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Well, OK, Professor Frey invited me to give the two lectures this week on first order equations, like that one, first order dy dt.
And the lectures next week will be on second order equation.
So we're looking for, you could say, formulas for the solution.
We'll get as far as we can with formulas, then numerical methods.
Graphical methods take over in more complicated problems.
This is a model problem.
It's linear.
I chose it to have constant coefficient a, and let me check the units.
Always good to see the units in a problem.
So let me think of this y, as the money in a bank, or bank balance, so y as in dollars, and t, time, in years.
So we're looking at the ups and downs of bank balance y.
The rate of change, so the units then are dollars per year.
So every term in the equation has to have the right units.
So y is in dollars, so the interest rate a is percent per year, say 6% a year.
So a could be 6%-- that's dimensionless-- per year, or half a percent per month if we change.
So if we change units, the constant a would change from 6 to a half.
But let's stay with 6.
And then q of t represents deposits and withdrawals, so that's in dollars per year again.
Has to be.
So that's continuous.
We think of the deposits and the interest as being computed continuously as time goes forward.
So if that's a constant-- and I'll take that case first, q equal 1-- that would mean that we're putting in, depositing $1 per year, continuously through the year.
So that's the model that comes from a differential equation.
A difference equation would give us finite time steps.
So I'm looking for the solution.
And with constant coefficients, linear, we're going to get a formula for the solution.
I could actually deal with variable interest rate for this one first order equation, but the formula becomes messy.
But you can still do it.
After that point, for a second order equations like oscillation, or for a system of several equations coupled together, constant coefficients is where you can get formulas.
So let's go with that case.
So how to solve that equation?
Let me take first of all, a constant, constant source.
So I think of q as the source term.
To get one nice formula, let me take this example, ay plus 1, let's say.
How do you find y of t to solve that?
And you start with some initial condition y of 0.
That's the opening deposit that you make at time 0.
How to solve that equation?
Well, we're looking for a solution.
And solutions to linear equations have two parts.
So the same will happen in linear algebra.
One part is a solution to that equation, so we're just looking for one, any one, and we'll call it a particular solution.
And the associated null equation, dy dt equal ay.
So this is an equation with q equals 0.
That's why it's called null.
And it's also called homogeneous.
So more textbooks use that long word homogeneous, but I use the word null because it's shorter and because it's the same word in linear algebra.
So let me call yn the null solution, the general null solution.
And y, I'm looking here for a particular solution yp, and I'm going to-- here's the key for linear equations.
Let me take that off and focus on those two equations.
How does solving the null equation, which is easy to do, help me?
Why can I, as I plan to do, add in yn to yp?
I just add the two equations.
Can I just add those two equations?
I get the derivative of yp plus yn on the left side.
And I have a times yp plus yn.
And that is a critical moment there when we use linearity.
I had a yp a yn, and I could put them together.
If it was y squared, yp squared and yn squared would not be the same as yp plus yn squared.
It's the linearity that comes, and then I add the 1.
So what do I see from this?
I see that yp plus yn also solves my equation.
So the whole family of solutions is 1 yp plus any yn.
And why do I say any yn?
Because when I find one, I find more.
The solutions to this equation are yn could be e to the at, because the derivative of e to the at does bring down a factor a.
But you see, I've left space for any multiple of e to the at.
This is where that long word homogeneous comes from.
It's homogeneous means I can multiply by any constant, and I still solve the equation.
And of course, the key again is linear.
So now I have-- well, you could say I've done half the job.
I've found yn, the general yn.
And now I just have to find one yp, one solution to the equation.
And with this source term, a constant, there's a nice way to find that solution.
Look for a constant solution.
So certain right hand sides, and those are the like the special functions for the special source terms for differential equation, certain right hand sides-- and I'm just going to go down a list of them today.
The next one on the list-- can I tell you what the next one on the list will be?
y prime equal ay.
I use prime for-- well, I'll write dy dt, but often I'll write y prime.
dy dt equal ay plus an exponential.
That'll be number two.
So I'm just preparing the way for number two.
Well, actually number one, this example is the same as that exponential example with exponent s equal 0, right?
If s is 0, then I have a constant.
So this is a special case of that one.
This is the most important source term in the whole subject.
But here we go with a constant 1.
So we've got yn.
And what's yp?
I just looked to see.
Can I think of one?
And with these special functions, you can often find a solution of the same form as the source term.
And in this case, that means a constant.
So if yp is a constant, this will be 0.
So I just want to pick the constant that makes this thing 0.
And of course, their right hand side is 0 when yp is minus 1 over a.
So I've got it.
We've solved that equation, except we didn't match the initial condition yet.
Let me if you take that final step.
So the general y is any multiple, any null solution, plus any one particular solution, that one.
And we want to match it to y of 0 at t equals 0.
So I want to take that solution.
I want to find that constant, here.
That's the only remaining step is find that constant.
You've done it in the homework.
So at t equals 0, y of 0 is-- at t equals 0, this is C. This is the minus 1 over a.
So I learn what the C has to be.
And that's the final step.
C is bring the 1 over a onto that side, so C will be y of 0 e to the at minus 1 over a e to at.
And here we had a minus 1 over a.
Well, it'll be plus 1 over a e to the at.
So now I've just put in the C, y of 0 plus 1 over a. y of 0 plus 1 over a has gone in for C. And now I have to subtract this 1 over a.
Here, I see a 1 over a, so I can do it neatly.
Got a solution.
We can check it, of course.
At t equals 0, this disappears, and this is y of 0.
And it has the form.
It's a multiple of e to the at and a particular solution.
So that's a good one.
Notice that to get the initial condition right, I couldn't take C to be y of 0 to get the initial condition right.
To get the initial condition right, I had to get that, this minus 1 over a in there.
Good for that one?
Let me move to the next one, exponentials.
So again, we know that the null equation with no source has this solution e to the at.
And I'm going to suppose that the a in e to the at in the null solution is different from the s in the source function, which will come up in the particular solution.
So you're going to see either the st in the particular solution and an e to the at in the null solution.
And in the case when s equals a, that's called resonance, the two exponents are the same, and the formula changes a little.
Let's leave that case for later.
How do I solve this?
I'm looking for a particular solution because I know the null solutions.
How am I going to get a particular solution of this equation?
Fundamental observation, the key point is it's going to be a multiple of e to the st.
If an exponential goes in, then that will be an exponential.
Its derivative will be an exponential.
I'll have e to the st's everywhere.
And I can get the number right.
So I'm looking for y try.
So I'll put try, knowing it's going to work, as some number times e to the st.
So this would be like the exponential response.
Response, do you know that word response?
So response is the solution.
The input is q, and the response is Y.
And here, the input is e to the st, and the response is a multiple of e to the st.
So plug it in.
The timed derivative will be Y.
Taking the derivative will bring down a 1. e to the st equals aY.
A aY e to the st plus 1 e to the st. Just what we hoped.
The beauty of exponentials is that when you take their derivatives, you just have more exponential.
That's the key thing.
That's why exponential is the most important function in this course, absolutely the most important function.
So it happened here.
I can cancel e to the st, because every term has one of them.
So I'm seeing that-- what am I getting for Y?
Getting a very important number for Y.
So I bring aY onto this side with sY.
On this side I just have a 1.
Maybe it's worth putting on its own board.
Y is, so Ys aY comes with a minus, and the 1, 1 over-- so Y was multiplied by s minus a.
That's the right quantity to get a particular solution.
And that 1 over s minus a, you see why I wanted s to be different from a. I If s equaled a in that case, in that possibility of resonance when the two exponents are the same, we would have 1 over 0, and we'd have to look somewhere else.
The name for that-- this has to have a name because it shows up all the time.
The exponential response function, you could call it that.
Most people would call it the transfer function.
So any constant coefficient linear equation's going to have a transfer function, easy to find.
Everything easy, that's what I'm emphasizing, here.
Everything's straightforward.
That transfer function tells you what multiplies the exponential.
So the source was here.
And the response is here, the response factor, you could say, the transfer function.
Multiply by 1 over s minus a.
So if s is close to a, if the input is almost at the same exponent as the natural, as the null solution, then we're going to get a big response.
So that's good.
For a constant coefficient problem second order, other problems we can find that response function.
It's the key function.
It's the function if we have, or if we were to look at Laplace transforms, that would be the key.
When you take Laplace transforms, the transfer function shows up.
Then when you take inverse Laplace transforms, you have to find what function has that Laplace transform.
So did we get the-- we got the final answer then.
Let me put it here.
y is e to the st times this factor.
So I divide by s minus a.
A nice solution.
Let me also anticipate something more.
An important case for e to the st is e to the i omega t. e to the st, we think about as exponential growth, exponential decay.
But that's for positive s and negative s. And all important in applications is oscillation.
So coming, let me say, coming is either late today or early Wednesday will be s equal i omega, so where the source term is e to the i omega t. And alternating, so this is electrical engineers would meet it constantly from alternating voltage source, alternating current source, AC, with frequency omega, 60 cycles per second, for example.
Why don't I just deal with this now?
Because it involves complex numbers.
And we've got to take a little step back and prepare for that.
But when we do it, we'll get not only e to the i omega t, which I brought out, but also, it's real part.
You remember the great formula with complex numbers, Euler's formula, that e to the i omega t is a combination of cosine omega t, the real part, and then the imaginary part is sine omega t. So this is looking like a complex problem.
But it actually solves two real problems, cosine and sine.
Cosine and sine will be on our short list of great functions that we can deal with.
But to deal with them neatly, we need a little thought about complex numbers.
So OK if I leave e to the i omega t for the end of the list, here?
So I'm ready for another one, another source term.
And I'm going to pick the step function.
So the next example is going to be dy dt equals ay plus a step.
Well, suppose I put H of t there.
Suppose I put H of t. And I ask you for the solution to that guy.
So that step function, its graph is here.
It's 0 for negative time, and it's 1 for positive time.
So we've already solved that problem, right?
Where did I solve this equation?
This equation is already on that board.
Because why?
Because H of t is for t positive.
That's the only place we're looking.
This whole problem, we're not looking at negative t. We're only looking at t from 0 forward.
And what is H of t from 0 forward?
It's 1.
It's a constant.
So that problem, as it stands, is identical to that problem.
Same thing, we have a 1.
No need to solve that again.
The real example is when this function jumps up at some later time T. Now I have the function is H of t minus T. Do you see that, why the step function that jumps at time T has that formula?
Because for little t before that time, in here, this is-- what's the deal?
If little t is smaller than big T, then t minus T is negative, right?
If t is in here, then t minus capital T is going to be a negative number.
And H of a negative number is 0.
But for t greater than capital T, this is a positive number.
And H of a positive number is 1.
Do you see how if you want to shift a graph, if you want the graph to shift, if you want to move the starting time, then algebraically, the way you do it is to change t to t minus the starting time.
And that's what I want to do.
So physically, what's happening with this equation?
So it starts with y of 0 as before.
Let's think of a bank balance and then other things, too.
If it's a bank balance, we put in a certain amount, y of 0.
We hope.
And that grew.
And then starting at time, capital T, this switch turns on.
Actually, physically, step function is really often describing a switch that's turned on, now.
This source term act begins to act at that time.
And it acts at 1.
So at time capital T we start putting money into our account.
Or taking it out, of course.
If this with a minus sign, I'd be putting money in.
Sorry, I would start with some money in, y of 0.
I would start with money in.
Yeah, actually, tell me what's the solution to this equation that starts from y of 0?
What's the solution up until the switch is turned on?
What's the solution before this switch happens, this solution while this is still 0?
So let's put that part of the answer down.
This is for t smaller than T. What's the answer?
This is all common sense.
It's coming fast, so I'm asking these questions.
And when I asked that question, it's a sort of indication that you can really see the answer.
You don't need to go back to the textbook for that.
What have we got here?
Yeah?
Is it the null solution [INAUDIBLE]?
It'll be this guy.
Yeah, the particular solution will be 0.
Right, the particular solution is 0 before this is on.
I'm sorry, the null solution is 0, and the particular solution, well, the particular solution is a guy that starts right.
I don't know.
Those names were not important.
And then the question is-- so it's just our initial deposit growing.
Now, all I ask, what about after time T?
What about after time T?
For t after time T, and hopefully, equal time T, what do you think y of t will be?
Again, we want to separate in our minds the stuff that's starting from the initial condition from the stuff that's piling up because of the source.
So one part will be that guy.
I haven't given the complete answer.
But this is continuing to grow.
And because it's linear, we're always using this neat fact that our equation is linear.
We can watch things separately, and then just add them together.
So I plan to add this part, which comes from initial condition to a part that-- maybe we can guess it-- that's coming from the source.
And how do we have any chance to guess it?
Only because that particular source, once it's turned on, jumps to a constant 1, and we've solved the equation for a constant 1.
Let me go back here.
I think our answer to this question-- so this is like just first practice with a step function, to get the hang of a step function.
So I'm seeing this same y of 0 e to the at in every case, because that's what happens to the initial deposit.
I'll say grow, assuming the bank's paying a positive interest rate.
And now, where did this term comes from?
What did that term represent?
The money that [INAUDIBLE]
The money that, yeah?
They had each of [INAUDIBLE]
The money that came in and grew.
It came in, and then it grew by itself, grew separately from that these guys.
So the initial condition is growing along.
And the money we put in starts growing.
Now, the point is what?
That over here, it's going to look just like that.
So I'm going to have a 1 over a.
And I'm going to have something like that.
But can you just guess what's going to go in there?
When I write it down, it'll make sense.
So this term is representing what we have at time little t, later on, from the deposits we made, not the initial one, but the source, the continuing deposits.
And let me write it.
It's going to be a 1 over a e to the a something minus 1.
It's going to look just like that guy.
When I say that guy, let me point to it again-- e to the at minus 1.
But it's not quite e to the at minus 1.
What is it?
t minus [INAUDIBLE]
t minus capital T, because it didn't start until that time.
So I'm going to leave that as, like, reasonable, sensible.
Think about a step function that's turned on a capital time T. Then it grows from that time.
Of course, mentally, I never write down a formula like that without checking at t equal to T, because that's the one important point, at t equal capital T. What is this at t equal capital T?
It's 0.
At t equal capital T, this is e to the 0, which is 1 minus 1 altogether 0.
And is that the right answer?
At t equal capital T is 0, should I have nothing here?
Yes?
No?
Give me a head shake.
Should I have nothing at t equal capital T?
I've got nothing.
e to the 0 minus 1, that's nothing?
Yes, yes that's the right thing.
Because at capital T, the source has just turned on, hasn't had time to build up anything, just that was the instant it turned on.
So that's a step function.
A step function is a little bit of a stretch from an ordinary function, but not as much of a stretch as its derivative.
In a way, this is like the highlight for today, coming up, to deal with not only a step function, but a delta function.
I guess every author and every teacher has to think am I going to let this delta function into my course or into the book?
And my answer is yes.
You have to do it.
You should do it.
Delta functions are-- they're not true functions.
As we'll see, no true function can do what a delta function does.
But it's such an intuitive, fantastic model of things happening over a very, very short time.
We just make that short time into 0.
So we're saying with the delta function, we're going to say that something can happen in 0 time.
Something can happen in 0 time.
It's a model of, you know, when a bat hits a ball.
There's a very short time.
Or a golf club hits a golf ball.
There's a very short time interval when they're in contact.
We're modeling that by 0 time, but still, the ball gets an impulse.
Normally, for 0 time, if you're doing things continuously, what you do over 0 time is no importance.
But we're not doing things continuously, at all.
So here we go.
You've seen this guy, I think.
But if you haven't, here's the time to see it.
So the delta function is the derivative of-- so I've written three important functions up here.
Let me start with a continuous one.
That function, the ramp is 0, and then the ramp suddenly ramps up to t. Take its derivative.
So the derivative, the slope of the ramp function is certainly 0 there.
And here, the slope is 1.
So the slope jumped from 0 to 1.
The slope of the ramp function is the step function.
Derivative of ramp equals step.
Why don't I write those words down?
Derivative of ramp equals step.
So there is already the step function.
In pure calculus, the step function has already got a little question mark.
Because at that point, the derivative in a calculus course doesn't exist, strictly doesn't exist, because we get a different answer 0 on the left side from the answer, 1 on the right side.
We just go with that.
I'm not going to worry about what is its value at that point.
It's 0 up for t negative, and it's 1 for t positive.
And often, I'll take it 1 for t equals 0, also.
Usually, I will.
That's the small problem.
Now, the bigger problem is the derivative of the-- so this is now the derivative of the step function.
So what's the derivative of this step function?
Well, the derivative along there is certainly 0.
The derivative along here is certainly 0.
But the derivative, when that jumped, the derivative, the slope was infinite.
That line is vertical.
Its slope is infinite.
So at that one point, you have an affinity, here, delta of 0.
You could say delta of 0 is infinite.
But you haven't said much, there.
Infinite is too vague.
Actually, I wouldn't know if you gave me infinite or 2 times infinite.
I couldn't tell the difference.
So I'll put it in quotes, because it sort of gives us comfort.
But it doesn't mean much.
What does mean much?
Somehow that's important.
Can I tell you how to work with delta functions, how to think about delta functions?
It's the right way to think about delta function.
So here's some comment on delta function.
Giving the values of the function, 0, and infinity, and 0, is not the best.
What you can do with a delta function is you can integrate it.
You can define the function by integrals.
Integrals of things are nice.
Do you think in your mind when you take derivatives, as we did going left to right, we were taking derivatives.
The function was getting crazy.
When we go right to left, take integrals, those are smoothing.
Integrals make functions smoother.
They cancel noise.
They smooth the function out.
So what we can do is to take the integral of the delta function.
We could take it from any negative number to any positive number.
And what answer would we get?
What would be the right, well, the one thing people know about the delta function is-- and actually, it's the key thing-- the integral of the delta function.
Again, I'm integrating the delta function from some negative number up to some positive number.
And it doesn't matter where n is, because the function is 0 there and there.
But what's the answer here?
Put me out of my misery.
Just tell me the number I'm looking for, here, the integral of the delta function.
Or maybe you haven't met it
[INAUDIBLE]
It's?
It's the one good number you could guess.
It's 1.
Now, why is it 1?
Because if the delta function is the derivative of the step function, this should be the step function evaluated between N and P. This should be the step function, , here, minus the step function, there And what is the step function?
You have to keep it straight.
Am I talking about the delta function?
No, right now, I've integrated it to get H of t. So this is H of P at the positive side, minus H of N. That's what integration's about.
And what do I get?
1, because H of P, the step function here, H is 1.
And here, it's 0, so I get 1.
Good, that's the thing that everybody remembers about the delta function.
And now I can make sense out of 2 delta function, 2 delta of t. That could be my source.
So if 2 delta of t was my source, what's the graph of 2 delta of t?
Again, it's 0 infinite 0.
You really can't tell from the infinity what's up, but what would be the integral of 2 delta of t, the integral of 2 delta of t or some other?
Well, let me put in the 2, here?
What's the integral of 2 delta of t, would be 2H of t. Keep going.
What do I get here?
2
It would be 2 of these guys, 2 of these, 2 of these, 2.
All right?
So we made sense out of the strength of the impulse, how hard the bat hit the ball.
But of course, we need units in there.
We have to have units.
And here, the value for that unit was 2.
Now, I'm going to-- because this is really worth doing with delta functions.
I didn't ask at the start have you seen them before.
But they are worth seeing.
And they just take a little practice.
But then in the end, delta functions are way easier to work with than some complicated function that attempts to model this.
We could model that by some Gaussian curve or something.
All the integrations would become impossible right away.
We could model it by a step function up and a step function down.
Then the integrations would be possible.
But still, we have this finite width.
I could let that width go to 0 and let the height go to infinity.
And what would happen?
I'd get the delta function.
So that's one way to create a delta function, if you like.
If you're OK with step functions, then one way to create delta is to take a big step up, step down, and then let the size of the step grow and the width of the steps shrink.
Keep the area 1, because area is integral.
So I keep this, that little width, times this big height equal to 1.
And in the end, I get delta.
Now again, my point is that delta functions, that you really understand them.
What you can legitimately do with them is integrate them.
But now in later problems, we might have not a 1 or a 2, but a function in here, like cosine t, or e to the t, or q of t. Can I practice with those?
Can I put in a function f of t?
I didn't leave enough space to write f of t, so I'm going to put it in here.
f of t delta of t dt.
And I'm going to go for the answer, there.
My question is what does that equal?
You see what the question is?
I got my delta function, which I only just met.
And I'm multiplying it by some ordinary function.
f of t gives us no problems.
Think of cosine t. Think of e to the t. What do you think is the right answer for that?
What do you think is the right answer?
And this tells you what the delta function is when you see this.
What do I need to know about f of t to get an answer, here?
Do I need to know what f is at t equals minus 1?
You could see from the way my voice asked that question that the answer is no.
Why do I not care what f is at minus 1?
Yeah?
Because you're multiplying by [INAUDIBLE]
Because I'm multiplying by somebody that's 0.
And similarly, at f equal minus 1/2, or at f equal plus 1/3, all those f's make no difference, because they're all multiplying 0.
What does make a difference?
What's the key information about f that does come into the answer?
f at?
At just at that one point, f at?
[INAUDIBLE]
0, f at 0 is the action.
The impulse is happening.
The bat's hitting the ball.
So we're modeling rocket launching, here.
We're launching in 0 seconds instead of a finite time.
So in other words, well, I don't know how to put this answer down other than just to write it.
I guess I'm hoping you're with me in seeing that what it should be.
Can I just write it?
All that matters is what f is at t equals 0, because that's where all the action is.
And that f of 0, if f of 0 was the 2 that I had there a little while ago, then the answer will be 2.
If f of 0 is a 1, if the answer is f of 0 times 1-- and I won't write times 1.
That's ridiculous.
Now we can integrate delta functions, not just a single integral of delta, but integral of a function, a nice function times delta.
And we get f of 0.
So can I just, while we're on the subject of delta functions, ask you a few examples?
What is the integral of e to the t delta of t dt?
It's 1
Yeah, say it again?
It's 1
It's 1.
It's 1, right.
Because e to the t, at the only point we care about, t equal 0 is 1.
And what if I change that to sine t?
Suppose I integrate sine t times delta of t?
What do I get now?
I get?
0
0, right.
And actually, that's totally reasonable.
This is a function, which is yeah, it's an odd function.
Anyway, sine, if I switch t to negative t, it goes negative.
0 is the right answer.
Let me ask you this one.
What about delta of t squared?
Because if we're up for a delta function, we might square it.
Now we've got a high-powered function, because squaring this crazy function delta of t gives us something truly crazy.
And what answer would you expect for that?
1
Would you expect 1?
So this is like?
I'm just getting intuition working, here, for delta functions.
What do you think?
I'm looking at the energy when I square something.
OK, so we had a guess of 1.
Is there another guess?
Yeah?
A third?
Sorry?
1/3
1/3, that's our second guess.
I'm open for other guesses before I-- OK, we have a rule here for f of t. And now what is the f of t that I'm asking about in this case?
It's delta of t, right?
If f of t is delta of t, then that would match this.
And therefore, the answer should match.
Do you see what I'm shooting for, yeah?
It'd be infinity?
It'd be infinity.
It would be infinity.
That's delta of t squared is that's an infinite energy function.
You never meet it, actually.
I apologize, so so write it down there.
I could erase it right away because you basically never see it.
It's infinite energy.
Well, I think you'd see it.
I mean, we're really going back to the days of Norbert Wiener.
When I came to the math department, Norbert Wiener was still here, still alive, still walking the hallway by touching the wall and counting offices.
And hard to talk to, because he always had a lot to say.
And you got kind of allowed to listen.
So anyway, Wiener was among the first to really use delta functions, successfully use delta functions.
Anyway, this is the big one.
This is the big one.
Now, so what's all that about?
I guess I was trying to prepare by talking about this function prepare for the equation when that's the source.
So dy equal ay plus a delta function.
Let me bring that delta function in at time T. So how do you interpret that equation?
So like part of this morning's lecture is to get a first handle on an impulse.
So let me write that word impulse, here.
Where am I going to write it?
So delta is an impulse.
That's our ordinary English word for something that happens fast.
And y of t is the impulse response.
And this is the most important.
Well, I said e to the st was the most important.
How can I have two most important examples?
Well, they're a tie, let's say.
e to the st is the most important ordinary function.
It's the key to the whole course.
Delta of t, the impulse, is the important one because if I can solve it for a delta function, I can solve it for anything.
Let's see if we can solve it for a delta function, a delta function, an impulse that starts at time T. Again, I'm just going to start writing down the solution and ask for your help what to write next.
So what do you expect as a first term in the solution?
So I'm starting again from y of 0.
Let's see if we can solve it by common sense.
So how do I start the solution to this?
Everybody sees what this equation is saying.
I have an initial deposit of y of 0 that starts growing.
And then at time capital T I make a deposit.
At that moment, at that instant, I make a deposit of 1.
That's an instant deposit of 1.
Which is, of course, what I do in reality.
I take $1 to the bank.
They've got it now.
At time T, I give them that $1, and it starts earning interest.
So what about y of t?
What do you think?
What's the first term coming from y of 0?
So the term coming from y of 0 will be y of 0 to start with, e to at.
That takes care of the y of 0.
Now, I need something.
It's like this, plus I need something that accounts for what this deposit brings.
So up until time T, what do I put?
So this is for t smaller than T and t bigger than T. So what goes there?
For t smaller than T, what's the benefit from the delta function?
0, didn't happen yet.
For t bigger than T, what's the benefit from the delta function?
[INAUDIBLE]
For t bigger than T, well, that's right.
OK, but now I've made that deposit at time capital T. Whatever's going there is whatever I'm getting from that deposit.
At time capital T, I gave them $1, and they start paying interest on it.
What's going to go there?
So if I gave them $1 at that initial time, so that $1 would have been part of y of 0.
What did I get at a later time?
e to the at.
Now I'm waiting.
I'm giving them the dollar at time capital T, and it starts growing.
So what do I have at a later time, for t later than capital T?
What has that $1 grown into?
e to the a times the-- right, it's critical.
It's the elapsed time.
It's the time since the deposit.
Is that right?
So what do I put here?
t minus capital T?
t minus capital T, good.
Apologies to bug you about this, but the only way to learn this stuff from a lecture is to be part of it.
So I constantly ask you, instead of just writing down a formula.
I think that looks good.
So suddenly, what does this amount to at t equal capital T?
Maybe I should allow t equal capital T. At t equal capital T, what do I have here?
1
1.
That's my $1.
At t equal capital T, we've got $1.
And later it's grown.
So we have now solved.
We have found the impulse response.
We have found the impulse response when the impulse happened at capital T. That was good going.
Now, I've given you my list of examples with the pause on the sine and cosine.
I pause on the sine and cosine because one way to think about sine and cosine is to get into complex numbers.
And that's really for next time.
But apart from that, we've done all the examples, so are we ready?
Oh yeah, I'm going to try for the big thing, the big formula.
So this is the key result of section 1.4, the solution to this equation.
So I'm going back to the original equation.
And just see if we can write down a formula for the answer.
So let me write the equation again.
dy dt is ay plus some source.
I think we can write down a formula that looks right.
And we could then actually plug it in and see, yeah, it is right.
So what's going to go into this formula?
We got enough examples, so now let's go for the whole thing.
So y of t, first of all, comes whatever depends on the initial condition.
So how much do we have at a later time when our initial deposit was y of 0?
So that's the one we've seen in every example.
Every one of these things has this term growing out of y of 0.
So let me put that in again.
So the part that grows out of y of 0 is y of 0 e to the at.
That's OK.
So that's what the initial.
So our money is coming from two sources, this initial deposit, which was easy, and this continuous, over time deposit, q of t. And I have to ask you about that.
That's going to be like the particular solution, the particular solution that comes from the source term.
This is the solution it comes from the initial condition.
So what do you think this thing looks like?
I just think once we see it, we can say, yeah, that makes sense.
So now I'm saying what?
If we've deposited q of t in varying amounts, maybe a constant for a while, maybe a ramp for awhile, maybe whatever, a step, how am I going to think about this?
So at every time t equal to s, so I'm using little t for the time I've reached.
Right?
Here's t starting at 0.
Now, let me use s for a time part way along.
So part way along, I input.
I deposit q of s. I deposit it at time s. And then what does it do?
That money is in the bank with everybody else.
It grows along with everything else.
So what's the growth factor?
What's the growth factor?
This is the amount I deposited at time s. And how much has it grown at time t?
This is the key question, and you can answer it.
It went in a time s. I'm looking at time t. What's the factor?
Is it e to the a t minus s
e to the a t minus s. So that's the contribution to our balance at time t from our input at time s. But now, I've been inputting all the way along.
s is running all the way from here to here.
So finish my formula.
Put me out of my misery.
Or it's not misery, actually.
Its success at this moment.
What do I do now?
I?
Integrate
I integrate, exactly.
I integrate.
I integrate.
So all these deposits went in.
They grew that amount in the remaining time.
And I integrate from 0 up to the current time t. So you see that formula?
Have a look at it.
This is a general formula, and every one of those examples could be found from that formula.
If q of s was 1, that was our very first example.
We could do that integration.
If q of s was e to the-- anyway, we could do every one.
I just want you to see that that formula makes sense.
Again, this is what grew out of the initial condition.
This is what grew out of the deposit at time s. And the whole point of calculus, the whole point of learning [?
1801 ?
], the integral equation part, the integrals part, is integrals just add up.
This term just adds up all the later deposits, times the growth factor in the remaining time.
And as I say, if I took q of s equal 1-- the examples I gave are really the examples where you can do the integral.
If q of s is e to the i omega s, I can do that integral.
Actually, it's not hard to do because e to the at doesn't depend on s. I can bring an e to the at out in this case.
That formula is just worth thinking about.
It's worth understanding.
I didn't, like, derive it.
And the book does, of course.
There's something called an integrating factor.
You can get at this formula systematically.
I'd rather get at it and understand it.
I'm more interested in understanding what the meaning of that formula is than the algebra.
Algebra is just a goal to understand, and that's what I shot for directly.
And as I say, that the book also, early section of the book, uses practice in calculus.
Substitute that in to the equation.
Figure out what is dy dt.
And check that it works.
It's worth actually looking at that end of what you need to know from calculus It's is.
You should be able to plug that in for y and see that solves the equation.
Right, now I have enough time to do cosine omega t. But I don't have enough time to do it the complex way.
So let me do as a final example, the equation.
Let me just think.
I don't know if I have enough space here.
I'm now going to do dy dt-- can I call that y prime to save a little space-- equal ay plus cosine of t. I'll take omega to be 1.
Now, how could we solve that one?
I'm going to solve it without complex numbers, just to see how easy or hard that is.
And you'll see, actually, it's easy.
But complex numbers will tell us more.
So it's easy, but not totally easy.
So what did I do in the earlier example if the right hand side was a 1, a constant?
I look for the solution to be a constant.
If the right hand side was an exponential, I look for the solution to be an exponential.
Now, my right hand side, my source term, is a cosine.
So what form of the solution am I going to look for?
I naturally think, OK, look for a cosine.
We could try y equals some number M cosine t. Now, you have to see what goes wrong and how to fix it.
So if I plug that in, looking for M the same way I look for capital Y earlier, I plug this in, and I get aM cosine t cosine t. But what do I get for y prime?
Sine t. And I can't match.
I can make it work.
I can't make a sine there magic a cosine here.
So what's the solution?
How do I fix it?
I better allow my solution to include some sine plus N sine t. So that's the problem with doing it, keeping things real.
I'll push this through, no problem.
But cosine by itself won't work.
I need to have sines there, because derivatives bring out sines.
So I have a combination of cosine and sine.
I have a combination of cosine and sine.
So the complex method will work in one shot because e to the i omega t is a combination of cosine and sine.
Or another way to say it is when I see cosine here, that's got two exponentials.
That's got e to the it and e to the-- anyway.
Let's go for the real one.
So I'm going to plug that into there.
So I'll get sines and cosines, right?
When I plug this into there, I'll have some sines and some cosines, and I'll just match the two separately.
So I'm going to get two equations.
First of all, let me say what's the cosine equation?
And then what's the sine equation?
So when I match cosine terms, what do I have?
What cosine terms do I get out of y prime, here?
The derivative.
Well, the derivative of cosine is a sine.
That that's not a cosine term.
The derivative of sine is cosine.
I think I get, if I just match cosines, I think I get an N cosine.
N cosine t equal ay.
How many cosines do I have from that term?
ay has an M cosine t. I think I have an aM, and here I've got 1.
That was a natural step, but new to us.
I'm matching the cosines.
I have on the left side, with this form of the solution, the derivative will have an N cosine t. So I had N cosines, aM cosines, and 1 cosine.
Now, what if I match signs?
What happens there?
We're pushing more than an hour, so hang on for another five minutes, and we're there.
Now, what happens if I match sines, sine t?
How do I get sine t in y prime?
So take the derivative of that, and what do you have?
Minus [INAUDIBLE]
Minus M sine t. That tells me how many sine t's are in there.
And on the right hand side, a times y, how many sine t's do I have from that?
You have N t's
N, good thinking.
And what about from this term?
None, no sine there.
So I have two equations by matching the cosines and sines.
Once you see it, you could do it again.
And we can solve those equations, two ordinary, very simple equations for M and N. Let's see if I make space.
Why don't I do it here, so you can see it.
So how do I solve those two equations?
Well, this equation gives me-- easy-- gives me M as minus aN.
So I'll just put that in for N. So I have N equals aM.
But M is minus aN.
I think I've got minus a squared N plus that 1.
All I did was solve the equation, just by common sense.
You could say by linear algebra, but linear algebra's got a little more to it than this.
So now I know M, and now I know N. So now I know the answer.
y is M, so M is minus aN.
Oh, well, I have to figure out what N is, here.
What is N?
This is giving me N, but I better figure it out.
What is N from that first equation?
And then I'll plug in.
And then I'm quit
[INAUDIBLE]
1 over, yeah
1 plus a squared
1 plus a squared, good.
Because that term goes over there, and we have 1 plus a squared.
So now y is M cosine t. So M is minus aN.
So minus aN is 1 over 1 plus a squared cosine t. Is that right?
That was the cosines.
And we had N sine t. But N is just 1-- I think I just add the sine t. Have I got it?
I think so.
Here is the N sine t, and here is the M cos t. It was just algebra.
Typical of these problems, there's a little thinking and then some algebra.
The thinking led us to this.
The thinking led us to the fact we needed cosines in there, as well as cosines.
But then once we did it, then the thinking said, OK, separately match the cosine terms and the sine term.
And then do the algebra.
Now, I just want to do this with complex.
So y prime equals ay plus e to the it.
To get an idea, you see the two.
And then I have to talk about it.
You see, I'm only going to go part way with this and then save it for Wednesday.
But if I see this, what solution do I assume?
This is like an e to the st.
I assume y is some Y e to the it.
See, I don't have cosines and sines anymore.
I have e to the it.
And if I take the derivative of e to the it, I'm still in the e to the it world.
So I do this.
I plug it in.
Uh-huh, let me leave that for Wednesday.
We have to have some excitement for Wednesday.
So we'll get a complex answer, and then we'll take the real part to solve that problem.
So we've got two steps, one way or the other way.
Here, we had two steps because we had to let sines sneak in.
Here, we have two steps because I could solve it, and you could solve that right away.
But then you have to take the real part.
I'll leave that.
Is there questions?
Do you want me to recap quickly what we've done
Yes
I try to leave on the board enough to make a recap possible.
Everything was about that equation.
We have only solved-- I shouldn't say only-- we have solved the constant coefficient, model constant coefficient, first order equation.
Wednesday comes nonlinear equation.
This one today was strictly linear.
So what did we do?
We solved this equation, first of all, for q equal 1; secondly, for q equal e to the st; thirdly, for q equal a step; fourthly for q equal-- where is it?
Where is that delta of t?
Maybe it's here.
Ah, it got erased.
So the fourth guy was y prime equal ay plus delta of t, or delta of t minus capital T. So those were our four examples.
And then what did we finally do?
So if we're recapping, compressing, we're compressing everything into two minutes.
We solved those four examples, and then we solved the general problem.
And when we solved the general problem, that gave us this integral, which my whole goal was that you should understand that this should seem right to you.
This is adding up the value at time t from all the inputs at different times s. So to add them up, we integrate from 0 to t. And finally, we returned to the question of cos t, all important question.
But awkward question, because we needed to let sine t in there too.
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Monday's lecture was all linear equations.
And I thought I would start today with nonlinear equations, still first order.
And we can't deal with every nonlinear equation.
That's too much to ask.
These are going to be made easier by a property called "separable."
So these will be separable nonlinear equations.
And let me start with a couple of examples and then you'll see the whole idea.
So one example would be the simplest nonlinear equation I can think of, with a y squared.
So how to get there?
Here's the trick.
This is the separable idea.
You're going to see it in one shot.
We can separate, put the Ys on one side and the Ds on the other.
So I write this as dy over y squared equal dt.
I put the dt up and brought the y squared down.
So now they're separated, in a kind of hookie way with infinitesimals.
But I'll makes sense out of that by integrating.
I'll integrate both sides.
I'll integrate time from 0 to t. And I have an initial condition, y of 0, always.
And since this one is about y, when t starts at 0, this guy starts at y of 0, up to, this ends at t, so this ends at y of t. OK. Now the point is, also the problem was nonlinear, we've got two separate ordinary integrals to do.
And we can do them.
We can certainly do the right hand side.
I get t. And on the left hand side, what do I get?
Well, that maybe I better leave a little space to figure out this one.
But the point is we can integrate 1 over y squared.
And I guess we get minus 1 over y.
So I get minus 1 over y between y of 0 and y of t. In other words, I'm getting let's see, so what the right, the derivative of the integral of 1 over y squared is minus 1 over y because I always check the derivative gives me that back.
So now I'm ready to plug-in those limits.
So I'll do the bottom limit first because it comes with a minus sign, canceling that minus, 1 over y of 0 minus 1 over y of t. Got it.
And that equals the other integral, which is just t. So that's the answer as it comes directly from integration.
And we can do more.
You can see that finding the solution when these things are separable has boiled down to two integrals.
And we could have a function of t here, too.
And that would be allowed, a function of t multiplying this guy, because then I would leave the function of t on that side.
And I would have to integrate that.
And I would bring the y.
You see, I've just separated the y.
In general, these equations look like dy, dt is some function of t divided by some function of y.
Maybe the book calls the top one g, I think, and the bottom one f. And everybody in this room sees that I can put the f of y up there.
I can put the dt up there.
And I've separated it.
OK.
So that's sort of the general situation.
This is a kind of nice example, nice example, dy, dt equals y squared.
Can we just play with this a little bit?
Let me take y of 0 to be 1, just to make the numbers easy.
So if y of 0 is 1, then I have, I'll just keep going a little bit.
You do have to keep going a little bit because when you finish the integral right there, you haven't got y equal.
You've got some equation that involves y, but you have to solve for y.
So I have to solve that equation for y.
Let me just do it.
So how would I solve it?
And let me take y of 0 to be 1.
So now, if I just write it below, I'm at 1 minus 1 over y equals t. Good?
So I'm going to put the 1 over y of t on that side and the t on that side.
So if I just continue here, I've got 1 over y of t on this side and, do I have 1 minus t on that side?
Yeah.
Looking good.
So solution starting from y of 0 equal 1 is y of t equal 1 over 1 minus t. You could do that.
You could do that.
And I can always, like, mentally I check the algebra at t equals 0.
That gives me the answer, 1.
But let's step back and look at that answer.
I mean, that's part of differential equations is to do some algebra, if possible, and get to a formula.
But if we don't think about the formula, we haven't learned anything.
Right there, yes.
Good.
So what happens?
I want to compare with the linear case that was like e to the t. This was y prime equal y, right?
And that led to e to the t. Y prime equals y squared leads to that one.
So first observation.
I haven't got exponentials anymore in that solution.
Exponentials are just like perfection for linear equations.
For nonlinear equations, we get other functions.
Professor Fry had a hyperbolic tangent function in his first lecture.
Other things happen.
OK. Now, how do those compare if I graph those?
It's just like, why not try?
So they both started at 1.
And e to the t went up exponentially.
E to the t. And, I don't know, we use exponential.
In our minds, we think, that's pretty fast growth.
!
mean, that's the common expression, grew exponentially.
But here, this guy is going to grow faster because y is going to be bigger than 1.
So y squared is going to be bigger than y.
That one's going to grow faster.
Faster than exponential.
This has the exponential growth.
Pretty fast.
Polynomial, of course, some parabola or something would be hanging way down here, left behind in the dust.
But this 1 over 1 minus t, that's going to grow really fast.
And what's more, it's going to go to infinity.
So that y prime equal y squared, the solution to that doesn't just-- e to the t goes to infinity at time infinity.
At any finite time, we get an answer.
Eventually, at t equal infinity, it's gone above every bound.
But this one, 1 over 1 minus t is what I want to graph now.
I believe that that takes off and at a certain point, capital T, it's going to infinity.
It's blown up.
So it's blow up in finite time.
Blow up in finite time.
And what is that time?
What's the time at which the y prime equal y squared has taken off, gone off the charts?
T equal--
1
--1.
Because when t reaches 1, I have 1 over 0, and I'm dividing by 0, and so that's the blow up.
Finite time blow up.
OK.
So this can happen for some nonlinear equations.
It wouldn't happen for a linear equation.
For a linear equation, exponentials are in control.
OK.
So that's one nice example.
Oh, another nice thing about that example.
Well, I say nice if you're OK with infinite series.
I just want to compare.
The book mentions the infinite series for these guys because that's an old way to solve differential equations is term-by-term in an infinite series.
It's sort of fun to see the two series.
Well, because they're the two most important series in math.
Actually, they're the two series that everybody should know.
The power series, Taylor series-- whatever word you want to give it for those two guys.
So let me do them.
E to the t, I'll put that one first, and 1 over 1 minus t. These are the great series of math.
Shall I just write them down and sort of talk through them?
Because this is not a lecture on infinite series by any means.
But having these two in front of us, coming from these two beautiful equations, y prime equal y squared and y prime equal y, I can't resist seeing what they look like this way.
So e to the t, do you remember e to the t?
It starts at 1.
What's the slope of e to the t?
At t equals 0.
So I'm doing everything-- this series is going to be, both of the series are going to be, around t equals 0.
That's my, like, starting point.
So this e to the t thing has a tangent.
It has a slope there.
And what's the slope of e to the t at t equals 0?
1
1.
It's derivative.
The derivative of e to the t is e to the t. The slope is 1.
So that tangent line has coefficient 1.
That's how it starts.
That's the linear approximation.
That's the heart of calculus, is this.
But we're going to go better.
We're going to get the next term.
So what's the next term?
That gave us the tangent line.
Now I'm going to move to the tangent parabola.
So the parabola has got another is still going to be below the real thing.
Can I squeeze in the words "line" and "parab," for "parabola?"
Parabola has bending.
I'm really explaining the Taylor series in what I hope is a sensible way.
Here is the starting point.
This has the slope.
The next term has the bending.
The bending comes from what derivative?
What derivative tells us about bending?
Second derivative.
Second derivative tells us how much it curves.
Well, the second derivative of e to the t is still e to the t. So the bending is 1.
The bending is also 1.
Now that comes in with a factor of a 1/2.
There is the tangent parabola.
And you will see what these numbers become.
Let me just go to, the third derivative would be responsible for the t cubed term.
And its coefficient would be 1 over 3 factorial.
So 2 is the same as 2 factorial.
3 factorial is 3 times 2 times 16.
So the numbers here go 1, 6, 24, 120, whatever the next one is.
720 or something.
They grow fast.
So that series always gives a finite answer.
It does grow with t. But it doesn't spike with t. Now compare that famous series.
And of course, this is 1 over 1 factorial, everything consistent.
Compare that with the series for 1 over 1 minus t. That's the other famous series that they learned in algebra.
I'll just write it.
That's 1 plus t plus t squared plus t cubed plus 1 so on, with coefficient 1.
This had 1 over n factorials.
Those make the series converge.
These don't have the n factorials.
This is 1, 1, 1, 1.
And, well, I could check that formula.
But do you remember the name for that series?
1 plus t plus t squared plus t cubed plus so on?
Algebra is taught differently in many high schools now.
And maybe that never got a name.
I guess I would call it the Geometric series.
Geometric series.
And you see, it's beautiful.
It's the other important series.
But it's quite different from this one because, what's the difference about this series?
Yeah?
It goes to infinity
It's a--
It goes to infinity
It goes to infinity.
But where?
At what time?
At what value of t is this sum going to fall apart?
Blow up?
At t equal 1.
When have 1 plus 1 plus 1 plus 1, I'm getting infinity.
So this blows up.
And of course, we see that it should because this blows up.
Left side blows up at t equal 1, the right side blows up a t equal 1.
Where the exponential series, which is the heart of ordinary differential equations, never blows up because of these big numbers in the denominator.
OK, I'm good for this first simple example, y prime equals y squared.
It has so much in it, it's worth thinking about.
I'm ready, you OK for a second example?
A second important separable equation.
I'm going to pick one.
So I'm going to pick an equation that starts out with our familiar linear growth.
This could be, you know, last time it was growth of money in a bank.
It could be growth of population.
The number of, to a sum first approximation, the rate of growth of the population comes from, like, births minus deaths.
And with modern medicine, births are a larger number than deaths.
So a is positive, and that grows.
But if we're talking about the, I mean, the United Nations tries to predict, everybody tries to predict, population of the world in future years.
And so this could be called the Population Equation.
But just to leave it as pure exponential is obviously wrong.
The world can't grow forever.
The population can't grow forever.
And the, I guess I hope it doesn't grow like 1 over 1 minus t. So this is at least a little slower.
But somehow competition for space, competition for food, for oil, for water-- which is going to be the big one-- is in here.
Competition here, of people versus people, a reasonable term, a first approximation, is a y squared, with a minus, is a y squared and with some coefficient.
That's a very famous equation.
A first model of population is it grows.
But this is a competition term, y against y.
And so, the same would be true if we were talking about epidemics.
That's a big subject with ordinary differential equations, epidemiology.
Or say, flu.
How does flu spread?
And how does it get cured?
So partly, people are getting over the flu.
But then y against y is telling us how many infected, how many new infections.
So we would like to solve that equation.
And it's separable.
I can do what I did before, dy over ay minus by squared equal dt.
And I can integrate, starting from year of 0.
Well, why don't we start from year 2014, with the population y at now-- the present population?
That would be a model that the UN would consider using.
That other people with very important interest in measuring population and measuring every resource would need equations like this.
And then they would put on more terms, like a term for immigration.
All sorts, many improvements have to go into this equation.
Let me just look at this as it is.
Well, I've got two choices here.
Well, it's this integral that I'm looking at.
That is a doable integral.
It's the type of integral that we saw in the Rocket problem.
The Rocket problem was more constant minus.
This was a drag term, when we were looking at rockets.
And this was a constant, say, gravity.
So it was still a second degree.
Still second degree, but a little different.
This has the linear in second degree terms.
If you look up that integral, you'll find it.
Or there's a systematic way to do it.
That's in 1801, I guess, called partial fractions.
It's not a lot of fun.
I don't plan to do it.
It's in the book.
Has to be because that's the way you can integr-- you can integrate polynomials over polynomials by partial fractions.
That's what they're for, but there's a neat way to do this one.
There's a neat trick that Bernoulli discovered to solve that equation, to turn it into a linear equation.
And of course, if we can turn it into a linear equation, we're on our way.
So the neat trick is let z be 1 over y.
You can put this in the category of lucky accidents, if you like.
So now I want an equation for z.
So I know that dz, dt if I take the derivative of that, that's y to the minus 1.
So it's minus 1 y to the minus 2 dy, dt.
That's the chain rule.
Take the derivative of 1 over 1, you get minus 1 over y squared.
Multiply by the derivative of what's inside.
That's the chain rule.
OK.
So I plan to substitute those in here.
So dy, dt, let's see.
Can you see me?
You can probably do it better than me.
So dy, dt is minus.
I'll bring that up.
Dy, dt I'm going to put-- I hope this'll work all right-- for dy, dt, I'm going to put in dz.
Using this, I'm going to put minus y squared dz, dt.
Did that look right?
I don't think I'm necessarily doing this the most brilliant way.
But dy, dt-- I put this up here and I got that-- equals ay.
So that's a over z. Oh, y Is 1 over z.
So get this, I want all Zs now.
So that's this part.
And ay is over z minus by squared is minus b over z squared.
Would you say OK to that?
I've got Zs now, instead of Ys.
I just took every term and replaced y by 1 over z. Y is 1 over z and dy, dt I can get that way.
OK. Yeah.
Now what?
Now look what happens, if I multiply through by z squared or by minus z squared.
Let me multiply through by minus z squared.
I get dz, dt.
Multiplying by minus z squared gives me a minus az.
And what do I get when I multiply this one by minus z squared?
Plus b
I get plus b.
Look what happened.
By this, like, some magic trick.
You could say, all right.
That was just a one time shot.
But it was a good one.
We ended up with a linear equation for z.
A linear equation for z.
And we solved that equation last time.
So let me squeeze in the solution for z, and then elsewhere.
So what was the solution for z of t?
It was some multiple of, no, yeah.
This is perfect review of last time.
We have a constant times z.
And so that's going to go into the exponential.
This will be the, it's a minus a, notice.
That will be the, what part of the solution is that one called?
That's the null solution.
The null solution, when b is o.
And now I add in a particular solution.
A particular solution.
And one good particular solution is choose the z to be a constant.
Then that'll be 0.
So I want that to be 0.
So what constant z makes that 0?
I think it's b over a, don't you?
I think b over a.
Does that work good?
That's every null solution plus one particular solution.
Let me say now, and I'll say again, looking for solutions which are steady states, b over a-- of this particular solution, that particular solution made this 0.
So it made this 0.
So it's a solution that's not going anywhere.
It's a constant solution.
It's a solution that can live for all time.
OK. B over a.
Let me put that word there, steady state.
OK. And now I would want to match the initial conditions using c. Yeah.
I'd better do that.
OK. And I have to get back to y. I have y is 1 over z.
So I'm going to have to flip this upside down.
I'm going to have to flip this upside down is what will actually happen.
Let me make it easy to flip.
Let me, I'll change c, which is just some constant to some constant d over a.
So then it's a is everywhere down below.
And I just write it here in the middle.
That makes it easier to flip.
So finally I get their solution.
Solution to the population equation.
But that's the famous word for it, the Logistic equation.
This is section 1.7 of the text on the differential equations in linear algebra.
It's a very, very much studied example.
It's a great example.
It fits the growth of human population with some, it's our first level approximation to growth of or other populations or other things.
It's a linear term giving us exponential growth, and a quadratic term of competition slowing it down.
And let's see that slow down.
So now that was a bit of algebra.
Much nicer than partial fractions.
The bit of algebra just came from this idea of going to z.
And now I want to go back to y.
So y is 1 over z.
So it's a over d e to the minus at plus b.
That's our solution.
A and b came out of the equation.
And d is going to be the number that makes the initial value correct.
So at t equals 0, I would have y of 0, whatever the initial population is, is a over d. T Is 0, so that's just 1 plus b.
So that tells me what d is.
D equals something.
It comes from y of 0.
So the answer, let me circle that answer.
That answer has three numbers in it, a, b, and d. a and b come from the equation.
D also involves the initial starting thing, which is exactly what it showed.
So you could say we've solved it.
But if you ever solve an equation like this, you want to graph it.
You want to graph it.
So let me draw its graph.
This is important picture.
So here is time.
Here is population.
Here's, maybe it started there.
This is times 0.
And now I want to graph this.
I want to graph that function.
Really, this is where we're going somewhere.
What happens for a long time?
At t equal infinity, what happens to the population?
Does it grow, like e to the t?
Just remember the examples here.
We had a growth like e to the t. We had a growth faster than e to the t that actually blew up.
What about this guy?
What will happen as t goes to infinity with that population?
It goes to?
A over b
A over b.
A over b.
That's the key number in the whole thing.
It keeps growing, but it never passes a over b.
This is y at infinity.
That's the final population.
So how does it do this?
If I draw this graph-- and what about negative time?
Let's go backwards in time.
What is it at t equals minus infinity?
Then you really see the whole curve.
At t equal minus infinity, what is this doing?
0 infinity
It's 0.
Good.
Good.
Good.
T equal minus infinity, this is enormous.
This is blowing up.
It's in the denominator.
We're dividing by it.
So the whole thing is going to 0.
So here's what the logistic curve looks like.
It creeps up.
And it's beautifully, there's a point of symmetry here.
The growth is increasing here.
And then, as a point of inflection you could say, growth is bending upwards for a while.
At this point, it starts bending downwards.
From that point on, ooh, let's see if I can draw it.
It'll get closer, and exponentially close.
That wasn't a bad picture.
The population here is half way.
Here, the population, the final population, is a over b.
And just by beautiful symmetry, the population here is a 1/2 of a over b.
At this point.
If this was the actual population of the world we live in-- I think we're pretty close to this point.
I believe, well, of course, nobody knows the numbers, unfortunately, because the model isn't perfect.
If the model was perfect, then we could just takes the census and we would know a and b.
But the model isn't that great.
But it's sort of, we're at a very interesting time, close to a very interesting time.
I believe that with reasonable numbers, this a over b might be maybe 12 billion.
And we might be, I think we're a little above six billion.
I think so.
So we're a little bit past it.
This is now.
This is halfway.
That's the halfway point.
It's perfectly symmetric.
It's called an S curve.
And many, many equations in math biology involve S curves.
So math biology often gives rise, with simple models, to a kind of problem we've had here with a quadratic term slowing things down.
Enzymes, all kinds of.
Ordinary differential equations are core ideas in a lot of topics, lot of areas of science.
OK. Do I want to say more about the logistic equation?
I guess I do want to distinguish one thing.
Yeah.
One thing about logistic equations and will of course come back to this.
OK. Let me look at that logistic equation.
Here's my equation.
So I've managed to solve it.
Fine.
Great.
Even graph it.
But let me come back to the question, suppose I just look at.
I can see two constant solutions, two steady states, two solutions where the derivative is 0.
So nothing will happen.
So in other words, I want to set this thing set to 0 equal to 0 to find steady solutions.
Steady means the derivative is 0.
So this side has to be 0.
So what are the two possible steady states where, if y of 0 is there, it'll stay there?
0
0.
Y equals 0 is one.
And the other?
A over b
Is a over b.
So steady equal to 0.
And I get two steady states.
Let me call them capital Y equals 0 because that's certainly 0 of, if we have 0 population, we'll never move.
Or setting this to 0, ay is by squared cancel y's divide by b a over b.
So the two steady states are here.
That's a steady state and that's a steady state.
Those are the only two in this problem.
You see how easy that was to find the steady states?
That's an important thing to do.
And then the other important thing to do is to decide, are those steady states stable?
When the population's near a steady state, does it approach that, does it go toward that steady state or away?
So what's the answer?
For this steady state, that steady state, y is a over b.
Is that stable or unstable?
So I'll write the word stable.
And I'm prepared to put in "un," unstable, if you want me to.
This is a key, key idea.
And with nonlinear equations, you can answer this stability stuff without formulas.
Without formulas.
That's the nice thing.
And then that comes in a later class.
But here's a perfect example.
So do we approach this answer or do we leave it?
We approach it, the solutions.
This is stable, yes.
And here's the other stationary point, capital Y.
The other steady state is that nothing happens.
So now if I'm close to that, if y is a little number, like 2, will that 2 drop to 0, will it approach this steady state, or will it leave it?
Leave it
Leave it.
So this steady state is
Unstable
Unstable.
Unstable.
Right.
Right.
With linear equations, we really only had one steady state, like 0.
Once it started, it took off forever.
Here, it doesn't go infinitely high.
It bends down again to that limit, that carrying capacity is what it's called, a over b. I guess I hope you think a nonlinear equation like got a little more to it.
Little more interesting, but a little more complicated, than linear equations.
Yep.
Yep.
Yep.
And similarly, the rocket equation, we could at the right time soon in the course, ask the same thing, a rocket equation was something like that.
What are the steady states?
Are they stable?
Are they unstable?
Can you find a formula?
Here.
This.
We got a formula.
And there are other nonlinear equations, which we'll see.
OK.
I could create more separable equations, but I guess I hope that you see with separable equations, you just separate them and integrate a y integral and a t integral.
Is that OK any question on this nonlinear separable stuff?
Differential equations courses and the subject tends to be types of equations as can solve.
And then there are a hell of a lot of equations that are not on anybody's list, where you could maybe solve them by an infinite series, but not by functions that we know.
OK.
I'm ready to do the other topic for today.
It's the topic that I left incomplete on Monday.
So I'm staying with first order equations, but actually this topic is essential for second order equations.
So I'm going to topic two for today.
So topic two will involve complex numbers.
So we have to deal with complex numbers.
And the purpose of introducing these complex numbers is to deal with what we met last time when the right hand side, the forcing term, was a cosine.
Typical alternating current, oscillating, rotating, rotation.
All these things produce trig functions.
Maybe rotation is more of a mechanical engineering phenomenon.
Alternating current more of an EE phenomenon.
But they're always there.
And what was the point?
The point was we had some linear equation, and we had some forcing by something like cos omega t. Or it could be A cos omega t and B sine omega t. Either just cosine alone, or maybe these come together.
And then the solution was y equals some combination of those same guys.
In other words, what I'm saying is cosines are nice right hand forcing functions.
Fortunately, because we see them all the time.
But they do lead to cosines and sines.
I emphasized that last time.
If we just have cosines in the forcing function, we can't expect that there's any damping, we can't expect only cosines.
We have to expect some sines.
In other words, we have to deal with combinations of them.
And the question is, how do you understand cos omega t plus 3.
Or let me take a first example.
Example-- cos t plus sine t. That's a perfect example.
So what is omega here in this example that I'm starting with?
1
1.
So I just read off the coefficient of t is 1, 1 hertz here.
But we have got this combination.
And the question is, how do we understand that cosine plus sine?
Two very simple functions, but they're added, unfortunately.
And there's a much better way to write this so you really see it.
You really see this.
That's called a sinusoid.
And the rule that want to focus on now is that everything of that kind, of this kind, of this kind, of a cosine plus a sine, can be compressed into one term.
One term.
Of course, it's got to have two constants to choose because that had an a and a b.
This had an m and an n. This had a 1 and a 1.
But the term I'm looking for is some number R times a pure cosine of omega t, but with a phase shift.
So you see there are two numbers here to choose.
It's really like going from rectangular to polar.
Say in complex numbers, let's just remember the first fact about a complex number.
If the real part is 3, and the imaginary part is, let's say 2, then here's a complex number, 3 plus 2 i.
So this was the real axis.
This was the imaginary axis.
I went along 3, I went up 2, I got to that number.
There it is.
I plotted the number 3 plus 2 i in the complex plane.
And for me, that number 3 plus and so on, really saying something important.
And maybe it's not entirely new.
I'm saying something important about complex numbers, this is their rectangular form.
Something plus something.
That form is nice to add to another complex number.
If I added 3 plus 2 i to 1 plus i, what would I get?
4 plus 3 i
4 plus 3 i.
But if I multiply, multiply, 3 plus 2 i times, let's say I square it.
I multiply 3 plus 2 i by 3 plus 2 i.
What do I get?
If I do it with this rectangular form, I get a mess.
I can't see what's happening.
It's the same over here.
This is like having a 1 and a 1, with an addition.
This is like a polar form where it's one term.
OK.
So let me answer the question here and then let me answer the question there.
And then you've got a good shot at what complex numbers can do, and why we like the polar form for squaring, for multiplying, for dividing.
What's the polar form?
Well, I'm using that word "polar" in the same way we use polar coordinates.
What are the polar coordinates of this point?
They're the radial distance, which is what?
So what's that distance?
That's the R you could say.
It corresponds to this R here.
So I'm just using Pythagoras.
That hypotenuse is what?
Square root of 13
Square root of 13.
Thanks.
9 plus 4, square root of 13.
And what's the other number that's locating this in polar coordinates?
The angle.
And the angle.
What can we say about that angle?
Let's call it phi is-- what's the angle?
Well, it's some number.
It's between 0 and pi over 2, I'm sure of that.
What do I know about that angle?
I know that this is 2 and this is 3.
So that's telling me the angle.
Well, what is that really telling me immediately?
It's telling me the
Tangent
Tangent of the angle.
So the tangent of the angle is 2 over 3.
And the magnitude is square root of 13.
OK.
So those beautiful numbers, 2 and 3, have become a little weirder.
Square root of 13, inverse tangent of 2/3.
You could say, well, that's not so nice.
What was I going to do?
I was going to try squaring that number.
So if I square 3 plus 2 i, or if I take the 10th power of 3 plus 2 i, or the exponential, all these things, then I'm happy with polar coordinates.
Like, what would be the magnitude of the square?
And where will the square of that number, so I want to put in 3 plus 2 i squared, which I can figure out in rectangular, of course-- a 9, and 6 i, or 12 i, or 4 i squared, stuff like that.
It's not pleasant.
What's the magnitude, what's the R for this guy?
What's the size of that number squared?
Yes?
Say that again
13
13.
Right.
I just have to square this square root so I get 13.
And the angle will be, what's the angle for the square there?
I don't want a number.
I guess I'm just doing this.
R e to the i phi squared is R squared.
And what's the angle here?
E to the i phi squared is e to the 2 i phi.
It's the angle doubled.
E to the 2 i phi.
The angle just went from phi to 2 phi.
The lengths went from square root of 13 to 13.
Squaring, multiplying is nice with complex numbers.
Maybe can I before I go on and on about complex numbers, I should ask you, how many know all this already?
Complex numbers are familiar?
Mostly.
Correctly, with a wiggle.
OK.
I won't go more about complex numbers.
Let me come back to my question here.
Let me come back to the application.
So here it is with complex numbers.
Here it is with sinusoids.
And the little beautiful bit of math is that the sinusoid question goes completely parallel to the complex number question.
So you have an idea on those complex numbers.
We'll see them again.
Let me go to this.
So I want this to be the same as this, OK. Maybe I'm going to have to use a new board for this.
Can I start a new board?
So I want cos t plus sine t to be some number R times cosine of t 1.
I can see omega's 1, so I just put to 1 minus some angle.
OK. And I want to choose R and phi to make that right.
You see what I like about it?
This tells me the magnitude of the oscillation.
It tells me how loud the station is.
When I see cos t and, separately, sine t, or I might see 3 cos t and 2 sine t. 3 cos t is a cosine curve.
2 sine t is a sine curve shifted by 90.
I put them together, it bumps, it bumps, bumps.
Not completely clear.
It seems to me just beautiful that if I put together a cosine curve that we know, that starts at 0, with a sine curve that starts at 0, the combination is a cosine curve.
Isn't that nice?
I mean, you know, that sometimes math gets worse and worse whatever you do.
But this is really nice that we can put the two into one.
But you see, it's going to-- well, let's do it.
What would R and phi be here?
So I'll use a trig fact here.
A cosine of a difference of angles, so this is R times cosine t, do you member this?
This was the whole point of going to high school.
Plus sine t sine phi.
So now, how do I get R and phi?
I use the same idea that worked last time.
I match the cosine terms and I match the sine terms.
So the cosine t has a 1.
1 cosine t is R cos phi.
That's what's multiplying cosine t. And the sine t has a 1.
And that has to agree with R sine phi.
So I'm in business if I solve those two equations.
And well, they're not linear equations.
But I can solve them.
Of course, the one fact that you never forget is that sine squared plus cosine squared is 1.
Right?
So if I square that one, and square that one, and add, what will I get?
1 squared and 1 squared will be 2, on the left hand side.
On the right hand side, I'll have R squared cos squared, R squared cos squared, and plus R squared sine squared.
And what's that?
What's R squared cosine squared plus r squared sine squared?
R squared
It's just R squared.
So all that added up to R squared.
In other words, it's just like polar coordinates.
R is the square root of 2.
That's telling us the magnitude of the response.
Square root of 2.
You see, it's just like complex numbers.
It's like the cosine gave us a real part and the sine gave us an imaginary part.
And R was the hypotenuse.
And that's really nice.
So R is the square root of 2.
OK. Now, the angle is never quite as nice.
But how can we get something about an angle out of there?
All we could get in this case here was the tangent of the angle.
And I'll be happy with that again here because it's a totally parallel question.
How am I going to get the tangent of the angle?
What do I have?
From these two equations, I want to eliminate R. So how do I eliminate R?
What do I do?
Divide.
Divide.
I guess if I want tangent sine over cosine, I'll divide this one in the top by this one in the bottom.
So I take the ratio.
That'll cancel the Rs perfectly.
It'll leave me with 10 phi.
And here it happens to be 1.
OK.
So what have I learned?
I've learned that when these two add up together, they equal what?
R square root of 2.
You see how easy it is.
Square root of 2 came from the square root of 1 plus.
It's like Pythagoras.
Pythagoras going in circles, really.
Times the cosine of t minus.
And what is phi?
Its tangent is 1, so what's the angle phi?
Pi over 4
Pi over 4.
Right.
So that's the sinusoidal identity when the numbers are 1 and 1.
But you saw the general rule.
Let me just take it.
Suppose this is the output, and cos omega t plus n sine omega t. What is the gain?
What's the magnitude, the amplitude, the loudness of the volume in this when I'm tuning the radio?
What's the R for this guy?
What's this R?
If we just follow the same idea.
So if we have m times a cosine and n times a sine, what's your guess?
What's your guess for R, the magnitude?
I'm guessing a square root of what?
Yeah?
You got.
What is it?
n squared-- [INTERPOSING VOICES]
Plus n squared.
Way to go.
M squared plus N squared.
And the angle is like the phase shift.
I'm not great at graphing, but let me try to go back to my simple example.
If I tried to add up on the same graph cosine t, which would start from 1 and drop to 0, go like that, right?
Something like that would be cosine.
And now I want to add sine t to that.
So that climbs up to 1 back to 0, down.
And now if I add those two, this formula is telling me that it comes out neat.
Neatly.
That one plus that one is another sinusoid with height square root of 2.
If I had different chalk, I've got at least a little bit different.
But does it start here?
Of course not.
It starts here, I guess.
But it goes up, right?
Because this comes down, but this is going up.
All together, it's up to, where is the peak?
Where is the peak on the sum?
So I'm adding, everybody sees what I'm doing?
I'm adding a cosine curve and a sine curve.
And it goes up.
And where does it peak?
What angle is it going to peek at?
What's the biggest value this gets to?
[INAUDIBLE]
At pi over 4, it'll peak.
At pi over 4, it'll be the cosine of 0, which is 1.
It's height'll be the magnitude, the gain, square root of 2.
So it'll peak at pi over 4, which is probably about there, right?
Peak at pi over 4 and, I don't know if I got it right frankly.
I did my best.
That's the sum.
That right there.
The first key point is it's a perfect cosine.
The second key point is it's a shifted cosine.
The third key point is its magnitude is the square root of 1 squared plus 1 squared, or n squared plus n squared, or a squared plus b squared.
So that's the sinusoidal identity.
A key identity and being able to deal with forcing terms, source terms, that are sinusoids.
OK. Now, I'm going to take one more step since we have just like 10 minutes left, and let the number i get in here properly.
Get a complex number to show up here.
OK. Before I start on this, let me recap.
Let me recap today's lecture.
It started with nonlinear separable equations.
And a great example was the logistic equation up there, with the S curve.
That took half the lecture.
The second half of the lecture has started with things real with sinusoids that are combinations of cosine and sine and has written them in a one term way.
And now I want to get the same one term picture from using complex numbers.
OK. OK. And everything I do would be based on this great fact from Euler that e to the i omega t. The real part is cosine omega t. And the imaginary part is sine omega t. That's a central formula.
Let me draw it rather than proving it.
Let me draw what that means.
I'm in the complex plane again.
Real part is the cosine.
The imaginary part is the sine.
That number there is e to the i omega t because it's got that real part and that imaginary part.
And what's its magnitude?
What's the R, the polar distance for cos omega t plus, for this number, which is for this number?
What's the hypotenuse here?
Everybody knows
1
1.
Hypotenuse is 1.
Cos squared plus sine squared is 1.
So e to the i omega t is on a circle of radius 1.
That's the most important circle in the complex world, the circle of radius 1.
And all these points are on it.
And their angles are omega t. And as t increases, the angle increases, and you go around the circle.
You've seen it.
Physics couldn't live without this model.
OK.
So that's basically what we have to know.
And now, how do we use it?
Well, the idea is to deal with the equation.
Like, the equation I had last time was dy, dt equals y plus cos t. That gave us some trouble because the solution didn't just involve cosines, it also involved sines.
Yeah.
So I want to write that equation differently, in complex form.
And this is the key point here.
So I'm going to look at the equation dz, dt equals z plus e to the i t. Well, I'll make that cos omega t just to have a little more, the units are better, everything's better if I have a frequency there.
Units of this are seconds and the units of this are 1 over seconds.
Now, question.
What's the relation between the solution z to that complex equation and the solution y to that equation?
Of course, they have to be related, otherwise it was stupid to move to this complex one.
My claim is that complex equation is easy to solve.
And it gives us the answer to the real equation.
And what's the connection between y and z?
So y's the real part
Y is, exactly, say it again
The real part--
Of z. Y is the real part of z.
So that gives us an idea.
Solve this equation and take it's real part.
If I can solve this equation without getting into cosine and sine separately and matching, I can stay real.
I solve the equation by totally real methods up to now.
Now I'm going to say, here's another approach.
Look at the complex equation, solve it, and take the real part.
You may prefer one method.
You may like to stay real.
In a way, it's a little more straightforward.
But the complex one is the one that will show us, it brings out this R, it brings out the gain, it brings out the important-- engineering quantities are important, if I do it this way.
Now, I believe that the solution to that is easy.
Actually, it is included in what I did last time.
It's a linear equation with a forcing term that's a pure exponential.
And what kind of solution do I look for?
I'm looking for a particular solution.
If I see an exponential forcing term, I say, great.
The solution will be an exponential.
So the solution will be sum.
Z is sum capital Z e to the i omega t. Plug it in.
What happens if I plug that in to find capital Z, which is just a number?
Right.
This is my method.
This is a linear equation, with one of those cool right hand sides, where the solution has the same form with a constant, and I just have to find that constant.
So I plug it in.
Dz, dt.
Take the derivative of this, z i omega will come down.
E to the i omega t. Z is just this, z to the i omega t. And this is just 1 e to the i omega t. So I plugged it in, hoping things would be good.
And they are because I can cancel e to the i omega t, that's the beauty of exponentials, leaving just a 1 there.
So what's capital Z?
What's capital Z then?
I've got a z here.
I better bring it over here.
And I've got the 1 there.
I think the z is 1 over.
When I bring that z over here, do you see what I'm getting?
It's all multiplying this e the i omega t. It's a number there.
Z times i omega and comes over as a minus z.
What do I have multiplying z here?
I see the i omega.
And what else have I got multiplying the z?
Minus
A negative 1 because it came over with a minus sign.
Done.
Equation solved.
Equation solved.
Complex equation solved.
So the point is, the complex equation was a cinch.
We just assumed the right form, plugged it in, found the number, we're done.
But there's one more step, which is what?
Take the real part.
So I have to take the real part of this.
So the correct answer is y is the real part of that number, 1 over i omega minus 1 times e to the i omega t. I'm tempted to stop there, but just with a little comment.
How am I going to find that real part?
And what form will it have?
What form will that real part have?
Yeah, maybe just to say what form will it have?
The real part, it's going to be a sinusoid.
But I have a complex number multiplying this guy.
The real part is going to be exactly of the form we-- well, of course, it had to be the form because that was another way to solve the equation.
It's going to be some number.
And I'll call it g, for gain, times the real part.
And so the real part will be a cosine.
Yeah, it's just perfect.
A cosine of omega t. And there'll be a phase.
Yeah.
i haven't taken that step fully.
I got to that fully.
And then I said that that, if I use some complex arithmetic, will come out to be this.
And you see the beauty of that answer, which was way better than a sum of sines and cosines.
We see the gain.
We see the amplitude.
And we see the phase shift.
Yeah.
So I don't know, that would be a good exercise in complex numbers.
Find g and find phi, in taking the real part of this thing.
Yeah.
It's a pure exercise in using complex numbers.
I don't feel like doing it today.
If we do it, you just see a lot of formulas.
Here, you see the point.
The point was that the complex equation could be solved in one line.
We just did it.
But that left us the problem of taking the real part.
That was the e to the i omega t there.
Left us the problem of taking the real part.
And that's a practice with complex arithmetic.
So you've got the choice.
Either stay real-- sign plus cosine.
And then use the sinusoidal identity, polar form.
Or get the polar form from here.
Same answer both ways.
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This week is my second pair of lectures.
Last week the two lectures were about first order differential equations, and this week second order.
Those are the two big topics in differential equations.
Let me start with most basic second order equation.
We see the second derivative and the function itself, and we don't see yet the first derivative term.
This is the nice case, when I just have y double prime and y.
In general, I-- I'm taking constant coefficients today.
Because if the coefficients depend on time, the problem gets much, much harder now.
So let's stay with constant coefficients, meaning we have a mass, for example, we have a spring.
The stiffness of the spring is k, the mass is m, and the y, the unknown displacement, tells us the movement of the mass.
The classical problem.
You will have seen it before.
Because you have an exam this afternoon, I wanted to start with things that-- they are about second order equations, but they're still close to the exam idea, particularly the idea of exponentials.
With constant coefficients, that's the fundamental message.
Exponentials in, exponentials out.
But it's not quite so clear when we had first order, y prime equal ay, we knew that the exponent was a.
The solution was e to the at.
Now we've got second derivatives coming in, and it won't be so much e to the at type thing.
Either the at was growth for a positive, decay for a negative.
Now we're going to see oscillation.
It's still exponentials, but oscillation.
Things going up and down, things going around.
Harmonic motion, you call it.
Sines and cosines.
And sines and cosines connect to complex exponentials.
So that instead of e to the at-- so now oscillations-- they're going to be coming from e to the i omega t. In other words, instead of an a, we're going to have an i omega.
Or, if we like to stay real, we can stay with cos-- cosine-- and sine.
And actually, I've written two real guys there, so I better have two complex ones.
And it will turn out to be plus or minus.
There are two frequencies there.
Plus i omega, minus i omega, and they turn into cosine and sine.
So in this case, with no damping term, we can stay entirely real without creating any problems.
We can work with cosines and sines.
The first question is, what's omega.
What is the frequency of oscillation.
And of course, another similar picture would be a pendulum, a linear pendulum, swinging side to side, keeping time, because that frequency will stay constant.
Always I'll start with zero on the right hand side.
Just look at these equations.
Constants there.
I'm looking for solutions.
And I'm looking for null solutions, looking for the natural motion of this spring, the natural up and down motion of this spring.
Classical problem.
Won't be brand new, but it's the right starting point for the full second order equation.
It'll get a little complicated on Wednesday, when damping gets in there.
The formula got a little messy, because you've got a mass-- you've got an m-- and a k, still, but you also will have a damping constant.
Then complex numbers really come in.
Here they're optional.
So this is my equation to solve.
Because we don't have a first derivative, a cosine will solve that.
So let me look for-- I could look for exponentials.
Maybe I should do that first, look for an exponential solution.
Yeah, that's a good idea.
And let me not jump ahead to know that the exponent has that i omega form.
Let me discover it.
So I look for no solutions-- because I have that zero there-- no solutions of the form e to the st, some exponent.
Plug it in.
That's the message with constant coefficients.
Look for exponentials, substitute them in, discover what s will be.
So let's just do that.
This is the most basic step.
For null solutions will be exponentials, I substitute into the equation.
I get ms squared from two derivatives.
It will bring s down twice.
This is just ke to the st, and I'm looking for null solutions.
Zero.
No forcing.
So this is undamped, unforced.
Undamped, unforced.
Natural motion.
What do I do now?
Plugged in an exponential, got this equation.
And the beauty is that the exponentials cancel.
An exponential is never zero, so I can safely divide by it.
So I cancel those, and I get ms squared plus k equals zero.
The key equation-- and it's so simple-- it's just we're doing algebra now.
The calculus, the derivative we took when we plugged it in, but now it's an algebra question.
And of course, solving that system is easy.
There's no s term, no damping term.
So the frequency, s, is-- put k on the other side, divide by n. s is-- s squared, let's say-- is k over m-- is minus k over m. Critical point.
That tells me, with that minus sign there, that s is an imaginary number.
A complex number has a real part and an imaginary part.
In this case, all imaginary.
No real part at all.
It's natural to think-- s is the square root of that, so I'm going to write-- everybody writes-- s, the frequency s, is i omega.
So if I plug that in, I have omega squared equal k over m. i squared and the minus 1 deal with each other.
So the frequency omega-- here is the great fact-- square root of k over m. That's-- yes?
What difference does having imaginary parts to answer affect the oscillation?
To-- OK. Oscillation, just pure oscillation-- which is what we would see here with no damping-- is the frequency, e to the-- the solution-- the displacement, I could write-- the displacement up and down, y, will involve e to the i omega t, and e to the minus i omega t. We've got second order equation.
Let me just go back to that key point.
When we have second order equations, we look for two-- we expect and we want and we need two-- solutions.
There will be two.
I didn't put it here, and I should.
s, the frequency, is plus or minus i omega, because in both cases when we square, it comes out right.
So we get two frequencies, and here they are.
So let's see how to answer your question.
The presence of this i is only telling me that, essentially, I've got sines and cosines.
That's really what-- when it's a pure imaginary number-- I would call that a pure imaginary number, there's no real part at all-- then equally cos omega t and sine omega t. I can now, if I want, go real.
I can say, OK, these were the general null solutions.
Let me put this down, then.
The null solution-- I'm looking only right now at null solutions-- is some combination of e to the i omega t and e to the minus i omega t. That's what we got from plugging in e to the st, discovering that s was an imaginary number, and we got these guys.
But equally-- equally-- yn is a combination of cos omega t and sine omega t. And maybe you'll like those better.
I think everybody practically likes those better.
Do you see that these guys are the same as these guys?
The c's are a little different because, well, we know that we can switch from one to the other.
We remember that basic fact that e to the i omega t is cos omega t plus i times sine omega t. You're used to maybe seeing that omega t as theta, e to the i theta is cos theta plus i sine theta.
And e to the minus i omega t, of course, is cos omega t minus i sine omega t. I hope you won't think I'm filling the blackboard with formulas, because I'm really just writing down-- well anyway, they're beautiful formulas.
So if I have these guys, then I have these and vice versa.
If I have cos-- how would you write cos omega t using the exponentials?
I want to just see totally clearly that I can go back and forth between complex imaginary exponentials and cosines and sines.
So how would I, I want to go in the opposite direction and write the cosine and the sine as combinations of these, just to show if I've got combinations of one, I've got combinations of the other.
Combinations of these are the same as combinations of those.
So what is cos omega t in terms of these guys?
[INAUDIBLE] some of them divided by 2?
Exactly.
If I add those two, this part cancels.
I've got two of these, so I have to divide by 2, as you say.
It's a half of the first plus a half of the second.
And how about sine omega t?
Sine omega t is always slightly more annoying, because it's the one-- it's the imaginary part that brings in an i.
What would be the same formula?
How could I produce sine omega t out of that?
Yes?
The difference divided by 2i
Yes.
If I take the difference, that'll cancel the cosines.
So I'm going to take e to the i omega t minus e-- minus e to the minus i omega t. Take the difference.
But then I've got 2i multiplying this sine.
Up here I had a 2, but now I've got-- when I take the subtract, these i's are in there, so I divide by 2i.
So this just tells me that I can go either way.
Next time, we'll see what happens when there is damping and there are complex numbers instead of pure i omegas.
We're golden here.
We've found the great quantity with the right units.
The right units of omega are 1 over time.
Actually the units are radians per second, would be the typical appropriate unit.
Radians per second.
I'll use the word frequency for that, but there's another definition of frequency, cycles per second.
I just want to think about steady motion around a circle.
So this tells me how many radians per second.
And if this is 2pi-- if omega happened to be 2pi-- then I would go once around the circle.
If omega was 2pi, then when t reached one, I would be around the circle.
Let me draw a circle in a minute.
So there's a 2pi here hiding behind the word radians.
And in many cases, you'll want also a definition in cycles per second.
So f is omega divided by the 2pi, and that's in cycles per second.
Full revolutions per second.
And that's hertz.
I think I misspoke last time in confusing these two, so let's get them straight here.
There's no complicated math in here, it's just a factor 2pi, but of course that factor is important.
So a typical frequency in everyday life would be like f, 60 cycles per second, 120pi radians per second.
So I'm going around in a circle.
Now I'm ready to have initial conditions.
This connects, again, to the afternoon exam.
We found the general solution with some constants, like here.
Let's keep that real form.
And now those constants get determined by the initial conditions.
Conditions plural, because we have an initial position, like I stretch it-- maybe I stretch it and let go.
Maybe I stretch the spring and then I let go.
What happens?
By stretching it, I'm giving it an initial displacement.
And I'm giving it zero initial velocity, because I stretched it and just let go.
Another possibility would be to strike it.
If I hit that mass, that would be a different initial condition.
What would be the initial condition if it's sitting there in equilibrium quietly minding its own business and I hit it?
Then I've given it an initial velocity, with initial displacement zero.
So those would be the two extreme possibilities.
Pull it down, let go, or strike it when it's sitting in equilibrium.
Anyway, we've got two initial conditions.
You see why-- y double prime is showing up because essentially we've got Newton's Law.
This is Newton's Law.
Mass times acceleration is equal to minus ky-- that's the, with the minus sign, and the all-important minus sign, that's the acceleration.
That's a force, sorry.
Mass, acceleration, this thing with a minus sign is the force, and the force is pulling back.
If y is stretching, the force is restoring.
Let me just go ahead with what you know.
The initial conditions.
And I want to solve my double prime plus ky equals 0.
So I'm still talking about the unforced with given y of 0 and y prime of 0.
Just think for a moment.
Could you do that?
This is the most basic second order equation.
We know what the solutions look like.
Let's do this one in a box, cosines and sines.
We know what omega is.
Omega had to be square root of k over m. Then the equation was solved.
All I've got left is to get c1 and c2.
All I have left is to match-- choose c1 and c2 to match the two initial conditions.
So let me just do that.
What are c2 and c2?
At time zero, I have to match an initial displacement.
So at time zero, this is a 1, cosine of zero is a one, and that's a zero.
So at t equals 0, I have y of 0.
The displacement matches c1 times cosine of omega t, which is a 1, plus c2 times 0.
I'll put it in there plus c2 times 0.
C1 cos 0 plus c2 sine 0.
I've learned c1.
And also-- what do I do next?
I want to get c2.
And where is c2 coming from?
Now I would like to know what's the coefficient of the-- the initial conditions are supposed to determine that coefficient.
It'll be that initial condition that determines it.
y prime of 0.
The initial velocity should match the derivative.
OK, so what's the derivative?
y prime.
So the derivative of the cosine will be a sine.
And that will disappear at t equals 0.
The derivative of this sine will be a cosine with a factor omega.
So I'll have y prime of 0 will be the c2 omega cos of 0.
Which is what?
That tells me c2.
You could do all this without my pointing the way.
I'm solving this equation.
I have the solution in general form with two constants.
Now I'm determining those constants, and cosine and sine just determine them perfectly because cosine is 1 and sine is 0 at the start.
So we've got the answer.
The solution is y of t is c1 is y of 0 cos omega t, and c2 is-- now you'll notice, c2 is y prime at 0 divided by omega.
y prime of 0 divided by omega sine omega t. There we go.
Finished.
Finished.
Unforced problem solved.
Everybody in this room could get to that point.
Let me make some comments about that.
It's a combination of cosine and sine.
They're both running at the same frequency, omega.
I'm going to give a special name to that frequency, omega, this famous formula, all-important.
Lots of physics in that formula.
I call that the natural frequency, because the next step will be to drive the system by a driving frequency, which would be different from omega.
So we need to-- we've got 2 omegas.
Actually when I first wrote the book I thought, we've got to keep these two separate.
Everybody has to keep them separate.
My first attempt was to use little omega and big omega for the two.
I concluded after looking at it for a while that it was better to be more conventional.
People had figured out a good way to do it.
And the good way is to call this the natural frequency and put a subscript, m. So all the omegas that you see on this board should be omega n. I can change them all, but let me just change it here.
I'll change omega n. So that's omega n. We've only got that one omega right now, because we don't have a driving term yet.
So natural frequency has the advantage, which kind of made me smile, that the n stands for natural, and everybody calls it the natural frequency.
And n also stands for null, and we're talking here about the null solution, because there's no forcing.
So I could have subscripts on all the y's.
Eventually I'll need subscripts on the y to separate what we've done.
This is really yn of t, the null solution.
Good?
Now we could take one more step.
This is a combination of cosine and sine.
And we learned last time that that could be put in a polar form, but I don't plan to do this.
Let me just say I could do it.
This would be some amplitude, some gain-- maybe g for gain-- no, a for amplitude is good-- times-- what is this second optional form, which I'm just going to write here, say that we could do it, remember a little about it, but not make a big deal-- what is it I'm after here?
I'm looking to write this combination of cosine and sine, which is two oscillations, a cosine curve and a sine curve, but with the same frequency.
Then I can combine them into a single cosine, a single cosine of omega nt.
But now what else have I got in this form?
There's a phase shift, minus phi.
Thanks.
So there's an a phi, two constants, or there's y of 0, y prime of 0, two constants.
Let me not write again the formula for a or for phi, I don't plan to do anything with it.
It just could be done.
In other words, what we've done so far is just to see that the single spring oscillates with the frequency omega n. That's really what we've done.
A single spring oscillates with a frequency omega n. Saying that makes me think, let me look ahead to the linear algebra part of the course.
So where is linear algebra going to come in?
It's going to come in for a system of springs.
When I have another spring.
Can I draw another spring and another mass and another spring?
Say six springs, six masses.
Then-- and they could be different k's, different m's, or not.
Then we've got six displacements-- six differential equations-- coupled together, because the whole system is coupled together.
So what happens at that point?
That's the point where linear algebra, where matrices are coming in.
You want to see what's the point of matrices.
It's not a separate course by any means.
It's a most necessary part, because a single spring happens in reality but also systems today are coupled.
Big, actually, there are many, many things.
You have an electric circuit with thousands or tens of thousands of elements.
You have a coupled system with many gears, many oscillations going on.
So we need matrices at that point.
Can I even just add one more word about the language?
When we had-- here we have a frequency of motion for one spring.
What are we going to have for two springs or six springs?
The motion will be a combination of six different frequencies.
And so you'll see that it's a much more interesting, much more not so simple motion.
A combination of six pure frequencies.
And those frequencies are determined from the six eigenvalues of the matrix.
I'm just using that word looking ahead.
We will have a 6x6 matrix to describe the coupled system.
That matrix will have six eigenvalues.
It will tell us six natural frequencies, and our solution will be a combination of all six oscillations.
Here, it's 1.
Here it's 1.
That spring is not there.
So the problem we've solved now is the fundamental, basic problem, and I have to-- next step is forcing.
I now want to add a force that drives the motion.
In general, it could be any function of time.
Calling it f of t. So that's what I'm going to put in now.
But in reality, very, very, very often f of t is also a simple harmonic motion.
It's also a cosine.
But at a different frequency, at a driving frequency.
So I'm going to-- the next equation to solve is to put in cosine-- let's stay real for now-- at another, driving frequency.
At a driving frequency.
And of course, it could have an amplitude.
But let me take that amplitude as 1 to keep things simple.
So now I'm talking about forced motion.
Can we solve it?
How can we solve this equation?
Let me take out the 0 or-- take out the 0-- equals cosine omega t. With a different omega.
If the two omegas were the same, if the driving frequency is the same as the natural frequency, the formulas have to be slightly adjusted.
There's still an answer, but it's a case of resonance and you have to look separately.
But let's say, no.
Let's say omega d is different from omega n. How are you going to solve this?
I have to think myself.
How do I solve that.
Let's start a fresh board.
my double prime plus ky equals cosine of omega dt-- or often, I won't put the d. I don't have to put the d anymore.
Omega will now represent the driving frequency, because I've got omega n, the natural frequency, as the square root of k over m. What am I looking for now?
I found the null solution.
I'm looking for a particular solution.
I'm trying to keep the whole thing systematic.
Null solutions are now dealt with.
Took a little more time than just ce to the at for first order equations, because we've now got a two-dimensional collection of null solutions, but we've got them.
Now I'm taking a forcing term.
So I'm looking for a particular solution.
I'm looking for any solution to this equation.
I'm looking for a particular guy.
What do you suggest?
Again, it's a neat problem because of that particular forcing term, a cosine, an oscillation.
So I'm going to look for yp is some gain times [INAUDIBLE].
This is the next and, fortunately, a highly, highly important case, in which the particular solution has the same form as the forcing term.
It's just a multiple of the forcing term.
That's best possible.
That's best possible, is to have the forcing term reveal to me-- the forcing term immediately reveals a particular solution.
Once I know what I'm looking for, what do I do?
Substitute it in.
So I substitute that particular solution in here.
And notice everything is going to be a cosine, mgy double prime.
So what do I get when I plug this in for that guy?
I want to-- you can do it quickly, but let's stay together and do it together, because we can with this case.
What happens when I plug that in and take its second derivative?
I get the g. And then what's the second derivative?
[INAUDIBLE]
We have a negative, because two derivatives of the cosine bring out a minus omega d, will come out twice.
And I'll keep writing omega d for a moment, but then I'll give up on the d. Cosine of omega dt.
And then k times this, g cosine of omega dt, equals the forcing term, cosine of omega dt.
It worked.
This is one of that small family of nice functions where the solution has the same form as the function.
Actually that list of what you could call best possible forcing functions, where the form of the forcing function tells you the form of the solution.
That's a small family.
But it's fortunately a very important one.
Cosines, sines are included, and we'll see all the other guys that are included.
Most forcing functions we couldn't just assume that the solution had the same form.
It's only these nice ones.
But cosines are nice.
So what do I do now?
Everything is multiplying cosines, so I just look at-- I have minus m omega squared g-- g is going to factor out-- minus m omega squared and a k times g. Let me remove that off for the moment.
I'm canceling cosine omega, so my right hand side is 1.
That's it.
We looked for a solution with that simple format, and we found it.
Now we know g, the gain.
So the solution is-- this is g is 1 over k minus m omega squared times cosine of omega t. And omega is omega d. Omega is omega d now.
Does that look good to you?
This is the periodic solution going at the driving.
This is what the-- this g is the gain, the driving force.
The driving force is 1 times cosine omega d, then that 1 gets multiplied by this number.
This is, you could say, the amplifying factor.
I guess frequency response would be the right word.
Can I bring in that word, response, again?
Response is a word for a solution.
It's what comes out.
When the input is this, a pure frequency, the output, the response, is a pure frequency-- same frequency, of course-- multiplied by that.
That is the frequency response factor.
Notice we could write that a cool way, by remembering that omega squared-- that's wrong as it stands.
What have I forgotten in writing k minus m omega squared in that denominator?
I forgot a subscript, which is n. Which is n. This is n. This is-- is that right?
No.
Is it?
Or is it d?
Maybe I didn't make a mistake.
Is it d?
You're seeing a kind of critical moment.
Which is it?
[INAUDIBLE]
It's d, isn't it?
Yeah.
It's d. Sorry.
It's d. But when I see this and remember what omega n squared is-- omega n squared is k over m-- I can see that I can get an omega-- I can use this in here to make it even more interesting.
So it'll be equals-- let me get this box ready-- cosine of omega dt divided by-- now I just want to rewrite that.
I want to take out an m. I'm going to write this as m times k over m. m times k over m. Safe to do that.
Now I have a factor, m, that I can bring out.
And what is m multiplying?
That's the neat thing.
What is m multiplying?
k over m is-- omega n squared.
And this is minus m omega d squared.
Minus omega d squared.
That's pretty terrific.
The gain is this multiplier, 1 over m, times that.
And we see that the gain is bigger and bigger when the frequency is near the natural frequency.
And of course everybody has seen the pictures of that bridge-- wherever the heck was that bridge?
Somewhere in the Northwest, I think.
You know the bridge I'm talking about?
[INAUDIBLE] Tacoma, Washington
Yeah, I think Tacoma, that's right.
The Tacoma Narrows Bridge.
Right.
Tacoma, Washington.
Where the natural-- when you build a bridge, you've built in a natural frequency.
And then when traffic comes, it's doing a driving frequency.
And if you haven't got those two well-separated, you're in trouble, as this shows.
Or similarly, when an architect designs a skyscraper, there's going to be a frequency of oscillation, a natural frequency, at which that skyscraper swings.
And then there's wind.
Actually I talked yesterday to the-- by chance, the math department is not a very party-going department, but once a year we let it out.
And so we had our party at Endicott house out in the suburbs, and all the usual people-- that's all the professors I know-- came, of course.
But also, there was a really cool person.
He's the key architect for Building 2.
You've noticed that Building 2 is under wraps and we're moved out.
And we move back in January 2016.
So we've been out a year and a quarter and we have another year and a quarter to go.
It's going to be cool.
And you may say, well, who cares.
But the key point is Building 1 is next, and Building 1 is going to have the same cool addition of a fourth floor.
We're putting in a fourth floor, which all the-- Buildings 3, 4, 5, 6 go up to four, but Buildings 2 and 1 stopped at the third floor.
But there's a lot of space up there under the roof.
And they've discovered they could put a fourth floor up there.
Here was one interesting thing, though.
These buildings that we're sitting in are sinking.
You know that MIT was built on marshy land, just the way the Back Bay-- which is like the greatest idea in the history of Boston, the Back Bay and the dam that makes the Charles River beautiful-- was built by bringing in trainloads of earth from Needham.
So whole mountains and hills in Needham have come into Boston and come here.
So anyway, we're sinking.
You may say something like 3/16 of an inch a year is not something to worry about, but now it's been more than 100 years that these buildings have been here.
Anyway, not good to sink faster.
So the weight had to be controlled.
So by putting in a fourth floor, that put in a lot a new weight, and faster sinking, probably by some formula here.
Probably there.
So the weight had to get subtracted out.
It turns out that the ceiling, the roof to Building 1-- Building 2 and no doubt to Building 1-- was more than a foot thick of concrete.
Really heavy.
And some more asbestos probably, which we don't want to think about.
That's much reduced.
A whole lot of weight came out of the roof.
I think they probably did the calculation right, so we won't get rain coming through, but it won't weigh as much and the fourth floor is acceptable.
All this was a big decision by MIT to pay for that, or to raise money and pay for the new fourth floor.
But it's going to be fantastic.
And it'll be fantastic in Building 1 also.
So all that is discussion of that formula.
That's the frequency response, this factor to frequency, omega d, or omega, is this factor.
I guess I should say something about resonance.
What happens when that formula breaks down?
When the driving force equals the natural frequency, then we're dividing by 0, and something is different.
The formula isn't right anymore.
What enters in the formula-- let me just tell you what enters, and then we'll see it in a simple example.
When I have this repeated thing, two things are equal, what tends to happen is a factor, an extra factor, t, appears.
So an extra factor, t, will appear in the case omega n equal omega d. The solution, y, will be some factor, I'll still call it g-- no, I don't want to call it g, let me call it a.
There'll be a factor, t, times cosine of omega t. So in this case, there's really only one frequency.
We're driving it.
So the oscillation grows.
As you know, when you push a child on a swing, the whole point of pushing that child is to push at the natural frequency.
You wait for this swing to swing back naturally and you drive it again with that-- at that-- maintain that frequency.
And of course you see the amplitude-- the child swing higher and higher.
Presumably you stop pushing before disaster for that child.
But that's a case of resonance.
And it's what happened in the Tacoma Narrows Bridge, and there was nothing to-- nobody stopped, traffic just kept coming.
The movie is amazing, because there's one car that shows up after it's already swinging wildly, some crazy person still driving across.
And you might think, OK, that's ancient history.
But you know the bridge in London, the pedestrian bridge, the Millennium Bridge-- it's just a walking bridge across the Thames-- a big feature of modern London, and it had the same problem.
It was swaying.
People could not walk across.
They couldn't keep their balance.
So they had change it.
So it's not trivial to anticipate.
So now we've solved it-- we've solved the the null equation with no force, and we've solved the driving force equal to a cosine.
And of course, we could do a sine.
What other driving force should we do?
I think we should do a delta function.
I think we have to understand the fundamental solution is the case when, if we can solve it-- there's always this general rule, if we can solve with a delta function, that will give us a formula for every driving force, because every function is some combination of delta functions.
So if we could do it with a delta-- really the great right hand sides are-- well, cosines and sines I'll include as great right hand sides.
Those are the exponentials in disguise.
So the great right hand sides are really exponentials at different frequencies and delta functions.
Delta of impulses.
So now I want to find the impulse response.
That's the next-- that's really a job.
At this point, in these last 20 minutes when I solve my double prime plus ky equal a delta function-- well, what I was going to say was I'm now taking you to something that you won't see on the exam this afternoon.
But maybe you will.
Delta function, right hand side.
I haven't seen it yet.
Or I haven't looked recently.
You won't see second derivatives, I guess.
So what is it?
So now this is of the form with an f of t, a very special f of t. And that very special f of t makes that extremely easy to solve.
That's really my point here, is it it's going to be a cinch to solve that, and we practically have done it already to solve that with a delta function.
And the reason is sort of physical.
We have here our spring.
And what am I doing with that force?
I'm hitting the mass.
I'm striking the mass.
Let me say, and I'll write it on the board, the point I want to make about this.
That point is that this equation with a delta function force starting from 0-- say, y of 0 equal y prime of 0 equals 0, let's give it starting from rest-- it starts from rest by hitting it.
And that hit, that impulse, is in no time at all.
It's not stretched out.
It's hit over one second.
So this has the same solution.
This is the beauty.
This is why we can solve it so easily.
Same solution as-- let me write it and see what you think-- as my double prime plus ky equal 0.
We know how to solve those.
With-- it's still, when I hit it-- when I hit it, what happens in that split second?
In that split second, it doesn't have time to move.
It doesn't move.
It still has y of 0 equals 0.
But in that split second, we've given it a velocity.
We've given it a velocity.
And that velocity will be y prime.
The initial velocity is 1-- because here I had a 1-- over an m. We have to have the units right.
So here's a point, and we will stay with it.
We'll come back to this point next time.
Maybe the first thing for you to take in is the fact that it's such a nice thing.
We have this equation with this mysterious delta function, and I'm saying that the solution is the same as this equation with no force, but starting from a mass.
I'm tempted to take an example to make this point.
Let me take an example where the whole thing is a lot simpler.
y double prime equal delta of t. I've taken the spring away, so the k is gone, the mass is 1.
What's the solution to y double prime equal delta of t?
If we concentrate on this example, we're good for today.
So my point is the same solution as-- now, what's the other problem?
I'm just repeating here, but making it simple by taking k equals 0 and m equal 1.
So the same solution as y double prime equals 0, with y of 0 equal what, and y prime of 0 equal what.
I just wanted to repeat here what I've said there, and then we'll solve it and we'll see that it's all true.
If I look for a solution to y double prime equal delta starting from 0-- this was starting from 0-- if I say that's the same as this, what should y of 0 be here?
Zero
Zero, right.
It hasn't had time to move.
It hasn't had time to move.
But in that instant, what happened to y prime?
It jumped to 1.
That's right.
That's right.
Exactly.
Now just solve that equation for me.
Solve this example for me.
Suppose y double prime-- yeah.
Here we go.
What's the solution if y double prime is 0?
What are the solutions to y double prime equals 0?
[INAUDIBLE]
Constant and linear.
a plus bt, right, have second derivative 0.
Now what's the solution that starts from 0 that kills the a and has slope 1?
What's the answer to that question?
[INAUDIBLE]
t. t. The solution to this equation is a ramp.
It's zero everything in this course, is zero up until time 0.
At time 0, in this example, all the action happens.
Everything happens.
And what happens is it gets a velocity of 1, and the solution is y equal to t. y is 0 here, of course.
At that point, that's the key point, t equals 0, right there-- it gets a slope.
We don't have a step function.
There's no jump in y.
The jump is in y prime, the y prime the velocity jumped from 0 to 1.
That's exactly-- I think when I introduced delta functions and drew a picture.
What is the derivative, the first derivative, y prime, for that guy?
Let's just review, because this is what we've seen already.
The first derivative is--
[INAUDIBLE] step
A step.
And the second derivative is delta.
The second derivative of this is the first derivative of a step.
The derivative of a step is 0 everywhere except at the step, at the jump when it jumps to 1.
So that's the solution in this example.
And now to end the lecture, let's solve it in this example.
Again, let me just say-- why do I like this forcing term?
Mathematically, I like it because, if I can solve that guy-- as we're doing, we are solving it-- if I can solve that one, I can solve all forces.
Over here, I could solve when I had a very happy f of t, a perfect f of t, where I could guess the answer and push through.
Now with a delta, I can build everything out of delta functions.
That's why I like it mathematically.
Why do I like it physically?
Because it's a very physical thing to have an impulse.
That happens in real time, in real things.
And by the way, let's just, before I write down any more formula, what would-- I would like to be able to solve it for a step function.
[?
Heavy thud.
?]
I would like to be able to do that one.
I'm going to have to erase something, or I'll write it right above just for the moment.
I would also like to solve my double prime plus ky equal a step function.
So I would call the solution, y, the step response.
And what would be a step function start?
A step function start would be like turning a switch.
Suddenly things happen.
That's forcing by a step, so I'm looking for the step response.
And how do you think these two are related?
I look at the relation at the right hand sides.
What's the relation of this step to the delta?
Yeah?
One's a derivative
One's a derivative of the other.
And we've got linear equations.
So the right hand sides.
The step response, y step, and the delta response, y delta-- I'll use a different letter for this because it's so important.
One is the derivative of the other.
The great thing about linear equations is we have linear equations, differentiation, integration.
Those are linear operations.
The step function is just like a steady-- anyway.
I was going to-- I won't-- is the integral of the delta.
Step function is the integral of the delta, so the step response is the integral of the delta response.
I guess to finish the lecture, why don't we solve this problem, which looks tricky because it's got a delta.
Instead, we'll solve this problem, which doesn't look tricky at all.
It's exactly what we started the lecture with.
Zero forcing and some initial conditions.
So let me just finally make space for the big deal from today's lecture, which would be the fundamental solution with a force by a delta.
I'm just going to write down the answer when you tell me what it is.
What's the answer to that?
What's the solution to this second order constant coefficient unforced equation with those initial conditions?
We probably had it here.
I may just have erased it.
But now let's get it.
So y is y delta.
This is the impulse response.
y of t-- and I'll give it later another name.
So here's a perfect review question.
What's the solution to this problem?
Everybody remembers-- what are the solutions, what's the general form for the solution to the equation?
I'm reviewing today's lecture.
The solution to that equation looks like what?
[INAUDIBLE]
It's a cosine and a sine, right.
And then how much of a cosine do we have and how much of a sine do we have?
The initial condition will tell me how much of a cosine we have.
And what's the answer?
None?
No cosine.
This condition, this initial velocity, will tell me how much of a sine we have, because the sines are the things that have initial velocities.
So it would be a sine of-- the sine of what?
Square root of k over m, right?
omega nt, right?
And what's the number?
What's the number so this has the right-- let me write again what I want.
I want y prime at 0 to be 1 over m. What's the number that I put in there?
I've got something, its derivative, at zero.
This is some number-- I'll call it little a for the moment, but I want to find out what it is.
Are we right?
Yeah?
I think we're right.
Yeah.
The derivative is at zero, so I just plug that into here, take the derivative at zero-- of course that makes it a cosine, which will be 1-- but it also brings out that factor.
So a times-- well, that factor will be 1 over m, and that tells me what a has to be.
Well, this is omega.
So a is-- this is omega a equal m. This is 1 over m omega.
And that's omega.
Sorry I'm erasing stuff which I-- this is the formula I'm after.
Sine omega t over omega.
I think we're good.
Are we?
Yeah?
Yeah.
I'll come back to this in-- Wednesday is my day to move to damping terms.
I've intentionally stayed with undamped equations here, because you're thinking about that level of equation.
Damping is going to bring in new stuff, and that should wait till Wednesday.
Shall I recap today?
I'll just recap today, and then we're done.
Today started with the unforced equation.
We solved it by assuming-- by not thinking ahead, just assume I have an exponential, because the beauty of exponentials is, when I plug it in, the exponential cancels.
And that told me that s was pure imaginary.
It told me that it had this form, e to the i omega t. And there were two s's.
Two possible s's, plus and minus.
I get to make a little comment about this example here.
What was omega?
What's the natural frequency in this problem?
What's the natural frequency here?
I guess this is a case where-- what's the natural frequency?
I guess this is a case where m is 1 and k is 0, is that right?
This does fit into that pattern, but it's a little special.
This is a case where m is 1 and k is 0.
So what's the natural frequency in this?
Zero
Zero.
Zero.
This is a crazy case of resonance.
It's a case in which the natural frequency and the driving frequency, say in this-- I'll have to do it here-- this simplest of all equations is, in a way, special.
It's a case when the natural frequency is zero and the driving frequency is zero and they're equal.
And what happens with resonance?
What's the new formula, the new term that comes in with resonance?
It's t. You saw it happen for this example, and we didn't have to use the word resonance.
We knew that we had a ramp.
We just used the word ramp, not resonance.
But this is a case of resonance.
When omega n is zero and omega d is zero.
And the factor t up here.
Anyway.
Just that small comment there.
And now, just going back to the recap.
The recap was, we tried exponentials.
We learned that they were pure oscillations.
We realized that we could do cosines and sines instead, and we did.
And we took off.
We got the formula.
Then of course the-- so this is section 2.1 of the book.
And it goes through all those steps carefully.
Section 2.2 of the book tells us about complex numbers, and section 2.3 brings damping in.
So that's what's coming next time.
So the recap again.
We found the null solution, we found a particular solution-- oh there's just one comment I want to make, and then I'm done.
Where was our particular solution?
Yeah.
This was our particular solution.
Here's my comment.
Here's my comment.
Suppose I want to solve this basic equation starting from a given y of 0 and a y prime of 0.
I'm going to do it in two parts, I think.
I've got the null solution, and I've got this particular solution.
Now here's my point.
If I want to get y of 0-- how shall I say this.
You can't just put together-- it's an easy mistake to make-- solve the null equation with the initial conditions and then add in the particular solution.
You'd think, I just followed all the rules.
But this particular solution that you added in has a-- at t equals 0, it's not zero.
So you have to change.
So the correct thing, the correct yp plus yn-- let me make that point.
Just a warning.
So in words the warning is, remember that the particular solution has some initial condition-- in that case, g-- and then that is going to affect the right null solution.
So again, y is y null plus y particular-- plus y of particular-- so it's some c1 cos omega nt plus some c2 sine omega nt plus this particular guy, g, cosine of omega dt.
All correct.
All correct.
But now, put in the initial conditions.
y of 0 is given.
And what do I get on the right hand side when I put in t equals 0?
I get c1 here.
What do I get when I put t equals 0 in there?
Nothing.
What do I get when I put t equals 0 in here?
g. So it's not c1 equal y of 0 anymore.
That's the easy mistake that I'm correcting.
When you put in this particular solution, it has an initial value.
That initial value is going to come in here.
So c1, then, the correct c1 is y of 0 minus g. End of story.
Just don't be too quick to just add the two pieces and think you can do them completely separately, because you're putting them together.
And then you have to put them together in the initial condition.
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So morning.
This is my fourth lecture on differential equations, that part of the course.
And I haven't said anything about the textbook.
That's Differential Equations and Linear Algebra.
I wrote it because so many courses like this one want to combine those two topics.
They're the two major topics of undergraduate math, after calculus, the two major directions, and the book connects those two directions.
And I just wanted to give you the website for lots of things connected with the book.
So it's the math website, dela for differential equations and linear algebra.
And so today is differential equations, second order, with a damping term, with a first derivative term.
So that in many engineering problems, those coefficients A, B, C would have the meaning of mass, damping, and stiffness.
Mass, damping, and stiffness.
And physically, we know what the mass comes from.
The stiffness comes from a spring, as I drew before.
And traditionally we describe the source of damping as a dashpot.
I guess in my whole life I've never seen a dashpot, but maybe it's-- think of a piston going up and down within a cylinder of oil or something, with resistance.
OK, so that's the left-hand side.
Linear constant coefficients still.
We don't have formulas, it's not easy to see what's happening when things are varying, when the equations non-linear, so this is the starting place.
OK. With some forcing.
So this is damped forced motion.
This is the ultimate within linear equations.
OK. And now, what's the forcing?
So now, always we solve first for the null solution.
With no force, what are the natural motions?
And we'll find a formula for those.
Then the big one, the special right-hand side is always an exponential.
So this is going to be y is going to be the null solution yn, and we'll get a formula for that.
This, the exponentials.
Always, the response to an exponential is an exponential.
That's like the most important fact in getting solutions.
So the response will be some ye to the st at the same frequency-- well, I say frequency.
If s is a real number that would be a growth or a decay, but very frequently s is an imaginary number, like last time.
And this is comes from a rotation or an oscillation.
And then, the complete picture comes from being able to solve it with an impulse.
So then y is the impulse response, which I write as g of t. Let me introduce just that letter g. That stands for growth factor in first order equations stands for Green's function, so that word Green is getting in here, his name.
And it represents the impulse response.
OK, one, two, three.
And then the point of doing this one is that then we can do it, then we can get a formula for any f of t. So this is the pattern that we followed for first order equations.
We followed it for second order equations that didn't have damping.
And now we're doing the big one with damping.
So what's the damping going to do?
What's going to be effect of damping?
Say if I had a right side of one, just a unit force?
The damping.
So I'll still have oscillation-- you'll see this-- I'll still have oscillation, but its amplitude damps out.
Like that like everything we know, like something swinging back and forth, but friction is it is eventually damping out that motion.
So it'll be oscillating with exponentially decaying amplitude.
Let's find this solution.
OK. How do we find the null solution?
I've got constant coefficients here.
So the nice, the right thing to look at is exponentials.
In first order equations it was e to the at.
Here, we'll have to see what it is.
So I'm going to try a particular-- I'm going to look for the right exponentials.
Certain exponentials will be null solution.
Can I do it?
So I take that equation and put in y equal e to the st.
Remember now, I'm doing 0 here so it's not the same s as the right-hand side.
OK. What happens if I put in-- so I'm looking for the null solution.
I'll try y equal e to the st. Plug it in.
I get m. Two derivatives gives me s squared e to the st, b.
One derivative gives me an s, e to the st, and k gives me k e to the st. And that's supposed to equal to 0 for the null solution.
OK. Have we made a good guess at what will work?
Yes, because I can cancel e to the st, and I come to the most important equation.
The equation that governs everything in this whole lecture is ms squared plus bs plus k equals 0.
That's called the characteristic equation, or it has many names.
It's obviously the big deal.
It's the thing.
It's an equation for s, for the special frequencies that solve the null equation with no force.
OK. And how many values of s do we expect?
Two.
So I expect solutions s1 and s2, and then my null solution is e to the s1t is a null solution.
E to the s2t is another one.
My equation is linear.
I can multiply that by any constant.
I can multiply this by any constant.
And I can superimpose, i can add, I can combine, because I can do linear algebra here.
This is the most important operation in linear algebra, multiply things by constants and add.
That's called a linear combination.
It's the basic operation in linear algebra and it's a basic operation here, because we're doing linear algebra with functions.
Do you see that that's it?
We've got it, except we should really write a formula and draw some pictures to show s1 and s2.
What would be the formula for s1 and s2, to the two solutions?
We remember that from school, it's the quadratic formula.
The two solutions to this.
Everybody remember this one?
Let's see if I do.
Does it start with-- a minus b.
And then there's going to be a denominator that I'll remember, which is 2a.
No.
I said a but I mean m. 2m.
And now this is plus or minus-- what goes into here?
B squared minus 4ac.
The key quantity in this whole business, b squared minus 4ac.
Ah, what is it?
Mk, thank you.
4mk.
Great.
And can I just remark, a little remark about units.
B squared has the same units as 4mk.
It has to or such a formula would be crazy.
So we will see that actually there's something called the damping ratio that involves the ratio of these guys.
OK, that's the formula.
But it's not like-- OK it's got a square root, and what's the point of-- the thing we have to remember with this square root is that if it's the square root of a positive number then we have a plus or minus, ordinary real numbers.
If this thing is negative, then what?
What's up if b squared is smaller than 4mk?
So if b squared is-- if there's not much damping, if it's underdamped, b squared would be smaller than 4mk.
And what then?
We've got a negative number here.
When we take its square root we have an imaginary number.
That's oscillation.
Under-damping is going to show oscillation.
Let me draw this.
Let me draw that curve for different choices of m, b, and k. This is a good picture to see.
So let me draw first of all one that has b equals 0.
OK.
So there's a curve.
What I'm drawing is, up on this curve is this ms squared plus bs plus k. Ms squared and bs and k. And here, this one is one with no damping at all.
This is just s squared plus 0 s plus 1.
That's what that curve is That's the example we saw before.
Y double prime plus y. Y double prime plus y equals 0.
This is coming from Y double prime plus y equals 0.
This is pure oscillation.
OK. Now let me bring in some damping.
Now, as I bring in damping the curve will move.
And it takes a little patience to see where it moves.
Let me have a little damping a little damping, so s squared plus 1 s plus 1.
I think the curve-- so this is s here.
And it's a parabola.
Everything I draw here is a parabola, is just that different parabolas come from different choices of b.
So I think that this choice, I think it goes a little bit like that.
That's the next one.
It goes down a bit.
Now, let me do s squared plus 2s plus 1.
Now, let's see, I really should stop for a moment and solve the equation, find the roots for each of these guys.
And then I'm going to have an s squared plus 3s plus 1.
So we're doing something very straightforward, parabolas.
But it shows us the different possibilities.
And we could give them names.
OK.
So I would call this one undamped.
And what are the roots of that equation?
S squared plus 0 s plus 1 equals 0.
What are the s1 and s2 for that guy.
So the roots of s squared plus 1 equals 0 are?
Stay with me here.
S squared plus 1 equals 0, that's the equation that has roots
I and minus i
I and minus i. I and minus i.
So this is undamped.
S1 is i and s2 is minus i.
That's the pure oscillation.
Pure oscillation, that's the case where b is 0 here, I have a square root of a negative number, and it gives me plus or minus 2i, and then the 2s cancel and I get plus or minus i.
So I can always go back to this but I'll try to choose numbers that come out nicely.
Now, what happens with some damping?
This guy.
What are the roots for this one?
Well, I better use this formula.
Now, I'm always keeping m equal 1 and k equal 1, in all those samples.
But now b has increased to 1.
So if b is 1, what do I have?
I have s is-- the roots are minus 1, plus or minus the square root-- And it's going to be just 2 down below.
What's in the square root, the all important square root.
When m and k and b are all 1.
Just do the calculation with me, so you see it.
It's negative?
Three
Negative 3.
So what's the point there?
The point is, this is going to be square root of 3i or minus i.
We have oscillation at a frequency square root of 3, and we have decay from s minus one-half.
The real part of this is giving us the drop off.
We didn't have any drop off at all in this case.
They were pure imaginary.
Now the s1 and s2 are whatever I have a minus 1, plus or minus square root of 3i over 2.
Those are the roots.
All right, I'm ready for this guy, and it's particularly nice.
It's particularly nice.
What do you see for s squared plus 2s plus 1?
When you see this parabola, now b has moved up to 2.
What's up with that one?
It's going to be-- let's see-- that looks like a perfect square to me.
Right, inside the square root is 0.
Right, exactly.
Inside the square root, b squared is 4, and 4mk is 4, so I have 0 inside this square root.
So now I have s equals minus 1 plus or minus 0 over 2.
That's when this was the case, when b moved up to 2
[INAUDIBLE] minus 2
I'm messing it up?
[INAUDIBLE] equals minus 2, plus or minus 0
Minus 2 plus or minus 0.
Thank you.
So what do I have?
What are the roots of this guy?
Negative 1.
What's the other one?
Negative 1.
A double root.
It's critical damping.
Critical damping-- it's not underdamped.
It's not overdamped.
It's right on the borderline.
And I see that, when you first saw quadratics, before anybody brought up that awful formula, you would have factored this into s plus 1 squared equals 0, and you would have discovered that s was minus 1 twice.
A double root.
So the picture there would be-- there's minus 1.
Yeah.
Everybody recognize that this, we're hitting 0, height 0.
We're hitting 0 twice at s equal minus 1.
So this is now the case.
This was b equal 0, no damping.
This was b equal 1, under-damping.
This is b equal 2, critical damping, just on the border.
And what do you think the s squared plus 3s plus 1 is going to look like.
Again, we can find it.
No, let me do the s squared plus-- let me take b equal 3 and find the roots and draw the picture.
If you're with, me that picture and these formulas tell you the difference between these four cases.
So what do I get?
Minus 3 plus or minus the square root of, 9 minus 4, is 5, over 2.
So I have two negative roots.
I have decay.
I have decay at a fast rate and a slow rate but both are giving decay.
So the curve now is coming down here and back up there, and it hits there, and it's got the two roots.
These two roots are x and x.
So these are s1 and s2.
Let me copy them over here.
S1 and s2 are minus 3 plus or minus the square root of 5 over 2.
Yeah.
Real.
Real roots.
So this is two real roots.
This is a double real root.
This is two complex roots.
And this is two pure imaginary roots.
The four possibilities.
The famous four.
OK. And for me, that picture is a-- so this was the b equal 3 curve, more overdamped.
It's a little interesting that overdamping has this root that's pretty near 0.
So overdamping doesn't mean that you go to 0 real fast.
Actually, the one that goes fastest to 0 is the critical damping.
Then, as b grows, one root gets closer to 0, so it's like slower decay, and another root is going off to fast decay.
I think you have to know those guys, because the physically that's very important, where's the damping.
But we've now found the null solution completely.
The null solution completely is-- let me write it again here-- is anything times e to the s1t and anything times e to the s2t.
And where do those constants, c1 and c2 get decided?
By the?
[INAUDIBLE]
Initial conditions.
Right.
These two conditions determine c1 and c2.
OK. Are we good?
So the null solution has already separated the different cases that depend on how much damping.
I'm ready for number two.
Number two now.
OK.
So the idea is I now have a forcing term, some frequency e to the st. And I will assume that it's not-- s not equal s1 or s2.
That's to make my life easy.
Just as last time, the formulas will break down.
And I'll have to put in something different if there's resonance, if the driving force is at the same frequency as a natural frequency.
In that case, there's a resonance.
And the way you spot a resonance formula is there's an extra factor of t. There's a growth of t. Up here, we're seeing no factors of t. Over there in the exponent, of course.
I mean, down below, there would be a t times e to the st. And the book does that carefully.
I don't see that it has a place in the very first lecture on this topic.
OK.
So this is saying no resonance.
All right.
What's the solution then?
The solution is a multiple of e to the st.
The input was e to the st.
The response is a multiple of e to st, so that's the frequency response.
Capital Y is telling us the response to s. And it's a very, very important function.
It's called the transfer function.
It's just the key to everything.
I probably say that so often, that this is the key to everything.
Well, it's partly because I just have one lecture to do a big part of differential equations, and it's got some key ideas.
And the first key idea is s1 and s2.
And the second key idea is Y.
And let's go for it.
All right, so this is the s1 and s2 picture.
I'll move that up, and now, in the equation, try Y e to the st. We hope it works.
What is this?
I'm trying, this is my solution, and it's a particular solution now.
I got the null solution.
I've moved to a particular And this is when the force is e to the st.
In other words, I'll just say again, if we have an exponential force, try an exponential response.
1803 would call this the exponential response formula.
So you could use the word exponential response, very appropriate.
You could use the word frequency response.
That frequency response is kind of the right word when s is an imaginary number giving an oscillation frequency.
OK, I'm going to plug that in.
So into the differential equation, m. Second derivative of that is s squared Y e to the st, right?
That's m Y double prime.
The beauty of exponentials.
Take every derivative, just brings down an s. The next term.
What's the next term?
Maybe do it with me, do it for me.
What happens when I plug in this as Y into b, so there'll be a b Y prime.
So what's Y prime?
[INAUDIBLE]
s, Y, this constant, e to the st. And then the final term is k times this Y itself, no derivative, so it's just Y e to the st, and that's matching e to the st. That's the force.
This is f. F, this is an exponential force.
Of course, I could and should have a constant here to give me the units of force.
Let me just keep the formula as clean as possible by taking units, so that's one.
OK what do I do?
[INAUDIBLE]
I cancel.
The nice part of 287 is canceling e to the st. That's the most fun you get.
And now, I have Y's on the left.
So Y is-- can I see what Y is?
Y is this 1, divided by the coefficient of Y.
And what's the coefficient of Y?
We know it.
We've seen that coefficient.
Y on the left is multiplied by ms squared plus bs plus k ms squared, again, ms squared, bs, and k. Multiplying Y, I divide by that, so I put it down here.
Ms squared plus bs plus k. And because s is not one of the roots, that's not 0, so we're golden.
That problem took five minutes.
The null solution took half an hour.
The exponential response is clear.
And you can see what it would be.
And let's give it a name.
This is the transfer function.
Widely used name.
Other names could be given but that's the best.
So it's a function of s. It's a function of s. And so, again, when f is e to the st, it sort of transfers the input into the output.
That's the way I think of the transfer function.
Here is the input.
The output is just multiplied by the transfer function.
And the transfer function is just that nice expression.
Just that nice expression.
So we are golden for a frequency, for a linear equation with an exponential forcing function.
What would be another example?
I'm using mass, dashpot, spring here.
If this was in electrical engineering, what three things would I be using instead of mass, dashpot, spring.
The three guys would be?
What would correspond to the dashpot?
So can I just draw here a little-- let me put on a low voltage and put on something that does this, and then something like this.
And then there's something like this.
And give me a break, tell me what these things are.
This guy is?
Inductor
An inductor.
This guy is a?
Resistor
Resistor.
Now that's the one that's like damping.
This resistor here is like damping.
Like the damping term, or maybe 1/b.
I'm not getting its units right because I haven't got any equation here at all.
Resisting is-- there's friction in that resistor.
It burns up heat.
And similarly, the dashpot slows things down.
And then this guy is a--
[INAUDIBLE]
Capacitor, right.
In other words, you can do the mechanical application and the electrical application with exactly the same ideas, just a change of letter, and of course, different units, but same problem.
OK, so that's a comment that you've seen before.
What else do I want to comment on?
Because this example was really so straightforward.
I think what I want to mention, and this is important, is that this is the central starting point for the Laplace transform.
So I can't do Laplace transforms all today by any means, and so Professor Fry will talk about the Laplace transform next week.
But what is the point of the Laplace transform?
The point of the Laplace transform is to get your money's worth out of the simple formula for exponentials.
Having an exponential there turns the whole differential equation problem into an algebra problem.
We just have quadratic equations.
We just have a division by a quadratic.
That's the great thing about the Laplace transform.
It turns the t domain, the time domain, where we have exponentials, into the s domain, the exponent domain, the frequency domain, where we just have quadratics.
And then first order equation's just linear.
And we can even get from second order of the quadratic to linear, because I can factor that guy.
If I factor into s minus s1 times s minus s2, I've two linear pieces.
And that's the first step in the Laplace transform, in the algebra.
So all of the algebra in the Laplace transform is this algebra for e to the st. And then the job of the Laplace transform-- and this is the tricky part.
So let me even take a little board space on that.
So this is like a heads up for next week.
So the Laplace transform.
OK.
So for any forcing function, f of t. That's the thing that we're going to take the Laplace transform of and the response and its response, y of t. OK.
So here's the idea of transforms in general.
I choose some terrific functions like exponentials.
So I want to convert my problem to exponentials.
e to the st for all s, s between let's say 0 and infinity.
So what do I do?
I take my function, and I figure out how much of every exponential is in it.
That's the Laplace transform.
I take my function f of t, and I go to what you'll see next week, its Laplace transform.
Let me call it capital F of s, or it's sometimes written the Laplace transform of F of s. Something like that.
I'm not going to do that.
I'm not going there.
So F of s is like the amount of a particular exponential in my function.
If my function is just the sum of two exponentials, then the Laplace transform just is a big bump on one and a big bump on the other.
But most functions, like some forcing function, has some e to the st is for all for a whole range of s. So I figure out how much of this is.
OK now second step.
Using linearity, I can solve the problem for when the right hand side is just that.
Then solve for right hand side, F of s, e to the st.
Solve for each frequency, each s separately.
And what does that mean?
That means just dividing by this.
So this was the easy step.
So this is step one, is take the Laplace transform.
The Laplace transform tells you how much of each exponential is in it.
Now step two is a cinch.
Step two is a cinch.
I just multiply by the transfer function.
I divide by this, bs plus k. And now, what's step three.
Step three, I have the solution for each separate exponential, but I've got a whole lots of exponentials, so I have to do an inverse Laplace transform, add up, figure out what function has this Laplace transform.
And that's often the place where the algebra gets harder.
In principle, we can do it for any F of t. We can take its Laplace transform, we can solve for each frequency in the transform, we can assemble them all together.
That's the inverse Laplace transform, so this is the inverse Laplace to get the solution y of t. So this is really the Laplace transform of the solution, and we have to get back to the solution itself.
Can I just let you think about those ideas?
I'm not up to describing the algebra here.
The point is the Laplace transform takes this into separate exponentials.
Each of those right hand sides is simple algebra, divide by that.
And then you have the job of okay, what function has this Laplace transform, to go backwards.
And that's usually the hard part.
So people make tables of Laplace transforms.
Everybody remembers the Laplace transforms of a few functions.
You could say that those few functions are the golden functions of differential equations.
The golden functions of differential equations are the ones where you know their Laplace transform and you can go back and forth easily.
And what are those golden functions?
Well, you might guess exponentials, simple polynomials, 1 t, t squared.
You can do Laplace transform for those.
What else do you think would be nice?
Cosine and sine.
And you could multiply those together.
We could deal with t times cosine t. So a little bit different.
But having said that, the reality is I've given you the whole list of nice functions.
Those functions show up in every simple method for solving equations.
There's a method called undetermined coefficients and what does it amount to?
I'm sorry, I don't have time to say it this morning, but it can come later.
Undetermined.
It just means if the right hand side is one of these nice guys-- shall I write down again the golden functions?
e to the st is like the platinum function.
And then some golden functions are like t and t squared and so on.
Of course, when we get that, we're close to cosine omega t, and sine omega t. All these guys, their Laplace transforms are nice.
We can deal with them completely.
Or multiply any of those together.
And when the right hand side is one of these golden functions, you can write down the answer.
We've focused on this one because it's the platinum one.
And we did these two too, because they come from s equal i omega.
And then these guys are a little bit of juice.
But that's it.
I'm sorry the list isn't longer.
It'd be nice to have--and of course, people for centuries have worked with the next hardest functions.
You know, the silver functions.
Famous functions have names like Bessel's function or Legendre's function.
Others where you can get pretty far.
Those are the best.
Then you have the famous ones where you get pretty far, in the web has the Laplace transforms, and then you get the general function, F of t. Oh, could you get anywhere with delta of t?
Oh, yes.
Does it belong on the list of golden functions?
Yes, it does.
I almost forgot it, and it's like-- I'll call it-- Yeah, delta of t. Yeah, that's a beauty.
That's a beauty.
The Laplace transform of delta of t happens to come out 1 over s. You can't ask for more than that.
Or maybe it's one.
Yeah, it's probably one.
Yeah, the Laplace transform of that is one.
Yeah.
It's got all exponentials in sort of, you could say, equal amounts.
OK.
So that's some thoughts about that about Laplace transforms, just sort of the big picture that takes the differential equation, turns it into an algebra problem, and then at the end, you have to get back, and that's the part that's not always doable.
OK.
So what's left for today is this guy.
Now this one now.
Have I got to that point?
So this will be the final ideas in this course, this four unit core elective.
What is the impulse response when there's damping?
What's the impulse response when there's damping?
OK. OK.
So that means that I would really like to solve m y double prime, b y prime, k Y equals an impulse, delta of t. Because if I can solve this, then I can solve everything.
And I can solve this.
It turns out to be easy.
Now why is it easy?
You might think, my gosh, we've got second order equations here, we've got a delta function there, where do we go?
And so my advice is go this way.
The solution to that is the same.
This with y of 0 equals 0 and y prime of 0 equal 0.
So you have a spring, so again we have a spring with a mass.
And that spring is in a damper.
So can I just, without knowing what I'm doing, draw a damper around it.
So the idea is I'm striking that mass at time 0.
Striking that mass at time 0.
What happens?
What happens immediately?
And then I don't touch it again.
I strike it and that's it.
I've set off, and what have I-- so my point is this has the same solution as m y double prime plus b y prime.
And Ky equals 0.
Nothing happens beyond time 0.
But what are y of 0-- Well, let me give this a letter g, just to emphasize it's special and deserves its own name.
So now, what is the starting position and the starting velocity for this picture?
So I'm saying that the y here is the same as the g there.
And really if you see it physically then you see it best.
So again, at the instant t equals 0, I'm hitting that mass with a unit impulse.
So what is the position of that mass, instantly after I've hit it?
0
Still 0.
Good for you.
Good for you.
And what is the velocity of that mass instantly after I've hit it?
1
1 is essentially the right answer.
1 is the right answer.
But the way I've set it up is, there is an m there.
If I put an m there, which would be nicer, then the answer would be 1.
But the way I've written it here, I have an m there, and I haven't fixed the units.
So that turns out to give me a 1/m there.
I could explain why it's a 1/m, but let me just for the moment say it's my fault.
It's because I didn't get any units right that we have 1/m.
No big deal.
But now, what's good here?
What's good?
This was a problem with a mysterious delta function.
This is a problem with 0.
And the only price we're paying is the impulse gave the mass a little velocity.
And you can imagine that the velocity gives it is 1/m because the strike didn't tell us about that.
So what I'm saying is we can solve that equation for g. We can find g, and in fact, we have.
So this really brings the lecture full circle.
What do I have here?
I have a null solution.
So this g here is a null solution.
So what form does it have?
What can you tell me about g of t?
And it's the same as y of t. So y is the same as g, and what's the form for it?
Yes?
Tell me?
I'm taking it back to the very beginning of the lecture where I solved it with a 0
[INAUDIBLE]
C1, thanks
[INAUDIBLE]
e to the?
C1, e to the s1, t, the two s's, s1 and s2, the special two s's, the special roots of the key equation.
That equation gives us s1 and s2, by this formula, or in a homework or an exam problem, we hope that it would come out easily.
So this is that part of it.
What's the rest of it?
C2 e to the s2 t. The impulse response is the particular null solution that starts with a shot, starts with an impulse, starts with a strike.
OK.
So I just have to find c1 and c2 here.
All right?
We've come back to the basic problem in differential equations.
We've got the solution.
We've got two constants.
We've got two equations.
We just plug that function into that equation.
It gives us one fact about c1, c2.
This gives us the second fact.
We solve them.
Why don't I just write down the answer?
It turns out to be e to the s1 t, and the other guy will come in with an opposite sign e to the s2t over s1 minus s2.
I think that gives us the c's.
Oh, there'd be an m. There'd be an m times this, because of my messing with units.
So can we just check?
At t equals 0, what do I get out of this?
0
0.
What I'm supposed to.
What's the derivative at t equals 0?
The derivative at t equals 0, this derivative is going to bring down an s1.
The derivative here will bring down an s2.
At t equals 0 the exponentials will all be one so I'll just have the s1 minus the s2.
It'll cancel that and it'll be 1 over m. So this is the neat formula for the impulse response.
That's the neat formula for the impulse response.
And then why-- can I use this little corner of the board?
Why do I want that impulse response?
What can I use it for?
It gives me the answer, not just for the impulse but for everything.
The particular solution is for any forces, force, I multiply by whatever the-- let me write the formula and I'll show you what it says.
Yes, there is the formula.
I'm sorry it's squeezed.
But really, the goal here was simply to get a handle on what is the response to any f. And again, I look at that this way.
F of s is the input at time s. G is the growth factor over the remaining time up until time t. So Y at time t, I take all the inputs up to time t, And each input gets multiplied by its growth factor.
It was e to the a, t minus s in the first order equation.
Now we've got two exponentials.
But that's the solution of the general problem.
So we have now in one lecture completed a solution to the second order constant coefficient differential equation.
Right.
Yeah.
By finding the impulse response.
Yes?
[INAUDIBLE] would we still be able to [INAUDIBLE] if s1 is equal to s2?
Ah, if s1 equals s2.
That's the case where formulas need a patch.
They need a patch.
If s1 equals s2, what do you think happens?
If s1 equal s2, everybody sees, I have 0/0.
And so this is like a technical question that I wasn't going to ask myself.
You asked it.
You're responsible.
What do we do for 0/0?
What did you learn in calculus?
Who's the crazy guy who figured out how to deal with 0/0?
In a way, calculus is all about 0/0, right?
Delta y over delta x, they're both headed to 0.
And suppose you have-- let me take the most famous example of 0/0.
It's like sine x over x, as x goes to 0.
So x going to 0.
Sine x goes to 0, x goes to 0.
What's the answer, by the way?
What happens to that ratio as x goes to 0?
This is maybe the most famous example.
Sine x over x, when x gets very small, is close to?
1
1, thanks.
And now just help me out with the name of the guy.
It's a crazy spelling name, and do you remember?
L'Hopital.
L'Hopital.
Everybody despises him.
Probably hi friends despised him.
But anyway, L'Hopital says in a situation like that, when you're going to 0/0, you're allowed to do something a little strange.
You're allowed to take the derivative of the top, so it has the same limit.
Instead of looking at this 0/0, you can take the derivative of the top, cos x, divided by the derivative of the bottom, 1.
And now you can let x go to 0, and you get the right answer.
So this gave a 0/0.
Unclear.
Fuzzy.
This gives-- what's the right answer then?
Just tell me again.
When x goes to 0 this becomes?
1
1.
Right.
So that's L'Hopital.
So that's what I would have to do here.
I would take the derivative of this, and the derivative of this-- the only sort of tricky part is it's the s derivative.
It's s1 going to s2.
Let me just tell you the result.
Since you asked.
A factor t comes out.
It's t e to the s1, or s1 is the same as s2, divided by the m. It actually looks simpler.
There's only one.
This is in the case s1 equal s2.
So s1 is the same as s2.
I just chose s1.
Yeah.
L'Hopital gives a simpler answer.
And it's got this suspicious and recognizable factor t. That came from L'Hopital.
OK, I won't do that stiff.
So let me say again, we've now done the second order constant coefficient equation I do just have 10 minutes of something to make it better.
And that is that the famous quadratic formula for s, for s1 and s2 is not beautiful.
It's correct.
It's correct.
But it's a little bit of a mess.
You've got three things, b and m and k playing around.
And we saw in this picture, we saw all the differences.
I guess in this example I kept m1 and I kept k1, and I increased b. I could do other examples where I increase the k, I make it stiffer and stiffer.
All these examples.
And engineers have worked for 100 years to see, out of this formula what are the important parameters, what are the important numbers, and hopefully, where possible dimensionless.
So I just want to- the final minutes would be-- back to high school-- playing with this formula, to get better numbers in there.
May I do that?
I just think, because then you'll see-- I learned this, actually, so this is like something math professors have no reason to do.
Look at that.
That's the formula.
End of story.
But the Web, 1803 website, has a class in which Professor Miller from the math department was teaching this subject, doing these, exactly these, but also Professor Vandiver from Engineering was putting in his suggestion of what are the good parameters?
What are the parameters that engineers look at?
So that would be my final comment, and I won't do it as well as Professor Vandiver did.
But can I just take that-- let me erase these two special examples, and look at this question.
Again, the book will do it.
So one nice-- b/2m is a pretty natural parameter to use.
Let me introduce that as one of them.
I'm going to, by taking ratios like b/2m, let me call that p. Let me call b/2m.
So that's a ratio of damping to mass.
And then this has got to come out simpler.
What does that come out?
If you'll allow me, I'm going to open the book so I don't write the wrong thing here.
This is in the book, on page 99.
The title is Better Formulas for s1 and s2.
Better Formulas for s1 and s2.
And here's my first better formula.
You can see that I get a minus b plus or minus the square root of something.
And that something will turn out to be p squared.
And then it'll be a minus something.
And that something will turn out to be the natural frequency squared.
Isn't that nice?
So what's the natural frequency?
Somehow, the natural frequency's coming in from this and this?
And just remind me what that second parameter is, omega n squared, the natural frequency of oscillation with no damping.
Tell me again what that is, because that was the fundamental ratio from last time.
It's central to all of engineering.
It's?
k/m
k/m.
Thanks, k/m.
So I believe-- and maybe Professor Fry could make this assignment a homework question, which is just algebra question.
Everybody sees that I have a minus p here.
And with a little care you get p squared, which is quite nice.
Which is quite nice.
And so we see that the decision between overdamping-- remember now?
Overdamping is when you damped so much that this became negative and you got an imaginary number in there.
Underdamping, it's still positive.
Overdamping, it's negative.
And so really that separation between overdamping and underdamping is the ratio of p to omega n. P to omega n is the damping ratio.
I think.
There may be a factor too.
Let me try to-- everybody sees that's the battle between these, if you accept that formula, and if it's in the book it's got to be right.
OK. And so the damping ratio is just that.
That's the damping ratio.
Now that's called zeta.
The Greek letter zeta.
I'm not Greek and not good writing zeta.
So I have unilaterally decided to use a capital zeta, which is a Z. Zeta is the Greek letter for Z. I could try to write it but you wouldn't be impressed.
So it's that damping ratio.
So now what does this mean?
Z smaller than 1, z equal to 1, c greater than 1?
Tell me what those-- obviously, when it's smaller than 1, p is smaller than omega n. Yeah, so what's going on here?
Which one is underdamping, which one is critical damping, which one is overdamping?
Because there's no difficult stuff here.
We're coasting in the last minutes here by just choosing words and notation that have turned out over a century to be more revealing than b squared minus 4mk, and this Z.
So z less than 1 will be what?
Z less than 1 will be p smaller than omega n. p smaller than this.
What's the story on that case then?
[INAUDIBLE]
That is underdamping, I guess.
p small has to go.
p is like b, the damping.
And small damping is underdamping.
So this is underdamp.
Underdamp.
And that's the case, in which we're going to have some imaginary stuff.
We're going to have some oscillation with the decay coming from there.
Now, what about z equal to 1?
z equal to 1 means p equals omega m, so that equals that so it's a big 0 in there.
What case is that?
[INAUDIBLE]
Critical damping.
It's this case in that picture.
It's that case with a double 0, equal s's.
Formulas that have to take account of that, this is critical.
And then finally, z greater than 1 is what?
Overdamped.
Overdamping.
Z bigger than 1 means p is big.
P big means b is big, damping is big, it's overdamped.
Overdamped.
So we've got it down to one parameter, the damping ratio, to tell us these things.
Rather than previously we had to say is b squared smaller than 4mk?
Is it equal to 4mk?
Is it bigger than 4mk?
Now we've got those words down to a single number z.
And let me just write next to us here that the z turns out to be the ratio of the damping to-- I think it's right-- it's the damping divided by the square root of 4mk, I think.
Can I just put a question mark there.
You couldn't mess around with the letters p and z.
But to get some variation from some other, but the point is, you see how much cleaner that is compared to this?
You're directly comparing that number to that number.
And that ratio is that number.
Yeah.
So all the formulas come out nicely.
Yeah, the formulas come out nicely.
And I guess what we see here-- final comment-- what we see here is what is the frequency of underdamped oscillations.
So I want to be in this underdamping case where there is oscillation.
There is an imaginary number coming out of that.
But there's also a real number.
Is the frequency of underdamped oscillation the same as omega m, the natural frequency?
No.
The frequency of that number-- so final comment, let me put it just here.
I would like this whole thing to be i, to give me oscillation, times omega d, the damped frequency.
And let me just say what that is.
So omega d, the damped frequency, squared, is this omega natural frequency squared minus the p squared.
If I had longer and we didn't have blackboards already full of formulas, I could-- it's the thing whose square root we're taking here.
So this is minus p plus or minus i, omega damped.
Omega damped is this square root.
There, we succeeded to fit in the better ratios, the good quantities to look at.
So again, the good quantities to look at are p, z, the damping ratio, omega d the damped frequency.
I think in a first lecture you could say, well, we already had correct formulas, we should just leave it there, and that's absolutely true, but anyway, this is what-- these are the letters people have introduced to make the formulas easier to understand in an engineering problem.
OK.
I'm all done except for questions.
Yes?
Don't ask me about resonance again.
Yes, OK.
Yes?
In the case of where we have the delta function, what is the velocity [INAUDIBLE]?
What is the what?
Why is the velocity equal to [INAUDIBLE]?
A-ha.
Okay.
You're right on the ball.
The question is where did this come from.
Where did that come from?
Can I tell you?
So if I integrate everything here, if I take the integral of everything, between 0, a little bit left of 0-- can I call that 0 minus?
Just a little bit left of 0.
This is crazy.
No math professor or whatever should ever do this.
To a little bit right of 0, just a real short time.
So what am I going to call a little bit right of 0?
0 plus.
OK now what is that integral?
Between a little left of 0 and a little right of 0, you know what the integral of the delta function is.
It is?
1
1.
Good.
Now, what are these ridiculous things?
Well, y is not changing in this tiny, tiny time.
So this is something, it's not getting big.
I'm integrating it over this tiny little time.
It's nothing.
Forget it.
Similarly here, y prime, the velocity is not climbing to infinity.
There's no-- and I'm just integrating over this infinitesimal little time.
Nothing here.
So this term has to be responsible for the 1.
And now you can tell me the integral of m y double prime.
What's the integral of m y prime?
m y prime
m y prime.
So m y prime plus, at 0 plus, minus m y prime at 0 minus.
But at 0 minus, it's 0.
You see what's happening?
And on the right side I'm getting a 1.
This is-- no person who had any skill with a blackboard would allow this to happen.
But that happened.
OK.
So these lower order terms are typical of math.
Lower order terms in the limit, forget them.
This is the top term, and it has to have something there, because it has to balance the 1.
And what it has is the jump in y prime.
So this is the instant jump in y prime in velocity, is times m gives 1.
So the instant jump jumped us from 0 to 1 over m. That's where I came from.
Well, that was a good question, and a kind of crazy answer but there it is.
OK, so we've got a mention of the Laplace transform as the algebra tool that works when you're staying with exponentials and nice functions.
And you'll see more of that.
So it's a frequently used tool to turn problems into algebra.
My name's Lorna Gibson.
I'm the professor for 3.054, it's a course on cellular solids.
And I've been working on cellular solids since I was a graduate student, since I did my Ph.D. And cellular solids are materials that are made up of an interconnected network of struts or plates.
And there's examples like engineering honeycombs and foams, and there's lots of examples in nature.
Things like wood and cork and there's a type of porous bone.
And there's lots of examples in medicine too.
Tissue engineering scaffolds, for example.
So my background is in civil engineering, and in civil engineering we study structures.
And typically people think of large structures like bridges or buildings.
But in fact when we analyze the cellular solids, we use the same kind of mechanics.
It's just the scale is very much smaller.
So we're looking at structures where the scale might be hundreds of microns or millimeters, things like that, but the same sort of mechanical principles apply to that.
OK, so I grew up in Niagara Falls, in Ontario.
And people always think of Niagara Falls as being the waterfall and all the tourist stuff, there's a casino there now.
But in fact, there's loads and loads of big civil engineering works in Niagara Falls, mostly associated with the hydroelectric power station.
So when they make hydroelectric power in Niagara Falls, the power station is actually about a mile downstream from the Falls.
And what they do is they have a big hydraulic gate that goes into the river and it diverts water from the river above the Falls into a whole series of canals and tunnels and there's a big reservoir where they store water.
And then the water from this reservoir goes into the penstocks, the tubes that go down to the turbines and then make the electricity.
Niagara Falls is not a big town, but if you drive around Niagara Falls, you see these canals, you see the reservoir, you see the big power station.
And so there's these really huge, impressive civil engineering works.
And my father worked for an engineering company in Niagara Falls and they specialized in the design of hydroelectric power stations, and I think that's how I got interested in engineering.
So I've been interested in bird watching for some time.
Mostly just because birds are beautiful and there's all sorts of interesting behaviors you can see with birds.
But since I started doing research on cellular solids and, in particular, teaching this course, I realize there's lots of examples of things about birds that have to do with cellular materials.
So for instance, some people had once told me that woodpeckers avoid head injury and brain injury by having a special cellular material in between their brain and their skull.
And that this acted kind of like a foam in a bicycle helmet.
That it would absorb the energy of the impact.
And I thought oh, well, I like bird watching and I study cellular materials, I should find out about this.
So I started looking into it and people had looked at the anatomy of the woodpecker skull and brain.
And, in fact, there is no special cellular material.
But by that point, I was kind of hooked.
And I actually did a project at one point looking at why it was that woodpeckers don't get brain injury.
And it's largely a scaling law.
It has to do with the fact that their brains are very small.
Another aspect of birds that has to do with cellular solids is how birds make themselves very light.
And here we have an owl skull.
This owl, unfortunately, had an accident with a car.
But somebody picked up its body and took it to Mass Audubon, and I got this from somebody at the Massachusetts Audubon Society.
And if you look at the skull-- I don't know if you can do a close-up here-- if you look at the skull, you can see there's a dense layer of bone on the outside and there's another dense layer bone on the inside, and there's a sort of foamy layer bone in between.
And that's called a sandwich structure.
And this foamy type of bone is called trabecular bone.
And that's one of the things that I study.
And it turns out that particular structure gives you a very stiff, strong, lightweight structure.
So you can see an example of how cellular materials are used in engineering but here sort of manifested in the owl's skull in making the skull very light.
Another part of the course is that I also have a project in the course.
And students do the projects in pairs.
Partly, because I just think it's nice for them to have somebody to work with.
And they've done a huge range of projects.
I let them do whatever project they want.
It has to have something to do with cellular solids.
And some of them have done, sort of, analytical things where they've done numerical finite element analysis of some cellular structure.
Some of them done very experimental things.
And some have been a lot of fun.
So, for instance, almost every year there are students who want to work on food foams.
And one year, for example, there was a group made bread.
And they looked at sort of the processing of the bread.
And they looked at how if you used more yeast, or you let the yeast rise, or the bread rise for longer, how that affected the cellular structures.
So they sort of changed the chemistry of how they made the bread.
And then they looked at the micro structure of the bread that they got.
I think they even did some mechanical tests of the bread.
So that was kind of fun.
And I think, probably, the most interesting project any student has done was on elephants skulls.
So you could imagine an elephant skull is huge, and it's bony.
So it would be very, very heavy, if it didn't have some pores in it.
And elephant skulls, it turns out, have some very large pores in them.
I think partly to reduce the weight of the skull, and the head, and the bone.
The neck has to, kind of, carry it all.
So these two students came to me.
And they said they had heard somewhere or they'd read somewhere that these pores in the elephant skull had an effect on how the elephants perceived sound and the acoustic transmission of sound waves through the skull.
And they wanted to do a project on elephant skulls.
So I was kind of intrigued by this.
I love all natural history kinds of things.
And I've worked before with people at Harvard's Museum of Comparative Zoology, where they have bones.
They have stuffed bodies.
They have all kinds of animals over there.
And I called up a colleague over there, and it turned out they had elephant skulls over there.
So I went with one of the students.
And it was an attic of the building, this kind of dingy place.
And there was this huge room.
And it was full of elephant bones.
And they had several skulls, which are like the size of this table.
They're huge.
They're this big.
And some of the skulls were cracked.
And you could see these big pores in the skulls.
And then the students found out that University of Texas at Austin has CT scans, Computed Tomography scans of all sorts of bones.
And sure enough, they had elephant skulls scanned.
And so they got a three-dimensional representation of the elephant skull through this University of Texas at Austin program.
And then they used that as input to a 3D printing set up.
So they 3D printed a sort of mimic of the elephant skull with some ceramic powder.
And they made a skull was about this big.
And then they wanted to look at the acoustic properties of it.
So what they did was they suspended the skull from a string.
And they had a speaker, and the speaker had a sound.
And they put an accelerometer on the skull.
And they measured the vibration of the skull.
And then to compare it with the skull that didn't have these pores.
And they got the CT image from Austin.
And they 3D printed this dolphin skull.
And they did the same thing.
They suspended the skull from as a thread.
And they measured the vibration.
And they could show-- and I've forgotten the details of their results-- but they could show, basically, that the two skulls had a different frequency response to the vibrations.
And they thought maybe part of it was because the pores.
So these two sections here are two sections of their 3D printed elephant skull.
I don't have the entire thing.
But you can see this is the orbit, where the eyes would have gone in here.
And if I turn it over and you look inside, you can see these pores in here.
And also if you look at this section lower down on the skull, you can see this whole porous structure here, as well.
And if we flip it over there's a little bit more over there.
So that was probably the most intriguing project that the students did as part of this course.
That was quite something.
So typically in the course we start off talking about the structure of these cellular materials.
We do some modeling of honeycomb and foam type materials, and that takes not quite half the term, but most of the first part of the term anyway.
And I give them more problem sets at the beginning of the term, and they don't really start on the projects at the beginning.
So I give them some background information to get them going.
And I assign the project at the beginning of the course, and I think there's three kind of deadlines.
One is they have to give me a proposal.
I think that's typically about a month into the course, and that can be fairly brief, but I want to at least know they've got a team.
They've got an idea.
They've got some idea of how they're going to carry out their project.
Then in about another month they have to send me a sort of update on the project, and by that point, I would have expected them, if they're doing a literature review, to start reading some papers and be able to tell me something about the background to their project.
If they're doing experiments, I would have expected them to at least gone into the lab and maybe made some materials, or bought some materials, and done some preliminary tests.
And I give them feedback at that point.
And at the end of the term, they hand in the final project.
And I see them twice a week in class, and I don't really have a formal recitation, but what I do do is every week that a problem set is due I have office hours.
And in fact, I don't actually have it in my office.
I book a room, and we do a little tutorial.
So there's times throughout the term they can see me and come and ask me questions about the project as well.
So the question is what kind of feedback do I give the students and how do they interact with me on the projects, and typically there's about 20 students that take the course.
So if they do it in pairs, there's roughly 10 projects.
Some of the projects students just do literature searches, and I don't really get that involved with those.
They're perfectly capable of going to the library and doing a literature search.
And some of the students who are taking it are graduate students, and they often do finite element numerical calculations.
And again, I give them some advice about how to set it up, but often they have experience doing this, and it's sort of applying what they already know to something new, but they kind of know what to do.
The way I get the most involved is if students do experimental projects.
So for instance, in this elephant skull project, I had this connection at the Museum of Comparatives Zoology, and I took the student up there.
They had this idea about 3D printing it, and we have a 3D printer in the department.
And one of the technical staff, Mike Tarkanian, is very, very good with the students and very helpful in getting them set up on the 3D printer, so he helps with that.
And I give them sort of general advice.
When they said now we want to measure some sort of acoustical response, I suggested maybe you could suspend it from a wire, or thread, or something and then put an accelerometer and measure the vibrations.
So I try to give them some general advice like that, but they then carry the experiment out themselves.
They have to figure out how to actually put it into practice and how to do it.
And I think they get a big kick out of that.
MIT students enjoy that kind of thing so that's kind of fun, and obviously I was very delighted too with this elephant skull project.
So there's different things I try to help with-- mostly giving general advice about how they can do their experimental projects.
So what are some of the challenges that students encounter in their projects.
So I think one thing is students sometimes are little over ambitious in what's possible to accomplish, because typically we have to cover some material before they can even start the project.
So typically they don't start the project until a month or six weeks into the term, and the term is only three months long more or less.
And so they really have a fairly limited amount of time-- maybe six weeks or eight weeks, something like that-- to actually do the project.
So if they want to make materials, if they want to do some sort of processing to make a foam, for example, they don't really have a lot of time, because often it takes some trial and error to be able to do that.
So the projects have to be fairly focused so that they can actually get something interesting out of it in a fairly short amount of time.
And sometimes students might want to do something where they would have to order materials, and it may take a few weeks for the materials to come in.
So again, I try to discourage them from doing projects where there's going to be a long lead time on getting some critical thing that they need for the project.
So there are some limitations, partly because of just the time we have for them to actually do it.
And they're not just taking my course.
They're taking other courses, so there's sort of a limited amount of time they can spend on it, too.
So really the way I set up the lectures is I write out notes for myself of what I want to cover, and the notes are pretty detailed.
And in the past I always just had the notes for me.
And even though they were reasonably neat and I could read them, I didn't hand them out to the students.
But because I've turned my fall course into an MITx course and I want to make the lecture notes available for that, when I was doing that course for MITx I made really nice notes.
And a friend of mine who lectures, I was asking her how she did it, and she said she actually goes and measures the chalkboard, and she measures the aspect ratio, how tall to wide the chalkboard is.
And then she sets up her notes so that they're the same aspect ratio, and she plans out exactly where she's going to put everything on the board on her notes.
And so I started doing that.
And when I'm doing the lecture I put it all on the board, and I find a lot of students-- even though I hand the notes out now-- they like to write their own notes.
And I think it helps them pay attention in class and helps them kind of focus on the material.
So another thing I do in the lectures, when I first started lecturing and for a long time I just focused on the engineering, you know, on the equations, this is the derivation, this is an example, this is how you use this.
And it was all just about kind of the engineering of whatever I was working on.
And I found over the last few years, actually in both courses, in the fall one on mechanical behavior and in the spring course on cellular solids, I look for more interesting examples and stories, kind of stories about the people who discovered some of the principles that we talk about.
I tell them stories about engineering situations that came up and there was some interesting thing happened.
And the students love it.
I mean, they really like having the kind of hard core mechanics broken up with some sort of stories.
So I do that a lot more now than I used to do.
So probably most lecturers have some kind of interesting example or historical thing that I talk about.
So for instance, when I teach the fall course, the mechanical behavior materials, one of the first things we talk about is stress.
So stress is a force per unit area.
If I take this piece a wood and I pull on it like this I'm pulling on it with a force that goes out like this.
Stress is just that force divided by that area.
And the unit of stress in the SI system is called the Pascal.
And it's named after Blaise Pascal who's a French mathematician.
And a couple of years ago I was in France for a conference and I was in a little town called Clermont-Ferrand, which is in the middle of France.
It's a pretty little town.
And I'm just walking around one day kind of seeing the square and the cathedral and all this stuff, and I see there is sign, there's like Pascal something or another.
And this was apparently the site of Blaise Pascal's house.
And right next door to it is Cafe Pascal.
So of course, I have to take a photograph of Cafe Pascal.
And in this other course, when we get to the bit about stress and I tell them about the Pascal I show them the picture of the Cafe Pascal.
And the other amusing thing that they kind of get a kick out of because of Boston, you know how in Boston there's the Freedom Trail and there's a red line goes around all these historical sites all this colonial and revolutionary stuff around Boston.
It's kind of cool, all the tourists to it.
Well in Clermont-Ferrand there's a Pascal Trail.
And there's little metal medallions of the portrait of Blaise Pascal put in the sidewalk and you can walk around Clermont-Ferrand doing the Pascal Trail.
So I kind of keep an eye out for stuff like that and I put that into the course now, and I never used to do that kind of stuff.
I have a picture from the Library of Congress, which I went to just for fun a while ago.
And one of the main buildings is this old, beautiful historical building for the Library of Congress.
And they have a marble staircase that goes up the middle.
And the staircase has all these little cherubs.
And there's an agriculture cherub and he's holding a sheaf of wheat.
And there's a wine cherub and he's holding a little thing of grapes.
Well, it turns out there's a mechanics cherub and he's holding a gear.
So at the end of the first lecture I show them the mechanic's cherub with the gear.
So there's these cute little things.
So I just keep an eye out for these cute little things.
And last fall, this past fall, one of the students came up at the end of the first lecture and he said, I really like art.
And he says, is there going to be more art in the class?
And I'm like, not usually.
This has kind of exhausted my art in mechanics.
And I said, but I'll keep an eye out.
And through the term there actually were different things that I saw that had to do with mechanics and art I showed the class.
So one of them was at the Peabody Essex Museum this fall there's been and display of the mobile sculptures of Alexander Calder.
And they also made these very large, I think they're called stabile sculptures, as well.
The big sail at MIT is one of his sculptures.
And these mobiles are actually a really nice example of free body diagrams in mechanics and balancing the forces.
So anyway, I got a picture of one of his mobiles and I went up to the exhibit.
And when I was at the exhibit I found that he did a degree in mechanical engineering.
He actually was a mechanical engineer.
And so the students were very, very tickled by the sculpture, the fact that he studied engineering.
So anyway, I look out for stuff like that.
So there's a lot of images in the class, partly because we're studying these materials and you can see just with the ones sitting in front of me, they have this porous, cellular structure.
So I show lots of images of materials.
We also look at how the images deform under load.
And I think, perhaps, that's something that might be a little unexpected.
So for instance, we have a stage in the electron microscope where we can actually deform things in the microscope.
And the stage is set up that it has a load cell on it.
And we could also measure how much it deforms, the materials.
So we can actually watch the materials as they're deforming.
So even though the cells may be 100 micron size, you can watch how they deform and how they fail, and you're going to learn a lot about the mechanics from these sorts of observations.
So we have both video of these deformations and also still photography at different time points that we show.
Another interesting little video clip that I show in the class is we look at the interactions between biological cells and tissue engineering scaffolds.
So for example, people who've looked at trying to heal, say, burns in skin where there's a large area of skin missing, one of the ways that you can do that is by using a collagen-based tissue engineering scaffold.
And when you have a burn in your skin and it just heals the normal way and you get all the scar tissue forming, that scar tissue is thought to form in conjunction with a process of what's called wound contraction.
So cells will actually migrate into the wound bed and actually mechanically pull the edges of the wound together, and that partly closes the wound and then scar forms as well.
And those two processes are thought to be related to each other.
So I've done a collaboration with Professor Ioannis Yannas here at MIT who developed one of these scaffolds for burn patients.
And he and I have been interested in this wound contraction problem.
And it turns out if you just put fibroblast skin cells into a dish of culture medium with one of these tissue engineering scaffolds, they will contract the scaffold.
And you can use an optical microscope to actually watch the cells do this.
So you can focus on an individual cell and you can see the cell elongating.
You can see the scaffold itself contracting.
And you can see this whole process.
So this is one of the little video clips that we show in class.
And the students always find that fascinating.
So I think one of the special things about this course and about my kind of research on cellular materials is that I spend a fair amount of time talking about materials in nature.
So I talk about wood, for instance, and what it is about the cellular structure that gives rise to the density dependence of wood properties and the anisotropy in wood properties.
I talk about trabecular bone, and I talk about the structure and properties of the bone, but also we do a little bit of modeling on how you might look at bone loss and osteoporosis.
So if you lose a certain fraction of the bone density, what residual strength would you expect the bone to have.
We have a project on bamboo right now.
We talk about the structure of bamboo.
Bamboo is actually a grass.
And this is a Chinese species of bamboo called moso bamboo, and you can see how big this one is.
They even get bigger, maybe six or eight inches across.
And what we're interested in doing is making something called structural bamboo products.
And this is an example of a bamboo oriented strand board.
So the same way people take wood, and they chop wood up into strands and make oriented strand boards for housing construction, you could, in principal, do the same thing with bamboo.
And we have a project that's in collaboration with some colleagues at the University of British Columbia.
They're the ones who are actually making the bamboo oriented strand board.
And with some architects in England, in Cambridge, England, we're looking at things like how you might modify wood building codes that talk about wood structural products, how you would modify that for bamboo structural products.
And what we're doing here at MIT is we're looking at the structure of the bamboo and doing some modeling of the mechanical properties of the bamboo itself.
So that's one example.
Here's another example.
This is a bamboo laminate.
So this is a little bit like a glue laminated wood member, but in fact, this one is made out of bamboo instead of out of wood.
So it's the same kind of idea.
Let's see.
We've also had a project in the past on cork.
So this is a cork from a wine bottle, and we've looked at that.
Cork has an unusual mechanical property.
You know, if you take a rubber band and you pull on it, if you can make it longer this way, it gets narrower that way.
Well, if you load cork in one direction, if you, say, pull on it or compress it, it doesn't get any wider or narrower in the other direction.
It just stays the same kind of size.
And you can show that that's related to the structure of the cells.
The cells are like little bellows, or like a little concertina.
So you can imagine if you have a little concertina, and you push on it this way, it doesn't really get any bigger that way or smaller that way.
It just stays the same dimension.
And the cork cells look a little bit like that, and that's why they do that.
So we have all these natural materials.
We talk a little about the hierarchical structure in plants.
The cell walls are fiber composites, and then there's a cellular structure.
And plant materials have a sort of hierarchical structure, with several different levels of hierarchy, and we talk about that in the class as well.
And we talk a little bit about why that makes these materials mechanically efficient and how you might look at designing engineering materials based on that.
So there's a little bit of biomimicking in the class as well.
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OK, so what we're going to talk about in this course are materials that have a cellular structure.
So they're all very porous.
And typically they have low volume fractions of solids, like less than 30% solid.
And we're going to talk about different types of cellular solids.
So one type are honeycombs.
And this would be kind of your standard hexagonal honeycomb, something like that.
We're going to talk about foams.
And I'm sure you've all seen polymer foams.
Here's an aluminum foam.
We'll talk about other sorts of foams as well.
We're going to talk about some medical materials.
And I brought in some bone samples, some trabecular bone samples here.
I brought in a little tissue engineering sample here.
And then, we're going to talk about materials in nature that have a cellular structure too.
So we're going to talk a little bit about wood.
And I brought in a piece of balsa wood.
I'll pass these around in a minute so you can play with them.
This is the lightest wood.
And I brought in lignum vitae.
This is the densest wood.
Lignum vitae is so dense that it sinks in water.
And I have a couple of projects right now on natural materials.
So I thought I'd talk a little bit about those too.
We have one on bamboo and making structural products out of bamboo.
So this is like a beam made out of bamboo.
And this is a piece of oriented strand board made out of bamboo.
So the same way you have like wood oriented strand board, you can do the same thing with bamboo.
And we have a project that involves this, so I thought talk a little bit about that later on in the course.
So we'll talk a little bit about the processing, how you make these materials, the structure.
We'll talk a little bit about how we do the modeling.
So we're going to start with the honeycombs, partly because they have a nice, simple unit cell that you can repeat and you can analyze fairly exactly.
And then we're going to talk about modeling the foams.
And then once we've got the modeling background, so we're going to get equations describing the mechanical properties, the stiffness and the strength, once we've got that, then we can apply it to lots of things.
So we can apply it to understanding the trabecular bone, the tissue engineering scaffolds, or how cells interact with scaffolds.
We're going to look at energy absorption in foams because they're good for absorbing energy.
We're going to look at a lightweight sandwich panels as well.
So we're going to look at all these different things.
And I have some slides I'm going to go over today that have more pictures of all of this.
And I'll pass this around to a sec, but let me go through the logistics first.
So this first hand out has a few of the kind of details.
There's two books that I've coauthored with colleagues.
The one on cellular solids, if you wanted to buy a book, is probably the most relevant one to buy.
If you don't want to buy it, you certainly don't have to.
I'm sure the library has multiple copies of it.
And the other more recent one is called Cellular Materials in Nature and Medicine.
And that's a little more specialized.
And again the library has that.
So you don't need to run out and buy that.
You can just get it there, but let me just mention there's that reference and it might be helpful.
OK, so let me talk a little bit about the projects.
What we do in this class is everybody does a project.
And I like for you to do it in pairs, just so that you have somebody to work with.
I think it's nice to have somebody to do it with.
And the project has to be something on cellular materials, but I really leave it pretty open ended what the project is.
It's really up to you to decide what you'd like to do.
And to give you some idea what people have done, people have done all kinds of stuff in the past.
So people have worked on negative Poisson's ratio honeycombs.
I didn't bring any of those in with me, but you can design these honeycombs so that they have this property that when you push it this way-- instead of if you make the strain smaller in this direction, it gets wider in this direction normally, but with negative Poisson's ratio materials, it will contract in that direction.
So I've had people do projects on that.
I've had people work on osteoporosis.
I had a group once who worked on elephant skulls and I brought part of their project with me.
So it turns out elephant skulls have a sort of porous layer to them.
And this is-- they have these large pores in the top of the skull.
I don't have a whole skull, but this is part of it.
And what they did was they had heard that elephant skulls have these pores and that the pores somehow affect sound transmission and how the elephant hears.
And so they wanted to do a project on that.
And that was really all they knew to start with, elephants have pores and we want to do a project on it because it's cool.
So I helped them put together this project.
We went up to Harvard.
We went up to the Museum of Comparative Zoology where they have elephant skulls, which are like about this big around, huge skulls.
And some of them had the outer part of the bone was broken and you could see these big pores.
So they got some nice pictures of elephant skulls.
And then they found that the University of Texas at Austin has computer tomography images of all sorts of bones of different animals.
And sure enough, they had elephant skulls.
So they got the file for the micro CT image, which gives you the sort of 3-D picture of the skull.
And then they use that to 3-D print small versions of the skull.
So that's what this is, they printed smaller versions.
And these were just a couple of slices that I got from them at the end project.
And then what they did with one of the skulls, they suspended it from a wire, and they took speakers, and they had sound that vibrated the skull, and then they put an accelerometer on the skull and they measured the vibration response of the skull.
And they also made, I think, it was a dolphin skull, which did not have these pores, and they kind of compared the vibration response from the sound for these two skulls with two different structures.
So that was a project for this class.
People have worked on tissue engineering scaffolds.
Often there's people who work on food foams.
So one year, people worked on bread.
They did bread processing.
They made bread by having different amounts of yeast, different rise times, different ingredients.
And they made bread.
And I have a little-- I like these historical things.
And I have a little thing here I wanted to show you.
So the people in 3032 last year, last fall, will know that I went to England in November.
And I went to the Royal Society for an editorial board meeting.
But I also went and looked at their archives.
And one of the things they showed me was this article, which is in the sort of an original, sort of archives of the Royal Society from 1600s.
And it's by a guy called John Evelyn.
He's famous for writing a book called Silva about trees and wood.
But he's written this article here.
And I love the title.
It's "The Several Manors of Making Bread in France, Where by General Consent, the Best Bread in the World is Eaten," by Mr. Evelyn.
So here we have in the Official Royal Society bread making science.
And in fact, the article was several pages long.
There was quite a lot on bread and how you make it in France, where the best bread is eaten.
So if you want to do something on food foams, people have done meringue before, various things.
They look at typically changing something about the recipe, or the composition, or the processing, or the baking.
And they look at the structure.
Sometimes they look at mechanical properties too.
So anyway, there's a whole list of things there.
You can think of other things.
If you look through the books, you might get some ideas as well for projects.
So what I'd thought I'd do next is I wanted to just kind of give an overview of what we're going to talk about.
And I've got a bunch of slides.
Let's see I forgot some things, because I do that.
Books, yeah, OK, let me pass some of these things around so you get to play with them too.
So here's some honeycombs.
I don't know if we can pass all these things around because it's a little unwieldy.
So this is an aluminum honeycomb.
I like bringing toys in.
These are little rubber honeycombs.
This is a little ceramic honeycomb.
This is a little paper honeycomb.
And let's see, we have some foams here.
So here's a metal foam and a ceramic foam.
And this is a sort of, not quite a foam, it's made of hollow spheres by centering hollow spheres together.
So you can play with that.
And what else do we have?
We have little lattice things here.
So here's a sort of 3-D lattice material.
So this has sort of a cellular structure, but it's very, very regular.
It's not like a foam.
So that's called a 3-D lattice material or sometimes a 3-D truss.
And what else should we pass around?
We need to do the bones.
Here's the wood.
You can feel how different the densities of the two woods are.
I would like these all at the end because I show these around for different classes.
Here's some bone, you can see of the bone looks like the foam.
And this is a tissue engineering scaffold for generating skin.
All right, so while those are getting passed around, I'll talk a little bit about what we're going to cover in the class with some slides.
OK, let's see, should I dim the lights?
Would that be a good thing, Craig, if I dimmed the lights?
They're kind of preset, you can try maybe two
OK.
Doesn't seem to-- there we go.
How about that?
That OK?
All right, so I like these historical things.
So this is a picture of Robert Hooke's drawing of cork from his book Micrographia.
And he was the first person who used the word cell to describe a biological cell.
And it comes from the Latin cella, which means a small compartment.
So you can think of the cells as small compartments.
That kind of makes sense.
And he kind of very modestly says, "I no sooner to discerned these, which were indeed the first microscopical pours I have ever saw, but me thought I had with the discovery of them perfectly hinted to me the true and intelligible reason for all the phenomena of cork."
So he's saying by looking at the structure of cork, he thinks he understands everything about the properties of cork-- very modest.
But in fact, this is kind of the foundation of material science.
So material science is all about looking at the structure of materials and trying to say something about the properties of the behavior of the materials.
And that sentence kind of sums that up.
So that's why I like that sentence.
So what we're going to do is look at different kinds of cellular materials.
We'll look at engineering ones.
And these we typically refer to as honeycombs or foams.
Honeycombs have two dimensional prismatic cells, while foams have three dimensional polyhedral cells.
And we'll look at applications for the honeycombs and foams in things like lightweight sandwich panels, in energy absorption devices, and things like thermal insulation.
We'll talk a little bit about the thermal properties.
We're also going to talk about cellular materials in medicine, so trabecular bone.
We'll talk about osteoporosis and how loss of bone reduces the strength, and how you might estimate that.
We'll talk about tissue engineering scaffolds, something about their mechanical properties.
You may think the mechanical properties aren't probably the most important thing.
But in fact, the mechanical properties do have some effect on how the cells interact with the scaffold.
So we'll talk a little bit about cell scaffold mechanics.
And then we're going to talk about cellular materials in nature at the end of the course.
So we'll talk a little bit about honeycomb like materials, like wood and cork, and foam like materials, like the trabecular bone.
There's also a type of tissue in plants called parenchyma that looks just like a foam.
And there's some sponges that have some interesting features.
And often, in nature the cellular material appears in combination with some solid material.
And so it's sort of a structural component.
And you can see sandwich structures in nature, leaves and skulls.
You can see materials that have density gradients, palm stems and bamboo are examples of that.
And you can see materials that have cylindrical shells with compliant cores, and things like plant stems and animal quills are like that.
So that's kind of the range of materials that we're going to talk about.
And one of the interesting things about cellular materials is that you can make cellular materials out of almost anything now.
And this is really a huge range of materials and lots of different applications for this.
So one of the fundamental things we're going to do is look at the mechanisms by which the materials deform and how they fail.
And we'll use a structural analysis to obtain the bulk mechanical properties, so things like the stiffness, the moduli, the strength, the fracture, toughness.
We'll look at how you can control the design of the microstructure to get the properties that you might want, and also how you might select for the best material for a given engineering application in engineering design.
So we're going to get those three things.
So let me start by showing you some more examples of engineering cellular solids, and in particular, some micrographs.
So that you can see what it looks like on a small scale too.
So these are the honeycombs.
These are the sorts of things that I'm passing around now.
The aluminum one, paper resin, and the ceramic ones.
The aluminum and the paper resin ones are typically used in the cores of sandwich panels.
And the ceramic ones are used in the catalytic converter in your car.
So what they do is they block off every other cell at one end and then the opposite cells at the other end and exhaust gas from your car is forced to go through those channels in the honeycomb.
And the walls, if you look at that triangular one, those triangular walls themselves are porous, and they're coated with the catalyst, which is platinum.
And that forces the exhaust gas through the wall in contact with the platinum, and then comes out the other end.
And they're ceramic, obviously, because the gas is hot and they need something that has a high thermal resistance.
So these are some examples of honeycombs.
These are some examples of engineering foams.
When I tell people I work on foams, they always think of polymer foams, like polystyrene or something.
And there's lots of polymer foams.
But you can actually foam any materials now.
There's metal foams.
There's ceramic foams, and glass foams, carbon foams, all sorts of foams.
So those are some examples.
You can see when you look at these images here that the foams have a low volume fraction of solids, like if you look at say this polyethylene one here.
Say we look at this guy up here, then you can see there's not much solid, there's a lot of gas.
So the volume fraction of solids is fairly low on that foam there.
So one of the things we're going to talk about is how the volume fraction of solids affects the properties.
You can also see on the top left and the top right, the top left one has what we call open cells.
There's just edges along the polyhedra, there's no faces over the membranes.
And the right hand one is a closed cell foam.
So there's like membranes that cover the faces of the cells.
So we're going to talk a little bit about the differences and the behavior of open cell and closed cell foams too.
These are food foams.
So I've already said you might want to do a project on food foams.
And these are just some examples of different kinds of foods that are in fact foams.
And it turns out the food industry spends quite a lot of time and effort thinking about the mechanical properties of food.
And it turns out if the texture of the food isn't right, then people don't like the way it feels in their mouth.
There's something they actually called mouth feel.
So it turns out if your cereals too soggy, it's icky.
If it's too crunchy, it's icky.
So it's sort of a happy medium.
And food companies spend quite a lot of time and money worrying about the mechanical properties of food.
This is an example of showing that the cells could be antisotropic, the cells could be elongated in one direction.
For instance, in the top one on the polyurethane foam.
And if they're elongated in one direction, it's not too surprising, you might have different properties in that direction from the plane perpendicular to it.
And then the bottom images is of pumice.
Pumice is a volcanic rock.
And you can see how the pores are kind of flattened out there.
And they're flattened out because that was once molten lava.
And the molten lava was flowing down a mountain side of a volcano.
And as it flowed, it got sheared.
And the shape of those pours reflects the shearing as the molten lava flow down the volcano.
And so this kind of sort of stretched out cell shape is going to give you antisotropic properties, different properties in different directions.
This is the 3-D truss that I'm passing around.
I don't know if it's exactly the same one, but it's a similar one.
And these trusses are triangulated structures.
And we'll talk a little bit about their properties too.
And then we also are going to talk about some applications.
So obviously, these materials are mostly air.
And that gives them a low weight.
And that means they're often used in structural sandwich panels as the core of the panel.
And these panels have stiff faces separated by a lightweight core.
And the idea is to make it a little bit like an I-beam.
So the way you have the flanges on the I-beam, the faces are like the flanges, and the porous core is like the web of the I-beam.
They can also undergo large deformations at relatively low stress.
And that means they can absorb a lot of energy.
So if you think of the energy as the area under the stress strain curve, if there's big strains and big deformations, then there's going to be a large area.
And that sort of energy absorption occurs at a fairly low stress.
So typically, when you want to absorb energy, it's not just how much energy you want to absorb.
You have to do it without actually breaking the thing you're trying to protect.
So you don't want to generate high stresses as you go along, and foams are good at this.
Foams are also good at being thermal insulators.
They have a low thermal conductivity.
And that's because they're largely made of gas and the gas has a lower conductivity than the solids.
So that gives them a lower conductivity.
And they have a large surface area.
And the smaller the pore size, the bigger the surface area per unit volume.
And that makes them good for things like carriers for catalysts.
And that's why they're used for these catalytic converters too.
OK, so here's some examples of cellular materials in medicine.
So here's some examples of trabecular bone.
Trabecular bone exists at the ends of your long bones.
So say in your hip or in your knee.
It also exists in your vertebrae in the middle of the spine in the vertebrae there.
And it also exists in your skull.
And you can see it's a porous type of bone.
It looks very similar to the foams and the sorts of mechanical models we make for foams can be applied to the bone as well.
And so that's one of the things we're going to do later in the course.
These are two slides showing what happens when people get osteoporosis.
The left hand slide is from a 55-year-old female to the same bone, the same slice.
And the right hand one is from an 86-year-old female.
This thing here, row star over row S, that's the relative density, the density of the bone divided by the solid that it's made from.
That's the same as the volume fraction of solids.
And so on the normal bone on the left it's about 17% solid.
And on the osteoporotic bone on the right it's, about 7%.
So you can kind of see the bone density has gotten lower, partly by thinning of the struts, but partly by resorption of the struts, as well.
And obviously the one on the right is going to have a lot lower strength than the one on the left.
These are micro CT images of bone.
And again, you can see how the structure looks different at different relative densities.
The one on the left is sort of in the middle at around 11% dense.
The one in the middle is the most dense 25%.
And the one on the right is 6% dense.
So it's not too surprising that the one on the right would have a much lower strength from the other ones.
And we'll look at how we can model that.
This is just showing some deformation in bone.
I have a colleague, Ralph Mueller who's got a micro CT machine, which allows you to do compression tests in the micro CT.
So he can make these sort of images where he scans it at zero strain.
He compresses it a little bit.
He scans it again.
He compresses it a bit more.
He scans it again.
And he these are stills from his images, but he makes animations from this.
And if you look at the top right up here, you see these struts here.
They're pretty straight in this one.
They're a little bent and starting to buckle here.
And then if you look at that one strut there, you can see how it's buckled right over.
So you can look at the deformation mechanisms by looking at the CT scans and things like that.
People are starting to think about using metal foams for coatings of orthopedic implants.
So one of the issues with implants is that say you have a hip implant or a knee implant, you remove the bone that's preexisting, and then you replace it with some sort of implant.
Typically, the implant has a stem that fits into the hollow part of the bone and then has a sort of joint piece to it that fits into the joint.
And you can get some loosening of the stem in the remaining bone.
And one idea is that you use porous coatings to minimize that.
And right now, typically what they do is center beads, metal beads on to the stem.
And another idea is maybe you could use a metal foam.
And these are some different types of metal foams that people are looking at.
Another type of cellular material in medicine is a tissue engineering scaffold.
And this just shows some different examples made by different processes.
And we'll talk more about these later on in the course.
This one here at the top left is a collagen based one made by a freeze drying process.
And I don't know if you saw MIT's website yesterday and today, Ioannis Yannas was the one who developed this.
And he's just being inducted into the National Inventors Hall of Fame.
And this is what he really was inducted for is he's invented a skin-- well, he calls it a dermal regeneration template for regenerating skin, mostly in people with serious burns.
Then these are some other sorts of scaffolds that are made by different processes.
This is by a sort of rapid prototyping process here.
The bottom two, these are kind of interesting.
These are actually the extracellular matrix in the body.
And they've had all the cells removed from it.
So these tissue engineering scaffolds are really designed to mimic the extracellular scaffold in your body or extracellular matrix in your body.
And you can see how when you remove the cells, the structure of those two things looks a lot like a closed cell foam.
So that's the kind of structure you're trying to replicate.
We'll look a bit at cell mechanics.
This is a cell contraction of a scaffold.
So here these sort of very thin transparent bits of the collagen based scaffold, and this is a fiber blast on it right here.
And I had a student, Toby [?
Fryman, ?]
who worked with me and Ioannis Yannas on this.
And you can see from the video the cell is actually contract in the scaffolding making it deform.
And you can calculate what forces the cell must be imposing on the scaffold by knowing something about the geometry of the struts and how the cells attached.
And then it's going a little bit more.
So we'll talk more about that.
And then there's a final picture down here, where you can see these two, the two points up here and down here have now been brought pretty much right together.
So we'll talk about that in more detail.
We've also done studies on cell attachment and how that attachment rate or the amount of cells that attach is related to the surface area, the surface area per unit volume.
So these are just some tests from that done by [?
Fergal ?]
O'Brien, a post-doc that worked with me.
We've also done some studies on cell migration.
And Brendan Harley was the student who did this.
And he stained the cells with one stain and he stained the scaffold with something else.
So red are the scaffold and green are the cells.
And then he used a confocal microscope to track the cells.
And he tracked the cells and where they moved versus time.
And if he has the location at different times he can get the velocity.
And one of the things he did was he changed the stiffness of the scaffold and he found that the migration speed depended on how stiff the scaffold was.
So he was looking at sort of interactions between mechanical properties of the scaffold and behaviors like cell migration.
And then we're going to look at materials in nature.
So here is wood.
So you can see the cellular structure of wood.
It's a lot like the honeycombs.
It has sort of a prismatic structure.
That one happens to be cedar, but other woods look similar.
Now, this is balsa wood.
And this is showing just how the balsa deforms.
I think this was loaded from top to bottom.
And this is at zero load.
And then this is more load, and more load, and more load.
And if you look at that cell there with this little kind of tear in it here, that's the same as the cell down here, and that's the tear there.
So you can see how the cell walls bend and how they deform.
And you can model that using the honeycomb models.
This is just another image showing actual failure of wood, buckling of the cell walls.
This is cork.
So these are modern scanning electron micrograph of cork.
And one of the interesting things is the cork cells have these little corrugations.
You see how they're not flat, they have little kind of wrinkles in them.
And that gives rise to sort of an interesting property of cork.
If you take cork and you load it.
So here we were pulling it in the direction of these arrows, pulling it like along this direction here.
And again, this is the same set of cells.
That tear there is the same as that tear there.
And you see all these little corrugations here, they've all straightened out when we're pulling on it.
And the Poisson's ratio of the cork is zero.
It's kind of like a bellows.
Like if you had an old camera, or you have an accordion bellows.
If you pull the bellows in and out, it doesn't get any wider this way or the other way.
You're just sort of opening the bellows and closing the bellows.
And the cork cells are doing kind of the same thing.
And it gives them this Poisson's ratio of zero.
Which it happens is one of the things that makes it easy to get the cork into your wine bottle, because as you're pushing on it, it's not pushing out in all directions.
This is only for this one direction, but it's not pushing out in all directions.
These are parenchyma cells in plants, in carrots and potatoes.
All those little blobs in the potato, those are starch blobs.
This is called a Venus flower basket sponge, and Joanna Aizenberg, at Harvard, has studied this quite a lot.
This has a hierarchical structure.
If you look at the overall sponge and then you look at each of the sort of struts that make up the lattice, that too has a hierarchical structure.
And she's looked at the optical properties of this glass sponge.
It's kind of a beautiful thing.
And then there's some cellular structures in nature as well.
There's sandwich structures.
There's density gradients.
And there's tubes with a cellular core.
So here's some examples of that.
Here's the iris leaf, you know the iris plant has these long kind of leaves that stand up like this.
And it's just like a sandwich panel.
The parenchyma are kind of like a foam in the middle here.
And there's very dense fibers called sclerenchyma that run up and down the length of the leaf.
And they're like fibers in a fiber composite.
And here's a bull rush or a cat tail leaf.
And they're like little I-beams almost.
It's like a whole series of little I-beams.
And again that's sort mechanically efficient.
These are examples of sandwich structures in bird skulls.
Some of you know I'm a birder.
So I sort of sneak in bird stuff from time to time.
But you can see how these birds skulls are all sandwich panels, and obviously birds want to minimize their weight for flight.
And this is one of the ways that they do that.
This is a horseshoe crab, sort of similar kind of thing.
This is from Mark Myers in San Diego.
He did a study on the crab in its shell as a sandwich.
And this is the ever so handsome cuttlefish.
And the cuttlefish has something called a cuttlefish bone.
This is the bone here and the bone is made up of these kind of sandwich type structures.
The cuttlefish is related to octopus and squid and things like that.
And it's hard to see in this picture here, but these little things here are actually like little tentacles.
There's several tentacles that it eats stuff with.
The cuttlefish is actually a mollusk.
All those things are mollusks.
It's called a fish, but it's not really a fish.
And here's an example of a natural material that has a radial density gradient.
Have you ever noticed if you look at a palm, like you see those pictures of Hollywood in LA, and the palm trees, you know, they line the street.
But they're all about the same diameter from the bottom to the top.
And when you think of like an oak tree, it's not like that.
It's big diameter at the bottom, skinny diameter at the top.
So when wood grows, the wood has more or less the same density in the bottom and at the top.
So as it's growing, the density is more or less the same.
And it resists the bigger loads from getting taller by adding circumference.
So it gets wider and wider as it gets older and older.
But palms don't do that.
They come out of the ground a certain diameter.
And most palms just grow that same diameter.
As they get taller and taller, you can imagine there's wind forces, and different kinds of forces are on it, the stresses get bigger and bigger.
And the way they resist those is that the cell walls get thicker, and they preferentially get thicker on the outside.
And if you think of moment of inertia, remember moment of inertia is increased more with the material on the outside of a beam.
And that's kind of what the palm is doing.
So if you look at young cell walls and old cell walls, here's some SCM pictures of young ones and it's sort of skinny.
And SCM pictures of older ones, and the cell wall has gotten thicker.
So we're going to talk a little bit about that.
And it turns out, this is an incredibly efficient way to deal with getting taller and needing to resist bigger loads.
Another material that has a radial density gradient is bamboo.
So this again shows these sort of dense sclerenchyma fibers.
Do you see these kind of dense parts here?
And you can see there's not very many of them here.
And there's more and more as you get towards the outside there.
So there's a density gradient there.
So we'll talk about that.
And some plant materials have a cylindrical shell with a compliant core.
Plant stems or commonly like.
This is a milkweed stem.
And you can see it's got these sort of fibers that are almost completely dense.
And then a sort of lower density core, cellular core, here, and a void in the very middle.
And you can show that that core helps prevent local buckling.
So if you take a drinking straw and you bend it, you get that kinking kind of failure.
You can show that having this sort of foam like core in the middle helps to resist that.
Imagine you have a drinking straw and now you put foam in the middle.
It's going to be harder to get it to kink like that.
So that's what the plants are doing.
Animal quills do the same thing.
That's a porcupine and a hedgehog quill.
And all of this stuff is in these two books.
So it doesn't really matter to me if you go out and buy the book.
I don't make very many much money on these books.
So this is not an income producing thing.
But those are the books that all these pictures have been taken from.
All right, so I think that's my sort of introduction to the class.
Are there any questions about how we're organized or what we're going to be doing?
Are we good?
It's OK?
OK.
So then I think what I'm going to do for the rest of the time is start the next section of the course which is on processing of cellular materials.
Now, I have another little hand out here.
So I don't know if I'll remember to do this for every lecture, but I like to have a little outline, that partly makes me be organized.
So it's just a little outline for the lecture
[INAUDIBLE]
You asked me to do what?
Put the room light back up
Put the what up?
Lights
Oh, the lights.
I'm going have another set of slides though.
So let me get out of the intro slides.
I know I have another set of slides.
So I'm going to just leave the screen up.
And kind of put stuff on the board and talk about the slides
That's fine
You're good?
OK. OK.
So I wanted to talk a little bit about processing of cellular solids and then, next time we'll start talking about the structures.
It seemed good to talk about the processing before we got to the structure.
So I'm going to talk a little bit about honeycombs and how they make honeycombs, and then foams, and then lattice materials.
Yeah?
The slide you're showing with the the shell with the foam inside it, are there techniques for analysis of it?
Well, I don't think we're going to get into all the gory details, but I can certainly give you references.
That's something that one of my students did at one point.
And in fact, I've been collaborating with Jennifer Lewis up at Harvard and she has a student who's making sort of cylindrical shells with foams out of ceramic foam.
So he's 3-D printing sort of with coaxial nozzles a cylindrical shell that's pretty solid.
And then a ceramic foam on the inside.
And that's one of the things he's playing around with.
So he's looking at ways you might make engineering versions of that.
So I wanted to start with looking at honeycombs and how they make honeycombs.
And I thought what I'd do is I've got some slides.
And I'm going to talk about the slides.
And then I'll put some notes on the board to kind of describe what we're doing, OK?
So this is the first sort of slide on the honeycombs.
And the two main techniques that they're made by, especially those aluminum honeycombs and paper honeycombs that I passed around, one technique is an expansion process.
So what they do is they take flat sheets of some metal, say aluminum, and they put little stripes of glue on it in different places.
So these little kind of specialty things are where the glue goes.
And then they stack those guys up in a sort of particular arrangement.
And then what they do is they pull it all apart, kind like a paper doll thing.
They pull it all apart and when they pull it apart, they get the hexagonal shape.
So let me just show on the board how you do the gluing and how that works.
So they would start with some sheets.
Say we start with two sheets like that.
They'd put some glue down, say there.
And then there's a gap.
And then they put glue on the opposite side over there.
And then there's another gap.
And then they put glue there.
And then they do the same positions but the opposite sides on the next sheet.
So they do that.
And then if you glue those together-- well, let me do another one.
Maybe do a couple more.
So then if I do one like that, it's glued there, and there, and there.
That guy gets there, and there and there.
And when glue that-- when you push that together and then take it apart, you've got something that looks like this.
So say I call that one, two, three, and four.
Then this would be 2 and 3.
OK, so this thing here is that.
Where it's not glued, you get them doing that.
And then it's glued again down here.
And so you get this kind of pattern.
And one of the things about these honeycombs that are made by the expansion process is these inclined walls have a thickness t. And because there's two sheets up here, the vertical walls have a thickness 2t.
So that's typically what you see in the commercial honeycombs that are made by this way.
And this process is used for aluminum honeycombs, for paper resin honeycombs, for Kevlar honeycombs.
And I'll just say note that the inclined walls have a thickness t, and the vertical walls are 2t.
So that's the expansion process.
And the process that's commonly used for honeycombs is called a corrugation process.
And for the corrugation process, it's just like the lower schematic here shows.
You take a flat sheet.
So you've got a roll of a flat sheet.
And you've got some rollers that have the right shape to give you the corrugated profile that you want.
You pass the sheet through the rolls and you get individual sheets out.
And each sheet is kind of a half hexagon.
And then you put the sheets together and that forms the whole hexagon.
So you have a flat sheet that's fed through a shaped wheel to form half hexagonal sheets, which you then bond together.
And it's the same kind of thing that the inclined walls have thickness t and the vertical walls have thickness 2t.
And this corrugation process, you can only really use it in materials that you can deform a fairly large amount to get the corrugations.
So typically, this is for metals.
And aluminum is probably the most common metal that this is used for
How are the corrugated sheets attached to each other?
I think they're just bonded with epoxy.
Yeah, so obviously if you wanted to use it for high temperature performance-- you know, all of these things are bonded with some sort of epoxy or some sort of resin.
So there's an issue if you wanted to use it higher temperatures.
So another process that's used to make ceramic honeycombs is an extrusion process.
And you just take a ceramic slurry and you pass it through a die.
And you can make a ceramic honeycomb by doing that.
And I believe that's how they make the ceramic honeycombs I passed around, the catalytic converter ones.
Other techniques involve rapid prototyping.
You can 3-D print honeycombs.
And Jennifer Lewis has a project on 3-D printing of honeycombs up at Harvard.
And one of the interesting things they're looking at is not just printing with an ink, but printing with a fiber reinforced ink.
So they're making cell walls of the honeycomb that are fiber reinforced.
And one of the tricks is trying to orient the fibers in the way that you want them to be oriented.
So there's rapid prototyping techniques as well.
You can use also selective laser centering.
Let's call it selective laser scanning.
So you can have a photosensitive polymer and use a laser to cure that and build up a honeycomb type structure.
And you can also cast honeycomb structures.
So those rubber honeycombs that I pass around, those are made by casting.
You take a liquid silicone rubber and you add a hardener and you pour it into a mold.
Another kind of interesting way that people have made-- well here's another example of the honeycombs that are 3-D printed.
And this is an example of-- or a couple of examples of looking at a bio carbon template.
So what that means is that these materials are based on the wood, but none of them are actually wood.
So what they do is they take wood, like they take pine, or they take beech or something.
They take some kind of wood and they carbonize it.
So that they do the same processes as used for making carbon fibers.
So you put the wood in an inert atmosphere.
And you pyrolyze it.
You heat it up to I think 800 degrees C. And all you're left with is the carbon.
And it preserves the structure.
And you replicate the structure.
You just get the same structure.
There's some shrinkage.
The shrinkage is about 30%.
But you get the same structure as the wood.
So this material up here is actually all carbon.
It's just replicating the wood that was used, the pine that was used.
An then what people have been doing is using that carbon structure and then infiltrating that with gaseous silicon.
And they form silicon carbide.
So these structures down here are all silicon carbide replicas of wood.
And they're thinking about using that for things like filters for high temperatures or for catalyst carriers.
And one of the attractive features is wood has fairly small cells.
The cells around 50 microns across.
And so you get a large surface area.
And this is a similar thing here.
These two are the carbon template.
And here they've used silicon and they've actually filled in the voids.
And so they've got silicon carbide where the cell walls used to be.
And they've got silicon where the void used to be.
So people are playing around with this is another way of making a honeycomb type of structure.
And they use other kinds of plants besides wood as well.
But that's the kind of general idea.
So the idea is that wood has a honeycomb like structure.
And the cells are fairly small.
the cells are in the order of 50 microns sort of in diameter and maybe a few millimeters long.
And this bio carbon template replicates the wood structure.
So the wood is pyrolized at 800 degrees C in an inert atmosphere.
So say an inert gas.
And that gives you the bio carbon template.
And you maintain the structure, although there's some shrinkage.
structures And then this carbon structure can then be further processed.
So for example, you can infiltrate it with a gaseous silicon.
And you end up with a silicon carbide wood replica.
So possible applications are things like high temperature filters, or catalyst carriers.
I think that's it on the honeycombs.
Are we good with all sorts of methods?
And my little talk here on processing is certainly not comprehensive.
I'm sure there's other ways people have developed.
These are some main ways.
All right then, I want to talk about foams as well.
People have developed different types of processes for different types of solids, so polymers, and metals, and ceramics.
So I just go through each class of solid and talk about that.
So the idea with polymer foams is that you want to introduce gas bubbles into either a liquid monomer or a hot polymer.
And then you want the bubbles to grow.
And then you want to stabilize them and solidify it by other cross linking or by cooling the hot polymer.
So there's a variety of ways of doing that, but let me just put that down.
So there's a few ways to get the bubbles in there in the first place.
One, is just by mechanical stirring.
So if you've ever made meringue, you know what that is, you just take a whisk and you beat egg whites and bubbles.
The air will get enveloped in the egg whites.
They also do that with polymers.
Or you can use a blowing agent.
And there's several varieties of blowing agents.
So the blowing agents are divided into physical and chemical blowing agents.
And the physical ones, they force the gas into solution under high pressure, and then you reduce the pressure, and the gas bundles expand.
So you can use physical blowing agents.
Or you can introduce liquids that, if you're using a hot polymer, that at the temperature of the hot polymer, they form a gas.
So that the liquid just turns into a gas.
And that would form vapor bubbles.
And then the chemical blowing agents.
There's a couple of different ways that those work.
You can either use chemical blowing agents where you have two parts that react together to form a gas.
And so that gas then blows the foam.
Or you can have a chemical blowing agent that reacts with the polymer to form a gas and that blows the foam.
So either way.
And you can have them decompose on heating.
So the same kind of thing.
Evolved the gas when it gets into the hot polymer.
So there's these different ways of blowing the foams.
And there's many, many different types of these blowing agents.
But, these are kind of the general techniques.
And whether or not the foam forms an open cell or a closed cell structure depends on the rheology of the polymers, so the viscosity of it, and also the surface tension.
Another way to make a foam is to make something called a syntactic foam.
A syntactic foam is made by taking thin walled hollow spheres and then using, say a resin, like an epoxy resin, to bond them together.
So you end up with something that's porous.
And you've got the void from the hollow sphere, but you don't foam it in the same way that you blow bubbles through it in some way.
One other thing about polymer foams is they sometimes have a skin on the surface.
So when you blow them, say you've got a mold, there will be a skin that forms against the mold, and sometimes the process is designed in a way that the skin is thick enough that it acts like a skin in a sandwich panel.
So they control the mold in a way and the blowing process so that they get a foam in the middle and thicker skins on the top and the bottom surface.
And that forms a sandwich panel.
Those are called structural foams.
Let's see.
So I think what I'm going to do next is the next section's on metal foams.
And I've got a few slides on that.
So I think I'm going to run through the schematics and just talk about it.
But, I'll put the notes on the board next time.
And there is one thing I forgot to do at the beginning.
I like to tell you a little about me.
And I want to hear about you.
So I wanted to leave a few minutes for that.
So let me just wait until people are finished writing stuff down.
And I'll go through these in a few minutes, and then we'll through it in more detail next time.
And I'll write notes down.
OK?
Are we good?
So there's a whole variety of ways of making foamed metals.
And most of them have been focused on aluminum.
But you could in theory do them with other types of metals.
So this was one of the first processes and it just involved taking a molten aluminum, so here's the aluminum down here in a crucible.
They added silicon carbide powder to it and then they just used a stirring paddle, like they just stirred it up and mixed gas in that way.
And they found that they got bubbles that rose up.
And then they used conveyor belts to just kind of pull the foam off.
And the thing about the silicon carbide was that if you didn't have that, then the bubbles wouldn't be stable enough that you could do this.
The bubbles would collapse before you got to be able to pull them up.
But the silicon carbide I think makes the aluminum melt more viscous and it helps prevent sort of drainage and collapse of the bubbles.
And so that's one way.
And there's a type of foam called Cymat, and this is an example of the foam that's made with that process.
Maybe I'll bring it next time and we can pass it around.
Another method is to use a metal powder and titanium hydride powder.
Then you can consolidate that.
So here's-- it's hard to see the writing, but this is a aluminum powder.
This would be the titanium hydride powder.
You mix them together and then you compact them.
You press them together.
And the titanium hydride decomposes and forms the hydrogen gas at a temperature at which the aluminum is not really quite molten but it's kind of viscous-y, kind of softening.
And so when the aluminum is soft like that and the titanium hydride decomposes and forms the hydrogen gas, you get a foam from that process.
And I think, somewhere, yeah, this foam here I think was made by that process.
Then in a similar thing, you can just put titanium hydride powder into molten aluminum, and again the titanium hydride powder evolves the hydrogen gas and you get foamed aluminum from that.
And I think this foam here was made with that process.
They all look kind of similar.
Another method is by replicating an open cell polymer foam.
So I think I passed an open cell polymer foam around.
And that's made by taking an open celled-- an open cell aluminum foam-- it's made by taking an open cell polymer foam, you fill up all the voids with sand.
You then burn off the polymer, but now you've got sand in all the sort of places where there were voids.
And then you infiltrate molten metal into that.
And then you get rid of the sand.
And then you're left with an aluminum foam that replicates the polymer that you started off with.
So this replication technique.
There's a vapor deposition technique.
And this was developed by Inco to make nickel foams.
So they take again an open cell polymer foam, that's kind of this thing here is, and they infiltrate into it nickel CO4.
The only teeny detail that's a problem with this process is that happens to be highly toxic.
So they put this gas through here.
And then they get nickel depositing on the polymer and they burn the polymer out and they center it.
So it is possible to do this.
They have done this.
But it's not that practical because of the toxicity of the gas.
Now another method is something called entrapped gas expansion.
And here, what you do is you take a can, like a metal can.
This one's titanium, a titanium alloy.
And then you put a titanium powder in here.
You evacuate the can.
So the can has a little valve on it, so you can evacuate it.
And then you backfill it with argon gas and you pressurize the argon gas.
So you've got a powder with sort of pressurized gas inside of a can.
And then you hot isostatically press it.
So you heat it up and you press it uniformly in all three directions.
And you compact it.
And then, if you want you can roll it.
Sometimes people roll it because they want to make sandwich panels, and they want to have a certain thickness, and they want to have faces on the panel.
But then you heat that up.
And as you heat it up, the gas evolves again.
And the thing expands and you get a foam that way.
So that's another method.
Another method is by making hollow spheres and then bonding the spheres together.
And in this process-- this was developed at Georgia Tech.
They used the titanium hydride again.
They made a slurry of the titanium hydride in an organic binder in a solvent.
And then they had a little kind of needle that they injected gas.
And so they had this slurry and they were blowing gas through this needle and they got hollow spheres of the titanium hydride.
And then they heated it up, and again evolved the hydrogen gas off.
But now they're just left with titanium spheres.
And then they bonded the spheres together.
And these aren't titanium.
This is an iron chromium, like it's not quite a stainless steel.
But this is the same thing.
I can pass that around next time too.
Those are the little beads that they make.
And then there's also fugitive phase technique.
So you can take say salt particles, put them in a mold and pour a liquid metal into that, and then, leach the salt away.
And I think that's it for the metal foams.
So I think I'm going to stop there for today in terms of the lecture.
I'll go over that again next time.
And I'll write stuff on the board.
But I wanted to tell you a bit about me.
So the people in 3032, they already know me because they had me in the fall.
But I see some unfamiliar faces.
So I thought I would tell you a little about me.
So I grew up in Niagara Falls in Canada, big power station.
Lots of big civil engineering works in Niagara Falls.
And my father worked at an engineering company that specialized in the design of hydroelectric power stations.
It was founded by the guy who designed the Niagara Falls power station.
And then I went to university in Toronto.
And I did a degree in civil engineering in Toronto.
And then when I finished my degree in civil engineering, I wasn't really sure what I wanted to do next.
And I applied for some jobs, and I applied to graduate school.
I applied to MIT.
And I didn't get in-- ouch.
But I did OK, it turns out.
And I ended up-- I had an advisor when I was in Toronto who had taken a sabbatical in Cambridge, England.
And he said he thought I might enjoy Cambridge, England.
And I ended up going to Cambridge, England to doing my PhD there.
And I worked on the cellular solids for my PhD.
And it was a nice combination because I was interested in material behavior and mechanics, but I had a background in me in civil engineering.
And these are just like civil engineering structures, but they're on a little teeny weeny scale, not like big buildings or bridges or something, like little teeny things.
And really all of this has come out of doing that PhD in Cambridge.
But when I was there, I never even thought about being an academic.
And I never applied for any academic jobs.
I didn't think I wanted to be an academic.
And I went and got a job in Calgary in the oil business.
And I was working at a consulting firm that did work for the oil business.
I hated it.
I just hated it.
It was like I had a boss.
I hated having this boss.
And, you know, the projects were too short term.
The winter in Calgary-- if you think this is bad, you've seen nothing.
Like less snow, but cold.
I mean like 30, 40 below, everyday, cold.
Real cold.
So I stayed there one winter.
And somewhere along the way there, I decided maybe the academic job thing would be good.
And I just sent my CV out to a bunch of Canadian universities.
And I ended up getting a job at the University of British Columbia in Vancouver.
And I lived in Vancouver for two years.
And I probably would have stayed there, except there was a gigantic recession, and it was all very depressing, and there was no money.
And that the universities in Canada are almost all run by the provincial governments.
And the government had no money.
It was all, you know, frustrating.
And I sort of thought, oh, I'll look around and see what else I can get.
And I answered a little ad in Civil Engineering Magazine for a job at MIT.
And I got the job at MIT.
And I was in the civil engineering department for about 12 years.
And then I moved over to the materials department.
Because my work started off on sort of sandwich panels and structural things.
And then it kind of became more biomedical stuff and had less and less to do with civil.
And I've been in the materials department since then.
And this is kind of what I do, this kind of work.
So that's kind of my little five minute story.
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I think, last time, we got as far as talking about processing of foams and we talked a bit about processing of polymer foams.
And today, I want to pick up where we left off.
And I was going to talk about processing metal foams.
We'll talk a little bit about carbon foams, ceramic foams, and glass foams.
We'll finish this section on processing today and then we'll start talking about the structure of cellular materials.
And hopefully-- we won't finish that today, but we'll finish it the start of tomorrow's lecture.
And then, we'll start doing mechanics of honeycombs tomorrow.
OK?
So that's the scheme.
I thought what I'd do-- I have a whole series of slides that show, schematically, a variety of different processes for making metal foams.
So I thought I would just go through the slides- and I think I did this really quickly last time-- but I would write down a little bit of notes on each of the slides so that you've got some notes on it, too.
This is the first method here.
And many of these methods were developed for aluminum foams.
But you could, in principle, use them for other types of foams, as well.
This first method here-- let me see if I can get my little pointer.
[INAUDIBLE] The idea here is that you take molten aluminum.
Down in this bath here, you've got molten aluminum.
And they put, into the aluminum, silicon carbide particles.
And the silicon carbide particles adjust the viscosity of the melt.
They make it more viscous.
And then they just have a-- I've got this thing here-- they've got a tube here that they blow gas in with.
And if you go to the bottom of the tube, there's an impeller, or a little paddle, that stirs the gas up.
And so the gas just forms in the molten aluminum here.
And then they have conveyors which pull off the molt-- or the metal foam, and then it cools.
The idea is that if you just had the aluminum, you couldn't really do this because the bubbles would collapse before they cooled down and became a solid.
But adding the silicon carbide particles increases the viscosity of the melt.
It helps prevent drainage of the foam.
Normally, if you have a liquid foam, just from gravity, you're going to have some drainage.
It's going to tend to-- some of the liquid is going to tend to sink, just from gravity.
By putting the silicon carbide particles in, you increase the viscosity.
It helps prevent the drainage.
And you can get the foam bubbles to be stable.
Let me just write one little note here.
The first method here involves just bubbling gas into the molten aluminum.
And that molten aluminum is stabilized by silicon carbide particles.
Sometimes people use aluminum particles and the particles increase the viscosity of the melt.
And that reduces the drainage and it stabilizes the foam.
That process was developed at Alcan in Canada and at Norsk Hydro in Norway.
And this foam here is an example of the foam that they've made.
And you can kind of see-- there's a density gradient in this foam.
I'll pass it around.
You can see it.
But the bubbles are smaller down here and there's fewer of them than there are up here.
And that's partly from the drainage.
The molten aluminum is draining down and you get more liquid at the bottom and then you get more solid at the bottom.
So that's the Alcan foam.
OK. Another-- there's half a dozen of these processes for metal foams.
Another version of the process involves taking metal powder and combining it with titanium hydride powder.
And then you consolidate it and you heat it.
So if I can show through the schematics here.
In the first step, you take the two powders, you mix them up.
So the second thing down here is mixing the powders up.
And then you press them together.
There's a dye here that you press it, and then you get pieces.
And then you can heat that up.
And the way that this works is that the titanium hydride decomposes at a temperature of about 300 degrees C. And if the other powder is something like aluminum-- aluminum melts at something like 660 degrees C-- the aluminum has become soft at 300 degrees C, but it's not molten.
And the titanium hydride-- when it decomposes-- forms hydrogen gas.
So the hydrogen gas forms the bubbles that you need to make the foam.
Are you riding your bicycle?
You are tough.
Very tough.
I ride my bike, but I gave up a couple weeks ago.
Well, three weeks ago, I guess.
When it started snowing, I gave up.
The idea with this process is you can use titanium hydride powder with aluminum and the aluminum becomes soft at the temperature that the titanium hydro decomposes.
And when it forms the hydrogen gas, that gives you the bubbles, and so you get the foam made by that process.
And that was developed by a place called Fraunhofer.
And I have one of their little foams here.
This is an example of their foam made by that powder metallurgy process there.
OK. Let me just write a little note about that.
So you can mix up titanium hydride powder with a metal powder and then heat that up.
When you do this, you need to have a metal powder that's going to be deforming by, say, high temperature creep at the temperature that the decomposition of the titanium hydride happens.
And for aluminum, that works.
So you need to have the metal material be able to deform fairly easily, in order to get these bubbles to form a nice foam.
But that's another way you can do it, is by consolidating these two powders.
Here's another way to do it.
You can also make use of this property of the titanium hydride-- that it'll decompose and form hydrogen gas-- by just putting it into molten aluminum.
And in this example here, you've got an aluminum melt in here.
And this time, they've added 2% calcium to it.
Again, to adjust the viscosity.
And then they add the titanium hydride in this step here and they've got a little mixer thing here that's spinning around and will mix it.
And then they'll put a lid on it to control the pressure and the titanium hydride will decompose, the hydrogen gas will evolve, and you'll get the foam made by that method, too.
So you can stir titanium hide right into a molten metal, as well.
OK.
So that's another method.
And I think that was made by something called the Alporas process.
And this is an example of one of their foams.
I'm pretty sure that's an Alporas foam there.
Yep
[INAUDIBLE]
They can't really control it perfectly.
In that first example that's going around-- maybe you might have missed it-- there's this drainage.
The first one's made by a molten process and you get drainage.
And you get different sized bubbles.
And you get different-- you get a density gradient in the thing, as well.
So you can't control these things perfectly.
OK.
Here's another method for the metal foams.
This one here involves replication.
In this method here, what you do is you start out with an open-cell polymer foam.
In this step up here, there's an open-cell polymer foam.
You fill that with sand.
So you fill up all the open parts of the cells with sand.
Then you burn off the polymer.
And so you've got a little channels where the polymer used to be.
And then you infiltrate those channels with the molten metal that you want to use.
So you replicate the polymer foam structure.
This is the infiltration process here.
This little thing here is your furnace.
And then you get rid of the sand and then you're left with a metal replica of the-- a replica of the original polymer foam.
And this example here is one of these things that's made by this replication process.
So that's an open-celled aluminum.
I think the-- if you look at the density of these things, they're fairly low.
And so there's quite a large volume of pores and they're all interconnected.
So it's not that hard to get the sand out.
This method would involve replication of the open-cell foam-- the polymer foam-- by casting.
So you fill the open-cell polymer foam with sand.
You burn off the polymer.
And then you infiltrate the metal into that.
And then you remove the sand.
OK?
Then, another process involves just using vapor depositions.
So you take an open-cell polymer foam again.
Here-- let me use my little arrow.
Here's the open-cell polymer foam up here.
And you have a furnace here with a vapor deposition system.
And they use a nickel CO4 system to do this.
You then burn out the polymer.
You're left with a metal foam with hollow cell walls, where the polymer used to be.
And then usually, what they do is they center it.
They heat it up again to try to densify the walls.
The only teeny weeny problem with this process is that the gas they use this incredibly toxic.
And so it's not cheap and it's got health hazards, as well, associated with it.
But you can do it.
That gives you a nickel foam when you're finished with that.
And you could also use an electrodeposition technique that's similar.
OK. That's another method.
Another method is shown here.
This is the entrapped gas expansion method.
And what they do in this method is they have a can and the can has a metal powder in it.
It's whatever metal you want to make the foam out of.
In this example here-- it's probably too small for you to read in the seats-- but it's titanium.
They've taken a titanium alloy.
They've got a powder of titanium alloy.
They then evacuate the can.
They take all the air out of it.
And then they back fill it with argon gas.
They put in an inert gas in there.
And then what they do is they pressurize and heat the thing up and so the gas is internally pressurized by doing that.
And then, sometimes when they do this, they want to have a skin on the two faces.
So in this next little image here, it's done where they roll the can and produce a panel that's got solid faces.
And then when you heat it up, the gas expands and you end up with a sandwich panel by doing this.
So this bottom figure down here-- I'm not having much luck with the pointer.
This bottom figure down here, they've heated it up.
The gas expands and you've got this solid skin on the thing, which is from where the can use to be.
So that's trapping of a gas.
OK?
There's a couple more methods for the metal foams.
One involves centering hollow metal spheres together.
And the trick there is to make the hollow spheres.
And the way that can be done is by taking titanium hydride again, if you want to make titanium spheres.
You put it in an organic binder-- in a solvent-- so you've got a slurry here.
And you've got a tube that you blow gas through.
And as you do that, you get hollow titanium hydride bubbles.
And then you can do the same thing where you heat that up.
The hydrogen gas evolves off, and you're left with titanium.
You're left with titanium spheres down at the bottom here.
And then you can pack those together and press those and form a cellular material that way.
You can also center hollow metal spheres.
And the last method I've got for the metal foams is that you can use a fugitive phase.
With the fugitive phase, you would take some material that you could get rid of at the end of the process.
Say, something like salt, that you could leach out.
Here, we have our bed filled with salt.
And then you would infiltrate that with a liquid metal, typically under pressure.
And then, after the metals cooled, you leech out the salt.
You get rid of that.
You can pressure infiltrate a leachable bed of particles, and then leach the particles out.
OK. We have a whole variety of methods that have been developed to make metal foams.
And most of these have been developed, probably, in the last 20 years.
Something like that.
Some of them are a little bit older than that.
But there's been a lot of interest in this recently.
Those are all methods to make metal foams.
I wanted to talk just a little bit about a few other types of foams.
People make carbon foams.
And they use the same kind of method as they do to make those bio carbon templates I told you about.
When you take wood and you heat it up in an inert atmosphere and it turns into a carbon template, you can do the same thing where you take a polymer foam, you heat it up in an inert atmosphere, and everything except the carbon is driven off.
And you're left with a carbon foam.
It's the same process they used to make carbon fibers.
There's carbon foams.
There's also ceramic foams that you can make.
I brought the little sample of ceramic foaming again.
You can pass that around.
And those are typically made by taking an open-cell polymer foam and passing a ceramic slurry through the polymer foam so that you coat the cell walls.
And then you fire it so that you bond the ceramic together and you burn the polymer off.
And you're left with a foam that's got hollow cell walls.
You can also make ceramic foams by doing a CVD process on the carbon foam that you could make by the previous process.
And people also make glass foams.
And to make glass foams, they use some of the same kinds of processes as people use for polymer foams.
I'll just say similar processes to polymer foams.
OK. That covers making the foams, and I think we talked about the honeycombs last time.
I wanted to talk a little bit about making what are called 3D lattice materials, or 3D truss materials, as well.
Let me [?
strip ?]
that up there.
Yeah?
[INAUDIBLE]
Chemical vapor deposition
[INAUDIBLE] use this for metal foams?
Well, people were quite interested in using them for sandwich panels-- the cores of sandwich panels-- lightweight panels.
There was interest in using them for energy absorption, say, car bumpers.
The automotive industry was quite interested in this, in terms of trying to make components with sandwich structures that would be lighter weight, or energy absorption for bumpers.
Or filling up-- if you take-- say you take a metal tube.
If you think of a car chassis and it's made of tubes, if you fill those tubes with a foam-- especially if you fill them with a metal foam-- you can increase the energy absorption quite substantially.
So when you have a tube in a chassis, if it's loaded axially, it will fold up and you get all these wavelengths of buckling.
And if you've put a foam in there, it changes the buckling wavelength and it increases the amount of energy you can absorb.
So not only is the energy absorbed by the foam itself, it actually changes the buckling of the tube so you can absorb more energy.
So there was a lot of interest in that.
There was an interest in using them for cooling devices for, say, electronic components.
The idea was you would take, say, an aluminium open-cell foam and you would flow air through that.
And say you have your device that's generating heat, you'd have a foam underneath it.
And the aluminum conducts the heat fairly well.
And then you would blow air through it to try to cool it off.
So there were a bunch of different applications people have had in mind for them
What about glass?
Glass foams, I think, are largely used for insulation in buildings, believe it or not.
I think, actually, one of the dorms at MIT-- maybe the Simmons dorm-- has a glass foam insulation
[INAUDIBLE]
Well, I think because the foam-- because the cells are closed, the gas is trapped in the cell.
Whereas with a fiberglass, gas could move through the fibers more easily.
So I think that's partly how it works.
OK. Well, let me talk a little bit about the lattice materials, too, and how we make those.
We're going to start talking about the modeling of honeycombs and foams.
And when we do that, we're going to see that if we have a structure that deforms by bending, the properties vary with the amount of material, in a certain way.
But if we have materials where the deformation is controlled by axial deformation, the stiffness and the strength are going to be higher at the same density.
People made these lattice-type materials to try to get something with a more regular structure, and especially a triangulated structure.
You see how these things are like little trusses?
Triangulated?
Triangulated structures, when you load them-- say I load this like this-- there's just axial components-- axial forces in each of the members.
And so, theoretically, this would be higher stiffness and strength for a given weight than, say, a foam would be.
So people were interested in these lattice material.
This one here is made of aluminum.
And I wanted to talk a little bit about how you can make these things.
One way you can do it is by injection molding.
And this here is just the centerpiece of something that would look like this.
So there would be a-- I didn't bring it, but there's a top face and a bottom face that fit onto this.
And they would be injection molded as three different pieces, and then assembled together.
So you can make a mold in this complicated geometry, and you can make a lattice material by injection molding.
We'll start with polymer lattices first.
One way is injection molding.
Another way to do it is by 3-D printing.
You can generate a structure like that by 3-D printing.
You can also make trusses in 2D and you can make them so that you can snap fit those together.
So you can make little 2D trusses.
Here's a little truss here and here's a little piece of a truss here.
And you can make a little snap fit joints.
Do you see how these ones have little divots in them?
And you can make it so that these things will fit together.
I think these guys-- can I do it?
No.
You'd have to take-- oops.
Wait a minute.
No, it's not that way.
There we go.
So you can snap them together like that.
OK.
I can't get it to-- there we go.
So you can do that.
And you do that over and over again.
And if you do it over and over again, you get something that looks like that.
OK?
You can make a snap fit thing.
Let me pass all these guys around and you can play with those.
That's the injection molding one.
This is the snap-fit one.
Got that?
Let's see.
I think I have a little picture here.
This is an example of the snap fit truss here.
It's the thing that's getting passed around.
And another clever way that was developed was by taking a monomer that's sensitive to light.
You take a photo sensitive monomer.
And you put a mask on top of it and the mask has holes in it.
And then you shine collimated light on it.
You shine, say, a laser on it.
And the light goes through the holes in the mask.
And it starts to polymerize the polymer because it's photosensitive.
And then, as the polymer-- as it polymerizes and becomes solid, it then acts as a waveguide and draws the light down deeper into the monomer.
And so the polymer acts as a wave guide.
It brings the light down.
And this is a schematic over here.
This is a schematic showing the set up.
And these are some examples of some 3D trusses that they've made using this technique.
And one of the nice things about this is you can get a very small size cell size.
So this is-- I think that bar-- it says 1,500 microns.
That's what?
One and a half millimeters.
So you can get a nice, small cell size if you want that.
Let me write that down.
You take a photosensitive monomer and then you have it in a mold beneath a mask.
And then you shine collimated light on it.
And as the light shines on it, it polymerizes the monomer.
So then it solidifies and then it guides the light deeper into the monomer.
OK. That's that.
And then finally, there's metal lattices, as well.
And so this is, obviously, a metal lattice here.
It's an aluminum alloy.
And the metal lattice is made by taking that polymer lattice that was made by the injection molding technique.
You coat that with a ceramic slurry.
You burn off the polymer, and then you infiltrate the metal where the polymer used to be.
OK. That's the section on processing of the honeycombs and the foams and the lattices.
So there's a variety of different techniques that people have developed for making these kinds of materials.
And I thought it'd be useful to just describe some of the techniques.
As I think I said last time, this isn't comprehensive.
This doesn't cover every technique.
But it gives you a flavor of what techniques people have developed for making these kinds of materials.
OK?
Are we good?
We're good.
OK.
The next part, I want to do on the structure of cellular materials.
And I have a little overview.
I don't think we'll finish this today, but we'll finish it tomorrow.
Yeah?
[INAUDIBLE]
Down here?
What happens to the ceramic?
They get rid of the ceramic.
Typically, the ceramic is not very strong and it's just fired enough so they can infiltrate it with the metal.
And then they-- yeah, I think with mechanical smushing around, you can get rid of the ceramic.
And the ceramic's brittle, so if you break the ceramic, you're not going to break the metal
I'm wondering if you could make a type of metal lattice [INAUDIBLE] with reducing [?
the ?]
oxides?
I guess you could, if you could-- but you'd have to then make the oxide in that shape, too.
You've always got to make something in that shape
[INAUDIBLE]
Yeah, maybe you could make a foam.
But to make these lattices, you need this really regular kind of structure and be able to control the structure.
OK. Let me scoot out of this set of slides and get the next set up.
OK. We want to talk about the structure of cellular solids.
And we classify cellular materials into two main groups.
One's called honeycombs.
This thing down here is a honeycomb.
And honeycombs have polygonal cells that fill a plane and then they're prismatic in the third direction.
So you can think of them as just being a prismatic-- and they can be hexagons, they can be squares, they can be triangles-- but you can think of them as prismatic cells.
And the cells are just in a 2D plane.
And then we also have foams.
All of these ones over here are foamed materials.
And they're made up of polyhedral cells.
The cells themselves are three-dimensional polyhedra.
And this slide here shows a number of different types of foams.
These ones are polymers up here.
These are two metals.
These are two ceramics.
This is a glass foam down here.
And this is another polymer foam down here.
OK?
[INAUDIBLE]
No.
I just know that

I took those pictures so I know that.
No, you can't tell by looking at them.
In fact, that's one of the things about how we model the cellular materials.
The fact that their structure is so similar is what gives them similar properties.
And they behave in similar ways because they've got similar structures.
OK. We've got 2D honeycombs, where we have polygonal cells that pack to fill a plane.
And then they're prismatic in the third direction.
And then we have what we call 3D cellular materials, which are foams, which have polyhedral cells.
And then they pack to fill space.
The properties of all of these materials depend, essentially, on three things.
They're going to depend on the solid that you make it from.
If you make the material from a rubber or from an aluminum, you're going to get different properties.
So they depend on the properties of the solid.
And some of the properties that we're going to use that are important for this type of modeling are a density of the solid-- which I'm going to call rho s-- a Young's modulus of the solid-- which I'm going to call es-- and some sort of strength of the solid-- which I'm going to call sigma ys for now.
And you could think of other things.
There could be a fractured toughness of the solid.
There could be other kinds of things.
One thing that the properties of the cellular material depend on is the properties of the solid.
Another is the relative density of the cellular material.
And that's the density of the cellular thing divided by the density of the solid.
And that's equivalent to the volume fraction of solids.
So it makes sense that the more solid you've got, the stiffer and stronger the material's going to be.
So it's going to depend on how much material you've got.
And it also depends-- the properties also depend on the cell geometry.
The cell shape can control things like whether or not the honeycomb or the foam is isotropic or anisotropic.
You can imagine, if you have a foam, and you've got equiaxed cells, you might expect to have the same properties in all directions.
But if you had cells that were elongated in some way, you might expect you'd have different properties in the direction that they're elongated relative to the other plane.
So cell shape can lead to anisotropy.
For the foams, you can also have what we call open-cell and closed-cell foams.
If you look at this slide here, and we look at this top right images-- these two up here-- the one on the left in the top is an open-celled foam.
There's just material in the edges.
There's no faces.
And so a gas can flow between one cell and another.
And then if you look at the one on the right, this is a closed-celled foam.
There's faces.
If you think of the polyhedra, you've got solid faces covering the faces of the polyhedra.
For an open-cell foam, you've only got solid in the edges of the polyhedra.
And the voids are continuous, so they're connected together.
And for a closed-cell foam, you've got solid in the edges and the faces.
And then the voids are separated off from each other.
So we'll say, the cells are closed off from one another.
Another feature of the cell geometry is the cell size.
And the cell size can be important for things like the thermal properties of foams.
It's important for things like the surface area per unit volume.
But typically, for the mechanical properties, it's not that important.
And we'll see why that is when we do the modeling.
OK.
Yes?
For the closed-cell foams-- because we can't really see it without cutting it, is it that all of the faces are closed?
Or is it like some fraction of the faces are closed?
If you look at this one on the top right here, they're pretty much all closed.
But the reason we have this little picture down here is some of them are closed and some of them are open.
So you can get ones that are in between.
But typically-- this is kind of unusual.
Usually, they're either all open or all closed.
If we look at the mechanical properties of cellular materials, typically the cell geometry doesn't have that much of an effect.
The relative density is much more important.
The relative density, we define as the density of the cellular solid.
And when I use a parameter like rho or e or something, if it's got a star, it's for the cellular thing and if it's got an s, it's for the solid.
So rho star is going to be the density of the cellular material.
And rho s is going to be the density of the solid it's made from.
And so the relative density is just rho star over rho s. And I just wanted to show you how this is the volume fraction of solids.
So rho star is going to be the mass of solid over the total volume.
Imagine you've got a honeycomb or a foam and you've got, say, a unit cube of it, the sum total volume of the whole thing-- the density of the cellular material is going to the mass of the solid over the whole volume.
The density of the solid is going to be the mass of the solid over the volume of the solid.
This is really just equivalent to the volume fraction of solids, how much solids you've got.
And that's also n equal to 1 minus the porosity.
Typical values for cellular materials-- I think last time I passed around one of those collagen scaffolds-- those tissue enineering scaffolds.
It was in a little plastic bag.
That collagen scaffold has a relative density of 0.005, so its 0.5% solid and 99.5% air.
And if we look at typical polymer foams, the relative density is typically between about 2% and 20%.
And if we look at something like softwoods-- wood is a cellular material.
And we look at softwoods, the relative density is usually between about 15% and about 40%.
Something like that.
OK?
As the relative density increases, you get more material on the cell edges, and if it's closed-cell foam, on the cell faces.
And the pore volume decreases.
And you can think of some limit.
If you keep increasing the relative density more and more and more, eventually you've got-- it's not really a cellular material anymore.
It's more like a solid with little isolated pores in it.
And so there's two bounds.
And the density has to be less than a certain amount for you to consider it a cellular material in the models that we're going to derive to be valid.
And if the relevant density is more than a certain amount, people model it as a solid with isolated holes.
If I have a unit square of material, if it's a cellular material, you might expect that you've got pores that would look like this.
And you've got relatively thin cell walls, relative to the length of the material.
And for a cellular material, typically, the relative density is less than about 0.3.
And when we come to the modeling for the honeycombs and the foams, we're going to see that the cell walls deform, in many cases, by bending.
And that you can model the deformation by modeling the bending.
And that the bending dominates the behavior if the density is less than about that.
And at the other extreme, you can imagine if you had just little teeny pores.
I have a little pore here and one here and one there and one there.
That's not really a cellular material.
It's just got a teeny weeny little bit of pores.
And that could be modeled as isolated pores in a solid.
Each one is acting independently.
And people have found that that is appropriate if the relative density is greater than about 0.8.
And then, in between, there's a transition in behavior between the cellular solid and the isolated pores in the solid.
OK?
Are we OK?
The next thing I wanted to talk about was unit cells.
Especially for honeycombs, people often use unit cells.
A hexagonal cell is an obvious one to use to model this kind of behavior.
For honeycomb materials, you can have unit cells and you can have different ones.
On the left here, we've got triangles, in the middle, I've got squares.
On the right-hand side, I've got hexagons.
And you can see, even if you have a certain unit cell, there's also different ways to stack it.
So the number of edges that meet at a vertex is different for, say, this example on the top left and this example on the bottom left.
Here, we've got six members coming into each vertex, and here, we've got four.
And again, this stacking for the two square cells is also different.
So you can have different numbers of edges per vertex.
Another thing to note that's kind of interesting-- if you look at the honeycomb cells here, this one on the top left-- this equilateral triangle one with the stacking-- and this one on the top right-- the regular hexagonal cells-- those two are isotropic for linear elastic behavior, whereas all the other ones are not.
So we have 2D honeycomb unit cells.
We can have triangles, squares, and hexagons.
They can be stacked in more than one way.
And that gives different numbers of edges per vertex.
And in that figure, a and e are isotropic, for linear elasticity.
OK.
When we come to modeling the honeycombs, we're going to focus on the hexagonal cells.
We'll talk a little bit about the square and triangular cells, as well.
And then, for foams, when you look at the structure of a foam, it's obviously not a unit cell that repeats over and over again.
But people started off trying to model the mechanical behavior of foams by looking at periodic repeating polyhedral cells.
And there's three cells here that are prismatic.
We're not really going to talk about those beyond this.
So they're not really physically realistic or interesting.
But people would use these two cells here in initial attempts to model foams.
And this one here is called the rhombic dodecahedra.
Rhombic because each of the faces has four sides and dodecahedra because each polyhedra has 12 faces.
I forget if I've bored you with my Latin already.
Hedron means face in-- oh, this is Greek, I think.
Hedron means face.
Do is two, deca is 10.
So dodeca is two plus 10.
It's got 12 faces.
OK?
So that's the rhombic dodecahedra over here.
And then this bottom one down here is a tetrakaidecahedra.
It's a similar thing.
Tetra's four, kai mean and.
Four and 10-- tetra kai deca-- it's got 14 faces.
OK?
And those two pack to fill space.
I think those are the only uniform polyhedra that pack to fill space.
Here is the 3D foams.
We have the rhombic dodecahedron and the tetrakaidecahedron.
And the tetrakaidecahedron packs in a bcc packing.
Initial models for foams-- they took these two unit cells.
And what they would do is have an infinite array of them to make up the whole material.
And then they would isolate a unit cell.
And they would apply loads-- some say compressive stress, for example.
And then they would figure out what the load, or force, was in every single member, and how much that member deformed.
And they would figure out the component of the deformation in the same direction that they were putting the load on.
And they would figure out things like a Young's modulus, or they would figure out when there was some failure of one of these struts, and they would figure out a strength for the foam.
But you can kind of imagine, geometrically, not that easy to keep straight.
A little bit complicated.
So one way to model foams is by using these unit cells.
But we're going to talk about a different way to do it, as well.
OK.
So those are unit cells.
When they make foams, as we just talked about, one way to make a foam is by blowing a gas into a liquid.
And if you blow a gas into a liquid, then the surface tension is going to have an effect on the cell geometry and on the shape of the cells.
And if the surface tension is isotropic-- if it's the same in all directions-- then the structure that you get is one that minimizes the surface area per unit volume.
And so people were interested in what sort of cell shape minimizes the surface area per unit volume.
And Lord Kelvin, in the 1800s, was the person who worked this out.
And this is called the Kelvin tetrakaidecahedron.
And it's not just a straight tetrakaidecahedron.
There's a slight curvature to the cells here, to the faces.
And you can kind of see it in some of the edges here.
Like if we-- let me get my little pointer.
If you look at that edge, it's not straight.
This edge here is not straight.
It's got a little bit of a curvature to it.
But this minimizes the surface area per unit volume.
And then more recently, in the 1990's, there were two people-- Dennis Weaire and Robert Phelan-- discovered that this structure here-- which isn't a single polyhedron, but it's made up of a few polyhedra.
That has a slightly smaller surface area per unit volume.
Smaller by 0.3%.
So, a tiny bit smaller.
OK. Let's see.
What I'll say here is that foams are often made by blowing a gas into a liquid.
And if the surface tension controls and it's isotropic, then the structure will minimize the surface area per unit volume.
OK. That's relevant if the foam is made by blowing a gas into a liquid and surface tension is the controlling factor.
Sometimes foams are made by supersaturating a liquid with a gas, and then you nucleate bubbles, and then the bubbles grow.
So there's a nucleation and growth process.
So that's a little bit different.
And if you have a nucleation and growth process, you get a structure that is similar to something called a Voronoi structure.
In an idealized case, imagine that you have random points that are nucleation points and that you start to grow bubbles at those nucleation points.
So you start off with these random points.
And the bubbles all start to grow at the same time and they all grow at the same linear rate.
If you have that situation, then you end up with this Voronoi kind of structure.
And I've shown a 2D version of it here just because it's easier to see in 2D, but you can imagine a 3D system.
And in order to make one of these Voronoi honeycombs, you can imagine-- if you have random points-- here, say that little point there is one of the nucleation points, and here's another point here-- you form the structure by drawing the perpendicular bisectors between each pair of points.
This is the bisector between these two points.
Here's a bisector between those two points.
And then you form the envelope of those lines around each nucleation point.
And that, then, gives you that structure.
And you can see, this structure here is kind of angular.
It doesn't look that representative of something like a foam.
But if you have an exclusion distance, where you say that you're not going to allow the nucleation points to be closer than some given distance-- your exclusion distance-- then you get this structure here.
And this starts to look a lot more like a foamy kind of structure.
So these Voronoi structures are representative of structures that are related to nucleation and growth of the bubbles, or nucleation and growth processes.
Let me write down something about Voronoi things.
And these Voronoi structures were first developed to look at grain growth in metals.
They weren't developed for cellular materials.
But you can use them to model cellular materials, as well, as long as it's a nucleation and growth process.
We'll say that foams are sometimes made by supersaturating a liquid with a gas, and then reducing the pressure so that the bubbles nucleate and grow.
So initially, the bubbles are going to form spheres.
But as the spheres grow, they start to intersect with each other and form polyhedral cells.
And you get the Voronoi structure by thinking about an idealized case in which you randomly nucleate the-- you have nucleation points at a randomly distributed space.
They start to grow at the same time and they grow at the same linear rate.
OK.
The Voronoi honeycomb, or the foam-- you can form that by drawing the perpendicular bisectors between the random points.
So each cell contains the points that are closer to the nucleation point than any other point-- or any other nucleation point.
And if we just do this process as I've described here, you end up with a Voronoi structure, where the cells appear kind of angular.
And if you specify an exclusion distance, where you say the nucleation points can't be closer than a certain distance, then the cells become less angular, and of more similar size.
OK.
So are we good with the Voronoi honeycomb nucleation and growth idea?
Alrighty.
All right.
If we think about cell shape-- if we start with honeycombs and we just think about it hexagonal honeycombs, if I have a regular hexagonal honeycomb so that all the edges are the same length and this angle here is 30 degrees, then that is an isotropic material in the plane in the linear elastic regime.
One of the things we're going to do is calculate-- if I loan it this way on, what's the Young's modulus?
If I load it that way on, what's the Young's modulus?
And we're going to find they're the same, in fact, no matter which way on I loaded it.
It would be the same.
But if I now have my honeycomb, and imagine that I stretched it out-- and I'm kind of exaggerating how much we might stretch it out.
But if we did something like that, it wouldn't be too surprising to think that the properties are going to be different if I loaded it this way on and that way on.
And in terms of the cell geometry, I'm going to call that vertical cell edge length h. And I'm going to call this one-- the inclined one-- of length l. I'm going to say that angle is the angle theta.
And the cell shape can be defined by the ratio of h over l and that angle theta.
OK.
When we derive equations for the mechanical properties of the honeycombs, we're going to find that they depend on some solid properties.
Say, the modulus of the honeycombs can depend on the modulus of the solid.
It's going to depend on the relative density raised to some power.
And we're going to figure out what that is.
And then it's going to depend, also, on some function of h over l and theta.
And that function really represents the contribution of the cell geometry to the mechanical properties.
OK. That's the honeycombs.
It's fairly straightforward to characterize the shell shape for the honeycombs.
It's a little more involved to do it for the foams.
And the technique that's used is called the mean intercept length.
At least, that's one technique that's used.
Let me wait until you've finished writing because I want you to see the picture as I talk about it.
OK?
OK.
Here's-- whoops.
My pointer keeps disappearing.
This top left picture here shows an SEM image of a foam.
And you can see, you've got some big cells and little cells and there's no obvious way to characterize the cell shape.
And what people do to calculate this mean intercept length is they would take an image.
They would then sketch out just the cell edges that touch a plane's surface.
So all these black lines are just the-- if you took your-- if you put ink on your foam and you just put it on a pad and put it on a piece of paper, you would get this outline of the edges of the cells, where they intersect that plane.
And then what people do is they draw test circles.
Here's the test circle here.
And they draw parallel equiaxed, or equidistant lines.
So the lines here are parallel.
They're all at, say, zero degrees.
And they're all the same distance apart.
And then they count the number of intercepts.
They count-- say we went out here.
The cell wall intercepts here.
There's one that intercepts here.
And then, it'd go up here.
Here's another intercept.
Here's another intercept.
So they count the number of intercepts of the cell wall with the lines.
And then they get a mean intercept length, which is characteristic of the cell dimension.
And then what they do-- because this is just in one orientation-- you would then rotate those parallel lines by, say, 5 degrees and do it all over again.
And get another length at 5 degrees and one at 10 degrees one at 15.
And so you get different lengths for the intercepts as you rotate your parallel lines around.
And then you make a polar plot, and that's what the thing is down at the bottom here.
And so you plot your mean intercept length as a function of the angle that you measured it at.
And you can fit it to an ellipse.
And if you do it in three dimensions, you fit it to an ellipsoid.
And the major and minor axes of that ellipse, or ellipsoid, are characteristic of how elongated the cell is in the different directions.
And the orientation of that ellipsoid is characteristic of the orientation of the cells.
Those of you who took 303, too, you remember Mohr's circles?
Is this beginning to look familiar?
It's the same kind of idea as Mohr's circles.
Same way we have principal stresses and orientation of principal stresses, now we have principal dimensions and the orientation of the principal dimensions.
So it's the same kind of idea.
OK?
Let's see.
I feel like I'm getting to the end here.
Maybe I'll stop there for today.
But next time, I'll write down the whole technique about how we get these mean intercepts and get this ellipsoid.
And I'm going to write the mean intercepts down as a matrix.
But you could also write it as a tensor.
And there's something called the fabric tensor, which characterizes the shape of the cells.
And as you might imagine, the same is with the honeycomb.
If you have equiaxed cells in the foams, you might expect you would get isotropic properties.
If you have cells that are stretched out in some way-- so you've got different principal dimensions for them-- then you've got anisotropic material.
And you can relate how much anisotropy to the shape of the cells.
OK.
I'm going to stop there for today.
I'll see you tomorrow.
Seems very sudden.
I'll see you tomorrow.
I'll pick up and I'll finish this section on the structure.
We've got a bit more to do.
And then we'll start looking at honeycombs and modeling honeycombs.
The honeycombs are simpler to model just because they have this nice simple unit cell.
So we'll start with that, and then we'll move from there to the foams.
OK?
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All right, so I guess I should start.
So I think last time we were talking about cell structure and cell geometry.
And I got as far as putting this image here up.
And I talked a little bit about how this works.
And I was going to go over it again and then write the notes down today.
So we'll start from there.
So we're going to do a little bit more on cell structure today.
We're going to talk about some topological laws for cellular materials for polyhedral cells.
And then we'll start talking about modeling honeycomb materials, and talk about how we look at the mechanical properties of honeycombs.
So I think where we left off last time was we started talking about mean intercept length.
So the idea is that with-- hello, hello-- if you have a honeycomb, it's fairly easy to define the cell shape in terms of the ratio of the cell edge lengths, h over l, and the angle that the inclined cell wall is to the vertical one.
But for a foam it's a little more difficult.
And what people do is they measure this mean intercept length.
So last time I talked about it briefly.
So the idea is you take an image of the structure you're interested in.
You sort of draw out the outline of a planar surface.
And then you put on that a test circle.
And you superimpose on the test circle equidistant parallel lines.
So say we start off at theta equals 0.
We then count how many times the cell walls intercept those lines.
And we get a number of cells per unit length.
And that gives us an intercept length for that orientation.
Then we rotate it around a little bit, say 5 degrees or something.
And we measure a new intercept length for that orientation.
And we keep doing it all the way around, 180 degrees.
And then you get-- then what you do is you plot those points-- if I can find my little pointer-- we plot those points.
And it will form an ellipse.
And the major and minor axes of the ellipse correspond to the principal dimensions of the cells on that plane.
And then you can do the same for perpendicular planes and form an ellipsoid.
And you can get the three principal dimensions.
You can also get the orientation of the cells by getting the orientation of this ellipse, or the ellipsoid in three dimensions.
So let me just write down some notes that kind of summarize all of that.
So I'm going to say that for foams, we characterize the cell shape and orientation by using these mean intercepts-- mean intercept lengths.
So we consider a circular test area on a plane section.
So you want to use a circle and not say, a square or a rectangle.
Because if you had a square, then you'd have different total lengths of line, depending on what orientation you took it.
So you want to use a circular test section.
And then you draw equidistant parallel lines.
So for example you might start at theta equals to 0.
And then you count the number of intercepts of the cell walls with the lines.
So if we say that Nc is the number of cells per unit length of line, then we can get the intercept length for that orientation.
Say for theta equals to 0, it works out to 1.5/Nc.
So it's not just 1/Nc, because you may not be cutting the cell-- you know, you're cutting the cell-- say you've got a three dimensional cell like this.
You're cutting it at different places along here.
So people have worked out the stereology of that.
And there's a constant that's 1.5 that fits into there.
So then you just repeat that process for different increments of theta.
So you increment theta by some amount, something like say, 5 degrees.
And then you repeat that whole process.
And then you plot a polar diagram of the intercept lengths versus theta.
And you then fit an ellipse to the points.
And if you did it in 3D, you'd have an ellipsoid.
And then the principal axes of the ellipsoid are the principal dimensions of the cells.
And the orientation of the ellipsoid is the orientation of the cells.
Hello.
And you can write the equation of the ellipsoid.
So it would be something like Ax1 squared plus B x2 squared plus Cx3 squared plus 2Dx1x2 plus 2Ex1x3 plus 2Fx2x3.
And that would all equal 1.
And you can write those coefficients as a matrix.
And if you do that, the first three, A, B, C, are the diagonals, and D, E, F are the off diagonals.
And you can also represent this is a tensor.
And if you do, it's called the fabric tensor.
So the fabric-- I think they use that term for other types of materials-- it says something about the orientation of the material.
And if you have this matrix here, if D, E, and F are all of 0, then A, B, and C are the principal dimensions of the cell.
So if all non-diagonal elements are 0, then A, B, and C are the principal dimensions.
Are we good with how this works?
So it's kind of a crank and churn kind of thing.
But it gives you a way of characterizing the cell shape and the orientation of the cells in the foams.
So the next thing I wanted to talk about was the connectivity of the cells.
So imagine we have an array of cells.
And if we have an array we have vertices that are connected by edges.
And edges surround faces.
And the faces enclose the cells.
And there's something called the edge connectivity, which is usually given the symbol Ze.
And that's the number of edges that meet at a vertex.
And there's a face connectivity, and that's the number of faces that meet at an edge.
And it's very common for honeycombs to be three connected.
So Ze is 3.
And it's common for foams to be four connected.
And there's some topological laws that are kind of interesting.
And so we'll get into those after this bit on connectivity.
So for the connectivity we have vertices, which are connected by edges, which surround faces, which enclose cells.
So we're going to talk about the vertices, the edges, the faces, and the cells.
So the edge connectivity, Ze, is the number of edges that meet at a vertex.
And for a honeycomb, say a hexagonal honeycomb, Ze is 3.
So I have a little honeycomb here.
If we look at the number of edges, there's three edges, connect at a vertex.
So Ze is 3.
And for a foam, Ze is typically four.
So I'll just say typically here.
And if I go back, if you look at the sketches here of the rhombic dodecahedra and the tetrakaidecadedra, if you look at the tetrakaidecadedra, you can see the number of edges.
So here's one edge.
Here's another one here.
There's another one here.
And there's another one here.
So that's the four edges that are meeting at that vertex there.
So if you look at arrays of cells, you can convince yourself that Ze is typically 4 for a foam.
And then we also have the face connectivity, Zf.
And that's the number of faces that meet at an edge.
And that's typically three for foams.
And so again if you look at these pictures you can sort of see how the face connectivity is three.
So if you look at this bottom one here, there's faces here.
And then there's another face-- well, let's see if we can-- it's kind of hard to show.
If you look at this one here, there's this cell.
There's that face there.
And then there's another one coming out of the page, I think.
OK.
So that's connectivity.
And the next thing are some of these topological laws.
So the first one I'm going to talk about is called Euler's law.
If you remember Euler from buckling, same guy.
So Euler's law relates the number of vertices and faces and cells and edges for a large array of cells.
So we can relate the total number of edges, so I'm going to call that E, vertices V, faces F, and cells C. That total number is related by Euler's law for a large aggregate of cells.
So in 2D Euler's law is that the total number of faces minus the number of edges plus the number of vertices is equal to 1.
And in 3D, you just add a minus the number of cells in front.
So minus the number of cells, plus the number of faces, minus the number of edges, plus the number of vertices is equal to 1.
So there's some interesting things you can figure out using Euler's law.
And one of the things I wanted to look at was if we had an irregular honeycomb that was three connected.
So say we have a honeycomb that's not just a regular hexagonal honeycomb, or not even one that's got repeating hexagonal cells.
But say we had a honeycomb where some of the cells had 5 sides to them, some of them had 7, some of them had 6.
You can ask, what's the average number of sides per face?
And you can use Euler's law to figure that out.
So we're going to look at an irregular three connected honeycomb.
Let's see, I can start that up here.
So when I say irregular, I mean we've got cells with different numbers of sides.
So we have an irregular-- oops.
You OK, Greg?
--that is three connected.
And what is the average number of sides per face?
And I'm going to call that n bar.
So we're told that it's three connected.
So that's saying that the edge connectivity is 3.
So there's three edges coming to each vertex.
So imagine that you had just a regular hexagonal honeycomb, like this.
Here's our vertex here.
And there's our kind of three edges that come into it.
And each vertex is going to have half of each edge, right?
Because the next vertex-- each edge is shared between two vertices.
So if Ze is equal to 3, then the number of edges per vertex is 3/2.
Because each edge is shared between two vertices.
And I'm also going to define something I'm going to call Fn.
And Fn is going to be the number of faces with n sides.
So I could have some number of the cells have six sides, some number have five sides, some number have seven sides.
So these are-- Fn is the number of faces with some particular number of sides, n. So if we have Fn is the number of faces with n sides, then if I sum up n times Fn and divide it by 2, that's the number of edges.
So imagine I have all these faces.
I take n times Fn.
So say we had four five-sided faces.
That would be four of those.
There's be 20 edges associated with that, 20 sides.
Then I have some number of six-sided, some number of seven-sided.
I add them all up.
And I have to divide by 2 to get the number of edges, because each edge separates two faces.
OK. And then I can use Euler's law.
So I can say F minus E plus V is going to equal 1.
And here I've got an F minus E plus-- I can put V here.
V's going to be 2/3 of E. So I've got that.
So that's the same as F minus 1/3 of E is equal to 1, like that.
And then for E I can substitute this thing here in.
So that gives me F minus 1/3 of the sum of n times Fn over 2 is equal to 1.
And then what I'm going to do is multiply everything by 6 so I can get rid of my denominator here.
So 6F minus sum of nFn is going to be 6.
And then I'm going to divide that through by F. So that's 6 minus sum of nFn over F is equal to 6/F.
And then if I let F go to a large number-- so say I've got a large aggregate of cell.
I've got lots of faces.
I'm going to let F become large.
So if F becomes large, then 6/F is going to tend to 0.
Let me get the other rubber
Professor?
Mhm?
That's-- F is like the total number of faces, not the total number of faces per cell
F is the total number of faces.
And Fn is the number of faces with n sides.
We've got some number with 5 sides, some number with 6, some number with 7.
So we're almost at the end here.
So 6/F goes to 0.
And so that says that the sum of n times Fn over F is equal to 6.
And that just is n bar.
That is the average number of sides per face.
This is your-- kind of your total number of sides in the whole thing.
And you're dividing by the total number of faces.
So that is the average number of sides per face.
So what this is saying is if you have a three connected honeycomb, the average number of sides per face is always 6.
So that if you introduce a cell with 5 sides, somewhere you have to introduce a cell with 7 sides, so that they compensate and you come back out to 6.
And I have a little soap bubble picture here that kind of illustrates that.
So here we have a soap honeycomb.
So you can make these just by putting two glass sheets close together with a soap bubble froth in between them.
And you can kind of see in this picture here, they've numbered how many sides each cell or each face has.
So here's one with 5, here's one with 7.
And I'm not going to add these up all together.
But you can kind of see that the 5's and the 7's kind of compensate for each other, and that the average works out to about 6.
And this is true for a large aggregate of cells.
So if you only have a few, it's not going to work out perfectly.
You need to have a large aggregate to have it work.
Yeah?
And that 5, 7 balance doesn't have to be touching, right?
No, no, no, overall.
Yeah, overall.
And, you know, you could have one with 4 sides, and you'd need two 7's or one 8 or something.
But you'd need to have that balance out and match up.
OK.
So that's the Euler law.
Now there's a couple more of these kinds of things.
There's something called the Aboav-Weaire law.
And so the Aboav-Weaire is sort of related to the Euler because it's looking at this idea that if you have a three connected honeycomb, if you introduce a 5-sided cell, you have to introduce a 7-sided cell to compensate.
And what Aboav noticed was that generally cells with more sides than average have neighbors with fewer sides than average.
Let's see.
So we'll say the introduction of a 5-sided cell requires the introduction of a 7-sided cell.
And I'll just put-- this is all for a 3 connected net, a 3 connected honeycomb.
So that generally, cells with more sides than average have neighbors with fewer sides than average.
And in 3D, you could say the same thing about the faces.
The cells that have more faces than average have neighbors with fewer faces than average.
So Aboav made observations of this.
And it was Dennis Weaire who made a proof of it.
And the equation that they came up with relates the average number of sides in the cell surrounding a candidate cell.
Let's see-- are we going to be able to put it in here?
Maybe.
So say you have a candidate cell and say it has n sides.
So you look at one particular cell and you say, count up, it's got n sides around it.
And then you count up the number of sides of all the cells surrounding it.
It has n neighbors because it has n sides, so then the average number of sides of the n neighbors is called m bar.
It has n sides.
So then the average number of sides of its n neighbors is m bar.
And m bar is equal to 5 plus 6 over n for a 2D honeycomb kind of cell structure.
OK, so that's the Aboav-Weaire law.
And then there's one more of these things.
The last one is called Lewis' rule.
And Lewis looked at biological cells and 2D cell patterns, and he found that the area of the cell varied linearly with the number of the sides.
And he found just an empirical relationship.
It looks like this.
We can say what everything is in a minute.
So he found that the area of a cell with n sides, that's A n, was linearly related to the number of sides, so this n over here, and naught's just a constant and A n bar is the area of the cell with the average number of sides.
So here, A n is the area of cells with n sides and A n bar.
And then n naught is just a constant.
And he found that for 2D cells n naught was equal to 2.
And if you look at voronoi honeycombs-- remember last time we talked about those voronoi honeycombs-- you can show that this holds for voronoi honeycombs.
And then you could write a 3D version of this as well.
And in 3D, it's the volume and faces instead of the areas and the number of sides.
So you can say that the volume of the cells with f faces relative to the volumes of the cells with the average number of faces, they vary linearly with the number of faces.
So there's sort of an exactly analogous expression here.
And here, this f naught is another constant.
And in 3D it's about equal to 3.
So these are all just kind of interesting topological rules that are nice to know about.
OK. Yeah?
Basic question.
In two dimensions, what is a face?
Like what does that refer to?
So-- and let me put my little slides again.
So say we have some honeycombs like this, then say we look at this guy.
So that's a vertex there, that's an edge, and this thing in the middle here is the face.
So in 2D, the face and the cell is kind of the same thing.
All right?
And then--
[INAUDIBLE] sides, what's different with the edge?
It's not quite.
Like if I count up the number of edges, I say one, two, three, four, five, and I count those up like that.
But if I say sides of the face-- see that's a face or it's a cell-- I would say that had six sides.
Right?
So typically when people say it has-- a cell or a face has so many sides, they count up how many all around it.
But because each side is-- or each edge is-- shared between two faces, the number of edges is actually half that.
Right?
Because there's one edge, but if I count up the number of sides for this face and I count up the number of sides for that face, I'm counting that twice.
So it's a little bit-- so I try to see edges when I'm talking about adding them all up and I try to say sides when I'm talking about, here's a face, how many sides does it have.
OK?
It's a little bit confusing.
Anybody else?
[INAUDIBLE] what is n bar naught?
n bar naught, did I put an n bar naught?
Oh, that's where I didn't erase this enough.
There you go.
It's just n naught.
OK.
So we've been talking about the structure of the honeycombs in the foams.
And what we ultimately want to do is be able to model the mechanical behavior or the thermal behavior, some sort of behavior of the cellular material.
And for looking at mechanical behavior, there's three main approaches that people take.
So you have to model the structure somehow.
So there's three main approaches.
And the first one is to use the unit cell.
So say, for example, for the honeycombs, if you have a hexagonal honeycomb you'd just use that unit hexagonal cell.
And that's what we're going to do probably starting later on today.
So for the hexagonal honeycomb, it's kind of obvious.
You would use a unit cell, the thing's periodic, you figure out how that unit cell behaves, you're all set.
For a foam, it was not so obvious a unit cell to use.
But the thing-- people use different things, but one of the common ones they use is a tetrakaidecahedron.
So the nice thing about the tetrakaidecahedron is that it packs to fill space.
So it's a repeating single cell.
Packs to fill space.
But, in fact, real foams aren't tetrakaidecahedrons, so this is a bit of an idealization.
So we'll just say foam cells are not tetrakaidecahedrons.
OK, so that's one approach.
And then a second approach is to use something called dimensional analysis.
So in dimensional analysis, what you want to do here, with this technique, is model the mechanisms of deformation and failure in the structure.
But you don't necessarily represent the cell geometry exactly.
And so what you do is, you say one thing is proportional to another and there's some constant of proportionality, and you just wrap all of those constants of proportionality up at the end.
So for instance, when people look at foams, the geometry of the foams is kind of complicated.
There's cells with different number of faces, there's different sizes of cells, it's kind of a mess.
So we could just say the geometry is complex and it's difficult to model.
And with dimensional analysis, instead what we do is we model the deformation mechanisms and failure mechanisms.
And you can get quite a bit out of just modeling those.
So when we look at modeling foams, we're going to do this, and you'll see how it works.
And then the third method is to use finite element analysis.
So this is a numerical technique and it's a very standard numerical technique.
And one of the nice things about this is you can apply it to random structures.
So for example, those voronoi structures we saw, if you want to try to see-- say you had a certain amount of material and you had a honeycomb that was a regular hexagonal honeycomb.
You had the same amount of material and you put it in a voronoi honeycomb, and you want to know how does having a random material affect the properties relative to the uniform material?
You could figure that out using finite element analysis.
So you can apply it to random structures.
And another thing you can do with finite element analysis-- and people in the orthopedics end of the world often do this-- if you're interested in trabecular bone, that porous type of bone I showed you the first day, people can take micro-computed tomography images of trabecular bone and they get a file which basically says, every voxel, says if it's solid or it's void.
And you can use those files as input to a finite element analysis.
And so you can analyze exactly how a piece of a trabecular bone would deform.
So it's nice for that.
So one of the things we're going to talk about later is we'll look at the structure of trabecular bone and the sort of mechanics of trabecular bone, and I'll show you some results that a former student of mine did, where we look at-- say you have a sort of intact structure of a certain density and then you reduce the density by fitting the cell walls or you reduce the density by removing cell walls-- because in osteoporosis sometimes the walls resorb all together-- and you can see what residual strength you would have for some given amount of density loss.
So it's good for looking at that kind of a thing.
You can also use it for looking at local effects.
So you can look at defects, for instance.
So if you think of the trabecular bone as having missing trabeculae, that would be a defect in the structure.
You can look at that.
You can look at size effects.
So when we were studying some of those metal foams, some of them have very large cells, and if you have a small sample relative to the cell size, you may only have four or five cells across the dimension.
Where you've cut the cells, you've got edges that are less constrained than in the bulk of the material because at the outer edge you've cut them.
They're not connected to anything else.
And so you can look at edge effects that relate to the size effects in foams as well.
So these are basically the three approaches that people use for modeling cellular materials.
And what we're going to do is we're going to start with looking at honeycombs and we're going to look at these hexagonal kind of honeycombs like this.
And one reason to start with them is that they have this unit cell structure.
And if you can analyze how that unit cell deforms and fails, you can say something about the whole structure.
And it turns out that the honeycombs, they deform and fail by the same mechanisms as the foams.
So if you can understand through this simple structure, it gives you a lot of insight into how the foams behave.
So if I deform this a little bit, the cell walls bend.
And you can show that if you deform this guy a little bit, the cell walls bend.
And so you can learn a lot by looking at the honeycombs and then sort of applying that to the foams.
So we're going to start off-- so that's the end of the section on sort of the structure of the cellular materials-- and now we're going to look at modeling the mechanical properties and we're going to start with the honeycombs and then we're going to do foams.
So when we talk about the honeycombs, we've got this hexagonal structure here and we're going to call properties in this plane the in-plane property.
So if I load it this way on or that way on, those are in-plane, and if I load it that way on, those are out of plane.
So think of the cells are in the plane and the prismatic direction is the out of plane.
And clearly, the honeycombs are going to have similar properties this way and that way, but they are going to have very different properties that way.
So we're going to start with the in-plane behavior.
So the honeycombs have these prismatic cells and they're widely available in different materials, polymers, metals, ceramics.
And they're used in a variety of applications.
So one of the most common is to use them in sandwich panels.
So I brought a couple of sandwich panels with me today.
So here's a couple of sandwich panels that have honeycomb cores.
This one's an aircraft flooring panel.
It's got carbon fiber faces and a Nomex core, honeycomb core.
And this is an aluminum honeycomb.
It's got aluminum honeycomb core and then aluminum faces, and it's kind of amazing how stiff that little panel is.
And each of the pieces is really not that stiff at all, but the thing put together is quite stiff, and we'll talk more about that when we get to sandwich panels.
So it's used in sandwich panels.
They're used for energy absorption.
So sometimes you'll see there's some natural disaster area and they fly in helicopters and they drop big crates of supplies, and they will have it like a pallet with a big crate thing.
Often they have a honeycomb, like a metal honeycomb, that is kind of oriented this way.
So the pallet would be like this and the honeycomb's like that.
And the idea is that when they drop it-- they sort of bring it down as close as they can-- and then the honeycomb absorbs some of the energy from the impact.
And they're also used, I think, sometimes in car bumpers.
So they're used for energy absorption.
And they're used as carriers for catalysts.
So the catalytic converter in your car looks like this, this is kind of the material that's used in the catalytic converter in your car.
And the way that works is, the cell walls here are actually porous and they're coated in the platinum, which is the catalyst, and half of the cells are blocked off on this end.
So every other cell on this end is blocked off and then every other cell over here, the opposite ones, are blocked off.
And so the gas is forced down a channel but then through the wall and then out the next channel.
And that's where the reaction actually occurs, is in the cell wall.
So they're used as carriers there.
And some natural materials also have a cellular structure, have a honeycomb structure.
So for example, things like woods and cork have a honeycomb structure, too.
So I said the mechanisms of deformation and failure in the hexagonal honeycombs parallel those in foams.
So we can learn a lot about foams by understanding the honeycombs.
And then, similarly, the mechanisms of deformation and failure in triangulated honeycombs parallel those in the lattice materials.
Remember I brought those lattice materials in?
The sort of trust type materials.
The triangular honeycombs have a behavior similar to those lattice materials.
OK.
So let me scoot over here.
OK.
So let me scoot out of this one.
OK.
So here is our kind of hexagonal geometry.
This is kind of an idealized geometry here.
And, as we talked about in this section on the structure, we're going to call these vertical walls, we're going to say they have a length h, the inclined walls have a length l, the wall thickness is t, and there's going to be an angle between the horizontal and that inclined wall of theta.
And I'm going to define three axes here, an x1, an x2, and an x3 axis.
So the x1, x2 plane is the in-plane and x3 is out of plane.
OK?
So if we load our honeycomb up, we get stress strain curves that look like this.
So the ones on the left here, over here these three are all in compression, and the ones on the right, these three are all in tension.
OK, so let's talk about the compression ones first.
And let's start at the top.
This is in elastomeric material, so like one of these rubber honeycombs.
This would be a material that yields plastically, so like an aluminum honeycomb.
And this would be a honeycomb that fails in a brittle manner, like one of those ceramic honeycombs.
OK?
So if we go up to the elastomeric one, up here, if I compress it I get a linear elastic part first, and then at some point, I get a stress plateau where the stress is almost constant for strains that are quite large.
And then finally, the stress starts to rise quite sharply at the end here.
So the strain here goes from 0 to 1.
So that's a strain of 100%.
You've completely flattened the thing there.
So that's a large strain.
So initially, when we're loading it up to smaller strains like this, we've got linear elastic behavior and these cell walls bend.
And you can relate the modulus here-- let's see where'd my little arrow go-- you can relate this Young's modulus here to the bending of those cell walls.
And we're going to-- I don't know if we'll finished that today-- but we'll start that today.
Then-- oops, lost my arrow, where'd my arrow go-- then at the stress plateau here, that plateau is related to collapse of the cells.
So if I have my little rubber honeycomb and I load it, at some point, the cell walls buckle.
So you see how they've buckled there?
And that stress plateau, I can smush that to quite large strains that are roughly constant stress.
OK?
So what's happening here is that we're buckling the cells.
And as we go along here, the buckling deformation gets bigger and bigger.
And then if I smush it and I've got it kind of like that, at some point it becomes much harder to press it together again because the cell walls are now touching each other and they're pressing against themselves, and to get a certain amount of strain, it gets much more difficult to do that.
And the stress required to do that gets much bigger.
And that's what leads to this last piece of the stress strain curve here, which is called densification because you've almost eliminated the pores.
It's hard to eliminate them entirely but the porosity has gone way down by the time you're up there.
So this type of stress strain curve where you've got linear elasticity and then a stress plateau and then the densification is classic for compressive behavior of cellular materials.
So elastomeric ones will have a plateau that's related to elastic buckling.
I lost my arrow again.
There we go.
If we had a metal honeycomb, we'd again have linear elasticity related to cell wall bending.
This plateau here would be related to yielding of the cell walls.
So say we had aluminum honeycomb, the aluminum could yield, and that would again cause this stress plateau.
And then we get densification.
If I had a brittle honeycomb like the ceramic one, we'd have initial linear elasticity.
Then you've got a stress plateau that's kind of a lot of up and down here.
And the serrated nature is related to the fracture of individual cell walls.
So it goes up and down because when you break a cell wall, the stress drops off, and then the other cell walls will try to pick the stress up again.
And so each one of those little up and downs corresponds to breaking a cell wall.
But if you kind of took an average of that, you can see there's a stress plateau and then there's the densification region.
So in compression, the shape of the curves is very similar and the mechanism of the plateau varies a little bit.
So let's see if I can write some of that down.
So in compression, we can say we have three regimes of behavior.
So we've got the linear elastic regime initially and we're going to see that's related to cell wall bending.
Oh, thanks.
And then we've got a stress plateau.
And for elastomeric materials, that's caused by buckling of the walls.
For metals, it would be caused by yielding.
And for ceramics, it would be caused by a brittle crushing.
And then we've got densification.
And that's related to the cell walls touching.
And if we increase the ratio of the thickness of the cell walls relative to their length, we're going to increase the stiffness, so the Young's modulus of the honeycomb.
The stress plateau I'm going to call sigma star.
And we decrease the strain at which that densification occurs, which I'm going to call epsilon D. So you see over on the right hand side of these plots here, that strain there is the densification strain, epsilon D. So that's in compression.
And these materials are very often loaded in compression.
In tension, we still get linear elasticity initially.
And that's going to, again, be related to the bending of the cell walls.
But if we look at the stress plateau, if you look at these three curves here on the right, the stress plateau only exists if the material has a yield point and you get some plastic yielding there.
If you have an elastomer, if I pull on this-- you don't get buckling in tension, so you're not going to get a stress plateau in tension.
And for a brittle material like one of those brittle honeycombs, if you pull that in tension, you would just get a crack propagate and the thing would break into two pieces and so you don't get a stress plateau there.
So the stress plateau in tension only exists if the material yields.
So you don't get any buckling in tension.
And for a brittle honeycomb, you would just get fracture.
OK?
These inclined walls are going to bend.
We're going to see that in more gory detail in one minute.
If you can wait one minute.
OK.
So are we good with a stress strain curves yet?
More of an abstract question, but is it possible [INAUDIBLE] to get collapse this way before the plateau?
I haven't seen that.
But maybe it it's possible, I don't know.
I don't think so, but maybe.
Anybody else?
OK. OK, so here's some photographs of honeycombs.
So these ones are white but these ones are just the same.
So most of them are just a regular hexagonal honeycomb and one of them is this funny shaped honeycomb here.
These two guys at the top, these two here, are unloaded.
And then the rest of them have some loading.
So if you look at this one down here-- so here I'm taking the honeycomb and I'm loading it.
This is the x1 direction, that way on.
And if you look very carefully, you can see what happens is these vertical walls just move sideways and that is going to zoot, with the sound effects.
And then the-- I can't help making sound effects-- and then these guys here bend.
OK, so this guy here-- it's maybe a little hard to see it on the image but it's actually a little bit bent there.
So that's loading it this way on.
And then similarly, if I take it and I load it that way on, that's the x2 direction.
And now these guys here, the vertical guys, are just going to compress a little.
But these guys here, the incline guys, are still going to bend a little bit.
And I've got some schematics that's going to show that a little bit better.
And then if I shear it, if I took it like this and I-- I can't do it because my hands aren't glued to the rubber-- but if I could sort of shear it this way on, then you would get this kind of deformation here and that also involves bending.
You can imagine these guys here would bend if I do that to them.
And then the buckling deformation looks like this.
If I scooch that like that, it should look kind of like that picture there.
OK?
So that's kind of what the deformation looks like for these sorts of honeycombs.
So these were elastomer rubbers.
These are just some images from an aluminum honeycomb.
So here's, the top one's the undeformed honeycomb, the middle one's loading it in the one direction from left to right, and the bottom one's loading it in the two direction from top and bottom like that.
OK?
So you can imagine that you've got little cell walls.
If you load it up high enough, those walls are going to yield and we're going to see that the yielding in the walls causes the formation of something called plastic hinges and the walls can rotate and they can produce these kind of shapes.
OK.
So then this is sort of a schematic stress strain curve here.
And this is showing what happens if you increase the thickness of the walls relative to the length, or you increase the relative density, you increase the volume per action of solids.
And this kind of shows the different regimes.
So over here-- well let's, first of all, start with the different relative density.
So this one here is the lowest relative density, then this is higher, and higher, and higher.
And not too surprisingly, the more you increase the density, the more material you've got, the stiffer it is, the stronger it's going to be.
And the more material you've got, the sooner it's going to densify.
If you've got more solid in here, you're going to reach that densification strain at a smaller number.
OK, so the shape of the curves looks like this.
And you can define these three kind of regimes.
So everything in here is linear elastic, everything in this big sort of envelope here is the plateau region, and everything up here is this densification region.
So this is just a bigger picture kind of plot.
OK, so let me skip through all of that.
All right.
OK, where are we?
Chalk?
Here we go.
OK, so I think I mentioned this before, but let me just go over it again.
So there's three types of things that affect the properties, the mechanical properties, of the honeycombs.
And probably the most important thing is the relative density of the honeycomb.
So remember, we said this was the same as the volume fraction of solids, and for a hexagonal honeycomb, you can show that's equal to the thickness to length ratio times h over l plus 2 divided by 2 cos theta times h over l plus sin theta.
And if you have regular hexagons, so h over l is equal to 1, all the sides are of equal length, theta is equal to 30 degrees, the relevant density is 2 over 3 times t over l. So it just goes linearly with t over l. The thickness to length ratio of the walls.
So it depends on how much solid you've got.
Depends on the solid properties.
So the Young's modulus of the solid, yield strength if it's a metal, some sort of fracture strength if it's brittle.
It also depends on the cell geometry, which we can describe with h over l and theta.
So if we think of a cell here-- that's our edge length h, that's our edge length l, that's our angle theta, here's the cell wall thickness t, and then we've got some set of solid properties here.
OK?
So that's kind of the set up.
And we're going to define x1, x2 axes like that.
And we're going to make a few assumptions just to make life a little bit simpler.
So we're going to assume that t over l is small, so that also means the relative density is small.
And what that means is that we're going to be able to neglect axial and shear deformations.
So you can imagine, if I have a thin wall and I'm applying loads that produce moments and produce bending, if the wall is very thin then the axial and the shear deformations are going to be small.
I'm also going to assume the deformations are small.
And what that means is that I'm going to neglect any changes in the geometry of the cell during the deformation.
And I'm going to assume that the cell wall is linear, elastic, and isotropic.
And we're going to start off with looking at in-plane behavior, and we're going to start with the elastic moduli.
And if we look at the elastic moduli, we're going to be talking about Hooke's law.
And Hooke's law and the elastic behavior of the material can be described by a set of elastic constants.
And if you recall, the number of independent elastic constants, how many constants you need to describe the material, depends on its symmetry.
And these materials are orthotropic.
So the regular hexagonal honeycomb is actually transversely isotropic, but imagine that h was not equal to l, then it would be orthotropic.
So remember, orthotropic means that you can rotate the structure 180 degrees about three mutually perpendicular axes and the structure looks the same.
So if I take this and I do that, it looks the same, and if I do that, it looks the same, if I-- no matter how I rotate this, about three mutually perpendicular axes, the structure remains unchanged.
So it's orthotropic.
So I'm going to write down Hooke's law for our orthotropic material, and then we'll talk about the constants that we're going to work out.
OK, so this is Hooke's law for our orthotropic material.
And let me just remind you what our notation is here.
So epsilon 1 is epsilon 1 1.
Epsilon 2 is epsilon 2 2.
Epsilon 3 is epsilon 3 3.
Epsilon 4 is gamma 2 3.
Epsilon 5 is gamma 1 3.
And epsilon 6 is gamma 1 2.
So these are the normal strains here, epsilon 1, 2, and 3, and these are the shear strains here, epsilon 4, 5, and 6.
And you remember this convention where the subscripts add up to 9.
So 4 plus 2 plus 3 is 9, 5 plus 1 plus 3 is 9, 6 plus 1 plus 2 is 9.
And then the stresses are a similar thing.
So sigma 1, sigma 2, and sigma 3 are the normal stresses.
And then sigma 4, sigma 5, and sigma 6 are the shear stresses.
And for the in-plane moduli, so we're dealing with the x1, x2 plane, there's four independent elastic constants.
So we could think of it as E1, E2, a Poisson's ratio 1 2 and a shear modulus in the 1 2 plane.
OK?
And the compliance matrix is symmetric, so there's the reciprocal relationship between the moduli and the Poisson's ratios.
And then the notation I'm going to use for Poisson's ratio, I'm going to say that mu i j is minus the ratio of the strain in the j direction divided by the strain in the i direction.
OK.
So what we're going to do next is we're going to calculate some of the elastic moduli.
I'm going to show you the derivation for E1 star and mu 1 2 star, and you can get the other two in a similar way.
So I'm not going to do all of them but next time we'll do the derivations for the Young's modulus in the Poisson's ratio.
OK?
And then we're going to talk about the out of plane direction later and we'll get the moduli for the out of plane direction as well.
OK?
So I think I'm going to stop there for today.
And then we'll start doing the derivations next time.
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OK, so it's five after.
We should probably start.
So last time we were talking about honeycombs, and I just wanted to quickly kind of review what we had talked about, and then today I'm going to start deriving equations for the mechanical properties of the honeycombs, OK?
So this is a slide of our honeycomb setup here.
These are the hexagonal cells we're going to look at.
We talked about the stress-strain behavior.
The curves on the left-hand side are for compression and the ones on the right-hand side are for tension.
And so what we're going to be doing today is we're going to start out by calculating a Young's modulus, this slope here.
We're going to calculate the stress plateaus for failure by elastic buckling in elastomeric honeycombs, by failure from plastic yielding in, say, a metal honeycomb, and by failure by a brittle crushing in, say, a ceramic honeycomb.
And if we have time, we'll get to the tension stuff.
I don't know if we'll get to that today or next time.
So we're going to start calculating those properties today.
And these were the deformation mechanisms.
Remember, we said the linear elastic behavior was related to bending of the cell walls, and then the plateau was related to buckling if it was an elastomer.
And the plateau was related to yielding if it was, say, a metal that had a yield point.
And then this was sort of an overview of the stress-strain curve showing those different regions, OK?
So what I'm going to talk about today to start is the linear elastic behavior.
And we're going to be starting with the in-plane behavior.
So in-plane means in the plane of the hexagonal cells.
And then next time we'll do the out-of-plane behavior, this way on.
So if I had to form my little honeycomb like this, what initially happens is the inclined cell walls bend.
So if you can see over here, we've kind of exaggerated it on this sketch.
So this wall here is bent.
This one here just kind of moves along, goes for the ride.
And this guy here is bent.
So this is for loading in the what we're calling the x1 direction, sigma 1.
And the same kind of thing happens when we load in the other direction, in the sigma 2 direction, these guys still bend.
Now the honeycomb gets wider that way.
It gets shorter this way, wider that way.
And we can calculate the Young's modulus if we can relate the load on the beam in the moments to this deflection here, right?
So the Young's modulus is going to be related to the stiffness and the stiffness is going to be related to how much deformation you get for a certain amount of load that you put on the beam.
So I'm going to calculate the modulus for the x1 direction, the thing on the left there.
And you can do the same thing for the x2 direction on the right, but I won't calculate that because it's exactly the same kind of process.
OK, so let me start here.
Get my chalk.
So I'm going to draw a one-unit cell here.
So here's my unit cell there, like that.
And this member here is of length h. That member there is of length l. That angle there is theta.
I'm going to say all the walls have equal thickness, and I'm going to call it t. And I'm going to define an x1 and x2 axis like this.
So the horizontal is x1 and the vertical axis is x2.
And I'm going to say that I apply a sort of global stress to it, sigma 1.
So there's a stress in the one direction there, sigma 1, OK?
And I'm going to say my honeycomb has a depth b into the page, but the depth s-- is because the honeycomb's prismatic, the b's are always going to cancel out of all the equations that we're going to get, because everything's uniform in that direction.
And we can think about a unit cell here, and in the x1 direction, we could say the length of our unit cell is 2l cos theta.
So in the x1 direction, that's our unit cell there, and that's 2l cos of theta.
And in the x2 direction, you might think that you go from this vertex up here down to that vertex there, but if you did that, then on the next layer of cells, you wouldn't have the same distance.
So the unit length in the x2 direction is actually from here to here, and then you can see the next cell, you would get the same thing.
You get this bit here from the inclined member, and then you would get h down here from the next member.
So this bit here is equal to h plus l times sine of theta.
So I can say in the x2 direction, the length of the unit cell is h plus l sine theta, OK?
So that's kind of the setup.
And then what we want to look at is that inclined member that bends, we want to look at how this guy bends under the load.
And if we can relate the forces on it to the deflection, and we need the component of the deflection in the one direction, then we're going to be able to get the modulus.
So I'm going to draw that inclined member again over here, and it's going to see some loads that I'm going to call p, and that's going to cause this thing to bend.
So I've kind of exaggerated it there, but there's the bending.
And there's some end deflection there, delta.
And there's moments at either end of the beam, or either end of that member, as well.
And this member here has a length.
That length is l. OK, are we good?
So it's just kind of the setup.
And I'm going to draw the deflection delta bigger over here.
So say that's delta.
That's the same parallel as this guy here.
What I'm going to want is the deflection in the x1 direction, and when I come to calculate the Poisson's ratio, I'm going to want the deflection in the x2 direction.
And if this angle here is theta, between the horizontal and the inclined member, then this angle up here is also theta, and so this bit here is delta sine theta.
And this bit here is delta cos theta.
Ba-doop-ba-doop-ba-doop.
So the Young's modulus is going to be the stress in the one direction divided by the strain in the one direction.
So I need to get the stress and the strain in the one direction.
So here the stress in the one direction.
If I'm applying my load, like this, sigma 1, the stress in the one direction is going to be this load p-- so this load, say, p on this member here, divided by this length here, the unit cell length, and then divided by b into the board, the width end of the board.
So sigma 1 is going to be p divided by h plus l sine theta times b.
And epsilon 1 is going to be the strain in the one direction, is going to be the deformation in the one direction divided by the unit cell length in the one direction.
So that's going to be delta sine theta divided by l cos theta.
So here, even though I've said this unit cell is 2l cos theta, there'd be two of the members here that would be twice that deflection in the one direction.
So it's, for one beam, it's delta sine theta over l cos theta.
Are we OK so far?
So far so good?
So for the hexagons, because we're going to figure out the equations more or less exactly, we're going to keep track of all the geometrical factors.
When we come to the foams, we're not going to keep track of all the geometrical factors.
So one of the things that makes us look a little kind of hairy is just the fact that we're keeping track of all these sines and cosines and all the dimensions and things.
All right, whoop.
So I need to be able to relate my load p to my deformation delta to get a stiffness out of this, to get a modulus, OK?
So the way I do that is, remember in 3032, we did those bending moment diagrams and we did the deflection of the beams?
This is where this comes in handy.
So I'm going to draw my beam a little bit differently now.
I'm going to turn it on its side.
So this is still my length l, but I'm going to turn on my side just so that you can see it the same kind of way we did the bending moment diagrams.
So this is still my length l across here.
And there's end moments, M and M here.
And P sin theta is just the perpendicular component of the load p. So p sin theta is just the component perpendicular to my beam.
So I could draw a shear diagram here and I could draw a bending moment diagram here.
And, if you remember, the shear diagram, if I have no concentrated load along here, and I have no distributed load along here, if this is zero, down here, it's just going to go up my P sin theta, and then be horizontal, and then come down by P sin theta, OK?
So that's the shear diagram.
And then the bending moment diagram, I'm going to draw down here.
So I've got some moment at the end here, and this would tend to bend like that.
So this would be a negative moment.
Remember, bending moments were negative if there was tension on the top, and they were positive if there was tension on the bottom.
So over here we'd have tension on the top, so that would give us a negative bending moment.
And then, if you also remember, the moment at a particular point is equal to the integral from, say, A to B-- well maybe I should write this another way.
And B minus Ma is the integral of the shear diagram between the two points.
A little sloppy.
OK.
So if I know how I have some moment here minus M, if I integrate this shear diagram up, then this is just going to be linear here, and then I'm going to be at plus M over there.
So if you look at this shear and bending moment diagram, it's really just the same as the shear and bending moment diagram for two cantilevers that are attached to each other.
So let me just draw over here what the cantilever looks like.
Let's see.
So imagine I just had a cantilever like this, and I have some force F on it like that.
And I call this distance here capital L, and I'm going to call that deflection capital delta, like that.
If I drew the shear diagram for that, there'd be a reaction here, F, there would be a moment here, FL.
Doot, doot, yup.
So this would look-- whoops-- it's a little too long.
Shear diagram here would look like this.
That would be zero.
This would be FL.
And the moment diagram would look like this.
Whoops.
A little too long again.
And that would be minus FL.
And that would be zero.
So do you see how the shear and the bending moment diagram here are really just like two cantilevers, OK?
So I know that the deflection for a cantilever, delta, is equal to F capital L cubed over 3EI.
It's kind of a standard result.
And so I can take this and apply that to this beam here.
So instead of working everything out from first principles, I'm just going to say that my beam here is like two cantilevers, and instead of F, I've got P sin theta.
And instead of capital L here as the length, I've got l/2 because l/2 would be the length of one of the cantilevers.
OK, so for the honeycomb, I've got two cantilevers of length l/2.
So delta for the inclined member on the honeycomb is going to be 2-- because I've got two cantilevers-- the force, instead of having F, I'm going to have P sin theta.
And instead of having capital L, I'm going to have l/2, all cubed.
So this is like F capital L cubed over 3.
And here, the modulus that I want is the modulus of the solid cell wall material, so I'm going to call that ES, and over the moment of inertia.
So you see how I've done it?
Is that OK?
So then I can just kind of simplify this thing here.
I've got P sin theta l cubed.
1/2 cubed is going to be 1/8.
So this is going to be 2, if that's 1/8, times 3 is 24.
So 2/24 is 12.
So I've got delta for my honeycomb member is P sin theta l cubed over 12 EsI.
And here I is the moment of inertia of that inclined member of the honeycomb.
And that's BT cubed over 12, OK?
So B is the depth into the board, and T is the thickness.
We'll cube that and divide by 12.
It's a rectangular section.
Yeah?
What was ES again?
ES is the Young's modulus of the solid that it's made from.
So clearly, if my honeycomb is made up of these members, whatever material the members are made of is going to affect the stiffness of the whole thing.
Are we good?
Because once we have this part, then we just combine these equations for the stress and the strain in the one direction.
And we have this equation relating delta and P, and we're going to be able to get our Young's modulus, OK?
We're happy?
OK. All right.
So I'm going to call the Young's modulus in the one direction E star 1.
So everything with a star refers to a cellular solid property, and 1 because it's in one direction.
So that's going to be sigma 1 over epsilon 1.
So if I go back up there, I can say sigma 1 is equal to P divided by h plus l sin theta b.
And epsilon 1 is equal to delta sine theta in the denominator over l cos theta.
And now instead of having delta here, I can substitute this thing here in for delta.
And then I'm going to able to cancel the P's out.
So delta was equal to P sine theta l cubed over 12 Es, and there was an I, a moment of inertia, and I was equal to bt cubed over 12.
And let's see here.
So that's delta.
And there's another sine theta here so I'm just going to square that sine theta there.
So now the P's cancel out.
The b's are going to cancel out.
The 12s are going to cancel out.
And I'm going to rearrange this a little bit.
So I'm going to write Young's modulus of the solid out in the front.
Then I've got a term here of t cubed and I'm going to multiply that by 1/l squared, and then everything else-- well, let's see.
We can take this l cubed here.
I can take that.
Put it underneath that, so that's going to give me t/l cubed.
And then I've got an h plus l sine theta here, and I've got an l there, so I'm going to take that to be h/l plus sine theta.
Boop-boop-da-doop.
So I've got this term with [?
h/l's ?]
in the thetas.
There's a cos theta from the numerator here.
This term here turns into h/l plus sine theta, and then I've got my sine squared thetas down there.
And that's my result for the Young's modulus in the one direction.
OK?
Let's make sure that seems right.
It seems good.
OK.
So one of the things to notice here is there's three types of parameters that are important.
So one is the solid properties.
So the Young's modulus of the solid comes into this.
So the stiffness of the whole thing depends on the stiffness of whatever it's made from.
There's this factor of t/l cubed-- that's directly related to the relative density or the volume fraction of solids.
So what this is saying is the relative density goes as t/l, so the Young's modulus depends on the cube of the relative density.
So it's very sensitive to the relative density.
And then this factor here really is just a factor that depends on the cell geometry.
Remember when we talked about the structure of the honeycomb, we said we could define the cell geometry by the ratio of h/l and theta, OK?
And since we often deal with regular hexagonal honeycombs, I'm just going to write down what this works out to be for regular hexagonal honeycombs.
So for a regular hexagonal honeycomb, h/l is 1.
All the members have the same length.
And theta's 3, and the modulus works out to 4 over root 3 times Es times t/l cubed, OK?
So do you see how we do these things?
So all the other properties work in a similar kind of way.
You have to say something about what the sort of bulk stress is on the whole thing and relate that to the loads on the members.
You have to say something about how the loads are related to deflections, or when we look at the strengths, we're going to look at moments and how the moments are related to failure moments of one sort or another.
But it's all just like a little structural analysis, OK?
Are we good?
You good, Teddy?
I thought you were going to put your hand up?
No?
You're OK?
OK. OK, so the next property we're going to look at is Poisson's ratio.
And I'm going to look at it for loading in the one direction.
So Poisson's 1 2, say we load uniaxially in the one direction, we want to know what the strain is in the two direction, it's minus epsilon 2 over epsilon 1.
And again, if I look at my inclined member, and I say that member's going to bend something like that, and that's my deflection delta there, and, say, got the same x1 and x2 axes.
And again, if I look at delta here, it's the same little sketch I had before.
That's delta sine theta.
And this is delta cos theta.
I'm going to need those components to get the two strains in the different directions.
So epsilon 1 is going to be delta sine theta over l cos theta.
And if I'm compressing it, that would get shorter.
And we get-- and epsilon 2 is going to be delta cos theta divided by h plus l sine theta.
And that would get longer.
So these two have opposite signs, and so the minus sign is going to disappear here.
Doodle-doodle-doot.
So then I can get my Poisson's ratio by just taking the ratio of those two guys.
So I could put a minus sign there and say that's the opposite sign to epsilon 2.
Then this would be delta cos theta divided by h plus l sine theta.
And epsilon 1 would be delta sine theta over l cos theta.
And the thing that's convenient here is that the two deltas just cancel out.
So the Poisson's ratio is the ratio of two strains.
Each one of the strains is going to be proportional to delta, and so the two deltas are just going to cancel out.
And so I can rewrite this thing here as cos squared theta divided by h/l plus sine theta times sine theta.
And so one of the interesting things to notice that the Poisson's ratio only depends on the cell geometry.
It doesn't depend on what solid the material is made from.
It doesn't depend on the relative density.
It only depends on the cell geometry.
Oops.
OK, and then we can also work out what the value is for a regular hexagonal cell.
And if we plug-in h is equal to l and theta's equal to 30, you get that it's equal to 1.
So one is kind of an unusual number for a Poisson's ratio.
When we think of most materials, it's around 0.3, so it's kind of unusual that it's that large.
The other thing that's interesting is that it can be negative.
So if theta is less than 0, then you can get a negative value.
If the cos squared is going to be a positive value, but you've got a sine theta down here, then that's going to give you a negative value.
So you can get negative values.
So let me just plug in an example.
So say h/l is equal to 2, and theta is equal to minus 30 degrees, then this turns out to be 3/4.
So cos of 30 is root 3/2, so square of that is 3/4.
h/l is 2, sine theta is 1/2, but it's minus 1/2.
So 2 minus 1/2 is 1 and 1/2.
And then the sine theta is minus 1/2.
And so it works out to be minus 1 for that particular combination.
And I brought my little honeycomb that has a negative Poisson's ratio in.
So this guy here-- let's see, I don't think there's an overhead here.
No overhead?
Guess not.
I'll just pass it around.
So if you take it, put your hands on the flat side and load it like this, and don't smoosh it like that.
Just load it a little bit, because you [?
want to be ?]
linear elastic.
If you load it just a little bit, you can see that as you push it this way, it contracts in sideways that way.
So don't smash it.
Just load it a little bit and you can kind of see it with your hands.
And if you put on a piece of lined paper, it's easier to see it.
OK, so that's kind of interesting.
So are we good with getting the Young's modulus in the one direction and the Poisson's ratio for loading in one direction?
OK, so you can do the same sort of thing to get the Young's modulus in the two direction and the Poisson's ratio for loading in the two direction.
And you get slightly different formulas, but it's the same idea.
And you can also get a shear modulus this way, and in-plane shear modulus.
It's a little bit-- the geometry of it's a little bit more complicated.
So all of those things are derived in the book, in the cellular solids book.
So if you wanted to figure those out, look at that, you could look at the book.
So let me just comment on that.
All right, so those are the in-plane linear elastic moduli, and remember we said that four of them describe the in-plane properties for an anisotropic honeycomb.
And you can use that reciprocal relationship to relate the two Young's moduli and the two Poisson's ratios.
All right, so the next thing I wanted to talk about was the compressive strength.
So let me just back up here a second.
So if we go back to here, remember we had for an elastomeric honeycomb, this stress plateau was related to elastic buckling.
So we're going to look at that buckling stress first.
And this plateau here is related to yielding.
And then we'll look at the yielding stress next.
And then this plateau here is related to a brittle sort of crushing, and we'll do that one third.
So we're going to go through each of those next.
And this is kind of a schematic for the elastic buckling.
So when you look at the elastic buckling, one of the things to note is that when you load the honeycomb this way on, if you load it in the one direction, you don't get buckling.
It just sort of continues to-- whoops, if I can keep it in plane.
It all just kind of folds up, so you just get larger and larger bending deflections.
You don't really get buckling.
But when you load it this way on, these vertical members here, the ones of length h, they're going to buckle.
So, see if I do that, my honeycomb looks like those cells up on the schematic there, OK?
So, whoops.
So we're going to look at the compressive stress or strength next.
That's sometimes called the plateau stress.
So we can get cell collapse by elastic buckling, if, for instance, the honeycomb is made of a polymer.
And then the stress-strain curve looks something like that.
And what happens is you get buckling of those vertical struts throughout the honeycomb And then you could also get a stress plateau by plastic yielding.
And what happens when you get plastic yielding is you get localization of the deformation.
So one band of cells will begin to yield initially, and then as the deformation proceeds, that deformation ban will propagate and get bigger and bigger, and you get a wider and wider band of cells yielding and failing.
So you get localization of yield, and then as deformation progresses, the deformation band widens throughout the material.
So if I go back and look-- if I look at this one here, when you look at this middle picture here, you can see how one band of cells has started to collapse and started to fail.
And as you continue to compress that in the one direction, this way on, then more and more neighboring cells are going to collapse and the whole thing will get wider until the whole thing has collapsed.
And that's kind of characteristic of the plastic failure.
And then the third possibility is brittle crushing.
And then you get these kind of serrated plateau.
And the peaks and valleys correspond to fractures of individual cell walls.
OK, so we're going to start off with the elastic buckling failure.
And I'm going to call these plateau stresses sigma star, for the sort of compressive strength.
And el means it's by elastic buckling.
And as I mentioned, you don't get it in the one direction.
The cells just fold up.
You only get it for loading in the two direction, so it's going to be sigma star el 2.
Oops, need a different piece of chalk.
So you get this elastic buckling for loading in the x2 direction, and the cell walls of length h buckle.
And you don't get it for loading in the one direction, the cells just fold up.
So again, let me draw a little kind of unit cell here.
And here is our stress sigma 2, like that.
And here's our little wall of length h that's going to buckle.
So if I load it up, initially it'll be linear elastic.
And then eventually, at some stress, it will get large enough that this wall here will buckle.
And we can relate to that plateau stress, or that compressive stress, to some Euler buckling load.
So you remember, if we have a pin-ended column, so just a single column, pins on either end, the Euler buckling load says you get buckling when the critical load is equal to some end constraint factor, n squared.
So n squared pi squared E, and here it's E of the solid, I over the length of the column, and in this case, the column length is h-- so h squared.
OK, so that's just the Euler formula.
And here, n is an end constraint factor.
And if you remember for a pin column, so if our column is pinned at both ends like that, and just buckles out like that, then n is equal to 1.
And if the column is fixed at both ends, something like that, then the column looks like that and then it's equal to 2, OK?
So if I know what the end condition is, I know what n is and I can use my Euler formula here.
So the trick to this is that it's not so obvious what n is.
Yes?
So, when you're loading in the x2 direction here, the first thing you're going to get is the incline members deforming?
Yeah
And then at some point, you hit a P critical that will cause the vertical members to buckle?
Exactly

Exactly.
That's exactly right.
Hello.
So the trick here is that we don't really know what this n is, initially.
They're not really pinned, pinned; they're not fixed, fixed.
And if you think about the setup with the honeycomb here, the constraint on that vertical member depends on how stiff the adjacent members are.
So you can kind of imagine, if I'm looking at one of these vertical members here, if these two adjacent inclined members were big honking thick things, it would be more constrained.
And if they were little thin, kind of teeny little membranes, it would be less constrained.
And you can think of it in terms of a rotational stiffness, that when the honeycomb buckles, you kind of see the member h goes from being horizontal to sort of it buckles over like this.
But that whole end joint, see the end joint at the top here or the end joint at the bottom, that whole joint rotates a little bit.
And so there's some rotational stiffness of that joint.
And that rotational stiffness depends on how stiff the member h is and how stiff those inclined members are.
So there's a thing called the elastic line analysis that you can use to calculate what n is.
And basically what that does is it matches the rotational stiffness of the column h with the rotational stiffness of those inclined members.
So we're not going to get into that.
I'm just going to tell you what the answer is.
But if you want to go through it, it's in an appendix in the book.
So you can look at it, if you want.
So here I'm just going to say that the constraint n depends on the stiffness of the adjacent inclined members.
And we can find that by something called the elastic line analysis.
And if you have the book, you can look in the appendix and see how that works.
But essentially what it does is it matches the rotational stiffness of the column h with the rotational stiffness of the inclined members.
So what you find is that n depends on the ratio of h/l.
And I'm just going to give you a table with a few values.
So for h/l equal to 1, then n is equal to 0.686.
For h equal to 1.5, it's equal to 0.76.
And for h/l equal to 2, it's equal to 0.806.
OK, so now if we have values for n, we can just substitute in to get the critical buckling load.
And if I take that load and divide it by the area of the unit cell, I'm going to get my buckling stress.
So it's pretty straightforward from this.
So my buckling stress is going to be that critical load divided by my unit cell area.
So it's divided by the unit cell length in the x1 direction to l cos theta times the depth b into the page.
So it's equal to n squared pi squared Es times I.
And I is bt cubed over 12.
Divided by the length of the column, h squared, and then divided by the area of the unit cell, 2l cos theta b, OK?
And I can rearrange that somewhat to put it in terms of dimensionless groups.
So if I pull all the constants out, it's n squared pi squared over 24 times the modulus of the solid, t/l cubed in the numerator divided by h/l squared times cos theta in the denominator.
So again, you can see that the buckling stress, the compressive sort of elastic collapse stress, depends on the solid property.
So here is the modulus of the cell wall in here.
Depends on the relative density through t/l cubed.
And then it depends on the cell geometry through h/l cos theta, and n depends on h/l as well, OK?
And then we can do the same thing where we figure out what it is for regular hexagonal cells.
And it's 0.22 Es times t/l cubed.
And then we can also notice that since E in the 2 direction, for a regular hexagonal cell, E is the same in the 2 direction and the 1 direction.
It's isotropic.
So E2 is also equal to 4 over root 3 Es times t/l cubed.
That's equal to-- whoops-- it's equals to 2.31 Es t/l cubed.
And we can say that the strain at which that buckling happens is just equal to a constant.
And for regular hexagonal honeycombs, it works out to a strain of 10%.
Are we good?
So we have a buckling load.
We divide by the area.
The only complicated thing is finding n. And you can find it by this elastic line analysis thing.
So each of these calculations is like a little structural analysis, only on a little teeny weeny scale of the cells.
So you see where my background in civil engineering comes in handy.
Yup.
OK.
So the honeycombs involve the most sort of complicated equations.
When we come to do the foams, we're going to use a dimensional analysis and all the equations are going to be much simpler.
So this is the most kind of tedious part of the whole thing.
So the next property I want to look at is the plastic collapse stress.
Say we had a metal honeycomb and we wanted to calculate the stress plateau for a metal honeycomb.
So we have this little schematic here, and say we load it in the one direction again.
So we're loading it here.
And we've got some load P, like that.
And if we have our honeycomb, we load it this way on, initially, the cell walls bend.
And you have linear elasticity and you have some Young's modulus.
But if you have a metal, if you continue to deform it and you continue to load it more and more, eventually you're going to hit the yield stress and the cell wall.
So the stresses in the cell wall are going to hit the yield stress.
And initially, the stresses are just going to be-- remember, if you have a beam, the stresses are maximum at the top and the bottom of the beam.
So initially you're going to hit the yield stress at the top and the bottom of the beam first.
But as you continue to load it, you're going to end up yielding the cross-section through the entire section.
So the entire section is going to be yielded.
And once the entire section yields, it forms what's called a plastic hinge.
Once the whole thing's yielded, then you can add more force and the thing just rotates.
And because it rotates, it's called a plastic hinge.
You know, if you take a coat hanger, and you bend it back and forth and bend it back and forth.
If you bend it enough, you form a plastic hinge because it just can bend easily.
So these little schematics here, if you look at the, say, one of these inclined members, the moments are maximum at the end.
So Remember when we had the linear elastic deformation and I looked at the little bending moment diagram?
The moments are maximum at the ends, and you're going to form those plastic hinges initially at the ends.
And so these little ellipsey things here, all the ends, those kind of show where the plastic hinges are.
So those plastic hinges are forming.
So here's for loading in the x1 direction, and here's for loading in the x2 direction, there.
So the thing we want to calculate is what stress does it take to form those plastic hinges and get this kind of plastic plateau stress?
OK, so we can say we get failure by yielding in the cell walls.
And I'm going to say the yield strength of the cell wall is sigma ys.
So sigma y for yield and s for the solid.
And the plastic hinge forms when the cross-section has fully yielded.
So let's look at the stress distribution through the cross-section when its first linear elastic.
So say that's the thickness t of the member.
And if the beam was linear elastic, the stress would just [?
vary ?]
linearly, like that, right?
And this would be the neutral axis, here, where there is no normal stress.
So that's what happens if it's linear elastic, and I'm hoping you remember something vaguely like that.
Sounds good?
But as we increase the load on it, and we increase the sort of external stress, this stress in the member is going to get bigger and bigger, and eventually, that's going to reach the yield stress, OK?
And once that reaches the yield stress, if we continue to load it, what happens is the yielding propagates down through the thickness of the thing here.
So we get yielding through the whole cross-section.
So let me scoot over here
Professor?
Yup?
When it starts to yield, does this curve change?
Yes.
I'm going to draw it for you
Oh,
That's the next step.
That would be the next thing.
OK, so once the stress at the outer fiber is the yield strength of the solid, then the yielding begins and it progresses through the section as the load increases.
So the stress distribution starts to look something like this once it yields.
OK, so that's sigma y of the solid.
Actually, let me rub that out because then I can show you something else.
So in 3D, this would be through the thickness of the beam.
That would be the thickness of the beam there.
And boop, boop.
It would look something like that.
OK?
And then this is still our neutral axis here.
And then eventually, as you load it more and more, the whole cross-section is going to yield.
Whoops.
And I'm assuming that the material is elastic, perfectly plastic.
So the stress-strain curve from the solid I'm idealizing as-- whoops.
That's not quite right.
I'm idealizing as that, OK?
So when you get to this point here, the entire cross-section has yielded, and that means you form the plastic hinge.
The idea here is that the section then just rotates like a pin.
All right.
So we can figure out the plateau stress that corresponds to this by looking at the moment that's associated with the plastic hinge formation.
So there's some internal moment associated with that.
And then equating that to the applied moment from the applied stress.
So doodle-loodle-oot.
Let me see me, maybe back up here.
So there's some-- if I have the stress distribution here, I could say this whole kind of stress block is equivalent to some force acting out like that and some force acting out like that.
It would be sigma ys times b comes t/2 would be f. And I can say there's some plastic moment.
If I think of the force here and the force there, they act as a couple and they have some moment, and that's called the plastic moment.
So that's like an internal moment when the plastic hinge forms.
So I'll say the internal moment at the formation of a plastic hinge.
I'm going to call that Mp, for plastic moment.
And we can work out Mp by looking at that stress distribution when the entire cross section has yielded.
The force F is going to be sigma ys times b comes t/2.
It's the stress times that area.
And then the moment arm between the two forces is also t/2.
And so that plastic moment is just sigma ys bt squared over 4, OK?
Are we good?
Sonya?
What's the second [INAUDIBLE]?
OK, so this is the force.
This thing here is the force F. And I have to-- if I'm getting a moment, I'm saying that that force, if I doot-doot-doot-- the distance between those two forces there is t/2.
So each force acts through the middle of the block, and so the distance between [?
it is ?]
t/2.
And I'm going to equate that moment to the applied moment from the sort of applied stress.
And then if I go back to my inclined member-- whoops, let's see.
Let me get a little more inclined.
That's my inclined member, there, of length l. I've got modes p that are applied at the end from sigma 1.
And I've got moments that are induced at the ends.
And that angle there would be theta.
This length here is l, like that.
And if I just use static equilibrium on that, I can say that I've got 2 times the moment, so I've got one at each end-- they're both the same sign-- minus P. And then the distance between these two P's, say I take moments about here, I've got M applied plus M applied, I've got minus P times l sine theta.
That's equal to 0.
So the applied moment there is just Pl sine theta over 2.
So now what I'm going to do is I'm going to equate this applied moment with this plastic moment, and I'm going to relate P to my applied stress sigma 1.
And then I'm going to get a strength in terms of the yield strength of the solid, there's going to be a t/l factor and there's going to be some geometrical factor.
So that's just the last step.
Boop-ba-doop-ba-doop.
So we get plastic collapse of the honeycomb.
And the stress I'm going to call sigma star plastic with a 1, because I'm going to look at the one direction.
And that happens when that internal plastic moment equals the applied moment.
So let's see.
I've got that.
Let me also write down over here, I've also got this sigma 1 is equal to P over h plus l sine theta times b.
So here I can write P in terms of sigma 1, in this thing.
And then write that, get the applied moment in terms of that, and then equate it to that.
So this term on the left-hand side corresponds to this expression for the applied moment where I've plugged in.
For P, I've plugged in sigma 1 times h plus l sine theta times b.
And that's my plastic moment on the right-hand side.
So if I just rearrange this, I can then solve for this plastic collapse stress.
So it's equal to the yield strength of the solid times t/l squared, and then times another geometrical factor.
2 times h/l plus sine theta times sine theta.
Doop-doop-doop.
So the same kind of thing, there's a solid property, a t/l, a relative density term, in then a cell geometry term.
And we can calculate with this for regular hexagonal cells.
And we can do a similar kind of calculation for loading in the other direction.
And you can get a shear strength if you want to do that, too
If you're going in the other direction, only the E or the M apply changes, right?
[?
Or ?]
like that section
Yeah, this thing here is the same
That stays
Right.
And this is-- there's a different geometry to it.
Because now you're loading it this way on.
OK, so we've calculated an elastic buckling plateau stress and a sort of plastic collapse plateau stress.
And if you have thin enough walled, say, even aluminum honeycombs, then the elastic buckling could precede the plastic collapse.
And so I'm just going to work out what the criterion would be for that to happen.
So the two stresses can be equated.
And then that's going to give us some criterion.
So the two are equal, I'm just going to write down the equations that we had.
So the buckling stress was n squared pi squared over 24 times E of the solid times t/l cubed divided by h/l squared times cos of theta.
And the plastic collapse stress for the 2 direction was sigma ys times t/l squared divided by 2 cos squared theta.
So I can write this-- because this has a t/l cubed term, and that has a t/l squared term, I can write this in terms of a t/l critical.
So if I leave it t/l here and I put everything else on the other side, I've got 12 over n squared pi squared, then h/l squared over cos theta times sigma ys over Es.
So if t/l is less than that, I'm going to get elastic buckling first.
And if it's more than that, I'm going to get plastic yielding first.
And we can work out an exact number for regular hexagonal honeycombs, so I'm going to do that.
So if I have a particular geometry, I can figure out what n is.
So for regular hexagonal honeycombs, t/l critical just works out to 3 times the yield strength of the solid over the Young's modulus of the solid.
So if we know that ratio of the yield strength of the modulus of the solid, we can get some idea of what that critical t/l would be.
So we'll do that next
And you said if t/l is less than that critical, then you're going to get the yielding first
No, if it's less, you get the buckling first.
If it's really skinny, it tends to buckle first.
So, for example, for metals, the yield strength over the modulus is roughly 0.002, like the 0.2% yield strength.
And so that means that t/l, the sort of transition or the critical value is at 0.6%.
So most metal honeycombs are denser than that.
That's a pretty low density.
But if we look at polymers, you can get polymers with a yield strength relative to the modulus of about 3% to 5%, and then that critical t/l is equal to about 10%, 15%.
So low-density polymers with yield points may buckle before they yield.
So we have one more of these compressive plateau stresses, and that's for the brittle honeycomb.
So I don't think I'm going to finish this today, but let me set it up and then we'll finish it next time.
So the idea here is that if you have a ceramic honeycomb-- remember I showed you some of those ceramic honeycombs-- that if you compress them, they can fail by a brittle crushing mode.
So ceramic honeycombs can fail in a brittle manner.
And again, initially there would be some cell wall bending, but at some point, you're going to reach the bending strength of the material.
And bending strengths are called modulus of rupture.
So you reach the modulus of rupture of the cell wall.
So I'm not going to write the equations down today because we're not going to get very far, so I'll do that next time.
But we're going to set this up exactly the same as we did for the last one, for the plastic yielding.
But instead of getting that sort of blocky, fully yielded cross-section stress distribution, we're just going to have the linear elastic stress distribution, and when the maximum stress reaches that modulus [?
rupture, ?]
the thing's going to fail.
So the form of the equations is going to be very similar to what we had for the plastic collapse stress, but there's a slightly different geometrical factor-- that's all.
So we'll do that next time.
And then next time we're also going to talk about the tensile behavior of honeycombs in-plane.
We'll work out a fracture toughness and then we'll start talking about the out-of-plane properties, as well.
So on Wednesday, we'll do the out-of-plane properties, OK?
So hopefully we'll finish the out-of-plane properties Wednesday.
And then next week, I was going to talk about some natural materials that have honeycomb-like structures, so things like wood and cork, OK?
All right, so this is the kind of most equationy lecture in the whole course.
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All right.
I should probably start.
Last time, we were talking about the honeycombs and doing some modeling of the mechanical behavior and we started off talking about the in plane behavior.
We're talking about loading it in this direction or that direction there.
And we talked about the elastic modulus.
I think I derived a Young's modulus for the one direction, a Poisson's ratio for loading in the one direction.
And then we started talking about the stress plateau and we went over the elastic buckling stress, for one of these elastomeric honeycombs like this.
And we went through the plastic collapse stress, for, say, a metal honeycomb that would yield.
And I think I started talking about a brittle honeycomb and brittle crushing.
The idea with a brittle honeycomb-- like a ceramic honeycomb-- is it could fail in a brittle manner.
And the failure is going to be controlled by the cell wall in bending.
And when that bending stress reaches the modulus of rupture, or the bending strength of the material, then you get wall fracture.
I think that's where we left it last time, right?
I had written down something about cell wall fracture.
Now, I wanted to do the little derivation.
Here's our little schematic up here.
Here's the honeycomb.
You've loaded it with sigma 1 here to such an extent that one of these cell walls has reached the modulus of rupture and has broken.
And this is the little free body diagram that corresponds.
I'm going to go through sigma 1 for loading in the one direction.
This is the same thing for loading in the two direction.
And the result for that's in the book.
OK.
If I have loading in the one direction, I can relate that horizontal force p to the stress in the one direction.
The little p is equal to sigma 1 times h plus sin theta times b.
And remember, b's the depth into the page.
And I'm going to define sigma fs as the modulus of rupture of the cell wall material.
It's the bending strength of the cell wall material.
And we're going to say that we get fracture of that bent wall when the applied moment is equal to the fracture moment.
From the plastic collapse stress from last time, we had the applied moment was equal to p times l sin theta over 2.
That was just using static equilibrium, looking at that free body diagram of the beam.
And if I write p, in terms of sigma 1 up here, I can just write that like this sigma 1 times h plus l sin theta times b.
And then I've got this other term of l sin theta and we divide that whole thing by 2.
That's the applied moment.
And we're going to get fracture when we reach the fracture moment.
I'm going to call that mf-- the moment at fracture.
Last time, we figured out a plastic moment to form a plastic hinge.
And this is an analogous thing.
But in this case, remember, if we have a beam and we have the stress profile through the cross section of the beam, it's going to look something like that.
So for our beam, that's going to be the thickness of the beam there.
So if it's linear elastic, we get the maximum stress at the top and the bottom.
And the neutral axis is here in the middle.
There's no stress there.
This is the normal stress distribution here.
And as we increase the stress for a brittle material that's going to be linear elastic till fracture, this is going to stay linear like this until we reach this modulus of rupture stress here.
When we reach that stress, then we're going to get fracture of the beam.
And we can say that there's some moment associated with that.
I could say that this stress block here is equivalent to some concentrated force and this stress block down here is also equivalent to the-- it's going to the same magnitude, but the opposite direction force.
And I can get the fracture moment by figuring out how big those forces are and multiplying by this moment arm between the two forces.
OK?
The magnitude of those forces is just going to be the volume, essentially, of this stress block here.
Imagine there's stresses there.
It's a triangle, so the area of it's going to be a half times t over 2 times of sigma fs.
And it's going to go b into the page.
So if you think of the force-- if this was the stress-- if that stress was constant, it would just be sigma fs times b times t. But it's not constant.
It's a linear relationship.
So I'm taking the area of that triangle.
That's the force.
And then I want to multiply that times the moment arm.
And the moment arm between those two forces-- each of these forces acts through the centroid of the area.
The centroid of the area is not in the middle for a triangle, and that total distance is 2/3 of the thickness, t. OK?
That's the moment arm that you get by figuring out where the centroid of these areas are.
I multiply that times 2/3 t and one of the 2's is going to cancel.
I can rewrite that and sigma fs times b times t squared over 6.
OK?
I think, last time when we talked about the plastic moment, we did a similar calculation and it worked out to sigma y bt squared over 4.
So now this is sigma fs, the modulus of rupture, times bt squred over 6.
The 6 is just slightly different because we've got a triangle here instead of a square shape like we had before.
And now I can get the brittle crushing strength and compression by just equating that applied moment to this fracture moment.
And if you do that, the result you get is this plateau stress for brittle crushing and compression.
In the one direction, it's sigma fs-- the modulus of rupture of the cell wall-- times t over l squared.
And then divided by a geometrical factor.
And her regular hexagons, it works out to 4/9 of the modulus of rupture times t over l squared.
OK?
Are we good?
We've got the in plane compressive properties now.
We've got the elastic moduli and we've got the three plateau stresses that correspond to the three mechanisms-- to the elastic buckling failure mechanism, the plastic yielding mechanism, and then the fracture mechanism for brittle crushing.
OK?
If you think of the stress-strain curve of these materials in compression, the stress strain curves all look something like that.
And now we've figured out equations that give us the modulus here and our collapse stress there.
OK?
So we can describe that stress-strain curve.
All right.
That's compression.
And the next thing I wanted to talk about is tension.
And if we think about the tensile behavior, the elastic moduli are going to be just the same.
So the moduli are the same in tension and compression.
And then, if we think about the stress plateau, we don't really have a stress plateau for an elastomeric material because there's no elastic buckling.
If you pull it in tension, you're not going to get buckling in tension.
You only get buckling if it's in compression.
If you have a material that yields like a metal, you can get a plastic collapse stress and a plastic plateau.
And that's very similar in tension and compression.
There's a very small geometrical difference, but you can, essentially, ignore it.
If you're loading the material in compression-- and imagine this was a metal-- if you load it in compression, the cell walls are getting a little further apart when I compress it.
And if you're loading in tension, like this, the cell walls are getting a little closer together.
So there's a small geometrical difference.
But if we ignore that, we can say that the plateau stress for plastic behavior is about the same in tension and compression.
And so really, the only property that's left is to look at a brittle honeycomb.
And for a brittle honeycomb, you can have fast fracture and we can calculate a fracture toughness.
So this next slide describes the fracture toughness calculation that we're going to do.
Here's our honeycomb.
I'm going to load it in the sigma 1 direction here.
I've turned the honeycomb 90 degrees, so this is still sigma 1.
And imagine now that we've got a crack here.
And I'm going to consider a situation where the crack is very large, relative to the cell size.
So it's not a crack in the cell walls.
It's a crack that goes through multiple cells.
I'm going to assume the crack is large, relative to the cell size.
I'm going to assume that the bending is the main deformation mode.
And what I'm going to do is look at-- if I have my crack tip here, I'm going to look at this cell wall a just ahead of the crack tip.
And I'm going to say, that cell wall is bent.
And I'm going to figure out something about the stress in that cell wall and look at when that fails.
And I'm going to assume that the cell wall has a constant modulus of rupture.
So the cell wall has a constant strength.
You can imagine the cell wall could have little tiny cracks in it, too.
And if a cell wall has a bigger crack, it's going to fail at a lower stress.
But let's imagine that the cell walls are all the same strength and they all have a constant modulus of rupture.
Let me write some of this down.
In tension, the elastic moduli are going to the same as in compression.
There's no elastic buckling in tension, so that's not going to happen.
The plastic plateau stress in tension is going to be very similar to that in compression.
As I mentioned, there's a small geometrical difference, but we're going to ignore that.
And then, if we had a brittle honeycomb, like one of those ceramic honeycombs I showed you, then we can have fast fracture.
What we want to calculate is the fracture toughness.
And I'm going to make a few assumptions here.
I'm going to assume that the crack length is large compared to the cell size.
And if I do that, I can say that I'm going to use the continuum assumption.
Hello.
We'll come back to that.
I'm going to say that axial forces can be neglected.
We're just going to look at bending forces.
And I'm also going to assume that the modulus of rupture is constant for the cell wall.
First, let's just think again about the continuum.
Imagine we just had a solid and we have a plate of the solid and it's loaded in tension with some remote stress-- some far away stress-- sigma 1.
And the plate has a crack of length 2c perpendicular to that normal stress.
And we're going to look at the stress-- local stress at the crack tip-- some distance r ahead of the crack tip there.
In fracture mechanics, it's been worked out what that local stress field is.
And it depends on the crack length, and then how far ahead of the crack tip you are.
So you can say that if you've got a crack length of 2c in a linear elastic solid, and the crack is normal to a remote tensile stress-- which I'm going to call sigma 1-- then that crack is going to create a local stress field at the crack tip.
And we're going to use this equation for the local stress field.
The local stress field is equal to the far away field divided by-- or multiplied by the square root of pi c and divided by the square root of 2 pi r. So there's a stress singularity at the crack tip.
And then the local stress decays as you move away from the crack tip
And what is r?
r is the distance from the crack tip.
So if that's the tip of my crack there, then r is my distance out.
OK.
In the honeycomb wall, if we look at the crack here, and then we look at that cell wall a that's just ahead of the crack tip, that cell wall is bent.
So in the honeycomb, we're going to be looking at the bent cell wall.
And that wall is going to fail when the applied moment equals the fracture moment.
If we look at wall a, we could say that the applied moment is going to be proportional to p times l. Getting ahead of myself there.
I'm going to do this-- because it's hard to say exactly where the crack tip is because there's a void there.
I'm going to use that argument here where I make everything proportional.
The moment's going to be proportional to p times l on wall a.
And the fracture stress is going to be proportional to sigma fs times bt squared.
Last time, we said it was sigma fs bt squared over 6.
It's the same thing.
I'm just dropping the 6 out.
And then I can also say that this applied moment, if it goes as pl-- p is just going to be my local stress times lb.
And then I multiply times l. So if you think of just thinking about-- if you got a load p on this member here, l, there's going to be some local stress there.
And p is just going to be that local stress times the cell wall length times the width into the page.
And then, that local stress, sigma l, I can replace with that equation over there.
So that local stress is going to go as sigma 1 times the root of c over the root of r. And I'm going to say the distance ahead of the crack tip goes as l. Instead of having r, I'm going to say it's l. It's not necessarily exactly l, but it's going to be some fraction of l. That's my local stress there.
And then I've got an l squared times b.
And if I set that equal to the fracture moment, that's going to be proportional to sigma fs bt squared.
Are we good here?
You have to think of the crack tip.
And there's some local stress field ahead of the crack tip.
And we're saying that the load p is equal to that local stress times a cell length times the depth into the board.
And then multiply it times l to get the moment.
And then I replace that local stress with this standard equation for the remote stress and the crack length and the distance ahead of the crank tip.
So here, the b's are going to cancel out.
And now I can solve for a fracture stress in the one direction.
And that's going to-- well, let me get proportional-- that's going to be proportional to sigma fs.
Then there's going to be a t over l squared.
And then, this is going to go as the square root of l over c like that.
And now, if I want to get a fracture toughness, the fracture toughness is just the strength times the square root of pi times the half crack length c. So here, my fracture toughness is just that strength times the root of pi c. So I can say that's equal to a constant times sigma fs times t over l squared.
And now, times the square root of l. These root of c's have canceled out.
So that's my equation for the fracture toughness.
And one of the interesting things here is that the fracture toughness depends on the cell size.
This is the first property that we've derived an equation for where it depends on the cell size.
OK. And here, c's just going to be a constant.
All right.
Now we've got a set of equations that describe the in plane properties.
We've got equations that describe the linear elastic moduli in the plane.
We've got three equations that describe the compressive stress for elastic buckling failure, for plastic yielding failure, and for a brittle crushing failure.
And we've got an equation for the fracture toughness, as well.
OK?
We've got a description of the in plane behavior of these hexagonal honeycombs.
The next thing I wanted to do was talk a little bit about in plane behavior, but for a different cell shape-- for triangular honeycombs.
Because they deform by a different mechanism.
And they can be used to represent the lattice materials that we looked at earlier, too.
If we have a triangular honeycomb with triangular cells, triangulated cells behave like a truss.
And you can analyze trusses by just saying that the joints are pin jointed.
There's no moments at the end of the joints or end of the members.
And the forces are all axial and so the behavior's a little bit different.
I wanted to show you how these triangular honeycombs work, too.
I can scoot this up.
Imagine that you've got a honeycomb that's an array of triangular cells like this.
And say we're applying some bulk stress sigma to it, like that.
And say it's got a depth b into the page.
When we have a triangulated structure like this, it behaves like a truss.
And we can analyze it as being a pin-jointed structure.
There's no moments at the joint.
And if it's pin-jointed and there's no moments at the nodes, then we just get axial forces along the members.
And even if the nodes were fixed-- as they are in these ceramic honeycombs-- you can show that if it's triangulated, even if you accounted for any bending, it really is a very tiny contribution to the deformation in the forces.
It's less than a couple of percent.
I'll say even if the ends are fixed-- I'll just say the bending contributes less than 2% to the forces in the deformation.
If I have a triangular cell like that, and say I pick a unit cell like this, and I say that the bulk stress produces a load of p on the top and p over 2 at each of the bottom nodes there, then the force in each member is going to be proportional to p. And for a given geometry of triangle, you can figure out exactly what the force distribution would be in each of the members.
But I'm going to use one of these proportional arguments again, just to get a general result.
Because I don't really care that much about the details of the geometry.
OK.
If I have a little set up like this, I can say that the overall stress is going to be proportional to p over lb.
And the stress in each member is going to be proportional to p over l times the thickness-- or b times the thickness.
This is the overall stress.
The overall strain is going to be proportional to some deflection of the triangle divided by the length.
So if I said, say, this length here was a length l. And then the deflection of each member is going to be proportional to p times l over es times the cross sectional area of the member, and that's just b times t. OK?
So this is the stress on the whole thing, the strain on the whole thing, and relating the delta to the p. And then, the modulus of the whole honeycomb is going to go as the stress over the strain.
So that's p over lb divided by delta over l. These l's here cancel.
And delta here is pl over es bt, and so the b's cancel and the p's cancel.
And the modulus I get for the honeycomb is just some constant related to the cell geometry times the modulus of the solid times t over l. And if you did an exact calculation for equilateral triangles, you'd find that that constant's 1.15.
The interesting thing to note here is that the modulus for these triangular honeycombs goes as t over l, not as t over l cubed.
For the hexagonal honeycomb, it went as t over l cubed.
And here, because the deformations are axial-- not bending-- it's much stiffer.
And it's much stiffer to have one of these triangulated structures.
I'll just say, here, that the modulus goes as t over l cubed for the hexagonal honeycombs due to the bending.
One of the reasons that people are interested in those lattice materials is that they, too, have moduli that go as t over l. That basically go with the relative density, rather than with the relative density cubed.
So they're much stiffer than, say, a hexagonal honeycomb.
OK?
Are we good with the triangulated honeycombs?
Yes?
What is c?
c's just a constant related to the cell geometry.
For equilateral triangles, it's 1.15.
You could work it out, but it just makes the whole thing a little more complicated to do that.
OK. That's the in-plane behavior.
And next, I wanted to talk about the out-of-plane behavior.
Remember, we said the hexagonal honeycombs are orthotropic and the orthotropic materials have nine elastic constants.
And we've figured out four so far.
We've figured out the four in-plane elastic constants.
There's five out-of-plane elastic constants to describe the elastic behavior completely.
And so we want to talk about these other elastic constants.
The honeycombs are also-- I should just back up a little bit.
The honeycombs are used in sandwich panels.
And when they're used in sandwich panels-- I brought a little panel in with carbon fiber faces and a nomex core.
If you bend that panel like that, you're going to get shear stresses in the core.
And the shear stresses are going to be going this way and this way on, and that way, that way on.
And so those shear stresses are out-of-plane.
They're in the x1, x3, or x2, x3 planes.
And so you need the out-of-plane properties for the shear properties in the sandwich panels.
Honeycombs are also sometimes used as energy absorption devices.
Not these rubber ones, but imagine there was a metal one.
And when they're used for energy absorption devices, they're typically loaded this way on.
Again, that's the out-of-plane direction and you need the out-of-plane properties.
And for the out-of-plane properties, the cell walls don't tend to bend.
Instead, they just extend or contract.
And you get stiffer and stronger properties.
Let me just write something down and then we'll start to derive some of those properties.
The cell walls contract or expand instead of bending, and that gives stiffer and stronger properties.
OK. OK.
There's five elastic constants in the out-of-plane directions.
We'll start with the Young's modulus.
And if I take my honeycomb and I load it in the x3 direction-- just taking this thing here and just loading it like that-- the cell walls just axially contract and the stiffness just depends on how much cell wall I've got.
So the modulus in the three direction is just equal to the area fraction times the modulus of the solid.
That's just the same as the volume fraction, or the relative density.
So it's quite straightforward.
The cell walls contract or extend axially.
e3 is just es times the relative density.
And that's just es times t over l. And then there's a geometrical factor here.
h over l plus 2 over 2. h over la plus sin theta times cos theta.
Again, a little bit like those triangular honeycombs.
The thing to notice here is that in the three direction, the modulus goes linearly with t over l, whereas in the in-plane directions, it goes with t over l cubed.
So there's a huge anisotropy in the honeycombs because of this difference.
Imagine a honeycomb might be 10% dense.
t over l might be something like a tenth-- 0.1.
So e star 3 is going to 0.1 of es, roughly, and in the other direction, it's going to be 1/1000th.
So there's a huge anisotropy because of this.
Let me just-- square honeycombs.
This just shows looking at the out-of-plane directions and the different stresses and properties that we're going to look at here.
The next one we're going to look at is Poisson's ratio.
And first, we're going to look at loading in the x3 direction.
And if we load it in the x3 direction, the cell wall's just strain by whatever the Poisson's ratio is for the solid times the strain in the three direction in the other two directions.
We'll say for loading in the x3 direction, the cell wall's strain by nu of the solid times whatever the strain is in the three direction in the other two directions.
If we load it in the x3 directions and everything contracts by that much in the other two directions, that just means that the Poisson's ratios-- nu 3 1 and nu 3 2 are going to be the same.
And they're just going to be equal to the Poisson's ratio of the solid.
So if each wall is going to contract by that amount, the whole thing's going to contract by that amount.
And that's going to give you that Poisson's ratio.
Let me just say, here, also-- and remember that I'm defining nu ij as minus epsilon j over epsilon i.
We're loading in the three direction here.
And then you can get the other two Poisson's ratios using those reciprocal relationships.
So nu 1 3 and nu 2 3 can be found from the reciprocal relations.
And remember, those relationships come from saying that the compliance tensor, or the stiffness tensor, is symmetric.
We can write, for instance, that nu 1 3 over e1 is equal to nu 3 1 over e3.
So I can write that like that.
And then I can say nu 1 3-- that is going to equal to nu 3 1 times e1 over e3.
And we just saw that nu 3 1 was equal to nu s. And we see, from before, the e1 is equal to some constant.
Let me just call it c1 times es times t over l cubed.
And e3 is going to be some other constant times es time t over l. The es's are going to go.
And if t over l is small-- even if it's say, a 10th-- and this is going as t over l cubed and that's going as t over l, then I can say this thing is about equal to 0.
It's going to be small.
It's not to be exactly [INAUDIBLE] 0, but it's going to be small so we're going to say it's 0.
So I'll just say for small t over l. And then, similarly, nu 2 3 is going to be close to 0, as well.
So there's the Poisson's ratios.
We've got the Young's modulus, the Poisson's ratios, and next we want to get the Shear moduli.
And the shear moduli is little more complicated.
The cell walls are loaded in shear but the neighboring cell walls constrain them and they produce some non-uniform strain.
I'm talking about shearing it this way on.
You can see on this figure here, we're talking about shearing it, like tau 2 3 or tau 1 3, this way.
And so each wall is going to shear, but the walls are attached to each other so they can't just do it independently.
They have to be constrained by each other.
And the exact solution is a little bit complicated.
And I'm just going to give you an estimate of what that modulus is.
And we're going to see that it depends linearly on t over l, as well.
I'll just say the cell walls are loaded in shear.
An estimate is g star 1 3 is equal to g of the solid times t over l times a geometric function.
It's cos theta over h over l plus sin theta.
And for regular hexagonal honeycombs, it's 1 over root 3 times gs times t over l. Again, just note the linear dependence of the modulus on t over l. And in the book, there's a method using upper and lower bounds that gives an estimate for g 2 3.
I'm not going to go into it.
I just want you to notice that the shear moduli go as t over l, just like the Young's modulus does.
And Sardar, who's sitting in the back there, has done even more involved calculations and analysis of the shear moduli of the honeycombs in this direction.
So I'm not going to go into all the gory details on that.
OK. That gives us the moduli now.
So now we've got all nine elastic moduli.
OK?
And the next thing to do is, then, to figure out the compressive strength.
So we're going to look at compression again, and then we'll look at tension.
If we look at compressive strengths, again, we've got different modes of failure.
And if I have an elastomeric honeycomb like this one here-- if these cell walls were a little longer, I might be able to actually do it.
If you compress this enough, you produce buckling in the cell walls.
And this is a schematic of this buckling pattern here.
And you can see there's a diamond pattern where it alternates up and down in the different cell walls.
We're going to do some approximate calculations, but you can see the idea of how the material behaves in this direction, just from these approximate calculations.
OK. Say we have our honeycomb like this, and here's the prism axis this way.
And now, we're going to load it up with some stress in the three direction.
I'm going to call this sigma star elastic 3 when it buckles.
And what we're going to do is just look at a single plate.
And look at the buckling of a plate.
We're going to analyze it just looking at a single plate and then adding up how many plates we have per unit cell.
It's actually more complicated than this because, obviously, the plates are attached together and there's some constraint by attaching the plates.
But we're not going to worry about that.
If you have a column-- just a, say, circular cross-section column-- and you apply a compressive load to it, it buckles at the Euler load.
And similarly, there's an Euler load for plates.
And that equation is usually written as a p critical is equal to some end constraint factor.
For plates, it's usually called k instead of n. So this is an end constraint factor.
It depends on the modulus of the plate.
It goes as t cubed.
Then, there's a factor of 1 minus mu of the solid squared and the length of the plate.
Say this plate here-- actually the width of the plate there is h and the length here is b.
And this thickness here is t like that.
Here, k is an end constraint factor.
And it's going to depend on the stiffness of the adjacent cell walls.
If I had a honeycomb, and say it was-- these walls here-- the adjacent walls-- were thicker, then you can imagine those thicker walls-- it'd be harder to get them to deform.
And the end constraint for the plate is going to depend on those thicker walls.
So that the end constraint, k, depends on these-- say I'm looking at this wall here of width h here, then how stiff these other two walls are is going to affect that end constraint factor.
What we're going to do is just do something very approximate.
We're going to say if these vertical edges here-- if this edge here and that one there-- if they were simply supported-- if they're just pinned to the next column, the next member-- then k has some value.
And if they're fixed, it has some other value.
And we're going to pick a value in between.
So we're going to do something very approximate.
I'll say if those vertical edges are simply supported-- that means they're free to rotate-- then k is equal to 2.0.
And this is if b is bigger than three times the length.
So this is h here, or we could say l. Either way.
It's really the-- it's the length when we look at the honeycomb this way on, but it's the width in that picture there.
And if the vertical edges are clamped, or fixed, then k is equal to 6.2.
These are values you can look up in tables of plate buckling.
And we're just going to approximate it by saying k is equal to 4.
We're just picking a value that's in between those two.
And then, the p total is going to be the sum of the p criticals for the columns that make up a unit cell.
For the unit cell, I have one wall of length h and two of length l. And if you just take that total load and divide by the area of the cell, you get that this compressive strength for elastic buckling is approximately equal to es over 1 minus nu s squared times t over l cubed.
And then there's a geometrical factor here.
And if you had regular hexagonal cells, this buckling stress works out to 5.2 times es times t over l cubed.
If you remember, for the loading in the two direction-- in the in-plane direction-- it has the same form.
And goes as es times t over l cubed, but it's much smaller.
This number here was, I think, 0.2.
It was much smaller.
So it has the same form, but it's a lot bigger.
OK?
Are we good with that?
The idea is we just use the standard equations for plate buckling.
We make some estimate of what that end constraint factor is.
And we just have an approximate calculation here.
OK. That's the elastic buckling.
If I had a metal honeycomb, then it might not fail by elastic buckling like that.
Instead, we'd probably get yielding.
If it was dense enough, we could just get axial yielding that-- if you just loaded it, you'd have axial forces.
And at some point, you'd reach the yield stress.
And so you can get failure by just uniaxial yield.
That's one option.
And if you get that, then it just depends on how much solid you've got again.
So it's just the yield strength of the solid times the relative density.
But usually, the honeycomb is thinner walled than that.
And usually, you get plastic buckling proceeding that.
In plastic buckling, you can think of it as-- say if you have a tube-- this is just shown for an individual tube here.
You can see how the tube folds up.
And you can get that same kind of thing with the honeycomb.
Here's a single tube.
It's been loaded along the prism axis of the tube.
And you can see, you get these folds, and the more you load it, the more number of folds you get.
And the more the folds concertina up.
To do an exact analysis for the honeycombs, you would have to take into account not just one tube, but the constraint of the neighboring tubes again.
And again, that gets to be a complicated, messy thing.
So again, we're going to do a more approximate thing.
What we're going to do is just say that we have members that are folding up like that.
So the same geometry.
But we're just going to look at a single cell wall and see what the single cell wall does.
And someone else has done the more exact calculation.
We'll just compare our approximate calculation to the exact one.
OK. We're going to consider an approximate calculation.
What we're going do is look at our isolated cell wall.
And if you look at the figure here, the wall is going to be vertical, initially.
And as we load it, eventually it's going to buckle and we're going to form one of those plastic hinges in the middle here.
And then, the thing is then going to rotate about that plastic hinge and just fold up.
So at the bottom here, it's completely folded up.
OK?
And we're going to do a little work calculations.
We're going to look at the internal work done and we're going to look at the external work done.
The external work is just going to be this load p times that deflection delta that the p moves through.
And if we say this is half of a wavelength-- if you think of this thing going through multiple wavelengths, just consider when it folds up like that, that's a half of a wavelength.
It would go two of those to get a full wavelength.
That's lambda over 2.
And so to go from this stage to that stage over here, the external work done is going to be approximately p times lambda over 2.
Say that it's thin and that 2t is small compared to lambda.
So it's going to be about p times lambda over 2.
And then, we're also going to look at the work done by the plastic moment.
And when we form the plastic hinge here, there's a plastic moment.
And that moment is going to rotate through an angle of pi.
So we start out straight here, we end up folded up like that, and we've gone from straight to that.
We had to go through 180 degrees to get there.
So it goes through an angle of pi.
And if you have a moment going through a rotation, the work done is the moment times the rotation.
We're going to equate those two works done.
We're going to look at the rotation of the cell wall by an angle of pi at the plastic hinge.
Our plastic moment-- it's going to be the yield strength of the solid again times t squared over 4, the same as when we were talking about the plastic moment before for the other loading direction.
But now, instead of multiplying this times b, we're multiplying it times 2l plus h. That's the length of the cell wall that's associated with one cell.
And now, it's not b because now we've turned the thing the other way on.
We're loading it the other way on.
And this plastic hinge-- if I think of-- if this was b before.
And now that b is l plus 2h-- or 2l plus h, rather.
That's the dimension of the-- let me draw a little hexagon so maybe you can see.
OK. Now we're forming a plastic hinge halfway down the board.
Imagine that this has some length b that way and we're halfway down the board.
And now, the plastic hinge has to form all the way around these members here for one cell.
Or you could think about it as this guy plus these guys is one cell.
You can think about the unit cell different ways, but it's one h plus two l's.
OK?
Are we OK with that?
OK. Then the internal plastic work is that plastic moment times the rotation pho-- or pi, rather.
Sorry.
Are we OK with this?
That the work done is m times our angle?
Imagine-- let me get rid of my honeycomb here.
Imagine you have a point here and you have some force over here.
Let's call that f. And say, the force is at distance r from f. And say that it moves through some distance.
The moment here would be r times f. And if that rotates, say, through some angle-- let's call it alpha-- and here is f here, then this distance here that the force moves through is just r times alpha.
So the work done is going to be r times alpha times f, or just the moment times alpha.
OK?
So that's all that we're doing.
OK. That's the internal plastic work.
And now we have to look at the external work done.
And that's equal to p times lambda over 2.
Here, lambda is the half wavelength of the buckling.
I'm going to say for these tubular kinds of things, it's in the order of l. So if you look at that last slide here-- oops.
Rats.
How'd that happen?
Let me scoot back down here.
There.
If we look at that guy again, the magnitude of the buckling wavelength is on the order of l. And here, below p, can be related to the stress in the three direction.
We'll just multiply it times the area of the unit cell.
And so if I equate the internal work and the external work, I can say p times lambda over 2 is equal to pi times my plastic moment.
And then, for p, I can write sigma 3 h plus l sin theta times 2l cos theta.
And then, lambda is l divided by 2 is equal to pi.
And then I've got my plastic moment over there.
And then if I solve for sigma 3, that's my compressor strength.
I've got pi by 4, the strength of the solid, sigma ys, times t over l squared.
Then h over l plus 2 divided by h over l plus sin theta times cos theta.
And for the regular hexagons, this works out to about 2 sigma ys times to over l squared.
And the exact calculation for regular honeycombs is equal to 5.6 times sigma ys times t over l to the 5/3 power.
This power here-- 5/3-- is a little less than 2.
And that's because the additional constraint of the neighboring cell walls.
But the main thing we're interested in, in these sorts of calculations, is the power dependence on the density and this simple calculation.
Obviously, it's not exact, but it gets you close.
OK.
I'm just going to wait for people to catch up a little.
OK.
The next property I'm going to look at is out-of-plane brittle fractures.
Say we loaded in tension, and if we had no cracks in the walls, we'd just see uniaxial tension and the strength would just be the strength of the solid times the relative density times the amount of solid.
We'll just say if defect free, the walls see uniaxial tension.
And then the fracture stress in the three direction is just equal to the relative density times the fracture strength of the solid.
If the cell walls are cracked, and if the crack length is very much bigger than the cell length, then the crack propagates normal to x3.
Then we can say the toughness gc-- or the critical strain energy release rate-- is just equal to the volume fraction of solid times gc for the solid.
And then the fracture toughness, k1c, is equal to the square root of the Young's modulus times gc.
And that's just equal to the relative density times the modulus of the solid.
And then the relative density times the toughness of the solid.
So it's just equal to the relative density times the fracture toughness of the solid.
It's just straightforward there.
Then we've got one last out-of-plane property.
And that's brittle crushing and compression.
And if we have some compressive strength of the cell wall-- say I call it cs-- then it's just the relative density times that strength.
And for brittle materials, that crushing strength is typically around 12 times the modulus of rupture, or fracture strength.
We could say that's about equal to 12 times the relative density times sigma fs, a fracture strength.
OK. That's the modeling of the honeycombs.
I know there's been a lot of equations and derivations, but that's the basis of a lot of the things we're going to do in the rest of the course.
The modeling we're going to do on the foams is based on this and the mathematics is just easier because we're going to use these dimensional arguments.
We're not going to figure out all these geometrical parameters.
Before we get to the foams, I wanted to talk a little bit about honeycombs in nature.
And we've only got a couple minutes left, so I won't really get that far.
But I wanted to talk a little bit about honeycomb materials in nature.
And the two examples we're going to talk about are wood and cork.
I'm going to talk a little bit about the structure of wood next time.
Then, we'll see how we can apply these models to understanding how wood behaves.
And we'll see how you can use these models to predict the density dependence of wood properties and also the anisotropy in wood properties.
And I guess we'll probably, maybe, start it Wednesday next week.
We'll talk about cork, as well.
Those of you who took 3032 know that I like cork because of Robert Hooke and his drawing of cork.
And I made a new video that I'm going to show you.
Remember in 3032, I showed you the video from the Bodleian Library, where they had the first edition of Hooke's Micrographia.
Well, it turns out Harvard has a first edition.
Harvard has three first editions.
Yeah.
Exactly.
MIT has zero first editions.
Gee, why does that surprise me?
And I have a friend who's a librarian at Harvard and she arranged for me to go and make a little video with the first edition of Micrographia.
So I can-- I don't if we'll play the whole thing, but I'll show you the first little bit of it.
And you can watch it at your leisure.
And Sardar came.
You came and saw it with me.
You came and saw the first edition with me, right?
Yes
Yeah.
Yeah.
It's very beautiful and you'll see some of the nice drawings.
And I talk about the cellular structure of some of the drawings.
So we'll talk about wood and cork next time.
But I think I'm going to stop there because that seems like enough equations for now.
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So I should probably get started.
So I just wanted to mention this Friday the libraries are having Furry Friday.
So they have therapy dogs come, and if you like dogs, it's kind of fun to go get cuddled by the dog.
The other thing I wanted to mention, last term, there was a student taking 3032 who was interested in art.
And I kept trying to find art pictures for him, and he's not here.
But I thought everybody else can like the art too.
So I belong to the Peabody Essex Museum in Salem, Mass.
And they have an exhibit right now on wood and on sort of using wood as a sculptural material.
And this is one of their posters to advertise it.
So this thing was carved out of a single piece of wood, I think.
And they've got lots of other sort of sculptural wood.
So I thought you might like to see that.
If you wanted to go to Salem, there's a couple of options, if you don't have a car.
You can take a commuter rail to Salem.
You can also take the ferry.
And if you take the ferry to Salem, it's like a five minute walk to where the ferry lets you off to get to the Peabody Essex.
And it's a kind of neat museum.
It's not too big.
It's kind of small.
But it's a beautiful building, and they have neat stuff there.
So you could go to the Peabody Essex Museum.
Hmm?
It's very nice
Yeah, you've been there?
Yeah, it's really nice.
So I was going to talk about honeycomb-like materials in nature today.
And I'm going to talk about wood today, and I might finish this today.
I might not.
And then I'm going to talk about cork for a little bit on Wednesday, and then we'll start talking about foams after that.
So I have a couple of sort of cute little language historically things.
And you know how I like that stuff too.
So I have two things about words that are related to wood.
So the word "materials--" you know where the word materials comes from?
It comes from the Latin.
So there's a Latin "materies materia."
And materies materia means "wood" or "the trunk of a tree."
So if you think of studying materials, in olden times that was like studying wood.
And another cute thing that I found was that in old Irish the names of the first few letters of the alphabet are named after trees.
So the letter A, that's called alem in old Irish, and alem is the word for elm.
And B is-- I don't know if I'm saying these right.
It's called beith, and that's the word for birch.
And C is call.
That's the word for hazel.
And D is dair, and that's the word for oak.
And so they sort of named the letters of the alphabet after different kinds of trees, different kinds of woods.
So I just thought those were kind of interesting historical things.
So I wanted to start by talking about wood structure.
And then we're going to look at how would deforms and fails, and talk about the data that people have measured for the wood properties, things like stiffness and strength.
And then we'll talk a little bit about how the honeycomb models can be applied to understanding the mechanical properties of woods.
So this is kind of a generic trunk of the tree here.
And we're defining three axes.
The radial axis comes radially out of the tree.
There's the tangential axis, so that's the x1 and the x2 axes.
And then there's the longitudinal or the axial axis, x3.
So if you think of the wood as being, in a very, very simple way, just like the honeycomb, the radial would be this way on.
The tangent would be that way on.
And that axial would be that way on.
So it's like that.
And if you neglect the growth rings, you can say that other woods orthotropic, and that's typically what people do.
They neglect the growth rings, and they say that it's orthotropic.
And the density of the woods, the relative density ranges from about 5% for balsa wood to about 80% for lignum vitae.
So I brought in some pieces of wood.
So this is balsa wood.
You're probably familiar, making different kinds of models with balsa.
So balsa's very light.
It grows in Ecuador.
And it's the lightest wood.
And this is lignum vitae.
You're probably not so familiar with that.
This actually grows in Florida, and it's the densest wood.
It has a relative density of 0.8.
And it's so dense, that if you put it in water, it sinks.
So it's a very dense wood.
And the way the wood cells grow is that if you look at the sort of structure here of a tree, there's the bark on the outside here, and then there's the kind of wood cells inside the bark.
And there's a layer of cells in between the bark and the wood called cambial layer.
And that's really the layer of cells that are alive and are dividing.
So if you think of the wood cells, they're living when they're in that little cambial layer there.
And they're dividing.
And that cambial layer, the cells have a plasma membrane and a protoplast.
And then they sort of exude the plant cell wall.
So a little like bone cells, like if you think of the bone in your body, there's osteoblasts and osteoclasis, different kinds of bone cells.
But the bone cells secrete the sort of collagen and the calcium phosphate that are the sort of hard mineral part of the bone that you think about in the bone.
And that's not a living thing.
The cells are the living thing.
That's like an extracellular matrix.
In the trees, it's a little bit the same.
So there's the living cells that are just under the bark, and they have this plasma membrane and the protoplasm.
And over a few weeks they excrete the plant cell wall, and then they die.
So the living cells die, and you're left with the plant cell walls.
And then as the tree grows, you're always having a layer of these cambial cells, and it forms bark on the outside and wood on the inside.
So there's sort of a layer of cells that are differentiated, such that on the outer layer they form the bark, and on the inner layer they form the wood.
And as the tree grows, that cambial layer is kind of expanding out radially.
So let me write some of these things down.
Let's see.
So let me just write down those two little word things because I think they're cute.
So the word materials is from the Latin materies materia.
And that means "wood" or the "trunk of a tree."
And here's the little old Irish thing.
It's not like I think-- I'm not going to put this on the test or something.
I just thought it was cute.
So the letter A is alem, which is elm.
And letter B is beith, which is birch.
Letter C is call.
That's hazel.
And D was dair, and that's oak.
So that's just for general interest.
So then the wood structure we can think of it as orthotropic, if we ignored the growth rings.
And if you have a sort of large diameter tree and you take a piece of wood not from the very center, but from somewhere near the outside, then that's not a bad approximation.
Now, the relative density of the woods ranges from about 0.05 for balsa to about 0.8 for lignum vitae.
Any Latin scholars here?
I took one year of Latin in high school.
Anybody take Latin?
No, no Latin.
So I think lignum vitae I think is "tree of life."
"Vitae" is the sort of life.
And when it has this ending A-E it means "of life."
So I think that's the "tree of life" is lignum vitae.
So trees have cambial layer beneath the bark.
And the cell division occurs in that cambial layer.
So the new cells on the outer part turn into bark, and the new cells on the inner part turn into wood.
And then we have the living plant cells that have the plasma membrane and the protoplast.
And those cells then secret the plant cell wall, which sort of surrounds them.
So in trees, the living cells lay down the plant cell wall over a period of a few weeks.
And then the living cells die.
Oops.
Back.
Here.
Now you always retain a layer of those cambial cells.
So you may have heard if you have a tree and you cut a ring around the tree through the bark, if you go into those cambial cells and you destroy them, you kill the tree, because you're killing that layer of living cells.
So then we want to look at the cellular structure of the woods as well.
And I've got a couple of slides here.
This one is of softwoods.
And softwoods have two types of cells.
That have tracheids, which are the bulk of the cells here, and the tracheids provide structural support.
And the tracheids also have little holes along the length of them at their ends called pits, and those pits allow fluid transport up and down the tree.
And then the softwood also has these ray cells here.
So those are examples of ray cells.
So this is a transverse section.
This is a longitudinal section here.
And the rays are parenchyma cells which store sugars.
So softwoods have tracheids and rays.
And then hardwoods, here's an example of a hardwood oak.
They have three types of cells.
There's cells called fibers, so these guys all in here would be fibers.
They provide the structural support.
They have vessels, these really large cells that provide fluid transport up and down the tree.
And they also have rays.
So here are some rays here.
And again, those rays are parenchyma cells that store sugars in the tree.
So let me just write down what all these cells are.
So in softwoods most of the cells are these tracheids, so they make up the bulk of the tree, something like 90% of the tree.
And they provide structural support.
They have holes in the cell wall for fluid transport, and those are called pits.
And to give you some idea of what size they are, they're are a few millimeters long, so something like 2 and 1/2 to seven millimeters long.
And then they're tens of microns in the other two directions, so they're something like 20 to 80 microns across.
And the cell wall thickness, t, is usually a few microns, so something between about two and seven microns.
So typically, the denser the wood, the thicker the cell wall's going to be.
Whoops, let's see if I can get the rays down here.
Put it on the same board.
So the rays are parenchyma cells that store sugar.
And then the hardwoods have three types of cells.
They have the fibers that provide the structural support.
And the amount of cells that are fibers varies, depending on the species, but it's usually somewhere around 35% to 70% of the cells.
And then they the vessels, which are the sap channels.
That provides for the conduction of fluids.
And that's between about 6% and 55% of the cells.
And then, again, there's rays that store sugars, and they usually make a boat 10% to 30% of the cells.
So there's the structure of this sort of cellular structure, at this kind of length scale of tens of microns.
And then there's also a structure within the cell wall itself.
And the cell wall itself is made up of cellulose fibrils in a matrix of lignin and something called hemicellulose.
So if you look at the cellulose structure, the cellulose has a regular structure, a sort of periodic lattice.
And it's crystalline for most of the length of the fibrils.
So this is the structure of the cellulose here, and this is showing it at a slightly larger length scale.
It might have a crystalline region here and then a non-crystalline region here.
And these macro fibrils, which are made up of bundles of micro fibrils, are about 10 to 25 nanometers.
And each one of the micro fibrils might be three to four nanometers across.
So you have these cellulose fibers.
And then the cell wall is made up of different layers.
So there's what's called the primary wall here, which has a random arrangement of the cellulose fibrils.
Then there's an outer layer here.
These are all called secondary layers.
This is S-- I think that's S1.
Yeah, it's S1.
And it has this arrangement of the fibrils.
Then there's a layer called S2, and it's generally the thickest layer in the cell wall.
And the cellulose fibrils are aligned not perfectly vertical, but a little off the vertical.
And the angle between the vertical and the orientation of the cellulose fibers is called the microfibril angle.
And then there's a third layer here, S3, with, again, a different winding of the fibers.
So because S2 layer is the thickest layer and because the fibrils are closest to the vertical axis, the S2 layer actually contributes the most to the longitudinal modulus and stiffness and strength of the cell wall.
So that's kind of the arrangement of the cell wall.
And then so that one cell would have that.
Another cell would have that.
And in between the two, there's a layer called the middle lamella that kind of glues them together.
So that's the arrangement of the cells.
Let me scoot over here.
So the cells are often modeled as a fiber reinforced composite that has four layers to it.
And in each layer there's different volume fraction of the fibers and different orientation of the fibers.
So the cell wall has this fiber-reinforced structure.
Here's the cellulose fibers in a matrix of lignin and hemicellulose.
And there's four layers, each with the fibers in a different orientation.
And then there's the middle lemella between the two cells.
So in doing the modeling of a material like wood, you need to know what the properties of the cell wall material are, because, obviously, the properties of the wood would depend on the cell wall properties.
And it turns out that they're similar.
They're not exactly the same, but they're similar in different species of wood, so we're going to call them more or less the same.
So the density of the solid is 1,500 kilograms per cubic meter.
The modulus of the solid in the axial direction is 35 gigapascals.
The modulus in the tangential direction or transverse direction is 10 gigapascals.
And the strength of the solid in the axial direction is 350 megapascals.
And the strength of the transverse direction is about 135.
So here A means Axial direction, and T is transverse.
And just for comparison, if you just look at cellulose, cellulose has some pretty amazing properties.
The modulus of cellulose is about 140 gigapascals, which is very high for a polymer.
And the strength of cellulose fibers run between about 700 and 900 megapascals.
So the cellulose fibers have very impressive properties.
And that's one of the things that gives wood very good properties.
So the next thing is I want to show you some stress-strain curves for wood.
And you'll see how similar they are to the honeycombs that we looked at before.
And then we'll look at how the cells are deforming as they're getting loaded.
And from that, we're going to do some modeling.
So let me just wait till people get caught up.
Are we caught up?
More or less?
OK.
So these are all compression curves, so I'm just going to talk about compression.
So these are curves for different types of woos.
And on the left, the wood is loaded in the tangential direction.
So in terms of the sort of honeycomb model, it's loading at kind of this way on, like that.
And on the right, are a set of curves for wood load it in axial compression.
So in axial compression loading, we're loading it that way on.
And we've got different species here.
So the lowest density is balsa, around about 100 kilograms per cubic meter.
The densest species on this plot is beech, which is around 700 kilograms per cubic meter.
And then there's pine and willow, some other species in between here.
So you can see the shapes of the curve look just like the curves that we had for the honeycombs.
So here there's a linear elastic bit.
There's a stress plateau.
And there's a densification bit.
And then, as the density goes up, it gets stiffer, and the strain at which the densification occurs gets smaller, and the strength gets higher.
And if we look at the axial properties, the shape of the curve is similar.
We get linear elastic stress plateau densification.
But if you look at the scale here, this scale goes from 0 to 100, whereas that scale went from 0 to 20.
And so the stiffness and the strength along the grain are much higher than they are across the grain.
And you probably already know that.
Wood is stronger and stiffer along the grain than across the grain.
So that's what the stress-strain curves looked like.
And the fact that we're getting the curves that look like that makes us think maybe the mechanisms of deformation and failure are similar to the honeycomb, too.
So here's a set of curves for balsa all plotted on the same scale.
And, again, you can see for loading across the grain, either in the radial or the tangential direction, the stiffness and the strength is a lot less than if you load it in the axial direction.
So a number of years ago, we had a project on balsa.
And the thing we were interested in doing was looking at how the cells deformed and failed.
And because balsa's a low-density wood, it was easier to see the deformation in the cells, because the cells were thin.
So that's why we chose balsa.
I actually have a project on balsa right now.
And [?
Sardar ?
], my postdoc, is doing more detailed kind of finite [INAUDIBLE] modeling, trying to represent the structure of balsa.
And I think I mentioned, the reason we're interested in it is the balsa's used as a core in sandwich panels in wind turbine blades.
It's actually the best material that they can find, it's better than any engineering material.
So that's comparing the three curves for balsa.
And then if you look at a specimen that's loaded in the [?
SCM ?
], with a loading stage, you can measure the stress-strain curve, and you can take photographs of what the cellular structure looks like at different stages of loading.
So here, this picture one, is unloaded.
And these four images here are looking at the same section of cells, the same area of cells.
And you can see there's a big vessel here, and that's the same vessel there.
So here, this image two, is at this point on the stress-strain curve.
Here's three, at that point.
And four is at that point.
So if you look at this carefully-- and I've got another higher mag picture I'll show you in a second-- you can see that what's happening is the cell walls are bending.
So it's kind of like taking my honeycomb like this, and I'm doing that to it, and the cell walls are bending, so just the same as the honeycomb.
And then eventually, if I load it enough, you get to this sort of densified stage.
And you're doing this, and the stress-strain curve increases sharply.
So here you can see how the cells have densified over here.
It kind of looks a lot like my honeycomb when I-- maybe I do it this way-- when I smush it up like that.
It looks kind of similar.
So if we look at the higher mag picture, again, these four images are the same area of the cells.
And if you look at that a little bit of crud, it's the same on all four of them there.
So this is the unloaded one.
And these were loaded from top to bottom.
And this is loaded to some extent/ that's loaded more, and that's loaded more.
So if you look at this cell here, it's got this little tear on it.
So you can sort of find it again.
If you look at that cell there, that's what it looks like unloaded.
And here you can see-- see that wall there?
You can see how it's bent up.
So it's bent like the honeycomb walls.
And here it's bent even more.
And eventually, it has this sort of a shape here, and it's deformed permanently.
It's formed one of these plastic hinges.
So it's like the aluminum honeycombs almost, that it's filled like that.
So in the balsa wood, when we load it in the tangential direction, we're getting bending of the cell walls and then yielding and plastic hinges forming, just the same as we would in an aluminum honeycomb.
Are we good with that?
Well, I'll go through all three directions.
And then I'll write down the notes.
So this is loading the balsa in the radial direction.
And these things here are the rays.
So we're loading it in that direction.
And here you see this also bending occurs, but the rays act a little bit like fiber reinforcement.
So the rays are a little bit stiffer, and they sort of reinforce the thing a bit.
And this is the loading platten here, and you can kind of see that the failure starts at the loading platten, and as you sort of load up more, it progresses in from the loading platten.
So we're going to look at the modeling of the balsa in the radial direction, and we're going to count for the rays, at least in a crude way.
And then when you load them balsa in the axial direction, initially you don't really see much happening.
So if one is unloaded, one's down here.
And two is at this peak stress up here.
And really, if you look between one and two, you just don't see an awful lot of difference.
And that's because what you're doing is you're taking the wood and you're loading it this way on.
And it's so stiff, you just don't see much deformation.
So there's not really much to see.
But then eventually, something starts to fail.
And in this case, what fails are the end caps.
So the balsa wood has these long cells here.
But then at the end of the cells, there's little caps on the ends.
And the cells kind of fit together like that.
So that eventually, if you keep smushing it, those end caps start to fail.
Here you can see how bright it gets, and the cells are starting to crush together and kind of fail those end caps.
And in fact, each one of these serrations here, if you look at, say, from that peak up to that peak, that corresponds to a length of about the length of the cell, or the length of the cell between the end caps.
So in axial deformation you're just actually deforming the cells until you break those end caps.
If you look at denser words, they fail in slightly different ways.
This is a Douglas fir, which is much denser.
This particular specimen, the whole thing is kind of buckled over.
So it's not really so representative of the structure itself.
This is Douglas fir in radial compression.
You can see this picture, it looks just like what I showed you for the balsa wood, that sort of propagation of the failure.
These long things here are the rays.
And this is a Norway spruce in axial compression.
And this is fairly common in denser words.
You get this buckling formation.
And what happens is, I think, you get some yielding of the cell walls initially, but that leads to buckling, like a plastic buckling.
And you can see on this higher mag picture down here, you get these really small wavelength buckles in the cell wall.
And the two-- you get a plane that kind of shears over itself.
And you can see in the top image, this top half has shared over relative to the bottom half.
And all the deformation is in this little band here.
So this stuff here is all going on in that band up there.
So let me write down some notes about how these things to form and fail.
And then we'll get to the modeling in little bit.
So we can say the stress-strain curves resemble those for honeycombs.
And I'll say the mechanisms of deformation and failure are most easily identified in low density balsa wood.
So for balsa, if we look at the tangential loading, we see bending of the cell walls and then eventually plastic yielding.
And for radial loading, the rays act as reinforcing.
And for axial loading, you get axial deformation and then the failure of the end caps.
And I'll just say failure by plastic buckling is also observed, say, in the denser woods.
[INAUDIBLE] So then we can look at some data for the properties of woods.
And these charts plot relative Young's modulus and relative strength against relative density.
So here the modulus of the wood is divided by the modulus of the solid cell wall material.
And here we've normalized everything by the modulus of the solid cell wall material in the axial direction, because the cell wall itself is anisotropic.
And so here's the relative modulus, and here's the relative density.
These are log-log plots.
And we see that when we load the wood in the axial direction, the moduli is just linearly related to the density.
And when we load it across the grain, it varies with the cube of the relative density.
So do you remember our little honeycomb models?
If I took the honeycomb and I loaded it this way, it went as the cube of T over L. And that's because the bending.
And so the wood doesn't lie perfectly on that cube line, but it's fairly close.
And then similarly, if we took the honeycomb and we loaded it this way on, it deformed axially.
The modulus depended linearly on the density.
So you get the same kind of relationships there.
And then if you look at the strength, the strength along the grain goes linearly, and the strength across the grain goes with the square.
And we'll see when we get to the modeling in a minute, that if we loaded, say, an aluminum honeycomb this way on, the strength would go linearly with the density, if we're just yielding the cell walls.
And if we loaded it this way on, it went as the square of T over L. So these things kind of correspond.
And you can see the structure of the wood is a lot more complicated than just a simple honeycomb.
And so these models are sort of first order, and they're fairly crude.
They don't try to capture every detail of the wood structure.
But they can give you a sense of where the wood properties are coming from.
So let me just write down some of these observations.
So the data for the wood-- the modulus along the grain goes linearly with density.
It goes more or less as the cube for loading in the tangential direction.
And the radial direction is somewhat stiffer different than that.
The strength in the axial direction goes linearly with the density.
And the strength across the grain goes with the square of the density.
And then there's data for the Poisson's ratios too.
So let me just write them down.
So the modeling based on the honeycomb is sort of a simplified model that gives you kind of a first-order description of the behavior.
And it doesn't really attempt to capture all the details of the softwood and hardwood structure.
And in the equations, I'm going to take the cell wall properties along the grain, or along the axial direction.
And we're going to have a bunch of constants that describe the cell geometry, and those constants are also going to reflect the cell wall anisotropy.
So we can model the wood structure as something that's a bit more of a simplified thing, just like this.
And we say we've got cells that are roughly hexagonal, and then we've got some cells that are more or less rectangular that are the ray cells.
And if you look at lots of micrographs, you can get some idea what the dimensions of the cells are.
And these dimensions were measured for a particular density of balsa wood.
So if we look at the linear elastic moduli, we can start off with a tangential loading.
And if we have the tangential loading, we can model it as a honeycomb loaded in the plane, and we get cell wall bending.
And from the cell wall bending in the honeycomb model, you would get that the tangential modulus varies with the relative density cubed.
And the structure's not quite that simple.
There's ray cells.
There's end caps.
And they act to stiffen it a little bit.
And the data lie a little bit above this line.
Then if we look at the radial loading, the rays kind of line up with the radial direction, and the rays act as reinforcing plates.
And so you can just use kind of an upper-bound composites idea to get the modulus.
And the rays tend to be a bit denser than the fibers.
So if I say a Vr is the volume fraction of rays, and R is the ratio of the relative density of the rays compared to the fibers, so it's rho over rho S for the rays divided by rho over rho S for the fibers.
And that varies a little bit from what one species to another, one specimen to another.
But it's something a little over 1, something between 1 and 2.
Then I can say the modulus in the radial direction is the volume fraction of the rays times R cubed times the tangential modulus plus 1 minus the volume fraction of rays times the tangential modulus.
And that works out to be about 1.5 times the tangential modulus.
I wanted to work this out in terms of the tangential modululs, so I've put this in terms of the tangential modules in the first term there.
So we get that the radial modulus is slightly larger than the tangential, but also goes roughly as the cube of the density.
And then for the axial loading, we just have axial deformation in the cell wall.
And the Young's modulus just varies linearly with the density.
So these are kind of simple models, but they kind of explain to first order the density dependence of the wood moduli and the anisotropy.
So it's kind of nice because they're fairly simple models, and it gives you kind of a big picture.
So if you wanted to know the modulus of a particular piece of wood, this probably isn't the best way to figure it out.
But if you wanted to kind of compare how do woods behave in general and how does the density affect the properties and why are they anisotropic, this is a pretty good way to do it.
We could also look at the Poisson's ratios.
And just because I didn't want to write them down again, I've just left on what the data were down here.
But let me just write what the model would give us for nu RT and nu TR, the model would give us one if we had regular hexagonal cells.
And these are the values we get here.
This might be 0.6, 0.7 would be a typical value, somewhere around 0.4 in there, so they're not quite one, but they're close to it.
And I think the reason they're a little less is because the rays in the end caps provide some constraint.
If you have the honeycomb, if I just had these cells, and I squeeze it like this, these guys can move out.
If it's a regular hexagonal honeycomb, the strain that I'm applying here is equal to the strain going out that way.
But if I have rays this way that sort of constrain it or end caps, it means that the Poisson's ratio is going to be a little bit less.
So I'll just say constraining effect of the end caps and rays-- constraining.
Then for nu RA and nu TA, the model says the value we would get would be zero.
And these are pretty close to zero.
They're not quite zero, but pretty close.
And then the last pair nu AR and nu AT, the model says that we would get nu of the solid.
And the data's close to 0.4, which we might expect would be about the nu of the solid.
So, again, there's some variation in the Poisson's ratios.
They're not all just one number.
But you can see these ones here are about zero, and that's roughly what the model says.
These ones here are closer to 1.
And then these ones here are closer to what you might expect for a solid material.
So it gives you the kind of general idea.
Are we good?
We're good, yeah?
So we can do a similar thing for the compressive strength.
So for tangential loading, we get plastic hinges forming and the bent cell walls, just like in an aluminum honeycomb.
Then we get that the strength over the cell wall strength goes as the relative density squared, so just like the honeycomb.
the radial loading, we can do the composites thing again.
So we can say the strengths in the radial direction is about equal to the volume fraction of rays times R squared times the tangential strength plus 1 minus the volume fraction of rays times the tangential strength.
And for balsa, I have some values here.
VR is about equal to 0.14.
R is about equal to 2.
And so the radial strength is about equal to 1.4 times the tangential.
And in higher density woods, the value of R is a little bit smaller, and in general, the radial strength is a bit larger than the tangential, and both depend on the density squared And then for axial loading, if the failure's initiated by yielding in the cell walls, then the axial strength's just going to depend linearly on the density.
So the idea with these models isn't that they kind of describe a particular piece of wood exactly.
It's more that it gives you a general picture of how the cells are deforming and failing, and how the properties scale with density and why the wood's anisotropic.
Are we good?
Yeah?
Caught up.
So there are a couple more sort of interesting things we can do with looking at the wood properties.
So we've been talking about how to model the cellular structure.
But people have also looked at how to model the cell wall as a fiber composite.
And this plot and the next one kind of show you how you can combine all of that together.
So remember, I said the modulus of the cellulose was around 140 gigapascals.
So here's the modulus of the cellulose, at least the crystalline part of the cellulose plotted in that little envelope there.
The lignin and the hemicellulose have a modulus around 2 or 3 gigapascals, so it's down there.
And if you made composites with cellulose fibers in lignin and hemicellulose matrix, those composites would have a modulus that fell in this envelope here.
They've got to be in between those two limits, right?
The modulus have to be between those two limits.
The density have to be within the densities of the constituents.
And if you look at the modulus of the wood cell wall, it lies in this envelope here.
Along the grain it'd be here, and then across the grain is further down here.
So the cell wall modulus is in here.
And then if you take that cell wall and you make it into the honeycomb-type material that wood is, if you load it along the grain, you're going to get this linear dependence of modulus on density.
And if you load it across the grain in the radial or the tangential direction, you're going to get this cubed dependence here.
So here's a set of data for different woods of different densities.
And that envelope kind of encompasses all of them.
But if you look at the slope of that data, it's roughly equal to a slope of 1.
And so it corresponds to that equation there.
And similarly, here's a set of data for different species of woods of different densities loaded perpendicular to the grain.
And they lie on a line that has more or less a slope of 3.
And this set of data here along the grain intersects the wood cell wall towards the top of that envelope, and this set of data here intersects closer to the bottom of that envelope for the cell wall material.
So this gives you a way of sort of putting everything together on one plot-- the cell wall as well as the cellular structure.
So that plot does it for the modulus.
And you can do the same kind of thing for the strength.
Here's the cellulose up here.
Here's the lignin down there.
Here's the wood cell wall, the composite made from those two.
And then here's data for different kinds of woods loaded along the grain and for load across the grain.
So it gives you a way of putting all this modeling into one set of plots.
So let me just write a couple of little things about that.
So we could say you could model the cell wall as a fiber composite.
And you can use the composition upper and lower bounds to give an envelope.
And then you can also show the cellular solids models on the same plot.
So overall, it shows you how the hierarchical structure fits together and can be modeled.
Now there's some more cute things we can see.
So another thing I want to talk about is material selection, because it turns out wood is very good compared to other materials in certain applications.
So we're going to look at, say, having a beam of a given stiffness at a given span, and say it's just a square cross-section beam of edge length T. And the question is, what material would minimize the mass of the beam?
So say we have some span we have to have.
It's got to have some rectangular cross-section, some given stiffness.
And the question is, what's the material that minimizes the mass?
So there's a little short calculation we can do to figure that out.
And then I've got another plot, and you can compare different materials on this other plot.
Then you'll see how good wood is compared to other materials.
So from beam of a given stiffness and given span, and say it's a square cross-section, then the question is, what material minimizes the mass of the beam?
So the mass is just going to be the density times t squared times l And if it's a beam, say it's got some central load on it, a concentrated load, the deflection's going to go as pl cubed divided by some constant and divided by the Young's modulus and the moment of inertia I.
So the stiffness, if I just rearrange this, the stiffness, p over delta, that's going to go as p over delta CEI, and I's going to go as t to the fourth over l cubed.
And then I can solve that for t squared.
And I want t squared because I'm going to plug it back into the equation for the mass.
So if I solve this for t squared, I've got my stiffness p over delta.
I've got l cubed divided by CE.
And then I take that whole thing to the 1/2 power like that.
And then I plug the t squared back into the little equation for the mass.
So I've got density minus p over delta times l cubed over CE.
And we'll take that whole thing to the 1/2 power-- [INAUDIBLE] another l. And so to minimize the mass, you want to look at the material properties.
And here, the material properties are the density and the Young's modulus.
And to minimize the mass, you want to minimize rho over E the 1/2 power.
Or conversely, you want to maximize E to the 1/2 over rho.
So if you just had a bar that you were just pulling on, you would just want to maximize E over rho.
But if it's a beam and bending, it works out that you want to maximize E to the 1/2 over rho.
And if we look at the next slide, this next slide then plots on a log-log scale, it plots the modulus on this axis and the density on that axis.
And here this plotted data for lots of different materials.
So there's engineering alloys.
Metals are up here.
Engineering ceramics are here.
Composites are here.
Polymers are down here.
Elastomer is way down here.
Foamy things, down here.
And this envelope here is woods.
And notice log scale here.
The lowest stiffness polymer foams here are 0.1 gigapascal, and diamond is up here at 1,000 gigapascal.
So there's like five orders of magnitude difference in the modulus here.
So then, if you look at the bottom right corner here, there's a bunch of old dashed lines.
And this red one here is E to the 1/2 over rho.
So if it's log-log plot, E to the 1/2 over rho's going to show up as a straight line.
And every point on that line has the same value of E to the 1/2 over rho.
And the material that would be the best for a beam of a given stiffness would be the one that has the biggest value of E to the 1/2 over rho.
And if I move the line up to the top left here, I'm increasing E. I'm decreasing rho.
It's got the biggest value of E to the 1/2 over rho.
So the materials that are on this line here, they all have the same value of E to the 1/2 over rho.
And they've got the biggest value-- well, virtually the biggest value.
So let's look at what those materials are.
There's things like engineering ceramics, like diamond that maybe are not the most convenient thing to make our beam out of, and tend to be brittle and might break.
So we have some issues.
There's engineering composite, so things like carbon fiber reinforced plastics.
And at this sort of tip of the composites, there'd be things like unidirectional fiber composites.
And then here's other woods down here.
So the woods have the same performance index, this is called, the same value of E to the 1/2 over rho as the best engineering composites.
And so they have very good properties for their weight.
And one of the interesting things is if you look at this performance index of E to the 1/2 over rho, this is the performance index for the wood.
This is for the solid cell wall material that the wood's made from, so E to the 1/2 of the solid over rho for the solid.
And from the modeling of the wood, just looking at the axial modulus, this thing here is equal to that times rho S over rho.
So if you look at this, this is the performance index for the wood.
This is the solid it's made from.
This number here is bigger than 1, right?
Because the density of the solid is bigger than the density of the wood, and so this is saying the wood is more efficient than the thing that it's made from, than the solid that it's made from.
And so that's the sort of plot for the stiffness.
And there's a similar plot for the strength.
That if you do the same little kind of calculation, you find that the performance index for the strength is some failure strength raised to the 2/3 power over a rho.
And again, here we're plotting strength versus density on a log-log plot.
And this red line here is the strength of the 2/3 over rho.
And again, if we scoot over here so we have a parallel line, every point on that line has the same value of the strength to the 2/3 power over the density.
And these are the materials that have the highest values.
And again, here's engineering composites.
These are ceramics.
But the ceramics, they have a high compressive strength, but they tend to be brittle.
So it's not really a practical strength.
These are metals in here.
And here's the woods down here.
So it's kind interesting just to see that the wood has such a good property.
Yes?
So I realize why this is valuable setting up the problem this way.
But if you're actually trying to design something, why would you want to fix your cross-section?
You could change your material and change your cross-section
So this is the starter version of this problem.
And there's another part two of the problem is to change the shape.
And you could look at what shape's efficient.
There's something called a shape factor that gives you the efficient shapes.
So you could take the material and turn it into a different shape and have a more efficient thing because it was a different shape.
So you can account for that
So then if you varied, like let's say you made your cross-section smaller, like even if it was still square, you could just still make it smaller
Yeah.
I'm saying we've got a given stiffness.
So if we're given a certain stiffness and a certain span, we would need a certain cross-section to get to that stiffness.
Are we happy?
OK.
So that's one thing.
Let's see here.
So let me just write a few more notes about the material selection, and then there's one more thing I wanted to show you about the woods.
Hmm?
C is just a constant.
So it's just a number.
So if you had a beam in three-point bending, then C would be three.
If you had a beam that was simply supported with a central load, C would be 48.
C is just a number
One more question.
So follow your line there, and the choice is really just about cost
No, it's not about cost.
There's nothing on cost here.
It's all really about the properties.
What's the best combination of properties to minimize the mass, and then which material has that combination of properties.
You can do charts like this that include cost.
You can make these charts with whatever property
[INAUDIBLE].
I guess maybe there's a difference off two or so of strength
Between pine and balsa?
Yeah, maybe more than that.
I think-- I can't quite see where-- pine's close to 100, and balsa's, I don't know, 20 or something
[INAUDIBLE]
Yeah, and it's not-- the point of this isn't so much looking at the absolute value of a strength.
It's looking at the value of this performance index.
And what you want to do is maximize that index to get the material that's going to minimize the mass.
So let me just read a couple notes about this.
So we have these-- these are called material selection charts.
So you plot the log of one property versus the log of some other property.
And then we have a line of constant E to the 1/2 over rho.
I'll just say it's shown in red because you're going have the same plots.
And the materials with the largest values are in the upper left.
So the woods have similar values to engineering composites.
And you can do a similar thing for strength.
So I have a few more minutes.
So I have a few minutes, and I want to talk about a couple of uses of woods.
So one is in old ships.
So I don't know if you know Professor Lechtman has this course Materials and the Human Experience, and they talk about sort of ancient uses of materials.
And I did a section, a module, on woods and the use of woods in old colonial ships, like The Constitution that's in Boston Harbor.
So this is kind of a schematic of an old ship.
And the thing that was interesting and the thing I talked about in this module was that people chose particular species for particular parts of the boat.
And they would choose a particular species depending on its properties.
And a lot of the hull was made of oak.
So oak's a very dense wood.
But they would get something they called straight oak, and they would get something they called compass oak.
And you can see this little thing down here, this little kind of schematic here, this little sketch.
This is straight oak, just a straight trunk.
And this thing here would be the compass oak.
And what they would do is they would use the straight oak for straight parts of the boat, so something like this, these pieces here.
And then they would actually look for trees that had the curve of the branches to match some part of the boat that they were looking for.
So, for instance, if you have the hull out here and the deck here and they had their cannons here, there's something called a knee, which is sort of a bracing piece that goes between the deck and the hull.
And that bracing piece is curved.
And they would actually look for trees in which the branches curved at the same kind of curvature as they were looking for in that piece.
And then they would use it for that piece.
And the advantage of this is they basically had the grain running along the curve, and so they got the best properties out of the wood by doing that.
So they had this straight oak and compass oak, and that was one cute thing.
And often they used white oaks.
And I brought a piece of white oak in.
You can see how dense it is.
And the US Navy often used something called live oak.
Live Oak grows in the South.
Anybody from the South?
You see these big trees with huge sort of spreading branches.
Those are the live oaks.
And apparently, the US Navy, I read somewhere, still has a forest somewhere with live oak for doing things like repairing The Constitution.
So let me just pass those guys around.
So those are a couple of oaks they would use for the hull.
Then they would use white pine for the masts.
And the reason they used white pine for the mast is because the white pine grows very, very tall and very straight.
And white pine was actually like a strategic resource in the 1600s, the 1700s.
And it turns out that when the British Royal Navy was doing all that colonial stuff in the 1600 and 1700s, Britain actually ran out of trees for masts for boats, and they would actually import masts from New England.
And there were these people called surveyors who would go around and they would mark certain trees that were supposed to be saved for these masts for the British Royal Navy.
And the thing was that the size of the boat and how many cannons you could put on the boat depended on how big the mast was.
So the size of the boat depended on the mast, because the mast height controlled how much sail area you could get.
So the taller the mast, the more sail/ the more sail, the bigger the ship.
The bigger the ship, the more cannons.
And so having these tall Eastern white pines was a sort of a strategic resource.
And I have a piece of white pine.
Unfortunately, my dog got to this one.
And be careful.
It's a bit splintery.
But you can see it's a lighter kind of wood.
And if you go around New England, if you go to the arboretum, you can see white oak.
You can see Eastern white pine.
The other wood they used is lignum vitae, that first dense one that I passed around.
And if you notice that lignum vitae has kind of a waxy feel to it.
And they used that in the block and tackle, so like pulleys and stuff like that.
And it was thought to be self-lubricating because of that kind of waxy layer on it.
And because it's very dense, if you think of like a block and tackle and you've got like a rope going over a pulley, you've got a pressure from everything sort of fitting together and the bits bearing against each other.
And the fact that was very dense made it very good for the block and tackle.
And so they used the lignum vitae for that.
And there's one other cute story about lignum vitae.
I don't know if any of you've ever read Dava Sobel's Longitude.
Anybody read that?
I'm a sucker for those history of science books.
So her book Longitude is about the development of an instrument to measure longitude.
Originally, they could get the latitude from the stars, but they were really bad at getting the longitude.
And so boats would go off, and they wouldn't really be able to figure out where they were, until they had a method to measure longitude.
And there was some British board of something or another.
They put forward a prize for somebody who could produce a way of measuring longitude accurately.
And there was a guy called John Harrison, and he built a clock.
He built a very accurate chronometer.
And if you knew when sunrise was and sunset was, and you knew the time and where you left, you could figure out where you-- it's kind of like time zones.
You could figure out where you are today.
And he built a chronometer, and one version of his chronometer used a lignum vitae for the same reason, because it was very dense, and it was very stable.
And the clock that he eventually won the prize with was in the 1700s, 1759.
I think they went on some trip with it.
It was 81 days at sea, and it lost five seconds over 81 days.
So that's pretty impressive for 250 years ago.
So that was the lignum vitae in the clock.
I have one more picture, and then I can finish up the thing on wood.
And we'll start the cork next time.
So this is another example of using wood.
And this is sort of a more modern use.
So this bridge here is made with a glue-laminated wood.
So this big beam here, the big arch, is made up of sections of wood which are glued together.
And you can glue the sections in a curved shape if you want.
They sort of have molds to do that.
And when they make this glue-laminated wood, they cut the defects out.
So they cut knots out, and they control the pieces of each laminate that they use to get the best quality.
And the glue-laminated wood actually has better properties than just two-by-fours or whatever you would cut down, lumber that you would cut from a tree.
So glue-laminated wood is kind of a nice kind of wood structure that's used now.
And you see it all the time in things like ice rink arenas, like large spans.
It's kind of beautiful.
You can see the wood grain in the curve in the wood grain when they make these things.
So that's the wood lecture.
I'm going to stop there.
So next time I'll talk about cork.
I just have a little bit about cork.
And then we'll start talking about foams.
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I wanted to talk a little bit about cork partly because cork is kind of interesting.
Cork has a structure that's a little bit like one of those honeycomb things.
What I'm going to do is I'm just going to talk and go through the slides.
I'm not going to write the notes on the board.
There's only a few pages of notes, and it's in the Stellar site.
I'm not going to write notes for this because it's just really for fun.
This first slide starts off with historical uses of cork.
Cork was used by the Romans.
They used it for the soles of sandals, the same as we do.
And they used it for stopping bottles of wine, the same as we do.
But they didn't realize that you could just use the cork.
They would take the cork, put it in the wine bottle, and then they would use pitch which is a tarry stuff from a tree.
They would use the pitch and seal the bottle with the pitch.
In the 1600s, there were some Benedictine monks that realized that you could just use the cork and not use the pitch.
They were the ones who really perfected the use of corks in wine bottles for sealing it.
So what is cork?
Cork is the bark of a tree called Quercus suber, the cork oak.
Here's a piece of the cork bark.
All trees have a layer of cork in them.
But the thing that's different about Quercus suber is that it's very thick and the corks are obtained from this thick layer.
Quercus suber is a Mediterranean type of tree.
It grows, Portugal is the main place that exports cork, but also places like Algeria, Spain.
You can grow it in California.
The cork is kind of unusual because-- let me scoot onto the next slide-- it's kind of unusual.
So here's a little picture.
I went to Portugal when I was a graduate student doing this project.
Here's the cork tree here.
Here's the little mini we rented.
Here's the cork being harvested here.
Cork's unusual because you can remove the bark from a cork oak tree and it regrows.
So most trees if you did this, you would kill the tree.
But cork doesn't get killed by doing this.
What happens is they plant trees.
You have to wait something like 10 or 15 years for the tree to get big enough to harvest the cork.
And then the first harvest is poor quality and they don't use that.
And then you have to wait another 10 years or so before you can actually harvest the cork.
So you can imagine if a cork orchard or forest gets chopped down to build a skyscraper, or apartment buildings, or something, it's not an economically feasible thing to plant cork trees these days.
So when the trees are cut down, they don't tend to get replanted at this point.
And there's a number of artificial substitutes for corks.
You've probably seen wine bundles with foam plastic corks in them too.
OK.
The reason it's called Quercus suber is that the cell walls in this particular type of cork oak are covered with a waxy substance called "suberine."
That's where the Quercus suber-- "Quercus" is "oak."
Every oak is Quercus something.
"Quercus alba" is white oak.
Quercus just means oak.
If we look at the structure of the cork, we can see that it's got these different views of the cork that are seen here.
This is a drawing by Robert Hooke in the 1600s from his book Micrographia.
He was the first person to really draw cork like this.
You can see he drew sort of boxy cells on this side.
And then, the other perpendicular plane, the cells have this structure here, sort of more rounded.
And here's a little sketch he's got of the cork tree.
Over here are SEM pictures.
These two planes here correspond to this plane in Hooke's drawing.
This plane here corresponds to this plane here.
This over here is Hooke's actual microscope.
I think the Royal Society still has that microscope that Hooke used in the 1600s.
Some of you know that Hooke wrote this book called Micrographia.
He got one of the first microscopes.
He looked at a lot of different materials, and he drew these beautiful drawings.
He wrote a page or two about each of the drawings.
Harvard has a first edition of Micrographia, and I made a little video on it.
So I thought I'd show you the video because the drawings are beautiful.
But the url's on the slides, and so you can watch it yourself.
There's more on this and on how he came to be so good at making scientific apparatus, how he came to do the Micrographia book.
At the end of it, there's a comparison of a number of his drawings with modern SEM images of the same thing.
He had this very famous picture where he draws a flea.
Don Galler, the person who runs the SEM for me, I had him put a flea in.
And he has essentially the same kind of image.
The thing that's spectacular is you can see how much of the detail that Hooke was able to capture in his drawings.
There's some really beautiful drawings.
OK.
So let's go back to cork.
Hooke was the first person to use the word "cell" to describe biological cells, and he described the cell in cork.
That's the structure looking at the SEM micrographs and his optical micrographs.
These are just some more higher resolution, higher magnification, images.
One of the things that you can see is that the cork has these little corrugations on the cell walls.
See those little wrinkles?
All the cell walls have those little wrinkles.
This plane here is the perpendicular plane.
If you look down into the cells, you can see those blurry things.
Those are the corrugations in the cell walls.
That's the structure.
Here's a schematic.
Here's the cork tree.
The cork is the layer just beneath the bark.
This is a picture of how the cells are oriented relative to their radial, and tangential, and axial directions.
You can think of them as roughly hexagonal.
They've got these little corrugations.
This is a schematic of an individual cell.
We measured the dimensions of the cells, and these are some average dimensions here.
Typically, the cells are tens of microns long, and the cell wall is about a micron thick.
Something like that.
OK?
One of the things we're going to see is that corrugated structure gives rise to some of the interesting properties of cork.
If we load cork and just do mechanical tests on it-- this is just a uniaxial stress-strain curve.
You can see the stress-strain curve looks like all these other curves we've seen for honeycombs.
There's a linear elastic part here.
There's a stress plateau here.
And then there's a densification part here.
Typically, the relative density of cork is around 0.15.
Something like that.
It densifies at a strain of something less than 0.85.
It's a stress-strain curve.
And when we did our little project on cork, we measured the properties in the three directions because in one direction, it's roughly hexagonal cells.
That plane is isotropic.
The e1,e2 plane is isotropic.
This compares the measured values of the properties versus what we calculated from the honeycomb model.
And really, we just used the sorts of equations that we talked about in class over the last couple of lectures, apart from loading in the x3 direction because in the x3 direction, you've got those corrugated walls.
And you have to take those corrugations into account.
So there's another complication that I'm not going to go into.
But there's a sort of modification you can do to account for that.
But you can see there's actually pretty good agreement here between the elastic moduli that we measured and what we calculated.
The compressive strengths down here are not quite so good.
They're off by a factor of two, more or less.
But they're in the right ballpark for the cork.
So those are some of the structures, some of the properties.
One of the interesting things about cork is what it's used for.
The uses of cork really exploit the mechanical properties.
Obviously, it's used for stoppers for bottles.
I brought a champagne cork along with me, and I brought a couple of other little pieces of cork here.
I'll pass those around in a minute.
One of the things to look at is just the still wine cork, which is the one on the right, and the champagne cork, which is the one on the left.
If you notice the still wine cork is just made of one piece of cork that's cored out from the bark.
And if you notice these little channels.
These little channels here are called "lenticels."
On the still wine cork, they go this way.
And on the champagne cork, they go that way.
They're oriented perpendicular.
It turns out that they are normal to the isotropic plane in the cork.
If you look at the champagne cork, this plane here is the isotropic plane.
If you think of that being put into a bottle, I think part of the reason they orient it this way is because it gives you a uniform compression against the neck of the bottle and gives a nice seal.
So that's one of the things about corks.
Another thing that's interesting is that cork has a Poisson's ratio equal to zero if you load it in that direction.
Let's see.
Did I not bring that picture?
Maybe I didn't bring that picture.
Hang on a sec.
Nope.
I guess I didn't.
Sorry.
I thought I brought a picture where I had the deformation of the cells when you load it in that direction.
When you load it-- so say the cells are corrugated this way on.
When you load it that way on, it's like having a bellows and folding up a bellows, or unfolding a bellows.
So when you load it that way on, there's no expansion or contraction this way on.
And so you get zero Poisson's ratio.
And if you think of trying to get the cork into the bottle, it's rather convenient to have zero Poisson's ratio because you don't get as much expansion.
As you're pushing it into the bottle, you don't get as much expansion that way out, pressing against it.
In fact, if you compare wine corks with rubber stoppers, this is a kind of typical rubber stopper.
Wine corks are always just cylinders.
In fact, even the champagne corks are cylinders when they start off.
When they put it into the bottle, it's just a straight cylinder.
It gets deformed like that from being in the bottle for some time.
They have straight sides and you can just squeeze them.
There's these funnel things that wine makers have for putting the cork in the bottle.
You can just squeeze them into the bottle top.
You can do that because the Poisson's ratio is zero and because the Young's modulus and the bulk modulus are both small.
But if you look at a rubber stopper, rubber stoppers always have these tapered sides to them.
And that's because the Poisson's ratio of the rubber is 0.5.
As you squeeze it in, it's trying to move out this way.
You couldn't get the stopper in unless it had those tapered sides.
So that's sort of an interesting thing about cork.
Let's see.
Another application of cork is it's used for gaskets for the same sorts of reasons.
It's relatively compliant.
It takes up any slack between two pieces that you want to press together.
It's often used for musical instruments that come in pieces.
Things like clarinets, there'll be a piece of cork-- you a clarinet player?
Yeah?
Yeah.
One of the interesting things about the clarinet is that, if you can see here at the ends, there's a piece of cork there.
And I think there's a piece of cork down there.
And the other pieces mate up with that.
The cork provides a seal.
And the way the cork is cut, it's cut in such a way that those lenticels go radially out like this, which means that the plane of isotropy and the direction that's got the zero Poisson's ratio is that radial direction.
When you're squeezing, say, the second part onto it, the cork does not expand circumferentially.
So as you're squeezing it this way, it doesn't expand that way.
It doesn't wrinkle or anything on your other part.
They use the cork in a particular orientation for that reason, I think.
So it's used for gaskets It's also used because it's got a good friction property.
It's got a property that is taken advantage of in things like flooring and shoes.
Cork has a high friction even if it gets wet.
Some sources of friction are from adhesion, from a surface effect.
Then if, say, the floor gets wet, then you break that, and it could be slippery.
But the source of friction in cork is from an energy loss and dissipation as you're deforming it.
Imagine you have a wheel here.
The wheel is rotating on this cork floor.
And here, a piece of cork is getting deformed as the wheel rolls over it.
As it gets deformed, there's some histeresis loop.
Cork has quite a lot of damping in it.
There's quite a lot of energy lost in that histeresis loop.
What that means is that's characteristic of the cork itself.
It's not a surface effect.
That means that if you use it for floors or for shoes, it doesn't lose that friction and damping when it gets wet.
Here's some measurements we did of the coefficient of friction for cork versus doing it dry and doing it with a liquid, soapy surface.
You can see the soap doesn't make any difference.
Cork is seen as a very attractive material for things like flooring.
It's actually not a cheap material to make your floors out of, but it's an attractive material for flooring.
Part of the reason it's used for floors and for the soles of shoes is because of these friction properties.
Another feature of cork is that it's got very small cells compared to a lot of engineering polymers.
The cells are on the order of tens of microns, whereas many polymer foams, the cells are hundreds of microns or millimeters.
We'll get into this later, but this plot here is really saying that the thermal conductivity of a cellular material depends, in part, on the cell size.
The cell size for foam plastics is in here.
And that for cork is down here.
Because it has a smaller cell size, it has a lower thermal conductivity.
Cork was at one point used to some extent for thermal insulation.
If you go to Portugal, where cork comes from, there's hermit caves.
There were these old hermit, religious people who had holed up in a cave.
And they would line the caves with cork to try to make it a little more insulated, a little more warm.
The other place you see this is if you look at cigarettes, you know cigarettes have that little brown tip on the part that touches your lips?
That's meant to look like cork.
And if you look, it has little dots on it.
The little dots are the little lenticels.
Apparently, they used cork originally in cigarettes as a sort of thermal insulation between the cigarette and your mouth.
So it was used for that too.
And then, one final thing.
Cork's also used, obviously, for bulletin boards.
If you push a pin into cork, then you get this local deformation here.
Here's our pin.
And here's cells locally deformed.
When you pull the pin out, you'll recover some of that deformation because the deformation is elastic.
So the hole will partly close.
So that's my little spiel on cork.
And that's just because it's interesting.
There's no test on cork or anything like that.
OK.
So are we good with cork?
Any questions?
We're OK?
OK. Let me scoot out of there.
Then the next part of the course, I wanted to talk about foams.
Let me just park the cork thing.
Let me pass these corks around.
So you can play with those too.
Oops.
There's little bits of cork.
There you go.
There's the champagne cork, the rubber cork, some little cork layers.
OK.
So the next part of the course, I wanted to talk about foams.
And I want to talk about how we model the mechanical behavior of foams.
If we look at the stress-strain curve for foams, these are some examples for foams made out of different materials with different characteristics.
The polyurethane and the polyethylene here are examples of elastomeric foams, really.
This one here is an open-celled foam.
This one's a closed-cell foam.
Polymethacrylamide is a polymer that has a yield point.
Mullite is a ceramic.
You can see the shapes of these curves resemble the shapes that we saw for the honeycombs.
Right?
They look exactly the same, in fact.
And the mechanisms of deformation in the foams are very similar to the honeycombs.
Even though the foams have a much more complicated geometry, we can use some of the ideas from the honeycombs to understand how the foams behave.
So that was part of the reason for doing the honeycomb analysis.
Let me back up.
These curves here were all in compression.
These curves here were all in tension.
So again, these ones in tension also look like the curves for the honeycombs.
Remember, in tension, we don't get any elastic buckling.
So if the foam was made of an elastomer, we don't see any stress plateau.
If the foam is made of material with a yield stress, then we get a very short yield plateau because of a slight geometrical difference between pulling and compressing the foam.
And if it's a brittle material, then we just get fracturing.
There's going to be some fracture toughness that's going to characterize it.
We can look at the deformation and the failure in these foams and look at the mechanisms.
And what we're going to do is model the mechanisms and not worry too much about the cell geometry.
So we're going to use dimensional arguments here.
Here's a foam in compression.
It was compressed from the top to the bottom.
And you can see this strut that's circled in red.
This is unloaded.
And then, this is after some load.
You can see this has bent somewhat.
And then, you can see this vertical strut here.
As the load gets larger, you can see that strut's buckled.
In an elastomeric foam, you get bending and buckling just the same as we did in the honeycomb.
Then here's a metal foam.
You form plastic hinges in the metal foam.
So here's a cell wall here.
And it's a little bent to start with at zero load.
But you can see it becomes more bent in this image over here.
And here's a cell wall that's more or less vertical.
And you can see that wall buckles.
It's a plastic buckling in this case.
There's a permanent deformation there.
Here's a brittle foam.
And you can see cell walls in this foam fracture.
So this region here is equivalent to this region here.
That little glitch there is the same as that little glitch there.
You can see there's a couple of cell walls here that are fractured.
So we get fracture.
The idea is is that the mechanisms of deformation in the foams parallel those in the honeycombs.
OK?
All right.
So let me write some of this stuff on the board.
We're going to start off by talking about open-celled foams, so foams where there's just solid in the edges, but not in the faces of the polyhedral cells.
Then we'll talk about close-cell foams where there's solid in the faces, as well.
But the open-celled ones are easier.
So we'll start with that.
In compression, we see the same three regimes as we did before for the honeycombs.
There's a linear elastic regime that corresponds to bending of the cell walls.
There's a stress plateau.
And for elastomeric foams, that corresponds to buckling.
For metal foams, that corresponds to the formation of plastic hinges.
And then, for ceramic or brittle foams, that corresponds to brittle crushing, so fracturing of the cell walls.
Then, if you load the foam up to higher strains and higher stresses, eventually you get to densification.
And in tension, just like the honeycombs, for the elastomeric materials there is no buckling.
We can get a stress plateau from plastic hinges if there's, say, a metal foam.
And for a brittle foam, we would get a fracture toughness and brittle fracture in tension.
So the idea is the mechanisms of deformation and failure just parallel what we've seen in the honeycombs.
So we'll start off with the linear elastic behavior.
And we'll start with open-cell foams.
The initial linear elasticity is due to bending of the cell walls.
And if the thickness of the cell edges relative to the length is small, the bending dominates the deformation.
But as the thickness to length ratio increases, then axial deformations can become important too.
What we're going to do is we're going to consider dimensional arguments.
We're going to set the dimensional arguments up so that we replicate the mechanisms of deformation and failure.
But we don't worry too much about the cell geometry.
What I'm going to start with is considering a cubic cell.
And I've arranged the cubic cell so that the cell edges are staggered.
That's going to give me the bending.
The edge length is going to be L. I'm going to say we have a square cross-section, t squared.
Here's our idealized model here with a cubic cell.
All the members have a length, l. All of them have a square cross-section, t squared.
That's an open-cell model.
We've got just solid on the edges and nothing on the faces.
The idea is that if I bend that, or if I load that in compression, so I apply, say, a stress out here that puts forces on those members there, because I've staggered these cell walls with these ones here, we're going to get bending in this cell edge here.
That bending is going to be what we model.
I'm going to set this up so that one thing is proportional to something else.
These relationships are going to be true regardless of the cell geometry.
So I could have picked a tetrakaidecahedra if I wanted to, and I would have had these same relationships.
I'm just picking a cubic thing because it's easier to think about.
So first of all, we look at the relative density.
Remember, the relative density is the volume fraction of solid.
So it's the volume of the solid over the total volume.
And that's going to go as t squared l over l cubed, or just t over l all squared.
You remember for the honeycomb, the relative density went linearly with t over l. For the open-celled foam, it goes as t over l squared.
The moment of inertia in this case is going to go as t to the fourth.
Remember, we have a square section, t squared.
So if it's bh cubed over 12, b is t, h is t. It's going as t to the fourth.
Then what I'm going to say is the stress is going to go as F over an area length squared.
OK?
So if I look at my little square thing here, I look at having my force here.
Here we have a force f. And it's acting over an area that's somehow related to l squared.
Right?
I don't know exactly what that constant is, and I'm going to not try to calculate that.
But it goes as F over l squared.
Similarly, I can write that the strain is going to go as delta over l. So the strain is going to go as this bending deflection here, that delta divided by the height of the cell.
And that's also l. Then I also know from structural mechanics that delta is going to go as Fl cubed over E of the solid and I.
Then I'm just going to put all these things together to get the modulus.
The modulus of the foam is going to go as the stress over the strain.
If I plug in what I have for the stress, it's F over l squared.
If I plug in what I have for the strain, it's delta over l. So this is F over l and delta.
I'm going to replace delta by Fl cubed over Es.
I'm going to use, instead of I, I'm going to use t to the fourth.
Then the F's are going to cancel out.
I've got that the modulus goes as the modulus of the solid times t over l to the fourth power.
Then I can put that in terms of the relative density.
It's going to go as the relative density squared.
So I can summarize all of this by saying that the Young's modulus of the foam is some constant C1, I'm going to call it, times the modulus of the solid times the relative density squared.
OK.
So this has the same kind of form as those equations we had for the honeycombs.
Right?
There's a solid-cell wall property.
The solid module's here.
For the honeycombs, I put it in terms of t over l. But the t over l was related to the relative density.
How much solid you've got is reflected in the relative density.
And then, this constant C1 wraps up all the geometrical constants that I've said, one thing's proportional to another, and something else is proportional to another.
C1 just wraps up all of those.
OK?
I'm just going to say here C1 includes all the geometrical constants.
We have to get C1 by looking at data.
If we look at data for the Young's modulus, we find that C1 is just about equal to one.
People have also done more sophisticated analyses than this.
There's a group of people who looked at doing a full-scale, structural analysis of an open-celled tetrakaidecahedral cell.
Remember, I said they pack to fill space.
So you can look at a unit cell.
They also made their cells such that the thickness along the length of the cell was not constant.
The thickness varied as something called a "plateau border."
If you have a foam that's made by surface tension, the edges will tend to have these plateau borders.
And the thickness will vary along the length of the edge.
So when they did all this whole, complicated thing, they could calculate a value for C1.
They calculated 0.98.
So it's very close to 1.
I'll say analysis of open-cell tetrakaidecahedron cells with these plateau borders give C1 equal to 0.98.
OK.
So that's the Young's modulus.
We can also look at the shear modulus.
The shear modulus is also going to be controlled by bending of the cell walls.
And so the shear modulus is just going to be some other constant times Es times the relative density squared, so a similar kind of relationship.
It's just a different constant.
And if the foam's isotropic, and if the Poisson's ratio is a third, then C2 is equal to 3/8.
Remember, if we have isotropy, then the shear modulus is equal to E over 2 1 plus nu.
And so you can get the C2 from that if you say nu is a third.
Then we can also get Poisson's ratio for the foam.
If the foam is isotropic, so we'll say for an isotropic foam here, nu is equal to E over 2G minus 1.
That's just rearranging this expression here.
And because E and G both depend on the relative density squared, they both depend on Es squared, that's all going to cancel out.
So this is going to be equal to C1 over 2 C2 minus 1.
So that's going to equal to a constant.
That constant's going to be independent of whatever material the foam is made from and the relative density.
The constant just depends on the cell geometry.
Remember, in honeycombs we found the same thing.
The Poisson's ratio for the honeycombs only depended on the cell geometry.
It didn't depend on the solid properties.
It didn't depend on the relative density.
So this is an exactly parallel thing here for the foams.
Yeah?
I have a silly kind of question.
What is the difference between foam and the honeycombs?
Oh.
The honeycombs have cells in a plane, and they're prismatic in the third direction.
And the foams have polyhedral cells.
You know what a tetrakaidecahedron, a 3D, polyhedral cell.
OK?
The honeycombs are prismatic.
And the foams have polyhedral cells.
OK?
Are we good?
OK.
So there's a couple more interesting things about Poisson's ratio.
The same way we can make honeycombs with negative Poisson's ratios, we can also make foams with negative Poisson's ratios.
They do it the same way as for the honeycombs, really.
The honeycombs had negative Poisson's ratios if the cell walls looked like this, this sort of arrangement.
So that sort of a thing.
We said that theta was negative for the honeycombs.
And if you invert the cell walls on a foam, you also get negative Poisson's ratios.
And the way they do that is they take a thermoplastic foam, and then, they load it hydrostatically.
So they compress it in all three directions.
And they smush the cells in on each other.
And then, they heat it up to a temperature above the glass transition temperature while it's still smushed.
And then, they cool it down.
So they end up with that structure frozen in.
And if they do that, they get a negative Poisson's ratio.
I'll just say they invert the cell angles analogous to the honeycomb.
They load the foam hydrostatically and heat to a temperature above Tg.
And then, they cool and release it.
I have a photograph here of a foam with a negative Poisson's ratio.
You can see how the cells have been smushed in.
It's the equivalent of the way it's done for the honeycomb.
OK?
Are we good?
OK. That's the linear elastic moduli for open-celled foams.
The next thing I wanted to do was closed-cell foams.
If we look at a closed-cell foam, we can idealize it in this kind of a way here.
I've set it up so that the edge thickness is different from the face thickness.
That's really representing the fact that in foams, many foams are made using a liquid, and the foaming is controlled by surface tension.
Often, the surface tension draws material into the edges and away from the faces.
So the faces tend to be thinner than the edges.
When we have deformation of the closed-cell foam, we've got bending of the edges the same as we did for the open-celled foam.
But the faces can stretch.
So they can have an axial stretching.
You can think of that as a cell membrane stretching here.
So imagine if I either pull on the foam or I compress it, there's going to be some axial load in the faces.
So when we analyze them, we have to account for both bending of the edges and axial deformation in the faces.
I'll just say we have edge bending as in open-cell foams.
And we also get a face stretching.
Another thing that can happen in the closed-cell foams is we can get compression of the gas.
In an open-cell foam, the gas can move from one cell to the next.
But in a closed-cell foam, the gas is trapped.
And as the volume of the cell changes, the gas gets compressed.
So we have another effect here.
So we'll say for polymer foams, surface tension tends to draw material to the edges during processing.
We define two thicknesses, one for the edge and one for the face.
And then we apply a force to this cubic structure.
And we can do an analysis a little bit like what we did for the open-cell foam.
Let me rub all this off.
OK.
I'm going to set this up a little bit differently than for the open-celled foam.
We're going to do a work argument.
We're going to look at the external work done by the force F going through a deformation, delta.
And that's going to have to equal the internal work done by the edges bending and by the faces stretching.
So let me set that up.
We're going to say the external work done, that's going to be proportional to F times delta.
So delta is how much the whole thing is going to deform.
Then, I've got internal work from bending of the edges.
That internal work is going to be proportional to F over delta times delta squared.
I'm going to end up with an expression where everything's in delta squared.
So I want to keep the delta squared there for now.
And F over delta is the stiffness.
That's going to go as E of the solid times I over l cubed.
I is going to go is Te to the fourth here because it's I of the edges.
I've also got internal work from stretching of the faces.
That internal work is going to be-- I'm going to run into my other equations here.
Let me put it down a little.
That's going to go as the stress on the face times the strain in the face times the volume of the face.
Or I could write that, instead of stress of the face, I can put it in terms of Hooke's law and say it's E of the solid times the strain in the face squared times the volume of the face.
And I can replace the strain in the face by delta over l. So it's delta over l squared.
And then, the volume of the face is going to be t of the face times l squared.
Then I want to balance the internal work and the external work.
I can say F times delta is going to equal some constant I'm going to call "alpha" times E of the solid times t of the edge to the fourth, that's this guy up here, over l cubed times delta squared plus some other constant I'm going to call "beta" times E of the solid times delta over l squared tf l squared.
So far, I've got this in terms of the force I'm applying.
But I want to get a modulus of the foam out of this.
I want to relate this force here to the modulus of the foam.
I can say the modulus of the foam is going to be related to F over l squared, that's the stress, divided by the strain, delta over l. I can write the force is proportional to the modulus times the deflection, delta, and times the member length, l. And then, I'm going to plug that guy into this expression up here.
And I'm going to get a delta squared on the left.
And I'm going to get delta squareds in each of the right-hand terms.
And so I'm going to cancel out the deltas.
If I put all that together, I have the modulus in the foam times delta squared times l. There's a delta here, and there's a delta there.
That gives me delta squared.
And that's going to equal alpha Es Te to the fourth over l cubed times delta squared plus beta times Es times delta squared tf, if I cancel out one of those l squareds.
So here I can get rid of the delta squareds in all these terms.
And if I just divide by l, I'm going to have the modulus of the foam.
I get a term here that it goes as alpha E of the solid times t of the edge over l to the fourth power plus beta times E of the solid times tf over l. Are we good?
OK. And what I'd like to do is instead of putting it in terms of te and tf, I'd like to put it in terms of the relative density.
I'm going to look at two limits.
I'm going to say if we just had open cells, if there were no faces on the membranes, if we just had open cells, and we had a uniform t, then the relative density would go as t over l squared.
If I just had closed cells, and I had a uniform t, then the relative density just goes linearly with t over l. The relative density is the volume fraction of solids.
It's the volume of the solid over the total volume.
For a closed-cell foam with a uniform t, the volume of the solid is going to be t times l squared.
And then, the volume total is l cubed.
So it's t over l. Now, I'm going to define one more thing.
If I say that phi is equal to the volume fraction of the solid in the edges, then I can say te over l is some constant times phi to the 1/2 times the relative density to the 1/2.
And I can say tf over l is equal to some other constant, c prime, I'll call it, times 1 minus phi, that's how much is in the faces, times the relative density.
Then I could combine all of this.
Can I put it in here?
Maybe I'll just stick it down here.
Hang on.
OK.
Put it up here.
This is my final expression here.
And this term here arises from the edge bending.
This term here arises from the face stretching.
So I think I should wait a bit for you to catch up.
The idea is the edges are bending.
The faces are stretching.
We're looking at the work done by deforming the edges and the faces and equating that to the work done by the externally applied load, f. OK?
That gives us two terms, one that accounts for the edge bending, and one that accounts for the face stretching.
To be comprehensive, we want to take into account the compression of the gas, as well.
So there's one more term I'm going to add on to that.
Typically the gas effect is only significant for elastomeric foams.
If you had a metal foam or a ceramic foam that was closed-cell, it wouldn't really contribute much.
But just to be complete, we'll go through this.
The idea here is we say we've got a cubic element of the foam.
Initially, it has a volume, v0.
If we apply a stress, a uniaxial stress, there's some change in the volume of the foam.
So there's some volume, v, after we compress it by some strain, epsilon.
If we can figure out what the volume is relative to the initial volume, we can figure out how the change of the amount of gas goes from the initial state to the compressed state.
Then we can use Boyle's law.
The idea is P1V1 equals to P2V2.
Then we can figure out, using Boyle's law, what the pressure must be.
And then, that pressure is what's contributing to the modulus.
When I do this, I'm going to write down some equations that just have the main results.
There's a whole bunch of algebra just to get from one to the other.
And I'm not going to write them all down.
When I write the equations, don't panic if it's not obvious how you get from one to the other.
In the notes that I'm going to put online, there's all the details of how you get from one to the other.
The idea is that we start with a cubic element of foam.
Initially, before it's loaded, it has a volume V0.
Then we apply a uniaxial stress.
And we say the axial strain in the direction of the stress, I'm going to call epsilon.
Just from the geometry, you can calculate the deformed volume.
So after you load it, the volume is V. And that volume on loading, V, divided by the initial volume, V0, you can show.
It's fairly straightforward.
It's 1 minus epsilon times 1 minus 2 times the Poisson's ratio of the foam.
Here, I'm taking compressive strain as positive.
And if you do this whole volume thing, you'll get terms in epsilon squared and epsilon cubed.
But because if it's linear elastic, epsilon is going to be small.
I'm going to ignore the epsilon squared and the epsilon cubed terms.
That's the total volume of the foam after and before the compression.
And then, what I want is the volume of the gas.
And the volume of the gas is just going to be the volume minus the volume of the solid.
And I can get that by just subtracting off the relative density.
So Vg over Vg0.
Again, Vg0 is the volume of the gas initially.
Vg is the volume of the gas when I'm compressing it.
Remember, the relative density is the volume fraction of solids.
So I'm essentially subtracting out the amount of solid to get the amount of gas.
Then we can use Boyle's law.
Here, p is going to be the pressure after the strain and p0 is going to be the initial pressure.
The building seems rather making unhappy noises for some reason.
I'm not sure what that's all about.
There'll be some initial pressure in the gas even before the strain, or before the stress is applied.
That's p0.
So the pressure we need to overcome is p minus p0.
Again, I'm missing out a bunch of steps.
But using that expression and this expression and this expression, you could find that p prime is equal to p0 times epsilon times 1 minus 2 nu divided by 1 minus epsilon 1 minus 2 nu minus the relative density.
Then the contribution of the gas to the modulus you can get by just taking the derivative of that pressure with respect to the strain.
So remember, modulus is stress over strain.
It's the same kind of idea.
So I'm going to call it E star g for the contribution from the gas.
So that's going to dp prime d epsilon.
And that's going to equal p0 times 1 minus 2 nu over 1 minus the relative density.
OK?
As I said, I'll scan the pages that have the details and put it online.
The final expression we get combines all of these things.
The modulus of the foam relative to the solid is phi squared rho over rho s squared plus 1 minus phi, that's the amount of material in the faces, times the relative density and then plus this gas compression term.
Like I said, for most foams, the gas compression is negligible.
So if p0 is equal to the atmospheric pressure, so about 0.1 megapascal, then the gas term's negligible, except for elastomeric foams.
So that's the Young's modulus.
Then we can do a similar thing for the shear modulus.
And the thing to notice in shear is that when you shear something, there's no volume change.
So if you shear it and there's no volume change, there's no gas compression.
There's no pressure built up because there's no change in the volume.
So for the shear modulus, you just have the first two terms.
And if the foam is isotropic and you use a third, then this constant here is going to be 3/8.
And then, Poisson's ratio is just going to be a function of the cell geometry, again.
And roughly, we could say it's going to be around a third.
OK. Are we good?
Let me wait a little bit.
So the idea is we're looking at the deformation of the bending in the cell edges, and the stretching in the cell faces, and the gas compression.
And we're accounting for those different terms.
The next thing is to compare these model equations with some data.
And here's data for Young's modulus.
Here, we're plotting the relative Young's modulus, so the modulus of the foam divided by the solid against the relative density.
Again, these are log-log scales.
On this plot, everything with open symbols is an open-cell foam, and everything with filled symbols is a closed-cell foam.
Here's our equation for the open-celled foams, the simplest thing that goes with density squared.
And that's that thick line there.
And you can see these are all open-cell foams.
There's some more up here.
So that gives a reasonable description of that data.
And then, these are two lines for closed-cell foams for different values of phi, so different values of the amount of solid in the edges.
And you can see all the filled symbols are the closed-cell foams.
And they're in between this line here and this line here, basically.
So that gives a fairly good description for the modulus of the foams considering how crude this model is.
We're not trying to account for the cell geometry.
We're just modeling the mechanisms of deformation and failure.
And then, here's a similar plot for the shear modulus.
Here's for open-cells, 3/8 times the relative density squared.
These are open-celled foams down here.
These are closed-cell foams up here.
If the amount of material in the faces is small, then you would just get the shear modulus varying with the relative density squared.
Then here's data for Poisson's ratio.
We don't really know what the constant is because we don't know what all these geometrical parameters are.
But here's a value of a third.
And you can see there's a lot of scatter here.
This is like less than 0.2.
This is more than 0.5.
And the scatter really represents differences in the cell geometry.
If the foams were, say, anisotropic, and the cells were stretched in one direction, then you would get different values of the Poisson's ratios.
So that's the Poisson's ratios there.
I'm thinking I'm going to stop there because that seems like enough equations for today.
And then, next time we'll start doing the stress plateau and we'll figure out the elastic collapse stress from buckling and a plastic collapse stress from yielding, and so on, and so on.
We'll finish doing the modeling next time.
And we'll probably start doing a little bit of other stuff on foams next time.
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All right.
And I really wanted to show you my little hook video and I downloaded it so I thought we'd start just by watching that and then I'll pick up about modeling phones.
So this takes like nine or 10 minutes, but I just thought it was cute.
And I made it and I want you to see it.
So let's do that to start.
[VIDEO PLAYBACK] We're here at the Harvard University Botany Library, looking at a first edition of Robert Hooke's Micrographia, published-- How do I get rid of the bar, Greg?
Oh, there it is.
show the microscopic structure of materials.
And it has a number of remarkable drawings in it.
Here we see drawings of silk.
These are two different silks.
On the top here, we have a fine-waled silk.
And in this more details drawing down here, you can see the patterned weaving of the silk.
The bottom image here is a drawing of watered silk.
And over here, there's another higher magnification image.
And you can see the pattern here is more sharply angled.
And it appears that this sharper angle here gives the different texture to the surface finish of the silk.
So here we see a drawing of charred wood.
And one of the things I find interesting about this drawing is how similar it is to modern electron micrographs which we've seen before.
And in this drawing, we can see two of the main features.
We see these small cells, which are fibers that provide structural support to the tree.
And we see these larger cells here, which are vessels which allow fluids to go up and down the tree.
And here we see a drawing of the surface of a rosemary leaf, with the unexpected, tiny, little bars.
And this is something that you can only see with the microscope.
You wouldn't expect to see those when you just feel the surface of the rosemary leaf.
So it's kind of interesting that with the microscope, you can see these features that are invisible to the naked eye.
One of the main themes of material science is that the property of materials are related to their structure.
And so being able to see the structure at a microscopic scale is very helpful.
And today, we can even see the structure at the atomic scale.
Robert Hooke understood this idea.
And in the description of the cork, Hooke states, "I no sooner discerned these-- which were the first microscopical pores I ever saw-- but methought that I had with the discovery of them, perfectly hinted to me the true and intelligible reason for all of the phenomena of cork."
So what he's saying here is that by looking at the structure and looking at the cells here in the drawing, he thinks he can understand the properties of cork or the phenomena of cork.
What was it about Robert Hooke that allowed him to make this book?
Why was it him and not somebody else?
Well, Robert Hooke had kind of an interesting history.
He grew up on the Isle of Wight.
And as a boy, he loved making drawings.
And he got quite skilled at making drawings.
The other thing was, he loved making models of things.
He made models of ships.
He made a wooden clock that was a working clock when he was a kid.
And as a teenager, he moved to London.
And he became an apprentice to Sir Peter Lilley, who was a famous painter of the time.
So his drawing was good enough that he would be working with a very well-known painter.
After he did that, he went to the Westminister School.
And he studied classics.
He studied mathematics.
But he also learned to use a lathe.
And this was also very helpful in him making various sorts of apparatus.
And as a student at Oxford, he worked in the lab of Robert Boyle.
And his job in that lab was to develop scientific apparatus.
And he did things like he built pumps that allowed Robert Boyle to do the experiments that led to Boyle's Law.
When he returned to London after Oxford, he became the Curator of Experiments at the Royal Society.
And one of the things he did was he got a microscope.
He improved that microscope, increasing their magnification, which was what allowed him to make the beautiful drawings that we see today.
And here in the preface of the book, we see that he even made a drawing of his microscope.
So this thing down here-- this is Robert Hooke's microscope.
The development of new microscopes with higher and higher magnifications continues to this day.
Scanning electron microscopes were invented in the 1960s.
And today, we have transmission electron microscopes and atomic force microscopes with even higher magnifications.
At these higher magnifications, we can see details that Hooke was unable to see because of the limitations of the microscope that he had-- the optical microscope.
But it's interesting to see today the images we see in a scanning electron microscope at a similar magnification to those that Hooke saw in his optical microscope.
And it's remarkable to see how many of the features that we see in these much more fancy microscopes that he was able to capture in his drawings with his simple optical microscope.
So here we have a picture of cork.
We have Hooke's drawings showing two perpendicular planes.
We also have this nice, little drawing of a cork branch here.
Cork is the bark from the cork oak tree.
And in Hooke's drawings of the microstructure, we can see these cells here are roughly box-like.
They're more or less rectangular.
And these cells here look more or less circular.
So there's these two different perpendicular planes in the cork.
And when we look at these scanning electron micrographs, we can see very similar structure.
There are some cells that are roughly boxlike, and others that are more or less hexagonal or roughly rounded.
One feature that Hooke was not able to see, though, that you do see on the scanning electron micrographs, is the waviness in the cell walls.
And that was because the resolution of his microscope was insufficient to see that level of detail.
And here in this illustration on the bottom here is a drawing of sponge.
And when we look at the scanning electron micrograph, we see that the structure is remarkably similar to what Hooke has drawn.
So here we have Hooke's drawing of feathers.
And we can see he's made several drawings at different length scales.
And if we look at this one here, we see the barbule.
And you can see these little hooked regions there.
And those hooks lock into the little feathers over on this side over here of the adjacent barbule.
And in the higher magnification picture, you can see on one barbule, there's hooks on one side but not on the other.
And it's this hooking of the two sections together that allows the feathers to maintain a smooth surface for the wing when the bird is flying.
And you can see the same sort of thing when you look at the electron micrograph.
So you can see the little hooks on one side of the barbules.
And you can see how they interconnect together with the next barb.
One of the most reproduced images from Hooke's book is that of the flea-- this image we see here.
And you can see why.
It's a gorgeous image.
And it shows details that people had never seen before.
People were amazed to see that the little flea that they might have found on their dog or something was actually made up of this compound body, with all these little plates and little hairs here.
And you can see these little tiny claws on the legs, and the legs have all these hairs.
Nobody had any idea that this is what a flea actually looked like.
And so it was an amazing drawing.
And it was something that people were just stunned by when Hooke's book came out.
And if we look at a modern electron micrograph, we can see it's remarkably similar if we look at the same magnification.
So Hooke showed many of the same details, showed some of the same hairs on the legs, showed the same sorts of plates, showed the claws at the ends of the legs.
And our modern image is probably from a different species of flea.
We don't know what species of flea that Hooke actually looked at.
But you can see there's a tremendous similarity between the two images.
And it's remarkable how many of the features that Hooke was able to capture in his drawing.
And here we have the compound eye of the fly.
And this, again, was astonishing to people in Hooke's day.
And even today, people look at this image, and they're pretty amazed at the detail in this drawing.
And again, we can compare this with a modern electron micrograph.
And again, you can see the similarities between what Hooke saw and what we see in a modern scanning electron microscope at a similar magnification.
In the 1980s, atomic force microscopes were invented, which have a resolution down to tens of nanometers.
And today, there's transmission electron microscopes, which allow you to see the atomic structure.
So for instance in a crystal lattice, you can see the individual atoms and the regular crystal structure.
Today, most experimental studies of materials include photographs of the microscopic structure of the material taken through some sort of microscope.
And the remarkable thing is that all of these studies really trace back to this book here that we're looking at today-- to Robert Hooke's Micrographia.
[END PLAYBACK] There you have it.
So I just thought that was kind of cute.
You might enjoy that.
So that was that.
All right, let's get out of there.
Stop.
So let's go back to the foams.
So I think last time, we got as far as talking about the linear elastic behavior of foams and modeling that.
But we didn't quite get to looking at the compressive strength of the foam.
So I think we got as far as comparing the models with these equations here, and these plots of the data.
And what I wanted to pick up with today was looking at the compressive strength.
And we'll look at the fracture toughness as well in tension.
So we're going to start with nonlinear elasticity and the elastic collapse stress.
So if we have an open-cell foam, the derivation for the elastic collapse stress is really pretty straightforward.
We say the elastic collapse occurs when the cell walls buckles.
So in this schematic here, you can see the vertical cell walls have buckled.
And so there's going to be some Euler load that's related to that buckling.
And that's just the usual Euler load-- n squared, the n constraint factor, pi squared E. This is going to be E of the solid, I over l squared-- the length of the member.
And then the stress that corresponds to that is just going to be proportional to that buckling load over the area of the cell, which is just l squared, so just P critical over l squared.
So that just goes as Es.
I is going to go as t to the fourth, because we have that square sectioned member.
And now this is going to be l squared.
And that's an l squared.
So that's l to the fourth.
And so if I combine all of that together, I can say that the elastic buckling stress is going to be some constant-- and I think we're up to C4 now-- times the Young's modulus of the solid times the relative density of the foam squared.
So that's our equation for the elastic buckling stress.
And if you compare this with data, you can make an estimate of what C4 is.
And we find that C4 is about 0.05.
And you can also say that 0.05 really corresponds to the strain at which the buckling occurs.
Because the Young's modulus goes as the constants 1 times Es times the relative density squared.
So the strain's just going to be the stress over the modulus.
So that does correspond to the strain.
So that's saying that buckling compressive stress occurs at a strain of about 5%.
So that's open cells.
And then if we look at closed cells, if you recall when we looked at the moduli we looked at a couple of extra terms.
One was associated with face stretching for the modulus.
And the other was associated with the compression of the gas.
For the buckling, the faces don't really contribute that much, because typically the faces are very thin relative to the struts.
And because they're so thin, they buckle at a much lower load, and they don't contribute too much.
So we're not going to worry about that contribution.
So I'm just going to say that the thickness of the face is often small compared to the edges.
And that really is from the surface tension in processing that draws material away from the face and into the edges.
There can be some contribution from the internal pressure.
So if the internal pressure is greater than atmospheric pressure, then the cell walls are pre-tensioned, and you'd have to account for that.
So the buckling would have to overcome that pressure as well.
So then you would have the buckling stress would just be what we have up there-- C4 times Es times the relative density squared.
And then we just add on that factor P0 minus P atmospheric.
The thing with the gas which tends to affect more than the buckling stress, though, is the post-collapse behavior.
So let me just show you a couple of things here.
So here's some data for the elastic collapse stress.
And you can see on the y-axis, we've got the stress normalized by the Young's modulus of the solid.
And on the x-axis, we've got the relative density.
And that solid line there-- sort of solid, dark line-- is that equation there, which is the same as this one up here.
And you can see the data lie fairly close to that.
But what's interesting is if you look at the-- why is this not working?
Maybe my batteries finally died.
If we look at the post-collapse behavior, you can see if these are the stress-strain curves, they're not flat here.
They have some rise to them.
And this is a closed-cell foam.
And you can imagine as you're compressing the closed-cell foam, you're reducing the volume of the cell.
And as you doing that, you're increasing the pressure inside the cell from the gas.
And you can calculate what that is.
And I'll do that in a second.
And if you subtract off that gas pressure contribution, that works out to this line here.
Then these lines will be more flat, like this.
And we already really pretty much worked out that gas contribution.
So I'll just say for the post-collapse behavior, the stress rises due to the gas compression.
And that's as long as the faces don't rupture.
So if you have an elastomeric foam, typically they don't rupture.
And what we had worked out before was that that pressure-- we called it P prime-- it was P0 minus P atmospheric-- that was equal to P0 times the amount of strain, epsilon, times 1 minus 2 times the Poisson's ratio divided by 1 minus epsilon times 1 minus 2 nu minus the relative density.
And once you get to the buckling stress, then the Poisson's ratio becomes 0.
So if you take a foam-- so I brought a little foam in so you can play around with this one-- so if you take a foam like this and you compress it, once you've buckled it like this, it's not getting any wider this way.
And part of the reason for that is you've got all these pores in here.
And the cells just collapse into the pores.
They don't really need to move out sideways.
So you can smush that yourself, and try to convince yourself that the Poisson's ratio is just 0.
Yes, Matt
[INAUDIBLE] I guess I want to measure [INAUDIBLE] the gas contribution?
Yes, so there is a strain rate effect with these things.
But I wasn't going to get into that here.
If you look in the book, it's described in the book.
So I think there's two things.
One is that the solid itself can have a rate dependency.
And then there could be something connected
[INAUDIBLE]
Yeah, I mean, I'm not going to go into that here.
But one could look at that.
So let me just write down one more thing here, because if we let nu be 0, then this thing here becomes simpler.
So we could say the stress post collapse as a function of strain would be our buckling stress and then plus this factor here.
So that curve on the bottom over here-- if this is the stress-strain curve-- this little dashed line here-- that's the gas contribution.
And that is this term here.
So you can kind of see how the shape of the curves reflects that gas contribution.
And when you subtract it out, you get pretty much a horizontal plateau over here.
Are we happy?
Yeah?
[INAUDIBLE]?
This is for the closed cell.
Because the closed cell are the ones that are going to have the gas pressure.
If it's open cells, the gas can just move out of the cells
[INAUDIBLE]?
Oh, sorry, that was to show you that the Poisson's ratio was 0.
And that's true for both of them.
So then we can look at the plastic collapse stress.
Say we had a metal foam.
And we do a calculation a little bit like the one we did for the honeycombs, too.
So we say the failure occurs when the applied moment equals the plastic moment.
And the applied moment is proportional to the applied stress times the length cubed.
So I'm going to call that applied stress-- our strength sigma star plastic times the length cubed.
So if you think of, say, the little schematic up here, the force is going to go with stress times the length squared.
And the moment's going to force times the length.
So it's the stress times the length cubed.
And then the plastic moment goes as the yield strength times the thickness cubed.
And then if I just combine those, I get that the plastic collapse stress in compression is another constant-- I'm going to call it C5-- times the yield strength times the relative density to the 3/2 power.
And if we look at data, we find that the constant is about equal to 0.3.
And if I go to the next slide, here's a plot of the yield strength or the plastic collapse strength of the foam divided by the yield strength of the solid, plotted against the relative density.
And that dark, bold line is this equation here.
And you can see the data lie pretty well on that line.
There's one data set that's a little bit above the line.
But you can see the slope of that data set is still about 3/2.
OK, and the same as in the honeycombs, we could say that we can get elastic collapse before the plastic collapse if we were at a low density.
You can get the same thing in the foams.
And you calculate out what the critical relative density is for that the same kind of way.
So we can say we can get elastic collapse precedes the plastic collapse if the elastic buckling stress is less than the plastic collapse stress.
So all we do is make those two things equal to figure out the critical relative density where you get the transition from one to the other.
So the relative density has to be less than 36 times the yield strength of the solid over the Young's modulus of the solid squared in order to get buckling before yielding.
And let's see, where can I put that?
So for rigid polymers, that ratio of the strength of the solid over the modulus of the solid is about one over 30.
And so the critical relative density for the transition is about 0.04.
So you'd have to have a pretty low-density foam, but it's possible.
And for metals, that ratio is about 1/1,000.
And then the critical transition density is less than 10 to the minus 5.
So essentially, it never happens for the metal foams.
And then for the closed-cell foams, we could include the terms for face stretching and for the gas.
But in practice, the faces don't really contribute very much.
And typically for foams like say metal foams or a rigid polymer that had a yield point, the faces rupture.
And then if the faces rupture, then you don't get the gas compression term, either.
So I'll just write the full thing down.
But typically, you don't need to use it.
So the first term would be from the edges bending.
And the second term would be from the faces stretching.
And this would be from the gas.
But in practice, the first term is really the only one that is significant.
So for closed-cell foam, this equation works pretty well, too-- the same one as for the open-cell foams.
OK, so if we had, say, a ceramic foam that was brittle, there'd be a brittle crushing strength.
And then we get failure when the applied moment M is equal to the fracture moment Mf.
And this works very similar to the plastic yield strength.
So we find the applied moment goes as the global stress times the length cubed.
And the fracture moment goes into the cell wall strength times the cell wall thickness cubed.
So the brittle crushing strength goes as another constant-- let's call it C6-- times the wall strength times the relative density to the 3/2 again.
And C6 is about equal to 0.2.
And typically, ceramic foams have open cells.
So I'm just going to leave it at the open-celled formula there.
So there's one last thing for the compressive behavior, and that's the densification strain.
And we just have an empirical relationship for the densification strain.
So if you compress the foam and you get to very large strains, then the cell walls start to touch, and the stress starts to rise steeply.
And there's some strain at which that occurs.
And we call that the densification strain.
And in the limit, the modulus at that point would go to the modulus of the solid.
If you could completely squeeze all the pores out, the stiffness of that would go to the modulus of the solid.
And you might expect that that densification strain is just 1 minus the relative density, but it actually occurs at a slightly smaller strain.
So in a large compressive stress, or strain, I guess we could say, cell walls touch, and we start to get this densification.
So the modulus in the limit would go to the modulus of the solid.
And you might expect that the densification strain was just equal to 1 minus the relative density.
But it occurs at a little bit less than that.
So empirically, we find that it's just 1 minus 1.4 times the relative density.
And then I have this plot here, which is really just fitting a line to that data for densification strain.
So those equations describe the compressive stress or the compressive behavior of the foam.
So we've got the moduli, we've got the three compressive strengths, and we've got the densification strain.
So what we're going to do later on in the course is we'll use those models to look at how we can use foams and things like sandwich panels and looking at energy absorption.
And we'll also look at these equations in terms of some biomedical materials-- looking at trabecular bone, and looking at tissue engineering scaffolds.
So there's one last property I wanted to go over, and that's the fracture toughness.
So if we were pulling the foam in tension, and we had a crack in the foam, we'd want to know what the fracture toughness would be for a brittle foam.
And this follows the same sort of argument as we had for the honeycomb.
So all of these equations really are just following the same kinds of arguments.
But you can kind of see how having the honeycomb calculations makes it easier to do the foam ones.
So we'll do the fracture toughness calculation, and then I want to talk a little bit about material selection and selection charts for foams.
So that's less equation-y.
OK, and we're just going to look at open cells here.
So imagine we have a crack of length 2a.
And we have some remote stress applied, so remote tensile stress, so I'm going to call that sigma infinity-- the far-away stress.
And then we have a local stress on the cell walls.
I'm going to call that signal local.
So I have a little schematic that kind of shows what we're doing.
So we're pulling on it.
There's some crack.
The crack length is large compared to the cell size.
And we want to know what the fracture toughness is.
So we can say from fracture mechanics the local stress is going to be equal to some constant times the faraway stress times the square root of pi a over the square root of 2 pi r. And that's at a distance r from the head of the crack tip.
And if we look at our little schematic here, we could say it's hard to say exactly where the crack tip is, but it would be somewhere in here.
And we'd say this next unbroken cell wall is a distance r ahead of the crack tip.
And that r is going to be related to l. It's going to be some function of l. So I can say the next unbroken wall ahead of the crack tip at some distance r is going to be related to l. And that's subject to a force, which is going to be the local stress times l squared.
So that force is going to go as local stress times l squared.
And the local stress-- I can substitute this thing here in-- that's going to be proportional to the faraway stress.
And I'm going to get rid of the pi's.
And I'm going to substitute for r. I'm going to put in l. So it's going to be proportional to the faraway stress times the root of a over l and times l squared there.
And then we're going to say, again, the edges are going to fail when the applied moment equals the fracture moment.
And the fracture moment is going to go as the modulus of rupture of the cell walls times t cubed.
And the applied moment is going to go as f times l. And I've got f from up there, so that goes as the faraway stress, sigma infinite, times the root of a over l. And now I've got l cubed, because there's an l squared there and there's an l down here.
And then if I just equate those, then this is going to go as sigma fs times t cubed, like that.
So then I can say the fracture strength is equal to my faraway stress.
That's going to go as my modulus of rupture times the root of l for a times t over l cubed.
And then my fracture toughness is going to be this tensile stress times the root of pi a.
So there's going to be some other constant here, which I'm going to call that C8.
We've got the modulus of rupture of the solid.
I've got the square root of l, and I'm going to multiply it by pi so it's like other fraction mechanics kinds of equations.
And then we multiply that times the relative density to the 3/2 power.
And here, if we look at data, we find that that constant is about equal to 0.65.
And here's another one of these plots.
So here I've normalized the fracture toughness of the foams by the modulus of rupture of the cell walls times the root of pi l. So I've taken the cell size into account here, and I've plotted against the relative density.
And that equation there is the same as this equation I've got down on the board.
And this is the only one of the properties that we've looked at that depends on the cell size.
There's a cell size dependence here.
All right, so I think that's all the modeling of the foams.
Are we good?
I gave you a lot of equations.
We're good?
All right.
So I want to talk about how we might design foams to improve their properties.
And then I want to talk about how we might select foams for certain applications and look at selection charts.
So when we've been talking about the foams, especially the open-cell foams, we've been saying their deformation is largely by bending of the cell edges.
And if we could do something to increase the stiffness of the edges or the strength of the edges, then that would increase the overall properties of the foam.
And there's a couple of ways to think about doing that.
So the foam properties-- if the foam is controlled by bending of the edges, and the edges have some flexural rigidity, EI, if we could increase that EI of the edges, we would increase the properties of the foam.
And one way to do that is by making the edges hollow.
So if we had hollow edges, and you had a tube, then that would increase the EI.
And we can work out how much it's going to increase them.
And I have a little example here of-- a natural example of hollow foam struts.
So this is a grass.
I don't know what kind of grass it is.
I just saw this grass.
And we picked some different grasses, and we took some SEM pictures.
And it has a really kind of common structure for grasses.
It's very common for grass stems to have sort of a solid outer part and then a foam-like inner part.
It's so common that botanists have a name for it.
They call it the core-rind structure.
And if you take one of these grass stems, and you look at the sort of foamy bit in the middle, and you do a SEM picture of that, you can see that the little cell walls are actually little hollow tubes.
So one of these things-- it's a little hollow tube.
So what I wanted to do is work out how much the modulus of the foam would increase if you could make all the edges into little hollow tubes.
So we're going to start by saying the foam behavior is dominated by cell bending, so edge bending.
And the foam properties can be increased by increasing the EI of the cell wall.
So there's a couple of ways we could do that.
So the first one is looking at hollow walls.
So imagine I have a thin-walled tube-- just a circular, thin-walled tube.
There's my little wall there.
It has some radius little r, and a wall thickness t. And then imagine I have the same amount of mass, but now I have a solid circular section.
And I'm going to say the radius of that is big R. So for our thin-walled tube, the moment of inertia is pi r cubed times the thickness, t, if it's thin.
And for our solid circular section, I is going to be pi big R to the 4th over 4.
And if I say I want to set this up so that the masses are equal, then the areas of the cross-sections have to be equal-- say it's from the same material.
So the masses are going to be equal if pi R squared is equal to 2 pi r t. So I'm going to solve here for R. So the pi's are going to cancel out.
So the masses are equal if R is equal to the square root of 2 times r times t. And then what we're going to do is see how the big is the moment of inertia of the tube relative to the solid.
And the tube is pi little r cubed t. And the solid was pi R to the 4th, divided by 4.
And I'm going to get rid of the R here, and get rid of the pi's there.
So R to the 4th is going to be 4r squared t squared.
So the 4s are going to go.
And this boils down to r over t. So if I had a thin-walled tube, the moment of inertia is going to be r over t bigger than if I had the same mass in a solid circular section.
So you can see for the little plant here, by making a thin-walled tube, you're increasing the stiffness of the foam with the same amount of material.
That's the idea.
And you can do a similar kind of analysis for other properties.
So that's if we have hollow tubes.
So another option is we could have cell walls that are sandwich structures.
So imagine if the cell walls themselves were little, tiny sandwich structures.
So when you have a sandwich beam, what you have is too stiff, strong faces that are separated by some sort of porous core, like a honeycomb or a foam or balsa wood.
And the idea with the sandwich structure-- if I draw a little sketch of the sandwich, here's my faces.
So imagine those are solid.
So they might be aluminum sheets, or they might be fiber reinforced composites.
And then we have some sort of cellular thing here as the core.
And the idea is, that's analogous to an I-beam.
So in the sandwich beam, we have two, stiff, strong faces separated by a lightweight core.
So the core is typically a honeycomb, or a foam, or balsa wood.
And the idea is, you increase the moment of inertia of the cross-section with little increase in weight.
And if you think of an I-beam, an I-beam has a large moment of inertia, because you're separating the flanges by the web.
And the sandwich beam works in the same way.
You're separating the faces by the core.
But the core doesn't weigh very much, because it's a cellular thing.
So the faces of the sandwich are like the flanges in the I-beam.
And then the core is like the web.
So the idea is to make something called a micro-sandwich foam.
So what you want to do is make the cell walls into sandwiches.
And one way to do that is to disperse a large volume fraction of thin-walled spheres into the foam.
And you have to get the geometry right to make it work.
So let me draw a little kind of sketch here of how it works.
So here's our thin-walled spheres.
And then you're going to distribute those in a foam.
Here's another sphere over here.
The spheres are not perfect.
Let's say there's another one in here.
And then the idea is this stuff in here would be the foam.
So these guys are hollow spheres.
And say the spheres have a diameter D. And say they have a wall thickness here of t. And say that the separation of the spheres I'm going to call c. You can see that there.
And then the cell size of the foam I'm going to call e. So there's a bunch of parameters you have to kind of play with to get this to work.
So you have to have thin-walled spheres so the faces are thin.
The sandwich panels work best when the faces are thin.
So you need the thickness of the sphere to be much less than D. You need the faces to be stiff relative to the foam.
So you need the modulus of the sphere material to be greater than the modulus of the foam.
And you need the volume fraction of the spheres to be relatively high to get the spheres close enough together for this to work.
So you want that volume fraction to be something like 50% to 60%.
And for the foam, you need to have the foam cell size less than the separation between the spheres.
You need to have a number of-- you can't just have one pore in here.
That's not really like a foam.
It won't behave like a foam as a continuum.
So you need to have a number of different cell sizes in between each sphere.
And so you need the cell size of the foam to be a lot less than the separation of the spheres there, c. But if you can control this geometry, you can get the sandwich effect.
And you can get improved properties by doing that.
So there's ways you can play around with the structure of the foams to improve their properties.
So that was one thing I wanted to say.
Another way to improve the properties of a foam-like material is to use one of those lattice materials.
So we've been talking about ways to improve the bending stiffness.
But if you could get rid of the bending altogether and have axial deformation in the cell walls, that would be much stiffer.
And you can get axial deformation by having those 3D truss kind of materials.
So I have a picture of this.
There we go, so there's one of those 3D truss materials.
So another alternative is to sort of get rid of the bending altogether, and to try to make a truss-type material.
So there's various ways to make these.
I think that we talked about a few of them earlier on.
And you can analyze them as truss-type structures.
And I can just run through a sort of little dimensional argument to get the modulus.
So the modulus is going to go as the stress over the strain.
The stress is going to go as a force over a length squared.
The strain's going to go as a deformation over l. So this is just like what we had before for the foams.
But in this case, the deformation is going to go with the force times the length over the area of the cross-section divided by Es, because we're pulling it or pushing it axially.
So that goes as Fl over t squared Es.
And if I just put that back in the equation here for the modulus, I get that we've got F over l. And I've got delta here, so that's F l t squared Es.
And you just get the modulus goes as the modulus of the solid times t over l squared.
And that goes as the modulus of the solid times the relative density.
So for the open-celled foams, the modulus went as the relative density squared.
So if it was 10% solid, the modulus would be 0.01.
And this is saying if it's 10% solid, the modulus is 0.1.
So it's much bigger.
So this is all sort of well and good.
The only difficulty is that when you look at the modulus, you can do reasonably well.
But when you look at the strength, some of the members are going to be inevitably in compression.
When you have these truss materials, some members are going to be in tension.
Some members are going to be in compression.
And the compression members tend to buckle.
And once the compression members buckle, then you're back to the same kind of strength relationship that you have for the foam.
So that's one of the difficulties of this.
So let me say that the strength-- so if the strength was controlled by uni-axial yield, it would go linearly with relative density.
But if it goes with buckling, it goes as the square.
So I'll just say the compression members can buckle.
And say you had a metal lattice.
Then there's some interaction between the plastic behavior and the buckling.
And you use what's called the tangent modulus instead of just the Young's modulus.
And the tangent modulus is lower.
And there's also what's called knock-down factors that can be large, too.
So the knock-down factor can be like 50%.
So the measured strength can be half of what you thought it was going to be.
This should be a squared over here.
Sorry.
So even though the stiffness of these 3D trusses can be quite good, the strength often isn't quite as good as one might hope.
So that's one of the issues with them.
All right.
So do you see the idea, though, with all these different micro structures, is that you can control the structure in a way to try to increase the bending stiffness or get rid of the bending stiffness and increase the axial stiffness?
So there's things you can do to play around with that.
And I wanted to talk a bit today about material selection charts for foams.
So when we talked about woods, we started talking about this.
Remember, I derived a little performance index.
We said if we had a material and we wanted to have a given stiffness, and we wanted to minimize the mass, we had that performance index that was E to the 1/2 over rho.
And we had a chart of modulus versus density.
And we saw that wood was really good.
You can do that for other sorts of properties, not just modulus.
So you can make-- depending on what the mechanical requirement is, you can work out different performance indices.
So I want to go into that in a little bit more detail.
So the question is, how do we select the best material for some mechanical requirement?
So in the wood section, we looked at the minimum mass of a beam of a given stiffness.
And we saw that the performance index was E to the 1/2 over rho.
So let me do another one of these little examples, and then I'll show you some more of them.
So another example would be what material-- minimize the mass of a beam of a given strength or a given failure load.
So we'll call the failure load Pf.
And we can see the maximum stress in the beam is going to be the moment in the beam times the distance from the neutral axis y, and divided by the moment of inertia.
So here, M is the maximum moment in the beam.
And y is the maximum distance from the neutral axis.
And I is the moment of inertia.
And I'm going to say i goes as t to the 4.
And I'm going to define a failure stress of the material sigma f. So sigma max is going to go as my failure load times the length.
That would be the moment.
The distance from the neutral axis is going to go as t. And the moment of inertia is going to go as t to the 4th.
And that's going to be the failure strength there.
So I can solve this for t. And then I'm going to write the mass in terms of t, and put that in there.
So here t goes as Pf l divided by sigma f. And that's going to be to the 1/3 power.
I guess I can scoot over here.
Then we can say that the mass M goes as the density of times t squared times l. So the mass M is going to go as rho times l times t squared.
So that whole thing goes to the 2/3 power.
So if we look at the material properties, the mass goes as the density times the failure stress raised to the 2/3 power.
So if we want to minimize the mass, we want to minimize rho over sigma f to the 2/3, or we want to maximize sigma F to the 2/3 over rho.
So that's the performance index for that case.
So we can obtain these performance indices for different loading configurations and different mechanical requirements.
And I don't want to go through a whole lot of them, but I'm going to put this up with the notes.
So this is from Mike Ashby's book on Material Selection in Mechanical Design.
And this is a whole series of these performance indices for different situations, for things loaded in torsion, for columns and buckling, for panels and bending.
So these ones are all for stiffness.
And they all involve a modulus raised to some power divided by a density.
So a tie in tension, c over rho, the beam in bending is E to the 1/2 over rho.
A plate in bending is E to 1/3 over rho.
So you don't need to memorize those.
But you can see you can derive these for different situations.
And here's another one for strength-limited design.
So the shaft is, depending on what the specifications are, it's the strength raised to the 2/3 power over rho.
The beam loaded in bending-- the top one there-- sigma f to the 2/3 over rho.
That's what we just did.
So there's all these different kind of performance indices.
So depending on what your situation is, you would pick one of these indices.
And then what you can do is use these material selection charts, which plot one property against another on log-log scales.
And because all of these performance indices involve a power, they always end up being a straight line on your log-log plot.
And here this one, I think, is the same as what I showed you for the wood.
This one's the modulus here plotted against density.
So foams are down here.
And other engineering materials are over here.
And these guidelines here are the different performance indices.
So this one's E over rho.
This one's E to the 1/2 over rho.
This one's E to the 1/3 over rho.
And for this case here, as you move the lines up to the top left-hand corner, E is getting bigger.
Rho is getting smaller.
And so the actual value of the performance index is getting bigger.
So you can use this to select a material.
So we've made these charts for foams as well.
So here's a couple of charts for foams.
And I think what I'm going to do is just go through them quickly.
And there aren't really that many notes, so I'll just put the notes on the website.
And you can come and write all the notes down.
And then we can finish this today.
So this one here is the Young's modulus versus density.
And these are all sorts of different foams.
So the low modulus ones tend to be flexible.
The higher modulus ones tend to be more rigid.
And you could use this to select foams, if you wanted.
You can also see what the range of values is.
So the values of the modulus here goes from a little less than a 100-- because this is two orders of magnitude here, I think, each one of these-- down to about 10 to the minus 4 or a little less than that.
So there's a huge range.
There's almost a range of a factor of a million in those moduli.
And the same with the strengths here.
The strengths go from 10 to the minus 3 mega-pascals up to about maybe 30 mega-pascals, something like that.
And you can see for the modulus and the strength, things like the metal foams are good.
The balsa's good.
Here's the balsa up here.
Metal foam's up there.
So you can kind of see the range of properties that you could get.
And then you could also-- need a drink, hang on.
You can also plot the specific property.
So here's the compressive strength divided by the density plotted against the Young's modulus divided by the density.
And here you want to be up at this end.
So you would have a high strength and a high stiffness.
So the balsa and the metal foams are good up here.
This next plot-- this is the compressive stress at 25% strain.
And this is the densification strain.
And if you think of having your stress-strain curve looks like this, something like that, so you could say that's a strain of 0.25 and that's the stress that corresponds to that.
So that stress times the densification strain, which is out here someplace, is an estimate of the energy underneath the stress-strain curve.
So you can think of this right-hand plot here-- those dashed lines-- these lines like this and this and this-- each one of those corresponds to how much energy you would absorb under the stress-strain curve.
So points that lie on here would have an energy of 0.001 megajoules per cubic meter.
And over here, we're at 10 joules per cubic meter.
So again, the balsa and the metal foams are good over here.
So you can use these plots to try to identify foams for particular applications.
And I think there's a couple more.
It doesn't have to be mechanical properties.
Here is thermal conductivity versus compressive strength.
So you can imagine if you wanted some insulation, you wanted to have a certain thermal conductivity value, you probably also need at least some minimal compressive strength.
You could also have something like a maximum service temperature, that maybe the foam is going to melt at some temperature.
You can't go beyond that.
So there's some property there.
And I think there's one more here.
You can look at things like the density in terms of the buoyancy of a foam, if you have some buoyancy application.
And you can look at cell size on this one here.
And cell size can be important for things like filtration and catalysis.
So the amount of surface area goes as 1 over the cell size-- the surface area per unit volume.
And so the cell size can be important for those sorts of applications.
So the idea is, you can make these material selection charts for foam.
And you can put data on there.
And you can compare foams.
And you can use these performance indices.
So I'm going to leave it at that.
There is a little bit more notes.
But I'll just put them on the website, and you can get them from there.
So I think we're good for today.
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So what I wanted to do today was talk about thermal properties of foams.
And foams are often used for thermal insulation.
And that's always closed cell foams that are used for thermal insulation.
And we'll see why.
And the foams tend to have a low thermal conductivity.
And that's largely because gases have lower conductivity than solids.
And if you have mostly gas, you're going to have a lower conductivity.
So they have a low conductivity because they have a high volume fraction of gas.
And they've got a low volume fraction of the solid.
They also have cells.
And the heat is transferred partly by radiation and convection.
And if you have small cells, you reduce the amount of convection and radiation.
And we'll see that.
So that, by having a cellular structure, and in particular, by having small cells, you can decrease the heat transfer.
OK, so let me write some of the stuff down.
So closed cell foams are widely used for thermal insulation.
And the only materials with lower thermal conductivity than closed cell foams are aerogels gels.
And I'll I talk a little bit more about aerogels later on today.
But the difficulty with aerogels is that they tend to be very weak and brittle, like ridiculously weak and brittle.
So we had a project on aerogels a couple of years ago.
And the students who I was collaborating with would make aerogels.
And they'd bring it up to my office.
And I would pick it up and like-- I would pick it up like this, and it would break.
So they have very low thermal conductivity, but they're very brittle.
And I brought a few of our samples of aerogels, just so you can see what they look like.
And I'll pass them around in that little tube, so you can kind of play with them.
OK, so we're going to focus on foams.
And whoops-- And we can say the low thermal conductivity of foam arises mostly from the high volume fraction of gas and that the gas has a low lambda, a low thermal conductivity.
So lambda is thermal conductivity, so I'm just going to put lambda there.
Then it has a small volume fraction of solid, which has a higher thermal conductivity.
And then the foams have a relatively small cell size.
So one of the things we're going to look at is how does the cell size effect the thermal transfer, the thermal conductivity.
OK, so there's lots of applications for foams.
And I guess one of the main ones is in buildings, insulating buildings-- also insulating refrigerated vehicles, things like LNG tankers.
So there's lots of the applications for using foams for thermal insulation.
Foams, in addition to having a low thermal conductivity, they also have good thermal shock resistance.
So thermal shock is if you have a material, you heat it up, and then you suddenly cool the surface of it, for example.
So say you takes something and you quench it in water or quench it in some fluid, then the surface, it wants to shrink because its temperature drops, but it's connected to everything underneath it and it can't really shrink.
And so it's constrained and you can get cracking and spalling.
And so, it turns out foams have a good resistance to that thermal shock kind of loading.
And we'll see why that is, too.
Roughly, you can see if the thermal expansion strain is the thermal expansion coefficient times the change in temperature.
And the stress that you might generate is just going to be related to the modulus times alpha times delta-t. And because we're going to see that alpha for the foam is the same as alpha for the solid, but the E foam is going to be a lot less that E of a solid would be.
So because the modulus is smaller, you would get a better thermal shock resistance.
OK, so I wanted to go over a couple of sort of laws of heat conduction, so we can talk about what thermal conductivity is and how we define it.
So the first one here-- --first one here is for steady state conduction.
So when we say steady state conduction, what we mean is that the temperature is constant with time, the temperature doesn't change with time.
So time's not going to come into the equation here.
And heat transfer for steady state conduction, where there is no change in the temperature with time, described by Fourier's Law.
And that says that the heat flux q is equal to minus lambda times the gradient in temperature.
And if you want to think about just a one diversion of that, it's equal to minus lambda times dt by dx.
So here, q is our heat flux.
So that would have units of joules per meter squared per second.
So how much heat transfer per unit area per unit time.
Lambda is the thermal conductivity.
And it has units of watts per meter k, so degrees kelvin.
And then delta or-- and then this is our temperature gradient.
OK, so that's Fourier's Law, and we're going to use that later on when we talk about the foams.
And then, just so that we have things a little more complete, if you have a non-steady heat conduction, if the temperature varies with time, then there's a difference equation that involves the thermal diffusivity.
So if we have non-steady heat conduction-- so t varies with time.
I'm going to call a time tau.
Then a partial differentiation, the partial derivative of temperature with respect to time is equal to the diffusivity, that's given the symbol a, times the second derivative of temperature with respect of distance, so with respect x squared.
So here a is the thermal diffusivity.
And it is equal to the thermal conductivity divided by the density and divided by the specific heat.
So here, rho is the density and cp is the specific heat.
The specific heat is the heat required to raise the temperature of a unit mass by the unit temperature.
And so, the density times cp is the volumetric heat capacity.
It's how much energy you would need to raise a certain volume by, say, 1 degree k instead of a certain mass.
OK, so on the table here, on the screen, we have different materials.
And we have the thermal conductivity lambda.
And we have the thermal diffusivity, a.
And I guess I should also say a has units of meters squared per second.
So this table is arranged in order of decreasing thermal conductivity.
So here's copper at the top, 384, watts per meter k. Here's, you know, different metals.
You've got aluminum.
Here's a couple of ceramics.
They're about a factor of 10 less than the metals.
Here's the polymers, another factor of 10 less than that.
And here's some gases.
Air is about 0.025.
Carbon dioxide is less than that.
Triclorofluoromethane, which used to be used as a gas in foams because it's got such low thermal conductivity, is 0.008.
But it's no longer used because it's a-- what you call it?
A fluorocarbon.
Anyway, it decreases our ozone layer.
So they don't use that anymore.
Now here's some wood.
So that's one sort of cellular solid.
And they're around 0.04-- something like that.
And here's a group of polymer foams.
And they're a little over 0.025.
So if you think of-- if you had the gas, air-- if the air was the gas inside the foam, 0.025 is lambda for the gas.
So you're not going to get lower than that.
And you have to use a low conductivity gas to get these values, like 0.025 here, 0.020, 0.017.
And then, hear some other sorts of sort of mineral fibers, glass foams, glass wools.
OK, so that's just a table so that you have some data there.
All right?
Yes?
[INAUDIBLE] --foams, if they are closed cell, with a different gas rate.
Because if they're open cell--
--Right.
The gas is going to--
--it would just always be air
It's going to go.
And in fact, one of the difficulties with using the lower conductivity of any gases is there's a phenomenon called aging, that if, you know, you've got your gas inside your foam, it's going to diffuse out into the air.
And air's going to diffuse in.
So over time, the thermal conductivity tends to increase because you're getting air coming and the local-- conductivity gas going out.
But I think, typically, that process takes a number of years.
It doesn't happen in a week.
But if you're designing a building and want the building to be there for 50 years, it occurs faster than that.
So it's not ideal from that point of view.
All right.
So let me talk a little bit more about thermal diffusivity.
Let me scoot over here.
So materials with a high value of that thermal diffusivity, a.
They rapidly adjust their temperature to their surroundings.
So if they have a high value of a, what it really means they've got a, say, a high value of lamda-- so high thermal conductivity.
And, say, a low value of this volumetric heat capacity.
So it doesn't take much energy to change their temperature.
And they also conduct heat well.
So they tend to adjust their temperature to their surroundings quickly.
OK, so then, let's talk about the thermal conductivity of a foam.
So I'm going to call that lambda star.
So the star is the foam.
And then we'll talk about-- lambda s will be the lambda for the solid that it's made from.
So if you think of the thermal conductivity of the foam, there's contributions from different types of heat transfer.
So you could have conduction through the solid.
I'm going to call that lambda s. You could have conduction through the gas.
You could have convection within the cell.
So convection has to do with having, say, within the cell, it might be a different temperature on one side of the cell to the other side of the cell.
And the warmer side of the cell, the gas is going to tend to rise to the warmer side and fall to the cooler side.
And you get a convection current set up.
So you can get heat transfer from that.
And you can also get heat transfer by radiation.
So radiation can cause heat transfer, as well.
So we're going to have contributions from conduction through the solid.
So the amount of conduction in the foam from the solid-- I'm going to call lambda star s. So lambda s would be the conductivity of the solid.
And lambda star s is the thermal conductivity contribution from the solid in the foam.
So we get kind of-- through the solid.
We have conductivity through the gas.
So it's lambda star g for gas.
And then we could have convection within the cells.
We'll call that lambda star c. And then we could get radiation through the cell walls and across the voids.
We'll call that lambda star r. And so, the thermal conductivity of the foam is just the sum of those four contributions.
So we're just going to go through each of those contributions, in turn, and work out how much thermal conductivity you get from each of them.
And it turns out most of the thermal conductivity comes through the gas.
So if we first look at just conduction through the solid, we've got that contribution to the conductivity of the foam from the solid, it's just equal to some efficiency factor times the thermal conductivity of the solid times the volume fraction of the solid or the relative density.
And here, eta is an efficiency factor.
And it accounts for that tortuosity in the foam.
So if you think of the solid in the foam, it's not like we have little fibers that just go from one side to the other like this and the heat just moves along those fibers.
You know, the foam cells have some complicated geometry and the heat has to kind of run along that complicated geometry.
And people have made estimates of what this is.
And it's roughly a factor of 2/3.
So I guess it would depend on exactly the foam cell geometry.
But typically it's around 2/3.
So that's conduction through a solid.
That's straight forward.
Conduction through gas is similarly straightforward.
It's just the conductivity of the gas times the amount of the gas.
And the volume fraction of the gas is just 1 minus the volume fraction of the solid.
So it's just 1 minus the relative density.
So the conduction through the gas is just lambda g times 1 minus the relative density.
So we can do a little example here.
And you can see how much of the conduction comes from the solid in the gas.
So for example, if we look at a foam that's 2.5% dense and say it's a closed cell poly-- what are we doing-- polystyrene.
So the total thermal conductivity of the foam is about 0.04 watts per meter k. And the thermal conductivity of polystyrene is 0.15 watts per meter k. And the thermal conductivity of air is 0.025.
So let's assume it's just blown with air.
And then if I just add up, what's the contribution of conduction through the solid and conduction through the gas-- so I just use those two little equations-- conduction through the solid-- it's going to be 2/3 of this value of lambda s times the amount of the solids-- that's 0.025 and then plus lambda g, which is 0.025 times the amount of the gas, which is 0.975.
And if I work those two things out, this is 0.003 and this is 0.024.
So that total is 0.027 watts per meter k. So you can see if the total is 0.04, most of it's come from the gas.
A little bit's come from the solid.
And the rest is going to be from convection and radiation.
And that's typical.
And that's the reason that they sometimes use low thermal conductivity gases to blow foams for thermal insulation because the gas makes up such a big fraction of the total conductivity.
If you can reduce that, you reduce the overall conductivity.
So, we'll say foams for insulation are blown with low conductivity gases.
But as I mentioned, you have this problem with aging that, over time, that gas is going to diffuse out and air is going to diffuse in.
Then the overall thermal conductivity of the foam is going to increase.
So that's the conduction.
And then the next contribution is from convection.
So imagine we have one of our little cells here.
And it's hotter on that side than it is on that side.
And hot air is going to rise.
Cold air is going to fall.
So you get a convection current set up.
And because of the density changes, you get a buoyancy force in the air.
So that's kind of driving the convection.
But you also have a viscous drag.
So the air is moving past the wall of the foam.
And there's going to be some viscous drag associated.
And how much convection you can get depends on the balance between this buoyancy force and the viscous drag.
So we'll say the gas rises and falls due to density changes with temperature.
And the density changes give rise to buoyancy forces.
But we also have these viscous forces from the drag of the air against the walls of the cell.
So air moving past the walls-- this is kind of a fluid mechanics thing-- so that air is a fluid.
And in fluid mechanics, they often use dimensionless numbers.
And there's a dimensionless number called the Rayleigh number.
And the Rayleigh number, you can think of it-- it's not quite the balance of the buoyancy force against the viscous forces.
But it involves those forces.
And convection is important if this Raleigh number's over 1,000.
And here's what the Rayleigh number is.
It's the density of the fluid times the acceleration of gravity times beta.
Beta's the volume expansion coefficient for the gas-- times the temperature change.
And we're going to look at a temperature change across a cell.
And then, times the length.
That's going to be the cell size.
And we divide that by the fluid viscosity and the thermal diffusivity.
So let me write down what all these things are.
So rho is the density of the gas.
So the g's gravitational acceleration.
Beta is the volume expansion of the gas.
And for a constant pressure that's equal to 1 over the temperature.
Then delta tc is the temperature difference across a cell.
And l is the cell size.
Mu is the dynamics viscosity the fluid.
And a is our thermal diffusivity.
So what I'm going to do is just work out, for a typical example, how big of a cell size do you need to get this Rayleigh number to be 1,000.
And we're going to see that, typically, that cell size is big.
It's like 20 millimeters.
So in most foams, the convection really isn't very important at all.
So it's typically-- people don't worry about convection.
And let me just show you how that works.
So for our Rayleigh number, which is ra-- for the Rayleigh number to be 1,000-- say we had air in the cells.
And say the temperature was room temperature.
Then the volume coefficient of expansion is just 1 over t. So it's 1 over 300, say.
degrees k to the minus 1.
Let's say our change in temperature across one cell was 1 degree k. Bless you.
The viscosity of air is 2 times 10 to the minus 5, pascal seconds.
The density of air is 1.2 kilograms per cubic meter.
And the thermal diffusivity for air is 2 times 10 to the minus 5 meters squared per second.
And if you plug all of these into that equation for the Rayleigh number and you solve for the cell size, you find that the cell size, l, is 20 millimeters.
So that says convection is only important if the cell size is bigger than that.
And so most foams have cells much smaller than that.
And convection is negligible.
So I have enclosed cells and the heat's not transferred so easily from one cell to another by the gas moving.
And by having small cells the convection drops out.
So you don't have to worry about that.
So the last contribution to heat transfer is from radiation.
And there's something called Stefan's law that describes the heat flux for radiated heat transfer from a surface at one temperature to another surface at a different temperature across a vacuum.
So we can say we have a heat flux qr not from a surface of one temperature.
So I'm going to call that t1-- to one at a lower temperature.
I'm going to call tnot-- with a vacuum in between them.
So this is [?
Stefan's ?]
law so this is the radiative heat flux is equal to the emissivity of the surfaces, which is beta 1 times a constant called Stefan's constant-- sigma times the fourth power of temperatures.
I'm taking the difference of the temperatures so here are the Stefan's Constant-- is sigma.
And that's equal to 5.67 times the 10 to the minus 8.
And that's in watts per meter squared per k to the fourth.
And beta is a constant describing the emissivity of the surfaces.
So it gives the radiant heat flux per unit area of the sample relative to a black body.
And that's a characteristic of the emissivity.
All right, so then, so if we-- yes?
[INAUDIBLE]
Now-- so right now, forget the foam.
We have no foam.
We just have two surfaces with a vacuum between them.
And now I'm going to stick a foam between the surfaces.
And we're going to see how that changes the heat flux, OK?
So the next step is we put the foam between those two surfaces.
And the heat flux is going to be reduced because the radiation is going to be absorbed by the solid and reflected by the cell walls.
And so we're going to characterize how much it's reduced.
So there's another law called Beer's Law, which characterizes the reduction in the heat flux.
Piece of chalk's getting to small OK, so Beer's Law gives us the attenuation, so the sort of reduction in the heat flow.
So qr is equal to qr not.
That would be the heat flux, if we just had the vacuum.
And then there's an exponential law.
And it's the exponential of minus k star t star.
And here, k stars in an extinction coefficient for the foam.
Talk a little bit more about that in a minute.
and t star is just the thickness of the foam.
And then this thing is called Beer's Law.
So we have very thin walls and struts.
And we're just going to consider optically thin walls and struts to make life easy.
Then we can say that, if they're optically thin, they're transparent to radiation.
They're optically thin if they're less than about 10 microns.
Then this extinction coefficient is just the amount of solid times the extension coefficient for the solids.
So it's just the relative density times the extinction coefficient for the solid.
OK, and then I can say, the heat flux by radiation.
I can use two equations to write that down now.
And then I'm going to let them be equal to get the thermal conductivity.
I can say qr is going equal to lambda r times dt by dx.
So that's the Fourier's Law that we started out with.
And then I've also got the qr that I'm going to get by combining the Stefan's Law with the Beer's Law up there.
So if I do that, I get that qr is beta 1 times sigma times t1 to the fourth minus t not the fourth.
So that's the qr not up there from down there.
And then I've got an exponential for the attenuation.
And instead of k star, I'm going to put the relative density of times ks.
and then I've got the thickness of the foam, t star, as well.
OK, so that's qr, but that has to equal lambda times dt by dx.
So I'm going to use some approximations.
Here and I'm going to end up with an expression for the contribution from radiation to heat transfer in the foam.
Yeah?
So when you say optically thin walls, where t is less than 10 microns, you mean like the walls of the foam?
Yeah, yea
So it's different t than the--
t star is the thickness of the whole thing, yeah.
So imagine we had our two surfaces.
And they might be like 100 millimeters apart or something.
t star is the sort of thickness of the foam in between the two surfaces.
And the optically thin is the cell walls, which are microns kind of thickness.
OK.
So I'm going to make some approximations here.
And that's going to allow me to solve for t star.
So I'm going to say that dt by dx x is approximately equal to just t1 minus t not over the thickness of the foam or I'll call that delta t over t star.
And then, the other approximation I'm going to use is that t1 to the 4th minus t not to the fourth is equal to 4 times delta t times the average temperature cubed.
So here t bar is the average temperature, t1 plus t not over 2.
So then, if I use those two approximations, I can write that qr, our heat flux from radiative transfer.
I got the beta 1.
I've got the sigma.
And instead of the difference of the fourth power, I'm going to write 4 delta t t bar cubed.
And then I've got my exponential.
Blah, blah, blah, blah, blah.
So then, here's the relative density times ks times t star, the overall thickness.
That's going to equal the radiative contribution to the thermal conductivity of the foam.
And instead of dt by dx, I'm going to have delta t over t star here.
So part of the reason for doing these approximations I end up with a delta t term on both sides.
Now I can cancel that out.
And if I just take this mess here and multiply it by t star, then I've got lambda r star.
That's our thermal conductivity contribution from radiation.
So one of the things to notice here is that, as the relative density goes down, then the contribution from radiation to the thermal conductivity of the foam goes up.
OK, so this chart here shows thermal conductivity as a function of relative density.
And it breaks down the contributions from the gas, g, the solid, s, and the radiation, r. And you kind of see the gas contribution doesn't change that much.
These are relative densities between a little over 2 and a little less than 5%.
So the amount of gas-- it's mostly gas in all of these things.
The solid contribution increases as the relative density increases.
So you'd expect that.
And then, as I just said, as the relative density goes down, the amount of radiation contribution goes up.
And so you can kind of see how that all fits together.
Another plot that shows the thermal conductivity versus the relative density.
These are for a few different types of foams.
You can see for this plot here, you reach a minimum in the thermal conductivity.
And that's because you've got this trade off between the contribution from the solid and the contribution from the radiation.
And those two kind of trade off and you get to a minimum.
So let me write some of this down.
So I'll just say that-- hang on.
Write this over again.
This is looking at the overall thermal conductivity.
And we can see the relative contributions of lambda, solid, lambda, gas, lambda, radiation.
I'll just say this shown in the figure.
I'm going to say the next figure shows a minimum in the thermal conductivity.
Then I'll just say there's a trade off between the conduction through the solid and I can direction from the radiation.
And then we also have a plot here that shows the conductivity versus the cell size.
And you can see that the conductivity increases with cell size.
And the reason for that is the bigger the cells get, the radiation is reflected less often.
And one thing I wanted to mention with the cell size is that if you look at aerogels, the way aerogels shells work is that they have a very small cell size, a very small pore size.
So typically, it's less than 100 nanometers.
And the mean free path of air is 68 nanometers.
So the mean free path is the average distance the molecules move before they collide with another molecule.
And if your pore size is less than the mean free path, then that reduces the thermal conductivity.
It reduces the ability of the atoms to pass the heat along between one another.
So the way the aerogels work is they have a very small pore size.
And what's important is how big the pores are relative to the mean free path of air.
OK, so that's the thermal conductivity.
I wanted to talk about a few other thermal properties of foams, as well, today.
So one is the specific heat.
And since the specific heat is the energy required to raise the temperature by a unit mass, then the mass is the same-- you know, if you have a certain mass of foam or a certain mass of solid-- the specific heat from the foam is the same as the solid.
So the specific heat for the foam is the same as the specific heat for the solid.
So that would have units of joules per kilogram per degree k. And the next property is the thermal expansion coefficient.
And it's a similar thing.
The thermal expansion coefficient for the foam is equal to the thermal coefficient of expansion for the solid.
So imagine you have-- say you had something like a honeycomb.
If you heat it up a certain amount, every member is going to expand by alpha.
And if every member expands by alpha, the whole thing expands by alpha.
And this is the same.
And it's the same idea with the foam.
So if every member just gets longer by alpha, then the whole thing gets bigger by alpha.
OK, so the last topic I wanted to talk about was the thermal shock resistance.
And thermal shock is the idea is that if you have something that's hot, and say you quench it in a liquid-- so you put it suddenly in a liquid-- the surface is going to cool down faster than the bulk of it.
And because the surface is trying to contract because it's cooling down, but it's attached to the bulk of it and it's constrained, it can't really cool down, then you generate stresses.
And if the stresses are big enough, you can cause fracture and have the thing crack and spall.
So we'll say if the materials is subjected to a sudden change in the surface temperature, that induces thermal stresses at the surface and can induce spalling and cracking.
So we're going to think about a material at one temperature that's dropped into, say, a liquid at a different temperature.
So the surface temperature is going to drop to the cooler liquid temperature and it's going to contract the surface layers.
And the fact that they're bound to the layers underneath that are not contracting as quickly, it means that you generate a thermal strain.
So the thermal strain is going to be the coefficient of thermal expansion times the change in temperature.
So you're going to constrain the surface to the original dimensions.
And then you're going to induce the stress.
So if it's a plane or thing, it's e alpha delta t. And then, there's a factor of 1 minus nu, just because it's a plane, in a plane.
And then you'll get cracking or spalling when that stress equals some failure, stress.
So I can rearrange this and solve for the critical delta t that you can withstand without getting cracking.
So I just rearranged this and say sigma's equal to sigma f. That would be sigma f times 1 minus nu over e and over alpha.
So that's the critical change in temperature to just cause cracking.
So now what I can do is I can substitute in there for what you would have for the foam.
And I'm going to do it just for the open cells just because it's easier to write the equations.
So for the foam, I would have some sort of fracture strength.
So when we did the modeling of the foams, we said that was equal to about 0.2 times the modulus of rupture times the relative density to the 3/2's power and 1 minus nu.
And if I divide by the modulus of the foam, that's es times the relative density squared.
And then we just had alpha for the foam was the same as alpha s. So then, I can rearrange this slightly and say it's equal to 0.2 over the relative density to the 1/2 power.
So I'm canceling out these relative densities here.
And then I can combine all the solid properties together.
And I'm going to say that nu for the solid is about equal to the same as nu for the foam.
So what I can do here is I can group all the solid properties together.
And this just is delta t critical for the solid, right?
So this is saying that the critical temperature range before you get cracking in the foam is equal to the range for the solid, but multiplied by this factor of 0.2 and divided by the square root of the relative density.
So if the square of-- the relative density is going to less than 1.
So this number here is going to be bigger than 1.
So it's saying that the temperature range that will give you spalling in the foam is going to be bigger than the temperature range in the solid.
So the foam's going to be better than the solid, OK?
And that uses our little models from before.
So I think I'm going to stop there, probably cause my throat is starting to get too sore.
There's a little case study in the notes.
And I'll just put that on the notes on the Stellar site.
It's like one page and it's really straightforward.
You can just read that, OK?
So this is the end of the bit on thermal conductivity.
That's just this one lecture.
And this is really the end of the whole section on modeling of the honey combs and the foams.
So that's kind of the first half of the term is modeling the honey combs and the foams.
And the second half of the term, we're kind of applying those models to different situations.
So next week, we'll have the review on Monday, have a test on Wednesday, week after that is Spring break.
I can't believe we're at Spring break already.
And then after that we'll start we'll do the trabecular bone for a week.
We'll do tissue engineering scaffolds and cell mechanics for two or three lectures.
We'll look at some other applications to engineering design, look at energy absorption and sandwich panels.
And then, I'm going to talk about plants a little bit at the very end, OK?
So we've already covered a lot of the kind of deriving equations part of the course.
The rest of the course is more applying the equations to lots of different situations, OK?
So I'm going to stop there just because my throat is giving out.
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So I was just going to be here to answer questions
Just clarifying, What was the material that we were covering?
In the test?
Yeah
So the test covers everything up to the end of the part on modeling foams, but not the bit on the performance indices, and the material selection charts for foams.
So I think up to the end of the fractured toughness of foams
I see.
OK.
So not covering past [INAUDIBLE]
Not covering thermal properties, no.
It doesn't cover thermal properties.
Here you go.
So you know I got this MacVicar award, and we had a lunch on Friday at the Catalyst restaurant.
And I had to get up and speak.
And just as I got up and spoke, there was a red-tailed hawk swooped by the window.
It was perfect.
It was perfect.
Yeah.
Hi.
So I'm just here to answer questions.
So come on.
Somebody must have questions.
It's all perfectly clear?
You want me to do--
Talk through test one from last year a little bit
You would have to give me test one from last year.
I didn't bring it with me.
I just brought the problem sets
I have it on my computer, and I could read you the problems or just hand you the laptop.
Whichever you prefer
Why don't you hand me the laptop and I'll try to do it.
Is that OK?
OK.
So the question is about the test for 2014.
OK.
So the first question was, describe four processes for making honeycombs, and comment on the type of material usually used for each process.
So I did post the solutions, right?
Did you look at that?
Yeah, I looked at them.
I guess I just feel like I don't fully understand why things are there.
But I can look at it some more
Well, I can go over it, if you want
I'll just look at it some more
No, I can go over it.
So four processes.
Let's see.
So there's the expansion process, where you take sheets and you glue the sheets together, and then you pull them apart.
So you can only really use that process for materials that are going to have large plastic deformations.
So you could use it for metals, some polymers.
But you couldn't really use it for ceramics.
You couldn't use it for glass because as soon as you yanked on it, you'd break the sheets, right?
So--
Is it rigid polymers that you can use it for?
Well, something like nylon you can use it for.
Something that's got a little yield, that will have some sort of yield point.
Not like if you had epoxy, you couldn't use it for epoxy.
So, OK.
So that's one process.
Another process is a corrugation process where you have a wheel that has little gear knobs on it.
And you run your flat sheet through that and it comes out with the half hexagonal profile, and you glue those together.
So again, you need something that's going to yield.
So that would typically be a metal that you would use that with.
Let's see.
Another processor making honeycombs is 3D printing.
You can 3D print honeycombs.
And there's different ways to do it.
One way is by having an ink.
So if you want to print in ink, typically that's some sort of polymer that you're printing.
I suppose physically it's possible to print glass or to print a metal, but you'd have to have some very high temperature setup to do that.
So typically a resin of some sort.
Let's see.
Other ways to make honeycombs.
You can extrude honeycombs.
So the ceramic honeycombs we saw were made by extruding a ceramic slurry.
And typically, you would do that with a ceramic that's a slurry and a powder.
You wouldn't necessarily do that with a metal.
I don't think I've seen any metal honeycombs that are extruded like that.
OK?
Are we good with number one?
Yeah, that makes sense
We now need your password
Oh, sorry.
So is there an actual difference between 3D printing and the extrusion process?
Yeah.
So the extrusion you have a die, and you squeeze the material through the die, right?
So extrusion's kind of like the toothpastey thing.
And 3D printing, you can have an ink or you can do 3D printing where you have, let's say, a powder.
And then you print the binder.
And then you heat it up some way to get the binder to cure.
And then you get rid of the powder that's not bound.
That's another way to do the 3D printing.
OK?
Can you also pour it into a mold?
Yeah.
Yeah, those silicon rubber honeycombs that I showed you, those are all made by pouring a liquid into a mold and then curing it.
Yeah.
Yeah, there's other ways, it just asked for four, so I randomly thought of four.
OK. OK, are we good?
So the next one is a hexagonal titanium alloy honeycomb has h/l is two, theta is 45, and t/l is 0.05.
It says, the end constraint factor for elastic buckling is n equals 0.806.
The titanium has a modulus of 110 gigapascals, and yield strength of 880.
And then you have to calculate some properties.
So would you like me to do that?
Yeah
Yeah, you would?
Does anybody else want that?
I don't see a lot of other people wanting anything else, so I might as well do that.
Let's see.
I would need a piece of chalk.
Here we go.
OK.
So it's a titanium alloy honeycomb, and we're told h/l is 2, theta's 45, and t/l is 0.05.
And we're told that n-- why don't I put the n over here-- n is 0.806.
And we're told the modulus of the solid is 110 gigapascals, and the yield strength of the solid is 880 mega pascals.
OK.
So it says calculate the value of and describe the mechanism of deformation failure for-- and the first part is the Young's modulus in the two direction, e star 2.
OK.
So I don't remember these formulas either, so I need to look at my notes.
And oh, I don't have the formula for e 2 in my notes.
Let me see.
Is it in any of the problems?
I have the formula sheet
You have the formula sheet?
I didn't bring the formula sheet with me.
You have it?
Yeah, I think it's there.
I'm pretty sure it's there.
OK, so this is just like substituting, and there's nothing complicated about this.
So it's equal to Es times t/l cubed times h/l plus sine theta divided by cos cubed theta.
So then you just plug everything in.
So this is 110 gigapascals.
And I'm going to put 110,000 mega pascals, because it's probably going to be less than a gigapascal.
Then t/l is 0.05.
So that's 0.05 cubed.
And then h/l is 2 plus sign of 45 is 0.707.
And then we divide by cos theta cubed, 0.707 cubed.
And then I'm not going to work it out, but that's OK.
I assume that's what's in the solution.
Are we good?
So it's just substituting.
That's all it is.
In the second part-- OK. You need to change the time on this, because it just keeps timing out
Sorry, I don't know how to change it, but I'll try
Oh, is that what this is?
OK, this is the test.
Oh, this is from 2013.
Oh, the next-- if we keep going.
Here we go.
He's got it, so you take your computer
That's probably better
That's perfect for everybody.
OK.
The second part is the plateau stress for loading in the x2 direction.
So that's that.
For loading in the x2 direction, it could either be an elastic buckling collapse stress, or a plastic buckling collapse stress.
So I'm going to calculate both, and then whichever one is lower, that's the one it would be.
So let's see here.
So that's going to be-- I'm missing my formulas again.
Here we are.
So here's the buckling one.
It's n squared pi squared over 24 times t cubed over lh squared times 1 over cos theta.
So you put 0.806 squared in here.
Pi squared 24.
So here you go.
This is t over l cubed times h over l squared.
So that would be 0.05 cubed.
And that would be 1 over 2 squared, and then 1 over 0.707, and then whatever that equals.
OK?
Are we good with that?
And then you'd want to calculate sigma star to plastic.
And that's equal to sigma ys, t over l squared, and 1 over 2 cos squared theta.
All right?
So then sigma ys was 880 mega pascals.
And t over l was our 0.05.
And that's 1 over 2 times 0.707 squared, and then whatever that equals.
OK?
And then whichever one of those would be less is the plateau stress.
So I don't-- yeah?
So I know-- I think I made this mistake in the problem set, but here, because it's titanium, we don't consider it brittle
Right.
Right.
I mean, you could.
But you don't need to
That was something with ceramics?
Yeah, or if it was a glass, or ceramic, or maybe an epoxy.
Something that was brittle.
And besides which, if you look at the question, I only give you a yield strength and a solid modulus.
So to get the brittle thing, I would have to give you a fracture strength
That's true with all the problems
I usually give you what you need.
Especially on the test, I'm going to give you what you need.
You're not going to be looking things up.
OK?
So we're good so far?
OK, let me go back to the question.
So then the third one is the out of plane Young's modulus in the x3 direction.
And that's just going to be Es times the relative density.
And the relative density-- let's see.
So it's Es times rho star over rho s. So that's 110,000 mega pascals.
And then there's also the very first equation on this is the relative density.
So it's t/l times h/l plus 2 divided by 2 cos theta h/l plus sine theta.
So I'm not going to substitute everything in, OK?
So that was it.
It was very plug and chug.
Are you good?
Can I ask a question about a specific question
I'm going to go through the rest of them.
She wants me to do the whole test.
You want me to do the whole test, right?
That would be great, but if other people have other questions it's fine
Why don't I do the whole test.
And then there's going to be time, I think
[INAUDIBLE]
Let me do the test from last unit.
OK, so that's number one.
Or that's one and two.
And then three is, a closed cell elastomeric polyethylene foam has a relative density of 0.05 and a volume fraction of solid in the edges of 0.6.
They give you the Young's modulus of the solid is 0.2 gigapascals.
The pressure within the cell walls is atmospheric, 0.1 mega pascals.
And the Poisson's ratio of the foam is 0.3.
And you're asked to get the Young's modulus of the foam, the compressive plateau stress.
And then there's a question about why does the Young's modulus depend on the solid modulus and relative density, while the Poisson's ratio does not.
So let me go through, then.
So the first one is, what's e star.
And we're told relative density is 0.05.
The volume fraction in the edges is 0.6.
So remember, that was what we called phi.
So phi's 0.6.
The Young's modulus of the solid is 0.2 gigapascals.
The initial pressure within the cells is 0.1 mega pascal.
and Poisson's ratio for the foam is 0.3.
And it's a closed cell.
So if you remember-- oh, let's see.
I think back here, was I supposed to-- I was supposed to say something about the mechanism of deformation and failure for the first one.
So the mechanism of deformation in the modulus in the 2 direction is bending.
The mechanism of failure here is buckling.
The mechanism of failure here is yielding.
And the mechanism of deformation here was axial deformation, OK?
So I forgot to say that.
OK, let me go back here.
So for this one, if it's a closed cell foam, remember there were three terms to the modulus.
There was one from bending of the edges, one from stretching of the faces, and one from the gas contribution if you've got gas inside the cells.
So again, I'm going to have to peek at the equation.
Foams.
Here we go, foams.
OK. And then this gas one.
Get rid of that.
OK, so this one is just the same kind of thing.
It's just plug and chug.
Do I need to put all the numbers in?
No.
I guess I'm a little bit confused why you need the pressure term.
Because you talked about faces bursting
Ah, so if it's the modulus, remember the modulus-- so the question is, why do you need to worry about the pressure because I talked about the faces bursting.
And remember, the stress strain curve looks something like that.
Maybe the slope of the curve is a little bit higher over here.
But this is the modulus down here, right?
The modulus is related to the initial stress strain relationship.
And initially, they're not going to burst.
You'd have to load it up to some amount of stress before the faces burst, right?
So when you're down here, the faces certainly down at the beginning, they're not burst, right?
You have to get some stress before they're going to burst.
And in some materials, when you get up around here, around the plateau stress-- let's just call that sigma star-- then they might burst.
And then the pressure term would disappear and the face term would disappear.
So if I don't tell you to ignore them, if I don't say they're going to burst, or I don't say they're negligible, I would calculate them.
And then if they're small, then you say, well, they're negligible.
OK?
Are we good with that?
You're good?
You're good?
Sardar, you don't need to be here.
But you can stay if you want, but you don't need to be here.
OK, are you good?
Everybody else good?
OK. That was A.
B, what's the compressive plateau stress of the foam.
So here, they want to know what sigma star is.
You're told it's elastomeric.
So if it's elastomeric, it's like a rubber.
It's rubbery.
So if it's rubbery, it's going to buckle.
It's not going to yield.
It's not going to be brittle.
So you can just calculate the elastic stress here.
And if we flip over to our handy dandy list of equations-- blah, blah, blah, blah.
Oh, pooh.
I'm realizing-- yeah, so I don't have the term here for the faces or for the gas, so here we could assume it's going to rupture.
So let's assume it's going to rupture.
So if we assume that it's going to rupture, it would be just like the open celled foams.
And then you can just use that.
And that's been found to work fairly well for the open celled and the closed cell foams
And we assume that faces rupture because--
Well, to be honest, I can't remember if last year I got to this point in the test and somebody said, we don't have the equation, and I gave them the equation with the other terms, or if we just assumed that the faces ruptured.
I can't remember.
To be honest, I think probably for elastomeric foam, you could assume that they'd probably don't rupture unless they're very, very thin
So what kind of foams do they typically rupture in?
So certainly if you had a metal foam, they'd probably rupture.
If you had, say, a polymer foam that was more rigid, like a rigid polyurethane.
So polyurethanes can be flexible, which means they're made out of an elastomer, or they can be rigid.
And the foams that are typically used for insulation, thermal insulation, are typically closed celled polyurethane foams.
And those typically have very thin faces, and they would rupture.
Yeah?
Are you looking for the Young's modulus in this problem?
The Young's modulus was the first part, right?
So part A was the Young's modulus
In B--
In B is the collapse stress, the compressor strength
So I think in my notes, I think it has this.
If the--
Right, if p0 is bigger than-- so what she's showing me in her notes is I had a little note in the class, in the lecture, that if the initial pressure in the cells is greater than atmospheric, then the cell walls are pre-stressed and you have to overcome that in the buckling
Is that atmospheric?
That is atmospheric pressure
So you don't need the [INAUDIBLE]
No.
Yeah.
OK?
OK. Yeah, you don't know that that's atmospheric.
So do you ever do things in PSI?
No, you don't.
Because when I was a student a long time ago, the thing I remember learning was atmospheric pressure is 14.7 PSI, more or less.
And the conversion between mega pascals and PSI is there's more or less 145 PSI to a mega pascal, so the atmospheric pressure is about 0.1 mega pascals.
Yeah?
I don't know if this is a silly question, but for part B, how do we get from the Young's modulus of the foam--
Oh, sorry, sorry, sorry, sorry.
I put the wrong thing down here.
Sorry, my mistake.
OK, now you happy?
Sorry.
OK, shall I move on to the next part?
Another question?
Well, I had a question about number two, but maybe we can come back to that--
OK, let me finish this, and then we'll go back to number two.
So this one here, the part C is why does the Young's modulus foam depend on the solid modulus and the relative density while the Poisson's ratio does not.
So when I write the equation for the Young's modulus, the solid modulus comes into it, the relative density comes into it.
And remember when we had the Poisson's ratio, it's just a constant that depends on the cell geometry.
So here's C, nu star just as a constant.
And that constant just depends on the cell geometry.
So if you think of the Poisson's ratio, it's the ratio of two strains, right?
So say I have my foam here.
So say that's just a block of foam.
Little cells in it here.
Little cells.
And say I press on it this way here.
And let's call this the one direction and the two direction.
If I press it in the two direction, the Poisson's ratio is then just nu would be-- let's see, this would be 2, 1.
It'd be the strain in the one direction over the strain in the two direction.
So it's the ratio of two strains, right?
And if you think of our model for the elastic behavior of the foam, each of those strains is going to be related to some bending deformation in the cell walls or the cell struts.
And this strain here is going to be-- let me make this proportional.
It's going to be proportional to a delta over l. And this one here is going to be proportional to the delta over l. So this might be delta in the one direction, and this will be delta in the two direction.
But those two things are both going to be related to the bending deflection of the beams, right?
And since both of those deltas are related to the bending deflection of the beams, we could write them-- if you want, I could write that as f l cubed over E of the solid times t to the fourth.
And then that's times 1 over l. And then this thing here is also f l cubed over E of the solid t to the fourth 1 over l. So everything cancels out except the geometrical constant.
And if you remember when we did the honeycombs, it looked exactly the same.
When we looked at the Poisson's ratio of the honeycombs, we had the strain in one direction over the strain in another direction.
And each of those strains was related to some component of delta.
There was a delta 1 and a delta 2.
But the delta 1 might be delta sine theta and delta 2 was delta cos theta.
So if the deltas are the same, then it all just cancels out.
And all you're left with is a geometrical constant.
OK?
Do you get physically why that is?
So I have a question about this.
Because we're given most of the equations that we need, is it only in conceptual questions that we should know how we actually derived that version?
I'm not going to ask you to derive that equation for a closed cell foam
I meant this last part
Well yeah, you should be able to explain that.
But I mean just at this level.
Nothing very mathematically involved
OK, cool
OK, are we good?
So that was the end of the test for the undergraduates, OK?
And then for the graduate students, just like the problem sets, I just have an extra question.
And that's what I did this year, too.
So the graduate students have one extra question.
So you and you.
Is anybody else a graduate student?
I think it's just the two of you.
You're post post-graduate.
OK. OK.
So let's see.
So this one says, the performance-- so this is on the performance indices which I told you you didn't need to know for this test, partly because, remember, we missed two lectures.
We're not exactly on the same spot as we were last year.
But I can do it if you want.
Do you want me to do it, or should we do other questions?
What do the grad students think?
Sure.
OK.
So the question is, the performance index to minimize the mass of a beam of a given bending stiffness, length and square cross-section is e to the one half over rho.
So you remember, we derived that e to the one half over rho in class.
In the section on wood, we saw that this performance index for wood is higher than that for the solid cell wall material in wood.
Do you remember that?
The e to the one half over rho for the wood was, I think, rho s over rho star to the one half times Es to the one half over rho s. So explain why wood has a higher value of e to the one half over rho than the solid cell wall material.
And then part B is, suggest a design for an engineering material based on wood that has high values of e to the one half over rho.
So one way to explain it is to say that if you're looking at e to the one half over rho, you can say for wood, E over Es is equal to rho star over rho s for loading in the axial direction.
So this will be for loading actually along the grain.
And that's what we were looking at.
So that's what I'm talking about here.
So I think-- let me just see if this is right.
Yeah.
So this equation here is exactly the same as that equation there, right?
And this is basically saying that this is the performance index for the wood.
This is the performance index for the solid.
And this factor here is bigger than 1, because the solid density is higher than the wood density.
OK?
So really, all you have to do is say that for the wood, the modulus in the longitudinal or the axial direction along with grain varies linearly with the relative density.
And it probably would be a good idea to say that this is a result of the cell walls deforming axially.
So when you take the cells, if you think of the wood cells as being something like that and you're loading it this way on, the cells just actually shorten, and the modulus depends on the-- it just is the volume fraction of solid times the modulus of the solid.
And that's where this comes from.
And once you have this, that basically gives you that.
OK?
Are we good with that?
So that's why it's higher.
Another way to look at it as sort of more of a hand-wavy argument is that if you have a certain amount of solid-- so say you have a certain mass of solid.
If it's solid, it takes up a certain cross-sectional area.
So say that your beam's a certain length, that's going to have a certain cross-sectional area.
And if you have wood, if you have a cellular material, if you have the same mass, you're essentially making the dimensions of that piece bigger.
So you're moving the material further away.
And as you're making it bigger, you're increasing the moment of inertia.
And so you're increasing the bending resistance of it.
That's another, more hand-waving way to talk about it.
And then the second part is to suggest a design for an engineering material based on wood that would have high values of e to the one half over rho.
So remember when we looked at those material performance charts, we said that wood was similar to engineering fiber composites.
But those data for fiber composites are assuming that it's solid, the fiber composite's a solid.
So if you could take fiber composites and make little tubes of fiber composites and assemble the tubes together so that it was like wood, would get something that would be even higher.
So if you could make, say, a fiber composite honeycomb material, and you'd want to have the fibers aligned along the prism axis of the honeycomb, then you would get higher values.
It would be the same-- it'd be this sort of argument again, but now with a fiber composite.
So you'd want-- if this was your fiber composite like this, you'd want the fibers-- well, in wood, they're at a little bit of an angle.
But say they were lined up like that.
You'd want them something like that, then loading it that way on, right?
And if one way to think about those charts is if you-- say we had a plot.
And say this was log of the modulus and that was log of the density.
And I think I'll just draw the envelope.
So foams were somewhere down here, metals were somewhere over here, and with elastomers we're somewhere in here.
I think ceramics were up here.
And then I can't remember exactly where composites were, but composites were around about here.
I'll just say FRC for fiber reinforced composites.
And I think woods were kind of in here.
Something like that.
And then we had our performance index, right?
So remember, there was a performance index, something like that.
And that slope of that was e to the one half over rho.
So every point on that line had the same value of e to the one half over rho.
And essentially, if you had the fiber composite and you made a honeycomb out of it, you would be taking the data from here and shifting them out that way.
You'd be pushing them out over here, so you'd get a higher value of that performance index.
OK?
So that's the test from last year.
That's the end-- yeah?
So for along right here or different it would be cubed
Yeah, so the thing about the honeycombs is--
The opposite
Right
[INAUDIBLE]
It'd be worse, that's true.
So the thing about the honeycombs is they're very stiff in the axial direction, but you pay for that in the other directions.
And it's the same for wood.
So the wood is very good when you load it along the grain, but you pay for it the other way.
But if you think of from the tree's point of view-- if you're a tree.
So here's my little tree.
So here, say we have a tree trunk, and we have some branch.
Branch over here, branches, tree.
So the grain is lined up this way.
And then when there's a branch, the grain turns around and goes that way, right?
So if you think of the tree as a whole, the whole tree blows in the wind like this.
So it's like a column like this, and everything's lined up that way.
And you're loading it this way.
So that is the stiff direction, right?
And if you're a branch, the branches are more loaded by gravity.
So they're loaded that way.
And then because the fibers, the grain turns around, they're also oriented in the good direction.
So from the tree's point of view, it's optimized things.
Then you remember when I talked about the old wooden sailing ships, when they made the old wooden sailing ships, if this was the deck here and that was the haul there, they would get pieces of wood to fit in here that were called the knees.
That was the knee.
And they would try to get a piece that was from a branch like this.
And they would try to match the curve of that joint with the branch with the curve that they needed in here so that the grain followed the pattern of what they needed for the boat.
OK. Other questions?
I'm not quite sure what the difference between tangential versus ray here
Oh, OK.
So in the wood, you mean?
Yes
OK.
So can I rub this stuff off?
We're happy?
Let's see.
Say again?
So tangential and radial.
OK.
So say the wood cells look something like this.
So these would be the fiber cells or the tracheids And then the rays typically are more rectangular cells.
So they might look something like that.
And then they would be some more fibers or tracheids, depending on if it was a soft wood or a hardwood.
So these would be either fibers or tracheids in a hardwood or a soft wood.
And then these would be the ray cells in here.
So they have a different structure.
They just look different.
They're different shape
This is the top?
From the top?
Yeah, this is looking from the top down.
And then if you think of the tree-- so the tree's going to have growth rings, right?
So the growth rings are going to look-- obviously I'm not making perfect circles, but you get the idea-- something like that.
And then the rays go this way.
They go radially.
OK?
So this would be the radial direction, and then those are the rays.
Are we good?
So which way's the tangential?
So the tangential would be this way on, OK?
So if I loaded this way like that, that would be loading it in the tangential direction.
And if I loaded it this way, that would be loading it in the radial direction.
The length of the rays runs in the radial direction.
The length this way on.
So this thing here corresponds to one of these lines I've drawn here.
And then these guys here are the stuff in between here
How do you know what the tangential, the Young's modulus and stiffness is?
So say we were loading it tangentially, we're loading it like that.
Then--
If you have a tree, how do you apply a tangential load on it?
Oh, well it's not the whole tree, right?
So say we have a piece of wood that we cut out like this.
So say I have that.
And if I loaded it this way on, I'd be loading it tangentially.
The tree's big, right?
So I'm not talking about loading the whole tree, I'm talking about taking a piece of wood out of the tree and loading it
OK, so you can't really load tangentially for the entire trunk
No, I'm talking about taking a piece out and loading that piece
How about a ray here?
Do you take the--
So the same thing.
You'd take-- say this was the piece of wood that you were looking at.
Now you would just load it this way on.
OK?
I think-- I brought my thing because I have the slides.
Let me see if I can find-- I think there was a slide that showed this.
OK. That was Furry Fridays.
That was the wood sculptor.
Here we go.
OK, so imagine that that cube is your piece of wood that you're loading, right?
So imagine this is the-- you cut a little piece out.
Then you're loading it tangent.
Can you see, then?
You can load it-- you're not loading the whole tree
Yeah, I was thinking about loading the entire tree and then applying the tangential load on it
Yeah, because then-- I see the problem
It's going to give us shears
Yeah.
Yeah, because I could say, well, if I was trying to load the whole thing.
Say I was loading it from here to there.
Well if you look at it one way, it looks tangential.
If you look at it the other way, it looks radial.
So think of cutting a piece out, because that is what you're going to do.
You're going to cut a piece out.
OK?
All right.
Are there other questions?
Yes?
With that formula sheet, do you only give that formula for the honeycombs, or also for the foams?
I'm going to give you-- if you look at, I think, problem set 2, I gave you a sheet that had three pages of equations.
And it looked exactly like this.
So there was one saying, properties of two dimensional cellular solids-- honeycombs.
There was a whole thing of in plane properties and out of plane properties.
That was one page.
The next page was properties of regular hexagonal honeycombs.
And then the next page was properties of three dimensional cellular solids foams
OK, excellent.
Thank you
OK?
Like I just said, I think there's maybe one or two equations missing from this.
But if it was something you needed, I would give it to you.
I would give it to you.
OK?
So I should have scrolled down for it?
What?
You should have scrolled?
So you're like me.
This happens to me all the time.
I have some website.
I'm looking at it.
I'm like, OK.
I got it.
I think I've got everything.
And then I realized I'm supposed to-- I missed something because I was supposed to scroll down
Problem set two was only the honeycombs.
That's why
Well, I think that was probably all we covered was the honeycombs on that problem set
So that's why I only got that part
OK. All right, yeah.
So I'm going to give you, this will be attached to the test.
OK?
So I think on the test that I posted it was attached, wasn't it?
Yeah.
[INAUDIBLE], did you have a question?
For 2 part D, I don't think we went over that.
I was confused as to how-- is it just that equilateral triangular cells always have-- is always truss behavior?
Let's see
Oh, sorry.
It's the 2014 test
2014-- oh, sorry.
I missed a part.
Sorry.
Yeah, so it says, the same titanium alloy is used to make a honeycomb with equilateral triangular cells.
And what is the in plane Young's modulus for loading in the x 2 direction of the triangular honeycomb?
So this is-- say you have cells that look like that now.
And that's x 1.
That's x 2.
OK.
So the Young's modulus for this-- so I happen to remember the formula.
So I guess I'm thinking you might have put this on your cheat sheets.
It's 1.15 times Es times that, times the relative density.
So I'm trying to remember.
Do I have that on here?
I don't have it on here
Are we just supposed to know that?
Are we just supposed to--
Well, I guess what I would hope that you would know is maybe not the constant, but that it should go as Es and linearly with the relative density.
Because it's a truss and because it deforms axially.
I don't really expect that you would remember the 1.15.
Yeah?
So does that mean for all the foams, of which there were like 10 constants, we don't need to write all down, like what C1 equals--
I think that's what this thing gives you.
Let's see.
It gives you all the Cs.
All right, then I'll make sure I give you the Cs.
I'll make sure I give you the Cs.
But who's got a pen so I can write that down to make sure that I do that?
Does somebody got a piece of paper.
Or I could write it down here.
Oh, here we go.
I can write it down this little sticky thing here.
OK.
So I'll stick that on there so I remember to do that
Will you also be writing what Cs?
Because you have, in your equation, C1, C2--
Well, I think I would say-- say I asked you for, I don't know, the yield stress for a foam in compression that yields plastically.
I would say, the constant for that is 0.3.
I wouldn't give you C2.
I wouldn't do it by numbers, I would tell you what the number was for the thing you needed, because I mean the way the numbers are, the only reason they're numbered is because that's the number they are in the book.
They're just ordered sequentially in the book.
But I don't expect you to remember which-- it's C6, or C5 or something.
So anyone else?
Can you explain the difference between uniaxial yield and plastic buckling?
Oh, OK.
So if you have something and it fails by uniaxial yield-- so say you have a honeycomb like this, and you're loading it this way on, OK?
So if you're loading it that way on, these walls of the honeycomb are just axially deforming, initially.
Right?
So the elastic behaviors, they just axially deform.
So it works out that if these cell walls are very thick, then you can reach a yield stress before any buckling occurs.
And then the strength would just be that yield stress of the cell wall material times the relative density.
OK?
But the cell walls have to be thick for that to happen.
So then imagine that the cell walls aren't thick.
Imagine that the cell walls are thin.
So say I have the same honeycomb like this.
If the walls are thin, and say the solid material itself-- so this is for the solid-- it has some stress strain curve.
And it may have a linear elastic part, and then a yield thing like that.
So say this is the yield strength here.
Say we compress that this way on the same thing in the three direction.
Then if a material's got a yield point, there can be an interaction between plastic yielding and elastic buckling.
And you can get plastic buckling.
And the plastic buckling, you're going to get the wrinkles that go along the length of it this way.
remember I showed you that tube that kind of collapsed and folded up kind of thing?
That's plastic buckling.
OK?
And typically, people use what's called the tangential modulus to calculate the buckling stress for plastic buckling.
And the tangential modulus would be something related to the tangent over there.
I don't expect you to be able to derive plastic buckling equations.
But the plastic buckling-- you know what elastic buckling is, right?
Yeah.
Yeah.
So one way to think about plastic buckling is, if you have-- and I'm trying to remember.
This is called the slenderness ratio.
And I'm trying to remember, is that l over r?
Imagine you had just a circular cross section and you had a length, l. So you have a column here, and it's got a length, l, and it's got a radius, r. Like that, OK?
So the longer it gets, the more slender it is, the higher the slenderness ratio is.
And this, I think, is some sort of stress.
If the slenderness ratio of just a single column is short-- if it's stubby, if you had a column that looked kind of like that, It's not going to buckle.
It's going to yield.
And so if you compress that, it would just yield.
And it's just going to yield at the yield stress, right?
It's just going to yield at sigma y of the solid, whatever the solid is.
If you have a long column, it would buckle elastically by an Euler buckling.
And Euler buckling-- let's see.
I'm going to run out of room here.
If you think of it in terms of a stress instead of a load, it's going to be in squared pi squared Es i.
Let's say i goes as r to the fourth.
And this is going to be l squared r squared.
Right?
This is going to be a pi in here.
There's going to be a pie in here.
I might have lost a factor of 4, but it's going to be-- let me make this proportional, OK?
So the slenderness ratio, there's going to be an l over r squared term here.
So I could cancel out the four there and put a squared.
So sigma Euler is going to go as Es times r over l squared.
Like that.
And so this is the Euler buckling stress here, OK?
So this would be elastic.
And right here at this little corner, it turns out life isn't quite that mathematically exact.
If you're near that corner, it's not like here it's buckling elastically, and here, it's buckling plastically.
What happens is, if you looked at data, data might do something like that.
So the sum interaction between the elastic and the plastic.
And that's kind of what's going on with this thing here.
Does that makes sense?
Plastic buckling can-- OK, so if you unload plastic buckling, you get some of the elastic part back?
You're not going to get much back
You're not?
No, because once-- to get the plastic buckling, you're very close to this.
By the time you get that deformation, you've got locally, it's yielded.
It's not all elastic everywhere.
It's going to yield in places.
And once it starts yielding, it's-- if you think of these buckles forming, it's not like you're at one spot on this curve throughout the whole thing.
Some of it's more deformed, and some of it's less deformed.
Let me pull up those plastically buckled columns, those tubes.
Get rid of that one.
Let me try and remind myself where they were.
I think-- honeycombs, I want honeycombs.
Out of plane, that's what I want.
It was this thing here.
So you see when you have-- that's just one tube, but the whole honeycomb would-- imagine that you have groups of tubes put together.
They would have to fail in some compatible way.
But the deformation and the stresses are not going to be uniform through this whole thing, right?
One part of it's going to be at one stress, and something else is going to be at another stress.
So parts of it are going to yield plastically, and you're not going to recover that.
OK?
So in fact, they use these sorts of things for energy absorption devices, like in cars and things like that.
To absorb the energy from the impact.
More questions?
Let him have a turn
Sorry, I have a question about I think 2013
2013,
The last question.
It's about the plastic
2013.
Let me rub some of this stuff off.
OK.
Here we go.
OK. Oh, we haven't covered this at all.
So the last question, this one here?
Right
Yeah, so this question's on energy absorption.
We haven't got there yet
How about--
And the third question's on sandwich structure.
So when I taught the course in 2013, I did the topics in a different order.
So I did honeycombs, and I did foams.
And then I think I did sandwich panels, and I did energy absorption.
And I left the stuff on the wood and the cork to the end.
So we haven't done that.
So don't panic if you haven't-- if you can't do that.
OK?
You should have known.
Come on, you should have known that if it talked about things we haven't covered yet, I'm not going to give it on the test.
OK, what else?
Can you explain more plastic hinges?
Plastic hinge, OK.
So let's just say we have a beam in bending, OK?
And say it just has a load p in the middle, all right?
Are we good?
So this load in the middle, then this reaction is p over 2, and that reaction is p over 2.
And if I drew the sheer force diagram, it'd go p over 2 up, we go over, go p over 2 down, like 2 p over 2 down.
Over and back up.
OK?
And then if I drew the bending moment diagram, it would go up and down like that.
And that would be zero.
And that would be zero.
And this would be PL over 4.
OK?
Are we OK with that?
We haven't got to the end of the answer, but--
I have a question.
We're not expected to--
No, no, you don't need to do this.
I'm just trying to explain it now.
You don't need to retain that information, for heaven's sakes.
No.
Come on, I'm so disappointed
For the moment, is it positive for the counterclockwise turns?
Oh, so you remember for the beam bending, there's a different convention.
It's positive if it's tension on the bottom.
Are you in mechanical engineering?
I though you were in mechanical engineering
No, I take physics
Oh, you do physics.
All I really want to say is the moment's maximum in the middle, OK?
So let me just say the moment's maximum in the middle, OK?
So then let's look at the cross section.
So say I look at a cross section here.
Let's just make it rectangular to make it easy for me to draw.
So it has width, b, and a height, h. OK?
So this would be h on this picture over here.
And remember, the neutral axis goes through the middle on the cross-section here.
And so one half of the beam is in tension, and the other half of the beam is in compression.
So for this situation here, this half of the bean is going to see compression, and that half of the beam is going to see tension, OK?
Are we happy with that?
We're happy with that.
OK. Now let me draw the stress distribution.
So if it's linear elastic and it hasn't yielded yet, the stress distribution is going to look like this.
So this is h again.
That's the height.
And now b is into the board.
And I'm plotting the stress this way.
So this thing here is my neutral axis.
It has no stress.
Remember, there was one plane that has no stress, and for a rectangular cross-section, it goes through the middle of the cross-section.
It goes through the centroid.
Is this ringing a bell?
I'm hoping this is ringing a bell.
Come on.
I know we did this in 302, too.
I know we did.
OK. OK, so this is all linear elastic, right?
So at some point-- so I'm going to get to the plastic hinge.
At some point, If you keep loading it and p gets bigger and bigger, the moment gets bigger and bigger, the stress gets bigger and bigger remember, the equation here for the stress is equal to My over i.
This moment, the maximum moment's going to be this moment here.
The maximum y is going to be h over 2, the distance from the neutral axis.
And i is going to be bh cubed over 12 for the rectangular section.
So if I keep loading it up, at some point, the maximum stress is going to equal the yield stress.
right?
And in our cellular things, is going to equal the yield stress of the solid.
So our beam is one of our edges in the foam, or struts in the honeycomb.
So at some point, is going to equal the yield strength of the solid.
So let me draw the stress distribution again, where we start to have plasticity.
So here, the stress is equal the yield stress.
And it's going to equal the yield stress at the bottom, too, because it's all symmetric, right?
And the neutral axis is still going to be in the middle here.
So it's going to 0 down there.
So initially, when it's just barely reached the yield stress at the outer part of the beam, then this stress distribution would still be linear in between.
But once you start to load it more than that, then the plastic region starts to seep in from the outside inwards.
And what we do here is we assume that the solid is elastic perfectly plastic.
And if you remember, when we said things were perfectly plastic, or if they were elastic perfectly plastic, they look like that.
The stress strain curve for the solid, I'm assuming, looks like that.
So I'm assuming the yield stress in the solid is just a constant.
That if I strain it more, there's no work hardening.
I'm neglecting work hardening
You said inward that the--
In board?
Inward, you said something like--
Inward.
So this is-- let's see.
I didn't bring a bean with me today.
No beam.
Do we have anything beam-like?
Ah, here we have a beam-like thing.
OK.
So say this is my beam, and I'm loading it this way on, OK?
And this is b, and that's h. So this picture here is looking at it that way on, OK?
And this picture here, I've drawn h, but now I'm just looking at the stress distribution across h. And b is into the board.
Is that OK?
Does that answer your question?
No, I mean for the plastic
For this part?
OK.
I'm working up.
I haven't finished it yet.
So this is the same kind of view as here.
I drew it a little bigger.
It shouldn't have draw bigger, I should have drawn it the same height.
But it's the same thing.
OK?
So you'll buy that at some point, we reach the yield strength here.
And if I keep loading it up and I assume that the solid is perfectly plastic, that there's no work hardening, then the stress distribution would look like this.
OK?
And then if it yields more, then it's going to look like that.
And if it yields more, eventually I'm going to get to the stage here, where it's-- let me redraw this.
That would be-- you get the idea, OK?
This will go over here, and down here, and like that.
OK?
OK.
So are we happy with this stress distribution across the cross-section?
Yeah, OK.
So that's when it forms the plastic hinge.
So when it forms a plastic hinge, the stress distribution looks like this.
So these are supposed to be the same size.
They're not quite.
Let's see.
So one of the things I talked about was the plastic moment that kind of characterized that plastic hinge.
And the plastic moment is just the internal amount of moment that the beam can withstand when it's yielded completely across the whole cross-section.
So when we're at this point here.
So you calculate that by saying, that stress there is equivalent to a force.
That stress there is equivalent to a force.
And you get the moment by multiplying those forces times that distance there.
OK?
Because you think of those two forces as being a couple, and the moments the force times the distance between them.
So the plastic moment was sigma ys.
And say we're talking about our honeycomb or foam or something.
That was our cell wall thickness t. So let me call it t instead of h for the foams and the honeycombs.
So this force here is going to be the stress time the area over which it acts.
And let's say we look at it for a honeycomb.
Then I've got the stress is acting over this distance here, and then times the depth into the page, right?
And if it's the honeycomb, that depth into the page is just b, OK?
And then this moment arm here is just t over 2 as well, because that's t over 4, and that's t over 4.
So it's t over 2 again.
So it's sigma ys bt squared over 4 for the honeycomb.
And say we have an open cell foam.
The edges aren't of thickness b, they're of thickness t. So then it's m p is just sigma ys t cubed over 4.
Are we happy?
And really, physically what that is is it means that the beam can't hold any more force.
You can't apply any more force to it.
It's just going to rotate like this once you've gotten to that plastic moment.
That's why it's called a hinge, because it just can rotate like hinge rotates.
Like a door hinge, OK?
How does it rotate?
Well, where's my original picture.
So if this was the beam, when you form the plastic hinge, your beam would just look like that.
And this would be your hinge point.
I'm a civil engineer originally.
We try to avoid this.
So that's why, in the foam and in the honeycomb, that's when it fails is when you get that plastic hinge forming.
OK?
All right.
We have a few more minutes
That kind of looks like plastic buckling
Well yeah, it's not buckling, but it's plastic, yeah.
It's permanent.
Anyone else?
OK. No other questions?
Should we call it a day?
Is that helpful?
All right, then.
It's what I do.
Come on.
It's what I do.
All right.
So I'll see you Wednesday.
And my plan is to grade the tests before spring break, so I shall have it back to you.
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So we're going to start talking about trabecular bone, and we're going to do a bone today and on Wednesday.
And I'm hoping we can more or less finish it on Wednesday.
So hello.
Hold on a sec.
So these are some images of trabecular bone, and you can see that it has a foam-like structure.
And trabecular bone exists in certain places in the body.
There's three main places that it exists.
So it exists at the end of the long bones.
And over here this is a femur.
This is the very top of the femur, and you can see this is all trabecular bone in here.
This is a tibia, and this is the top of your knee there.
And you can see how the bone get more bulbous at the ends, and it's filled with a trabecular bone.
This is a vertebrae.
So vertebrae are actually mostly trabecular bone, and they have a really thin shell of what's called cortical bone, the dense bone, on top of it.
So trabecular bone exists at the ends of the long bones, it exists in the core of the vertebrae, and it also exists in sort of shell or plate-like bones.
So in your skull, for example, there's a layer of trabecular bone in between two layers of the compact dense bone.
And in your pelvis, it's the same thing.
So those of you who took 3032, you remember when I passed around the bird skulls, there was that very porous kind of trabecular bone.
So trabecular bone is of interest medically in three main kind of medical situations.
So the first one is osteoporosis.
So I want to talk a little bit about osteoporosis now, and then we'll talk about it in more detail later on.
I guess, we'll probably start today.
Another medical issue is osteoarthritis, and the properties of the trabecular bone are important in arthritis, and the third issue is in joint replacements.
And so we're going to talk a little bit about osteoporosis, osteoarthritis, and then joint replacements.
And then I'll talk a little bit more about modeling bone like a foam and how it deforms and how it fails.
And then we'll talk a little bit how we can model osteoporosis.
So let me write down some of these things.
I guess we'll start here.
So trabecular bone has a foam-like structure, and what we're going to see is that we can use the models for foams to describe the mechanical behavior of the bone.
It exists at the ends of the long bones.
And at the ends of the long bones, the bones become more bulbous.
And really what that's for is to increase the surface area so that there's cartilage between the ends of the two bones.
So there would be a bone here and a bone there, and there's cartilage in between them that sort of lubricates that joint and makes low friction at the joint.
And the bone gets larger to decrease the stresses on the cartilage.
So if you have the same force and you have a larger area, you're going to have a smaller stress.
So that's why the bone gets bulbous like that.
And then by having the trabecular bone, because it's so porous and lightweight, you're not having a big, dense hunk of bone at the end of the long bones.
So it exists at the ends of the long bones.
And I'll just say the ends have a larger area than the shafts.
And that's to distribute the loads on the cartilage or to reduce the stress on the cartilage.
And then the trabecular bone reduces the weight.
So it also exists in the core of the vertebrae, and in fact, it makes up most of the vertebrae and then in things like the skull and the pelvic bones in shell and plate-like bones.
And so it's the core of a sandwich structure there.
So it's of interest in osteoporosis, in osteoarthritis, and in joint replacements.
So if we start by thinking about osteoporosis, you probably know that osteoporosis is a disease where the bone mass becomes reduced and there's a greater risk of fracture, so there's especially a greater risk of hip fractures and vertebral fractures.
And it turns out in both of those sites, if you just look at these bones here, typically if you have a hip fracture, what happens is the neck of the femur breaks.
So this is called the neck here.
This is called the head, this spherical bit there.
So the neck has a fracture, and you can see most of the bone there is trabecular bone, so it's really carrying most of the load and the same with the vertebrae.
This sort of cylindrical part of the vertebrae here carries most of the load.
It has a shell of really thin cortical bone, but it's mostly trabecular bone.
And when the loads are vertical like this, that's really the trabecular bone that's carrying most of the load.
And people get sometimes what are called wedge fractures where instead of having a sort of a cylinder with parallel faces like this, the trabecular bone fails, and the bone ends up like that so that there's-- yeah, I know.
You make that wincing expression.
It's like ouchy.
And in fact, it's very ouchy for people who get that.
And when you see little old ladies who are all hunched over, that's why.
The bone has actually failed.
It's actually been crushed into these wedge fractures, and there's no way they can straighten it out, and it's quite painful.
So people who look at osteoporosis are quite interested in the mechanical properties of trabecular bone for this sort of reason.
The hip fractures are particularly serious because people become immobilized and then sometimes because they're immobilized, they get pneumonia, and in elderly people they sometimes die.
So something like 40% of elderly patients who are over 65 die within a year of having hip fracture.
So it's not that the hip fracture kills them, it's that they become so immobile, and they can't move, and they can't walk around, and they end up getting pneumonia.
So it's quite a serious thing.
And there's something like 300,000 hip fractures a year in the US, and the cost of treating these hip fractures is something like $19 billion.
So yeah, it's a huge problem.
And as the population is aging, as baby boomers like me get older and older, there's going more people having hip fractures.
So it's a huge deal, osteoporosis.
So we'll say bone mass decreases with age, and osteoporosis is extreme bone loss.
And a little later today I'll show you some pictures of what it looks like when people have osteoporosis.
So the most common fractures are of the hip and the vertebrae, and at both sites, most of the load is carried by the trabecular bone.
And the hip fractures you are the most serious.
40% of elderly patients pass away within a year.
So that's sort of a little introduction to osteoporosis.
The next issue that people are interested in is osteoarthritis.
And in osteoarthritis, there's a degradation of cartilage at the joints, and the stress on the cartilage is affected by the modulus of the bone that presses against the cartilage.
You can kind of magic if you have a fiber compass, for instance, most of the stresses, if you're loading it along the fibers, is carried by the fibers because they're stiffer.
So if you have like say trabecular bone that has varying density, the denser bits are going to have higher moduli, and it's there's going more stress associated with that.
And so the modulus of the trabecular bone can affect how the loads are distributed in the cartilage, and that can affect the damage in the cartilage.
And the shell, as I mentioned before, this sort of shell of cortical bone or the dense bone at the joints can be quite thin.
It can be less than a millimeter.
So I brought my little bones along with me again.
So this is the head of a femur here, and this is a piece of a knee joint here from a tibia.
And you can see just looking at these how thin the cortical shell is.
So you can get an idea of how thin that is.
So osteoarthritis involves a degradation of the cartilage at the joints.
And the stress on the cartilage is affected by the moduli of the underlying bone, and the cortical shell, the totally dense bone, can be quite thin.
So the mechanical properties of the trabecular bone can affect the stress distribution on the cartilage.
And if osteoarthritis gets particularly bad, then sometimes people have joint replacements.
So when it gets really bad, the cartilage is degraded completely, and the bone is rubbing on bone, and that's quite painful.
And when it gets to that point, people generally have a joint replacement.
And so the way the joint replacements are done is say somebody who is going to have a hip replacement, what they do is they chop off the top of the femur.
So they would chop the femur off somewhere around here, and then they have a metal implant that has a spherical ball.
That's like the head of the femur.
And then it has a sort of stem and a shaft here that goes into the hollow part of the long part of the shaft of the femur.
And so they use a number of different metals for this, titanium and stainless steel, and there's a cobalt-chromium alloy are also used.
So you need metals that are biocompatible, aren't going to corrode, aren't going to have degradation products.
And then the bone grows around that implant, and the bone grows in response to mechanical loads.
So the density of the bone depends on the magnitude of the load, and the orientation of the trabeculae depends on the orientation of the principle stresses that are applied.
So let me write that down.
So they cut off the end of the bone, and they insert the implant into the hollow shaft of the remaining bone.
And the metals they use are titanium, stainless steel, and a chromium-cobalt alloy.
And then the bone grows into that implant.
And the bone grows in response to mechanical loads.
So the density of the bone depends on the magnitude of the stresses, and the orientation of the bone depends on the principle stresses.
So one of the issues that comes up in joint replacements is that there's a mismatch in the moduli between the metal and the bone.
So if you think the metal, like something like stainless steel, has a modulus of around 200, 210 gigapascals.
And the cortical bone has a modulus of about 18 gigapascals, and the trabecular bone has a modulus between about 0.01 and 2 gigapascals, depending on its density.
So you're taking the bone out, and you're replacing it with something that's much, much stiffer, and that changes the stress distribution around the remaining bone.
And one of the things that can happen is you can get a loosening of the implant.
So the bone can grow in initially, but over time, you get a different stress field in the bone.
And if you have a different stress field, then the bone can resorb away from the implant and cause what's called loosening.
So if the implant becomes loose, that's clearly not a good thing.
It's a bad thing.
And often orthopedic surgeons don't like to do these joint replacements in young people partly because they don't always loosen, but occasionally they do.
And if they loosen they can go back and do a revision.
But you can kind of imagine after they've chopped the head of the femur off and they put one implant in, it's not that easy to go back in and replace that with another one.
You would need one with a longer stem, and the whole thing becomes a little bit more complicated.
So this issue of stress shielding is what it's called when you have something much stiffer that's shielding the stresses in the bone.
The issue of stress shielding means that they don't like to do the replacements on younger patients unless you can get stress shielding.
And if we just compare-- if we look at the cobalt and chromium alloy, the modulus of that in gigapascals is about 210.
If we look at the titanium alloys that are used, the modulus is about 110.
If we look at the stainless steel-- it's 316 stainless steel-- it has a modulus of around 210.
And then if we look at the bone, the cortical bone has a modulus of about 18, and the trabecular bone has a modulus 0.01 to 2 gigapascals depending on the density.
So after the joint replacement happens, the remodeling of the bone is affected.
So the idea is that the stiffer metal carries more of the load, and then the bone carries less load, and then it resorbs.
And that can lead to this thing called loosening, which is not desirable.
Now, this typically doesn't happen till about 15 years after you've had the implant, so it's not something that would happen right away, but it can happen later on.
So these are all sort of medical reasons why people are interested in trabecular bone because of osteoporosis, osteoarthritis, and joint replacements.
So I wanted to start by talking about the structure of trabecular bone.
And then we'll talk about what the stress-strain curves look like in compression and tension, what are the mechanisms of deformation and failure, and how we can apply our models for foams to the trabecular bone.
So the idea is that the structure of the bone resembles a foam, and here's some SCM images of trabecular bone.
And you can see that the bone has a varying structure.
If it's relatively low density, this is a bone that's almost like an open-cell foam if I didn't tell you that was a bone, you might actually think it was an open-cell foam.
And here's a denser piece of bone, and you can see there's still interconnections between all the openings, so it's not exactly like a closed-cell foam, but it's much denser, and it's almost like there's perforated plates in the structure.
And then as I said the bone can grow in response to loads.
So if you have loads that are more or less vertical, the trabeculae tend to line up and be more or less vertical with some sort of horizontal bracing.
So this is a piece of bone from a knee, the condyle is sort of towards the top of the knee.
And you can see these are sort of plate-like pieces of bone.
They're almost parallel, and not too surprisingly in your knee, the loads are typically vertical, and then there's a little bracing bits that go horizontally here.
So you can get different structures depending on the loading on the bone, and the density of the bone corresponds to the magnitude of the load, and the orientation of the trabeculae corresponds to the orientation of the load
[INAUDIBLE]
Resorb.
So when the bone density goes down, when you lose bone mass, that's called resorption.
So the idea is that the trabecular bone resembles a foam.
And in fact, the word trabecular comes from Latin, and in Latin, it means little beam.
So the foams to form by bending.
They act like little beams, and so the trabeculae are like little beams, even in Latin.
There's a range of relative densities, and you can see in that image up there, you can see that there's a range.
And they range typically between about a 5% in dense and 50% in dense.
So something like 0.1 or 0.2 might be typical.
And the low-density bone resembles an open-cell foam.
And the higher density, it becomes more like perforated plates.
And the structure can be highly anisotropic depending on the stress field.
And then I've got another image here of the trabecular bone.
These images are using what's called micro computed tomography.
So you've probably heard of computed tomography.
Say somebody has cancer, they get put in a CT machine, and they do a scan.
The micro CT is more of a research tool.
It's the same kind of technology, but it's got a much finer resolution, and typically, you put a small specimen into a machine to do this.
So the specimen might be half an inch in diameter and an inch tall, something like that.
So these are done by a colleague, Ralph Muller, who's in Zurich, and this is one of his bread and butter things that he has these images, and he looks at osteoporosis.
And you can see here the difference in the structure for the different densities.
So here's a 26% dense piece of bone in the femoral head.
It looks pretty sturdy and substantial.
Here's an 11% dense piece from the lumbar spine, and here's a 6% dense piece.
And you can kind of see when you go from 26 to 11, the struts get a little bit thinner.
And when you go from 11 to 6, the struts get very thin, and in fact, if they get too thin, the struts resorb altogether, and some of their struts can just disappear.
So when people get osteoporosis, what happens is they first lose bone mass by thinning the struts, but then at some point, the struts just resorb altogether.
And if you think of the struts as a biological material, they have bone cells in them.
So there's little osteoclasts and osteoblasts and osteocytes that live in the bone, the mineral thing, the bony thing.
And those cells have dimensions of 10s of microns, so maybe 20, 30 microns, something like that.
So the struts can't get any thinner than that.
If they get thinner than that, then the cells can't live, and the thing just disappears altogether.
And you can think of from a mechanical point of view, if you lose density by thinning the struts, you can use our sort of foam equations.
And say the density went from 0.2 to 0.1, you could make some estimate of how the modulus and how the strength would vary depending on our foam models.
But if you lose density by resorbing the struts, the struts just disappear altogether, then it's as if you had a steel scaffold or a steel structure of a building.
And now you're starting to remove columns and remove beams.
Yes, I know.
That's not good, not good.
And so we'll talk a little bit more about that when we talk more about osteoporosis, and you can see the consequences of that.
But this image here kind of gives you a little bit of a picture of what the bone structure looks like as it gets less dense.
So I want to talk a little bit more about the bone growing in response to load.
Let me rub off the board.
So you're probably already a little bit familiar with this idea.
So when astronauts go up into space, they often do exercises where they have a treadmill, and they've got springs, and they're pulling on the springs to try to exercise themselves.
And the reason they do that is when they're in microgravity, if they were doing some kind of exercise, they would lose bone mass.
And they will get back to Earth where we have Earth gravity, and they would have a problem.
So you see it in astronauts, in microgravity.
The other place you see this just in everyday life is in professional tennis players.
People have done like x-rays of the bones of professional tennis players, and obviously, they have one arm that they hit the ball with their racquet.
The bones in that arm actually get bigger because they're loading that bone over and over again pretty much every day when they're playing tennis, and they're not loading the other arm.
So their two arms are not symmetrical because of this loading from hitting the racquet over and over.
And the people in 3032 have already seen this, but I couldn't resist bringing up the Guinea fowl experiments again.
So obviously, you can only do x-rays on human.
You can't sacrifice the humans and look at their bones, but you can with Guinea fowl.
And so people have done experiments where they run Guinea fowl on treadmills, and they have one set of Guinea fowl that they run on the treadmill that's horizontal.
They have another set of Guinea fowl that they run on a treadmill that's inclined to 20 degrees, so one would think they might have more stress on their bones from that, and then they have a control group that they don't run on the treadmill at all.
And then what they do is they have a forced plate on the treadmill so as the Guinea fowl is running, they measure the maximum force in they're taking high-speed video.
And then they measure the angle of the knee at that point at which the force is maximum.
And they can see there's a change in the angle of the knee when they put them on the inclined treadmill, not too surprisingly.
And then these are juvenile Guinea fowl that haven't completely matured their bones.
And then after about six weeks of this, they sacrifice the Guinea fowl, and they do scans on the bone, and they look at the orientation of the bone, and they measure what's called the orientation of the peak trabecular density, which is a way of characterizing the orientation of the bone.
And they find that the angle of the knee when the Guinea fowl are running changes by about 14 degrees.
And it turns out the angle of the bone, the orientation of the bone also changes by about 14 degrees.
So the bone has remodeled to match that change in the forces that are applied as the Guinea fowl are running on a treadmill.
So this is all a demonstration just to show that bone grows in response to load.
So let me write down some of this stuff.
So I will say astronauts-- did you see Michael Collins is going to come to the talk at MIT?
When I was a kid in '60s, he was one of the Apollo astronauts.
He was like one of the first NASA astronauts.
Anyway astronauts, so in microgravity, they would lose bone if they don't exercise.
And tennis players, the bones get larger in the arm that they hold the racket with.
And then I'll just write a little bit of notes about the Guinea fowl experiments.
So this was done by-- it's in a paper, Ponzer et al 2006.
So they have one set of Guinea fowl that run on a level treadmill, they have another set that run on a inclined treadmill, and it's inclined at 20 degrees.
And then they have a control group that doesn't run on the treadmill.
And then they measure the angle at the knee at the moment of peak force on the treadmill.
And after six weeks, they sacrificed the Guinea fowl, and they measured the orientation of the peak trabecular density.
And they find that the knee flexion angle changed by 13.7 degrees.
And if you compared the inclined versus the level treadmill, and they found the orientation of the peak trabecular density, which they called OPDD, also changed by 13.6 degrees.
So the idea is that the orientation of the trabeculae changed to match the orientation of the loading.
Then I have a little video here.
Do you like video?
So I have a colleague who's at Harvard who studies animal locomotion, and they didn't do this set of experiments, but they do do experiments on Guinea fowl running on treadmills, and thought you might find this amusing.
So let me see if I can make this work.
[VIDEO PLAYBACK] -Sometimes you walk into a lab and you just think this is what science is all about.
-I just put the Guinea fowl on the treadmill, and this is something that we commonly do.
-Welcome to the Concord Field Station, a defunct Nike missile base turned scientific menagerie.
It's owned by Harvard, and biologist Andy Biewener is the director here.
So think of it as-- -A research lab facility for doing comparative biomechanics and physiology of largely animal movement.
-And the birds are just the tip of the iceberg.
-So do you want to see the baby goat and the emu?
-Obviously.
-OK. We keep it because it's sort of like a mascot.
There used to be a lizard colony.
You can hear the African greys.
Then the jerboas are housed in this room here.
This is where we originally did our pigeon flight studies.
So usually the ones with claws and sharp teeth and aggressive behaviors, you want to watch out for them.
-As you might expect.
But did you know that Guinea fowl-- -They're really lovely to work with.
-Sometimes.
Or that-- -Rats are not very good on treadmills.
-Yes.
That's what a rat treadmill looks like.
And this?
-And this was historically a treadmill of note, the treadmill that they first taught kangaroos on and showed that kangaroos stored energy in their tendons enough that they don't actually increase their metabolic rate when they hop at faster speeds.
-These are the kind of discoveries made here with the use of high-speed video and x-ray machines and semi cooperative animals.
But beyond the basic biology, Biewener says engineers are using this research to build better robots, and it can help improve medical treatment for people with movement disorders.
Today the big excitement at the lab is happening here.
Ivo Ros is studying how heart rate changes when cockatiels fly at different speeds.
So this is a way to look at how much energy it takes to fly, and that cord is measuring heart rate.
But instead of the birds flying faster, the wind changes speed.
-I'm going to turn it on then.
-It's hard to fly fast, and it's hard to fly slow, Ros says.
So the expectation is that the heart rate should be shaped like a U.
But so far Ros is finding that it's a flat line.
It's like the bird goes into a stress reaction when it takes off.
Is that just because of the wind tunnel?
What Ros wants to know is-- - --whether or not they need to be stressed to fly in the first place.
-That's something that Ros is looking into, but today is mostly about training.
-Keep going.
Come on.
-Imagine you're a cockatiel.
A wind tunnel is kind of a strange experience.
-A bird in a wind tunnel has to confront the fact that the world is not moving past, which defies its normal sensory cues.
-Which pretty well sums up the Concord Field Station generally.
For Science Friday, I'm Flora Lichtman.
[END PLAYBACK] And if read The New York Times, Flora Lichtman used to work for NPR and would make these Science Friday videos for them.
But now she has a gig doing things for The New York Times, and she does science videos still.
I don't know if you quite call them videos, but what they do is they have these little paper puppets, and the paper puppets are animated and re-enact different episodes in science.
And it's kind of amazing how they do these little science videos.
So if you Google Flora Lichtman, you'll see more Science Videos with all sorts of things.
I guess the other interesting anecdote is I went to the Concord Field Station once.
And I had done a study on quills and animals that have quills because the quills have a foamy structure in the middle.
So they're carrot, and they have sort of like carrot-like structures.
And they have an outer shell that's dense, and then they have a foamy thing in the middle.
Anyway I did this paper on quills and how they work mechanically.
And Technology reviewed a little article about it, and they said they wanted to take a picture of me with a hedgehog.
The hedgehogs are little European-- they're like little small, cute things.
And I said, well, if you can find a hedgehog, I'm happy to have my photograph taken.
And they had a hedgehog at the Concord Field Station.
So we went out there, and we had these big leather gloves and took a picture.
And I don't if it was Andy or I don't know who it was, but I said one of the people there, what did you do with a hedgehog?
And he said, well, we tried to do the treadmill study.
But hedgehogs are like porcupines.
When they get scared, they curl up into a little ball.
And so they said they would put the hedgehog down under the treadmill, and they would start it up, and it would make a noise, and it would get scared it.
And it would just go into a little ball and kind of slide along to the end, and then it would kind of get flopped off.
So that was the end of the hedgehog experiments.
But they did have wallabies there the day I went.
So they have all sorts of animals that they put on to treadmills, birds that they fly, so it's kind of interesting to go there.
But the main idea here is that there was this set of experiments with Guinea fowl that showed just how precisely the orientation of the bone matches the orientation of the loads
Was there a difference between the control groups?
Ah, so I have slides.
I have slides.
Hang on a sec.
I got distracted by my video.
Sorry.
So here's the sort of schematic of Guinea fowl on treadmill.
And where's the little doo-da here?
So on the level, the knee flexion angle was whatever this is, 76.3, and here was the 62.6, so the difference is 13.7.
And then this kind of table here summarizes these results.
So this is the maximum trabecular density, and this is the angle.
And here we have the incline.
Let's see, the control was the yellow, and the level was the blue, and they've got the values for that peak trabecular density orientation.
So they've got that for the level.
And then they looked at the difference between the incline and the level in the knee angle.
That's what this thing here is.
And then between the level and the control, there wasn't really any difference in the knee angle because the control ones, they were just walking around.
So that's the sort of slide that has the actual data on it.
All right.
And then I showed you the video.
All right.
So we need to do a couple more things before we get to the modeling.
Let me get a drink.
So if we want to use the models for foams to try to describe the trabecular bone, we need to know something about the properties of the solid.
Remember we used the properties of the solid in the models.
So we want to get the properties of the solid in the trabeculae, and there's a couple of ways you can do this.
To get the moduli, you can use an ultrasonic wave propagation method, and you can measure a modulus that way.
And if they do that, they measure a modulus between about 15 and 18 gigapascals.
Another way to do it is to take a piece of bone, do a compression test on it, measure the modulus.
Before you do the test, you put it in the micro CT machine, and you get a picture of the structure, and then you use that as the input to a finite element analysis.
So the finite element analysis is a computer numerical analysis to do mechanical calculations.
And if you know what the modulus of the structure is, you can back out what the modulus of the solid must have been from the fine element thing.
And those sorts of experiments also showed that the modulus was around 18.
And it turns out that moduli is about the same as cortical bone, and the properties of the solid trabeculae are very similar to the solid cortical bone.
So let me scoot over here.
So if you use an ultrasonic wave propagation, people have measured a modulus for the solid in trabecular bone of 18 gigapascals, or you can do a finite element calculation based on micro CT data for the structure.
And then you measure the overall modulus for the trabecular bone, and then you back out the modulus of the solid.
And people who've done that have gotten values of around 18 gigapascals too, and that's very similar to what the cortical bone is.
And so we're going to use the following properties for the solid in the trabecular bone.
We're going to say the density is 1,800 kilograms per cubic meter.
The Young's modulus is 18 gigapascals.
The yield strength has different values in tension and compression.
It's about 182 megapascals in compression.
And it's about 115 megapascals in tension.
So those are the solid properties.
So then if we look at the compressive stress-strain curves, they have the shape shown on the screen there.
And you can see how similar the stress-strain curves are for those for a foam.
So there's the same three regimes that we see for the foam.
There is a linear elastic regime over here, there's a stress plateau here, and there's some densification regime here.
These are three curves for three different relative densities.
As the relative density goes up, the stiffness goes up, the plateau stress goes up, and the densification strain goes down.
So is the same as the foams that we've looked at before.
And if we look at the mechanisms of deformation and failure, people have looked at this.
These are on a whale vertebrae.
So these are tests that are done in a micron CT machine, again, by Ralph Muller's group.
And here the specimen is unloaded, and here's the same specimen loaded.
So you can see this platen has come down a little bit.
And if you look at this column here, this trabecular here, you can see it's bent out and bowed out more.
And people have found that usually the linear elastic behavior is controlled by bending of that trabeculae, and the plateau stress is usually controlled by some sort of buckling.
But it's not elastic buckling.
You don't recover it.
If you take a piece of bone and you compress it, it's going to have a permanent deformation.
So it's inelastic buckling.
And I think we have some more pictures.
This is another example from whale bone from Ralph Muller's group.
So here's the bone unloaded.
Here it's loaded to 4% strain, here it's to 8%.
And you can start seeing right in this area here if you compare with up there, it's starting to form.
And if you go up here to 12% strain, you see that strut right there.
That was this guy up here, and you can see that it's buckled right over.
So people have made measurements like this in observations, and you can actually see the buckling.
And people have also done finite element modeling.
They can take a micro CT scan and input that to the fine element model.
And then if they do the compression and they input the properties of the solid, they can see that they get a buckling kind of failure.
If you have trabeculae that are very aligned-- we have more.
Here's one more in the buckling.
So this is one of Ralph's little movies.
So when it unloads, it looks like it recovers, but this is all just an animation.
He takes several stills and puts them together, and it doesn't actually recover.
It's just the way that it shows.
But again, these are two different specimens of different densities.
You can see how the struts deform.
They bend and then they buckle.
Let me stop there, and I'll put some stuff on the board.
So we'll say the compressive stress-strain curve has the characteristic shape of cellular solids.
And the mechanisms of deformation and failure, usually there is bending followed by, usually, inelastic or plastic buckling.
And sometimes if the trabeculae are aligned like that knee that I showed you, or if the trabecular are aligned or if they're very dense, then the actual deformation is important.
And I'll just say people have found this by making observations using micro computer tomography or by finite element calculations.
And this is a stress-strain curve and tension here, a tension you get failure at small strains, then you get micro cracks in the bone.
And these next plots just show some data for the bone.
So we're plotting the Young's modulus here.
So this is a relative Young's modulus, the modulus of the bone divided by the solid cell wall material.
Here's the relative density.
Here's data for lots of different specimens of bone.
So some of this data is for human bones, some is for bovine bone.
Sometimes the data is taken where the orientation of the trabeculae doesn't line up with the direction of the loading.
So you might have trabeculae that are oriented this way, but you're loading it this way.
There's sometimes different strain rates.
Let's see.
There's different groups, and so there's a huge scatter in the range of the data.
But you can see if you look at it broadly and you look at that whole cluster of data, the data lie close to a line of a slope of 2.
And if you think of the open-celled foam model and you had bending of the cell walls, you'd expect that the modulus would vary [?
to the ?]
density squared.
So that's kind of the limit of how we do the modeling.
We're really just interested in seeing how the properties vary with density.
If you had a particular piece of bone, I don't think you could use the models to exactly predict what the modulus of that bone would be.
And here's the compressive strength here.
So this is the relative compressive strength.
Here we've normalized it with the yield stress of the solid bone, and here's the relative density.
And you can see, again, this line is of slope 2, so that kind of speaks to the buckling-type failure mode.
And I think I have another one here.
This is the tensile strength.
So if you pull the bone in tension, you wouldn't expect to get buckling, you'd expect to get plastic yielding.
And if you got yielding and you use the open-cell foam model, you'd expect a slope of 3/2, so this line has a slope of 3/2.
And this line is sort of towards the upper bound of that set of data.
You can imagine a line that went through it a little bit lower but the same slope.
And so these open-celled foam models, they don't predict the properties of a particular piece of bone because the bone can have some anisotropy to it.
The orientation of these things may not be perfectly lined up with the loading.
But overall the models give you a sense of how the bone is deforming and failing.
So let me write some of this down.
So that's data for the modulus, the compressive strength, and the tensile strength.
And those have been on those plots.
Those values are normalized by data for cortical bone.
I thought somebody was talking.
It's just the chair squeaking.
And as I said the spread in the data is large, and that's due to anisotropy in the bone and misalignment between the bone orientation and the loading direction.
So when people first started doing tests on trabecular bone, they typically were orthopedics labs.
And the orthopedics labs tended to initially cut the bone specimens on anatomical axes.
So they would do you know the superior-inferior, or the medial-lateral, or the posterior-anterior.
But the bone orientation didn't line up with those directions.
So the bone might have been this way, but they were loading it this way, and so that gave this misalignment.
And there could be some variation in the solid properties too.
So you could imagine some solid might have more micro cracks than another.
So if you took say human bone of different ages, you might expect the older bone to have more micro cracks in it.
So these plots put a lot of data together, and then the lines are based on models for open-cell foams.
So the relative modulus goes roughly as a relative density squared, and the cell walls are bending.
And the compressive strength goes roughly as the modulus squared, and that's related to this plastic buckling.
And then the tensile stress or tensile strength depends on the formation of plastic hinges, and it goes roughly as the density to the 3/2 power.
And one observation that people have made is that in compression, if the modulus and the strength both go as a density squared, then the ratio of the strength to the modulus is just a constant, and that, in fact, is just the strain at failure, or the strain for that say, the plateau.
And that's a strain of about 0.7%, and that's pretty consistent in trabecular bone.
Let's see.
And we said sometimes the bone was relatively aligned.
So here's that picture of the femoral condyle again in the knee, and you can see the bones lined up.
If you have bone that's lined up like that and you load it along the direction of alignment, then you can get axial deformation in the trabeculae.
And then you would expect the moduli would go linearly with the density.
And here's some data for the Young's modulus and the compressive strength of bone that was fairly aligned.
So this was selected to be aligned.
So here's the modulus here.
And the square data points are the longitudinal direction, and the little diamond, these little stars, are transverse.
So here's a line of slope 1, and again, they don't all lie perfectly on that line, but roughly the slope is about 1 there.
And then similarly here, this is the compressive strength.
Now the little squares are the longitudinal data, and they're not exactly on a slope of 1, but they're more or less on a slope of 1.
So I'll just say in some regions, the bone may be aligned.
And then axial deformation is important.
And then you would expect the modulus to go linearly with the density and the strength to go linearly with the density in the longitudinal direction.
Then finally I wanted to finish up the bit on the modeling by making one of these plots a little bit like we did for wood.
So here's the Young's modulus of bone plotted against the density.
The trabecular bone is down here.
It's sort the lowest density.
And then this is the collagen that's in the solid part of the bone, and this is hydroxyapatite, the mineral.
So the modulus of hydroxyapatite is around 120 gigapascals, and the modulus of collagen is somewhere around 5.
And if you make composites of collagen and hydroxyapatite, their moduli are going to be in this envelope here, and compact bone, the modulus fits in around here.
Remember I said it was around 18 gigapascals.
So then if you take a compact bone and you turn it into trabecular bone, you'd expect the modulus would go down along a slope of 2.
So here's our little slope of 2, and more or less that's what you see with the trabecular bone.
So the idea is that the models give you kind of a general idea of how the bone is behaving, but it's not really meant to predict a particular piece of bone.
Because a particular piece is going to have a particular geometry.
Typically they're not equi ax and isotropic.
All right.
So are we good with the general overview?
Are we good with how fewer equations there are now that we got past the first part of the course?
So I'm going to talk a bit more about osteoporosis, and I'm going to talk about some modeling that my group did to look at the consequences of osteoporosis.
And then later on we're going to talk a little bit about using metal foams as a possible replacement material for a trabecular bone as well.
And I have a little bit of a talk on using trabecular bone in evolutionary studies to see whether or not a species was bipedal or quadrupedal.
So I think I talked about this a little bit in 3032, but I have more slides and more stuff I'm going to talk about.
So let me get myself organized.
So osteoporosis comes from the Latin, and it actually means porous bones.
So osteo means bone, and not too surprisingly porosis means porous.
So this next slide gives you some idea what osteoporotic bone looks like.
So the top slide is normal bone in a 55-year-old woman.
These are sections from the lumbar spine.
And that bone up here is 17% dense, so the relative density is point 0.17.
And this is a section from the same area of bone in an 86-year-old woman, and it's 7% dense.
So you can see there's a huge difference in the density, and you can start to see what happens when you lose bone mass.
So if you look at this bone up here, it's all well connected.
Each little trabeculae is connected to its neighboring friends.
And you can see down here, I mean, you look at this and you kind of go ouch just looking at it.
Because this piece of bone here is just kind of dangling off, not connected to anything.
And you can see the struts have gotten thinner, so they've lost bone mass by thinning.
And then as I said when the thickness gets less than they are roughly equal to the size of the cells, then the cells can't live anymore, and the bone strut just disappears altogether.
So it's not too surprising that if you lose this much bone mass, there's mechanical consequences, and there's a greater risk of fracture.
And as I said the two most common types of fractures are hip fractures and vertebral fractures.
So let's see here.
So as people age, everybody loses bone mass.
And happily for you and not so happily for me, the bone mass peaks at about 25 years old.
So you're probably either not at the peak or just barely at the peak.
And then it decreases after that every year.
I'm considerably older than you.
And in women, when you go through menopause, the cessation of estrogen production increases the bone loss.
And so typically, osteoporosis is most common in post menopausal women.
And osteoporosis is defined as a bone mass 2.5 standard deviations or more below that of a young, normal mean.
So it's not like you fall and break your hip and they say you have osteoporosis.
It's based on the bone mass.
And as I said, the trabeculae thin and then they resorb completely.
So anybody here take Latin?
Yes.
I did Latin for one year in high school.
So trabeculae, with an E on the end here, trabeculae, I suppose, is the plural.
Trabecula with an A is singular.
And that comes from Latin.
So you don't say trabeculas.
That's a no-no.
All right.
Let me get rid of these.
So if we saw that the strength of the bone varies as the density squared, you can begin to see how sensitive the strength is going to be to this bone mass loss.
So say you went from a density of 0.2 to a density of 0.1, then the densities changed by a factor of 2 so that the densities gone down by 1/2, but the strength is going to go down by a factor of 4.
You're going to have the strength to be a 1/4.
And so you're going to have a big change in the strength.
And you can imagine is the trabeculae thins, this buckling gets easier to happen.
And then once the trabeculae begin to resorb as they disappear altogether, it's like I said.
it's like having a building's framework.
Now, you're removing beams and columns, and the strength is going to go down even more dramatically.
And the way we model the osteoporotic bone is we use finite element analysis.
So before we talked about using the unit cell for the honeycomb.
But to use the unit cell, you have to have repeating unit cells, and obviously, you don't have that.
You've got local variations in what the structure looks like.
And we also used a dimensional analysis.
And the dimensional analysis relies on the geometry being similar from one specimen to another, and you can't really rely on that either for the osteoporotic bone.
And so what we've done is-- and this is what other people do as well, is use finite element modeling to represent the bone.
So initially what we did was we used a 2D Voronoi model.
So remember we talked about Voronoi honeycombs and Voronoi foams.
So I like to start out with simple things, so we started out with 2D Voronoi model for a honeycomb.
Then we did a 2D representation of vertebral bone.
And then we had a 3D Voronoi.
And I had a couple of students who did this.
Matt Silver was the one who did the first two, and Sereca Vagilla was the one who did the last one.
So let's see.
I think that's probably a good place to stop there for today.
So next time I'll talk about the modeling of osteoporotic bone, and we might talk a little bit about metal bones as a substitute for trabecular-- metal foams as a substitute.
I don't think we'll get to the evolution stuff.
We probably won't quite finish next time.
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So last time we were talking about trabecular bone and that it's this porous kind of foam-like type of bone.
And we talked a little bit about the modeling.
And I think I got as far as starting to talk about osteoporosis.
And I wanted to talk today about how we can model osteoporosis using those voronoi honeycombs that we talked about a while ago when we were talking about the structure.
So Bruno has a project on trabecular bone for the class.
And he needed bone samples.
And so we talked about different kind of bone samples we could get.
And the thing is if you get human bone-- well, there's all sorts of issues about just handling human bone and permissions, and it's complicated.
So that was too complicated.
We've used bovine bone before.
You just go to the slaughterhouse and get bovine bone.
But one of the things with trabecular bone is because it grows in response to loads, the geometry of it can be different, sort of architecture can vary from one spot in the bone to another.
And I have a colleague who started doing tests on whale bones, just because it's a way of getting nice, uniform bone.
So I was on a Ph.D. Committee a few years ago for a student who was in the Woods Hole program.
She was at MIT, but was doing Woods Hole thing.
And I don't know if you've heard of the Atlantic right whales, the North Atlantic right whales.
They're endangered.
There's about 500 of them left in the world.
And they migrate between like typically the Bay of Fundy and off the Florida coast.
So they go up and down the coast.
And they sometimes get hit by ships.
And then bones break and that kills them.
And her study was on ship impacts on right whales.
And so I got to know people at Woods Hole who worked on whales.
So Bruno, I called up my friend at Woods Hole.
And the Woods Hole guy didn't have any bones.
But he put me in touch with somebody at the Mass Fish and Wildlife Department.
And he had a couple of whale vertebrae that he was willing to give up.
So I got one of them for you.
So here it is.
So I went out to the Mass Fish and Wildlife yesterday.
And they produced this bone for me.
I could either pass it around.
It's not too heavy.
Or you could come up and look afterwards.
Shall I pass it around?
Or do you want came up afterwards?
Maybe come up-- pass it?
OK.
So one of the things is our like vertebrae, there's these things that stick up like this.
There's one that's missing off of this bone.
There should have been one down here too.
But this is sort of what's called the body of the vertebrae.
And in human vertebrae it's about that big.
But it's the same kind of general structure.
There's these kind of bony plates that come off.
And the body is almost all trabecular bone.
So you can see on this side, this is a growth plate here.
But on this side, there normally would have been a thin shell of the cortical bone.
And when I pass it around, if you look up at this point here, you can see there's just a little bit of that left.
But it's kind of gotten worn out.
And the rest of this, if you look, you can see it's the trabecular bone.
And you can see it's pretty uniform, which is why I thought this might be good for your tests.
So I thought you should get in touch with Mike Tarkanian.
And I talked to him a little bit about cutting it with a water jet cutter.
So I think we could use the water jet cutter.
And I think I emailed you.
If you could cut it in half, I'd like to have some pictures of it cut in half.
And he's got a diamond corer, cylindrical corer.
So you could make little cylindrical specimens.
And I know if you want to do compression tests or beams
Well, I was planning to do compression tests
Yeah, so you could probably use some kind of a bands or something to cut them up.
So, and this is where the spinal column goes through in the whale.
So there you have it.
Anybody want to ask me anything about the whale bone?
OK, so let me pass that around.
And I guess I have to tell you a couple of other stories.
So while I was there, they have a brand new building.
And it's all got solar panels and geothermal heat.
And it's all very groovy.
And the guy who I was talking to about the bone, he wanted to give me a little tour of the building.
And he said, oh, you've got to see your freezer.
OK, so the freezer.
So he opens the freezer door.
The freezer's like a room.
And there's like a bear.
I'm serious, like a bear on the floor, like a dead bear on the floor of the freezer.
And he said it was like a two-year-old bear that had, I guess, just come out of hibernation a couple weeks ago.
I think it had gotten hit by a car or something.
And somebody must have called them up.
And they have it.
So they had this bear.
They had like several deer.
They had a coyote.
They had like boxes full of all kinds of animals.
So anyway it was kind interesting to see all these animals there.
And you know what he said about the deer?
He said, normally deer in Massachusetts don't starve.
Like if you live in the suburbs, you actually have a problem with deer eating your vegetable garden because there's deer all over the place.
But he said, this winter, you know how much snow we've had and how cold it's been?
He said, deer have been starving this winter.
And I think a couple of the deer they'd had had actually starved to death.
And people had called up.
And they had come and kind of collected the carcasses.
So anyway that was my little trip out to Mass Fish and Wildlife yesterday.
OK, so let's go back to talking about the bone.
And I think last time we kind of left off more or less here.
So this was a slide of what osteoporosis looks like.
And you can see the bone loss is a combination of thinning of the struts and resorption of the struts.
And we wanted to try to model this, sort of an engineering sense.
And the way we did that is we used our voronoi honeycombs.
So the bone has an irregular structure.
And we wanted to look and see if we could model something that had an irregular structure.
So we used the voronoi honeycomb.
And if you remember when we talked about the structure of cellular materials earlier on in the course, we said that these are generated by putting down random seed points and then drawing the perpendicular bisectors.
And if you have a constraint that the seed points can't be closer than some exclusion distance, then you get structure where you have cells that are not all exactly the same size, but roughly the same size.
So that's what we did here.
We got this structure here.
And I had a graduate student, Matt Silva, who did a lot of these studies.
And then he used this and analyzed it using finite element analysis.
So the first thing he did was he calculated the elastic moduli of the structure.
So he applied loads.
We calculated deformations.
Figured out elastic moduli.
And so this plot here shows a comparison between the analytical equations that we derived at the first part of the course and what he calculated for these voronoi honeycombs for the final element analysis.
So here's Young's modulus down here, for example.
In the closed form, the line, that's just the analytical equation we had originally.
And the little dashed line is his finite element.
Here's the Shear modulus down here.
And here's the possum's ratio up here.
So you can see, there's a pretty good agreement between these two things here.
So let me write some of this on the board.
And then I can keep going after that.
So for the 2D voronoi honeycomb, we have the random seed points.
And we used the perpendicular bisectors.
And we used a minimum separation distance between the points.
So we generated the structure.
And then we did a finite element analysis.
And from that, we calculated the modulus.
And what we found was the finite element analysis results were pretty close to our closed form analytical model.
And if we think about the modulus, the modulus is the average stiffness over the whole honeycomb.
And when we look at the strengths next, the strength is going to be related to the weakest few struts.
And we're going to find that the strength doesn't work quite the same way.
So first, we got the modulus.
That's sort of the simplest thing to calculate.
And then after that, we wanted to calculate the compressive strength.
So we did a similar thing.
We set up the voronoi honeycomb in the finite element analysis.
We had a few more elements along the length of each strut.
And then we modeled the elastic buckling and the plastic failure behavior.
And we looked at honeycombs that had different densities.
And the lowest densities failed by buckling.
And the higher densities failed by a sort of plastic yielding.
And we assumed that the cell walls were elastic, perfectly plastic.
Remember we said if we have a material where-- this is for the solids-- so say this is the stress in the solid and that's the strain in the solid.
If it behaves like that, we say that's elastic, perfectly plastic.
So we modeled the walls as elastic, perfectly plastic.
And we made the ratio of the solid modulus to the yield strength similar to what it would be for bone, which is about 0.01.
And for that particular value, the transition between the elastic buckling and the plastic yielding failure was at a relative density of about 0.035.
So then what we did was we analyzed structures that were a little lower than that were equal to that and then a little higher than that.
So we would try and see what happened with these different failure modes.
And then we found that if we look at the compressive stress strain curve, this model here had a relative density of 15%.
And the transition occurred at a relative density about 3.5%.
So this one here failed by a plastic failure.
You can see, if we unload it, there's some plastic deformation.
We get a little strange softening here, which is kind of characteristic of the plastic failure.
When we look at the overall deformation of the honeycomb, we saw local kind of failure, like we do in aluminum honeycombs.
So that was the kind of stress strain curve for a relatively dense honeycomb that failed by yielding.
And then this is one for a much lower density.
This is now point 1.5%.
And this one fails by elastic buckling.
And if we load it up and unload it, we recover most of the deformation.
Barry, do you think you could make that stop?
Yeah
There's a chance it could be--
Below.
Just because we're recording it.
It just doesn't seem very good.
OK, so what we did was we made five different voronoi honeycomb.
So we had five sets of seed points.
And we had five slightly different geometries.
And then we averaged the results of those.
So if we make that calculation, this is the strength of the voronoi honeycomb here divided by the strength of the periodic regular hexagonal honeycomb.
And this was plotted against relative density.
And you can see the strengths for the voronoi structure are a little bit lower than for the regular periodic hexagonal honeycomb.
And they reach a minimum here.
And the minimum's around about 0.05 relative density.
So this was the 1.5.
That was a 3.5.
That's 5% and 15%
Why is there like a wide gap between 0.05 and--
Well, I think because we felt-- I think this one failed by some combination of-- there was a limit to how many of these we were going to do.
And this is what we chose to do.
We wanted one that we knew was going to fail by plastic yielding.
And that was this one.
And we wanted one that we knew it was going to fail by elastic buckling.
And we wanted a couple in between.
So we didn't do-- I guess we were lazy is the real reason there's not another point in the middle there
It just seems like there might be a variable--
You think it might go [CAREENING NOISE] like that.
Except there's a physical reason why this happens.
And I'm going to get to that in a minute.
So let me I'll finish explaining this, and then I'll put the notes on the board.
So first of all, the strength is less than the regular hexagonal honeycomb.
And then there's also this minimum here around about 5% density.
And I think the reason that the strength is not the same as in the voronoi in the periodic honeycomb is because if you look at the distribution of strains-- or you can think of the distribution of stresses-- so these were the normal strains at the nodes in the honeycombs.
And the distribution here is for the voronoi honeycomb.
So the voronoi honeycomb, you have lots of members of different lengths.
There are different orientations.
And so there's some distribution of stresses and strains in each member.
And that gives you the distribution.
In the regular hexagonal honeycomb, if you look at just the nodes, there's really just the vertical member, which has a certain strain.
And all the vertical members are going to have the same strain at the nodes.
And then the obliques members, if you just look at the nodes, there's just going to be a maximum tension and a maximum compression at the nodes.
Because there's a unit cell and it repeats, all the oblique ones are going to have the same maximum and the same minimum.
So the dashed lines here are for the regular hexagonal honeycomb.
So the thing to observe here is that the voronoi has some strains and correspondingly some stresses that are outside the range of the regular periodic honeycomb.
And so if there's parts of it that are seeing higher strains and higher stresses, it's going to fail at a lower load.
So I think it's this distribution because you've got this random structure, and you've got different lengths and different orientations of the members.
So that's one reason why these strengths are less than the periodic structure.
I think there's a minimum here, because-- I think before the test I mentioned there's some interaction between elastic buckling and plastic yielding.
And when you get that interaction, that also reduces the strength.
And so there's a minimum near where the crossover is between the elastic buckling and the plastic yielding.
So let me write some notes of the compressive strength.
So we'll say the cell wall-- so the cell wall elastic, perfectly plastic.
And the yield strength relative to the modulus for the solid was 0.1, which is pretty much what it is for bone.
And we assumed that the possum's ratio of the solid was 0.3.
And for this value of sigma Ys over Es, the transition between elastic buckling and plastic yielding is at about 3.5% relative density.
So then we made models with densities that were a little bit less than that and a little bit more than that.
And then the strengths we got from the voronoi were between about 0.6 and 0.8 times what we got from the periodic.
So then what we looked at were these maximum strains of the nodes.
And we found that because the voronoi honeycomb had a much broader distribution of those strains, that led to the lower strengths.
And then we found the minimum strength was at a density of about 5%.
And if you think just about a pin ended column, and you make a plot of the strength-- so I'm just going to say the strength of that columm-- against l/r, the slenderness ratio.
So say it's a cylinder, l would be the length.
r would be the radius.
If you just had Euler buckling, you get a curve that looked like that.
The Euler buckling load is pi squared EI over l. squared.
So I goes as r to the 4th.
And if we get the stress, then it's going to be that divided by the area of the column.
So it's pi squared E. Moment of inertia is pi r to the 4th over 4 l squared.
And the area is pi r squared.
So it goes is as r over l squared, or 1 over l over r squared.
So this would be the Euler buckling.
So that's elastic buckling.
But at some point, the column is going to yield.
If I make it really, really short, then it's going to yield before it buckles.
And at some point, this would be the real stress here.
And this would be failure by the plastic yielding.
And in practice, there's not like a sharp corner here.
You know, if you made columns of progressively longer length, and you tested them, the little short ones would yield.
So they'd be along here.
But they don't kind of yield and the next one buckles.
In fact, you get something like that.
So that when you're near that transition, when you're near this point here, the actual failure stress is a little bit less than that.
And I think that's partly what's going on over there.
It's the minimum.
OK, so those two studies, we just looked at the modeling and the strength of honeycombs.
And we were kind of looking at how does the random structure change the properties.
So the randomness of the structure didn't really change the modulus much at all.
But it did affect the strength.
And then the next thing we did was we looked at putting defects into the bone.
So we knew that the bone, partly the density is reduced by thinning of the struts and partly by resorption when there's a lot of density loss, a lot of bone loss.
So we wanted to look at putting defects where we actually removed some of the struts.
So we did another series of voronoi models.
So we did another series of tests where we looked at it if we have a certain density that we start with and then we look at losing the same bone mass or relative density in our honeycomb, and we look at what happens if we thin the struts versus if we remove struts.
So these plots here show that.
And Matt Silva also did this.
So this is the residual modulus plotted against the reduction in relative density.
So residual modulus is the modulus after we've reduced the density by some amount relative to the initial modulus.
And if we have an intact honeycomb where we just thin the struts, we don't remove any of the struts, if we have an intact honeycomb, as you reduce the density, the modulus just goes down like that.
So that's really just the same as those analytical formulas that we had.
But if you start removing struts, not too surprisingly, the modulus goes down substantially more.
And at this value here, I think it was 30% or 35% density reduction, you reach what's called the percolation threshold.
At the percolation threshold, you have two separate pieces of material.
So obviously the mechanical properties are going to go down to zero when you reach that percolation threshold.
And then this plot over here is the same sort of thing, but now for strength.
So it's the strength of the bone, or the strength of the honeycomb, after you've either thinned or resorbed the wall, divided by the strength of the intact honeycomb.
And again, you're reducing the density here.
And this is for the intact model where you're just thinning the struts.
And this is for the models where you're removing the struts.
So you can see that if you think of-- this is kind of a simple model, but if you think of the bone, if you first thin the struts, you're going to lose a little bit of strength.
But then if you start removing the struts, you're going to lose a lot of strength.
So that's really where people run into-- there's much higher risk of fracture once you get to the point where you might be resorbing struts.
OK, so let me see, what's next thing?
OK, let me just finish the slides, and then I'll put the notes up.
So this is the same sort of thing, just plotting the strength and the modulus on the same plot.
So you can see the shape of the curves is very similar.
The modulus is a little bit more sensitive than the strength.
And here we are thinning, and here we are removing the struts.
And then the next step was that we made a model that was more similar to bones.
So let me write down the notes for this.
And then we'll do that one that's more similar to bone.
Thought it didn't makes sense.
OK, so then we were interested in trying to model something that was a little closer to the structure of bone.
And so we set up this model here.
So we started with just a square voronoi.
So you just force the points into a square, voronoi, or a square pattern, you get a square voronoi.
And then what we did is we just perturbed the points a little bit.
And we got this perturbed voronoi array.
And so we made this model here.
And we took a piece of vertebral bone.
And we measured the angle of orientation of a lot of the struts.
And we matched our voronoi model to that distribution in the bone.
So our model looked something like this.
So you can kind of see how it's more or less vertical and horizontal, but not exactly.
And here's a sort of comparison with a slice of vertebral bone.
And again, because the loads are more or less vertical in the bone, the trabeculae tend to orient that way.
And then have some horizontal struts.
So here you can see on the left, we've got a voronoi model that's more or less representative of the bone.
And we've removed some of the struts.
So we're going to the same thing with this model.
We thin the struts.
And we remove the struts.
And we try to see what the residual strength is.
And you can see there's for of at least a 2d model this isn't a bad representation of the vertebral trabecular bone.
And this was the stress strain curve for both the specimen of the vertebral bone that was tested and then the honeycomb model that we made to kind of match it.
So a similar kind of behavior.
This is how our model failed, this sort of a local band of struts that fail.
Let's call it local deformation band.
And then what we did was we thought about changing the density.
And what we did was we removed either horizontal or vertical struts, or we thinned either the vertical or the horizontal struts.
So these are the same kinds of plots as I showed before.
This is the residual modulus.
This is the residual strength.
This is the density reduction.
And here we're reducing the density by making struts thinner.
So it's still intact.
We haven't removed any struts.
But each of the struts gets thinner.
And this top line is for thinning the longitudinal or the vertical struts.
And this was for thinning the transverse or horizontal struts.
And then this is for removing struts here.
So we're removing a bigger number.
So the more we remove, the more the density changes.
And then this plot here is for the strength.
So again, this is thinning.
So we're moving either the horizontal or the vertical ones-- I'm sorry, thinning either the horizontal ones or the vertical ones.
And then this is for removing struts.
And again removing more to reduce the density more.
So you can kind of see we're kind of working our way to more complex models here.
So this one here was looking at the bone.
Let me write some notes for this.
And then we did a 3D voronoi model.
So I'll do that one next.
Oop, over here.
So I'll just say the model was adapted to reflect the trabecular bones study in the vertebrae more.
So we perturb a square with array of seed points to get a structure that was more like the bone.
And then we looked at the reduction in the thickness and the number of strides in the longitudinal and transverse directions.
OK, and then the next model we did was the 3D version.
So we made a same kind of thing.
But with the 3D model, we had fewer cells in that model, because once it goes to 3D, you've got more struts in each cell.
But this was the same idea.
We uniformly thinned the struts, or we removed the struts.
And again, you can see removing the struts is a much bigger effect.
And also the other thing to look at here is for 3D the percolation threshold is 50%.
So it kind of makes sense that if it's in 3D, and you've got struts in all directions, you're going to have to remove more of them.
You're going to reduce the density more to break it into two separate pieces at the percolation threshold.
So that was the 3D model there.
And then this is just a comparison of the 3D with the 2D for the modulus.
We just did the modulus for the 3D structure because it sort of gets computationally more involved.
So the 3D, these two lines here corresponded to removing the struts and the change in the modulus.
And the little triangles were the 3D voronoi calculation that we did.
The little crosses here were the same sort of calculation done for tetratridecahedron.
One of my former students had loaded that.
And then at the bottom here are the lines for the 2D structures, for either a regular hexagonal cell or for a 2D voronoi cell.
And you can see, there's not a huge difference whether or not you take a regular structure or a voronoi random structure.
But there's a fairly significant difference between the 2D and the 3D.
So the 3D, just you have to remove more material before you get the same reduction in modulus.
So let me write some notes for that.
So it's the same kind of analysis, but just with a 3D model.
So do you see the idea with these models?
It was an attempt to look at a way that you could computationally estimate how much modulus loss or strength loss you get by either thinning the struts or removing the struts.
So I gave you a way to model the osteoperotic bone.
Yes
Can you clarify the percolation threshold?
So the percolation threshold is say you have some network and you start removing things.
If your remove enough, you have two separate pieces of things.
If you remove enough struts, you have two separate pieces.
That's called the percolation threshold.
So I think it's-- you want to know why it's called that?
So I think that originated because it wasn't used in this kind of context.
I think instead it was used in a context where imagine you have 2D plate.
So you put 2D holes in it, little circular holes.
And they we're looking at flow of a fluid through the plate.
So you can imagine, if you put enough holes, eventually they line up or they-- line up is not the right term-- but there's enough of them that they connect.
And you end up with two separate.
You have a path through for the fluid.
That's called the percolation threshold.
But in mechanics it's sort of two separate pieces.
OK, does that makes sense?
So you know, at the percolation threshold, the stiffness is zero because you have two separate things and the strength is 0
So the two doesn't have to be like separated by--
It could be-- yeah, yeah, it doesn't have to be a straight line
Does it make a threshold between whether everything is interconnected?
Yes.
Or there's some path that separates them.
Yeah, that's what it is.
OK, so that's the end of the bit on osteoporosis.
I had a couple more things I wanted to talk about on bone.
So the next part is on the idea of using metal foams as a bone substitute materials.
So when they make hip implants-- so they're typically metals.
They're titanium or stainless steel or something.
Often what they do is they coat the outside with some sort of porous coating.
And the idea is the porous coating allows the bone to grow in.
So you can kind of imagine, like especially on the stem and around the head of a femur, you'd like the bone to grow into that to attach.
And having a porous coating helps.
And there's a couple of ways they do it currently.
They use porous sintered beads.
So they put little beads of metal on the outside.
And the idea is that the bone grows into the little gaps between the beads.
Or sometimes if they have-- not so much for hip implants but sometimes when people have say car accidents and their face get sort of smashed up, and they have to have, say, a plate put in their face, and they need like a flatter plate, they use like a wire mesh.
And they have sort of a wire mesh that goes on the outside.
And it's the same thing.
It's the idea to try to get the bone to grow into that plate.
So some people are thinking instead of using the porous sintered beads, or instead of using one of these wire mesh plates, that you could use metal foams.
And so there's been some interest in using metal foams in coatings of implants.
And longer term, there's been some interest in trying to make sort of a vertebral body that would involve using a metal foam from the vertebral body.
So you know that whale bone that we just passed around, that vetebral body, that cylindrical part, it's almost all trabecular bone.
So there's some interest in trying to use metal foams for spinal surgeries.
Maybe not to replace the whole body.
But to use in part of the surgery.
So I have a little slide here which shows a bunch of metal foams that people have made with the idea in mind that perhaps some of this could be used in orthopedic surgery.
So these are some different kinds of metal foams.
And these ones are made from titanium or tantalum.
So typically, the metals that they use in orthopedic implants are the cobalt chromium alloys.
Titanium, they sometimes use tantalum or stainless steel.
And they use those because they're biocompatible.
And they're very corrosion resistant.
So these ones here are mostly titanium.
And this is one that's a tantalum.
So let me just go over how they make them.
And then I've got another slide that goes over it in more detail.
And I'll write a few notes down.
So this guy on the top left up here, that's made by taking an open cell polyurethane foam, like a seating foam, like a cushion.
And then what they do is they heat that up in an inert atmosphere, so that they are left just with the carbon.
So it's a sort of vitreous carbon.
And then what they do is they coat that carbon by a CVD process with tantalum.
And so they end up with a tantalum foam with a very thin layer of carbon at the core.
The carbon makes up something like 1% of the final composition and the tantalum is the 99%.
So that's sort of a replica process there.
This one here is made by another replica process.
They take an open cell polyurethane foam.
They infiltrate it with a slurry, which has the titanium hydride particles in it.
Remember when we talk about processing of the foams at the beginning, I said, if you heat up the titanium hydride, eventually the hydrogen would be driven off, and you'd be left with the titanium.
So they do that, and then they sinter it, and they get a titanium foam.
This one is made by a fugitive phase process.
So the idea with a fugitive phase is you burn off some phase.
So you could mix powders, consolidate the powders, then you burn off one of the powders.
And you're left with the other one.
And then you need to sinter that together to make it have some reasonable mechanical properties.
This one here is made by using a foaming agent.
This one's made by expansion of argon gas.
I think when we talked about the metal foams, we talked about the idea of packing, say, titanium powder in a can.
And then you evaporate the can.
And you then pressurize the can with argon gas.
And then you heat treat it.
So you heat it up.
And as you heat it up, the argon gas expands.
And you're left with the pores.
This one's made by a freeze casting process.
I have a slide.
And I'll talk about that in a minute.
This one's made by a selective laser sensory.
So it's like a 3D printing.
But instead of printing in ink, this time you have a bed of a powder.
And you've got a laser that selectively sinters.
So you turn the laser on and off.
And where it's on, it's going to bind the material.
When it's off, it's not going to bind the material.
And then you raise the thing.
You make a little bit more powder.
You do it all over again.
And you can get different patterns.
And then this is made by a sort of process in which they take powders and press them and ignite them.
But I think that's not very commonly used.
So this is some more details about how they might do it.
So this is the fugitive phase process here.
You could take a titanium and a filler powder, pack them together.
You know, you'd mix them up, pack them together.
You would raise it to a certain temperature to decompose the filler.
So typically the filler decomposes at a lower temperature than what you sinter the powder, the metal powder that's left.
So you decompose the filler.
You drive that off.
And then you sinter the metal powder.
And you're left with porous titanium.
This one's the expansion of the foaming agent.
So you take your titanium powder.
You might have a binder and then a foaming agent.
Mix those all together.
They heat them up until typically the binder becomes a liquid.
And the foam foams up the liquid binder.
Then they drive off the binder.
And then they sinter at a higher temperature the titanium.
So you you've got a porous titanium left.
This is the freeze casting or the freeze drying process.
So here they would take titanium powder and put it in agar.
And the agar's in water.
So the agar is like a jelly, like a gel.
But it's mostly water.
And then if you freeze it, what happens is the water freezes.
And it drives off the titanium agar into the interstitial zones between the frozen ice crystals.
So these little dots here are the ice crystals.
The ice crystals are growing as it gets colder.
And as the ice crystals get bigger and bigger, you're left with the titanium and the agar in between those ice crystals.
And then if you sublimate the ice off, you're left with a porous thing.
And you can sinter the titanium if you want to make it a little more dense.
And this is a rapid prototyping thing here.
So you spread the powder.
You could either print a binder or you could use a laser to sort of almost weld the particles together.
Then you would drop the piston, put more powder down, have the binder go again until you made your product.
And then you would get rid of all the unbound material.
And you'd have your final product.
OK, so these are some of the methods they can use for making metal foams.
I don't know, should I write notes?
Or are you good if I just put the notes on the website?
I'll put the notes on the website.
And then this is a stress strain curve for a titanium foam.
So it looks like all these other kinds of foams and bone and wood and everything else that we've looked at.
And this is some data for titanium foams that are made by different processes.
And we've just taken different data from the literature and put it together.
So this is the modulus here.
This is a slope of 2.
You can see some of the data is sort of near that slope, but below it.
And there is obviously a large spread in the data too.
But if you go back and look at the different types of structures, then not all these have this kind of typical foam-like structures.
The structures aren't all quite like a foam either.
Yeah?
So is there any objective in this to make the foam similar in structure to the bone that will be growing into it?
Or does it just need to be scaffolding?
I think for this, they just want to make a porous thing.
And they're thinking about coatings.
So the coatings aren't necessarily similar to the bone.
When we finish the section on bone, we're going to start talking, I think, about tissue engineering.
And when we talk about tissue engineering, people have made scaffolds to try to make them the same shape as the anatomical part that they're trying to mimic.
And then they make a cellular kind of core.
So they have made some more of an attempt to do that.
I could give you a sneak preview.
Would you like a preview?
So these are some scaffolds that are generated from a making different kinds of tissue.
These aren't all from bone.
But this one here, for example, is for regenerating bone.
And they've printed it in this exact geometry, because that's going to replace some anatomical piece.
And they want it in that geometry.
So I'll talk more about that.
But so for the scaffolds, they sometimes do that.
But not so much for these bone coatings.
Let's see.
Se we did that.
We did that.
So these are the data.
And then over here, there's the compressive strength.
And again, you know, this is our line with a slope of 3/2.
Again, there's a lot of scatter, because there's a lot of variation in the structure of these things.
But it's in the ballpark.
So are we good with metal foams?
And there's just like a page and a half of little notes.
Should I just put that on the website?
You're good?
OK. OK, so the last topic I wanted to talk about on bone has to do with how people look at the structure of bone in evolution and in evolutionary studies.
So the idea here is, in particular in looking at sort of pre-human species, sort of hominid species, people are interested in when primates went from being quadruplets to bipeds.
So obviously, bipedalism, walking on two legs, is characteristic of us.
And people would like to know if they find some fossil-- you can't just tell from the fossil directly is it a biped or a quadraped.
You can't see the species moving because it doesn't exist anymore.
So you'd like to have some way of making some estimate of whether or not it was a biped or a quadruped.
Let me get my notes together here.
So I wanted to kind of look at the big picture a little bit and look at the evolution of different species.
And this is a phylogenetic sort of chart.
And this is kind of-- have you heard of the tree of life?
This is like the tree of life.
So this piece of it is-- metazoa is for animals.
So not plants, animals.
And this goes back about 1.2 billion years.
So these are millions of years ago.
And then these are different sort of eras and ages that are defined.
And when we have a branch here, this is a common ancestor.
And then this is a branch in one direction, and that's a branch in another direction.
So these points here, like 1 and 2 and so on, the implication there is that there was a common ancestor to everything that traces back to there.
So this point here, 1, is 1.2 billion years ago.
So this was sort of very early kind of species.
And the very first things were, well, multicellular things.
I mean, there were little amoeba type things.
But the more sort of sophisticated animals were sponges.
And there's three kind of classes of the sponges.
There's calcarea.
And they're called calcarea because they're mineralized.
And they're mineralized with a calcium carbonate.
And then there's another branch of them called-- I don't know if I'm going to say this right, but hexactinellida.
And those have glass.
Those are called glass sponges because they have SiO2 is the mineral in those.
And then there's these guys here the demospongiae.
I think some of those have calcium carbonate and some of them don't.
Oh, no, some of them have silica.
And some of them don't.
So these things here are sort of very early multicellular structures.
And the mineral in them is either calcium carbonate or silica.
And then if we move up, I've sort of highlighted a few of them.
The cnideria-- I know there's a C, but that's actually-- when I say s-nideria, my biologist friends laugh at me.
And they say no, no, no, it's nideria.
The cnideria are the corals and the jellyfish.
And corals are also mineralized with calcium carbonate.
And you can see they branched off something like a billion years ago.
Then we get up here.
These are the mollusks.
So the mollusks are things like bivalves, like if you like to eat claims, things like that.
So bivalves, snails, and things like octopi, octopus.
So those are all molluscs.
And molluscs, when they're mineralized, also are calcium carbonate.
So those are the calcium carbonate kind of shell.
So we haven't got up to anything bony yet.
Bone is collagen plus a calcium phosphate.
So we haven't gotten to anything that's a calcium phosphate yet.
Then another large class is arthropoda.
That's insects and spiders and crustaceans.
Those all have a chiton skeleton.
So if you think of like the exoskeleton of an insect or a spider, those are chiton.
And crustaceans, things like lobsters, those also have a chiton shell.
And in crustaceans, it might be mineralized.
But again, the mineral is a calcium carbonate.
So all the way up here, most of these things, if there is any mineral, it's calcium carbonate.
And if we get up finally to the vertebraes, the vertebraes have the calcium phosphate and have a bone, like what we think of as real bone.
So the vertebrates obviously involve things like mammals, birds, snakes, and fish.
So this is kind of the big picture going back.
And sort of one of the interesting things is that bone doesn't come along until you get somewhere over here.
And I have one more little, nice slide here.
So when I talked about the sponges, they were these guys here with the glass sponges.
Joanna Aizenberg at Harvard did a nice study on glass sponges.
And she looked at this one here.
It's called the Venus flower basket is kind of the common name.
And it has this hierarchical structure.
And it's remarkably stiff and tough.
And what she did in her paper was look at optical and mechanical properties.
But they looked at the structure at different length scales.
So there's kind of a cellular structure at this length scale.
It's kind of a tube.
This was just a picture I took in a natural history museum.
But if you look at higher magnification, each little strut is made up of sort of fibers and has a hierarchical structure itself.
So that's just one of the sponges.
And here's a similar chart for the vertebrates.
So this point here is where the other chart kind of branched off.
And if we start with the earliest things again, the earliest ones with the most common ancestor back here is something called cyclostomata.
And those are things like jawless fish, so things like lamprey and hagfish.
Do you know what a hagfish is?
It's this kind of eely thing.
And I have the video for you if you don't know what a hagfish is.
So let me get out of here because it's just an amazing thing, the hagfish.
OK, so let's see, I got my sound on.
[VIDEO PLAYBACK]
Here, at the University of Guelph, about an hour outside Toronto, materials scientist Atsuko Negishi and biologist Julia Herr think that these lovely creatures, called hagfish, may revolutionize how we make strong materials
These are specific hagfish.
They're well known for their unique defense mechanism
So if I wanted to see this, what would we do?
Like could we poke at it with a stick?
I think the best way to do it is to reach in there and grab one
Oh, my gosh, look at that disgusting-- oh, no, I've been slimed.
I feel like an outtake from Ghostbusters.
Look at the quantities of this stuff.
[END PLAYBACK]
He used to do this for The New York Times.
And I think he's got his own going on, but he used to make these little videos.
And he had a show on PBS a year or two ago all about materials.
And there were like four different episodes.
And he talked about different kinds of materials.
And he went to different labs.
But he's quite a character.
But anyway the hagfish have this defense mechanism where they make the slime.
And I don't know if you know Professor McKinley over in mechanical engineering here at MIT, but he collaborates with those people of Guelph.
And he studies what he calls non-Newtonian fluids.
Well, a lot of people call non-Newtonian fluids.
A Newtonian fluid is a thing like water, where the viscosity is a constant no matter what sort of strain rate you shear it at.
And a non-Newtonian fluid does not have that property.
The viscosity changes.
And some of them have a strength, like in the hagfish one has a strength.
So Gareth's studies things like this kind of hagfish slime.
I don't know if he still has them.
He used to have hagfish in his lab over him building, whatever it is, 1 or 3 or something over there.
OK, so that's what the hagfish are, just because that's amusing.
And they don't have bone.
So they and the jawless fish don't have bone, even though they're called vertebrates.
Then the next sort of most recent thing is the chondrocytes.
Those are the cartilaginous fish, so things like shark.
So sharks don't actually have bones.
They have cartilage.
And they have a little bit of mineralization in the cartilage, but they don't actually have bone.
And the first thing that actually has a bone is the ray finned fish, which are these guys here.
And that occurred about 450 million years ago.
And then everything in the vertebrate since then is bony.
So there's coelacanth-- I don't know if you've ever see those.
Every now and then they find one of these things in Florida or something-- lung fish.
There's these guys here, which are the frogs and the toads and the salamanders.
So it's finally getting to be spring after the winter from hell.
And the salamanders are going to come out into the vernal pools and mate.
And it's cute.
So anyway, they come out this time of year.
And then there's the amniota, things that have eggs of one sort or another.
So that includes us, the mammals, the birds, the snakes, and the turtles.
And so those would have branched off about there.
So the last thing I wanted to point out here is that there's this huge kind of diversity of animals that have evolved over millions and millions of years.
And that the first ones that were mineralized used the calcium carbonate and that the bony type materials didn't really evolve or didn't appear until about 450 million years ago.
And then these are the vertebrates that have these kind of bony things.
So as I've said many times now, the bone grows in response to loads.
And the bone structure reflects the mechanical loads and the function.
And evolutionary studies have looked at both cortical bone and trabecular bone architecture to try to say something about the locomotion of the animal or of the species.
So there's ones that look at cortical bone.
But I'm just going to talk about one that deals with trabecular bone.
So this study was done by a group of peoples, the first author was Rook.
And what they studied was the ileum.
So this is a pelvis.
And there's the ileum is one of the bones in the pelvis.
And they found fossils of a hominid species that was about 8 million years old, called Oreopithecus bambolii.
And bambolii refers to the place in Italy where these fossils were found.
And so they found two-- or at least somebody found two pieces of an ilium.
And they took x-rays.
And they made a digital reconstruction so that they would get one ilium-- it turned out that the two pieces were two different parts-- so they could make a whole one out of the two pieces.
And then they looked at the structure of the trabecular bone.
And they compared that structure to the structure of the trabecular bone in humans and other primates.
And they wanted to see is it more like the humans, which are obviously biped, or is it more like some of the primates, which are quadrupeds.
So this just shows for a human and a non-human primate what the structure of the ilium is.
And these little black boxes with the letters are what are called anatomical landmarks.
So they're sort of comparable spots on the bone of different species.
And what they were doing was looking at the trabecular architecture at these different spots.
So you can kind of see how they're more or less analogous in the two different species.
And this is the digitally reconstructed ilium that they put together from their fossil species.
And again, these little letters refer to these anatomical landmarks.
And then what they did was they compared the Oreopithicus, the fossil, with the human and then three non-human primates.
So these four images are all from the fossil.
These are the human.
These are champi, siamang and baboon.
And this square here corresponds to that one there.
This is B, corresponds to that one.
And C and D are those two there.
So they had two fossils, they made the digital reconstruction.
They looked at certain areas.
And then they looked at the same or analogous areas in these other species.
And they tried to look for similarities and differences in the bone structure.
So let's look at this first box A.
You can see there's a very white bit here.
And the white corresponds to more dense.
So there's a white bit that's more dense in their fossil.
And in the human bone, you see is a similar sort of band right there.
And if you look at the other non-human primates, that band is missing.
So that says to them, OK, this feature, this one feature at least, is more similar to the-- sorry, in the fossil here is more similar to the human than it is to the non-human primates.
And then they had three other landmarks that they looked at like that.
So this one here again is a sort of dense regions.
So you can see that white dense region.
There's some there.
There's some here.
So those are the fossil and the human.
But there's a teeny bit here and a teeny bit there.
But it's not as pronounced in the non-human primates.
Yes
From I get at least so far, the portions of the bone that are dense versus not dense seem less relevant to the direction of loading than the orientation of the foamy parts of the trabecular bone
So the density reflects more or less the magnitude of the stresses.
So if the stresses are higher, it's going to be denser.
And the orientation of the bone, like whether or not which way the struts are oriented, that reflects the sort of ratio of the principal stresses.
So if the principal stresses go in a certain direction, the bone's going to tend to line up in that direction.
That's what that Guinea fowl study was kind of about.
OK?
Are we good?
OK. And then these other two, so in B-- and you can't really see it from here, but they looked at the sort of architecture of the trabecular bone.
And they said that it was more similar in the fossil in the human than it was in these other three primates.
And this region here, this looks very similar to that bit there to me.
But I think there was some other feature about this region that they were looking at.
And again, it was more similar to the human than it was to the non-human primates.
So by just looking at the pattern of the bone and the density of the bone and comparing it to these other species, they said that the-- and you know the hip, because we're standing like this, you would kind of expect to see differences in the hip.
That's why they wanted to look at the ilium.
So the conclusion they made was that these observations suggested that the species was bipedal, or at least spent some of its time as a bipedal animal.
And I think that might be it.
Do I have more I have one little summary here.
So just to summarize what we've done on bone this week.
We talked about the structure of the bone.
We talked about mechanical properties in the foam models.
We talked about osteoporosis in the voronoi models, how you can try to represent the loss of bone and the loss of bone strength using those models.
We talked just a little bit about the idea of using metal foams as bone substitutes or coatings as implants, and then this little bit on trabecular architecture and evolutionary studies.
I have some notes, but I think we've got just a couple minutes left.
So maybe I'll just scan those and put them on the website?
Are we good with that?
I have a question about what we just looked at about the different species.
We always consider on the density changes.
Can there always be changes in the solids?
So it changes a little bit from one species to another.
So the question is, does the solid properties of the bone, the solid bone itself change?
So if you look at cortical bone in different species, it changes a little bit, but like 10%, not a huge amount.
So the two most common things people have compared are bovine bone and human bone.
And there is a slight difference between them.
But it's not a huge difference.
One of the other things people have looked at in osteoporosis that I didn't really talk about was there's another whole set of things that can go on that reflect what you're talking about.
So they talk about bone quality as well.
And when they talk about bone quality, they're talking about are there micro cracks in the solid.
So you might imagine as you get older, it's not just that you lose the struts or that the struts get thinner, but the solid bit itself has more cracks in it.
So you can imagine if the solid bone itself had little micro cracks, then it too would be weaker.
And then you think what I put up was bad.
It gets even worse if you put that in as well.
So, yes, people do look at bone quality, which is sort of looking at with age.
And typically it's fairly elderly people that the bone quality is an issue.
I guess there are certain diseases where it's an issue.
But in osteoporosis, it's sort of elderly people.
Any thing else?
Should I stop there for today?
So there were hardly any equations in this.
Did you know that?
So we got to the part where there's lots of equations.
So next week I'm going to talk about tissue engineering.
I think I'm going to talk a little bit about different kinds of scaffolds, how they make scaffolds, how the scaffolds fit sort of into an anatomical things, what is that supposed to represent, how they use them clinically a little bit.
And we had a research project on osteochondrol scaffold, so scaffolds for replacing bone and cartilage.
And I'm going to talk a little bit about that as sort of a case study in tissue engineering scaffolds.
And I have some stuff on cell mechanics and how biological cells interact with scaffold.
I don't if we're going to get to that next week or not, but somewhere close to that.
So the next bit is on tissue engineering scaffolds, osteochondrol scaffolds, cell mechanics.
And there's not that many equations in that part either.
So OK.
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So last time, we finished talking about trabecular bone.
And what I wanted to talk about this week was tissue engineering scaffolds.
So the idea with tissue engineering is you want to be able to repair damaged or diseased tissue.
And typically, that's done by regenerating the tissue in some way.
So in your body, many types of cells, like maybe not blood cells, but most types of cells for sort of structural tissues are attached to an extracellular matrix.
And this is sort of a schematic of the extracellular matrix here.
And the composition depends on the type of tissue that's involved.
For example, in skin, it would be collagen and something called glycosaminoglycans and elastin.
In bone, it would be collagen and a mineral-- a calcium phosphate mineral, hydroxyapatite.
So the composition varies.
But the idea with the tissue engineering scaffold is that you want to make a material that essentially substitutes for the extracellular matrix.
And it does so on a sort of temporary basis.
So the idea is you put something in the body.
The cells attach to that, whatever scaffold you put in.
And the scaffold has to be made in such a way that the material-- that the cells, they can migrate through it.
They can attach to it.
They can differentiate.
They can proliferate.
So the cells can do all their normal function.
And the idea is that as the cells are doing their normal function, they then secrete the natural extracellular matrix.
And the engineered thing that you put in is resorbed.
So there has to be a balance between the rate at which the scaffold you've made resorbs and the rate at which the cells are depositing the new native extracellular matrix.
So that's one of the key things about this.
So this is just an example of sort of a schematic of an extracellular matrix.
In this case here, there's collagen fibers.
So these guys here collagen fibers.
And these kind of hairy-looking things are proteoglycans.
So they have a core of protein with sugars kind of hanging off of them.
And there's different kinds of GAGs, they're called, that hang off of them.
So one's chondroitin sulfate.
The CS here stands for Chondroitin Sulfate.
There's one called dermatan sulfate.
And there's one called heparin sulfate.
So there's different of these glycosaminoglycans.
So let me write down some of this, and then we'll kind of get into this.
So what I wanted to do today was show you some examples of tissue engineering scaffolds.
Show you some of the sort of design requirements.
So you have to have, obviously, a material that's porous so the cells can get in there.
And that's where the cellular solids comes in.
So we'll talk about some scaffolds, some of the sort of design requirements for them.
And also, we'll talk a little bit about processing of the scaffolds and mechanical properties of the scaffolds.
So I'm hoping I can finish most of that today.
And then next time, I have a little case study on osteochondral scaffolds.
So osteo means bone.
And the chondral means cartilage.
So this was sort of a two-layer scaffold that we developed in collaboration with some other people at MIT and some people at Cambridge University.
And it went from a research thing to a startup thing and being used clinically.
So I was going to talk about that Wednesday.
OK.
So let me just get started here.
So the goal of tissue engineering is to regenerate diseased or damaged tissues.
And in the body, the cells attach to the extracellular matrix.
And that's sometimes called the ECM.
Sometimes, the scaffolds are called scaffolds.
And sometimes, they're called matrix because the extracellular matrix is called matrix.
So then, the composition of the ECM depends on the tissue, but it usually involves some sort of structural protein.
So something like collagen or elastin.
It also typically involves some sort adhesive proteins.
So something like fibronectin or laminin.
And it involves these proteoglycans, which are the core of protein with a sugar hanging off them.
Whoops.
And the sugars are typically glycosaminoglycans.
And for short, people call them GAGs, just because it's easier to say.
So some examples of the GAGs are chondroitin sulfate.
And we're going to talk more about that a little later because that's one that we've used in making collagen-based scaffolds with [INAUDIBLE].
So for example, if you look at the composition of extracellular matrix in something like cartilage, it has a collagen component and a hyaluronic acid component and some GAGs.
And this hyaluronic acid is a proteoglycan.
If you look at bone, extracellular matrix in bone, it's made up mostly of collagen and hydroxyapatite.
And if you look at skin, skin is made up of collagen, elastin, and proteoglycans.
And the idea is that the cells have to be attached to this extracellular matrix in order to function.
Or, they have to be attached to this or to other cells in most cases.
So next week, I'm going to talk a bit about cell mechanics.
And we have some video where we watch a cell deforming a scaffold.
And I don't have videos to show you, but I had a student who took videos of cells migrating along with scaffolds.
We'll look at how the stiffness of the scaffold affects how the cells migrate along the scaffold.
So that's kind of the native ECM in the body.
And the idea with tissue engineering is that you want to make a scaffold that's porous that mimics the extracellular matrix in the body.
So, say, there's some tissue that's damaged.
Say there's damaged cartilage.
You want to provide sort of an extracellular matrix that's a sort of synthetic thing that's going to provide the same function as the ECM in the native tissue.
So people have been working on scaffolds for regenerating all sorts of different tissues.
And probably, the most successful one has been used to regenerate skin.
And there's been scaffolds available for regenerating skin for probably almost 20 years now.
And one of the first ones was developed by Professor Yannas in mechanical engineering here at MIT.
And it's actually still sold by a company called Integra.
But this research to develop scaffolds for lots of different tissues, orthopedic tissues, things like bone and cartilage, cardiovascular tissues, nerve-- like Professor Yannas works on peripheral nerve these days.
People have looked at trying to make scaffolds for gastrointestinal tissues.
So all sorts of different tissues.
And at MIT, there's quite a lot of interest in this.
There's a lot of people working on it.
So Bob Langer works on it.
Linda Griffith.
Sangeeta Bhatia.
There's really quite a number of people at MIT who work on it.
Those are just some of the people at MIT.
So the idea is in the body, the cells are constantly resorbing the extracellular matrix and depositing more.
So if you think about bone, for example.
Remember we said bone grows in response to load?
Even in healthy bone, the bone is constantly being resorbed and deposited.
And in normal bone, the rate of resorption and the rate of deposition is roughly the same.
When people get osteoporosis, the thing that happens is that that balance gets out of whack.
And so it's not being deposited at the same rate it's being resorbed.
And the idea with the tissue engineering scaffolds is that they degrade over time.
And that the cells that were attached to them are forming their own extracellular matrix.
So there's kind of a balancing act between the cells depositing the native ECM and the tissue engineering scaffold that was provided, say, by the clinician being resorbed.
So the scaffolds are actually designed to degrade.
And controlling that degradation rate is one of the design parameters of the scaffolds.
OK.
So that's kind of the overall, kind of big picture.
And what I wanted to talk about next was some design requirements for the scaffolds.
So if you think about it, there's different sort of ways you can think about what the requirements are.
So you have to make the scaffold out of some solid.
And there's some requirements for the solid.
So obviously, you want a solid that's biocompatible.
That's one kind of main requirement.
Another requirement is that not only the solid has to be biocompatible, but when the solid decomposes, if it decomposes into other components, they have to be biocompatible too.
So you don't want the solid to degrade during this resorption process into toxic components.
That would be a bad idea.
And then, the other thing is the solid itself has to promote cell attachment, and cell proliferation, and cell migration, all these kinds of things.
So we're going to talk about some different materials for the scaffolds.
And some of them are sort of native proteins.
So some of them are things like collagen.
Collagen already has binding sites for cells to attach to it.
Obviously, it's one of the proteins in the native ECM.
There's also a number of synthetic polymers you can use.
And with the synthetic polymers, they don't have natural binding sites for the cells.
And so you have to coat them with something else.
So you have to coat them with, say, adhesive proteins so that the cells will attach to them.
So we're going to talk about the requirements for the solid.
Then, you make the solid into some sort of porous, foamy thing.
I think I have some slides here.
Here's an example of a collagen GAG scaffold.
And this is one of the ones that is made in Yannas' lab.
And you can see it looks a lot like a foam.
It's very, very porous.
And there's some requirements for the sort of cellular structure, the foamy structure of the scaffold as well.
So typically, you want interconnected pores so it's easy for the cells to migrate in.
Typically, you want pores to be within a certain range.
It turns out if the pores are too small, it makes it difficult for the cells to get in.
Sometimes, they can by eating away at the material.
But typically, the pores-- you want them to be bigger than a certain size.
You also want them to be smaller than a certain size because how much specific surface area, how much surface area per unit volume you've got, depends on the pore size.
The smaller the pores, the more surface area per unit volume you have.
And then, the number of binding sites you have for cells to attach to it depends on that specific surface area.
I'm going to write all this down, so I'll do that.
So there's requirements for sort of the pore structure.
And then, there's also some requirements for the whole scaffold itself.
So for instance, it's got to have some minimal mechanical integrity.
So there's sort of some requirements for that.
So there's requirements for the solid.
There's requirements for the sort of cellular structure, the porous structure.
And there's requirements for the overall scaffold.
So let me write some of these things down.
So there's some requirements for the solid.
So it must be biocompatible.
It must also promote cell attachment and proliferation in the cell functions.
And then it must degrade into nontoxic components.
So there's some requirements for the cellular structure, too.
And what you want to have is a large volume fraction of interconnected pores.
And so you want that to facilitate the cell migration.
And also, the transport of nutrients into the cells.
So you also want the pore size to be within a critical range.
So you need the pores bigger than a lower limit so the cells can migrate through easily, can kind of get in there.
And you want the pore size to be less than an upper limit to have enough surface area to have enough binding sites to actually attach cells.
And for different tissues, there's different critical ranges of the pore size.
So for example, for skin they found that you want to have a pore size between about 20 microns and about 150 microns.
And for bone, the pore sizes that people tend to use are between about 100 and 500 microns.
So there's the pore size.
And one other feature is that the pore geometry should be conducive for the cell type.
So lots of cells are somewhat equiaxed.
Maybe a little elongated.
But if you look at something like nerve cells, like peripheral nerve, they're incredibly elongated.
So you want to have pores in the scaffold that are also very elongated.
And then for the overall scaffold, it needs to have some mechanical integrity.
You guys OK?
Yeah?
We're good.
Oh, achy.
Yeah.
I know.
So you want to have some overall mechanical integrity.
I mean, the thing has to be put into the body in surgery.
And people are going to be pushing and poking at it.
And so it has to have some just overall mechanical integrity.
Also, it turns out that if you put stem cells into scaffolds, the types of cells they differentiate into depends in part on the stiffness of the scaffold.
So you want to be able to control the stiffness of the scaffold
How do they do this research?
Do they use animals?
Or they do it in vitro?
Yeah.
So the question is, how do they do the research?
So they do a sort of series of different things.
So at one level, you could have the scaffold and you'd put cells on it.
Say you're making a bone scaffold.
You'd put osteocytes onto it.
So one level, you just put cells onto it.
And you want to see, are the cells attaching?
Are they dying?
Are they proliferating?
So sometimes, what people will do is put the-- they'll seed a certain number of cells at a certain time.
Say, time 0.
Then, they'll look at how many cells are attached at 24 hours or 48 hours.
And you kind of see the cell attachment.
You can measure relatively easily.
Another thing people do is animal studies.
So for instance, Yannas does research on peripheral nerves and scaffolds for peripheral nerves.
And they cut a piece out of the sciatic nerve of rats.
So obviously, they have a surgeon and do a surgery thing with it.
You can't just kind of do this in the lab.
You've got to get permissions and stuff to do it.
And then, they put in the scaffold.
And the scaffold's actually in a tube.
And so they put the two stumps of the nerve end at either end of the tube, and then the tube's filled with a sort of porous scaffold that we're talking about here.
And then, they wait some period of time.
And they take video of rat running, things like that.
They then sacrifice the rat and they do histology.
And they look at the sort of cross-sections and see what it looks like.
And so I'm going to talk next time about this osteochondral scaffold we worked on.
So we did cell studies.
We did goat studies.
We put into goat knees.
There was a longer term sheep study.
And then, the student who's in Cambridge, England, started up a company and he ended up getting approval in Europe to start clinical trials.
And then he worked with an orthopedic surgeon who started putting it in people.
But typically, they're looking at cells, looking at animals before you get to the people stage.
And one of the things people do when they're making these scaffolds is you want to use materials that already have some sort of regulatory approval.
So say FDA approval or approval in Europe.
So typically, people don't start with a brand new material from scratch.
Because to get approval for that would just take a very long time.
So typically, people start with-- the solid material is already approved for some other sort of use.
OK.
So one requirement for the overall scaffold is it has to have sufficient mechanical integrity.
Sufficient.
And then also, as I mentioned, the stiffness of the scaffold can affect differentiation of cells.
And the other thing that is really a factor for the overall scaffold is you want to control the rate of degradation of the overall scaffold.
So you want that rate to be matched to the rate at which the new tissue is forming.
So it has to degrade at a controllable rate.
OK.
So I want to talk about the materials that people use.
And you can kind of break them down into a few classes.
So one class is natural polymers.
So things like collage.
So you can get collagen.
And that's an example up there of a scaffold that's made with collagen.
Another class of materials is synthetic biomaterials.
And if you've had stitches or surgery, you may know that some of the sutures they use are resorbable.
So some of those polymers that they use for resorbable sutures are also used for tissue engineering scaffold.
And then, there's also hydrogels that people use as well.
So those are probably the three main groups are sort of natural polymers, synthetic biopolymers, and I guess the hydrogels are sort of a subset of the sympathetic biopolymers.
So collagen is probably the most common kind of natural polymer that's used.
They also use GAGs.
And this scaffold up here is made by making a coprecipitative collagen with a GAG chondroitin sulfate.
People also use alginate.
I think one of the project groups-- you guys are going to make some sort of alginate scaffold, right?
No?
[INAUDIBLE] foamy thing?
Yeah.
And people also use something called chitosan, which is a derivative of chiton, which is what's in the exoskeleton of insects and things like lobsters.
So those would be all examples of natural polymers that can be used and people have tried.
I'm going to talk a bit more about collagen, just because it's the most common one.
So collagen is a major component in the natural extracellular matrix.
And not surprisingly, it has binding sites for cells to attach to it.
So if you use that, that kind of takes care of that issue.
Let me put it down here.
So collagen exists in many types of tissues.
Exists in skin.
Exists in bone, cartilage, ligament, tendon-- cartilage.
So it's very common.
It has surface binding sites for cells.
It has a relatively low Young's modulus.
So the Young's modulus is a little less than a gigapascal.
But you can increase the modulus by either cross-linking or by using it in conjunction with some synthetic polymers.
And I'm going to talk a little bit about how you make these scaffolds up here.
And the first step in making those scaffolds is you put the collagen in acetic acid, and then you add the glycosaminoglycan and it forms a coprecipitate.
And the fact that it forms a coprecipitate with the glycosaminoglycan, the GAG, means that you can use a freeze-drying process.
And that's how that's made.
Collagen is one option.
So then synthetic biopolymers is another option.
And as I said, typically they use the materials that are used for resorbable sutures.
So there's several of those.
There's something called PGA.
That's polyglycolic acid.
And something called PLA.
That's polylactic acid.
And then you can combine those two and make something called PLGA polylactic co-glycolic acid.
And you can control the degradation rate of these things by controlling the molecular weight.
And in this case, you can also control it by controlling how much of each of those things you put in.
And there's another one called polycaprolactone.
So those are several synthetic biopolymers that people use.
There's lots of different materials, but these are just some typical ones.
And then another class are hydrogels, which are produced by cross-linking water soluble polymers to form an insoluble network.
And those are typically used for soft tissues.
Sometimes, they're used for things like cartilage.
And again, there's a few different materials that are commonly used.
One's PEG, Polyethylene Glycol.
One's PVA, Polyvinyl Alcohol.
And another one's PAA, Polyacrylic Acid.
So for these synthetic polymers, there's many different processing techniques available.
And I'll talk a little bit about some of the processing techniques.
But one of the limitations is they don't have natural binding sites.
And you have to coat them with some sort of binding agent, like an adhesive protein, to get the cells to attach to them.
And then, as I mentioned before, you have to make sure that whatever material you choose, if it's a synthetic material that when it degrades, it's not toxic to the cells.
Because you don't want to have some sort of toxic reaction or inflammation.
OK.
So there's a couple more things about materials.
So those are all polymer-based materials.
When people are trying to make scaffolds for bone tissue engineering, they also include a calcium phosphate mineral.
And there's different versions of the calcium phosphate.
So they can include-- you can buy, for instance, hydroxy powders now.
There's another calcium phosphate called octacalcium phosphate, which will, with water, turn into hydroxyapatite.
So typically, there is this mineral, some sort of calcium phosphate, is combined with either collagen or with one of these synthetic biopolymers.
And one other option is something called an acellular scaffold.
And what that is they take some natural tissue and they remove all the cell material from it.
And so when they remove all the cell material, what they're left with is the native ECM.
And that's called an acellular scaffold.
So it's a native ECM with all the cell matter removed.
And they remove the cells by-- they can use sort of a physical agitation or chemical, or using enzymatic methods.
Using something like trypsin to get rid of the cell.
So there's ways that they can get rid of the cells.
OK.
So are we good so far?
So there are some requirements for what materials we kind of use, what the cell structure should be, and these are some examples of typical materials.
So I wanted to talk a little bit about the processing of the materials.
Let me wait until people catch up a little.
Oh, and I have some scaffolds I was going to pass around.
So this big sheet is a piece of the collagen GAG scaffold that I showed a minute ago.
And then this little piece is a mineralized version of that.
So this has the collagen plus calcium phosphate plus hydroxyapatite in it.
OK.
So this slide shows some examples of different scaffolds that people have made.
And I was going to talk a little bit about some of these methods.
And why don't I talk about them, and then I'll write some notes on the board.
So this top one here on the top left, that's the collagen GAG scaffold that's made in Yannas' group.
And that's made by a freeze-drying process.
So you put the collagen in acetic acid, then you put in the GAG.
The GAG and the collagen form a coprecipitate.
And then you can freeze that.
And if you freeze it, what happens is-- it's just like if you freeze saltwater.
The water freezes.
The pure water freezes.
And you've got increasingly higher brine content in the bit in between the water grains, or in between the ice.
So you get the sort of solid ice forming.
And the collagen and the GAG are kind of squeezed into the interstitial bits between the ice crystals.
And then if you sublimate the ice off, you're left with this porous kind of structure that looks like a foam.
So that's made by a freeze-drying process.
And I'll go over it in a bit more detail when I write the notes on the board.
You could also foam some of these polymers.
So just blowing a gas.
The same way you can blow a gas through engineering foam.
You do the same thing with some of these polymers.
This one's made by foaming.
You can have a fugitive phase process.
So this is made by salt leaching, the second row on the left there.
So you could imagine you could take a polymer powder.
You could mix it with salt.
You can sort of mix them up, combine them together.
You heat it up to get the polymer to melt and to sort of form a connected mass.
And then you leech out the salt.
And then you get pores where the salt was.
This one here is made by an electrospinning process.
So you have a nozzle.
You feed the polymer through the nozzle.
Then you have plates that are charged and you get fibers forming and kind of scattered in different directions by the electrospinning process.
Then, this one here represents scaffolds that are made by things like 3D printing, selective laser centering.
You can have laser-sensitive polymer.
And you can produce scaffolds that way.
I think the geometry of this one matches some part in the body.
I think it was a knee or something like that.
And then, these two examples down here are the acellular scaffolds.
That's what I was talking about at the very end there.
So those are from porcine pork heart tissue.
You know, pig heart tissue.
And those are mostly elastin.
And they've had all the cell matter removed from them.
OK.
So you can kind of see that these synthetic scaffolds here have a structure that's not so different from these native ECM scaffolds down here.
OK.
So let me write some of the things about the processing on the board.
Let me rub this off.
Start over here.
So these freeze-dried scaffolds are used for skin regeneration.
And I think I have some more slides here.
So it's kind of a two-step process.
In the first step, you make what they call a slurry.
So you make the slurry by taking the collagen.
And for skin, I think you want type 1 collagen.
Yeah, skin is type 1 collagen.
There's different types.
You put it in acetic acid, and then you add the GAG.
And we use chondroitin 6-sulfate is just the particular GAG that we use.
And one of the things that the acid does is that it swells the collagen.
And collagen has a sort of periodic structure in it, sort of periodic banding.
And the acid destroys that periodic banding structure.
And that helps increase the resistance to having some host immune response.
So that you remove the immunological markers and it makes it less likely that the scaffold's going to get rejected by the body.
So then when you put the GAG in, you form a coprecipitate.
So this next step just shows kind of mixing the whole thing up.
And then you've got kind of a little slurry that you can store.
And then, this is the freeze-drying step here.
So you put the slurry, the suspension, into a pan.
Kind of just like a cookie sheet, really.
And then you freeze it.
So if you think of this phase diagram here, where you have temperature and pressure.
So here we have liquid, solid, and vapor.
So if you start off at this point here, you freeze it.
So you've reduced the temperature.
So that forms the ice.
And the ice is surrounded by the collagen and the GAG fibers.
And then if you do the sublimation step, you reduce the pressure and increase the temperature a bit.
And then you get over to the vapor end of the world.
And then you're left with this porous scaffold.
And then, let me see.
Let me do one more step here.
And then you can control the size of the pores by controlling the freezing temperature.
So the size of the pores is exactly related to the size of the ice grains that are forming.
And the faster it freezes, the smaller the grains are going to be.
And then, the smaller your pore size are going to be.
So you can control that.
The type 1 collagen is mixed with acetic acid.
And it then swells the collagen and disrupts periodic banding.
And it removes immunological markers.
And then you add the GAG, the chondroitin 6-sulfate.
And then that cross links with the collagen and forms a coprecipitate.
And then you can freeze dry that to get the porous scaffold.
And typically, the relative densities of these scaffolds is very low.
So typically, they're 0.5% dense.
The relative density is 0.005.
So they're 99.5% air and the rest of it's the collagen and the GAG.
And the pore sizes are typically between about 100 and 150 microns.
And Yannas uses the same scaffolds for the nerve regeneration.
And he uses a directional cooling.
And that then elongates the pores, so that-- the idea is that they elongate so the nerves kind of grow along that length.
So that's one way.
Another way is leaching a fugitive phase.
So let's see.
I think-- yep.
Here we go.
Back to there.
So if you look at the one on the second row on the left, that's done by using salt as the fugitive phase.
People use other things.
You can use wax.
Paraffin wax works as well.
So it doesn't have to be salt.
There's different things you can use.
So you combine a powder of the polymer with your fugitive phase Say, salt.
Then, you heat it up to get the polymer to bind.
And then you leach out the salt.
So you can control the porosity by the volume fraction of the fugitive phase, and then the pore size by the size of whatever the fugitive phase is.
Another technique is electrospinning.
The idea is you produce fibers from a polymer solution that you extrude through a nozzle.
And then you apply a voltage across some plates to spin the fibers.
And then you get a network of these fibers.
And typically, their micron-scale diameter.
And the last method I'm going to talk about is rapid prototyping.
So you can think of using 3D printing.
Or you could use selective laser centering or stereo lithography using a photo-sensitive polymer.
So the idea is-- you know how this works.
You just build up layers of solid, one layer at a time.
And then you can make complex geometries with that.
So that's one of the advantages.
If you wanted to make a part to fit a particular place in the tissue, then it's convenient that you can control the geometry of the whole part.
OK.
So that just kind of summarizes very briefly, kind of how the tissue engineering scaffolds are meant to work, what kinds of materials people make them from, and a few of the processes.
And there's many, many processes.
These are just sort of a few common processes.
I wanted to talk a little bit about the mechanical behavior because that's kind of what I do.
And so this next plot just shows a stress strain curve in compression for a collagen GAG scaffold.
And I'm hoping that by now you're getting the idea all of these cellular materials have this kind of shape of a curve.
So there's the same kind of linear elastic regime, and then a collapse plateau.
These collagen things, as you can imagine just pressing them in your hand, they fail by buckling, by inelastic buckling.
And then there's a densification regime.
So they look like all the other kinds of curves that we've got.
One of the things we've done in the modeling is typically in the model, we want to be able to calculate or measure the modulus of the solid or the strength of the solid from which the thing's made.
So [?
Brendan ?]
Harley was one of my PhD students.
And he took a little microscope, cut a little strut.
The struts are very small.
The pore size is 100 microns.
So the struts are on that order.
He glued one end of the strut to a glass slide, and then he used an AFM probe to do a little bending test on that little strut.
If he could measure the deflection, he knew the length.
He knew the geometry of the strut.
He could figure out what the modulus was for the solid.
So he backs out what the modulus is.
So he did these measurements on a dry strut.
And then by comparing the overall modulus of a dry scaffold with a wet scaffold, he estimated what the modulus of the wet scaffold or the wet strut would be.
So the modulus of the dry collagen GAG was 672 megapascals.
A little less than a gigapascal.
And wet it was about 5.
So there's a huge difference between the wet and the dry.
OK.
So let me just write a few notes about that.
So in compression, there's the three regimes that we see for all these cellular materials.
And so you can estimate the modulus by using the model we have for the foam for the modulus.
The modulus of the foam divided by the modulus of the solid goes as the relative density squared.
And that's related to bending in the cell wall.
And then, the collapse plateau is related to elastic buckling.
And so that's equal to some constant times E of the solid times the relative density squared.
And that's related to elastic buckling.
And then we measured E of the solid doing this little AFM beam bending test.
OK. And for one of these low-density scaffolds, we measured the modulus.
We measured the buckling strength.
And we got pretty good agreement by using these equations here.
And the good agreement was if this constant was 0.2.
That was the strain, that buckling.
Yeah
So what does it mean to be wet in here?
And why is it so much lower?
Well, we make the scaffolds by this freeze-drying thing.
And like this thing I passed around, that was dry.
We just immerse it in water.
We just put it in water.
And then, he does the test.
So he does the test on the whole scaffold dry, and then he does the test on the whole scaffold wet.
And I assume there's some sort of bonding that gets disrupted by having the water.
I don't know the details of how that works, but there must be some change in the bonding to make that happen.
So let's see.
I'm trying to see.
What else should we do today?
Oh, yeah.
So we get pretty good agreement with this sort of simple foamy model.
One of the things we did find was that when we tested higher-density scaffolds, the agreement wasn't so good.
And I think this was because when you get higher density, it's just hard to get the collagen GAG mixture to mix in with the acetic acid.
And so we ended up getting inhomogeneous scaffolds with sort of large voids in them.
So in order for the modeling to work, you have to have a scaffold that's relatively homogeneous.
You don't have kind of big defects in it.
I think that's all I'm going to say about that.
I have one more slide here.
So those are sort of three-dimensional scaffolds we've been talking about.
Sort of foam-like scaffolds.
People have also made honeycomb-like scaffolds as well.
And this just shows some examples of some honeycomb-type scaffolds.
So this one here, I think was made in Sangeeta Bhatia's lab.
You can sort of think of it as a hexagon, but it's also kind of triangulated as well.
These two here were made by George Engelmayr.
He worked with Bob Langer at one point.
And these two scaffolds here, they're both sort of rectangular cells.
And they were designed to look at how the cell geometry or the pore geometry affected the sort of morphology of the cells that attached.
So if you have different, say, [?
porous, ?]
you get different morphology in the cells that you're trying to attach.
And then, George Engelmayr also made these scaffolds here.
And those were designed to be anisotropic and have different mechanical properties in different directions.
And I think what they had done was try to match the anisotropy in the mechanical properties to anisotropy in heart tissue, in cardiac tissue.
So these are some examples of honeycomb-type scaffolds.
So let me just write down a few things about those.
So I think these were more-- obviously, the scaffolds are used-- the ultimate goal is to use them clinically.
But sometimes, people make scaffolds just to study cell behavior.
And some of these, I think, were made just to study how the cells would behave on them.
So they're sort of idealized to do that.
So that triangulated.
It kind of looks like a hexagon, but there's also sort of triangles in there.
I think the thing they were looking at with that was transport of nutrients to cells.
And from a mechanical point of view, if it's triangulated you'd expect, say, the modulus of that to go linearly with how much solid there is there.
So the rectangular honeycomb and the diamond shaped pores, they were used to study the effect of pore geometry on the cell orientation.
They used fibroblasts.
And I'm going to call it the accordion-like honeycomb.
That's mechanically anisotropic.
And the mechanical anisotropy is matched to the cardiac tissue.
OK.
So I think I'm going to stop there for today.
That's sort of the end of this part.
And next time, I'm going to talk about the osteochondral scaffolds.
I don't think it really makes sense to start it for two or three minutes.
So I'll start that tomorrow.
Or Wednesday.
And we should be able to finish that on Wednesday.
So one of the things that these honeycomb-type scaffolds kind of suggests is that the scaffolds are used both to try to regenerate tissue in the body in clinical applications, but they're also used as sort of an environment for cells, in order to study cell behavior.
So next time, I'm going to talk about an osteochondral scaffold.
And the idea with that was to try to use it clinically.
But next week, I'm going to talk about cell mechanics a bit.
And when people study cell mechanics, or look at the mechanics of biological cells, not the cellular structures.
So when people look at trying to study how cells behave, they need some environment to put them on.
Typically, people started by just using flat 2D substrates.
But the flat 2D substrates are kind of easy to study, but they don't really represent the tissue in the body.
And so people are now using tissue engineering scaffold as an environment that they can control to study how cells behave.
So they study cell attachment, cell proliferation, cell migration, and cell differentiation all by using the scaffolds as kind of a controlled environment.
So we'll talk about that, probably not-- I don't know if we'll get to it on Wednesday.
But we might start it on Wednesday and finish it next week.
OK?
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What I thought I would do today is two things.
I wanted to give this talk about osteochondral scaffolds, and this is just slides.
I'm not going to write anything down.
So there aren't going to be any notes.
And I'll put the slides on the Stellar site, I don't know, this afternoon or tonight or something.
And this isn't going to take the whole time we have today, so the other thing I thought I'd do is I wanted to walk you through this little booklet that I handed out about how to write a paper.
So this is by Mike Ashby, who you know, I've written books with.
He was my PhD advisor and we wrote the books on cellular solids together.
He has written many books.
I looked him up on Amazon this morning, and even though I know he'd written many books, I was shocked how many books.
They had 58 listings for just books with him as a co-author and some are second editions and third editions.
He's written a lot of books.
He's written a lot of materials, papers.
And he's, in the materials community, he's seen as a very clear and lucid writer.
So he's written this thing for his students and I thought we could just walk through it and I can talk to you a little bit about writing.
Because I know you're not quite there with your projects, but later on in the term you're going to want to write up your project, and it's not too early to start thinking about writing.
And there's some really good advice and it's short and it's to the point.
It's really helpful, that little brochure.
So we won't read every single thing, but I wanted to kind of walk through some of the main points in it, and that pretty much should take the hour.
So last time we talked about tissue engineering scaffolds, and today, I wanted to do a case study, and this was a project we had here at MIT and in collaboration with some people in the materials department at Cambridge University.
And we made what we called an osteochondral scaffold.
So osteo means there was a part for regenerating bone and chondral means there was a part for regenerating cartilage.
And really the point of the scaffold was to try to repair small defects in cartilage.
And I'll explain why we did the bone thing as well.
So this is just a little outline.
I'm going to talk a little bit about what the structure of cartilage is.
We have a little schematic of that.
Then talk about how small defects in cartilage are currently treated.
So these are things that, say you're an athlete, you might tear some cartilage, that kind of thing, not like you have osteoarthritis and you need a new knee.
It's not going to do that.
So I'll talk a little bit about cartilage and the current treatments.
I'll talk about some of the things we thought about in making an osteochondral scaffold, like what parameters were important.
And we based it on that collagen-GAG scaffold that I talked about last time.
So the collagen glycoseaminoglycan scaffold that we talked about last time.
And what we did was we had a layer that was a collagen based scaffold for the cartilage and we had another layer that was a mineralized version of that.
And one of the main things we did in this project was figure out how to mineralize that scaffold.
And then we made this two-layer osteochondral scaffold.
So I'll talk about that, OK?
So are we good?
So this is a schematic of articular cartilage.
Articular just means it's in a joint, so between like, say, two of your long bones, and there's several regions.
So this shows both the cartilage and the bone underneath it.
And so this top layer is called the superficial layer, and then there's a transition zone, and then there's this deep cartilage here.
And the little white lines represent how the collagen is oriented in the cartilage.
So the collagen is oriented, more or less, vertically here and then it becomes more kind of woven and horizontal towards the top.
And so those three different zones, the collagen is oriented differently.
Then there's a region down here called the tidemark.
Everything above here is just cartilage and everything below is calcified to some extent.
So this next layer down here it's more cartilage-like, but it's got some calcification in it as well.
And then here's the compact bone.
They call it subchondral bone.
It's the bone below the cartilage.
So chondral is cartilage.
And then here's the trabecular bone.
We talked about trabecular bone a couple of weeks ago, OK?
So one of the things is there's different types of collagen and the different types may have slightly different fiber structures or slightly different compositions.
They're all related, but they're slightly different types.
And bone has what's called type I collagen and cartilage has type II collagen.
So when we made the scaffold, we wanted the bony layer to be made with type I and the cartilage layer to be made with type II.
So I think with the 3032 people have probably seen it.
I think I did a version of this for you guys, didn't I at the end of term, something?
Yeah, but there's other people, so it's really for them.
So if you think of articular cartilage, it has difficulty repairing itself and one of the reasons that it's difficult for the cartilage to repair itself is that there's no blood supply in it and another reason is that there's not very many cells.
So chondrocytes are the cells in cartilage and if there's a low volume fraction of the cells, then it's not so easy for that small number of cells to actually produce the extracellular matrix, which is kind of what you think of as the cartilage.
And it can be damaged, either as I said, from sports injuries, typically that's what young people get, they tear their cartilage in some sporting injury accident, or from osteoarthritis.
So these scaffolds we're talking about, they're not really meant to repair large amounts of cartilage that are damaged.
And as I said, the cartilage has a poor capacity for self-repair.
So there's several treatments, there's three treatments that are given currently and the most common one is called marrow stimulation.
And some of these orthopedic treatments are fairly crude when you look at it.
So this marrow stimulation, what's involved is they take a drill.
So here's the drill.
And they basically drill through the cartilage.
So this would be the cartilage layer here.
And they drill down into the bone and they want to get down into the trabecular bone.
So they want to go below the cortical or the subchondral bone and they want to go down into the trabecular bone.
And the reason they want to do that is the trabecular bone has bone marrow in it and the bone marrow has mesenchymal stem cells, and they want those mesenchymal stem cells to move up into the cartilage layer.
So that's what this red glob is here.
The idea is that you've made a hole and now that hole is going to fill up with a blood clot, which will have these mesenchymal stem cells, and that those will form cartilage.
So it does work to some extent, but it's not really a great result.
There's something like 75,000 of these done a year.
As I said, this is the most common kind of repair.
The next most sophisticated type of repair, what they do is they take plugs of bone and cartilage from another place.
So say this is where the defect is and they want to repair that.
They take bone and cartilage from this spot up here.
So they sort of drill out little cylindrical cores and then they plug them into this bit here.
And that's called an osteochondral autograft.
Sometimes it's called mosaicplasty because they build up a little mosaic from all those little pieces.
And then they just leave these donor sites, where they took the plugs from, they just leave those empty.
So they try to take the bone and cartilage from regions where the loads are lower, but they end up leaving holes, which is not so desirable.
And again, those holes may fill a little bit by the previous mechanism.
Then the most fancy method that they use now is called autologous chondrocyte implantation.
So what they do is they harvest cartilage cells from the patient.
They then take them to a lab and they culture them for two or three weeks and they get them to multiply and proliferate and grow and then they re-implant the cells.
So this works fairly well, but the difficulty is it involves two surgeries, so one to get the cells out and one to put them back in, and there's the cost of the cell culture.
So this is a much more expensive procedure because of the two surgeries and the cost of doing the cell culture, but I think it's the method that works the best.
And when I talked about this in 3032, I mentioned Dara Torres.
Dara Torres is an Olympic swimmer.
She's 47.
She began swimming in the Olympics in 1984, more than 30 years ago.
And she has swum in the Olympics, not every one, but up until 2008.
And even 2008, she won three silvers that year.
She was, I think, the oldest person who's ever won a medal in the Olympics.
And she had this surgery done at the Brigham a few years ago.
And it used to be, they've stopped running these commercials but the Brigham for a time was running these commercials, that featured Dara Torres and saying basically that she had this surgery done there.
I'm assuming she was happy with it.
OK, so that's what they do currently and what we were thinking of and what other groups have thought about is could you repair damage in the cartilage by using a tissue engineering scaffold?
So what we were thinking about when we tried to do this project was we wanted to use a healthy articular cartilage joint as a model for our scaffold.
We wanted to have a layer that would go down into the bone so that you would have access to those mesenchymal stem cells.
And that's why we wanted an osteochondral scaffold, so that we would have a layer for the cartilage but also a layer that would go down into the bone.
We wanted to be able to control scaffold parameters, things like the mineral content and the pore size.
Remember, I said these tissue engineering scaffolds, the pore size is one of the parameters, that you want to have the pore size in a certain range.
And we wanted to use materials that would be appropriate for approval from things like the FDA.
So typically, when people make tissue engineering scaffolds, they don't start with some material that's never been approved before.
They start with something that already has approval for something else and that's what we wanted to do too.
So this was our idea of what we wanted the scaffold to look like.
We wanted an unmineralized type II collagen scaffold up here that would be for the cartilage.
We wanted a mineralized type I collagen scaffold down there for the bone.
And we wanted some region that had some gradient in mineralization because it would be like that layer of cartilage that was slightly mineralized.
So we wanted to duplicate that whole structure.
So that was our picture of what we wanted to do and we had a pretty good idea of how we were going to make this.
We were just going to use that same process that [INAUDIBLE] developed for the skin scaffolds.
He was involved with this project but just used type II collagen instead of type I.
The challenge was really figuring out how to make the mineralized collagen scaffold and then how to get this gradient in the mineralization between the two layers.
So this, I think, I showed you last time.
So this is just the method we used to make the collagen-GAG scaffold.
So we take type II collagen.
We put it in acetic acid.
Remember, that destroys the periodic banding of the collagen and it improves the immunological response.
We add the chondroitin 6-sulfate, the GAG, to crosslink it.
We then just make a slurry out of that.
So we mix that all together.
We keep it as a slurry.
And then the second stage is you put the slurry or the suspension into a pan, and then you do the freeze drying process.
So you go through this process where you start at this room temperature and atmospheric pressure.
You cool it down to freeze it, and then you sublimate it, and you're left with a very porous collagen-GAG scaffold.
So these are pictures.
I think this was with a type I collagen, but it looks the same with the type II collagen.
So that's what the structure looks like.
I think I showed you this last time.
We can control the pore size by controlling the freezing temperature.
So the way the freeze dryer works is there is shelves that you put these pans on, and these cooling elements go through the shelves, and you can set the temperature of those shelves.
So we would set the temperature of those shelves to different values and we got different pore sizes.
So the colder the shelf temperature was, the faster the freezing, and the smaller the size of the ice grains, and then the smaller the size of the pores.
And this is just the mechanical response again.
So we did mechanical tests on it.
These are some of the numbers for the properties.
So we tested it dry and wet, and we measured a modulus and a buckling collapse stress for it.
So those are just some values there.
And then it came to making the mineralized collagen-GAG scaffold, and one of the students in Cambridge, England was really the main person who did this, Andrew Lynn.
And he worked with Brendan Harley, who was our student here, and after they graduated, I had another student, Biraja Kanungo, who worked on this too.
So Andrew Lynn really developed this technique.
And this involved taking the collagen and the GAG, so the same as for the other scaffold, but this time, if we want to make a mineralized scaffold, we somehow have to get calcium phosphate into it.
So it's got to have sort of a hydroxyapatite-ish type of component to it.
So this time we used phosphoric acid instead of the acetic acid and we put some calcium salts into the mixture as well.
And Andrew was really the one who figured out how to do this and what salts to use.
And then the process was very similar.
So we have this slurry.
We mix the slurry up.
We did a freeze drying process, and then we crosslinked it with a chemical crosslinker called EDAC.
So it's just a chemical that you put into this and it crosslinks it all.
So that was how we made the mineralized scaffold.
The mineral we got was something called brushite, which is a calcium phosphate, but it's not exactly the same as hydroxyapatite.
And we could control the amount of brushite by different weight fractions or volume fractions by controlling how much of the calcium salts we put in and what the molarity of the phosphoric acid was.
If you take brushite and you put it in water, it then converts to octacalcium phosphate and then to apatite by a hydrolytic conversion.
So the apatite is related to the hydroxyapatite in bone.
And this was the structure of the mineralized scaffold that we got.
So you can see it looks a little different from the collagen-GAG scaffold.
It's much denser.
Typically, the densities were like 5% or 10% dense.
And remember, the collagen scaffold was 0.5% dense.
So it's a lot denser.
But if you notice, there's a few things to notice here.
So one is the pore size, that's a 500 micron bar.
And we could make pores between about 50 and 1,000 microns, depending on the freezing conditions.
And the range of pore sizes we were shooting for was somewhere between 100 and 500, so we could get in the right ballpark with that.
And you can see, just looking at that picture, if that's 500 microns, those pores are somewhere of that order.
Another thing to look at is this image here, and this just shows, the white little dots are the calcium phosphate mineral, and it just shows that the mineral is uniformly distributed throughout the thickness of the scaffold.
So some people had tried to make scaffolds for regenerating bone where they take, say, one of those polymers for resorbable sutures or they take collagen and they coat it with hydroxyapatite, but the shortcoming of that is that the scaffold is going to resorb over time.
The cells are going to secrete enzymes, which are going to eat away at the scaffold.
And if the scaffold resorbs, you're eating away at the hydroxyapatite first and then you're left with whatever polymer is underneath that.
Whereas this gives you a more uniform composition throughout the thickness of the scaffold, and you've got calcium phosphate everywhere throughout the thickness of the struts in the scaffold.
So that's the structure of that.
We wanted to make sure that the mineral was uniformly distributed throughout the scaffold, and we did some micro-CT imaging, some micro computer tomography.
And this is our sample here, and this red line just says that is the plane at which this image was taken, and the black is the mineral.
And then here's a lower plane and here's another image.
And you can see the calcium phosphate's pretty uniformly distributed throughout that specimen there.
And this was just another way of looking at the same thing using EDX in an SEM to look at where the calcium was and where the phosphate was.
So again, this is uniform distribution of the mineral.
We did mechanical tests on these scaffold as well.
So we measured moduli and collapse stresses.
We get the same kind of stress strain curve as all these other cellular solids.
This was done by- Biraja Kanungo was the student who did this bit.
One of the things we found was that with these mineralized scaffolds you could manually compress them.
If you pushed them down and you hydrated them, they would recover all the deformation, but we were increased in improving the mechanical properties of the mineralized scaffold for improved handling during surgery.
And there's another reason that I'll get to as well.
So the second reason is that, if you look at how bone itself forms, it forms from a collagen precursor, so bone in your body.
And there's something called osteoid, which is this collagen-based precursor to bone, and it has a modulus of about 25 to 40 kilopascals and Angler showed that if you have mesenchymal stem cells and you put them on substrates of different stiffnesses, they differentiate into different kinds of cells, depending on the stiffness of the substrate.
So the substrate stiffness can affect what kind of cells you get.
And the idea here was we thought, well, if we could get a stiffness that was close to this osteoid, what the natural bone formation has, then that might help the mesenchymal stem cells differentiate into the bony cells that we want them to.
So we wanted to try to reach a stiffness of this in the wet state.
And if I back up here, you can see the stiffness we had was around four in the wet state, four kilopascals and we want to get to 30 or 40, something like that.
So these are our equations for the modeling of the mineralized scaffold, the open celled foam models for the modulus and for the collapse strength.
So we could change different things.
We could change the solid properties or we could change the relative density.
The geometry of the thing is probably not going to help us too much.
So basically, that's what we did.
We first started off trying to increase the mineral content.
We thought if there was more mineral content that would make it stiffer.
So Biraja made these more highly mineralized scaffolds and these are just some SEM images of those.
But the thing he found was that the properties actually got worse when he had the more highly mineralized scaffold.
The modulus went down and the strength went down, so that wasn't very helpful.
And when he looked into it more detail, he found that he had more voids in the cell walls and he had more disconnected walls.
This shows some of the micrographs, so this isn't really very quantitative, but you can see there's holes here.
There's a few holes here, but there tended to be a bigger volume fraction of holes in the more highly mineralized scaffolds and more walls that were disconnected.
So we realized increasing the mineral content wasn't going to work very well.
And then the second thing he tried was increasing the relative density.
So we started off at this density of 4 1/2% dense and he developed a method of increasing the relative density up to about almost 20% here.
He did this by, first of all, he tried to just mix more of the constituents into the slurry, that's kind of the most obvious thing.
But as you add more constituents, it gets harder and harder to mix the thing up and have it homogeneously distributed.
So in the end, that's not how he made these things.
He started with the starting mixture and then he had a vacuum system for sucking water out of it.
So he would reduce the amount of water, which essentially increased the amount of solids that was in there.
And you can see in this last one here, the most dense scaffold, he was sucking the water in one direction, and he sucked it so much that he was starting to get the cells collapsing.
So this one here, these cells that have this sort of elongated orientation, that's because the cells are starting to collapse because of the vacuum that he was applying.
But he could get a pretty good difference.
This was almost 5%.
That's almost 20%, so roughly a factor of four.
There are, yeah, four difference between them.
So then he did mechanical tests too, and he measured the relative density.
Here's the moduli wet and then the strength dry and the strength wet.
So you can see here, if you look at the dry moduli, for instance, when you go from this relative density to that relative density, from about 14% to 19%, the modulus actually drops down.
And if you look at the structure, I think, it's because you've got this flattened structure here.
You've collapsed the cells a bit already.
So there's a maximum density that you might want to go to, probably somewhere around 14% or 15%.
But the wet modulus we've got here is around 35 kilopascals, so that's close to the target that we had.
It's close to what we wanted to have.
So one way you could get the right or the appropriate modulus is by increasing the density to that value.
Another way is by playing around with the crosslinking.
So those values were for non-crosslinked scaffolds.
So here, this is the 14% dense scaffold wet, non-crosslinked it was around 35 kilopascals.
This is a dehydrothermal treatment, just basically heating it up.
It gives you a higher modulus.
And then this is a chemical crosslinking technique that increases the modulus again.
So if you put all of this together, you can show these results for the modulus on one table.
So these are all the wet modulus.
So it's pretty clear by playing around with the relative density and the crosslinking, that you can get a scaffold in that range.
And the idea is that that would help get the mesenchymal stem cells to differentiate into these osteoblast-like cells.
So then we wanted to also see if our models for the cellular solids could be applied to this scaffold, and we needed the solid properties.
So Kristyn Van Vliet helped us with this, and we isolated a single strut, bonded it to a glass slide, and, I think, I mentioned this last time, we used an AFM tip to then do a little bending test.
So I think the one I mentioned last time was for the collagen-GAG scaffold.
Then we also did the same thing for the mineralized scaffold.
And here we measured a modulus of about seven gigapascals, so that's a dry modulus.
And just for comparison, the modulus of the solid and trabecular bone is something around 18.
So it's lower, but it's in the same ballpark.
And by nanoindentation, we measured a strength of about 200, and that's similar to what you would get in trabecular bone.
So the solid in the struts themselves is not exactly like trabecular bone in mechanical properties, but not too far off.
And then here's a plot of the scaffold modulus divided by the solid modulus against the relative density.
And this line here, the curve, is a squared relationship.
So that's what we'd expect for the foam models.
And this is the strength.
These things fail by a plastic or a brittle failure, and this curve is a three halves power with relative density.
And again, that's what you'd expect from the cellular solids model.
So that gives you a reasonable description of the behavior of the mineralized scaffold.
So again, these were the considerations in trying to make the osteochondral scaffold.
So now we have a collagen scaffold and we have a mineralized scaffolding, and we want to put the two of them together.
So we wanted to use the joint as a model, and we wanted to have some intermediate layer that had some gradation in the mineralization.
So that was the next step.
And the way we did that was we used what we fancily called liquid-phase co-synthesis.
This just meant that we took the mineralized collagen-GAG slurry, we poured that into a mold, and then we poured the non-mineralized slurry into the same mold.
And then we just allowed the two slurries to interdiffuse for some time period.
And then we did the freeze drying step.
So the idea was, that if you poured the one on top of the other, the mineralized one is denser and then you put the less dense one on top, but over some period of time, there will be some diffusion between the two.
And then we just did the freeze drying step.
So this is the scaffold we ended up with.
This is a micro computer tomography image.
So here's the collagen-GAG scaffold for the cartilage on top, and here's the mineralized collagen-GAG calcium phosphate scaffold on the bottom for the bone.
And that just kind of shows what it looked like.
The porosities and the pore sizes we got, the collagen-GAG was about 98% porous and had a pore size of around 650 microns.
The mineralized scaffold was 95.5% porous and had a pore size of 400, roughly, microns.
And then this is what the structure looked like in the EDX.
You can see, this is the collagen-GAG layer, this is the mineralized layer, and there's some zone in between that's a little bit mineralized, not as much as the bony layer but more than the cartilage layer, and the same with the phosphorus.
So we have this collagen-GAG slightly mineralized layer, and then a more mineralized layer And then finally, there was a student, Scott Vickers, who worked with Myron Spector at the Brigham.
And oops, oops.
No, I don't want the weekly updates, thank you.
Sorry.
Well, let me just get rid of this.
Oop, where's my little mousy mouse?
There we go.
OK, so Scott Vickers worked with Myron Spector, and Myron had a surgeon who could do animal studies.
So we did some animal studies on goats.
I think there were six goats at the Brigham.
And they took a plug out of the knee of the goats, and they put our scaffold in.
So this is one of the surgeries as they're about to poke the scaffold in.
And then, Scott waited, I think it was four months, and then sacrificed the goats, and then did the histology.
And this is one of the images from his PhD.
So this staining shows that you've got tissue growing in.
The scaffold was where these little black dotted lines are.
So you had bony tissue in here, and there was a cartilage-like tissue formed at the top.
It wasn't perfect articular cartilage, but it was something similar to that.
And really was as far as we took the project with the funding that we had.
We had a-- I don't know if you remember that-- well, it's beyond before your time.
But Cambridge and MIT had a big research collaboration and the student exchange was part of that.
And this was done through that research collaboration between Cambridge and MIT.
So this was as far as we took it with that research funding.
Andrew Lynn, who was the student in Cambridge, who developed the mineralized scaffold, he then started up a company called Orthomimetics, and he had longer term animal studies done, and he took it a little further.
In Europe, there's something called CE Mark approval, and CE Mark approval means you can start doing clinical trials.
So he never got FDA approval for it, but he got approval to have clinical trials in Europe.
And the first clinical use was in February of 2009.
And they started off using it for the donor sites for the mosaicplasties.
Remember the second method I talked about?
They take plugs out of one region and put them into the region with the damage.
So what they were doing was using our scaffold to fill up these donor sites here and they found that worked quite well.
And then eventually, they started using it for the primary sites as well.
And as of about April a few years ago, April 2012, they had treated about 200 people with this
Wait, so why not just directly start putting it in the-- why start with the--
I think because these were supposed to be sites that were less loaded, weren't as highly stressed, and they thought that was a more, not as critical a place to put them in.
Yeah?
So what is different about this scaffold?
Is it the lack of the dense bone that allows all of the marrow cells to migrate up to the cartilage?
Well, I think, there's not that many people that have made osteochondral scaffolds, so it's good that we've got these two layers.
And the idea that you try to get the stem cells up into the cartilage.
The stem cells will differentiate into the bone or the cartilage, I think, partly depending on the stiffness of the surrounding tissue
But they don't do that in normal bone because of the dense layer?
No, I think they would.
But the thing is, I mean, in some ways, that very first technique where you just drill holes in, I mean, in some ways, that's what it's counting on, right, is that the marrow mesenchymal stem cells are going to differentiate either into the bone or into the cartilage.
But it's just got a hole to differentiate into.
So this gives the cell something to attach to and I think, gives a better result.
Yeah?
So is part of the scaffold and this bone, are they removing the original bone?
Yeah, they remove-- yeah.
Let me back up a step.
So when they do this thing here, so this is in a goat, but they would do the same thing in a person.
So when they drill the hole to put that in, they go through the cartilage, they go through the compact bone, the dense bone, and they go into the trabecular bone, because you don't really get into the marrow until you're in the pores of the trabecular bone.
OK?
Are we good?
How do they attach it?
How do they attach it?
Yes
I think it's just a press fit.
I think they just drill a hole and stick it in.
And these are all in joints, right?
So there's always another bone pressing against it.
I mean, in the surgery, they kind of peel things apart so they can do the surgery but there's another bone pressing down on it.
So I don't think there was any glue or anything.
So there's that.
So this is just a summary.
So we were able to make this two-layer scaffold with a gradient interface, and we tried to make it so that it mimicked the osteochondral tissues.
And this freeze drawing process allowed us to control things like the mineral content, the porosity, the pore sizes.
And then we used materials that had already been approved for medical devices.
And this was funded by a number of places.
So the Cambridge MIT Institute was that collaboration I mentioned.
I have a chair and I used some money for that.
Brendan Harley was one of the students involved and he got a fellowship from MIT.
And Andrew Lynn got a fellowship through the Cambridge Commonwealth Trust and through St. John's College in Cambridge, that's his college there.
So he had had funding through that.
So I think that's the end of that talk.
So are we good with scaffolds?
Yeah, you're good?
So obviously, this probably whole course is on tissue engineering.
This is just scratching the surface and giving you an introduction to it, but I wanted to show you how a lot of these scaffolds look a lot like the foamy materials that I work on, and that you can use the same models for trying to understand the mechanical properties of the scaffold.
And even though the scaffolds are acting in a biological way, there's actually a connection between the mechanical behavior of the scaffolds and the biological response.
So I'm going to talk on Monday about cell/scaffold interactions, so how the environment biological cells are in can affect how they behave.
So we're going to talk about things like cell adhesion and cell migration and cell contraction, contractile behavior.
So I've got another talk a little bit like this and that means I'll have a few notes I'll put on the board on Monday about how the environment that the cells are in affects how they behave, OK?
So this is leading up to that.
It's sort of a similar thing.
So I thought for the rest of the time today, because I knew this was going to end early, I thought what I would do is just switch gears and talk about writing.
So normally when I give a class, I don't talk about writing all that much.
I talk about the technical stuff.
But it turns out writing is actually a huge part of what scientists do and engineers do and whether or not you end up with an academic job or a job in industry or working for government, no matter where you are, you are going to have to write things.
And the better you write, the better off you're going to be.
So I made copies of this brochure and I have a few little slides.
And you're going to have to write up your project report for me, and I thought it might be helpful to just go through this little brochure.
So I'm not going to write stuff on the board.
Everything is pretty much in this brochure, but I thought I'd just walk you through some of the main parts of it.
Oh, did you get one?
Here, have one.
So you might want to think about this when you're writing up your project report for this class, but this really is, it's general.
It's really for any time you have to write something.
This is helpful.
So Mike Ashby put this together.
And as I mentioned, he's done a lot of scientific writing and especially in material science and engineering.
He also makes paintings.
This is one of Mike's little paintings of him writing a paper.
And what I was going to do is just walk through it.
So there's just a few figures, and mostly it's text, but let me go through some of the figures.
So I'm going to just turn to what's page three, OK?
So one of the things that he says to start, and I think this makes a lot of sense, is that you can think of writing a paper the same way you can think about designing something in engineering.
So when you think about design, there's different stages of design, right?
There's a conceptual design, where you just decide roughly what the thing's going to be.
There's embodiment, where you work out a lot of the more details.
You design one version of it.
And then there's the detailed design, where you do all the fine-tuning.
You do all the final design things.
And before you really start your design, if you were going to design an engineering thing, you would first of all think about what's the market.
And if you're writing, the market is your readers.
And so when you think about what you're going to write, you have to think about who's going to read it.
And it's the same thing when I give a talk.
When I give a talk, before I make a single PowerPoint slide, the first thing I think about is who am I talking to and what do they already know?
What's the audience?
What are they looking to get out of the talk?
What am I trying to convey in the talk?
And the writing is the same thing.
You have to think about your audience.
So for instance, I have technical talks.
Obviously, I go give technical talks at meetings, but I do other sorts of talks too.
I've got the woodpecker talk.
I go give the woodpecker talk at the Mass Audubon.
And people come to listen to that talk who are interested in birds but they're not engineers.
So when I do that talk, I have to make it so somebody could understand it who's intelligent but they're not necessarily an engineer.
So I give a different kind of talk when I do that than when I give an engineering talk.
And I got invited to a student dinner.
I'm going out to somewhere with some students on Friday.
I'm going to give that how I became a professor talk, and when I do the how I became a professor talk, it's more general, and so it's a different audience that I'm thinking about.
And in some ways, when I do these talks, it can be the same group of people, but depending on what I'm talking about, the way they look at it is different, OK?
So the market thing is something to think about.
So if you're writing a thesis, your market is your PhD committee, who's going to read the thesis and examine you on the thesis.
If you're writing a paper, you've got to think about the market as being other people in your research field, some of who are going to be the reviewers, who are going to be reading it and criticizing it.
If you're writing a general popular science book, it's a different kind of audience.
If you're writing a research proposal, the audience is going to be the funding agency.
Are they interested in what you're talking about, but also reviewers, who are going to be deciding whether or not to give you the money and what the criticisms are.
So you have to think about who the market is.
So that's one thing.
And one thing that think about in doing the writing-- you know Mike and I have written several books together, and when I tell my neighbors I've written books, they somehow think-- their first idea is that we start on page one and we start writing the book from page one and then we work our way through to page 500.
And that's not how we do it at all and that's not how I write papers.
That's not how most people write papers.
You've got to think about the big picture and think about, roughly, what goes where.
And then maybe think about a draft that gets the scientific facts right.
You put the information down, and then you try to figure out about how do you make the style really nice?
How to make it read well?
How do you make it easy to understand?
How does one paragraph lead into the next paragraph?
So it's an iterative thing.
It's not like you start at page one or line one and you just start writing.
So it's an iterative process, the same as engineering design is an iterative process.
If you were going to design, I don't know, a skateboard or something, you wouldn't just think you were going to make one and that would be it.
That's how writing is, and often, students don't see it as this iterative thing.
OK, so that's that page there.
Let's see, I already talked about market.
I think I have another little slide about-- here's another slide here about markets.
So this is what I just said.
Who are the readers?
How are they going to use it?
So you've got to think about who you're writing for or if you're giving a talk, who's going to look at that.
OK, now the next phase is to make what Mike calls a concept sheet.
I would just think of this as an outline.
And I always make some sort of outline before I try to write anything.
And there's different ways you can do that.
The thing that Mike's got here and there's a nice little figure on page six, which is what I've got up here, is he takes a big piece of paper.
In Europe, it would be known as the A3.
Here it would be known as the eight and a half by 17.
You take a big piece of paper.
This stationery company actually makes paper with big blank space in the middle and little note space on the outside or you could just use the back of it.
So you take a big piece of paper and you may think this is douffy, but it actually helps.
So you take your big piece of paper, and you just make boxes about each topic.
You're going to have an introduction, you're going to have methods and materials, you're going to have a result section.
And you think about what should go into each of those boxes.
And what you're trying to do here is think about the whole paper and how it all fits together and what goes where and what are you going to include and what are you not going to include.
So people think about writing as sitting at the keyboard and typing, but that's just the-- how when you get a problem on a problem set, you turn and crank, that's the turning and cranking part.
The thoughtful part is figuring out what to put in, what to leave out, what figures you want, how you organize the whole thing, how you put it all together, and this helps you do that.
And you can make this-- it's just for you.
It doesn't have to look great.
You can make this messy if you want.
It doesn't really matter.
But it's good to have some sort of an overview of how you want to put the thing together, and that's what this stage helps you do.
So Mike's put other little things here.
So see papers by so-and-so and so-and-so.
So you've got an introduction.
You want to talk about something.
You know there should be some references go here.
Maybe you think there's some extra references you haven't got.
Maybe something in the method, there's some analogy you can use here.
Maybe you need a figure, needs a good figure.
You don't have to actually have the figure.
You just say I need a figure and you have a vague idea what the figure looks like.
Here there's some discussion point.
Discuss this with collaborators, Ed, all this stuff.
So you just put down roughly what needs to go where and you think about the whole thing and how it's all going to fit together.
And then as you work through it, it looks more like this.
So this would be on page seven.
And this is filled in, and if fact, this particular one, what he's written down here is the overview he made for making this booklet, OK?
So the things refer to this book.
So here's the introduction.
Here's the little chart we went through and through, the concept design, embodiment, and detail, blah, blah, blah.
Here's the need, the market need.
We just talked about that.
Here's the concept thing that we we're just talking about now.
Then we're going to talk about each of these different phases.
And his initials are MFA, Michael F. Ashby.
Think out, that means think about this more.
I haven't figured this out yet.
I need an example here, all these kinds of things.
So this is the way he does it.
There's a couple of other ways you can do it.
One way is to just make a bullet outline, that's typically what I do.
You know you're going to have an introduction, these different headings.
And you might put in the bullet outline these same sorts of things, like I need this figure or we need to get one more set of data or I need to look up this reference.
So it doesn't have to be finished, but it's a thing that tells you what have you got, what do you need to do to make it all come together.
So that's one way to do it.
Another way to do it is by thinking about what figures you want in the paper.
So some people, before they write any words, they say, well, I know I want to have these figures, and then they build the words around the figures.
And they can even sketch out what the figures are.
Some people even start before they start the project, they say what kind of figures do I want at the end of the project?
And they don't know if the data is going to do this or that.
They don't know which way the data is going to go, but they say I want to have a plot of one thing versus another thing.
Oh, you're looking like this is not OK?
No
No?
OK, yeah, some people do this.
So even when you're ready to write, you could say to yourself, well, in the methods, do I need a figure that's a schematic of some apparatus?
In the results, how am I going to present the results in the discussion?
Do we need some other kind of figure?
So if you have a set of figures, you can work the text around those figures.
So that's another way to do this.
So those are just three options, but all of them have in common that you think about the whole paper and how it all fits together and what goes where and what do you have already and what else do you need to get, OK?
So that's that.
And then what I was going to do-- I think that's probably the last figure that's-- yeah, OK.
So there's just this little thing here.
So what I was going to do is just talk about some of the other things in this little booklet that work through each of these stages of the writing.
So that's the concept, and then the next phase would be what's called the embodiment, if you think of the design language, and that would be the first draft.
And I think most people find the most difficult thing is to write the first draft.
I mean, I find that the hardest.
Once you've got something, editing it is relatively straightforward.
You go, oh, this piece should go over there or there's something missing.
But getting the first draft down is the hardest thing.
And one of the things-- it's like when I talked about writing the book, you don't write the draft sequentially either.
I typically tell students to start with the materials and methods because that's the most straightforward and it's the easiest to write.
There's not a whole lot of mystery to how you actually did something you've already done.
So I tell people don't write the introduction first.
That's a bad idea because it's actually often not so easy to write the introduction.
So I tell people to write the materials and methods first.
Write the result section because you've got your results and you know what the results are going to be.
So those are the two easiest things.
And often, I think, what people find is if they find it hard to write it's because they don't know what they want to write or they haven't thought things through.
They haven't got their thoughts together.
And if you've got your thoughts, if you know what you want to say, then the writing becomes easier.
So one of the pieces of advice Mike gave me, and it's in this booklet too, is in the first draft what you should try to do is just get the facts down.
Just get the information down and don't worry about if it doesn't sound quite right or if this sentence doesn't lead into that sentence.
Don't worry about the style of it at all.
Just try to get the facts down.
Just try to get the information down.
And once you've got the information and you've got some kind of framework-- I'm not saying that you don't want to make the style good, you do, but you don't need to do that the first thing.
The first thing is just to get the facts down, and then you can edit it later on.
It's more of this iterative process.
OK, so the first draft, the most important thing is just to get the facts down.
Then, I'm not going to read all of these things because you can just read them yourself, but I'll just comment on a few things.
So one thing he's got-- I'm on page eight now, is the abstract.
So the abstract should be concise.
It should be fairly short and you want to tell people why you're doing what you're doing, what you did, what the key result is, and what the conclusions are.
So it can be pretty short.
There's a section on the introduction.
You can just read that.
Let's see here.
Yeah, I think, you can read yourself these other things.
I don't want to go through the whole thing.
All right, so that's that.
OK, so you get to the end of the first draft, there's a little section on figures here.
Let me just make some comments on that.
People who are busy and may not want to read every word that you've written will look at the figures, and they'll look at the figures to make some judgment about does this look useful?
Does this look like you've got something I want to spend more time on?
So it's important to make the figures easily understandable and to be fairly self-contained.
So you want to make the figures clear.
You don't want to have too much information on it so that people can't follow what's going on, and you want them to illustrate the points that you're trying to make.
So the figures are very important.
And then when you've got the first draft finished, the next step, which I find tremendously useful and students often don't get that this is a step.
The next step is you put it somewhere and you don't do anything with it for a while.
And this is why it's important not to write it up at the last second.
Because there's something about, you work on something, you put it to one side, you don't do it.
You just leave it and you go do something else.
And then when you come back to it, all of a sudden, you see things that you didn't see the first time through.
So the next step is you just put it to one side for a couple days.
But you can only do that step for a class if you started before the night before it's due, OK?
So that's why you have to start earlier.
That's why I'm telling you this now, OK?
So you put it aside and then you go have a cup of coffee.
You see there's several cups of coffee that Mike has happily drawn for us here.
OK, then the next part of this is all about grammar and sentence structure.
I'm not going to go over that.
You can read all that yourself.
But there is one amusing thing I want to tell you because there's a cute story.
I'm on page 16 now.
There's a thing about spelling.
Mike is a terrible speller.
When I was his student, he used to-- there were no word processors.
And we used to write things and he would write things and I would say, I can't read your writing.
And he says, ah, well, I make up for my bad spelling with my bad writing.
So I couldn't tell if he had spelled something right.
The other thing we found was that, we wrote the cellular solid book, the first edition was in the 1980s.
And I grew up in Canada, then I lived in England, then I moved here.
He grew up in Australia, then he moved to England, then he moved to the States then he moved back to England.
And words that end, that we spell I-Z-E, like normalized, like normalized variables.
They're I-Z-E here.
In England, they're all I-S-E. OK, and then some of the words, it turns out, Canadians spell with an I-Z-E and some with I-S-E.
So anyway, we wrote the entire fricking book.
We were almost ready to send it off, and we have all those graphs where the axes are normalized, and we must have used the word normalized like 100 times, and we realized half the time we had spelled it I-S-E and half the time we had spelled it I-Z-E. And then we had to go back and find them all and fix it all.
So anyway, that's just our little story about spelling.
OK, there's a thing about punctuation.
You can read that.
That's kind of boring.
I wanted to move on to the bit about style, because style is important too.
And people remember papers that are well written.
Obviously, papers that have valuable scientific information are memorable too, but if it's well-written, it's more memorable, and the style really is important.
And it's the same with giving a talk.
You can convey the same information, but if you don't have it presented in a clear way, people aren't going to remember it.
So the first rule-- so I'm on page 20.
The first rule, now, is to be clear.
And being clear, it doesn't mean having long-winded sentences with lots of jargon.
It really just means make it short, make it concise, make it clear what you want to say.
So Mike has several examples here of headlines from the newspapers, which were not clear.
So I'll just read these out because I find them amusing.
"Red Tape Holds Up New Bridge."
So, OK, maybe the red tape temporally held it up, but not spatially held it up.
OK, and then here's another one.
"Something Went Wrong in Jet Crash, Expert Says."
In fact, you hear this all the time on the news now.
Obviously, something went wrong.
You don't have to be an engineer to figure out something went wrong in a jet crash.
Then, "Chef Throws Heart In To Help Feed the Hungry."
OK, well, maybe that's a little too much.
This is my favorite one, "Prostitutes Appeal to Pope."
OK, there's different ways you can appeal.
You don't have to appeal to the Pope in that way.
And then, "Panda Mating Fails, Vet Takes Over."
I don't think the vet really took over with the mating, but you know what they mean.
OK, so one thing is clarity.
Another point is don't waffle.
You want things to be concise and get to the point.
So he's got this example here from what he cites as a well-known but anonymous materials text.
"The selection of the proper material is a key step in the design process, because it is the crucial decision that links computer calculations and the lines on an engineering drawing with a real or working design."
But what does it say?
It says material selection is important.
So don't say something in some long-winded way that you could say in a short way.
OK, let's see.
So let me move on to the next page.
The next step that I-- let me emphasize one more time, is 8.5, revise and rewrite.
So writing really is iterative.
Those books that you've seen that Mike and I have written, I cannot tell you how many drafts of each chapter we went through over and over and over again.
So to make it good, you have to do it over and over again.
OK, there's that.
Let me see.
Is there anything else I have here?
Ah, here's a couple things that are not so obvious.
So one is, for each paragraph you should have a good first sentence.
You should tell the reader something they don't already know.
So there's some examples here, and you can read through them.
But it is really helpful for each paragraph to have a good opening sentence.
Another thing is at the end of the paragraph it's good if there's a sentence that links it to the next one.
So it hints at what the next paragraph's going to be about, so it leads the reader through it, and it is a logical progression.
So an opening sentence for the paragraph and the final sentence, you might want to look at in a bit more gory detail.
OK, and then there are some references at the very back, at page 25.
And I brought a couple of books in that I have found helpful.
So you may have seen these.
One is called The Elements of Style and this is by Strunk and White, so William Strunk, Jr. and EB White.
You guys know EB White, Charlotte's Web, same guy.
He's written this small, small, not too long to have a look at book about style.
So there's some rules of usage, which is grammar stuff, but there's principles of composition and there's a chapter called "An Approach to Style."
And so here's some of the headings.
Place yourself in the background.
Write in a way that comes naturally.
Work from a suitable design.
These people are not engineers, but they're saying work from a suitable design.
It's like that concept thing.
Revise and rewrite, did I mention that, revise and rewrite?
Do not overwrite.
Don't make it too complicated.
Avoid the use of qualifiers.
If you say something's very stiff, what does that mean?
You can just say it's stiff.
You don't have to say it's very stiff.
Very compared to what?
Let's see.
Use orthodox spelling, blah, blah, blah, blah, blah.
Avoid fancy words.
OK, be clear, all this kind of stuff.
So Strunk and White is old but good.
And then, there's this book here by Bill Bryson.
Bryson's Dictionary of Troublesome Words.
So this is, it is like a dictionary.
It goes by letter and has different words, but it talks about how people sometimes use a word thinking it means one thing when it doesn't, it actually means something else.
And so it's just words that people use commonly that they sometimes confuse with what they really mean.
So it's a very handy thing too.
Do you know Bill Bryson?
Very funny travel writer.
If you ever go somewhere that Bill Bryson has been you should get the book he's written about it because he's very funny.
OK, so I think that's it.
Oh, and then, I think, the very end of this booklet has some examples of good writing and bad writing.
So you can have a look at that too.
All right, so are we good on how to write a paper?
Do you have questions on how to write?
Writing is important.
So I have one more writing story.
So when I first got this job at MIT, I was living in Arlington, and my mom comes to visit me from Toronto.
And I'm work, work, work, work, working, working, and my mom says to me, she says, my, you spend a lot of your life writing.
I'm like, Mom, that's what I get paid to do.
I get paid to write.
Like she always thought I was should be spending maybe 30 hours a week lecturing or something like that.
Like she, well, you're a teacher.
You only teach three hours a week, what is this?
I'm like, I write, Mom.
That's what I do.
So anyways, so writing is important and giving presentations is important.
And no matter what you end up doing, you're going to end up having to do those two things.
And if you do it well, it's going to make your life better, and if you do it badly, you're not going to be so good.
So that's my message.
So I'm going to stop there because I haven't got the next lecture ready, and I can start that on Monday.
So Monday we'll do the cell mechanic stuff.
And then let me just-- because we're getting ridiculously close to the end of term.
Like four weeks from today is the last test.
I know, it's shocking.
So there's one lecture on cell mechanics, and then there's some more engineering applications of foam.
I'm going to talk about energy absorption in foams and foams in sandwich panels, that kind of stuff.
And then, we're going to talk about some natural materials again.
We're going to talk about natural structures, so like natural sandwich panels and natural cylindrical shells with foamy cores.
So I've got lots of nature things at the last part of the course, and you know how I like those nature things.
So there you have it.
Bob's your uncle, as my mother would say.
You know what Bob's your uncle is?
Bob's your uncle is an English expression.
It just means there you have it.
There you have it.
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All right.
So last time we were talking about tissue engineering scaffolds.
And what we're going to talk about today still has to do with tissue engineering scaffolds, but we're going to look at it from a different perspective.
So last time we were looking more at sort of a clinical perspective, and looking at those osteochondral scaffolds for repairing small defects in cartilage.
And today what we're going to talk about are how cells-- how biological cells, interact with the scaffolds.
And there's various kinds of interactions.
So we're going to go through a bunch of these.
So the first one I'm going to talk about is degradation of the scaffolds.
Then we'll talk about the cell attachment.
Cell morphology-- so the shape of the pores in the scaffold can affect the way the biological cells-- what shape they have.
Biological cells could also contract the scaffold and apply mechanical forces.
So we're going to talk about that.
The stiffness of the scaffold and the pore size can affect the speed of cell migration.
And the stiffness of the scaffold can affect the differentiation of cells, so from one cell type to another.
So I thought today I'd talk about that.
This probably won't take the whole hour.
The next topic is on energy absorption in foams.
And so we'll probably start that towards the end of the lecture.
OK.
So the idea here is that we're looking at how scaffolds are being used, really, to provide a 3D environment to characterize the behavior of cells.
And in particular, how the cells interact with their environment.
So let's write that down.
So how the cell behavior is affected by the substrate it's on.
OK.
So the first thing we're going to talk about is scaffold degradation.
And if you think of the native extracellular matrix, the cells secrete enzymes which resorb that matrix and then they also deposit new matrix.
So it was kind of like what we were talking about with the bone.
The bone is always being resorbed and deposited.
And if there's a balance between that those two, then the density of the bone stays the same.
And if one of the rates gets out of whack, then you get osteoporosis and you lose bone mass.
So the idea is that in just the native extracellular matrix, the cells are producing enzymes that degrade the scaffold.
And those enzymes are also going to degrade the tissue engineering scaffolds as well.
And you want to be able to control the rate of degradation, versus the rate at which the native extracellular matrix gets deposited.
Excuse me, sorry.
So you can kind of imagine if the tissue engineering scaffold did not resorb quickly enough, you'd have scaffold there.
And the cells would be trying to put down their own extracellular matrix, and there wouldn't be a place to put it.
And if it resorbs too quickly, then the cells don't have something to attach to.
So there has to be a balance between the rate at which the enzymes are resorbing the tissue engineering scaffold, versus the rate at which the cells are depositing their own extracellular matrix.
So in the native extracellular matrix, the enzymes produced by the cells are resorbing the extracellular matrix.
And then the cells are also synthesizing so they synthesize ECM to replace it.
So the cells are also going to degrade the tissue engineering scaffold that you put in.
And the length of time that the scaffold is insoluble, or so that it remains in the body as a solid, is called the residence time.
And so then we require the scaffold degradation to occur over a time that balances with the new ECM synthesis.
And so the scaffold residence time must be about equal to the time required to make new native extracellular matrix.
So the degradation rate depends on the composition of the scaffold, on how much cross linking there is, and on the relative density.
Obviously, the more scaffold there is, the longer it's going to take to degrade it.
And with synthetic polymers you can vary the molecular weight of the polymer.
And sometimes if you have copolymers, one may degrade faster than the other.
And you can control the balance of how much of each copolymer you have.
And for natural proteins, like collagen, you can control the amount of cross linking.
So you can do the cross linking by various techniques.
That's what's called physical methods.
There's something called dehydrothermal treatment, where you heat the collagen up to 105 degrees C in a vacuum, in a dry environment.
And that eliminates water and causes more cross linking.
There's a UV treatment, ultraviolet light treatment, you can use.
And there's also chemical cross linkers you can use.
So there's different chemical methods you can also use to cross link the collagen.
OK.
So the next thing I wanted to talk about was cell adhesion.
And let's just wait a minute for people to catch up.
Are we just about there?
So this next slide shows a sort of schematic of how a cell would adhere to a substrate.
So down at the bottom here, all these little squiggly lines are representing the extracellular matrix in the native tissue.
Or you can think of it as a, say, a collagen scaffold.
But here we have the ECM.
And this little blob here is our cell.
This is the nucleus of the cell here, the little green blob in the middle.
And the cell is attached to the ECM through something called focal adhesion points.
And this schematic here is a blow up of that focal adhesion.
And at the focal adhesion there's proteins called integrins.
And integrins pass across the cell membrane.
So the idea is the integrins attach to ligands on the extracellular matrix.
And then they also attached to the sub membrane plaque within the cell.
And then that plaque attaches to the side of skeleton.
Things like actin filaments within the cell.
So this is what attaches the cell as a whole to the extracellular matrix, is these focal adhesion sites here.
And different kinds of cell behaviors-- obviously, things like cell attachment, but also things like cell migration, are affected by those focal adhesions there.
So we have that the cells attach to the ECM at focal adhesions.
And sometimes you see those referred to just as FA.
And at the adhesion point the cell has integrins.
And the integrins are transmembrane proteins, so they go across the membrane.
And they bind to like ends on the ECM.
And then the other end of the integrin is attached to the submembrane plaque within the cell.
And then that connects to the cytoskeleton.
And then different kinds of cell behaviors-- so for example, things like adhesion, and proliferation, and migration.
And that cell contraction, we're going to talk more about that in a minute.
They all depend in part on this adhesion between the cells and the extracellular matrix.
And the biological activity depends on how many binding sites there are.
So if you think of the extracellular matrix, it's got these ligands and it depends on the density of binding sites, how much interaction you can get.
So things like how much cell attachment you can get, depends in part on just how many of these binding sites you've got for the cells to attach to.
And that density of the binding sites or the density of the ligands depends on the composition of the scaffold.
But also on the surface area per unit volume of the scaffold.
So if you think of first, just the composition, if you have native proteins, like collagen, they have binding sites themselves.
They have native binding sites.
But if you think of synthetic polymers, like the resorbable sutured type of polymers that we talked about, they don't have binding sites and you have to coat the scaffold with some sort of adhesive protein.
And then the surface area per unit volume of the scaffold is related to the pore size and the relative density.
Let's call the specific surface area, surface area per unit volume.
And if you think of having some scaffold that's like an open celled foam, you can roughly calculate what the surface area per unit volume is.
So say each strut was a cylinder, then the surface area of each cylinder is going to be 2 pi rl.
If each one has a radius r and a length l. Say we had n of them, that would be your surface area.
And the volume of the whole scaffold, or one cell, would go as l cubed, the length of each strut cubed.
So if we just forget about all the constants here.
Forget about n. This just goes as r over l times 1 over l, and that goes as the relative density to the 1/2 power times 1 over the pore size.
So the specific surface area depends on the relative density and on the pore size.
And if you have a tetrakaidecahedron cell, you can work out exactly what that relationship is.
It's sort of a model.
And that gives you the relationship there.
And in this particular case, I think the relative density was 0.5%.
And so it's a constant over the cell size.
So one of the things we did in my group was look at how cell attachment varied with this specific surface area.
So we seeded cells onto scaffolds of different pore sizes.
We kept the relative density constant and we changed the pore sizes.
Remember I said, when we make these scaffolds by freeze drying we can control the pore size, by controlling the freezing temperature.
And we see that it's just a linear relationship between how many cells attach, or the percentage of the cells that were seeded that attach, and the specific surface area.
In here we used MC 3T3 cells.
It's sort of a standard cell one that you can get.
So Fergal O'Brien was the post-doc in my group who did that.
So I'll just say we find cell attachment is proportional to the specific surface area.
OK.
So that's the cell attachment.
So you can see how the scaffold design is going to affect how the cells attach.
So there's some relationship between them there.
Another thing people have looked at is cell morphology.
And so if you change, the sort of, orientation of the pores, how does that change the orientation of the cells?
So this was a study done in another group.
So here we have randomly oriented fibers that make up the scaffold.
And here they're not perfectly oriented this way, but more or less.
And then these are cells that have been seeded onto them, so that the green staining is the cells.
And you can see if the scaffold is random, the cells themselves line up with that fiber structure and become more or less random.
And if the scaffold has fibers that are aligned, then the cells, they also line up and be aligned.
So the morphology of the cells can be affected by the orientation of the scaffold pores.
Also the cell morphology can be affected by the stiffness of the cells.
Or the stiffness of the substrate.
So this is a substrate.
Here this was a PEG-fibrinogen hydrogel.
And they varied the cross linking of this hydrogel.
So they got different modularly for the hydrogel.
So these numbers here, are all the stiffness of the four different hydrogels.
And you can see the cell morphology changes from being a spread out thing on the least stiff substrate, to being just a little spherical or circular blob on the most stiff substrate.
So the cells respond to the substrate.
And so how the cells behave, depends in part on their environment.
So I wanted to also talk about womb contraction.
And talk about how cells contract scaffolds as well.
So one of the things people have found when they look at say, skin and regeneration of skin-- so say you had somebody with a burn and the surgeons will clean the burnt out.
And then what will happen as it heals, is scar tissue will form.
And the scar tissue forms in conjunction with the wound contracting.
So cells will actually migrate into the wound bed and they'll pull the edges of the wound together to try to close the wound.
And they won't close it completely, but they'll partially close it.
And that's called wound contraction.
And that is thought to be associated with the formation of scar tissue.
So the cells can actually apply mechanical loads.
And they can contract the wound.
And one of the things that Professor Yannas found was that if you use one of his collagen and gag scaffolds, you can inhibit that wound contraction.
And if you can prevent the wound contraction from occurring, you also prevent the formation of the scar tissue.
And that allows normal dermis to form.
So you get normal skin.
So this photograph here is of somebody who had burns over their entire torso.
And they put this tissue injury scaffold on this part at the bottom, but not on that part at the top.
And you can see these lines here are contracture lines from the scar formation.
And you can see this skin down here is relatively normal.
And in fact, when people look at the histology of the skin the forms using these scaffolds, they find that it is pretty much the same as normal dermis.
It doesn't have sweat glands and it doesn't have hair follicles.
So you can't sweat from that skin and you don't grow hair.
But apart from that, it's more or less normal dermis.
So this observation that if you can inhibit the womb contraction, you can prevent scar formation and you can get normal dermis to form.
That's led to some interest in just seeing how is it that the cells do this contract l process.
I think hitting the thing and my battery is dead.
So one of the things people have done, is they've just taken what's called, free floating scaffold.
They've just taken little disks of scaffold and put it in a cell culture medium in a Petri dish.
And they find that if you put, say fiberblast on it, the fiberblast will contract that scaffold.
And people have measured how much the diameter of the scaffold changes.
And so they've kind of measured this contraction just by-- it's almost like measuring a strain.
And what we wanted to do is we wanted to try to measure the forces that were involved.
So we first developed something called a cell force monitor, and I'll show you that.
And then we tried to calculate how much an individual cell could apply in terms of the force.
So we used this scaffold here.
This is the same collagen GAG scaffold I showed you before.
And here's the cell force monitor.
So that's just a schematic of holding a piece of the scaffold between two clamps.
So here it is in elevation view.
And then I'll just build the whole thing up, so you can see how it works.
So it's on a base plate.
It's attached to a horizontal stage that's adjustable.
Then there's a very thin beam here.
So this is another adjustable stage here, and this very thin beam here.
And that's attached to one end of this clamp.
And here's the matrix.
And this is attached to this other adjustable stage here.
And then when we have a proximity sensor-- so what's going to happen is, this is fixed over here.
The scaffold is going to contract with the cells applying these contract l forces.
This beam here is going to bend and the proximity sensor is going to tell us how much it's bent.
So we can measure how much that's bent.
If we know how much that's bent, and we calibrate the beam, we can figure out the force in the beam.
OK.
So we can figure out how much is the total force that the cells are contracting with.
And then this just is a little silicone well with some culture medium.
So that's the whole setup there.
Toby Fryman was a student who did that, who's married to Professor Van Vliet.
And I have a very big soft spot for both of them.
So anyway, that's the set up.
And the thing that Toby measured was the force, by measuring how much that beam deflected.
And he measured the force over time.
And he found that if he put say, a certain number of fiberblasts onto the scaffold, the force would increase and then reach an asymptotic point.
And you could describe these curves by this equation here.
Here's the asymptotic force.
And it's a 1 minus exponential of minus time over a time constant tao.
And then this number here is the number of fiberblast that were attached at 22 hours.
So he ran these tests for 22 hours.
And when he was finished, he could count the number of cells that were attached in the scaffolds.
So you would just wash off any cells that weren't attached and you can do accounting of how many cells are left.
And one of the things that he found was that if you plot that asymptotic force-- if you plot through this force over here, against the number of cells that were attached, you just get a linear relationship.
And the slope of that is roughly the force per cell.
And that's about one nano neutron.
Now this is a little deceptive because not all the cells are contracting.
And not all the cells are lined up in one direction.
So there are cells in different orientations.
But just as an order of magnitude the cells are applying something like one minute per cell.
So that's the effect of the cell number.
Another thing he did was he looked at what happens if you change the stiffness of that beam if.
You make that beam in the device different stiffnesses, how do the cells react.
And so the stiffness here are the stiffness of the system.
So there's 0.7 newtons per meter up to ten, so it's a factor of a little over ten difference.
And you can see the displacement per cell changes.
The stiffer the system is the less the cells can displace it.
But if you then plot the force per cell, you find that the force per cell is about the same.
So you develop about the same force.
So that suggests the cells are capable of applying a certain amount of force, and not any more force.
No larger force.
So he did that.
Then we were interested in what was the mechanism of this.
How were the cells applying this force?
Because I was kind of surprised to find out the cells even could apply forces.
So we were interested in understanding the mechanism of this.
And one of the things we knew that we didn't quite figure out how this all worked together was, we knew that the cells elongated.
If you just take a substrate, like even just a 2d substrate, and you put cells on it they'll be rounded to start out with.
And over time, over a few hours, they'll spread.
And that's pretty standard.
Many types of cells will do that.
So we knew the cells were starting off as rounded and they were spreading.
So the cells are getting longer, but our whole scaffolds getting shorter.
And so it wasn't obvious how was the cells going longer, but the scaffold's getting shorter.
And so the next thing we thought we would do is just watch the cells and see what they did.
And so we measured the aspect ratio of the cells at different time points.
And we did this by just impregnating the scaffold in the cells at different time points with a resin, and then using a stain, and then using digital image analysis.
So what we found was that the fiber of the fiberglass morphology looked like this.
So the long thready things of the scaffold, and these little blobs here are the fiberblast of the cells.
So here at time 0 you can see-- like I said, the cells are pretty rounded they're not very spread out.
Here at eight hours you can see-- here's a cell that's gotten longer.
Here's another one.
This guy here is still rounded, it's not doing much.
22 hours, again, some of the cells are quite elongated.
Some of them are still not that elongated.
So they don't all become active.
But one of the things we noticed, if you look at this image here, you can see these cells are attached at one end, and at the other end.
But they're not attached in the middle.
There's sort of a gap between the cell and the strut.
And this is another example here.
Here's a cell here, and this is the collagen GAG strut that it's attached to.
And you can see it's attached to the two ends, but not in the middle.
And this starts to explain how it is that the cells are elongating but the scaffolds getting shorter.
It's that the cells are just attached at two ends.
And the cells are moving along a strut and they're attached to the two ends.
And if you think of the cells attached through those focal adhesion points, they're applying tension to the cell.
And the actin filaments in the cell are in tension.
Obviously, filaments can't be in compression.
They're only going to be in tension.
And what happens is that puts the stress into compression.
And if the struts in compression, at some point it's going to buckle.
And you can see this strut here has basically buckled under that cell.
And so if the cells are getting longer, and they're buckling the struts, then that's going to shorten the struts and the whole scaffold is going to get shorter.
And so then Toby plotted the aspect ratio of the cell, so that is a measure of their elongation against the time.
And again, he found one of these curves with the same kind of form as the curve for the forced development.
And he found the time constant here for the change in the aspect ratio was about five hours.
And for the development of the force it was about 5.7 hours.
So the time constant for the elongation of the cells, more or less matches up with a time constant for developing the force.
So that's what that says.
And that suggests there's a link between the elongation of the cell population and the macroscopic contraction of the population.
So then we wanted to take it one step further.
And we wanted to look at what the cells were doing live.
Like as they were doing it.
So Toby devised this little schematic thing here.
So he had just an optical microscope.
He had a microscope slide with a fairly thick well in it, so that we could put culture medium in the well.
We put a cell seeded matrix in here.
And he had a heated stage here.
And then he took little videos of what the cells were doing.
And this required some patience because as you could see not all the cells did anything.
Some of them just sat there and did nothing.
So he would set this up for a day, and watch a cell, and it would do nothing.
And then he would have to find another cell.
But he did find some cells that were responsible for the contraction.
And that was it was kind of neat.
So here's the scaffold again.
All these little bits here are the scaffold.
This is a strut of the scaffold.
And this is a fiberblast parked on the scaffold.
And this has a little video here.
And you can see what's happening is the strut here is starting to buckle.
And you can see these two sides here, those two things are coming closer together.
So they originally were this piece here, and that piece there.
And now they're at that point there.
And then if I let it go a little bit longer, it continues to do that process.
And then the final thing-- this kind of smushed up mess here is these two things having me brought completely together.
And this strut here is some strut down over here.
So you can see how the cells are elongating and causing contraction of the scaffold.
Here's a series of stills taken from another video that he did.
So this sort of square thing is the scaffold.
So b is the scaffold.
And a, this little blob here, is the fiberblast.
And you can see, even from this image to this one, you can see that the fiberblast has spread a little.
Do you see how it's kind of oozed out along the scaffold there.
And eventually it attaches over here.
And you can see that it's buckled this strut underneath it.
And here it's a little bit more deformed.
It then grabs on down here somewhere and deforms it even more.
So you can see that's more deformed.
And then Toby put alcohol on the whole thing, which kills the cells and the cell let's go.
And you can see you recover some of the deformation.
You don't recover all of it, but you recover some of it.
This was another example.
And this was kind of interesting.
Here there was a scaffold junction where there were three struts that came together, a little bit like a strut.
And there was a little cell right there.
And you can see the cell elongates.
You see how this elongated and its grabbing on up here somewhere.
But the amount of force the cell was kind of pulling with must have been less than the-- or rather must been more than the force of the focal adhesion.
Because what happens was eventually the focal adhesion let go.
And the cell kind of bounces back and ends up over here.
So the cell was kind of snapped back on to the other focal adhesion over here.
And here it's rounded again.
And here it elongates again.
And then this focal adhesion lets go and now it's moved back over to there.
So these struts here are so stiff.
They're much stiffer, I think, partly because they're triangulated.
And it looks like they're just shorter and a lot thicker.
The cell isn't being able to deform those.
But it's elongating and then focal adhesion was letting go.
So this is a little schematic of what we thinks going on.
So the cell starts out-- it's some elongation here.
It's attached at that point.
It's attached at that point there.
And the cell is getting longer.
And if you think about it as the cell's getting longer-- if you think about the Euler Buckling formula, the buckling load goes as 1 over l squared.
So the longer the length of this piece of the strut of the scaffold underneath the cell is, the smaller the load it takes to actually cause it to buckle.
So at some point it buckles like this.
And this is just a little force diagram.
So the actin fibers are in tension and the matrix strut is in compression.
Sometimes we saw some bending.
So you could see if a cell was spanning between two struts, you could get the cell bending the struts as well.
That was another possibility.
And so we think that the cell elongation was related to the contraction.
The time constants for the two things were almost the same.
And as the cell elongates there's a gap between the cell and the matrix on the central portion.
And then the cell is adhered at the periphery of the adhesion points.
And then the tensile forces in these act.
And filaments inside the cell induce compression in the strut, and that causes buckling.
And then Toby graduated.
And then I got another student, Brendan.
And Brendan saw what Toby did and he wanted to do a little more with that.
Brandon was also involved that osteochondral project that I talked about last time.
And Brendan this other thing as well for his project.
So he wanted to measure the force of an individual cell.
So when we had that cell force monitor, that was the total force of all the cells in that one direction.
But Brendan wanted to know if he could measure the force of a single cell.
And now that we knew that the contractal process was related to buckling, We thought, well, we could just use Euler's formula.
If we knew what the modulus of the solid was, and we knew what the dimensions of the struts were.
So that would allow us to calculate the contractile force of a single fiberblast.
So I think I've shown you this thing here.
So Brendan was the one who did these experiments.
He cut a single strut out of the scaffold.
And the single strut is about 100 microns long.
He used a microscope to do this.
He then glued it onto a glass slide and he used the atomic force microscope probe to bend the strut like a cantilever beam.
And he measured this displacement here.
And from that he could back out what the modulus of the solid was.
He did these tests in the dry state.
But we could extrapolate to the wet state from looking at the behavior of the whole scaffold.
So he had a modulus for the wet scaffold solid.
And then this is our formula for Euler buckling here.
So that's just the standard formula.
I had a student from civil engineering, who looked at hydrostatic loading of a tetrakaidecahedral cell and he looked at buckling.
If you had a tetrakaidecahedral cell and you load it in all three directions.
He looked at the buckling.
And he had calculated that the n constraint factor-- the n squared was point 0.34.
So we have some idea of what that n squared value should be.
Although it's somewhat of an estimate.
I had a UROP student who took Toby's images and measured the dimensions of the struts.
So he measured the diameter and the thickness of the struts.
And from that, we just plugged everything into the Euler formula.
And we found that the average single cell force is somewhere between about 11 and 41 nano neutron.
It was something like 26 nano neutrons.
So it would make sense that it's more than the one nano neutron per cell because not all of those cells were active and they weren't all going in the same direction.
So Brendan Harley and Matt Wong did that part of the project.
OK.
So that's the contraction.
Are we good with contraction?
So it's kind of interesting that cells will contract and we can measure some forces.
So the next type of interaction between the cells and the scaffolds that I wanted to talk about is cell migration.
And these are some studies from the literature.
These are two different studies.
But the top one here, they've measured migration rate as a function of the cross linking treatment of a scaffold.
And the decreasing stiffness goes this way.
And so they're seeing that the speed of migration-- this is in millimeters per day.
Cells don't move too quickly.
They go millimeters per day.
But you can see that the migration speed, the speed at which the cells can move, depends on the stiffness of the scaffold that they're attached to.
And in this study on the bottom here, what they did was they had just a flat 2d substrate.
Just a flat polymer.
And what they did was they cross linked one part of the polymer more than the other part of polymer.
So over here, this was the less cross linked.
That was the soft part.
And this was the more highly cross linked.
This was the stiffer part.
And they found that if they put a cell on the soft part it would migrate onto the stiff part.
But if they put a cell on the stiff part, it would start going this way towards the soft part.
But when it got to the interface it would just spread out along the interface.
And it wouldn't go into the soft part.
So the cells were somehow sensing the stiffness of the substrate.
And for some reason, I don't know what, but for some reason these particular cells seem to prefer being on the stiff substrate.
So this is just really showing that there's some interaction between the substrate stiffness and the way the cells are behaving and migrating.
And then Brendan also wanted to study this.
And he got some of the collagen GAG scaffold.
He made some of the scaffold.
And he stained that with a stain that made it turn red.
So these lines here are all red struts in the scaffold.
And then he put fiberblasts on to the scaffold and stained them green.
So all these little blobs here that are green are the cells.
And then he used confocal microscopy.
And the confocal microscopy allowed him to look at a certain volume of the scaffold.
And he had some software that would track the centroid of each cell as it moved through the scaffold.
And so he had a thing he called spot tracking.
So each of these little spheres here corresponds to a cell.
And the white box is the volume of material that you could see in the scaffold.
And this color scale here really corresponds to time.
So I've forgotten which round.
I think blue is the original time 0, and then red is maybe five seconds, and yellow was 10 seconds.
The different colors correspond to different times.
So he could track the path of each cell and also what the position was at different time points.
So he knew what the position was at different time points.
And obviously from that, he could get the speed of the scaffold.
And he did these experiments on scaffolds of different stiffnesses, as well as, different pore size.
And here you can see the cell speed.
He's measuring it in microns per hour now.
The cell speed increases at first and then decreases with the strut stiffness.
So we don't know exactly why this is.
But there is an effect between the stiffness of the scaffold and the migration speed.
And another thing he did was he looked at how the cell speed varies with the pore size.
And as the pore size gets smaller, the speed goes up.
And we're not entirely sure why that is.
But I think that might be related to this binding site thing too.
As the pore size goes down, the number of binding sites is going to go up.
And if you think of the cells migrating by having these adhesion sites, and the adhesion sites are just at the ends of the cells, and the cells kind of putting out a little extension, and then looking for somewhere else it can bind.
The more binding sites there are, the faster it's going to find a binding site.
And the faster, I think, it's going to move on.
So I think that the cell speed depends on pore size, at least in part because of the increase in the binding sites with smaller pore sizes.
So pore size and the migration.
And then the last thing I wanted to talk about was cell differentiation.
And this is a study study by Engler.
And one of the things he found was he put mesenchymal stem cells on 2d substrates.
Just flat 2d substrates of different stiffnesses.
And again, he could control the stiffness by cross linking.
And what he's showing up here in the first bit is that he's looking at the stiffness of tissues of different kinds.
So here's brain type tissue.
Something like one kilo pascal.
Muscle might be something like 10 kilo pascal.
And collagenous bone-- this is sort of the osteoid that is the precursor of bone, not the bone itself.
Is about 100 kilo pascals.
And what he did was he put these mesenchymal stem cells-- so here's his cell onto his substrate.
And he varied the stiffness of the substrate.
And then he looked at the shape of the cells.
So here's the least stiff substrate, so between point 1 and 1 kilo pascals.
And here's 4 hours, 24 hours, 96 hours.
And these cells formed long processes extending beyond the cell body.
And they looked kind of like neurons.
So they he called those neuron like.
Then there's an intermediate stiffness of substrate here.
And these cells became even more elongated.
And became something like a muscle cell, myoblast like.
And then cells that were put onto a substrate that was between about 25 and 40 kilo pascals, they developed a shape that was something like an osteoblast, like a bone cell.
So one of the things he was looking at here, was how the stiffness of the substrate affected how a stem cell might differentiate into different cell types.
And another thing that he did was he looked at different cell markers.
And he found that the cells were expressing markers that were corresponding to the types of tissue.
So I couldn't tell you the names of all these things and what they are.
But I think the red here is expressing more of a particular marker.
And I think these wounds were related to nerve tissue.
These wounds here, were related more to muscle tissue.
And these wounds here were related more to bone tissue.
So the things the cells were expressing also seemed to correspond to the different types of tissue that they were corresponding to.
So I'm just going to end this part by going through a little summary here.
So what I've tried to show you today is different types of cell behavior that are affected by the scaffold.
And they're affected by things like the number of binding sites, by the pore size, by the stiffness of the scaffold.
So we started with a cell attachment.
We saw that the cell attachment increases linearly with a specific surface area.
We saw that the cell morphology depends on the orientation of the pores.
And that kind of makes sense, they got to line up with the pores.
We talked about the contraction behaviors.
So the cells bind at the periphery, the cells elongate, and that causes this buckling.
And you can calculate the buckling forces.
It's around 10 to 40 nano neutrons.
We looked at the cell migration speed.
That increases with the stiffness of 1D fibers.
And we looked at cell migration in the collagen gag scaffolds.
So that depends on the stiffness of the pore size.
And then there was this final study on the cell differentiation.
So I wasn't going to write any notes on this because the slides I think pretty much explain it.
So I was just going to put the slides on the website at the end after today's lecture.
So are we good with how cells and the scaffolds of the environments interact?
Because I think it's not so obvious that this actual mechanical environment makes a difference.
People think of the chemical, the biochemical environment.
That obviously affects the cells.
But people don't think at first that something like the sort of structure of the pores, the pore size, or the orientation of the pores, or the mechanical properties are going to affect how the cells behave.
But in fact, they do.
So that's it.
And this is all various people who worked with me on these projects.
So it was a lot of fun.
OK.
So hang on a sec here.
What's this all about?
I'm going to get rid of that.
Go away.
Here we go.
OK.
So are we good with cells and substrates?
Yeah?
OK.
So let's just take a little moment and I'll rub the board off.
And then we can start the next bit.
OK. OK.
So that's the end of the medical material stuff.
So we talked about the bone.
We talked about the tissue engineering scaffolds.
And then we talked about the cell scaffold interactions.
So now we're going to go back to more engineering topics.
And the next thing I wanted to talk about was energy absorption in foams.
So foams are very widely used for energy absorption applications, things like bicycle helmets, different kinds of helmets.
You buy a new computer, it comes in foam packaging.
And the reason foams are used so much is they're extremely good at absorbing energy from impact.
And in fact, they're better than the solid that they're made from.
So let's just look at this curve here for a minute.
So here's a stress strain curve in compression for the foam.
And the material that it's made from would have the stiffness something like this.
It would be much, much stiffer than the foam.
And if you think about how much energy you can absorb, the energy you can absorb is just the area under the stress/strain curve.
That's the energy you can absorb in a given volume of foam.
And so when you're thinking about these energy absorption problems, it's not just that you need to absorb a certain energy.
You need to absorb it without exceeding a certain peak stress.
So whatever it is you're trying to protect, at some point it's going to break.
This is what you want to avoid.
You want to avoid it breaking.
So you don't want to have a stress bigger than the stress that's going to break whatever it is, your computer, or your head, or whatever.
So say you have a given peak stress that you can tolerate here.
And we've normalized things by the solid modules.
But just say that's a peak stress here.
The foam is going to absorb this amount of energy up here, this whole little shaded region.
And the solid is going to absorb that little, teeny weeny bit in there.
So what you want to do is absorb the energy without exceeding a certain peak stress.
And the foam is always going to be better than the solid that it's made from.
There's a couple other things that make the foams good because they're more or less isotropic, maybe not perfectly.
But roughly, they have the same properties in all directions.
Sometimes you don't know what direction the impact's going to come from.
And so if you've got the same properties in all directions or roughly the same, that's a good thing.
You also want the protective thing to be light.
If you're paying for shipping for your computer or whatever, the fact that the packaging is light makes the shipping easier.
If you have a helmet for your head, you don't want some big heavy thing.
You want something fairly light.
And foams are cheap.
So the fact that they're roughly isotropic, they're light, they're cheap, this all helps as well.
But from a mechanical point of view, foams are very good at absorbing energy.
And so what we're going to do in the next-- the rest of this lecture and on Wednesday-- we're going to see how we can convert these stress/strain curves into what are called energy absorption diagrams.
We're going to look at some energy absorption diagrams that we just measure from the stress/strain curves.
And we're going to look at how we can predict the energy absorption diagrams as well.
OK.
So the main idea here is that the impact protection has to absorb the energy from the impact but without exceeding a certain peak stress.
So the direction of loading may not be predictable.
And foams are good because they're roughly Isotropic.
And they would have the same energy absorption capacity from any direction.
And foams are also light and cheap.
We can say for a given peak stress the foam is always going to absorb more energy than the solid it's made from.
So other things that make foams good are that they have a capacity to undergo large deformations.
And they do that at roughly constant stress.
So that if you look at the stress strain curve for the foam, you're going to be able to absorb all this energy under here.
And these strains that the foam might go to might be 0.08 to 0.09, so huge strains on an engineering scale.
And then this is your energy-- would absorb is that area under the stress/strain curve.
So I wanted to say something about strain rates too.
So typically we're going to be talking about problems of impact.
And in impact, the strain rates are typically on the order of 10 to 100 per second, something like that.
We're not going to talk about things like blast.
If you have a blast loading, then you have to take inertial effects into account.
And blasts involves strain rates, which are 1,000 to 10,000 per second, much, much higher.
So we're going to talk about strain rates that are about 10 to 100 per second, maybe a bit more than that.
And for instance, you can roughly estimate what one of these impact rates would be.
So you had something that you dropped from a height of 1 meter.
Then the velocity on impact is just if you just equate the potential energy with a kinetic energy.
The velocity and impact is just the square root of 2gh.
So g's the gravity acceleration.
And h is the height.
So that's the square root of 2 plus 9.81 meters per second times 1 meter.
And that comes out to 4.4 meters per second.
And say you had some foam packaging that was 100 millimeters thick.
Then you could say roughly that the strain rate would be approximately equal to that velocity over the thickness, so 4.4 per second over 0.1 meters.
That' would be 44 per second.
So it's somewhere in that range.
Obviously, the thickness could be a little bit smaller, it could be bigger.
But it's in that ballpark.
And if you do tests on servo controlled instrons or you do a drop hammer test, you can get strain rates in that ballpark.
OK.
So we're talking about impact and not blast.
OK.
So most of the energy that's absorbed is really absorbed in that stress plateau.
So if you think of the stress/strain curve, most of the area under the stress/strain curve comes from the area from underneath the stress plateau.
So the mechanisms of absorbing the energy are going to be mechanisms that are associated with a plateau stress.
So for elastomeric foams, we've got elastic buckling of the cells.
And one of the advantages or disadvantages-- depending on what you want-- of this is that the deformation is recoverable and you got to have rebounds.
So if you have an object and you drop it onto elastomeric foam, it's going to bounce around like that.
So the elastic deformation is going to be recovered, and you're going to get rebound.
If you have a foam that has a plastic yield point or is brittle, then the deformation is going to be largely from dissipating plastic work or work of fracture.
And in that case, there's no rebound.
But once you've loaded it, you've crushed the thing, and you've permanently deformed it, and you can't use it again.
So sometimes if you ride your bicycle like I do, if you have a helmet, you should wear your bicycle helmet.
If you have a problem, if you have an accident, and your helmet get smooshed, that's it.
You have to throw your helmet away.
You can't use it again.
And this is why.
[INAUDIBLE], even if it doesn't get smooshed, if you hit your head at all, [INAUDIBLE].
Exactly.
[INAUDIBLE] Yeah.
You need a new helmet.
Yeah.
Go ahead.
Talk about helmets because I'm on a helmet conversion thing.
Yes.
You've got to wear your helmet.
And you should change it every now and then.
Anything else you'd like to add about bicycle helmet safety?
No, absolutely.
You've got to wear your helmet.
So I know several people who would have had their head smooshed had they not been wearing their helmet.
So you have to wear your helmet.
Let's see.
OK.
If you think about natural cellular materials, things like wood, they often have cell walls that are fiber compensates.
And you can dissipate energy by mechanisms related to the fiber nature, so by things like fiber pull out fracture.
And then you can also have open cell foams with fluids.
You can have fluid within the cells.
And if the cells are open cells, the fluid effect is really only going to be important if the cells are extremely small or the fluid is particularly viscous, or the strain rates are very high.
So in most cases, the fluid effects aren't important in open cell foams.
But, for example, you could try to make an open cell foam that had more energy absorption by putting a fluid into it.
So you could put glycerin into the fluid, and that would increase how much energy it would absorb.
Or, you can put this honey into it.
That would make it more energy absorption.
And enclosed cell foams, you may have an effect of the gas within the cells.
But it's really only going to be significant if you have elastimeric foams where the cell faces don't rupture.
The cell faces rupture, then the gas is just going to flow out of them, and that's not going to do much.
So the next step is I want to go from having the stress/strain curve that we've become very familiar with, and make something with that that is a little easier to see graphically that shows how much energy we can absorb.
Remember, what I said what we're really interested in is absorbing a certain amount of energy without exceeding a certain peak stress.
So what I'm going to do is plot another plot that's based on that.
It's going to be the energy absorbed.
So w is going to be energy absorbed per unit volume.
And I'm going to plot that against the peak stress.
OK.
So we're going to look at three different regimes here.
We're going to look at what happens in the linear elastic part, what happens in the stress plateau, and then what happens in the densification part.
So let's think about the elastic regime first.
And if I moved up-- say I moved up to some point right there where the little x is on the stress/strain curve.
Then the amount of energy I absorbed would just be equal to this little bit here.
And if I moved up, and then the peak stress would be this peak stress there.
We'll call that sigma p1 and w1.
And if I moved up over here, I'd be at w2.
And that would be sigma p2, right?
And if I know the modulus, I know what that relationship is.
And I get a relationship.
And these are going to be-- I'm going to do this on log scales here.
There's going to be log, and that's going to be log.
I'm going to get in that linear elastic regime.
The energy is going to go as the peak stress squared over 2 times the modulus of the foam.
Remember, energy is a half stress times strain.
And I can say strain is sigma p over e. So it's 1/2 sigma p squared over e. So on my log1 plot here, this is just going to be a straight line like that.
And then I'm going to get to this value here.
I'm going to get to my collapse stress here.
So let's call that single star.
And at that point, the more I go along here, every point I go along, like that, I'm going to absorb more and more energy.
But the stress isn't going to go up at all.
So then this thing here is going to go like that because I'm absorbing more and more energy.
But the stress just stays the same.
So this is good news if we want to absorb energy.
And then once I get to the densification point, then it's going to do the opposite thing.
As I go along here, at each increment I'm not absorbing that much more energy.
But the stress is going up.
So at some point it turns and starts to look like that.
So this part here corresponds to linear elasticity.
This bit here corresponds to the stress plateau.
And this bit here corresponds to densification.
And the point where I would like to be is right here, because here I'm going to absorb the most energy possible through the peak stress.
So you can think of that as sort of an optimal point.
And I'm going to refer to that as a shoulder because it's the shoulder between where the curve bends over again.
So I've only got a couple minutes left.
But let me just show you one thing and then we'll talk about this more next time.
So I've just done this for one relative density.
But if you look at the screen, you can imagine I would have stress/strain curves for lots of different relative densities.
And let's say these are all at the same temperature and all at the same strain rate.
And I could draw a curve that looks like that for each stress/strain curve.
And if I did that, I'd get a family of them.
So this is our energy absorbed here.
I've normalized it by dividing by the solid modulus.
This is our peak stress here.
And I've normalized that by dividing by a solid modulus.
And I've got a sort of family of these things, right?
They all have the same shape.
But they shift depending on the relative density.
And then the thing that makes life good is that these shoulder points you can connect with a line.
And you can mark off the relative density for those shoulder points on each line.
And then the last step you can do is you can just plot these lines.
And you can repeat this for different strain rates.
So this would be a family of these guys here.
There's a family of those lines at different strain rates.
And then you would join up the points that correspond to each relative density.
So you can make a drawing that looks like this that summarizes the most energy you can absorb for a certain peak stress for foams of different relative densities tested at different strain rates.
You could do it for different temperatures if you wanted to.
So next time, we'll talk about that.
But I'm going to stop there for today.
OK?
Are we good?
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All right, so last time we started talking about energy absorption, and I wanted to try to finish that up today.
And then next week we would talk about sandwich panels and using honeycombs and foams in sandwich panels.
So I think we got as far as the idea of introducing what these energy absorption diagrams are.
So let me run through this little sequence again, and then I'll put the notes up on the board.
So the idea is that you have your compressive stress strain curve-- so here's a series of curves for different densities of a foam.
And we would do these all at the same strain rate and temperature, so that those aren't variables.
And then what we would do is we could turn those into energy absorption diagrams.
So notice here, this is a log plot.
So here's that energy absorption here.
Here's the peak stress here-- so that's the peak stress up to some level of energy that you've absorbed.
And we've normalized both of those by the solid modulus.
So say we look at one density here, say we look at the lowest density-- 0.01.
So here's our curve here for the test.
If I go up to some stress level that's still in the linear elastic region, then that's going to translate into somewhere along here on the energy absorption curve.
And one of the things we're going to do today-- I'm going to show you how you draw these, and how you can set them up.
And you can either get them from experiments-- so this is, say the top curve were experiments, doing it from experiments-- or you can use the models for the foams to do it, as well.
So we'll see how you can do from both ways.
So this would be the linear elastic bit here.
This vertical part-- where the energy is increasing, but the peak stress isn't increasing-- that corresponds to the plateau here.
And then this part here, where the energy is not increasing very much, but the stress increases a lot-- that corresponds to the densification part over there.
So what you would do is you would do tests on foams of different densities, and from the test data, you could draw an energy absorption curve for each density.
So you plot-- there's four different densities here, so it forms a family of these curves.
And then what you do is you say-- and I think we talked with this last time-- that the place you want to be is at this shoulder point.
You want to be absorbing as much energy as possible at that plateau stress.
So you want to be at the point just before it turns around to the densification regime.
So you can mark that little shoulder point for each density.
So here's like 0.01, here's 0.03, and so on.
And then those points can be connected by a line-- so that heavy line then connects those shoulder points, or those optimum points.
And then what you can do is then repeat this whole process for different strain rates.
So this family of lines here is at different strain rates-- so this set that goes this way, corresponds to this line here.
And if you do them at different strain rates, where the shoulder appears at a slightly different point-- and so if you mark those shorter points, you can draw these lines here that connect up for one relative density.
So this line on the left hand side here, these are all relative densities of 0.01, these ones are all relative densities of 0.03.
So this diagram down here, doesn't really look like this at all-- it doesn't look like the basic energy absorption diagram.
But it actually has information for what the optimum would be for a range of densities and a range of strain rates.
So typically, these foams are viscoelastic and they have some strain rate sensitivity.
So you'd like to get the strain rate sensitivity into it.
And it's not shown here, but you could do the same thing at a constant strain rate and varying the temperature, too.
So if you had things at different temperatures, you could do the same kind of idea.
OK-- so are we good with how this works?
Because now I'm going to write some notes on the board so you have it in your notes.
So the idea is that you turn your stress strain curve to look something like that, into an energy absorption diagram.
And you can plot on log log scales, the energy versus the peak stress.
And you get something like that.
And this point here, I'm going to call the shoulder point.
And that would use the material in the most efficient way-- you get the most energy absorption without getting higher than that plateau stress.
So we can say at this stress plateau, the energy increases without much increase in the peak stress.
And then as the foam densifies, then you get an increase in that peak stress, with little increase in the energy absorbed.
And so ideally, you want to be at that shoulder point.
So to construct these energy diagrams, you can test the series of foams at different relative densities and constant strain rate and temperature.
And then you make that plot of the energy absorbed normalized by the solid modulus, versus the peak stress normalized by the solid modulus for each curve.
And typically what people do is they would take the solid modulus at a constant strain rate and temperature, so you don't have to introduce that, as well.
Then you would mark the density for each of those shoulder points, and then you would connect them.
And then you could repeat this whole process for different strain rates.
And then you would draw the final diagram at the bottom, where you have this family of lines that describes the shoulder points for different densities and different strain rates.
And you could treat different temperatures in the same way, if you wanted to.
You would hold the strain rate constant, and vary the temperature.
so it's kind of a nice way of just putting a lot of information in one diagram.
So one of the things about this is that, because we're normalizing by Es, and if you think of elastomeric foams, elastomeric foams, both the Young's modulus depends on Es, and the plateau stress depends Es.
So the Young's modulus depends on the stiffness of the solid, and also because the plateau stress is related to elastic buckling, it also depends on the modulus of the solid.
So for elastomeric foams-- it's because you normalized it with respect to Es-- one of these diagrams will represent all elastomeric foams.
So that's rather a nice thing-- so you can have different elastomeric foams, but one of those diagrams is going to represent all of them.
So we can say, elastomeric foams can all be plotted on one plot, or one curve, since both the modulus and the plateau stress are related to Es.
So if we look at this next figure here-- maybe I'll just wait a minute for people to stop writing.
Some people write faster than others.
So if we look at this next plot here, here's a compressive stress and strain.
These tests are done for one density, but at different strain rates-- so it's kind of the other version of this.
But here's the stress strain curves here, and here's the energy absorption diagram that's derived from those.
And then here's a summary diagram that has the different strain rates, and that would have different densities here.
And the idea is this diagram here could represent all elastomeric foams.
So this has been put together for polyurethane, but it should be able to represent other sorts of flexible elastomeric foams.
Sorry?
In those diagrams, then, the intersection of this strain rate line and your relative density line should be the shoulder position?
Yeah, exactly.
And so this line here is for 0.01, and that one's for 0.03.
So 0.02 is going to be-- you'd have to interpolate somewhere in between there.
So we've just put on certain values, because we're not going to put on a million values.
We just put on certain ones, and then you can estimate where other densities would appear on there.
OK?
So here's another example here-- these are curves for two different foams.
So here's a polyurethane a polyethylene, so these are both elastomers.
And one is the dashed line, and one is the solid line.
And you can kind of see how the lines mesh up.
So here's a density of 0.01 for the polyurethane.
Here's 0.05 for the polyurethane.
And here's 0.06 for the polyethylene.
And you can see how the 0.06 and 0.05-- they're not quite on top of each other, but they're pretty close to coming on top of each other.
And then 0.1, 0.12-- and so you get a family of them for the different densities.
And you can also do this for materials that have a yield point, so polymethacrylimid has a yield point, so you do exactly the same kind of thing.
So here's the energy absorption diagram that's been developed from the stress strain curves.
But now this curve here, or this set of curves here, is really just valid for one ratio of sigma [?
ys ?]
to Es-- so the solid yield strength of the solid modulus.
So it's valid for whatever it was for that particular type of foam-- the polymethacrylimid.
So I'll just say, if we have foams that are made from a material with a yield point-- and so they have a plastic collapse stress-- then the curve will be valid for foams with the same ratio of [?
sigma ys ?]
to Es.
So in that case there, for the polymethacrylimid, that ratio is about equal to 1 over 30.
So that plot would probably give not a bad description of other foams, with the same value of sigma ys over Es.
So the idea here is we can generate these diagrams either from data-- the way those ones have been done-- or from the models.
So another way to generate these is to think about the models that we have for the foams, and the foam behavior.
So we have an equation that describes the Young's modulus, we have equations that describe the plateau stresses, we have an empirical equation for the densification strain.
And we can use those to generate these diagrams.
And they're kind of useful, because they show you what's going on a sort of mechanistic basis.
So this is a diagram here that's been generated for open cell elastomeric foams.
Here's our energy absorbed, here's our peak stress down here.
This little inset is sort of a schematic of the idealized foam behavior.
So here's the Young's modulus, here's the elastic [?
collapse ?]
stress, here's the densification over here.
So obviously it's kind of a very idealized set-up.
But we can generate this diagram-- and I'm going to go through the equations that will let us do that.
So one thing to note is here's the curves for each density.
Here is this line that connects the shoulder points.
There's a couple of other lines on here that I just wanted to mention something about.
If you think about the densities-- like that's 0.01, this is 0.03, that's 0.1-- if you had a fully dense solid that was made of the same elastomer, you could plot the curve for that, and that is going to show up over here.
So this is kind of an upper bound.
It can't get any more from that.
And we've also got a dotted line here, which takes into account fluid flow within the cells.
So there can be some fluid flow, and because we haven't talked about that, we're not going to go into that.
So we can just ignore that dotted line for now.
So let me go through how we can do the modeling.
So we're going to divide the stress strain curve up into bits, and I'm going to write equations for the energy absorbed for each bit.
So we're going to start with a linear elastic behavior.
And I'm going to-- let's see-- yeah, so let me just note here that this is the densification strain out here.
And this strain, epsilon naught, corresponds to the strain at which we first reach the stress plateau.
So here, for the linear elastic part, I'm going to say that the strain is less than that.
So the strain is less than absolute naught.
And then I can say, the energy absorbed-- so if you just remember from Hooke's law on linear elasticity, energy under the stress strain curve for the linear elastic part is 1/2 of sigma squared over E. So I'm going to call sigma-- whatever the stress is going to be, the peak stress that we get to.
And now we're going to divide by E of the foam-- because these were our foams here.
And I can use our model here to say-- and now I've got to divide that by Es, because I've normalized here.
So because I know from my modeling that the Young's modulus of the foam is just equal to Es times the relative density squared for the open celled foam, that means I've now got an Es squared in the denominator, so I've got a sigma p over Es squared.
And then I've got a 1 over the relative density squared term.
So that factor there, that equation there, gives you these first set of lines here.
Gives you that bit, and this bit, and that bit.
So it gives you those first parts of the energy absorption diagram.
And then if we look at the stress plateau-- so here we're going to say that epsilon naught is less than epsilon is less than the densification strain-- so we're on the plateau somewhere.
And now the energy absorbed is just going to be our plateau stress times the amount of strain we've got.
So that's epsilon minus epsilon naught.
And if I normalized with respect the solid modulus, I can write down that my plateau stress is 0.05 times the relative density squared, and then multiply that times epsilon minus epsilon naught.
So that equation then corresponds to these vertical parts-- so this part here, that part there, this part here.
It corresponds to those vertical lines on the figure.
Vertical lines on the diagram.
And then the plateau stress is going to end at the densification strain.
And at that point, the energy diagram is just going to become vertical.
OK, and then the last part-- I'll try to rub this off a little better.
And then the last part is when we're at the end of this stress plateau, and the strain is equal to that densification strain.
And so the amount of energy we absorb here is really going to be the maximum.
So this is the energy that's going to correspond to that shoulder point that I've been talking about.
So I'm going to call that W max, and normalize that with respect to Es.
And that's then going to be our plateau stress times the densification strain.
And the densification strain was just 1 minus 1.4 times the relative density.
So here I'm assuming that the densification strain is very much bigger than the strain at which the buckling first occurs, and I'm neglecting that linear elastic part.
So then we could say that the optimum foam is at that shoulder point.
And I can say that the peak stress at that point is just equal to the plateau stress.
And what I want to do is get an equation for that solid line up here that connects all those shoulder points.
So I want an equation, in terms of the energy, and the peak stress, instead of in terms of the density.
So what I'm going to do a solve this for the density, and then plug that back into there.
So I get that the relative density is, then, 20 times the peak stress over the solid modulus, and I take the square root of that.
And now I can substitute this up here for the relative density.
I think I need another board.
So then I've got-- this is my peak stress, or my plateau stress, over Es, and this is all just the densification strain.
And that's just 1 minus 1.4 times the relative density.
But now I'm putting the relative density in terms of the peak stress, or the plateau stress.
If I just simplify that slightly with the constant, it's 1 minus 6.26 times sigma p over Es to the 1/2 power.
So that equation there describes the-- oops, no more updates.
No updates.
Go away.
Ah, so that equation there describes this line here that connects those shoulder points.
And that's the line you're the most interested in, because each of those shoulder points is a point where the foam is being used in the most efficient way, or the optimum way.
And then, let's see-- is that going to fit?
No-- let me try the other board.
And then the last thing we can do is calculate that line that corresponds to the dense solid.
Yeah?
On those ones over there, it says it corresponds to the vertical lines on the diagram.
And then later it says then it becomes vertical.
Is that referring to two different diagrams?
So this stress plateau equation here corresponds to these vertical lines here.
So for relative density of 0.01, it corresponds to that part.
For 0.03 it's this part.
And your 0.01 it's that part
But then some of the [INAUDIBLE] it says, then w [?
versus ?]
sigma becomes vertical.
Should that--
So, well the plateau stress ends at the densification strain.
So the plateau stress ends here.
Oh, let's see-- and then it becomes-- should be horizontal.
Sorry.
OK, sorry.
Happy?
And now I'm going to rub that all off.
OK, did everybody get this?
I can rub it off?
OK, so then the last part is what happens when the foam is densified.
And if it was fully dense-- and you never really can get to this point-- but if it was fully dense, you would get rid of all the pores, and it would just be a solid.
And then the energy absorption curve would be the curve for the solid.
So I'll just say, when fully densified, I'm going to say a curve approaches that for the solid.
And for the solid, you would just have that W over Es is equal to 1/2 the peak stress squared over Es.
So this model curve, the curves have the same shape as when you get the diagrams from the experiments.
And you can see how the different mechanisms of deformation and failure contribute to the diagram, where the diagram comes from.
And I guess one other point is you can see that the foams are always going to be a lot better than the solid.
And remember this is a log log curve, so that a foam that has a density of 3% here, there's a huge difference in the peak stress.
So say you wanted to absorb this amount of energy up here, for a foam that was 0.03 dense, the peak stress would be a little less than 10 to the minus 4, normalized by the modulus.
And for the solid, it would be 10 to the minus 2-- so it's orders of magnitude better to have the foam rather than the solid.
All right, now let's see what else we have.
So we could do a similar thing for closed-cell foams, and you get diagrams that look like this.
One of the differences with the closed-cell foams, if you assume that the faces don't rupture, the plateau stresses and horizontals-- remember we had that gas contribution, and you can take that into account?
So I'm not going to go over the details of that.
The next one I wanted to talk about was looking at foams that have a yield point.
And again, you can generate a similar kind of diagram, but now, instead of having an elastic failure here, you've got a plastic failure-- you form plastic hinges.
And then again, this diagram is less general than the one for elastomeric foams, so this diagram would be valid for whatever ratio of sigma ys over Es you've the calculation for.
So this one here is for 0.01.
So let me just run through the same kind of calculation for the plastic foams.
Can I rub this off, and then I can use this board to start here?
OK, so the linear elastic part is just the same as for the elastomeric foams.
So you get W over Es is 1/2.
Sigma p over Es squared times 1 over the relative density squared, so it's just the same thing.
And the stress plateau-- you get w over Es is just the plastic collapse strength times the strain range that you go up to.
So if you remember the plastic collapse strength was 0.3 sigma ys times the relative density to the 3/2 power, and then times the strain range.
And then at the end of the stress plateau, you've got the maximum energy absorbed.
So normalize that by Es.
And that's going to be your peak stress over Es times the densification strain again.
So this bit here is the densification strain.
And then you can do a similar thing to figure out the equation of that line that joins the shoulder points.
So the first step is to solve for the relative density there.
So if this part here-- 0.3 sigma ys times the relative density to the 3/2 power is the plastic collapse stress, then at the densification point, the relative density, you just rewrite that and it comes out to the 2/3 power, because you turn the power around.
And then you just substitute this up in here.
So you get that the maximum energy absorbed, normalized by the solid modulus, is your peak stress, times 1 minus 1.4 times this thing in brackets to the 2/3 power-- the 3.3 times the peak stress over the yield strength of the solid.
And then if I just rearrange that, and get the constants, it's 1 minus 3.1 times our ratio of the stresses there.
So maybe I'll just put over here-- the curves are less general than for elastomeric foams.
So each family of curves would be for a particular ratio of the solid yield strength to the solid modulus.
So you get the idea?
It's fairly straightforward.
So I wanted to finish up this topic by giving you a few examples of how you can use these curves.
So the next thing is to look at the selection of foams for impact protection.
And typically, you're given some information about the objects-- so typically you want to protect some object.
It could be a computer, it could be your head, some part of your body.
So typically you know something about the object you want to protect.
So you might know its mass, you might know the area of contact, you might say, well, if it's my computer, I want to make sure it doesn't break if I drop it from a height of a meter.
Or you could say whatever, [?
maybe it's ?]
2 meters.
But you pick something.
And so there's a certain amount of energy you know that you need to absorb.
And you may know that whatever the component is, or the body part, or whatever-- there's some maximum acceleration you can tolerate.
So you might say, well, I want to make sure my computer doesn't break under acceleration of so much.
So you're given the acceleration.
And so if you know the mass and the acceleration, and you know the area of contact, you can figure out a force over an area, and that gives you the peak stress.
So typically, you know those things in the problem.
And typically the problem involves choosing a material, or choosing-- like choosing what kind of material do you want to make the foam out of, and what density of foam do you want to use, what thickness of foam do you want to use?
So I've got a couple of examples just to show you how this works.
So let me just write down a couple notes, and then the rest of it I think I'm just going to take from the slides.
So typically, you know what it is you want to protect, and you know something about it.
So if you know what it is, typically you know what the mass is.
You might know what the contact area would be.
Say a maximum drop height, maximum tolerable acceleration.
So say if you're worried about brain injury-- and making a helmet-- you might know what the maximum tolerable acceleration would be.
So typically, you know what the peak allowable stress is, just from whatever the object itself is.
And the variables that you have to play around with-- variables-- are things like the foam material, the foam density, and the foam thickness.
So I have a couple of examples that have different setups here.
And we'll just see how the thing works out here.
So the first example, we're told the mass of the packaged object is 1/2 kilogram.
We're told the area of contact between the foam and the object is going to be point.
0.01 of 1 meter squared.
And we're told that it's supposed to be designed to withstand a drop of 1 meter-- so if the drop height is 1 meter, then the velocity is just the square root of 2gh.
So g is gravity, so you can work out the velocity on impact would be 4.5 meters per second.
And if you have the velocity on impact and the mass, you can figure out the energy to be absorbed.
You can say mv squared screwed over 2, or you could say it's mgh-- either way.
So here it's going to work out to 5 joules.
And in this case, we're told that the maximum deceleration is 10g-- so then if that's 10g, the maximum force is the mass times the acceleration.
That works out to 50 Newtons.
And then that gives us a peak stress, the maximum allowable peak stress of the force over the area it's 5 kiloNewtons per meter squared.
And in this case, we're told that the foam is going to be a flexible polyurethane, and it has a solid modulus of 50 megapascals.
And so we can calculate this normalized peak stress.
So the normalized peak stress here is 10 to the minus 4.
So in this problem here, we need to figure out what's the foam density, and what's the thickness of the foam to protect the object?
And so that last slide is just summarized here.
And I realized that when I was going to put this up, the font is going to be kind of small, so I blew it up a little.
So we have figured out that we're at sigma p over Es-- the peak stress normalized by the solid modulus is 10 to the minus 4.
And we know that we're going to use a flexible elastomeric polyurethane, so we pull out our diagram for elastomeric foams.
And this sort of hashed band here corresponds to all the different strain rates, and all the different densities.
So I haven't plotted each individual line, we've just got this band that represents the whole thing.
So we know that our normalized peak stress is going to be somewhere along this line here.
That's from everything that's given, and what we can calculate.
And we want to know what density of foam to use, and what thickness.
So the way you approach this is the thickness is going to affect the strain rate.
So the strain rate's just going to be the velocity on impact, divided by the thickness-- or it's an approximation for the strain rate.
So we don't know if we're at this point down here, or if we're at this point up there, because we don't know where we are in the strain rate end of things.
So the way you solve this is you just guess a thickness.
And if you guess a thickness, you can calculate a strain rate.
Then if you calculate the strain rate, you know where in this band you are.
You can figure out a value of W over Es.
And you can use an iterative process.
And the way this is set up is we've chosen two very different initial thicknesses, and the point of doing that is to show you that it converges very quickly.
So the first iteration on this side here, we've chosen the thickness of a meter, which is probably unlikely that we need a meter of foam.
And on the side here, we've chosen a millimeter-- 0.001 meters.
So probably, we need more than that.
So we probably need to be somewhere between those two bounds.
So if the thickness was a meter, then the strain rate turns out to be 4.5 per second.
And then we know where we are in this diagram, and we can read off a value of W over Es.
So we know we're on this line here, and for a particular strain rate, we can read off the W over Es.
So here's the value-- 5.25 times 10 to the minus 5.
And if we know Es, which we do, we can then calculate the actual energy absorbed per unit volume.
We get W-- so W is [?
2620 ?]
joules per cubic meter.
And we can use that value-- because that's an energy per unit volume-- we know the area of contact, and we can use that to get another value of the thickness.
So we use that to calculate the next iteration of the thickness.
So U is the total energy in joules, and W is the energy per unit volume.
So U, the energy in joules, is going to equal the energy unit volume times the area times the thickness.
So we know what the area is, too.
We can then calculate a new thickness.
So now our new thickness is 0.19 meters.
We can use that to get a new strain rate-- that's 24 per second, and go through the whole thing again.
And we end up with a revised energy of 3,300 joules per cubic meter.
Now if we started at the other end, if we started with the first guess was a millimeter, then the strain rate is 4.5 times 10 to the minus 3 per second.
That gives us a different value that we read off here for W over Es, and a different value for W. And then we use this value here for W to make another guess for the thickness.
So that value is 0.14.
And then we go through the whole thing again-- we get a revised strain rate a revised W over Es, and a revised W. And you can see after just two iterations, these two things are almost exactly the same.
And then on the third iteration, they both would give you a thickness of point 0.15 meters.
So even though you can pick wildly wrong first iterations, it converges very quickly, and it's a fairly simple calculation to do.
So you know the thickness that you want is in here, and then you can get the optimum density here.
So you know that your sigma p over Es is along this line of 10 to the minus 4--we're somewhere in here.
These two strain rates here-- one was 24, one was 32-- so the final value is going to be somewhere around 30.
And if you look on this thing here, you can see there's a line that corresponds to 10, there's a line that corresponds to 100.
We're going to be right around in there.
So the relative density is going to be right around 0.01.
So you can use the diagram to get the thickness and to get the density.
OK, are we good?
We're good?
So I've written some notes, and I'll just scan those and I'll put them in the Stellar site.
So here's another example here-- and in this example, it's set up a little bit differently.
So in this example here, we're not told the material, but we're told the thickness of the foam.
So this time we want to get the material, and we want to get the density of the foam.
So here we've got the specification.
We're told the mass is 2 and 1/2 kilograms.
The area of contact is 0.025 meters squared.
We're told the thickness-- here thickness is 20 millimeters.
We've got a drop height of 1 meter again.
And the velocity of impact is then going to be 4.5 meters per second again.
And since we know T, we know that the strain rate is going to be around 225 per second.
We can calculate the energy absorbed-- [?
MGH-- ?]
25 joules.
We can calculate the energy absorbed per unit volume, because we've got the area and the thickness-- so I'm just going to divide that by the area and the thickness.
Now we have 5 times 10 to the 4th joules per cubic meter.
And we're told that we're supposed to design it so the package can withstand a deceleration of 100g-- and so that gives you a maximum force.
And here we've got a maximum allowable peak stress of 10 of the 5 Newtons per meter squared.
So here we have our curve for the elastomeric foams again-- let's assume it's going to be an elastomeric foam, but we don't know what kind of foam.
So the way you solve this problem is that you make a guess for what Es is.
So remember, these diagrams are all normalized by Es.
So to plot a point on there, we need to know what Es is.
So here, we're going to make a guess, and we're going to assume that Es is 100 megapascals to start.
And if we had that value of Es, we know W, up here, and we know sigma p there.
So we just divide those values of W and sigma p by Es, and we get these two values here.
And we plot those two values on the thing here.
So here's our point A-- and that corresponds to that first guess of 100 mega Pascals for the modulus of the solid.
So that's not necessarily the final answer, that's not the right answer-- that's just somewhere to start.
So the thing to notice is, if we have that point there, the strain rate there is probably not quite right.
This upper bound here is-- let's see.
Got to get closer.
Oh, let's see-- that's 10 to the minus 2, that's 10 to the 2.
So the strain rate we want to be is closer up to here, it's not quite down there.
So we're not at the right strain rate.
But the thing to notice is that if we draw a line of slope 1-- so there's this dash line of slope 1-- when we move up and down that line, we're just changing Es.
Because everything is normalized with respect to Es-- if that line has a slope of 1, then we're just moving up and down with respect to Es.
So what we do is we scoot up the line to get to the point that's on the right strain rate.
So if this was a strain rate of 100, or around 200, we want to be at point B.
And then from point B, we can read off what's the value of W over Es, and sigma p over Es.
And from that, we can back out what the solid modulus we want is.
So these are the values that we read off the chart.
This gives us an Es of 28 mega Pascals.
And again, you can go to this more detailed diagram here, and if you read off the sigma p over Es, and the W over Es, the density you want is about 0.1.
So it tells you the modulus of the solid, and from that you can pick a solid, and it tells you the density.
Are we good?
OK, so I have one more.
So there's a slightly different way you can do it, too.
So now I want to talk about bicycle helmets-- you're my bicycle helmet person.
So this is another little case study that involves a slightly different way to do this.
And this involves a slightly different diagram.
So the idea here is to choose a material for a bicycle helmet.
And as you probably all know, the bicycle helmets have a hard shell, and they have some sort of foamy liner.
And the foam's usually around 20 millimeters thick.
And you want something that's light, because you don't want your helmet to be too heavy-- but you want something that will absorb the energy from the impact.
So I've set this up here-- so we assume the mass of the head is about 3 kilograms.
And we assume that it can withstand a deceleration of something like 300g-- and so then you can get a force mass times the acceleration, because you have the force.
And I've assumed an area of contact of something like 0.01 meters squared, so that gives you a peak stress.
So this method here is based on the idea that you have these material selection charts for foams.
Remember we talked about that earlier.
And this is the compressive stress at 25% strain.
So the idea is that axis there is meant to represent the plateau stress, and this axis here represents the densification strain.
And these dashed lines here correspond to basically a value of W-- and energy absorbed per unit volume.
So this is an energy absorbed per unit volume of 0.001 megajoules per cubic meter, 0.01, and so on.
So if you know the peak stress that you can tolerate, for the numbers I gave you it works out to be 0.9 mega Pascals-- so here it's just a little less than 1.
So that's the peak stress that you can tolerate there.
And you can see that the material that's going to absorb the most energy-- so you're absorbing more energy as you move over this way on the curve on the plot-- so the material that's going to do the best is something like an expanded polystyrene that's 5% dense.
So I've highlighted that in red-- expanded polystyrene that's 5% dense.
And then this is the densification strain here.
And you know that the lines of the energy absorption are just the stress times the strain.
So you can basically just read off from here that expanded polystyrene that was 5% dense would be a good choice for a bicycle helmet foam.
Would you like to add anything about bicycle helmet foams?
I think that is exactly what they use, yes.
So this is just another way to do this kind of thing.
And I did one more little calculation here-- if you know the thickness of the foam is, say, 20 millimeters, and we've estimated the area of the contact, you can figure out what the energy absorbed is per unit volume, and from that you can back out the energy in terms of joules, and from that you can back out the velocity that's the maximum speed that one would want to get dinged at on your bicycle.
And the maximum speed works out about 22 miles an hour.
So that's just another example.
Are we good?
OK-- yeah, exactly, your head would be hitting the ground at 22 miles an hour which, would be ouchy.
Which is why you want to wear your helmet, because your skull will not be happy if that happens.
And your brain will not be happy.
And you will not be happy.
And your mom and dad will be really, really, really unhappy.
So you don't want that to happen.
so I have a few minutes left-- and I know I've done this for the people in 3032, but I was going to buy woodpecker talk, because it's about energy absorption.
So if you don't want to watch it, if you've already seen it, you can go.
But there are some people-- you guys haven't seen it, and you haven't seen it.
There's a few people that haven't seen.
And it's cute, involves birds.
And it involves energy absorption-- so I thought I would do this woodpecker talk.
So Barry, this is all just slides.
I don't know if you want to do the lights differently.
I'm not going to write anything on the board, I'm just going to talk
Well, if it will help them see the slides better then certainly
Yeah, I think maybe we could turn the lights down a little, please
Let's try this one
Oh, there we go.
That's good.
Yeah, now I don't feel like I'm looking in the spotlight so much.
That's good.
OK, so you guys know that I like to watch birds.
And you know that I work on foams.
And if you look at bird books, sometimes the bird books say that woodpeckers can withstand the impact from pecking because they have a special material between their skulls and their brains.
And I thought, oh, well I study foams, and I'm interested in woodpeckers-- I should find out what this special material is.
So I started looking into this.
And it turns out there is no special material.
People have looked at the anatomy of woodpecker brains and skulls, and there is no special material.
But it turns out there were also a group of neurologists in the late 1970s who got interested in why woodpeckers don't get brain injury, and they took high speed video of a woodpecker pecking.
And it was kind of amazing what they found out.
So it turns out the woodpeckers can withstand incredibly high decelerations-- much higher than our brains could withstand.
And so I kind of got interested in this, and I decided to try to figure out how it works.
So here's an acorn woodpecker.
And let's see-- I got a little video here.
Wait a minute-- there we go.
There we go.
So here's a little acorn woodpecker.
These live in California.
Anybody from California?
Yeah-- have you seen them?
No, OK. Well you have not been looking carefully.
So here's our little acorn-- and you can see they peck-- oh?
Like around San Francisco
[INAUDIBLE].
[LAUGHS]
So they're pecking, and when they're pecking-- they don't just go bonk.
They do this repeatedly-- they can peck at 10 or 20 times per second.
So the question is, why don't they get brain injury?
So, first of all, why do they peck?
So as we in that little video, that woodpecker was foraging-- it was trying to get little things out of the bark.
And woodpeckers eat insects, and so the bird books say that they can actually hear insects scurrying around under the bark, and they'll peck at the bark and forage to try to get insects.
And there's one other anatomical feature of woodpeckers, which is kind of amazing-- and you can see it in this picture here.
So this is the woodpecker tongue, and the tongue is connected to something called the hyoid process.
And the hyoid process wraps around their eyeballs.
And then when they peck-- I mean, the idea is they're making a hole in the tree, and they've got to get their tongue into the hole to get the bug.
And the end of their tongue has little barbs on it.
And when they contract this thing here, their tongue scoots out and gets the little bugs.
So they pick partly to forage, but they also build something called cavity nests.
So they'll find a tree that's started to rot, and they'll drill a sort of horizontal hole, and then they'll drill a cup underneath that, and they lay the eggs and have their nest at the bottom of the cup.
And then they also-- especially at this time of year, in fact, today I heard woodpeckers drumming-- so it's one of these mating things, one of these courtship things.
So woodpeckers will peck on a hollow branch or a hollow tree, just to make a big loud noise to say, here I am, looking for sex, I'm ready, this is my territory.
And so in the spring-- so this time of year you hear woodpeckers drumming.
And I think naturally they do this on hollow trees to try to make a big sound, but they have adapted to metal downspouts, which are also very effective for making this loud noise.
And people who've-- I wrote a paper on this-- people who have read my paper sometimes email me and say, oh, I heard you know about woodpecker pecking.
How can I get them to stop drumming on my downspout?
Because it's kind of annoying, if you're the human inside the house.
So anyway, they peck for these reasons.
And then acorn woodpeckers are special-- acorn woodpeckers are what I think of as the champions of pecking.
And they do one more behavior-- so here's from David Sibley's beautiful Guide to Birds, here's the acorn woodpecker.
They store acorns in what's called a granary, and they look at old tree trunks that are beginning to rot, and they pick holes in the tree trunk and they store the acorns in the holes.
So see all those little dark dots on that trunk?
Those are all holes that the acorn woodpecker has pecked.
And they'll do this-- they live in social groups, so there might be like 10 or 20 of them living together, and they'll peck like 10,000 holes into their granary.
There will be thousands and thousands of these holes.
And here we have the acorn woodpecker with the acorn in its beak, and you can see some of these holes have acorns, and some of them are empty.
So here is the acorn woodpecker in action.
And here's the little video again-- I won't play the video.
So I often give this talk for non-engineers, so I'm going to explain some things-- there's going to be some writing on the slides that you guys already know.
So the impact force depends on the deceleration-- how quickly the brain stops when the beak hits the tree.
And I should mention, these videos are from the Cornell Lab of Ornithology-- they have an amazing collection of bird audio, like bird calls, bird songs, and bird photographs and videos.
It's incredible, the collection that they've got.
And then I explain what acceleration is-- so I'm going to put acceleration in terms of gravity.
And when the beak hits the tree, there's going to be a deceleration on impact.
And just as a comparison, human brain injury occurs roughly at about 100g.
And so the question is, how much deceleration can the woodpecker brain take?
And that's where the neurologists in the Bay Area come into the picture.
They found out that there was a park ranger-- I think maybe at Point Reyes, just north of San Francisco.
And he had an acorn woodpecker that, I don't, had an injured wing or something.
Anyway, he had this acorn woodpecker that he kept.
And they were able to use this acorn woodpecker, and they took high-speed video of the woodpecker pecking.
So from the high-speed video, the video that they took went at something like 2,000 frames a second.
So they have a picture of where the head is every 2,000th of a second.
So if you know the position at these times, you can get the velocity, and you can get the deceleration.
So they measured the deceleration, and they measured some amazing things.
So they measured that on impact, the woodpecker's bill was going something like 15 miles an hour.
And the decelerations were up to 1,500g-- so many times more than what we can withstand.
And they also measured the stopping time-- and they thought it was between about 1/500th to 1/1,000th of a second.
And that's going to be important later on.
So one of the interesting things about this was how they got this whole thing set up.
So I don't know how many of you do UROPs-- but you know, part of the thing in doing UROPs, and doing experiments in the lab is just how do you do the experiments?
So you know, the park rangers got the acorn woodpecker, that's all very good.
But they have to get the woodpecker to peck in front of a camera, and they have to get the camera to turn on as it's pecking-- so there's some experimental challenges.
So the important thing, the critical thing, is the date of the paper.
This was written in 1979, which meant they probably did the experiments in 1978.
And I was a graduate student in 1978.
And I can report there were no Apple computers in 1978-- there were no laptops.
You couldn't just kind of do your little PowerPoint slides.
And most offices had something called an IBM Selectric typewriter-- I don't know if you've ever seen the old IBM typewriters.
But the typewriters, when you typed on the typewriters, they made these noises, and it sounded kind of like a woodpecker pecking.
And the ranger had one of these typewriters in his office, and he had discovered that if he typed on the typewriter, the woodpecker thought, oh, there's another woodpecker.
I'll start pecking.
And so they used the typewriter as a way to get the woodpeckers to peck.
And I think they had some old stump, and I don't know if they put nuts or peanut butter or something into the stump to get the woodpecker to peck at it-- because they needed to peck at a particular spot, so they have the camera all set up.
So anyway, they had this whole arrangement to do this high-speed video of the woodpecker pecking.
OK, so it goes that up to 1,500g, which is kind of amazing.
And then, remember also, they do this repeatedly-- they do it at like 10 or 20 times a second.
Then this is just explaining what stress is-- but you already know what stress is, so I'm going to skip over that.
And so the way you can think about this is to think about it in terms of a scaling argument.
So imagine there's the brain and there's the skull, and when the head hits the tree, the brain's going to accelerate, and the brain and the skull are both going to decelerate.
And you can think of the stress as the force over the area.
So the force is the mass times the deceleration over the area.
So that value for the woodpecker, you can say, is roughly equal to the same values, but for the human.
So what that argument relies on is the idea that the stress to cause damage in the woodpecker brain is similar to the stress to cause damage in the human brain.
And that's not totally unreasonable.
So if you look at bone, for example, like when you measure the strength of the whale bones, it's not going to be that different from what you would measure from human bones.
So when you look at a particular type of tissue, and you look at the strength of it in different species, the properties aren't that different from one species to another.
So let's say the brain tissue gets damaged at the same stress in the two species.
So we can write this sort of equation down.
And the mass is going to depend on the density of the brain tissue times the volume-- and the volume goes as the radius cubed.
And the density is going to be the same for the woodpecker, or for the human-- so assume the brain tissue has the same kind of basic stuff.
So the mass goes as the radius cubed, and the area of contact is going to go as the radius squared.
And so there's a radius term.
So you can say the radius times the deceleration in the woodpecker should be equal to the radius times the deceleration of the human.
And obviously, the woodpecker radius is going to be a lot smaller, so the woodpecker deceleration is going to be a lot bigger.
So part of the thing is the brain is just a lot smaller.
Then there's another factor that comes into it-- these are photographs of an acorn woodpecker skull and a human skull that I got from the Museum of Comparative Zoology at Harvard.
So this is looking down on the top, and this is an elevation view.
And if you think of the brain as roughly a hemisphere-- I mean, obviously it's not a perfect hemisphere, but let's say it's roughly a hemisphere-- the orientation of the brain in the skull is slightly different in the birds and in the human.
So if you think of it as being this way on, and in the woodpecker it's turned roughly this way on, and the contact area-- think of this as the contact area.
The projected area would be a full circle, and in the human, the brain is more this way on.
And then the projected contact area is a semicircle.
So there's a factor of 2 difference, just because of the orientation of the brain.
And so I've put the factor of 2 that accounts for that.
So then I could say the deceleration that the woodpecker can take is twice the ratio of the radii of the human and the woodpecker brain, times the deceleration the human can take.
And this was around 100g, remember.
So then I wanted to know what the ratio of the sizes was, and I thought, oh geez-- I'm going to have to mess around with skulls, and try to make some sort of measurements, and this is going to be a drag.
And then I found this paper-- I couldn't believe it-- I found this paper called "Brain Size in Birds."
And this guy had table after table after table of the mass of the brain in different birds-- and he had the acorn woodpecker, lucky for me.
So if you just assume that the brain is roughly a hemisphere, if you have the mass, you can work out roughly what the radius is.
So it turns out there's a factor of 8 difference between the radii.
And so the human brain is about eight times the size of the woodpecker.
And then if you take into account this factor of 2, that means the woodpecker should be able to tolerate a deceleration of 16 times what the human can.
So remember, I said the human can take about 100-- so this would get the woodpecker up to 1,600.
But they measured 1,500.
And I'm a civil engineer, originally, I like big factors of safety-- that's a little too close for comfort for me.
And so it turns out there's one more factor that matters.
People have studied human brain injury pretty extensively, and one of the things they've looked at is how much acceleration you can tolerate without injury, relative to the duration of the impact.
So this is from a car crash conference-- so here's the tolerable acceleration, and here's the duration of the impact.
And typically, for human head impacts, the deceleration occurs over a few milliseconds-- like over 3 to 10 milliseconds.
So if this is the duration of the typical head impact for a human, here's the range of the tolerable accelerations between about 80g and 160g-- so you know, I said around 100.
So we can take this curve, and now we have this factor of 16, and we can just scale it up by our factor of 16 for the woodpecker.
So if I scale it up by the factor of 16, we're there.
But the duration of the impacts was more around 1/2 a millisecond, to a millisecond, so I've extrapolated this a little bit.
And the duration of them is up in here.
So this is saying the woodpecker can take these sorts of decelerations here, and that adds on another factor of 4.
And these were the measured decelerations-- 1,500 was about the biggest, and I think it went to about a few hundred g. So there's really three factors-- one is the small brain size, so there's this sort of scaling factor.
One is the orientation of the brain, that was another factor of 2.
And then there's this duration of impact, which is a factor of 4.
And so that's how you can get this huge decelerations that they've measured in the woodpecker when they're pecking.
So now I have just a couple more slides.
So woodpeckers have various adaptations to pecking, too.
So one of the things is they have amazingly stiff tail feathers-- you see the tails here?
You can't quite see it because this didn't reproduce quite properly, but the woodpecker is perched on a tree here, and the tail is pressed up against the tree.
And if you think about pecking, like imagine if its tail was not pressed up against the tree-- it would be kind of grabbing on with its feet, and it would be trying to peck, and it would be hard to push.
You know, you need something to push against.
And so it's got these stiff tail feathers, which make it easier to push against the tree.
And so here's the stiff tail feathers here.
It's also got kind of unusual feet for birds-- it's got two toes forward and two toes back.
And again, if it's grabbing with its feet, that helps it get some purchase to push against.
That's called zygodactyl in the bird world.
So it's got these adaptations, the pecking.
And then finally, I just like this slide here from The New Yorker, because it's a woodpecker pecking out a woodpecker, so it seemed kind of cute.
And several people helped me with this project.
Trey Crisco studies head injury at Brown.
He actually does work for the NFL on football brain injuries, and looks at football helmets.
Sharon Swartz is a friend of mine at Brown, who's a biologist.
She studies bat flight.
And she just thought this was kind of a cool project.
So, in fact, I was walking the dog last week, a few days ago, and I saw bats flying around overhead at night.
And it was so great to see the little bats.
So I had to immediately email Sharon, and say, bats-- there's bats in my neighborhood.
Andy Biewener runs the Concord Field Station, and studies animal locomotion, and I talked to him about it.
Jeremy Trimble was the one who gave me the skull pictures, and Matt Dawson was a student helped me do some of the images.
So that's my woodpecker talk.
So that's the end of energy absorption, I just thought that was kind of amusing.
So next time, we'll start talking about sandwich panels.
So, let's see-- Monday's a holiday.
I will be in Toronto on Monday, seeing my family.
In fact, I'm going to see one of my old professors-- I'm going to my old fluid mechanics professor on Monday and have lunch.
I'll see my family on the weekend.
And so we'll meet Wednesday next week, and I'll start the bit on sandwich panels.
So I think there's two lectures on engineering sandwich panels, and then there's a lecture on natural sandwich panels-- so bird skulls, for example, are natural sandwich panels.
So I'll talk about that.
And then I think the last lecture-- there's only a handful of lectures left-- the last one I think I talk about natural materials, because you know, I like that.
So I just do that for fun.
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OK, so we should probably start.
So last time we finished up talking about energy absorption in foamy cellular materials.
And today I wanted to start a new topic.
We're going to talk about sandwich panels.
So sandwich panels have two stiff, strong skins that are separated by some sort of lightweight core.
So the skins are typically, say, a metal like aluminum, or some sort of fiber composite.
And the core is usually some sort of cellular material.
Sometimes it's an engineering honeycomb.
Sometimes it's a foam.
Sometimes it's balsa wood.
And the idea is that what you're doing with the core is you're using a light material to separate the faces, and if you think about an I-beam-- so if you remember when we talk about bending and we talk about I-beams, the whole idea is that in bending, you want to increase the moment of inertia.
So you want to make as much material as far away from the middle of the beam as possible to increase the moment of inertia.
So if you think about an I-beam, you put the flanges far apart with the web, and that increases the moment of inertia.
And the sandwich panels and the sandwich beams essentially do the same thing, but they're using a lightweight core instead of a web.
And so the idea is you use a lightweight core.
It separates the faces.
It increases the moment of inertia.
But you don't add a whole lot of weight because you've got this lightweight core in the middle.
So I brought some examples that I'll pass around and we can play with.
So these are some examples up on the screen, and some of those I have down here.
So for instance, the top-- turn my little gizmo on-- the top left here, this is a helicopter rotor blade, and that has a honeycomb core in it.
This is an aircraft flooring panel that has a honeycomb core and has carbon fiber faces.
So that's this thing here.
I'll pass that around in a minute.
This is a downhill ski.
This has aluminum faces and a polyurethane foam core.
And that's the ski here.
And it's quite common in skis now to have these sandwich panels.
This is a little piece of a small sailing boat.
It had, I think, glass fiber faces and a balsa wood core.
And I don't know if any of you sail, but MIT has new sailing boats.
Do you sail?
I do not much these days
OK, but those little tech dinghies that you see out in the river, those have sandwich panel holes to them.
So those are little sandwich panels.
This is an example from a building panel.
This has a dry wall face and a plywood face and a foam core, and the idea with panels for buildings is that usually they use a foam core because the foam has some thermal insulation.
So as well as sort of separating the faces and having a structural role, it has a role in thermally insulating the building.
The foams are a little less efficient than using a honeycomb core.
So for the same weight, you get a stiffer structure with a honeycomb core than a foam core.
But if you want thermal insulation as well as a structural requirement, then the foam cores are good.
And these are a couple examples of sandwiches in nature.
This is the human skull.
And your skull is a sandwich of two dense layers of the compact bone, and you can see there's a little thin layer of the trabecular bone in between.
So your head is like a sandwich, your skull is like a sandwich.
And I don't know if I'll get to it next time, but in the next couple of lectures, I'm going to talk a little bit about sandwich panels in nature, sandwich shells in nature.
You see this all the time.
And this is a bird wing, here.
And so you can see there's got the dense bone on the top and the bottom, and it's got this kind of almost trust-like structure in the middle.
And obviously birds want to reduce their weight because they want to fly, so reducing the weight's very important.
And so this is one of the ways that birds reduce their weight, is by having a sandwich kind of structure.
So I have a couple of things here.
These are the two panels at the top there.
This is the ski, and you can yank those around.
I also have a few panels that people at MIT have made.
And I have the pieces that they're made from.
So you can see how effective the sandwich thing is.
So this was made by a guy called Dirk Moore.
He was a graduate student in ocean engineering.
And it has aluminum faces and a little thin aluminum core.
So you can see, if you try and bend that with your fingers, you really can't bend it any noticeable amount.
And this panel here is roughly the same thickness of the face on that.
And you can see how easy it is for me to bend that-- very easy.
And this is the same kind of thing as the core.
It's thicker than that core, but you can see how easy it is for me to bend this, too.
So each of the pieces is not very stiff at all.
But when you put them all together, it's very stiff.
So that's really the beauty of this.
You can have lightweight components, and by putting them together in the right way, they're quite stiff.
So here's another example here.
This is a panel that one of my students, Kevin Chang made.
And this has actually already been broken a little bit, so it's not quite as stiff as it used to be.
And you can kind of hear, it squeaks.
But you can feel that and see how stiff that is.
And this is the face panel here.
And you can see, I can bend that quite easily with my hands.
Doodle-doot.
And then this is the core piece here.
And again, this is very flexible.
So it's really about putting all those pieces together.
So you get this sandwich construction and you get that effect, OK?
[?
Oop-loo.
?]
All right, so what we're going to do is first of all look at the stiffness of these panels, calculate their deflection.
We're going to look at the minimum weight design of them.
So we're going to look at how for, say, given materials in a given span, how do we minimize the weight of the beam for a given stiffness?
And then we're going to look at the stresses in the sandwich beams.
So there's going to be one set of stresses in the faces, and a different kind of stress distribution in the core.
So we'll look at the stress distribution.
And then we'll talk about failure modes, how these things can fail, and then how to figure out which failure mode is dominant, which one occurs at the lowest load.
And then we'll look at optimizing the design, minimizing the design for a certain strength and stiffness.
So we're not going to get all that way through everything today, but we'll kind of make a start on that.
OK.
So let me start.
So the idea here is we have two stiff, strong skins, or faces, separated by a lightweight core.
And the idea is that by separating the faces, you increase the moment of inertia with little increase in weight.
So these are particularly good if you want to resist bending, or if you want to resist buckling.
Because both of those involve the moment of inertia.
And they work like an I-beam.
So the faces of the sandwich are like the flanges of the I-beam, and the core is like the web.
And the faces are typically made of either fiber reinforced composites or metals.
So typically, something like aluminum, usually you're trying to reduce the weight if you use these things, so a lightweight metal like aluminum is sometimes used.
And the cores are usually honeycombs, or foams, or balsa.
And when they use balsa wood, what they do is-- I brought a piece of balsa here-- what they do is they would take a block like this and chop it into pieces around here.
And then they would lay those pieces on a cloth mat.
So typically the pieces are maybe two inches by two inches.
They lay them on a cloth mat, and because they're not one monolithic piece, they can then shape that mat to curved shapes.
So it doesn't have to be just a flat panel.
They can curve it around a curved surface if they want.
So we'll say the honeycombs are lighter than the foams for a given stiffness or strength.
But the foams provide thermal insulation as well as a mechanical support.
And the overall mechanical properties of the honeycomb depend on the properties of each of the two parts, of the faces and the core, and also the geometry of the whole thing-- how thick's the core, how thick's the face, how dense is the core?
That kind of thing.
And typically, the panel has to have some required stiffness or strength.
And often what you want to do is minimize the weight for that required stiffness or strength.
So often these panels are used in some sort of vehicle, like we talked about the sailboat, or like a helicopter, or like an airplane.
They're also used in like refrigerated trucks-- they would have a foam core because they'd want the thermal insulation.
So if you were going to use it in some sort of a vehicle, you want to reduce the mass of the vehicle and you want to have the lightest panel that you can.
Yup?
So if you saw the base material and you'd have the [INAUDIBLE] sandwich panel, that piece [INAUDIBLE] the sandwich panel with something [?
solid ?]
in the middle?
Well--
So, as we're getting [INAUDIBLE] aluminum piece that's was as thick as a sandwich panel
Yeah
[INAUDIBLE]
Oh, well, if you have the solid aluminum piece that was as thick as the sandwich, it's going to be stiffer, but it's going to be a lot, lot heavier.
So the stiffness per unit weight would not be as good.
OK?
So we're going to calculate the stiffness in just one minute.
And then we're going to look at how we minimize the weight, OK?
OK.
So what I'm going to do is set this up as kind of a general thing.
We're just going to look at sandwich beams rather than plates, just because it's simpler.
But the plates, everything we say for the beams basically applies to the plates.
The equation's just a little bit more complicated.
So we're going to start with analyzing beams.
And I'm just going to start with a beam, say, in three-point bending.
So there's my faces there.
Boop.
And I've made it kind of more stumpy than it would be in real life, just because it makes it easier to draw it.
And then if I look at it the other way on, it would look something like that.
So say there's some load P here.
Say the span of the beam is l. Say the load's in the middle, so each of the supports just sees a load of P/2.
And then let me just define some geometrical parameters here.
I'm going to say the width of the beam is b.
And I'm going to say the face thicknesses are each t. So the thickness of each face is t. And the thickness of the core is c, OK?
So that's just sort of definitions.
And I'm going to say the face has a set of properties, the core has a set of properties, and then the solid from which the core is made has another set of properties.
So the face properties that we're going to use are a density of the face.
We'll call that row f. The modulus of the face, Ef, and some sort of strength of the face, let's imagine it's aluminum and it yields, that would be sigma y of the face.
And then the core similarly is going to have a density, rho star c. It's going to have a modulus, E star c. And it's going to have some strength, I'm going to call sigma star c. And then the solid from which the core is made is going to have a density row s, a modulus Es, and some strength, sigma ys, OK?
So the core is going to be some kind of cellular material, a honeycomb, or a foam, or balsa.
And typically, the modulus of the core is going to be a lot less than the modulus of the face.
So I'm just going to say here that the E star c is typically much greater than Ef.
And we're going to use that later on.
So we're going to derive some equations, for example, for an equivalent flexural rigidity for the section, an Ei equivalent.
And that has several terms.
But if we can say the core stiffness is much less than the face thickness, and also if we can say the core-- the stiffness is less and also the thickness of the core is much greater than the thickness of the face, a lot of the expressions we're going to use simplify.
So we're going to make those assumptions.
So let me just draw the shear diagram here.
So V is shear, so that's the shear diagram.
We have some load P/2 at the support.
There's no other load applied until we get to here.
Than the shear diagram goes down by P, so we're at minus P/2.
Then there's no load here, so this just stays constant, and then we go back up to 0.
And then let me just draw bending moment diagram.
The bending moment diagram for this is just going to look like a triangle.
Remember, if we integrate the shear diagram, we get the bending moment diagram.
And that maximum moment there is going to Pl over 4.
OK.
So initially, I'm going to calculate the deflections.
And I don't really need those diagrams for that, but then I'm going to calculate the stresses, and I'm going to need those diagrams for the stresses.
So just kind of keep those in mind for now.
So to calculate the deflections, sandwich panels are a little bit different from homogeneous beams.
In a sandwich panel, the core is not very stiff compared to the faces.
And we've got some shear stresses acting on the thing.
And the shear stresses are largely carried by the core.
So the core is actually going to shear, and there's going to be a significant deflection of the core and shear as well as the overall bending of the whole panel.
So you have to count for that.
So we're going to have a bending term and a shear term-- that's what those two terms are there.
So we're going to say there's a bending deflection and a shear deflection.
And that shearing deflection arises from the core being sheared and the fact that the core, say, Young's modulus or also the shear modulus, is quite a bit less than the face modulus.
So if you think of the core as being much more compliant than the face, then the core is going to have some deflection from that shear stress.
OK, so we're going to start out with this term here, the bending term.
And if I just had a homogeneous beam in three-point bending, the central deflections-- so these are all the central deflections I'm calculating here-- with Pl cubed over it turns out to be 48 is the number, and divided by EI.
And because we don't have a homogeneous beam here, I'm going to call that equivalent EI.
And to make it a little bit more general, instead of putting 48, that number, I'm just going to put a constant B1.
And that B1 constant is just going to depend on the loading geometry.
So any time I have a concentrated load on a beam, the deflection's always Pl cubed over EI, and then the sum number in the denominator and that number just depends on the loading configuration.
So for three-point bending, it's 48.
For the flexion of a cantilever, B1 would be 3.
So think of that as just a number that you can work out for the particular loading configuration.
So here we'll say B1 is just a constant that depends on the loading configuration.
And I'll say, for example, for three-point bending, B1 is 48.
For a cantilever end deflection, then B1 would be 3.
So it's just a number.
So the next thing we have to figure out is what's the EI equivalent.
So if this was just a homogeneous beam, and it was rectangular, E would just be E of the material and I would be the width B times the height H cubed divided by 12.
So here we don't quite have that because we have two different materials.
So here we have to use something called the parallel axis theorem, which I'm hoping you may have seen somewhere in calculus, maybe?
But, yeah, somebody is nodding yes.
OK, so what we do, what we want to do is get the equivalent EI-- I'm going to put it back up, don't panic-- of this thing here, right?
So I want-- this is the neutral axis here, and I want the EI about that neutral axis there.
So, OK, you happy?
There.
OK, so I've got a term for the core.
OK, the core, that is the middle of the core, right?
So for the core, it's just going to be E of the core times bc cubed over 12.
Remember, for a rectangular section, it's bh cubed over 12 is the moment of inertia.
And here our height for the core is just c, OK?
And then if I took the moment of inertia for, say, one face about its own centroidal axis, I would get E of the face now times bt cubed over 12.
So that's taking the moment of inertia of one face about the middle of the face.
And I have two of those, right?
Because I have two faces.
And the parallel axis theorem tells you what the moment of inertia is going to be if you move it, not to the-- you don't use the centroid of the area, but you use some other parallel axis.
And what that tells you to do is take the area that you're interested in-- so the area of the face is bt, and you multiply by the square of the distance between the two axes that you're interested in.
Oop, yeah.
Let me change my little brackets.
Boop.
So, oop-a-doop-a-doop.
Maybe I'll stick this, make a little sketch over here again.
OK, all right.
So this term here, Ef bt cubed over 12, that would be the moment of inertia of this piece here, about the axis that goes through the middle of that, right?
Its own centroidal axis.
But what I want to do is I want to know what the moment of inertia of this piece is about this axis here.
This is the neutral axis.
So let's call this the centroidal axis.
And the parallel axis theorem tells me what I do is I take the area of this little thing here, so that's the b times t, and I multiply by the square of the distance between those two axes.
So the distance between those axes is just c plus t over 2, and I square it.
And then I multiply that whole thing by 2 because I've got two faces.
Are we good?
[INAUDIBLE]
Yeah?
The center [INAUDIBLE] and the [INAUDIBLE], are those Ed's or Ef's?
These are Ef's because this is the face now, right?
So this term here is for the core.
So here the core is E star c. And these two Ef's are for the face up there, OK?
Because you have to account for the modulus of the material of the bit that you're getting the moment of inertia for.
Are we good?
OK.
So now I'm just going to simplify these guys a little bit.
Doodle-doodle-doodle-do-doot.
OK?
So I've just multiplied the twos, and maybe I'll just write down here this is the parallel axis theorem.
Doot-doot-doot.
Yes, sorry?
So for the term that comes from the parallel axis theorem, why do we only consider Ef and not [INAUDIBLE]
Because I'm taking-- what I'm looking at-- so the very first term, this guy, here--
Yeah, [INAUDIBLE]
Accounts for this, right?
And these two terms both account for the face
Oh, OK, so the face acting--
Yeah, about this axis.
So the parallel axis theorem says you take the moment of inertia of your area about its own centroidal axis, and then you add this term here.
But it's really referring to that face, OK?
Let me scoot that down and then scoot over here.
And this is where we get to say the modulus of the face is much greater than the modulus of the core.
And also, typically c, the core thickness, is much greater than t, the face thickness.
So if that's true, then it turns out this term is small compared to that one.
And also this term is small compared to this one.
And also this term, instead of having c plus t squared, if c is big compared to t, then I can just call it c squared, OK?
So you can see here, if Ec is small, then this is going to be small compared to these.
If t is small, then this guy is going to be small.
So even though it looks ugly, many times we can make this simpler approximation.
OK, so we can just approximate it as Ef times btc squared over 2.
So then this bending term here, we've got everything we need now to get that bit there.
So the next bit we want to get is the shearing deflection.
So what's the shearing deflection equal to?
So say we just thought about the core, and all we're interested in here is what's the deflection of the core and shear?
And so say that's P/2, that's P/2, that's l/2.
We'll say that's-- oops.
That's our shearing deflection there.
We can say the shear stress in the core is going to equal the shear modulus times the shear strain, so we can say P over the area of the core is going to be proportional to the Young's modulus times delta s over l. And let's not worry about the constant just yet.
So delta s is going to be proportional to-- well, let me [?
make it ?]
proportional at this point.
Delta s is going to equal Pl divided by some other constant that I'm going to call B2, and divided by the shear modulus of the core, and essentially the area of the core.
And here B2 is another constant.
So again, B2 just depends on the loading configuration.
Yeah, this is a little bit of an approximation here, but I'm just going to leave it at that.
OK, so then we have these two terms and we just add them up to get the final thing.
Start another board.
OK.
So that would give us an equation for the deflection.
And one thing to note here is that this shear modulus of the core, if the core is a foam, then we have an equation for that.
We also could use an equation if it's a honeycomb.
But I'm just going to write for foam cores.
Whoops.
This is for-- that will be for open-cell foam cores.
Oops, don't want to-- and get rid of that.
We won't update just now, thank you.
OK, so the next thing I want to think about is how we would minimize the weight for a given stiffness.
So say if we're given a stiffness, we're given P over delta, so I could take out the two P's here.
If I divide it through by P, delta over P would be the compliance, P over delta would be the stiffness.
So imagine that you're given the face and core materials, and you're told how long the span has to be, you're told how wide the beam is going to be, and you're told the loading configuration.
So you know if it's three-point bending, or four-point bending, or a cantilever-- whatever it is.
And you might be asked to find the core thickness, the face thickness, and the core density that would minimize the weight.
So I have a little schematic here.
I don't know if you're going be able to read it.
So I'm going to walk through it and then I'll write things on the board.
Whoops, hit the wrong button.
OK, so we start with the weight equation here.
The weight's obviously the sum of the weight of the faces, the weight of the core, so those two terms there.
So I'll write that down in a minute.
And then we have the stiffness constraint here.
So this equation here is just this equation that I have down here on the board, OK?
Then what you do is you solve that stiffness constraint for the density of the core.
So this equation here just solves-- we're solving this equation here in terms of the density, and we get the density by substituting in this equation here for the shear modulus of the core.
So you substitute that there.
It's kind of a messy thing, but you solve that in terms of the density.
Then you put that version of the density here in terms of this weight equation up here.
So then you've eliminated the density out of the weight equation, now you've just got it in terms of the other variables.
And then you take the partial derivative of the weight with respect to the core thickness c, set that equal to 0, and you take the partial derivative of the weight with respect to the face thickness, t, and you set that equal to 0.
And that then gives you two equations and two unknowns.
You've got the core thickness and the face thickness are the two unknowns.
And you've got the two equations, so then you solve those.
So the value you get for the core thickness is then the optimum, so it's going to be some function of the stiffness, the material properties you started with in the beam geometry.
And similarly, you get some equation for the optimum face thickness, t. And again, it's a function of the stiffness and the material properties in the beam geometry.
Then you take those two values for c and t, those two optimum values, and plug it back into this equation here, and get the optimum value of the core density.
And so what you end up are three equations for the optimal values of the core thickness, the face thickness, and the core density in terms of the required stiffness, the material properties, and then the loading geometry.
So I'm going to write down some more notes, because I'll put this on the Stellar site.
But it's hard to read just here.
So let me write it down and I'll also write out the equations so that you have the equations for calculating those optimum values.
So before I do that, though, one of the interesting things though is if you figure out the optimal values of the core thickness and the face thickness and the core density, and you substitute it back into the weight, and you calculate this is the weight of the face relative to the weight of the core, no matter what the geometry is, and what the loading configuration is, the weight of the face is always a quarter of the weight of the core.
So the ratio of how much material is in the core and the face is constant, regardless of the core-- of the loading configuration.
And this is the bending deflection relative to the total deflection.
It's always 1/3.
And the shearing deflection relative to the total deflection is always 2/3.
So regardless of how you set things up, the ratio of what weight the face is relative to the core and the amount of shearing and bending deflections is always a constant at the optimum.
OK, so let's say we're given the face and the core materials.
So that means we're given their material property, too.
And say we're given the beam length and width and the loading configuration.
So that means we're given those constants, B1 and B2.
If I told you it was three-point bending, you would know what B1 and B2 are.
So then what you need to do is find the core thickness, c, the face thickness, t, and the core density, rho c, to minimize the weight of the beam.
So there's two faces, so the weight of the face is 2 rho f g times btl.
And then the weight of the core is rho c g times bcl.
So I'm going to write down the steps and then I'll write down the solution.
So you solve.
So you put this equation for the shear modulus of the core into here, and then you rearrange this equation in terms of the density of the core here.
So you have an equation for the core density in terms of that stiffness, and then you solve the partial derivatives of the weight equation with respect to the core thickness, c, and put that equal to 0.
And then the partial of dw [?
over ?]
dt and set that to 0.
And if you do that, you can then solve for the optimal values of the face and core thicknesses.
Yes?
[INAUDIBLE] for weight, what is g?
Gravity

Just density is mass, mass times gravity-- weight.
That's all it is.
And then you've got a version of this that's in terms of the core density.
You can substitute those values of the optimum face and core thicknesses into that equation and get the optimum core density.
And then in the final equations, you get, when you do all that, and I'm going to make them all dimensionless, so this is the core thickness normalized by the span of the beam is equal to this thing, here.
So you can see each of these parameters here, the design parameters that we're calculating the optimum of.
I've grouped the constants B1 and B2 together that describe the loading configuration so you'd be given those.
C2 is this constant-- oop, which I just rubbed off-- that relates the shear modulus of the foam core.
So you'd be given that.
These are the material properties of the-- you know, say, it's a polyurethane foam core, this would be the density of the polyurethane.
Say it's aluminum faces, that would be the density of the aluminum.
so you'd be given that.
You'd be given the stiffnesses of the two materials, the solid from which the core is made and the face material.
And then this is the stiffness here that you're given, just divided by the width of the beam, B.
So the stiffness, you'd be given the width B.
So you're given all those things, then you could calculate what that optimum design would be.
So the next slide here just shows some experiments.
And these were done on sandwiches with aluminum faces and a rigid polyurethane foam core.
And here we knew what the relationship was for the shear modulus.
We measured that.
And what we did here was we designed the beams to all have the same stiffness, and they all had the same span in the width, B, then we kept one parameter at the optimum value and we varied the other ones.
So here, on this beam, this set of beams here, the density was at the optimum.
And we varied the core thickness, and we varied the face thickness, and the solid line was our model or our sort of optimization.
And the little X's were the experiments.
So you can see there's pretty good agreement there.
Then the second set here, we kept the face thickness at the optimum value and we varied the core thickness, we varied the core density.
So the same thing, the solid line is the sort of theory and the X's are the experiments.
And here we had the core thickness of the optimum value, and we varied the face thickness and the core density.
So you can kind of see how you can see this here.
And over here, just because I forgot to say it, this is the stiffness per unit weight, over here, OK?
So these are the optimum designs here, all right?
So there was pretty good agreement between these calculations and what we measured on some beams.
Do I need to write anything down?
Do you think you've got that?
Yeah?
I was just going to ask, for the optimum design column that you have there, do those numbers like fall out of these equations if you do the math?
They do, yeah.
I mean, it's-- yeah, exactly.
So if you remember the equation we had for the weight, so the weight is equal to 2 rho f gbtl plus the density of the core, bcl, so if you plug these things into there, then-- so this is the way to the face, that's the way to the core, then it drops out to be a quarter.
So it's kind of magical.
I mean, you have this big, long, complicated gory thing, and then, poof, everything disappears except a factor of 1/4.
And the same for the bending deflection.
So we had those two terms, so there was the bending and the shear.
If you just calculate each of those terms and take the ratio of 1 over the total, or the one over the other, everything drops out except that number.
So that's why I pointed it out, because it seemed kind of amazing that everything would drop out except for that one thing.
OK, so then the next thing-- so that's the stiffness in optimizing the stiffness.
Are we happy-ish?
Yeah?
OK.
So the next thing-- oh, well, let's see.
I don't think I need to write any.
I think if you have that graph, I don't really need to write much down.
So the next thing then is the strength of the sandwich beams.
So let me get rid of that.
You guys OK?
Yeah?
Yeah
Yeah, but you're shaking your head like this is very, very helpful for me
[INAUDIBLE]
Oh OK, that's
[INAUDIBLE]
That's OK, you can do that.
I don't mind.
But as long as you don't have questions for me.
OK, and so the first step in trying to figure out about this strength is we need to figure out the stresses in the beams.
So we need to find out about the stresses.
And we're going to have normal stresses and we're going to have shear stresses.
So I'm going to do the normal stresses first and then we'll do the shear stresses.
So you do this in a way that's just analogous to how you figure out the stresses in a homogeneous beam.
So we'll say the stresses in the face-- normally it would be My over I. M is the moment, y is the distance from the neutral axis, I is the moment of inertia.
So this time, instead of having a moment of inertia, we have this equivalent moment of inertia.
And we multiply by E of the face.
So you can think of this as being the strain essentially.
And then you multiply by E of the face to get the stress.
The maximum distance from the neutral axis, we can call c/2.
So that's y.
Then EI equivalent we had Ef btc squared over 2.
And then I have a term of Ef here.
c squared.
So one of the c's goes, the 2's go, the Ef's go.
Then you just get that the normal stress in the face is the moment at that section divided by the width, b, the face thickness, t, the core thickness, c. And I can do the same kind of thing for the stress in the core, except now I multiply by the core modulus.
So if I go through the same kind of thing, it's the same factor of M over btc, but now I multiply times E of core over E of the face.
And since E of the core is a lot smaller than E of the face, typically these normal stresses in the core are much smaller than the normal stresses in the face.
So the faces carry almost all of the normal stresses.
And if you look at an I-beam, the flanges of the I-beam carry almost the normal stresses.
So I want to do one more thing here.
I want to relate the moment to some concentrated load.
So let's say we have a beam with a concentrated load, P. So for example, something in three-point bending, typically we're interested in the maximum stresses, so we want the maximum moment.
So M max is going to be P times l over some number.
And this B3 is another constant that depends on the loading configuration.
So if it was three-point bending, B3 would be 4.
If it was a cantilever, B3 would be 1.
So if I put those things together, the normal stress in the face is Pl B3 divided by btc.
OK, so that's the normal stresses.
And then the next thing is the shear stresses, and the shear stresses are going to be carried largely by the core.
And if you do all the exact calculations, they vary parabolically through the core.
But if we make those same approximations that the face is stiff compared to the core, and that the face is thin compared to the core, then you can say that the shear stress is just constant through the core.
So we'll say the shear stresses vary parabolically through the core.
But if the face is much stiffer than the core and the core is much thicker than the face, then you can say that the shear stress in the core is just equal to the sheer force over the area of the core, bc.
So here, V is the shear force of the cross-section you're interested in.
And bc is just the area of the core.
And we could say the maximum shear force is just going to be V over-- actually, let's make it P, P over yet another constant.
And B4 also depends on the loading configuration.
So if I was giving you a problem, I would give you all these B1, B2, B3, B4's and everything.
So the maximum shear stress in the core is in just the applied load, P, divided by this B4 and divided by the area of the core.
OK, so this next figure up here just shows those stress distributions.
So here's a piece of the cross-section here.
So there's the face thickness and the core thickness.
You can think of that as a piece along the length, if you want.
This is the normal stress distribution, here.
So this is all really from saying plane sections remain plane.
These are the stresses, the normal stresses in the core.
And you can see they're a lot smaller in this schematic than the ones in the face.
And then this is the parabolic stress in the core.
And similarly, there'd be a different parabola in the face.
And these are the approximations.
Typically these approximations are made so the normal stress in the face is just taken as a constant.
The normal stress in the core is often neglected.
And here the shear stress in the core is just a constant here.
So the two things you need to worry about are the normal stress for the face and the shear stress for the core.
Are we good?
We're good?
Yeah, good-ish.
OK, so if we want to talk about the strength of the beam, we now have to talk about different failure modes.
And the next slide just shows some schematics of the failure modes.
So there's different ways the beam can fail.
Say it's in three-point bending just for the sake of convenience.
One way it can fail is, say it had aluminum faces.
This face here would be in tension, and the face could just yield.
So you could just get yielding of the aluminum.
That would be one way.
It could be a composite face and you could have some sort of composite failure mode.
You can get more complicated failure modes for composites, but there could be some sort of failure mode.
This face up here is in compression, and if you compress that face, you can get something called face wrinkling.
You get sort of a local buckling mode.
So imagine you have the face, that you're pressing on it, but the core is kind of acting like an elastic foundation underneath it.
And you can get this kind of local buckling, and that's called wrinkling.
That's another mode of failure.
You can also get the core failing in shear.
So here's these two little cracks, denoting shear failure in the core.
And there's a couple of other modes you can get, but we're going to not pay much attention to those.
The whole thing can delaminate, and, as you might guess, if the whole thing delaminates, you're in deep doo-doo.
Because, remember when I passed those samples around, how flexible the face was by itself and how flexible the core is by itself.
If the whole thing delaminates, you lose that whole sandwich effect and the whole thing kind of falls apart.
We're going to assume we have a perfect bond and that we don't have to worry about that.
The other sort of failure mode you can get is called indentation.
So imagine that you apply this load here over a very small area.
The load can just transfer straight through the face and just kind of indent the core underneath it.
We're going to assume that you distribute this load over a big enough area here, that you don't indent the core.
So we're going to worry about these three failure modes here-- the face yielding, the face wrinkling, and the core failing and shear, OK?
So let me just write that down.
And then you also can have debonding or delamination, and we're going to assume perfect bond.
And then you can have indentation, and we're going to assume the loads are applied over a large enough area that you don't get-- So you can have different modes of failure, and the question becomes which mode is going to be dominant?
So whichever one occurs at the lowest load is going to be the dominant failure mode.
So you'd like to know what that lowest failure mode is.
So we want to write equations for each of these failure modes and then figure out which one occurs first.
So we'll look at the face yielding here.
And face yielding is going to occur just when the normal stress in the face is equal to the yield stress of the face.
So this is fairly straightforward.
So this was our equation for the stress in the face.
And when that's equal to the face yield strength, then you'll get failure.
And the face wrinkling occurs when the normal compressive stress in the face equals a local buckling stress.
And people have worked that out by looking at what's called buckling on an elastic foundation.
So the core acts as elastic support.
You can think that as the face is trying to buckle into the core, the core is pushing back on the face.
And so the core is acting like a spring that pushes back, and that's called an elastic foundation.
So people have calculated this local buckling stress, and they found that's equal to 0.57 times the modulus of the face to the 1/3 power times the modulus of the core to the 2/3 power.
And here, if we use our model for open cell foams, we can say the core modulus goes as the relative density squared times the solid modulus.
And so you can plug that in there.
So then the wrinkling occurs when the stress in the face, the Pl over the B3 btc is equal to this thing here.
OK, so one more failure mode that's the core shear, and that's going to occur when the shear stress in the core is just equal to the sheer strength of the core.
So the shear stress is P over B4 times bc, and the shear strength is some constant, I think it's C11, times the relative density of the core to the three halves power times the yield strength of the solid.
And here, this constant is about equal to 0.15, something like that.
So now we have a set of equations for the different failure modes, and we could solve each of them, not in terms of a stress, but in terms of a load P. The load P is what's applied to the beam, right?
So we could solve each of these in terms of the load, P. And then we can see which one occurs at the lowest load, P. And that's going to be the dominant failure mode.
So one way to do it would be to, for every time you wanted to do this, to work out all these three equations and figure out which one's the lowest load.
But there's actually something called a failure mode map, which we're going to talk about.
So let me just show you it and we'll start now.
I don't know if we'll get finished this.
But there's a way that you can manipulate these equations and plot the results as this failure mode map.
And you'll end up plotting the core density on this plot, on this axis here, and the face thickness to span ratio here, and so this will kind of tell you, for different configurations of the beam, different designs, for these ones here, the face is going to wrinkle, for those ones there, the face is going to yield, and for these ones here, the core is going to shear.
So I'm going to work through these equations, but I don't think we're going to finish it today.
So this is just kind of where we're headed is to getting this map.
So we'll say the dominant failure mode is the one that occurs at the lowest load.
So the question we're going to answer is how does the failure mode depend on the beam design?
And we're going to do this by looking at the transition from one failure mode to another.
So at the transition from one mode to another, the two modes occur at the same load.
So I'm going to take those equations I had for each of the failure modes, and instead of writing this in terms of, say, the stress in the face, I'm going to write it in terms of the load, P. So using that first one over there, the load for face yielding, I'm just rearranging that.
It's B3 times bc times t/l times the yield strength of the face.
And similarly for face wrinkling, I can take this equation down here and solve it for this P here, OK?
And then I can take that equation at the top and solve that for P2 for the core shear, and that's equal to C11 times B4 times bc times sigma ys times-- oops, wrong thing-- times the relative density to the 2/3 power.
OK?
And then the next step is to equate these guys.
So you get a transition from one mode to the other when two of these guys are equal to each other, right?
So there's going to be a transition from face yielding to face wrinkling when these guys are equal.
And I'm not going to start that because we're going to run out of time.
But let me just say that I can pair these two up and say there's a transition between those two.
And that transition is going to correspond to this line here, OK?
So at this line here, that means you get face yielding and face wrinkling at the same load, OK?
And then if I paired up-- let's see here.
If I paired up face wrinkling and core shear, these two guys here, I'm going to get this equation here on that plot.
And then if I paired up these two guys here, the face shielding and the core shear, I would get that line there, OK?
So once I have those lines, that tells me, you know, anything with a lower density core and a smaller face thickness is going to fail by face wrinkling.
Anything with a bigger density is going to fail by face yielding.
And anything with a larger face thickness and a larger density is going to fail by core shearing.
And so you can start to see that it-- I'll work out the equations next time, but you can start to see that it kind physically makes sense.
Intuitively, this face wrinkling, it depends on the normal stress in the face, in compression.
So obviously the thinner the face gets, the more likely that's going to be to happen.
So it's going to happen at this end of the diagram.
And it also depends on that elastic foundation, on how much spring support the foundation has, right?
So the lower the core density, the more likely that is to happen.
Then if you, say you have small t, so the face is going to fail before the core, as you increase the core density, you're making that elastic foundation stiffer and stiffer, and you're making it harder for the buckling to occur.
It can't buckle into the elastic foundation, so then you're going to push it up to the yielding.
And then as you make the face thickness bigger, as t gets bigger, then the face isn't going to fail and the core is going to fail.
So you can kind of see just looking at the relative position of those things, they all kind of make physical sense.
So I'm going to stop there for today and I'll finish the equations for that next time.
And we'll also talk about how to optimize for strength next time.
And we'll talk about a few other things on sandwich panels.
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All right, then, I guess we may as well start.
So what I wanted to talk about today was natural sandwich panels and sandwich beams.
So there's lots of examples of sandwich structures in nature, and we've been looking at the engineering sandwich structures.
And we've seen that you can get a lightweight structure by having this sandwich construction.
And so there are several examples I was going to talk about today.
And I think because this isn't really on the test, I'm not going to write a lot on the board.
So there's some notes.
I'll just put them on the website, and you can look at that if you want.
Because we have kind of a shorter time today.
I'll just try and talk and explain what's what.
Hey, Bruno.
How are you?
So this is the first example.
So many leaves of Monocotyledon plants have a sandwich structure.
And this is an iris plant and iris leaves.
And for those of you in 3032, I think you know that these are glass flowers.
So the Harvard Museum of Natural History has a glass flower collection that was made in the 1800s.
And there was a botany professor there who made these as sort of a lecture demonstration vehicle.
And so he would bring then to class, and he would show different things about the plants with the glass flowers.
But now they're just in the museum, and they're very realistic.
So I just wanted to show you those.
So let's see, it's not working.
Turn it on.
There we go.
So if we look at a cross-section of an iris leaf, it looks like the diagram on the left.
So here's the iris.
And you can see there's these kind of solid fibers, and those solid fibers are called schlerchyma.
And they only exist at the top and the bottom of the leaf.
So I went out this morning.
And if you look outside of the Stata building, there's that little kind of river-y thing, and there's some iris leaves growing there.
So I went and got some iris leaves.
And you can tell we had a horrible winter because usually when I give this lecture in the spring, the leaves are like twice as big.
But this year, they're just little, short, wimpy ones.
But I'm going to pass it around.
And if you just like move your thumb over the top, you can feel little ridges, little bumps.
And those little ridges that you can feel are these little schlerchyma fibers.
So you kind of see they kind of stick up a little.
And so when you move your thumb over it, you can feel that.
And then you can see that the middle of the iris leaf has this kind of foamy-type structure here, and that's called parenchyma cells.
So you can think of the leaf as very much like one of the sandwiches.
This is like a fiber-reinforced composite at the top and at the bottom.
And then this is kind of like a foam core in between, separating the fiber-reinforced faces.
And so the iris leaf behaves mechanically like a sandwich beam.
So I'm going to talk a little bit about how we can actually demonstrate that using the equations that we developed in class.
This is another example.
This is, I guess, what Americans call the cat tail, but Canadians and English people call it a bull rush.
And you can see this is a slightly different construction, but there's the same sort of idea.
So instead of having a foamy core as in the iris leaf, you've got these kind of webs here that go in between the top and the bottom, and that forms like a series of I-beams almost.
And you can think of that also like a sandwich panel or a sandwich beam.
So you've got two stiff top and bottom pieces, and then you've got these kind of webs that separate them, kind of like a honeycomb core would be.
So that's another example of a leaf that has the sandwich-type structure.
And this is very common in these Monocotyledon leaves.
So if you think of a cat tail or you think of an iris, they tend to be kind of narrow at the base, maybe an inch or two wide at the base, and they can be quite tall.
The iris leaves can get two or three feet tall.
The cat tails can get five or six feet tall.
And they stand up more or less straight.
They bend over a little, but they stand up more or less straight.
And this sandwich structure is one of the things that lets them stand up straight at a fairly low weight.
And from the plant's point of view, there's a sort of metabolic cost associated with making more material.
So it we can minimize the amount of material, it's a better thing for the plant.
These are some other examples of grasses that are sandwich-type constructions.
This is from some papers by Julian Vincent.
And the black little circles here are the schlerchyma, are those sort of dense fibers.
Then you can see in both of these cases, the dense fibers are on the outside, and the parenchyma cells, which is the white, are on the inside.
And so this is sort of another set of micrographs of the iris.
So this is just showing the outside, and these are the ribs viewed from the outside.
And this is the core, just sort of viewed along the length of it.
And so you can idealize the structure as being like a sandwich that's got sort of fibers on the top and on the bottom.
So the top and the bottom are like a fiber composite.
And the middle part, with the parenchyma cells, is kind of like a foam.
And so we did a little project on iris leaves, and we wanted to see if you could show that they behave mechanically, like a sandwich beam.
So you remember that we had that equation for the deflection of the sandwich beam.
There were two terms.
There was a bending term, and then there was a shearing term.
And so we took some little sandwich beams.
We cut little kind of rectangular beams.
We hung little weights.
We measured how much they deflected, and we wanted to see if we could use this equation to predict their stiffness and how much they deflected.
So to do that, we needed to know a bunch of things.
We needed to know some of the geometrical parameters.
So we needed to know what volume fraction of the face is those solid ribs, how thick's the core, how thick's the face?
And so we measured a bunch of these geometrical parameters.
We tested it like a cantilever so we knew what B1 and B2 were for the cantilever.
We knew how long the beam was, so we know what l is.
We knew what loads we applied, so we knew what P was.
But we needed to make some estimate of what the face modulus was and what the core shear modulus was, too.
And so we made some estimates of that.
So this table here just shows some of the dimensions of the leaf.
The leaf tapers, and this is at the thin end, so here's the face thickness.
Here's the sort of length of this.
Some square cells in the face.
This is the core thickness here.
This is the dimensions of the core cells.
This is the diameter of the ribs, the spacing of the ribs, the volume fraction of solids in the ribs.
And we did that at different lengths along the different positions along the length of the rib, or length of the leaf.
So we had the geometrical parameters, but we needed to get this E of the face and G of the core.
And to do that, we looked at the literature.
And people had done tests on the fiber parts of leaves.
They'd done little tensile tests, and they'd measured modulii between about two and 20 gigapascals.
And then we did some tension tests on the iris leaf.
And in tension, those ribs are going to take most of the stress.
And if you know the volume fraction of the ribs, you can back out what the stiffness of the ribs must have been.
If you know the stiffness of the ribs, you can figure out the stiffness of the face.
So we calculated that, and then we looked at the literature.
And people have done tests on parenchyma cells and different types of tissue on things like apples and potatoes and carrots.
And these are the values for the Young's modulus they get.
They're between about 1, and the highest one was 14 megapascals.
But most of these values for the Young's modulus are around about four.
And the shear modulus is roughly about half of the Young's modulus.
So we said the shear modulus was around two.
So we have these values we could plug it in and then calculate what the stiffness would be for the iris leaf.
And so this was a little analysis we did.
So this was the measured beam stiffness up here.
We had four beams, and they were different stiffnesses.
They all had the same length.
They all the same face thickness.
The core thickness varied.
They all had the same width.
We cut them to have the same width so we could calculate a flexural rigidity.
That's the EI equivalent.
We could calculate the bending deflection term, the shear deflection term.
And this is the calculated beam stiffness.
And then this is the ratio of the calculated over the measured.
So it's not exactly right.
Obviously, there's some difference here.
But it's in the same order of magnitude.
It's in the same ballpark.
And one of the complications that we didn't really try to take into account was that the leaf isn't a nice rectangular structure.
The leaf has this kind of curved cross-section to it.
And we made a bit of an approximation to that, but it wasn't that close, really.
We could have probably done better on that.
But I think the idea that the iris behaves like a sandwich is a reasonable one.
So that was the iris leaf.
And then I wanted to show you some other structures in nature that are sandwiches.
So this is a seakelp, help like a seaweed thing, in New Zealand.
This is the largest intertidal seaweed.
The fronds, the sort of long pieces of it, are up to 12 meters long.
So that's almost 40 feet.
So 40 feet is probably like from one side of this room to the other side of the room.
It's quite long.
And you can see, if you look at this section here, this is all like a honeycomb-type section here.
And the honeycomb is like a honeycomb in a sandwich, and the top and the bottom faces are like the face of the sandwich.
So this would be like the face here.
That would be the honeycomb core.
And that would be the other face on the other side over there.
And those honeycomb-like cores, apparently, have some gas-filled pockets that then provide buoyancy to keep the whole thing floating.
So it photosynthesizes.
So one of the things about these leaves is that they have multiple functions.
It's not just that they have to have a certain stiffness so they don't fall over.
The plant wants to photosynthesize, so you want to maximize the surface area as well, and you want to have exposure to the sunlight.
So there's a number of things that the plant's trying to do in having this structure.
So that seakelp is one example.
These are skulls from birds.
And so this is a pigeon here.
This is a magpie.
If you come from the West you see magpies out West.
You see them in Europe as well.
And this is a long-eared owl.
This long-eared owl's around here.
And I brought in a couple of bird skulls as well.
And you can see that all of those birds skulls are sandwich structures.
The one for the pigeon has sort of a foam-like core here.
And you can see that the two faces aren't sort of concentric for the pigeon skull.
They sort of not following each other.
But here, this would be, say, on the top shell of the magpie, where the two, the inner and outer face, are sort of concentric.
Then you get these kind of little ribs of trabecular bone in between them, and then the same with a long-eared owl.
You get these little ribs in between them.
And so you can see that there's a sandwich structure there.
And obviously, birds want to be light.
They have to be light to fly, to take off, and so they want to be light.
So I've got two skulls here.
And I'll pass them around.
Please be careful because they're kind of delicate.
This one is from a screech owl, and you see screech owls around here.
This was a screech owl that had an intersection with a car.
Yeah, so the skull fractured, but you can see the sandwich right there.
You see the two little bits?
So you can see the inner plate and the outer plate and the foam, the trabecular bone.
So that's the screech owl.
And this is a red tail hawk.
So you can't really see the shell and the sandwich structure here.
But I want to pass it around just so you can see how light it is.
So it's amazingly light.
So a red tail hawk is probably about this big, something like that.
And this is one of the things that makes them very light.
So those are the bird skulls.
Oh, yes, so now I have to tell you about the owl.
So I think the people in 3032 have heard this before.
But the other people haven't.
So one of the things about the owl is if you look at the whole skull, if you look at this picture here, one of the things is that this bone here is not symmetrical with that bone there.
Normally, when you think of a body, you think of the bones being symmetrical.
But those bones are not symmetrical, and those bones are near where the ear is.
And it turns out on owls, at least on some owls, the ears are at different heights on their heads.
And people think that one of the things that allows the owls to do is it allows their hearing to sort of pinpoint where something is.
And owls can catch little creatures at night, but they can also catch little creatures underneath the snow.
So they can catch things that they can't even see.
And they have a number of adaptations to improve their hearing, but this is one of them.
So here's a little owl Allison Curtis is a Canadian friend who lives in northern Ontario, and this is looking out of her living room window.
And that's a barred owl.
And you can see the barred owl has caught this little vole here.
And you can see in the background it's winter in Canada.
and there's snow all over the place.
So this owl has probably caught that little vole underneath the snow.
And then it's come to eat it.
And this is another picture of-- you can see this is where an owl landed in the snow.
It's wings hit the snow, trying to catch something underneath.
And this is another kind of beautiful print of the owl's wings hitting the snow in the winter time.
So did I show you the fox video?
Should I show you the fox video?
You saw it, right?
I think I showed it last time in 3032.
But you guys haven't seen it.
Let me show you the fox video because foxes do the same kind of thing.
Their ears are the same as ours.
They're in the same position.
But they have this-- let me see.
Where's the sound thing?
We don't really need the sound for this, but there's BBC sound.
So we get this music, even though the fox can't hear the music.
Here we go, fox no drive.
Check this out.
Is it going to come up?
Is that going to play?
OK [VIDEO PLAYBACK] -It listens for the tiny sounds of its prey moving about below
So you see how it cocks its head, and it does this with its head?
It's putting its ears at different heights when it does that.
So check this out.
And look carefully, you can see the little animal it's got in its mouth when it comes out.
There's a little tail.
So part of the reason dogs and foxes and coyotes do that thing, I think, is because they put their ears at different heights, and it helps them pinpoint where something is.
[END PLAYBACK] You know I love these Nature videos, right?
So that's the fox video.
Let me see if I can stop that.
So that's one of the interesting things about owls.
Let me go back to my little PowerPoints.
So here's another example of a creature that has a sandwich-type structures.
So here's the sandwich here.
Here is the ever so charming looking cuttlefish.
And the cuttlefish is not actually a fish.
It's a mollusk.
So it's related to things like octopus, things like that, and squids.
It's a cephalopod.
And you can't see it so well in this picture, but I'm going to show you something else and you see it.
It's got like little tentacles.
These things here are actually separate little tentacles.
And because it's not a fish, it doesn't have like fins that can kind of swim with.
And it's got this thing called the cuttlefish bone.
And this is a cuttlefish bone here.
And that bone has the sandwich structure here.
And it's not actually a bone.
It's really a shell.
It's a calcium carbonate thing, not a calcium phosphate thing.
But the cuttlefish can control how much air goes into those little pockets.
And it can control its buoyancy by controlling how much air goes into those little pockets.
And I brought with me a cuttlefish bone.
Have you ever owned like, I don't know, like a parrot or a pet bird?
Apparently, pet birds love to sharpen their beaks on this cuttlefish bone.
So if you go to a pet store, you can buy this stuff.
So you won't be able to see the little sandwich structure because it's a very small length scale.
But you can kind of see there's a sort of different material on the inside than there is on the outside of that.
So do people know the other thing that cuttlefish are famous for, besides the bone?
Change colors.
Can I show you a video of cuttlefish changing color?
Yeah, of course.
So let me get rid of this again.
Go back to this.
Let's see, somewhere-- where's the cuttlefish?
Here we go.
Did I do it?
Is it thinking?
Here we go.
Where's the cuttlefish?
So this is another one of these Science Friday videos from National Public Radio with Flora Lichtman.
[VIDEO PLAYBACK] -OK, let's play a game.
[GAME SHOW MUSIC PLAYING] [APPLAUSE]
See it?
-Biologist Sarah Zielinski took these shots.
And if you needed a helping hand to find the cuttlefish, don't feel bad.
-I've certainly taken photos in the past then come back to look at them and gone, I'm sure there was a cuttlefish in there somewhere!
-These cephalopods are master camouflagers.
But while they're hiding their body, they're revealing something about their mind, or at least their visual system.
-In very simple terms, they can tell us what they can see by the body patterns they produce on their skin.
-They produce these body patterns by expanding or contracting chromatophores, these little ink sacks on their skin.
And they use different displays for different reasons, like for male-to-male combat.
-Two males will turn into each other and pass these kind of waves of dark chromatophores over a really bright sort of iridescent stripey body pattern and somehow solve these combats.
Eventually, one male gives up and goes away.
-And then there's this unsolved mystery.
It changes color when it grabs a snack.
-That doesn't make perfect sense because it seems to make it very conspicuous.
So one theory is that it's just a happy signal of how excited it is to have caught something, some response that it doesn't have any control over.
-But most of the time they seem to be using their chromatophores more intentionally, primarily to blend in.
-Because otherwise they're more likely to be eaten, so it's very important they don't make mistakes about ambiguous visual information.
-And ambiguous visual information is specifically what Zielinski's interested in.
So here's the experimental setup.
Print out laminated patterns, like this checkerboard, and stick them in a tank.
-And we place the animals in the tank.
And we record the body patterns that they produce.
-You're seeing them on squares, but they do the same thing on top of circles.
They produce-- - --the disruptive pattern, where you get these blocky components of high-contrast components.
-But when you put a cuttlefish over squiggles, it produces-- - --a sort of mottley pattern, where you get these little groups of dark spots showing across the body.
-So what happens when you put a cuttlefish on something in between, when you put them on incomplete circles?
When we see something like this, our visual system likes to fill in the blanks, something we do constantly, Zielinski says.
-The reason why cartoons and sketches work is because we can recognize objects based on their edges alone.
-And we can identify objects even if they're broken up or-- - --have an object that is occluded by another object.
That's no problem for us.
We can still work out what the object is most of the time.
And I was interested to know whether cuttlefish can solve similar problems.
-And Zielinski and colleagues report this week that cuttlefish do seem to-- - --fill in those gaps and interpret those little segments as a whole circle.
-Or anyway, the broken circles prompted the same camo pattern as full circles.
So if you're wondering, uh, I see these as circles, too.
What's the big deal?
The weird thing here is that there's no reason why cuttlefish, which are-- - --invertebrates, and they're in the same group as slugs and snails.
- --should see the world the way we do.
-Yes, it's like they're alien, but we also seem to have so much in common with them.
-So the next step?
-Because we can't share the perceptive experience of a cuttlefish, it's hard to know exactly what it is that they're doing to fill in that missing information.
And I want to try to get a better grasp on that and also see whether they actually respond to true illusory contours.
-So you're going to show optical illusions to cuttlefish?
-(LAUGHING) That's what I'm hoping to do, yes.
[END PLAYBACK]
So let's go back to sandwiches.
I think I have-- do I have one more?
There we go.
So horseshoe crab shells, so different sorts of arthropods, the shells are sandwiched too.
This is from Mark Myers' work.
So we're looking at the cross-section of a horseshoe crab shell.
So again, it's the same idea-- the animal wants to minimize the amount of material or minimize the weight, and this is a way of doing that.
And I went to the Galapagos about a year ago.
And there was a place where they had these giant Galapagos tortoise shells.
And one of them was broken, and you could see there was a sandwich structure in the Galapagos tortoise shells.
These Galapagos tortoises, their shell is like this big.
They're gigantic.
They're huge.
So those are my examples of sandwich panels and beams and shells and whatnot in nature.
So the idea is that nature too wants to minimize weight and minimize the amount of material, and the sandwich structure is a way of doing that.
So I have one more thing I wanted to talk about today.
So this isn't quite sandwich structures, but it's looking at another kind of natural structure that is designed to reduce the weight of plant stems, in this case, palm stems.
And there's a couple of interesting things about this.
So when you look at palms, like let's pretend we're not in Boston.
We're in California, where they have palms.
And we're in LA, and they don't have winter.
And if you look at the palms growing, when the palm's short, it's about this big in diameter.
And as it gets taller and taller, the diameter doesn't really change.
It gets taller and taller and taller, but the diameter doesn't change, at least in some species.
Whereas if you think of a tree, a tree starts out with a little skinny diameter.
And as the tree gets taller, the diameter gets bigger.
And it sort of tapers and does that whole thing.
So palms don't do that.
And palms are not trees.
They're a botanically different thing from trees.
So here's a coconut palm.
And so the question is, as the stem gets taller and taller, how does it resist the bending loads that get bigger and bigger?
So probably, the main load on these sorts of things is from the wind.
And often these plants are in areas where they have hurricanes.
And you see them in hurricanes, you see the pictures of the palm stem blowing way over.
And so how do they resist the larger internal stresses as they get taller and taller, if the diameter doesn't get bigger and bigger?
And the way they do that is that they deposit additional layers of cell wall as the plant ages.
So if you think of a tree, when a tree grows, it just deposits more and more cells.
And the cells have roughly the same thickness.
So there's ones that are deposited in the spring have thinner walls.
Then the summer and the fall have thicker walls.
But more or less, it's similar.
Whereas the palm, it deposit cells, and then as the trunk of the palm gets taller, as the stem gets taller, it deposits more layers on the cell wall.
So this is an example in an SCM.
You can see here this is a young cell, and it's got-- this one that's not marked is a primary cell wall, and then this is the first layer of the secondary cell wall.
And then this is an older palm.
And you can see here it's got more layers, and so the cell wall itself has gotten thicker.
So that means that the density of the tissue changes as the palm ages.
And it does so in a very kind of clever way.
If you think of the palm as being like a cantilever that's vertical and it's bending in the wind, when we have a cantilever beam or any kind of beam, the stresses are going to be biggest on the periphery, right?
They're going to be biggest on the outside.
And if you think of the palm as having a circular cross-section, that outer periphery is going to see the biggest stresses.
So it would make the most sense if that was the densest tissue.
And that's exactly what the palm does.
So there was a nice study done by Paul Rich quite a number of years ago.
And he studied palms in Central America and looked at the density and measured the mechanical properties.
And I'm going to talk about his stuff today.
So the white is the low density.
The gray's the medium, and the black's the high.
So you can see the low density's on the middle of the young stem, and just at the very base and then the periphery is the dense tissue.
But as the stem gets taller and gets older, then stuff that was low density is now high density.
And only the very middle here is the low density.
And that some stuff that was low density has turned to middle density.
And some stuff that was low density has turned to high density.
So it's done this by adding more and more layers to the cell wall, making the cell wall thicker and making the cells themselves denser.
So this is looking just at a single palm.
So each one of these lines is a single palm.
And this is looking at how the density changes from the periphery to the center of the palm.
So if you cut the palm down and, say, we take a little sample radially from the middle to the outside or from the outside to the middle, he then measured the density.
And it's probably easiest to think about the dry ones because that's kind of what you would compare wood to.
So the dry densities varied from about one gram per CC, that's about 1,000 kilograms per cubic meter, down to almost zero in this particular species here, probably like 50 or something like that.
And if you compare this with woods, this little arrow here is the density of most common woods.
So if you looked at pine and spruce an oak and maple and ash and hickory, they would all be in that little range there.
So a single palm stem can have a bigger range of densities than many different species of wood.
So it has this kind of profile of the density.
And the thing I was interested in is seeing how mechanically efficient that was to put the denser material at the outside.
So I looked at the stiffness of the palm, and I also looked at the strength.
So I just replotted that data on this slightly different axes here.
So this is the radial position relative to the outer radius, and this is the density.
And I subtracted off the minimum and then took the range.
And for this species here, the minimum density was almost zero.
So this expression simplifies to something like that.
And just because it's mathematically simpler, that's what we're going to look at.
So the density goes roughly as the radius squared.
And Paul Rich also did a lot of mechanical tests on the palm, and he took out little beams of different densities.
And he measured the stiffness and the strength of the beams.
So he measured the modulus of elasticity here versus density.
And he measured the modulus of rupture here.
And these are all along the grain.
And he found that the Young's modulus varied with the density to the 2.5 power, and the strength varied as the density squared.
And if the-- [BUZZING SOUND] Oh, hello.
[LAUGHTER] So these were just sorts of empirical findings that he made.
If you have prismatic cells and you deform them axially, and the cell wall was the same in the different specimens, then the solid modulus would be a constant.
And you would expect that the modulus of the beam would go just linearly with the density, sort of like a honeycomb loaded [?
at a ?]
plane.
But what he measured was that the modulus and the strength varied with some power of the density.
And the reason for that really was that the cell walls of the denser material had more layers.
And in the additional layers, the cellulose microfibular angle was probably different, so that the different layers had different stiffnesses.
And if you have layers of differences, then you're going to get this power relationship.
So what I then did was I took his data, and I tried to see how efficient that would be in bending.
So he had found that the density varied with the radius raised to some power.
This power n was 2, but I wanted to do it just for a general case, so I said I was just n. And he said that he found that the modulus varied with the density raised to some other power m. And for him, m was 2 and 1/2.
And so I could write just another equation saying that the modulus goes as the radius to the mn power.
And then you could do a little calculation where you work out with the equivalent flexural rigidity is.
So you have to integrate up.
You kind of say you have a little band at a certain radius.
That radius has a certain modulus.
And you can figure out the moment of inertia that goes with that particular radius.
And then if you integrate it up over the whole thing, you can say that the flexural rigidity for the gradient density is some constant times pi times the outer radius to the fourth power divided by those two powers mn plus 4.
So m was the power here for the modulus.
And n was the power there for the density.
And then you could compare that with having the same mass just uniformly distributed over the whole cross-section.
And then if you take the ratio of the flexural rigidity for the density gradient versus the flexural rigidity for the uniform density, you can show that it's this equation here.
And then if you plug in these measured values for those exponents for n and m, you find that the flexural rigidity with the gradient density relative to the uniform density is a factor of 2 and 1/2.
So the stem is 2 and 1/2 times stiffer by having that density profile.
So there's a huge sort of mechanical advantage to doing that.
And just sort of physically, if you know the stresses our biggest on the outside, it would make sense to put the denser material on the outside.
And then the other thing I looked at was the strength of the palm.
So imagine this is our very schematic palm here, and then there's a circular cross section.
So I wanted to compare the bending stress distribution with the bending strength distribution.
So the stress goes as the modulus times the strain, just Hooke's law.
And here we're assuming that plane sections remain plane, like that's the standard assumption of bending.
So if you assume plane sections remain plane, then the strain goes with the curvature times the distance y from the neutral axis, the distance from the middle.
So this distance here would be the dis-- [?
same ?]
at loaded with a loaded p here.
That distance would be y there.
And then I can plug in some things here.
So instead of E, I'm going to plug-in my relationship with the radius to that mn power.
And here's my curvature, and instead of y, if I say that some radius, I'm going to say y is our r cos theta.
And so I'm going to say that the stress goes-- [SNEEZE] Bless you.
Goes as radius raised to some power mn plus 1.
And again, for the species I know what n and m are, so the stress goes as the radius to the sixth power.
And then I can also compare with what Paul Rich had found for the strength.
He found that the strength-- so sigma star is the strength-- was proportional to the density raised to some power q, and that power was 2 in the measurements that he made.
And so I can say that the strength goes as the radius to this power nq, so to the fourth power.
And then if I plot the stress distribution and the strength distribution-- so imagine, this is through the cross-section here.
So this is the diameter of the stem.
And this is the neutral axis here in the middle.
The strength goes as that solid line there.
It goes as the fourth power.
And the stress goes as that dashed line there, as the sixth power.
So they're not exactly on top of each other, but they're very close to being on top of each other.
So basically what the palm has done is it's arranged the material in such a way that the strength matches the stresses that are applied to it.
So if I just had a constant density, my stress profile would look like that.
And if I had a constant density, the strength profile would kind of like that.
So the strength here would be a constant, and this would be the stress here.
So the stuff in the middle, it's much stronger than it needs to be.
Whereas the palm has arranged things so that it's got just the right amount of strength for the stress, as a function of the radial position.
So it's kind of a clever thing.
So that's kind of a beautiful thing.
And I think that is it.
I think that's-- yeah, that's the end of it.
So all these images came from this other book that we wrote.
And if you wanted to get the sources, you could get them from there.
So that all I wanted to talk about today was some examples of sort of efficient mechanical design in nature and the sandwich panel structures as one, and these radial density gradients is another.
We have a project on bamboo right now, and the bamboo also has a radial density gradient, and it's the same thing.
The densest material's on the outside, and the least dense is on the inside.
So I think I'm going to stop there for today.
So what I was going to do on Monday is talk a little bit about bio-mimicking.
And that won't take the whole class at all.
And I thought we could spend the rest of the class on Monday just doing a review.
So the test's on Wednesday.
So if you want to bring questions, that would be a beautiful thing.
I can't really can I review the whole last six weeks or something in an hour and a half or something.
So if you want to bring questions, I'll be here and we can just go over questions.
Does that sound good?
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All right, well, I guess I may as well start.
I don't know if anybody else is going to come.
So I wanted to finish up by talking a little bit about the biomimicking.
And some of these examples, you've seen before.
But I just thought I'd put them all together, and we could look at them as one thing.
So if you remember, when we talked about wood, one of the things that I showed you was that people have taken wood and pyrolized it.
So they get a carbon template of the wood cells.
And then they infiltrate that with silicon carbide-- or, with a silicon vapor infiltration.
And they make a silicon carbide ceramic.
So they can get a replica of the structure.
So sometimes when people say "biomimicking," some people think of it as replicating something.
But it doesn't have to be just replicating something.
It can be also just some design inspired by the biological material.
But this thing really is a replica.
And this was another version where there was the-- they took the silicon carbide material and then infiltrated that with liquid silicon to get a fiber-reinforced material, really.
So there was the wood composites we talked about.
Jennifer Lewis's group up at Harvard is doing 3-D printing of honeycombs.
And one of the things they've been interested in is not just printing of the pure resin, but having a fiber-reinforced resin.
And the first thing they did was they made these honeycombs like this.
And they had small silicon carbon fibers-- or, were they silicon carbide?
Maybe it was carbon fibers-- in the ink.
And so they would just-- if the ink was being laid down, the fibers would just line up in the direction of the ink.
And so the fibers tended to be in the plane of the honeycomb.
And if you think of things like wood, you want the fibers to normal to that plane.
And more recently, they've-- hello.
Oh, hello.
Oh, look.
Oh, look, almost everybody is here.
So more recently, they've got a technology now where they're rotating the nozzle as they print the honeycomb.
And as they rotate the nozzle, they get some change in the orientation of the fiber.
So they're beginning to be able to make honeycombs that are fiber-reinforced.
And they can get the fibers aligned with the prism axis of the honeycomb, which is more or less what the wood does.
So I wasn't going to write anything on the board today.
I was just going to go over some of the slides and do the review.
So they're beginning to make honeycombs that have the same sort of structure on the cell wall level, or at least a similar structure, as to what the wood composites have.
So that's another example there.
This is just another close-up of their fiber-reinforced walls in honeycomb specimens.
We talked about trabecular bone, and we talked about the fact that people are starting to look at using foamed metals as coatings on orthopedic implants.
And there's been some interest in looking at using foamed metals for more permanent parts of the body, more permanent bone parts, things like vertebral cages, stuff like that.
And so this is just an example here, with the trabecular bone on the left and the tantalum foam that's made by replicating an open-celled polyurethane foam on the right.
And you can see the similarity in the structures of those two things there.
Then we talked about tissue engineering scaffolds.
And if you remember, these two scaffolds here on the bottom were made from pig heart tissue.
And they're made by just removing all the cells.
So that actually is the natural extracellular matrix.
And then, these other structures up here, these were all engineered tissue engineering scaffolds that are made in a synthetic way.
And the idea is to try to mimic the extracellular matrix in the body.
So you can see the similarities there.
And then more recently, we were talking about sandwich panels.
So this is the example from the helicopter rotor blade.
That's from an aircraft flooring panel.
And this was the Irish leaf, and these were the bird skulls.
So the same sort of idea, that there's these engineering lightweight structures, and there's also similar things in nature.
And then, I think, last time, we also were talking about palms.
And the palm stems had density gradients in them.
And one of the things we showed was that by having that density gradient, the stress distribution across, say, the radius of the palm was almost matched by the strength distribution.
So it was a very efficient way to use the material.
And at MIT, that was a student in architecture who was looking at doing this with concretes with aerated or foam concretes and making a radial density distribution.
So he made beams.
He made columns.
He made different kinds of things.
With the concrete, there's a little bit of a limitation, because the concrete is much stronger in compression than it is in tension.
So if you had a beam loaded in tension or a concrete column that might buckle, you'd still have to have some reinforcing bars in there to take the tensile loads.
And one thing I think I didn't really talk about was animal quills and other sorts of plants stems.
And many of these have a structure that's made up of a cylindrical shell with a foam or a honeycomb core.
So here are some examples in nature.
If you look at grass stems, there's a dense layer on the outside, and then a foamy layer on the inside, and then just a hollow layer in the middle.
And if you look at porcupine quills-- this is a porcupine quill-- these are made of keratin.
They're like modified hairs.
So it has a dense layer on the outside, and then this foamy stuff on the inside.
This is a hedgehog spine here.
And again, you can see there's a dense layer on the outside and these ribs on the inside.
And this is the toucan beak-- you know, the toucans that live in Central America.
And the beak has a foam core.
Again, they're keratin structures, and yet the outside is solid.
But the inside has a foamy structure.
And Mark Myers did a paper on this a while ago.
So we got interested in these structures that have a solid shell on the outside but a foamy thing on the inside.
And we wondered if there was a mechanical reason for that.
And you can show that, at least in some of them, if, say, you have a grass stem-- it's really common in plant stems.
Do I have some more plant stems?
Yeah, here we go.
Here's a milkweed stem.
So it's got these dense fibers with this foamy core here.
And blue jay feathers-- feathers have this as well.
So they have an outside layer on the quill that's solid, and then an inside layer that's foamy.
If you look at things like plant stems, they blow in the wind.
And you can look at the buckling resistance of the stem.
And because there's this shell with the foam-like core, it's not just the overall buckling of the whole thing.
You can get that local face-wrinkling mode again.
So the outer shell can wrinkle.
And you can show that having a foam-like core helps prevent that wrinkling from happening, the same as with the sandwich panel.
You remember we talked about the face wrinkling on the sandwich panel?
Well, on these stems, you can get wrinkling of the outer shell.
And the foam helps prevent that from happening.
And you can show that you can actually-- for the same buckling load, you can reduce the weight of the plant stem, or the bird feather quill, or whatever by having that foamy core.
So people have looked at this too.
And there's a group in Germany who had looked at the idea of mimicking the horsetail stem.
So this is a plant stem here, the horsetail plant.
And they made something they called a technical plant stem, where they made this structure here.
And they made it out of fiber-reinforced composites.
And you can see, the little holes here represent those holes there, in the plant stem.
So the idea was to try to get something that was good at resisting the buckling, but at a lower weight.
And they were doing that with this thing here.
And there was a group in Japan that did a similar thing.
They used-- I think they took a copper tube, and then they took copper and aluminum wires and filled the copper tube with the wires.
They extruded that.
And then they melted out the aluminum, which, I think, also helped to soften and make the copper bond together.
And they got these structures here.
And you can see, that's similar to some of the plant stems as well.
So these are all examples of cellular structures that have-- mechanically efficient structures.
They're lightweight, and they're strong, and they're stiff.
And these natural structures have been mimicked in engineering applications.
So that's really all I wanted to talk about today.
But I think we wanted to use the rest of the class as a review.
So I haven't made a one-hour summary of the last six weeks, because that's not really possible.
So I thought I would just answer questions.
If you have questions, I'll try and answer them.
So for the test-- so, the test is on Wednesday.
You can bring one 8 and 1/2 by 11 sheet.
I wasn't going to give you all those honeycomb and foam equations, partly because-- the only thing I would really want you know is open-cell foams.
And I was hoping, by now, that you might have registered those equations somewhere in your brain.
So I'm not going to ask you for-- some equations are obscure.
I might ask you-- expect you to know what the Young's modulus of an open-celled foam is by now, or the axial modulus of a honeycomb or something.
But I don't think I'm going to ask you anything like, calculate air pressure contributions to the modulus of the closed-cell foam.
I don't think we're going to do that on this test.
So I think you should know what the modulus of a-- the Young's modulus of an open-celled foam is, the shear modulus of a foam, because you need that for the sandwich panels.
But you don't need reams and reams of those equations, so I wasn't going to give you those this time.
OK?
So Jenny, did you have-- so I finished the biomimicking thing.
We're just going to do a review.
I don't know if you want to stay or if you want to go.
You want to stay?
Well, whatever.
So do you have questions, Jenny?
I do.
I just wanted to have question 4 from the last pset reexplained.
Because I know that you explained it to in office hours, and I know that the solutions are online, and I looked at them.
I'm still confused
OK.
So, you're going to have to remind me what problem 4 is, because I don't remember
Question 4 says, polymethacrylate foam at solid strength of 3.0-- or, solid Young's modulus, rather, of 3.0-- gigapascals is being considered for the energy absorption layer in a bicycle helmet
OK.
I'll tell you what, I think I have it on my little disk here.
Maybe that's easier, because then I can read it
[INAUDIBLE]
Yeah, let me just see if that's going to come up.
There it is.
OK.
So this one here about the foam, about the--
The one with the graphs
--energy absorption?
I'm just a little bit confused by the graphs
OK. Let me see if I can make this bigger.
Hang on a sec, my computer's thinking.
OK. OK. And I think I gave you-- right, I gave you this graph here.
Right?
Yes
OK.
So can you read what I've got here, or is that-- should I make it bigger?
I can't really read it, [INAUDIBLE]
Does that help?
OK.
So I have to admit, when I put this together, somebody-- I can't remember who it was-- told me you thought it was overconstrained.
And it turned out it was overconstrained.
And then I said, forget the thickness.
Forget that I've given you the thickness, right?
So disregard the thickness
No, the velocity
Oh, speed-- the speed?
The speed-- all right, OK, the speed.
No, I don't want to register.
OK.
So from what I gave you, you can figure out the normalized peak stress, right?
So you can get-- so if I do this, is this good?
You can see what I'm pointing at?
So you can get, the peak stress is just the mass times the acceleration over the area-- are we good with that-- divided by Es, which I gave you.
So I think most people probably got the peak stress here.
And then, because I had given you the velocity and the thickness, I was thinking you could calculate the strain rate.
But if I don't give you the velocity, say we're not calculating the strain rate at this point here, OK?
So here, we're-- did I put this-- I don't know if I have the graph on this solution.
There we go.
So this point here is the 2.5 times 10 to the minus 4 for the peak stress.
And if you're not given the velocity, I think what I thought you would do then would be just assume a velocity.
And then you can check it at the end.
So since I had given you this velocity of 12, let's just say that's what we assumed.
OK?
So then you get a strain rate of 480 per second.
So let's just say we assumed that velocity.
Then we could scoot back over here.
So we know we're on this line here for the sigma p over Es.
And we want to be up towards the top of these different strain rates.
So if you look at the strain rates, see, the very last one at the top is 1,000 per second, and the next one's 100 per second.
So we're halfway in between those.
And they're so close together, you can't really read the difference.
But we're up here somewhere.
So then we read off a w over Es for that.
OK?
Are we good?
Then, if we know the w over Es, this is the number, here, that I read off.
You know what the Es is, so you can get w. If you can get w, w is in joules per cubic meter.
It's in energy per unit volume.
But you know the area, and you know the thickness.
So you can get the energy in joules.
And then-- oh, did I-- I must have rubbed that off.
Didn't have it on this version.
The version in my notebook, I think, calculated what the velocity would be that corresponds to this.
And I think it turned out to be 8 meters per second or something.
So I had assumed 12, and I think it worked out to 8.
And on those log log graphs, whether or not it's a strain rate of 480, or a little bit less, or a little bit more, you can't read the difference on these things.
OK?
Can you explain how this graph relates to the other graphs from lecture that were simpler?
Because we had ones that were all density, and ones that were all the same strain rate.
And I guess I'm just confused why
Oh, OK.
Hang on.
So let me see if I can-- hang on.
No, I think I know what you mean.
Let me see.
I want to pull up the lecture notes.
I think I'm finding the right thing.
Here we are.
So in the lecture notes, there was a thing that looked like this.
Is that what you're talking about?
Yeah.
So the top set are the stress strain curves.
So those are OK. You do a compression test, you measure that.
Then, the middle set, you take, say, for one density, for one stress strain curve-- say that's your curve, the middle one there, 0.03.
Say you loaded it up to some point, or say you looked at some point here, on the curve.
You would figure out, for that stress, what's the area under the curve up to that stress.
So you'd have a stress and an area under the curve up to that stress.
And you could then-- say we know what this foam is.
It's-- I don't know-- a polyurethane or something.
So say we know Es.
Then we could divide those two numbers by Es.
And we would plot that one thing.
Let me just walk over here.
So this is the 0.03.
So if it's in the linear elastic reading, it would be somewhere in here.
Then I would scoot along, say, to here someplace.
I would do a whole bunch of points all the way along there.
And at every point, I would say, what's the stress, and what's the energy absorbed up to that stress.
OK?
And then I would plot-- doot, doot, doot, doot, doot-- up here, I would plot all those points.
OK?
And then when I got to this part here, that corresponds to that part over there.
OK?
Are we all good with that?
OK.
So then we repeat-- so we get one curve on the middle chart.
Then, for the different densities and the different stress strain curves, we plot a different curve for each of the different densities doing the same process.
And the thing we notice is that these points here are really the optimum point.
Because at that point there, you absorb as much energy as you possibly can for that stress.
OK?
And we notice, happily, that those points lie on a line, basically, on a straight line.
And we tick of what the different densities are-- so 0.01, 0.03, 0.1, 0.3.
OK?
And that line there, and all of these stress strain curves, and all these lines here, curves here, they all correspond to one strain rate.
All right?
So now I could take that line there for that first strain rate, and it would be one of these thinner lines here for a particular strain rate.
And I would mark off-- just the same as I've got here, 0.01, 0.03-- I'd go 0.01, 0.03, 0.1, 0.3.
All right?
And then I would repeat this whole process again for a different strain rate.
So I'd go back to doing some mechanical tests at, now, a new strain rate.
And typically, the new strain rate is going to be bigger or smaller by a factor of 10 or so, because you're not going to see much difference in the behavior unless you get big changes in the strain rate.
So you're going to change the strain rate significantly.
You get a new series of these stress strain curves.
Then you get a similar-- very similar, but not quite the same-- series of these curves here.
And what you'd find is you'd have another line here that would be offset a little from the first one.
And the positions of where these densities were would also be offset a little from the first one.
And then you'd draw the second one.
So the second strain rate line would go here.
And the little density positions would be offset a little bit, and you would mark them off.
And you basically repeat it for different strain rates, and then you build this thing up.
And then the density lines connect too.
Is that OK?
So that's how you generate that.
So the idea is, this bottom diagram is really summarizing all of those shoulder points.
The bottom diagram is really summarizing all of these points where the stress starts to scoot up at the densification machine.
Yeah?
So in question 3, we were constructing the graph in the middle for this particular one.
What confused me is that on the graph, the variable on the x-axis is stress, whereas in the question, we were given the stress.
So I drew something that looked as it should have looked.
And it turned out to be right, but I'm very confused as to why
OK. Let me try and get rid of that.
And I'm going to have to remind myself what the question is again.
OK.
So we have an open filled aluminum foam.
I asked you to write the equations for each of the different regimes.
And those were straight out of the notes, I think.
Right?
Yeah, I think the three inputs were different relative densities
Yeah, and then you had to construct the energy absorption curve based on those equations.
And I give you three relative densities.
So this was my solution.
So all this stuff here, I think, was straight out of the notes.
So there was the-- oops, let me back up so we start at the beginning.
So there was the linear elastic part.
There was the stress plateau.
That was just as it starts to densify.
And then, there's that line joining up the points.
So you were OK with all of that?
Yeah.
No, the only part that confused me is because once I simplified-- I inputted all of what we were given such that I had equations in terms of the relative density so I could just apply it to the different ones given.
But then I wasn't very sure how-- because I got constants, I wasn't really sure how-- the slopes should look like, because they were constants
So you got-- I'm not sure what you mean about you got constants.
So what I did was I said, well, I know that the diagram has to have this basic shape, right?
In the first part here is where it's linear elastic.
And this part here is the stress plateau.
And that part there is the densification.
OK?
And I said, well, if I can find these two points that correspond to the change between the linear elastic part and the stress plateau and the point that corresponds to the stress plateau and the densification, then I've got the diagram.
Right?
OK?
Yeah, I think that--
Oh, can I stop talking now?
I mean, if you wanted to--
Well, it's for you.
So you're OK?
I think
You
So what confused me about this question was that in order to find the two points, you'd need to know the strain at which that [INAUDIBLE].
So I wasn't sure how you would find that strain
Yeah, I think I didn't tell you.
Let's see.
Well, let's see.
Do you have to have the strain?
You have that.
I think if you have-- I think you don't have to have it in terms of the strain.
I suppose you could put it in terms of the strain.
I mean, that would be a different way to do it
But isn't the equation for the stress plateau already in terms of the strain?
Yeah, but this assumes that the stress plateau is just perfectly flat, right?
So this thing here would be useful, the strain at which the plateau starts.
But I think that's the same as saying-- that's the same point as saying you're at that point there, because that's where the plateau starts.
OK?
So I think there's different ways you could try to approach this.
But I think-- let's see, I'm just trying to remember what I did here.
Yeah, so what I did here-- so to get point a, I said, well, this is the equation for the linear elastic bit, right?
And this is the equation for the stress plateau.
Where I'm at point a here, this stress is the stress plateau.
So that's why I've put sigma star plastic there.
And then I said, this is-- the sigma star plastic is-- and this works out to some number here, some number times the relative density to the 3/2 power.
And then I just made this little table here, where I said, OK, these are the densities I need to get the curve for.
This is going to be my sigma star plastic for those densities.
And from that, then I can get this column here.
It's basically just that equation with this substituted in.
And do you see that that corresponds to the point a?
So I mean, you could do it by figuring out what that strain is and then figuring out all the points along that line up to that strain.
But you don't have to do it that way.
Do you know what I'm saying?
So you're saying-- if I had-- so say I have my idealized stress strain curve like that, right?
You're saying, well, I had to know that strain there so that I know where this stops, right?
And I'm saying that corres-- and you could do it this way if you wanted to.
You could say that this is the energy absorbed up to that point.
And if you know that the modulus goes as the relative density squared times Es, and you know this, if you know those two things, you can figure out what that strain is.
So you could do it that way if you wanted to, but I just did it a slightly different way
Yeah, so I don't understand how you just were able to get rid of the epsilon minus epsilon on that curve
Oh, let's see.
So I think, in the first part, where it's linear elastic, I just go up to epsilon 0, right?
So this part here doesn't really involve the epsilon, because I've gotten rid of it by taking the square of the stress.
And then here, the other point I'm looking at is this point here.
And I've assumed that epsilon 0 is much smaller than epsilon d, and I've ignored it.
And that's, I think, how I did it in the notes in the class.
OK?
So when you get to that point, this strain here is usually a few percent at most.
And this strain here is typically 80% or 90%.
So it's pretty common to do that.
OK?
Other questions?
So don't forget, the test covers everything from the thermal conductivity of the foams, it covers the stuff on trabecular bone, sandwich panels, and then, the energy absorption stuff.
Yeah?
Are you a little exhausted already?
Yeah.
So I'm guessing that this week and next week, you have everything due.
You have papers, and projects, and tests.
Do you have very many exams left, like final exams?
Yeah, you've got finals?
[INAUDIBLE]
Oh.
All right.
Anybody else have any questions?
Because I think we can just go and do other things if nobody has any other questions.
So the test is on-- so let's just review where we're at for the rest of the term.
So the test's on Wednesday.
And you can bring one cheat sheet, but I am not giving you all those equations with honey combs and foams.
So I think, on your cheat sheet, you would want to put Young's modulus of an open-celled foam, shear modulus of an open-celled foam, compressive strength of an open-celled foam, shear strength of an open-celled foam.
But I don't think there's going to be anything more complicated than that.
And Monday, I was going to do the how I became a professor talk.
So if you've seen it and you don't want to see it again, you're welcome to not come.
But for you guys, I just talk about how I got here.
And it's about my life.
It's not about cellular solids or anything.
And I don't know, you guys liked it.
You like it, don't you?
Yeah
Yeah.
Yeah.
So if you want to come, I'll do that.
And then Wednesday, I thought, I just need to collect the projects.
I wasn't going to do anything on Wednesday.
And so I've been thinking about the how I became a professor talk.
And I think-- so I have this idea that students would like to hear more of these talks from other faculty.
Would that be true?
Yes
Ah.
Because I've been in touch with Cindy Barnhart, and we're going to try and organize something for the fall.
So I'm going to approach some other professors-- not just in our department, across all of MIT-- and see if I can get other people to do the how I became a professor talk.
So you would like that?
OK, yeah.
So we'll see if we can make that happen.
The following content is provided by MIT OpenCourseWare under a Creative Commons license.
Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu
--not of the vindictive sort.
You skip class, you've skipped a lot of important stuff.
But I'll get you on the quiz, that's all.
Kidding aside, there were a number of things that I passed out.
I think nobody needs a set of problem set number 13.
That was the one that had something on symmetry constraints and working with second-rank tensors.
If anybody missed that, you can see me during break.
Some people, I think, missed problem set 14.
And that's the one where you are invited to diagonalize some tensors, either using the method of successive approximations or the direct diagonalization- by-an-eigenvalue procedure.
Anybody need one of that?
And I'm sure that nobody has a copy of problem set number 15, which deals with piezoelectricity.
And I know you don't have it, because I just put it together.
So I'd like to hand it out and invite you to explore things that deal with third-rank tensors.
And I hope that, even though doing the problem sets is optional, particularly at this juncture in the semester when things have come to a set of successive crunches.
But if you don't know how to do it, for goodness' sakes, come see me.
Or raise it in our next class.
You know, some question like, I haven't the foggiest idea how to do problem number two.
Could you say a little bit about that, please?
And I'd be happy to oblige.
All right.
I will have for you next time the quizzes and also all the problem sets which will have been turned in up to that point.
And what I spent my time doing instead is writing out notes for those people who missed the last lecture, and also notes covering what we're going to do today.
Because it is very exquisitely intensive, algebraically.
It's not hard, but there are a lot of variables with a lot of subscripts.
So let me pass this around.
I'll split it up into packs.
These are notes on some basic relations in electromagnetism which you may or may not have forgotten.
Take one off the top.
And it's coming at you from either side, so you're going to pass it back.
And the notes also cover everything that we're going to do on piezoelectricity.
Most of it will take place today.
And I can zip along a little more rapidly if you have notes to follow.
I would like to ask you when you get a set of the notes-- I could really kick myself-- the introductory discussion reminds you of the definition of a dipole.
And down in the middle of the page, on the cover sheet, two different types of polarizability are defined.
And one of them involves the separation of charge on an individual atom.
And that is called, quite appropriately, the electronic polarizability because it involves polarization of the electrons and protons on the individual atoms.
And then there's another type of induced dipole moment that comes when the structure is ionic.
And then an electric field will pull positive ions in one direction and negative ions in the opposite direction.
And that is very often referred to as the ionic polarizability.
And it's easy to keep them straight.
One involves electrons, which all atoms have.
The other involves ions.
And not all structures and materials have ions.
So the second one is unique to ionic structures.
And then, these are sometimes also referred to as the dielectric polarizability.
And I meant to purge that from the notes.
And as I put this together to xerox it, I grabbed the uncorrected sheet.
So please, just below you see the displaced positive and negative ion on the middle of the page, cross out "dielectric" polarizability.
And change that to "ionic" polarizability.
And I didn't catch that.
There's one other little typo as we go partway through.
In any case, we'll talk about third-rank tensor properties.
We'll introduce some other ones other than piezoelectricity later on.
But piezoelectricity is one of the primary examples of a third-rank tensor property.
And it's one that has a lot of applications in devices-- pressure sensors, audio equipment, electronic devices.
It's a very important property in terms of devices and present-day technology.
But there are others.
And we'll cover those in due course.
Some of them are rather exotic.
OK.
But to review these basic concepts in electromagnetism, we remind you again of the definition of a dipole.
We mentioned this last time.
But to quickly review, a dipole is a pair of charges of opposite sign but equal magnitude, separated by a separation, d. And then one defines a dipole moment, which has a vector character, as the product of one of the charges of magnitude, q.
And the vector that separates the two charges in the sense of the vector is defined as going from the negative charge and pointing towards the positive charge.
It's purely a definition.
But the reason it's convenient is that the vector sense and the product of charge and separation comes up again and again in all sorts of problems, among them definition of the piezoelectric effects.
Also a dipole in an electric field is going to experience a torque because the electric field will pull on the positive charge in the same direction.
It will pull on the negative charge in the opposite direction from the field.
And that's going to create a torque on this little gizmo.
OK. Now it's important, too-- and interesting to note-- that there are three different kinds of dipole moments.
There are some molecules-- and water is the primary example.
Water has seen an asymmetrical arrangement of hydrogen, relative to the oxygen ion to which they are connected.
And that gives water a permanent dipole moment, which is what makes water such a darn good solvent.
And its ability to dissolve primordial juices probably accounts for our fact, intelligent design notwithstanding, of why we are here today.
On the other hand, dipoles, as I was just saying, can be induced when you impose an electric field on matter.
And these are of two kinds.
One is the dipole moment that is induced on an individual atom.
And that results in displacement of the positive nucleus relative to the negative electron shell.
It's found that the dipole moment is proportional to the magnitude of the electric field.
And the proportionality constant, alpha, is called the polarizability.
And for an individual atom, as I said a moment ago, it's defined as the electronic polarizability.
We write it as a scalar quantity, but actually, by now you're probably sensitized to being a little bit skeptical when things that relate to vectors are described as a scalar.
And in fact, the electronic polarizability is not a scalar, it's a tensor.
And one should really write that the i-th component of the dipole moment is given by alpha i,j times the j-th component of the electric field.
Second type of induced dipole moment involves, again as we said a moment ago, the effect of imposing an electric field on an ionic structure.
And again, the field will pull the positive ions in one direction and negative ions in the other.
And here quite clearly, if this pair of ions is in a structure, and that structure has some symmetry, we really have to consider this second origin to induce dipole moments as a tensor.
And this is referred to as the ionic polarizability.
So both of these types of dipole moments will be present in matter, in general.
The relative importance of each depends on whether the electric field is a static field or an oscillatory electric field.
And then the frequency dependence of these two polarizabilities has a consequence on the magnitude for fields of different frequencies.
And not surprisingly, the ability of the ions in the structure to polarize is going to poop out with high-frequency electric fields a lot quicker than just the displacement of the light electrons about a positive nucleus.
So there is a frequency dependence of the net polarizability.
We won't go into that.
But just keep in mind that at very high-frequency electric fields, the ionic polarizability will damp out.
OK. Then we went through a rather simplified but amusing model for the electronic polarizability.
And there's some rather severe assumptions that are made.
But making those assumptions let's you get a rigorous result, which tells you something about the electronic polarizability.
So we model the electron distribution on the atom as a uniform charge density in a distribution that goes up to some radius, r, and then quits.
So there's a sharp cutoff to the distribution of electrons, which is obviously ridiculous.
You know there are a collection of wave functions that give you charge probabilities that tail off slowly to large distances, getting progressively smaller and smaller.
So this is not terribly realistic.
And then the other thing we assume to make this model, something that we can solve exactly, is that the nucleus and the electron distribution displace as units.
In other words, we start with a sphere of electrons, the center displaces, but it stays a sphere of uniformly distributed electrons.
And then having made those assumptions and having lost any credibility for the model, if we carry through to see what the model predicts, it's rather interesting.
We use a fact, again, known to freshman and sophomore students of electromagnetism, that a charge inside of a uniform distribution of charge experiences no force.
And that's surprising, but it's something that you are very often invited to do on problem sets.
So therefore, if the center of the electron distribution is displaced from the nucleus, then the restoring force between the electron sphere and the nucleus is simply the coulombic force between a nucleus of charge plus ze, and a fraction of the total number of electrons, namely that fraction of the electrons which are contained within a sphere that has a radius equal to the displacement.
And that geometry and that algebra's carried out for you on the bottom of the first page.
If you set that up and ask what the dipole moment will be, it comes out beautifully simple.
It comes out to be equal to whatever proportionality constant you use in Coulomb's law.
I use rationalized MKS units from force of habit.
So there's a 4 pi epsilon 0 in there, then times the cube of the radius of the electron distribution, times the electric field.
So the two items of note that come out of this simplified treatment is first of all, the induced dipole moment is proportional to the applied electric field, which is what we assumed.
And so therefore, the polarizability, which is the quantity that relates the dipole moment to the magnitude of the field, is a constant.
And it turns out to be equal to 4 pi epsilon 0, times the radius.
So not only does this tell us that the electronic polarizability is something that relates dipole moment in direct proportion to the magnitude of the field.
And secondly, the electronic polarizability involves the radius of the charge distribution, cubed.
And even on a qualitative basis, this is interesting.
It tells you that high-atomic-number, big, fat ions are going to have a very, very large polarizability.
And things way down in the periodic table, like beryllium and lithium, and other low-z atoms, are going to be tough little nuts that don't display much polarization at all.
And in point of fact, several individuals have tabulated empirical sets of electronic polarizabilities.
One of the earliest ones are the so-called TKS values published a long time ago by Tessman, Kahn, and "Wild Bill" Shockley.
And I give you the reference to those.
There is another set of values that were assembled by a fellow at DuPont named Bob Shannon.
And I'll give you a citation to those values.
But in any case, if you look at these values, you find that the cation that has highest electronic polarizability is thallium, way down on the bottom of the periodic table, next to lead.
And that's just a big, fat, flabby atom that can be deformed very, very easily.
How do you get these polarizabilities?
Well, the equations at the bottom of page two-- which we won't make any use of but, nevertheless, will tell you where they come from-- there's a relation between the square of the index of refraction and the sum of the polarizabilities of the individual species, times the number of those species per unit volume.
And that is something that's called the Lorentz-Lorenz equation, equation.
I can't resist saying everything twice, Lorenz and Lorentz.
You can put this in another form that involves the molecular weight of a molecular structure and the polarizability per molecule.
It's the same equation, but for organic compounds.
It's a useful form.
And it turns out that the dielectric constant of the material is directly related to the square of the index of refraction.
So you can write those two equations in terms of either the dielectric constant or the index of refraction.
And if you substitute dielectric constant in place of n squared in those equations, the equations get new names.
And they're not called the Lorentz-Lorenz equation, equations.
They're called the Clausius-Mossotti equations, which shows you that sometimes fame and immortality can be gained simply by a trivial substitution of variables.
Nice to keep in mind if you can find something like that.
Finally, we don't see dipole moments on individual atoms or molecules.
We see evidence of polarization in bulk.
And on page three is a little model that reminds you of what the polarization of a material is.
And that's defined as simply the dipole moment per unit volume.
And that's represented by a capital P rather than a small p. And this is also a vector quantity.
And we can see how this is related to the individual dipole moments in the solid by dividing the solid up into individual cells.
And I use the term of "cell" loosely.
Is this a unit cell?
Is this a box around each of the atoms?
It really doesn't matter.
Because all of these little dipoles on the individual unit cells are packed together back to front in the solid.
So internally, the negative end of one induced dipole moment is always adjacent to the positive end of the neighboring dipole moment.
And everything cancels out internally, except for the two surfaces of the solid.
So macroscopically, if you impose an electric field and it induces dipole moments, what happens is you see that there is a charge on the surface of the piece of material.
And that is a real charge.
It's a bound charge.
You can't draw that charge off as a current.
Because the minute you reduce and eliminate the electric field, those dipoles disappear.
And the charges all go back to where they came from.
So this is the thing that keeps the exact model that we use for one of the cells in the solid [INAUDIBLE] of no importance whatsoever.
You could say that the dipole moment on each atom is the charge times the separation of the positive and negative charge.
And qi is the induced charge.
The number of cells per unit volume, if we say that they're little cubes of the same edge length, delta, is unit volume one divided by the volume per cell, which will be delta cubed.
And the dipole moment per unit volume is the number of cells per unit volume times the polarization on each.
And you find that everything drops out.
And the polarization indeed turns out to be equal to induced charge per unit area, whether that's induced charge per unit area of one of our cells or on a square centimeter of material.
So polarization, dipole moment per unit volume is numerically equal to, and physically equivalent to, an induced charge per unit area.
OK.
So much for freshman electromagnetism in fast forward.
And now I'd like to turn to something that involves tensors and materials.
As I mentioned a moment ago, piezoelectricity, literally "pressure electricity," is a very technologically important property, useful property, but also provides a nice example of a tensor property that has to be defined in terms of a sensor of third rank.
There are a number of different piezoelectric effects.
The first one is the so-called direct piezoelectric effect.
And it refers to the fact that if you subject a material to an applied stress, it will develop a surface charge.
Remove the applied stress, the surface charge goes away.
The surface charge can be described in terms of a polarization, a dipole moment per unit volume.
And it will be in direct proportion to a stress tensor sigma j,k.
And so what we'll assume-- and this is an assumption.
And it's followed, except for very extreme conditions, that each component of the polarization, p sub i, is given by a linear combination of every one of the nine elements of stress, sigma i,j.
Oops, not i,j.
I have to use a different index.
I'm used to writing i,j.
So each component of the polarization, where i ranges from 1 to 3, is given by a linear combination of all nine of the elements of stress, sigma j,k.
And the proportionality constants, di,j,k, are known as the piezoelectric moduli.
And this is called the direct piezoelectric effect.
OK. That's simply an assumption that the polarization is proportional to the applied stress.
We know that stress, however, is a field tensor of second rank.
We know that the polarization has the character of a vector, charge times the length.
We know how second-rank tensors transform.
We know how first-rank tensors transform.
And therefore, knowing that, we can say that the coefficients, the array of coefficients, di,j,k, will transform like a third-rank tensor.
And therefore they are a tensor.
So we will have three components of polarization.
And we will have nine components of stress.
And so there will be 27 piezoelectric moduli, di,j,k.
So things go up rapidly in terms of number of coefficients as the rank of a tensor increases.
Now let's take a look at the nature of these equations.
We'll have, for example, the x1 component of P being given by d1,1,1 times the element of stress sigma 1,1.
The next term will be d1,2,2 times sigma 2,2.
I'm putting down the tensor components first.
But this is just an equation.
I can write the terms in any order.
Next will be a d1,3,3 times sigma 3,3.
And the next one would be a d1,2,3 times a shear component of stress sigma 2,3 and other terms.
I don't want to write out all nine of them.
Because I think I have written enough to make my point.
The point is that the value of i is always tied to the component of the resulting vector that we're defining.
But the second pair of subscripts, j and k, always go together as an unseparable pair with the component of stress that they modify.
So the 1,1 here goes with the 1,1.
The 2,2 goes with the 2,2.
The 3,3 goes with the 3,3.
So why in the world, if they're always going to go together, do we have to use two indices to define the piezoelectric moduli?
Why don't we use a single symbol?
Well, there is a good reason for not doing it.
But we'll issue that caveat later on.
We could use an a, a b, and a c, or an alpha or a beta or a gamma, or some other esoteric symbol.
But what makes sense since we're using Arabic numerals to represent subscripts, let's write the pair of indices in terms of an index that's related to the stress tensor, which is sigma 1,1; sigma 2,2; sigma-- whoops.
Sigma 1,2; sigma 1,3.
And then comes sigma 2,1; sigma 2,2; sigma 2,3; sigma 3,1; sigma 3,2; sigma 3,3.
We know that this is a symmetric tensor.
So these off-diagonal terms are always equal to one another.
So we're numerically going to have to enter each of those twice.
So to define a single index that represents the components of stress.
And I'll describe it in this fashion so you can remember, as a mnemonic device, what makes this work.
We'll go down the main diagonal of the stress tensor, this fashion.
And then having reached the bottom, we'll go up along the right-hand side and then jog over to the left to pick up the last term.
And we'll define a number associated with these indices that goes in the form 1 to 2, to 3, to 4, to 5, to 6.
So we'll just use these three integers, ranging from 1 to 6, to represent pairs of integers in the stress tensor.
So things tidy up quite nicely then.
This would be P1 is d1,1 times sigma 1; plus d1,2 times sigma 2; plus d1,3 times sigma 3; plus d1,4 times sigma 4.
And now, uh-oh, Houston.
We've got a problem.
Because there's another term in here that is d1,3,2 times sigma 3,2.
And sigma 3,2 is required, because the stress tensor is symmetric, to be numerically equal to sigma 2,3.
So this is really d1,2,3 times sigma 2,3.
And then we've got a d1,3,2 times a sigma 3,2.
I know that this is equal to this.
But is there any reason d1,2,3 has to be equal to d1,3,2?
I can't see any reason why.
OK. Let me confide in you that, yes, they are equal.
But we can't show it or claim it on the basis of what we've got before us right now.
It's going to come later.
But we can show that they are equal.
OK.
But equal or not, this means we'll have a term d1,4 sigma 4, and we're going to get it in there twice.
So when the subscripts go up to 4, we're going to get a 2 out in front.
And then we'll get for the term sigma 1,3 and 3,1 we'll have a d1,5 times a sigma 5.
This would be sigma 1,3.
And then we'd have another d1,5 for sigma 5 again.
And this would be sigma 3,1.
So what are we going to do?
We're going to say, well, it's nice we're getting rid of an unneeded subscript.
P1 is the d1,1 times sigma 1; plus d1,2 times sigma 2; plus d1,3 times sigma 3.
Are we then going to say 2 d1,4 times sigma four, and say that the coefficient here is di,j when i is equal to 1, 2, 3.
But the coefficient is 2 di,j when j is 4, 5, or 6?
Hell of a matrix that would be.
That's going to be a bother.
Since it doesn't seem that we can measure these two coefficients, d1,2,3 and d1,3,2, independently anyhow, let's just write our relation in reduced subscripts with the two added together as the element d1,4.
So we are going to define d1,4 equals d1,3,2 plus d1,2,3.
So that's a definition.
And d1,5 will be defined as the sum of d1,1,3 plus d1,3,1.
And we're going to define d1,6, finally, as d1,1,2 plus d1,2,1.
OK.
So if we do that, then and only then are we entitled to write a nice, simple, compact little nugget that says that in our redefined form, Pi is equal to di,j times sigma of j.
And i goes from 1 to 3, and j goes from 1 to 6.
And this is kind of a neat compact, easily managed outcome.
So the moral of this story is, I like to think, is you can have your cake if you eat its 2.
Oh come on, this is a tough, tough crowd.
You're just all worn out from the MRS meeting.
That'll do it to anybody.
So why do we worry about the fact that to the direct piezoelectric effect should be a third-rank tensor?
The answer, my friends, is that this is no longer a tensor relationship.
It's a matrix relationship.
A tensor is a matrix, but it's a matrix with a difference.
It's a matrix for which a law of transformation is defined, if you change axes.
There's no such law defined for the di,j's.
So they qualify as a matrix, but they are not a tensor.
So to attempt to say-- as you might be inclined to do when we raise the issue of what the symmetry restrictions are on these tensors-- if you attempt to say, well, I'm going to find d2,1 prime.
And that's going to be c2,i c1,j times di,j.
Wrong.
You go down in flames because that's just nonsense.
That's just nonsense.
So this is a matrix relation.
And if you want to do exercises such as slice a piezoelectric wafer out of a piece of quartz, and then having done so, refer the properties to axes taken along those edges of the plate that you've cut, you've got to-- in order to find the piezoelectric moduli for that plate in that new coordinate system-- you've got to be prepared always to go back to the full tensor notation.
If you want to derive symmetry restrictions, which we're going to do whether you want to or not-- but we won't do them exhaustively-- you've got to go from this matrix notation back to the full three-subscript tensor notation.
OK?
Yes, sir?
Couldn't you extend c sub i,k's to be [?
3,6's ?]
and extend most others [INAUDIBLE]?
Oh, yeah.
No, if we would write all these down, we'd have-- I don't know if that's what you're asking-- but we'd have P1 is equal to d1,j times sigma sub j; P2 is equal to d2,j sigma sub j; and P3 is equal to d3,j times sigma sub j.
So I just did that for the line with i equal to 1, because I was too lazy to write down all three relations.
Is that what you're asking?
No, I'm saying you could have a transformation model if you were to extend a c sub i,j matrix to be 3,6
No.
It just won't do it.
I mean, think of what these indices are.
They're 1, 2, and 3 standing for the x1 direction, the x2 direction, the x3 direction.
What's the x5 direction?
It's just not defined.
Good try, but you just can't do it
I'll think of something next time
Oh, I'm sure you will.
Whether it will work or not is an absolutely different question.
OK.
So beware.
Do not try to transform tensors of third rank expressed with two subscripts to other coordinate systems.
All right.
What I would like to do now, then, is to examine the symmetry restrictions that are imposed on a third-rank tensor by crystallographic symmetry.
And these are embedded in the notes that you have.
But it's going to be bothersome to leaf through those.
So I separated out the pages separately.
And I'm going to look at a few transformations so that you can see how they work, and then point out some very, very curious consequences of this, which are non-intuitive.
So first of all, how would we do it?
If we want to get the form of di,j, the matrix for-- let's do a very, very simple one.
Well, let's do one that we did last time, for those who were here, for inversion.
For inversion, the direction cosine scheme is c1,1, 0; 0, 0-- I'm sorry.
For 1 bar, the transformation of axes is c1,1, 0, 0; 0, c2,2, 0; 0, 0, c3,3.
So we'll take di,j, change it to the full three-subscript notation, di,j,k.
And the law for transformation of the third-rank tensor is di,j,k prime is equal to ci, capital I; cj, capital J; ck, capital K; times d, capital I, capital J, capital K. We've not done this much.
But we've mentioned quite some time ago that this is the way a third-rank tensor would transform element by element.
The only terms that are non-zero in this array, when the symmetry transformation is inversion-- where x1, x2, x3 goes to minus x1, minus x2, minus x3-- is whatever i is, minus 1; whatever j is-- only the diagonal terms are there, and they're all minus 1-- whatever k is, it's going to be minus 1, times di,j,k.
So we find that for inversion, di,j,k prime is always, regardless of what the three indices are, is going to be minus di,j,k.
And if inversion is a symmetry operation which the crystal possesses, we're demanding that the transform index be identical to the original index.
But there's a minus sign in there.
So we can say that every single element vanishes, has to vanish.
So any crystal that has inversion in it is not going to be able to display the piezoelectric effect or any other third-rank tensor property.
The electro-optic effect, piezoresistance, and there are a whole slew of them, which are examined for you in part on this sheet.
One third-rank tensor property is the direct piezoelectric effect, which we've been discussing as our example.
There is something called the converse piezoelectric effect, which describes the phenomenon where an applied electric field creates a strain.
So the direct effect is you [?
scush ?]
your material, you develop charge.
The converse piezoelectric effect says if you apply an electric field that's going to move the atoms around and induce polarizations, you are going to create a strain.
And the absolutely mind-boggling thing is that the same array of 3 by 9 coefficients describe both the direct piezoelectric effect and the converse piezoelectric effect.
So in one relation, we have that each component of polarization is di,j,k times the nine elements of stress.
And that means we have to write three equations in nine variables, the elements of stress.
And the converse piezoelectric effect, we're developing a strain.
And there are, therefore, nine elements of strain.
And we're applying a vector, e sub i, a field.
So there are three components of that vector.
27 elements, 3 times 9.
27 elements, 9 times 3.
So both of these are third-rank tensors.
Strain transforms like a second-rank tensor.
Field vectors transform like a first-rank tensor.
Therefore, the coefficients are elements of a third-rank tensor.
But what boggles the mind is that the same 27 numbers describe these two seemingly disparate phenomena.
You don't prove this by symmetry.
You don't prove this by the nature of stress and strain.
This hinges on the thermodynamic argument, which I'm not going to go into.
But what you do if you're willing to stretch tensor notation a little bit-- you can't have a 3 by 9 array when you've got a tensor of second rank on the left and a vector on the right.
You have to write this in this fashion, that di,j,k times ei gives you the element of strain, epsilon j,k.
And that's not proper tensor notation.
But we're not going to quibble.
Because if you allow this little departure from convention, then you can write both the converse piezoelectric effect and the direct piezoelectric effect in terms of the same matrix.
But again, that is not intuitively obvious.
It does not have to be the case.
And it hinges on an argument in thermodynamics.
Some other examples of piezoelectric relations that I've mentioned in the notes.
If you apply a stress and you get a polarization, that stress produces a strain.
So you also have to get a polarization if you're applying a stress, and write it in terms of the strain that's produced.
Similarly, if you apply an electric field and it produces a strain, the crystal must be in a state of stress.
So there must be relation between stress and applied electric field, which will also be a third-rank tensor.
Those two relations are represented by coefficients given the symbol e. And again, the same array of 27 elements describes both effects.
And these are not dignified with any special names.
They're just tensor relations that have to be true because of the fact that stress and strain are coupled by elastic relations.
Not surprisingly, then, the coefficients e must somehow involve the piezoelectric moduli di,j and the elastic properties of the material.
Another effect which is a very interesting one is the electro-optic effect.
If you apply an electric field to a material, you change the birefringence of the field.
The birefringence is defined as the difference in index of refraction for light polarized in two orthogonal directions.
Another tensor effect-- and I'll give you a note defining the terms next time we meet-- there is a piezoresistive effect, which you don't see talked about very much.
But that's a property that Texas Instruments was interested in at one time.
And I have a set of sheets defining those relations that the presenter of a paper at a meeting, one time, kindly gave to me.
OK.
So what other symmetry restrictions are there?
Having shown that inversion will not permit any third-rank tensor property, we have gone from 32 point groups, lost interest in the 11 centrosymmetric point groups.
And so there are only 21 piezoelectric point groups.
And the way we would plod through all 21 of them would be to simply define, starting with a twofold axis.
Let's say a twofold axis parallel to x3 would correspond to a change of axes ci,j that describes the new axes in terms of the original ones that consisted of minus 1, 0, 0; 0, minus 1, 0; 0, 0, 1.
So if we look at an element in the piezoelectric matrix-- something like d1,6-- d1,6 actually corresponds in tensor notation to d1,3,2 plus d1,2,3.
And d1,3,2 prime is going to be c1,i c3,j c2,k times all of the original tensor elements di,j,k.
The only term of the form ci, something that is non-zero is c,1,1.
And that has a value, minus 1.
The only term of the form c3, something which is non-zero is c3,3.
And that turns out to be plus 1. c2, something, the only form that's non-zero is c2,2.
And that has value, minus 1.
And this should be times d. And the only value of i that stayed was 1.
The only value of j that stayed was 3.
And then only value of k that stayed was 2.
So this says that d1,3,2 should be equal to d1,3,2, which I don't like, unless it's supposed to be negative.
I did something wrong here.
Well, you see how easy it is, even if it didn't turn out right.
And the form of the tensor for monoclinic crystal of symmetry 2, with a twofold access parallel to x3, has as shown in the lower left-hand corner of the handout on symmetry restrictions, it has eight non-zero terms.
If you do the same thing for a mirror plane perpendicular to x3, you find that there are 10 non-zero terms.
So we don't have the situation where all of the point groups that are able to show the property within a given crystal systems like monoclinic have exactly the same form of the property tensor.
In fact, you'll notice a curious correspondence between the restrictions for symmetry 2 and symmetry m. All the terms that are 0 in symmetry 2 are non-zero in symmetry m, and vice versa.
All the non-zero terms in symmetry 2 are 0 in symmetry m. And the reason for that is simply that this is the form of the direction cosine scheme for symmetry 2.
The form of the direction cosine scheme for symmetry m, where the m is perpendicular to x3, would have the form 1, 0, 0; 0, 1, 0; 0, 0, minus 1.
So ci,j, for a mirror plane perpendicular to x3, is exactly the negative of the direction cosine scheme for a twofold axis parallel to x3.
And since the number of direction cosines is odd, this means that everything that has an equality between the di,j,k's for m would have the transformed element be the negative of the original one for 2, and vice versa.
So that's why there's this complementary form of the tensors for the two monoclinic symmetries.
Point out a couple of curious things in the tables.
You really have to go through 21 of the symmetries independently.
And you find that some of them do come out the same.
Symmetry 4 bar 3m and symmetry 2:3 have restrictions of the same form.
But for the most part-- Yes?
I know why this is wrong
Why is that wrong?
Because your notation of d1,6 is in fact not d1,3,2 but d1,2,1.
Since you have two same indices, [?
d3 ?]
and minus [?
d1.
?]
OK. d1,6; d1,6.
Ah, of course.
Of course.
d1,2,6 is the one up here.
And that's 1, 2 and 2, 1 for the strains.
OK.
Thank you.
So this is d1,1,2 plus d1,2,1.
And so we would have 1 by 1,k and 2,k.
1, 1, and 2.
And 1, 1, 1.
1,i; 1,j; and 2,j for this one.
So we'd have c1,1 c1,1 c2,2.
c c1,1 is minus 1; c1,1 is minus 1; c2,2 is minus 1.
So d1,1,2 is minus d1,2,1, which means they have to be identically 0.
Thank you.
I'm sure nobody cares at this point.
Very good.
OK. Another curious thing that happens is that for cubic symmetry 4:3:2, it's acentric.
But every single modulus is 0.
And the reason, the explanation, is there's so many different transformations which have to leave the tensor invariant that the poor tensor just can't do it.
It gives up, packs up, and goes home, leaving all the elements zero.
Just no way you can get all the qualities to be satisfied.
Another curious result that I point out for some of the hexagonal symmetries for symmetry 3:2, for symmetry 6:2:2, and for 3 over m and 6 bar 2m, all the elements of the form d3, something are identically 0.
Which says you simply cannot create a polarization that has a component P3.
You just cannot create a polarization perpendicular to the axis of high symmetry.
Yeah.
Got a question?
So all of the cubic that has 4:3:2 symmetry or [?
4:1:0 ?
], that means you can't get--
You just don't have any piezoelectric effect, even though the crystal is not centrosymmetric, requiring that all those transformations leave the tensor invariant; simultaneously, require that everything has to be 0.
One final thing, and then I'm running a little bit over.
But I'd like to go on to other aspects of piezoelectricity during the next hour.
Remember something that we've shown and which I've asked you to look into again on one of the problem sets, and that is that the trace of the strain tensor gives you the change in volume.
The first 3 by 3 blocks of this tensor that we've been looking at, were we to write the elements of strain, epsilon i,j, in terms of d-- epsilon j,k-- in terms of di,j,k times e sub k, it is this block in here which enters into the terms epsilon 1,1; epsilon 2,2; and epsilon 3,3.
These terms give you the fractional change in volume.
And the elements that are involved in the piezoelectric matrix are these terms in here.
So if all of those terms are 0, it turns out that when you apply a stress and look at the electric field, or apply an electric field and look at the strain, if you apply a field and all of those nine terms are 0, you cannot create a strain.
So there's no volume change.
There can be a strain, but no volume change.
The only deformation you can create is pure shear.
So if you drive a piezoelectric oscillator with an electric field, for those property tensors for which that first 3 by 3 block are all zero, the thing can shear-- in the water, for example-- but it can't pulse.
It can't have a volume change.
And if you were to create a transducer for sonar applications, what you would want to do is to have piezoelectric device which expanded this way, to create a sound wave going through the water.
If it just goes back and forth, it's going to slosh back and forth in the water and not create any sonic wave that could be used in sonar.
Now that is true for elements being identically 0.
It turns out it's also true for elements such as 4 bar, where two of those terms in the 3 by 3 block are the negative of one of the other.
These will also have no volume change.
So that's another very curious property of piezoelectric response that follows from these symmetry restrictions.
OK. That's enough for our first session.
When we come back, we'll look at the converse piezoelectric effect.
And we'll also look at representation surfaces for specific piezoelectric devices and ask if it's possible for a third-rank tensor to have a representation surface that's analogous to the representation quadric for second-rank tensors.
So I'm sure you'll all want to come back and hear the answer to that question.
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The final quiz is scheduled for a week from today, on December 8.
Following that weekend, we'll have one class on the 13th.
And the last days of classes are Monday, Tuesday, and Wednesday, the 12th, 13th, and 14th.
So I know a number of you spoke up last time and said that you have a conflict on Thursday, the eighth, with a quiz in another class.
So could I see a show of hands again of how many of you have this conflict?
So three, six people, seven people, you're going to enroll in the other course just so you can put off taking the quiz.
That's unfortunate, I don't think anybody should have to take two quizzes in one day.
We can't move it up.
We'll have to move it back.
So I don't know if I am violating any Institute rule, but I know that it is strictly illegal to give assignments that are due after the last day of classes, let alone have a quiz after the last day of classes, which has not been previously scheduled as a final examination.
So we can't give it after Wednesday, the 14th, the last day of classes.
So, do any of the six impacted people have a strong preference for when we should schedule the quiz?
Monday?
Due it Monday?
It doesn't have to be that soon.
Monday?
Why not Tuesday?
Because Tuesday we have a class
Or Wednesday?
Well we would have a Thursday class instead.
Correct?
No.
If we're doing it after the quiz next week, there's a class.
We have a last meeting here on the 13th.
And I'm not going to shift it for other people, just for the impacted.
So of the six who are entitled to vote, how many would prefer to have it Monday, the 12th?
Three and I know how it's going to turn out.
And for Wednesday, the 14th?
Three
I can't take the test because-- not because of another test-- I'm out of town.
And I might come back Saturday.
And Monday is just a little quick to take it after we get back
I think I'm going to have to make an executive-- yes
Professor, when you mean by conflict, do you mean having an exam at the exact time?
No, it's on the exact same day.
Considering this is a two hour examination, to have another examination exactly the same day, if you're not brain dead after an hour or two, you will be pretty close to it.
So I think that is an unfair penalty to pay.
I think, to keep it as close as possible to the quiz that the rest of the people will be taking, why don't we make it-- since it's a tie vote-- on Monday, the 12th.
Do it sooner, rather than later.
And I'll let you know next time of where we will hold it.
I'll have to arrange a room for it.
So that will not be a make-up quiz.
It would be a made-up-- invented-- quiz.
And, as I say on Monday, I should have all of the homework and the previous quiz to return to you.
Reason you don't have it now is all the little crabbed handwriting that you see before you in the form of these notes, which takes forever.
All right, so since we've satisfied the unpleasant aspects of the end of the term, let's get back to discussing some of the other piezoelectric effects that we have defined.
And then also ask the question rhetorically, is there anything like a representation surface for a third-ranked tensor property?
And it doesn't look promising.
But there are some things that we can do to discuss variation of properties with direction, and we'll see directly what those are.
But first let's look at the converse piezoelectric effect.
And this again is a third-ranked tensor property, but what we do is to have the elements of strain, epsilon ij, a second-ranked tensor.
And to take advantage of this curious relation between the directed converse effects, we define the converse piezoelectric effect as giving you nine elements of strain in terms of a third-ranked tensor dijk times e sub i.
So what is not standard is our convention for the order of the subscripts on the moduli.
And we make up the rules, we can do it any way we like.
And the advantage of defining it this way, in nonstandard tensor notation, is that we can use the same coefficients for both the direct and the converse effects.
So let me-- to illustrate what these equations look like-- write out a few examples.
Epsilon 11 would be d1jk times e sub 1.
I'm writing it in simple fashion, and not expanding fully, just to save space and time.
Epsilon 22 would be d2jk times e2.
Epsilon 33 will be d3jk times epsilon 3.
And let me stop after these six terms, and write, at least for these, an expansion.
Because this gives us some interesting information.
I'm not doing the equal signs in here.
So this would say that the element of strain epsilon 11 is d111 times e1 plus d112 times e2 plus d113 times e3.
And I want to say this is 11k.
And I want to say that this is 22k.
The next term would be epsilon 22, and that would be d122 times epsilon 2 plus d212
Wouldn't the 2's be balanced?
Hm?
Wouldn't that be d2's?
Yeah, you're right.
This should be epsilon jk equals dijk times e sub i.
So this is all e1.
This is e1.
And this one goes with this one.
This 2 goes with this one.
You're right.
d2 and this is the d311 times e11.
This would be 122 times e1 plus 222 times e2 plus d322 times e3.
And the fourth one would be e33 equals d133 times e1 plus d233 times e2 plus d333 times e3.
OK, these elements here are the three that appear in the box that I had indicated for the terms djk.
When we write the direct piezoelectric effect, this would be the box of coefficients.
When we write the converse effect, this is the box of coefficients.
And now what I wanted to point out is that delta v over v is equal to episilon 11 plus epsilon 22 plus epsilon 33, which is the trace of the strain tensor.
And we're going to get one set of terms which depends on the x1 component of the field, another set of three terms that depend on the x2 component of the field, and another one on the x3 component.
If these expressions sum to 0, then there would be no volume change.
So this first set of 3-- 3 of the first 6 equations-- give you an indication of when the application of a field will result in no volume change in the sample.
And that again, I remind you, can be for two reasons.
It could be because all of these six of the nine piezoelectric moduli are 0.
Or alternatively, if you examined the form of the piezoelectric moduli matrices that is required by symmetry constraints, you'll find that there are a substantial number of point groups for which some terms are 0.
But then there is, in addition, an equality between some of the other terms, which make the volume change zero.
Even though, not all of the elements within this 1/2 box of moduli are zero.
So that is an interesting effect.
And it turns out that the majority of the non-centrosymmetric point groups do not have a volume change, when you apply in a field in any way you choose.
A few do, but it's a minority.
Alright, but this is not the main point of writing this.
I want to-- at this point-- point out that this tells you about the volume change.
And then we would have additional terms, we would have a term of the form e1 something like e123 or e132.
And this would be e14.
This would also be e14 since we replace both of those subscripts by a single subscript.
And the tensor elements that would go in here would be d123.
We just want one of the them.
d123 times e1, and then we want d223 times e2, and then d323 times e3.
We're writing one of the specific equations for the shear strings.
If we write the expression for d132, this is going to be d132 times e1 plus d232 times e2 plus d332 times e3.
There are two interesting consequences of this.
The strain tensor is symmetric.
And this element of strain-- why have I got three subscripts in here?
Don't want that one in there.
These two strains are equal.
And therefore, if we would apply just an e1 for example, let e be equal to just a component of field along x1.
The strains have to be equal.
But we have two different tensor elements here.
And the only way that strain can be symmetric, and it's defined as such, is that d123 be identical to d132.
And that resolves the issue that came up in connection with the direct case electric effect.
We said the direct piezoelectric effect depends just on the sum of the elements dijk and dikj.
And since they're lumped together, all we can measure is the sum.
And, so, we'll just have to call that a single matrix element.
The converse piezoelectric effect tells us these tensor elements have to be equal, if the strain tensor is to be symmetric.
So that says that, since we defined d16 as the sum of d123 plus d132, this says that d132 is equal to d123 is equal to 1/2 of d16, just making the equality in the reverse direction.
The converse effect let's us say that 123 has to be 132, that any ijk has to be equal to a dijk.
So if I tried now to write this first expression in the reduced subscript notation, e23 is what we let e5 be.
And now we have this equal to and in our reduced subscripts, we have this as 1/2 of d16.
And the field that's multiplying this is piezoelectric modulus is e1.
And that messy factor of 2 has come back to haunt us again.
It's like trying to stuff a jack-in-the box back in the box.
It keeps popping up.
We ate the factor of 2 in defining the matrix representation of the piezoelectric electric modulus.
And now when we try to go to a reduced subscript notation for the converse piezoelectric effect, we've got a 1/2 in there.
And similarly, the second equation would be e5-- same result epsilon 5-- and it's 132, but 132 is 1/2 of d16 times e1.
And we have a similar mess for the other coefficients here.
So what do we do?
Do we say that the relation between strain epsilon j equals dij e sub j has 1/2 in front of several of the coefficients and not in others?
Well, we can't really absorb the factor of 2 in the definition of the piezoelectric moduli, because we've already done that.
So the only thing we can do is to say that we will have to take the off diagonal strains, and define them as having a 1/2 in front, and we add these up.
So we will have to write matrix strain in this reduced subscript notation.
We'll have to take e11, epsilon 12, epsilon 13, epsilon 21, epsilon 22, epsilon 23, epsilon 31, epsilon 32, and epsilon 33.
And in converting this to matrix form, we'll call this epsilon 1, analogous to what we did for the tensile stresses.
We'll call this epsilon 2 and this epsilon 3.
And then for all of the off-diagonal elements of strain, in order to avoid the factor of 2 popping up in front of the matrix representation of the piezoelectric moduli, we're going to have to put in here 1/2 of epsilon 4, 1/2 of epsilon 5, and 1/2 of epsilon 6, and same for the off diagonal terms 1/2 of epsilon 5, 1/2 of epsilon 4, and 1/2 of epsilon 6.
So the moral of this story is that you can't win, but if you play it right, you can come out even.
So only if we define the reduced subscript strains in this fashion, can we write an expression of this form.
So this algebra is carried through for you for the other elements in the notes.
But this is the way we are forced, unless we want to have a factor of 2 in some terms, and not in others, is the way we have to define matrix strain.
All this is formalism and definition, but I'd like to now do two things.
First of all, give you some examples of real numbers for piezoelectric moduli, and then ask the question about representation surfaces.
Once again, these numbers are in the handout for you, so you don't have to make note of them.
But one of the very important piezoelectric materials is the quartz form of SiO2.
SiO2 has many polymorphic forms.
Quartz is the form that's stable at room temperature, and it has point group 32 asymmetric.
And there are higher temperature polymorphs of SiO2.
There's a phase transition in quartz to a more symmetric form, and then there are cubic forms at the highest temperatures.
Now, quartz is not the material that displays the largest piezoelectric moduli.
But it has the following advantages.
One is it is a naturally occurring material that is very inexpensive.
So it's not an exotic expensive material.
Very stable.
It's not water soluble.
Extremely tough.
You can take a thin wafer of quartz, for example, if you want to make a monochromator for a diffraction experiment.
You can take a thin wafer of quartz, and bend it like this, and it does not break, very elastic.
And if you want a material that's going to earn its living by being squished, you want something that doesn't plastically deform and something that is very hard and resistant to stress.
So quartz, even though the moduli are not the largest, is a very attractive material, and is used in a variety of devices.
There was a time when CB radios were very, very popular.
Everybody had to have one in their car.
I guess so they could pretend that they were truck drivers.
But anyway, you don't have them anymore now that cellphones have come in in existence.
But for your CB radio, you needed something called a crystal.
And they were fairly expensive.
And the number of channels on which you could communicate dependent on the number of crystals that you could plug into your CB radio.
The so-called crystal was exactly that.
It was a little black box that looked almost like a transistor.
And there were two leads coming out of it.
If you ever got curious and broke this thing open, what you found was a nice wafer of quartz.
And on the wafer of quartz-- brazed onto it-- were two wires.
And that's all there was in the box.
The crystal really was a crystal.
And the crystals had been very precisely ground to thicknesses such that when a field caused these wafers to hit a resonance, that resonance would be at exactly a particular frequency.
And that was the frequency of that channel.
One of the crises, during the Second World War, is that the highest quality natural crystals of quartz come from Brazil, and during the conflict the sea channels were essentially blocked.
And so, people-- in order to make all these communication devices-- had to learn how to synthesize crystals of quartz synthetically.
And there are a number of companies, such as Sylvania up on the North Shore, that developed entire buildings devoted to growing single crystals of quartz.
And they're big tanks like something out of the aquarium.
And in the center of the tank is a rod, and seeds of quartz are placed on the rod.
And the thing very slowly rotates around in this solution.
And on the rod, eventually, are single crystals of quartz that are this size.
And its a very spectacular thing to see.
In any case, symmetry 3 2, and the moduli that are 0 and non-zero.
If we refer the reference axes to a set of coordinates with x1 in this direction, and x2-- since it has to be orthogonal to x1-- in between the two-fold axes, and x3.
And for all materials of commerce that are anisotropic, there has to be some standard for defining the choice of axes.
For example, crystallographers would say the unique axis should be along the z direction, the x3 direction.
But why isn't x1 and x2 in between the two-fold axis?
There's some professional society that is responsible for giving standards for representing property measurements in some mutually agreed upon form.
And this is the standard set of axes for the quartz and symmetry 3, 2.
The moduli, dij, not dijk, but dij, these two are constrained to be equal.
This one is 0.
This is minus 0.67, 0, 0, 0, 0, 0, 0, 0.67, 4.6, 0, 0, 0, 0, 0, 0, 0.
One of the strange materials for which no field can create a strain-- field along x3 cannot create a strain.
And these are all in units of 10 to the minus 12 coulombs per Newton.
An example they give you here, if you apply a field of 100 volts per centimeter, which is not terribly large, but would be comparable to what you have in some electronic device, perhaps.
So this, since our units are MKS, this would correspond to 10 to the 4 volts per meter.
The strain epsilon 1, which is d11 times e1.
It turns out to be minus 2.3, which means it contracts.
That's the significance of the negative times 10 to the 4.
And that turns out to be 10 to the minus 12 times 10 to the fourth.
That turns out to be a strain of minus 2.3 times 10 to the minus eighth.
10 to the minus eighth is not exactly a large point strain.
You're not going to see the crystal wafer twitch and jump, if you apply a field of 100 volts on it.
But yet, even a strain of this sort is more than enough to be useful.
But this is just to illustrate that quartz is not the most sensitive of piezoelectric materials.
Another one that I give you data for is so-called ADP.
And this is a widely used material.
This is ammonium dihydrogen phosphate.
And this is one of the family of salts that have very large piezoelectric responses.
The nice part about it is that it's water soluble, so you can grow very, very large crystals easily from solution.
The nasty part about it is that it is water soluble, so you have to be careful to protect this material from moisture if you're going to use it in any sort of device.
But if we look at the moduli, relative to the standard axes, and this has point group 4-bar 2m.
So x3 is taken along the 4-bar access.
Then there are two-fold axes and orientations like this.
And since they are orthogonal, you can take both x1 and x2 along the two-fold axes.
And the numbers here for dij, are 0, 0, 0, 1.7, 0, 0, 0, 0, 0, 0, 1.7, 0.
This is one of the interesting tensors where there's a diagonal row of non-zero terms off on the right hand side.
And finally, the big surprise is the third modulus this is 51.7.
So you can see this has a very strong effect.
This is over 10 times the maximum piezoelectric modulus in quartz.
So this is a material that's very commonly used in transducers.
This is again times ten to the minus 12 coulombs per Newton.
One of the very, very exciting developments in recent years is a class of materials that are perovskites.
And they are very, very new.
I gave you the reference to the first one that was reported, and that was just in the spring of 2000.
And these are perovskites.
Perovskites are materials that in the type form are cubic.
But depending on composition and temperature, they can transform to a distorted version of this very simple cubic structure that is tetragonal.
And this material can exhibit piezoelectric effects.
And whether it distorts or not depends on the relative sizes of what goes into the perovskite.
Perovskite has a composition ABO3 like barium titanate is one example.
And the material has two different cations.
And they have different valences so they have different sizes.
And I won't bother to describe the structure, but it is only a very restricted locus in the field RA versus RB, where both the A and the B can remain in contact with the oxygen without distortion.
And it turns out to be a line that does something like that.
Any other perovskite-- and there are lots of them in this field of radii-- has to have one of the ion sort of flopping around.
And if that gets too serious, the structure distorts so that all these ions can remain in contact with the oxygen.
OK in order to do that, many of them distort to tetragonal forms.
Others distort to super structures, which have very, very large unit cells.
But in any case, when you're right at the phase boundary between the distorted structure and the true perovskite structure, these materials sometimes have very, very complicated x solutions of the two phases, on a very sort of fine scale.
And it's not known exactly why they have this property.
But they have absolutely enormous piezoelectric moduli, very close to this phase boundary.
And the references that I give you-- here-- is a compound that is a lead titanium zinc niobate.
And the complicated composition is to get you close to this phase boundary.
This has a d333 that is greater than 2,000 picocuries per Newton.
And pico is 10 to the minus 12.
So this is a piezoelectric electric modulus that is 10 to the 3 times d11 for quartz.
So 3 orders of magnitude stronger than this very commonly used piezoelectric material.
Some of these materials have strains getting close to 1%.
So this is something that will actually twitch on the lab bench, when you apply a field to it.
So these are entirely new.
People still don't know the origin of this behavior.
And it's still under considerable study.
So this is a new family of materials.
It's very exciting, and undergoing a lot of investigation and development work at the moment.
Alright, we are almost out of time.
Time goes fast when you're having fun.
Let me raise the question that we'll consider next time, which will be one of our last lectures.
And that is, is it possible to create representation surfaces that tell you how the piezoelectric properties of a particular material will vary with direction?
Well, vary with what?
Well, we talk about the direct piezoelectric effect.
This gives us components of-- well let's look at the simpler one, in terms of what we apply.
We have the converse piezoelectric effect that says that epsilon ijk is going to be dijk times e sub i.
So we've got a piece of material, and we apply a field, e sub i.
So we can vary this in space relative to a coordinate system x1, x2, x3.
But how in the world are we going to show what happens?
So I always do an extra thing in here.
There are 9 components to the strain tensor, which is symmetric.
So they're really six responses that are unique.
So yes, we can define the direction of the applied field.
But they're going to be 6 different strains.
So we're going to need six representation surfaces.
One for each of the three tensile strains, and one for each of the three shear strains.
So you can't do it with a single surface.
So you can't do much other than say, there are certain responses which are intended to emphasize one particular sort of strain or one particular sort of polarization.
So one of the things we might do is to cut a very thin plate of something like quartz, and subject it to a uniaxial stress.
So let's say sigma along the x3 axis.
So we're looking at a very restricted strain tensor.
That's 0, 0, 0, 0, 0, 0, 0, 0, sigma 3.
And in response to that strain, there are going to be three different components of the polarization.
Polarization is manifested as a charge per unit area.
So if we make a very thin plate-- to be sure there will be charges induced on these thin edges-- but if it's got a surface area that's a large compared to the area of these thin edges, we are going to be measuring primarily p3-- the component of polarization that's normal to this surface-- and that might have a charge per unit area that is comparable to these other two charged surfaces.
But because the area by design of our specimen is so large, the response that would be most easy to detect, and which would be the largest response, by design, would be p3, which is the charge per unit area on this surface.
So this is an effect we can define for a particular sample, and for a particular special form of the generalized force.
And we then can ask, what is the value of the single modulus that relates p3 to sigma 3?
And that's a question we can ask.
And we can plot that response as a function of direction of a plate that we consider as being cut out of a single crystal, and different orientations, and then ask how this modulus-- which connects the two-- changes with the orientation of x3.
So that is a question we can ask.
And these surfaces are absolutely wild, highly anisotropic, can be identically zero in certain special directions, and they are very interesting, and a lot of fun to look at.
So we'll take a quick look at a couple of those, which won't come as a surprise because they're already worked out for you in the notes.
So we'll take a look at one of those.
And in the problem set, I invite you to amuse yourself by looking at such representation surfaces for two other point groups.
And with that, having kept you til five after the hour I will quit.
Today, we are going to change gears after the first hour to a very different set of topics.
What I wanted to do today was to take a look at notation for space groups.
And this is a question that is far from trivial.
There are conventions, and we'll see the reasons for them with a few examples.
And then I would like to pass around representations of some non-trivial space groups.
First, as they appear in the old International Tables for X-Ray Crystallography, Volume 1.
And then second, to give you a look at the larger, heavier, and much more expensive Volume A, which gives much more information which is both its liability and its advantage.
There are things in there that are not in the international tables.
But there's so much information and so cluttered that it's really overwhelming.
But I'll give you a chance of what their representations of the space group look like.
Let me begin by passing around an example of the information in the International Tables, Volume 1 for one of the orthorhombic space groups.
And it all fits on one page.
It's not terribly complicated.
But it is considerably more complex both in the arrangement of atoms and in the cluster of symmetry elements, compared to the very simple monoclinic space groups which we looked at in their entirety and in considerable detail.
So first of all, some indication of what's on the lower part of the page.
This is exactly analogous to what you've seen and presumably are familiar with for the two dimensional plane groups.
There are many possible origins that could be selected for the 0, 0, 0, position relative to which the atomic locations are expressed.
So the first thing, just below the picture of the arrangement of atoms and the depiction of the arrangement of symmetry elements is a statement of where the origin is.
So it says, origin at the inversion center 2/m.
And if you look at the arrangement of symmetry elements, there is nothing but the representation for screw axes normal to the page.
But there is, off on the right-hand side, a solid chevron, and that is the indication for a mirror plane.
Something again that is nowhere stated, but always assumed is that the two variables x and y or the axes a and b are assumed to run from upper left to lower left for a and the coordinate x and from upper left to upper right for the axis b.
And therefore, plus c comes out normal to the plane of the projection.
Here, for the first time, we're seeing a complex space group that really requires the full range of symbols for representing the atomic locations.
You will see, for each circle, a vertical line dividing it in half.
These are two atoms then that are superposed in projection.
One of them, corresponding to the little comma inside the circle, is the enantiomorph of those circles which are unadorned with that little comma.
And they are, therefore, of opposite handedness.
Alongside of each circle, you will see two symbols giving the elevation of that atom within the cell.
The one at the furthest upper left corner says 1/2 plus.
So this says that the coordinate is z, whatever that is for the representative atom, plus 1/2.
The half of the circle that has the comma inside has a minus sign next do it.
That means that it is at an elevation minus z.
Well, where is the representative atom?
This is tucked just inside the corner of the cell where the vertical line divides the circle into 1/2.
That is a plus, so that's an atom at plus z.
The right-hand side of that circle has a comma.
It's an enantiomorphic motif, and that exist at 1/2 minus z.
So you find labels on all of the atoms in the cell.
There are a total of four clusters of eight.
So are a total of 16 atoms within the unit cell.
If you look at this pattern and reflect on it for a moment, so to speak, you can see that the same cluster of eight sits at the corners of the cell as in the middle of the cell.
Exactly the same.
So therefore, this is a lattice that is side-centered.
There's a lattice point at the corner of the cell and one lattice point, in addition, at 1/2, 1/2, 0, in the center of the cell.
The c-axis comes out of the plane of the paper.
This then is a side-centered c-lattice.
And that is the capital letter symbol that appears first in the symbol for the space group.
It is a side-centered lattice, and the extra lattice point is in the middle of the face out of which c comes.
Then come, in the space group symbol at upper left-- this is the abbreviated form-- mcm.
And what I wanted to go over this example specifically for, there are three directions in an orthorhombic symmetry, no one any more or less special than any other.
So if you look at the arrangement of symmetry elements, there's a 2 sub 1 screw axis coming out in the direction of c. The symbols running left to right, arrows and barbs, the arrows stand for two-fold axes.
The barbs stand for 2 sub 1 screw axes.
And running up and down parallel to the a-axis, again, arrows that indicate two-fold axes and barbs that indicate 2 sub 1 screw axes.
So which is the special direction?
Which one should come first?
It's not like a cubic symmetry, where the direction of the four-fold axis is always along the edge of the unit cell, or a tetragonal symmetry, where the four-fold axis is always taken to be the direction of c. Here, that's not specified.
So again, the first convention is that the magnitudes of the translations determine a, b, and c in orthorhombic.
Because there's nothing more or less special about the symmetry of these three orthogonal directions.
And the convention which we've mentioned earlier is that the magnitude of b is greater than the magnitude of a.
And then seemingly logically, c is the smallest.
So if this is the direction of a and this is the direction of b, then c comes up.
And this is the intermediate length translation.
This is the maximum translation.
And normal to the blackboard is the shortest translation.
What symmetry goes first?
And here again, for orthorhombic, we need a convention.
And the space group symbol has-- I [INAUDIBLE] I use lambda for this, but-- a symbol for the lattice type.
And we've seen that the capital C that comes first indicates that there is a c-centered lattice.
And then what one does for orthorhombic is give the symmetry elements in the order abc.
So the axis that is parallel to a over the plane that is perpendicular to a.
Next, the axis that's parallel to b over the plane that is perpendicular to b.
And then the axis that is parallel to c over the plane, if any, that is perpendicular to c. So then this is a convention for orthorhombic.
And again, this is the only crystal system where conventions this elaborate are necessary.
If this were tetragonal, 4 or 4/m would always come first because that's the direction of c. So even the order of the symmetry elements is different.
But we've got another problem.
If we look at the axes, if any, that are along a, we've got this symbol that stands for a two-fold axis and this single barbed arrow that stands for a 2 sub 1 screw.
So what kind of axis is along a?
There are two different kinds of axes.
So in here, we could put either a 2 or a 2 sub 1.
And if we look at the planes that are perpendicular to a, there's two of them.
There's a dashed line, and this is a glide where the direction of tau is in the plane of the paper so that we're looking edge on in the glide plane and perpendicular to tau.
And then there is a solid line.
That's the symbol for a mirror plane.
This would be a glide in which tau is a long b, so this is a b-glide.
So those are the two sorts of symmetry planes that are perpendicular to a.
Again, there are two of them.
So we could put in either m, or we could put in a for the plane that is perpendicular to a.
So already, there are four different possibilities for the first character.
So obviously, what I'm leading up to is the fact that we're going to have to have conventions of necessity to come to a symbol on which everyone agrees.
I think you get the idea now, so let me move along quickly along the direction of the arrow for a two-fold axis, a barb for 2 sub 1 screw axis.
So again, there are two kinds of symmetry axes that could go in here for the axes that are parallel to b, either 2 or 2 sub 1.
If we look at the planes that are perpendicular to b, there's a dotted line.
And this is a glide for which tau is directed upwards.
We're looking directly along the orientation of tau.
The axis that is normal to the board is the c-axis.
So this is a c-glide.
And then appears a dash-dot line alongside of it.
The dash-dot line is a glide plane where you're looking neither perpendicular to tau nor parallel to tau, but halfway in between.
So this is a glide plane where tau is equal to 1/2 of a plus 1/2 of b.
And that's a diagonal glide represented, just to confuse the innocent, by the symbol n. So two choices for axes along b.
Two choices for the glide plane, either c or n. And finally, mercifully, along the direction of the c-axis, which is easiest to visualize, because things are either along it or parallel to it, we've got only a 2 sub 1 screw axis.
So this is the only symbol which is not ambiguous.
And then perpendicular to that, we have the symbol for a mirror plane, the solid chevron off to one side.
And then the chevron adorned with a diagonal arrow and that is, again, a diagonal glide n. We notice that the center of the cell at z equals 0 is the same as the grouping of atoms at the corner.
So that is a side-centered lattice.
The centered lattice point is coming out of the face which c emerges from.
And so this is a c-lattice.
You have a question?
Professor, for the first term [?
x of the ?]
plane perpendicular to a, why is it a and not b?
I'm sorry.
You're absolutely right.
I wrote it down as a b-glide.
Perpendicular to a is a b-glide.
Good.
Congratulations.
So we need a convention.
And the convention for establishing a hierarchy of symbols is easy to remember.
It's if you have both a two-fold screw axis and a two-fold axis, the order of preference is that the 2 is chosen in preference to the 2 sub 1 screw symbol.
So by this greater than, I mean the preference for choosing it is greater than the symbol to the right.
So we get rid of the 2 sub 1 here.
We pick the 2.
Get rid of the 2 sub 1 here, and that stays at 2 sub 1.
For planes, you pick m in preference to an a-glide, in preference to a b-glide, in preference to a c-glide, in preference to a diagonal glide, in preference to a diamond glide, assuming you have a choice of two different planes.
So in this case, we picked the m in preference to b.
Here, we've got a c-glide and an n-glide.
We picked the c, and it's in preference to the n. And here again, for the plane that is perpendicular to c, we picked the m in preference to the n. So the full symbol would be C, 2 over m, 2 over c, and 2 sub 1 over m. Now I'll peek and see if I got this right.
Yes, I did.
How about that?
And the short symbol would be simply Cmcm, which indeed is what we see listed in the upper left-hand corner of the page.
This is a space group based on the point group 2/m, 2/m, 2/m, which is D2h.
And there is the large number 17 in the superscript, which is the 17th orthorhombic space group, which Schoenflies was able to derive in the order in which he did them.
Any questions?
Yes
If you look at the other symbols to describe the space group, would we construct the same space group using that function [INAUDIBLE]?
Yes, you could.
And let me say, yes, it's possible.
But also to answer a question, you might have about the short form of the symbol, we're throwing away information.
We've thrown away the fact that there are two-fold axes, and a pair of directions, and 2 sub 1 only in the third direction.
Is it possible that two space groups could have exactly the same three symbols in the short form?
And the answer is no.
Because if you think about it, you start with a lattice type.
If you put a mirror plane in one orientation, a mirror plane in the other orientation, that really is all you have to specify in order to determine every other symmetry element that's present.
You simply combine those operations, and you find that when you've taken these symmetries and rotations and combine them with the lattice translation, it comes out to one and the same unique result.
So your question was what happens if you take another three symbols.
You end up with the same result.
Three of them, really, are enough to specify what the group will turn out to be.
Now the other thing I wanted to do was to show you how remarkably the symbol for the space group will change if you take a different set of axes simply because the relative lengths of the cell edge for this arrangement of symmetry elements have changed.
This takes dominance.
This determines the labels that go to the axes.
And then the symbol for the plane group, and what is a b-glide, and what is a c-glide follows from this choice of axes.
So let me take another example.
And I think if you follow what I've done so far, I can zip through this a little more quickly.
Somebody pick a new direction for this axis.
Let's not pick a.
We have used that for this.
So somebody pick a b or a c. Do you want to pick a b or a c?
[INAUDIBLE]
b.
So we're going to change this direction to b.
Somebody want to pick a label for this axis?
c
c. And nobody gets to pick this axis.
If you think you can, you're out of here.
Because that now has determined the coordinate system.
So we've got b this way, c this way.
We want a right-handed abc convention.
So a goes up.
So abc right-handed system.
So exactly the same arrangement of symmetry elements except that we've now changed the labels on the axes.
So we have changed the name that we have to apply to the glide planes.
So let me now proceed to write down here the new collection of symbols which would appear for this setting of exactly the same space group.
First of all, the centered lattice point is now in the middle of the face out of which a comes.
So this is an a-lattice, side-centered a.
If we look along the direction of a, all that we have for an axis is 2 sub 1.
Then we have these two planes.
One is an m. One is a diagonal glide.
So that is m or n. If I look along b, I have the choice of either 2 or 2 sub 1 as both are present.
Perpendicular to b, however, is a glide for which tau is along the direction of c. So that is a c-glide.
And then the other symbol for an axis that's perpendicular to b is a mirror plane.
So it's either c or m. Then finally, along the direction of c, I once more have two-fold axes and 2 sub 1 screw axes.
And then perpendicular to c is a diagonal glide.
And this is a glide.
The dotted line indicates that the direction of tau is along a.
So I have a choice of either a or n in here.
So putting down the symbols in the order of preference, this would be A, 2 sub 1 over m, 2 over m, 2 over a.
Or the short form of the symbol would be Amma, which I submit doesn't look anything like Cmcm.
And when you start out in diffraction, many people have a first disconcerting experience.
You spend probably the better part of a month taking single crystal diffraction patterns.
You come up with a trial space group for your particular material.
And then you go to the International Tables, and you say, oh, it's not in there, what did I do wrong?
And you go back, and you check all your calculations.
You look once more at all of the systematic absence of reflection.
See, I have it right.
Why is it not in there?
Well, the answer is more than likely that you have a different setting for the labels that you applied to the axes.
And you have, therefore, metamorphosed the symbol for the space group from one form to a symbol that is very, very different.
How do you tell, when you have a particular space group, what all of the possible symbols for one and the same symmetry might be?
So let me now pass out two things.
If you have a question like that, you can bet your bottom bippy that it's in the International Tables if you look in the right place.
So first of all, let me do something I should have done early on.
I meant to hand out a copy of the space group that I drew up here.
If I can find it again.
It's buried with all of the stuff that I brought in.
So I do have a copy for you, but it's gone at the moment.
Let me pass out a listing of all of the possible variants of the space groups for the 230 three-dimensional space groups.
And it starts out simple for triclinic symmetries.
There's only two possible symbols period, P1 and P1 bar.
And then when you come to monoclinic, the Schoenflies symbol has the marvelous property that it is independent of the orientation and labelling of axes.
It depends only on the point group that the crystal has.
For monoclinic, either C2 or Cs-- that's a mirror plane-- or C2h-- that's 2 over a horizontal mirror plane.
But the labels can still change depending whether in the first setting either a, b, or the diagonal of the oblique net has to be labeled a and c. So at the top of the column, you'll see the permutation of the two symbols a and b or b and a.
And again, the change is significant.
Number eight, as indicated in the left-hand column, can be Am or Bm.
And for the groups that are based on 2 over m, you can see, again, there are different symbols.
A, 2 over b, A, 2 over a.
Again, not at all similar.
Next, come the orthorhombic space groups.
And you can amuse yourself by looking through those.
Again, what is called the standard setting under the heading symbols for various settings, this is the one that's listed in the tables.
But then for various permutations of a, b, and c, you can have five other possibilities.
And you can marvel at how the symbol for the space group changes, as you do nothing but change the labels a, b, and c that goes onto the axes.
So there are tons of orthorhombic space groups because you have three different axes to play with, and three different planes, and four different lattice types.
So there is an absolutely mind boggling collection of orthorhombic space groups.
It is the most densely populated crystal system.
For tetragonal, no real alternative.
Something like P4, 4 is in the direction of the c-axis.
There are two-fold axes parallel to that, but the symbol that is used for the symmetry part of the space group symbol looks very, very much like the symbol for the point group.
So these are pretty much self-explanatory.
Notice that there are two kinds of lattices for tetragonal crystals, either primitive or body-centered.
So you see families with P for primitive or I for body-centered.
But notice how many you can get out of a single type of axis and a single lattice.
Numbers 75, 76, 77, and 78 are a four-fold axis and a primitive lattice, a 4 sub 1 screw, a 4 sub 2 screw, or a 4 sub 3 screw.
Then you do the same thing with I.
And so it goes.
Then it picks up at the top of the page again.
And you see, again, different possibilities for the symbols.
Continuing still on-- and I think you have the idea now-- same for trigonal crystals.
They chose to list those separate from hexagonal.
And then hexagonal symmetries.
The number is fairly large here because when you take an axis and add it to a primitive hexagonal lattice-- for number 168, for example-- you get P6.
Change the six-fold axis to a screw axis.
There are five different types of six-fold screw axes.
So there's a P6 sub 1, a P6 sub 2, a P6 sub 3, a P6 sub 4, and a P6 sub 5.
And then finally, cubic.
A fair number of cubic symmetries.
But the edge of the cell is always along the direction of the four-fold axis or the two-fold axis, depending on whether the symmetry is based on 2, 3 or 4, 3, 2.
So there are really no alternative symbols for different settings of the symmetry elements relative to the edges.
So there you have them.
All 230 of our cast of characters.
But in forms of the very different mantles in which they can be cloaked.
What else did I want to do?
I want to contrast what is done for us in the old Volume 1 of the International Tables.
And I will pass out an example for one of the tetragonal space groups.
This is the one with maximum symmetry 4 over m, 2 over m, 2 over m. Sorry.
Can you pass those back?
And this is something that is analogous to P4mm in two dimensions with the four-fold and two-fold axes extended parallel to the edge of what is now a tetragonal cell.
And then there are two-fold axes perpendicular to the four-fold axis.
So actually 4 over m, 2 over m, 2 over m has been dropped in at a lattice point of a primitive tetragonal lattice.
The chevron in the upper right of the arrangement of symmetry elements is the mirror plane that's perpendicular to the four-fold axis.
The other mirror planes are the same as in the plane group P4mm.
And the glide plane, which now would be called a diagonal glide, is exactly the same as in the two-dimensional symmetry.
Notice, however, the exquisite number of general and special positions that are present in the space group.
The general position has the letter U.
You've gone through almost the entire alphabet before you've labeled all of these positions.
And again, either on the mirror planes, but now there's a vertical mirror plane, a diagonal vertical mirror plane, a horizontal mirror plane, mirror planes halfway along each of the axes, mirror planes halfway along the diagonal translation.
So there are lots of positions of point group m. Lots of positions with point group 2mm.
And then finally, 2 over m, 2 over m, 2 over m. And the highest symmetry, 4 over m, 2 over m, 2 over m, which occurs at the origin, at the center of the cell, and at the positions 0, 0, 1/2 and 1/2, 1/2, 1/2.
Then I'd like to pass around one more example of a space group from the International Tables.
And I know I copied it, and I don't have it with me.
I left some stuff behind, which we'll get after we take our break.
No, here it is.
This is an example from Volume 1, the older edition, for one of the cubic point groups.
And this is a very high symmetry.
This is symmetry 4 over m, 3 bar 2 over m, the highest symmetry cubic point group dropped into a body-centered lattice.
So this turns out to be, in the long form, I, 4 over m, 3 bar, 2 over m. Let me pass these back.
So the first thing you'll notice is no picture of symmetry elements.
How do you draw something where the symmetry elements are not merely parallel to the plane of the depiction or perpendicular to it?
Here, you've got mirror planes that are inclined to the paper, axes that are inclined to the paper.
How do you represent them?
You'll notice also the enormous number of atoms in the general position.
96 atoms.
Drop in one atom at xyz, and all hell breaks loose.
You get 95 other atoms.
And again, when the lattice is centered, they do not list all 96 sets of coordinates.
They denote at the top of the page 0, 0, 0, 1/2, 1/2, plus.
That means because the lattice is body-centered to those 48 positions that are listed, you add 0, 0, 0, which is easy to do, and then add 1/2, 1/2, 1/2 to x, y, and z.
And these are the atoms that hang at the centered lattice point.
Again, because the symmetry is so high, surprisingly, there are not that many special positions.
Because all of the symmetry elements can serve as special positions are related by the symmetry that's there.
So this is all that you see for the body-centered lattice with 4 over m, 3 bar, 2 over m dropped into it.
And again, you can describe some very, very complicated crystal structures with a very brief set of notes, give the value of the single lattice constant a.
And then say, you've got an atom in the general position 96l with x equals something, y equals something, z equals some number.
And another atom in position 16f x, x, x, and just the value of x would be given.
So you've described a structure that has roughly 125 atoms in it with just that modest specification of atomic positions.
Let me now pass out for you some samples of what the space groups look like in Volume A of the new International Tables for Crystallography.
Not X-ray crystallography, but just crystallography in general.
And I think you will be blown away because the amount of information that's there is almost suffocating in its detail and its density.
The first sample space group that is given here is, not by coincidence, one of the tetragonal space groups, the one that we just discussed, P, 4 over m, 2 over m, 2 over m. They've done some very useful things.
They have specified the asymmetric unit within which you need specify coordinates of atoms.
And here, they say that you have to tell what's in the range x equals 0 to x equal 1/2, y equals 0 to y equals 1/2, and z equal to 1/2, but greater or equal to 0.
Then they give a representative set of symmetry operations which generate all of the atoms within the unit cell.
The symmetry operations are not independent, but they list 16 symmetry elements.
And now if you look to the next page on the list of 16 atomic coordinates in the general position, you will see a number in parentheses in front of each one.
The 15, for example, tells you that you get this atom from the atom at xyz by the operation of symmetry element 15, which turns out to be a mirror plane parallel to the x minus xz direction.
And this is very, very useful if you want to describe a structure and you've labeled your atoms silicon 1 and silicon 2 if they're two different kinds of silicon ion in the structure.
And then you're talking about a silicon, oxygen, silicon bond, and it's not at all clear which of the symmetry related atoms are involved in that bond.
Well, the presentation of a standard numbering of atoms related by symmetry lets you say that, for example, this is the bond between silicon superscript 5, oxygen, silicon superscript 14.
And you can identify the coordinates of the atoms that went into that particular bond angle or bond distance.
And that's very nice.
All sorts of information a la group theory.
The maximal non-isomorphic subgroups, the maximal isomorphic subgroups of lowest index, the minimal non-isomorphic supergroups.
At one time, I may have known what all that meant, and I've long since forgotten and have never regretted having forgotten.
So this is very exotic, higher-level information about the subgroups that exist for the space group.
Notice how much more information is present though, compared to the depiction of the same space group in the earlier International Tables for X-Ray Crystallography.
Well, more or less at random, I picked out some other space groups.
I, 4 sub 1, amd.
So this is 4 over m, 2 over m, 2 over m in which the 4 has been replaced by a 4 sub 1 screw axis.
The mirror plane perpendicular to the edge of the cell replaced b an a-glide.
The mirror plane for the second mirrors that's perpendicular to the cell edge, the a-glide is diagonal.
And then perpendicular to the diagonal, two-fold axis is a d-glide.
Again, the different operations are identified by a number, the glide plane, and the inversion centers, and two-fold axes that are present.
And then in the coordinates of the general position, you are given the nature of the symmetry element that produces the atom at a particular set of coordinates from the atom that's in the general position.
Now again, the maximal non-isomorphic this, the minimal non-isomorphic supergroup, and so on.
Lots of information.
Then some examples of still higher symmetries.
P6 sub 3 over mmc based on 6 over m, 2 over m, 2 over m with the 6 replaced by a 6 sub 3 screw axis and one of the mirror planes replaced by a c-glide.
If you go on, here comes the really exciting part.
The next page gives you a depiction of a cubic space group.
Heroic.
What they do is show stereographic projections at locations, such as 0, 0, 0 and 0, 1/2, 1/2, I believe it is.
And then the best part of all, down at the bottom of that page, you see the arrangement of atoms in stereo.
And if you're one of these people who can stare at the thing cross-eyed and let your eyes sort of blonk out, you can watch the two halves merge.
And all of a sudden, zing, the thing leaps out of the page at you in three dimensions.
If you don't have that ability to cross your eyes and see stereographic projections, you have to get a little viewer.
But nevertheless, if you want to see it in three dimensions badly enough, you can do it.
So there you are.
For the first time, pictures of the general position in cubic space groups.
I think that's kind of fun.
Sometimes I gaze at it for so long I'm afraid my eyes are going to get stuck.
And then I'll be in real trouble.
Here's another one.
P 4 over m, 3 bar, 2 over m. You notice the problems that they have in indicating the orientation of three-fold axes which are not parallel to or perpendicular to the paper.
You have to do that with a little stereographic projection.
And then finally, after we've done a couple of more cubic space groups, you come to the one that I gave you the handout from the International Tables for X-Ray Crystallography.
I, 4 over m, 3 bar, 2 over m. And look at all the information that is there that no attempt is made to give you in the earlier tables.
So this is for better or for worse what space group tables look like.
This is the information that hopefully you'll be equipped to use.
If nothing else, if you can't derive these things or reproduce the arrangement of symmetries from the symbol, if you ever have to construct the atomic arrangement from a material from the crystallographic notation in which the atomic arrangement is provided, you can hopefully go to the International Tables, know where to look, and how to go about identifying the coordinates of all of the atoms within the unit cell.
Timed myself to finish just at five of the hour.
So let me, before you leave, give you one final problem set on symmetry, problem set number 10.
And this is related to identifying what the symbols used to give atomic locations represent.
First problem asks you to look at a pair of atoms and determine the symmetry element which has related them.
The second problem on the second sheet asks you to do what we did here for this orthorhombic space group.
Determine the symbol when a, b, and c are particular translations of the three that are unique.
And then change the orientation, and see what the space group symbol morphs into.
Let me finish with one final handout.
And that is could you please summarize what we spent the last month and a half doing?
Gladly.
I have summarized everything that we did and the theorems that we used to derive it on one piece of paper.
So all of symmetry theory is here admittedly written in a rather tight hand.
But here is an indication of everything we did to get from a definition of basic operations, a flowsheet that gets us down to 230 three-dimensional crystallographic space groups.
So I'll pass that around for your awe and amazement.
Sorry I gave a big pack out on the right-hand side of the room.
They'll come around to you.
So let us take our usual 10 minute break.
All right, now for something completely different.
Before beginning though, I would like to raise a procedural question.
It was my intent that quiz number two, which is like a little more than a week away, was going to be part symmetry and part tensors.
I don't want to have your fortunes based by weight of 2:1 on symmetry theory as opposed to properties.
What I would suggest, and you can express your opinion either way, is that we postpone quiz number two by maybe as much as 10 days, so that it can be half symmetry and half the introductory discussion of tensors.
And if that does not create a conflict with some of your other classes, I would propose that as a suggestion.
If not, we can just have on the quiz what we're going to cover of tensors in the next couple days and have the rest symmetry.
So what I am suggesting is the quiz was scheduled for a week from November 8, which is a Tuesday.
What I would suggest is putting that off to the 18th, and that will give us four extra lectures-- six extra lectures-- on tensors and we can make the quiz 50-50
What does that do to quiz number three?
Quiz number three is going to stay in place because it's right at the end of the term
What about the subject in three?
It will be all tensors.
It will be all tensors and properties.
And the third quiz is set up for longer than you think.
It's December 8, so it'll be about three weeks
So everything from here on?
Yeah.
Does that make sense?
Or is somebody going to have a big, traumatic responsibility at that point?
I don't know, but I think it would be probably be easiest for us not to have it three days before close of exams
OK, OK, that's what I wanted to hear.
How about if we went just a little bit further then and did it not on the-- when was the suggestion?
The 18th.
Let's say we did it on Tuesday the 15th?
Does that conflict with anything?
That's a little bit
We were doing it on the 18th
Is that Friday?
[INTERPOSING VOICES]
I'm sorry, 18th.
I said it would be 17th.
Excuse me
That definitely would be a little better
OK, let me allow you to think about that, and that would be pushing it back one week and maybe it'd be-- we'll see how much material on tensors we cover.
So let's table it for now but let's say, tentatively, let's consider seriously moving it to the 15th.
OK?
But you guys have final say if there's a preference that you have.
OK let's talk about properties in first a general and rather philosophic way.
When we discuss properties we very commonly lump together a set of behaviors which have some common phenomenology.
So for example, we will talk about mechanical properties.
And that's a basket in which we place many different sorts of behavior.
We place things like fracture toughness, yield strength, elastic constants, a whole variety of things involving strength deformation and so on.
Or another thing we very often lump together in a basket is something called electrical properties.
And we talk about electrical conductivity, ionic conductivity, electronic conductivity.
We talk about dielectric constants, we talk about permittivity and permeability, magnetic properties-- a whole bunch of other classes of behavior, which have some sort of root or description in terms of electromagnetism.
There are though a lot of strange aspects to properties.
There are some properties that are determined for material which no longer exists when you're through determining the property.
You have a sample you want to know what the yield strength is.
So you put the gradually increasing load on it until finally, POP, it breaks.
And now you know what the yield strength is for a material that no longer exists.
Another example is an old one.
Back before the days of X-ray diffraction when people would try to characterize material, anything that is very intensely colored usually looks black, and that's not a very definitive property.
The color of something is a very prominent characteristic.
So for mineralogists, in particular, anybody going out into the hills and wanting to be prepared to determine what a particular rock that they tripped over was had something on a rope around their neck called a streak plate.
And what the streak plate was was a little rectangle of porcelain.
And they would take this black mineral and rub it on this piece of porcelain and it would leave a mark-- that was called the streak.
And what this was was actually little bunch of fragments that rubbed off on to the porcelain because the porcelain was white and the fragments were very small.
A black piece of rock could give you a streak that was brown or green or deep orange.
And so you really were determining the color of the material when it was in a form finely divided enough that light could pass through it.
So the streak test was a very important diagnostic for determining minerals.
There are some properties that we refer to as structure sensitive.
In the sense that their value-- conductivity is a good example-- the value of ionic conductivity depends on the impurities and point defects that are present in the material.
So to say that the ionic conductivity of a certain material is such and such, you have to specify the purity and perfection of the material or the property's meaningless.
There are some properties that are not even single-valued.
And the best example there are dielectric or magnetic properties.
If you plot the magnetization of a material, the magnetic moment per unit volume as a function of the applied magnetic field B, the material is initially non-magnetized.
When you apply B, you get a magnetic moment per unit volume that eventually saturates.
And then if you remove the magnetic field, the material keeps some of its magnetization.
It doesn't go to zero when you reduce the field to zero.
You have to reverse the field to get the property to go to zero and then magnetize it in another direction.
And if you continue to cycle the magnetic field, you get a behavior like this.
This is hysteresis, and this type of behavior is called generally hysteretic.
Same thing would be a relation between the displacement vector and an applied electric field.
But clearly you can have any value of the magnetization in this range.
And if you relate the magnetization as a function of the applied field, you can get any value of the proportionality constant you like between a negative maximum and a positive maximum including zero.
So there's a property that's not even single-valued.
Depends on the past history of the sample.
And then there are even more peculiar properties.
There are properties that are-- we call them composite properties and very often qualitative.
What do they mean by a composite property.
Let me give you one example-- The property fuzzy.
If I say fuzzy, you know exactly what I mean.
It means something that has a diffuse reflectivity, something that has a surface texture that's yielding.
It's not like rubbing a wire brush.
It's soft and giving, a whole collection of different properties.
But yet when I say fuzzy, you know exactly what I mean.
Let me not be so facetious as talking about a fuzzy property.
People such as ceramists or powder metallurgists very often will take a powder of a material and consolidate it by compacting it and heating it, very often subject to a compacting stress.
And when you do that little necks grow between the particles and they hold together and the material is centric.
So you refer to a property of a powder as being centerable or the centerability of a powder.
You know exactly what somebody means by that.
But what does it depend on?
It depends on the surface energy, it depends on vapor pressure, depends on bulk diffusion coefficients.
It depends on surface diffusion coefficients.
And all of these things have to be just right to make the powder something that easily densifies upon heating in the application or not of pressure.
So you know exactly what I mean by centerability, but it is a very complex property.
And was one that really was not understood until probably the late 1950s.
Until then, it was an art that was entirely empirical.
And it was somebody here at MIT, a fellow named Robert Coble, who developed a theory of centering that was the first really workable theory that described densification by heating and compaction.
OK, composite property that depends on many different individual properties of the material.
OK, what we are going to examine here are equilibrium properties that can be rigorously defined and measurable.
So we're going to leave out of the picture things like fuzzy and centerability.
And let me give a very nice, obscure definition of what I mean by a property.
And what I mean by a property, in terms of a formal statement, is the response of a material to a specific change in a given set of conditions.
So the response of the material to a specific change in a set of conditions that relates independent properties and dependent properties for a particular process.
So I'll state that again because I think it's terribly elegant.
So a property is the response of the material to a specific change in a given set of conditions that relates independent and dependent quantities in a particular process.
Now there are a lot of properties that have their roots solidly embedded in thermodynamics.
So let me give you a few of those.
What we'll talk about when we specify a property is something that we will refer to as a displacement.
We'll talk about a generalized displacement in response to a generalized force.
So some of the thermodynamic quantities that are related in this fashion-- if we list some forces and some displacements, the thing that happens as a result of that stimulus that's applied to the material.
Temperature can be regarded as a force that results in all sorts of processes as a result-- thermal energy flow, thermal expansion, all sorts of things.
But one of the things that will happen in response to a temperature change thermodynamically is an entropy.
So you can view entropy as a generalized displacement resulting from the application of temperature as a generalized force.
Another example is electric field and the thing that happens there, and electromagnetism, is the quantity, D, which is displacement.
Still another example, stress, and the result of applying a stress, among other things, is a strain.
What is special about these forces in this displacement is that their product is, in each of these cases, energy, the change in internal energy of the particular body.
And when that is the case, these are said to be conjugate-- a conjugate force and displacement.
And there are other examples that one can come up .with.
Now to talk about a specific set of properties.
Very often, and this crept into the earlier discussion, very often the thing that we do to a material, as a generalized force, is a vector.
So very often the thing that we do to the material, apply an electric field, apply a magnetic field, apply a temperature gradient, apply a tensile stress, it has the characteristics of a vector, magnitude and direction.
And in many cases, the thing that happens as a generalized displacement, we may call this in general, q, is also a vector.
An example is if we apply an electric field as a generalized force, one thing that might happen is a current flow, which is also a vector.
And we are accustomed to writing, q, the generalized displacement, as a proportionality constant, sigma, which is in the case of electric field and current flow, the electrical conductivity.
Let me call it, in general terms, proportionality constant, a, times the applied vector, p. Is this something we would like to stick with as a general relation?
Well let me submit that writing an expression of this form makes an inherent assumption.
Namely that this will be true for small, and we'll have to define for each property what we mean by a small, applied vector.
Let me give you an example.
If p were-- the applied vector was electric field, and the resulting vector was current flow, and the relation between those is the conductivity-- a relation of this sort says that if you double the field, you double the conductivity.
You triple the field, you triple the conductivity.
Obviously this can't go on indefinitely because all a sudden, POW, dielectric breakdown.
The sample evaporates at the smoke.
And again, you have the property of a material which no longer exists.
So a lot of properties, we inherently assume that the applied vector is small in order to write something in an expression of this form.
So for conductivity, dielectric breakdown is going to destroy the nice linear relation between current flow and electric field.
That's not always the case.
Let me give you an example of another property.
And this is a property, magnetic susceptibility, which relates the magnetization, which is magnetic moment per unit volume, and relates that to an applied electric field-- an applied magnetic field, H. And the proportionality constant, represented by a Greek chi, is the magnetic susceptibility.
So let me give you a specific example.
Suppose we have a chunk of glass and the glass contains a dilute concentration of iron.
And iron carries a permanent magnetic moment.
And the reason I want to make it a dilute concentration of iron is I don't want these magnetic moments close enough that they can interact with one another.
I want this to, therefore, be a dilute system.
So we have different iron atoms in this.
Each iron atom has a magnetic moment.
And then we put this in a magnetic field, H. And the magnetic field acts on each of these little magnetic moments just as though it were a compass needle.
And so it will try to take each of these moments and drag it into coincidence with the magnetic field.
But at a finite temperature, temperatures making these magnetic moments jiggle around, so at a finite temperature, the magnetic moments will just not simply zing into coincidence with the magnetic field.
You'll have to increase the magnetic field to make it larger.
When that happens, more and more of the magnetic moments will come into alignment.
And if you put on a really strong magnetic field, then every single magnetic moment will be dragged into exact alignment with the magnetic field and the system has saturated.
There's no way you could squeeze further magnetic moment per unit volume out of it.
So again, you would expect, for this particular property, magnetic susceptibility of a material, if you plot it as a function of the applied magnetic field-- linear maybe be at low applied fields-- but eventually if you make the field strong enough, the system is going to saturate.
And then again, no longer will the direction of the magnetization, and that magnetic moment, be parallel to H. But the property becomes nonlinear if the applied vector is strong enough.
This is an example of a property where you would not go wrong at all by stating direct proportionality.
For ordering to occur, temperature and applied field go hand in hand.
Increased temperature tends to create more disorder.
Increased magnetic field tends to align the moments.
In the days when MIT had a national magnet laboratory, the magnet laboratory held the record for the strongest magnetic field ever produced artificially by man.
And if you took that magnetic field and applied it to this system of dilute iron in a glass, you would have to lower the temperature of the sample to about 3 Kelvin before you would begin to see saturation.
So even the strongest field that you could produce in a laboratory environment would not succeed in producing non-linearity until you cooled the sample down almost to absolute zero.
So here would be a case where under any practical consideration whatsoever, assumption of strict proportionality would be right on the money.
You'd be absolutely correct.
But there's another assumption built into this statement.
P, the generalized force, is a vector in the cases we're discussing now, and q is also a vector.
So when we write an expression of this form you're making another assumption.
And that is that the vector displacement that results is always exactly parallel to the vector that you apply.
Do I make a big deal out of this?
Isn't that always going to be the case?
I mean whoever heard of taking a piece of metal and putting an electric field on it in this direction and having the current run off in this direction.
It's absurd.
Or maybe it isn't so absurd.
So let's think of some of the atomistics of this process.
Now, since I know I'm among friends, I will not hesitate to display my ignorance, total ignorance, of polymer chemistry.
So suppose we had a piece of polymer.
That's what a polymer molecule looks like.
It's a more or less linear molecule, and so these might, in a very highly ordered polymer, be chains that lined up like this.
And suppose we now put an electric field on this polymer and asked how the current will flow.
Again, it would not be absurd to say that an electron sitting on this polymer chain in response to this field would be constrained only to flow in the direction of the chain and would find it rather difficult to hop from one chain to another.
So maybe, just maybe, we could apply an electric field to a polymer and it would have a flow in this direction, that would be pointing in this direction, and the current flow, J, would be in that direction.
So should we maybe rethink this idea of the direction of the generalized displacement being not parallel to the direction of the generalized force.
OK, these are hypothetical examples.
Let me now give you an example of a real property for a real material, for which the thing that happens, the vector that happens, is decidedly not parallel to the direction of the applied vector.
OK, thermal conductivity is something that relates a heat flux, usually represented by the symbol K, and relates that to a temperature gradient, dt dx, which is a vector.
The thing that gives rise to thermal conductivity can be either propagation of radiation as in propagation of light traveling through a transparent material.
But the other mechanism for thermal conductivity is modes of lattice vibration.
When a material is hot, the atoms are jiggling around and you can make the displacement of an individual atom be represented by a sum of waves moving in all different directions with a variety of wavelengths in a variety of amplitudes.
Take all those waves, add them together at a particular time, and you get the displacement of the atom.
Propagation of heat by this mechanism, by modes of thermal vibration, can be very, very anisotropic and probably the best example of this is-- get a single crystal of graphite.
And have the layers, the graphite sheets, which are hexagonal rings in which each carbon has three neighbors, and have the sheets be parallel to the surface of an extended two-dimensional slab.
Now we can't do that.
Single crystals of graphite don't occur in sizes like that.
But what you can do is make a material called pyrolytic graphite that you make by having a reaction in the vapor phase and having the soot that's formed settle down onto a substrate.
And what happens is you nucleate a graphite crystal in one orientation, and that grows preferentially with its layers parallel to the surface on which it's nucleated.
And these nuclei occur at random.
So you get a bunch of single crystals of graphite, which all have their layers, their three coordinated layers at parallel, but they are oriented at random about their c-axis.
And that's a material that's very easy to prepare.
And it's called pyrolytic graphite, and it has interesting applications and properties.
OK, now I'm going to suggest an experiment to you and I'll caution you, please do not try this at home.
Get yourself a piece of pyrolytic graphite.
Put a Bunsen burner underneath it.
Play the Bunsen burner on the bottom of the sheet of pyrolytic graphite.
And to determine temperature, we'll take a ball of cotton and put it directly above the flame.
And then take another ball of cotton and put it down at the end of the sheet.
If you do that and bring up the Bunsen burner, this piece of cotton will sit there and sit there and sit there and sit there.
And this piece of cotton will instantly burst into flame.
So here's a case where dt dx, the thermal gradient, points in this direction.
And the heat flow goes off in a direction at right angles to the temperature gradient, almost exactly at right angles.
It turns out that the thermal conductivity in this direction, the value of K in the-- I want to put crystallographic directions on this.
But K parallel to the layers is equal to 10 to the third times K, perpendicular to the layers.
So there's an anisotropy of the property that amounts to a factor of 1,000.
Very dramatic.
So what I'm leading up is that in a general relation between a generalized displacement, p, and a generalized force-- I'm sorry, I'm using q-- p and q-- we just can't simply put some constant in front.
Because that assumes inherently that the direction of what happens, the vector that happens, is exactly parallel to the direction of the applied vector and that generally is not true.
The difference may be small but it has to be taken into consideration.
All right, let me now suggest a way of patching this up.
What I'm going to do is assume-- and it is an assumption-- we're going to assume that each component of the vector that happens is given by-- in fancy terms, the vector that results as the generalized displacement is given by a linear combination of every component of the vector that's applied.
And that's what we've defined as the generalized force.
So how would we express this analytically?
I am going to use as my coordinate system, not x, y, and z as we usually do, I'm going to call it x1, x2, and x3.
The reason is we'll see some unique properties of these indices that are very useful algebraically.
So I'm going to now take my applied vector, p, and I'm going to assume that it has three components, p1, p2, and p3 along x1, x2, x3, respectively.
My coordinate system, although I did not state it explicitly, is going to be a Cartesian coordinate system, not a crystallographic coordinate system.
So what I'm proposing here is that we take each component of the resulting vector q-- let's say q1-- and we'll assume that that's given by a linear combination of all three components of the vector p. So it'll be some number times p1 plus some number times p2 plus some number times p3.
And those numbers in general will be different.
So let's say I call the coefficient a, and now I'm going to define a convention that will stay with us.
I'm going to define each of these coefficients, a, in terms of the index of the component of the generalized displacement which is being computed, and the coefficient modifies the component of the generalized force for that particular term.
So I'm going to call this a11, where the 1 goes with this and the 1 goes with this.
I'm going to call this a12, so that the 1 again says it's a contribution to q1.
The 2 says that this term modifies p2.
.
And this similarly would be a13.
For the term component of the generalized displacement, q2, I'll use the coefficients a21 times p1 plus a22 times p2, plus 23 times p3.
And q3 similarly will be a31 times p1, plus a32 times p2 plus a33 times p3.
So I've got three simultaneous equations then, one for each component of the vector that results, the generalized displacement.
So I can sum up this set of three equations by saying that the i-th component of q, where this is some particular component, q1, q2, or q3, is given by the sum over j from 1 to 3 of aij times p sub j.
So I've got ai1 times p1, ai2 times p2 plus ai3 times p3.
So this is in a nice, compact little nugget the expression that we are assuming will apply for all sorts of physical properties in which this is a vector and this is a vector-- electrical conductivity, magnetic susceptibility, thermal conductivity, and so on.
OK, this is a bad point to introduce something as hairy as what comes next.
But I've got one minute left, and I think I can do it.
I'm going to introduce something called the Einstein convention after old Albert Einstein himself.
I'm sorry, but nobody said everything this term had to be easy.
So let me introduce, if you're ready for it, the Einstein convention.
Old Albert, I think, was just as lazy as anybody else.
And he said it is going to be a bloody pain in the butt-- I don't know if he put it exactly this way.
It's going to be a pain in the butt to write this summation every time we want to combine three terms in a linear combination.
So the Einstein convention is let us throw out the sigma and write this expression just as qi times aij p sub j.
And whenever we see a subscript repeated, summation over repeated subscript is implied.
OK so that's the Einstein convention.
It will save us the trouble of writing in a lot of summation signs.
But I would caution you that this convention, compact and convenient as it is, can define some polynomials that are nightmares.
So let me give you one example of this, and this is actually a physical property.
Let me say that Cijkl is given by ai capital I aj capital J ak capital K, al capital L times let's say D, capital I, capital J, capital K, capital L. That actually, believe it or not, means something physically and we're going to get to that in due course.
But what this is, taking the Einstein convention into account, this is a quadruple summation over capital I, capital J, capital K, and capital L. These are very often referred to in this business as dummy indices, meaning that they don't really mean anything physically.
They're just indices of summation.
I would caution you that this is a term that does not permute.
If I say those are dummy indices, it means one thing.
If I say those are indices, dummy, it means something completely different, and you're apt to get a poke in the nose.
OK so these are dummy indices.
Now, what does this represent?
I won't say what it represents physically, but this consists of terms in four variables times a coefficient.
So there are five quantities in each term.
If I sum over capital I, J, and K from 1 to 3, there are going to be 81 such terms, each with five elements in each term.
So it's going to be on the order of 405 terms in this summation, and we've collapsed the whole thing down, 405 terms in this nice summation.
So it is a great facility for writing down expressions explicitly, but these terms can hide a terribly, terribly complex polynomial.
All right, I think that's a good place to quit.
It continues to amaze me, though, how some absolutely trivial convention, when first proposed by a great man, will carry that man's name no matter how stupid it is.
So this is called the Einstein convention by everybody who works in this field.
Another one, just briefly to finish up, when you discuss dislocations, you talk about a circuit of steps around the dislocation.
That's called a Burgers circuit.
And if you do the same circuit in a corresponding piece of material that doesn't have a dislocation, if this circuit closes, this circuit fails to close by something that's called the Burgers vector.
Every book on dislocation theory says, Shockley called this good material.
I'm sorry, Shockley called this good material, and Shockley called this bad material.
Big deal-- good material, bad material because it's got a dislocation in it.
But that is identified with Shockley's name in every discussion of dislocation.
I'm just jealous because nobody has ever said according to me, such and such is-- all right, see you later in the week for more great things.
All right, I would like to then get back to a discussion of some of the basic relations that we have been discussing.
We didn't get terribly far, but I'd like to start with the Cartesian coordinate system that we set up.
Rather than using x, y, and z, I'm labeling the axes x1, x2, and x3.
And we'll see that the subscripts play a very useful role in the formalism we're about to develop.
Now, the first thing we might want to specify in this coordinate is the orientation of a vector and its components.
So let's suppose that this is some vector P. And what I will do to define its orientation is to use the three angles that the vector makes, or the direction makes with respect to x1, x2, x3.
And we could define these angles as theta1, that's the angle between the direction and x1, theta2, the angle between our direction or our vector and x2, and finally, not surprisingly, I'll call this one theta3.
So the three components of the vector could be written as P1, the component along x1 is going to be the magnitude of P times the cosine of theta1.
The x2 component of P would be the magnitude of P times the cosine of theta2.
And P3, the third component, would be the magnitude of P times the cosine of theta3.
Now, we will have so many relations that involve the cosine of the angle between a direction and one of our reference axes that it is convenient to define a special term for the cosines of these angles.
So I'll define this as magnitude of P times the quantity l1, magnitude of P times l2, magnitude of P times l3, which is a lot easier to write.
And we will define these things as the direction cosines.
With these equations it's easy to attach some meaning to the direction cosines.
Suppose we had a vector of magnitude 1, something that we will refer to as a unit vector.
And if we put in magnitude of P equal to 1, it follows that l1m l2m l3 are simply the components of a unit vector in a particular direction along, obviously, x1, x2, and x3, respectively.
Trivial piece of algebra, but it attaches a physical and geometric significance to the direction cosines.
Now, the vector is something that could represent a physical quantity.
In any case, it is something that is absolute.
And it sits embedded majestically, relative to some absolute coordinate system.
The magnitudes of the components P1, P2, and P3 will change their values if we would decide to change the coordinate system that we're using as our reference system.
So the next question we might ask is, suppose we change the coordinate system to some new values, x1 prime, x2 prime, and x3 prime?
And I'll illustrate my point with just a two dimensional analog of this.
This is x1, and this is x2.
And this is my vector P. And I change x1 to some new value, x1 prime, and change x2 to some new orientation, x2 prime.
Then clearly the component of P on x1 has changed its numerical value if I refer it to x1 prime instead.
And similarly, this value would be the component P2.
If I change the direction of x2 and draw a perpendicular to x2, this would be P2 prime.
So if I change coordinate system along the fashion I suggested, the three components of a vector, P1, P2, P3, are going to change to some new values, P1 prime, P2 prime, P3 prime.
OK.
So the question I'd like to address next is given the change of coordinate system, and given the three components of P in the original coordinate system, how do I compute the values of the new components P1 prime, P2 prime, P3 prime?
I'll say it in words, and then we'll define a mechanism for specifying the change in coordinate system.
What I'll say is-- and this was apparent in the sketch that I just erased-- the new component of the vector P1 prime is simply going to be the sum of the components of P1, P2, and P3 along the new x prime direction.
So I'm saying that this is going to be the sum of the component of P1 along x1 plus the component of P2 along x1 prime and the component of P3 along x1 prime.
So in short, I'm doing nothing more complicated than saying, I can get the values of the new components if I take the vector P, split it up into its three parts, and then find the component of each of these three parts along the x1 prime access, then do the same thing for the x2 prime axis, and then the same thing for x3 prime.
So I'm going to need, now, a notation for a change in a three-dimensional Cartesian coordinate system.
So here is x1, here is x2, and here is x3.
And I will change them.
And again, I'm always keeping the coordinate system Cartesian.
So here's an x1 prime, here's an x2 prime, and then x3 prime will point out in some direction like this.
I don't want a prime on that.
So I'm going to say now that the component of P along the new x1 prime is going to be the magnitude of P1 times the cosine of the angle between x1 and x1 prime, plus P2 times the cosine of the angle between x1 and-- am I doing this right?
P1 onto x1 prime.
And I want P2 onto x1 prime.
So this is going to be the angle between x2 and x1 prime plus P3 times the cosine of the angle between x3 and x1 prime.
Well, we used C's or l earlier on to represent a direction cosine.
Let me define Cij as the cosine of the angle between x1 prime, xi prime, and x sub j.
So that means I can write this expression here in this nice compact form.
With our definition of direction cosines, I can say that P1 prime is going to be equal to C1 1 times P1 plus C1 2 times P2 plus C1 3 times P3.
Can write that P2 prime in the same way.
It's going to be the cosine of the angle between P2 prime and P1 and x1, plus the cosine of the angle between x2 prime and x2 times P2 plus C2 3 which is the cosine of the angle between x2 prime and x3, times the component P3.
And in a very similar fashion, P3 prime will be C3 1 P1 plus C3 2 times P2 plus C3 3 times P3.
So here is the way a vector will transform.
And we can write this compactly in matrix form.
We can say that P sub i prime, where this is a column matrix, one by three, is going to be equal to Cij, a three by three matrix times the original components of the vector P sub j.
And just to cement the notation that we're using, if I put my old axes up here, x1, x2, x3, and put the new axes, x1 prime, x2 prime, x3 prime, down this way, then the cosine of the angle between the quantities that are in this column and the quantities that are in this row would be C1 1, C1 2, C1 3, C2 1, C2 2, C2 3, C3 1, C3 2, C3 3.
Nothing fancy except the notation.
It's the description of some very simple geometry.
This array, Cij, is something that I will refer to as a direction cosine scheme.
Let me pause here and see if that's all sunk in, whether you have any questions on this.
One of the nasty properties of what we're going to be doing for the next month or so is that the notions are really very, very simple, but the notation is horribly cumbersome and complex.
So it takes a bit of getting used to in application to actual real cases before you feel fully at home with it.
OK. Just a matter of definition so far.
Let me note that this direction cosine array is going to be useful for defining how a vector changes as we go from the original coordinate system to a new coordinate system.
But this direction cosine array will also tell us how the axes in one coordinate system are related to the axes in the new coordinate system.
It follows from the fact that the axes themselves can be regarded as unit vectors.
And we said that the components of a unit vector are the direction cosines of that vector, relative to a coordinate system.
So let's ask, what are the new components of x1 in terms of the original axes unprimed.
Well, x1 prime is going to be the unit vector x1 times the cosine of the angle between x1 and x1 prime, plus the unit vector along x2 prime times the cosine of the angle between x1 prime, and x2.
And that's C1 2.
Plus x3 regarded as a unit vector times the cosine of the angle between x1 prime and x3.
So we can actually write an equation for unit vectors along each of our new axes.
And they will go as C1 1 times x1, C1 2 times x2, C1 3 times x3, plus C2 2 times x2 plus C2 3 times x3 times x3 prime.
And x3 primal will be C3 1 x1 plus C3 2 x2 plus C3 3 x3.
So this, then, is an equation between the unit vectors along the three reference axes in the new coordinate system relative to those in the original coordinate system.
And the direction cosine scheme does the job.
OK. We could, using the same argument, give the array that specifies the reverse transformation.
If we would change our mind, for example, and say we don't like what we've done, let's write the original coordinate system x1, x2, and x3 in terms of the unit vectors along the new axes.
And we can use exactly the same array.
We can say that the original x1, in terms of the three new axes, x1 prime, x2 prime, and x3 prime, is going to involve the cosine of the angle between x1 prime and x1, and that is C1 1, plus the cosine of the angle between x2 prime and x1, and that's C2 1, plus the cosine of the angle between x3 prime and x1, and that C3 1.
If we continue on this, if you have the idea, the angle x2 is going to be given in terms of x1 prime, x2 prime, and x3 prime, as the cosine of the angle between x2 and x1 prime, and that is C1 2.
Here we want the cosine of the angle between x2 and x2 prime, and here the cosine of the angle between x3 prime and x2.
And you can see the way this is playing out.
C1 3 times x1 plus C2 3, x2 prime plus C3 3 times x3 prime.
So there's the reverse transformation, using the same array of coefficients as we did the first time.
So it turns out if we write this symbolically in a compact form, xi prime is given by Cijx sub j.
And the reverse transformation using the same direction cosine says that xi is going to be Cji times x sub j prime.
In other words, the reverse transformation, let's write it as Cij minus 1, the inverse transformation, turns out to be simply Cji.
And that, in matrix algebra, is written as the transpose of the original array of coefficients.
And transpose is either given by a squiggle, a tilde on top of the matrix.
Some people like to use a superscript T. But we'll use this particular notation.
But you can see either notation used to indicate the transpose.
The array Cij, which has this property, and it also has another property which I won't bother to prove, but the determinant of Cij is equal to 1.
And this is something called a unitary matrix.
Unitary matrix has the property that the inverse matrix is the transpose.
We will very, very shortly start writing down numbers for some specific transformations.
And then I think that will give us a little facility in doing these manipulations.
Comments or questions?
Is this old stuff or old stuff for which the notation is still confusing?
All right.
Let me point out something that is perhaps apparent to you.
And that is that not all nine of these numbers are independent.
There are relations between them.
And let's point out some of these relations.
C1 1, C1 2, C1 2 represent the components of a unit vector along x1 prime, in the original coordinate system of the elements in any row is equal to 1.
Because these are the components of a unit vector.
And the magnitude of a unit vector is 1.
In the same way, if we look at any column of terms in this matrix, for example, C1 1, C2 1, C3 1, these are terms that represent the cosine of angles between x1 in the original coordinate system and x1 prime, x2 prime, x3 prime, our new coordinate system.
So this gives us the magnitude of x1, but x1 is a unit vector.
So it follows that the sum of the column C1 1 squared plus C2 1 squared plus C3 1 squared also has to be unity, because that gives us the magnitude of a unit vector along x1, So the sum of the squares of elements in any column of the direction cosine is unity.
These expressions are useful.
But they have one ambiguity.
That is the cosine of an angle can be either positive or negative, depending on whether the angle is less than 90 degrees or greater than 90 degrees.
These relations involve the squares of direction cosines, and therefore we can't tell whether the direction cosine itself is positive or negative.
So let me put down a limitation here.
And that is we cannot tell the sign.
Every time I point this out to people I wince inside.
Because I once spent two weeks trying to debug a computer program, and it wasn't working.
And it turns out the reason it wasn't working properly was that I didn't realize that you cannot tell the sign when all you know is the squares of the direction cosines.
So I remember this as a rather pointed observation.
Happily, there are other relations among this array of coefficients.
This row of terms represents the components of x1 prime in the original coordinate system x1, x2, x3.
This row immediately below it represents the components of a unit vector x2 prime relative to the original coordinate system x1, x2, x3.
Our coordinate systems are Cartesian.
Therefore, this unit vector has to be perpendicular to the unit vector along x2 prime.
And that means their dot product has to be 0.
So let me indicate that this way.
The unit vector along x1 prime dotted with the unit vector along x2 prime has to be 0.
And that dot product is going to be C1 1 times C2 1, that's the product of these two terms, plus C1 2 times C2 2 plus C1 3 times C2 3.
And that has to be 0.
And this involves only the first product of the direction cosine.
So to make it come out 0 when we add up the magnitudes, we will get the sign of the direction cosine.
So this is a much more powerful relation.
And similarly, the product of the coefficients in the first and the third row have to add up to 0.
And the second and third row have to be 0.
So there are three different relations we can write between products of corresponding coefficients in the rows.
So to sum up in words, the sum of the corresponding elements-- of the product of-- in any pair of rows of Cij must be 0.
But we're not done yet.
If we look at the columns in this array, this represents the components of x1 relative to x1 prime, x2 prime, x3 prime.
And these terms here represent the components of x2 relative to x1 prime, x2 prime, and x3 prime.
And for similar reasons, the dot product of those two vectors has to be 0.
So we can say that in addition, the sum of any sum of pairs of corresponding coefficients in any pair of columns must be 0.
So we're working here on the direct matrix of the transformation.
We've seen that the reverse relation, the inverse matrix of Cij is Cij transpose.
And therefore the inverse matrix has to have this same relationship that the products of terms and rows or columns, any pair of rows or columns has to be 0.
Now, there's one other pair of relations among the coefficients, which is not quite so geometrically obvious.
And I won't attempt to prove it.
I'll just state it.
I said a moment ago that these are unitary matrices.
The determinant of the coefficient Cij then has to be unity.
But interestingly, it will be plus 1 if one goes from a right-handed system to a right-handed system.
That is to say the set of axes x1, x2, x3 might be right-handed.
And if the new coordinate system x1 prime, x2 prime, x3 prime is also right-handed, then the determinant of coefficients is plus 1.
On the other hand, if one goes from a right-handed system to a left-handed reference system or from a left-handed one to a right-handed coefficient, then, interestingly the determinant of coefficients is minus 1.
So the determinant of the matrix of the transformation is plus 1 if you go to coordinate system of the same chirality.
It's equal to minus 1 if you go to a coordinate system of changed chirality.
All right, so to repeat something I said at the outset but which you now probably truly believe, the elements in the direction cosine scheme that get you from one coordinate system to another have lots of inter-relations.
And all of these coefficients are not independent.
There are these relations that couple them.
How are we doing on time?
We mentioned last time that a large collection of physical properties of materials are properties that relate a pair of vectors.
So let me, to make this specific, talk about a particular physical property, electrical conductivity.
And electrical conductivity relates a current density vector, and that it charge per unit area per unit time to an applied vector, and that vector is the electric field vector.
And the electric field has units of volts per unit length, so volts per meter in MKS.
And provided the electric field that's supplied is not too strong, it turns out that every component of the current flow is given by a linear combination of every component of the applied electric field.
So the flow of current along x1 will be given by a proportionality constant, an element sigma 1 1 times the x1 component of the electric field.
Let me write it out the first couple of times we do this.
Sigma 1 1 times E1 plus sigma 1 2 times E2 plus sigma 1 3 times E3.
J2 will be sigma 2 1 times E1 plus sigma 2 2 times E2 plus sigma 2 3 times E3.
And J3 will be equal to sigma 3 1 times E1 plus sigma 3 2 times E2 plus sigma 3 3 times E3.
Looks formally like the relation between unit vectors that define a coordinate system.
Number of subscripts is the same, but actually this is something that's completely different.
It's dealing with vectors that have some physical significance.
So in compact reduced subscript notation, this is the definition of electrical conductivity.
This matrix that relates the electric field vector to the current density vector is said to be a tensor of the second rank.
OK, tensor.
First thing you might say, why do you call it a tensor, dummy?
It's a matrix.
It's a plain old matrix.
There's a subtle but very important difference.
A tensor is a matrix with an attitude.
And I'll make the distinction clear a little bit later on.
But there are tensors also of higher rank.
These expressions where summation over repeated subscripts is implied can hide, as I indicated last time, some absolutely horrendous polynomials.
But tensor at very least is a term that makes the faces of all who hear it pale, and makes the knees of even the very strong to weaken.
And in case you don't believe that, I'll show you what I have to wear whenever I give these lectures.
And consequently it's kind of scuzzy and worn out.
But I have to put on these knee braces from wobbling braces.
And you can see what it says on here.
"Tensor."
So that's a consequence of this frightening definition that we've just made.
Let me next set the stage for what we ought to do next.
E sub j represents the components of an electric field, x1, x2, x3, in a first coordinate system.
ji represent the components of the current flow in a coordinate system, x1, x2, x3.
If we were to change coordinate system for any reason, these three numbers would wink on and off.
Some might go negative.
The magnitudes would change.
And as a result, the components of the current flow would have to do the same thing.
Because the components of these vectors, without changing anything physically, have to change their numerical values if we refer them to a new set of reference axes.
If we change coordinate system and these numbers change, and if we change coordinate system, these numbers change, we're still applying field in the same direction.
The current still flows in the same direction.
But the components we use to define these two vectors change.
And it follows just algebraically, the elements of the tensor have to change and link into different values.
It follows automatically.
So a question, then, is that if we have a coordinate system, x1, x2, x3, and we change it into a new coordinate system, x1 prime, x2 prime, x3 prime, then j sub i changes to some new values, j sub i prime.
E sub j changes to some new values, E sub j prime.
And therefore, of necessity, sigma i j, the conductivity tensor, has to change to new values sigma ij prime.
So I'll let you rest up to brace yourself for this.
The question is, how can we get sigma ij prime, the nine elements of the tensor in the new coordinate system, in terms of the direction cosine scheme that defines this transformation and in terms of the elements of the original conductivity tensor?
And this, my friends, is what makes a tensor a tensor and not a matrix.
I can write a matrix for you, a really lovely matrix.
Let's put in some elements here.
Let's put in 6.2, square root of minus 1e, and 23.
And as other elements, I'll put in pi 23.4, 6, and 0.
It's a perfectly good matrix.
It's just an array of numbers, any numbers, real or imaginary, or whatever I like.
So this is a matrix.
What a tensor is, is a matrix for which a law of transformation is defined.
And that's what makes a tensor a tensor.
What does it mean to take this two-by-four matrix that I just wrote down?
How do I transform that to a different coordinate system?
It's meaningless, just an array of numbers.
It's an array of numbers that has some useful properties, like matrix multiplication and the like.
But to talk about transformation of this set of four ridiculous numbers to a new coordinate system is something that's absolutely meaningless.
Not so for something like conductivity or the piezoelectric moduli or the elastic constants.
These change their values.
There's a law of transformation when we go from one Cartesian reference system to another.
So what we will do when and if you return is to derive a law for transformation for second-rank tensors, and then, by implication, look at higher-rank tensors and decide how they would transform.
But why would you want to do this?
Why would you want to muck things up and have to worry about transforming these numbers?
Well, let me give you just one simple example.
Suppose we had conductivity of a plate, of a crystal.
And what would you do?
You'd measure it relative to a set of axes, which, if you have a little fragment of crystal, you have no reference system.
So say that the axes x of i are taken relative to the lattice constants of the material, so relative to the edges of the unit cell, possibly.
Then you decide that this material really has some useful properties, and you would like to cut a piece out of it so that you get a plate for which the maximum conductivity in that plate is in a direction normal to the plate.
So you know just what sort of plate you want to cut out.
You know what the direction cosines are.
But once you've cut a plate from the crystal, the tensor relative to the old axes, x1, x2, x3, is not going to be terribly useful.
You're going to want to find the tensor relative to this as one set of axes, and these perhaps as a new set of axes within the plane of the plate.
So there's a good example.
Cut a piece from a crystal and cut that piece so that the extreme values are along x, y, and z for the new coordinate system.
Then you will be faced with the necessity of transforming the tensor from one coordinate system to another one.
Or you might measure the thermal conductivity tensor.
You might want to cut a rod out of the material so that the maximum conductivity or the minimum thermal conductivity is along the direction of the rod.
You might want to use that as a push rod to hold a sample in position and not have it be a big heat sink for the temperature that's inside of your sample chamber.
So I've hopefully convinced you that there are lots of cases where it would be necessary and convenient to transform the tensor that describes a property to a new coordinate system.
All right, so let us take our break now.
Some internal clock always tells me when it's five of the hour, unless I get really excited about something.
And it is indeed that time now.
So let's stop.
OK, ready for more?
I defined a problem for you.
Now let's address it.
Said that if we have one coordinate system, and if we have some vector, q, that's defined as a second-rank tensor, aij times some other vector p sub j-- and let me digress in passing.
I am very careful to say "a second-rank tensor" and not a "second-order tensor," because higher-order order terms means negligible and non-important.
And when I say "second-order tensor," I don't mean to say it's not important and negligible.
It's very important, so I say "rank," which has some sort of dignity to it.
So I don't like the term "order," because it has another meaning.
OK, so here is a tensor that relates a vector pj to give us the components of a vector qi.
If we change coordinate system, the components of p, representing exactly one in the same vector, wink on and off and take different values.
The values for q take on different values, and therefore, of necessity, the three by three array of coefficients, which relates these different numbers, must also change its numerical values.
And I hopefully convinced you at the end of last hour that there's times when you might actually want to do this when you're cutting out a particular sample from a single-crystal specimen.
How do we get the new tensor in terms of the direction cosine scheme that specifies the change of axes and the original tensor?
So I am going to refer to my notes quite closely here, because I want it to come out pretty and not have to redefine variables when I'm done.
So let's start with the original tensor relation, q sub i equals aij times p sub j.
Now, what we want is something of the form q sub i prime equals aij prime times p sub j prime.
And we know the relations forward and reverse in terms of the direction cosine scheme cij So let's begin by writing qi prime in terms of qi And we know how that is going to transform.
It's going to be elements of the direction cosine scheme cim times q sub m. So that will give me the i-th component of q in the new coordinate system.
I know how q arises from the applied vector p. So let me write qm in terms of the applied pj.
And this is going to be aij times-- be careful of my variables here-- this is going to be a ml times p sub l. And that is going to, from my definition of a second-rank tensor, give me the m-th component of q.
So far, so good.
I've got two different repeated subscripts here, so this is a double summation.
Now, I'll have what I want to have, namely a q sub i prime on the left-hand side and a p sub j prime on the right-hand side if I can express the original components of p in terms of the new components of p. And I do that by the reverse transformation.
So let me now write a ml.
And then in place of p sub l, I will write cjl times p sub j prime.
Notice the inverted order of the subscripts.
That is the reverse transformation that's going to give me p sub l. So you really have now what I'm after.
This is a triple summation in m, l, and j. I can write the terms that are in what is going to be a triple summation over a product of terms.
I could write these terms in any order.
So to simplify it, let me write q sub i prime is equal to cim cjl times a ml, times p sub j prime.
And now, hotcha, I've got an expression that has q prime on the left and p prime on the right, and paying close attention to my notes so that the subscripts all came out the way I would like them to.
So what this says is that the transform tensor aij prime is going to be equal to, by definition-- or what we've shown here, it's going to be equal-- just picking off terms-- it's going to be cim cjl times a ml.
Just picking off these terms.
m has no physical meaning, because m simply is an index of summation.
l has no specific meaning.
That is just an index of summation.
But the i and the j do have meaning.
They go with the i and j on the particular tensor element that we were attempting to evaluate.
So in other words, to be specific, if we want the new value of the tensor element a1 2 prime, it's going to be c1 something, c2 something, and those somethings m and l would vary from 1 to 3.
So if I write this out not in the reduced subscript notation but put a summation sign in there, so a12 prime is going to be the sum over m and the sum over l of terms c1m-- the first index is always 1-- c2 something-- first index is always 2-- and then m and l take on all possible values.
OK, we've got two results.
We learned how to transform a vector.
And a vector, if you will, is simply a tensor of first rank.
And transforming the vector, we summed over all three components of the original vector.
And the coefficients in that summation were one direction cosine.
Now we're transforming a second-rank tensor.
Again, each new element is a linear combination of all nine of the elements in the original tensor, and the coefficients are a product of two direction cosines.
So we've got two points.
Let's draw a line through them, and we can say that, in general, any new tensor element of any rank-- and you could prove it through exactly this method by going up, now, to the third rank, fourth rank, and so on, and writing substitutions of this form-- it turns out that a new tensor element aijkl however far you want to go, is going to be given by a linear combination of direction cosine element ciI, cj capital J, ck capital K, cl capital L, times however far you have to go times aijkl, and so on.
So these are the true indices.
These have physical meaning.
These have relevance to how a particular property will behave.
The capital I, capital J, capital L, and so on, are what we referred to last time as dummy indices.
These are indices of summation.
But the first index on the direction cosines has specific meaning.
They are tied to the indices on the subscript of the term that you would like to evaluate.
So we've got now a very profound relation for a tensor of any rank.
And really, it just involves substitutions using the reverse or the forward transformation until you get one element on the one side related to another element on the right-hand side.
And this is a specific element in the new tensor, in our case, of the second-rank tensor, aml.
This is all very abstract.
It is something that we'll have to do a couple of times for a real problem before you see how it works out.
The number of elements that figure into these transformations is really astronomical.
Suppose, for example, the tensor involved were something like the elastic stiffness tensor, which is represented by c. And we need four subscripts.
This is a tensor or fourth rank.
If we wanted to transform a particular stiffness to a new coordinate system, we would need a summation ci capital I, cj capital J, ck capital K, cl capital L, times all of the elements in the original tensor, aijkl.
A fourth-rank tensor consists of an array of 9 by 9 terms.
So there are 81 of these.
We'd have four direction cosines out in front, and there would be a total of 81 times 5 characters that we would have to write.
So to do the complete tensor transformation, we would have to write on the order of 400 quantities to get just one of the 81 elements in the new transform tensor.
So the total number of elements we'd have to write to do this would be 81 squared times 5.
That's a lot of elements.
We'll do a few of these transformations directly, but let me assure you that if we do a transformation that is going to involve symmetry, a lot of the direction cosines, if we're lucky, will be 0.
So it's not quite as onerous as it seems.
So we would make use of this sort of formalism if we wanted to go from one set of reference axes to a new set that might represent a special specimen that we cut out of a crystal.
But there's another formal way in which we could make very profound and non-intuitive use of these relations.
Crystals, except for the abominable triclinic crystals, have symmetry.
If a crystal has symmetry, you can transform the solid physically by that symmetry operation.
And you have to measure the same property before and after.
So suppose we have a crystal that has a twofold axis.
And this crystal is something that looks like this.
So this is side A, and this is side B.
We could move the crystal by a 180-degree rotation.
Put it down.
I won't draw it, because it's going to look exactly the same, except now this thing-- I won't draw, and I do draw it-- this is face A, and this is face B.
If we had electrodes on the crystal before and after that transformation, we have to measure, let's say, the same electrical conductivity for both orientations of the crystal.
Now, moving a crystal relative to some coordinate system, relative to a pair of electrodes that we're fastening onto the crystal, is exactly the same thing as doing the reverse transformation of the coordinate system.
That's a vague, strange-sounding term.
So suppose we have a crystal with a fourfold axis with four faces, A, B, C, D. And here are our electrodes.
To move the crystal relative to the electrodes by a 90-degree rotation would involve rotating face D up to this location.
A would move to this location.
C would move to this location.
B would move to this location, and we'd fasten electrodes on the crystal again.
If the crystal originally had a coordinate system such that this were X1 and this is X2, moving the electrodes onto a different direction on the crystal is the same as moving the crystal in the opposite sense.
So we could either envision moving the crystal relative to the electrodes like this, or we could move the electrodes relative to the crystal by the reverse transformation.
And the result is the same.
So what I'm saying is that if a crystal has symmetry-- and let me be specific and suppose that our crystal has a twofold rotation axis along X3.
Let's ask how that twofold access would change the coordinate system relative to the crystal.
That's the same as moving the crystal relative to the coordinate system.
It's going to take X1 and move it to this location X1 prime.
It's going to take X2 and move it through 180-degree rotation.
This is going to be X2 prime.
And if the twofold axis is along X3, X3 prime is the same as X3.
So what is the direction cosine scheme for this change of axes?
You might immediately start working and saying, well, C11 is the cosine of the angle between X1 prime and X1.
That's 180 degrees.
Cosine of 180 degrees is minus 1.
But let me remind you that the direction cosine scheme, c ij, simply gives us the relation between the new axes x sub i prime and the old axes, x sub j.
So let me, just by inspection, write down the relation between these two sets of coordinate systems.
So X1 prime is equal to minus X1.
X2 prime is equal to minus X2.
X3 prime is equal to X3.
So the direction cosine scheme for this particular transformation is simply minus 1 0 0, 0 minus 1 0, 0 0 1.
So I just evaluated a nine-element direction cosine scheme by inspection, if I can write the relation between the coordinate system before and after the transformation.
OK?
And as we examine higher symmetry, the same is going to be true for the threefold axis, let's say, along the 1, 1, 1 direction of a cubic crystal, for a sixfold axis, and so on.
So we'll be able to write the direction cosine schemes for a symmetry transformation simply by inspection.
So for a twofold axis along X3, this is the form of the direction cosine scheme.
So now let me transform the elements of a second-rank tensor term by term and see what we get.
Suppose I want-- let's stick with conductivity as an example.
Suppose I want the value for the conductivity element sigma1 1 prime.
That's going to be c1 something, c1 something, because these are the elements that go in here, times every element of a conductivity tensor sigma lm.
The only element of the form c1 something that is non-zero in this row c11, c12, c13, is the term c11.
In the same way in the same row, the only term of the form c1 something is c11.
So this is going to be simply c11 times c11 times sigma11.
That's the only term that survives.
c11 has a numerical value of minus 1, and that says that sigma11 prime is equal to sigma11.
So is there any constraint, any restriction on sigma 11?
No, sigma11 could be anything it likes.
So there's no constraint.
Let's do another element.
Let's see what sigma12 prime would be.
This will be c1 something times c2 something times sigma something something.
The only element of the form c1 something that is non-zero is c11.
So I'll put in a 1 for the l. The only direction cosine element of the form c2 something which is non-zero is c22.
So I'll put in c22 and let n be equal to 2. c11 is minus 1, c22 is minus 1.
So again, this gives us something not terribly interesting.
Sigma 12 prime is equal to sigma 12.
So there's no constraint, at which point you're probably getting very restive, say, this is not telling us anything.
So let me shake you up by doing one further transformation, and that is to find the value for c13 prime.
And that would be c1 something, c3 something times sigma lm.
The only form of this term of the form c1 something, that's non-zero is c11, as we've seen.
So I'll put in just that single term and replace l by 1.
The term of the form c3 something that is non-zero is c33.
And I'll put in a 3 for m, and this is then c11 times c33 times sigma13.
1, And c11 is minus 1. c33 is plus 1.
And that says that sigma13 prime is minus sigma13.
But if this is a symmetry transformation, the tensor has to remain invariant.
And if we're to have sigma13 equals minus sigma13, there's only one number that can make that claim, and that's 0.
So sigma13 is identically 0.
And that places a rather severe constraint on the way in which the crystal is going to relate an applied electric field to a current flow.
Sigma11 is anything.
Sigma12 is anything.
Sigma13 has to be identically 0.
Now, let's cut to the bottom line.
The direction cosine scheme is diagonal, so we can say that for any element that we pick to transform sigma ij prime is going to be cii, the diagonal term which has the second subscript equal to the first, times cjj times sigma ij.
And this says that if we have i or j is equal to 3, then sigma ij has to be 0, because we're going to have a minus 1 times a plus 1.
If neither i or j is equal to 3, then again, we will have a minus 1 times a minus 1.
There will be no constraint.
And the only final possibility is that both i and j are equal to 3.
That would be the single element c33.
Then we would have plus 1 times plus 1 as the product of direction cosines, and there will be no constraint.
So for a crystal that has a twofold axis, and in which that twofold axis is along the direction of x3, the form of the tensor will be sigma11, sigma12, 0, sigma21, sigma22, 0, sigma31, that's going to be equal to 0, and sigma33 has no constraints.
So rather than having nine elements, there are only five independent elements rather than nine.
And there is now another relation that can occur in a second-rank tensor.
The off-diagonal terms sigma12 and sigma21 do not have to be related.
But for most-- but not all-- most second-rank tensor properties happily have sigma ij identical to sigma ji.
In other words, the tensor is symmetric across its principal diagonal.
That is a condition that does not arise from symmetry That depends on the specific physical property.
So let me emphasize that this depends on the tensor property.
And for a great many physical properties-- conductivity, diffusivity, permeability, susceptibility-- you can show that the tensor has to be symmetric.
But there are a lot of tensors, particularly for the more obscure physical properties, where to my knowledge, this proof has never been given.
And along the same lines, it is well known that there is one physical property for which this is not true.
This is the thermal electricity tensor.
So for at least one property, you can show for sure that the tensor does not have to be symmetric and that for a crystal of symmetry 2, this term and this term are definitely not equal.
All right, let us do another transformation for another symmetry, and we can see that it goes fast when the direction cosine scheme is relatively sparse.
Let's ask the restrictions, if any, that are imposed by inversion.
So what is the direction cosine scheme here?
Here's x1, here's x2, here's x3.
Then operation of inversion at the intersection of these axes will invert the direction of x1 prime to here.
It'll invert the direction of x2 prime here.
It will invert the direction of x3 prime to here.
So the relation between the reference axes is that x1 prime is equal to minus x1.
x2 prime is equal to minus x2.
x3 prime is equal to minus x3, so that the form of the direction cosine scheme crj is minus 1, 00, 0 minus 1 0, 00 minus 1.
Slightly different from that for a two-fold axis for which the first two diagonal elements were minus 1, the third one was 0.
Well, let's jump right to it and see if we can generalize this.
It's a diagonal direction cosine scheme once again.
And this says that if we transform a particular element sigma ij, it's going to be given by cil, cjm, sigma lm, where l and m are variables of summation.
The only ones that survive are the ones for which i equals l and for which j equals m. So it's going to be cii, cjj times sigma ij.
Regardless of the values of i and j, the diagonal terms are always minus 1.
And therefore, sigma ij prime is always going to turn out to be equal to sigma ij, so there's going to be no constraint on any element.
And this shortens the job that's facing us immeasurably.
So let me write that down, because that's important.
Inversion imposes no constraint on any second-rank tensor property.
So if we stay with monoclinic crystals, we looked at symmetry 2.
Symmetry 2 over m is equal to 2 with an inversion center put on it.
But inversion doesn't require anything, so the symmetry constraints for 2 over m have to be the same as for symmetry 2.
We look at the constraints that might be imposed by a mirror plane.
A mirror plane plus inversion is 2 over m. 2 over m has to be the same as 2.
So this will be the same as 2.
And now we've shown that for any monoclinic crystal, regardless of whether the symmetry, the point group, is 2m or 2 over m, the form of the tensor has to be exactly the same.
So for any monoclinic crystal, namely 2m or 2 over m, this is the form of the tensor where the twofold axis, again, is along x3, the mirror plane would have to be perpendicular to x3, and for 2 over m, both of the preceding conditions.
So how about that?
Let me issue a caveat, because we're almost out of time.
There are five independent elements.
And that's true.
But elements sigma12, sigma21, sigma13, and sigma31 are 0 only for this arrangement of axes relative to the symmetry elements.
If you wanted to take a different set of axes, you know how to get the tensor for that set of axes.
Each tensor element is going to be given by a linear combination of these five non-zero elements.
And if the orientation of the axes relative to the symmetry elements is quite general, all nine elements of the tensor will be non-zero.
There will be only five independent numbers, which composes each of those nine elements, and they will be given by a product of two direction cosines, sometimes each of these five non-zero elements.
But there will be no zeros in this array at all for an arbitrary set of coordinate systems.
Something that I think I'll ask you to do as a problem, because it's really easy to do, if the tensor is symmetric, which most of them are, one thing that you can show quite directly is that a symmetric tensor remains symmetric for any arbitrary change of axes.
And that, again, is something that's fairly easy to prove, and I'll let you have the fun and exhilaration of doing that for yourself.
OK, so this means that for everything except thermal electricity, you really have to transform, at most, only six elements if you go from one coordinate system to another.
That's still a lot, but it's considerably better than transforming all nine.
So if the tensor originally is symmetric in one coordinate system, it stays symmetric in any other coordinate system.
Now, one thing that I should mention-- I passed over it rather quickly-- we said that a property of a direction cosine scheme is that it is what's called a unitary transformation.
And it has the property that the determinant of the coefficients is plus 1 if the axis retains the same chirality.
The determinant is minus 1 if you change the handedness.
That is only true for what is called a measure-preserving transformation.
That's what it's called.
And when it's a measure-preserving transformation, then the direction cosine scheme is a unitary matrix.
What is measure-preserving transformation?
If it's a right-handed system beforehand, there's no squishing.
It doesn't go to an oblique coordinate system.
Cartesian stays Cartesian.
If the reference axes are of equal lengths, they don't stretch upon the transformation.
You can define transformations like that if you like, where the angles between them go from orthogonal to oblique after the transformation, and the units of length along the three axes change dimension.
But then the determinant of the coefficients is not unity, and a lot of the nice, convenient properties that we've seen here do not hold.
All right, that is a good place to stop, I think.
Next week, no quiz.
If you came in late, we're going to postpone the quiz for a week.
And let's see, look at all I can ask you just after one lecture on tensors.
What we will do next is explore the form of the tensors for other crystal symmetries.
It goes fairly quickly.
And then having done all that, I'll show you how you can determine the symmetry constraints by inspection for a tensor of any rank.
And you're going to despise me for that, but this was useful, because we can get used to manipulating the notation.
But there is a method called-- appropriately enough-- the method of direct inspection where you can very quickly and very easily do the symmetry transformations.
So all this and more will be revealed next time, which is going to be a lot more beneficial than taking a quiz.
Do them individually so I can continue to put names and faces together.
I'm happy to announce that the registrar has now got everybody's photograph online for registration in the course.
So anonymity is a thing of the past, so you have to watch your step from now on.
I handed back, to those who didn't get it, problem set number four, which asked you to tackle some patterns, nontrivial patterns.
And actually, that was a dirty trick, because we hadn't, at that point, derived the plane groups, and you really didn't know what to do or what to look for.
But nevertheless, it got you thinking about patterns and some of the symmetry elements which we had discussed up to that point.
At this point we have derived exhaustively every last one of the 17 plane groups.
So now you are armed with this new-found power, and when faced with a pattern, you should know exactly what to look for and how to go about deciding what plane group it is.
At the very least, you'll have the drawings of the arrangement of symmetry elements in the plane groups before you, and you can work by the process of elimination.
For example, high symmetry usually hits you right between the eyes, and if something is square-ish, you can pretty quickly guess that it's based on a square lattice.
And if it has a square lattice, there jolly well better be a 4-fold axis in there that makes it square.
If you can find the 4-fold axis, then you have to ask yourself only three questions.
So 4-fold axis, fine.
Is there a mirror line in there?
Yeah.
Does the mirror line go through the 4-fold axis?
Then it is P 4 MM.
And you know just where to look for everything else, including these very subtle glide planes that are hard to spot.
If there is a mirror plane there, but it doesn't go through the 4-fold axis, then it's P 4 MG. And if there is no mirror plane, then it's P 4.
So just by asking one or two simple questions, you can narrow it down to what the plane group has to be.
This is another indication that the informed intellect is always more than a match for sheer, raw native intelligence.
If you know what to look for, it's a lot easier.
Because you really didn't have much practice with patterns, we're having a quiz, as you know, next Thursday, that will cover up through completion of the plane groups and not the material we've been doing now.
So I think it might be of use to you to have some practice analyzing a few more patterns.
So there are four additional patterns in this problem set.
As always, it's optional, but if you would like to try them, and you want to see if you've got them right, come in and see me tomorrow or on Thursday morning.
I'd be happy to go over them with you on the spot.
So this is for practice.
And some of you did extraordinarily well on the first try.
Others, I think, could use the additional practice.
There were a few people who identified the plane group correctly, but got the name that's assigned to it wrong.
And one other very confusing thing is that there is one plane group that has the symbols G and M, and another plane group which has the symbols M and G. So if you found a mirror plane and a glide plane is an independent symmetry plane, when is it MG, and when is it GM?
I have a little mnemonic device.
GM, general manager, is the guy who sits on top of the organization.
So GM should be the plane group that has the highest symmetry.
P 4 GM, P 4 general manager.
Now that is-- hey, it works for me.
But another way of saying it is that there are two plane groups, one has MG and the other has GM, and I like cars, and I think an MG is much classier than anything that General Motors, GM, puts out, so MG should be the one of highest quality, highest symmetry.
And that's just the reverse.
But whatever works for you.
You could keep them straight through that simple algorithm.
And as they say, if it works for me, but if it doesn't work for you, don't use it.
All right.
What I will bring in during intermission, for those of you had trouble identifying translations in the patterns that we handed out earlier, I've taken these and put them on overhead transparencies.
And I'll have two of each.
So if you don't see the symmetry or translations that are present, you can actually take one pattern and physically move it and lay it on top of the other one, and that's a good way to convince yourself what a translation looks like when it occurs in a pattern.
So I'll bring those in at our break between class.
All right, any questions before we move on?
Any questions that have arisen as you have gotten ready for the quiz?
You haven't gotten ready for the quiz yet, so there are no questions.
That's OK.
I know how things work at MIT.
You deal with one crisis at a time.
Any questions?
Anything you want to go over?
There was one interesting wrinkle in a problem that I had not encountered before, and this was the one that asked you to look at, in two dimensions, a plane with indices h and k. And then, when h and k were mutually prime, to move that plane by the translations plus T1 and plus and minus T2, and then show that the number of intervals between the origin and the intercept plane, the one that hit lattice points on both translations, was equal to h times k if they were mutually prime.
That is true only if the lattice is primitive.
And what the problem said was to pick one of the cells that you used in the first problem, number one.
Well, that problem asked you to begin with identifying different primitive cells.
If you take a multiple cell, this operation of going plus and minus T1 does not put a lattice line through each of the lattice points.
If you picked a double cell, that process decorated only half of the lattice points with planes, and the other half sat there with nothing hanging on them at all.
The key to the difference, if you looked at a double cell, was that if h plus k was even, then you automatically got a plane on every lattice point.
If h plus k was odd, as it would have been for the plane 2,1 for example, 2 plus 1 is 3-- hey, this isn't even one of my good days-- then half of the lattice points did not have planes hanging on them.
Now, there's great relevance of this observation to diffraction, and you probably are all familiar, if only vaguely, with the magic rules that say, if h plus k is equal to 3 pi plus 4, then the intensity is identically 0.
Well, for a double cell, the lattice planes that are repeated by translation diffract x-rays, and there is no reason why the intensity should be something other than 0.
So here comes, a la Bragg, an x-ray beam coming in at angle theta, and then you say you get diffraction when scattering from this lattice plane is exactly in phase with this one.
And this gives the familiar relation that an integral number of wavelengths is equal to 2 d sine of theta when the crystal diffracts.
So there is exactly-- this says there's exactly 2 pi phase difference or n lambda path difference between these two planes.
Now, if the lattice would be a double cell, then there is an additional lattice point in here that does not get a plane hung on it.
So if the lattice is a double cell, there's another plane that has to hang on this lattice point, and that one is exactly out of phase with this plane, and the intensity is 0.
So this observation that a non-primitive lattice has a interplanar spacing that is a sub-multiple of that of a primitive lattice gives some insight into why certain reflections-- certain diffraction maxima-- are identically 0 in intensity for a crystal that has a non-primitive lattice.
So that is something I had not noticed before.
I should have, but I will phrase the problem a little bit more precisely in the future.
All right.
So to conclude my preamble, I hope you'll try playing with some of the four additional patterns that I handed out, just to give yourself some practice.
And the implication of this is that you're going to see a pattern on the quiz, and I will tell you that you will.
So if you want to see how you did on the patterns that I distributed, please come in and talk to me about them.
All right then.
Let me remind you where we were last time.
We started to begin to build a framework of symmetry elements in three dimensions.
And we asked the question, what would happen if we take a first rotation axis, A alpha, combine it with a second rotation axis, B beta, in such a way that they intersect at a point.
This means that their operation and reproducing atoms or motifs is going to leave at least one point in space unchanged, and that will be the point of intersection.
We ask ourselves, what will be the combined effect-- we have two operations in space-- what would be the combined effect of rotating alpha degrees about A followed immediately by beta degrees about B.
So what we're going to do then is to take a first motif-- and let's say it's left handed.
Being a left-handed person, I like to give right handed motifs and left handed motifs equal time.
If we rotate that through an angle alpha, and this is number 2, it will stay left handed.
Then if we rotate that by B beta, it'll move it over here to number 3 and it will stay left handed as well.
And the question is then, what net operation is equivalent to the combined operation of these two transformations?
And to specify the type of operation is really a no-brainer.
All of these motifs are of the same corality so the only thing that can relate them is translation or another rotation.
And clearly the first and the third have no reason to be parallel to one another, and the distance between them is going to depend on how far they are away from the axis, so translation won't do the job, a and the only thing that's left as a net operation that's equivalent to those two steps is rotation about a third axis C. And what we're going to do today is answer the question, where is axis C located, and what is the angle of rotation, given the value of alpha and beta and the angle between these two axes?
And let's define that as a lowercase c. So clearly the location of the axis and the amount of the rotation is going to be a function of alpha, beta, and the angle between them.
We want this to be a combination of operations that exists in a symmetry operation.
And if this is to be a crystallographic symmetry, these will be restricted to the angular throws of a 1-fold, 2-fold, 3-fold, 4-fold, a 6-fold axis.
We can take these two at a time, ask what the net effect is if we combine at a given angle c. And what comes out here must be a rotation which is also crystallographic.
So there are going to be severe constraints on this combination.
Two rotations about an intersecting point will always be a third rotation, but if this is to be a set of operations in a symmetry group, the result must be crystallographic.
That's a tough problem, and how will we undertake it is going to be non-intuitive.
OK, the problem is most readily treated with spherical trigonometry.
So on the surface of a sphere, I'm going to map the point at which A alpha protrudes-- and I'll call this point A-- and then I'll mark out the point where axis B beta exits the sphere, and I'll mark that point B.
And this is the angle C. And we said that in spherical trigonometry, the measure of the length of the arc separating A and B is given by the angle subtended at the center, so the length of this distance between A and B is the angle c. Again, it sort of boggles the mind when you measure lengths in terms of degrees rather than some metric unit.
All right.
So I will now not bother to show the sphere on which the geometry is taking place.
I'll just draw A and B, and this is the arc between them, c. And somewhere or other there will be some third axis, C, which is going to be the combined effect of the rotation about axis A and axis B.
So what I would like to do is to locate the position of this axis C. In order to do that, I'll have to know what the angle between A and C is, and I'll call that, by analogy to what I've done here, I'll call that b.
And I'll want to know what the angle between B and C is, and I'll call that angle a.
So again, going back to three dimensions momentarily, if this is the rotation operation C gamma, and this is A alpha, and this is B beta, the axis c is this, the angle b is this, and the angle a is this.
OK.
It's a non-trivial problem and it is not by accident that the solution to this problem was first given by a very, very famous mathematician, Leonhard Euler, and this construction that we're about to go through is called Euler's construction.
All right.
Let me find where these different locations are going to be.
We've specified the location of point A and the location of point B, and we know that the angle between them is the length of the arc ab, which is the angle between A and C. So let me now do some constructions.
Let me find a great circle that by design is alpha over 2 away from the arc ab, and that is by construction.
And I'm going to say, then, that if A alpha works in this direction, the operation of A alpha is going to take this great circle and move it over to a great circle which is alpha over 2 on the other side of the arc ab.
Fine, you say, so what?
Well, just going to leave those there for now.
I'll have B beta work in the same sense.
And I'm now going to create a line here that by construction is beta over 2 on one side of the arc ab, and if I let B beta go to work, that will map this great circle over to a new location beta over 2 on the other side.
What has this done for me, other than perhaps confuse me and clutter the diagram?
Well, now I'm going to determine unequivocally the location of the axis C, and where it emerges from the reference here.
And how will I do that?
I'm going to use a definition that may have seemed trivial the first time we made the observation.
I said that a symmetry element is the locus of points that is left unmoved by an operation.
OK?
I rotated by A alpha from here to here, that took everything along this line and mapped it to a new location here.
I took this line and rotated it by B beta, and that took everything along this line and moved it to a new location.
So my question now is if I rotate by A alpha and then rotate in the same direction by B beta, what point is left unmoved?
It's only one point that can make that claim, and that is where these two great circles intersect.
The rotation A alpha will take this location-- and I'm going to call it C because I've identified now what it is-- it's going to take C and move it over to here, call that C prime, and then B beta takes that point and only that point, and restores it back to its original location.
So this, then, ladies and gentlemen, is where the rotation axis C gamma pokes out of the sphere of reflection.
Still don't know what this angle is in here, and I would dearly love to know what these arcs b and a are, and then I will have specified all three of the interaxial angles between A, B and C. OK, let me do something quite similar to what I did before.
I'm going to again let A alpha work on a particular point, and then let B beta map it.
So here's A alpha, here's B beta.
And now I'm going to look specifically at how these two rotations transform point A, where A is the point at which axis A alpha pokes out of the sphere.
A alpha, when it acts on this point, does nothing to it.
It leaves it alone.
B beta is going to map A to a new location, A prime.
Now, doing A alpha and following up by B beta is supposed to be equal to the rotation C gamma.
So that says that this point and this point must be related by the rotation gamma.
So say that again.
We're doing exactly what we did here except we're starting with an initial point, not this arc, but we're starting with the specific point A, operate on it by A alpha, it twirled around but stays put.
Rotate that by B beta, it goes through a total angle beta to this location here.
The net effect of getting from A to A prime is supposed to be the rotation C gamma, so this angle is then gamma and this is the location of C. OK?
OK, one other step that's a fairly easy one.
This length is equal to this length, because they were produced by rotation.
This side is common to these two triangles, and this angle then is beta over 2, this is beta over 2.
And if these two triangles, A, B, and C, that triangle is similar to A prime BC, and therefore I can say that angle A prime CA is identical to ACB, so therefore this angle has to equal this angle, and if the total angle is gamma, this is gamma over 2, and this is gamma over 2.
So now let me extract from this the information that I would like to use.
Here are three axes, A alpha, B beta, and C gamma.
This angle in here is gamma over 2.
This angle in here is beta over 2, and this angle in here is alpha over 2.
And let me emphasize that in this magic triangle, out of which we're going to extract some dazzlingly profound stuff, it is half the angular throw of the rotation axes that appear in here as these spherical angles, and not the entire angle of rotation.
So here's how properties of the three rotation axes are related one to another.
And now, we introduced without proof last time something called the law of cosines in spherical trigonometry.
And I not only do not want to prove it, but I have no idea how I would go about doing so, but that doesn't prevent me from using it.
So if here are three edges, a, b, c, and three angles in there, A, B, and C, we said that the law of cosines in spherical trigonometry, analogous in a way to the law of cosines and plane geometry, except since the lengths of the triangles are measured in degrees, there are trigonometric functions of these angles that appear in the law of cosines.
This says that cosine of b cosine of c plus sine b sine of c times cosine of a is equal to cosine of a.
So this now is an interesting relation that we can apply to this spherical triangle, which connects together the three rotation axes.
Let me apply it to find the angle c which we have picked as the angle between the initial two axes a and b.
That says that this should be equal to cosine of b cosine of c, the angle between the other two axes, plus sine of b sine of c times the cosine of angle a, and angle a is cosine of alpha over 2.
So all these quantities that we'd like to determine are hooked together by the law of cosines.
And this is a lovely relation, but it doesn't do us a bit of good, because in this relation we know only one quantity, and that is the rotation angle of a.
We can pick the angle between a and b, that's this, but I have no idea what these other angles are.
That's what I'd like to find out.
I'd like to find out the angles at which three rotations have to be combined in order that rotation about one followed by rotation about the second be the third.
So this equation is a beautiful equation, but it involves everything that I don't know and only one quantity that I do know.
So it looks as though we're up the creek.
Yes, sir?
So you're looking for cosine c, so shouldn't it be cosine b cosine a?
Oh, I'm sorry.
Yeah, I did that wrong.
Yeah.
You're absolutely right.
This should be cosine of a, and this a goes with this alpha over 2.
Absolutely.
Sorry about that.
So anyway, the point still stands that what this equation involves is the three interaxial angles, and I would like to know how I could combine a and b to get it to come out to a crystallographic rotation c, and where that location is relative to the first two axes.
So it involves everything I don't know, and only one thing that I do.
But now we introduce another curious aspect of spherical triangles, which I mentioned last time.
You may have thought that that's interesting, but who cares?
Here are the three points, A, B, and C, and these are the three arcs little c, little a and little b.
And then we said we could construct something called the polar triangle of ABC.
And what we would do, we would find the pole of arc b, and that will be some point B prime.
We'll find the pole of arc a, and that will be some point A prime.
We'll find, similarly, the pole of arc c, and that will be some point C prime.
And now we can connect together A prime, B prime, and C prime, and get something that's called the polar triangle.
And now comes the useful part.
We said that a curious property of the polar triangle is that the side of the polar triangle plus the angle opposite it add up to 180 degrees.
In my original triangle, this is beta over 2, this is gamma over 2, and this is alpha over 2.
So the length of this side is going to be 180 degrees minus beta over 2, the length of this side is going to be 180 degrees minus alpha over 2, and the length of this side is going to be 180 degrees minus gamma over 2.
And now let's use these angles and these lengths in the law of cosines, and I'll leave out the little bit of intervening algebra.
And what we will get out of this is that cosine of c-- and I'll solve for that-- is equal to cosine of alpha over 2 cosine of beta over 2 plus cosine of gamma over 2 divided by sine alpha over 2 sine beta over 2.
And that is something we can sink our teeth into and run with, because now I can ask the question, suppose I want a to be a 2-fold rotation axis, b to be a 3-fold rotation axis, and c be a 4-fold rotation axis?
Then the value of alpha over 2 is half of 180 degrees or 90.
Well, you can see I put in half the value of the rotation axes.
And then I solve for c, and that is the angle at which I have to put axis a and b together to get c to turn out to be whatever angle gamma over 2 is.
So I can do this systematically now without thinking.
And I can set up the problem by taking the crystallographic rotation axes and combining them together three at a time in all possible combinations.
Right?
In addition to this relation, I have two other relations.
And let me assemble them off to the left, because we have to solve three equations to find out the nature of the combination that is required.
So just permuting terms, the angle between A and B, c, has to follow from cosine of c equals cosine of alpha over 2 cosine of beta over 2 plus cosine of gamma over 2.
Notice that the single term by itself is the cosine of half the angle of the opposite rotation axis c. Then in the denominator is the sine of these two angles.
And so just permuting terms, one can see that cosine of b is going to turn out to be cosine of alpha over 2 cosine of gamma over 2 plus cosine of theta over 2 divided by sine of alpha over 2 sine of gamma over 2.
And a third analogous expression will give me the angle that will be the one between B and C. And this will be cosine of beta over 2 cosine gamma over 2 plus cosine of alpha over 2 divided by sine beta over 2 sine gamma over 2.
OK?
So now we don't have to think anymore.
It's just plug and chug.
And I'll pause to suck in air and let you catch up, and then we'll set up the problem and look at a few solutions.
And all this, in the event that you're thoroughly bewildered, is in the set of notes that I handed out last time.
So you can read it over at your leisure
Will this stuff be on the quiz?
No.
Quiz will go up to the end of the two-dimensional plane groups and stop.
We won't say anything three dimensional.
OK, let's, then, if there's no objection or complaint, look at possible values for-- let me do it the same way that I did it in the notes so that it's consistent-- let's put down the value for axis b, the rank of axis b and the rank of axis a.
And A could be a 1-fold axis, B could be a 1-fold access, and we could take a 1 with a 1 with a 1, a 1 with a 1 with a 2, a 1 with a 1 with a 3, a 1 with a 1 with a 4, and a 1 with a 1 with a 6.
This is clearly impossible.
If I did nothing, and followed it by doing nothing, and wanted it to come out to be a 6-fold rotation, you'd all be spinning on your axes like tops right now.
So you can't do nothing and follow it by doing nothing and have it come out to be a net rotation.
So these are impossible.
So we don't have to consider those.
A could be a 2, though, and I don't want to do 2, 1, 1, because I've got a 1, 1, 2 here.
The order doesn't make any difference.
So I'll start with a 2, 1, 2, a 2, 1, 3, a 2, 1, 4, and a 2, 1, 6.
So those are four combinations that I should be examining.
I could look at a 3 with a-- 2 with a 1, I have here in the form of 2, 1, 3, so the next one I would want to look at is a 3, 1, 3, a 3, 1, 4, and a 3, 1, 6.
And let me put in a couple more here.
If B were a 2, I should look at a 2 with a 2 with a 2, a 2 with a 2 with a 3, a 2 with a 2 with a 4, 2 with a 2 with a 6.
And then A 3 with a 2-- and I've got 3, 2, 2 up here, so I'll start with 3, 2, 3, 3, 2, 4, 3, 2, 6, run out of room here, but there should be a similar entry with a 4 and a 6.
So this sets up the problem.
If you count up the number of ways one can do this, we only have to consider the off-diagonal boxes here, because interchanging a and b, for example, looking at 3, 1, 3, that's going to be the same as 1, 3, 3 up here.
So it's just the off-diagonal boxes that we have to consider.
So there should be a 3, 3, 3 in here, a 3, 3, 4, and a 3, 3, 6.
So what we would have to do in order to determine the unique combinations is to look at all of these combinations in turn, and I'm going to not try all of them.
I will do one that's going to be clearly impossible.
So let's look at 2, 1, 4.
So here A corresponds to a 2-fold axis, B corresponds to a 1-fold axis, and C would correspond to a 4-fold axis.
So could we combine these three axes at appropriate angles such that a 2-fold followed by a 1-fold is equivalent to a 4-fold?
This clearly isn't going to work.
If I do a 180-degree rotation then don't do anything and ask is that equivalent to a 4-fold rotation, that is saying that the 2-fold axis should be identical to the 4-fold axis, and that is not going to work.
So let me do now a generic family that I know turns out to be possible.
Let me look at an n-fold axis with a 2-fold axis with a 2-fold axis, and I can show that this combination is possible for any integer n whatsoever.
So this will include a lot of non-crystallographic symmetries.
So let's say that this is C, this is A, and this is B.
But make it C, B, A if you'd like.
OK, so my first equation says that cosine of c, the angle between A and B, should be equal to the cosine of alpha over 2 times the cosine of beta over 2 plus the cosine of gamma over 2 divided by sine of alpha over 2 sine of beta over 2.
B is a 2-fold axis, so alpha is equal to 180 degrees.
A is a 2-fold-- I'm sorry.
B beta, A alpha.
Beta is equal to 180 degrees, the angular throw of a 2-fold axis.
Alpha is equal to 180 degrees, the angular throw of a 2-fold axis.
And gamma is equal to whatever 2 pi over n would be.
That's the throw of the n-fold axis than I'm letting be equal to C. So the cosine of A and B, to get the result of rotation A followed by rotation B being equal to the net rotation of an n-fold axis is that the cosine of c should be the cosine of 180 degrees over 2 times the cosine of 180 over 2 plus the cosine of gamma over 2, and gamma is whatever the rank of the axis determines, and that's divided by sine of alpha over 2, and alpha is 180 and sine beta over 2, and that's 180 over 2.
So the cosine of c is going to be the cosine of 90, which is 0, times the cosine of 90, which is 0, plus the cosine of gamma over 2, whatever that might be, over the sine of 90 which is 1, sine of 90 which is 1.
So this says that cosine of c is equal to cosine of gamma over 2.
So the angle between axis A and axis B ought to be equal to one half the angular throw of rotation axis C. So let's start putting down some of this information.
This says that if this is axis C gamma, then A pi and B pi, the 2 180-degree rotations, should be at an angle gamma over 2.
We still need values for b, and we still need a value for a.
So let's find out what those are.
And let me start over here at the left again, because I've got the relation that I need for b sitting here.
The cosine of b is equal to the cosine of alpha over 2, and that is the cosine of pi over 2, plus alpha is a 90-degree rotation.
Then cosine of gamma over 2, whatever gamma happens to be, plus the cosine of beta over 2, and that's cosine of pi over 2, and this is all over sine of pi over 2 times the sine of gamma over 2.
So this is going to be cosine of pi over 2, which is 0, plus cosine of pi over 2 divided by sine of pi over 2, which is 1.
This gamma over 2-- no, cosine of pi over 2 is 0.
So this is 0 plus 0 times sine of gamma over 2, whatever that turns out to be.
So cosine of b is 0, and that says that the angle b between axis A and axis C turns out to be 90 degrees.
So in order to get pi followed by c gamma to be equal to b pi, I've got to make b be equal to pi over 2.
And if I put it in this orientation, then a pi followed by c gamma is going to be equal to b pi.
One final angle, and that's the value for a. OK, cosine of a should be equal to the cosine of beta over 2, that's pi over 2 times the cosine of gamma over 2, whatever it is, plus the cosine of beta over 2, and that's cosine of pi over 2, over sine pi over 2 sine of gamma over 2.
And this, as for b, turns out to be 0 plus 0 over sine pi over 2, which is 1 times sine of gamma over 2.
So cosine of a turns out to be 0, and this says that the angle a is also pi over 2.
So we've got a whole-- Yeah
Did you really need to go through all three equations?
Yeah, because I had to show that all three work.
And in general, if I combine, let's say, a 4-fold with a 3-fold with a 2-fold, which is something I want to do, all three angles a, b, and c, will be different.
OK?
So the answer is yes.
And there will be a few cases where a value for one angle will exist and the value for the one or two others will be impossible.
And that's also something that I have to know.
So repetitious as the exercise might be, the answer is yeah, you do have to do all three
[INAUDIBLE]?
Oh, you don't have to do different permutations.
That's just a question of labeling.
OK?
So n with a 2 with a 2 is the same as a 2 with an n with a 2 is the same as a 2 with a 2 with an n, that's just labeling.
So that's why my boxes, when I filled them out, the list got shorter and shorter, until finally for a 6-fold axis, it would be just 6, 1, 1, 6, 1, 2, 6, 1, 3, and so on.
Just that one entry in the box where I enumerated what should be considered.
Well, here is a whole slew of possible solutions and a lot of them are non-crystallographic, but still possible.
This says that a combination of a 2-fold axis-- but remember now that these are equations and operations.
But the only operation that's present for a 2-fold axis is a rotation a pi, b pi or c pi.
So I could combine three 2-fold axes that are mutually orthogonal, and that is an allowable combination.
And what we are obtaining here is a sort of scaffolding, a framework, based on pure rotation operations, that by themselves will be an allowable 3-dimensional point group, but which also provides a framework, a Christmas tree, that we can decorate with mirror planes and inversion centers to get still additional groups of higher symmetry.
So here's one possible crystallographic o combination of rotation axes.
What we will use to denote this combination is the same rule that we use for our other notation.
We will make a running list of the independent operations that are present, and what we have combined here are three distinct independent 2-fold axes.
A solid that would have this symmetry plus some other symmetry would be an orthogonal brick with one 2-fold axis coming out here.
This is the operation c pi, another 2-fold axis coming out the front, and this would be the operation a pi, and another 2-fold axis coming out of this face, and this would be the operation b pi.
Now let me show you-- it's rather amusing-- that what we have done really works.
We've shown supposedly that a pi followed by b pi should be equal to a net rotation c pi about an axis that's orthogonal to the first two.
So let's pick a motif, and for convenience I'll put it at one corner of this brick.
Here's object 1, I rotate it by 180 degrees about a.
Here sits object number 2, same corality, and then I rotate it 180 degrees about B beta, and that's going to give me number 3.
What is the net way of getting from 1 to 3?
Holy mackerel.
It's a net 180-degree rotation about c pi.
It really works.
Or I could do the operations in a different order.
I could rotate by d, rotate by c, and the way I get from the first to the third is a rotation a pi.
So that is a self consistent set of rotation axes.
That is 2, 2, 2.
Let me do one more.
Well, no, let me take a break here and let you absorb all this, and then we'll look at some remaining ones, and this will include some that are non- crystallographic.
And that's perfectly OK.
But they're lovely groups, they constitute groups, but they won't be groups that can occur in combination with a lattice.
Any questions about where we left off-- up to where we left off?
OK, what I'll do then is give you a few more examples of the combinations in 22 to show which ones we have to retain as frameworks for crystallographic point groups and which ones exist as groups but which involve rotational symmetries that are not permitted to a lattice.
So we've seen a combination of three orthogonal twofold axes and then projection that would look like this.
And the international symbol for that point group is just a running list of the different axes that are present, 222.
The next group in the sequence would be 3 2 2, where we took a 120 degree rotation.
We combine that with a twofold axis perpendicular to it and the new twofold axis comes out and reminds you again of things that are quite clear but which are easy to forget-- that this angle here is 1/2 of 2 pi over 3.
Don't forget that 1/2.
So the neighboring twofold axis is 60 degrees away and then if we allow these axes to operate on each other, the net symmetry consistent set of axes looks like this.
But let us look at a solid that has this symmetry.
And such a solid would be a trigonal-- triangular prism.
And again we can use the corners of this polyhedron as the reference locations of our motifs.
So let us put a first motif here, number 1.
Let's rotate it by 120 degrees to get a second motif here.
And then let us rotate that one down by a twofold axis coming out of one of the edges.
That will give us number 3 that is down here or of the same chirality.
And how do we get from 1 to 3 directly in one shot?
And the answer is about a twofold axis that comes out of the face of the prism.
So again the prism that we've used as our reference is a prism that looks like this.
We've got one twofold axis coming out of the face.
The next twofold axis is the one that is equivalent to a threefold rotation followed by a twofold rotation.
Now here we hit a situation in deciding on the name for the combination that is analogous to what we found in two-dimensional plane groups for 3mm.
We saw that only one mirror plane was distinct.
We look at this arrangement of axes-- this was axis a, alpha, this was axis b, beta.
But if I repeat one of these twofold axes by 120 degree rotations, this one is the opposite end of this one, this one is the opposite end of this one, and this one is the opposite end of this one.
So there are only three kinds of-- there are three, twofold axes and they are related by the 120 degree rotation.
Another way of saying that is that if we look at a trigonal prism, each of the twofold axes comes out of an edge and out of the opposite face.
So there they all do the same thing in this regular prism-- twofold axes extend between corners of the opposite face.
So there are only-- there's only one kind of twofold axis present in terms of being symmetry independent.
So just as we called 3mm, 3m, and we call this one 3 2 because all the twofold axes are symmetry equivalent.
The next one that is crystallographic would be a combination of a 90 degree rotation with a pair of twofold axes that are normal to it and separated by 1/2 of pi over 2, the [?
throw ?]
of a fourfold axis.
And the symbol for this one would be a fourfold and now there are two different kinds of twofold axes.
And if we look at a regular square prism, we can again show that what we've been demonstrating for these other prisms is true.
One twofold axis would come out of the face, the other twofold axis would come out of the midpoint of an edge, and the fourfold axis would come up here.
And a 90 degree rotation, from here to here, followed by a twofold rotation about this axis, gives us as a net effect a 1, 2, 3.
And that rotation from 1 to 3 about this twofold axis gives us the combined mappings.
Notice that the order in which we do them is unimportant.
For example we could do the same thing but do two 180 degree rotations about the twofold axes.
Let's say we start by doing a rotation about this twofold axis.
From here down to here.
And then we do a twofold axis about the-- twofold axis that comes out the face and that would take number 2, and rotate it up to number 3.
And the way we get from 1 to 3 directly is by a rotation of C pi over 2 about the square face of the prism.
So do them in any order you like-- two 180 degree rotations, or a 90 degree rotation, one of the two full rotations with 90 is the other twofold-- the other type of twofold axis.
The 90 degree rotation plus the second type of the twofold axis is the same as the first.
So the result can be permuted and turns out to be the same combination.
The next one that we would hit if we proceed systematically is non-crystallographic.
And this would be a fivefold axis.
And I'm foolhardy to even start trying to sketch this in three dimensions.
Nevertheless, nothing ventured, nothing gained.
Start with a twofold axis out of one of the edges and rotate from 1 down to 2.
Follow that by a twofold rotation about the twofold axis that comes out of the face.
And that gives us one number 3 up here, and lo and behold, the way you get from 1 to 3 directly is by a rotation through 1/5 of 2 pi.
So this would be the non-crystallographic point group, 5- well it's not going to be 5 2 2.
And that has a whole bunch of twofold axes separated by 1/10 of 2 pi.
And just as in 3 2, twofold axes here all come out of the face and out of the opposite edge.
So this would be called 522.
A nice symmetry but not crystallographic, so we can promptly forget about.
So what comes out of this is a family of groups that are all of the form n 2 2.
The crystallographic ones are 222, 32, 422 which we've looked at in detail, and one that I won't draw because there's so much symmetry it get's messy.
This is 622.
There is a Schoenflies notation.
You remember the language that we encountered for our two dimensional point groups.
We used m for mirror in the international notation.
We used C standing for cyclic group, subscript s standing for Spiegel in the Schoenflies notation.
The Schoenflies notation for all of this family of symmetry is Dn, and the D stands for dihedral.
And the reason for that name is that the difference between all of these groups, besides the n-fold axis, is this angle between adjacent twofolds and this is a dihedral angle.
I have a set of planes passing through a common axis; the angle between those planes is termed a dihedral angle.
So this is called D for dihedral and then a subscript that gives the rank of the axis.
So Dn generically.
This is D2, this is D3, this is D4, and this is D6.
Comments or debate?
Yes, sir
[INAUDIBLE]
[INAUDIBLE] --out of this face.
So I got from here down to the diametrically opposed axis.
And I rotate it about the adjacent one which comes out of an edge and-- what did I do here.
Here I did A pi over 2 from 1 to 2, and then I did B pi, where this is B pi, and they turned out to be C pi, which is this one here.
So going from here to here down to number 3 is the same as going from 1 to 3 in one shot about a twofold axis normal to the face.
And actually I'm courageous to try to do this in three dimensions.
We could do it in projection and then things are used-- this for a point that's up, and use this for a point that's down.
And then what we've done is to go from 1 that's up, to 2 that's up, and then we rotate it about this twofold axis.
That was 3, that's down.
So when we get into complicated symmetries where you just can't do a proper job drawing them in an orthographic drawing, we'll do them in projection and use a solid dot for something that's up and an open circle for something that's down.
Is there any other way we can combine things?
Well what you would have to do is use these three relations, and plug and chug your way through all of the combinations which were enumerated in the handout.
And I'll save ourselves a lot of work by saying that there are only two more combinations.
And these are combinations of axes at angles that have relevance to directions in a cube.
One of them is a twofold axis with a threefold axis with a threefold rotation.
Again remember these are not equations in symmetry elements.
This is really A pi, combined with B 2 pi over 3, combined with C 2 pi over 3.
And the angles that fall out of this are all crazy things like 109 point something degrees and they make no sense whatsoever.
They're not nice things like some multiples of 2 pi, 90 degrees or 120 degrees.
And they make no sense at all until you refer them to directions in a cube.
And this one, 2 3 3, consists of a twofold axis coming out of the face of the cube, a 120 degree rotation, so this is B 2 pi over 3, this is A pi.
And the other one, C2 pi over 3, corresponds to a threefold axis coming out of another body diagonal.
I don't know if I want to be gutsy enough to try to illustrate that that really works.
But one thing that we should do is to let these axes go to work on one another, and see what comes out.
First thing we can say is that this threefold axis-- if we extend it, it comes out of the bottom diagonal of the cube.
This one, if we extend it, will come out of this diagonal of the cube.
So there's a threefold here, and a threefold here.
The twofold axis gives us a threefold axis that will come down this way.
So there's a threefold axis here and a threefold axis coming out here.
And then the twofold axis will rotate this threefold axis over to the remaining pair of corners.
So we have created this by looking at a twofold rotation, combined with the threefold axis, combined with the rotation of another threefold axis.
But in point of fact, if you let the twofold axis operate on the threefold axis coming out of one body diagonal, you get threefold axes automatically out of all body diagonals.
So this is given the international symbol 2 3, because if you start with one twofold axis out of a face normal and one threefold axis out of-- along a body diagonal, the twofold axis, acting on that threefold axis gives you one along every body diagonal.
And the threefold axis-- let me point out that a cube standing up on its body diagonal, with a threefold axis coming out here and these faces sloping down into the blackboard.
If I have a twofold axis coming out of one face, the threefold axis puts a twofold axis on this face and rotates again 120 degrees and puts a twofold axis on this face.
So the threefold axis relates all twofold axes coming out normal to the cubed face.
And they were really just two independent axes in this combination.
So 2 3 is the international symbol-- one kind of twofold axis, one kind of threefold axis-- inclined at these crazy angles that are in a cube.
The Schoenflies notation for this is T, and that stands for tetrahedron.
And let me try to convince you that if I look at a tetrahedron, that that is the arrangement of pure rotation axes in a tetrahedron.
And the way to show a tetrahedron is to inscribe it in a cube.
So if I connect these two faces together and these two faces together, that will define for me a solid that has four triangular faces.
Easier to recognize it when we put it up on one face.
So this is a tetrahedron.
Schoenflies symbol is T. And I love to get to this part of the semester and be at this point at the end of the hour, because if I draw a stereographic projection of the twofold axes, in this symmetry, I can take a couple of more twofold axes and add it to this combination which destroys the group.
But lets me wish everybody a happy Halloween and exit to a stunned silence at the end of the hour.
So this is a nice point group for October.
There is one more and that is the highest symmetry of all.
And this is a combination of a rotation-- A pi over 2, a 90 degree rotation, with a rotation B 2 pi over 3, rotation through one third of the circle, and the rotation C pi.
And the directions between the axes that come out there don't come out some multiples of 2 pi.
Again they are crazy angles that make no sense at all unless you refer them to directions that occur in a cube.
The fourfold axis is in the direction that corresponds to the normal to a face.
The twofold axis corresponds to a direction out of one of the edges.
And the threefold axis corresponds to a direction that is a body diagonal.
So again this is a mess.
If we let those axes work on one another however, we will produce, more readily appreciated in a stereographic projection, fourfold axes along the directions that correspond to face normal.
So the cube, threefold axes coming out of the body diagonals and twofold axes in between all of the fourfold axes in directions that correspond to the lines from the center of the cube out through the edges.
So this is a group which would be called 4 3 2.
There's one kind of fourfold axis related to all the others by the other rotation axes that are present.
One kind of threefold axis that is related to all of the other threefold axes along the body diagonal by other rotation axes that are present.
And twofold axes, one kind, all coming out of the edges.
So this is the group that, in international tables, is called 4 3 2.
And the Schoenflies notation, this is called O.
And that is the general reaction when one sees this lovely combination.
You go ooh and O is what it's called.
But the O doesn't stand for a gasp, it stands for an octahedral.
This is the symmetry of an octahedron-- rotational symmetry of an octahedron.
So that's it.
That is the bestiary of ways in which you can combine crystallographic rotation axes in space about a fixed point of intersection.
And there are eleven of them.
There are the axes by themselves-- 1, 2, 3, 4, and 6.
There are the dihedral groups, 222, 32, 422, and 622.. And then the two cubic groups, T and O.
In the international notation, I'm mixing metaphors.
These are 23 and 432.
I call to your attention the insidious similarity of the two combinations of axes, 32 and 23.
When the 3 comes first, this is a group of the form n22.
When the 2 comes first, that is the tetrahedral group.
So if you count them all up, there are 4, 2 is 6, and 5.
There are eleven axial combinations.
Quite a few more than the situation in two dimensions where we just had single rotation axes 1, 2, 3, 4, 6.
Now we have those as in two dimensions but the dihedral groups and the two cubic arrangements of axes as well.
So we don't have to stretch our vocabulary too much more to be all inclusive here.
OK, comments?
Takes your breath away, doesn't it?
All right, let me indicate the next step in outline.
What we will next do is introduce our remaining two symmetry operations into the picture.
We have the eleven axial combinations.
As I said, we can regard these as a framework that we can decorate with mirror planes and or the inversion center, which has not appeared until this point because inversion is inherently a three dimensional transformation.
So what we're going to do is to take these axial combinations and add another symmetry operation to the group.
And this as-- we used the term earlier, this is an extender.
We have something that constitutes a group by itself then we muck things up by adding another operation.
Remember in all of this we are combining operations, not symmetry elements.
So we'll take an axial combination and add the reflection sigma, a reflection operation sigma.
Or take a rotation and combine it with the operation of inversion as an extender.
So let's itemize the sorts of extenders we should consider.
And the ground rules are that the extender should leave the arrangement of rotation axes invariant.
Because if it doesn't, we are going to create a rotation operation that does not conform to the constraints that we used in Euler's construction.
So for example, if we take 222, which contains the operations A pi, B pi, C pi and identity, that's the group 222.
If we would add to the arrangement 222 a mirror plane that snaked through some arbitrary fashion like this, that mirror plane is going to reproduce the twofold axis over to here.
And that is either going to not constitute a group because this twofold, this twofold, and this twofold don't conform to Euler's construction.
Or alternatively, if we put it in carefully at 45 degrees with this twofold axis, we're going to get twofold axes that are 45 degrees apart and that's going to change this into a fourfold axis.
So if the addition of a mirror plane does not leave the arrangement of rotation axes-- rotation operations-- invariant, we're either going to get something that's impossible and does not constitute a group.
Or else we're going to get a combination of rotation operations of higher symmetry which we've already found because we went through that process of combination using Euler's construction in an exhaustive fashion.
So the rule then is that if we add the reflection operation sigma, then the arrangement of rotation operations must be left invariant.
OK let's look at the single axes.
If there's an n-fold axis, the ways we can add a reflection operation to that axis is to pass the reflection operation through the axis.
And this is going to look very much like the two dimensional point groups of the form nmm except that rather than having a mirror line, imagine the whole works as extending upwards along the rotation axis and space.
So this extender is called a vertical mirror plane.
The other way we could add a mirror plane to an n-fold rotation axis is to put the mirror plane in an orientation that's perpendicular to the rotation axis.
That didn't exist in two dimensions because that plane that's perpendicular to the axis is the plane of our paper.
And unless we wanted to have a two-sided group that was on both the top of the paper and the bottom of the paper, and we make up the rules since it's our ball game.
And that could be a group and these would be the two-sided plane groups, a plane point groups, but we didn't do that here.
Adding the mirror plane, the reflection operation sigma, in a fashion that is normal to the rotation operation, A 2 pi over n, is another distinct combination.
And this is called, very descriptively a horizontal mirror plane.
That looks like about all you can do except for the cases where we have more than one kind of rotation axis present.
So let me use 422 as an example.
We could add a horizontal mirror plane perpendicular to the principal axis of symmetry, and that would be the horizontal sigma.
We could add a vertical mirror plane.
And now there are two ways we can do it.
We could put the mirror plane, the operation sigma, through the fourfold axis and in a fashion that was perpendicular to the twofold axis.
And we will retain the term of vertical sigma for that addition.
But the other thing that we could do would be to put the reflection operation interleaved between the twofold axes.
That's going to take this one and flip it into this one, this one flip it into this one, flip these back and forth, and that doesn't create any new reflection.
And this is referred to as a diagonal reflection plane.
Diagonal to what?
Diagonally interleaved between the twofold axes.
And these are distinct additions and they will lead to different groups.
In as far as addition of reflection operations is concerned, that's about all we can do that's distinct.
And notice that this is for the groups Dn, and tetrahedral, and octahedral only.
The distinction here is not defined for just a single axis.
So there are three possible extenders here-- a vertical mirror plane, a horizontal mirror plane, a diagonal mirror plane, added to each of the eleven arrangements of rotation axes.
And then the final extender that we could add is to add inversion.
And the symbol for the inversion operation is 1 bar.
And obviously if one point in space is going to be left invariant, you either add this on a single axis, and that is what you'd have to do for the groups Cn, or at the point of intersection.
And that would be the case if more than one axis, and that's the groups of the form n22, T, and O.
That's it.
That's the job.
So we should consider each of these possible additions of an extender systematically.
I don't propose to do every single one independently.
If we do a couple, you'll get the general idea.
And I think because I have an honest face, and you've come to trust me, I can just describe the remaining results to you and we won't grind through every single one.
Now the enormity of what I've proposed becomes apparent when I say that we now are going to have need of a number of different-- what I call combination theorems, that let us complete the group multiplication table.
And deduce, as a consequence, which symmetry operations must come into being because of these additions.
So we'll want to know what happens when you add a vertical sigma to a rotation operation, A 2 pi over n. We'll want to know what happens when you add a horizontal sigma to a rotation operation, A 2 pi over n. And we're going to want to know what happens when you add a diagonal mirror plane, this really is a special case of the vertical mirror plane.
And we'll want to know what happens when you add an inversion center to a rotation axis.
So let me do a few of these and then next time we can start off and start driving the three dimensional symmetries.
So let's just repeat the ones that we've already done.
We said that if we have a rotation operation A alpha, and we put a reflection plane through it, that A alpha followed by a reflection plane passing through it-- let me call this sigma V-- because this is the so-called vertical, up orientation.
We've already seen that in two dimensions.
This is a vertical mirror plane, sigma prime, that is going to be alpha over 2 away from the first.
So that is something that we've already seen in it's entirety in the two dimensional point groups.
There was m sigma, there was 2mm, and that was C2V, 3m, that's C3V, 4mm, and that was C6V, and 6mm, and that's C4V, and that's C6V.
So that is the result that we obtained for two dimensions.
And you can see now the reason for distinguishing the mirror plane by saying it is a vertical mirror plane because this is in the three dimensional sense.
It's vertical parallel to and passing through the rotation axis-- no longer a rotation point but a rotation axis.
So we've got those theorems.
What happens if we take a rotation operation, A pi, and put a horizontal reflection operation through it normal to that axis?
So I'll call this sigma h, a horizontal operation.
OK what we have to do is draw it out once and for all.
Here's the first one, let's say it's right-handed.
We'll rotate by A pi to get a second one which stays right-handed.
And then we'll reflect it down in the horizontal mirror plane to get a third one which is left-handed.
And now the question is, how is number one related to number three?
Anybody want to hazard a guess?
We've got to go from a right-handed one to a left-handed one.
But these two guys are oriented anti-parallel to one another.
So how do we relate the first one to the third one?
I heard somebody mumble softly enough to remain anonymous.
Inversion.
Right.
So as we go along making these combinations, if we had not been bright enough to think of the operation of inversion as a general transformation where the sense of all three coordinates is changed, we would have stumbled over headlong right here.
Combine a rotation operation, A pi, with a mirror reflection that is perpendicular to the axis.
The way you get from one to three in one shot is by inversion through a point that is at the intersection between the rotation axis and the mirror plane.
So rotation followed by rotation in a vertical mirror plane that's perpendicular to the axis is the operation of inversion at the point of intersection.
So again, if we had not been clever enough to invent it or tell you about in advance, there it is.
When we start forming the group multiplication table, we would have had to have defined this operation to describe this relation
Is that sigma sub h?
Ah yeah.
Let me write that, sigma h. I thought that V didn't look like a V, so I changed it so it looked like a V but it shouldn't be a V. That's a horizontal mirror plane.
OK this is a new combination and if we see what operations are going to be present in the group, we've got the two operations of the twofold axis, 1 and A pi.
And what we have added is sigma h as an extender.
So 1 and A pi, this little box here is the subgroup that we know and love as the twofold axis.
And then we'll write sigma h here and now let's fill in the group multiplication table.
Doing the identity operation twice is identity.
Doing the identity operation followed by A pi is A pi.
Identity followed by sigma h is sigma h. Identity followed by A pi, sigma h, lets me fill in those boxes.
Do a rotation twice, that's the identity operation.
Do a rotation of A pi and follow that by a reflection, A pi followed by reflection.
This is the inversion operation.
And so I should add inversion to my list of operations since it's come up.
So 1 followed by inversion is inversion.
A pi followed by inversion is the horizontal reflection.
Horizontal reflection followed by A pi is the same as inversion.
Horizontal reflection followed by horizontal reflection is the identity operation-- brings me back to where I started from.
And a horizontal reflection followed by inversion is the same as A pi.
So I'll have another object down here to complete the three dimensional arrangement.
And this is the group for operations-- A pi, inversion, horizontal reflection, and the identity operation.
So what do we call this one?
The operation A pi plus a horizontal reflection operation gives rise to a group that is a twofold axis of its operations, perpendicular to a mirror plane.
And this written as a fraction means that the 2 is perpendicular to the mirror plane rather than being parallel and in the plane of the mirror plane.
So two codes for writing symbols.
A 2 followed on the same line by the m means that the m is parallel to 2.
We know that when we write them as a fraction, that will be our way of designating that this mirror plane is perpendicular to a twofold axis.
OK it's the witching hour, 4 o'clock exactly.
That's when we ought to quit, so let's stop there.
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All right.
The quiz on Thursday will cover up through piezoelectricity.
You've had a set of notes in your hands that cover just about everything that I wanted to say about piezoelectricity.
There are other modulae that one could talk about.
But these follow quite directly from the one or two that we will do.
So we will not have anything to say about elasticity until the last lecture of the term, which is a nice outcome because you certainly don't want to take a quiz on forthright tensors.
You'll spend the entire hour just writing out all these cumbersome equations.
All right.
So the quiz will cover up through piezoelectricity, including a few of the representation surfaces that we'll examine today.
The thing that we'll be looking at is-- I'm sorry.
You have a question?
Yes.
A question.
Well, you said [INAUDIBLE] example
I'm sorry.
Third-rank tensors

Also, we had just a little bit of that going into the second quiz.
And we didn't really ask anything about that on the second quiz.
So it'd be third-rank tensors.
But you have to use second-rank tensors to define third-rank tensors.
So it really will be not the emphasis, but certainly you should be familiar with the early part of what we did with second-rank tensors.
All right.
So today we'll look at some representation surfaces to your wonder and delight at how incredibly anisotropic the variation of these third-rank properties are with direction.
One of the things that I love to do for problems when we get to these different piezoelectric effects is make up hypothetical devices just for fun.
So among those that I've invented, the first one is a earthquake sensing device because it's well known that California is going to split in half, and half will fall into the sea any day now.
So if this is the San Andreas fault, I have developed large, prismatic monoclinic crystals that I embed into the San Andreas fault at regular intervals.
The bottom of these crystals is grounded.
There's an electrode at the top.
And if the San Andreas fault starts to move, and there is shear on these crystals, there will be a charge developed on the top.
And I ask you to relate the charge on the top in terms of the piezoelectric modulae to the shear along the San Andreas fault.
So there's a very clever little device that clearly is going to be lucrative because these crystals will have to be huge in size.
And they'll cost more even than silicon.
Another device that I've invented is used down at the Boston fish pier.
This is a crystal on which I hang a pan, and you put fish in it.
That creates a tensile stress in this direction.
We measure the charge on this face.
And we put a little meter that measures the charge that's accumulated.
So this is how the fishermen can weigh their fish after they haul them in off the boat.
So you see this is really practical material we're dealing with.
This has applications in all realms of life.
So today I'd like to show you, also, another problem that sets up.
And I pass this out just so you can, again, have a look at some of the questions that you should be equipped to answer.
So here-- not fully expecting anybody to do it, but you can see the sorts of problems one might ask.
Here is problem set number 16, which asks you, should you be so inclined, to think about manipulations of third-rank tensors.
But then the second question is an idea for a device which the Center for Material Science and Engineering is considering employing here in building 13.
And you can read all about that..
This is the famous soup cell.
I think probably you will see some sort of for fun modulus for a particular device on the quiz.
I haven't made one up yet.
But I think you'll have a look at something like that.
And when we do the longitudinal piezoelectric modulus for quartz, which we'll do momentarily, this will give you an idea of how to set up these expressions.
The problem, basically, is that the direct piezoelectric effect measures a polarization in terms of all of the elements of applied stress.
So there is a modulus dijk times all of the elements of stress, sigma jk.
So this, until you use the condensation of subscripts, contains nine terms going this way and three equations going this way.
As we discussed earlier, since only six of the nine elements of stress are independent, you can condense this down into a 3 by 6 array of terms.
The problem in doing that, though, even though one has only 18 different modulae to work with, instead of 27-- that's a considerable economy in notation, and an elimination of a great deal of redundancy-- the matrix form of this relation-- and I emphasize that it is a matrix, and it's no longer a tensor-- the matrix form cannot be transformed.
Again, there are only six terms in the matrix elements of stress.
And there are three equations, again, one for each component of polarization.
So instead of having 27, one has only 18.
But if you are considering changing the reference axes-- and that is one of the things that it's interesting to do for these various modulae that one can define-- change the orientation of a particular rod-shaped specimen that you cut out of a crystal to different crystallographic orientations and then ask how the scalar modulus changes as you change the direction in which you've sliced out the wafer or the rod of material.
In order to do that, you want to transform the piezoelectric tensor to a new set of reference axes.
And you cannot transform the dij's because they are a matrix and not a tensor.
And no law of transformation is defined.
So what you have to do in any generic problem of this sort is, for a particular single crystal, look up the form of the dij matrix that will have the equalities between tensor elements written in and the 0's, those modulae which are identically 0, entered into the array.
And then you have to work your way backwards to get to the full tensor notation.
So we'll see this when we look at the modulae for symmetry three two.
You'll have to write in the exact matrix subscripts without absorbing the equalities in the notation.
And then you'll have to expand the matrix terms into full three-subscript tensor terms.
Then if you want to transform the axes, which is to say you want to cut out your specimen, be it a plate or a rod, in a different orientation, you have to transform the full three subscript tensor elements to a new setting.
And then if you want to continue to work in that setting in the compact matrix form, collapse it back down to matrix form, insert the equalities, and then you're back to where you started from.
So this problem of seeing how modulae that describe different phenomena vary with crystal symmetry, you have to go through this problem of expanding the compact form and then collapsing back down when you've got it as a function of some orientational angle or in terms of a coordinate system that you want to work in.
The problem is exactly the same for elasticity.
And we'll look at some of these modulae next term.
You've heard these names before, I'm sure-- Young's modulus, shear modulus, and so on.
We'll take a look on next Tuesday, a week from today, at Young's modulus, which is one of the more important ones.
And probably are used to seeing this in the form of information for a polycrystalline material, which is essentially isotropic.
The modulus is much more interesting for single crystal materials.
And then the surfaces that are defined particularly for the lower symmetries are absolutely wild things with lumps and wiggles and lobes and things of that sort, nothing like the dumb old, uninteresting ellipsoids that we encountered for second-rank properties.
So let's, then, take a look at how we had set up and defined the direct piezoelectric effect.
We set this up as a proper tensor relation, saying that P1 is equal to d1.
V1 And then, you'll recall, we have nine elements of strain-- sigma 11, sigma 12, sigma 13, sigma 21, sigma 22, sigma 23, sigm a 31, sigma 32, and sigma 33.
But the tensor is symmetric, so we really only need to enter into our relation six of these nine terms explicitly.
And what we did was to replace the two subscripts on the elements of strain, which, again, you need if you want to refer to those elements of stress through a different coordinate system.
You have to know the elements of stress in tensor form.
But we convert it to six terms by going and replacing the pairs of subscripts with a single one, two, and three, marching down the diagonal of the tensor this way and then marching up the right-hand side, calling two, three, four, and calling one, three, five, and finally ending up in this slot here.
And we call that six.
So that was the notation we used to get to a matrix representation.
But the place where all this started is-- and I'll write just one line of this to be merciful-- d 111 times sigma 11, so this pair of subscripts goes with this pair, plus d 122.
I'm putting that one in next because we're going to number the terms for stress in this order, one through six.
Times sigma 22 plus d 133 times sigma 33 plus d 123 times sigma 23 plus d 132 times sigma 32 plus d113 times sigma 23 plus d 132 times sigma 32.
And finally we end up in slot number six.
And we have a d 112 times sigma 12 plus a d 121 times sigma 21.
Look at that.
There's an equation that covers two whole blackboards.
So if we now condense this down to matrix form we would say that P1 is d 11 times sigma 1 plus d 12 times sigma 2 times d 13 times sigma 3 plus-- and now we have this messy problem with the 2's-- we have a d 14 times sigma 4 plus, again, a d 14 times a sigma -- 2 3 is equal to 32, so I can call this 4-- and then these terms become 15 sigma 5 and, again, a 15 times sigma 5 plus a 16 times sigma 6 plus d 16 times sigma 6.
Now we have to make a choice.
Either we are going to have, in a general matrix relation, that P sub i is equal to dij times sigma j, if j is equal to 1, 2, or 3.
But it's equal to 2 dij times sigma j if j is equal to 4, 5, or 6.
And that is something we like to avoid, if possible.
That's ugly.
That's ugly.
And it's going to be a hell of a matrix if we have factors of two in front of some of the matrix elements but not in terms of others.
So this is something we could do.
Hey, it's our ballgame.
We make up the rules.
But that's going to be an ugly thing to have to deal with.
So instead, as we mentioned last time, what we will do is to lump together these terms and define d 14 not as these individual tensor elements, but define those matrix elements as the sum of these two elements.
And then we saw before that-- we saw last time in our earlier meeting-- that from the converse piezoelectric effect, which expresses strain in terms of an applied field, where the elements of strain 1, 23, and 32 appear in separate equations.
And knowing that the same array of piezoelectric coefficients amazingly describes the converse piezoelectric effect as well as the direct piezoelectric effect, we know that d 123 equals d 132.
And that is from the converse effect.
So equivalent to saying this is to define d 14 as twice d 123 because these two elements are equal.
So we're eating the factor of two here so that we can write a nice matrix relation that doesn't involve a factor of two.
So making this combination of terms for the shear stresses, we would have simply d 14 sigma 4 plus d 15 times sigma 5 plus d 16 times sigma 6.
And we can say, in general, for the other two equations, by analogy to this one, that P sub i is dij times sigma j-- a nice, neat matrix relation but one for which, unfortunately, there's no law of transformation for the matrix modulae dij.
If you want to change to another coordinate system, we have to be prepared to resurrect this full three subscript notation on the piezoelectric modulae.
OK.
Comments or questions at this point?
All right.
If not, let me remind you that in the notes which I distributed last time, there are summarized all of the constraints imposed on the piezoelectric modulae for single crystals.
And, again, these constraints, these requirements that the tensors remain invariant for the change of axes produced by a symmetry element that the crystal possesses, these transformations show that no third-rank tensor property can exist in a crystal that has inversion.
So the 11 [INAUDIBLE] group, so-called, that possess inversion have absolutely no property and can be not considered further for third-rank properties.
And then one must consider all of the 32 minus 11 21 point groups that lack conversion separately.
There's no reason why they should behave the same way.
Remember that for second-rank tensor properties we pulled the argument that inversion imposes no restrictions or constraints whatsoever on second-rank properties so, therefore, two different symmetries that differ only by the presence or absence of an inversion center.
That is to say, you change 2 to 2 over m if you add inversion.
But the argument was inversion requires nothing, so the constraints or symmetry, too, look exactly like those for symmetry.
And here you've got to plod through every single one of the non-centrosymmetric point groups separately.
And they all have tensors that have different forms.
So what I'd like to do is look at one specific one.
And that is the matrix for symmetry 32.
And that is a point group that you'll recall has a threefold axis and twofold axes at intervals of 60 degrees.
And in your list of the qualities and absences, 32, where the 3 is parallel to the axis x3, and the twofold axis is parallel to x1.
So we're defining this as x1, this as x3, and x2 comes out halfway between the two.
So let me draw this looking down along the threefold axis.
These are all twofold axes.
And we'll take x1 in this direction.
x2 pokes out in between two twofold axes.
And x3 comes straight up along the threefold axis.
With that coordinate system, the form of the piezoelectric modulus matrix has this form-- d 11 minus d 11 0 d 14 0 0 0 0 0 d 15 minus d 14 0 and in the bottom row d 31 d 31 d 33 0 0 0.
So this matrix with the equalities put in is obtained by looking at the full three-subscript tensor and requiring that it look the same before and after any of the rotations involved by these three distinct axes-- the threefold and the pair of twofolds.
OK. Now, there are many different scalar modulae that one could define, some of them serious and worth the consideration because of the application in devices or other practical situations.
The modulus that I'd like to examine is something called the longitudinal piezoelectric effect.
And let's emphasize, again, that it is impossible to come up with one representation surface that fits every need because, again, the direct piezoelectric effect relates the components of a vector to a tensor, sigma ij.
There are six independent tensor elements.
So how can you describe how this vector is going to change as you change orientation of a crystal whose behavior is described by all of these modulae?
But I would point out, however, that there are only two distinct-- oops, I'm sorry.
This is d 14.
I don't know how I made that d 15.
There are only, in this array-- and I slipped a notch.
Excuse me.
Last couple of days are such that I am not able to even read from my notes.
So these bottom lines, my apologies, are all 0.
And there are two modulae, d 11 and d 14.
So there are two independent numbers, but they appear as different matrix elements and, therefore, different tensor elements, as well.
And this should be minus 2 d 14.
I'm sorry.
I slipped down a notch, and I got some for 32 and some for 6.
And I'm glad I found it at this point, or I'd really be in deep trouble.
OK.
So two numbers and that is, in fact, the form of the matrix for symmetry 32.
So the longitudinal piezoelectric effect is one of the representation surfaces that gives you the way in which the polarization will change for one very, very specific type of stress.
In particular, what we'll do is set this up as a coordinate system.
And we will look at a coordinate system where this is the reference axis, x1.
We'll come down with a compressive stress, sigma 11, along that axis.
And, therefore, we're applying a uni-axial stress, which has just one component of stress.
So that's very specialized.
In general, there would be six different components of stress.
But we're looking at one specific stimulus applied to this crystal plate.
In response to that sigma 11 there are charges induced on all of these surfaces.
And these charges, this charge per unit area, is proportional to the component of polarization P1, the component of polarization P2 along the surface out of which x 2 comes, and the charge per unit area or the polarization along x3.
So this would be P3.
And this would be the direction of x3.
Now I deliberately tried to show this sample as a thin wafer, which has a much larger surface area here than it does on the other two surfaces-- a much smaller area there.
Therefore, since polarization is charge per unit area, if this area normal to x1 is a very large area, there's a lot of charge accumulated there.
It's going to be easy to measure.
If we make the wafer vanishingly thin, then the charge per unit area is high, but the total area is small.
So there's going to be a negligible accumulation of charge on these two side surfaces.
So the longitudinal piezoelectric effect and the longitudinal piezoelectric electric modulus is an effect, where we look at the component of polarization, P1, in response to an applied stress, sigma 11.
So it's that simple.
Look at all the terms we've thrown out.
We've thrown out a whole bunch of elements of stress, which we could impose if we wanted to.
And we've thrown away two of the three components of polarization by designing a specialized sample.
So all that's left then is that P1 equals sigma 11.
And the relation between those two parameters is the 111.
So all this is going to hinge on one single piezoelectric electric modulus, d 111, and how that changes with direction.
So this is for one orientation of a plate.
And I had not specified how the orientation of this plate is related to the symmetry axes.
So let's do that now.
What I'm going to assume is that this is a crystal.
It doesn't look like it's hexagonal.
But imagine that this is a crystal of quartz.
And we could look at an x1 that's in this direction.
And imagine that we have cut out of this crystal a wafer that has a normal along x1.
And then relative to this coordinate system, if this is x1, the modulus d 111 would tell us what charges accumulated on these two surfaces.
But now, what we could do if we wanted to know how this modulus changed with direction would be to cut out a plate, a thin plate, in another orientation, where this is x1 prime.
And this has changed relative to the orientation of the cell edges in the crystal.
The crystal is fixed.
We're just cutting a wafer out in a different orientation.
So this is x1 prime.
We're going to, again, squeeze it with a tensile stress sigma 11 prime.
And we'll ask how the polarization P1 prime is related to sigma 11 prime.
And the answer is that P1 prime will be a tensor element d 11 prime times sigma 11 prime.
So what we are asking, essentially, is how does d 11 transform when we take the direction of x1 in a different orientation and, thus, change the value of d 11 prime?
It's going to change all of the piezoelectric matrix elements.
But we're looking at an effect in a sample that is deliberately prepared such that we will measure only the surface charge given by P1 prime.
And, therefore, the way in which the properties of this plate change as we vary the way in which we've sliced it out of the single crystal is going to be simply the variation of d 11 prime with direction.
So this is the general nature of what we will do when we define any of the scalar modulae related to the piezoelectric modulus tensor.
We can change our notation a little bit in that we have a modulus which I'll define as a scalar modules d. And that d is going to be d 111 prime.
And I know how to evaluate that.
d 111 priime will be C 1l, C1m, C 1n, where these are direction cosines, times all of the elements in the original tensors, dlmn, in the tensor referred to the original coordinate system.
So even though this looks simple-- it's just one modulus-- when we transform it, we've got a product of three direction cosines out in front at every single one of the 27 tensor elements in the original tensor.
So it's not as trivial as it seems.
So this is how this modulus that relates compressive stress to induced surface charge will change with orientation.
But what are these direction cosines?
These are the direction cosines-- not the full direction cosine matrix.
These are the direction cosines for x1 prime.
OK?
So we can get rid of this two-subscript notation if it's understood that these are the direction cosines of x1 and simply call these l l, l m, and l n, just as we did for the direction cosines of a vector because we're only concerned about the orientation of one of the axes, namely x1 prime.
We don't care diddly-bop about x2 prime or x3 prime because these don't enter into the modulus that we have defined.
And this will be times dlmn.
OK. Is what we're doing clear?
So we have defined this particular effect in terms of those of the 27 piezoelectric modulae which are necessary to describe it.
And then we've established how they will change with a change of the direction of one particular direction.
And we don't care anything about x2 or x3.
All right.
Now we go through this process of inserting for matrix notation with the equalities built in.
The proper matrix notation in the first term is d 11 in matrix notation.
That's this term up here in the upper left-hand corner.
The second term, the term that we've written as minus d 11, that's not d 11 at all.
This is, by definition, d 12.
And we need the true subscripts, if we're going to transform this.
So this really is not even a matrix because the subscripts have lost meaning, and we're just using them to identify equalities.
Then comes a 0.
And next comes something that we've labeled d 14.
And the subscripts there are correct.
That is, indeed, the fourth term in the first row.
But we're going to want to convert d 14 into a tensor element momentarily.
Now let's get the rest of the terms that are non-zero.
This is really d 25.
So the next term that is non-zero is d 25, which just happens, because of symmetry, to be equal to minus d 14.
But this is the true matrix subscripts, and this is the true matrix subscript here.
The next term over to the right is the fifth and final non-zero term.
This is minus 2 d 11.
And this is really d 26.
Those are the true matrix elements.
So we put in the proper matrix subscripts.
And now the next, final, step in the expansion is to convert these terms into actual tensor elements.
So this is d 111.
And these are tensor subscripts, so this is something we can transform.
This is d 12.
In tensor notation this is d 122.
And that's something that has a law of transformation.
d 14 is really d 123 plus d 132.
We lumped two tensor elements together to define this modulus.
Minus d 14 that appears in the next to the last non-zero spot is d 25. d 25 is really d 231 plus d 213.
Then, finally, d 26 is d 121 plus d 112
Shouldn't that be d221?
Sorry.
d 26, you're right.
That's down in the second row.
d 221 and d 212.
Now we've got something we can transform.
The law for transformation is l sub l, l sub m, l sub n, dlmn.
So And these are the direction cosines of x1.
So this term will transform as l 1, l 1, l1 times d 111 .
1 The next term will transform as l 1, l 2, l 2 times d 122.
And that will be d 122 prime for different orientation of x1.
This will be two terms.
This will be l 123.
And I can write them in any order.
So this is l 123 times d 123 plus d 132.
And these terms prime, when we change axes, are going to be equal to l 2, l 1, l 3 times d 231 plus d 213.
And this last term will be l 1, l 2 squared times d 221 plus d 212.
All right.
So this now is our new tensor element, d 11 prime.
And that's given by this sum of terms.
So we'll have a first term l 1 cubed times d 111.
And if I look through these other terms, that's the only term in l 1 cubed that I'll have.
The next term will involve the product of three cosines-- l 1 and l 2 squared times d 122.
And if I go down here, here's an l 1, l 2 squared again.
So I have plus d 221 plus d 212.
And then, finally, the other coefficient that I have is plus l 1, l 2, l 3-- which is what this should be.
And that will be times the sum of terms d 123 plus d 132.
Up here, you've got the same thing again-- plus d 231 plus d 213.
And I have a total of 1, 2, 3, 4, 5-- 1, 2, 3, 4, 5 terms.
OK, that is how the longitudinal piezoelectric modulus will change as we change the direction of the normal to the plate that we have cut out of the crystal.
So these are direction cosines relative to the crystallographic axes.
l 3 is the angle between the normal to the plate and the threefold axis.
l 1 is the angle cosine to the angle between the normal to the plate and one of the twofold axes.
And l 2 is the direction cosine for the normal to the threefold axis and the twofold axis.
OK.
So we've got it now in terms of tensor elements.
And now-- yeah?
Is it at all reasonable to assume instead of taking those sums in d 123, d 122, just saying 2 d 123?
Is that OK in assuming?
Or not necessarily?
Well, we could do that.
But I did it the long way to not obscure what we're doing.
OK?
This is a well-defined summation over subscripts.
And we're going to collapse immediately down to the sums.
And we're going to replace the equalities.
So let's see what comes out of this, if we now, having reached the zenith, having transformed the tensor elements, go down and replace this with a consolidation of terms and an insertion of the equalities between the matrix elements.
OK The first term is d 11.
So I will have l 1 cubed times d 11.
Notice I'm getting third powers of direction cosines, which is going to be what causes the exotic nature of these anisotropies.
And then I have a product of l 1 and l 2 squared.
And this is d 12.
And this second term is-- where did it go?
This is d 21 plus d 212.
And that is what we call d 26.
And then, finally, this product of three different direction cosines-- l 1, l 2, l 3.
And we have d 231 plus d 213.
And this is d 25.
And, again, an l 1, l 2, l 3 times d 14.
And the second term here is d 25, if I've done it correctly.
d 25 -- this is d 24.
And this one is d 24.
2/4 OK. Let's now insert the equalities-- back to where we came from.
d 11 is d 11. d 12, however, for symmetry 32-- I'm going to my handy-dandy chart of symmetry restrictions.
I don't want to do that.
That's fourth rank
It's still on the board
It's still on the board?
Yes.
Thank you.
When your nose is in it, it's hard to see.
d 12 is minus d 11.
1 And d 26 is minus 2 d 11.
So these two terms can be consolidated.
l 1 cubed plus l 1, l 2 squared times d 11 minus d 11.
So these two terms die.
And I have a minus 2 d 11 that's left.
If I insert the equalities here, I'll have l 1, l 2, l 3. d 25 is minus d 14.
And here's a d 14 itself.
So these two terms die.
And then I had d 25.
And that is--
You don't have d25
I don't have d 25.
Where did I get the extra one?
[INAUDIBLE]
OK.
I'll take your word for it.
And I know how it has to turn out.
OK.
So these two terms kill each other.
And I'm left with, then, an l 1, l 2, l 3.
Or have I left something out?
This is d 1-- this is d 15 and d 231 and d 23 -- uh, this is d 2 -- and the combination of 13 and 31 is d 25.
Right?
And if I look at my equalities, this is l 1, l 2, l 3 minus d 14 plus d 14 plus d 25.
And d 25 is minus d 14
Why would you add this d25?
Let me check my notes and see what I've got here
[INAUDIBLE]
OK. See what I -- d 25 is minus d 14.
And then I have just a d 14.
I don't know -- I see what I did.
I put it in the wrong slot.
This is d 14.
And d 25 is the one that's minus d 14. so this term dies, which is nice because that cross term is messy.
So what I'm left with, then, if I check against my notes, is d l 1 cubed plus l 1, l 2 squared times d 11.
And then I have minus d 1 minus-- this doesn't belong in here.
This wants to end up being l 1 cubed minus 3 l 1, l 2 squared times d 11.
The first term is l 1 cubed.
That's correct.
The second term is l 1, l 2 squared.
We've got a minus d 1 plus minus 2 d 11.
So I have minus 3.
It should be a minus.
Yeah, that carries down to a minus.
So I have minus 3 l 1, l 2 squared all time d 11.
OK.
So what we have ended up with is an expression for the longitudinal piezoelectric modulus as a function of orientation.
The surprising thing is that l 3 does not appear here at all.
It doesn't depend on the angle between the normal to the plate and the threefold axis.
It depends only on one modulus, and that is a remarkable thing.
This says that the shape of this surface is independent, essentially, of the property, any property that relates the one one prime to a uni-axial stimulus, sigma 11.
And you measure a vectory component in the same direction is always going to have this universal surface.
And it involves just a geometric term and then one modulus that changes the magnitude of the longitudinal piezoelectric modulus but does not change the asymmetry.
So let's see what this function looks like as a function of direction.
Maybe we better wait for that until we come back because that's going to take a few minutes.
So this is what we found.
That is correct.
And we have to now decide what this looks like, which will take a few more minutes.
But let's stop here rather than run late.
All right.
Let's take our 10-minute break as usual.
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OK. Let's resume.
I cut things off at a time when we had the final answer.
And I left you hanging because we don't know what the final answer is telling us.
This says that as we change the orientation of the normal to the plate, relative to x3 and, secondarily, as we change the angle between the threefold axis and x2, we get this strange third-rank trigonometric function.
Let's convert these cosines into the appropriate angles.
This was the twofold axis.
And this was x1.
This was the direction of the threefold axis.
This was x3.
And this is the direction x2.
And that comes out in between a pair of twofold axes.
And we may want to look at this from up above, relative to these twofold axes that occur.
This thing here looks like a trigonometric identity, doesn't it?
Let's let this angle here be theta.
And l1, then, is cosine of theta.
And so our geometry is that this is l1 cosine of theta.
This is l2, which is the cosine of the angle between our direction and x2.
And that is the cosine of pi over 2, minus theta.
And that, then, is equal to sine of theta.
So this identity, as you all know-- I'm not telling you anything that you don't already know-- this is d is equal to d1,1 1, times the cosine of 3 theta, right?
You knew that.
Believe it or not, it is.
It's one of these obscure trigonometric identities, which wouldn't occur to you in a million years unless you go digging through some handbooks.
So this is rather astonishingly simple.
It's simply a cosine function.
But the interesting thing is it goes as a function of 3 theta.
So this goes through one cycle between a pair of adjacent twofold axes.
So it starts out as d1,1.
And it finishes up at d1,1.
And in between it is minus d1,1.
And so it goes down.
And now we come to something that I do differently than most people do.
Most people will say it goes as minus d1,1.
But minus d1,1 I like to show as a lobe going off in this direction, with a minus sign
How do you get that cosine of 3 theta?
Hmm?
How do you get d equals d1 cosine 3 theta?
How do I get that?
Yeah
That is just a trigonometric identify, believe it or not, a well-known trigonometric identify.
Actually, it's an exceedingly obscure trigonometric identity.
So the way this is going to go from x1 is it's going to be a positive lobe around x1.
There's going to be a negative lobe along x2 and then a positive lobe again about the twofold axis that's 120 degrees away.
And then a negative lobe, and then a positive lobe opposite this negative lobe, and a negative lobe opposite this positive lobe.
So it's a six-membered-- six lobes.
There is always a positive lobe opposite a negative lobe.
And what this means is that the charge is of opposite sign on opposite ends of the twofold axes.
So the response peaks up on twofold axes.
Now the thing that I don't like is that what Nye does is to say, OK, this is a negative value.
So you should plot the radius in a negative direction.
And that puts it over here, right on top of this positive lobe.
So what Nye shows is the polar plot of this result, in the plane of the twofold axes, is simply this.
This is x1.
And he shows a lobe here.
And he shows a lobe here.
And he shows a lobe here.
And if you interpret that as a polar plot you say, well, I know what the value is in here.
It's decreasing.
And there's nothing going on in here.
So the response must be 0.
And then it starts coming up again.
And there's a response in different directions here.
And then it goes back down to 0.
And in this range, there's nothing going on.
And that's not true.
What's going on is a negative value of the modulus.
And to me, that becomes abundantly clear if you just put a sign that labels the sign of the modulus within those lobes.
Now I submit, that's pretty anisotropic, isn't it?
Yes, Steve?
Did you just arbitrarily choose where your positive and negatives go, is it [INAUDIBLE] out here?
No.
This comes out here.
When l1 is 0, that's-- excuse me.
When the angle is 0, l1 is plus 1.
Remember that l1 is the cosine of this angle.
When that angle is 0, then the value is plus 1.
So I was careful to put the label x1 on this.
Now the other thing that we should examine is how these lobes vary in a direction perpendicular to x, in a direction that includes x3.
So let us look at how the function varies in a direction perpendicular to-- that includes x1 and x3.
OK.
Call this angle phi.
And in the x1, x3 plane, d1,1,1 prime turns out to be l1 cubed d1,1.
Before you go to our general expression.
And l2 is cosine of 90.
This thing drops out.
We're left with simply d1,1 prime equals l cubed times d1,1.
So this then goes as cosine cubed of phi, which is a lobe that starts out at plus 1.
And then because it's cosine cubed, this dies out very, very rapidly.
So these lobes are very flat in the x1, x2 plane, die out very rapidly as the cube of phi, where phi is the angle between the twofold axis and x3.
So the interesting thing about this surface is that if you decided to pick up the random fragment of crystal and determine what its Piezoelectric Modulus is.
One way of doing this, a poor man's test for piezoelectricity is to just clamp a fragment of crystal between two electrodes and then hook this up to a variable frequency generator that sweeps through a range of frequencies, changes the frequency of a voltage across these plates, and then comes back and sweeps again.
Or alternatively, have a knob one on that lets you change frequency.
Then put a pair of earphones across the crystal.
And if we would do that, by having this variable frequency generator, as we turned it, when we hit a resonant frequency that set up either a full wavelength of a vibrational wave in the crystal-- or half wavelength, or one wavelength, or 3/2 wavelength-- there'd be a resonance.
And the capacitance does something crazy.
It does like this as you go through the resonant frequency.
And then your simple detector-- a pair of earphones-- as you tuned the frequency, you would hear static as you went through this discontinuity in capacitance.
Or alternatively, you could put this on a cathode ray tube and just have a sweep frequency and put the voltage across the electrodes on the oscilloscope screen.
And you would see something like this, and then this discontinuity, and then maybe second harmonic.
And it would work just fine.
This is a poor man's way of detecting piezoelectricity.
But what if-- what if-- you happen to have your piece of quartz with the electrodes directly on the c-axis?
The modulus would be 0, and you wouldn't detect anything.
Or if you put on the probes such that they were in a direction that was in between these lobes where, again, the Piezoelectric Modulus has gone to 0.
You wouldn't find anything.
So measuring the Piezoelectric Modulus for a random chunk of material is dangerous, if you just look at one direction and say nothing's going on; it's not piezoelectric.
The other thing is the material could have a piezoelectric response that's so weak you just can't measure it.
Yeah, OK?
This stuff's only for single crystals, right?
Because if you [INAUDIBLE] polycrystal material, you're getting [INAUDIBLE] averaging
You're averaging over all directions.
So this is for a single crystal.
And that's the only time you get these exotic, very anisotropic surfaces.
Another method that I've seen in a rather old book is really nice.
What you do is you cut your crystal into a little plate.
And then you drive the plate by means of putting electrodes on it and hook this up to a variable frequency.
And if the material is transparent, you will set up-- depending on the velocity of sound in the material-- you will set up standing waves when you hit the right frequency.
And these lines that I've drawn are places where the displacement is always 0.
And these things that I've indicated at little squares is a region where the displacement goes up and down between its maximum and minimum extreme.
Now if this crystal is transparent and you shine a light through it, these little regions bounded by lines of zero displacement act like little lenses.
So if you pass a beam of light through it, you get a bunch of right maxima, little focused spots on the sheet.
And you can determine for a particular frequency if this is the 1 and 1/2 wavelengths along a dimension that you know, you can again find the Piezoelectric Moduli.
So there are lots of ways of detecting this effect.
But when you have a single crystal, you've got to really look at this as a function of direction, to be absolutely sure.
One other Piezoelectric Modulus is in the problem set.
4 bar 3m, this is the structure of sphalerite and a lot of the compound semiconductors.
This is not in the notes.
But if you try the problem, what you find is, again, a highly, highly anisotropic behavior.
And this is the direction of the unit cell of the crystal.
You find that the piezoelectric response is a series of very sharp lobes, positive and negative, going along the directions of the body diagonal, namely the 1, 1, 1 directions in the crystal.
And then there's another lobe that goes like this, negative, positive.
Another lobe that goes up here, positive and negative.
So very sharp lobes along the body diagonal, alternately positive, negative, positive, negative as you go around 0, 0, 1 plane.
OK. That's what the longitudinal piezoelectric effect is like.
You can invent single-crystal devices for weighing fish.
And then you have an interesting question.
If you hang the weight on the crystal, and it's a flat plate, how would you orient the crystal to get the optimum sensitivity?
So you define a scalar modulus for the crystal in the particular direction you're exerting a uniaxial stress.
And then you express this module in terms of theta, the angle of rotation, within the crystographic plane.
In the case of the sup cell if you looked at that geometry, that is a flat plate subject to compression, where you measure the surface charge perpendicular to the direction of the uniaxial stress, not parallel to it as we've done here.
And then there's a question if you rotate the plane of the crystal about the imposed stress, what orientation gives you optimum, maximum response?
So there are fun things that you could do with that.
We have a quiz next time.
There's a lame-duck period after the quiz will be over.
And I want to say a little bit about the tensor aspects of elastic moduli.
Mechanical behavior is something that you have or will cover in great detail in a graduate-level class on mechanics, but I think very few people who deal with things other than cubic materials or with single-crystal materials where the elastic constants are functions of crystal symmetry and direction.
So let me, at least at the time available, set up a definition of stress in terms of strain.
Stress is a second-rank tensor.
There are six unique components.
So we could write an expression for sigma 1,1; sigma 2,2; sigma 3,3; sigma 4,4; and sigma 5,5.
And sigma--
--1,2
No, I want to go down like this.
So this would be sigma 2,3; sigma 1,3; and sigma 1,2; eventually to be known as sigma 1, sigma 2, sigma 3, sigma 4, sigma 5, and sigma six.
But actually, if we're writing out a full tensor relation, we should write down all six elements here.
So let me do that and put in a 3,2.
Put in all nine of them, because before we condense to matrix form, that is what we're going to have.
So we'll have a sigma 2,3; we'll have a sigma 3,2; we'll have a sigma 1,3; a sigma 3,1; a sigma 1,2; and a sigma 2,1.
And we can express each of the stresses in terms of the elements of strain.
So I'll have an epsilon 1,1; an epsilon 1,2; an epsilon 2,2; an epsilon 3,3; an epsilon 2,3; an epsilon 3,2; an epsilon 1,3; an epsilon 3,1; an epsilon 1,2; and an epsilon 2,1.
Nine terms, nine terms.
And in between is going to be a tensor consisting of 9 by 9 elements.
There will be 81 different coefficients in here.
If we want to derive symmetry constraints, we're going to have to, for every one of those 81 elements, do a transformation.
And each transformed element is going to be a sum of four direction cosines times one tensor element, repeated 81 different times.
Not something to be undertaken on a short afternoon.
OK. What are the tensor elements in here?
They are represented by the symbol c. This would be c1,1,1,1.
This would be c1,1,2,2 times epsilon 2,2; a c1,1,3,3; a c1,1,2,3; and so on.
And these c's, in one of the great perversions of scientific notation, are called stiffnesses.
The other thing we could do is to write strain in terms of stress.
And really, stress is something you can do to the crystal.
So stress is an independent variable.
And this is something that I feel more at home with.
So I'll have an epsilon 1,1; 2,2; 3,3; epsilon 3,2; 2,3; epsilon 1,3; epsilon 3,1; epsilon 2,1; and an epsilon 1,2.
Again, nine elements of strain.
And these will be given by coefficients times each of the nine elements of stress, sigma 1,1; sigma 2,2; sigma 3,3; sigma 2,3; sigma 3,2; and so on for nine different components.
Andy the coefficients here are designated with the symbol s. These are the tensor elements.
So this is s1,1,1.
This is s1,1,2,2.
And the s's stand for compliances.
Now English is a very strange language.
But to call compliance s and to call stiffness c is surely a perversion that can be designed for no other purpose than to confuse the introductory student and give the instructor some feeling of superiority.
I don't feel superior, I feel embarrassed that I have to explain this.
Stiffness, c; compliance, s. It's got be one of the atrocities of scientific notation.
To remember which goes with which, I'll tell you what works for me.
If it doesn't work for you, forget it.
The s goes with sigma.
And the little c goes with the epsilon.
That's a c with a bar in it.
That works for me.
If it doesn't work for you, use your own mnemonic device.
English does a lot of things strangely, but nothing is as perverse as this.
Now what's some examples?
Consider this.
You park your car in your driveway, but you drive your car on a parkway.
Why don't you drive your car on a driveway and park your car in a parkway?
No, it's the other way around.
It's almost as bad as this.
What are some other ones?
In my basement, I have something that's called a hot water heater.
You don't heat hot water, you heat cold water.
Why isn't it called a cold water heater?
Even better, I once worked for a year and a half in Switzerland.
And in the kitchen of my home I had, in German, an elektrowarmwasse rsheissungsapparat.
That's one word that's 11 syllables.
The Germans have a knack with the language that we have never approached in English.
But there's another one you've probably heard.
You know what the prefix "pro" means.
That means for, and "con" means against.
And so progress is moving forward, and I leave it to you to decide what Congress means.
OK.
So I could go on and on.
I probably shouldn't, unless you egg me on.
No, I won't.
But anyway, there are-- OK, I'll give another one.
Why is what a doctor does called "practice"?
This is not reassuring at all.
Let me invite you over to my practice.
And why is it when you take an airplane your flight ends with a terminal?
That's doesn't sound very encouraging either.
So anyway, language is silly.
And I guess if you want to call c stiffnesses and s compliances, it's OK.
The verbal description of these terms means something.
Because if the stiffness is very high, that says that a little bit of strain requires that you haven't posed a very large stress.
So stiff is when the material is resistant to stress.
So you have to have a very, very large strain to get yourself a given level of stress.
Conversely, if the material is very compliant, a very small stress for a compliant material should give you a big strain.
And that's exactly what a large value of s will do.
It will give you a large strain for a relatively small stress.
So the words describe what's going on.
OK.
There is great utility in reducing these equations to a matrix form, which is going to cut us down from a 9 by 9, or a tensor with 81 elements, to a smaller number of subscripts and a smaller number of matrix elements, so if we let sigma 1 be sigma 1,1; if we let sigma 2 be equal to sigma 2,2 and sigma 3 be equal to sigma 3.
And then if we write the inequalities we've had before, and we let epsilon 1,1 to be replaced by epsilon 1, then we would have out in front here a term simply c1,1, a 1 for the sigma and a 1 for the epsilon.
The next term would be c1,2 times epsilon 2, where obviously epsilon 2 has replaced epsilon 2,2, and 1 stands for the two indices, 1,1.
So this will go c1,3; epsilon 3.
And then you hit these messy factors of two.
You'll have a pair of terms.
You'll have c1,4.
And that will stand for a term c1,1,2,3.
And then there'll be another term c1,1,3,2.
And if you want to write this as a matrix, c1,4 has to be equal to the sum of these two things, times epsilon 4.
That's the only way you can write it as a matrix.
And if you think it was bad worrying about a factor of 2 in front of half of the terms on the right-hand side of the equation, we're going to be dealing with factors of 2 for terms in the upper 3 by 3 array.
We're going to be worrying about a factor of 4, somehow or other, down in the lower 3 by 3 array.
So the way you handle the factor of 2 really is a nightmare for relations in elasticity.
But continuing on with the first line, c1,5 times epsilon 5 would be a combination of c1,1,3,1 plus 1,1,1,3.
And then finally, when you get to c1,6 times epsilon 6, that would represent a combination of c1,1,2,1 plus c1,1,1,2.
You have to make a decision where to swallow the 2.
And probably the best thing I can do is to give you conventions for relabeling stress, strain, stiffness, and compliance, which I'll pass around to you.
No big deal, except that this is a case where you can eat the 2's early on, but you can't even break even.
So very, very briefly, what you do is you convert the tensor elements of stress to matrix elements in exactly the same way we did with piezoelectricity.
For strains, we do exactly the same thing that we did with the converse piezoelectric effect.
The factor of 2 that you swallowed in definition of stress pops up to haunt you when you deal with the strains.
And you cannot write a nice matrix relation unless you divide the off-diagonal elements of strain by 1/2.
And that was exactly the same thing we encountered with the converse piezoelectric effect, which related strain to an applied electric field.
In defining the matrix stiffnesses, the c's, you forge straight ahead, and you let ci,j,k,l, which is identical to ci,j,l,k-- that's the term that relates the two equal shear stresses-- and you define that as a matrix term with two subscripts.
No factors of 2 or 4 are involved.
Then when you hit the s's, there's a nightmare.
si,j,k,l is sm,n if m and n are 1, 2, or 3; that is to say, not 4, 5, or 6. si,j,k,l, which is the same as s,i,j,l,k, has to be defined as one half of sm,n, where m or n is 4, 5, or 6.
And then finally, you have to throw in a factor of 4 when both m and n are 4, 5, or six or, in other words, m and n are not 1, 2, or 3.
So it is a mess.
And these are rules that you have to bear in mind if you're ever going to go from matrix form, which works absolutely lovely in a fixed coordinate system.
But if you have a single crystal and you want to refer it to different axes, you have to be prepared to resurrect the full tensor form.
And then and only then can you work symmetry transformations.
I don't want to go through the simple algebra of expanding and contracting these terms.
So the next two sheets show you how you condense from tensor to matrix notation, and then go back from matrix notation to tensor notation.
This is for the compliances si,j,k.
And the next page does the same thing, if you care to write stress in terms of strain.
And there are a lot more factors of 2 that appear there.
But again, it shows you how you go from tensor notation, for subscripts on the c's, down to matrix notation with 2 subscripts, and then go back up again to tensor notation.
So it's a tedious business.
And you've got to keep careful tracks of your factors of 2 and your factors of 4.
How about symmetry restrictions?
Holy mackerel.
Transformation of 81 different elements.
Well, in order to do the tensor transformations, you have to go to the full-force subscript notation.
Only then is a law of transformation defined.
Fourth-rank properties are even tensors.
So mercifully, there are not as many different possibilities.
All point groups that differ by presence or absence of inversion have exactly the same form of the tensor.
So symmetry 2, m, and 2/m look alike.
Symmetry 2, 2mm, and 2/m, 2/m, 2/m look alike.
So there's only one orthorhombic tensor, only one kind of monoclinic tensor, only one kind of triclinic tensor, and so on.
And the only place that you have more than one form of the tensor for a particular set of point groups is looking at the point groups that are based on single-rotation axes, something like 4, 4 bar 4m, and those that are based on the axial arrangements 4, 2, 2.
So they're different, just as they were for second-rank tensor properties.
For cubic, a rather remarkable result.
There are three independent compliances.
But cubic crystals are not elastically isotropic.
They are anisotropic.
And it's just the nature of the tensor that requires that.
So rather than letting you hang by your thumbs wondering what's on these pages, let me pass around the summary of symmetry constraints.
If you understood what we did for third-rank tensors or even second-rank tensors, you how to do this.
And fortunately, the terms almost over.
So I can't make you do it, which is probably an enormous relief to you.
Some additional bits of information.
The transformations for hexagonal crystals are complicated because threefold and sixfold axes do not change the reference axes x1, x2, x3 into one another.
And the equations that relate the individual tensor elements are more complex.
For the threefold axis, for example, 2 s1,1 minus 2 s1,2 is equal to s4,4.
So they're not simple relations between them.
And that's simply because you're not changing one reference access into another.
You're changing x1, for example, into a linear combination of x1 and x2.
Two things down here of interest.
I said that cubic crystals are not isotropic.
What would have to be the case if the material were to be isotropic?
Well, it turns out that if you do this for the compliances, s, if 1/2 of s4,4 is equal to s1,1 minus s1,2, then the material is elastically isotropic.
If you do this in terms stiffnesses, the c's, then 2 c4,4 is equal to c1,1 minus c1,2 if the material is supposed to be isotropic or going to be isotropic.
Then there is one other inequality that depends on structure.
And this is something known as the Cauchy relation.
And it depends on the interatomic forces.
The first condition that has to be met is that the forces between the atoms should be central forces.
What is a central force?
Well, this is a case where the attractive force between the atoms is directly along the line joining their centers.
Isn't that always the case?
Don't crystals hold together because there's an attractive force between atoms?
Well actually, for metallic crystals and ionic crystals, maybe that's not a bad assumption, particularly for ionic crystals.
But if you had a covalent material, where the bonding was due to overlap of orbitals like this, then the thing that holds the crystal together is overlap between these orbitals.
And the force holding the atoms together is a force that goes through this shared electron pair.
And that's not a central force.
And so the Cauchy relation generally fails rather badly for covalently bonded materials.
Second assumption is that each atom is at a center of symmetry.
And the reason for that is so the force in the plus-x direction is the same as the force in the minus-x direction.
And that is not true for any material that has tetrahedral coordination.
So it's not true for any of the forms of SiO2.
It's not true for any of the compound semiconductors that are based on tetrahedral units.
So there, the atom inside of these tetrahedra is decidedly not at an inversion center.
And finally, you assume that the crystal is under no state of initial stress.
Because if it is under initial stress, you've squished it, you've stretched the bonds.
And the forces-- not to a major degree, to be sure, but-- the forces will not truly be perfect central forces.
So if all three of these assumptions are satisfied, then you have an additional equality c1,2 is equal to c4,4.
What I'll do next time is I'll bring in some examples of data for stiffnesses and compliances and, in particular, show you how some of these elastic tensor elements change with temperature.
And we can examine them to see how well the Cauchy equality or the isotropy condition is satisfied.
All right.
That's about all that we'll do about the definition of fourth-rank tensor properties.
On next Tuesday to wrap the term up, we'll look at the variation of some elastic moduli, such as Young's Modulus or the shear modulus, for different symmetries as a function of direction in the crystal, OK?
But you all knew that.
That is quite a mouthful.
So if you like, you can refer to it as SST.
And today, we're taking off.
So my name, for those of you who don't know me, is Bernardt Wuensch.
My room number is 13-4037, The office that's become a legend in its own time.
And my extension number is 3-6889.
And hey, just to get you sensitized and thinking about the right things, let me point out that my extension number has a point of 180 degree rotational symmetry right in the middle.
You can pick it up, turn it head over heels by 180 degrees.
And it's mapped into coincidence with itself.
Just happened to get it.
You might think I would have had to have fought for years to get an extension number like that.
But no, it just happened to come my way.
OK, some words about the formalities of the subject.
First of all, the format of the class is unusual.
We meet four hours a week, but because most if not all of you are graduate students anxious to get some work done in the laboratory, we do this in two two-hour chunks.
So we meet Tuesday and Thursdays for two hours.
Two hours is a lot of time for anything, however good.
So what we do is to take a long intermission halfway through and let you go out and enjoy what's left of the lingering summer for 10 or 12 minutes.
And then, come back refreshed and we will resume.
Most graduate students like this arrangement because it gives them a chance to duck out and make a setting on a furnace or turn something off in the laboratory.
And it works out better for them than having a one hour time chunk every day of the week or four of the five days of the week.
In any case, nobody's complained about it.
So I assume that will work satisfactorily for you as well.
The other question that one immediately asked at the beginning of the term, how many quizzes?
And we're supposed to tell you that straight up.
There will be three quizzes.
No final examination-- do I look like the kind of Scrooge that would prevent you from getting a good flight home at Christmas time or have you working, cramming for a final examination a few days before Christmas?
And for my part, I can remember the good old days-- or not so good old days-- when I did give a final.
And there I would be, lying down on my stomach under the Christmas tree grading final examinations.
And every time one of my little kids would come near, I'd lash out with my foot and say, get out of here, kid!
Can't you see Daddy's got papers to correct?
Well, no Scrooge.
No final examination.
We'll have three quizzes.
The quizzes will also be a little bit unique.
Since we have two hour chunks of time, I found by experience that if I give the quiz during the first hour everybody is sitting around glassy-eyed, absolutely brain dead and pay no attention to the lecture that follows.
If I give the quiz in the second hour, everybody is pretending to pay attention and then sneaking surreptitious looks at their notes just so they can have everything packed away before they have to write on paper.
So my quizzes are two hour quizzes which lets you not work for two hours, but gives you all the time you could possibly want.
And people start leaving after about an hour and a quarter.
But you can stay for the entire two hours if you want.
Even at that, I found by experience that if I give you two hours for the quiz, after about an hour and a half, everybody's looking out the window, looking back at the ceiling, not a single pencil is moving.
Then I will say, OK, you all done?
Everybody starts writing again and going through their papers once more.
And even after two hours I find that in order to get the quiz papers, I have to plant one foot on the edge of your table, grab hold of your quiz with both hands and drag it out of your clutching fingers.
So you can take the full two hours.
But it should not be necessary for you to consume that much time.
The quizzes will come-- and this is something else we're supposed to explain to you-- they will come at one third of the way through the term, two thirds of the way through the term, and 2.983/3 of the way through the term.
And if you're wondering where that number comes from, this lets me put the quiz just before the final week of the term when we're not supposed to give examinations.
So there'll be three quizzes.
You will have opportunity for lots of practice with problems.
We will have on the order of 15 problem sets.
And for the most part, they will be very short, or modestly short, and designed to give you some practice in working with the material.
Because as the nature of this subject begins to unfold, you'll see that it involves a type of mathematics that you've really perhaps not had much practice with.
It involves geometrical relations.
And to really master it, you have to work with the material and get some practice.
Another question that's perennially asked if not outright raised in private, what do the quizzes count?
What do the problem sets count?
The answer to that is that the problem sets will count a lot towards your understanding of the material.
But I'm not going to grade them.
And I'm not going to factor them in along with the quizzes to decide your final eventual fortunes in this class.
The problem sets, moreover, there are a lot of them.
But they will be optional in the sense that if you do them, I will carefully correct them, add words of inspiration and advice, correct things where you've gone wrong and then return them to you as quickly as possible.
But if you how to do the problem, you say, ah!
Why does he wants us to do this and waste our time with a silly problem like this?
Don't do it.
Don't do it, because if you know how to do the problem set, that's fine.
And you've got better things to do with your time.
And I've got better things to do with my time if the feedback is not going to be of benefit to you.
So I hope you will do them.
And as I said, if you do them, I will correct them promptly and thoroughly.
And if you haven't got the foggiest idea how to do the problem, do what you can.
And then, write down a plea of help-- I don't understand what's going on here!
OK, and then I will take the time to write out what's going on, hopefully to your benefit and use.
Will I turn out solutions?
Only on an individual basis in the fashion that I've just described.
I find that if I write out a solution to each of these problems, you don't do them.
And you say, oh, so that's how you do that, throw it in a file, and not look at it until the night before the quiz.
So there will not be solutions handed out other than correction on an individual basis on your papers.
Is that all that I wanted to say?
I think that's about all for the formalities.
Actually, I should add one postscript to say the problem sets don't count anything toward your final grade.
They do in one minor sense.
When you have a large class and you plot up the grades and there are no lumps with gaps in between, there comes a point where you have to separate one grade from another.
And if you've done well on the quizzes and you've done well on the problem sets and there's just one quiz that's a little bit, I say, OK, he or she had a bad day that afternoon.
And I'll give you the benefit of the doubt.
And even though I'm not a vindictive sort, if you're right on the fence and you haven't done any of the problems, then without malice I say, gotcha!
And you go down [INAUDIBLE] on the low side of the barricade.
And I think that's only a natural indication because there are some cases where, with Solomonic judgement, you have to decide who gets what grade.
OK, let me say a little bit about the texts.
There are a number of books that deal with crystallography.
For the most part, though, they consists of an introductory chapter.
Every single book on the solid state feels compelled to write some sort of half baked chapter on crystal structure or crystallography.
And usually, these chapters consist of big tables.
And they say, there are 14 of these.
There are 17 of these.
There are 32 of these.
There are like 230 of these.
There are 1,170 of these.
And then, that has all the excitement and stimulation of reading the telephone directory.
It's a crazy cast of characters, but it's awfully hard to see the plot.
So what we will do all the way through is derive everything.
So you can not only see how it turns out, but why it has to be that way.
And that's the way, in my opinion, one really learns this material.
A couple of other very pedantic comments about the material.
In the early part of the term, the first half in fact, we're going to use plain old geometry.
Now, geometry really doesn't cut much mustard around the Institute.
If you can't integrate it or take its Fourier transform, that's a mathematics you don't have to take seriously.
Well, geometry is a perfectly valid branch of mathematics.
And one can do what we're going to do in more complex terms using the language of group theory.
And we will, in fact, use a little bit of that later on.
But for the most part, just diagrams with simple geometry are going to be one of the principle tools in the initial part of the class.
About halfway through, we'll switch over to something that is much more mathematical in the traditional sense.
Here, a little bit of linear algebra and matrix algebra will help you.
If you haven't had that or haven't looked at it for a while, we'll build it up from ground zero so that you'll be able to fully understand it.
We'll hit a few eigenvalue problems towards the end of the term.
If that doesn't get your adrenaline pumping, that will be developed in a physical context so that you're doing the sort of problem before you even know what it's called.
So it's going to be a user friendly course that doesn't rely on something that you may have had two or three years ago.
OK, but the other thing that I wanted to say was that this class is not like many classes in that you talk about something for one week.
And then, you put it aside and you talk about something completely different the next week.
Our first half of the course will be one long process of synthesis.
We're going to start out very, very simply with little mapping transformations.
This is picked up and rotated and slid over to here.
And you'll say, ho-hum, let's get on with it.
Come on, go faster.
But we'll build on this and then build on what we've just done to what comes next.
And unlike most of the classes in science that you take where you start with general terms and you zero in on some little nugget like f equals ma, e equals mc squared, lambda equals 2d sin (theta), a little nugget like a bullion cube that you can drop in your pocket.
And then when you need it later on, you pull it out and add hot water.
And then, you have a tool that you can use.
We will do something that's completely different in its structure.
It will start out simple.
It will grow.
It will blossom like an elegant [?
Filigree ?]
structure that gets more and more complicated and diverges rather than converging to a nice, tight, little nugget.
It's going to get very, very complicated.
And the reason for doing this gradually and thoroughly is so that you can understand the complexity and where it comes from.
OK, so my moral here is keep up.
It may seem easy when you start.
But we're going to assume that you've got that down cold before we go on to the next step.
OK, texts.
Apart from these half baked treatments which I just keep [INAUDIBLE] on, one of the very best books is by an old MIT guy, Martin Buerger, who was one of MIT's most distinguished faculty.
He was the very first faculty member to be honored with the title Institute Professor, the very first one.
Chairman of the Faculty, all sorts of awards from professional societies-- he has a book called Elementary Crystallography.
This is published by Wiley.
There's some who dispute the term "elementary."
But he really has a book which uses, at the outset, nothing more than geometry.
He doesn't throw in comments like, "It can be shown that," or "By further work, it turns out--."
He does everything for you.
Everything is down there so you can see how it's done and what the results are.
To me, it is the best book on the subject.
That's the good part.
The bad part is that it's been out of print for about 15 years.
So what I am going to do is to make-- now that I know how many of you are going to be present-- I'm going to make a Xerox copy for you of the first half of the book.
What a department!
What a class!
You're going to get a classic text, 50% of it, without spending a nickel.
And that'll be the text for the first part of the class.
We will be doing some derivation that are not in this book.
And for that, I will have notes that I have written out.
And you'll get Xerox copies of that.
So we'll have lots and lots of handouts during the course of this semester.
I'd like to call your attention, though, to two other books.
These are not textbooks.
These are reference books.
And you can see from the shape of this one that this is one of my favorite volumes.
It's thoroughly worn out.
This is something that is called The International Tables for X-ray Crystallography.
And it is published by an organization called the International Union for Crystallography.
The funny sounding term, "International Union for Crystallography," sounds like an organization under which diffractionists go out and strike for higher pay.
But no, this is actually a federation of all of the national societies of crystallography from all over the world.
And among the useful things that they do, besides having a splendid conference every couple of years, is to publish these tables.
And volume one is called Symmetry Tables.
And everything that we will derive and all of its properties-- physical and geometrical-- are tabulated in this book.
It is, however, a reference book and not a textbook.
You don't learn it for the first time from this book.
But in terms of generating atomic arrangements from the data that's present in the literature, looking at the arrangement of symmetry elements in space, and how they move atoms around, it is the code book that tells you how to crack the arcane language in which diffraction and structural results are recorded and find out how to unravel it.
I call also to your attention, although it will not be germane to this class, there are four other volumes.
Volume two is called Mathematical Tables.
And this has all sorts of useful stuff.
If you've ever done diffraction, you know that depending on the symmetry of the crystal, there are some planes for which h squared plus k squared plus l squared divided by 2 pi is not a reflection [INAUDIBLE] if the crystal is green, and other arcane rules like that.
All of these are summarized in these books.
There are quantities that you need to calculate, things like interplanar spacings.
Tables are available there.
So this is a handy thing primarily for diffraction.
Volume three is called Physical Tables.
And this is where you find things like absorption coefficients for x-rays and for neutrons.
It's where you find the latest values of absorption coefficients, neutron scattering length.
And since these things are derived experimentally, the values improve and change from time to time.
So this is where you find the most up to date values of physical constants and items that are necessary for diffraction.
It never ceases to amaze me how somebody who has the good fortune of having to use the diffraction for a thesis will labor carefully over making the measurements and reducing the data.
And then when it comes to using a wavelength, which is how the final numbers will be determined, goes to an appendix of a book on diffraction that was published 20 years ago.
And that's not the most up to date value.
Scattering powers of x-rays by the electrons on the atoms are calculated from wave functions, which constantly get better from year to year.
And the value of the scattering powers of the function of angle gets better from year to year.
So this is where you want to go if you need any of that physical data.
And finally, volume four is-- it's not its title, but it's essentially an update of the Physical Tables, giving later values which came out about 10 years later.
OK, this series was getting out of hand.
So I have to bend my knees and use two hands when I pick up this one.
This is a continuation of the series, essentially.
But this one is called International Tables for Crystallography, period, no x-rays because neutrons and electrons are just as important today for doing scattering experiments.
And this is International Tables for Crystallography.
No x-ray in there.
And there are now something like six volumes out.
They're not called one, two, three, and four, but they're called A, B, and C to avoid confusion.
And volume A is one called Space Group Symmetry.
And then, there are a whole series of other ones.
As I say, I think there's six of them that give physical data and all sorts of useful guides.
I have mixed feelings about the new series.
You will see that it is about three times as large and three times as heavy, which means it's nine times as expensive.
And to me, it's almost the case for most people of a situation where if it wasn't broke, you shouldn't fix it.
And what they've done is that they've put in all sorts of esoteric theory which probably is going to be of interest and use to perhaps 5% of the readers.
But nevertheless, if you wanted, you'll find it there, which is something that could not be said before.
They've added a few things which are useful, but a lot of additional information which you don't really need.
And you pay for that whether you want it or not.
Nevertheless, it's been done.
You can't buy the old volumes any longer.
You have to buy the new volumes.
So anyway, this is what you'll find in the library now.
Maybe they do still have the old volumes, one through four.
This, we will make reference to in the course of the term.
I will give you some copies of certain pages in here as handouts when we need them for purposes of illustration or for use.
But I spent the last five minutes just to make you aware of the existence of these books.
And these are really the penultimate source of information and numerical quantities that will be used in diffraction, one of the principle applications of crystallography.
I think I have just enough enough-- to start things off, I have a syllabus for the course that is, in very dense form, exactly what we will be covering this term.
And I'd like to lead you by the hand through this.
All right, what we will be doing in the first half of the term is something that is known as crystallography.
OK, the meaning of the word is almost self-explanatory.
The first part is crystal.
We're going to be dealing with the crystalline state of matter.
To me, amorphous materials, although they may be important, have all the interest of a piece of steak before it's been cooked.
The atoms in amorphous materials are fine.
But they really get interesting when they organize themselves into an ordered fashion.
So the name is self-explanatory.
The first part, crystal, means we're going to deal with the crystalline state.
What does the graphy mean?
That means mapping or geometry.
And let me give you an example of a few other words that have the same sort of structure.
Geo-- the Earth-- followed by graph, geography, is the mapping of the Earth.
And there are many other terms that involve these two separate parts.
Crystallography, though, is very often subdivided into different flavors.
There is something well defined called x-ray crystallography.
And this is the experimental determination of the crystallography of a material using diffraction, usually x-rays because they're relatively inexpensive and they're widely available.
But increasingly, neutron scattering or electron scattering is used for this purpose.
And there are a number of very powerful, very exciting sources of neutrons, either from reactor sources of unprecedented intensity or from what's called a spallation source, where an entire synchrotron is built just to direct a beam of particles onto a heavy metal target.
And those high energy particles split off neutrons from the nuclei of the target material.
Doesn't really matter what the material is.
It helps if it's a heavy metal.
The nice thing about these sources of neutron radiation is that they're so expensive they are all national facilities.
And the consequence of that is that anybody with a good idea and a project worth doing can apply for beam time.
And if it's a good problem, you get it.
So you're using a facility that cost $1 billion.
You have people whose sole function in life is to help you do the experiment and make sure you're doing it properly.
And this is a very, very exciting time to be somebody working with diffraction using these neutron sources.
There's another branch of crystallography which is called optical crystallography.
And this is the characterization and study of crystalline materials using polarized light.
You can identify unknowns using their optical properties if they're transparent about 10 times faster than you can do with x-ray diffraction.
It's a technique that today is little used.
But it's a very powerful technique.
And all it takes is a microscope, and you're off and running.
Some other flavors of crystallography, well, I'll mention the one that we're going to use.
What we're going to talk about is something called geometrical crystallography, to distinguish it from these other branches.
And this is synonymous with symmetry theory.
So that's what we'll do for the first month and a half or so.
All right, let me introduce now some basic concepts.
Geometrical crystallography is the study of patterns and their symmetry.
So let me give you an example of some very simple patterns that extent in one dimension.
And let me put in a figure.
The thing that is in the pattern is something that's called the motif.
And let me use a plump, little fat comma.
And I'll make a chain of these things extending in one dimension.
The nice thing about this fat little comma is that it is a figure which, in itself, has no inherent symmetry.
So it is asymmetric, without symmetry.
And imagine this is being repeated without limit in both directions, both to the left and to the right.
We then draw another pattern with a different sort of motif.
And let me use a rectangle with one concave side.
OK, and I think you get the picture of this one.
And imagine that as extending without limit indefinitely to the left and to the right.
Then, I'm getting tired of inventing new motifs.
So let me use the same motif the second time, but arrange it in a slightly different way.
And again, imagine that as extended indefinitely.
OK, having now generated these three patterns in two dimensions but extending periodically in two dimensions.
Let me ask the question now.
And even if you have not the foggiest idea, you have a 50% chance of being right.
Are any of these patterns the same?
Or are they all different?
Are any of the patterns the same?
Or are they different?
Well, that's a-- yeah?
[INAUDIBLE]
OK, that is an answer that's right because the bottom two involve the same sort of figure.
They have the same sort of the motif.
They both have the same rectangle with one concave side.
And that's a valid answer.
Do you have a different answer?
The first and third are the same
First and third are the same.
Why do you say that?
They both have [?
rotational symmetry.
?]
OK.
This is the point I was trying to introduce.
And that is your choice of answering the question, one is the nature the motif.
And you're absolutely correct.
This pattern and this pattern are both based on the same motif.
But in patterns, we are less concerned with the motif that is in the pattern than we are with the relations between one motif and all of the others.
And in that context, the first and the third pattern, although they look entirely different, are really exactly the same sort of pattern.
So let's begin to analyze what sort of operations are in these patterns that take one motif-- and obviously, they're all the same-- and relate it to all of the others.
First of all, there is an operation which I'll call translation for obvious reasons.
And I'll represent that by a vector, T, since a translation has magnitude and direction but no unique origin.
I could take this pair of objects sitting nose to nose, pick them up, slide them over by T, put them down again.
And I have the relation that gives me this neighboring pair.
Pick it up again, move it to the right by the same translation in the same direction, put it down again.
And I've got this pair.
So that is one operation that can exist in patterns.
This is the operation of translation.
So let me call that by a vector relation.
And it has magnitude.
It has direction, but no unique origin, just like a plain old vector.
So in other words, I can't say that the translation moves us from here to here or from here to here.
It's all the same thing-- magnitude and direction, no unique origin.
In fact, all of these patterns have translational periodicity.
There's a translation in this bottom pattern and another translation from here to here in the middle pattern.
The thing that makes a crystal a crystal is that it is an arrangement of atoms or molecules which is related one part to another by the operation of translation.
If you don't have translational periodicity, you do not have a crystal.
So that comes to the essence of what crystallography is about.
You can imagine, in one sense, the generation of this pattern by a rubber stamp sort of operation.
Suppose I have a rubber stamp.
And I put on the rubber stamp the pair of motifs like this.
Pick it up, move it over, chunk.
Pick it up, move it over, chunk.
And I can stamp out the pattern in that fashion.
Notice that my statement about no unique origin in these terms can be stated that it doesn't matter where the two motifs are on the stamp.
they could be up in the upper left hand corner, right in the middle, down in the bottom.
As long as I move the stamp through the same distance and the same direction, I get the same pattern.
Now, that's not bad for an introduction.
But I want to be more general than this because when I deal in terms of a rubber stamp operation, that is a transformation that involves taking one little chunk of a two dimensional space, picking it up, and putting it down in another location to another unique location in space.
So I'm going to now make another generalization that operations, which we've begun to define, act on all of space.
So I don't want you to think of this repetition in terms of a rubber stamp, although we could get the pattern that way and it's conceptually appealing.
But I'm going to say now that this string of motifs has translational periodicity if, when I pick it up, move it by T in a particular direction, and drop the whole infinite chain back down again, it is mapped into congruence with itself.
Which leads me to another definition-- an object or a space possesses symmetry when there is an operation or a set of operations that maps it into congruence with itself.
In other words, in plain words, you can't tell that it's been moved.
OK, is there anything else that is a transformation which leaves the set invariant?
OK, if we look at the first pattern, there are [?
rho ?]
sides such as this one here, or this one here, or this one here, about which I can rotate one motif into its neighbor or, for that matter, pick up the entire chain and flip it end over elbow through 180 degrees.
And it will be mapped into coincidence with itself.
And that is an operation, and another sort of distinct operation of transformation.
And this is one that I could call rotation for obvious reasons.
And there are two things I have to tell you about a rotation operation.
The first one is the point about which the rotation takes place, and that's going to be some point.
And let me call this point here A.
So this will be some labelled point that is the location of the rotation axis.
But then, the other thing that I have to tell you is the angle through which I'm going to rotate.
And I'll append to the A as a subscript the angle of rotation.
So this particular operation, called a twofold rotation because it rotates through half of a circle, would be the operation A pi.
This point is A.
We rotate through an angle pi.
This pattern here has also rotational symmetry.
In addition to the translation, there is a rotation operation, A pi, in the lower pattern.
So the follow who is unfortunate enough not to have a seat-- and I should have given you this one a long time ago.
I'll give that to you as your reward for giving the best answer.
And you get a seat wherever you would like to place it.
The first and the final pattern are the same in the sense that they contain two operations, translation and rotation.
This pattern is a much more interesting one.
This also has a rotational symmetry, A pi.
It also is based on a translation.
But now, there's another operation that we can do to leave the pattern invariant.
There exists [?
rho sides ?]
that pass through the center of this rectangular figure across which I could flip an individual motif, or for that matter the entire pattern, from left to right.
It's a reflection sort of operation.
So this is a new type of transformation.
So we'll add that to our list.
And the symbol that's usually used to indicate the locus of this operation is m, standing from mirror.
And that does it for these particular patterns.
Three sorts of operations-- translation, rotation, and reflection.
And in fact, that is all you can have in a two dimensional space-- not necessarily a rotation that's restricted to 180 degrees.
If these patterns are translational periodic in more than one direction, you can have higher symmetries.
One of the things I would like to suggest to you is that you look around you in everyday life at the sort of patterns that enrich your environment.
I see a one dimensionally periodic pattern there, the black and white stripes.
It's translationally periodic, going up and down.
It also has mirror planes running through the black stripes and the white stripes.
I see another two dimensional pattern back there.
That has translation.
But you could rotate-- no, you can't do anything.
That just has translation, nothing else.
Get a new shirt.
That's not terribly interesting.
There's another one there that's so complex I don't think I can look at it without climbing all over him and drawing some translational vectors and things like that.
But that's a nice periodic pattern.
That's a good one.
But there's lots of stuff like that.
Look at the grills in the ventilators.
They have mirror planes.
They are translationally periodic in one direction.
We've got floor tiles.
These are lovely because these have examples of 90 degree rotational symmetry.
Same is true of the tiles up in the ceiling, same sort of pattern.
And there's another pattern with four-fold symmetry in the grills that are underneath the fluorescent fixtures.
So symmetry is everywhere.
It surrounds us.
We wear it.
We walk on it.
We sit on it.
And think how much richer your life will be when you can understand this part of your environment.
Hey, that's a good, chauvinistic note, overstated, on which to end.
So why don't we take our break?
I'll hang around if you have any questions or get you a copy of anything that came around that you missed getting one of.
And it is now, according to my Timex watch, about three minutes before the hour.
So let's take a break and stretch for 10 minutes.
Ya'll come back because I've got your name's on a list.
OK, let us resume.
I had no idea how many people would be here today, and I think I made 25 copies of the handout.
And I see 25 names on the list.
And that means that two people did not get a copy of the syllabus.
Does anybody need a copy?
That's strange.
OK. All right.
We covered some introductory material, and I think we've covered enough that you can do a problem set.
So it gives me great pleasure to hand out problem set number one.
OK, you can think about that.
It is the sort of problem that will either take you two minutes or two hours or infinity.
So don't spend too much time on it, but I would like you to put your name on it and turn it in either at the end of the hour or next time so I can make comments if there's something that's mostly right but not quite right.
Let's return to these three simple patterns that we put on the blackboard.
And let me make another point about symmetry.
The people who sensed that this pattern and the one on the bottom were the same because they had the same motif in them, that they had the same rectangle with one concave side.
And I drew a mirror line in here because that locus, when I review this is a reflection from left to right, left the motif, as well as the entire pattern, unchanged upon making that transformation.
Is there not also a mirror line there?
Worked for this motif.
Why not for this motif?
Well, the answer is no, that this is not a mirror line because the symmetry transformations acts on everything, and not just one little bit of space.
And if I would take this chain of objects that's translationally periodic with a translation running this way, and I reflected that, I should have another chain running like this.
So the direction of the translation vector is not left invariant by this reflection.
So the conclusion here, and it's a subtle one, matter of definition, almost, is that the transformation, the symmetry transformation, if it's to be a symmetry transformation, acts on all of the space, and not just on one local domain.
So let me give you an example of a pattern that doesn't involve translational periodicity.
So let me try to make a star as carefully as I can.
What sort of symmetric does that have, or would it have had I drawn it more perfectly?
Well, that would be a five-fold rotation axis in the middle because I could rotate through one fifth of the circle.
And any of those rotations twice or three times, or just 2 pi over five, would be something that maps the pattern into congruence with itself.
There are also mirror lines that go from one tip of the star to the other end.
So that is an example of a pattern, non-periodic, but one that has five-fold rotational symmetry and mirror symmetry.
Now, if I put that star in a box and ask, what is the symmetry of that space?
There's only one operation which is common to the star and to the enclosing rectangle, and that's this mirror plane.
So the symmetry acts not just on one little part of the space, but it has to leave everything invariant.
So in that sense, going to this pattern here, this is not a mirror plane because it doesn't leave the entire pattern invariant.
That plane would reflect this one up to here, and we don't have anything there.
So the space is not left invariant.
One further definition.
We defined what we mean when we say a space or an object has symmetry.
We said an object or a space possesses symmetry when there is an operation, or set of operations, that maps the space or the object into congruence with itself.
Let me make another definition, and that a symmetry element-- another bit of terminology-- is the locus of points that's left unmoved by the operation.
Left unmoved or left invariant.
So for some specific examples, this vertical line is a locus which is left invariant by either the five-fold rotation or any of the mirror planes passing through the points that would be true of the star.
So for the star, these are all symmetry elements.
For the net combination of the star and the enclosing rectangle, the only thing that leaves a space invariant is this line.
The locus that's left and moved is this line, so we refer to that as a mirror plane.
Now, these may be seeming kind of definitions.
Nice to have, but what use are they?
We will use some of these definitions to answer a question which may seem tricky.
If we do a couple of things in sequence, for example, what is the net consequence of doing, let's say, a rotation combined with a reflection?
You can answer that question by saying, what has been left unmoved?
And that is the locuses of whatever net transformation results from a combination of two or more.
So that, again, is abstract, but we'll use that later on.
Any question on this?
Let me summarize very quickly what we have found for two dimensions.
We found that there are, in a two-dimensional space, three kinds of operations.
There is the operation of translation, which we'll call by the vector, T, corresponding to that transformation.
And there is an operation of reflection.
And the locus of the plane, in which the reflection occurs, we'll call a mirror plane.
And that's a linear locus.
And then, we've seen in these two patterns here an operation of rotation.
In particular, in these two-dimensional patterns, we saw the rotation operation A pi.
Now, I would put forth for your consideration something that is a profound conclusion.
These are the only single-step transformations that can exist in a two-dimensional space.
These are the only ones that can exist as single-step operations.
We can view these as operations that result in a transformation of coordinates.
And in two dimensions, if we have some position, x, y, in the space, what the transformation of a translation does is to take x and add a constant to it.
It takes y and adds, perhaps, a different constant to it.
Do the operation a second time, and we'll go to x plus 2a and y plus 2b.
So analytically, we can look at these symmetry operations in terms of the transformation of a representative coordinate.
If we have a reflection plane, and let's set up a coordinate system where this is y and this is x.
We have an object here at the location x, y, and we reflect it across this locus.
It goes to minus x, y, if the mirror line runs through the origin and is perpendicular to x.
So one example of a transformation by reflection is that x, y goes to minus x, y.
And this is a case where the mirror plane is perpendicular to x and passes through the origin.
If we do the operation a second time, minus x, y would get mapped back into x, y again.
It comes back to where it started from.
So this would be the first reflection.
This would be the second reflection.
And we saw in the patterns an example of one other transformation.
Let's suppose there was an operation, A pi, at the origin, and this was x, and this was y.
We started out with a motif here at x, y.
If we rotated that by 180 degrees, it would go down to a location minus x, minus y.
So the operation of a 180-degree rotation is going to analytically correspond to a transformation of going from x, y to minus x, minus y.
If we perform it again, it would go back to x, y. OK, let's look at this in more general terms.
In a two-dimensional space, we've got two dimensions to diddle with.
We can change the sense of no coordinate.
That's translation.
We can change the sense of one coordinate.
That's going to be reflection.
We can change the sense of both coordinates.
That's going to be a rotation.
That's all we can do.
So these are the three basic operations in a two-dimensional space.
I gave you special cases to make things easy, but regardless of where the mirror plane is, parallel to or perpendicular to an axis or not, and whether it passes through the origin or not, a mirror plane has the operation of reversing the sense of one direction.
Just the sense of one direction that is reversed.
And the rotation, be it a rotation through 60 degrees or 90 degrees or 180 degrees, is always taking both coordinates.
It's making a transformation of both coordinates.
If you only have two coordinates with which to play, that's all you can do.
Let's do some giant extrapolations.
If we have a strictly one-dimensional pattern where there's x and nothing else, than they're only going to be two coordinates.
And there are going to be only two ways we can transform coordinates.
So in a one-dimensional space, we can change the sense of no coordinate, and that's going to be the operation of translation, or we can change the sense of one coordinate, and that's going to be the operation of reflection.
No rotation in a one-dimensional space.
Now, let's extrapolate in the other direction.
In a three-dimensional space, the sort that we're going to be concerned with when we want to describe the symmetry of real crystals, you've got three coordinates to permute.
So it follows then, without saying what they are, in 3D, there are going to be four distinct one-step operations.
And then five dimensions?
Hey, that's a nice thing about mathematics.
You could play any game you like.
Not only that, but you make up the rules.
In a five-dimensional space, there's going to be six transformations.
Would we ever want to worry about five-dimensional crystallography?
Well, let me hang out a teaser and not answer the question.
Yeah, there are crystals for which as many as six-dimensional symmetries are necessary.
Wow.
Doesn't that blow the mind?
We'll return to that, and I'll explain why later on.
Another thing you might ask, why did I sneak this in?
Why did I say one-step operation?
Well, it's something we should worry about, and unfortunately, we will.
What if you take a motif, translate it, rotate it around a couple of times, reflect it, bounce it up and down three times, and then put it down?
How do you get from the first motif to the final one there?
Is there an infinite number of operations?
Mercifully, no.
The number is small and very finite.
And we will systematically, in another week's time, examine specifically two-step operations.
And as with many things that we'll encounter, we might not be clever enough to think them up.
But when we start putting things together into a synthesis, suddenly we're going to stumble over something we don't know how to explain, and we will have arrived, like it or not, at a new feature which we perhaps hadn't anticipated.
OK, any question at this point?
All this has been in a way of general introduction.
We're going to now take things more slowly and proceed one step at a time.
I would like to confine our attention for the moment on two-dimensional symmetries and examine the sorts of patterns that can exist in two dimensions, fabric patterns, floor tile, grillwork, and so on.
And we've seen that, basically, there seem to be three operations, three kinds of operations, translation, reflection, and rotation.
That's an infinite number of operations because we are not specifying whether or not the rotation angle is restricted to any particular value.
No reason why it should be.
There are lots of rotational symmetries that are absolutely lovely.
But let's build things up.
And I would like to first look at the operation of translation, which we've said a great deal about to this point.
Translation has magnitude.
It has direction.
So it acts like a vector.
But just like a vector, it has no unique origin.
Perform the operation twice, and you have a position that is two translations removed from the origin.
Do it three times, you have a translation that's three times out.
If a motif sits here, the motif must sit at the end of this translation in the same orientation parallel to itself.
It must exist at the end of two translations.
And if the operation acts on all of the space, if we say a translation is present, we really imply that there's an infinite row going to plus infinity and back to minus infinity.
And there is a motif hanging at the terminal point of every vector.
Now, we can summarize this periodicity with a convenient device.
Let's take some fiducial point and summarize the translational periodicity by saying that something that is hung at one point, either here, or maybe hung also off in some other direction relative to the translation, that something hung on one of these points is automatically reproduced for us at every point.
So what we have done through this device is defined something that is called a lattice point.
And this is an abstraction of the translational periodicity.
There is an array of points, geometric fictions, which we have constructed.
And we ascribe to this geometric fiction the property that anything hung at one of these points, be it a benzene ring or be it a Santa Claus on Christmas wrapping paper, is understood to be automatically reproduced at every other one of these points.
It is this array of fictitious points that is the proper designation of what we refer to as a lattice.
So a lattice is an array of fictitious points that summarizes the translational periodicity of the crystal.
It has a property to repeat that something hung at a particular disposition relative to that point and with a particular orientation is understood to be hung at every other lattice point in exactly the same way.
So that is a lattice.
And this is one of the most abused terms in crystallography.
We talk about the sodium chloride lattice.
The sodium chloride lattice is a set of points that are placed at the corners of a cube and in the middle of all the faces.
This is the NaCl lattice.
If I choose to decorate that lattice with one sodium and one chlorine, then I have atoms sitting at these lattice points.
And that is the sodium chloride structure.
That is the proper term for the atomic configuration.
So lattice is a geometrical term, and it's an abstraction.
Structure is the actual atomic arrangement.
Now, since I realize already that I am among friends, I can confess that I very often recklessly abuse the term lattice.
If I talk about lattice energy, lattice diffusion, lattice vibration, I'm not talking about abstract points bobbling around or something going through this array of little points.
I mean, I should talk about structure diffusion, structure energy, structure vibration.
But man, that just doesn't have the established terminology, and it doesn't have the zing and music of something like lattice vibrations.
So I do it all the time.
Don't tell anybody else that I said this to you, frankly.
But it's never going to be stamped out.
But now you perhaps are informed enough to at least blush slightly when you talk about lattice energy or lattice diffusion, realizing you're using the term incorrectly and that you know better, but everybody else does it, so you do the same thing.
So that is the definition of lattice.
Now, suppose I take this space, to which I've added a first translation, and I'll call it T1, implying that I'm going to add something else to this space, which I'm free to do.
I can put in a second translational periodicity because this is a two-dimensional space.
How do I do this?
And the answer is very carefully because the second translation could not go in the space parallel to the first one if I put in a second translation, T2, which is totally incommensurate with T1.
The things blow up in my face.
I don't have a lattice.
I will get lattice points all over the place.
So this is impossible.
So if T1 is not equal to T2, this space self destructs.
If T1 is a multiple of T2, then if I say a translation exists of length T1, and I add a second translation twice T1.
I've already got those lattice points.
And that's nothing new.
So if I want to say there's a second translational periodicity in the space, the only thing I can do is pick a T2 which is not parallel to T1.
And then this T2 will pick up everything in the space.
It's going to take these lattice points and generate them at equal intervals, T2.
But for that matter, it acts on everything in the space.
So we could think of this translation, T2, as moving this entire infinite string of lattice points separated by T1 and giving me a whole string of lattice points.
So now, having taken two noncollinear translations, those translations will imply a two-dimensional space lattice in which motifs will be hung at translations nT1 plus mT2 where n, m are integers that go from minus infinity to plus infinity.
OK, so this is a two-dimensional space lattice, or sometimes it's referred to by the term a lattice net.
Good term.
It looks like what fishermen throw in the water to snag fish.
So it is a net, in terms of something that we're familiar with in everyday life.
All right.
So we've specified a space lattice, but it is a highly redundant pattern.
We've got a doubly infinite set of lattice points.
And the unique nature of the pattern, the structure, is going to be whatever is associated with one lattice point.
So if we specify what's going on in the vicinity of one lattice point and establish that at every other lattice point, we have the entire infinite two-dimensional structure.
So let's ask now, how we can define the area that is unique to one lattice point.
And there are several ways of doing this.
We can specify T1, and then specify T2.
We'll repeat T1 up to here.
T1 will repeat to T2 over to here, and we will have defined the area that is uniquely associated with one lattice point.
So if I can tell you what's going on within the confines of this parallelogram, then I have given you the unique part of what is hung at a lattice point, and which is reproduced only by translation.
And this is a very important construct.
It is something that is referred to as the unit cell, or sometimes just cell for short.
And now we encounter a curious ambiguity.
T1 and T2 imply an array of lattice points.
And this particular choice of T1 and T2 define a cell.
But the reverse is not true.
If I give you-- and what do I want to say?
That a particular lattice does not specify a unique unit cell.
Or, stated another way, there are many different choices for T1 and T2 that would specify the same unique area.
I could take this as a T1 prime, and then I would have a cell that looks like this.
And that would also define the area associated with one lattice point.
It's not clear this oblique thing with one very long T1 prime would have very much to commend it, but there are many ways, many choices, for T1 and T2, to find exactly the same lattice.
We could take this as T1, this as T2, same lattice, same array of lattice points.
Take this as T1, this as T2, same array of lattice points.
Take this as T1, this as T2, that's yet another choice.
So there are an infinite number of translations.
Special name for this, to introduce a bit of jargon again, these are very often called conjugate translations.
So all this is still nothing more than simple geometry, but if you invent some fancy words, you really have to do that to impress your friends.
Yeah, you had a question here?
Yeah.
So you can define magnitude for T1 and T2, all those constants.
But you're changing the directions of T1 and T2, and you're saying, even though you're changing those directions, it's still the same unit cell?
Yeah, provided I have some new translation like this one here, which is really this T1 plus this T2, this would define a very, very oblique cell that looks like this.
But yet, the terminal points of T1 prime and-- I need a term for this.
I'll call this T2 prime.
The terminal points here are going to be exactly the same as the nodes that are defined here.
So they are two choices for one in the same lattice.
OK, so the implication of this is we're going to have to have rules.
And some of these make common sense.
You could pick, in a two-dimensional lattice, some absolutely ridiculous unit cells defined in terms of very long vectors that define a cell that is a very, very oblique cell.
So it's the lattice that's defined by this translation here.
And the next translation parallel to this one would go way up to something like this.
So there's a T1.
There's a T2.
This crazy cell here works.
That's the area that's associated with one lattice point.
But clearly, it has absolutely nothing to commend this choice.
There's nothing to be gained by using these long translations that make very extreme intertranslation angles.
Your intuition would say, why would you want to do that, you dummy?
Let's take these as the translations, which is something I sort of naturally did all along.
And what are we doing?
We're picking the shortest translations.
So there's one very common sense rule.
Another rule, getting a little bit ahead of the game, but suppose I examine the lattice that describes the arrangement of four floor tiles.
If I take a lattice point right at the point of intersection of the joins between the tiles, that is a cell that is exactly square.
And it's exactly square because there's a four-fold axis in that pattern that leaves things invariant after a 90-degree rotation.
So if that's the nature of a lattice, if it in fact is constrained because of the symmetry that is there to have two translations identical in length, in fact, identical in every way.
Pick those as the choice of the cell to emphasize that special key feature of the lattice.
So a second row, which is a second and final one, is to pick a T1 and T2 that displays the symmetry, if any, of the lattice.
Which introduces us to a feature which we'll elaborate much more later on, that the translational periodicity and the symmetry of the lattice are two things that go together.
That the fact that there is translational symmetry drastically reduces the number of symmetries that you could have, the fact that there are symmetries possible for presence in a lattice restricts the number of different kinds of cells.
So these are two aspects of the pattern, the symmetry that's in it and its periodicity.
OK, but these are the only two rules that we really need to pick what's called the standard cell.
Take the shortest translations that are available to you, and pick translations that display the symmetry that may be present in the lattice.
Any questions or comments?
Any comments?
OK, I think I have time for one last major point of discussion.
And what we are going to embark on now is a process of synthesis, which will occupy us for a couple of weeks.
What I'm going to do is start with a translation.
And this defines an infinite string of lattice points.
Now, I know that in two dimensions, I have two kinds of symmetry operations that are present, either rotation or translation.
So now, I'm going to ask the question, what happens if I define a lattice, or at least one translation in a lattice, and now I add to that lattice an operation of rotation, OK?
I can do that.
We've seen examples of translationally periodic patterns that have rotational symmetry.
So let me suppose I add to this space a rotation operation, A alpha.
And there's no unique origin to the translation.
There is no unique location for a lattice point.
So I can put the operation A alpha in at my designated lattice point.
Now, if I do that, all hell breaks loose because now I have a rotation operation A alpha.
This has a translation coming out of it.
That translation will be repeated up here, an angle alpha away.
A alpha acts on everything, so it's going to take this translation and move it over here to a location for another translation.
This is a lattice point.
This is a lattice point.
And this business is going to go on until it comes around full circle.
Let me focus my attention on just one of these translations, and this will be this one up here, the one that is alpha away from the first in a counterclockwise direction.
So here sits another translation, and that means this is a lattice point.
At this end of the translation, the same thing is going to happen.
The operation A alpha is moved to this location at the end of the translation.
That means that anything coming out of this lattice point must also be repeated at angular intervals, alpha.
And now I'm going to focus my attention on this translation here.
And there will be a translation that goes up like this, and this is a lattice point.
And now, in the words of that famous musical, there's big trouble in River City.
Because we started out by saying that everything in the space was periodic at an interval, T, a translational interval, T. This is T. This is T. This is T. Here we have a lattice point.
This jolly well has to be T as well, or we've contradicted the basic assumption of our construction.
Well, that's over restrictive.
This doesn't have to be T, but it has to be some multiple, p, of that translation.
p could be 0. p could be 5.
But it has to be an integral number of translations because this translational periodicity has to work everywhere, including up on the top of this trapezohedron.
So that's a constraint.
This angle is alpha.
We cannot let alpha be arbitrary because the only way we can add a rotation operation A alpha to a lattice is for a value of alpha which makes this translation be a multiple of the original one.
Now, let me take this geometry, and I'm going to extract the basic constraint from it.
This is some integer, p times T. This is T. This is T. This is T. This is alpha.
Let me lickety split drop down a perpendicular to the original translation.
This is T times the cosine of alpha.
This is T times the cosine of alpha.
This total length is T. This length in here is p times T. And now I can go away from the geometry to an equation, something you probably prefer to deal with.
And what this constraint is expressed analytically is that my original translation, T, minus twice T times the cosine of alpha has to come out equal to an integer, p times T. And there's my constraint.
Alpha has to satisfy that condition.
Well, I can immediately cancel the T and write this as one minus 2 cosine of alpha is equal to an integer, p. And it figures that that has to be the case because none of this construction depends on the size of the original translation that I took.
And now, let me solve for the values of alpha which are compatible with a lattice.
This says that cosine of alpha is 1 minus p over 2.
And unless that condition holds, my combination is incompatible.
So I'm going to let that stew with you until next time.
But what we've set up is something where we can just plug and chug, put in different values of p. And if I start out with a value of p, and let's let p be equal to 4, and then find one minus p over 2, which is supposedly the cosine of an angle, alpha.
If that's 4, I will have minus 3/2.
And the value of alpha obviously does not exist.
Cosine of alpha cannot get greater than 1.
If p is equal to 3, then 1 minus 3 over 2 is minus 2 over 2, or minus 1.
You like the way I do that arithmetic in my head just like that?
And the angle whose cosine is minus 1 is 180 degrees.
And what that says is that a two-fold axis works.
So I can drop a two-fold axis into a net.
And what that's going to do is take my original translation, rotate it 180 degrees, and the second translation is going to sit here.
Rotate it 180 degrees in the reverse direction, and then the second lattice point sits here.
And lo and behold, just as advertised, the distance between the first lattice point and the final lattice point is three translations.
So I can put a two-fold axis in any lattice whatsoever because this is compatible simply with a lattice row.
So one possible combination of rotation in a lattice is going to be any lattice whatsoever, and what we can add to this is a rotation operation A pi.
And we'll have two full rotation operations which are translationally equivalent.
All right.
Several integers to go.
We would want to try p equals 2.
That's going to work.
p equals plus 1 is going to work.
p equals 0 is going to work.
And we will find a very limited number of rotational operations that are compatible with a lattice.
And this is going to give us a small number of the possible combinations of lattice and rotational symmetry in two dimensions.
So we'll pick up from there next time, and we'll very quickly determine the remaining possibilities and take a look at what the arrangement of symmetry elements look like in these lattices.
OK, once again, I have some extra copies of the syllabus if somebody did not get one.
And I'll have extra copies of the problem set.
OK, why don't we you get started again?
I can understand why there is an air of excitement in the room since tomorrow's a holiday.
But we still got two hours before the day is over.
Any questions on what we have done up to this point?
No, that's good.
Let me erase some of this art work then and ask why one would indulge in this rather bizarre exercise of taking the elements of something that represents a physical property and constructing a surface out of them.
Well, the name suggests that maybe these surfaces, be they hyperboloids of one or two sheets or imaginary ellipsoids, have something to say about the property that the tensor that formed the coefficient of these functions is doing.
So let me take the case of an ellipsoid.
That would be the case where all the coefficients are positive.
And let's let this be x1 and this x2.
And let me ask a first question.
If this is to be a well deserved function, does the surface transform in exactly the same way as the tensor elements?
In other words, if we take the tensor in one coordinate system and then change the coordinate system to a different coordinate system, are the coefficients in front of x1, x2, and x3 still the elements of the tensor in that new coordinate system?
So let me show you that this is, in fact, the case.
Suppose our original relation, sticking to conductivity, is that x sigma ij xi xj equals 1.
And then for some reason or another, we change axes.
So the new equation will be some different coefficients.
Because as we change axes, they're going to wink on and off, still the same surface, but now referred to different axes xi prime and xj prime still equal to 1.
And let me now use the reverse transformation to put xi prime in the terms of xi.
If we do that, we can say that xi prime is cli x sub l prime.
That's the reverse transformation.
So notice the inverted order of the direction cosine subscripts.
And then xj prime is going to be equal to cmj times x sub ep prime.
To find this, let's just rearrange these terms in a trivial fashion.
Then I will have cx sigma ij cil cmj times xl prime xm prime equals 1.
And what is this?
This is a summation of direction cosines where the first subscript goes with the subscript on sigma prime.
What did I do wrong?
This is mj.
Why is that not coming first?
[INAUDIBLE]
Well, what I would like to get this in the form of is a summation of sigma ij prime over two dummy indices l and m. What did I do wrong?
Ah, yes, right, right, right, right, this is li.
Thank you somebody said it.
And my nose was right in it.
And I didn't notice it.
OK, this is a summation of all the original elements sigma ij prime times dummy indices l and m. And this actually is sigma ijn.
So it's the same set of coefficients.
And that is a technicality, which probably you wouldn't want to worry about.
So anyway, the elements of the tensor transform in the same way that the equations for the surfaces transform.
So if you change coordinate system, the coefficients that you should use should be the elements that are in the transform tensor.
OK, now I'm going to ask a question.
Suppose I define some direction by a set of direction cosines li and I ask the value of the radius of the surface in that direction.
So this is some radius vector.
The radius vector will have components, R1, R2, R3 or Ri.
And each of those components will be equal to the magnitude of R times the direction cosines li.
Yes, sir?
Could you briefly rego over what you just stated before you erased it [INAUDIBLE]?
Oh, OK.
I'm sorry.
I said aij xi xj equals 1 is the original equation for the quadric.
If we change the coordinate system, we're going to have some new elements aij prime times xi prime xj prime.
Because we've got a new coordinate system.
And therefore, the coefficients in front of the equation for the quadric have to change.
And now what I did was to express xi prime and xj prime in terms of xi and xj using the reverse transformation.
And xi prime in terms of the original x's will be c sub l sub i x sub l prime.
Usually, we write it xl equals cil x sub l. This is the reverse transformation where the order of the subscripts is reversed.
So this will give me x of i prime.
The expression for x sub j prime will be given by cmj x sub j where m is a dummy index.
And that's equal to 1.
And if I just rearrange this, then I will have cli cmj times aij prime equals 1 times xl xj.
And this is going to be alx lxm.
And this will be alm times xl xm equals 1, which is the equation for the quadric in the original coordinate system.
I think that was better when I was standing in front of it so you couldn't see it.
It's just that the surface transforms formally to the surface that we would create if we used the new tensor elements for the same change of coordinates system, which you would probably will be willing to take my word for
[INAUDIBLE]
This is a direction cosine for the--
[INAUDIBLE] right below
Right below?
[INAUDIBLE]
cli cmj aij prime xl xm, and then I collected the cli cmj times alm.
And that is the transfer.
Yes?
In the previous [INAUDIBLE], you defined the derivitave side as being xi prime equals cil xl.
[INAUDIBLE]?
You can say that x prime is cil times x sub l. And if we want to use the same direction cosine scheme but do it in reverse, we would say that x sub l is equal to c-- yeah, OK, x sub i is cli times x sub i, right, no, times x sub l-- is that right-- x sub l prime
On the second line to the third line, you say that xi prime equals cli [INAUDIBLE].
It's either xi equals xcli xl prime or xi prime equals cil
OK, you want to say this.
xi prime and cil becomes xl, yes, yeah.
And then mj times x-- that's x sub m. OK, and then this says that cil cjm times aij prime should be alm, right, OK, OK.
So it's OK then.
OK, that was supposed to be a small point that everybody would accept.
But now it's done correctly.
OK, let's get back to this, which is more hair raising and exciting.
What is the radius of the quadric in a given direction that we specify by three direction cosines l1, l2, l3?
So there's some radius from the center of the quadric out to the surface in that particular direction specified by three direction cosines.
And we know what those three components are, sub i, are going to be in terms of the direction cosines, magnitude of R times li.
And those points at the terminus of the radius vector go to one point on the surface of the quadric.
So these values of R are coordinates that satisfy the surface of the quadric.
So we can say that just substitute Ri for the different values xi in the equation for the quadric.
And this says that sigma ij magnitude of R times li, that would correspond to x sub i times magnitude of R times l sub j.
That's the magnitude of xj.
That should be equal to 1.
And if I rearranged this slightly, this says that the magnitude of R squared is going to be equal to 1 over sigma ij times li lj.
So the radius of the quadric gives me not the value of the property in that direction.
This is the value of the property that will occur in the direction specified by the direction cosines lj.
The radius of the quadric is going to be equal to 1 over the square root of the value of the property in that direction.
So this is why it's called the representation quadric.
If you construct the surface from the tensor elements, you will have defined a quadratic form, which has the property that, as you go in different directions look at the distance out to the surface of the quadric in that direction, that that distance squared is going to be equal to 1 over the property in that direction or, alternatively, the radius is going to be 1 over the square root of the value of the property.
There is an enormous implication here.
This says that the value of the property, if the quadratic form is an ellipsoid, the value of the property as we go around in different directions in the crystal is going to be a smooth, uniformly varying function.
They're going to be no lobes sticking out.
They're going to be no dimples, no lumps.
It's going to be an almost monotonous property, not very interesting.
It's going to change in a uniform way.
And in fact, we could put the two surfaces side by side.
If this is the value of the quadric as a function of direction, if we make a polar plot of the property as a function of direction, it's going to look sort of like the reciprocal of this.
The minimum value of the property will be in the direction of the maximum value of the radius of the quadric.
And the maximum value of the property is going to go in the direction of the minimum value.
So the value of the property as a function of direction is going to be a quasi-ellipsoidal sort of variation but not really an ellipsoid.
It's going to go as this inverse square of the radii in ellipsoid.
But the thing is it's going to be something that varies uniformly between extreme values of the maximum and minimum value of the property.
We are, I assure you, for higher ranked tensor property going to look at some absolutely wild surfaces with surfaces that do have lobes and extreme values and very, very irregular variation of properties with directions but not for second ranked tensor properties.
There will be a few variations on this theme, which are also interesting to touch upon.
In principle, we can get other quadratic forms from the tensor if some of the elements of the tensor are negative.
And in particular, what would we see-- and I'll draw this relative to the principal axes of the surface-- what would we see for an hyperboloid of one sheet?
This is the sort of surface that might result if one principal value of the tensor had a negative value.
Well, this is the quadric.
And this is a radius in one particular direction.
What would this say about the property?
Well, let me, to make it clear, look at one of these sections through the quadric that our hyperbola.
In directions like this, we have a radius that is a minimum out to the surface.
In other directions, the radius gets progressively larger.
And then if we plot the reciprocal of the square root of that, that says that, in this direction, we get the maximum value of the property.
And as the direction approaches the asymptote, the reciprocal will go down to 0.
But how did I get two y-axes here?
OK, so the maximum will be in the direction of the minimum radius.
And then it will go down to 0 for the asymptote.
And that's going to be symmetrical on either side of this principle axis.
So what then are the radii in directions outside of the asymptotes of this hyperboloid?
Within this range, the radius is imaginary.
But if you square it, you get a negative number.
So within these two lobes, the value of the property is positive.
As you go away from the asymptotes, you'll get another lobe like this and another lobe like this where the value of the property is negative.
How in the world could you get anything that looked like that?
Well, a good example of a property that has this behavior is thermal expansion.
And let me give you two examples.
The structure of selenium and tellurium is a hexagonal structure.
And there are chain-like molecules that are pairs of bonds that rise up around a threefold screw axis.
So this is two coordinated atoms in the structure just spirals up in this triangular spiral around the threefold screw axis.
So this is a material in which the bonding is very, very anisotropic.
The bonding within these covalently bonded spirals, which are like springs that you might put on your screen door in the summertime, the bonding is very strong.
Between these individual molecular chains, the bonding is very weak.
So what happens when you heat this stuff up?
It expands like the dickens in the direction of these weak bonds.
But the spirals are, in part, held in this extended form by repulsive interactions in like chains.
So when these spirals move apart as a result of large thermal expansion in the plane normal to the chain, the chains relax a little bit.
So you have a large positive thermal expansion here.
But in one direction along the normal to the hexagonal in the structure, the thermal expansion is negative.
And that gives you a variation of property with direction that looks exactly like this.
Negative value of the property, that means the structure, contracts in that direction, positive values this way.
The structure expands.
So there's a good example of this.
There's another example of a material, which, again, has a negative thermal expansion in one direction.
And this is calcium carbonate, which is a complicated hexagonal structure.
But it looks very, very much like the structure of rock salt in a distorted form.
These are the calcium atoms.
And calcite is CaCL3 And the calciums are in a face-centered cubic arrangement just like in rock salt.
The carbonate groups are very tightly bonded little triangles with the carbon in the middle.
And these things are arranged on the edges of the cell.
And this is going to be very schematic.
These triangles are normal to the body diagonal of the cells.
So you have one family of triangles that are all parallel to one another.
On the next edge, the triangles are anti-parallel to this first orientation.
So think of rock salt.
Tip it up on its body diagonal.
Every place you have a chlorine anion, place a triangle in an orientation that is perpendicular to the body's diagonal of the rock salt structure.
Again, these tightly bonded little triangles don't do much as you increase temperature.
But the bonding between the calcium and the triangles is rather weak.
So again, you find that the structure expands in one direction so strongly that it contracts to the other direction so calcium carbonate.
The calcite form also has the distinction of having a negative thermal expansion coefficient.
We'll talk quite a bit about expansion coefficients as one example of a tensor property.
Interestingly, and we'll say more about this and I'll give you some references when we come to this point, can you get a material that has a negative thermal expansion coefficient in all directions?
No, that seems as though it would violate in a flagrant way some vast law of thermodynamics that things have to increase their volume when you increase temperature.
There were, however, a few odd ball materials that over a very, very limited temperature range would contract but only by a tiny amount in all directions.
And then one of the very interesting discoveries of recent years done primarily by a crystal chemist named Art Slate, who's out at the University of Oregon, he discovered a family of materials that have large negative expansion coefficients in all directions over a considerable range of temperatures, like 100 degrees.
So these are materials, when you heat them up, they contract amazing as that seems.
And there's a structural reason for it.
These are tetrahedral frameworks in which one corner of a tetrahedron is dangling.
And as you heat it up, this tetrahedral corner can get closer to other tetrahedra, which takes energy to do.
But in so doing, you actually are changing the net volume occupied by the solid so very, very anomalous set of compounds.
More of these have been found.
There are probably a dozen materials now that have negative thermal expansion coefficients in all directions.
So is the imaginary ellipsoid a viable representation quadric for real materials?
Yes, in the case of thermal expansion coefficients.
OK, so the representation quadric then is a dandy device for seeing with one function how a property will vary with direction.
For an ellipsodial quadric, the variation of the property itself with direction is not really ellipsodial but quasi-ellipsodial in that there's a small principal axis, a large principal axis.
And the large value of the property goes with the short axis of the quadric.
So we have a nice device for representing the value of the property as a function of direction.
It looks as though we've lost all information about the direction of the resulting vector, the generalized displacement.
It looks as though that's not in here at all.
Well, there was an ad years ago for a bottle spaghetti sauce, which offends many cooks who are very fiercely proud of their own spaghetti sauce.
So here's a wife who's using the canned stuff.
And her husband comes home and looks at it very, very skeptically and says, where's the basil?
And the housewife says, it's in there.
It's in there.
And he sniffs again and says, where is the basil?
And she says finally, it's in there.
It's in there.
Well, this is a similar situation.
Where is the direction of the generalized force?
And I say, it's in there.
It's in there.
Not obvious, but it's in there.
So let me now show you where it is lurking.
I think we have time to carry this through.
Let me describe how it's embodied in the quadric and then prove to you that this is indeed the case.
And I'll use a general quadric in the form of an ellipsoid.
That looks more like an egg than an ellipsoid.
It has a pointed end.
OK, this is something that's called the radius normal property.
And what it says, in words, is that, if you pick a particular direction relative to the quadric, we know that its length is going to be inversely proportional to the square root of the value of the property.
If you want to know what the direction of the resulting vector is, for example, in our relation for conductivity now getting very warm, it says the current flow is given by a linear combination of every component of the electric field.
The radius normal properties is that, if you want to know where J is for this particular direction, look at the point where the radius vector intersects the surface of the quadric and, at that point, construct a perpendicular to the surface, which, in general, will not be parallel to R. And this will be the direction of J.
It won't give you the magnitude.
But it'll give you the direction of it.
OK, let me now prove to you that the quadric does have this property.
In our conductivity relation, the direction of J is going to be given by the tensor relation.
Namely that J sub i is equal to sigma ij times E sub j, which can be written as sigma ij times the direction cosines of E sub j times the magnitude of E. I'm going to want to compare this expression, with which you're very familiar term by term, with the normal to the surface that we can compute from its derivative at that point.
So this will say that J1 is equal to sigma 1, 1 l1 times the magnitude of E plus sigma 1, 2 times l2 times the magnitude of E plus sigma 1, 3 times l3 times the magnitude of E. And I don't need to write much more.
J2 will be sigma 2, 1 times l1 magnitude of E plus sigma 2, 2 l2 times the magnitude of E and so on.
What is going to be the normal to the surface as a function of direction?
OK, to do this I'm going to say that the normal to the surface is going to be, if I have some function of xyz, the normal to that function is the gradient.
Let me say that that's G of xyz.
And we can say then that the x1 component of the normal is not going to be equal to but proportional to the gradient of the equation for the quadric with respect to x1, dx2, and dx3.
So it's going to be proportional to the differential of the function that gives us the surface with respect to x1 times i plus the differential of the function with respect to x2 times j plus dF dx3 times k. So this is the normal entirely.
So if we split this into components, N1 is going to be equal to dF dx1.
And if I differentiate the equation for the quadric, that's going to be 2 sigma 1, 1 times x1.
Remember the equation for the quadrant is sigma 1, 1 times x1 squared plus sigma 1, 2 times x1 x2.
So if I differentiate-- let me write it down.
If I take those terms and differentiate with respect to x1, I'm going to get 2 sigma 1, 1 times x1.
Here I'll get sigma 1, 2 times x2.
But then down in this line where I have a sigma 2, 1 x2 x1, if I differentiate with respect to x1, I'll have another term sigma 2, 1.
And if I differentiate with respect to x3, I'll have sigma 1, 3 plus sigma 3, 1 times x3.
And a similar thing for the x2 component of the normal, that's going to be proportional to the gradient with respect to x2.
And that's going to be equal to sigma 1, 2 plus sigma 2, 1 times x1 plus 2 sigma 2, 2 times x2 plus sigma 2, 3 plus sigma 3, 2 times x3 and similarly for N3.
OK, I've made my case if I can demonstrate that each component of J, J sub i, is proportional to each component of the gradient to the function that defines the quadric.
If you look at them, they're close but no cigar.
The first term is OK.
I've got a 2 out in front of sigma 1, 1 for the x1 term.
Here, though, I have the sum of two off-diagonal terms.
And for this expression, I have just sigma 1, 2.
And down here I've just the single term sigma 2, 1.
So the conclusion we're forced to draw is that these two expressions would have components of a vector that are parallel to one another if sigma 1, 2 was equal to sigma 2, 1.
Because if these were equal, I could write just twice sigma 1, 2.
And down here I could write twice sigma 2, 1 and do that all the way through these two expressions.
And then this factor of 2 would just be part of the proportionality constant.
So the radius normal property will be true or valid only for symmetric tensors.
That is it's going to be true only if sigma ij is equal to sigma ji.
We mentioned when we first started talking about second ranked tensors that a tensor does not have to be symmetric.
And showing that the tensor is symmetric has nothing to do with symmetry.
Because it's symmetric in an algebraic sense not in the literal sense of geometrical symmetry.
And there are some properties, the thermal electricity tensor is one notable example, where the tensor is a second ranked tensor but it decidedly is not symmetric.
So this is OK for most properties but not all.
OK, so here are properties of the representation surface that we can construct from the representation quadric.
And what I would like to do following this and after the inevitable unpleasantness of a quiz is to look at some specific properties that illustrate second ranked tensor properties.
And in particular, I would like to look at some other sorts of second ranked tensor properties that represent generalized forces, namely stress and strain.
You've seen these before in other contexts perhaps not actually defined rigorously in terms of tensors.
Because, obviously, stress and strain can be regarded as generalized forces.
A stress is something that you can apply to a solid.
And that will cause various things to happen.
In addition to mechanical behavior, different properties and effects can result.
So that is going to involve a generalized force, which is a second ranked tensor.
And a thing that might happen could be a vector.
It could be another second ranked tensor.
And this is going to introduce us to the nasty world of higher ranked tensor properties and their representation surfaces.
So this is a nice place to quit and pause for a quiz.
When we resume, we'll look at stress and strain in terms of our tensor algebra.
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This, as you know, is the last lecture of thermal tensor properties of crystals for the semester.
Come on, can't somebody go, aww, just to make me feel good?
At least nobody said yay, so that makes me feel good, too.
So I guess on balance I come out feeling pretty happy.
What I'm going to do today is to talk a little bit more about forthright tensor properties, which we've not said much about.
And I'm going to look at some scalar moduli and look at them for a couple of different symmetries and restrictions on the compliances.
The surfaces that you might expect turn out to be really, really weird.
Because these will be fourth order variation with the direction cosines.
So when you take trigonometric functions and raise them to the fourth power, you get severe anisotropies for a great many of the scalar moduli.
I'm sorry to say, years ago when I made the acquaintance of a computer programmer in course six who was looking for something challenging to do in connection with material science and engineering, I said, wow, have I got something for you.
How about doing some computer graphics?
And this was a few years ago when such products were fairly rare.
And how about making a program that would let us see visually how some of these scalar moduli will vary with direction and provide the provision so you can turn them over in space, just as you would pick up a model in your hands and rotate it around till you finally came to appreciate it?
And then for something like a triclinic crystal, where there are almost a couple of dozen different moduli, how about allowing people to scale up or scale down the value of one of the elastic constants and let them see how the shape changes?
And oh, that was-- he went to work on it for a semester and came back.
And it was a gorgeous thing.
You could take a monoclinic representation of three dimensions of Young's modulus.
Say you want to change S31.
And it would change by a factor of 10.
You'd see a big nose blow out on the surface and then contract down again.
And you could turn around.
Anyway, the program having told you how marvelous it is, it's no longer supported because of the operating system that it operated under having done defunct.
But anyway, that was a fun thing to play with
What operating system?
I don't even remember anymore.
It was a fairly obscure one.
Was not only the system, but it worked through a software package which operated on that system.
All right.
So we're going to talk today about fourth rank tensor properties.
The most important-- in fact, probably the only ones you've ever heard of, are the elastic, stiffnesses, and compliances, which are ways of representing strain in terms of stress or stress in terms of strain.
I think I'll sort of see how the time plays out.
If I finish just about everything that I'm hoping to say by on the hour, I think I'll just go an extra five or 10 minutes.
And then comes a great, grand, and traditional event at MIT.
All semester long I've been grilling you with problem sets and quizzes, and you've had to dance through your steps just because I told you to.
Now how's this for reciprocity-- you get to evaluate me.
You get to grade me.
MIT, at the end of every semester, has a course evaluation.
It's submitted anonymously by you, so you could really vent your spleen without ever having called to task for it.
So when we finish, either for our break or at the end of a slightly extended first session, I will bow, to thunderous applause, I hope, and exit.
And then you will be left alone to fill out the course evaluation.
Corinne will collect them from you, and she will deliver them over to department headquarters.
The results will be tabulated, and so will the comments.
But they will not be in your hand, all of which I recognize, having spent hour after hour grading your quizzes.
So they will be submitted to me in a completely sanitized fashion.
I won't know who said what.
A final question that I will answer before you ask it, what about the quizzes?
How are you doing on them?
Well, I have come to hate them.
I think I will give one-question quizzes from now on the future.
It takes me about five to seven minutes to grade each question.
And there are about 25 of you.
So it takes me, to grade one question, somewhere between two and 1/2 and three hours.
And there are 10 questions on the two quizzes remaining to be graded.
And so I let you do the arithmetic, and you can see what sort of torment I've been living in the last few days in return for the torment I put you through for two brief hours.
So wow.
I will leave it to you to decide who's getting the worst of this deal.
OK, fourth rank tensors.
Elasticity's probably the only one you can think of.
Let me give you an example in this handout, an example of another forthright tensor property.
And this is the piezoresistive effect, the way electrical resistance of the material changes in response to an applied stress.
It's a pretty exotic property.
I had never heard of it until I was at a meeting of the American Crystallographic Association and somebody, believe it or not, from Texas Instruments-- so you better bet that this property is useful in technology-- fellow named Ahmed Amin got up and gave a talk on the piezoresistive effect.
And he had marvelous slides at the outset that defined the effect So what you have here on these few sheets is a few pages that define the piezoresistance effect.
We're not going to do anything with it.
It's a fourth rank tensor, so it will proceed to transform like other forthright tensors, such as elastic stiffness and compliance.
I should warn you about these following pages, which were slides that he used at his talk.
He seems to have done the lettering with a magic marker.
Because all of the I's and J's are indistinguishable, which makes interpretation of what he's saying here a little bit challenging.
And then the other thing that throws one off is that he uses x to represent stress.
I don't know what field or in what discipline the elements of stress are called xij, but this is what he does.
And once you figure that out, the interpretation of what he has said here is fairly apparent if you think about it.
Anyway, he takes a state of zero stress and strain and expands it as a series, and then picks off different coefficients.
And one of them is a set of moduli which he calls pi, pi subscript ijkl, which relate stress, xkl in his notation, to a change in the density, delta row ij.
So here's a fourth rank tensor property that will indeed have to conform to all the restrictions for the moduli that we have defined for the elastic stiffnesses and moduli.
All right.
Let me then turn to stress and strain.
We introduced the relation between stress and strain, but didn't really go into detail on the bizarre absorptions of factors of two or four that have to be done in order to make this come out in a nice, clean matrix form.
So let me remind you of where these silly factors came in.
We took our stress tensor, sigma 1 1, sigma 1 2, sigma 1 3, 2 1, 2 2, and 2 3, sigma 3 1, sigma 3 2, and sigma 3 3.
The tensor had to be symmetric.
The off-diagonal term sigma ij had to be equal to the term sigma ij.
And this was for the reason of mechanical equilibrium.
These off-diagonal terms, sigma ij and sigma ij, we saw, exerted a torque on a body.
And unless they were equal, the body would undergo an angular acceleration.
And then we renumbered these according to the convention that as we would march down the diagonal of the tensor and take pairs of subscripts, 1 1, 2 2, 3 3, represent them by a single 1, 2, and 3, then march up the side to define a sigma 4, sigma 5, and a sigma 6.
If you remember that little algorithm you can always keep straight what these reduced subscripts represent.
So with that definition, then, the stress tensor reduced to sigma 1, sigma 6, sigma 5, sigma 6, sigma 2, sigma 4, sigma 5, sigma 4, sigma 3.
And that takes a tensor and degrades it to a matrix.
Because there is no law of transformation for this representation of the elements of stress.
OK, then we do something very similar with the strain tensor, epsilon 1 1, 1 2, 1 3, epsilon 2 1, epsilon 2 2, epsilon 2 3, epsilon 3 1, epsilon 3 2, epsilon 3 3.
And we saw that for physical reasons this tensor also had to be symmetric.
We had to have epsilon ij identical to epsilon ji, the reason being that if these off-diagonal shear strains were not equal, the state that we would be defining was one of actual deformation combined with rigid body rotation.
So unless this was the case, the definition performed by the tensor epsilon iij would define rigid body notation as well.
OK. We use the same process of renumbering to convert this from a tensor to a matrix.
And we got this into a form, epsilon 1, epsilon 6, epsilon 5, epsilon 6, epsilon 2, epsilon 4, epsilon 5, epsilon 4, epsilon 3.
But in order to do that, we saw we had to consume a factor of two.
So there is a factor of two built into the off-diagonal terms.
And what we have put in here is we converted this to actually epsilon 1, 1/2, epsilon 6, 1/2, epsilon 5.
And I should have written that this way to begin with.
So all these factors of two appear so that when we combine these epsilons we get a nice, clean matrix relation between stress and strain that doesn't involve factors of two.
And that's a great convenience if we're going to be doing all of our deformation within the framework of one coordinate system.
OK.
So if we now use these definitions to write the stress in terms of strain, we will set up our fourth rank tensor property.
If we do this first for the stress in terms of strain, you would have a tensor element of stress sigma 1 1, which would be equal to C1 1 1 1 times epsilon 1 1 plus C1 1 2 2, epsilon 2 2 plus C1 1 3 3, epsilon 3 3.
And then we would have, in addition, off-diagonal elements of strain.
We'd have C1 1 2 3 times epsilon 2 3 plus C1 1 3 2 times epsilon 3 2 plus C1 1 2 1, epsilon 2 1 plus C1 1 1 2, epsilon 1 2 plus C1 1-- and I should have made this 1 3 so that the other integers come out right.
And C1 1 1 2 times an epsilon 1 1 1 2 plus a C 1 1 2 1 times an epsilon 2 1.
So that is one line of the relation between stress and strain.
And when we condense this to a matrix form, it is surprising, once we expand it once more, at how cumbersome this expression is.
Now, why do I remind you of all this?
Isn't it nice to work in the form of a fixed coordinate system where we can use a matrix for the Cijkl's?
The answer is, fine, unless you want to change the coordinate system.
And you might want to do that for practical reasons such as cutting out a specimen which makes it convenient to define the stiffnesses and compliances relative to a coordinate system taken along the logical directions in the specimen.
And the other reason for doing it, and that is what I want to do a little bit of this afternoon, is to derive the symmetry restrictions on the stiffnesses and compliances.
You cannot transform a matrix representation of the stiffness or the compliance.
You can only do this for the full-blown tensor arrangement.
So having collapsed down, which we'll do momentarily just to remind you of how it goes, we will have to expand again to derive symmetry restrictions.
Do not exit at this point.
We're not going to do any of these calculations in their full glory detail.
We'll just set up the problem and then jump immediately to the outset.
So how does this pay out if we try to make the condensation?
I remind you again that the symbol C stands for stiffness, and the symbol S, which we'll see later, the Sijkl's are call compliances.
And the perversity of that semantic description of these tensor elements is perverse for reasons that I have never really been able to understand.
So if we go down to matrix form, we'll write instead of sigma 1 1, simply sigma 1 1.
We'll call this C1 1 times epsilon 1 plus C1 2 times epsilon 2.
So far so good.
C1 3 times epsilon 3.
And now we have problems, because epsilon 2 3 was defined as 1/2 of epsilon 5.
Now we'll have a C1 5 here, and then a C1 3 2 is C1 5 again.
And this also is 1/2 of epsilon 5.
And then similarly, this is C1-- I'm sorry.
This is C1 4.
This is C1 5 times 1/2 of epsilon 5 plus C1 5 times 1/2 of epsilon 5 plus C1 6 times 1/2 of epsilon 6 plus C1 6 times 1/2 of epsilon 6.
So this, given the way in which we had defined matrix strain, initially in connection with piezoelectricity when we entered the realm of third rank tensors, this will play out OK.
This says that sigma 1 is simply C1 epsilon 1 plus C1 2 times epsilon 2 plus C1 3 times epsilon 3.
And now the 1/2 gets absorbed, and it's simply C1 4 times epsilon 4, and so on.
If we look at another line, one that involves some of the off-diagonal terms and strain, if we, for example, look at sigma 2 3 and we write this down as sigma 2 3 times 1 1 times epsilon 1 1 plus C2 3 2 2 times epsilon 2 2 plus C2 3 3 3 times epsilon 3 3 plus C-- and I'll just write down a few of these additional terms.
This would be C2 3 3 2 times epsilon 3 2 plus C2 3 2 3, epsilon 2 3, and then other terms for additional terms, and epsilon ij with i not equal to j, which are going to behave the same way.
So if we convert this, we go to sigma, and we would call this sigma 4.
This would be C4 1 in matrix notation times epsilon 1 plus C4 2 time epsilon 2 2 plus C4 3 times epsilon 3.
And now this matrix element would be C4 4.
And in place of the tensor element of strain epsilon 3 2, we would write 1/2 of epsilon 4.
And the next term is again a C4 4 times a 1/2 epsilon 4 and the terms in epsilon 5 and epsilon 6 would behave the same way.
So you can see that the factor 2 is absorbed, and this becomes simply C4 4 times epsilon 4.
So all the factors of 2 have disappeared.
And we can say, provided we remember the way we have condensed the elements of tensor strain, we can say with complete confidence that in our matrix notation, sigma i is Cij times epsilon j where i goes 1, 2, and 3, and j goes 1, 2, 3 all the way up to six
[INAUDIBLE]
No.
No, no, these are the-- yeah, I'm sorry.
Yeah, inj.
So it's a six by six matrix, right.
Right you are.
And this is a six by six.
And in here are 36 businesses stiffnesses.
OK, any comments other than, yuck?
And again, if you stay in one coordinate system it's not so bad.
In fact, instead of having nine by nine 81 terms you have 36 stiffnesses.
And that's a great convenience.
Unfortunately, things are not quite so simple if we work with the compliances represented by the symbol S. And if we attempt to write strain in terms of stress, we'll have an S1 1 1 1 times sigma 1.
We'll have an S1 1 2 2 times sigma 2 2 plus an S1 1 3 3 times a sigma 3 3 plus an S1 1 2 3 times a sigma 2 3 plus an S1 1 3 2 times a sigma 3 2, and an S1 1 2 3 3 1 times sigma 3 1 plus an S1 1 3 times a sigma 1 3, and then other off-diagonal terms which I won't bother to mention.
So if we convert this to matrix form, epsilon 1 1 would be replaced by the matrix term epsilon 1.
S1 1 1 1 would become S1 1, and this would be sigma 1, and an S1 2 times a sigma 2, plus and S1 3 times sigma 3.
And now we have a problem, Houston.
Because we would have an S1 4 times a sigma 4 plus an S1 4 times sigma 4.
And now in the sigmas there is no factor 1/2 in the matrix representation of stress as there was with strain.
Because we ate the factor of 2 in defining strain.
But now there is no factor of 2 and stress.
So we're going to have to do something with that.
So this would give us some additional terms, S1 5 times sigma 5, and then an S1 5 times a sigma 5 again.
So what are we going to do?
Again, we're stuck.
We can either say that epsilon i is equal to Sij times sigma j if j is not equal to 4, 5, or 6.
Or we can absorb, again, a factor of 2 in the definition of the compliances.
And that, again, since we will usually be working in one, and the same coordinate system, is the convenient thing to do.
So what we will do is to define this as S1 4 times sigma 4.
And in so doing, we have to combine the factor of 2 into the definition of S1 4.
So S1 4 would be equal to 1/2 of S1 1 2 3 plus S1 1 3 2.
So it's going to be 1/2 of one of these equal terms that involve one on-diagonal subscript and one off-diagonal subscript.
So this is the only way we're going to be able to write a nice, clean matrix representation.
Things get worse
Sir?
Yes?
You're sure of your [INAUDIBLE]?
Yeah, this would be equal to 2S1 4.
I'm sure it's 1/2.
I'm not sure of what term would go in front.
So I would have 2S1 4 times sigma 4.
And if I want to write this-- I'm sorry.
Let me go back to the full matrix expression.
I would have S1 1 2 3 times sigma 2 3 plus S1 1 3 2 times sigma 3 2.
I know that sigma 2 3 is equal to sigma 3 2, so I could write this as S1 1 2 3 plus S1 1 3 2 just simply times sigma 2 3, one of them.
Now what I would really like to do is to write this as S1.
Let's change this to sigma 4.
I would like to write this as S1 4.
So it follows then that S1 4 is equal to S1 2 3 plus S1 1 3 2.
And that is the other way around, isn't it?
Thank you.
I knew there was a factor of 1/2, but it goes in here
And then there's no 2 at all.
If you wanted to say that S1 1 2 3 is equal to S, 1 1 3 2, then you would have S1 4 equal to 2S1 1 3 2
OK, you're right, you're right.
So it's this.
Let's leave it at that.
But if these are equal, we could say-- and we did show that the compliance tensor is symmetric, so we could say that this is equal to 2S1 1 2 3 or 2S1 1 3 2 as they're equal.
I told you it was going to get bad, but I didn't think I would be contributing to it the extent that I am.
I don't know if you really want to see this, but this gets even worse when we deal with something like epsilon 2 3, where this is a term that would be replaced by epsilon 4.
And then all of our absorbed factors come back to haunt us.
Let me go through this quickly.
This would be S2 3 1 1 times sigma 1 1 plus S2 3 2 2 times sigma 2 2 plus S2 3 3 3 times sigma 3 3.
And then we'll have these terms, off-diagonal terms S2 3 3 1, sigma 3 1 plus S2 3 1 3 times sigma 1 3, and so on, other terms.
OK, and going from epsilon 2 3 to epsilon 4, we have to put on a factor 1/2.
And that says that that 1/2 is going to create problems in the first term in this expansion.
We'd call this sigma 1 1.
We'd like to call this S4 1.
But what is S4 1 from what we have defined here?
It's the sum of two tensor elements.
So we will have to define, now, again, S4 1 equals the same as we did before.
It's going to be equal to the term S2 3 1 1 plus S3 2 1 1 1.
And therefore if I put a 1/2 in this definition, then that will cancel, so far.
But then in the terms that come down in the lower quadrant of the matrix-- and I will just right in two terms.
We've got an S2 3 31 that we would write as S4 4 times sigma 4.
And then I would have another term, S4 4 times sigma 4.
But the S4 4 really is a sum of two tensor elements.
So my definition of an S4 4 is that it should be--
Isn't that sigma 5 [INAUDIBLE]?
You're right, you're right.
Yeah, this is 5.
So this was right, 3 1, and this is sigma 5.
You're right.
And I go to S4 5, yeah.
No, sigma-- OK, yeah.
OK, let me cut to the chase.
What we're going to have to do is to put in not the factor 1/2 that we had here, but there's going to be a factor 4 that-- something like S-- which one are we dealing with?
Things like S2 3 3 1, something like S4-- and we're dealing with 5.
S4 5 is going to be defined as S1, S2 3 1 3 plus S2 3 3 1 plus S3 2 1 3 plus S3 2 3 1.
So we're going to have to put in a factor of 1/4 in front of the S4 5 in order to accommodate for those terms.
So our definition, then, in going from tensor compliances to matrix compliances is much more complicated.
And the rule-- and just to summarize this, I gave you a handout last time.
The necessary conventions to absorb these factors of 2 and 4 is that-- and I didn't bring it with me.
OK, I just simply did not bring it with me.
So I wouldn't attempt to summarize it.
All right.
So hopefully I've convinced you of nothing other than that these conversions are messy.
But the summary is that for the compliances, the Sijkl, you take this equal to Slm if l and m are equal to 1, 2, or 3.
Then you have to replace Sijkl by 1/2 of Slm if i, j, or kl is 4, 5, or 6.
And then you have to write Sijkl as 1/4 of Slm if i, j, and kl are 4, 5, or 6.
But once you're into the coordinate system, which will be fixed in matrix notation, things are simple, since you don't have to worry about these factors of 2 or 3.
The reason I did this is I would like to now talk a little bit about how one can define important scalar moduli that describe a particular phenomenon.
If you ask the question, how do mechanical properties vary with direction, we've got six unique elements of stress.
And we've got six independent elements of strain that can work.
And if you wanted to show how each one of those elements varied with the other one, you'd need 36-- 6 times 6, 36-- representation surfaces and this is not going to be fruitful.
So just as we did for piezoelectricity where we could define a scalar modulus which represented the component of polarization normal to a very thin plate-- making it a thin plate means you're going to measure primarily the charge on the large surface, because polarization is charge per unit area.
So if you looked at the surface of a plate with x1 normal to the plate, you are going to have a specimen that primarily gives you a measure of P1, or the charge on a surface normal to x1.
And then you could apply any of six different states of simple stress.
And one thing that you could do is squeeze it with the tensile stress in the same direction as the normal to the surface.
And that was the longitudinal piezoelectric modulus.
Then on the quiz you looked at another modulus that related a component of polarization on another surface in response to a tensile stress that was not parallel to it.
There are a number of such moduli and mechanical properties.
Probably the most important one is something called Young's modulus, which I'm sure you've all heard of.
And Young's modulus involves taking a very long rod of the material and hanging a weight on it.
And that weight will induce a strain.
And if we take this as the direction of x1, we will, with Young's modulus, relate the change of length to the initial length of the rod.
And my question now is, rhetorically, do we want to do this in terms of stiffnesses or compliances?
Does it make any difference?
Yeah, makes a lot of difference in terms of the simplicity of the result that you get.
Suppose we wanted to do this in terms of the compliances.
What we'd be then applying would be a sigma 1.
And this would be given by the compliance C1 1 times the strain E1.
That looks like exactly what we want.
This delta l over l for this one-dimensional specimen with a one-dimensional directional applied stress, this would be simply epsilon 1.
And that's force per unit area here.
This would be sigma 1 1.
And it looks as though what we want is a C1 1 1 that relates a sigma 1 1 to an epsilon 1 1.
Is this going to be a definition that I would want to make for describing this?
My colleague here shakes his head very seriously.
No.
Do you want to share your reservation?
Because as I recall, Young's modulus, once strain is weaker output [INAUDIBLE] in terms of stress you're going to want a compliance
That's one answer
You could theoretically get either one [INAUDIBLE]
If I could restate your objection, we can't impose a strain to get a stress.
We really stresses the independent variable, practically speaking.
We can stress it, but we can't instantly say, [INAUDIBLE] develop a epsilon 1.
Yeah?
[INAUDIBLE]
You betcha.
That's the reason.
Added onto this is not only this term, but there will be a C1 2 times epsilon 2 plus a C1 3 times an epsilon 3, and so on.
So we're not going to be able to get a nice, tidy relation between the tensile strain that we're measuring and the tensile uniaxial stress that is produced there.
So the relation of stress in terms of strain that involves the stiffnesses is just not going to work.
But if we would instead express epsilon1 in terms of a compliance, a matrix compliance S1 1 times sigma 1, that's all that she wrote.
We're measuring this by design by selecting an elongated specimen for which we'll primarily be seeing epsilon 1.
This is not to say these other strains exist.
There will be lateral strains here that will be epsilon 2 and epsilon 3.
There will be shear strains.
Sounds mind-boggling.
You pull the sample in this direction, and what it does is shear.
Well, if this were a single crystal, that would happen.
And these would be other components of deformation.
But if we measure a pi sigma 1 and measure epsilon 1, then this is the way we have to define it.
And the definition of Young's modulus is that 1 over S1 1 is sigma 1 over epsilon 1.
And this is Young's modulus.
I would hasten to observe that it is unfortunate that this very practical modulus involves the S's, which have all these complicated factors of 2 and 4.
And therefore, if we try to go to a particular symmetry and ask the question, how does Young's modulus change as we cut this rod in different orientations from the single crystal.
So I would like to illustrate for you how we would do this.
I'm going to take a very, very simple example.
If you look in the table of symmetry restrictions that I passed out a couple of times ago, prior to the quiz, for an isotropic the tensor has a different form than it does for a cubic crystal.
Cubic crystals are elastically anisotropic.
But the form of the stiffness tensor for an isotropic material was S1 1, S1 2, S1 2, 0, 0, 0, S1 2, S1 1, S1 2, 0, 0, 0.
Then the diagonal terms, if the material was isotropic, was 2s1 1 minus S1 2 for this term, 0, 0, 0, 0, 0, 0, and again, a 2S1 1 minus S1 2 plus a 0, and then 0, 0, 0, 0, 2S1 1 minus S1 2.
OK. Let me now derive Young's modulus as a function of direction and show that, in fact, this says that regardless how you orient the rod, the value of Young's modulus will stay the same.
Let me also do something else first, though, which is something we should have scratched our head over when we first encountered it.
Here's something curious.
When we looked at symmetry constraints on single crystals, we found that for a cubic crystal, C6 6 had to be equal to 1/2 of C1 1 minus C1 2 if the stiffnesses were to be independent for all symmetry transformations.
For the compliances, however, we found that S6 6 had to be equal to 2 times S1 1 minus S1 2.
Two very different equalities between tensor elements to make the elastic behavior be invariant to the symmetry transformations of a cubic crystal.
How come?
How come these are so different?
Well, they have to be, in terms of the tensor, the same sort of a quality.
And the reason they don't look the same is because of our different absorption of the factors of 2 and 4 in defining the stiffnesses an in defining the compliances.
Remember that for the Cijkl's there was no factor of 2 or 4 introduced in defining the matrix elements.
So this one is OK.
This is a true equality between tensor elements, for any tensor whatsoever of fourth rank for a cubic crystal.
For these terms, these are off-diagonal.
This is an off-diagonal compliance.
And our definition is that Sijkl is equal to Smn, for m and n not equal to 4, 5, or 6.
Here's this crazy thing again.
It's equal to 1/2 of Smn for m or n equal to 4, 5, or 6.
And it's equal to 1/4 of Smn for m and n equal to 4, 5, or 6.
So S6 6 here is actually 4S1 2 1 2.
And that's supposedly equal to 2S1 1, which is really S1 1 1, minus S1 1 2 2.
And this says that 1/2 of S1 2 1 2 is equal to S1 1 1 1 minus S1 1 2 2.
So this is exactly the same constraint on tensor elements, when we take out these factors of 2 and 4 that have been absorbed.
So this, in fact, although it looks very different with those factors absorbed, is exactly this same relation.
So the symmetry constraint is the same
How'd you get 1/2 there?
Hmm?
How did I get 1/2 here?
It's 4S1 2 2 had to be equal to 2S1 1 minus S1 2.
And this was the equality.
I'm working up this way.
So this was the statement in the matrix forms of the compliances that I handed out.
Looks different from the term on the left.
So I wrote this now in terms of the matrix elements.
And the factor of 4 comes in here.
The 2 was here in the equality that had to be held that the tensors stay invariant.
And if I bring this 2 over on the left-hand side, it's exactly the same as when--
Shouldn't it be 2 now?
[INAUDIBLE].
[INAUDIBLE] the right side
No, this is just replacement of the matrix terms with the definition of the tensor
Right, but [INAUDIBLE]
You're just dividing 4 by 2, right?
Yeah, exactly
So wouldn't that make [INAUDIBLE]?
Oh, ha,
You want to be able to [INAUDIBLE] by 4, so it'd be S12 1 2 equals 1/2 the difference to be equal to that
OK, OK.
Thank you.
OK, let me then go back to what I started out to do.
And I will write just one line of how we would go about expanding this just to indicate how intricate it is, and then I'll give a few examples for two point groups of how Young's modulus varies with direction.
OK. What we're going to say, then, is that one of these compliances, S1 1 1 1 prime is going to be equal to l 1/4 times S1 1 1 1 plus l 2/4 times S2 2 2 2 plus l 3/4 times S3 3 3 3 .
And just now doing what we know, that any Sijkl prime is going to be Cii, Cj capital J, Ck capital K, Cl capital L-- these are direction cosines now, not stiffnesses-- times Sijkl.
So it's a quadrupole summation over all of the non-zero tensor elements.
So what I'm doing now is taking these terms and expanding them to the full tensor form.
So S1 1 1 1 prime, that's the first term in the transformed tensor.
That's going to be these terms plus l1l2 squared times S1 1 2 2 plus S2 2 1 1 plus l1 squared l3 squared times S1 1 3 3 plus S3 3 1 1, and then still another term, l2 squared l3 squared times S2 2 3 3 plus S3 3 2 2.
And then there will be terms of the form l2 squared l3 squared times, again, four terms, S3 2 3 2 plus S1 2 3 3 plus S2 3 3 2 plus S2 3 2 3, and then similarly, terms in the squares of l1 and l2, and these would involve four terms of the form S1 2 1 2 plus permutation 0, and then l1 squared l3 squared times, again, four terms, S1 3 1 3 and permutation 0 for four terms.
OK.
So what do we have now?
We have an expression for S1 1 1 1 prime.
And this is, in fact, 1 over Young's modulus E. And we've done this summation over the supposed non-zero tensor elements in the matrix for an isotropic material.
And if I simplify this, and this will be just one more tedious and then we can see, that is for these equalities, we ought to, for l1, l2, l3, get the same value, S1 1.
And that is indeed what happens.
So the summation, if I simplify, is l 1/4 times S1 1 plus l 2.4 times S2 2 plus l 3/4 times S3 3.
And then a collection of terms plus l1 squared l2 squared, S1 2 plus S2 1.
And then similar terms in l1 and l3, l1 squared l3 squared, S1 3 plus S 3 1 plus l2 squared l3 squared times S2 3 plus S3 2.
And then some terms that stand by themselves, l2 squared l3 squared times S4 4 plus l1 squared l2 squared S6 6 plus l1 squared plus l3 squared times S5 5.
And consolidating this, this is going to be equal to l 1/4 plus l 2/4 plus l 3/4 times S1 1.
And then combining terms in the second power of direction cosines, l1 squared l2 squared plus l1 squared l3 squared plus l2 squared l3 squared.
And this is all times 2S1 2 plus 2S1 1 minus 2S1 2.
So S1 2 drops out.
And all this will be simply l 1/4 plus l 2/4 plus l 3/4.
And then I'll add in the other terms here that involve products of squares of direction cosines.
And all this is times S1 1.
But this turns out to be equal to simply l1 squared plus l2 squared plus l3 squared, the sum of the squares of the direction cosines of our rod, quantity squared times S1 1.
And this term inside the parentheses is 1.
So indeed, S1 1 prime is equal to S1 1.
So the value of Young's modulus has not changed with direction if the form of the compliance tensor is like so.
And I think you would probably have taken my word for that, but it was intended primarily as an example of how, when S1 1 prime is the reciprocal of Young's modulus, how you would set up a transformation for S1 1 prime.
And the form of this polynomial would be exactly the same, even for a triclinic crystal, if you included all the non-zero terms in this fashion.
All right, so let me wrap things up in another couple minutes for some cases that are real symmetries, where Young's modulus is anisotropic.
And working in exactly the same way for the tensor that is the appropriate one for a cubic crystal, we would lift out the form of the stiffness matrix.
The reciprocal of Young's modulus is S1 1 prime, so what we're saying is we have a long, skinny rod.
This is x1.
And what we're doing, if this is a single crystal, is examining how Young's modulus changes as we change the direction of the rod to a new orientation, x1 prime, that's described by direction cosines l1, l2, L3, relative to x1.
The expression that results by exactly the process that we muddled through a moment ago is that for a cubic crystal, the form of S1 1 prime as a function of direction does indeed give anisotropy.
Isaiah It's S1 1 minus 2 times S1 1 minus S1 2 minus 1/2 of S4 4.
Remember, for a cubic crystal, 1 1, 2 2, and 1 4 are the only non-zero terms.
And then this is times a polynomial l1 squared l2 squared plus l2 squared l3 squared plus l3 squared l1 squared.
So this is, again the reciprocal of Young's modulus.
And it turns out to have a constant term, S1 1 , which is what we found for the isotropic material.
But from that, as the direction cosines change, we subtract off a term that is a linear combination of 1 1, 1 2, and 4 4.
So the question is, is this positive or negative?
The direction cosines are all squared, so this term here is always going to be positive.
So are we going to take the thing that we found for an isotropic material, which was a constant, Young's modulus, which was 1 over S1 1 prime, and that's equal to the Young's modulus E. Are we going to add or subtract something to it?
Well, it turns out that if you look at real materials, it can be either positive or negative.
It's usually positive.
So that says we are going to subtract off a term which goes as products of squares of direction cosines.
And this is something that's going to be zero along the direction 1 0 0.
So along the reference axes x1, x2, x3, which if you remember, where this form of the matrix came from was taking the axes along the edges of the cubic crystal.
It turns out that this is going to be equal to 1/3 along the direction 1 1 1.
So if this term is plus, nothing gets added onto the surface in the directions that correspond to the four-fold axes or the twofold axes of a cubic crystal.
Nothing gets added on if it's plus.
But along the body diagonals, this takes on a value of 1 3, so a positive thing gets added on.
And the best way I can describe this surface-- it's not a simple surface-- it looks like a cube with fuzzy edges.
So we had a cube and started to dissolve it.
So this is how the reciprocal of Young's modulus varies with direction.
If, on the other hand, this term is negative-- so this is what you get if it's positive.
If it's negative, again, you start with the basic isotropic variation of 1 over S1 1 1.
If it is negative, and there's one metal, that's molybdenum-- molybdenum has a negative value of these compliances.
It looks like a surface that has indentations along the 1 1 1 direction.
So it looks like something with a dimple in it.
There is a final case, which is not realized for any material that I know of, if that term is 0, then it's isotropic.
Let me give you one other variation that comes straight out of [INAUDIBLE].
And this is for a hexagonal crystal that might be a hexagonal close-packed crystal.
For zinc specifically, which is a hexagonal close-packed metal, the form of the matrix is S1 1, S1 2, S1 3, 0, 0, 0, S1 1, S1 2, 0, 0, 0, S3 3, 0, 0, 0, S4 4, 0, 0, S4 4, 0, and S4 4.
In the values of these specific compliances are 8.4 for S1 1, for S1 2 1.1, for S1 3, minus 7.8, for S3 3, 28.7, for S4 4, 26.4.
And these are all in units of 10 to the minus 12 meters squared per Newton.
That's good old MKS units.
It turns out that there are 10 to the 2 meters squared per Newton per 1 centimeter squared per dyne, if you like CGS units.
The form of the reciprocal of Young's modulus, it turns out to be a surface of revolution.
This is x3, which is the direction of C. And that's the surface of revolution and the value of S1 1 prime as a function of the angle theta.
That's the only parameter we need examine the variation with since it's a surface of revolution.
S1 1 prime is equal to S1 1 times the sine of theta to the fourth power plus S3 3 times the cosine of theta to the fourth power plus S4 4 plus 2S1 3 times sine squared theta, cosine squared theta.
So it's a fairly exotic surface, even as a surface of revolution.
And what this looks like, it's something that peaks out at 3 times 10 to the minus 12 centimeters squared per dyne along the direction of C. It's something that comes down very sharply as you approach the normal to the C axis.
And then there's a cute little wiggle just near the axis.
So it looks something like that.
Not a terribly isotropic surface, and a variation of Young's modulus of 3 to 1 in the direction parallel to C and perpendicular to C. So there are lots of exotic surfaces of this sort.
All right.
I did run a little bit over.
And I will absent myself and let you express yourself candidly on the questionnaire.
Except for these numbers that I just put up on the blackboard, I haven't really said anything about, much about, numbers.
So let me pass around some examples, not for metals, which we just looked at, but for oxides.
And this is interesting, because you'll remember there was an additional equality between the stiffnesses called the [?
Kowshi ?]
equality, which was supposed to hold for physical reasons, not for reasons of symmetry, if the forces were central, if the crystal was under a state of no stress, and if all of the atoms were situated at a center of symmetry.
The first set of data that you have on the top sheet is for MGO, which is cubic.
It's predominantly ionic compounds.
And the atoms are all octahedrally coordinated.
So all of the requirements for the [?
Kowshi ?]
equality should hold.
And you can see they're not really exact.
And that's probably due to the covalent character.
The two different sets of data here are for measurements by two separate observers.
The second page is for aluminum oxide.
And again, you see the order of magnitudes of the numbers on the order of 10 to the 12 dynes per centimeter squared in general, and variation from one to another for a lumen of a factor of about four.
I meant to bring the book in, but I left it behind in my rush to come off.
Where do you get these numbers?
And it's hard to really find.
And the book from which I took these data are by two workers, Simmons and Wang.
And it's called Handbook of Elastic Constants, I think that is.
And this is a book that was published by MIT Press.
Valuable repository of data, but it's all numbers.
And there are many, many materials in here.
Simmons and Wang, interestingly, are geophysicists.
Why should geophysicists care about stiffnesses?
Why?
Because these guys are going all over the face of the earth setting off little bits of explosive that send off seismic waves.
And they probed the interior of the earth by the propagation of elastic waves.
And the elastic waves depend on the square root of stiffnesses over the density of the material.
So they're very, very interested in the elastic properties of particularly rock-forming minerals.
Another interesting comment on this book is that there is no set of data for a single triclinic crystal.
Too many stiffnesses that are independent, I guess, and too few materials that, fortunately, are anisotropic and triclinic.
OK, I am going to quit.
And thank you for your attention.
You've been a good class.
And we've covered all sorts of exotic aspects of the behavior of crystalline materials.
And you might think with some justification that you know all there is to know about symmetry and tensor properties of materials.
But nevertheless, I would caution you that you don't know everything.
So I have a final handout fill in the chinks, the cracks that we've not been able to fill.
And this is a set of data that is titled "Think You Know Everything?"
And this will fill in some things that you really don't know.
And it has a lot of very useful items on here.
For example, there are 293 different ways to make change for $1.
You didn't know that.
2/3 of the world's eggplant is grown in New Jersey.
I'm a native of New Jersey, and even I didn't know that.
The longest one-syllable word in the English language is "screeched."
OK. And so it goes.
In most advertisements-- this is one I never checked out-- in most advertisements the time displayed on a watch is 10 minutes after 10:00.
Don't know why.
How many ridges are around the edge of a dime?
118.
And how many little dimples are there on a regulation golf ball?
If you count them up, you'll find that there are 336.
So I'll end with one for the benefit of some of our visiting students.
In England, the Speaker of the House is not allowed to speak.
There is my oxymoron of the day.
And with that I shall leave you, and I hope you enjoy these.
It's one of these things that circulates around on the internet.
So as the sheet finishes, now you do know everything.
So I'll leave you, and again I thank you for your faithful attendance.
And you shall see me when you come to call for your quizzes.
And I will, as most of you requested, send out an email to let you know when you can come and pick those up.
So again, thank you.
Enjoy the mid-term break.
And take advantage of some of the really interesting things that go on extracurricularly around the institute.
So that's it.
Thank you, and au revoir.
[APPLAUSE]
All right, I feel almost as though I should introduce myself all over again.
It's been a week and a half since we had a lecture.
So let me begin by reminding you of what we were doing.
We had derived all of the plane groups, and for better or worse had put them behind us.
And then we moved into three dimensions, where things get a lot more involved and a lot more complicated, and to say the least, a lot more numerous.
And the first question we asked was to say, when we're in a three-dimensional space, we can combine a first rotation operation with a second rotation operation, B beta If we're to begin by deriving point groups-- that is to say, the least [INAUDIBLE] point in this three-dimensional space is not going to move.
[INAUDIBLE] two axes to intersect a point.
For a space group, they could be parallel to one another.
But that's gonna be an infinite set of symmetry elements and operations that extends through all space.
Then we asked the question-- rhetorically, because you knew I was going to answer-- what is the net result of a sequence of rotating from a first object to a second, and then picking up the second and rotating it to an angle beta about second axis.
Begin the third one here.
What is the net effect?
And again, we could do that by the process of elimination.
It has to be either translation or another rotation, because these are the only two generic sorts of operations which leaves the chirality unchanged.
And I think I convinced you that indeed there was some third axis, C, which rotated directly from the first to the third by some different angle, gamma.
So that is the consequence of combining the first two rotation axes.
What they would anticipate is that the location and also the value of the rotation [INAUDIBLE] depends on alpha and beta, and also [INAUDIBLE] at which we combine them.
And that's what we'll see.
You could do this in a number of ways that if you don't [INAUDIBLE] that gamma would turn out to be a crystallographic rotation.
And then your result would be true, but it would not be a rotation operation which could exist in a three-dimensional point group.
So using the genius of Leonhard Euler and a construction known as Euler's construction.
We set up a little spherical triangle, which we could analyze.
Let me tell you a little bit about Euler, because he's a remarkable individual.
The first remarkable feature of Euler is that he's Swiss, and there are not many world-class famous people who are Swiss, simply because the population is so small.
The probability of somebody rising to heights is constant among all populations.
If you have a small population, there are not going to be very many.
And to demonstrate that point, can somebody identify some other citizen of Switzerland who rose to great heights, as world-famous as [INAUDIBLE]?
Think of one other person?
I'm fairly pressed to do so myself.
There's my uncle, but he actually didn't amount to much.
But there's an artist, Paul Klee, who is world-class.
He was one of the early modern artists.
And Switzerland has just finished constructing a marvelous museum on the outskirts of the capital city, Berne.
It's a structure that is supposed to mimic the rolling countryside of the central part of Switzerland.
So it's a series of cylindrical structures, glass in front, glass on top.
And it divides the area into three.
Two of them are exhibit spaces and one is a space for scholars and researchers.
And it's an absolutely marvelous structure.
It appeared in the pages of Time Magazine when it opened [INAUDIBLE] about two months ago.
Anyway, Euler was born in 1707, so he operated a long time ago.
And he died in St. Petersburg on September 18, 1783.
That is exactly 350 years and one day after my birthday.
That's another remarkable thing about him.
He was 76 years old when he died.
And that time of primitive medicine and plague, not many people got to live to their 60s and 70s.
Euler studied at the University of Basel under the Bernoullis.
I think you've all heard of the Bernoullis.
There's a very famous principle of physics known as the Bernoulli effect, which stated in its simple practical form says that if you have the Sunday paper on the front seat alongside of you, and you drive your car with the windows down, the paper will blow out the window.
That's Bernoulli's principle in action.
Euler got his doctorate from Basel at age 16.
It sort of leads one to the rhetorical question, how come you guys have been spinning your wheels for so long?
But then I said, he was an unusual individual.
The Bernoullis went to St. Petersburg in Russia, under Catherine the First.
Russia was trying very hard at that time to enter the ranks of the Western world as a full member.
Euler followed them a little bit later on.
And he succeeded one of the Bernoullis as a professor of mathematics in 1733.
Then unfortunately, two years later in 1735, he lost the sight of one eye.
And why?
Because at that time, astronomy had been using this newfangled telescope which had recently been perfected.
One of the hottest things going was studying the heavens looking through a telescope.
And if you wanted to look at the sun, people knew nothing about the damaging effect on retinas of the sun's rays.
So he lost the sight of one eye.
1741, he went back to Europe again, to Berlin.
Why?
Because the reigning monarch in Russia at that time was called Ivan the Terrible, and that says reams about why Euler would want to get out of Russia.
But then 1776, he went back to St. Petersburg under the next monarch, who was Catherine the Great.
And that says why one would be interested to go back.
Finally, in 1766, he went fully blind.
Did that slow him down?
Not one bit.
He published in his lifetime 800 papers.
You talk to some big cheese around MIT, they've published maybe 200 or 300 papers, and that with the assistance of an army of graduate students, and also, one might add, the assistance of Xerox machines and word processors.
So back in the days when you wrote everything out by hand with a [INAUDIBLE] quill pen, 800 papers is an absolutely unbelievable accomplishment.
It took 35 years after he passed away to publish everything that he'd written.
People had kept busy publishing what he did.
And among his accomplishments, he was one of the first people to apply real hardcore mathematics to astronomy, to make it quantitative.
He was one of the first to suggest that light was a wave form, and that color was a function of wave length.
That was astonishingly precocious.
And then, lest he seem like an egghead who spent all his time staring through telescopes and working out theorems to use in crystallography, he also wrote a popular account of science for the general public, which was published in 1768.
And that book was published for 90 years, three generations of people kept gobbling up [INAUDIBLE] pretty good.
He impinged upon our own language and activities in several important ways.
He was the one who used lowercase i to define the square root of minus 1.
We can thank Euler for that.
He was the person who used e to define the constant, 2.71828182845904523536.
And he was the person who first used f to stand for function.
So he contributed not only a lot of good mathematics, but a lot [INAUDIBLE].
So this does not have to be easy.
Euler was a great guy.
And this geometry of rotations about different axes is something that also survives in a mechanism that involves achieving angular core rotation on a axis by rotation on two orthogonal arcs.
And that's something that's called an Euler Cradle.
And that is geometry that;s used in a great number of mechanical devices.
In any case, back to instruction for our purposes.
The thing that we would like to do is let alpha and beta take on all possible crystallographic values, namely 360 degrees or onefold axis, although we know that that's not gonna work.
Twofold, 180 degrees.
Threefold, 120.
Fourfold, 90.
Sixfold, 60.
And let that give the values to alpha and beta, taking two at a time.
And then let us ask the question, at what angle should we combine these two axes to get gamma to be a crystallographic rotation axis?
And if it is crystallographic and not something like 37.9234 degrees, what are the remaining axes with interaxial angles B and A?
So this is the problem that Euler's construction solved.
And I won't go through all the arguments that we need to set this up.
But what we found after some f sleight of hand when we were working on the polar triangle with spherical trigonometry, what we found was the result that said that if we want to combine two axes, alpha and beta, so that the third one turned out to be a rotation of gamma, then the cosine of the angle between A and B should be the cosine of alpha/2, cosine of beta/2 plus cosine of gamma/2, divided by the sine of alpha/2, sine beta/2.
So if you pick your alpha and beta, and you decide what you would want these first two rotations to turn out to be.
And generally it's not gonna work.
But there are a surprising number of cases where it does work.
So you specify the combination you had.
You also determine the angle between A and C. And you have to also determine the angle between the axes B and C. And there are similar sorts of expressions that one obtains simply by [INAUDIBLE] alpha and beta again.
So then we set it up just by looking at all possible combinations of twofold, of two different rotation axes, and a third, which the net effect might be.
We're not interested in permuting A, B, and C. And A equal to C equal to B is just as interesting or not as A equal to B equal to C. We don't care about permutations.
And we generated-- just as we [INAUDIBLE] a week and a half ago-- a set of combinations that we should consider, what the axis A would be, what the axis B would be, and them different choices for the axis C. So A could be 1.
B could be 1.
And we could look for 1, 1, 1; 1, 1, 2; 1, 1, 3; 1, 1, 4; 1, 1, 6.
Those are legitimate combinations?
Those are absurd combinations, because doing nothing about the first onefold axis, doing nothing about the second onefold axis could hardly result in the net effect of the 90-degree rotation by the third axis.
And [INAUDIBLE] suggested is that sitting around and doing nothing twice was equal to a rotation [INAUDIBLE] its junctures [INAUDIBLE] we'd find ourselves spinning on our axis like tops.
Twofold axis.
1, 1, 2, we have here, so we don't have to consider 2, 1, 1.
But we should consider 2, 1, 2; 2, 1, 3; 2, 1, 4; 2, 1, 6, and so on.
If we filled out this whole table, last time you got a copy of it and some notes, and all that remains then is to quote [INAUDIBLE] and see where we get allowable axial combinations.
And not surprisingly, there are so very, very few.
And we showed-- again, when we finished up last time-- that you can always take any n-fold axis that has a C gamma that's equal to C 2 pi over n and combine it with twofold axes at right angles, provided you make the angle between the twofold axes equal 2 of gamma over 2.
So the crystallographic possibilities for C are, first of all, C could be a twofold axis, in which case you could combine with it a pair of twofold axes, and this 1/2 of 180 degrees would also be a right angle.
We could let C be a threefold axis, in which case the twofold axes have to be at right angles, two- to threefold axes.
And they should be combined at half the angular throw of the threefold axes, which is 60 degrees.
Two more possibilities are four [INAUDIBLE] pair of twofold axes at right angles.
Of the angle between them, half of the throw of a fourfold axis would have to be equal to 5 degrees.
And the last one is sixfold axis with a pair of twofold axes.
Add angles to it.
And a third [INAUDIBLE].
So notice the insidious fact that the angle between the twofold axes is always a half, 1/2 the rotation angle in principal axis C are not equal to this rotation axis.
The other thing we saw that is that these twofold axes are different, distinct, symmetry-independent axes.
They're different in that the principal axis C never rotates this axis into the second one, and therefore demands that whatever's going on around one twofold axis be identical to what's going on at the other twofold axis, different in the sense that they function in different ways in the pattern, or if you're describing the symmetry of an object.
So probably the best demonstration of this is a regular prism with a triangular shape or with a square shape or with a hexagonal shape.
And the adjacent twofold axes for these prisms would come out of the normal to a face.
And then if the second twofold axis is going to be 45 degrees away from the first, the other one has to come [INAUDIBLE].
So yeah, they function in different ways in the space.
One is normal to the face of a regular prism, if that's what's in our space.
The other one comes out of the edge.
Similarly for a sixfold axis, a hexagonal prism has one twofold axis coming out normal to a face, the adjacent twofold axis coming out to an edge.
And as advertised, that angle is 33.
So they function in different ways.
The only exception to that, again, is the [INAUDIBLE] threefold axis.
And the twofold axes there, which were 60 degrees, come out of one side from a corner of the triangular prism.
On the other side, that was [INAUDIBLE].
So all of the twofold axes were the same thing.
There's only one independent kind of twofold axis, just as there was only kind of mirror plane in the combination of a mirror plane passing through a twofold axis.
The names for these are always, as we've done in the past, a running list of the independent symmetry operations that are present.
So this general one, n, 2, 2, with the n-fold axis for some generic sort of a twofold axis.
This one would be 2, 2, 2.
This one would be 4, 2, 2.
And this one would be 6, 2, 2.
This [INAUDIBLE] with a threefold axis is, again, called not 3, 2, but 3, 2, because there's only one kind of twofold axis, just as there's only one kind [INAUDIBLE].
We pause [INAUDIBLE], see if you have any questions.
These are the crystallographic combinations of this [INAUDIBLE].
There is no reason why you should not in something that doesn't have to be compatible with a lattice, combine an n-fold axis of any sort with twofold axes or right angles.
And indeed, if I look at my old friend, the saguaro cactus, we can add anything like 28- up to 32-fold symmetry.
This cactus stem had a 28-fold symmetry.
It would be a twofold axis coming out of the string with one of the ribs, another twofold axis coming out of the crevice between the pair of these ribs.
And if I took that thing up, very carefully because of the spines, and with great effort because it weights several tons, and rotated it about one axis, and then rotate it again about a second axis, turning it upside-down, [INAUDIBLE] axis would be rotation to 128 [INAUDIBLE].
Valid symmetry, but not crystallographic.
OK, any comments or questions?
Get to know these results, because the exercise that's going to occupy us for the next week is going to be asking how we can decorate these frameworks with mirror planes and with inversion.
If you want orientation, we could add another operation to the collection of axes while it pops up.
Where [INAUDIBLE].
Comments or questions?
OK, there are only two other combinations that are crystallographic that involve directions that are not simple.
And one of them involves a combination of a threefold axis with twofold axes that come out of directions at a normal to the face of a cube.
So these turn out to be very, very strange angles which make no sense at all until you refer them to directions in the cube.
The direction of the threefold axis turns out to be correspondent with the angle of a cube.
The direction of the twofold axis corresponds to the normal two faces.
You can show-- I did show, and I don't think anybody really followed me, so I had to hand out that [INAUDIBLE]-- that if you start with one twofold axis and one threefold axis, what you're going to get is a threefold axis coming out of all of the [INAUDIBLE] diagonals, but they're all equivalent to this threefold axis.
And twofold axes come out to normal for all the faces of the cube.
So there's only one kind of twofold axis and one kind of threefold axis, so even though we got this by combining a pair of twofold axes-- I'm sorry, a pair of threefold axes-- that's what happens when you stay up late.
You forget about this one.
A pair of threefold axes at the diagonals and one twofold axis.
And this is the combination that is called 23, because there's only one sort of twofold axis and one sort of threefold axis.
And I'd like to point out and I'd like to warn you of traps when we come across them, make sure we don't [INAUDIBLE] across them.
Notice the insidious relation of this pair of integers to the symmetry that we label [?
232.
?]
32 is a threefold axis with a twofold axis normal to it.
23 is this combination that involves corrections in a cube.
Now, let me pause parenthetically with an aside.
You might say, how can this be?
Here is a cube.
That cube has got a fourfold axis about it.
Don't call that a twofold axis.
A cube has a fourfold axis coming out of it.
OK, let me give you an example in real life.
There is a fairly common mineral, iron disulfide-- pyrite.
This forms nice, shiny cubes, but the cube faces have striations on them.
If you look at them, there's a set of lines running this way.
And what those lines are if you look at this crystal face with a magnifying glass, is that these are little steps of a second face.
And this is a face of the form hk0.
And this oscillates back and forth, and there seem to be lines scribed on the surface.
So this sort of a [INAUDIBLE] crystal growth of one face which never really develops is not that uncommon.
There's a threefold axis coming out of the corner here, but this is really a surface that is left at variant only by 183 [INAUDIBLE].
You cannot rotate that surface 90 degrees.
The orientation of the lines have changed.
But there is a bona fide threefold axis coming up there, so these striations, if I rotate them to this face, will go in a way like this.
This edge turns into this edge, and therefore the lines will run down like this.
And if I rotate [INAUDIBLE] n by 90 degrees, the striations on the adjacent face will run down.
So there's a decorated cube.
And if you rolled it up and say how can I move this cube around and leave its appearance totally unchanged?
the answer is, rotate it by 120 degrees [INAUDIBLE] diagonal.
But we can only rotate it 180 degrees around the face.
So there's an example of a crystal [INAUDIBLE] on the arrangement of rotation axes 23.
The final one, the highest symmetry of all, involves a fourfold axis coming out of a direction normal to a face, a threefold axis coming out of the [INAUDIBLE] diagonal, and a twofold axis coming out normal to an edge.
And if you let these axes work on one another, there's a twofold axis that comes out of every edge, and a fourfold axis that comes out of every face.
And this one is named 432, because it's a combination of a fourfold, a threefold, and a twofold.
That is the symmetry to the cube.
And for crystallographic symmetries, that's about as complicated as it gets.
Now, if we look at the regular solids that we've encountered here, with symmetry 23, there is a regular [INAUDIBLE] consisting of four triangular faces.
That's a tetrahedron.
And for 432, one of the polyhedra that can form from the crystal [INAUDIBLE] is an octahedron.
These were the lovely solids called Platonic solids, after Plato, that we used as our trophies early this afternoon.
So this is an octahedron.
Let me finish up before our break by asking is there any other regular polyhedra that can result from a combination of rotation axes that are not crystallographic?
Now, that's a tough question to ask.
You instantly scan your knowledge of geometry.
Clearly, there are a lot of prisms [INAUDIBLE] infinite number of prisms [INAUDIBLE] n22.
But there's only one other combination of axes non-crystallographic which results in a regular polyhedron.
And this is a combination, believe it or not, of a fivefold axis with a threefold axis.
This is a fivefold axis and a threefold axis and a twofold axis.
And these result in a regular solid called an icosahedron.
And that is so complicated that I won't attempt to draw it.
But having said so, it looks like this.
It has diamond-shaped faces.
And there are five of these that come together in a [INAUDIBLE] form.
So here's one of the [INAUDIBLE] diamond-shaped faces, another diamond-shaped face, and then the two other ones that come in like this.
So there are 1, 2, 3, 4, 5 faces, so this is the orientation of a fivefold axis.
The twofold axis comes out of a place where two of these edges are shared.
And the threefold axis-- [INAUDIBLE] triangular faces.
[INAUDIBLE].
So here's the fivefold axis.
These are the twofold axes [INAUDIBLE].
And these are threefold axes.
But you know all this.
Is there anybody who [INAUDIBLE] show me that they've never seen an icosahedron?
Anybody ever who has not seen-- have you seen it?
[INAUDIBLE].
Here-- I spared no expense-- is a live icosahedron for those who would like to look
How many sides does it have?
He'll count them for you.
I don't remember.
I know the number of faces and the number of edges, but not the [INAUDIBLE]
[INAUDIBLE] faces
I think it's-- I'm not sure [INAUDIBLE].
Not sure
But you said you know the number of faces
Yeah, but I can't tell you everything I know [INAUDIBLE]
Yeah, yeah
There's another figure which also has this symmetry, and it's a regular solid.
And this is called a rhombic dodecahedron.
And, wise guy, this has 12 faces.
And the faces are pentagonal.
And there are three pentagonal faces that come together at the threefold axis, a fivefold axis out of each of the pentagonal faces, and a twofold axis out of the edges.
The face that this has 12 faces at regular intervals leads an entrepreneur who was familiar with injection molding to cast these little things as a plastic, and then puts a month of the year on each of the 12 faces and made a nice little desk calendar.
[INAUDIBLE] something that reminds you of symmetry as well [INAUDIBLE].
So this has fivefold faces, 12 of them, and that's the rhombic dodecahedron as opposed to the normal dodecahedron which is [INAUDIBLE]
So are those both [INAUDIBLE]?
I'm sorry?
Are those both [INAUDIBLE]?
The icosahedron, [INAUDIBLE]?
Yeah, this has a fivefold axis coming out of the pentagonal [INAUDIBLE].
Is that what you're asking?
And the threefold axis, there are three of them that, come together, and they do something like this.
So here's the threefold, here's the fivefold, here's the twofold.
Here we have to see it [INAUDIBLE].
Look for somebody who's got one of these desk calendars.
[INAUDIBLE] used to sell them.
[INAUDIBLE].
OK, this sets up the next stage of our game.
We've got these arrangements of axes.
And if you count them up on the fingers of your hands and one toe, there are 11 of them.
There are the axes by themselves-- onefold, twofold, threefold, fourfold, sixfold.
There are these so-called dihedral combinations, where the only thing that changes from one to the other is the symmetry of the main axis, the [INAUDIBLE] symmetry, and therefore the dihedral angle between the twofold axes.
And these are 222, 32, 422, and 622.
And then the two cubic arrangements, 32 and 432.
So when we return, we'll ask the question, how can we add mirror planes for an inversion center to this combination of axes?
And these are going to give us new symmetries involving not only rotation, but [INAUDIBLE] help with rotation inversion as well.
And the constraint in doing this is that we have to add the reflection plane and the inversion center in such a way that it doesn't create any new rotation axes, because we have systematically derived all of the possible combinations of crystallographic rotation axes.
So if the addition of a mirror plane, creates a new axis, that's going to be something that you already have with a combination of a greater number of rotation axes.
For example, if you take a single twofold axis and a mirror plane of an angle, that generated another twofold axis 90 degrees away.
But we've already got that.
If a mirror plane moves an axis to an angle that doesn't correspond to one of these arrangements, it's going to be impossible, because we have systematically derived, using Euler's construction, all the combinations of rotation operations that are possible.
So that's going to be the constraint.
We want to add mirror planes or an inversion center in all possible combinations.
And this means we're gonna need a theorem.
What happens when you add a mirror plane to a rotation operation?
We're already familiar with one of them.
You take an axis and you pass an mirror plane through it, you get another mirror plane that is rotated about the axis, away from the first, by half the rotation angle of the axis.
So let's take a breather, and let us resume in about 10 minutes.
To mention, most everyone did extremely well on the quiz.
But I sense that there's still some of you who have not yet come to terms with crystallographic directions and planes, and you feel a little bit awkward in distinguishing brackets around the HKL and parentheses around HKL.
And there are some people who generally get that straightened out, but when I said point group, suddenly pictures of lattices with fourfold axes and twofold axes adorning them came in, and that isn't involved in a point group at also.
Again, a point group the symmetry about point.
A space group is symmetry spread out through all of space and infinite numbers.
So let me say a little bit about resources.
I don't know whether you've been following what we've been doing in the notes from Buerger's book that I passed out.
That was hard to do is because we did the plane groups, and he doesn't touch them at all.
So now we're back following once again Buerger's treatment quite closely.
So read the book.
And if you like, I can tell you with the end of each lecture, this stuff is on pages 57 through 62.
The other thing.
As you'll notice, this nonintrusive gentleman in the back is making videotapes of all the lectures.
These are eventually going to go up on the website as OpenCourseWare.
We were just speaking about that, and it takes a while before they get up, but I have a disk of every lecture.
So if there's something you didn't follow or a place where I chewed my lines and you want to go back over it again-- not that that happens very often-- you are more than welcome to ask me to borrow and borrow the disk if you want to review it.
So that's another resource.
And then I will regularly throughout the term give hard-copy handouts of some of the things that we're doing, particularly when it involves geometry.
And when we move on to three-dimensional geometry, unless of the graphics is really tight and precise, it's hard to follow what's going on.
So in that vein, one of the first things I wanted to pass out-- the only other one for today-- is a demonstration that in fact in the Group 23 all you need is a single twofold axis oriented along the normal to a face and a single threefold axis coming out of one body diagonal.
And that gives you all of the axes you are going to get into 23.
So when this comes around, there are number of steps that you can perform letting the axes work on each other.
And if you start with just the single twofold axis and the single threefold axis-- which the symbol suggests is all you need, there's only one kind of each-- if you look at a cube along its body diagonal, the twofold axis that's coming out of one face gets rotated into directions normal to all the other faces if you rotate by 120 degrees.
So the little diagram in the upper right hand corner of the sheet hopefully convinces you of this.
Now we've got three mutually orthogonal twofold axes and one threefold axis coming out of a body diagonal.
So the vertical twofold axis swings that around by 180 degrees to give you another threefold axis along a body diagonal.
And I labeled that one 3 prime in the middle diagram at right.
And then if you take the axis 3 prime and repeat it by the second of the twofold axes that we have along face normals, that in the middle diagram on the right hand edge takes 3 prime and repeats it to 3 double prime.
Then finally, the third twofold axis that we generated repeats the threefold prime axis to the remaining threefold axis along the fourth body diagonal.
So the results start with one twofold axis oriented along the normal to a cube and one threefold axis along the diagonal of the cube, you get axes coming out of all faces and all diagonal.
And there's a staple on that sheet for reasons that I don't understand, but it there was, probably on the surface of the Xerox machine when I went over there.
So let me now return to our next step, and that is to add what in the language of group theory is called an extender, a new symmetry operation that can be added to a preexisting group that will generate new operations.
And let's see what sort of theorems we need to describe what we should look for and in which particular orientation.
We've got these 11 axial combinations, and these are frameworks that we can hang mirror planes on.
So let's look at a first simple combination.
Suppose we have a twofold axis, and the only nontrivial operation there is a rotation through 180 degrees.
So this an axis A pi.
And again, the ground rules are that if we're to add a mirror plane to this axis, which along with identity constitutes a group, we can't create any new axes.
So there are two ways we can do this.
One is the three-dimensional analog of something that we have already done, namely to pass a mirror plane through the twofold axis.
And this is the group that we found as a two-dimensional Point Group, and we called it 2mm.
And to do that, we used the theorem that says that if you take a rotation operation A alpha and combine it with a reflection operation that goes through it, you get a new reflection plane, sigma prime, that's at an angle alpha over 2 to the first.
So what gave us the second mirror line in two dimensions, in three dimensions that would give us another mirror plane that's at right angles to the first.
And this will give us a three-dimensional symmetry, which is also called 2mm.
So it's nothing more than taking the two- dimensional Point Group and letting it come out at the board at you-- Man, that'll give you nightmares when these things are coming out of the paper at you-- and that is a valid group 2mm.
There's another way that we can add a mirror plane to a rotation axis though which will not create any new axes, and that's to add the mirror plane-- the reflection operation sigma I should say since it is a combinations of operations-- exactly perpendicular to the locus of the rotation operation.
And that reflection operation sigma then just flips the rotation axis end to end.
And the rotation operation just swirls the locus of the reflection plane around within its own locus and doesn't create any new reflection plane.
So in the particular combination A pi, if we add a mirror plane perpendicular to that as the operation sigma, this would be then all of the operations of a twofold axis over all of the operations of a mirror plane.
So we've got now a combination of a twofold axis with a mirror plane.
We've got the same thing here.
To distinguish these two combinations, we'll write this as a fraction 2/m.
And that literally in words is the way we've added the twofold axis.
It is sitting over the mirror plane and gets reflected down into its other end when the mirror plane act on.
So it's just language.
It's nice though, as we said some weeks ago, if our language have some descriptive content to so when we look at it we can remind ourselves of what it means.
So two 2mm means the mirror planes are parallel to the axis that contain it.
2/m means the twofold axis is over the mirror plane.
It goes through and pierces it.
What I would like to ask though is have we got a group yet?
And let's take a first object-- let's call it right handed-- rotate it by 180 degrees.
You get a second one, which will stay right handed.
And then repeat it by a reflection operation in the mirror plane, and we'll get a third operation.
Reflection changes chirality, so the third one is left handed.
So we are performing the sequence of operations A pi followed by sigma.
And let me append just so we make no mistake and not confuse it with 2mm that the reflection operation is normal to the mirror plane.
So question, what is the net effect?
And lo and behold, we have stumbled over-- if we had not been clever enough to invented it and suggested it early on-- the only way you can get from 1 to 3 in one shot is to turn it inside out and change its chirality by projecting it through a point which is the location where the twofold rotation pierces the mirror plane.
And what this is going to do is to change the sense of all three coordinates.
This is going to take the coordinates XYZ of object number 1 and change them into minus x, minus y, minus c. And what we're doing is inverting the motif, turning it inside out as it were, into an enantiomorph.
And we have discovered as soon as we combine a pi with a perpendicular reflection plane, a new operation which we'll call in words inversion.
And the symbol that used to describe this is a one with a bar over it, and we'll see why later on.
But this is a onefold axis with an inversion center sitting on it, and that's what an inversion center by itself is.
The implication is that they're going to be other axes that we can abbreviate such as 3-bar, 4-bar, and so on.
So this is analogous to a situation that will come later where we really need a new notation.
So we've fell headlong over a new type of symmetry operation, and we should consider taking inversion and adding that to the rotation axes by themselves as an extender.
Obviously, if we take inversion and add it to a mirror plane, we're going to get A pi.
If we take a twofold rotation, add it perpendicular to a mirror plane, we get inversion.
If we put inversion and put it on a 180-degree rotation, we'll get the mirror plane back.
So these things all permute one to another.
You may even remember some time ago we asked in general terms when do two operations permute without changing anything.
And the answer is if these operations to leave the locus of the other one alone.
And the mirror plane obviously leaves the locus of the rotation operation unchanged.
The rotation operation spins the mirror plane around in its own plane and doesn't create a new axis.
And the inversion center leaves the mirror plane alone and takes the twofold axis top to bottom.
So those three operations, inversion A pi, sigma are the three nontrivial operations that exist in the space.
The fourth one is the identity operation.
So here is the set of elements in the group that we will call 2/m.
2/m implies three operations inversion, a 180-degree rotation, reflection in a plane perpendicular to the axis, and the identity operation.
And I'll leave it to yourself for you to convince yourself that I can rotate and then reflect or I can reflect and then invert.
And all of these operations do not create any new motifs in the set.
The group multiplication table, in other words, contains just the four elements, 1, 1-bar, A pi, and sigma.
So the full pattern that corresponds to 2/m consists of four objects.
It'll be a fourth one down here.
The twofold axis tells you have this fellow is related to this one.
Inversion tells you how this one is related to this one.
And the mirror plane tells you how this one, number 1, is related to number 4.
And the identity operation tells you how 1 is related to itself.
So, again, as we've seen in the set of operations that constitute a group, there's a one-to-one correspondence between the transformations that are elements of the group and the number of objects in the pattern.
So we've made one combination, and what we found from this is a new transformation inversion that involves changing the sign of all the coordinates in a space through a point which is called the inversion center.
Questions?
If not, let me quickly rattle off other Point Groups in this family.
We could take the operation A pi/2 in 90-degree rotation and add this perpendicular to mirror plane.
Let me now say something that I've said again many times before.
The pattern of objects that will result is the pattern of objects that's produced by the initial group-- let's say a fourfold axis-- repeated by the extender.
And the operation sigma perpendicular is the extender.
So the pattern, without making any big deal about it, is going to look like this square of objects reflected down below the mirror plane.
So it'll be one going down like this.
One like this.
One like this.
The operation A pi sits perpendicular to the operation of reflection, so there will be an inversion center that arises at the point of intersection.
And indeed the square above can be inverted through this point-- little hasty repairs there-- and every one up above gets inverted down to an enantiomorph below.
So all of these guys up on top are right handed.
All these guys down below are left handed.
So this is another group.
This is 4/m in international notation, a fourfold axis perpendicular to a mirror plane.
There is also, unfortunately, a Schoenflies notation.
The international notation tells you what you've got, a twofold axis or a fourfold axis perpendicular to a mirror plane.
The Schoenflies notation tells you how you derived it.
And the symbol for a twofold axis is C2, so this is a twofold axis.
And what we did was to add an extender consisting of a horizontal mirror plane.
Schoenflies calls this one C2h; Group C2, which is a twofold axis; a horizontal m is the extender.
So Schoenflies tells you how you make it.
The international notation tells you what you get as a result.
Schoenflies notation here would be C4, that's the symbol for a fourfold axis, and the extender is an h. And then without making any big fuss about it, if I do the same with a sixfold axis, I will have six objects related by a sixfold rotation axis.
If I take that sixfold axis and put a mirror plane perpendicular to it, these will be reflected down to a hexagon of enantiomorph equidistant below the mirror plane.
That's not terribly good, but it's not terribly bad either.
So this would be called 6/m, Schoenflies notation C6h.
And the operation A pi exists in a sixfold axis, so there is an inversion center but also arises as a new symmetry element at the point of intersection.
So with reckless abandon, you can continue on here and derive noncrystallographic groups for all the even-fold axes and derive an 8/m and a 12/m and 16/m.
Lovely symmetries.
I wouldn't want to draw them, but they're still symmetries.
They all have an inversion center in them, but they're noncrystallographic.
So we don't have to worry about them for prism purposes.
I left one out because it introduces a complication that is kind of curious and interesting.
Any questions on what we've done here?
Yes?
The 2mm, it's just the same-- Schoenflies notation is just C2--
Schoenflies notation for three dimensions is exactly the same as in two.
So the three-dimensional version where this extends in a direction that is normal to the two-dimensional space of our two dimensions.
Two dimensions it was this.
Now just imagine them coming out of the blackboard at you.
The symbol for this one was 2mm.
The symbol for this one is also 2mm, the same thing.
The mirror plane is a vertical mirror plane.
So the Schoenflies notation is exactly the same as what we used per two dimensions.
It's called C2v.
We've added a vertical mirror plane.
And again, horizontal and vertical.
Horizontal is horizontal with respect to the axis of higher symmetry.
Vertical is vertical with respect to the two-dimensional space of the two-dimensional Point Group, parallel to the axis in three dimensions.
So that's the vertical indication.
So let's, though, tuck that away for future reference.
We've got two kinds of extenders.
We've got a horizontal mirror plane, and we've got a vertical mirror plane, and these are extenders that we should consider adding.
So we've taken care of 2/m, 4/m and 6/m.
2mm, 3m, 4mm, and 6mm are just the extensions into a third dimension of what we've seen and come to love in the two-dimensional space.
Let me now turn to the threefold axis.
And this is a curious one.
Threefold axes require fewer symbols to indicate the vertical mirror planes because there's only one independent one.
But let's see what would happen.
And now I'm not going to attempt to draw these in three dimensions anymore.
I'm going to use a stereographic projection.
And what I'll do is use a solid dot for a point that's up above the equatorial plane and an open dot for one that's down below the equatorial plane.
So my stereographic projection of 4mm would look like.
This is the fourfold axis.
This is the mirror plane.
And I've got one up that gets reproduced by the axis to give me a set of four.
All of these are, let's say, right handed.
And then directly below them is another set of four repeated by reflection, and these are all left handed.
And then there's an inversion center at the point of intersection.
And I'll indicate that by the little open circle sitting right on the fourfold axis.
So there is a projection of what 4/m looks like.
So let me now do the same thing for a threefold axis.
And I'll add to the triangle of points that a threefold axis would generate.
So these guys are all of one chirality.
Let's say right handed.
Then I'll reflect them down, and I'll get three objects that are down.
And that's what 3/m looks like.
Is there an inversion center here?
No.
No, because the operation of A pi is missing.
And it was the horizontal mirror plane combined with A pi that gave us the inversion center with all of the even rotation axis.
So one of the things we have to say here is that there is no 1-bar that's present, which means in this instance, unlike the other ones, we have another option.
So we can use the operation of inversion as an extender too.
So we're going to get another group out of the threefold axis besides this one.
And this one we will name 3/m or C3h in Schoenflies notation.
They're six objects here, so there should be six operations in the group.
So let me number these guys up on top as number 1, number 2, number 3.
And 1 is related to itself by inversion.
There's an operation A 2 pi/3.
And that tells me how the one that's up is related to the second one that's up and how the left-handed one that's down is related to the one that's directly below number 2.
There is an operation A 4 pi/3.
And that's the same as saying 8 minus 2 pi/3.
And that tells us how the things that are separated by 240 degrees are related, both up and down.
I know how 3 up is related to 3 down and how 1 up is related to 1 down and 2 up is related to 2 down.
This is all with the horizontal mirror plane, which I'll call sigma h. That is a total of one, two, three, four-- whoops-- one, two, three four operations.
I need six.
Let's ask how is this one that's up related to this one that's down?
I just got rotation operations and reflection.
The only way I can get from this one number 1 to this one number 3 left that's down is to take two steps to do it.
I can't get from this one up here to this one down here unless I rotate 60 degrees and then invert.
If I move over to 3-bar.
This is A 2 pi/3 with 1-bar as an extender.
The pattern would, again, look like what are threefold axis does.
But then if I repeat this set of three by inversion, the two triangles above and below are skewed.
The ones down below are enantiomorphs.
The three that are up are of opposite chirality.
And this is a new type of pattern.
And in international notation, what do we call this?
It's a threefold axis.
But how do we indicate a symbol for three with inversion sitting on it?
Let's ask if we know how each of these objects is related to each of the other.
So here's 1, 2, and 3; 1 goes to 2 by A 2 pi/3; 1 goes to 3 by A minus 2 pi/3.
Let's put some numbers on here for the ones down below.
Let's call them 4, 5, and 6; 1 goes to 6 by an version.
How do I get from 1 up to 4 that's down?
I can do that only by taking two steps.
Rotate 1 from here to number 2.
Don't yet put it down.
First invert it.
So 1 to 4 involves the operation A 2 pi/3 followed immediately by inversion.
And I go from 1 up to 5 down by doing the operation A minus 2 pi/3 followed immediately by inversion.
And then 1 goes to 1 and itself by the identity operation.
So I have six objects, one, two, three, four, five, six operations.
Yes?
In the cases where you're rotating and inverting, does it matter which way to the other?
No, it shouldn't because they leave each other alone.
So I can rotate from here to here and invert.
Or I can invert from here to here and then rotate.
It's the same transformation.
Again, they permute if the two loci of the two operations leave the other locus alone.
Maybe the enormity of what we've shown here has not sunk in.
This is a new two-step operation.
We can't describe it any simpler than saying, rotate and not put it down yet, follow up by inversion.
Yes, sir?
Couldn't we express that in another way by sort of extending the three-directional glide plane by saying invert, then transform by some vector that's parallel to the glide plane?
Maybe they do in space group, but as soon as we introduce a glide plane, you've got an operation that's half a lattice translation.
And that means you've got to have a lattice translation and double the lattice translation, so--
Oh, we don't have to worry about that
--when we're in a space group, yeah.
That could be present.
But not for a point group because the ground rules are at least one point has to remain immutably fixed in space.
So this is a two-step operation, and what we're going to call it is rotoinversion.
It consists of as a first step an operation by rotating alpha from 0.1 to a virtual point number 2.
But before you put it down, you will invert it to a new object number 2 which is of opposite chirality.
So here then are the operation of the group that results when you combine a threefold rotation axis and add to it an version center as an extended; A 2 pi/3; A minus 2 pi/3; a rotoinversion operation through 2 pi/3 and then inverting; a rotoinversion operation of A minus 2 pi/3 followed by inversion.
And the symbol that is used to represent that new two-step operation is putting a bar over the top of the symbol for the axis.
And then, finally, we have inversion by itself.
So that's a group rank 6.
The Schoenflies notation is called C3i because we got this group by adding an inversion to C3, the threefold axis.
The international notation picks up on putting a bar over an axis to indicate a rotoinversion operation.
So this is called 3-bar in the international notation.
So there is a new group, and it is an oddball.
It sort of stands alone from the other groups of the form C3h.
This we derived by using the rotation of the threefold axis and adding 1-bar as an extender.
So there's no mirror plane in this
That's not the same as--
That's the same as 3/m, no.
3/m is C3h.
3-bar is C3i, different extender added to the same subgroup 3
What was the definition of 3/m?
Oh, we never really finished that.
That if we need the six operations that control the group, we'll have a sixfold rotoinversion axis.
But this pattern looks just like the triangle produced by 3, and we add an reflection operation as an inversion, and the 3 go down.
So if we ask how every one of the top is related to one underneath, that's by this horizontal mirror plane.
If I want to know how I get from this one that's up to this one that's down, then I've got to rotate through 60 degrees and invert.
So that would be a rotoinversion operation.
Let us to extend this idea of a rotoinversion operation.
And we would find this eventually in adding different extenders and falling headlong over this rotoinversion operation as we did here with 3-bar.
But let me in this case start by defining a 4-bar operation.
And this would contain the operation A pi/2 followed immediately by inversion.
And we'll call this step A pi/2-bar.
So let's try to do that and see what we get.
Start with a first point, number 1.
And that's up, so it's a solid dot.
And let's say it's right handed.
If we combine that with a rotation of 90 degrees.
Not yet put it down.
That's a virtual motif.
Before putting it down, we inverted it.
We would get one that's down, and it would be left handed.
Do the operation again.
I'll put the little tadpole inside.
Do the operation again.
Rotate 90 degrees and invert.
We're back up again.
So this was 1.
This is 2.
This was 3.
And that's up.
Do the operation again.
Rotate and invert.
And here is number 4, and it's down.
Do it a fifth time, and we're back to where we started.
So this is a crazy pattern.
It's a pair of objects that's up and a pair of objects that's down.
So there's a twofold axis in there.
That twofold axis A pi leave the pattern invariant.
But there is no way of specifying the relation between the two that are up and the two that are down other than doing this two-step process of rotating 90 degrees and then inverting.
So there is actually in this pattern a new type of operation analogous to 3-bar, and it's called a 4-bar axis.
And it's indicated geometrically by drawing a square because there's a 90-degree angular symmetry to this.
But a twofold axis inscribed inside of it because this is a pattern that has a twofold symmetry.
So something that has this symmetry is the symmetry of a tetrahedron.
And if we draw a line from the upper edge to the lower edge, this is the locus of a 4-bar axis.
International notation this is called 4-bar.
That's how we generated the pattern.
The Schoenflies notation is an S, little bit of exotica.
This geometric solid is something that's called a sphenoid.
And sphenoid.
Is the Greek word for axe.
And you can imagine a handle put onto this thing, and it does look kind of like an axe.
You could splits firewood with a thing like that.
It looks like a tetrahedron, but in a tetrahedron, it's either elongated along the 4-bar axis are squished.
It doesn't have to be regular.
So this is called S4, and the S stands for sphenoid
Is it part of a regular tetrahedron?
No, no.
A regular tetrahedron would be something where all of the edges had equal length.
And what we're doing is taking one edge and the edge that's opposite it and either stretching it or squishing it.
So there are two edges.
This one, and this one, which are the same length.
And then these four inclined edges have a different length.
It could be either elongated or squished.
But it's not a regular tetrahedron.
If those three distances were equal, then geometrically it would be a tetrahedron.
But strictly speaking, a tetrahedron is not a tetrahedron, just as a square prism with eight sides approximately equal can't claim to be a cube unless there's symmetry present that demands that this be true.
In this case, the 4-bar requires that these four edges inclined to the 4-bar axis be of one length.
And this have to have the same length.
But there's nothing that constrains all six to the edges to be of identical length.
So it's not a tetrahedron, so squished or deformed tetrahedron.
So there is another two-step symmetry element that we would not have been clever enough to think of had we not discovered this sort of rotoinversion operation when we added inversion to a threefold axis.
A 3-bar axis is a step that's present when you add inversion to a threefold axis.
So 3-bar, what we call it for short, is identical to a threefold axis plus inversion sitting on it.
A 4-bar is not equal to a fourfold axis with inversion added to it.
A 4-bar is something that you cannot describe any more simply than saying there is a two-step operation in there, and it's a group of rank 4.
Let me finish by setting up the task of going through this systematically.
We have 11 axial combinations 1, 2, 3, 4, 6, 222, 32, 422, 622, 23, and 432.
So there are 11 of those.
We want to examine as extenders a vertical mirror plane that would be one extender.
We should add that to each of these symmetries.
We already done a lot of these.
We've done pretty much up here.
We could add a horizontal mirror plane.
Or we've encountered inversion when we added a mirror plane perpendicular to an even-fold axis.
We could add inversion as an extender.
And to be complete, we should add to our list of axes in quotation marks, the 4-bar axis, having discovered it.
And does that do it?
Is there anything else we could do to these axes that would leave them invariant?
What about 2-bar?
2-bar; 2-bar would be rotate 180 degrees and invert.
So 2-bar is identical to a horizontal mirror plane.
So that's nothing new.
We're already running a little over time so-- Yeah?
3-bar?
3-bar is 3 plus 1, and we call at 3i, so this one down here.
This is 3-bar.
We describe it for short as that, but it really is a threefold axis with an inversion center sitting on it.
This thing is distinct because it's not a fourfold axis with inversion sitting on it.
A 4-bar is a 4-bar is a 4-bar.
You can't decompose it as a twofold axis is a subgroup.
That's only half the story.
I don't want to keep you anxious, not anxious to find out what the answer is but anxious to get on your way and go home.
So let me submit that when we have more than one rotation axis, such as 222 or as in 32, if we put the mirror plane in normal to the principal axis, we'll call that a horizontal mirror plane.
If we add the mirror plane through the principal axis, we could pass it through the threefold axis and the twofold axis, pass it through the vertical twofold axis and the horizontal twofold axis.
We will call this a vertical mirror plane.
And that's all we could say for a single axis, the mirror plane was perpendicular to the axis or passed through it.
But when there's more than one axis, another thing we could do would be to snake the mirror plane in between the twofold axis.
In that case, this twofold axis gets reflected into this one.
But I haven't created any new axes.
So that is going to leave the results of Euler's construction unchanged.
I can't similarly put a vertical mirror plane through this first twofold axis but in between the other two.
In that case, this is no longer 222 because these two mirror planes are equivalent by reflection, so I want to drop that at very least.
So in any case, without belaboring the point, it's late.
I could do for each of the groups that involved more than one axis I could add a diagonal mirror plane, or I should try to add a diagonal mirror planes.
And this means interleaved between axes that are present in these combinations, added such that no new axis is created.
But the addition clearly is going to be a new disposition of symmetry elements arranged in any different fashion and space.
So the game's afoot.
This is what remains to be done next.
What I'll do for next time is prepare a chart that looks like this that has the results of all of the unique combinations shown and then hand out pictures of stereographic projections of all the results.
I think once you know how to play the game to go through and do every single one in detail is probably not necessary.
If you know how to do some of them and you know all the tricks for adding extenders, you could do it if you had to.
All right.
So, again, sorry we started late and sorry that we last long as well.
Good afternoon and welcome back.
I'm glad you all came back even though there was a problem in finding a seat for every one last time.
I would like you to turn in the problem set if you've been able to do it.
If you haven't been able to do it, that's no great problem, but I hope you find it mildly amusing.
I have given that problem set out a couple of times over the years, and I can recall one very pale student who appeared at my door the next morning saying it's the first problem set, you've only talked an hour and I can't do it.
And then there was another group of three who formed a consortium to attempt to solve the code using a computer.
They didn't get very far either.
This is an interesting sort of problem because it requires you to think in a slightly different direction than you're accustomed to thinking, and therefore it's a little bit amusing.
This is not my creation, it came from a book by a man named Polya, the title of the book is Mathematics and Plausible Reasoning.
And he gives this as an example of a problem that can be solved only if you think in a slightly different direction.
I'll give you another example of a problem from his book.
Suppose, not that the problem arises in this era when you do all of your graphics on a computer console, but suppose you had to, in solving a problem in short notice, draw a circle that had a diameter of 4 inches.
And you went looking for your pair of compasses which you never use very often, and when you found them, they were rusted solid, and they were open to a distance of 5 inches.
OK, so there's the problem.
You have a pair of compasses that can only draw a circle that's 5 inches in diameter, you must draw a circle that's 4 inches in diameter.
What sort of construction, what sort of mapping out of arcs that you connected together could you do to create the circle of smaller diameter?
Anybody have an idea?
It seems impossible doesn't it?
Well suppose you got yourself a little block of wood that had a height that was equal to the square root of 5 squared minus 4 squared.
And so you have one end of the compass is up on top of the block of wood the other end of the compass traces out a circle that has a smaller diameter.
It's fairly obvious.
But what you have to do is to think of a problem that has poked your nose into two-dimensions and think of it in terms of a three-dimensional problem, and then the answer is easy.
So that was the sort of thought provoking thing that Polya presented in one portion of this book.
OK, if you enjoyed that problem, I have another one for you in a similar vein.
And perhaps you'll enjoy this one as well.
So while I'm talking, let me pass this around, I think there's enough for everyone.
Last time I got so caught up with the displaying the heft of the International Tables for X-Ray Crystallography that I forgot to mention entirely that there is another text that we will use in the class.
And this we will not need until halfway through the semester, and therefore I did not feel terribly remiss in not mentioning it.
It's a book by somebody named Nye, and it's called the Physical Properties of Crystals.
And this is published by Oxford University Press and the publication date of the original addition was in 1967.
This is a book that is really, I don't think I'm being extravagant in calling it a classic.
It is a beautifully written book.
The first 2/3 of it deal systematically with tensor properties, the particular sort of mathematics that is used to set them up, transformation of axes, and then looks at specific physical properties and numbers that have to be described in terms of tensors.
This is actually the third book in a sequence.
It was a book by Wooster that covered things very similarly, but the notation that was used for the tensors was not the modern current notation.
And then the subject started in the form of a third book, Woldemar Voigt.
And the title of this book is Lehrbuch der Kristallphysik.
And this was published back in 1910.
This is the first time anybody had anything to say on the matter.
It in fact is a big fat book that contains some topics that are not covered in Wooster and Nye.
Nye's book is beautifully written.
The first 2/3 of it concerns tensor formalism and the last 1/3, which we will not touch at all, deals with thermodynamic relations between different properties that are represented by tensors.
So I don't recommend you go out and buy this book until you determine whether or not you need it because I'll try to make the lectures self-contained and I'll have lots of notes and handouts.
The problem with Nye's book, and any book that is intensely mathematical, is that you can't jump in on page 73 to find the answer to a specific question.
Because when you go there, it will say as we showed back in Chapter 4, now what is he talking about?
So you go back to Chapter 4, and Chapter 4 says, starting with our definition of Chapter 2, and you have to go back and read Chapter 2.
So it's awfully hard to pick something out to answer a specific question.
You have to really go all the way through it.
The good news is that Nye's book has been published in paperback.
And it is available at the COOP and paperback means cheap, cheap, or relatively inexpensive.
No books are really cheap these days.
So we will cover material that's in there, we'll have a slightly different emphasis, but the notation and the general mathematics that's involved is in Nye's book.
The second thing that I mentioned last time is that there is a very, very nice and thorough and geometric treatment of crystal symmetry.
And I said that's the good news.
The bad news is that it's out of print, so I promised, what a guy, that I give you a Xerox copy of the first half of the book.
So here is the text that we'll use in the first part of the term.
I included at the beginning, the table of contents, so you can see each other topics that are covered in the book.
We will not go through all of the material that's covered.
There are a lot of different symmetries to be derived, and it turns out that if you get the general idea and you can summarize the results, there's no need to derive every single one of them.
It's nice to know that there's a place where you can find out how it is done if you really have a particular question.
Did everybody get a copy or are there a few who did not?
I made extras, OK, nobody in need of one, great.
OK, let me now start with a general rhetorical question.
Crystallography, as we mentioned last time, is the geometry of crystals.
It's the geometry of patterns and the sorts of symmetries that are in those patterns.
Now you might ask yourself, why should I as a material scientist or a physical scientist of some sort, worry about this stuff?
I'm not training to be a wallpaper designer, I'm going to do physical things.
I'm going to heat things and measure properties, and that sort of thing.
Well there are at least three answers to that question.
First of all, whether you like it or not, the arcane language of symmetry is the language that's used to describe crystals.
It's the language that's used to describe structures.
The normal thing that you do when you're trying to describe verbally a ball and pin model of the geometrical arrangement of atoms in a crystal is to say the red balls are at the corners of the cube, the green balls are in the middle of the edges, and the chartreuse balls are sort of tucked up inside one of the corners, but slightly closer to 1 face than to the other 2 faces.
The point I'm trying to make is that is a language that has limited utility.
It deals, it's capable of dealing only with the simplest sort of atomic configurations.
So there is a general language based on symmetry theory, based on group theory, that is universally used to describe atomic arrangements.
So instead of saying red balls at the corners of the cube and green balls in the middle of the faces, I can say space group 4 over m3 bar 2 over m, atom a in position for b, m3m, atom b in position for c, m3m.
That's what rock salt it.
And that is the way, not only it, but especially more complicated structural arrangements are described in the literature.
So this is the language of describing such arrangements.
And finally, sooner or later, I bet you that every one of you will be involved with some crystalline material, and the first question you will answer is what is its structure, what is the atomic arrangement.
That's where properties start.
And you'll go a book or a set of volumes that describe structural data.
There's a big long compendium of books that fill about that much of a bookshelf which are called structure reports.
They started a number of years ago to compile all of the structures that had been determined within a given calendar year.
They did a pretty good job of staying caught up back in 1915 and 1920.
And then as it became easier to obtain such results, partly due to the advent of rapid large computers, they fell further and further behind and I think now they are about 5 or 10 miles-- 5 or 10 years, miles as well.
But this is one of the places to go to look up, without going to the original literature, whether the material that you're interested in has had its atomic arrangement determined.
When you go there you're going to find the atomic arrangement, not in terms of red balls at one position on the cell, but you're going to find it in terms of the language of symmetry theory.
So one of the things I hope you'll be able to do by the time we finish this time together, is to be able to go to such literature and know exactly what to do and where to go to reconstruct the geometrical arrangement of the atoms.
OK, so hopefully you're at least mildly convinced that this exercise is going to be worthwhile.
Before we continue where we left off last time, I would like to say a little bit about the language in which these geometries are described.
And we mentioned last time without thoroughly demonstrating why that in a 3-dimensional space there are 4 basically different kinds of operations.
And one of these is something that all crystals must by definition display, and this is the operation of translation.
Analytically it can be described as a mapping in which every coordinate in a space xyz is mapped to a location x plus some constant, y plus some constant, z plus some constant.
And if you do the operation again, this would go to a location x plus 2a y plus 2b, z plus 2c.
A feature of translation that is unique to this particular symmetry transformation is that it has no origin.
If I have a pair of motifs that are related by translation, we could think of them as being related by a vector, magnitude and direction, that takes this motif and moves it to this location.
We said that more generally we should view these operations, not just acting on one little domain and space, but acting on everything.
So this implies that there be a infinite chain of motifs if the operation of translation is to be present.
Because only that infinite, doubly infinite string is consistent with all of space being mapped into itself.
Like any vector, there's no unique origin.
You could say it extends from here to here, or from here to here, or any other choice of translation, provided the direction and the magnitude are the same in every choice.
As a result, it's not really possible to specify the locus of this particular operation.
It has magnitude and direction, but no unique origin.
What we can do through a very neat device is to nevertheless, take some reference point and have each of these reference points separated by T, and have each motif lurking off in space in exactly the same location and distance from this point that we've constructed.
And this array of abstractions, so these geometrical abstractions, these points, are what are called lattice points, and they are a very neat summary of the translational periodicity of the crystal.
It is absolutely essential not to mix up these lattice points which are a construct that we have created, and the atoms themselves that are present in a crystal.
The atoms are atoms and they're not necessarily the lattice points.
Another way of saying that not all atoms of the same chemical species need be translation equivalent.
We'll see some examples of this later on, so do not mix up the atoms and the lattice points.
When I talk about the sodium chloride lattice, I mean an array of points in space that are located at the corners of a cube and in the middle of the faces of the cube.
If I talk about the arrangement of sodium ions and chlorine ions, that is the sodium chloride structure and not the sodium chloride lattice.
And then last time I apologized for usage so as not to appear hypocritical.
Everybody talks about lattice vibration, lattice energy, lattice dynamics, and so on, but that's a misuse of the term.
But nevertheless, it is much more musical than saying structure energy, structure vibration.
So we'll go on misusing the term lattice I'm afraid.
OK, let's look at another operation.
Here we change the sense of no coordinate.
Let's next look at an operation that might take xyz, and map it into minus xyz.
This would be a situation where if I set up a coordinate system, here's x, here's y, here's z.
What I've done is to take an object that sits off here, at a coordinate plus x, and I've changed the sign of x so that this object now sits off here.
This is exactly what happens when I take something and reflect it in a mirror.
And if that's not immediately obvious, it just so happens, not at all by accident, I brought along with me a mirror.
OK, here is one hand, and if you look in the mirror, there is the other hand.
It's the same distance behind the plane of the mirror, two coordinates have been left unchanged.
The two coordinates within the plane of the mirror, if that is my choice of the reference system, and one of them has been reversed.
Now I'll take the second motif out of the geometric construct, and I'd like to point out one very curious feature of the pair of motifs that's generated by this transformation.
They're both the same thing, clearly.
But no matter how I try, I cannot move one so that it coincides with the other.
And we intuitively appreciate this difference by saying we actually use our hands by analogy.
We say one is left-handed and one is right-handed, and they are not congruent.
The fancy name that's used to describe this relation is to say that they are enantiomorph.
Another term that's used, particularly in chemistry, is to say that they are chiral.
Which one is the left-handed one, which is a right-handed one?
This is what I call my left hand, this is what I call my right hand.
But can we distinguish them physically, any other way?
No, these are terms that have come in to regular use in both our everyday language and also in science because we use our hands instinctively as readily available examples of enantiomorphs, readily at hand, I might say to almost make a pun.
One of the really brilliant figures in physics was a man named Richard Feynman.
Recently deceased, Feynman gave a very famous series of lectures on science at Cornell University.
And he has one entire vector that was devoted to the difference between right and left.
And he comes up with a funny story, he pretends that the hero of the story is someone who's trying to communicate with beings in outer space and suddenly he gets lucky and he gets a response to his message.
And they work out a way to communicate and eventually they try to describe each other to the other individual.
Well what they look like?
Well we're bipedal, and we have two organs related by reflection that let us sense light and form images.
And we have an aperture through which we ingest things that can be metabolized.
And our circulation and body works because we have a pump on the left hand side that circulates fluids through our-- wait, I don't understand, what's left?
So then there follows along this course on how to define left in an absolute sense.
How do you describe to someone what makes your left hand left, and your right hand right without being anthropomorphic about it.
So it goes on and on and on and he gets into physical phenomena which are objective and independent of a human being.
And finally he comes to the anisotropic emission beta particles in radioactive decay.
And that depends on direction relative to the magnetic moment and that defines an absolute sense of right and left.
So that's something physical, it doesn't depend on the nature of the human being.
And then Feynman wraps up his story by saying, if finally our extraterrestrial being travels to space, gets out of his spaceship and he walks forward to greet you and you put out your right hand, and he puts out his left hand, get out of their fast because it means he is made out of anti-matter.
And nuclei of matter in this anisotropic emission of data rays shoot out the beta particle in one sense, anti-matter shoots out the beta particle in the chiral sense.
So it's a cute little story that emphasizes the problem in defining absolutely left from right.
But they're of opposite handedness that we can say.
Mirrors are interesting things and I brought along a couple of mirrors and they have very, very peculiar characteristics.
And I would invite you to come up and look at these in private because if I hold them up in front of you, you're not going to be able to see what I'm doing at all, although I kid myself that you can, and I walk around and show you what I'm doing.
Here is something scientific, it's a chemical compound carbon dioxide, and-- OK, and if I hand this down you can see carbon dioxide in the mirror.
Can you see that?
Uh oh, it's not reflecting-- sorry, I didn't turn it on.
Now I think we can get it.
And is it working now?
OK, now it's working.
You can see why this is a very special kind of mirror because it reflects only red letters and it leaves the black letters unchanged.
You're going to have to come up, I see some of you straining your necks, you'll have to come up and look at that in person.
But the black letters are completely unchanged, the red letters are reflected into letters of an opposite chirality.
It's a very special kind of mirror.
I've got another kind of mirror that works in a different way.
Where Is my other piece of paper?
OK.
This is an interesting mirror because it reflects only male names are not female names.
This was a very topical sort of mirror a few years ago when there was a lawsuit.
There was a college down South called the Citadel which would only admit male applicants and not female applicants.
So I claim that this was a mirror that I got from the Citadel because it doesn't change the male names, but does change, does reject or reflect, female names.
So you can play with this during our break.
But if I look at myself in a mirror, I take a look at myself.
If I wink my left eye, the in there winks his right eye back at me.
So I'm not really seeing myself, what I'm seeing is my enantiomorph.
Doesn't that shake you up?
You have never ever seen yourself in exactly the same way as other people see you.
You are only familiar with your enantiomorph.
Does that make a difference?
Well I'll bring in something that I put together and I couldn't put my hands on.
We are very, very sensitive to the symmetry in our faces.
And if they are reflected left to right, you surely are going to look different to the other person.
And the way to see that is to take a photograph of somebody and cut it down the middle, and put the two different sides reflected left to right.
And the expression on the person's face, and the general spirit that that image conveys is entirely different if you use the one half of the face reflected left to right, and the other half reflected left to right.
So think of this, when you look in the mirror, you see your enantiomorph and other people see you differently.
Let me ask you to scratch your head now.
Is there any time when, in point of fact, you may have seen yourself without being reflected into the enantiomorph?
Yeah
Picture?
Absolutely.
A Photograph or a TV monitor.
When you look at a picture on television, you can read all the signs, and they didn't make up special signs in reflection so that they'd look right when they photographed you.
So photography or a video camera does not change the chirality.
But I've got another way in which I can see myself in the exact same chirality.
And this I can't really convince you of, you'll have to try it.
If I put two mirrors together at 90 degrees, and then adjust them so that I am looking right the point of intersection so that my two images coincide, then I see myself in the normal way.
If I now blink my left eye, this guy blinks his left eye at me too.
This is really astounding, two mirrors at 90 degrees, if you look at yourself right at their point of intersection, give you a non-chiral image of yourself.
So I invite you to come up and try that, that's truly astounding.
So why should somebody in material science or chemistry care about chirality?
Does it really make any difference?
Let me give you a little experiment that you can try.
Suppose you have a little cell on a couple pieces of Polaroid, and the cell has a glass front and a glass back and you fill it with sugar solution.
And then you pass a beam of polarized light through the sugar solution, and what happens is that the sugar solution rotates the direction of polarization in proportion to the thickness of solution at the light is passed through, in proportion to the concentration of sugar.
The point of polarization gets rotated.
Now that's pretty curious, so you scratch your head about that.
Why does that happen?
Well, maybe I'd better go back and try it again.
And a day or two later, you go back and you repeat the experiment.
And once again, the point of polarization rotates, but it rotates in the opposite direction.
The reason for this is that if I was not careful in cleanliness and there were some little bugs lurking in the corners of that cell and when they sense the sugar solution, they said wow, free lunch, and they crawled out And gobbled it up.
It turns out, those guys can gobble up just the sugar of one chirality.
Sugar is a chiral molecule.
And in fact there is a product that's called invert sugar and this is sugar that is all of one handedness.
But everything in the world around us, everything from sugar beats to sugar cane to other things that make sucrose, manufacture sugar of one chirality, not mixed.
All chiral molecules that are produced by living organisms are all of the same kind chirality.
If we make them synthetically, there's no reason to favor synthesis of one molecule or the opposite handedness, so synthesized molecules are of equal proportion in the left-handed chirality and the right-hand chirality.
This means that in the case of pharmaceuticals at the very best, you are going to use only half of the product that you've made.
There is a pharmaceutical product that is prescribed for attention deficit disorder, this is called Ritalin, and only one chirality of the Ritalin molecule does anything for you.
The other part is just metabolized and doesn't do anything.
But there are other much more sinister cases.
There was a serious problem about 20 years ago, primarily in Europe, where a particular pharmaceutical thalidomide was prescribed for pregnant women, it was to act as a sedative.
Only one chirality of the molecule did this, the other chirality tragically caused birth defects.
So you have to be very careful about the chirality of the pharmaceutical molecule that you synthesize.
Another example, there is-- I don't remember the name of it.
This is something that is taken to-- this is something called Ethambutal which is used to treat tuberculosis.
Only the molecule of one handedness does this, the other one causes blindness.
That's really a sinister and antiomorph.
Then there's some even crazier examples.
Ibuprofen is a chiral molecule, and this in a most remarkable situation is a molecule which you're body converts to the molecule of the chirality that has the intended purpose.
So here your body is clever enough to change ibuprofen into the molecule which is the one that you need for its pharmaceutical effect.
OK, so mirrors are interesting things.
I would invite you to come up and play with the special mirrors that do strange things and see yourself as others see you.
And now I would like to continue on in this discussion to mention the ways in which we can represent a mirror plane in a graphic language.
This is what a mirror plane does, it changes the sense of one coordinate.
If there is a locus across which that transformation is performed, we would like first of all, an analytic symbol.
Some way of indicating the presence of that particular operation in the pattern, and a mirror is very descriptive so the symbol m is used to represent the presence of a mirror plane in a particular symbol.
We might want to indicate a specific operation.
There are only two operations in the case of a mirror plane reflecting left to right and reflecting right to left.
But there are other operations such as rotation.
If we have a 16-fold rotation axis, there is one operation that consists of rotating 1/16 of 2 pi, another operation that will also leave the space invariant that's rotating 2/16 of 2 pi, and so on.
So an individual operation is something that we will want to designate.
And for a mirror plane, something that is used commonly in physics is to use an operation sigma for a particular reflection.
This is not done in Buerger.
If you get into reading it, he uses m for both.
And then finally, it's going to be convenient when we have a pattern before us to use a geometric symbol to indicate in the pattern the locus of this particular operation.
And what we use in the case of a mirror plane is a bold line.
And that if this were the pattern and we wanted to indicate where the mirror plane was, or the mirror line in 2-dimensions that relates those two motifs, we draw it in thusly.
We began last time to examine the properties of rotation, but that's another sort of symmetry and that is a rotation which takes place about a rotation axis.
The symbol that is used to represent the collection of operations, the analytic symbol, is based on the fact that the angular rotation, alpha, has to be equal to some sub-multiple of 2 pi.
2 pi over n, where n is some integer.
And the reason for that I think is quite clear, if I take a particular motif and rotate through an angle alpha, if I am not rotating by some sub-multiple of 2 pi, I'll just go round and round and round and I will never get a finite set of objects that is separated from its neighbor by the same angular interval alpha.
This will only happen if alpha is an integral sub-multiple of 2 pi.
And the symbol that is used for the collection of operations that is usually embodied in a rotation axis is n, the same n that is in the denominator.
The symbol for individual rotation, so we mentioned last time we have to specify the location of the point about which we rotate, and we have to indicate the angle alpha through which we've rotated.
So A alpha will be an individual operation, and the geometric symbol will be an n-Gon which has the symmetry of the rotation axis.
So for a sixfold axis we would use a hexagon.
For a fivefold axis we will use a pentagon, for a fourfold, a square, for a threefold, a triangle.
Now an n-Gon with 180 degree rotation is a line segment.
And that would be easily overlooked and it's not very aesthetic, so here we indulge in a little bit of artistic license and fatten out the middle of the line segment to get an oval with pointed ends.
And that's the symbol for a twofold axis.
What about a one-fold axis?
One-fold axes exist anywhere, so you can sprinkle them around with reckless abandon.
A one-fold axis has no symmetry at all, but that is a very nice symbol to use for no symmetry at all.
So symmetry 1 is the absence of symmetry.
So it does come up occasionally in notation.
Now if you look at what we've done so far, we have a transformation that changes the sense of no coordinate.
We have a transformation that changes the sense of two coordinates, one coordinate, no coordinate, one coordinate.
Rotation is interchanging the sense of two coordinates in a plane, and in a 2-dimensional pattern, that's all there is.
But for a 3-dimensional space, we have the option of changing the sense of no coordinate, the sense 1, the sense of 2, or change the sense of all 3 coordinates.
So if this is x and this is y and this is z, and up here in space lurks my initial motif, if I change the sense of x, the sense of y, and the sense of z, namely take xyz and map it to minus x, minus y, minus z, what I'm going to do is to essentially turn the object inside out.
And if my initial one was right-handed, I will produce a chiral object, a left-handed object.
This is an operation which is called inversion.
In this operation of turning the object inside out if you will, is inverting it to a new location.
And this analytically is the exchange in coordinates provided the point of inversion is at the center.
The analytic symbol for inversion is 1 with a bar over the top, pronounced 1 bar.
And I'll have to leave to later indication of exactly where that notation comes from.
The individual operation is also called 1 bar, and the geometric symbol that is used to indicate the location of an inversion center is a tiny little open circle large enough so that you don't miss it, but not so large that it might be confused with an atom in a drawing of an atomic arrangement.
So in this case, we would adorn our sketch was a little circle at the origin if that was the point through which the space was being inverted.
So that, ladies and gentlemen, is our basic bag of tricks in 3-dimensions.
Let me point out that inversion can exist in 3-dimensions only because I have to have 3 coordinates to play with or else I cannot define the operation.
Suppose I have a mapping operation xyz that goes to minus x, minus y, minus z, and I get rid of z to make it 2-dimensional.
Then my transformation is xy going to minus x, minus y and that's exactly what a twofold axis does.
So inversion, when you throw out the third coordinate, looks like a 180 degree rotation.
So you need 3 dimensions in order to define that transformation.
If we really wanted to go crazy, we could go on to say what happens in 4-dimensions?
There should in principle be five different operations and yes, mathematically you can define them.
They're very difficult to draw because we have to have some sort of operation that take something and pulls it out of our 3-dimensional world.
We have no idea where it went and then all of sudden, [POP], it pops back into our space.
But mathematically there are cases when you need a fourth variable to describe the symmetry of an arrangement.
And this generally occurs in something called a modulated structure where there's a periodic change in some variable other than the atomic positions.
And let me give you two quick examples without going into it exhaustively.
One characteristic of an atom besides its location and its atomic mass and things like that, is perhaps a magnetic atom that has a magnetic moment attached to it.
There are magnetically ordered structures.
One of them looks exactly like rock salt.
And I'll draw just the magnetic cations which sit in locations like this.
And the magnetic moment here is up, the magnetic moment here is up, the magnetic moment here is down, the magnetic moment here is down.
So what I've drawn here is no longer the lattice and in fact, the lattice constant of this material looks like a rock salt as far as the atomic positions are concerned, but the magnetic moments have to be continued on in another direction and some extra distance.
Actually some examples of this sort of behavior is FeO, cobalt oxide, nickel oxide.
All of these cations are magnetic, they have magnetic moments which are ordered and the unit cell turns out to be when you take magnetic moment into account, a larger cell, a super cell.
There's a more interesting type of magnetic structure though in which the magnetic moment is inclined relative to some translation in the structure.
And the magnetic moments all lie on the generators of cones, but as you walk along the chain of atoms, the orientation of the moment rotates to different orientations.
There is a family of materials that are said to have cubicle spin structures in which the periodicity of the march of the magnetic moment around the surface of the cone occurs with a period that is in commensurate with the spacing of the chain of atoms.
So strictly speaking, this material does not have a lattice in this direction, so it's not a crystal unless you use a fourth variable to describe the periodicity of the orientation of the moment.
And one final one at the risk of carrying this too far, here's a pattern that is based on a square lattice.
It has a fourfold axis in it unless I make the pattern out of squares that are black and white, make a checkerboard.
This is now no longer a fourfold axis because I can't rotate 90 degrees and leave the pattern invariant.
So this is an example of something called a black-white symmetry, or a color symmetry.
And it requires more than just 4 operations to describe the relation between one motif and another.
We need a fourth operation, switching of a color from black to white, or switching it from white to black, that's a forth operation.
Again, within the confines of a pattern that exists in our space.
Yes?
Does that actually have-- so you're saying it doesn't add value to the [INAUDIBLE]?
I did say no rotational symmetry if it has a fourfold axis here but this used to be a fourfold axis, and that now changes into a twofold axis.
And then I have the problem of describing how this square is a square exactly like this square except for its color.
So I need then an operation which rotates 90 degrees and switches from white to black.
And then rotates 90 degrees again and switches from black to white.
So there's a fifth operation, a color change that is necessary in a 3-dimensional space or a forth operation, a color change in a 2-dimensional checkerboard for example.
So there are lots of nuances to symmetry theory, it's mathematics and the nice thing about mathematics is it's your ballgame, you could make up the rules and as long as you play according to those rules consistently, then you've got something that people can't quarrel with.
OK, I think my internal clock has just told me that it's five minutes of the hour and it's time to take our break.
Come up by all means and play with the mirrors if you'd like and we'll resume the lecture part of our discussion in 10 minutes.
Another strange observation about mirrors that I've never really understood-- the mirror plane is reflecting me left to right, so it looks as though a mirror has a grain to it.
It knows what line to reflect me back and forth.
But if I kept the mirror in fixed orientation and I lay down, the thing should reflect me side to side, but it doesn't.
It still reflects me from top to bottom.
So how can that be?
Why does a mirror plane, if I hold it in one orientation, appear to have a direction across which I'm reflected but it doesn't follow me if I move?
Know what I mean?
You have any explanation of that?
Hmm?
Rotate your eyes, too
Rotate my eyes.
I can roll them around.
I can't rotate in any other fashion.
That's strange.
I mean, you look at yourself every morning-- several times, perhaps, and you're reflected always left to right.
And if you turn the mirror, it doesn't reflect you top to bottom.
Or conversely, you can leave the mirror alone and you can-- why does a mirror plane just reflect you left to right?
Ah.
I'll let you stew about that one for a while.
And I can tell you that I know of three papers in scientific journals that tried to explain this.
I can give you literature citations, but you think about it until our next meeting.
Why does it know which way I'm oriented?
Ah, I can't think about it.
Alright, back to more straightforward questions.
We are now about to embark on a grand process of synthesis which will take us the better part of half a semester.
And we've identified our four basic operations-- four basic one step operations-- namely translation, reflection, rotation, and for the time being I'm going to look just at two-dimensional symmetries, so I'll leave inversion out of the picture.
It's only defined in three dimensions, and the logic which I will follow will be to first build up two-dimensional symmetries and then we'll turn them into three-dimensional symmetries by picking another translation that's not coplanar with the first two.
So we're going to look for the time being at just two-dimensional symmetries.
The nice thing about doing this is that the number of two-dimensional symmetries is relatively small, so we can derive them rigorously and exhaustively.
To do so in three dimensions is a much more time consuming and elaborate exercise.
It's no different in principle, there's just more work.
So if we do two-dimensional symmetries first, it's an easy case.
We can do it rigorously and completely, and then what we'll do is just look at a few examples in three dimensions and look at how the results are designated and tabulated.
So the first combination I will choose to make is one that I set up last time and should not have started when there was no time to finish it.
So let me take an initially pristine space and say that to this I'm going to the operation of translation.
That immediately implies a string of translations and a string of lattice points, but I'm just going to focus my attention on the first one.
Then what I said I'll do is add to the space a rotation operation A alpha.
I'm going to for convenience put it right at the lattice point, but as there is no unique origin to the translation, I can start and stop the translation anywhere I like.
So here's my translation.
The rotation operation is going to take that translation and repeat it at angular intervals alpha so that I get a radiating porcupine-like sheaf of translations all coming out of a common point.
Alpha, we observed, has to be a submultiple of 2 pi, so I will have a cluster of translations separated by the equal interval alpha-- angular interval alpha-- and I'm going to choose to focus my attention on just the first one.
So this, since it's repeated by rotation, has the same length T. And then I'm going to look at the other end of the translation and say that similarly I must have a set of equally spaced angular-wise translations all separated by alpha, and I'm going to choose to focus my attention just on this one.
So this angle is alpha, and this angle here is alpha.
And now there's big trouble in River City.
This is a translation, here's a lattice point, here's a lattice point.
This is a translation-- a lattice point sits up here.
This is a translation-- a lattice points sits up here.
Now I've got two lattice points eyeball to eyeball, and they jolly well better be separated by either the interval T or some multiple P times T, where P is some integer.
It would be quite all right if there were two translation separation or five translation separation, but it must be some multiple of translation or I have violated my initial premise that everything in this space is periodic at a translational interval T. So let me take this geometrical constraint and very quickly convert it into analytical form so that we can proceed to systematically find out what the possibilities are.
Let me drop a perpendicular down to the original translation.
So this distance in here is PT, this distance will be T times the cosine of alpha, and this distance in here will be T times the cosine of alpha.
So in analytic form, then, I can say that my original translation T is equal to T cosine of alpha plus T cosine of alpha.
That's 2T cosine of alpha plus PT.
And the first thing we can see is that the magnitude of T drops out because this construction and the constraint it embodies in no way depends on the magnitude of the translation.
So this says, then, that 1 is equal to 2 cosine of alpha plus the integer P. And if I solve for the value of cosine of alpha, cosine of alpha will be 1 minus an integer P divided by 2.
So there is the constraint that must be followed if my construction is to be self-consistent.
So we've got this now in a plug and chug situation.
So what I'm going to put down is the value of P, taking all possible integers for which a value of cosine of alpha exists.
I'm then going to evaluate cosine of alpha, which is 1 minus P over 2, and then I'm going to identify the n-fold axis that corresponds to that particular value of alpha.
Let's put in P equals 3, and this is as far as we got last time.
Well, if we put 4, then cosine of alpha is minus 3/2-- it's not defined.
If we drop P down to 3, then cosine of alpha is minus 1, 1 minus 3 over 2.
And the angle whose cosine is minus 1 is a-- let me put down the value of alpha and then the n-fold axis.
The angle whose cosine is minus 1 is 180 degrees, and that would describe quite nicely the rotational throw of a twofold axis.
If I let P drop down to the value 2, then I have minus 1/2 of the cosine of alpha.
The angle whose cosine is minus 1/2 is 120 degrees.
And guess what?
That's the angular throw of a threefold axis.
Let P drop down to 1, and then cosine of alpha is 0.
The angle whose cosine is 0 is 90 degrees, and that would be a fourfold axis.
Looks like we're done, except P could be equal to 0.
In that case, cosine of alpha is plus 1/2.
The angle whose cosine is plus 1/2 is equal to 60 degrees, and that's a sixfold access.
What about negative integers?
Minus one, will that work?
This says that cosine of alpha is 1, and the angle whose cosine is one is 0 degrees or 360 degrees.
And that would be no rotational symmetry all.
It's rather amusing that the trivial case of no symmetry at all also falls out of this construction.
So this is a momentous result.
We've shown that if you're going to have a pattern that has a repetition by translation in it, the number of rotational symmetries that can be added are either no symmetry at all, a twofold rotational symmetry, threefold, fourfold, or sixfold.
In other words, the axis-- no symmetry at all, twofold, threefold, fourfold, or sixfold, nothing else.
This tells you a lot about the shapes that you can see macroscopically on crystals.
You could have a crystal that had the shape of the trigonal prism, and that would be perfectly fine.
You could have a crystal that had the shape of an orthogonal brick that would have twofold axes coming out of the faces.
You could have a crystal in the shape of a hexagonal present, or you could have a crystal in the shape of the square prism.
But something thing like a crystal with a pentagonal cross section that would be a fivefold access-- that is strictly forbidden, because the external shape of the crystal has to reflect the internal symmetry among the arrangement of atoms.
There are lots of things in nature that have crystallographic symmetry.
There is a little cactus that looks like this with some little spines coming out like this on the top, and its proper name is astrophytum myriostigma, also known as the ornamented bishops cap.
Beautiful example of fivefold symmetry, but the cells inside of that cactus can not have the same size and shape and be repeated by translation.
There are flowers that have very common examples of fivefold and even sevenfold symmetry.
There's one little purple flower that looks like this that comes out in the spring, and that's called periwinkle.
Fivefold symmetry-- fine for a plant, but if you got down inside the stem of this flower, the cells cannot be repeated by translation.
There are astronomically high symmetries.
These big giant cacti in the Southwest, the saguaro, they have rotational symmetries that run from 18-fold, 19-fold, 23-fold, so that if you picked up one of these guys very carefully, because they're covered with spines, and if you could lift it-- which would be very hard because they weigh a couple of tons-- take that guy and rotate them through 1/27 of a circle, and if he had 27-fold symmetry, you could plop them down and you couldn't tell that it had been moved.
There is no crystallographer who can resist cacti.
All sorts of symmetry, even symmetries that violate crystal graphic symmetries-- wonderful textures and colors.
And they have another really remarkable property, which commends them as house plants.
If they die, this sheath of spines stays intact until someday when you're watering it, you brush against it and you poke a hole right through the spines and there's nothing inside.
So a house plant that dies and you can't tell for two years or so is a very good plant to have as a companion.
So cacti have all sorts of strange symmetries.
Fine, but you can say something about the internal structure of that flower or that cactus.
But this little almost trivial proof has told us something else about crystals, because the presence of translation imposes a constraint on the rotational symmetry that can be present, and the rotational symmetry tells us something about the nature of the two-dimensional lattice which can accommodate that symmetry.
So let's go through this list once more, and let's pay attention to the value of P. For a twofold rotational symmetry, what we would do would be to take this original translation, we put the twofold axis here, and that takes the translation and rotates it around.
So here's a lattice point here, one here, one here, and the twofold axis here, takes the translation and we rotate in a counterclockwise sense.
Here's another lattice point here.
P was equal to three translations, and I'll be darned if that isn't exactly what we have.
Three translations from this lattice point A-- let me put some labels up here.
This was lattice point A, this was lattice point B, and this was our translation PT.
Three translations, just as advertised.
What constraint does this put on a lattice?
None whatsoever, because all this says is that if you have a translation that translation must be repeated into an extended one-dimensional row.
So you can put this an any 2D lattice whatsoever.
In other words, the magnitudes of the two translations-- let's call them T1 and T2 are under no constraint to be related one to another-- and this angle between them, alpha, can be anything that it likes.
You could have a twofold axis, but that requires simply a lattice row parallel to T1 and a lattice row parallel to T2.
The next integer we hit was minus 1/2.
That was cosine of minus 1/2, this was P equals 3, and that corresponded to something that could accept a threefold symmetry.
So let me put a guide to the I in here and make an equilateral triangle.
If we translate and rotate up by 120 degrees using a threefold axis and then rotate minus 120 degrees about the other lattice point, that puts another translation up here, and now-- I'm sorry, this was P equals 2 And as required, this is PT, and that's equal to two translations.
This has put a constraint on the sort of lattice which can exist in this space because we have two translations, T1 and T2, which are equal in magnitude, identical in magnitude, because they're related by symmetry.
And consequently, we have defined a space lattice, a two-dimensional space lattice.
We'll call this T1 and call this T2.
This lattice has the constraint that magnitude of T1 must be identical to the magnitude of T2, and the angle between T1 and T2 is again identically 120 degrees.
Not 119.9, but exactly 120 degrees because there is a threefold axis in here that demands that that be so.
So this is a very specialized kind of lattice, restricted to have two translations identical in magnitude.
And if there's a threefold axis there, you should be able to find these two.
And if there's a threefold axis there, then the angle between these two specialized translations is 120 degrees.
Our next magic integer was 1, and that corresponded to a fourfold axis.
And if we do what we claimed we did in that construction, we'd put a fold axis here.
That takes T1 and rotates it exactly 90 degrees to a translation T2.
Once again, two noncollinear translations, so we have defined a two-dimensional net.
If we complete the cell, this is one translation.
PT is equal to one translation in here, and this angle is 90 degrees because it's produced by a fourfold axis.
So we have a very special lattice.
Again, the magnitudes of two translations are identical-- not approximately the same, they're identical-- and the angle between them is identically 90 degrees.
Only a couple to go.
P could be equal to 0, and that was the case for a 60 degree rotation, a sixfold axis.
So again, let's draw what came out of this particular special case.
Here's T1, here is T2.
This angle is 60 degrees exactly because there's a six-fold rotation axis here, a sixfold rotation axis here, and the rotation of 60 in the opposite sense gives us another translation here.
These two lattice points coincide and there is PT equal to 0T, and these two points coincide.
Now if I complete a standard unit cell with T1 T2 as I've done in other cases-- this was T1, this was T2, this was a translation which I'm not going to use.
So this is the shape of the lattice and these now are lattice points with a sixfold axis on them.
The dimensional specialization is exactly the same as I found for a threefold axis, T1 identical to T2.
If I pick this cell, T1 to T2 can be described as an angle of 120 degrees.
Exactly the same lattice that we found for a threefold axis.
So with this simple minded little construction we've found two profound things-- that there are five kinds of rotation axes, including the onefold no rotational symmetry at all.
And it turns out that there are one, a general lattice, a hexagonal lattice, and the square lattice.
There are three kinds of two-dimensional lattices that are required by these symmetry elements.
So these guys require that there be three lattice of different specializations that are able to accommodate them.
So let me call this a parallelogram net, and that has T1 not equal to T2, the angle between T1 and T2 general.
And this is a lattice that can accommodate either no symmetry at all or a twofold rotation access.
Then there was a net that I'll call the hexagonal net, and this had T1 identical to T2 in magnitude, and it had the angle between them, the angle between T1 and T2 as identically 120 degrees.
And this could accommodate either a threefold or a sixfold axis.
And then finally, the general net, which I call a parallelogram net, and that has magnitudes of the two translations not equal to one another.
They can have any values they like, and the angle between T1 and T2 is completely general.
And that's exactly what I had up here for the-- oh, we did that once already.
The one that I'm missing, the third one, is the square net.
Square net has T1 identical to T2.
In magnitude, the angle between them is exactly 90 degrees, and that is required by a fourfold axis.
Let me pause here to see if there are any questions.
Yes, sir
When you write on the last one with the sixfold, it's only 120.
You could have also written 60, correct?
Yes, I could have.
And this lattice is actually the same as what I found for a threefold axis.
I could pick either this or this as the cell, but the two translations in those two cells are equal.
And again at various stages along the way, we'll need a convention.
And if I have a net that looks like this, a parallelogram, whether a specialized parallelogram or not, I have a choice of two angles that I could use.
We call that alpha.
This is going to be 180 degrees minus alpha.
Which do I pick?
You need a rule, and the rule is that for labeling the cell, pick alpha so that it's greater or equal to 90 degrees.
That's pure convention, but you want to have a rule just like a language so people use the same words to describe the same thing.
Here we want to use the same geometry to describe one and the same thing.
The other convention is that there are no unique translations that define a net.
We could take linear combinations of these vectors and define the same lattice, so we need another rule for selecting the standard translations-- again, so that two people can do an x-ray diffraction experiment and report the results in terms of the same lattice, and this is a fairly reasonable thing.
This is pick the two shortest translations, and that clearly makes sense.
There's absolutely nothing at all to commend a cell that has this as T2 and this as T1 so you get a long, skinny oblique things.
Your natural inclination would pick the two shortest translations in the net.
OK, so these are conventions.
This has nothing to do with the nature of the symmetry or what makes it unique, but just so that people have one defined way of labeling things
Right.
[INAUDIBLE] my question was actually that--
It was a good answer, even if was not to the question that you asked
I said the sixfold and the threefold are exactly the same, but then I realized they are because there is no crystal that can have threefold symmetry without sixfold
You were doing fine.
You should have quit just before that last statement.
How can you have a hexagonal lattice that sometimes has sixfold symmetry and sometimes has threefold symmetry?
Well, let me give you an example for that, and it makes a very useful point.
Let me draw two hexagonal nets, and in this one I'll put a threefold axis.
So I'll have one motif here, I'll go 120 degrees away.
Here is a motif here, and I'll go 120 degrees away, and here is a third motif.
So these three guys form a triangle about this lattice point.
And we would have about the other lattice points at the corners of the cell exactly the same triangle of motifs, and the same thing over here.
So there is a pattern.
It has a hexagonal lattice, and it has a triangle of objects related by a threefold axis.
And now let me take exactly the same lattice, and now I'll put in instead a sixfold axis.
And that means I'm going to have a hexagon of objects, and it'll do something like this.
And I don't want to push my luck and try to draw that twice, but there would be a hexagon here and a hexagon of motifs here, another one at this lattice point, and another one at this lattice point here.
Same lattice, same shape, same dimensions-- although that's not critical-- but one of them has only a threefold axis.
One of them has only a sixfold axis in it.
Why?
Because I decided to put a threefold axis in this pattern, and I decided to put a sixfold axis in this pattern, but they both end up being contented and happy with a lattice with the same degree of specialization.
Now we will come as we progressed a little bit further that we have to go in two dimensions to the reverse situation.
We have a particular symmetry and it's happy with two different sorts of lattices with different shapes and different specializations, and that's going to come up directly.
Any other questions?
Yeah
In this example, the sixfold [INAUDIBLE] axis [INAUDIBLE].
It's just that [INAUDIBLE]
Yes.
You're saying that here hanging at this lattice point is something that has sixfold symmetry.
Here is something has only threefold symmetry.
The nature of the lattice and the symmetry that is in that lattice are two inseparable aspects of the pattern.
But often, as we've seen here, there's more than one possibility for a given lattice.
And as we'll see very shortly, for some other symmetries, for one given symmetry, there are two kinds of lattices.
But nevertheless, the thing to keep fixed in your mind is that we call this a hexagonal lattice.
Why?
Because this translation is equal to this one, and they are exactly 120 degrees apart.
That lattice can have that specialness only if there's either a threefold or a sixfold axis in it.
So the specialization of a lattice is inseparable from the symmetry that is in the lattice that demands that specialization of the lattice.
Conversely, a lattice can have the specialization only if you place in a symmetry which demands precisely that specialization.
So if you measure a lattice, and this turns out to be 119.99 degrees and these turn out to be 3.21 angstroms and 3.21 angstroms, that is not a hexagonal lattice, because there's no symmetry in there that demands that this angle be 120 degrees.
And that may seem to be an academic fine point, but we'll see that in due course the properties of a crystal depend on the symmetry of that crystal.
If the crystal has symmetry, the property also has to have that symmetry.
And it is the atoms inside the cell which determine the symmetry of the property, and the properties is one aspect of the symmetry that goes along with lattice dimensions and lattice angles.
Is that clear?
So let me say it again, because this is an important point.
The specialness of a lattice is inseparable from the symmetry that is existing in that lattice that demands that specialness.
So if you have a crystal with three orthogonal translations that is equal in length as you might care to measure, that crystal is not cubic unless there's symmetry in that lattice that demands that the edges of the cell conform to the geometry of a cube.
Any other questions?
Let's then in the time that's remaining look at the other symmetry element that could be present in a two-dimensional crystal, and that's the mirror plane.
So here's a mirror plane.
That's the one remaining symmetry element in two dimensions, and let's ask how we might combine in this space along with the mirror plane a translation.
If I just pop in a translation, and call this T1, and for convenience, I'll take the lattice point on the mirror plane.
Here's another lattice point that sits here.
The mirror plane acts on everything.
It's going to take this lattice point and flip it over to here, and it's going to take the translation that goes from the origin lattice point to this lattice point and give me a T1 prime that sits here.
And now I have two non collinear translations, so these have defined for me a cell that looks like this.
This is T1, this is T2, and this is some angle alpha between them.
That's a special lattice.
This is a lattice which has, just as in the hexagonal or square net, two translations that are identical in magnitude.
Why?
Because they're repeated by a mirror plane, and the angle between them-- the angle between T1 and T2-- is completely general.
It can be anything it likes.
It can close up to a very narrow angle or it can open up to an almost flat angle.
That's a new kind of lattice.
None of the preceding lattices could make that claim.
Now let me point out that this is for the first time a case where it would be much to our advantage to not choose a cell that contains one lattice point.
Let me put some dotted lines in here.
And let me submit-- these will also be lattice points-- that I could pick a larger cell that would have this as T1, it would have this as T2, and the angle between them now would be identically 90 degrees because of the geometry that gives us a rhombus here.
This would have two translations that are not equal in magnitude, and it would have an angle between these two translations that is identically 90 degrees, but it is no longer a cell that contains one lattice point.
It now catches a second lattice point that's in the middle.
So this is a double cell, and it has a rectangular shape.
It's a centered rectangular lattice.
Being a double cell that's redundant has twice the area that is unique in the pattern, and anything that's hanging up here is going to be hanging down here, so it has a twofold redundancy.
But the thing that you get in return for paying the price of that redundancy is a cell that has a right angle in it.
And as we'll see, we're going to use the edges of the unit cell as the basis of a coordinate system for describing what goes on at positions xy within the cell.
And the advantages of an orthogonal coordinate system, whenever you can take advantage of it, far outweighs the price of pain-- twice the area to describe the same pattern.
So this is what is generally taken as the standard cell.
It's a double cell.
But I think you're used to that sort of compromise, because you've all heard of face centered cubic lattices.
The primitive cell in a face centered cubic lattice is a rhombohedron.
But, oh, that Cartesian coordinate system is so great to use rather than something that has an oblique coordinate system.
So is this is a definition of convenience.
Notice the curious duality-- either special relation between the translations angle general or special relation between the translations, an angle special.
General translation, special angle, relation between the translation, general angle-- so, it's a curious sort of duality.
You can have one but not the other or vice versa.
So this is a fourth sort of lattice.
Number one, number two, number three, and now we have number four, which is a centered rectangular lattice.
And this particular lattice has T1 not equal to T2 in magnitude, but it has the angle between T1 and T2 exactly 90 degrees, and it's a double cell.
It's centered.
Are we done?
The reason I asked that silly rhetorical question is that obviously I suspect that we're not done, and what else might we do?
We got our centered rectangular net by starting with a translation-- starting with an mirror plane, really-- and then we added a general translation, and that reflected it across and gave us a diamond shaped net which we could define as a rectangular double shell.
Does that always happen?
Do I always get that oblique diamond shaped cell?
No.
Suppose I put in my first translation deliberately in a fashion such that it was at exactly right angles to the mirror plane.
That mirror plane will then reflect the translation and change its direction, and now I have generated a one-dimensional lattice row with translational periodicity T1.
And I've got a translation that's exactly perpendicular to the mirror plane.
How do I now make a space lattice, a two-dimensional space lattice?
And the answer is very carefully.
Suppose I throw in a second translation T2 and the mirror plane reflects it across to here.
This interval between lattice points up here is totally in commensurate with the first translation, and that won't work.
It's violated my initial choice of the translational periodicity unless I do one of two things, and let's try them both and dispose of this quickly-- this is straight away.
Here's my one-dimensional lattice row.
I do not violate this periodicity T1 if I do two things.
I could pick T2 so that it fell exactly along the mirror line, and that's going to generate for me a new type of lattice in which this is 90 degrees and these two translations are unequal in length.
So this is a lattice that has a rectangular shape.
It's a primitive rectangular cell, and it has T1 not equal to T2 in magnitude, and it has T1 and T2 exactly 90 degrees apart.
The second choice that would not result in any contradiction would be to have this as T1, and then pick T2 very carefully so that it spanned the mirror line with one half of T1 exactly up here and one half of T1 exactly here.
And this would be compatible with the separation T1 down at the start of this translation.
I think you can see that what this is going to give me is the centered rectangular net right back again, so this is the centered rectangular net-- nothing new.
But we did pick up one additional two-dimensional lattice of distinct character.
Number 5 is a primitive rectangular network, and it has the characteristics T1 is not equal to T2 in magnitude.
Just as in the centered rectangular net, T1 is at exactly 90 degrees to T2, but this is a primitive cell.
And that's it.
We really set up the ground rules for the geometry of a periodic two-dimensional space.
There are five kinds of rotation axes-- one, two, three, four, and six.
Each one requires one or more of the specialized two-dimensional lattices.
We have a case where interestingly two different symmetries are compatible with a lattice of the same specialization.
In the case of the [?
hexagon ?]
on that, either three or sixfold symmetry could require that.
In the case of the mirror plane we have one symmetry element, M, that can fit into two different kinds of lattices.
So in one case, the same lattice can take two different symmetries.
In this case, two different lattices can accommodate the same symmetry.
So that's the story for two dimensions, and we have just one final thing to do.
That is to add to the lattices that we have found, and there are five of them, and add to the lattice point the symmetries that we have found require them.
And there are a limited number of these-- one, two, three, four, or sixfold rotational symmetry in a mirror plane.
And when we have finished these additions, we will end up with a combination of lattice and symmetry, which is something that is called a plane group-- group because the operations that are present in the space follow the requirements of the mathematical entity called a group, plane because this is a distribution of symmetry elements throughout a plane and not just fixed at one particular point.
Let me finish with some general observation, and we will obtain some of these rules later on.
Suppose I take a pristine space-- and this blackboard is no longer pristine-- and I put in the space a first operation, and there's a motif in there.
This first operation moves around this first motif and gives me a second one, number two.
Then I say let me put in another operation.
I'd like to combine these things and see how many different combinations I have, so I put in operation number two.
Operation number two will take the second object and repeat it to a third object.
Number two is identical to number one, so this gives me a third object reproduced from the second by operation number two.
Now I have a space, and sitting in it are two different operations and three different motifs.
Motif number one and motif number three are the same darn thing because they've been repeated by symmetry steps, so there must automatically be some third transformation that is equal to the combined effect of going from one to two and then from two to three-- going from one to three directly.
So this is another truth about these symmetries.
Whenever you take two operations and combine them in a space, the net effect of those two operations is equal to a third operation.
So a question we're going to ask all along the way until we are at the end of the month-- if you take a translation and combine it with a mirror plane, what new operation has to arise?
If you take two rotation operations and put them together, what third net operation has to arrive?
If we have some general rules, then we can automatically say, OK, I'm going to take a square lattice and I'm going to put in a fourfold axis.
What else is going to pop up elsewhere within the cell?
We'll be able to do this systematically but fairly automatically.
So that's where we're going from here.
When we're done, we will have derived systematically and rigorously the sorts of symmetries that can combine symmetries such as rotation and reflection with a lattice, and we will know completely the different sorts of patterns that exist around us in two dimensions-- in floor tiles, brick work, wrapping paper, and plaid shirts.
We'll pick up at this point on Thursday.
Let me caution you we are going into territory that is not covered in Berger's book.
I've just passed it out to you.
What we've said in the early parts of the term are in there, but we're going to do things in a slightly different way and then return to his text later on.
All right, my trusty $5 Timex watch says it's five after the hour.
So I guess we might as well begin.
I have corrected all the problems that were turned in-- the puzzles, not problems, puzzles.
And I just yesterday got hold of the photographs of you few people.
And I'd like to hand out the papers individually so I can start to put names together with faces.
So I'll turn them back individually next time.
I have one additional problem set that for you.
And again, this is in the form of a puzzle.
Be assured that this is the last entertaining problem set.
From here on, we move into drudgery-- tedious, drawn out things that hopefully will teach you something.
But this is the last one that's done with a little bit of tongue in cheek humor.
This one is a little more subtle than the other two.
And for this one, I will hand out a solution because I would like to use these problems to make a couple of basic points about the nature of symmetry theory in crystals.
So see how you do on this one.
OK, also before we begin, he gets a little bit uneasy before a crowd.
But I brought a little friend in with me today.
And could someone hold the box, please, so I could take him out?
Sorry.
He gets a little bit nervous before a crowd.
Come on, take it easy.
Take it easy.
This is Rodney.
He's an old friend of mine.
Rodney is a crystallographic abomination.
Rodney has five-fold symmetry-- impossible in crystallography, an abomination.
Why did Rodney decide to do this?
Anybody have any idea?
Why not six fold symmetry?
He's got five-fold symmetry.
Yuck!
But why?
There's always a reason for things if you look carefully enough and know what questions to ask.
Yeah?
He had narrow symmetry
Yeah, he had narrow symmetry.
But he could do that-- what if you were square?
Why not a square starfish?
Why five?
He did this for a reason.
He's been doing this for probably a couple of million years, he and his ancestors.
There has to be a reason.
OK, let me turn him over.
And maybe that will give you a clue.
The backside is not nearly as nice as the front side.
And in fact, if you look at Rodney carefully, Rodney has sutures between these five arms.
OK, this is what he looks like from the top.
But if you look at him from the bottom, there are sutures between these arms that do this.
And these sutures are weak points in his structure.
And by having five-fold symmetry, he puts the suture between this pair of arms opposite the midpoint of the opposite rib.
And that gives him a great deal more structural integrity.
So when a large fish comes along and gloms onto one of his arms and shakes, he doesn't split in half and end his career prematurely.
What happens instead, at very worst, the tip of the arm breaks off.
And then, Rodney gets to scamper away and make more starfish or do whatever his purpose in our environment is.
If he had an even-fold symmetry, that wouldn't happen.
He would really be done in by a fish grabbing hold and shaking
What if [INAUDIBLE]?
That's a very good question.
If he had only three arms and a fish glomed on and one arm broke off, his mobility would be greatly impaired.
In a way, he would go hip, hop, flop.
Hip, hip, flop.
Hip, hip, flop.
Hip, hip, flop.
It'd be a pathetic thing.
So five is much more advantageous than three
Why not seven?
Why not seven?
That's a good question, too.
Would you believe that on the Australian Barrier Reef, there are starfish with nine-fold symmetry?
Quite odd.
That's really a creepy thing with all those arms flopping around.
But as many things in our environment in our natural world, there's a reason for things if you can only think what it might be.
So we gave you another example of a non-crystallographic symmetry, the large cacti of the Southwest, the saguaros.
And there are lots of other ribbed casts.
They have symmetries that range between 19 and 23 fold.
Why do you think they have this rib structure?
Why can't they have a nice, smooth trunk like a birch tree?
Any idea there?
I'll give you a hint.
They live out in the desert.
There's not much water
[INAUDIBLE]
Hm?
More surface area for the [INAUDIBLE]
Yeah, but that's bad.
That means they would lose their water.
So things that don't want to lose water through their surface-- you're right-- try to minimize their surface.
But you're onto something.
And it is exactly the reverse of that.
Water is scarce.
So when it rains, this cactus wants to soak up the moisture and store it.
If he has this pleated structure, he's able to expand.
If he had a smooth surface with minimal area, he would just pop.
And that would be the end of him or her, as the case may be.
"It" is probably the appropriate term.
But anyway, there's the reason for these many, many ribs.
It gives them the ability to expand, store water, and then contract when the water evaporates.
So again, it's always interesting to look at the world around us and see examples of things that are strange because it's the strange things that very often can lead to penetrating conclusions.
All right, last time we had talked about chirality in finite atomic structures, molecules.
And I had pointed out-- but did not make the obvious observation-- I cannot think of a more dramatic connection between structure and properties, which is what material science is all about, than the very, very different properties of molecules that are of opposite handedness.
We pointed out, for example, that many pharmaceuticals are of chiral molecules that exist in right handed and left handed forms.
And very often, one form of one handedness has a very different set of properties, and in particular pharmacological action, then it it's enan enantiomorph.
So in the case of thalidomide, one handedness of the molecule acts as a tranquilizer.
The other handedness acts as a source of birth defects.
One of the other ones less familiar to me, but this is [?
ethambutol.
?]
It's used as a drug to treat tuberculosis.
The molecule of opposite handedness creates blindness.
The drug Ritalin that is used to treat hyperactivity, one handedness of the molecule is absolutely useless.
So you only use half of what you pay for.
So it's of great interest to pharmaceutical companies to be able to synthesize a molecule of one particular handedness with little if none of the other one.
And I looked up a number since we met last time.
It turns out that in 1997, enantiomorph pure drugs were a $400 billion dollar industry.
So again, this is structure property relations in action.
So let me ask you another question.
It turns out that products that are derived from nature-- a good example being sugar, which is produced by sugar beets or sugar cane-- produces a molecule only of one handedness.
They are [?
enantiomer ?]
pure.
But every synthetic compound that can exist in right handed and left handed forms occurs with equal probability.
So how do you make a pharmaceutical of one particular handedness?
And this is an interesting question because I think it was two years ago the Nobel Prize in chemistry was given to three people for their work in being able to synthesize molecules of a given handedness.
Anybody an idea how you do it?
I don't have an idea how you do it, but what do you mean by one handedness versus--?
OK, left handed or right handed.
So my left hand is exactly the same as my right hand, but I cannot match them into congruence with one another.
And we don't know how to describe this other than we used the term derived from our physiology.
We say left handed and right handed.
So when I was referring to a molecule of one handedness-- and I'll bring in a Xerox copy of a couple of examples.
There are many examples of fairly common molecules which can exist in one form in which the hydroxyl group points out this way, then another form where the hydroxyl groups points out this way.
Sorry, you should have asked that earlier
So if it's not the same molecule, it would be that molecule but--?
Chemically, it is exactly the same concept.
But the way in which those component models are configured has the same number of side groups, the same number of appendages.
But they, don't have a better word.
One is left handed.
One is right handed.
I'll give you some examples next time.
Somebody else, you had a question?
I was just going to see if I could venture a guess how they do it.
I was going to guess they use an electric field that has its bearings to set the right handed course [INAUDIBLE] that would make [?
enantiomorphs ?]
or more advantageous than [INAUDIBLE]
That's not a bad idea.
Anything that could bias the energy of one configuration over the other would do it.
But actually, no reason why you should know how to do it.
If you knew how to do it, maybe you could have gotten the Nobel Prize instead of these other guys.
But actually, you do it through catalysis.
And you use a catalyst which itself has chirality.
And that chirality favors the formation of one molecule over the other almost exclusively.
And that's why living plants make-- or animals-- make chiral molecules in just one handedness as well.
So it's using a catalyst that also has some chirality
How do you get the one hundred catalysts [INAUDIBLE]?
How do you have it--?
OK, there are lots of materials.
Any crystalline material that does not have a mirror plane in it can exist in chiral forms.
And one of the classic examples of this is quartz.
Quartz is SiO2.
And quartz has actually only a threefold symmetry.
But it forms crystals that look prismatic, as though they were hexagonal prisms.
And if I draw two of these prisms side by side, you could not tell which prism was left handed and which was right handed.
But quartz ends up growing with a couple of little facets on it that look like this.
They spiral up this way in a crystal of one handedness and they spiral up in the opposite direction for a crystal of opposite handedness.
So the external faces on the crystal show you that this crystal shape cannot be mapped one into another.
They are of opposite handedness.
And they have very different properties.
And so here's an example.
I'm not answer your question.
I'm talking around it very adroitly in the hope that if I talk enough you will forget what you asked.
But how do you make a chiral catalyst?
Well, there are some materials if you prepare them in a single crystal form, once you nucleate one enantiomorph, that will continue to grow.
So I would presume one could do exactly the same with a catalyst.
Actually, the properties of quartz are strikingly different for these two different forms.
If you would subject the crystal to compression, this crystal would develop a positive charge on the surface.
This crystal would develop a negative charge on the surface.
That's a property that we'll discuss in detail later on.
That's something called piezoelectricity, literally pressure electricity.
So the nature of the charge induced on the crystal is different for the crystals of the two chiralities.
OK, any other comment or question?
OK, last time we had taken the first small steps in what will turn out to be a long and protracted process of synthesis.
We've covered by now some of the general notions of operations that can exist in a pattern, two dimensional pattern of wallpaper or fabric or a three dimensional pattern that constitutes a crystal.
And we said that in three dimensions, there are four basic different operations that can exist that tell you how one part of the system is related to another.
There is an operation of translation.
And this has all the characteristics of a vector, magnitude and direction.
We summarize that translational periodicity by a set of fiducial markers that are called lattice points.
And analytically, the transformation takes the coordinate x, y, z.
And if you do the operation once, maps this to x plus a, y plus b, z plus c where the translation has components a, b, and c in x, y, and z directions.
Another operation is the operation of reflection.
And what this operation does is to take the coordinates of an initial part of the space atom or location and maps it into some location where one direction is reversed in sign.
Which direction that is or whether it's a pair of directions will depend on how the reflection locus is located relative to the coordinate system.
Next, there was an operation of rotation.
And this involves repeating things at angular intervals.
Take the space or the location of the atom and rotate it through some angle, alpha, about some rotation, a.
And in a sense, this is a periodicity which is reminiscent of translation except translation strings things out in a line.
Rotation wraps up the directions about some central axis.
But it's also periodic.
Then finally, in three dimensions, there is another operation called inversion.
And what this does is-- to could describe it as such in literal terms-- take something and turns it inside out to a point.
So if the inversion point were at the origin, it would take x, y, z, and map it into minus x, minus y, minus z.
And then, we concluded last time with a definition of a vocabulary for indicating analytically a single operation, the set of operations that constitutes the symmetry element, and then a geometric symbol for the locus of that operation.
And for translation, we could represent a single operation by t. The set of operations could be-- and I'll use this customary set of braces to indicate the set-- this would be a one dimensional lattice, a set of operations, p, where p is an integer times t. The geometric symbol, it's convenient to use an arrow extending from one lattice point to another.
But I emphasize that there's no unique origin to the translation.
We cannot be entitled to say it sits here as opposed to down here or down here.
There are an infinite number of parallel directions that could specify a given interval between lattice points.
For rotation operations, the notation for a single operation in as much as we must specify two things, the point about which we rotate and then the angular interval about which we perform the rotation.
And so we need to specify two things, the location of the axis, a point a, and the angle, alpha, through which we rotate.
The set of operations, the rotation axis, is indicated by an integer, n, where the rotation part of the operation is 2 pi divided by the integer, n. So we would use, for example, 2 to represent a rotation axis where things were moved through 180 degrees, 3 for a symmetry where things were mapped through intervals of 120 degrees, and 22 for my good friend, the Saguaro cactus, which is left invariant perhaps by a rotation through 1/22 of 2 pi.
For the geometric symbol for a rotation axis, we will use a little polyhedron.
Triangle for a threefold axis.
Square for a fourfold axis.
For a twofold axis, the polyhedron that had that symmetry would be a line segment.
So we'll take a little artistic license and let the middle of the line segment bulge out a little bit.
Finally for inversion, for reasons that will not become clear for another meeting or two, the single operation is denoted one bar.
The set of operations of which there are only two, inverting and coming back again, also indicated by the symbol one bar.
And the geometric symbol for the locus is a little, tiny circle that's open large enough to be noticed, but not so large that it looks like an atom in an atomic arrangement.
So there's our basic bag of tricks.
And last time, we had already used some simple geometry to come to a rather astonishing conclusion.
Yes, sir?
Reflection?
Reflection, that's this middle one.
Whoops, not going to help.
Thank you very much.
I get excited and carried away.
OK, reflection, the second operation.
Single operation, although crystallographers don't often use it in condensed matter physics, a single operation is represented by the symbol sigma.
Symbol for a mirror plane is m, easily appreciated.
And the geometric locus across which things are reflected left to right and vice versa is done with a bold, solid line when you're looking down parallel to the surface of the reflection locus.
Then, I will now be caught up.
And I can set off on something new.
We had said that in a pattern, very often more than one of these operations are combined in a single space.
And one of the things that must be present-- if the symmetry of the entity we are discussing is a crystal-- one of the things that must be present is a translation.
And so I asked last time the question, what if we want to say that our space has a translational periodicity described by t. And then, I add to one lattice point a rotation operation, A alpha, that has to get translated to another location where that operation, A alpha, exists.
And if A alpha exists and I repeat that translation by a rotation alpha, I'm going to have a sheath of translations all separated by a distance alpha.
If I single out the one that is removed from the first by an operation of rotation A alpha in a counterclockwise direction and single out from that sheath another one that is alpha away in the clockwise direction.
Then, I have lattice points at the end of all of these translations.
But these two guys up here are also lattice points.
And therefore, I have mucked up to the entire construction unless I can claim that this distance between these two lattice points is either T or some multiple P times T, where P is an integer.
And then lickety split, before you could catch your breath in wonder, I dropped a perpendicular down to the original translation.
This distance in here was PT.
This distance on either side was T times the cosine of alpha.
And that led me to a constraint that alpha, that rotation operation A alpha, if it were to be combined with translation, would be restricted to values such that the cosine of alpha was equal to 1 minus an integer P divided by 2.
Wow, simple, but so incredibly profound.
What came out of this was that alpha could either be 0 or 360 degrees.
That was a one-fold axis.
Alpha could be 180 degrees.
That was a two-fold axis.
120, which would be the angular rotation of a three-fold axis.
90, that would be the operation of a four-fold axis.
Or 60, that would be the operation of a six-fold axis.
It's incredible.
Anything, any pattern, any crystal that is based on a lattice can contain only these five rotational symmetries including no symmetry at all.
So this says a lot about what the morphology of crystalline materials are permitted to be.
Then, we looked at the nature of the lattices that were described by these angular rotations.
And we found that there were only three types of lattices that were required by rotation.
These were two dimensional lattice nets.
One was a general oblique net which had two different translations, some arbitrary angle between them, two translations, T1 and T2, which were not equal to one another.
We would call this the oblique net.
Then, there was another net that was square in the sense that two translations, T1 and T2, were identical to one another-- not close, but identical.
And the angle between them was identically 90 degrees.
This oblique net could accept either a one-fold axis or a two-fold axis.
This was the specialized shape that was required by a four-fold axis.
And then finally, there was another net that looked oblique except it was a specialized oblique net.
It had two translations, T1 and T2, which were identical just as in the square net and an angle between them, which was exactly 120 degrees.
And this same sort of net could be compatible with either a three-fold axis or a six-fold axis.
Then, we took the only other operation that could be present in a two dimensional pattern, and this was the operation of reflection, a mirror point, m. And we saw that this by itself could require two different sorts of specializations in a lattice, a lattice in which two directions were exactly orthogonal but the lengths of the translations were unequal.
And then, a diamond shaped net-- and this was a new wrinkle.
This was a net that had two translations that were Identical in length and an arbitrary angle between them.
Or alternatively, if we chose to pick a redundant lattice, one that had two translations, T1 and T2 prime which were different but an angle between them that was exactly 90 degrees, so this one we would refer to as a rectangular lattice, a rectangular net, and this one as a centered rectangular net-- same shape, but an extra lattice point in the middle.
Why pick a double cell, which is redundant, takes up more, and doesn't convey any more information?
And the advantage comes when you want to analytically describe the relation between different directions or the positions of atoms within the cell.
In that case, an orthogonal coordinate system, even if it's not Cartesian, is of immense advantage as opposed to an oblique system.
OK, slightly out of breath, but that is everything we've covered to this point.
And you can find this material discussed in the first chapter, in the introduction, and then half of the second chapter of Berger's book if you want to read over it.
OK, any questions or comments?
All right, I wanted to say a little bit more about lattices and something with which you are no doubt familiar, but perhaps have some questions about why one does it in such an obscure, difficult way.
And let me introduce what's done by an analogy which we have around us in everyday life.
Let me draw a system of streets and avenues for a city that might be one like Boston where the streets were laid out by cows wandering around in search of forage.
And suppose somebody met you on the street and said, may I stop you here at point A and ask you a question?
How do I get from point A to point B?
That's not a vector.
That's a point.
How do I do that?
You would not say, go 342 meters to the east and 26.4 meters to the south.
That's perfectly logical that is admirably Cartesian but you wouldn't do it that way what we just said you would say go one block straight ahead, and then turn one block to your left, and then go down the street to your right and turn to your right again, and you'll be right there.
You can't miss it.
Of course, the person in Boston would miss it and would have to stop a second time and ask again and by a method of successive approximations eventually get to where he or she would like to be.
I submit if a system gives you a grid, it makes sense to use that grid as the basis vectors of coordinate system even if the intervals are different, in two different directions, and even though the intervals might be not vectors that are orthogonal to one another.
If the system is periodic and has this grid work to it, it makes sense to use that as the basis of a coordinate system.
So if we have a lattice that is oblique and there are lattice points defined by these two translations, T1 and T2, it makes great sense if you want to specify the vector that runs from this lattice point to this lattice point to do that by saying that this is a lattice point that is at the end of the vector 1 T1 plus 2 T2.
And that, just as a system of streets and avenues, is a lot more efficient than saying go 12.2 angstroms in the x direction and go 15.3 angstroms in the y direction.
What we have defined here is something that is called a rational direction.
And a rational direction is something that is going to be a direction that extends between lattice points.
And what's rational about it is that the coefficients in front of T1 and T2 that would appear would be integers.
And an integers is sometimes referred to as a rational number.
That would be quite different from a vector that extended from a lattice point to, let's say, some atom that is within the boundaries of this cell.
And this might be a radial vector from the origin to one lattice.
And that could be used to specify the atomic coordinates.
This is going to complete the dichotomy, a feature that is referred to as irrational.
Why?
Because the components of the steps along T1, x, and along T2, y, are going to be fractional numbers.
So you'll hit irrational notation for directions that go to different locations with a unit cell.
You'll use rational vectors in directions that extend between integral numbers of lattice points.
These directions occur all the time when one discusses the physical properties of crystals.
I'm sure in any class that you've had that's discussed mechanical properties, one of the favorite topics is to discuss the ways in which close packed metals can deform.
And this can be discussed easily in terms of one layer of close packed spheres sliding over another.
And one of the ways that the layer can slide to go from one set of close packed hollows to another is in a direction like this.
And that turns out to be a rational direction.
And these are directions in which a close packed metal will easily deform.
So directions of easy plastic deformations almost always involve rational directions.
So there is a unique application of these sorts of features in the discussion of mechanical properties.
Another thing that can happen is that a particular structure, if it is weakly bonded or relatively weakly bonded in one direction, if you come down with a chisel on that plane and give it a little tunk with a hammer, the crystal will fall exactly in half and give you two very smooth surfaces on either side of some plane.
And the reason is if you look down into the guts of the crystal and look at the atoms and how they're bonded together, there may be some direction of weak bonding between atoms that is very easily separated by some sort of mechanical force.
This is true even for crystals that are very strongly bonded.
The way in which people begin to facet diamonds, for example, is to sit for a month and study the diamond and then decide how you're going to put a straight edge to it and give it a little tunk so that it falls neatly into two pieces at the sides that you want to facet.
Doesn't always work, so you sit for a long time to figure out just what you're going to do.
I think diamond cutters burn out early and have a useful career that last perhaps 3 and 1/2 years.
And then, they're burned out.
I don't know that for sure.
But it must be a nerve wracking way of making an existence.
Not a far-fetched story.
At one time, I was doing a diffraction study on a material which was a low temperature face, very interesting, but formed stable only at such low temperatures that you could not make it by cooking the components.
But it had been found as a mineral in the Swiss Alps, all sorts of exotica attached to this particular material.
There was a particular place up in the Alps which had a very unusual chemistry involving lead and arsenic and thallium and sulfur-- not what you find in your usual backyard rock pile.
And there was something like 37 unique minerals that were found in this little spot that were known nowhere else on the face of the Earth.
And I was very interested in the atomic arrangements in one of them.
So I wrote to the British Museum that had found the sole supplier of this material that-- back in about 1901 or so-- I wrote and I said, do you have any crystals of this stuff?
Yes.
Would you like to study it?
And I said yes, I would.
Well, we'd be happy to send it.
But please be careful not to damage the morphology because it is a very rare material.
So I said fine.
I'm as careful as the next meticulous guy.
They sent two crystals.
The crystals measured-- one of them was 2/10 of a millimeter in diameter.
One of them was about 0.15 millimeters in diameter.
And I was supposed to take off a piece for study without damaging the morphology.
Man, I felt just like one of these Dutch diamond cutters.
I sat and I looked at those two little suckers under the microscope for, I think, a week.
And then finally, the moment of truth, and I took a needle and popped off a corner.
And it went into two pieces.
One went for a microprobe analysis to determine the composition.
One of them went to a single crystal x-ray diffraction study to determine the atomic arrangement.
Both analyses can be done on a very small piece of material.
So my story about the diamond cutter agonizing before looking for a cleavage or a fracture surface is not just made up fiction.
It is something that, when you work with crystals very often, is encountered.
OK, cleavage planes are also rational directions.
The cleavage plane of rock salt is legendary.
And it's nice, really, to spend an afternoon splitting it up just to see how neat it falls apart.
OK, how are we going to denote rational planes?
Talked about rational and irrational directions, and that's easy.
How are we going to describe rational planes in crystals?
And this is rather bizarre.
So let me tell you what you do, which is something you've probably heard before, and then explain why one does it this way.
Let's suppose we have three vectors that are the vectors we will use to define the lattice.
And if there is a plane hanging on one of these lattice points, everything is translationally periodic, so there must be a similar plane in the same orientation that passes through all the lattice points.
So let me look at a very special plane, namely one that cuts a lattice point at an integral number of translations, T3, and an integral number of translations, T2, and an integral number of translations, T1.
And I don't want any common factor here, so let's look at this point here.
And now, let's connect these lattice points together.
And that will define a plane uniquely.
And what is special about this point is that it hits a lattice point on each of these three axes.
And this is a plane that I'll call the intercept plane.
There will be a similar plane hanging on all of these other lattice points.
But those planes will not hit a lattice point on the directions of T1, T2, and T3.
This is the first one out from the origin that does that.
Now, we will use this oblique set of vectors, T1, T2, and T3, as the coordinate system for specifying direction and angles.
So let me ask a question now that's going to blow you away.
What is the equation for the locus of this plane If I use this oblique coordinate system based on T1, T2, and T3 as my coordinate system?
Holy mackerel.
What a way to wrap up a Tuesday.
Actually, it's very, very easy.
First of all, it has to be a linear equation.
So it's going to be linear in x, y, and z.
So something times x, something times y, something times z equals a constant.
That's going to define, if I use the proper coefficients, the equation for this plane.
And let me determine quite easily what those coefficients are.
What is the coordinate of this point, which sits on the plane and therefore must satisfy this equation?
Well, y is 0. z is zero.
So this is the point-- if I am a translations out from the origin, this is the point x equals A.
So if I'm three translations out from the origin, that integer is the coefficient in front of x because when y and z are 0, if I make the constants on the right hand side equal to 1, this is the point x equals a.
And it's nice to have one on the right hand side.
That's about as nice and neat and tidy a constant as you could have.
We do the same thing for this point at the end of two translations, T2.
let's say in general that for this intercept plane, we are two translations out along T2.
So for the point 0-- this point, 0-- if I put the number of translations, b, in front of y, This is the point y equals b when x and z are 0.
And in the same way, if we are C translations out along T3, C times z-- that integer-- is equal to 1
Can you explain it again?
[INAUDIBLE]
I'm sorry.
What?
[INAUDIBLE]
Yeah, I'm saying I'm going to [INAUDIBLE] my definition at a plane which cuts the direction of T1, which is what I called the variable x, at a translations out.
In this case, this is 3.
This is 2.
And this is 1.
So for this particular example, I would have 3x plus 2y plus 1z equals 1 because when y is 0 and z is 0, the point on the surface is-- 1/3.
Sorry.
Good question.
OK, that's a lot better.
I would've gotten in much deeper trouble if you had let me persist in that.
So thank you for the correction.
But now, I would like to have integers out in front here.
So let me multiply both sides through by A times B times C. And now, I do get something with integers in front of x and integers in front of y.
And this would be A times C, and integers in front of z.
And that's A times B equals-- and now on the right side, I would have a times B times C. So everything, every coefficient and the term on the right hand side are products of integers.
So they are all integers themselves.
OK, let me next ask the question these planes passing through all of the other lattice points that are contained within this little tetrahedron of volume are going to be equally spaced.
How will I claim that these planes are all equally spaced?
Well, that's easy just in general terms.
Here is a lattice point.
There's a plane passing through it.
There's some lattice point neighboring it that's out at distance, D. Here is another lattice point.
It has a plane passing through it.
There must be a plane out here on D. And I'm not going to be able to have the same environment for every lattice point unless all of these planes have the same spacing from one another.
So to say that they pass in lattice points in the environment of everybody lattice point is by definition identical is possible only if the planes are equally spaced.
Next question-- and I'm not going to be gutsy enough to try to demonstrate this in three dimensions, so let me do it in two.
Let us ask the question, how many planes are there hanging on all the lattice points between the origin and the intercept plane?
And let me demonstrate that for a two dimensional case.
And I will refer you to a drawing in Berger's book because this is not convincing unless you do it absolutely exactly.
And to do that in front of a live audience with chalk is not always the easiest thing in the world.
So here is T1.
Here is T2.
And let me look at this plane.
This is the intercept plane.
A is equal to 3 and B is equal to 2.
Now, what I'm going to do is to use plus and minus the translation T1 to repeat this intercept plane.
OK, I go minus T1.
And that's going to give me a plane in here.
I'm going to go minus T1 again.
And that's going to give me a plane in here.
So I have split the interval between the origin and the intercept plane into A parts.
OK so far?
Let me now use the translation T2.
And I will use the plus and minus the translation T2 to take this stack of A planes and repeat it B times.
And that's going to take the plane at the origin and move it up to here.
It's going to move the planes into stacks like this.
I will map the stack of A planes B times.
And that is going to give me A times B intervals except if a special condition applies.
And then, I will go back to it in a moment.
This is the equation of the intercept plane.
I will argue that the three dimensional analogy of what I've done is that there will be, if I let the translations T1, T2, and T3 go to work, there will be A times B times C intervals between the origin and the intercept plane.
So if this is the intercept plane and this is A times B times C times the spacing between the planes out from the origin, then the first plane from the origin is going to happen to intercept which is 1ABC to the first.
And ABC divided by ABC turns out to be 1 for any value of A, B, and C. So the first plane from the origin is going to have the simple equation be BC times x plus AC times y plus AB times z equals 1.
And the second plane out from the origin will have the same coefficients except the term on the left will be 2.
Third plane will have 3 on the right and finally the integer ABC when we get to the intercept plane.
OK, now I'm ready to make a momentous definition just as we get five of the hour.
B times C is an integer.
Let me define that integer by a single integer.
And just for the heck of it, I'll call it H. A times C is an integer.
Let me call that integer K. And AB is an integer.
Let me call that integer L. So using T1 and T2 and T3 as the basis vectors of my coordinate system, the first plane from the origin has the equation Hx plus Ky plus Lz equals 1.
And these three integers, H, K, and L, are said to be the Miller indices of this plane
Yes, sir?
Just on that figure, [INAUDIBLE] connect to the lattice points.
I'm trying to make sense of why did you put the planes in between them?
OK, what I did was I started with just these three hanging on the lattice point separated by T1.
And then, I move this by T2.
OK, so T2 has a plane in the middle of it.
And it's going to move that up to here and give me another plane.
If you just draw this out-- and you're going to get a problem that asks you to do this-- since these integers are mutually prime, the second translation is going to interleave planes between the first set.
And it gets even worse in three dimensions because the third translation, which is the axis C translations out, is going to interleave that set of AB planes C times.
And they have to be at equal distances.
Otherwise, the definition of the lattice point having identical environment is violated.
OK, these are the infamous Miller indices that are the downfall of all freshman who take 309.1.
It is not necessarily a very straightforward definition.
But the advantage of it is that it lets you analytically describe the locus of the lattice planes in a very, very convenient way.
Actually, let me tell you something maybe you have not heard before.
These indices were not invented by a fellow named Miller who was an Englishman who wrote a book, an early book on crystallography that incorporated this notation.
The guy who invented them and first proposed them was Auguste Bravais, a very famous French crystallographer-- mathematician, actually-- who gave his name to the three dimensional space lattices.
These are universally known as the Bravais lattices, 14 of them.
It's a great name.
It's music.
it rolls of the tongue-- Bravis, Bravais.
Actually, Bravias wasn't the first guy to try to derive them.
The first guy who tried to drive the lattices was somebody named Frankenheim.
He blew it.
He didn't get 14.
He only found 13.
So there's a moral here.
If immortality is your game, get it right the first time because everybody has heard of Bravais, Bravais.
Nobody has heard of Frankenheim.
And for that, I say thank god.
What a name!
It makes you think of somebody who's got plugs in his neck and walks around like this and has a green face.
Frankenheim-- bah!
Bravais, Bravais, it really sings to you.
OK, I'm getting silly so it's time to quit and take a break.
Come up and say hello to Rodney.
All right.
The end of this lecture generated so much excitement that we've continued on into the start of the next segment.
So I think we'd better begin.
What I told you, as several people were clever enough to observe, that the number of planes between the origin and intercept plane is equal to A times B times C, where the intercept is A T1 out plus B T2 out equals C T3 out.
That is true only if the numbers A, B, and C are mutually prime.
If it's something like 4, 2, 1, you won't get that number of planes.
You'll get a submultiple.
And the reason for that is that some of the planes, when you generate them by these three different translations in different directions, some of them are mapped on top of existing planes.
And I'm not going to attempt to prove this, but let me just tell you the result, that if A and B contain some common factor, p, and B and C contain some common factor, q, and in the worst possible case, A and C contains some common factor, r, then the number of intervals is equal to A times B times C divided by p times q times r. And probably the best way to show that is let you show, on a problem set in the very near future, to actually map out the planes in the two-dimensional situation, for a and b that are mutually prime, and then do the same thing for a and b that contain some common factor.
So this is the so-called Miller indices, the Miller-Bravais indices, and everybody hears about them, but it is not at all clear why one takes this very indirect and non-intuitive way of defining the orientation of a plane.
And the reason for this, ultimately, is that in x-ray diffraction, you want to integrate up the scattering of all atoms in a crystal.
So you want to integrate over a plane that contains the atoms, and this locus has a convenient form only if you choose these indices to find the orientation of the plane.
And what comes out of this, as I said to a couple of people during intermission, is that the amplitude of the scattered wave has some summation over all of the atoms.
And in that phase of that ray, that is e to the plus 2 pi i, hx plus ky plus lz.
Very, very simple form.
And it has that form only because the locus, the way this plane is defined was done very, very carefully.
So it's something beyond just simple geometry that forms a reason for it.
Let me say a couple more things in general about the Miller-Bravais indices.
They are very, very intuitive, counter-intuitive.
Let's suppose that this is a, and this is b, and this is c. And I put down here, again, the equations of the entire set of planes in the stack.
And let's look at the first planes and ask what the intercepts are.
The intercepts on x, y, and z are going to be 1 over h on x, 1 over k on y, and 1 over l times z.
So the first plane from the origin is going to be 1 over h of the a translation, 1 over k of the b translation, and 1 over l of the c translation.
So this gives us one way of determining, given a particular rational plane, show it relative to the three translations.
From the first plane from the origin, h is going to be 1 over the intercept on T1.
I'll call it a, which is what we usually do.
k is going to be 1 over the intercept on b, and l is going to be 1 over the intercept on c for the first plane from the origin.
So if you have a drawing of the set of planes relative to the translation, such as the one that I obliterated from just a moment ago, reciprocals of the intercepts of the first plane from the origin give you h, k, and l very easily.
The thing that is counter-intuitive is that, if you look out of context at the three indices, h, k, and l, you have the feeling that as l gets larger, the intercept of the plane on c should gradually creep up from here, to here, to here, and so on, as l gets larger.
Actually, the reverse is true.
As l increases, the intercept on c gets smaller and smaller and smaller and smaller, until finally, when l is infinity, the intercept is 0.
So it's counter-intuitive.
The intercept on the axis gets larger as the Miller index gets larger.
Let me give you a proprietary Uncle Bernie's secret method for determining Miller indices without stress and strain.
This works every time.
Find the intercepts of any plane.
And we know from these equations what those are going to be.
For the nth plane, the intercepts are going to be n over h, n over k on b, and n over l on c. Take the reciprocals.
And the reciprocals are going to be h over n, k over n, and l over n. And then find whatever integer, n, you must multiply those reciprocals by in order to convert them to mutually prime intercepts, and you've got the Miller indices in a no-brainer, automatic fashion.
Another bit of jargon.
We will also use the basis vectors to define locations and to define directions in a lattice.
To distinguish three integers that indicate a plane, and you all know this from previous experience, the parentheses says plane, an individual plane.
There are other times when you would like to use the indices of one representative plane to indicate the entire set of planes that are related by the symmetry of the crystal.
So let me give one specific example.
Suppose we had a cubic crystal, where these are the three translations that form the edge of the unit cell.
And let's say that this is a1, this is a2, and this is a3.
I'm using a1, a2, and a3, not a, b, and c. But let's not go there.
OK, these are lattice points.
And if I look at this plane that's perpendicular to a1 and cuts it at one lattice point, l, this would be the 1, 0, 0 plane.
And I would indicate that that is an individual plane that we're talking about.
But if this crystal is cubic, all six faces of the set, 1, 0, 0; 0, 1, 0; 0, 0, 1; and then the plane that's in the opposite direction, minus 1, 0, 0; 0, minus 1, 0; and 0, 0, minus 1.
These six planes define the surfaces of the cube.
So there are times, since these are equivalent to planes and will have the same properties, the same hardness, the same [?
edge ?]
bits and everything else, there are times when we may want to refer to the entire symmetrical set.
And we do that.
We call that set a form.
It is a symmetry-related set.
And you can pick any one you like.
Generally, you pick one with simple indices, like the one that is perpendicular to the translation a.
And then you indicate by braces, which is the symbol for set.
This means the set 1, 0, 0, and that would correspond to all six of these faces.
The number, if you do this in reverse, and I'm going to give you the indices, what are the separate faces?
That depends on the symmetry.
So to give you an example, if I have a unit cell that has three orthogonal translations, and this is a, and this is b, and this is c, so that there is no cubic symmetry to it.
This face, which would be 1, 0, 0, and the one that cuts the axes at minus a, minus 1, 0 0, those are going to be the only faces with a single 1 and a pair of 0's which are equivalent by symmetry.
So in this case, for this brick shaped unit cell, 1, 0, 0 would mean this pair of two faces, rather than six.
So to go from a representative face to the set of planes, you have to know the symmetry.
If you can write the planes on the basis of symmetry, take one representative and let that be the form.
So it leads one to observe that crystals undergoing plastic deformation have bad form.
Yeah?
Is that different from when you talk about a direction?
Yes, yes.
You will see in just a moment that we use the same system of three integers, but in order to distinguish which one he's talking about when one's thinking about integers with a different set of numbers.
Again, if we use the lattice as the basis of a coordinate system, another thing we might want to specify is the orientation of a rational direction.
And let me go to a two-dimensional case.
Assume that c goes straight up.
Directions of easy plastic deformation will very often involve integral number of translations.
For example, this direction here is obtained by going one translation in the a direction plus one translation in the b direction.
So we could use the pair of integers 1, 1, and if we are normal to see we'd use a 0 for the third integer.
But how are we going to tell whether we're talking about a direction or a plane?
And the way one does that is to use square braces to indicate one particular direction.
You may want to also indicate the location of a particular atom within the unit cell.
In order to do that, one could really use the location of a vector, the components of a vector, xa plus yb plus zc, the vector from the origin to that atom as coefficients which correspond to the atom location.
So if that is the case, the numbers are usually not integers.
And one simply uses the three integers in an unadorned fashion.
So the lattice translations give you the basis of a coordinate system, and we use those to specify features, rational or not, of planes and directions and coordinates.
OK, any questions here?
OK, we are next going to take the first small step in a process of synthesis.
We have found that there are, in two dimensions, five distinct kinds of lattices.
The oblique, the rectangular, the centric rectangular, the square, and the hexagonal.
Five kinds of lattices.
We've shown, to this point, that each of these lattices can accommodate a certain rotational symmetry.
1 or 2 here.
m and m for the rectangular and centric rectangular.
4, the fourfold axis for a square.
3 or 6 for the hexagonal lattice.
So what I would like to do next, after we make a brief direct diversion, is to combine with each of these nets one of the two symmetries which can be accommodated in that net.
And what we will have determined then are the patterns' lattice and symmetry, which can exist in two dimensions.
The sort of things we look at actually, like the pattern of square tiles on the floor, or the two-dimensional pattern that is on a fabric design.
So we'll derive those exhaustively.
The thing that results is something that is called a plane group.
And I'm going to take this through in detail because there are not that many combinations that can be made.
And the number is enough to count on your fingers and toes.
You've got to use both, so they're a fair number, but not a staggering number.
If we were to do this exercise in three dimensions, there would be 230 distinct combinations.
And that's too much for anybody to go through.
I know of very few people who work professionally in this area who've actually sat down and derived every single one.
If you know the principles of the derivation and know how you could go about doing it if you were forced to, that's sufficient.
The results are tabulated in an admirable compilation.
We mentioned that at the beginning.
This is the International Tables for X-ray Crystallography, Volume 1.
So what we should end up with is a way for you to go to this reference data and know how to interpret it and make use of it.
But this is a nice microcosm.
There are not many plane groups, and to do this exhaustively, will let us see the sort of reasoning that goes into it.
I have to tell you that this is not in Buerger's book.
We're going to do two-dimensional crystallography first, and then extend it in a third direction.
In fact, I know of no book that does this.
And so you're going to get a unique, handwritten document by none other than yours truly that will do this in detail every step of the way.
And I don't know of any other place where this is done.
Now, what is the origin of this name?
Plane, well, these are symmetries confined to a plane.
And then there's this mysterious term, group.
And the corresponding thing in 3D is something called a space group.
This is a language derived from a branch of mathematics that is called group theory, and it is exactly the language that describes things that we've mentioned conversationally, such as if you take two transformations and perform them in succession, they're going to give you a third that has to be present.
If you've got two present, there must be a third.
We talk about symmetries being self consistent.
If you have a rotation operation that is not an integral submultiple of 2 pi, it's not going to be self consistent because you go around, and around, and around, and you will never come exactly full circle.
So group theory is a branch of mathematics that deals with concepts of this sort.
So it's not very hard to develop, and it gives its ideas to the notation that is used in crystallography.
So let me say a few words about group theory.
What is a group?
A group is a set of elements-- And what are elements?
Elements are things-- For which a law of combination is defined.
And the nature of the elements can be very, very different things.
They could be numbers.
They could be complex numbers, and the law of combination here might be multiplication.
They could be matrices, for which the law of combination is defined as matrix multiplication.
Or they could be pots of different color paint, for which the law of combination is defined as mixing the two colors 50-50.
I don't know any practical application of that sort of group, but nevertheless, it follows the definition, a set of things for which a law of combination is defined.
And in addition, which satisfies three so-called group postulates.
And the group postulates are the combination of any two elements is also a member of the group.
The second group postulate is that the identity-- I'm going to call it just identity-- and let me define, identity is doing nothing-- Is also a member of the group.
And if we define the identity operation as I, this means that for every element, I followed by a or a followed by I, is the same as a by itself.
If the law of multiplication were just arithmetic multiplication, then number 1 would be the identity operation because 1 times any number gives you the number.
Any number times 1 gives you the same number back again.
And the third postulate is that, for every element, an inverse exists-- let's call the elements a and a minus 1-- such that a followed by a minus 1 is equal to the identity operation, and the inverse element followed by the original element, a, is also the identity operation.
So if the law of combination were defined as multiplication, you would have to find, for every number, another number, which multiplied by it, gives you the number 1.
OK, let me now illustrate with a couple of simple examples.
Let's consider the set of numbers 1, n minus 1.
The number of elements in the group is said to be the rank of the group, or some people like to use the term order.
And I don't like that term because second order or fourth order is a term applied to quantities which are not terribly important.
So we expand this except for higher order terms, which are negligible.
And I don't like that connotation.
So I will not use order, although it's sometimes used to talk about the rank of a group.
The rank or order of the group is simply the number of elements contained in a set.
So let's show that the numbers 1 and minus 1 constitute a group of rank two.
Is the combination of any pair of elements also a member of the group?
Well, to do this, it's convenient to set up an array called the group multiplication table.
And what you do is you simply put the elements of the group along a row.
And in this case, it's 1 and minus 1.
And then you put the same elements along a column, and then you combine them.
1 times 1 is 1.
Minus 1 times 1 is minus 1.
1 times minus 1 is minus 1.
Minus 1 times minus 1 is plus 1.
Sometimes, as I'm giving this part of the lecture, somebody passes down the hallway, and they hear 1 times 1 is 1.
1 times minus 1-- and they go in reverse to see what in the world could be going on in this high-powered mathematical lecture.
That's why I close the door.
OK, so any combination of elements is also a member of the group.
Is the identity operation present?
Yes because minus 1 times plus 1 or plus 1 times minus 1 gives you the same element back again.
So for every element in the group, the combination of any two elements is present.
For each operation, we've shown that an inverse exists and for every element there's an identity operation, and the inverse exists.
So the pair of numbers, not terribly exciting, 1 and minus 1 where the law of combination is defined is a group of rank 2.
I'll give you another example.
I won't bother to carry out all the terms, but the numbers 1, minus 1, i, and minus i, where i is equal to the square root of minus 1, and where the law of combination is defined as multiplication, constitute a group of rank 4.
With this simple introduction, I think one can see that symmetry transformations can be regarded as operations that are elements in a group.
And let me give you a simple example.
Let's look at a combination.
We still haven't considered the mind-boggling possibility of having more than one symmetry element present in a space at the same time.
But let's suppose we have two mirror planes that are completely orthogonal, and see what sort of pattern they would generate.
Here's a first motif.
If I call this mirror plane m1, and this mirror plane m2.
m1 is going to take this object and reflect it across to here.
m2 will take this object and reflect it down to here.
Take this one, and reflect it down to here.
And now I know how this is related to this, how this is related to this, how this is related to this.
But this one and this one are exactly the same thing.
And those two objects are not related by reflection.
If this is a right-handed one, this is a left-handed one, this is a left-handed one, and this is a right-handed one.
So if something relates these two, it has to be something that does not produce an enantiomorph.
The only thing that's possible then would be translation.
And these two guys are not parallel to one another.
You can't get this one by just sliding this parallel to itself.
The only thing that's left is rotation.
And, lo and behold, we can get from this one to this one by a 180-degree rotation.
And we can get from this one to this one by a 180-degree rotation.
So here's an example of a way in which we've combined without showing why this is possible, or how I got there.
Here's a combination of symmetry elements that gives us a possible arrangement of objects in space.
It's going to be convenient to have a notation to indicate such combinations.
And we call them Harry, George, and Sam, or some popular, affectionate name.
Or call them number 1, number 2, number 3.
But a nice notation is going to be a descriptive one which tells you what you have.
And what is done in crystallography is to indicate these possible combinations by a running list of the different kinds of symmetry elements that are present.
And here, we've got a twofold axis, for which the symbol is 2, and two mirror planes that are different and function in different ways in the pattern.
So this combination is called 2mm.
OK, that's getting ahead of our story.
But what I wanted to do now is to show you that this set of symmetry elements contains four operations that constitute a group.
There's the operation of a one-fold axis.
That's the identity operation.
There is a reflection across the first type of mirror plane.
And notice, now, the utility of having a symbol that indicates a specific operation rather than a symmetry element.
There's a second reflection operation, sigma 2, and there is a rotation operation through a 180 degrees, A pi.
So I'd like to show that these four operations constitute a group of rank 4.
So how do we show this?
We first set up the group multiplication table.
And then we simply combine these four operations pairwise.
Put the same 1, sigma 1, sigma 2, and A pi down this side of the array.
Doing nothing, doing nothing, is the same as a one-fold axis doing nothing.
Doing a reflection sigma 1 followed by doing nothing gives you sigma 1 back again.
Doing a reflection sigma 2 followed by doing nothing gives you sigma 2 back again.
And similarly, rotating A pi followed by doing nothing is the same.
If I go along the columns, doing nothing followed by sigma 1 is sigma 1.
Sigma 1 and 1 combined is sigma 2.
A pi combined with sigma 1 is A pi.
So far so good, but not surprising.
Doing sigma 1 and following it by sigma 1 is reflecting left to right across mirror plane 1, and then reflecting right back again.
So that is going to be a one-fold axis doing nothing.
Doing sigma 1 following up by sigma 2 would reflect across, and then follow up by reflection in the second mirror plane.
I get from the first one to the final one by the rotation A pi.
So this is A pi.
And if I do the first reflection, sigma 1, and follow that by A pi, that gets me from number 1 to number 2 to number 3.
And I get from number 1 to number 3 directly by the reflection operation, sigma 2.
OK, you notice what's happening?
I get the same four elements back again in every column or row.
Sigma 1, 1, A pi, sigma 2.
And if I rattle these off quickly, this as A pi.
This is identity, and this is sigma 1.
And this will be sigma 2.
This will be sigma 1.
And this will be the identity operation.
So the first group postulate is satisfied.
These four elements combined pairwise always give you nothing but one of these elements back.
An identity operation is present because the one-fold rotation axis is the same as leaving the thing alone.
And we've seen that for every operation, an inverse exists because 1 occurs once in each row and column.
So sigma 2 is its own inverse, and for sigma 1, sigma 1 is its own inverse.
For A pi, A pi is its own inverse.
So an inverse exists.
So we've got a group.
And this is another way of saying, in general terms, if we put two mirror planes and a twofold axis together in this specific fashion, these operations, when they go to work on an initial motif, reproduce a finite set of objects and not a clutter that just fills space and never closes upon itself.
Yes, sir?
I see why you have sigma 1, sigma 2, A pi in that group, but why do you have the 1, which is a conversion?
OK.
If I didn't put it in-- no, that's the identity operation.
This is a one-fold axis.
Just this 2 was a twofold axis.
So one-fold axis is a nice symbol for identity because it doesn't do anything.
It picks something up and puts it right back down where it came from.
OK. Yeah.
That's deceptive because 1, when we're talking about multiplication of numbers, also functions as the identity operation.
Yeah?
If all these symmetries are commutative, [?
that is, ?]
if they all commute, do you really [INAUDIBLE] bottom diagonal [INAUDIBLE]?
In general, yes
Generally?
Yeah.
Actually, this is a special group because if I have two elements, a, and take b and follow it by a, that is the same result as a followed by b.
And a group that has this character is said to be abelian.
There's a standing rotten joke in mathematics that asks rhetorically, what is purple and commutes?
The answer is an abelian grape.
I don't think it's terribly funny either, but if you had a gaggle of mathematicians here, they would chortle and double over in laughter.
Ha, ha, ha, abelian grape.
Well, I don't feel badly.
You react about the same way to some of my funny things, too.
But you get used to that sort of thing.
All right.
The feeling I want to leave you with at this point is the idea that group theory and some of its concepts are exactly what we're meaning when we grope for a definition of the fact that a certain collection of symmetry operation should be finite.
That is to say, it's finite.
The operations combined on each other have to give a finite number, must constitute a group.
And the main reason for being familiar with this is that group theory forms the basis of a lot of the words that are used to describe the combinations of symmetry elements.
For example, what we have derived here, the symmetry that's called 2mm, is something that's called a point group.
Point because there is at least one point in space that's left unchanged, and that's this point of intersection of all three symmetry elements.
And it's called a group because the individual operations of these symmetry elements, when combined according to group theory, can be shown to constitute a group.
The patterns that we get when we add symmetry elements to a lattice are groups in the sense, if you regard the pattern as extending infinitely in all directions, that the combination of any two elements gives you something that's also in the group.
But the number of elements is infinite.
For example, in the lattice, here's T1.
Here's T2.
Every other lattice point can be described as some combination of T1 followed by T2.
But the numbers in front of those translations can be infinite.
If we put a symmetry element in this lattice, like a twofold axis, for example, the translations reproduce this twofold axis to an infinite number of locations.
But yet, all of the group postulates are satisfied.
The number of elements in the group does, however, not have to be finite.
So that is another distinction that's worth making, that there are infinite groups and there are finite groups.
And the number of elements can be infinite.
Collection of symmetry elements that leave one point in space unmoved is called a point group.
If we do something like this, put a twofold axis in combination with a pair of translations, that is an infinite set of operations that acts on all of space.
This is something that's called a space group.
Another designation, distinction, that's sometimes useful to make are crystallographic point groups, such as this one, 2mm, as opposed to combinations of symmetry elements, which are perfectly lovely and which are valid groups, such as a combination of rotations of 2 pi over 5, or a combination of a fivefold axis with a mirror plane.
Perfectly lovely symbols, constitute groups, but these are non-crystallographic These would be non-crystallographic point groups.
Satisfy all the requirements of the group, but if it's to be in a crystal, the fivefold axis must be combined with translation, and that's impossible.
So you can have non-crystallographic point groups.
There are no non-crystallographic space groups because they are, by definition, something that involve translation yet constitute groups.
All right.
Set the stage for next time, so you all come back excited.
We have shown that there are a limited number of symmetry elements that are possible in a lattice, rotation 1, 2, 3, 4, and 6 in a mirror plane.
And these are now examples of what we would call crystallographic point groups.
These would be the sort of symmetry elements that are candidates for two-dimensional patterns.
But are these the only symmetries that can be added to lattices?
Could we not combine a mirror plane with a twofold axis?
The answer I say to that is, sure.
You bet you can.
You just did it for us.
But could you combine a mirror plane with a threefold axis, a mirror plane with a fourfold axis, a mirror plane with a sixfold axis?
The answer to all of those questions are yes.
So there are going to be spaces here, where there will be additional groups.
And when we're done, we will have the two-dimensional crystallographic point groups.
But now I have to be able to complete something.
If I take A pi and combine it with a mirror operation sigma 1, what do I get?
If I take A 2 pi over 3, a threefold rotation, combine that with a mirror plane that passes through it, what do I get?
Again, a characteristic of symmetry theory is that whenever I take a motif and repeat it by operation number 1 to get a second motif, and then I repeat that by operation number 2, I have three things that are identical.
And there must be some way, some operation, that automatically arises that tells me how number 1 is related to number 2.
Number 1 is related to number 3.
Tied in with group theory, this says that if you combine operation 1 in a space, if it's to be a group, if operation number 2 is present, you must combine those two steps, and whatever pops up must be a member of the set of operations that are present.
So we're going to have to come up with something that I call combination theorems.
And this is simply expressing the result of a combination of two elements in a group multiplication table.
It goes with a caveat.
If you're so excited about this stuff, you're going to run home and start reading Buerger's book as soon as you get there.
You can regard these symmetry transformations as operators.
And you're familiar with other sorts of operators like d by dx, d by dy.
And when you apply them in succession, you actually write it as a product.
d by dy followed by d by dx you write as d squared dxdy.
Or you write it as d squared dx squared when you differentiate twice.
You don't mean you take the differential of the function with respect to x and then square it.
You mean you do this operation twice.
So these are all operators.
And what we usually understand is that if you do d by dy first and then follow that with d by dx, the sequence of operation goes from right to left.
And I think that will come right quite naturally to you.
You do that with differentials and other sort of operators.
That is what we'll do.
Not Martin Buerger.
A very strong-minded person, and he did what he thought was sensible.
And he said we read things from left to right.
We read everything from left to right.
So I will be damned if I'm going to write d by dx, d by dy to be a sequence of operations that goes from right to left.
So all throughout Buerger's book-- and it's easy to get adjusted to, if he writes A dot B, he means B followed by A.
Whereas normally, in the calculus of operations, we will use, as the rest of the world does, this is A followed by B.
Trivial point, but one that can throw you for a loop the first time you see it in Buerger's book if you follow it.
Distinction does not matter if the group is abelian.
So that the order of operations doesn't make any difference.
OK, that's enough for one day.
Beginning next time, know we will start to begin this process of synthesis.
And the first thing we'll ask is, how do you combine a rotation operation with a reflection operation?
See you next week.
Having looked at the quizzes in a preliminary fashion, I come to the conclusion that some notes on tensors would be useful for the remainder of the term.
So at great pain and personal sacrifice I will endeavor to [INAUDIBLE] All right, let me remind you of where we were a week ago-- before a slight unpleasantness intervened-- and we had just begun to look at the properties of a very useful surface, the representation quadric.
And we said that to define it we would take the elements of a second rank tensor, something of the form A11, A12, A13, A21, A22, A23, A31, A32, A33.
And we'd use those elements as coefficients in a second rank, a quadratic equation, of the form Aij Xi Xj equals 1.
OK, so the Xi and Xj are the coordinates in our three-dimensional space.
And the set of coordinates that satisfy this equation-- setting the right hand side equal to a constant-- will define some surface in space.
And it's going to be a surface that is a quadratic form, so it is going to be one of four different surfaces that can be defined by such an equation.
One would be an ellipsoid, and in general this would be an ellipsoid oriented in an arbitrary fashion with respect to the reference axes.
And it would have a general shape, so the three principle axes of the ellipsoid would be of different lengths.
Then we saw we could also get a hyperboloid of one sheet.
One sheet means one surface-- continuous surface.
This was the surface that had an hourglass like shape that pinched down in the middle, and the cross-section, in general, would be an ellipsoid.
And the asymptotes to this surface would define a boundary between directions in which the radius was real, which meant a positive value of the property, and a direction in which the radius was imaginary, even though that may boggle the mind.
And if you take an imaginary quantity and square it, you're going to get a negative number.
So this surface would describe a property that was positive in some directions and negative in magnitude over another range of directions.
Then the third surface was a hyperboloid of two sheets.
And this was a surface that looked like two [INAUDIBLE] with an elliptical cross-section that were nose to nose, but not touching the origin.
And in this range of directions between the asymptote to those two lobes the radius was imaginary, the property negative and then there were a range of directions that would intersect the surface of either of these two sheets in there in those directions the property would be positive.
Then the fourth one-- which is encountered only rarely, but as we pointed out does exist in the case of thermal expansion as a property-- and this would be an imaginary ellipsoid.
This would be an ellipsoid in which the distance from the origin out to the surface was everywhere imaginary.
And this would be a property then that was negative in all directions in space.
And that doesn't seem to be a physically realizable thing, we pointed out that for the linear thermal expansion coefficient there are a small number of very unusual materials that when you heat them up contract uniformly in all directions.
Very remarkable property, but that does indeed, thankfully, provide me with an example of an imaginary ellipsoid quadric that does represent, indeed, a realizable physical property.
OK, this surface, we said, had two rather remarkable properties.
The first property was that if we looked at the radius from the origin out to the surface of the quadric, a property of the quadric is that the radius out in some direction-- that would be defined in terms of the three direction cosines-- the radius has the property that it is given by 1 over the square root of the value of the property in that direction.
Or, alternatively, the value of the property in the direction is 1 over the magnitude of the radius squared.
OK, so this is what gives us the meaning of the imaginary radii, the imaginary radii 1 squared would give you a negative property.
But again, I point out, it's obvious here-- but we have to keep it in mind-- that the value of the property and the magnitude of the radius are related in a reciprocal fashion, not only reciprocal, but reciprocal of the square.
So if this is the quadric A, a polar quad of the value of the property A as a function of direction would have its maximum value along the minimum principal axis and, correspondingly, the minimum value of the property along the maximum radius of the quadric.
So that is sort of counter-intuitive, you think big radius, big value of the property, no, goes not only as the reciprocal, but as the reciprocal of the square.
So the value of the property with direction is not a quadratic form any longer, it's based on a quadratic equation, but when you square the radius it's no longer quadratic.
Looks as though we have the value of the property as a function of direction, but looks as though we've lost information about the direction of the resulting vector.
The direction that we're talking about is always the direction in which we're applying the generalized force, the electric field, or the temperature gradient, or the magnetic field.
But the direction of the thing that happens is defined by the basic equation that gives us the tensor.
But as I demonstrated with you last time, that information is in the quadric also, and this is the so-called radius-normal property.
Which doesn't involve exactly what it sounds like, but what it tells us to do is a way of determining the direction of what happens is to go out in a particular direction, along some radius that will intersect the surface of the quadric at some point.
And then at the point where the directional vector emerges, construct a vector that is normal to the surface.
And so if this then were the direction of an applied electric field, the direction to the-- normal to the surface of the quadric where that direction intersected-- the surface of the representation quadric-- this would be in the case of conductivity the direction in which the current flows.
So everything that you care to know about the anisotropy of the property that's described by a given tensor, and about the direction of what happens-- the generalized displacement as you apply a vector-- is contained within the quadric.
But one caveat, this holds only if the tensor is symmetric.
And I think this is where we finished up just before the quiz, only if Aij equals Aji does the radius-normal property work.
OK, any comments or questions on this?
OK, let me now turn to a practical question of interpretation.
If you were indeed to measure a physical property as a function of direction, and you get a tensor that is a general tensor, A11, A12, A13.
A21, A22, A23.
A31, A32, A33.
All the numbers are non-zero.
You know that the diagonal values in the tensor are going to give you the value of the property along X1.
Or, in general, the diagonal terms give you the value of the property along Xi.
Aii gives you the property along Xi.
You have not the foggiest idea, if the quadric has a general orientation, what the maximum and minimum values of the property might be, and these might concern you.
Given the tensor that you've determined, what are the largest values, what is the largest value of the property, what is the smallest value of the property?
And if the crystal has any symmetry these directions might be the direction of symmetry axes in the crystal.
So for many reasons you might want to know the maximum and minimum value of the property.
And then for some applications for a real chunk of crystal that you've examined relative to an arbitrary set of axes, you might want to ask how should I orient that crystal so that I can cut a rod from it that has, let's say, the maximum or minimum thermal conductivity.
If you're using a piece of ceramic let's say, as a probe into a furnace, you wouldn't want that probe to conduct much heat, so you'd like the minimum thermal expansion coefficient, for example.
If you were making a window out of a transparent single crystal, it'd be a very small window, but you might want the smallest thermal conductivity normal to the surface, and therefore you might want to make your single crystal window oriented in a particular direction.
So I hope I've convinced you with enough straw questions that we can knock down easily that, yes, this would be an interesting thing to know.
So how can we do this?
Let me show you some simple geometry that let's us set up an equation for finding these directions right away.
And it'll be based on the following observation, that when the direction of, let's say, the applied electric field is a long one of the principal axes, then and only then is the direction of the generalized displacement exactly parallel to the radius vector.
We go off in any other direction other than a principal axis, the generalized displacement is not parallel to the radius vector out to the surface of the quadric.
We look at the equation of the property A11 X1 plus A12 X2 plus A13 X3.
This is the X1 component of the generalized displacement.
Similarly, A21 X1 plus A22 X2 plus A23 X3 is going to be the X2 component.
And A31 X1 plus A32 X2 plus A33 times X3, guess what?
That's going to be the X3 component of the displacement.
Now we said that these X's give us the direction of the displacement as we let the coordinates of that point X1, X2, X3 roam around the surface of the quadric.
When we land at a point on the surface of the quadric which is where a principal axis emerges, then, and only then, is the resulting vector parallel to the applied vector.
And this means each component of the resulting vector has to be proportional to a vector out to the point at that location, X1, X2, X3.
So what I'm saying then is that when we are on a principal axis the X1 component of what happens is going to be proportional to X1.
And the X2 component of the vector that happens has to be proportional to X2, and this is the radial vector out to that point.
And, similarly, the X3 component of what happens has to be parallel to the coordinate X3, the vector out to the surface of the quadric along X3.
So let me make this an equation, now, by putting in a constant for the unknown proportionality constant.
And what I'll do, just for arbitrary reasons, is call that proportionality constant "lambda."
So this, then, is a set of equations that will give me, if I solve for X1, X2, and X3, the coordinates of one of the three points that sit out on the surface of the quadric at the location where a principal axis emerges.
So let me rearrange these equations to a form that I can solve.
And I'll make the equations homogeneous by bringing lambda X1 over to the left hand side of the equation, and let me point out that this is the same lambda in all three equations.
There's a proportionality constant between the radial vector out to the surface of the quadric and each component of the vector that results.
So my set of equations will be A11 minus lambda X1 plus A12 times X2 plus A13 times X3, and that's now equal to zero on the right.
Next equation would be A21 X1-- and notice I'm not combining A12 and A21 into numerically the same constant, representing it by the same quantity, I'm just leaving them be separate and independent at this point.
And then the middle term here is A22 minus lambda times X2 plus A23 times X3 equals 0.
And the third equation, you can anticipate how that turns out, it's A31 X1 plus A32 times X2 plus A33 minus lambda, and that's equal to zero.
So this, then, is a set of linear equations and they're called homogeneous equations.
And that has solved our problem for us, all we have to do is solve for X1, X2, X3.
Except that if the nine coefficients Aij are all arbitrary and lambda is a constant this set of equations has only one solution, and that solution is X1 equals X2 equals X3 equals 0, and that indeed will wipe out every one of these lines.
And that is not a very interesting solution.
The only case in which this is not the only solution is that if the condition that the determinant of the coefficients is equal to zero, then we can find a real set of X1, X2, X3.
So the condition that a solution exists is that A11 minus lambda A12 A13, and A21 A22 minus lambda A23, A31 A32 A33 minus lambda, this determinant has to be equal to zero.
All right, so we know how to expand the determinant, we'll do a little number crunching, and I'm not going to write it down explicitly, but we'll have a term A11 minus lambda and this will be among other terms in the expansion A22 minus lambda A23 A32 A33 minus lambda and then there'll be another determinant that involves this term A12 plus its cofactor plus A13 times its cofactor.
Unfortunately none of these terms are zero in general, so I'm going to have three terms here.
If I expand this term, I'm going to get something in A11 minus lambda A22 minus lambda times A33 minus lambda and a bunch of other terms that I won't bother to write down explicitly.
The only point I want to draw at this particular juncture is to note that this is a third rank equation.
In lambda.
It's a cubic equation.
And what this means is that there are going to be three roots.
And lets call them lambda 1, lambda 2, and lambda 3.
And that's the way it should be.
There are three principle axes, so we should have three values of the constant lambda that we could put into this equation that will let us solve explicitly for the coordinates X1, X2, and X3, which sits on the surface of the quadric where our principal axis pokes out.
Three different principal axes, so this should be three solutions to the third rank equation that we've defined here.
OK I think you've all seen this sort of problem before.
The lambdas are called characteristic values.
Or, on the basis that things sound more impressive in German, these are called the eigenvalues, meaning the same thing.
So using nothing more than the properties of the quadric and some linear algebra we have stumbled headlong into an eigenvalue problem, which is a well known sort of mathematical problem associated with a large number of physical situations and a lot to do with physical properties in particular.
OK, now I have a question for you.
How we going to solve this silly thing?
I know-- well the answer is I poke it into my laptop, and out comes the answer.
But that's not satisfying, we should know how to do this if our batteries were dead and we needed the answer in a hurry.
Well, I know how to solve a second rank question.
If I have equation of the form ax squared plus bx plus c equals 0, I know the solution to that.
It's etched indelibly in my memory and it says that x equals minus b plus or minus the square root of b squared minus 4ac all over 2a.
Impressed you, didn't I?
You know why I remember that?
I learned high school algebra from a throwback-- a teacher who was a throwback to the Elizabethan period.
You know how when people go to a remote island they suddenly find an example of a species that everyone thought was extinct?
Well, to secondary education [?
Ms. Dourier ?]
was a species that had long gone extinct, but happened to be preserved at the high school that I attended.
[?
Ms. Dourier ?]
was a very tall woman, ramrod straight, with a pile of snow white hair on the top of her head.
She invariably dressed in a long, black skirt, and a white blouse that had puffy sleeves, and a collar with frilly things on the top, and, invariably, a black ribbon around the frilly lace collar top.
Her mode of enlightened education was to have all of the students in the class have a notebook that was open.
And one hapless member of the class was called upon to solve a problem in real time, and as the student recited we all wrote down what the student was saying in our notebooks.
All the while [?
Ms. Dourier, ?]
holding a ruler like a swagger stick, strode up and down the aisles, and woe be unto the poor student who faltered slightly, let alone get something wrong.
And that would bring a slap of the ruler down on the desk and a sigh of disgust, and then, all right, you take it, Audrey.
And then Audrey would begin to recite until she screwed up and we would all copy in our notebooks.
That tyrant so terrorized and intimidated a bunch of kids in a redneck, working class high school that we really learned algebra.
So to this very day, as a reflex action, minus b plus or minus the square root of b squared minus 4ac all over 2a.
But I don't have the foggiest idea how to solve a third order equation until I look it up.
You probably never had to do it, so let me, for your general education and amusement, pass around a sheet that tells you how to solve a third rank equation.
OK the first step is to convert an equation in, let's say, y squared plus y cubed plus py squared plus qy plus r equals 0 to a so-called normal form where you get rid of one of the terms.
So if you make a substitution of variables and let y be replaced by x minus the coefficient p/3, then you get an equation that has a cubic term, linear term x, and a constant.
And the constants a and b are combinations of p and q and r in the original equation.
And then the solutions, not well known to be sure, are there's a first solution x is equal to a plus b, and then, something messy, x2 and x3 are minus 1/2 of a plus b plus or minus an imaginary term-- those are the second and third roots-- i times the square root of 3 over 2a minus b where A and B are complicated functions of a and b.
Capital A and capital B are functions of little a and little b.
And then if the coefficients in the original equation are real, then you get one real and two conjugated imaginary roots.
If this combination of terms b squared and a squared are greater than 0, it is b squared over 4 plus a cubed over 27 equals 0.
If it's that, then you get three real roots, two of which are equal.
In the most general case that we would be encountering here is b squared over 4 plus a cubed over 27 is less than 0, and then there are three real unequal roots.
Unfortunately, if b squared over 4 plus a cubed over 27 is negative that term appears inside of a square root sign in the solutions.
So the only case that we'd really be interested in doesn't work for the solution.
But fortunately there's another solution and that's given at the bottom of the page, and if you really wanted to solve a third rank equation by hand, these solutions would do it for you.
But I think you'd much prefer to have your computer solve the equations for you, but let me show you a way of doing this without any computer, without solving any equations.
And it is a very clever method of successive approximations that's based on the properties of the quadric.
So let's suppose we have a tensor, and that tensor has some set of coefficients Aij, all of which are non-zero.
And I'll assume although this will work for other quadratic surfaces that the quadric has the shape of an ellipsoid.
All right let me now pick some direction at random, actually I don't have to pick it at random, but I'll show you what the shrewd first guess would be.
So let's say we picked this direction.
And let us find the direction if this is-- let's do it in terms of current and conductivity.
Let's let this be the first guess for the applied field.
That's clearly not going to be one of the principal axes, and let this be the first resulting direction of current flow J1.
So what I'm finding then is a J sub i in terms of an Aij times an E sub J, and this is my first result, and this was my first assumption.
Let me now let the direction of J be taken as the direction of the applied field for a second guess.
And we could normalize to a unit vector, but we don't even have to do that.
So let's simply say that my second guess for the electric field that I hope will point along one principal axis is E2, and E2 I'll take as identical-- let's say proportional-- to J1 from my original guess.
So this then would be E2, the second guess.
And if I find the direction to the normal-- to the quadric in that direction, this would be my second result for the current flow J, I should have put this in parentheses to indicate that these are not components of E and J.
And you can see what's happening, look at where I am, I started out here after just one iteration I have defined this as the potential direction of one of the principal axes.
So I'm going to find E1-- the direction of the field-- E1 that comes from sigma ij times Ji, I'll take my second guess E2 either as identical to J1 and this is going to give me then a new value for my second iteration, this would be J2.
And this process is going to converge very, very rapidly on the shortest principal axis.
As you can see in two shots I'm pretty close to being parallel to this direction, which would be X2 in my illustration.
Now if I wanted to-- if I wanted to see if I was close to convergence-- this is going to be awkward because if I don't normalize the magnitude of the resulting vector, J is going to get larger and larger and larger.
So periodically I would want to normalize-- take the magnitude of J, divide that into the components, and then I'd have a unit vector.
If I wrote a computer program to do this, I would do the normalization each time to test convergence.
Now, this is my kind of procedure, because if I'm doing this by hand and I screw up and I make a mistake and my answer is thrown off a little bit, if I continue to iterate, the thing will proceed to continue to converge to the shortest principal axis.
So I can make a mistake and it will correct itself and come back again, and that's my kind of solution.
Yeah, Jason?
It's going to give you the minimum value of the property, right?
No.
It's going to give you the minimum principal axis, that's going to be the maximum value of the property.
So that's one out of three, hey, that's not bad.
What do I do for the others?
Let me tell you without proof to find the maximum principal axis.
And that would be the minimum value of the property.
What you would do is exactly the same procedure, and you would operate on not the tensor, but on-- using as a matrix of coefficients-- the matrix of the tensor inverted.
And I assume you know how to find the inverse of a matrix, except that you don't have to even find the inverse, the inverse is going to be a collection of functions of the original elements divided by the determinant.
You don't have to worry about normalizing, all that's important is the relative value of these coefficients.
Yes, sir?
If your tensor is symmetric, can't you say that they'll just [INAUDIBLE] on to another?
That's what you're going to do.
But when you know just one, that won't work.
The other two are floating around somewhere and you don't know in what direction.
So if you do this, then you have two out of the three and now very shrewdly, as you point out, if we have two of the principal axes and the tensor is symmetric, then we can automatically get the direction of the third.
If it's not symmetric, then you have to use the set of equations, and you have three principal axes as eigenvectors, as they're called in eigenvalue problems.
And they do not have to be orthogonal, but you have the components in a Cartesian coordinate system so you can find the angle between those axes.
OK?
Yes, sir
With respect to whether the matrix is symmetrical or is not symmetric, the quadric is always-- has three principal axes at right angles right?
Only if the tensor is symmetric.
Only if the tensor is symmetric.
And let me put your question aside for about five minutes, and I want to take an overview of what we've learned about the geometry of the quadric and the symmetry restrictions that we can impose on the tensor in a formal fashion, and see how they compare and how one allows an interpretation of the other.
So I'm going to answer your question, I'd like to wait for about three minutes.
Other questions?
OK, again this is all very symbolic, I'm not giving you an explicit answer, I can't do it.
But what I can do you-- do for you is, ha ha, do to you.
I almost slipped and said that, what I can't do to you, is give you a problem that asks you to solve for the principal axes, for example of a property tensor.
I will be merciful and perhaps not have all nine coefficients non-zero, I will have one of them zero, or maybe two of them zero.
So again to try this it is very instructive and it will cement what the individual steps that I performed here actually involve.
Now, a question that I thought you were going to ask about principal axes is that in order to do what I did here I am using the radius-normal property, and the radius-normal property only works for a symmetric tensor.
So sounds like I'm swindling you, except that if I use this procedure, the radius-normal won't actually be identically parallel to the generalized displacement.
But it's not going to be terribly different from it, and I'm going to get something that again will converge towards the shortest principal axis.
And once I'm close to the principal axis it is true-- symmetric or not-- that the only three directions before which the generalized displacement is parallel to the radius vector are the principal axes.
But anyway this is a dicey situation if the tensor is not symmetric, and it probably comes as a great relief to learn that there's really only one property tensor of second rank that's known for sure to be non-symmetric.
So in most cases-- 99 out of 100, if not more-- we will be dealing with property tensors that are symmetric, so we could use this procedure.
So again this property converges remarkably well, and as I say it has the admirable quality of being self-correcting if you make a mistake, if you're grinding this out by hand.
But a computer program that you write that just simply takes the direction of J, normalizes it to a unit vector, applies that as the generalized force, finds a quote "J" as a second iteration, normalizes that, and then you can check to whatever degree of convergence you wish to have in your solution.
And it's a simple matter to set up a program to do this.
All right, I think we still have some time, so let me now put everything together.
And look at what we've seen of the geometric properties of the quadric, and the tensor arrays that we found for different symmetries.
For triclinic crystals, for which the recommended procedure is to leave by the nearest exit and work on something different instead, you would have nine elements altogether that are necessary to define the property.
I'll discuss this in terms of tensors that gives you an ellipsoid.
The argument is not different for hyperboloids of one or two sheets, and we can't draw imaginary ellipsoids, but an ellipsoid is a much easier thing to draw.
OK, triclinic symmetries, either one or one bar, nine elements in the tensor.
No constraints whatsoever on the shape of the quadric or on its orientation relative to our coordinate system.
So, let's total up now looking at the quadric how many degrees of freedom there are in this situation.
We have three principal axes, so those are three degrees of freedom.
The orientation of the quadric is arbitrary, so there are three orientational degrees of freedom.
And that comes out to six.
Supposedly nine degrees of freedom, what are the other three?
Again, if this is a general tensor, which is non-symmetric, the eigenvectors-- the principal axes that you have to choose to squeeze this thing into a diagonal form-- become non-orthogonal.
OK?
So you have another three degrees of freedom that specify the mutual directions of the eigenvectors-- the angles between them.
OK?
The directions in which you have to pick your coordinate system, X1, X2, and X3, that force this thing into a diagonal form is going to involve a coordinate system in which these three angles are not 90 degrees and are fixed if you're going to squeeze this thing into a diagonal form.
And that's it.
So we add these three interaxial angles in, they're a total of nine degrees of freedom for a general non symmetric tensor.
To convince you that these interaxial angles for the eigenvectors really are variables, let me tell you something that we may prove later as a recreational exercise, or I may give it to you on a problem set, it's not difficult to prove.
And that is the result that a symmetric tensor remains symmetric for any arbitrary transformation of axes.
That is from one Cartesian coordinate system to another.
Now a very salutary effect of this is that the tensor has nine elements.
And if the tensor is symmetric-- if you want to transform that tensor-- you only have to do the off-diagonal terms, because whatever it turns into when you change the axes, these off-diagonal terms are going to be equal to the off-diagonal terms in the new tensor.
OK?
So this means that only six-- if the tensor was symmetric in one coordinate system, only six elements have to be transformed.
Actually, it's better than that, you don't have to do six, you only have to do five.
And this is the last diagonal term that can be obtained from the trace of the tensor.
And that would be the trace of the tensor T prime after transformation because A11 prime plus A22 prime plus A33 prime has to be equal to the original trace T, which was A11 plus A22 plus A33 in the original system.
So if the tensor is symmetric, you really have to crank through a transformation for only five of the nine terms.
Now what is the relevance of this to what I just said?
If we have the tensor diagonalized, to a new form A11, 0, 0, 0, A22 prime, 0, 0, 0, A33 prime, that is a special case of a symmetric tensor.
So if you decide to go from the coordinate system that produced this diagonalized form to some other coordinate system, the new tensor that you're going to get is going to be symmetric.
But the thing was not originally symmetric, so how can that be?
The answer is you can put it in diagonal form only in a non-Cartesian coordinate system.
And, therefore, that is evidence that the reference axes-- the principal axes-- cannot be orthogonal.
Otherwise you could take the diagonalized tensor and crank it back to some arbitrary Cartesian coordinate, and suddenly it would go to a symmetric tensor when it was not symmetric to begin with.
OK?
QED.
So you can diagonalize a general tensor only if you take the axes along the eigenvectors, which cannot be orthogonal.
OK, I saw you raise your hand, I wasn't ignoring you.
That was it?
Good
I wanted to make sure that Cartesian [INAUDIBLE]
It's always nice to see the class one step ahead of what you're doing.
All right, let's look at a couple of more crystal systems in the form of tensors in those systems and show that that in fact does correspond to the geometric constraints on the quadric as well.
For monoclinic crystals-- and this is symmetry 2, symmetry M, and symmetry 2/M.
The form of the tensor when we took the coordinate system along symmetry elements was A11, A12, 0-- terms with a single three vanished-- A21, A22, 0, 0, 0, A33.
So this was the case where X3 was along the two-fold axis and or perpendicular to the mirror point.
OK, number of independent variables to describe this property is five.
And if we look at this in terms of a constraint and shape in orientation of an ellipsoidal quadric, says that one of the principal axes, if the quadric is to remain invariant, has to be along the two-fold axis and or perpendicular to a plane that contains the mirror plane, if a mirror plane is also present.
So what are the degrees of freedom here?
This has to be along one reference axis, x3.
And, indeed, this form of the tensor occurred only when the two-fold axes were parallel, 2 X3.
And this ellipsoid then was left with one degree of freedom.
If this is the direction of the reference axis X2, we have a degree of freedom in-- shouldn't have called these X1 and X2, these need not be the coordinate systems.
Looking down on the quadric along the two-fold axis, the section of the quadric perpendicular to the two-fold axis can have a general shape, and it can have one degree of freedom in the orientation of the principal axis relative to the coordinate system.
So we have three parameters to specify the principal axes, since this is a general quadric.
We have one degree of freedom in orientation, and that adds up to four, but we said five.
So what's going on here?
Again this is a question of where the eigenvectors point.
If this term is not equal to this term, in order to force the tensor into a diagonal form, you have to pick eigenvectors in the X1, X2 plane, which will force the tensor into a symmetric form.
That is, to make this term equal to this term.
It's only a pair of axes involved, so there's only one angle between these two eigenvectors, which is a variable.
So for a nonsymmetric tensor plus one angle between the eigenvectors.
So that comes out, a-ha, five.
And again the way to see that is I will never get the tensor into a diagonal form making all the off-diagonal terms zero when I'm starting with a tensor which is not symmetric, and I know that a symmetric tensor, even in the special case where the off-diagonal terms are zero, is going to go to a symmetric tensor in another coordinate system.
So I know that there has to be an additional degree of freedom.
OK, the rest come very, very easily, so let me take just a couple of minutes to finish up this stage of the discussion, and we will shortly go on to something different.
The next step up in symmetry is orthorhombic.
And if we pick arc-- and this could be 222, 2MM or 2/M 2/M 2/M.
We found that when we took X1, X2, and X3 along two-fold axes that the tensor took a diagonal form A11, 0, 0, 0, A22, 0, 0, 0, A33.
That says that the quadric if it's an ellipsoid-- or any other quadratic form-- has all three of its principal axes along the reference system X1, X2, X3.
So the only degree of freedoms here-- and things are again starting to set up like supercooled water-- the only degrees of freedom are the three principal axes.
And that's it, the orientation is fixed.
And, moreover, the tensor is diagonal in this reference system-- it's going to be symmetric in this reference system, it's going to remain symmetric in any other coordinate system.
So the tensor is always symmetric, the eigenvectors then are always orthogonal, even if the coordinate system stays Cartesian.
And there are three variables, three degrees of freedom.
And finally two remaining cases can be disposed of quickly.
We saw that for an arbitrary theta that is a symmetry element-- so let me not say a symmetry-- for an arbitrary rotation about X3.
This was the ingenious little proof in which we transform the tensor by an arbitrary rotation theta about X3.
We found that the form of the tensor was A11, A12, 0, and now I get to use tensor notation by calling this term A12 again and not the proper indices A21, so I'll put quotation marks.
"A22" was equal to "A11."
So going to an arbitrary rotation theta this must be the form of the tensor we discovered, and this will cover then fourfold axes, three-fold axes, and sixfold axes.
These two elements are constrained to have the two values, these two are constrained to have the same values, so the tensor is always symmetric.
Symmetric relative to these axes, symmetric in all coordinates, and any choice of coordinate system, then.
And that means that if the quadric has this property, it's going to have one principal axis along X3, it's going to be circular in the plane that contains X1 and X2, so there are only, count them up, there are only three degrees of freedom.
In terms of tensor elements, these degrees of freedom are A1, A12-- in the diagonalized form-- and A33.
And there are three degrees of freedom in terms of the independent elements as well.
So this is for anything that involves rotational symmetry other than 180 degrees.
And, finally, for cubic crystals, if a symmetry operation other than 180 degrees gives us this form of the tensor, two off-diagonal terms are zero, two diagonal terms equal.
If we impose this along all three reference axes, then the form of the tensor is going to go to a diagonal form with all diagonal terms equal.
The quadric that corresponds to that form is a sphere that says that there's one quantity, one degree of freedom in the form of the tensor.
There are zero degrees of orientational degree of freedom.
Thus the spherical quadrics stay spherical for any orientation.
So again, one independent element in the tensor, just one degree of freedom in the quadric-- namely its radius-- and, again, the formal constraints that we obtained by symmetry transformations agree with those that are the same degrees of freedom when you want the quadric to coincide with the axes.
Yes?
Speaking of the circle and the exponents to play only if A12 equals zero?
These two were equal, but non-zero.
These two are equal, this one was independent.
And that would be for the specific case of a fourfold axis.
Now if we put a fourfold axis along X1, that's going to involve these two being equal and the non-zero elements become-- let's see-- along X2 it would be A13 and A23.
No.
OK, we put the four-fold axis along X2, then this one would be A22, A11, and A13 would have to be equal, and 21 and 12 would have to be zero, 23 and 32 would have to be equal.
I think that's the way it'll go, and these three have to be zero.
So when you impose all three constraints, here we've had these two equal, now we have these two equal, put the four-fold access in the next-- in the remaining direction, and again it has to be diagonal, and pairwise all of the off-diagonal terms will be required to be zero
I mean, when you diagonalize your matrix the diagonal term are going to be all different in this case if A12 is different from zero
This is not a final result, this is an intermediate step.
So we impose this constraint, plus this constraint, and that's actually going to give us all the qualities we're going to get, but we'd want to do the same thing for a four-fold axis along X3.
You put them all together, the fact that these are zero will wipe out all of the off-diagonal terms, and these things-- for another choice of axes, these two would have to be zero.
And for another choice of the orientation of the four-fold axis, this and this would have to be equal.
And, finally, this and this would have to be equal.
So this is what we found, in fact, when we cranked through the symmetry transformations, and this is what we would get when we said that the quadric has to have a shape that conforms to cubic symmetric
My question was in the case of the arbitrary relation of those three.
All three degrees of freedom, are they, in fact, the three principal axes?
For this case here?
Yes
For a symmetric tensor, they are two principal axes.
OK?
If the tensor is nonsymmetric, then these two don't have to be equal any longer, and they give you the extra degree of freedom.
OK?
What are the three degrees of freedom in this case?
I just want you to know I appreciate your question
May I make a suggestion?
Yeah
I think you need two degrees of freedom to specify the first one, and since it's symmetric and since it's orthogonal you'd only need one more to find the second one and then another to divide the third.
So that's three degrees of freedom to find three principal axes in an orthogonal system
Not sure I like that, because this has to be the shape of the quadric for an arbitrary rotation angle
Does that angle between X1 and X2-- does it have to be 90 degrees if it's not connected?
That's correct, that's correct.
OK, let's not tie up the whole audience, let me-- let's talk about this and I'll think about it too.
But you make a very, very good point, but let's not settle it in real time.
We'll throw everybody out, we'll close the door, and you can come back and see who won the scuffle, OK?
OK, so let's take our break, I think you're more than ready for it.
OK, its always disappointing when, just as you're ready to tie everything up in a nice, neat little package and leave the room to a round of applause, you screw up the last little detail.
Anyway, what's wrong here is, I said for an arbitrary rotation around X3, and I'm comforted only by the fact that nobody else caught this little glitch either, the answer elements are a11, a11, a33, a12, not 0, and -a12, 0, 0, 0, 0.
So there's a skew-symmetric form when there's just a single rotation axis.
So this would be asymmetry n and this would be one of a family of two different symmetries, the dihedral groups n22, the two-fold axes would require that these be equal, so, 0, 0, 0, a11, 0, 0, 0, a33.
So this is not the form of the tensor for a four-fold axis, it's the form for a crystal with symmetry 422.
And same for a three-fold axis.
This is the form for 32 and 622.
For just the single rotation axis of any sort other than 2, these two terms are equal in magnitude, but non-0.
Now if you look at the equation for the tensor, the term in a12, x1, x2, and a21, x1, x2 are opposite in sign and disappear.
So the quadric is, in fact a surface of revolution So the question is why are there two degrees of freedom rather than three?
And I think the answer to that is going to lie in what is required of the eigenvectors when the off diagonal turns are skew-symmetric, and that is something I've not thought about before.
All right, so that one minor loose end sticking out of the lid as we try to close the box on this segment of our discussion.
I'd like t move on to a different set of considerations.
One of the things that we can subject a crystal to as a generalized force is not merely a vector force, but stimulations that are more complicated.
And I would like to in particular now, as a prelude to get into getting into property tensors of higher rank.
I'd like to .
first discuss stress, and then define strain, and you've all heard of these questions in other contexts.
I'm going to introduce it, though, in terms of the tensor notation that we've used for other sorts of tensors, just so that we have everything in terms of the same language.
So let me introduce stress by supposing I have a volume element in a solid, and it will not surprise you that I'm going to define the coordinate system in the solid by three axes, X1, X2, X3 in a Cartesian.
system, and I will look at a surface on the exterior of the solid that cuts axis X1 at A, axis X2 at B, and axis X3 at C. Then I'll assume that I have a force density, force per unit area applied to the surface.
And if this is force per unit area, why don't I call it a pressure?
Well, a special case of what we normally think of as a pressure is a hydrostatic pressure, and that is something that's always exactly normal to the surface.
So I don't want to call it a pressure to mislead you into the assumption that this force per unit area, which is a vector, has to be normal to the surface.
So I'll call it a force density, which becomes a hydrostatic pressure when that force vector is normal to the surface.
And I'll call that force density vector K, and it will have three components, KI.
Now, the solid is in equilibrium, I will assume.
And if that external force is not balance on the other side by balancing forced, the volume element in the solid would undergo a linear acceleration, or an angular acceleration at the very least.
So I will assume that this force on the outside, KI, which is a force per unit area.
So to make that a force, I will have to multiply that force density by the area of the surface ABC, on which it acts.
And I'll say that this has to be balanced for each of the three directions, X1 and X2 and X3 by a force per unit area acting on these internal surfaces.
So I'll assume that there is a force acting in the x1 direction, and I'll call those a sigma IJ.
And I'll say that there's a first force acting in the X1 direction on-- now I need to somehow specify the orientation to the surface, because on this internal surface there's also a force acting in this direction.
Now on this back surface there's also a force acting in this direction.
So the forces acting on all three internal surfaces in the X1 direction have to balance exactly the X1 direction, the X1 component of the force density.
So let me turn this to K1, and I'll say this has to be a force per unit area acting on area.
First of all, I'll call the origin O on area AOC.
Plus a force acting in the X1 direction on area A-- whoops, what did I do?
I wanted to do this first on area BOC, excuse me.
And a force in the X1 direction acting on area AOC.
And finally a force acting on area AOB.
And these are all acting in the X1 direction, and I now would like to introduce some notation that indicates the direction of the direction orientation of the surface on which that force acts.
And I will use the normal to these internal surfaces to define that.
So the force per unit area on area AOB, that is a surface whose normal is X3.
And I will call this component sigma 13.
And I'll call this one area AOC is normal to X2.
That's why I did that little sleight of hand as I began this, and finally area BOC has as its normal X1.
So my use of subscripts here is that this first subscript is the direction in which the force acts.
And the second subscript will be the normal to the internal surface.
So I've set up algebraically a system of subscripts that works.
We'll see that these have some physical significance in just a moment.
So what will I call these?
Well, let me get this in a slightly different form.
Let me take the left hand side and divide area ABC into all of these other areas.
So this then will have sigma 11 times area BOC over area ABC.
And not surprisingly, sigma 12 will then have multiplying it a term area AOC divided by area ABC.
And this is all looking very cumbersome, but you'll be amazed at how this cleans up in just a moment.
And this will be area AOB over finally area AOC.
ABC again.
OK, so how can we tidy this up?
Well, let me imagine that I have some surface, a planar surface, and its normal points in this direction.
And I'm going to take that area and project it onto another surface whose normal points in this direction.
And let's suppose that the angle between these two directions is phi, and the angle between the two surfaces is phi.
The area of this original surface is A, and its normal points in this direction.
And I project it onto a surface whose normal is an angle phi away.
This new area, A prime is equal to A times the cosine of phi.
So if we accept that, then the first line of this equation, which will be eventually a set of three equations, this is the ratio of area BOC over area ABC, and what is that?
That is the angle between the direction of the force density and area ABC.
And here is area ABC.
That's the normal here.
And the first term involves area BOC, and that is the direction of X, the normal to that is X1, and therefore the ratio of these two areas is the same thing as the cosine of the angle between K and X1.
And this second ratio of areas is going to be sigma 12 times the cosine of the angle between the force density vector K and X2.
And the third term is going to be sigma 13 times the cosine of the angle between K and X3.
What these cosines are are just the cosines of the angle between K and the three reference axes, X1, X2, and X3.
So this is simply the direction cosine L1 of K. And this is the direction cosine L2 of K. And this is the direction cosine L3.
So the X1 component of the force density vector is going to be equal to a term sigma 11 times L1, plus the term sigma 12 times L2, the term sigma 13 L3.
And if I would write down similar balances, forces for the X2 component of the force density vector, not surprisingly these things are going to give me coefficients that I'll label sigma 121, sigma 22 times L2, plus sigma 23 times L3.
And in general then the requirement that this body be in equilibrium states that each component of the force density vector should be given by a coefficient sigma IJ times L sub J.
These are the components of the force density vector.
This is the direction cosines of the direction of K. This is a unit vector in the direction of K. This is the direction of a vector.
These are the components of K. And therefore these coefficients sigma IJ relate to vectors, and they must be a second rank tensor.
And these are called, as you all know, it comes as no surprise as the components of stress.
So the reason that these are really tensors, and not just a vector force per unit area essentially comes down to the fact that the behavior of the body, dynamically and mechanically, is going to depend not only on the direction of the force, but on the direction of the surface on which it acts.
Quite clear here, if I shove, is a force density, and that's a vector.
But the way in which this desk responds to my pushing on it with a force vector of that magnitude is going to depend whether I do this, which is going to compress it, or whether it takes the same force and direct it this way, which if I don't move the thing is going to result in a shear.
So the behavior of the body will depend on the direction of the force per unit area that I apply, and the direction of the surface on which I apply it.
In this case, that would be the surface ABC.
Yes, sir?
You say that the force density is not parallel to the normal of the surface?
That's correct
So when you do the ratio of [INAUDIBLE] isn't it supposed to be [INAUDIBLE] of the normal to the [INAUDIBLE] and X1?
Another way of defining it.
This is a vector, and what I'm doing is splitting that vector up into three internal forces that act on surfaces whose normal are here.
So actually, what I'm going to do is to take the way in which the internal force pushes back.
I see, that's what's confusing.
If the force that I'm putting on the body, I didn't mean to indicate any particular direction, but if this component of K points in this direction, this force that balances it has to go in the opposite direction
So you're saying that there were surface ABC is defined as normal to K?
ABC is not normal.
No.
ABC is not normal to K. All I'm saying is that it if it were normal to K, then the balancing force would be just a function of this orientation of the surface.
But if I allow this to be a general force per unit area, then the balancing force on the inside depends on the angle between K and the angle between this internal surface
I don't see the relation between K and the ratio between other ends.
BOC and ABC [INAUDIBLE]
OK.
The reason that there are three areas, is that what's troubling you?
Because I'm really now splitting the balancing force up into a force per unit area on this surface, and a force per unit area on this surface, and a force per unit area on this surface.
And that's so that I can look at these different components of resistance in terms of a Cartesian system.
Clear that the ratio of these two areas does go as the-- I erased it now, that is going to go as the-- This is A prime, and this is the original A, the relation between those areas is given by the cosine of this angle phi.
If I [INAUDIBLE] this down in this projection, and this is the area of A.
This is the direction of the normal to the surface ABC.
And I'm going to let this be the direction of the normal to one of the internal surfaces, and these internal surfaces are going to have as their normal either X1, X2, or X3
So [INAUDIBLE] between the normal to ABC and X1
And X1, yes.
And that's what I'm saying its, and that is going to be the same thing.
As the cosine of that angle is just going to be the direction cosine of the normal to ABC relative to X1, X2, X3.
OK?
So in the end, you're saying that K is parallel to the normal of ABC?
No, I'm not.
No, I'm not.
This is a relation between the areas.
And these are the forces acting on those internal surfaces.
And I'm saying there are three of them, and those three components that are all acting in the X1 direction will depend on the ratio of these areas.
The ratio of this area is the cosine of the angle between X1 and K. OK.
This you're OK with?
Yeah
OK. And what I'm saying now is that there is acting on this internal surface something that has to balance the component of K, which points in the direction of X1.
And the component of K before we divided out the area, the component of K1 times the area gave me the net force.
So this K1 now is a force per unit area, the component of K per unit area that points along the X1 direction.
Now as a force per unit area I multiply it by an area that gave me force.
There are three forces internally that balance this force, and they all act in the X1 direction
You're just [INAUDIBLE].
Shouldn't it be the cosine of the angle between the normals of [INAUDIBLE]?
Yeah.
That's what I'm saying here.
That for X, for this surface, yeah, that's what I'm saying.
One of these surfaces here, if we look at the first term, this is the direction of X1, and the force here will be the force along K. And if I take this force and put it down in X1, this is going to be the force times the cosine of the angle.
And the two areas are going to go as the cosine of the angle
[INAUDIBLE] cosine of the normal [INAUDIBLE]?
OK.
This is indeed the ratio between the internal area and the external area.
Since the internal area has a normal that's either X1, X2, or X3 the angle between the normal to these internal surfaces and K is going to be the direction cosine of K. I suggest since we're getting out of time quickly we either resolve this after the end of class or leave it until next time.
That's a bad note on which to end, too.
The point I want to leave, still subject to resolution and debate is that the coefficient sigma IJ I would like to claim relate to vectors.
One is the direction of the external surface, and if it relates to vectors, then this set of coefficients qualifies as a second rank tensor.
And if you accept that, however grudgingly, everything that we've said about second rank tensors holds for the quantity sigma IJ, which are called the elements of stress.
In particular, we can talk about a stress quadrant.
We can talk about the value of a stress in a particular direction.
If we have a stress tensor in which show all the terms are non-0, it can be diagonalized.
We can also say that if the crystal has symmetry, for example, if the crystal is cubic, you can say that the form of a second rank tensor for a cubic crystal has to be diagonal.
And these diagonal elements represent compressive stresses, so you can say that you can only subject a cubic crystal to compressive stress.
On the other hand, if the crystal is a triclinic crystal, you can say that there's a sigma 11, there's a sigma 12, a sigma 13, a sigma 21, a sigma 22, sigma 23, sigma 31, sigma 32, sigma 33.
So for a triclinic crystal, there appear to be nine elements, but next time we will show that the nature of stress is the body has to be in equilibrium requires that the stress tensor be by definition symmetric if there's to be no net couple to force on it.
Going to buy this?
You didn't buy the things that I said that perhaps were slightly incorrect.
This is massively incorrect.
Jason?
You can rotate a cubic crystal [INAUDIBLE]
No.
Because say a tensor of this form acts just like the scalar.
Just like a cubic crystal, you don't suddenly get off diagonal properties if you transfer it to a different axis.
So unless you tell me what's wrong here, I'm going to leave this on your plate until next week.
And I don't want you fretting over the weekend.
This is clearly nonsense, and it is a contradiction that arises only if you carry too much of what we've been doing for the last month with you into a discussion of stress and strain.
Stress is something that you impose on a crystal.
It has nothing to do with what kind of crystal it is and what's going on in the interior of the crystal.
There may be constraints on the coefficients that relate stress and strains, and there surely are.
But I can take a cubic crystal, and I can squeeze it, and I can twist it and share it anyway I like.
So this is an important distinction between the tensors that represent physical properties, which we've been discussing up to this point.
And the properties such as conductivity, diffusivity, susceptibility and so on and all their many varieties, these are something which we'll refer to as property tensors.
And they are tensors demonstrably, but they are tensors which describe the physical behavior of a crystal.
Nye uses a term that I don't think is self explanatory, so I don't care for it.
He calls these field tensors.
That sounds like some description of a corn patch that requires a tensor.
But no.
We probably don't really think of what we mean when we say a field.
When we say an electric field, you think of an e-vector.
You think of a vector.
But what a field is is just a set of coordinates, XYZ, and that describes an area or a volume.
And just like the things that corn grow in are referred to as corn fields, it's an area that has corn on a lattice very often also.
So a field is just a set of coordinates in space to which a value of something is a sign.
So if we have an electric field, and we express that electric field as a function of X, Y, and Z, that is a field of vectors.
If we take something that has intrinsically, that intrinsically needs to be described as a tensor, then we have to every point in space XYZ a tensor sigma IJ, whose value and whose components value vary with position within the space.
So talking about the field as assigning values of something to every coordinate in the space that could be a tensor.
What we refer to as an electric field really is a field vector.
So that's Nye's term, but the fact that it took me three minutes to explain why he uses it explains as well why I don't care for it.
So we'll talk about these as property tensors.
And property tensors are something that's innate to the material.
Something like stress, strain, and other second rank tensors are external stimuli, just like an electric field.
And they have value that is imposed on the crystal, and the stress field is defined as a function of position.
So that's the distinction between a property tensor, which is subject to symmetry constraints, and a tensor that represents a stimulus, an external stimulus applied to the crystal, and it's defined as a function of position.
And that is a field tensor.
OK, so that's the difference.
Since this is something that's imposed externally, it can have any form of the tensor that you wish to oppose.
And there are no symmetry restrictions on it.
The only restriction is, as we'll show next time, that it must be symmetric if the body is to be in equilibrium.
All right.
It is time to quit, and that I think is a good place to leave things.
And we'll say more about stress next time.
We are thundering rapidly to the midpoint of the term.
I think a week from this coming Thursday will be exactly halfway, and the halfway point is when I usually try, give or take a day, to change gears very abruptly and start looking at the symmetry of physical properties and tensors and actual discussion of physical properties that are anisotropic.
So we'll be finishing up our discussion of symmetry fairly quickly.
We still have, however, some major derivations to perform, and we'll get into one of these today.
Last time we had begun the process of deriving what are called the point groups as opposed to the plane groups.
The plane groups were combinations of symmetry with a lattice that extended throughout a plane in space.
A point group we've seen in the form of two-dimensional point groups.
We're now going to today derive the three-dimensional point groups.
And, again, the name stems from the fact that these are clusters of symmetry about a common point so that at very least that one point in space stays put.
So, hence, point groups-- these are symmetries about a point.
And to put some embellishing adjectives on here, these will be the three-dimensional crystallographic point groups.
Because there are an infinite number of point groups, we're considering just those symmetries that can be combined with a lattice and therefore are permissible to crystals.
I'd like to point out in case you've not been making much use of Buerger's book that we actually are getting back to Buerger's treatment.
We'll do some of the derivations over the next few days with a little bit more detail and with a slightly different procedure than Buerger.
But the bases covered are the same.
So let me just point out where some of this material is in Buerger's book.
Buerger does Euler's Construction, and he does this in a once over lightly.
I think this is not the best part of the book.
This is on pages 35 to 43.
In particular, he sort of leaps ahead quickly and just looks for one angle between rotation axes and doesn't get the-- do specifically the ones that are 90 degrees and are not as interesting.
Some of the combination theorems are in Chapter 6, and there you'll find the statement that two reflections at an angle mu is equivalent to a net rotation of mu, except Buerger doesn't call them sigmas.
He calls them m's.
He uses the same symbol for individual operations and for the symmetry of them, and I think that's a little untidy.
Also, the theorem that says that if we have a mirror plane that's perpendicular to a twofold rotation axis, that gives rise to inversion center at the point of intersection.
And then he launches into a rather condensed version of the derivation of the point groups on pages 59 to 68.
Then, in Chapter 7, derives the two dimensional lattices or the plane nets.
This is on page 69 to 83.
We've done that differently and much more exhaustively by deriving the two-dimensional plane groups and the lattices are part of the plane groups, and we've gotten them automatically.
There is a place where in the derivation of the space lattices, he's a little bit incomplete and actually does something that's wrong A-ha-- I showed Buerger was wrong at one place in his book.
But anyway, this is where the material that we're covering now is mentioned in Buerger's book.
Last time, we had hit a new surprise.
We had asked if we take the basic rotational symmetries and add an inversion to something like three-- so, three is a subgroup, the operation of inversion is an extender.
We found that there was a new operation that came up, and this was a rotoinversion operation.
And then the resulting point group, which we called 3-bar-- the symbol itself, 3-bar, is a threefold rotoinversion.
And if we were not clever enough to invent a rotoinversion operation, we would have stumbled headlong over it in adding inversion as an extender to a threefold axis.
And a rotoinversion axis is something that involves in general rotating through some angle alpha to get from a first object to a second object of the same handedness and then not yet putting it down, first inverting it through axis-- through a point on the axis-- to get a second object.
I shouldn't call this second because this is not rotation with inversion simultaneously but a two step operation.
So we rotate from 1 to this virtual object also of same handedness and then before putting it down, we invert it to get number two, which is left handed.
And the operation of getting there in two steps but in one repetition is the operation of rotoinversion.
That operation was one of the members of the group that results when you add inversion as an extender to 3.
But, really, this was not something that was a new sort of symmetry element.
It was an operation that came up, but we could regard the point down that we've labeled 3-bar as simply a threefold axis with an inversion center sitting on it.
But then we looked at another one.
And by anticipation, I said we ought to look at these.
And we looked at what happened when we took a fourfold rotoinversion operation where we would take the first object number one, rotate it through 90 degrees, and not yet put it down.
We'd first invert it through a point on an axis, and we got one that was 90 degrees away but pointing down in of opposite handedness.
If we repeated that operation several times, we found that we got two objects up above a plane normal to the rotation part of the operation and passing through the point, which we inverted.
So there's one right handed one up here, a second right handed one up here, and then two down below, which were of opposite chirality, number 3 and 4.
And that is an entirely new beast.
We have no way of describing the relation between these four objects other than saying rotate, don't put it down, first invert.
Rotate, don't put it down, first invert again.
So this is something entirely new.
It's a two step operation.
You cannot describe it any more simply than saying take two steps to do the repetition in the same sense that the glide plane that we discovered in the plane groups had a step of translation followed immediately by reflection, and that was a new sort of operation.
So here's another example of a two-step operation.
The symbol for this point group is 4-bar, and the symbol for it-- if you want to indicate its locus, is, it's got a squareness to it because these objects as mutually are separated by 90 degrees, but it really has only a twofold symmetry, so the symmetry for a 4-bar axis.
It's not really an axis at all.
It's a 4-bar point, because only a point is left invariant.
But the symbol for a 4-bar "axis" is this square with a little twofold axis inscribed in it.
Alright.
We ought to really, then, step back and ask how many different rotoinversion axes there might be, and add these to our bags of tricks.
And what I'm going to invite you to do is to examine this systematically on a problem set which I will pass around, along with several other problems for your consideration, as well.
And having stumbled onto that combination, another thing we might ask to broaden our bag of tricks still further is to say is there such a thing as a rotoreflection axis?
And we called n-bar a rotoinversion.
I wrote a reflection axis is indicated by an n with this little squiggle on top.
Anyone who has studied Spanish realizes that the proper name for the squiggle is a tilde, T-I-L-D-E. And this would be an operation where there's a locus in which we'll perform the reflection part of the operation, so this would be a 2 step operation that involves taking a first one and rotating it to a virtual object number two but not putting it down.
Before putting it down, we will reflect in a planar locus.
So this is number two, and it will be of opposite chirality.
So that's another operation that involves rotation coupled as one part of a step that involves R, the second of our three translation free operations, namely reflection and inversion.
So are these something we should consider?
What I've invited you to do is to draw out for n equals 1 to 8 the patterns produced by rotoinversion and the patterns created by rotoreflection, and then see if you regard these as new operations or whether they can be broken down into the simultaneous presence of more than one of our old friends such as 3-bar being a plain old threefold axis with an inversion sitting on it.
The fact that I'm asking you take your time to do this suggests that yes, you are going to find some rotoinversion and rotoreflection axes, which are inherently 2 step operations and can't be decomposed.
What I'd like to do now, then, is to systematically look at the axial combinations.
These are of the form n one, two, three, four, and six and the dihedral symmetries.
These are of the form n22, or just 32 in the case of the one with the threefold axis.
And then the two cubic arrangements 23, twofold axis out of the face normal to a cube, threefold axes out of all of the body diagonals, and 432.
And what I'm going to do with you is to use these as worksheets to guide the logic of what we're doing, and we won't do every single one of them.
I think once we do a few, the general principle is clear and the results there are before you.
But what we'll take, then, is the axes n, the axes n22, and then the two cubic symmetries 23 and 432.
Then, finally, we found 4-bar as a different rotation axis-like symmetry element, so I'll add that to the list.
Then consider what we can add to these symmetries as extenders, and these are an additional element that we can add to what is already a self-contained nice little group.
We muck things up by introducing another operation, and then we'll have to take combinations of that operation with all the symmetry elements that are there in the parent subgroup and see what new symmetry elements arise.
The ground rules in these additions are fairly simple but nevertheless profound.
The addition of the extender cannot create any new symmetry axes by its operation, and the reason for that should be quite clear.
We obtained these combinations of rotation axes using Euler's principle, and that was thorough and orderly, systematic and complete.
These are the only arrangements of crystallographic rotation axes that are possible.
So if we add an extender that creates, for example, a new twofold axis in n22, it's either going to be something that we already have in this list, and therefore not interesting, or something that just cannot constitute a group.
We'll take combinations of operations and we will never find a closed finite set.
So if we add a mirror operation, a reflection operation as an extender, the addition of the reflection operation sigma must be either perpendicular to the axis.
And what that's going to do is just flip the top part of the rotation axis and reflect it down to the bottom axis.
Nothing new was created.
And it could alternatively be parallel to the axis and passing through it.
In that case, we've already seen the consequences of this addition in deriving the two-dimensional point groups, namely to 2mm, 3m, 4mm, and 6mm.
No new axis is created.
One of the additions that is something we mentioned last time is that in the case of the dihedral groups where we have an n-fold axis and then twofold axes arranged in some fashion-- not in some arbitrary fashion.
It's going to be 2 pi over n times 1/2, depending on the n-fold axis-- we could pass the mirror operation through the twofold axis.
And that's what we'll call a vertical mirror plane.
But when there's more than one axis present, we could also put the reflection operation diagonally in between the twofold axes.
So we're going to refer to this as a diagonal mirror.
It's snaked in between the axes that are present.
So, perpendicular reflection operation as an extender, one parallel to the axis and passing through it, or one that is diagonal passing in between them.
And finally, that exhausts what we can do with reflection, but we could add inversion.
And if we're not going create new axes, and we're going to get a point group-- a finite set of objects-- the inversion center has to be on one of the single axis, namely these groups n, or at the point of intersection, if there's more than one.
So there's the job laid out for us.
And we've got two theorems to aid us in quickly finding the new operations that arise-- in effect, taking a shortcut to establish the group multiplication table, and the two theorems are as follows.
We saw that if you take a rotation operation A pi, that takes a first object that's right handed and rotates it to a second object that is also right handed.
Then follow that by reflection in a mirror plane that is perpendicular to the axis.
The net effect of going from 1 to number 3, which is left handed, is to create an inversion center at the point of intersection.
So there's a theorem that we can write down once and for all and say that A pi followed by reflection in a mirror plane that's perpendicular to axis A has, as a net effect, the operation of inversion.
And if this is not a twofold axis, it's something like a fourfold axis or a sixfold axis, that contains the operation A pi as part of the operations contained in that axis, and that's going to create an inversion center as well.
So for any evenfold axis, add a perpendicular mirror plane, and automatically the inversion center comes up.
And we can permute the order of these operations if we rotate by 180 degrees and then invert the net effect is reflection in the perpendicular mirror plane.
So this is something that you permute around into three different combinations.
Does the order of the operation make a difference?
And the answer is no, the order is not important.
For example, rotating and then reflecting is the same as reflecting and then rotating it.
It went up at the same point, number 3.
Remember a fancy word for this when we define what's meant by a group, that these groups are going to be what's called Abelian, and an Abelian group is where any combination of operations a followed by b is identical to b followed by a. I told you that great joke among mathematicians, what is purple and commutes?
And the answer is an Abelian grape.
Ha, ha, ha-- I think that's stupid, but it drives mathematicians into guffaws of laughter.
A second theorem, and this is one that we've already seen in two dimensions, is that if you take an operation A alpha, pass a reflection operation sigma 1 through it, that to the effect of that is going to be another mirror plane that is alpha over 2 away from the first.
Notice I'm a little bit sloppy in talking about reflection operation and mirror plane because if a reflection operation is there, that's all I need to have to say that this is the locus of a symmetry point, a mirror plane.
So this is another one.
Do you think that this an Abelian combination to say that A alpha followed by sigma 1 is equal to sigma 2.
Is this Abelian?
No
No, it's not.
And how do you answer that?
Not through any means more profound than just drawing it out and seeing what you get.
Let's take 3n as an example.
So let's do the operation.
In this order, do the rotation A 2 pi over 3 to go from this one up to this one, and then let's reflect across this horizontal mirror plane.
So this is 1 right handed, this is 2 right handed reflect, here's number three and it's left handed.
So that's A 2 pi over 3 followed by reflection in sigma 1.
And if we do the operations in reverse order, reflect in sigma 1 to go up to here, so this would be 2 prime and then rotate by 120 degrees.
We would go to here to here, and that is not the same location as this one.
So this is not equal to sigma 1 followed by A 2 pi over 3.
How can you tell when is the combination of operations is going to be Abelian and when is it not?
It sounds rather clumsy to state it in words, but whenever the two symmetry operations leave each other unmoved, then you can permute the order.
For example, in this combination here the rotation leaves the mirror plane unchanged.
It just spins around in its own plane.
The mirror plane leaves the rotation unchanged, it just reflects it end to end.
Here this rotation axis and these mirror planes obviously do not leave each other unchanged because the threefold axis rotates the mirror plane to two other new locations.
So whenever that's the case, if the operations do not leave each other unchanged, the order makes a difference.
So that's a useful thing to keep in mind.
Let's now turn to these sheets that I passed out, and you might want to unstaple them because they're designed to go side by side, and there are three pages.
What I've done across the top of the pages is to give the different combinations of crystallographic rotation axes that are possible.
So going from the first sheet across to the last, there's the rotation axes by themselves-- one, two, three, four, and six.
Then there are the groups of the form n22, 222, 32, 422, 622, and then the two cubic combinations 23 and 432.
So there are the 11 possible combination of crystallographic rotation axes, and stuck off by itself at the end is the oddball 4-bar.
For the axes by themselves, the diagonal mirror addition is not defined.
The diagonal mirror position by definition snakes in between axes that are there in the axial combination.
If there's only one axis, that's not defined.
But you could add a horizontal mirror plane.
A horizontal mirror planes adds a onefold axis.
It's just a mirror plane.
So that's called m in the international notation for mirror.
And it's called C sub S, C because it's a cyclic group, and the S stands for Spiegel.
That's the Schoenflies notation.
If we add a-- take two and add a horizontal mirror plane-- this is the thing that we sketched out here-- this gives a pattern of four objects, two that are up above the mirror plane of one chirality, two down below the mirror plane of opposite chirality.
These projections of the patterns are down along the rotation axis, and when you see a dot in a circle, the circle represents the point that is down and the dot is one that's up on top.
So there are two pairs of objects-- two right handed ones that are up and two left-handed ones that are down.
And that, in international notation, is 2/m the twofold axis is over a mirror plane.
So that's the way to make sense of the symbol and remember what it means.
In the Schoenflies notation at C2.
You've added a horizontal mirror plane as an extender.
Going down the remaining ones in that list because they're all very, very similar-- add a mirror plane to a threefold axis, the triangle is reflected down to a triangle of motifs of opposite chirality, so that's 3 over m. For sure, it's called 6-bar, but let's forget about that.
The Schoenflies notation C3H, and similarly there's a 4 over m, a square above, and a square of opposite handedness below the mirror plane.
The horizontal mirror plane in all of these cases is shown as a bold line.
They've added fainter vertical lines to give you an angular reference.
Don't be confused by that in the Xerox copy-- the weight of the lines is not distinguished.
So the only symmetry element in 4 over m, for example, should be shown by this solid bold circle that is in the plane of the paper.
The two crossed lines are lighter and those are just as mutual orientations.
And finally, there's another one of this form which is 6 over m. On all of the evenfold additions-- that is, to say everything but 3m-- if you add that horizontal mirror plane, it's going to be perpendicular to an operation A pi.
Therefore, inversion pops up as one of the operations in the group, and 2 over m, 4 over m, 6 over m, those regular polyhedra of objects-- can be inverted through the point of intersection of the axis in the mirror plane, and that leaves the set unchanged.
So if we go down to the bottom of the list and say, can we add inversion to these axial arrangements, in the case of 2, 4, and 6-- no, adding inversion doesn't give you anything new, because that's already there in the groups of the form CNH, or 2/m, 4/m, 6/m.
Diagonal is not defined, so the only additional one that we could pick up by adding inversion is adding inversion to a threefold axis, and that's one we did the other day.
Adding inversion to a threefold axis gives us three that are up of one handedness, three that are down of opposite handedness, and the orientation of the triangle is skewed.
That is called 3-bar.
But it's just nothing more than a threefold axis with an inversion center sitting on it.
Schoenflies' notation is C3i.
Then the only other additions to the single axes are adding the mirror plane in a vertical fashion passing through the axis, and these we've already seen in two dimensions so we don't have to spend much time on them.
There's 2 mm, which is C2V.
3m-- all the mirror planes result upon adding a single mirror plane to the threefold axis-- 4 mm, and 6 mm.
These are just three-dimensional extensions parallel to the axis of the symmetry that we already derived for a plane in space in the two-dimensional point groups.
So we're almost half done.
It's easy, isn't it?
Any questions at this point?
So the only ones that are really new that we haven't seen already in two dimensions are the rotation axis perpendicular to the mirrored plane.
That brings us to roughly the middle of the second sheet, and the arrangement of axes 222 represents three mutually orthogonal twofold axes.
If we add a horizontal mirror plane, that's going to put an inversion center at the point of intersection because that horizontal reflection is going to be sitting normal to a twofold axis.
And then since the other twofold axes see an inversion center sitting on it, there's a mirror plane perpendicular to those other twofold axes.
So adding again to go through that again, adding inversion to the point of intersection of the twofold axes creates a mirror plane perpendicular to each of those twofold axes.
So it becomes 2/m 2/m 2/m-- each of the twofold axes acquires a different sort of mirror plane perpendicular to its orientation.
And 2/m 2/m 2/m, even though I love saying it, is kind of a mouthful.
So that's very often abbreviated to just mmm, which really is a nice exclamation when you see what a lovely symmetry it is.
The pattern shown in the diagram to the left is just the pattern of 222 with the objects reflected up or reflected down, and you get a total of 8.
That means that there are eight operations in the group, and you can add up quickly what they are.
They are the three operations of rotating by pi, the operation of identity, then three mirror planes because the mirror planes all indistinct, and then the operation of inversion.
So there are the eight operations that are present that generate the objects from a single one.
Add a horizontal mirror plane to 32 and you get a threefold axis perpendicular to the threefold axis, so you write that as 3m.
That mirror plane now passes through the horizontal twofold axes, and a twofold axis with a mirror plane passing through it wants another mirror plane 90 degrees away, so a new mirror plane pops up passing through each of the twofold axes and perpendicular to the horizontal mirror plane.
Those mirror planes are in the same direction as the twofold axis.
They're not perpendicular to it, and when symmetry elements are parallel to a common direction you write them out on the same line, so here's a case of a mixed metaphor.
One mirror plane that's perpendicular to an axis, the threefold axis, so you write that as 3 over m. The other twofold axes have mirror points passing through them, so you write that as 2m and the threefold axes makes all of those 2m symmetries equivalent to one another.
Pattern, and this is true for all of these symmetries, is nothing more than the pattern of the subgroup repeated by the one operation that you've added as an extender .
So if you look at the pattern of 32, object all of the same chirality, all apparently up and down, and add the horizontal mirror plane.
The one that is up goes down, the one that's down goes up, and you get a set of four around each of the twofold axes.
422 behaves very similarly.
The pattern is just the pattern of 422 reflected in a plane, so you have a set of 16 objects-- eight of them up of one chirality, those are the solid dots, if you will, and then another two, four, six, eight that are down of opposite chirality.
You have a mirror plane that you added as the horizontal mirror plane perpendicular to the fourfold axis.
All of the horizontal twofold axes see a mirror plane passing through them.
So they've got to have a mirror plane that's 90 degrees away, and those are the vertical mirror planes.
Either the point is obvious now or you're not following me at all, so I'll simply say on the left-hand column of the final sheet, 622 with a horizontal mirror plane goes to 6/m, 2/m, 2/m, called D6h in Schoenflies.
I'll leave the cubic ones until last.
They're not nearly as bad as they seem.
But obviously they have a lot of symmetry all over the place, and we'll want to take a closer look at the patterns.
The next extender is a vertical mirror plane, and we've already got the groups of the form cnv, because they are just extensions in a third dimension of one set we've seen in two dimensions.
If we try adding the vertical mirror plane, which is defined as passing through the principal axis of high symmetry and perpendicular to the twofold axes, we've already encountered those in every group except 32.
2/m, 2/m, 2/m already has the vertical mirror plane.
The same is true of 3/mm2.
The same is true of 4/m, 2/m, 2/m and 6/m, 2/m, 2/m.
The vertical mirror plane comes in automatically when we add the horizontal mirror plane as the extender, So nothing new there at all.
And now we come to one that is interesting.
This is the diagonal mirror plane.
And I'll do this one slowly and then in some detail because it's another example of how we would trip over something if we were not clever enough to think of it.
Let me begin by drawing the three orthogonal axes of 222.
So we have one object that's up and one object that's down.
The twofold axis perpendicular to the board will take this one that's down and rotate it to here and take this one that's up and rotate it to here.
So that is the pattern of 222.
We can snake a mirror plane in between the twofold axes, and a mirror plane passing through the vertical twofold axis has to be accompanied by one that's 90 degrees away, half the throw of the axis, which comes from our theorem that says A pi dot sigma vertical has to be equal to a sigma prime that's vertical and pi/2 away from the first.
In terms of the pattern, this second mirror plane that comes up is going to reflect this one across to here, it's going to reflect this one across to here, it's going to take this one and reflect it over to here, and take this one and reflect it to the right, as well.
So now we've got a total of eight objects.
This one is right handed, and this one is also right handed because we repeated it by rotation.
Then we reflected that pair, so these two are left handed, and reflect it across the other diagonal mirror plane.
They're back to right handed again, and these have to be left handed as well, so there's the pattern.
Is there an inversion center in this pattern?
The answer is no, because there is no mirror plane that is perpendicular to a twofold axis, so this point group has no inversion in it.
It's said to be acentric, without a inversion center.
Do we know how everything is related to everything else?
We know how this one is related to this one by twofold rotation.
We know how this one is related to this one, and that's by reflection across the mirror plane.
How is the first one related to this one?
How do you get it through 90 degrees and then pop it down and change the chirality?
[INAUDIBLE]?
You can't do that in one step.
You've got to do just what I said in words.
Rotate it 90 degrees, don't put it down yet, and reflect it down in a horizontal mirror plane.
This horizontal mirror plane is not a symmetry element, it's part of the operation that's necessary to get us from this guy over to this guy of opposite chirality.
So this is an example a new type of operation, a 2 step operation.
This is a 90-degree rotoreflection axis.
And again, I feel compelled to put the axis in quotation marks, because really it's a pair of operations that leaves only a point unmoved, so it's really a point symmetry element.
There's another way of describing the same thing, and that would be rotoinversion.
We could rotate by 90 degrees in the reverse sense, not yet put it down, and invert, and we get, again, the relation between-- what did I do?
Rotate it here, yes-- rotate to here and then invert, and we got that one.
So that's another way of defining the relation.
A 90-degree rotoreflection is a minus 90-degree rotoinversion.
So you pays your money and you takes your choice, and what the groundbreakers who went before us did was to take rotoinversion.
as the operation to survive.
So here is, if we were not smart enough to invent it, our 4-bar rotoinversion axis.
So if we hadn't been clever enough-- and it would take a pretty clever, devious mind to invent a 2 step rotoinversion operation-- as soon as we added this sort of extender, a diagonal mirror plane to a group of the form n22, we would have found that there was a new sort of transformation that could not be described any more simply than to say take two steps to do it.
Rotate 90 degrees and then invert.
So we have to add the operation of A pi over 2-bar to our basic bag of tricks for generating patterns.
The group 32 is also a group to which we can add a diagonal mirror plane.
And if you do that, again, the bold lines that are mirror planes in that group and the axes that are faint lines are easy to mix up.
But if we examine that group without the confusing lines, here are the twofold axes.
And remember that they are all equivalent to one another by the threefold axis.
So this axial group is the group 32.
If we put down a mirror plane interleaved between the twofold axes, notice that each of those mirror is perpendicular to a twofold axis.
So we can write this as 2/m.
2/m means there's an inversion center at the intersection of the twofold axes with the mirror planes.
And a inversion center sitting on a threefold axis makes it a 3-bar axis, so this group is called 3-bar 2/m.
And the pattern is what a twofold axis would do.
These would be up-down-right ones.
Reflect those and you get an up-down pair of left ones.
Reflect again and you get an up-down pair of right-handed ones.
Reflect yet again, and you get an up-down pair of left-handed ones.
Reflect still once more, and you get a down-up pair of right-handed ones.
One more time, up-down left-handed ones.
So you get a total of 2, 4, 6, 8, 10, 12 points.
So this is 3-bar 2/m in Schoenflies notation D3-- that's the symbol 32-- with a diagonal mirror plane added.
You indicate that by a d in the subscript.
OK.
I think that's probably a point where you're more than ready for a break.
We've got very few yet to do.
One surprise is that if you add diagonal mirror planes to 422 and 622, you get perfectly lovely exquisite groups but they're not crystallographic, so we don't include them in our list.
But we'll discuss them when we come back, and then take a look also at the cubic symmetries.
Impossible to draw, but really not all that difficult to understand.
So let's stop at this point and take a stretch.
People are stretching already.
They need it.
We'll resume in 10 minutes' time.
--questions about what we've done.
I think it's been fast, but hopefully if you understand the principles, we didn't go too fast.
But any questions on what we've done?
I guess you haven't had a chance to think of questions yet.
Let me take care of our oddball symmetry at the end of the chart that I handed out-- and this 4 bar-- and ask what we can do there.
Having discovered the four bar operation in 2, 2, 2 with diagonal mirror planes, we can consider that as a new type of symmetry element.
And this is the symbol for it.
And this would take a pair of objects and repeat them by a 180 degree rotation.
And then another pair of opposite handedness, opposite chirality would be rotated 90 degrees and inverted, rotated 90 degrees and inverted.
And I mentioned last time that a solid that has this shape is something called a sphenoid.
And the 4 bar axis takes a pair of faces that are up and a pair of faces that are down and skewed by 90 degrees.
Now what can we do?
If we add a mirror plane that's perpendicular to the 4 bar axis, the right-handed one goes down, the left-handed one comes up.
And it becomes simply 4 over m, which we've already got.
If we add a vertical mirror plane through the 4 bar, if you look a bit earlier at 4 bar 2m, now we've got the 4 bar, now we've got the mirror plane.
Not surprisingly, we get the two mirror planes.
And so S4v is going to be the same as 4 bar 2m.
And that's something we already have D2d.
So that exhausts the possibilities.
4 bar just stands by itself.
At a vertical mirror plane, it's D2d 4 bar 2m.
At a horizontal mirror plane, it becomes 4 over m. The diagonal mirror plane is not possible.
There's nothing to place the mirror plane diagonal to.
And if you add inversion, it changes into 4 over m. So this one is an odd group.
It sits by itself.
Except that we could start there, and add a vertical mirror plane, and get 4 bar 2m once more.
That leaves the cubic ones, which are really easy to deal with because a cube has such high symmetry.
We all know what cubes look like.
But when you show arrangements of motifs, it gets a little bit confusing.
So let's take a look at the tetrahedral groups.
T is a combination of twofold axes coming out in directions corresponding to face normals to a cube.
So these are the twofold axes.
And then there are threefold axes coming out of directions that correspond to the face diagonal.
So this is the jack-o'-lantern stereographic projection, which is something I love to draw at this time of year.
What will the pattern look like?
Well, it's going to look like 2, 2, 2.
So that's a subgroup.
So let's draw in a pair of objects on either side of the twofold axis, and that's the pattern of 2, 2, 2.
Now this guy here is lurking off of a threefold axis.
So that threefold axis is going to repeat it three locations that are 120 degrees apart.
And so there's going to be a triangle of objects up above.
This threefold axis is going to reproduce this into a triangle of objects that are down below.
The twofold axis will take this triangle of objects and move it over to here.
Also, up.
And the twofold axis that's vertical will take this triangle and move it down to three that are below.
So it looks very complicated in projection.
But it's simply a planar triangle of atoms here rotated 180 degrees.
So you've got one like this, one like this.
And then down below, two other planar triangles of objects.
Remembering what the symmetry of a tetrahedron looks like, you can draw this arrangement of symmetry elements relative to a tetrahedron.
And the threefold axes now are coming out normal to the faces.
The twofold axes coming out normal to the edges.
And imagine that we put a triangle on this face, a similar triangle on the face behind, and a triangle pointing in the other direction down this way.
These are the three below.
So that's the pattern of 2, 3.
Take a tetrahedron, smack a triangle on the two upper faces, and an equilateral triangle on the two lower faces.
And the threefold axis that comes out here comes out of the center of this triangle and out of the center of this triangle.
The next one that you can get from this is Th, the tetrahedral arrangement of axes with a horizontal mirror plane.
If you put in a horizontal mirror plane, it's going to take this triangle and reflect it down.
It's going to take this triangle and reflect it up.
And they will overlap in projection.
And so that's the pattern for Th.
There's a mirror plane perpendicular to a twofold axis.
That creates an inversion center.
So we can call this a 3 bar axis.
So that's Th 2 over m 3 bar.
So just have triangles on the upper and lower faces as well.
The one remaining one is Td.
Here are the twofold axes.
Here are the threefold axes.
And now if we put a diagonal mirror plane in, diagonal with respect to what?
Well, diagonal with respect to these twofold axes.
So this mirror plane goes down like this, passes through a twofold axis.
There must be a mirror plane 90 degrees away.
And the threefold axis is going to repeat these mirror planes so that we get mirror planes that are at an angle with respect to the vertical twofold axis.
And this is Td.
And this, in the international notation, is called-- if we can figure it out-- the diagonal mirror planes with respect to these twofold axes have changed them into 4 bar axes.
So this is called 4 bar 3m, which doesn't look cubic at all.
So that's a little bit deceptive.
I am foolhardy for even trying to do this.
And so I don't think I'm going to do it.
4, 3, 2, we know what that looks like.
That is fourfold axes coming out in directions that correspond to the face normals to a cube.
A threefold axis coming out of body diagonals.
Twofold axes between all of the fourfold axes that are normal to the edges of the cube.
So all sorts of rotational symmetry.
That, if you want a pattern, has a triangle that's on about each of the threefold axes.
But the one that is up, points in the opposite orientation of the one that's on the threefold axis coming out of the other end of the cube.
So you have one triangle that's like this, up.
And another triangle like this that's on the other end of the threefold axis that points down into the blackboard.
What can you add as other extenders?
You can put in a mirror plane that is perpendicular to the fourfold axis.
And that leaves everything unchanged.
Now we've got a mirror plane passing through a fourfold axis, so we have to have mirror planes at 45 degree intervals.
And that will create mirror planes there.
This horizontal mirror plane goes through this fourfold axis, so there must be mirror planes at 45 degree intervals.
So there's another one like this and 90 degrees away as well.
The fourfold axis is perpendicular to a mirror plane, as are these other twofold axes, so there's an inversion center.
Here's a fourfold axis with a vertical mirror plane going through it.
And it has to have a vertical mirror plane going like this at 45 degrees away.
So that's the symmetry.
This is the regular symmetry of a cube or an octahedron.
And I will not, even if pressured, try to draw a pattern that conforms to that.
We've got the fourfold axis perpendicular to a mirror plane.
And this is called O sub h, O with a horizontal mirror plane.
The fourfold axis in 4, 3, 2 has got a mirror plane perpendicular to it.
The twofold axes all pick up mirror planes perpendicular to them.
There's an inversion center at the point of intersection.
So we label this axis a 3 bar axis.
So that's O sub h, 4 over m, 3 bar, 2 over m. And this, as I say, is the symmetry of a regular cube or an octahedron.
Nothing else we can do here.
There's so many symmetry axes all over the place that there's no way we can snake in any diagonal mirror plane.
Because there is no second axis of the same kind adjacent to either the twofold, the threefold, or the fourfold axis.
So we can't put a mirror plane in here between the threefold and the twofold.
We can't put a mirror plane in here between the fourfold and the twofold.
So we're done.
And this is the final and 30 second combination.
So there are 32 crystallographic point groups
Professor?
Yes, sir
So if m3m is a regular operation, are there symbols [INAUDIBLE]?
O is the symbol for 4, 3, 2.
And what we added as an extender is a mirror plane perpendicular to the fourfold axis.
That is just one way of adding an extender.
This is also Oi.
If we add an inversion center, we get all this.
And if you look at the ones have been honored by being designated by a symbol, the horizontal mirror plane takes the precedence.
So for example, for 2 over m, 2 over m, 2 over m, you've got two different vertical mirror planes, and you've got one horizontal mirror plane.
But it's called 2 over m, 2 over m, 2 over m. You can call it 2, 2, 2, m, m, m. That's also a possible symbol, but much more of a mouthful.
There's some arbitrariness to the symbol because the arrangement of symmetry elements is what's real.
And we decide how we want to devise a notation to label them.
And as we've seen, there are two different people who adopted a different code for giving them names.
So if it's rational and informative, it's a good notation.
Interestingly, the international notation, on the one hand, and the Schoenflies notation, on the other, complimentary.
The international notation tells you unambiguously what you have.
So 2, 2, 2 over m, m, m tells you three orthogonal twofold axes if you remember what came out of [INAUDIBLE] construction.
And perpendicular to each of them is a mirror plane.
And that's what you've got.
Schoenflies tells you how you derived it.
You took D2, and that's the dihedral group 2, 2, 2, and you added an h, a horizontal mirror plane.
And all hell broke loose.
And you got mirror planes perpendicular to all the twofold axes.
So there's a certain complementarity to the two different notations.
And the people who do diffraction and crystallography for the most part follow the international notation, the Hermann-Mauguin notation.
The people who do condensed matter physics use the Schoenflies notation.
Because it's more inscrutable.
And condensed matter physicists like to be inscrutable because it's how you gain respect.
So both notation survive, but are more prevalent in some disciplines than in others and vice versa.
Let's take a brief look ahead.
What remains to be done?
We now have the 32 crystallographic point groups.
These are the way things can be arranged by symmetry about a fixed point in space.
If we were to proceed in the same way that we did for the two-dimensional space groups, what we should do next is decide what sort of three-dimensional space lattices these symmetries will require.
And then proceed to drop each of the point groups into each of the lattices that can accommodate them.
And then use the tricks that we've used in two dimensions, take mirror planes and replace them by glide planes.
And then having done that, we would take the mirror planes and the rotation axes and interweave them.
And lest that job seem too daunting, let me point out that we already have 17 of the space groups.
All we have to do is take the plane groups, take a translation that's perpendicular to the plane of the plane group, and let all the rotation axes and mirror planes extend up indefinitely in three dimensions parallel to that translation.
So we already have 17 of the three-dimensional space groups.
They look just like the plane groups except they extend in a direction that's perpendicular to the plane of our original two-dimensional group.
So we're already a long way towards deriving a three-dimensional space group without really knowing it.
But what I would like to consider next is the lattices that are required for three dimensions.
And I've already given you how one can approach this problem.
Take the lattices that have been required in the two-dimensional plane groups.
And if the presence of a threefold axis and the base of the cell require, say, net that is hexagonal with a1 equal to a2 identically in magnitude and exactly at 120 degrees with respect to one another.
Let's let the third translation be perpendicular to the base.
If it's hexagonal, we have to have at least a threefold axis here.
And we've found that adding a threefold axis to that net gave rise to two other threefold axes in the center of the triangle.
So one lattice would be one in which the third translation, which we'll label c, is straight up.
In other words, perpendicular to the net that we derived in two dimensions.
What we will have as a constraint is that if we pick a certain translation that is not normal to the plane group, and let's say it terminates here in projection.
So the third translation goes up and over.
If there's a threefold axis at this end of the translation, there must be a threefold axis at the end of that translation.
And that mucks everything up, and we no longer have a group.
We can't have another threefold axis poking down through the plane of this two-dimensional plane group.
But it is perfectly OK if the third translation moves this threefold axis and puts down a lattice point and another threefold axis directly over this one.
Or alternatively, another choice for T3 would be this one.
That would put the lattice point in a threefold axis directly over this one.
So again, we have threefold axes extending normal to the plane of my drawing.
And I've created no threefold axes.
So with a threefold axis, there are three potentially different space lattices, one in which the third translation is normal to the plane of the plane group, another one where the translation terminates over the other two twofold axes.
And so this is the way in which one would proceed to derive the space lattices.
Buerger does not do it this way.
He doesn't look at the plane groups.
He looks just at where the axes sit in the nets.
No mirror planes whatsoever.
And I'll return to the point.
And the place where I think he's wrong is that if you add a twofold axis to a centered net, you get twofold axes in all of these locations.
These are the twofold axes in C2mm.
And if that's all you look at, there's no reason why this should be the plane group.
And it looks as though you can make T3 terminate over this twofold axis.
And that would give you a peculiar triple cell with three lattice points along the long diagonal of a rectangular net.
And you say, wow, 15 space lattices.
We're going to be famous, if not rich.
You can't do that because this plane group can exist only if there are mirror planes in here like this.
And then through these twofold axes, these have to be glides.
So you can't take 2mm and put it on top of 2gg.
It's impossible.
So there are 14 space lattices.
The 15th doesn't exist.
But it looks as though it's possible in Buerger's treatment when he looks just at the placement of the rotation axes alone and not the location of any mirror planes that are in the plane groups.
So that is where we're going to go next.
And that is a process that actually is surprisingly simple.
And we will get the space lattices very quickly.
We'll get a number of space groups along the way, as we've just seen.
And then we will cut to the bottom line and just look at how this information is tabulated in tables for you in a fashion that's analogous to the representation of the plane groups.
So we're pretty close.
We'll be wrapping things up in another two or three meetings.
What I would like to do in the time that remains though, the point groups are very difficult to visualize unless you look at real crystals.
So what I'm going to do is pass around a collection of models of actual crystals.
I'm not sure I can tell you what every one of these models represents.
These are curious things.
I would beg you not to drop them on one of the sharp corners.
These are made out of wood.
They're made out of pear wood in the Black Forest by elves.
No, that last part is not true.
But they are incredibly expensive.
Because the angles in these things have to be exactly right.
So they are made on the same sort of machine that you use for faceting diamonds.
It's something that has two degrees of freedom, so you can get a surface that you want on the material exactly parallel to a grinding surface.
And then you have a provision for advancing it normal to that direction by a controlled amount.
If it were not precise, you would not see sharp edges and sharp corners.
So these are incredibly expensive.
I, therefore, ask it that you hold them tightly with both hands.
Don't drop them on the floor.
And I'll take around a couple of handfuls of these.
I don't know if I have enough for everyone.
With the tables of the point groups in front of you, why don't you try to identify the point group that's possible in each of these models.
This is a big one.
So I'm sure you won't drop that one.
I don't think I have enough for everyone, so I'll start passing them out to every other person.
Maybe you can look on with the person adjacent to you.
Take one of these.
Pass those over.
Here you go.
I gave you that just to be mean.
That is an example of something that's called a twinned crystal.
You can't have reentrant faces.
Yeah, that's actually two crystals that are intergrown
So we can't identify the [INAUDIBLE]?
You can look.
Block one of them out in your mind, and try to imagine what it would look like
[INAUDIBLE]
Yeah.
Here's an example of another one.
That's actually gypsum.
And that's a very common twin in gypsum.
And if you could take this and rotate it-- this one doesn't rotate-- rotate it 90 degrees, then you would have a crystal that had a parallelogram shape.
And actually that's been rotated by 90 degrees about an axis.
That's within the base
No, no, I think that's just 4m or 4 over m
Yeah, there's no mirror there.
That's a 4 bar, I think
Oh, is that the inversion?
There's no-- neither.
I am a touchy-feely guy.
If I see something up here, I look down here and try to feel something that's parallel to it.
But you're right, you're right.
If I rotate 90 degrees-- remember rotoinversion is the same as rotoreflection.
So I could rotate to here and then reflect down, and I get this one
So it's 4 bar
It's 4 bar, yeah.
And you see in the cross section, it is square.
There's a little square on top here, so there is a fourfold aspect to it when you look at planes in a special orientation
I have a question about 4 bar.
4 bar in planes are twofold rotation
A twofold pure rotation.
Probably the best example of 4 bar is a tetrahedron.
Two faces on top, two faces twisted 90 degrees, and inverted down
So that means an 8 bar would be a fourfold proper rotation.
So why cannot I put an 8 bar in the point group?
Because it's only a fourfold rotation, and that's allowed
The reason is that there's no origin to translations.
So if you had translations in an 8 bar crystal, you would have four translations that were repeated by rotation and then another translation 45 degrees and inverted down.
But you can assemble those at this same point.
So you would have eight translations 45 degrees apart.
And that's impossible in a lattice.
Need another one?
I'm sorry
Do you need another one to look at?
No, I'm good
OK.
I can't have you just sitting there doing nothing
So is this half of this [INAUDIBLE] material?
Yeah, that's twinned by a 180 degree rotation on the 1, 1, 1 plane
Oh, yeah, the 1, 1, 1 [INAUDIBLE]
Yeah.
That is actually half of an octahedron.
This is a-- whoops, no, it's not.
Yes, it is.
No, I think that's just the threefold-- 3, 3, 3.
3 bar, 2 over m. And it's been rotated by 180 degrees, which is not a symmetry transformation.
So the 3 bar is rotated 180 degrees
That's a 3 bar, 2 over m?
I think so.
I think it's a 3 bar here, a face here, and a face down below, inverted, 60 degrees away.
And they're mirror planes.
And there two full axes perpendicular to those mirror planes, I do believe
I don't see the twofold axes
You're only seeing half the crystal, and I think that's the reason why.
This is the mirror plane here.
And this edge is parallel to that mirror plane.
And there's a twofold axis that comes out of the middle of that edge.
You can't see it.
That edge.
And the mirror plane is exactly parallel to that.
It's tough to see because you're only seeing half of it
[INAUDIBLE]
Yeah
This [INAUDIBLE]?
Yeah.
This is a fourfold axis perpendicular to a mirror plane.
There's a threefold axis coming out here and a twofold axis coming out here.
And there is a mirror plane perpendicular to the fourfold axis and perpendicular to the twofold axis also under the mirror plane.
So this is 4 over m, 3 bar, 2 over m
[INAUDIBLE]
That's the rotational symmetry.
But you get all sorts of mirror planes coming through here.
So really, it's this one.
Mirror plane this way, mirror plane this way, mirror plane this way.
So if we set it up relative to this, there's the fourfold coming out here, here's the fourfold coming out here.
And there are mirror planes this way and also this way, 45 degrees away.
Nothing to do?
Can I steal one that you're not working with?
There's a--
This one is [INAUDIBLE]
That's 2, 3 with no mirror planes.
Oh, no, no, that's not true.
No, that is 4 bar, 2m, believe it or not.
Look at the three different directions.
These two corners are the same.
There's a little, tiny line segment there
Yeah, so there is a mirror plane right here?
Yeah, but there's no mirror plane going through the other edges.
So there's a mirror plane here, mirror plane here.
And then there is twofold axes coming out of this little straight line segment here
Really?
Yeah.
And that's different from this.
So these two edges are the same
Where is the principle axis?
The principle axis would be this one
This one?
Yeah
So basically, I have this mirror plane right here.
Then twofold axis in the middle of the circle.
So now here, where is it?
Well, it's hexagonal.
So you want to back up a little bit.
Right there
This one?
No, sorry.
This one.
Threefold axis, twofold.
That's perpendicular to a mirror plane
So here's my mirror plane
Coming right out of this little edge here
Let's say like this.
No mirror plane?
No
Which one is this one?
It's like this one
Let's put it right here.
Here is--
Like this?
No, you've got to get it up like this.
OK.
There is a mirror plane
No, this is not.
No, the mirror plane is right here
Yeah, OK.
It goes up like-- oh.
Let me get it set up here.
OK.
There is the mirror plane.
This face is different from the others
And the twofold axes are on the edges
Yeah
So basically, there are three for one mirror plane
Here are the mirror planes coming through this way, this way, this way.
They're 60 degrees apart
You say there are three mirror planes?
Yeah, this one, this one, and this one
I don't agree.
This one is no mirror plane
You're right.
You're right
There is only one when you look from above.
But there are two for just [INAUDIBLE] 45 degrees-- or, not 45 degrees.
That's a triangle
It's a terrible one
Because if we decide that the top is one, the principle--
There's a twofold axis.
That's a mirror plane, and that's a mirror plane.
And it looks like it is 2nn.
I think it's 2nn
OK.
There's nothing further?
Nothing further.
Got that one?
4 bar 2m?
I had to look to find a twofold axis.
That was the tricky part
No, no threefold axis
Well, twofold
Yeah
Finding those was [INAUDIBLE]
That's it
I can see the mirror planes
That's the twofold axis
Yeah
And this two on top and two underneath skewed by 90 degrees, that's a 4 bar.
So that's 4 bar 2m.
You have nothing to do?
Oh, you're talking with them,
What is this one actually?
There is 4, 4, 2, 4, 3, 4, 3
Right, so it's cubic.
And mirror planes are going down through the twofold axis.
So that's enough to tell you that it is this.
So let's set it up here.
Fourfold axes are coming out of the points.
So you've got mirror planes this way, mirror point 45 degrees.
Here is this twofold axis.
Here is this twofold axis.
And here are the twofold axes that are in the same plane.
And here's the other fourfold axis
I see.
That one's not fair.
This is actually two crystals that are grown together by an operation that is not a symmetry element of the crystal.
And actually these two crystals have symmetry 2 over m. And they are rotated relative to one another by a 180 degree rotation.
It's not a symmetry element.
So this is something that's called a twinned crystal
Twinned?
Twinned crystal.
So it's really two of them.
And if I could twist this one around so that the crystal continued on in that direction, it would be something that had a lozenge-like shape.
So I should put that one away.
That's only confusing people
Professor Wuensch?
Yes, sir
Is that a 6 over mmm?
Yes, that's 6 over m, 2 over m, 2 over m. Six folds.
Two kinds of twofold, one out of the edge, one out of the corner.
And a mirror plane perpendicular to each of those.
6 over m, 2 over m, 2 over m
So that's the same thing [INAUDIBLE].
So the 6 over m comes from this way.
And then the 2 over m is there
Yeah, so it's [INAUDIBLE]
Yeah, exactly
Let me put that one away.
It's just confusing people
Oh, that one was easy
That's a twinned crystal.
It was easy?
You found it easy?
Yeah, so 3 over m, right?
Yeah, I guess.
This looks as though it might be the corner of an octahedron because you don't see much of it.
But these two faces and these two faces are not related by a fourfold rotation
Well, either way, there wouldn't be anywhere to put a fourfold rotation
Yeah, right.
So if you look at the whole thing, this is actually two intergrown crystals when you see this sort of reentrant angle.
This is two crystals grown together by an operation which is not a symmetry operation.
So this is something that's called a twin, a twinned crystal
But the symmetry is nevertheless 3 over m, correct?
Of the whole thing?
Yeah, yeah, of the whole thing
And this is just 3
No, this is 3mm.
There's a mirror plane down there and a mirror plane down here
You're right
No twofold axis because this end is different from this end
So this is a regular solid, right?
What is it called?
This is a rhombic dodecahedron
Oh, the one you were talking about the other day
Yeah.
This is 4, 3, 2.
And then mirror planes too all over the place
So all of these solids do not have to be members of this list, right?
Because these are finite objects.
They can be [?
twelvefold ?]
and--
They could be.
These actually, in fact, are models of real crystalline materials.
And they're made with the faces and the relative sizes that these minerals actually have

So somewhere floating around is something that's very characteristically quartz.
And there's one that's a flat one with a reentrant angle in it, that's gypsum
[INAUDIBLE]
Yeah, that's fourfold.
Twofold here.
This goes into this.
No mirror planes because these things are inclined.
So it's 4, 2, 2.
Good.
It's easy when you know what to look for.
When you say, they are only a few possibilities.
And if I see a fourfold axis in it, that narrows it down to two or three
Like that.
That's an inversion center running through here.
And there's no rotation there [INAUDIBLE]
[INAUDIBLE] here?
Here through this way?
No, because this doesn't map [INAUDIBLE].
Does that map that by a mirror?
The mirror [INAUDIBLE]?
It cuts across this way
Yeah, I could buy that
You've reached a consensus?
We've got a mirror, and we've got an inversion.
But we're not sure what else we've got
While I like the inversion, I'd also like the mirror.
I don't see the mirror.
All I see is a really an inversion axis there.
I think I'm missing that mirror that you're saying
There's a mirror that goes down this way.
And there's a twofold axis here
OK, that was the one we missed
So that's 2 over m
Over m. And the inversion just falls out
Yeah, that's in there too.
Inversions are nice.
You can feel inversions.
You put your finger on one face and your finger on the other face.
[INTERPOSING VOICES]
Get it?
So this is this one, right?
Yes, very good.
And that is a silicate mineral called garnet
What's that?
Garnet.
It's actually a silicate mineral, which is used as a gemstone sometimes
So this should be a 4, 3, 2 face
Yep
But I don't know if it's the end of the [INAUDIBLE].
Could be m3m
Where are the-- it doesn't an inversion center
Hmm?
It doesn't have an inversion center here
Oh yeah, it does
Oh yes, it does.
Yes, it does.
Actually, the thing to look for are the high symmetry things.
Your eye spots them.
That's got a fourfold axis in it.
OK So are there other fourfold axes?
Yes, yes.
So that has to be based on 4, 3, 2.
Because you have three orthogonal fourfold axes.
Threefold is here.
Twofold axis is here

OK, thanks
There are-- oh, I already saw those ones
You want to look at some more?
All right
Enough for one day?
Professor Wuensch, I've been staring at this one for such a long time.
And I couldn't match it with anything on here
This looks like a tetrahedron except the face is puckered up into this little thing.
So the way I would start with saying, OK, this is tetrahedral in which case if I make these things flat, that's a perfect tetrahedron.
Four sides.
But what's happened is that this face is not quite 1, 1, 1.
The 4 bar axes come out these three directions.
And they have got twofold symmetry, mirror planes running through the twofold axes.
This is the 4 bar axis with mirror planes running this way, this way.
So this is actually 4 bar.
This is still further.
It's a cubic crystal.
This is based on the tetrahedral symmetry.
Here is the 4 bar axis.
Mirror plane this way, mirror plane this way.
Another 4 bar coming out of these two edges.
And then these are the diagonal mirror planes.
So this is it
I think I was looking for a threefold axis in the center
No, the 4 bar comes out of the edge of the tetrahedron.
And this is the other 4 bar coming out here.
Threefold like this and mirror planes going through the threefold axis like that
I see it
It's hard when you look at this thing.
I never saw this thing before.
What's here?
But if you know the results have to be 1 of these 32 possibilities, and this is obviously cubic, and it's based on the tetrahedron, that means there are only three possibilities.
No mirror planes.
If you find one mirror plane, you ask yourself, does the mirror plane go through the threefold axis, or does it miss the threefold axis?
It goes through the threefold axis, so it's got to be this.
And then you know just what to look for
Is it just [INAUDIBLE] m?
This is a dirty one.
This is a dirty one.
If you look at this very carefully, this face is the same as this face, but it's not the same as that one.
It's a little bigger.
So if you say, what's in here, these two things do not come together at a common vertex like these two.
There's another little line segment between these faces.
So there are no threefold axes coming out of this.
Looks like it might be tetrahedral.
Clearly, a mirror plane going this way.
And then there is also a twofold axis coming out here.
So there's a 2 over m and another 2 over m. I did this once before with somebody, and this is the hardest one in the whole set.
So a mirror plane this way.
There's got to be another mirror plane because there's a twofold axis.
4 bar.
Up, down, up, down.
So this is 4 bar, 2 over m. 4 bar, 2 over m.
I think it's time that we got started.
I haven't given a problem set out for a couple of days.
So I have one that will cover some material that we'll go over with and develop today and some new material that we're not going to introduce for a while
[INAUDIBLE]?
Yes, this is Thursday.
And this is for the near future.
I've given you two-- two a week before.
So it's the adjective you're objecting to not the problem set, of course.
Today we're going undertake another major step in our development of three dimensional crystallographic symmetries.
We have just completed derivation of the crystallographic point groups, so these are the macroscopic symmetries that crystals can have.
And then, by analogy, to what we did in two dimensions, the next step would be to take each of these point groups and hang it at a lattice point of some three dimensional lattice that can accommodate them.
We have not as yet developed the three dimensional lattices, and that will be our agenda for today.
Seems like we've already done that, haven't we?
We showed that in two dimensions, a cell can be oblique, it can be rectangular, it can be centered rectangular, it can be square, or it can be hexagonal.
So these are the different kinds of bases that one can have for the cell, right?
So there should be the same number of space lattices.
Actually it's more complicated than that.
And let me remind you by another handout what the two dimensional space groups, the plane groups, look like assembled in all of their glory on one sheet.
And there you see the different ways symmetry can be placed in these five different kinds of lattices-- oblique, rectangular, centered rectangular, square, and hexagonal.
Now a lattice in two dimensions, a potential base for a unit cell, can be rectangular if and only if there is symmetry within that net.
That demands that it be exactly rectangular.
In other words, if a lattice is to be rectangular in the base of the cell, it has to have in it one of the symmetries that correspond to those in the plane groups that give and require, indeed demand, a rectangular lattice.
Similarly a lattice can be truly square, identically square, only if there's a fourfold axis in it.
That demands that it be square.
The same could be said for either threefold or sixfold symmetry.
One of the groups, P3, [?
P3, ?]
mP3, 1m, P6, and P6mm, has to be in the base of the cell if that cell is to have the specialization.
So let's give an example with a twofold axis.
Let's suppose we have a base of the cell that has an unspecialized shape-- two translations that are unequal in length and an angle between them which is a general obtuse angle.
That's about as general as you can get.
But suppose that this net has as well a two dimensional-- a twofold axis in it.
And suppose we pick our third translation such that it picks up this net which is going to be the base of our three dimensional lattice and translates it over to this point in the base.
Well that translation picks up everything.
It picks up not only the two translations that we've been calling T1 and T2 to this point, but it also picks up all the twofold axes that are hanging on these lattice points.
So if this is the terminus of T3 and projection, it will have picked up a twofold axis and plopped down a twofold axis here where no twofold axis exists.
And we can take one of the theorems that we've seen, namely that the operation A pi, followed by translation, is equal to a new rotation operation through pi located halfway along the translation.
And turn this around and ask what happens when we combine two rotations through 180 degrees about different points.
And let's just draw it out to see what happens.
Let's say there's a twofold axis here.
It takes motif number 1, moves it to number 2.
And then here's a second twofold axis and this takes number 2, rotates it over here to number 3.
I is number 1 related to 3?
But in terms of this first theorem, not surprisingly, we have introduced a translation into the space which is equal to twice the spacing between the twofold axes, which I"ll label as delta.
So in other words, if we put a twofold axis down in this place in addition to the ones that we already have, we have created, in the base of the cell, a new twofold axis in this location.
That just mucks everything up because it's going to rotate the translations that we have, it's going to rotate the twofold axes that we have.
And we will not any longer have a group.
We will not have a set of operations that closes upon itself.
So the conclusion then is that if we're going to safely pick a third translation, T3, and combine it with a net that has twofold axes in it, in order to get a group, we are going to have to ensure that the twofold axes line up in every net in the stack.
And there could be several ways of doing that.
We could pick T3 so that it goes straight up.
And then the twofold axes are just all slit up parallel to themselves, and they remain in coincidence.
Or we could pick T3 so that it terminates over the point that's halfway along T2.
And then this twofold axis gets picked up and put down over one that's already there.
This one gets picked up and put down over one that's already there and so on.
So that's an OK choice.
That's a safe choice as well.
Similarly we could pick T3 such that, within the plane of the net, it had a component that was one half of T1, so that it picked up this net and slid it over to this location.
So once again all the twofold axes lined up.
Or, in the same fashion, let it terminate over the center of the cell.
So those are the four ways we can combine an oblique net that contains a twofold axis in such a way that the lattice points in the existing twofold axes remain invariant and are not moved into locations that create new twofold axes.
OK so let's back off a little bit and start with plane group P1-- no symmetry at all.
So this is T1 and T2.
If there's no symmetry whatsoever in the base of the cell, we are free to pick the third translation in any orientation that we choose.
So trying to sketch it in three dimensions.
If this is T1, and this is T2, T3 can be in any orientation such that not only is this angle general, but this angle is general, and this angle general as well.
And if I would try to carefully complete the parallelepiped so that its geometry looks reasonable.
This is going to be the general oblique parallelepiped with three translations that are unequal in magnitude, and three angles between these translations.
Let me write them as the angle between T1 and T2-- not equal to the angle between T2 and T3 necessarily, not equal to the angle between T3 and T1.
And all of these angles are completely general and can assume any values that they like.
So this then is the lowest common denominator.
This is a totally unspecialized space lattice.
It has the shape of the general parallelepiped-- no special relations between any of the translations whatsoever
[INAUDIBLE]
I say they're-- Oh I'm sorry, magnitude are not equal.
[INAUDIBLE] I meant to put not equal signs there.
Yeah, they can have any length they wish.
They're not equivalent.
Any other questions?
OK let's go back then to the case of an oblique net to which we've decided to add a twofold axis.
And that gives twofold axes in the locations that we've become familiar with in the plane group.
And if I pick the translation so that it goes directly normal to the base of the cell, I'm going to have a new kind of lattice.
The lattice is going to have two angles between axes that are exactly 90 degrees, not approximately so, but exactly so, because only if that translation, T3, is exactly normal to the base of the cell will the twofold axes in all of these nets lineup exactly.
So here we have a degree of specialization.
The translations remain unequal in magnitude.
They can be any length they wish and not change things, but two of the angles-- axial angles, T1 to T2, is identically 90 degrees.
The angle between T2 and T3 is identically 90 degrees.
And the third angle, the angle-- I'm sorry, T1 and T2 is my general angle.
So this can be anything.
So that can be any general angle.
T2 and T3 and T1 and T3, want to be exactly 90 degrees.
So we're getting space lattices that have a degree of precise specialization.
And the reason is that the geometry of the base of the cell is inseparable from symmetry in the base of the cell that demands the degree of specialization that makes the space lattice unique.
So let us look at a couple of the other choices for the third translation that will leave the twofold axes in coincidence.
Another choice would be to pick the third translation such that it terminated exactly over the midpoint of T2.
So here's T1.
Here's T2.
Now what I'm going to do is something I have not done to this point.
Having this translation T3 terminate exactly over the midpoint of a translation [INAUDIBLE] was clearly a specialization.
If I were to go up two translations, T3, I would be directly over the end of T2, and that is a much more convenient way of demonstrating that special feature.
So let me pick a new T3 prime.
It goes up exactly normal to the base of the cell, and in so doing, I will have defined a cell which is a double cell.
And the catch is an extra lattice point in the middle of one of the pair of faces.
So I'll refer to this general sort of situation-- this is a double cell.
And I'll call this, in words, a side-centered cell.
And it has, with this redefinition of translations, exactly the same degree of specialization as before-- magnitude of T1 not equal to the magnitude of T2, not equal to the magnitude of T3.
And it has the same specialization of the angles.
The angle between T1 and T3 is exactly 90 degrees.
The angle between T2 and T3 is exactly 90 degrees.
And the angle between T1 and T2 is general.
General, but by convention we will assume the obtuse angle rather than the acute angle.
OK so by picking this double cell, we've defined a cell that has a twofold redundancy.
But in doing so, we have gained the advantage of showing that this specialized geometry is exactly the same as this one-- in the same sense that in two dimensions, the primitive rectangular cell and the centered rectangular cell were both cousins.
They were both had orthogonal translations.
But the other thing that we gain is that even though we can't have an orthogonal coordinate system, if we use the three translations as the basis of a coordinate system, at least we have two 90 degree angles on our coordinate system.
And operationally, that is going to be an enormous advantage.
Yes, sir
So is T3 [INAUDIBLE]
T3 was selected so that we picked it as 2 T2.
I'm sorry, 2 T3 minus T2.
That was my new T3 prime.
I went up two translations in T2 and then back T2, and that put me directly over the lattice point from which my translations emanate
So your T3 is actually [INAUDIBLE]
It goes to the next level up in the stack of nets.
If I would draw this in projection, which is not nearly as clear in some respects, this would be the base of the cell.
Let me use x's for the next level up.
So here's the net offset by half of T2.
And then if I go up two of these T3's and then back by T2, my next layer up is directly over the first one-- the third layer up is directly over the first
Shouldn't it be T3 prime then?
Yes if I were to be consistent, I should call this T3 prime, which I did here.
But now I should call that T3 as well.
I got a little bit careless because we're going to forget that we ever had this T3-- to find this selected as a stacking vector.
OK well in P2 there are four different kinds of twofold axes.
So we could have the third translation terminate over either of the remaining pair of twofold axes as well.
I think that it's apparent that if I pick T3, the original T3, as one half of T1, so that the next layer up terminates in lattice points in this orientation.
And then I pick a T3 prime, which is equal to 2 T1 minus-- 2 T3 minus T1, I will have gotten again a side-centered lattice.
The only difference is that it's going to be a different pair of faces.
It's going to be the O1Oa faces that have the extra lattice point.
So that is not fundamentally a different sort of lattice.
It is also side-centered.
And differs from our second result only in whether it is the side face with the shortest translation or the longest translations of the base that bears the extra lattice point.
One final choice though does result in a new lattice with special features.
And this would be one where I pick the third translation such that it terminates over the center of the net below.
So let's let this be T1, this be T2, and pick T3 so that it moves the origin lattice point directly over the center of the parallelogram face below.
And then if I go up two translations, T3, and then subtract off T1, and subtract off T2, I will again have a T3 prime, which is exactly normal to the base of the cell.
OK this is a type of lattice that you've come to know and love.
In the cubic system, this is referred to as a body-centered lattice
So then, T3 is just [INAUDIBLE]
T3, the original choice, would pick this original net up and move it so that that net was moved-- such that it's corner was over the center of the original net.
So if we go up two translations, we have lattice point over lattice point, and we redefine T3 by going back 1 T1 and minus 1 T2.
And that makes it exactly normal to the base
Not on the sides?
Nothing on the sides .
Now in fact, if we try to do that, we would have something that's not a lattice.
If we tried to have some additional lattice points in the center and the edges of this upper level, we would have destroyed the original T1 and T2 in the base of the cell.
Let me do one or two more and then I think the way in which one proceeds should be quite clear.
If you take this list of all of the plane groups, if we would like to have a three dimensional cell that has a rectangular base, the base can be rectangular only if there is a mirror plane in that net or a twofold axis with the mirror plane.
The two situations are different.
So let's examine first the case of a rectangular net that has either a mirror plane or a glide plane in it.
And they're both going to have the same sort of constraint.
If I have a mirror plane, that-- let me use a slightly wiggly line so the mirror line is not confused with the edges of the cell.
So here are two translations, T1 and T2.
And if I pick a T3 that moves this rectangular net up and over itself, we have to do so in such a way that the mirror planes coincide.
Because if I have a mirror plane in space that passes in proximity to a lattice point, I can turn my theorem around and say that a translation, T1, combined with a mirror plane that is a reflection operation that's removed from the lattice point, I would have to introduce a new translation to delta which is going to be incommensurate with T1.
So I have to pick my third translation in one of two ways.
Let me indicate that by drawing a circle.
T3 can terminate anywhere over this mirror plane.
I could pick this net up, slide it parallel to T2, and as long as I put it down so that this is the lattice point in the next layer up, that will be a legitimate placement of the net.
Because all the mirror planes are simply slid into coincidence with one another.
So if I sketch that thing in three dimensions-- this is the base of the cell and it has a rectangular shape.
And the third translation goes up at an angle.
That's going to be exactly something that I've made before.
And the only difference is that the oblique angle is now on the vertical plane but that's a right angle.
And this stays a right angle.
So that's exactly a lattice of the shape that we had last time with twofold axes, except this cell is sitting on one of its rectangular sides rather than sitting on its oblique base.
So that is something that's not new.
And if I would pick the translations so that T3 had a component that was one half of T1, plus some amount z that is perpendicular to the base, I can redefine that in terms of a cell that has one pair of faces inclined to one another.
And the difference is going to be that the lattice point would be caught in the center of the cell.
This would be my original translation T3 and I'll redefine that as a T3 prime.
And that gives me two right angles and one general angle.
So those are exactly the same lattices that I obtained from stacking a net with a twofold axis in it.
Let me introduce now another piece of jargon as we go along.
The relative angles between the translations are going to determine the shape of the coordinate system that we pick along those translations to specify the geometry of features within the lattice.
So we've had a first case, and this was just a general oblique net.
So we had a T1 not equal to a T2, not equal to T3, and all three angles were general.
All three pair of axes are inclined.
And this coordinate system, if we want to refer to it in terms of words, is called a triclinic.
This is the triclinic system.
And the triclinic system is short for coordinate system.
So this is as general as things get.
And if this is the shape of the lattice there is no symmetry.
Which is the same as saying that the point group, the only point group that can fit into such a lattice, is point group 1.
And then we hit another sort of geometry.
And that is where the three translations had a general angle-- a third translation that was exactly normal to the plane of the first two.
And one pair of axes is inclined.
And you can see that this notation is a little strange, perhaps, but consistent.
And it says something about the nature of the arrangement of cell edges.
So one pair of axes inclined is called the monoclinic system.
And this can contain lattices that are either primitive, side-centered-- with either the longest or shortest translation in the parallelogram that's coming out of the face that's centered.
Or we saw there was one final possibility.
And that was body-centered.
So we'll refer to these lattices subsequently as monoclinic primitive, monoclinic side-centered, or monoclinic body-centered.
These lattices were compatible with point group 2 or point group m. We obtained a lattice of the same shape with point group m or with point group 2.
If it works for 2 and it works for m, why not both at the same time.
It would also work for 2 over m-- twofold axis perpendicular to a mirror plane.
Any question at this point?
Yeah?
Also 2m, right?
[INAUDIBLE] Yeah
No,because as soon as you got a mirror plane, then you have to have a rectangular net.
So it's going to be-- it's going to fall into this family.
You're right, a mirror plane by itself and we've got that here.
That's where the mirror plane falls in the side of the thing.
You're right.
But if it has a rectangular shape, then we have to have either orthogonal mirror planes and a twofold axis.
The base, if the base-- yeah.
OK. Part of the amusing part of this game is that you can, upon appropriate redefinition of cells, come up with a given coordinate system that defines more than one lattice.
And I'll give you-- because you get a little space doing this-- I'll give you a two dimensional example.
Here is one cell.
This is T1, this is T2.
This is a primitive cell.
Nobody should bother to write this down.
Here's a T1, here's a T2.
So anybody know what that is?
This is a primitive cell.
This is a jail cell.
Couple sniggers, thank you.
Still another one-- this is T1, this is T2.
Anybody guess what that is?
That's a soft sell.
I got a million of them.
Know what that is?
It's an underground cell.
Here's one that's kind of dated-- I don't think anybody will get this one.
Does anybody know what that is?
That was a misadventure of the Ford Motor Company that was named the Ed cell.
Well, I'll give you equal time.
Anybody who, after intermission, wants to respond in kind, I'll let you have an opportunity to do so.
But obviously these are not standard crystallographic cells, but they make for a smile in an otherwise dreary business.
I don't know how to proceed from here.
I have a set of notes for you which carries all this out in incredible thoroughness.
So let me pass this around.
Let you take a copy of this and we won't go over every single thing, but I think just to outline quickly the sort of choices that we have.
We haven't yet finished with the rectangular cell, but if it's got something like 2mm in it.
with twofold axes here and mirror planes running this way-- P2mm in the base, P2gm, P2gg, can be stacked only in such a fashion that T3 brings twofold axes and mirror planes into coincidence with one another.
So T3 could be 0T1 plus 0T2 plus some amount z which is straight up.
Or it could be equal to one half of T1, 0T2, plus z.
That would be making the origin twofold axis fall over this location.
Or T3 could be equal to 0 plus one half of T2 plus z.
And they're both going to, upon redefinition, give the same result.
Or T2 could be equal to one half of T1 plus one half of T2 plus z straight up.
And upon redefinition, the first one that would be a brick-shaped unit cell.
That's going to be a primitive cell.
Picking T3 with a component of either one half of T1 or one half of T2 is going to be something that gives us a cell, which can again be redefined.
This was the original T3 in terms of a T3 prime, which is equal to 2T1 or 2T2 minus [?
2T3 ?]
minus T1 or T2.
This is going to be side-centered.
And the final choice where it is a T3 that's one half of T1 plus one half of T2 can once again be defined in terms of-- redefined in terms of a body-centered lattice.
OK this is a coordinate system where all interaxial angles are 90 degrees.
The translations, however, have arbitrary magnitude with respect to one another.
All of these cells could be described as rhombuses.
In this case all angles in the rhombus are 90 degrees, so this is called the orthorhombic system-- orthorhombic coordinate system.
All the angles between the axes are 90 degrees.
So we can get that out of any one of the rectangular groups that has twofold axes and symmetry planes in it.
And conversely, take one of these lattices and let the symmetry elements that we found in the plane group extend through them and you've got a three dimensional space group.
So without emphasizing the fact is we go along, we're picking up a healthy number of three dimensional space groups as we go along
[INAUDIBLE]?
This is primitive.
It's also orthorhombic -- same dimensionality, three translations, and they're all mutually orthogonal.
That's true of these other cells except they're double cells in the case of the side-centered and double as well for the case of the body-centered.
It's time we introduced some other conventions for notation in describing lattices.
We've been using T1, T2, and T3 to designate the three translations that are used to define the cell.
One has to have some sort of rules for picking a standard unit cell so that two people can do an x-ray diffraction experiment, let's say, on a particular material and end up defining the lattice in exactly the same way.
So let me now list, so that we can use these labels as we go along further, conventions for selecting cell edges.
OK one convention is that one should select the shortest translations in the lattice.
And these shortest translations, if there's no symmetry, will be [?
dignified ?]
by calling them a, b and, c-- standing for the x, y, and z directions.
One picks always, if there's nothing special about the translations in the triclinic system, the magnitude of b greater than a.
And in a rational world-- every so often you come to a point like this and it feels almost the silliest talking about underground cells and jail cells-- in a logical, rational thinking world, you would do this.
This is not what one does-- for reasons that have to do with the perverse nature of crystals.
Take b greater than a, one does that always, but you pick c as the shortest translation.
Why?
It has something to do with the peculiar behavior of crystals.
If you were working with a crystal that had the shape of a needle, in other words, greatly elongated in one direction so that the crystal looks something like this.
How would you draw a picture of that?
Would you put it straight up and down like this or would you put it on its side.
Takes up a lot of room this way.
You'd almost certainly instinctively put it this way.
Now if you didn't know anything about the translations down inside the innards of that crystal-- and this was the position that the people who did crystallography solely on the basis of crystal shape and morphology-- and had no idea if there was even a lattice inside of the crystal that caused this regular shape.
When you draw a coordinate system x, y, z, z always goes up.
Who draws a Cartesian coordinate system with z going this way, x going this way, and y going this way?
It's obscene.
We always put z going straight up and down.
You do the same if you don't know what the lattice translations are with the labels on the axes of the crystal.
So this is a, this is b, and this is c. And there were tons of pages published with drawings of crystal morphology on them.
And then along came x-rays.
And when they begin to determine lattice constants of a the material that looked like this, almost invariably they found the translations inside of a needle-like crystal corresponded to a unit cell in the shape of the pancake.
Crystals, when they have a very, very short lattice vector, almost always grow most rapidly in that direction.
So this is a terrible pickle.
Here are reams of books that published descriptions of crystals that had a needle-like-- an acicular shape, if you want to use a fancy term, which had c as a direction which corresponded to the shortest translation in the crystal.
So what do you do?
You don't want to have to redraw all these figures and redefine all these directions in the crystal.
So what you do is cop out.
And you make c the shortest direction in the crystal.
You think this field is rigorous but people wimp out and they do something that's contrary to the logical thing to do.
Interesting question is why crystals should grow most rapidly in the direction of the shortest lattice vector.
And I think I know the answer to that-- I've never seen anybody express it in the writing.
The clue comes from some early work that looked at the growth of the crystal from solution.
And this was done first with-- I think it was some sort of iodide.
I can look it up-- something like cadmium iodide.
And this material in solution grew very, very slowly in the form of hexagonal plates.
And continued to grow most rapidly in this direction.
This was the direction of the shortest lattice translation.
Then all a sudden something happened.
And suddenly the rate of growth and the nature of the morphology changed.
The crystal took off like a bat in these directions and finally ended up in a needle-like fashion like this.
This was an observation that led to the dislocation theory of crystal growth.
The interpretation was that in the early stages of growth, the crystal is growing by accretion of atoms on to these planar surfaces.
And the probability of an atom sticking is less high when the atom is on the flat surface than when an atom is on the surface that might have a little crevice in it so it can bind to two planes at once.
So the original postulate-- and now we all take this for granted-- is that if there was a flaw such that the crystal developed a dislocation on this surface, and an exposed ledge, atoms would accrete and stick more tenaciously to this exposed ledge of the dislocation.
And this is the so-called dislocation-- screw dislocation mechanism-- for crystal growth.
And after the crystal has grown for a while, you can see spiral steps on the surface of the crystal that correspond to this dislocation winding around.
OK so why should crystals grow most rapidly along the direction of the shortest translation?
It's going to be a lot easier to create a screw dislocation with a Burgers vector for a translation that is short, rather than the translation that's 10 or 20 angstroms that the Burgers vector would have to involve-- many, many layers of atoms.
If the translation is very small, just two or three atoms, it doesn't cost you much work to make a screw dislocation.
So I think that is the reason why crystals with very short lattice translations grow most rapidly in that direction, because it's easier to make the screw dislocation with a Burgers vector that's parallel to that translation.
So here's the first flagrant bit of logic that flies in our face.
Let me give a few more definitions and then we'll return to this.
If we have a crystal in which two translations are equivalent by symmetry-- and we would have such a situation if we picked a translation that was perpendicular to a square net.
And would we call the base of the cell something defined by translations a and b?
If this is a fourfold axis, what goes on in this direction is identical to what goes on in this direction by symmetry.
And if these two directions are identical, why call one of them a and one of them b?
So if two directions are equivalent, and this is true of the square net, what you do is call one a1 and you call one a2-- because they're the same thing.
So let's call them both a.
And the third direction then is c. And in such a situation here, c is always defined as the unique direction.
What do I mean by a unique direction?
This is the direction of high symmetry in crystals that are based on square bases and hexagonal bases.
Yes sir?
You pretty much contradict yourself if c is the longest [INAUDIBLE]
Not always, not always
Well, I mean in that picture
Well I drew it that way.
You know why?
Because if I draw it like this, the top of my cell doesn't get in the way of the bottom of my cell.
And if I want to draw a small c translation, then I get into geometrical complications that doesn't look nearly as nice
Well, then I'm just saying two [INAUDIBLE]
No, no, no.
It's different for different systems.
This is what is true for a triclinic system.
And in the case of a brick-shaped unit cell-- orthorhombic.
So this is done for triclinic.
It's also done for orthorhombic.
Not for monoclinic.
For monoclinic, there is a special direction because there is a direction of a twofold axis or maybe a twofold axis perpendicular to a mirror plane.
In that case, if we were rational, we'd call that 1c.
OK I'll close with an opinion poll.
I will invariably, and I'm sure most people do, draw a crystal with a square base, with the c axis the longest.
And if we have someone drawing a sketch of a cell in which the base of the cell is a hexagonal net, we'd again call one a1, one a2.
And perpendicular to that would be the same direction.
Say look, there I did it again.
I made c longer than a1 and a2.
All right.
Let's look at hexagonal crystals.
Let me ask for a poll.
What percent of all crystals with threefold and sixfold axes do in fact have the length of c greater than the length of a?
50/50?
How many would say equally probable to have it the longest or the shortest?
No opinion.
Let me tell you the astounding fact.
Seventy percent of all hexagonal crystals have c greater than a. Astounding.
Let's look at the next-- haven't really defined these terms-- tetragonal crystals.
These are crystals that have a fourfold axis in them.
What percentage have c greater than a?
This is almost a first cousin of the cubic crystal or all three translations are equal.
How many would say more have c greater than a, fewer have c greater than a?
One vote.
More?
That's a good answer, but it's only about 60 some percent.
So let's go down to orthorhombic crystals.
And orthorhombic crystals are the ones that have a brick-shaped unit cell.
How many have c greater than a?
It's declining.
So if you extrapolate 50/50?
Zero?
Who said that?
Are you a wise guy?
You weren't supposed to give the right answer.
It's 0 percent.
It's 0 percent because there's nothing that makes one direction more special than any other.
So you decide on the labelling of axes on the basis of their relative lengths.
And you, by definition, make c the smallest axis.
Congratulations.
To the rest of you, gotcha.
So that's a good note on which to quit.
And a few remaining mysteries of coordinate systems and lattices will be expounded in the second half of our lecture.
Since z, that's straight up.
And that is going to provide for you a cell that has the shape of a square prism.
This would be a1.
This would be a2.
And this would be a3-- I'm sorry, this would be c. And if you took the choice for the third translation as 1/2 half of a1 plus 1/2 of a2 plus some amount z up above the base of the cell, and then redefined a T3 prime that would be equal to-- sorry.
What am I doing here?
Yeah.
This is right.
1/2 of a1 1/2 of a2, and z straight up.
Define a T3 prime as 2T3 minus a1 minus a2.
And that will define for you, again, a cell in the shape of a square prism with a1 and a2.
And now the translation that went up directly over the center of the base of the net below.
Twice that minus a2 minus a1 brings you back to a third translation T3 prime that's normal to the base.
So both cells have the same shape.
This is a primitive.
This is a body-centered tetragonal lattice.
And the symbol that's used to represent the body-centered lattice is I.
One of the few cases, again, where one has to be bilingual.
The German word for body-centered is innenzentriert.
So there's Schoenflies at work again.
So primitive body-centered represented by I.
And that brings us to one final case, and that is the plane groups that have a hexagonal shape.
And here we have a funny situation in that the same choices of the third translation do not work for all of the plane groups.
If you have a base to the cell-- and again, the two edges of that net are identical by symmetry-- if it has a three-fold axis in it and nothing else, than there are two choices.
You can either have T3 be equal to 0a1 plus 0a2 and an amount z straight up.
And that will give you a cell that is a primitive cell.
This would be a1.
This would be a2.
And just as we did for tetragonal, we'd call the third axis c, This will work for a three-fold axis.
So there is a space group P3.
This will also work for all of the other groups.
So there's P3M1 and a P31M.
And a P6 will work.
And a P6mm will work.
So look it here.
We've got five additional space groups in addition to the lattice type.
This is the primitive hexagonal lattice.
For a three-fold axis, we have another choice.
We could pick T3 so that it took the origin three-fold axis and put it directly over this one.
Looks as though you could have a different distinct lattice if you move the origin lattice point over the three-fold axis that sits at a location 1/3, 2/3.
This one is at 2/3, 1/3.
Let me demonstrate, I think convincingly and clearly, that those two choices are exactly the same thing.
And the way that I can show that is, once again, to draw this net a1, a2.
And suppose I pick the part of the offset of T3 that is within the plane of the net as going from this lattice point to this one here.
Then if I go twice that translation T3, that's gonna put me directly over this three-fold axis.
And if I go three translations T3, I'll be directly over the lattice point on the ground floor.
So putting the original lattice point over this three-fold axis or this three-fold axis is exactly the same thing.
And I can change one into the other just by flipping this thing 180 degrees.
So it's obviously the same choice.
This description of T3, if I define the direction of c now as 3T3 defined in this fashion, minus a1 minus a2, is gonna be a funny situation.
It's gonna be a peculiar sort of double body-centered cell.
Of course, along the long diagonal of the hexagonal cell, I'll have one lattice point that's 1/3 of the way along that long diagonal, and another interior lattice point that's 2/3 of the way along that long diagonal.
OK.
This double body-centered cell, so to speak, actually can be redefined in terms of a primitive cell, not a primitive cell that is this particular description of a primitive hexagonal lattice.
This is not a primitive hexagonal lattice.
But suppose I go from this interior lattice point along the long diagonal of the cell to this location and call that a1.
Remember that there is a three-fold axis sneaking down through the center of this triangle, goes through the lattice point, and comes out the center of the triangle below.
If I go up to this lattice point and up to this lattice point, I have three translations that are straddling the three-fold axis.
And that three-fold axis rotates one into the other.
So this new definition of a cell that is in fact primitive a1, a2-- and they're all equivalent by symmetry.
I always get in trouble when I try to draw it, but that actually defines a cell with a shape that we've not seen before.
That is a cell that has the shape of a rhombohedron.
And as I say, I always get in trouble when I try to draw it.
It's got three translations like this and three translations skewed by 60 degree sitting on top.
Everybody can see that's a rhombohedron, can't you?
Say yes.
Be nice.
So this is another primitive cell that's a choice for this triple cell.
And this consequently gives its name to this lattice.
This is called a rhombohedral lattice, regardless of whether you pick the primitive cell in the shape of the rhombohedron or this peculiar double body-centered cell.
So this is a rhombohedral lattice.
And this is represented in a space group symbol by the letter R, standing for rhombohedral.
So just the three-fold axis with this choice of translations, that would be space group R3
[INAUDIBLE]?
Yeah.
a1 like this.
a2 like this.
a3, I'm taking a right-handed system, comes from the centered lattice point.
So let me draw it in projection.
It's easier to see.
Here's the outline of the hexagonal net.
Here's the three-fold axis.
And I've taken a1 coming up like this.
I've taken a2 coming up out of the board like this.
And I've taken a3 coming out like this.
OK.
So these are the three edges of the rhombohedron.
Notice that I cannot have that lattice with a lot of the hexagonal plane groups.
I cannot do it for six-fold axis.
Because a six-fold axis-- P6 and P6MM have a six-fold axis here and three-fold axes in the middle of the cell.
So I cannot have a rhombohedral lattice for P6 or P6<M.
The primitive hexagonal lattice is the only thing that works.
And then we have this curious situation with the two alternative settings of point group 3M in a hexagonal net, P3M1 and P31M.
For P3M1, I have symmetry 3M at both the origin lattice point and in the center of these two triangles.
So for P3M1, in addition to a three-fold axis, I could put in 3M1, stack up plane group P3M1.
But I cannot do this for 31M.
That's impossible.
Because if you look at P31M, there is symmetry 3M at the origin and only symmetry 3 in the center of the triangles.
So if I try to pick this as a third translation for plane group P31M, I'm taking symmetry 3M and plopping it down on top of symmetry 3, and that wrecks the plane group.
So for the rhombohedral lattice, I can only have 3M1 or P3.
P6 has a six-fold axis at the origin lattice point.
The two centers of the triangles are three-fold axes.
So primitive for the six-fold axis, either P6 or P6MM, is the only choice.
So if anybody's counting-- and I wasn't-- we've got 11 lattices so far.
And the only ones that we haven't been able to get by simple stacking of the plane groups are the systems that you all know as your favorites.
The cubic or isometric stems for the fact that the translations are identical in all three directions.
Let me quickly dispose of those by starting with a four-fold axis coming out of one face of a tetragonal cell and saying now if this is a cubic symmetry, I have a three-fold access that comes out, sort of, in the direction of a body diagonal.
But this three-fold axis is gonna rotate that four-fold access to this direction and to this direction.
We've seen that if I look down along the diagonal of a equally axed tetragonal prism, the three-fold axis takes this face, rotates it into this face, and rotates it into this face.
So I have to have a four-fold axis coming out of every one of the faces.
And having that arrangement of four-fold axes means that not only will the two translations be identical in the base of the cell, as is the case for tetragonal, but if I put in that three-fold axis as an extender, all three edges of the cell have to be identical by symmetry.
So I can do this for a primitive tetragonal lattice.
And that's gonna give me a primitive isometric lattice.
I can do this for a body-centered tetragonal lattice.
So that would be indicated by the symbol I.
So it's a body-centered cubic lattice.
And if you add up everything we've done to this point, there are 13 space lattices, a very unlucky number.
And as you all know, there are 14.
We've missed one.
So what have we not done?
We stacked up all the plane groups in all possible ways.
We've taken the four-fold axes out of the face normals, all three face normals of a tetragonal cell.
And that forced it to become a cubic lattice.
What have we done wrong?
Actually, we can get the fourteenth space lattice by a little bit of sleight of hand that is not obvious.
Let's say that this is the base of a tetragonal lattice that we used as the parent for our cubic lattice.
Let me change that primitive square in the base of the tetragonal cell and change it into a centered square.
There'd be no reason for doing that for a plane group.
But let me notice now that if this is the direction of a four-fold axis, rather than having the three-fold axis come out of this direction relative to this super double square, I can make the three-fold axis come out in this direction.
Ah-ha.
So what I have now is a cell that is centered, has a center in the base and has translations that are perpendicular to the base like this.
And if this is the direction of the four-fold axis and this is the direction of the three-fold axis, that three-fold axis is going to put a centered lattice point on the remaining faces of the cube.
So it's only if I notice I can make a double square base to the cell again, put the three-fold axis of a cubic symmetry along the direction of the body diagonal, I generate lattice points in the middle of the other faces as well, and this gives me the lattice known as F, the face-centered cubic lattice.
And the edges of the cell here are all identical, equivalent by symmetry.
So they're labeled a1, a2, a3.
Have you ever seen a lattice constant for a cubic crystal a0?
Has anybody ever seen that?
A lot of people do it.
And if you see that notation in a publication, you will be entitled to sneer at it knowingly.
It's logical, three, two, one, zero.
By extrapolation, if you've got an a3, an a2, a1, you gotta have an a0, right?
Well, not necessarily.
That sort of algebra doesn't apply to labeling of cell edges.
It comes from a curious historic precedent.
Come back, again, to the fact that for literally more than 100 years, people developed symmetry theory and had a pretty good idea that crystals were based on packing of units, molecules, whatever they were, that were on the nodes of a lattice.
But they couldn't measure the size and shape of the lattice.
And that was the brilliance of the famous experiment by Max von Laue because that, in one fell swoop, show that the mysterious radiation x-rays were just electromagnetic radiation of a very short wavelength.
And that's what he really gained notoriety for.
But at the same time, he proved for the first time, unequivocally, that crystals were based on a lattice.
And that was the other side of his double-edged accomplishment.
Up until that time, people learned about crystals by studying the morphology and measuring the angles between crystal faces and recording different faces on a crystal.
I mean, I can just see one of these old guys working a little bit before his wife made him come down and sit for supper because it was getting cold.
And he'd let out a, whoop, huh?
A 13 27 face.
What the hey.
It's a new face on potassium tartrate, something like that.
This was the way these guys got their jollies.
And on the basis of the angles between faces, as I said on several occasions, they could determine a ratio of axes, a ratio of a to b to c, if their assignment of indices to faces were correct.
And how did you know?
Well, along came x-rays, a and you could measure for sure.
So what people started doing to distinguish a to b to c ratios determined from crystal morphology, they put the zeros on it to indicate real values from diffraction.
And when little people came afterwards and they didn't realize the reason for doing that-- nobody's done this for over 100 years because everybody determines lattice constants using diffraction-- but some people have seen these zeros.
So they say, I'm not doing proper notation.
And you know how furious crystallographers get if you don't use proper notation.
So let me write a0 as the lattice constant of my cubic crystal.
You see this commonly.
And there's been no need for it for 100 years.
So anyway, that's the origin of the subscript zero.
There are a lot of dumb things that take place among normally rational people.
I said a moment ago-- and I crossed my fingers when I said it-- that c is the label that is assigned to the unique axis, the axis that is unique because of symmetry.
Sometimes you see monoclinic crystals, which have a two-fold axis and/or a mirror plane perpendicular to the two-fold axis, labelled with this as b, this is a, and this as c, and the angle between them-- that's the general obtuse angle-- as beta.
There is a story behind this lapse of standardness in notation.
How do people decide on what proper notation is?
Well, there is an organization called the International Union of Crystallography, which is the equivalent of the International Union of Pure and Applied Physics and the International Union of Pure and Applied Chemistry.
And they meet every three years.
And there are special commissions for people who like to fight and argue out conventions and nomenclature.
And this is such a silly abomination that there was a move made when they came out with their new edition of international tables to change this and do it the way you do for all of the other crystal systems, namely a, b with gamma between them, and c coming out this way as the direction of the two-fold axis.
And this would be the mirror plane.
And that would put this in accord with tetragonal and hexagonal and everything else.
And it was logical and seemed the right thing to do.
So how do people decide on nomenclature?
This is something that people fight with with more passion than anything else in science.
So you make a proposal.
The proposal goes to the national committees of crystallography in countries over the world.
They discuss it.
They fight about it.
They vote on it.
Then it comes back to the International Union.
And finally, if everybody seems agreed, they present it at the next meeting of the International Union Of crystallography for a vote.
And that's exactly what happened with the proposal to standardized notation.
This took place maybe, I don't know, in the 1950s.
Everybody was in favor.
The discussion proceeded rationally.
Then this little guy jumps up in the front row, short of stature, balding of head, with a very piercing nasal voice.
And he gets up and he enters into this diatribe.
b-axis unique was good enough for Curie.
b-axis unique was good enough for Mauguin.
B-axis unique was-- and all the names were French.
Actually, he wasn't French.
He was Belgian.
So he went on and on and on.
And he so cowed this group of 300 rational delegates, that they didn't cave in entirely.
What they did was they entered every entry for monoclinic crystals in the international tables twice, once with this set of labels, once with this set of labels.
And to indicate that this was the preferred notation, this is labeled the first setting.
And to indicate second-class citizenship no matter what Curie, Mauguin, Bravais, and all of these other guys felt, this was called the second setting.
And I'm not kidding.
I'll bring in a copy of the international tables for x-ray crystallography.
Every entry for a monoclinic crystals is in there twice
And no one in the past 50 years has tried to get rid of the second set?
No.
Actually, the reason is it is expensive to revise these tables.
And I brought in on the first day, you don't remember, my copy of international tables for x-ray crystallography, which is about this big and cost $100.
And then the new version which is twice as large dimensionally, four times as heavy in terms of weight, and I think about eight times as expensive in terms of actual cost.
So you don't do this casually.
If there's something actually new comes out, you make a little supplement that people can stick in the pages somewhere
Don't they still have PDFs nowadays or something?
The last volume came out before PDF, before very intensive use of computers for such data.
The journals now are all done online.
You submit papers online and they're archived online as well in addition to hard copy.
But that's something that was before the time of the publication of the last set of volumes.
Well, I am almost done with talking about lattices.
But I have a thought question.
You've all heard of this castle in Ireland that has this famous rock in the side of the castle called the blarney stone?
And you, to get the gift of gab, have to lean over this rather dangerous wall and somebody holds your ankles and you go down and you kiss the blarney stone.
And that gives you supposedly the gift of gab.
But kissing is a dangerous occupation.
You've all heard about kissing disease mononucleosis?
And with all those people smooching the blarney stone, I've always been afraid the blarney stone might come down with mono.
And if the blarney stone became ill in this session, would it have to be transferred to the Mono Clinic?
I don't know.
All right.
Here in all their glory are the 14 space lattices, or Bravais lattices as they're finally called.
The labels a, b, c, et cetera, are not on them, but the c axis is vertical.
You'll see that the hexagonal rhombohedral lattice is shown referred to hexagonal axes.
In the international tables, you will find all the information both referred to rhombohedral lattices and to the triple hexagonal cell as well.
Now, into these lattices one should drop those of the 32 point groups that the lattices can accommodate.
That's gonna be a lot of additions.
Besides that, there is the option of having the symmetry elements of the point group not intersect at a common point, but interleave them as we did, for example, with the glide planes in P2GG, and in some of the other two-dimensional plane groups as well.
Then we would do what we also did in two dimensions, take the mirror planes and replace them by glide planes, one or the both.
And you might think that we really have an enormous amount of work to do.
We haven't considered any symmetry combinations that involve a horizontal mirror plane.
For example, we look at what symmetry 2 requires and we found the primitive monoclinic lattice, and the side-centered monoclinic lattice, and the body-centered monoclinic lattice.
So there's more than one lattice type into which we can drop a given point group.
But what would you have if you had a mirror plane perpendicular to the two-fold axis?
What kind of lattice will that require?
OK. Let me, in one masterful swoop, convince you that we have done almost the better part of the work.
Let us ask what does inversion require of a lattice?
What does the symmetry of inversion, which didn't appear in any of the plane groups at all because that's inherently a three-dimensional point group operation?
Well, think about it.
All inversion says is that, if, hey, there's a translation that goes up this way, there should be another translation minus T in the opposite direction.
And any lattice whatsoever can make that claim and satisfy that requirement.
So let me now just quickly say that if we have derived the lattice types, and by implication the space groups as well, for a two-fold axis, for a four-fold axis, for a six-fold axis, and so on, we have automatically satisfied the requirements for any point group that results when you add inversion to a two-fold axis, when you add inversion to a four-fold axis, or add inversion to a six-fold axis.
So it turns out that the requirements of a six-fold axis, which gave us the primitive hexagonal lattice, are also met by 6 over M. The requirements of a four-fold axis, which gave us the lattices primitive tetragonal and body-centered tetragonal, are also satisfied for 4 over M. So, any one of the point groups that differs from the 11-- from the 10 two-dimensional symmetries-- namely 1, 2, 3, 4, 6, M, 2MM, 3M, 4MM, and 6MM-- any symmetry that you get by adding inversion to those two-dimensional point groups is automatically going to be happy in the space lattices that we've derived to this point.
So we have done an enormous fraction of the work that's required to derive space lattices, and the space groups as well.
So there is not too much more to be done.
And what I will do for probably the next meeting is to derive a few of the space groups for the lower symmetries which are not too complicated, just to show you how the game plays out.
And then I will mention a little bit about the special idiosyncrasies of the orthorhombic brick-shaped cells where there is nothing more distinct about one direction and the other.
All the inter-axial angles are 90 degrees.
The translations are general.
So there are peculiarities in notation there that are unique to the orthorhombic system.
And then I want to spend, perhaps, one class talking a little bit about crystal chemistry.
Have any of you had a class in solid state chemistry or crystal chemistry?
OK. One.
One of the main uses of space group theory is the description of crystal structures.
So I think I might, as a minor digression, take an hour to talk about packing and some of the concepts used to describe the close-packed structures and ionic structures and give you a problem that ask you, as I already did, to generate a structure, simple-minded two-dimensional structure, and do some interpretation of it.
So, I think this will be worth saying a little bit about just so that one of the main applications of symmetry theory does not escape without any discussion at all.
We will, after about two more lectures, have a complete change of direction and start talking about properties of crystals and how these properties are impacted by the fact that crystals have symmetry.
So we're gonna use some of our results in symmetry, but not all that much, but develop remarkable mileage in predicting anisotropy in crystals just on the basis of the symmetry which that crystal possesses, that certain physical properties as a function of direction are weird surfaces that have lumps and noses and bulges in them.
And certain properties have this anisotropy universally, regardless of the chemistry or composition of the crystal.
Given the symmetry of the crystal, the anisotropy has to be in this fashion.
We'll be able to show that some properties simply cannot exist, period, regardless of composition and structure in materials that have certain symmetries.
So that'll be the direction we take off in.
I mentioned at the beginning of the term-- you've probably forgotten-- there is a lovely book on tensors and physical properties by a gentleman named J. F. Nye, N-Y-E. And the title of the book is Physical Properties of Crystals.
This is a nice place to go for further reading.
We will go through roughly half of Nye's book with a little more emphasis on the consequences of symmetry than what he does.
But if you want to do some extra reading, I asked the Coop to stock this book.
But I suggest that you not rush out and buy it.
It's available in paperback, which means it's cheap, but only relatively so.
There's no such thing as a cheap technical book, paperbacked or not, these days.
It's also on reserve in the library.
So, see if you need the extra help.
See if you would like to do some reading in Nye.
And make sure you want to before you go out and put your money down at the Coop.
The last part of Nye is an elegant, beautiful part of the book.
It deals with the thermodynamic relations between different properties.
And it's a unique treatment that is developed by Nye.
We won't do any of that, but it's a nice thing to have.
So don't rush out and spend your money.
But if you want additional reading and just going to the reserved book collection is not enough, then Nye is not at all a bad copy of a book to have.
I think most of you have probably seen Nye, not the book but the man.
How many of you have seen the famous Bragg bubble raft movie on disk locations, where he floats sheets of bundles and sheers them and you see his dislocation go zipping across?
At one stage in that movie, there's an assistant who tiptoes, trying to be unobtrusive, across the background.
But the camera catches him scurrying behind Bragg.
That's Nye.
So if you've seen the movie, you've seen what Nye looks like.
OK.
I'll call it quits there for today.
And we'll see you next week on Tuesday.
Everyone seems to be here.
At least all the seats are filled up, so why don't we begin.
One thing that I did finally, to be specific, I told you that we would have three quizzes and that they would be uniformly spaced one third, and 2/3 and 2.89 thirds of the way through the term.
And I figured out where those dates would be.
So since I have them, let me give you the date.
So Quiz 1 is going to be on Thursday, October 6.
Quiz 2 will be on Tuesday, November 8, and Quiz 3 will be on Thursday, December 8.
And this, as advertised, is exactly one third of the way through the lectures.
This is exactly 2/3 of the way through the lectures, and this is the next to the last meeting that we'll have.
This is the week before the final week of the terms.
so I deliberately kept it a little bit away from the usual end-of-term crunch.
Being person who is sensitive to symmetry and order, I find it highly regrettable that we cannot have the quiz on October 8.
Then you would on the 8th of October, November, December, you have a quiz.
Unfortunately, October 8, much as that would beautify the schedule, is on a Saturday, so we can't do it so.
I could do it, but I don't think anybody would come.
I've been, believe it or not, going through these problem sets in great detail.
I didn't want to hand them out until I had the list-- and you may or may not know-- that is made up for the registrants in each class, And we get a nice array of photographs with names and department numbers and years numbers under them.
That something, by the way, that only the instructor in a class requests, not even the secretary can do it.
So this is not widely broadcast.
I found to my dismay though when I finally got it that about one third of the pictures are missing.
And I don't know whether you folks have been photographed fairly late on the term, but a third of the pictures are missing.
I already been putting some faces in my memory banks, but to match up names and faces so that I can hand out the problem sets individually, I need more photographs.
So I'll do it one way or another next time, and we'll see some photographs, additional photographs, have appeared in the interim.
I've given you three problem sets so far, and they were really just for fun as well as to get you thinking about the right sorts of things, but there were little puzzle.
I'm, unfortunately, going to have to tell you from now on no more cutesy little puzzles.
You're going to get real problem sets, long, tedious, drudgery.
They're optional though, so that's the saving grace.
So I'm going to pass it around the problem set 4.
And this asks you to demonstrate some of the things that we talked about last time and convince yourself that they really work and ask you to make some simple applications of Miller-Bravais indices to two-dimensional lattices.
And then finally, the last problem is to get you thinking about patterns.
And I've asked you to identify the lattice points and the symmetry elements in two patterns.
I'm almost willing to wager that not one of you will get it completely right, that there's going to be some little thing that you missed or did wrong.
At one time, I bet a bag of potato chips for everybody in the class against the class wagering against me one bag of potato chips.
Well, I'm going to watch my weight, and I've won that invariably, and I really can't eat that greasy sort of stuff.
In any case, you'll see that in the case of symmetry and patterns it is immensely simplified when you know exactly what to look for, when you know the number of possibilities was finite,.
And when we're through with this, you can ask just one question about a pattern and then know exactly what the symmetry of that pattern has to be and exactly where to look for everything else.
I can guarantee you that by the time you finish this class you will never sit in a bathroom and stare at the tile floor and see it in exactly the same way again.
I think that's a good observation because I like to think that that's what education is all about, not sitting in the bathroom, but seeing things differently after you've had the experience than you did before.
That's what education is all about.
So have fun with the patterns.
And again, it will become-- one seat up here.
Today then we're about ready to embark on an adventure.
As I said last time, we'll develop two-dimensional symmetries first because there are relatively few of them, and one could do this rigorously and in great detail.
We're going to have to touch on things more lightly when we get to three-dimensional symmetries so we can finish at a decent point in the term.
Let me remind you that we have so far identified two types of symmetry that can exist in a lattice, onefold, twofold, threefold, or sixfold rotation axes.
Any number of different rotational symmetries are possible, but if you're going to want them to be compatible with a lattice, you must restrict the rotation axis to one of these five, including a onefold axis, which is no symmetry it all.
And in two-dimensions, besides translation, we saw the operation of reflection.
And now we're going to begin to put things together and make elaborate combinations.
The first thing I will ask is can we have more than one of these symmetry elements present and operating about the same locus at the same time.
The answer to that would be, why not.
Because if we look at a sixfold rotation axis, that's really what a twofold rotation axis does combined with what a threefold rotation axis does.
And we plot these on top of one another, and what we end up with is the arrangement of motifs that is generated by sixfold axis.
So in a sense, a sixfold axis is a twofold axis superimposed on a threefold axis.
That kind of a naive way about thinking of this though.
What we're really saying is that what a sixfold axis has is the set of operations 1 A 2 pi/ 6, A twice 2 pi /6 -- that would be 120 degree rotation-- and A pi, A 5 pi/6.
And there's 1, 2, 3, 4, 5.
And we're missing one.
60, 120, 240.
So that would be A twice 2 pi/3, 4 pi/6 So what we're really saying is a sixfold axis consists of these six elements, and these elements constitute a group because I can combine any two of these rotation operations and find something that's already a member of the set.
By saying that a sixfold axis consists of a threefold axis sitting on top of a twofold axis, what we're saying is that there's one collection of elements here, namely identity, or same thing as A 2 pi and A pi, that is a subset of these, or we could say a subgroup.
So what we're doing in saying that these two sets of rotation operations exist simultaneously, we're saying that the twofold axis is a subgroup of the sixfold axis.
And similarly, we can separate out three other elements, a onefold axis, same as A 2 pi; a 120-degree rotation, A 2 pi/3; and a 240-degree rotation, A 4 pi/3.
And these three operations, a group of rank 2, is what constitutes another subgroup, a subset of elements which by themselves satisfy all of the requirements of a group.
So this is one example of how we can have more than one symmetry element, a set of operations existing about the same locus and space.
This is not how one would go about the deriving these higher symmetry however.
Because what we have to do is to add something to the set, something that's called an extender, and then show that all the requirements of a set being a group is satisfied; namely, that a combination of any two elements in the group is also a member the group, that for every operation an inverse exists, and the identity operation is a member of the group.
So you can build up more higher symmetries, more complex symmetries by adding some operation called an extender and then taking all of the products of these elements and see what new operations arise.
So we can have more than one symmetry element operating about the same locus and the set of individual operations that it embodies.
And here sitting all by itself is a mirror plane.
Why don't we combine a mirror plane with the rotation axis?
And we've already seen examples of that.
For example, in the square pattern that these tiles make up, there's a fourfold axis in the middle of each of the squares that are also mirror planes that pass through that location.
So let me look at a combination that is a little simpler to handle.
And what I'm going to say is here sits a mirror plane with a reflection operation that I'll call sigma 1 for the individual operation.
And let me say that I combine now in that space a second reflection operation about a line that intersects the first one.
And the question I'm going to ask now is happens when I take a first motif, number 1, reflected in the locus of the first mirror plane to get number 2, and then reflect that a second time in the locus of the second mirror plane to get one that sits up here.
So the question is now what is the operation sigma 1 followed by the operation sigma 2 equals to?
We can almost answer by the process of elimination.
If this one is left handed, reflection changes the polarity, so number 2 is right hand.
And if we reflect a second time, the right-handed one goes to a left-handed one.
So what we're asking is how do I get from one left-handed motif to another left-handed motif?
Of the three operations that exist in a two-dimensional space could be translation.
But, clearly, I can't take the first one and slide it parallel to itself and make it coincide with the third.
So translation would not change chirality, but that won't work.
What's left?
Rotation.
Yes.
You want to clarify something first?
That third one, it looks kind of left-handed.
[INAUDIBLE]
Oops.
It does.
It does indeed.
Sorry about that.
When you've got your nose poked right in these things, it's easy to overlook it.
Yes.
Absolutely right.
That should be over canted over a bit more like I had it the first way.
Very good thank you.
I don't want to proceed much.
Yes?
What did you [INAUDIBLE] axis [INAUDIBLE]?
We could do that.
And let's do it, and we're going to encounter this sooner or later.
So we have a first mirror plane reflect from a first one, which is right handed, to second one that's left handed.
And then define a second locus, and that one would take the second one and move it over to here, the same distance on the other side of this mirror plane.
And I'll let you answer the question since you asked it.
How is the first one related to the third one?
Translation
Translation.
Yes
It's translated
Exactly So if this distance is delta, the first and the third are related be a translation which is twice delta.
So that would be in addition where we'll eventually want to consider when we have patterns that are based on a lattice.
But right now, I want to consider symmetry operations that leave at least a point in space and move perhaps a line in the case of the mirror plane.
But back to the original question, what is this first reflection followed by a second reflection?
We've just thrown out translation.
That won't work unless the mirror planes are parallel to one of those.
The only thing left is rotation.
And in point of fact, even looking at this diagram, you can get from the first one to the third one by a rotation.
So the answer to the question is that a combination of two reflection operations is a rotation operation about the point of intersection of the mirror planes.
But we can go further and say exactly how large this rotation is.
Let's draw an extra line in here.
And let's say that this angle is alpha.
And if I repeat this by reflection, that angle is also alpha.
Let's label this angle beta.
Repeat that by reflection, and that angle must also be beta.
The angle between the two mirror planes is the angle mu.
And clearly mu is equal to alpha plus beta.
So the answer to the question quite generally is have two mirror planes intersect at an angle mu, and their successive operation is going to be equivalent to the net result of rotating through twice of the angle between them mirror line.
So we don't have to think about that anymore now.
We know if we combine to mirror planes we get a net rotation.
So that is a basic result, and that's a first example of something that I'm going to call by my own pet term.
I'm going to call them combination theorem, the theorem that tells you what you get when you combine two operations.
And what this is going to give us is a way of filling in one box in the group multiplication table.
So knowing what two mirror planes combined at an angle should be, I can now combine the mirror planes at angles which gives me a rotation operation which is one of the ones that's allowed for a lattice.
In general, if I'm not yet putting these symmetry elements in a lattice, I could combine them in such a way such that I got a 17-fold rotation as the angle.
And then the angle between the mirror planes would be half of 2 pi/17.
Could be a lovely symmetry, nice for a pendant or a ring, but not for a lattice and not for a crystal.
So I'm going to want to combine mirror planes at angles that correspond to rotations that are compatible with the lattice.
So what I could do is take a first reflection locus, sigma 1.
And if I want the rotation angle to be 180 degrees for a twofold axis, I would want the second mirror plane with the operation sigma 2 to be at one half of pi with respect to the first.
And for this specific combination then, reflection in the first mirror planes followed by reflection in the second mirror planes to give me a third one, is going to be equivalent to-- Oops.
Back here-- it's going to be equivalent to a twofold axis.
So sigma 1 followed by sigma 2 at an angle one half of pi is going to be equivalent to the rotation operation A pi.
This is not yet a complete pattern because here's a mirror planes that wants to reflect this motifs as well.
So let's just reflect number 3 across the mirror planes 1.
And then we're going to get a fourth object.
Now let's show that this constitutes a group.
We've got the identity operation doing nothing.
We've got the operation sigma 1.
We've got the operation sigma 2, and we've got the operation a pi.
I'll put the same operations again in the vertical column.
Here's operation 1 doing nothing, sigma 1, sigma 2, and A pi.
So doing nothing and following it by one of these four operations just gives you the same thing back again.
And doing these operations followed by 1 will give me the same operation back again.
If I reflect twice across sigma 1 from left to right and right to left, I'm back to where I started from.
So sigma 1 followed by sigma 1 is the identity operation.
Sigma 1 followed by sigma 2 is what we just did.
That turns out to be the operation A pi.
And sigma 1 reflection and following that by A pi is the way in which I get to the fourth one.
So that is going to be equal to sigma 2.
Well, let me rips through the last ones fairly quickly.
Sigma 2 followed by signal 1.
Sigma 2 followed by sigma 1 is the same as A pi.
Sigma 2 followed by sigma 2 gets us back to where we started from.
That's the identity operation.
Sigma 2 followed by A pi reflect from here to here and then rotate.
That's the same as sigma 1.
And the final sequence, rotation followed by reflection would give me sigma 2.
Rotation followed by sigma 2 will give me sigma 1.
And doing the rotation operation twice is the same as the identity operation.
So are the group postulates satisfied?
Yes.
The combination of any two elements is a member of the group.
For every operation, an inverse exists.
And we can answer that question very easily by merely looking at the column under a particular element, and somewhere we find the identity operation.
So sigma 1 is its own inverse.
So, yes, an inverse exists for every operation.
And then finally, identity is a member of the group.
So everything's lights up.
Bells ring.
This set of four elements is entitled to call itself a group.
And in particular, it is a group of rank 4 because there are four elements in the group.
I show you a cool thing.
I noticed that the number of objects in the pattern, the number of motifs in the pattern, is exactly the same as the order of the group.
Why?
Because these four operations tell you how to get from any operation to the remaining three and tell you how to get it into itself.
So identity relates it to itself.
Sigma 1 reflects it to here.
A pi rotates it down to here.
Sigma 2 reflects it down to here.
So there's always a one-to-one correspondence between the elements that are in the group and the rank of a group and the number of objects that are in the pattern.
Yes, sir?
How do you show if the inverse exists again?
The inverse exists if I can find something that combined with an element in my basic set gives me the identity operation.
So it's the operation followed by its inverse has to be the same as doing nothing.
So if I look under each element and find the identity operation, then sigma 2 is its own inverse.
If I look under A pi, A pi is its own inverse.
This is generally not the case.
This is a very simple symmetry.
So really we can work with commas and little figures, and that's one way of getting these symmetries.
But group theory I think you can see is a very nice, very elegant language for describing some characteristics of these patterns.
Now little bit of jargon.
As if this material were not confusing enough, there are two different languages that are used to denote these unique combinations.
This notation where we simply have a number for a rotation axis or an m for mirror plane is something that was first proposed by two mathematicians, Hermann and Mauguin.
And this was adopted as the preferred notation in the international tables, that hefty tome that I brought it on the first day of classes.
So this is also referred to synonymously as the international notation.
Let me put these things out horizontally.
For rotation axes by themselves, the symbol in Hermann notation is simply n. And for lattice, n-- as we've seen many times-- is restricted to what in Hermann-Mauguin notation would be either 1, 2, 3, 4, and 6.
There's another type of notation that is use quite frequently by condensed matter physicist, and this is called the Schoenflies notation after one of the mathematicians who was the first to derive the three-dimensional symmetries.
This was a curious thing.
There were three people in different parts of the world about the turn of the century-- and this was before email or even air mail-- who were trying to derive the three-dimensional symmetries.
Schoenflies was one of them.
And all three of them got to the same place at about the same time, and the results were obtained.
But each one used his own method and had his different notation.
Schoenflies was one of the people who did this.
Schoenflies was a mathematician and based his notation on group theory.
Something called a cyclic group is what the C stands for.
And a cyclic group is one in which all elements are power, so to speak, of some basic operation.
For example, a fourfold axis consists of the set of operations A pi/2; a 90 degree rotation; A pi, which can be written as A pi over 2 squared; A pi over 2 squared; [?
then we need ?]
A pi/2; and A pi/ 2 again; A 3 pi/2, which can be viewed as doing A pi/2 three times; and finally, the identity operation, which is the same as A 2 pi, which is equivalent to doing the basic operation A pi/2 four times.
So the group of rank 4, there are four elements in the group, each one is a power of the basic operation A pi/2.
So a rotation axis by itself as a cyclic group.
Schoenflies indicates these generically by C subscript n, where n is the rank of the axis.
Hence, this is C1.
This is C2.
This is C3.
This is C4.
And this is C6.
A mirror plane by itself is indicated m in the international notation.
A mirror plane is also a cyclic group.
There are two operations, reflection and reflection back to where you came from.
Doing the reflection operation twice is the same as doing nothing.
So Schoenflies also call this one C. And the one thing that is without meaning in English is his subscript S. And Schoenflies was German, and the S stands for spiegel, which is the German word for mirror.
We have in many cities a paper that's called the Daily Mirror.
In Germany, there's a paper called Der Spiegel.
So they use it the same way in everyday life as well as in notation.
So C sub S in the Schoenflies notation is a mirror plane.
Cyclic group the S stand for spiegel.
For the combinations of which we have seen only one, in the Schoenflies notation these are all of the form Cnv, a rotation axis Cn with a vertical mirror plane passing through it.
So this one, for example-- we've yet to look at the others-- would be C2v.
So we'll have to draw the first, and then I'll give you the complete set.
So this is a rotation axis with a mirror plane passing vertically through it.
Let's do a few more.
And I think having done a couple in great detail, we'll see what the others will look like quite readily.
Let's take a fourfold axis and add to the rotation operations A pi/2, a first mirror plane that I'll label sigma 1.
I can permute the order of operations and say that sigma 1 followed by the rotation operation A pi/2 is going to be equal to a second reflection operation that is equal to one half of pi/2 away from the first.
So I'm claiming that if I reflect in the first mirror plane and then rotate by 90 degrees that should be another mirror plane at 45 degrees to the first.
So let me do exactly what I advertised.
Here's the first, one right handed.
Reflect across to get a second one, which is left handed.
Reflect, then rotate, and that would bring my left-handed one up to this location here.
So here's 3, and it's left handed.
And lo and behold, just as advertised, I get from the first one to the third one by reflecting across a mirror plane, which is one half of pi/2.
If let these symmetry elements operate on each other, what I'll end up with is a set of mirror planes at 45 degree intervals.
And what I'll have is a pair of objects hung in the same fashion at every mirror plane.
Is that too fast?
You want to go through that a little slower?
Oh, go ahead say, go through it more slowly.
Everybody's afraid to say, yeah, yeah, and been seem like a class dummy.
Do you want me to do it more slowly?
I see people still writing, so I think what you'd rather have is me be quiet for a bit while you catch up to where I am.
In the Schoenflies notation, this lovely thing here is C4.
It's a rotation axis of rank 4 with a vertical mirror plane added to it.
And the international notation, now the Hermann-Mauguin notation, is just a running list of the individual symmetry elements that are present.
And now I really are going to have a mouthful.
This is a fourfold axis, so the symbol for that is 4.
This is a mirror plane.
This is a mirror plane.
This is a mirror plane.
This is a mirror plane.
This is a mirror plane.
This is a mirror plane.
That's a mirror plane, and that's a mirror plane.
So it looks as though I should call this 4mmmmmmmm, which comes 4mmmmmmmm.
Well, that is a typical reaction.
It's a lovely symmetry.
But that is a mouthful.
So it isn't really necessary to give all these m's So the international notation is a running list of the independent symmetry elements, and that's the new wrinkle that I'm introducing.
It's a running list of the independent symmetry elements.
And all these mirror planes are just different sigma that exist as operations in the group.
But how many different kinds of mirror planes are they?
Well, there are two different kinds of mirror planes, both in terms of the way in which they function in the pattern.
The two motifs related by reflection hang close to this mirror plane, but they're widely separated for this mirror plane.
So the motifs do different things relative to those two kinds of mirror planes.
Another way of asking what's independent is if I start with this 1 mirror plane and repeat it by 90-degree rotations, I'll get these 4 mirror planes 90 degrees away.
So they are not independent in that these mirror planes are all related by the rotational symmetry that's present.
You don't get this mirror plane in any fashion other than saying, if I combine the rotation operation with this reflection operation sigma, the net result is this reflection plane.
So there are two mirror planes that are distinct in this arrangement of symmetry elements, distinct in the sense that they function in different ways in the pattern; distinct in the sense that no other operation that is present will throw these two operations into one another.
Another example, and this is in fact symmetry 4mm, is the square tile.
If you look at the mirror planes there, they are 45 degrees apart.
But one of those mirror planes comes out normal to the edge of the tile.
The other mirror plane comes out of the vertex of the tile.
So they are different in the way they function in the space which has this symmetry.
So, mercifully, we don't have to call this 4mmmmmmmm.
We drop the last 6 m's, and this one is called simply for 4mm, 2 kinds of mirror planes with a fourfold axis.
If you're familiar with the way in which these were derived, the symbol tells you exactly what you've got.
So it's a very useful notation for these symmetries.
Let me pause here, give you a chance to catch up.
Yes, sir?
So wait, all you did here is take two separate mirror planes at an angle of 45 degrees between, and you ended up with a fourfold symmetry
Exactly
But you didn't put on the fourfold symmetry, it was just what fell out of it?
No.
I Got this mirror plane to begin with by combining this reflection operation sigma with the operation A pi/2, and that's where this operation came from.
But a general theorem that says if I have a rotation operation A alpha and combine it with a reflection operation sigma, the combined effect is a reflection operation sigma 2, which is alpha/2 away from the first.
So, again, showing you for this now rather messy diagram, if I start with the reflection operation that takes 1 of them throws of the 2 and then rotates 2 up to location 3, the way I get from 1 to 3 in one shot is to reflect in this locus sigma 2.
Yes, sir?
You could have started with any one of those two and got the third
Absolutely.
Absolutely.
The operations that are present operate everything in the space.
So when I say that there's a mirror plane here that relates this one to this one, it also relates this one to this one, this one to this one, this one to this one, that mirror planes operates on everything.
You can't say that a symmetry operation grab this little packet of space and moves it to another packet removed from it.
To say it's a symmetry of a pattern or of a crystal, it has to leave everything invariant.
OK.
But, again, in terms of the language of groups theory, if you combine this pair of operations, then when you combine everything that you get pairwise, you will get in this case a total of, how many operations?
How many operations are in this pattern?
I said a moment ago the rank of the group is the number of operations that are present.
It's the number of objects that are in the group because each operation in a group tells you how to get from anyone to all of the other.
So, 2, 4, 6, 8, this is a group of rank 8, and there are 8 operation that are present.
And I can rattle them off quickly.
Four operations, identity, 90, 180, 270 for the fourfold axis.
That's 4.
This mirror plane.
This mirror plane.
The mirror plane.
And this mirror plane.
That's 8.
Four for rotation.
Four for reflection.
Other questions?
All right.
Let me then wrap up this quickly and get onto something new.
Did I hear the hiccup or a question?
No.
It was a hiccup.
Not a yawn I hope.
Let me do the highest symmetry of all in two dimensions, and this is if I take a sixfold axis and combine it with a mirror plane.
So here's the first operation sigma.
This one is something of a bear to draw.
This is sigma 1.
Sigma 1 followed by the operation A 2 pi/6 should be equal to a reflection that is an angle one half of 2 pi/6 30 degrees away from the first.
So this will be sigma 2.
So if I reflect from 1 to 2 and then rotate by 60 degrees, I'm going to get one that sets up here.
And the way I get from number 1 to number 3 in one shot is by a reflection sigma 2.
Now if I draw in all of those are mirror planes when they are repeated by the sixfold axis, I shouldn't have given this my 6mmmmmm treatment because there are 12 mirror planes.
Two objects hanging on this one in one fashion spaced in a different way that are belly to belly.
They're back to back here, so it does something different in the pattern.
A pair hanging here.
A pair hanging in here.
A pair hanging and here.
A pair hanging in here.
And finally, a pair hanging in here.
So there are a total 2, 4, 6, 8, 10, 12 motifs in this pattern.
Pairs hanging on mirror planes that are 30 degrees apart and hanging disposed and pointing in different ways on the adjacent mirror planes.
So this is one that we would call in international notation 6mmm and in Schoenflies notation C6v.
So we're making extraordinary progress here.
The one that I did for you initially was C2v, and that's 2mn in international notation.
The only one that I left out to this point is a threefold axis compared with a vertical mirror planes.
And let's start with the operation A 2 pi/3, combined with that a first reflection operation sigma 1 that passes through it.
And for reference, I'll drawn in some lines that are separated by intervals of 60 degrees.
So this reflection from 1 to 2 followed by a rotation 120 degrees should give me this one as number 3.
The first is right handed.
The second is left handed, and the third stays left handed.
And lo and behold, the way I get from 1 to 3 directly is by a reflection sigma 2 across a mirror line that is-- I'm sorry.
This is number 1 down here-- across a mirror line that is 30 degrees away from the first.
If I would complete the pattern, reflect this one across to here, take this pair and reflect it or rotate it up here, and I have six objects.
So this is a group of rank 6.
And it has characteristics that are quite analogous to those we did for the other rotation axes in all respect except one.
The international symbol for this combination is C3v.
We've got a threefold axis.
We've got the mirror plane that we added.
And then we've got a mirror plane that is 30 degrees away from the first-- I'm sorry-- 60 degrees away from the first, one half of 120 degrees.
I claim that this is not the proper symbol because the mirror planes that are listed should be the symmetry-independent, distinct sort of mirror planes.
And is that the case with this one?
The answer is no.
It isn't.
Here is one mirror plane.
It's got a pair of motifs hanging on it.
This mirror plane here is a mirror plane that has a pair of motifs hanging it.
Same is true of this mirror plane.
So all six of these mirror planes are doing the same thing in the pattern.
Each one has a pair of a motifs on either side of it in the same fashion.
And I can get one mirror plane and the motifs hanging on it by a rotation of 120 degrees.
So there's only one kind of mirror planes, so we don't need that m. So all of the other rotational symmetries, m, a onefold axis with a mirror plane; 2mm, 4mm, 6mm have two kinds of mirror planes.
C3v has only one kind of mirror planes.
And another way of showing that is if I look at a trigonal prism.
It's got a mirror plane coming out of a corner and out of the opposite face.
Mirror plane coming out of the corner and out the opposite face.
Mirror plane coming out of a corner and out the opposite face.
And that's exactly the rearrangement of symmetry elements here.
So each mirror planes when drawn in relative to a trigonal prism, which is a body that has this symmetry, each mirror planes does exactly the same thing, as opposed to the square tile or square prism that has one kind of mirror plane that comes out of faces and one kind of mirror plane that comes out of corners.
So there are two different types of mirror planes.
Let's add them all up, summarize, and we've got the cast of characters that we can be use in deriving two-dimensional symmetries.
We've got C1, which is a onefold axis; C2, twofold axis; C3, that's a threefold axis; C4, and that's a fourfold axis; C6, that's a sixfold axis.
Then we've got the additions that involve adding a mirror plane to a rotational axis.
So this is simply m and CS in Schoenflies notation.
Then we added a mirror plane to a twofold axis.
We've got C2 with a vertical mirror plane, 2mm; C3v; and not 3mm, but only 3m; C4v, and that's 4mm; and C6v, and that's 6mm.
Add them all up, and there are 10 unique possibilities, no more, no less.
Yes, sir?
I'm missing a point.
Why is there the mirror plane sigma 2 rather than rotating sigma 1 is not different?
How do you?
The same thing is going on, on this mirror plane as on this mirror planes, on this mirror plane.
Or if you like to see it in terms of a geometric solid, the mirror planes here in a trigonal prism all of them do the same thing.
They come out of one corner and out of the opposite edge.
Each of these has one pair of objects hanging on it.
And that is true of all three of them.
And if you like, this end of the mirror plane here is nothing more than one that's related to the first one by one of the rotations of the threefold axis extended back in the opposite direction.
So the mirror plane doesn't just work up here.
It works down here as well.
It works all through the space.
So any way you want to look at it and whatever terms work for you, each of these mirror planes is the same, but they have a different type of end to them.
There's something different hanging at one end than at the other end.
Think of it in terms of actual physical object
[?
What about the ?]
case of the fourfold [?
there?
?]
In the fourfold, one comes out of the face.
One comes out at the edge.
The motifs are oriented and spaced at a different distance from each of those neighboring mirror planes.
So these two are different.
The fourfold axis never rotates this one into this one.
These two are not different because the threefold axis rotates, if you do it twice, this end into this end of what is one in the same mirror plane as what's down on here.
So what's confusing you I think is that there's a polarity to the mirror planes.
Both ends of the mirror planes do not have the same disposition of motifs on them, but there's nothing that says that this has to be the case.
The motifs could just be at one end.
And that, if you think about it a little bit if it help, figure out-- not on company time though, but on your own time-- what a fivefold axis does.
And a fivefold axis is a non crystallographic symmetry.
But it turns out that C5v, which is a 5 with an m, has only one kind of mirror plane in it as well and for exactly the same reason.
A regular figure that has the symmetry is a pentagon.
A mirror plane comes out a corner and out of the opposite face.
And that's true for all of the mirror planes that are in there and separated by one half of 2 pi/5
That's just whether or not the [INAUDIBLE] axis is [INAUDIBLE]?
Yeah.
That's what I said.
Yeah.
So before I go through all of the rotation axes with vertical mirror planes, the international notation would be 2mm, 3m, 4mm, 5m, and 6mm, 7m.
For the odd symmetries, there's only one kind of mirror plane in a pattern, in the tile, or whatever you want to have.
So there are 10 distinct possibilities, and these are called the Point Groups.
Why?
Because they are clusters of symmetry elements about at least one point that's fixed and embedded immovably in space.
They're called groups because, as we've seen for one simple example, the collection of operations follows the postulates for the set of elements which we have defined as a group.
More specifically, we could call these the 10 two-dimensional crystallographic graphic Point Groups, which is more of a mouthful.
But it emphasizes the fact that there are lots of two-dimensional Point Groups, but the two-dimensional crystallographic Point Groups are those that involve rotational symmetry that are compatible with a lattice.
Hence, they are crystallographic.
But the number of Point Groups is actually infinite if you include the ones that are not compatible with translation.
All right.
I think that's probably a good place to quit.
And guess what?
My internal clock has told me that this is the time for our break.
This is one crossroads in our development.
And what we'll do next for the faint hearted who may not want to have more of this stuff for one day than what we've just done, we're now going to do the final penultimate combination, and say we have also shown that there are 5 two-dimensional lattices.
And the final step will be to say if we have a pattern that has symmetry and is based on translation, we can obtain these by taking each of the 10 crystallographic Point Groups in turn and dropping them into each of the five lattices that can accommodate them.
We would not try, for example, to take a Point Group like 4mm and try to drop it into the hexagonal lattice.
It's not going to fit.
But there will be two, maybe three ways in which we can add a given Point Group to one of these lattice types.
And what we've done when we finished is we have exhaustively derived the symmetries' translational lattices and symmetry operations of reflection and rotation that are possible for two-dimensional crystal.
You might say, well two-dimensional crystal, I've heard that people who do thin film work to make monolayers and really make a two-dimensional crystal.
That's not something you could say only a few years ago.
But why do we worry about two-dimensional symmetries if we're not to be wallpaper designers?
Well, actually, I'll give you one example.
One of the difficulties you had in conveying the nature of a crystal structure is taking something that's fairly complicated in three dimensions and getting it onto a two-dimensional sheet of paper.
And what you invariably do is you project the contents of the unit cell down along one of the cell edges.
And what you have then is a two-dimensional crystal structure with a lattice and symmetry.
So you'll see two-dimensional representations of translationally periodic arrangements of atoms quite frequently when you look at projections of actual crystal structures, which is the only reasonable way to convey the information of any structure once you get on beyond the baby stuff of sodium chlorides, zinc, sulfide and body-centered iron.
OK.
I'm going to pause, suck in air.
I am happy to turn you loose for 10 minutes and will meet here again, let's say, at 10 after 3:00.
That I know of that derives two-dimensional symmetries.
And so, what I will give you instead is a set of notes-- nothing fancy, just handwritten-- but which will carry out each of the derivations in detail, so you'll have something to fall back on if it seems overly complicated.
Any questions on what we've done to this point?
We've really set forth the ground rules and components of our final step.
And that, as I mentioned earlier, is to combine one or more of these collections of symmetry elements around a fixed point in space, with one of the five two-dimensional lattices that can accommodate them.
And when we find a successful, feasible combination, we will have derived something that we could call the two-dimensional crystallographic space groups.
These are collections of operations which really pick up everything-- the whole space-- move it, tumble it end over end, and throw it back down into coincidence with itself.
So hence the term space group, rather than point group, where it's only a point that's left invariant at worst.
OK, so what we're going to do then is to take each of the point groups in turn, and combine each of them, when possible, with one of the five lattice types.
And I remind you that those are the oblique lattice or parallelogram lattice.
And this was what was required by two and one, no symmetry at all.
Then there was the primitive rectangular net, and the symmetry element that required that was a mirror plane, or we could have, depending on how the mirror plane was aligned relative to the translation, we could also have a lattice that was non-primitive, a double cell.
And that was a centered rectangular net.
And either of those was compatible with a mirror plane.
Then we had a square cell-- square net or square cell-- and that was what was required by a 4-fold axis.
And then the hexagonal net, and that could be required by either a 3-fold or a 6-fold.
Well, if you add up the number of point groups that we've listed here, there are only six, but there are 10 crystallographic point groups.
So what about the ones that we have not yet considered adding to a lattice, and these are the nets of the form Cnv.
So let's consider, in turn, each of those point groups and see what they would require of the lattice that could accommodate them.
But let's do 2MM first.
2MM, the 2 is compatible with an oblique net.
The mirror plane wants either a rectangular net or a centered rectangular net.
So all we can do is to add this to a rectangular or centered rectangular net.
Let's put a 2-fold axis at the lattice point.
The mirror plane has to be aligned along the edge of a rectangular net.
If there are two of them, you just put one along each of the edges of the rectangular net.
A mirror plane in a net could either go in like this, or could go, for the centered net, in like this.
So what we're saying, 2-fold doesn't care, add a mirror plane, it means that the parallelogram has to become a rectangle.
You can put the two mirror planes of 2MM in along the two edges of the rectangle, or do this equally well with a centered rectangular net, put the mirror plane in parallel to the edges, just like we did in CM, put the 2-fold axis in at the lattice points.
So there's a lattice point at the corner, and also a lattice point in the middle of the cell.
So for the rectangular net, we could put in either M or 2MM in both of these rectangular nets, the primitive and the centered.
4-fold access requires that the net become still more specialized and have the shape of a square
Professor?
Hmm?
We can't see on this side
Oh, I'm sorry.
I'll pace back and forth, which I usually do.
Then everybody will get equal time.
OK, so I'm glad you spoke up.
4-fold axis requires a square net.
There are two mirror planes, 45 degrees apart, and those mirror planes have to be along the edges of a rectangular net.
Put one of them in this orientation, and that surely is along the edge of a rectangular net, along the edge of a net that's not only rectangular but square.
But it has a higher symmetry, and that's fine.
Looks as though this diagonal mirror plane might cause troubles, but let's note that if we look at the diagonal translations in this lattice, that they will define a centered rectangular net, so putting the other mirror plane in this way puts it along the edge of a centered rectangular net.
So this mirror plane is perfectly happy as well.
So in the square lattice we could pop either symmetry 4 or 4MM.
3-fold is an interesting one.
That's the odd symmetry that does strange things.
Here is a hexagonal net, and that's what a 3-fold axis requires.
Two translations, identical in magnitude, and in terms of what goes on along them, in exactly 120 degrees apart.
Now, I don't readily see any rectangle along the edge of which I could align the mirror plane in 3M.
But it's the oblique shape of this net that throws you off.
Let's notice that if I draw several unit cells side by side, here is a double cell that has a rectangular shape.
OK?
Take 2T1 plus T2 as one edge of the cell, and leave the other one as T2, and we have defined a cell where that is identically a 90-degree angle.
So one of the mirror planes of 3M could go in this orientation, and the other one would be 30 degrees away.
So this would be the orientation of the two mirror planes.
So we're putting in 3M with M perpendicular to the translations that define the primitive hexagonal cell, T1 and T2.
But there's another way you can do it.
Let me draw the same net, the same hexagonal net, and look at a different kind of double cell.
OK, here's my hexagonal net, two translations, T1 and T2, with a 3-fold axis at the lattice point.
And now I will again draw a couple of extra cells-- double T1, and let's double T2.
And then let me point out that I can put in the mirror plane this way and along this way, and here is a double cell, and I can put the mirror plane along this direction and this direction.
So this is crazy.
This is the same lattice and the same point group, but there are two different ways I can align the point group in the lattice.
I can put the mirror plane in such that it's perpendicular to a cell edge, or put the mirror plane in 30 degrees away from that orientation such that it is along the edge of a centric rectangular net.
So I could add 3M as well with the mirror plane aligned parallel to the translations in the net.
So lots of funny permutations here.
Here I have one symmetry element, two different lattices that I can combine with it.
Here I have one point group, 3M, but two different ways in which I can set it relative to the edges of the cell.
Yes?
Why are you putting two mirror planes in each of those figures when you really only could do one mirror plane?
Actually, I need-- well, there are going to be a radiating sheaf of mirror lines coming out of that 3-fold axis
An angle of 60 degrees
They are at an angle of 60 degrees.
It has to be 60 degrees, and that is indeed what I have here.
I've just drawn part of 3M, and if I draw the remaining mirror planes, they would do this.
OK, I think that's fairly intelligible.
Another reason for notes is to draw these diagrams when they're reasonably complex.
On a blackboard, when you're rushing to get through it, is something that's easier said than done, even though you may have done it a number of times.
So there is the job that we've laid out for ourselves.
Five lattices, 10 point groups, and we can drop each of the point groups into one or more of the lattices of the five that we have determined.
As soon as we combine things, we're going to have the situation that we've encountered before.
In general, if I have an operation number one that produces one transformation of coordinates, and combine it with an operation number two which has another change of coordinates, there must automatically arise some third operation.
And we can write it in terms of the language of operators by saying operation number one followed by operation number two is equivalent to an operation number three.
And in terms of the pattern, this says that whenever you throw two symmetry transformations together, a third one pops up.
You can't push it down.
It's going to be there once you add the first two combinations.
You may have to be really clever to see where it is and what it is, but it's going to be there.
It's going to be there.
It comes in automatically.
So this general class of statement, this equality in operations, is something that I like to call, for short, a combination theorem.
And we've seen several examples of this already, notably the one in the last hour that said whenever you combine two mirror lines at an angle mu, you automatically create, like it or not, a rotation operation of a 2 mu.
So as soon as we start making these combinations, we're going to find a new operation arising.
So let's start with the simplest one.
Add a 1-fold axis to a general oblique net.
So there is T1, there is T2, we don't put anything in it, it's just a simple array of translations with lattice points that we can put at the terminal point of these translations.
There's no symmetry.
If we put in one atom, that atom hangs on every lattice point, and that's all there is to it.
It's so simple it's not only uninteresting, it's ugly.
It's just a simple atom arranged in a periodic array.
As with everything we've done so far, it's convenient to have a notation to indicate which particular plane group one has, and the symbol here is to give side by side the type of lattice as the first part of the symbol, and then the point group that you have added.
So what do we have here?
We've taken a primitive oblique lattice, and we've added no symmetry at all, a 1-fold axis.
And here people who have derived symmetry theory assume that the reader knows something.
And so you're going to get a lot of respect when you show your colleagues that you know what this notation means.
All you have to do, if you know the symmetry, is to state whether the lattice is primitive, as it has to be for an oblique net, or whether it's centered, in the case of the rectangular net.
Three dimensions is a lot more complicated.
If the lattice is cubic, it could be primitive, it could be face-centered, it could be body-centered.
So you need a symbol that tells the type of lattice.
In this case it's primitive, and the point group that we've added is 1, so this rather boring ugly thing here is called P1.
Another bit of code.
We're going to do eventually the same thing in three dimensions.
We're going to take just a general oblique three-dimensional lattice, and not adorn it with any sort of symmetry at all, and that will also be called P1.
But to distinguish the two, capital P1 means a three-dimensional lattice, and we won't be working with those for a while.
We haven't even enumerated them.
A lowercase p, that is the symbol for a two-dimensional net.
In other words, a plane group rather than a space group.
So here's the first, P1, and it really doesn't give shivers running up and down my spine to look at it.
So let's do one that's a little more interesting.
Let's take a 2-fold axis and combine it first with a primitive rectangular net.
I'm sorry, what am I saying?
I want an oblique net.
So the symbol-- and I'm assuming this is going to be unique, and we haven't seen anything with a 2-fold axis in it, so it is.
So this would be P for a primitive lattice, 2 for the symmetry that you've added to the lattice.
And since you are all cognoscenti, I don't have to tell you it's an oblique lattice, because you know, don't you, deep down in your heart, that a 2-fold axis requires only that the lattice be an oblique lattice.
Nothing special is required.
So what we will do is to take this general oblique lattice with two translations, T1 and T2, and to this-- remember, we don't deal with symmetry elements.
We have to do this operation by operation.
So what we're doing is taking the operation A pi, the only non-trivial operation contained in a 2-fold axis, and putting that in at a lattice point.
OK. Bells ring and whistles go off.
We don't know what happens when we make this combination.
And let me do this in more general terms, so we can derive the theorem once and for all.
Here's a translation, and let me add to the end of that translation a rotation operation A alpha.
In this case, it would be the operation A pi.
What's the net result of rotating and then translating?
That's a non-trivial question.
So if I'm doing this for a general rotation of alpha, let me take a translation, and I'm going to make my construction in the following way, because I'm really clever.
And I'm going to put this translation T at one half of alpha on one side of the perpendicular to T. And then I'll let that rotation go to work.
And it'll take the translation and move it to alpha over 2 on the other side of the perpendicular.
And now I'll let the translation move to this rotated translation, and if it does so, it's going to move the translation over to here.
Here's a motif hanging on this translation.
It's rotated over to here, and then gets slid over to here.
This one is right-handed, this one is right-handed, this one is right-handed.
How do I get from this one to this one?
Only two ways, translation or rotation.
If we rotate, about what point are we rotating?
Now I'm going to use a definition which seemed trivial.
I said, a symmetry element is the locus of points that is unmoved by the operation.
And if we decided, because these two motifs are not parallel to one another and have the same corrality, that the rotation-- has to be a rotation that relates the two.
When asked finally the question, what point is left unmoved?
Ha.
It's this point where these two translations come together at a single point.
That's the point that's left unmoved by rotating from here to here and then sliding over by the translation T. And I can say exactly where that rotation is.
It's going to be along the perpendicular bisector of my original translation.
And where is it going to be located?
This line makes an angle of alpha over 2 with respect to the perpendicular to T. So that angle is also alpha over 2.
This line and this line both make an angle of alpha over 2 relative to the normal to the translation, so this angle is also alpha over 2.
And look how this has turned out.
The combination of a rotation with a translation is another rotation, B, about a different point but through the same net amount alpha as the original translation.
On top of that, really to nail this down, where should we look?
This distance in here is T over 2, and the tangent of alpha is T over 2 over this distance x.
And if I just solve for that distance, x is equal to T over 2-- I'm sorry.
I got this garbled.
T over 2 over x is equal to the tangent of alpha.
That's what I wanted to do.
So if I solve for x, that distance x is T over 2 over the tangent of alpha, and I can write that as T over 2 times the cotangent of alpha over 2.
So here it is, done properly at last.
Go up a distance x along the perpendicular bisector of T, and you go up a distance T over 2 times the cotangent of alpha over 2.
So this is a general result for any rotation operation.
Combine a rotation operation A alpha with a perpendicular translation-- perpendicular to the rotation axis A-- and the result is a new rotation operation through the same angle, but at a different location.
It's at a distance x, which is equal to T over 2 times the cotangent of alpha over 2 along the perpendicular bisector of T. So this is a general theorem.
This is a fact of two-dimensional life that is always going to be true regardless of alpha.
And the only constraint is that the translation has to be added perpendicular to the locus of the rotation axis, and that has to be the case in two dimensions, because the rotation axis has to always be normal to the space, the two-dimensional space.
In three dimensions, we'll have to generalize this to make the translation have arbitrary orientation relative to the rotation axis, but we've got enough to chew on at the moment.
Yes
[INAUDIBLE]
The distance x is the distance up along the perpendicular bisector.
I'm sorry.
I said it but I didn't write it in.
OK?
Let me go through it again since we've got everything on the board.
We start with the original object, rotate it by alpha, and then we then further map it to a new location by the translation T. Both motifs are right-handed, but they're not parallel, so therefore, they have to be related by a rotation.
If we ask what is the locus about which this location has occurred, we use the fact that the locus of a rotation axis is the only point that's left unmoved by the net transformation, and the place where this translation intersects the translation that has been shifted over by T is a point that sits up along the perpendicular bisector of the translation by a distance T over 2 times the cotangent of alpha over 2.
So that's in the abstract.
Let's now look at a particular application of this to our addition of a 2-fold rotation to a translation.
So here's the translation T. We're going to put a 2-fold axis and the only operation which it possesses, namely A pi.
And according to our theorem, up along the perpendicular bisector of T by distance x equals T over 2 times the cotangent of alpha over 2 should be a new operation B pi.
So in this case, this would be T over 2 times the cotangent of pi over 2.
Cotangent of 90 degrees is 0.
So this says that you go 0 of the way along the perpendicular bisector, which means you stay right at the midpoint of T. So this says that if you rotate through 180 degrees, follow that by this translation, you should find that the net effect is a rotation of pi over 2 about this point.
Boggles the mind, but let's show that it works.
Here is an initial point, initial motif number 1, and it is right-handed.
We rotate by 180 degrees.
Here's number 2, it stays right-handed, and I pick it up and I move it by this translation, and it moves from here to here.
Here's number 3, it stayed right-handed, and the way I get from 1 to 3 in one shot is to rotate 180 degrees about the midpoint of the translation.
How about that?
People are so astounded that they're stunned, except for the people who couldn't see what I was doing, and they're moving back and forth frantically trying to see what I wrote.
I'll let you catch your breath, and then let's do this, use this to derive the two-dimensional space group, the plane group that results when we add a 2-fold access to an oblique net.
And it's turned out very simply, because if I've got the operation B pi here, that implies that I've got a 2-fold axis, because that's the only operation that exists within a 2-fold axis.
So I'll start by putting in a 2-fold axis here.
I'll combine A pi with this translation T1, I get a new rotation operation, B pi, in the middle of this translation, so there's a new 2-fold axis there.
The 2-fold axis here hanging at this lattice point as well, and at these other lattice points.
Combine the rotation A pi with this translation T2, I get a new operation, B pi, sitting in the middle of this translation, so I've got all I need to say that there's a 2-fold axis there.
That 2-fold axis will get translated down here by T1.
This 2-fold axis will be translated over to here by T2.
Then the only other thing that I have to combine this translation with is the translation T1 plus T2, and that combined with A pi says I should have B pi in the middle of that translation as well, so I'll get another 2-fold axis in the middle of the cell.
And that's it.
We've got our first nontrivial two-dimensional space group.
And what do we call this?
Confusing as hell, some might say, but crystallographers would say that it's a primitive lattice combined with a 2-fold axis, this is P2.
I don't have to tell you that it's an oblique net, because you know that that is all that a 2-fold axis requires, that the net be oblique.
What does the pattern look like?
I'll say another thing that's so simple, it takes a while for it to sink in.
The pattern of a plane group is the pattern that is generated by the point group merely hung at every lattice point in the net.
So if this is what a 2-fold axis does, relates a pair of motifs like this, a pattern for P2 is this pair hung at every lattice point.
So all of the objects are of the same pirality, all either left-handed or right-handed.
I've got the same pair at every corner of the cell.
And notice that these new symmetry elements that popped up just like what we found in the point groups, the new mirror planes that came arose as a consequence of the combination that we had made.
And they were independent in most cases, 3M being the one case where they weren't.
These additional three 2-fold axes are independent 2-fold axes.
They're different kinds of 2-fold axes than the ones at the corner.
Not only are they not equivalent by translation, but they do different things relative to the motifs that are in the pattern.
This 2-fold axis has a different relation relative to the two closest motifs than does this one, than does this one.
They relate different objects pairwise in the pattern.
And no two of these 2-fold axes describe the same pairwise relationship.
So these are independent 2-fold axes.
Independent in the sense that they do different things relative to the motifs in the pattern, different in the sense that there is no other operation that throws this one into this one, and therefore, of necessity, would require that the pair about this 2-fold axis be the same as the pair about this 2-fold axis.
OK.
So there is a first nontrivial two-dimensional space group.
I should mention-- probably an obvious fact-- that this also has a cousin in three dimensions.
If I just imagine all of these 2-fold axes extending out through space perpendicular to the plane of the blackboard, and take another translation that is also perpendicular to the blackboard, I would have the three-dimensional space group which has the same symbol except capital P, indicating that it's a space lattice.
So almost at no extra charge, we've done all the work to determine one of the three-dimensional space groups.
Yes?
If you change the order of the combination of the translation rotation, do we get the same operation?
OK. That is a good question.
I really meant to mention it earlier in connection with the point groups.
Does interchange of the order of the operations in a combination change the result, or change, in other words, the symmetry element that is equivalent to the net combination of the two?
How many think yes, it does change?
Could you say that again?
OK, if we have two operations A and B, is operation A followed by B the same result as the operation B followed by A?
Does the order make a difference?
And how many say yes--
The order makes a difference?
The order does make a difference.
How many say no?
Well, this is one of these happy occasions where I can say you're all right.
Sometimes yes, sometimes no.
So let's ask when does it make a difference and when does it not.
Let's look at one of our point groups.
Let me take a non-trivial point group.
This would be point group 4MM.
OK?
So those are all mirror planes of two different kinds.
And let me quickly sketch in the pattern.
The pattern is just a pair of things hanging on the mirror plane, and that pair is rotated by 90 degrees.
OK.
Suppose we look at the operation A pi over 2, a 90-degree rotation in a counterclockwise direction, and let the other operation be sigma 1, this reflection plane here.
Suppose we reflect and follow that by A pi over 2.
We reflect to here and then we rotate by 90 degrees, and that's going to give us the number 3 sitting up here.
So we reflect it and then rotate it.
How about if we first rotate that same object by A pi over 2 and follow it by sigma 1.
So we'll rotate number 1 to here, so this is 2 prime, and you can see we're already in trouble.
If we now reflect along sigma 1, this is 3 prime.
So in one case we ended up here, in the other case we ended up here.
So the order did make a difference.
Let us look at another example.
Let's look at 2MM.
So here's an operation sigma 1, here's an operation sigma 2, so let's first do sigma 1 followed by A pi.
So here's object 1, reflect to here, and then we rotate by 180 degrees, and this is number 3.
Now let's do the operation A pi and follow it by sigma 1.
So we'll take number 1 and rotate it up to here.
So here's number 1 prime.
And then we reflect by sigma 1, and 3 and 3 prime are exactly the same.
So it didn't make a difference.
So the answer is sometimes yes, sometimes no, which is interesting, but even more interesting is how can you tell?
Yeah, you have a question?
In that first identity you drew--
Yep
The right side, it seems like your commas are drawn backwards and that just affects--
Could be.
Mm, no, they're-- ah, these are drawn backwards.
They should all have their tails pointing in.
No, it doesn't work, actually, because even if I screwed up on the orientation, the positions are different.
Again, first time I reflected, and then rotated, so this is 3.
If I rotate and then reflect, I'm way over here.
So this is one location, this is another location, and even if I screwed up by having the tails pointed the wrong orientation, still not the same.
But here, clearly, reflecting and rotating is the same as rotating and reflecting.
I end up in the same place.
So how can you tell?
It's a very difficult thing to prove that this is the case, but you can interchange the order of operations when the two symmetry operations that are involved leave each other untransformed.
And if the two symmetry operations move one another, then changing the order changes the net result.
It doesn't change the nature of the operation, but it does change the locus about which it operates.
Let me show you what I mean.
Here's 2MM.
The 2-fold axis takes this mirror plane, turns it upside down.
It takes this mirror plane and flips it over.
The mirror plane is left unchanged in both cases.
The mirror plane has the 2-fold axis sitting right on it, so it doesn't change the location of the 2-fold axis when it acts on it.
And similarly, this mirror plane takes this mirror plane and reflects it, and it doesn't change anything.
So all of these operations-- the three independent operations, sigma 1, sigma 2, and A pi-- all leave their locus of operation unchanged.
Whereas that, clearly, is not the same for a 4-fold axis and a mirror plane.
The mirror plane, when it acts on the 4-fold axis, leaves it unchanged, but the 4-fold axis, when it acts on the mirror plane, moves it into a new, entirely different location.
And when that is the case, the order does make a difference.
So I can take that corny joke of the mathematicians, what is purple and commutes?
An Abelian grape.
I can say, what is purple and commutes?
It's a purple 4-fold axis in a mirror plane, a purple 4MM.
OK, what other aspect of the plane groups and three dimensional space groups is an analytic representation of the way the plane group maps atoms around?
And I see that it is five of the hour, so let me just make one final statement to indicate what I'm going to talk about next time.
Here's the arrangement of 2-fold axes, and if you have an atom sitting in here, and this is T1 and T2, and we use this as the basis of a coordinate system, so this atom sits at a location x along T1 and y along T2.
And then they ask, what other atoms are related to it?
Well, that 2-fold axis will move x to minus x, and move y to minus y, and those are the only two atoms I get.
So a characteristic of this plane group, if I were describing to you a crystal structure which had this symmetry, I might say something like a lithium in a position with coordinates xy, and therefore minus x minus y would also have to occur.
I have an oxygen at a different number, x prime y prime, and that must then also occur at minus x prime minus y prime.
And it looks as though any time you throw in an atom at a location xy, you get a symmetry related companion to it with negative coordinates.
And this is what is called the general position of the space group.
There's nothing special about it, and you put in an atom here, you get another atom here.
This is going to be a nice economical way of describing an atomic arrangement, particularly if you realize that there are almost 150 atoms that appear in the unit cell for some of the cubic symmetries.
Drop in one, wow, all hell breaks loose, 150 other ones.
But what you can do is codify the coordinates of all these symmetry-related atoms.
And that is a great utility in describing crystal structures, because I only have to give you the coordinates of one representative atom.
But this is not the only thing that happens.
This is the general position because it's general relative to its location with respect to the symmetry elements.
If x and y were both 0, I would have one atom moving to the origin to coincide with the second one.
And if I put one in at 0, 0, that's all I'm going to get.
The 2-fold axis is just going to twirl it around, and the translation will repeat it to the other lattice points.
So if I put in an atom at 0, 0, that's all I'm going to get.
And the same thing would occur for any of these four independent 2-fold axes.
If I let x and y migrate to 0, 1/2, this atom will join together with this atom, and I would get just one atom sitting there.
So there's a position also of the form 0, 1/2, and a position 1/2, 0, and a position 1/2, 1/2 that are the location of these four distinct 2-fold axes.
OK, to have a shorthand way of referring to the general position in these positions, which are called special positions-- what's special about them is that they're right on top of a symmetry element, and whenever that happens, the number per cell is a sub-multiple of the number in the general position, because coalescence has occurred.
And so they're three independent 2-fold axes, so there are four locations where you get only one atom per cell rather than two.
And then to give you a way of referring to these positions readily, they are labeled starting with the most specialized, just going through the alphabet, position a, position b, position c, position d, position e, which is the general position.
And then another piece of information that's almost trivial here, but is not for more complicated symmetries, what's called the rank of the position, and that is the number that you get per cell, one per cell, one per cell, and for the general position you get two.
So the rank of the position is just the number per cell.
And then, if it's a special position, the atom has to sit at a site of some symmetry, and in this case, the atoms all sit at 2-fold axes for the special position.
The general position always has site symmetry 1 always.
So that's the characteristics of the way in which this symmetry can move atoms around.
And this is of great utility in interpreting structures.
For example, suppose our motif was not an individual atom, but was a molecule.
And suppose you had one molecule per cell.
That would tell you the molecule has to fit in a special position.
And the special positions are all 2-fold axes.
So if the molecule has to sit on a 2-fold axis, the configuration of that molecule has to have at least a 2-fold symmetry, or else you can't fit in one per cell in this particular plane group.
So you can use this information to determine structures.
I'm going to give you a problem a little bit later on when we get to three-dimensional crystallography, and I'm going to give you the magnitude of the cell edge and the density of rock salt.
And from that, in 10 minutes, you can find out that there's only one possible structure for rock salt.
The Braggs fiddled around for the better part of a year doing x-ray experiments and trying to calculate what the intensities would be for different atomic positions.
You, in another two weeks, can get the same answer in 10 minutes knowing only the lattice constant and the density and the space group.
So a lot of physical consequences of the nature of the structure, the confirmation of the molecule, the extent or absence of solid substitution of one species for another, can be dredged out of the properties of the space group.
OK, I've run 2 minutes over, but we took our break two minutes later, so let's call it a day and I shall see you on Thursday, hopefully armed with pictures so that I can personally present each one of you with your corrected problem sets.
We had begun to discuss, really what is, although you would not gather it from last time, a fairly straightforward matter.
Set up our usual Cartesian coordinate system.
And then look at the surface of the solid, and applied to this solid a force per unit area, which we'll represent by a vector k. And then put on some labels here.
Let this be O.
Let this be A, let this be B, and this be point C. Then we proceeded to-- I will say, attempt to-- set up a balance of forces between the external force per unit area and the internal forces per unit area, which have to push back on the external force if the solid is to remain motionless.
The thing that I did not carry through is that there are two directions involved.
One is the direction to the normal, to the particular surface that we're looking at.
We'll call that N. And there is the direction that k, the vector k, makes with respect to the three reference x axes, x1, x2, x3.
Let's take k and immediately split it up into the three components, k1, k2, k3, because throughout, we're going to be dealing with the balance of those three forces per unit area in the directions of x1, x2, and x3.
We want to take this surface area and resolve it into three separate areas that are normal to x1, x2, x3.
And what I did at the end of last hour was to look at k and try to express these internal areas in terms of the direction cosines of k. Forget that.
We've split up k into its three components, and I didn't draw in the direction of the normal to the surface N last time.
So I kept saying, direction cosines of k, and people in the audience said, no, no, no, no.
And I said, yeah, because you see, that's the ratio of these two areas.
And everybody said, no, that's wrong, and I said, ha.
And that's where we ended my most inglorious hour.
And they've got this all down on tape.
My reputation's going to be ruined.
So, anyway, let's now do it properly.
And what I will say is that in the x1 direction, we have to have internal forces that push back, and we have to balance those forces in all three directions.
So I'll say is that in the x1 direction, and this as you saw last time, gets very cumbersome because we have to deal with all these clumsy areas, that the force per unit area on ABC has to be balanced for each of the three components of k by internal forces acting along x1, x2, and x3.
So we can say is that the first in the x1 direction, the component of k that acts on the surface ABC in the x1 direction has to be balanced by internal forces.
And let's look first at the x1 direction.
So there are three internal surfaces.
They are the area BOC.
That is this back surface here, and the normal to that surface is x1.
So let me define some forces per unit area that are internal, and I'll use the term sigma for that.
And I will use a first subscript that gives the direction in which that force per unit area acts, and I'll use a second subscript that indicates the normal to the internal surface.
So area BOC obviously, as we observed a moment ago, has x1 as its normal.
But there are three internal surfaces.
There will also be an internal area, AOC, on which a force per unit area acts in the x1 direction.
So I'll call this one by analogy to what I did a moment ago.
I'll call this sigma 1, because it acts in the x1 direction, and it acts on the surface AOC, which has x2 as its normal.
Then finally, there will be a third force per unit area, sigma 1 3, acting in the x1 direction, on a surface whose normal is perpendicular to-- along x3.
The surface is perpendicular to x3.
And that's area AOB.
OK, so this is just a simple balance of forces, external and internal balancing forces, which act in the x1 direction.
Then we'll write a similar thing for the x2 direction.
And we'll say that the x2 component of force per unit area external to the volume element acting on area ABC is going to be balanced by, again, forces now acting in the x2 direction.
So first, there will be a component of 4 sigma 2 acting on the area that's normal to x1, and that's the same area BOC, plus another force per unit area acting in the x2 direction on the surface, whose normal is a long x2.
And that is, once more, area AOC.
And that would be a force per unit area that I'll define, because it's my ball game and I own the ball.
So again, that would be sigma 2 2, and then finally a sigma 2 3 times the area whose normal is x3.
And I think I need not write the third term that follows automatically from what I've written so far.
So now comes the geometry part, where I'm going to get rid of these messy areas, and I'm going to divide through by the external area on the surface of the solid.
I will then write accordingly that k1 is equal to sigma 11 times the ratio of area BOC divided by area ABC plus sigma 12 times area AOC divided by area AOB-- ABC, excuse me.
Then finally sigma 1 3 times area AOB divided by area ABC.
So now I'll introduce a little bit of geometry that caused such difficulties last time.
If I have an area and project it onto another area that makes an angle C with respect to the first, this area A prime is equal to the area A times the cosine of Phi and you could define that Phi in a different way, namely take the normal to one of the two areas and let M prime be the other normal, so C then is exactly the same thing as the angle between the normals to these two surfaces.
So let's now look at what the ratio of these two areas that we have is.
What we're dealing with here is an angle between the normal to the surface, not k, but the normal to the surface, and either x1, x2, or x3.
So the ratio of-- let's look at one of these specific terms.
The ratio of area BOC through the ratio of the area ABC should be the cosine of the angle between the normal to the surface and the normal to area BOC.
Well, the normal to area BOC is x1, so that first ratio of areas is just going to be the cosine of this angle.
And that cosine is just the direction cosine of N1 along the x1 of M along the x1 direction.
So this is just the direction cosine between N, the normal to the surface, and x1.
And let me say then that, by definition, I'll let the direction to the normal be its three direction cosines, L 1 plus L 2, plus J plus L3 times k. These are the direction cosines of the normal to the surface, and not k. So this then simply comes down to sigma 11 times L1, the ratio of the area AOC to ABC is going to be the angle between N and x2.
And that's the angle whose direction cosine is L2.
And similarly, the ratio of these two areas will be cosine minus 1 of L3.
So this cumbersome construction with ratios of areas just reduces to sigma 11 L1 plus sigma 12 times L2 plus sigma 13 times L3.
And the other two equations will follow a similar form.
K2 will be sigma 21 times L1 plus sigma 22 times L2 plus sigma 23 times L3.
Or in general, case of I will be given by sigma I J times L to J.
Easy.
Nice and compact.
Nice and simple.
What can we say about this?
K is a vector which has components k1, k2, and k3.
The direction cosines can be, as we said a moment ago, can be regarded as the components of a unit vector pointing along the normal to the surface.
So k is a vector.
These three direction cosines are the direction cosine to a unit vector normal to the surface.
If we were to be so foolish to try to change the coordinate system., having just gotten it right for the first time in one coordinate system, we know exactly how the two vectors will change because we have a law for transformation of a vector.
If we were to change the coordinate system, the components of k would wink on end of and take new values and we know exactly how the components of a vector will change as we change coordinate system.
OK, if this transforms like a vector and this transforms like the vector, then the 3 by 3 array of coefficients which relate those to quantities must be a tensor.
So we've proven rigorously that sigma IJ is a tensor, in particular, it's a second rank tensor.
And were we to change coordinate system, we know from what we've done in the past, exactly how the numbers that make up this 3 by 3 array will transform.
These quantities, sigma IJ, are something that you've probably encountered, introduced to be sure, probably in a little different way.
These are the elements of stress, and sigma IJ is the stress tensor.
We can immediately, on physical grounds, established that sigma IJ must be a symmetric tensor.
So let's set up two axes, x1, and x2, and x3 obviously is normal to the board.
And let us look at some off-diagonal terms like sigma 12 and sigma 21.
Sigma 1 2 would be a force acting on the x1 direction on a surface whose normal is x2.
So this is going to be a sheer stress and that, to prevent any translation of the volume element, must be balanced by a sigma 12 on the other face that acts in the opposite direction.
We're going to need a convention for defining sign, S-I-G-N, of what constitutes a positive element of stress and a negative element of stress.
And I'll define a plus sigma IJ as a force per unit area acting in the plus xi direction on a surface whose normal is plus xj.
So if this is a force per unit area acting on this surface, and it acts in the direction of x1 on a surface whose normal is x2, what I've shown here is a positive sigma 12.
A sheer force going in the opposite direction would be a negative sigma 12.
A positive sigma 11 would be a stress that's tensile, a negative sigma 11 would be a stress that's compressive.
So here's a sigma 12, a sigma 12 on the opposite face.
If we went away and turned our back, this volume element would start accelerating, getting an angular acceleration, would start furiously spinning about the x3 axis like a top.
So something, if this volume element is, by definition, an equilibrium, something has to balance this couple.
And if it's to stay put, we must have a sheer force like this and a sheer force like this.
And that's the only thing available from the nine elements of stress that will give a net torque on the volume element.
So what are these?
This is a stress that acts on the surface whose normal is x2-- whose normal as x1 in the x2 direction.
So this is sigma 21.
It acts in x2 direction, and its normal is x1.
An x, and then [?
plus ?]
x2 direction on [?
I ?]
surface, which has a normal that is along x1.
So the only way we can avoid an angular acceleration is to say that sigma 12 has to be identical to sigma 21 if there's no rotation about x3.
And looking at the couple's that act about x2, and x3, we could say in general that sigma IJ is going to have to be equal to sigma JI for a volume element in equilibrium.
So stress then is equilibrium is in place, has to be a symmetric tensor.
It is, however, a plain old second-rank tensor like all of the other second-rank tensors that we've spent the last month talking about.
And I think I tried to swindle you on this one before.
If this is a second-rank tensor, the stress tensor must be symmetric and diagonal for a cubic crystal.
In other words, the stress tensor sigma IJ must have the form sigma 0 0, 0 sigma 0, 0 0 sigma because we've shown that that is what is required of a tensor if the crystal has cubic symmetry.
So surprisingly, you cannot subject a single crystal with cubic material to sheer stress.
Any quarrel with that?
Should we move on?
We better not, but somebody better raise an objection.
That's not true, obviously.
I can take a cubic crystal and shear the stuffings out of it.
So what's going on here?
The tensor property that you are applying to [INAUDIBLE] property of the [?
crystal.
?]
But if it is an outside force, it doesn't [INAUDIBLE]
This is one of your good days
[INAUDIBLE] society [INAUDIBLE]
That makes it an especially good day because you're paying attention to what I said for a change.
The distinction here, and I did make it last time-- it was one of the things that I did correctly-- that there are really two kinds of things that have to be described by second-rank tensors.
There are the things that we have been spending most of our time on, and these are property tensors and-- I mentioned last time, too, so I know I said this before-- Nye calls them field tensors.
I'm sorry.
Nye calls them property tensors, also.
And these are something that depends on the structure and bonding and other characteristics of the material.
The stress is something that we we're taking and imposing on an innocent, unsuspecting crystal, and that is something we do.
And we can, by design, twist it, shear it, stretch it, as hard as we like, provided we don't break it.
And so this is something that is externally imposed on the crystal, and this is something that's called a field tensor.
We mentioned last time the meaning of field.
The field is a volume of space and a field tensor is a space in which a value of the tensor is defined as every point, x1, x2, x3, in the space.
Just as an electric field is a vector field and you define the electric field we as a function of every point x1, x2, x3, in the space.
So there's no reason whatsoever that there need be restrictions on the tensor that we apply.
And when we can cross out that interesting, but incorrect, supposition.
But they can be a lot of different forms, special forms, of stress tensors which have special forms simply because of the nature of what we do.
And these have restrictions and equalities that have no counterpart in property tensors.
For example, we can have the cancer sigma 00000000, which is something we could never have identically so for a property tensor.
And what does this represent?
This is a tensor for which sigma IJ is 0, unless I equals J equals 1.
So this is the only thing that's going on, is a force per unit area along the x1 direction.
So this is a uniaxial stress.
And that is a stress field which we could apply on a piece of material very easily.
Just screw a hook into the material and hang a weight on the hook and you've got uniaxial stress.
We can also have something that's a little harder to visualize, a biaxial stress.
And this would be a stress that had elements sigma 11, register 0-- at least when we refer to principal axes-- 0 sigma 2 2 0 0 0 0.
When it's in diagonal form, two of the diagonal terms are 0.
That doesn't seem to make sense because, if this is a force per unit area along x1, and this is a force per unit area along x2, can't we add up those vectors and say that that's a uniaxial stress?
Depends on your coordinate system.
If you want an example of a biaxial stress, imagine some firemen underneath a burning building, and they're holding out a blanket for somebody to jump into.
You have one pair of firemen pulling in this direction to keep the blanket taut, and another pair of firemen are not quite so vigorous, and they're pulling in an orthogonal direction.
And those two forces, sigma 11 and sigma 22, create a biaxial stress event material.
You could change the form of the tensor by rotating the coordinate system, but this is one example of a biaxial stress.
The final general form, a triaxial stress, obviously by extension, is going to be where of all three diagonals, sigma 11, sigma 22, and sigma 33, are non-zero when you put the tensor in diagonal form.
This would be a general triaxial stress.
And an example of that would be when the jumping person lands on the blanket, there's going to be a force that his or her mass exerts on the blanket, and that would be a third sigma 3 3.
So that would be a triaxial stress.
Let me give you a couple of special cases of specialized stress tensors.
And suppose I had a triaxial stress, but all the diagonal terms were equal.
That's something that I tried a moment ago to hoodwink you into believing was required by a cubic crystal, but suppose that was the form of the stress tensor.
What would that represent?
[INAUDIBLE]
That is something that's going to be-- yeah, very good.
That's not a pressure.
It's a dilation.
But if I make it into a more familiar form, put it here, minus P, minus P, minus P. It's a force prevented area.
That's where the stress is.
If it's a hydrostatic pressure, it's squeezing the volume element so it has to be, by our definition, acting in a negative x1 direction on a surface whose normal is plus x1.
So minus P, minus P, minus P, represents a hydrostatic pressure.
And now, unless you've seen it before, or you're reading the notes instead of listening to me, let me give you a really strange-looking one.
What is this stress tensor?
Where sigma 22 is opposite in sign but equal in magnitude to sigma 11
[INAUDIBLE]
Yeah.
Good.
The way to see that geometrically is the following.
What we're doing is, we're pushing with a force per unit area in this direction.
That would be sigma 11, and then we squeeze in the orthogonal direction.
And that is a sigma 22, which is negative.
If we rotate the solid by 90 degrees, then we've got this, plus this, acting on this face, same forces per unit area as we have here.
And that is a net force per unit area like this.
On the opposite surface, we've got this force per unit area and this force per unit area, component going in this direction.
And that is going to be a force that goes like this.
That's going to be a force per unit area like this, and similarly, this combined with this is going to be a force acting in the reverse direction.
That would be, again, this with this, and this would be this with that.
So clearly, that is a body that's being subjected to pure sheer, and if we rotated the axes x1 and x2 by 45 degrees to an x1 prime and x2 prime like this, and then transformed the tensor, it would take the form 0 sigma sigma 0 0.
And then, what do we want now?
This would be the only thing we get.
Except I should be careful to put a sigma prime on this because, when we change axes, these sigmas will not be of the same magnitude as the original ones.
There's going to be a square root of 2 coming from cosine of 45 degrees in there.
But, anyway, just rotating the axes 45 degrees of as I've indicated schematically here with vector sums is going to give me two off diagonal terms equal in magnitude that would correspond to sigma 12 and sigma 21.
So those are some special forms of a stress tensor when you place it in a diagonalized form.
But let's conclude this part of our discussion by observing that everything that we've said about second-rank tensors holds for the stress tensor.
For example, you can always select new axes that put the tensor in diagonal form.
So that's one general property.
You can define a stress quadric, and we'll do that with the equation sigma ij xi xj equals 1.
Why not?
It's a second-ranked tensor.
We can take the elements and construct a surface from it.
This quadric will have the properties of any other quadric for a second-rank tensor.
In particular, if we construct the quadric, it's now for sure going to be able to be either an ellipsoid, an hyperboloid of one or two sheets, or in this case, even an imaginary ellipsoid for a stress that's everywhere compressive.
The quadric could have its diagonal elements all negative.
The radius of the stress quadric in a particular direction is going to be the 1 over the square root of the property in that direction.
What does that mean, stress in a given direction?
Well, let's again remember how we have defined the stress tensor.
A force per unit area has three components, k1, k2, k3, and they're equal to sigma ij times lj.
So the value of the property in this direction is going to be the value of the property in the direction defined by [?
elsa ?]
j and the values of [?
elsa ?]
j are components of the unit vector that point in that direction.
So if this is the direction of k, [?
off ?]
like this, the value of this quadric in this direction is going to be the value of the part of k that's parallel to the radius vector over the magnitude of the radius vector.
But the [?
radius ?]
vector is a unit vector, so it's simply going to be equal to k parallel.
And if k parallel is the part of the vector that is along the surface normal, that is going to be the component of k which is purely tensile.
This is a force per unit area, has parts that are, components that are normal to the surface normal.
Those are shear components.
The component that's directly along the normal to the surface is going to be a purely tensile component.
And, therefore, the radius of the quadric in a given direction gives us the tensile component of the force density vector k that is transmitted across the face-- so let's not say face, but interface-- whose normal is in the direction of R. So this is the radius vector.
That's the direction of the normal to a surface and 1 over the radius of-- The radius of the quadric in that direction is going to be 1 over the square root of the tensile component of stress that's transmitted across an interface in this direction.
How are we doing on time?
Any questions at this point?
Can you explain how you know that's [INAUDIBLE]?
Remember what our radius normal property said.
If we have a tensor like good conductivity, where a current density is related to an applied electric field, [?
essa ?]
j and we've-- unfortunately, that's segment 2.
This is the conductivity tensor.
What we said is that, if we take the components of the tensor, sigma ij, and construct the surface sigma ij xij equals unity, that will be some sort of quadratic form that has the property that the radius in any given direction is 1 over the square root of the value of the property, sigma conductivity in this case.
Bad case.
I shouldn't have picked things where the sigma is also involved.
And the value of the property in the given direction is going to be the parallel component of what happens over the magnitude of what you do.
So in the case of electrical conductivity, if we apply a field in this direction, the value of the property is going to be that part of the current flow, which is parallel to the electric field, divided by the magnitude of the electric field.
So that is a relation that you know by heart and have come to not only know, but to love.
So what are we saying here?
For the stress tensor, we can take the elements of the stress tensor and construct a quadric.
And the thing that's related here is a force per unit area that we're applying to the surface of the crystal, and that's transmitted through the volume of the crystal by the relation that we have defined as the stress tensor.
So, unfortunately, again, sigma Ij.
And now what is involved are the direction cosines of the normal to the surface.
So what is the applied vector?
The applied vector is a unit vector, and this is the normal to the surface.
And the surface in question would be a surface like this.
The thing that results is a force per unit area k. So what we're doing is taking k, the value of sigma, in that direction.
A scalar quantity will be the part of k that's parallel to the normal vector, divided by the magnitude of the unit vector.
And that's, a magnitude of n is 1, so that's just k parallel.
So this is the surface.
This is its normal.
And what the radius of the quadric is going to give us is that part of the force density, which is the force density vector resolved onto the normal vector.
And this is a tensile.
This is the part of the tensile stress that's transmitted.
This is the total stress that has a sheer part and a tensile part.
The radius of the quadric is going to give you the part that's transmitted across the surface in this orientation.
That is purely a tensile force.
That make sense?
I should have done it without sigmas being present in both tensors.
So everything you want to know about stress that's applied to a solid is contained in a second-rank tensor that is symmetric and obeys all of the characteristics that we derived earlier for property tensors, even though this is a field tensor, and not a property tensor.
Let us then turn to the next thing that I want to discuss, again something you've heard before.
But we'll do it within the context of the tensor algebra that we've developed before.
And this is the concept of strength.
And let me introduce the subject by a one-dimensional analog.
Suppose we have a very thin elastic band fastened to some immovable surface, and this will be the direction x.
And then we pull on it.
And here again, we have to be complete and note that there are several different kinds of behavior.
One type of behavior is a case where the elastic band stretches, and if there's some point P on the initial state that point P will move to a point P prime that is displaced from the original location by a vector U.
There's another point Q part way along the elastic band.
If the thing that we have fastened to the elastic band slipped, and there's no deformation at all, Q would be displaced by the same amount U if there was just translation of the elastic because it wasn't fastened down solidly.
But if it's stretching here between this point and this point, it's going to stretch up here, too.
And, in general, it will move over to some point Q prime, and this displacement vector is going to be the original U plus delta U.
And what we care about in defining strength is not the absolute coordinates, but the interval between these points.
This separation is U, and this separation is U plus delta U.
Now this is a common, but not a necessarily unique, sort of behavior.
This is something where the displacement does not change along the length of the elastic bands.
This is something that's called homogeneous deformation.
And this is characterized by a displacement view that varies linearly with position along the elastic band.
Does something like this.
It's possible, in a [?
grade ?]
of material, if you'd like a real example, that U, as a function of distance along this one-dimensional object, does something like this.
Why not?
That's what might happen for example in-- that's almost an oxymoron, a one-dimensional elastic band whose cross-sectional area changes with distance in that one dimension.
So this is homogeneous deformation, as opposed to inhomogeneous deformation.
There's another type of behavior, and this is apparent if you place a thin piece of metal or plastic, let's say, between the pair of grips in the Instron machine, and after deformation, it does something like this.
There's an inhomogeneous deformation, which is to referred to colloquially as necking.
A very good example of that is what happens when you grab hold of some cheap plastic shrink wrap that has been shrunk around a package that has been mailed to you.
And you want to get it off, so you grab it.
And it doesn't deform elastically.
What happens is that it necks in places where there's a stress concentration.
And you get something exactly like this.
And if we were to indicate that sort of behavior, U as a function of position would be one U, and then it would increase very rapidly.
The [?
UDX ?]
would be very much larger than the places which had uniform elastic deformation, so it would be a nonlinear behavior of this sort.
And that would be necking.
If you've never noticed that sort of behavior before, I give you an assignment tonight of going home and getting a piece of Saran wrap from your kitchen drawer and grabbing hold of it and pulling it.
And it deforms non-uniformly in a fashion that's indicative of necking.
OK, it's just about five of.
Why don't I stop there.
And what I would like to do next is to cast this relation, which is a one-dimensional stream, into a three-dimensional situation.
And we'll look at some of the properties of the description of a body that's been deformed in three dimensions.
Resume by going back to our one-dimensional body that has undergone some elastic deformation.
And what I would like to do now is to distinguish between displacement of an object and fractional change of length, which turned out to be measured by the same thing, that thing that we're going to name strain when properly defined.
OK, here is our one-dimensional case.
And we said that originally some point, P, at a location x gets mapped to a point P prime that is at x plus some displacement U.
So this is the displacement vector U.
Our point Q, which is originally at some location x plus delta x, where delta x is the original separation between P and Q, gets mapped to a point Q prime, which is going to be equal to a whole collection of terms.
It's going to be equal to x plus delta x, the original location, plus the linear variation of U with x that has to go U times x plus delta x.
And if we simplify this a little bit, Q prime is going to be at a location, factoring out x plus delta x, x plus delta x times 1 plus e
Is e represented here?
e is the linear relation between displacement U and position along the body.
OK, so what has happened here?
The relative change of length is going to be P prime Q prime minus PQ.
And if we just substitute in our two expressions, the relative change of length is going to be equal to delta x times 1 plus e minus delta x.
This is the relative change of length.
And I should label this such.
That's going to be equal to delta x times 1 plus e minus delta x divided by delta x.
And that simply going to be e, which is equal to delta U over delta x.
And that is what we define as strain.
And that is a dimensionless quantity.
Because it has units of length over length.
All right, how can we patch this one-dimensional sort of behavior up to a three-dimensional situation?
What we are saying in this one-dimensional case is that the displacement of a particular point P varies linearly with position in the body.
So let's generalize that into three dimensions by saying that the component of displacement U1-- and this is going to look like an algebraic identity.
It's going to be the way in which U changes with x1 times the position x1.
So this is saying that the displacement will vary linearly with x1.
And the rate will be du1 dx1.
But also U1 is going to change with the coordinate x2.
And the coefficient there will be du1 dx2.
And there's going to be a third term that will be the way in which U1, the x1 component of displacement, changes with x3 times the position in the body x3.
So this is saying exactly the same thing that we did in one dimension, that displacement depends linearly with position, except that there's now three coordinates for position.
And the displacement will change with an increment of a change in position along each of those three axes.
Similarly, we can say that U2 is going to be equal to the way in which U2 changes with x1 times the position x1, the way in which U2 changes with x2 times the coordinate within the body x2, and the same for U2 dx3 and the actual positional coordinate x3.
And so, in general, we're going to propose that the i-th component of displacement is given by the way in which displacement changes with x sub j times x sub j.
Or you could do the same thing, not in terms of actual displacement, but differences in displacement, the same way we could define strain as the fractional change of length of our line segment on the elastic band or, alternatively, as the shift in position.
So we could define it also as the difference in distances or displacements let me call them.
And that really is a trivial change.
We'll say the change of a length along x1 delta U1 is going to be the way in which U1 changes with x1 times delta x1 plus the way in which U2 changes with x1 times the way in which U1 changes with x2 times delta x2 plus the way in which U1 changes with x3 times delta x3 or, in general, that the fractional change in length is going to be equal to dui dxj times delta xj.
All right, so what we're going to define now is dui.
Dxj will be defined as something like strain.
It's not exactly strain yet.
And this is going to be a measure of deformation.
I'm hedging my words because of something that we'll want to impose upon the proper strain tensor.
But first, let's see what these three-dimensional terms mean.
And to see that, let me look at a place within the body.
And I'm going to look at a line segment that goes between a P and Q that is oriented along x1.
And I'm picking that deliberately.
Because I want to keep things simple and isolate one of these terms so we can identify its meaning.
So let's say here we have two points P and Q separated by a distance delta x1 prior to deformation.
And now we squish the body.
And what happens is that P will move to a position P prime.
Q will move to a position Q prime.
And this will be the new line segment, P prime Q prime.
This will be the original delta x1.
To that we're going to tack on an instrument which is delta U1.
But then there will also be a delta U2.
And clearly what has happened is that the length of the line segment P prime Q prime has changed.
But also the orientation of the line segment has changed by an angle, which I'll define as phi.
So again, we have a line segment.
We deform the body.
P is displaced.
Q is displaced.
The component of that line segment along x1 has changed in length by an amount delta U1.
The coordinate along x2 will change by an amount delta U2.
So we not only change the length.
But we rotate the line segment.
We can express these in terms of our general relation that I proposed a moment ago.
Delta U1 is going to be equal to du1 dx1 times delta x1.
And that is going to be given by the element epsilon 1, 1 times delta x1.
The value of x2 was going to be equal to du2 dx1 times delta x1.
And that's going to be by definition e2, 1 times delta x1.
And the reason this is so simple is that I initially picked the line segment which was parallel to x1.
So not all of the four terms that would be present in the x1, x2 system have come in.
So there's no contribution of delta x2 because of the fact that I've picked this special orientation.
OK, the fractional change of length resolved on x1 is going to be, well, it's going to be exactly delta x plus delta U1 quantity squared, it's going to be this horizontal line segment, plus delta U2 quantity squared all to the power 1/2.
And now I'm going to say that delta U2 is negligible compared to this big delta x plus delta U1.
And I'll say that this is approximately equal to delta x plus delta U1, OK, just taking the whole works outside of the square root sign.
So delta U1 over delta x1 is going to be equal to the term e1, 1 from this expression here.
So the term e1, 1 represents the tensile strain along x1.
So we can see how these derivatives are going to enter into changes of length.
The P prime Q prime has also been rotated by phi.
And we can say exactly what that is that the tangent of phi is going to be given exactly by delta U2 divided by the original length of the line segment delta x1 plus the little increment of displacement along x1.
And this is-- I'll walk down here so I can see it.
This is delta U1.
This is delta U2 over times delta x1-- sorry, I can't see what I've got down here-- delta U1 plus delta x1.
OK, so it's the amount of displacement along x2 over the original line segment delta x1 plus the change in displacement delta U1.
And clearly delta U1 can be claimed to be small with respect to delta x1.
So this is approximately equal to delta U2 over delta x1.
And tangent of phi is going to be tangent of a very small angle.
So this will be delta U2 over delta x1.
And so this angle phi is, for small strains, going to be equal to delta U2 over delta x1.
And that is the definition of our element of strain e1, 2.
So e1, 2 corresponds to a rotation of a line segment that was originally parallel to x1 in the direction of x2--
[INAUDIBLE]
e2, 1, I'm sorry, e2, 1, yeah, along x1 in the direction of x2.
And that is counterintuitive as I have just demonstrated.
e2, 1 is a rotation of a line segment along x1 in the direction of x2, which is just the reverse of the subscript.
So eij, a general off-diagonal element of the array eij, is going to be a rotation of a line segment initially along x sub j in the direction of x sub i.
And for very small strains, numerically that term eij will give an angle in radiants
For the equation you have over there, should it be delta x plus delta x2?
No, I would say this--
Because aren't you saying delta U1?
Oh, OK, you're right.
Yeah, I didn't take the difference here
Delta U1 is basically negligible?
Yep, yep, so I'm saying that that [INAUDIBLE] should be delta x1--
Plus delta U2?
--plus delta U2.
You're right.
No, I'm throwing this out.
And that's right.
So I'm saying that this is essentially delta x1 plus delta U1.
I'm saying that this is negligible.
So strictly speaking the distance between P prime and Q prime is the square root of this squared plus this squared.
But if this thing is tiny, P prime Q prime is essentially going to be this distance, delta x1 plus delta U1.
So that's right.
OK, so we have something that looks like it measures deformation.
But I would like to ask if this is a suitable measure of deformation.
And I would like to show that, unless the tensor eij is symmetric, that we have included in our definition of strain rigid body rotation as well as true deformation.
So in order to demonstrate that, what I'm going to look at is a case where we have actually, by the way in which we apply a stress to the material, actually done nothing more than rotate x1 to x1 prime and rotate x2 to x2 prime.
And in general, for a real amount of deformation, that is something that is going to happen.
Let's say I decide to deform this eraser in shear before your very eyes.
And I try shearing it.
Nothing much has happened.
And I squeeze a little harder.
And finally, I've got it wrestled down into an orientation like this.
And when I finished, this was the original position of the eraser.
Here's x1.
Here's x2.
And by the time I've wrestled it around to a deformed state, it sits down like this, maybe deformed a little bit.
But would you say that this angle here is a measure of deformation?
No, that's clearly a rigid body rotation.
And what I would intuitively do, if I wanted to measure true deformation, if this were the original body and after deformation it went to a location with some deformation to be sure but this has been rotated to here.
This has been rotated down.
And I wouldn't want to say that this is a measure of deformation.
This would be e1, 2.
What I would want to do intuitively would be to position the body symmetrically between the axes and then say that this is a measure of shear strain.
So let me show you now in a more rigorous treatment that, if there is a component of rigid body rotation, let me show you what a rotation of the coordinate system to a new orientation x1, x prime would do.
So let's say that this is an original point Q1.
And after deformation, it moves by a rotation phi that's equal to e2, 1 to a new location Q1 prime.
Let's suppose that this is a position Q2.
And after what we think is the deformation, this moves to a location Q2 prime.
And this would be e1, 2.
That's equal to minus phi.
This is e2, 1.
And that's equal to plus phi.
So what we would do is like to get rid of that rigid body rotation.
And what I'll do is to show that, if we have a point that's out at the end of a radial vector, R, which has coordinates xi, x1, x2, x3, and if we have a displacement, which is absolutely perpendicular to that radial vector, that this would be characteristic of rigid body rotation.
And I will say that, if Ui is perpendicular to the radial vector xi, then U.R should be equal to 0 for every position xi.
So if we just multiply this out, this is saying that Ui Ri should be equal to 0 for rigid body rotation.
And let's simply carry out this expansion.
And I will have then eij xi xj forming this dot product.
And that is going to be 0 for a rigid body rotation.
And if I expand this, this will contain terms like e1, 1 times x1 x1 plus e2, 2 times x2 x2 plus e3, 3 times x3 squared.
And then they'll be cross-terms e1, 3 plus e3, 1 times x1 x3 plus e1, 2 plus e2, 1 times x1 x2 plus e2, 3 plus e3, 2 times x2, x3.
And my claim now is that, if this represents rigid body rotation, then that should be 0.
That is to say the displacement is absolutely perpendicular to the radius vector for small displacements.
This must be 0 for all xi.
And the only way that's going to be possible for any value of x1, x2, x3 is that each of these six terms vanish individual.
It's the only way I'm going to be able to get the whole thing to disappear for any value of coordinate x1, x2, x3.
So we're going to have to then have e1, 1 equals e2, 2 equals e3, 3 equals 0.
All the diagonal terms are going to have to be 0.
In order to get the fourth term to vanish, I'm going to have to have e1, 3 equal to the negative of e3, 1 and e1, 2 the negative of e2, 1 and e2, 3 the negative of e3, 2.
So what is this going to look like for the part of the e tensor that corresponds to rigid body rotation?
The diagonal terms are all going to be 0.
And the off-diagonal terms are going to be the negative of one another.
So this is e1, 2. e2, 1 would have to be equal to minus e1, 2. e3, 1 would have to be equal to the negative of e1, 3.
And so we will have something like this.
So this suggests that our definition of the true deformation, which will be a tensor epsilon ij-- I bet you kind of guessed I was going to call it epsilon and not E. I can write this as a sum of two parts.
I'll turn that around and say I'll write epsilon ij plus another three by three array omega ij is equal to the tensor eij.
That's a novel idea.
This is addition of two tensors element by element.
And if I do that and define epsilon ij equal to 1/2 of eij plus eji, so for the diagonal elements that is going to take 1/2 of e1, 1 plus e1, 1.
And that's just going to give me e 1, 1 back again.
And I'm going to define the terms omega ij as 1/2 of eij minus eji.
And if I do that, the resulting tensor will be symmetric.
OK, so from tensor eij we can split it up into two parts and a part omega ij.
And I won't bother to write it out.
But you can see that omega ij plus eij is going to be equal to eij minus 1/2 of 2eij minus 1/2 of 0.
And that's just going to be eij.
So this plus this is indeed going to give me the tensor eij.
So given a tensor eij, which is not symmetric, I will define eij as the sum of two parts, a tensor epsilon ij, which is true strain, and a part omega ij, which is rigid body rotation.
And now at last we have a satisfactory measure of true deformation in terms of the displacement of points within a deformed body where that displacement can arise from either rigid body rotation and or deformation.
So finally, we have a tensor epsilon ij, which is the strain tensor.
And now I can finally make my claim that, if this really is true deformation, that for cubic crystals-- you can see the same [INAUDIBLE] window coming again-- since second ranked tensors have to be symmetric for cubic crystals, the form of strain for a cubic crystal can only be this.
It has to be diagonal if the tensor is to conform to cubic symmetry.
And you say, ah, ah, you tried to pull that on us with stress, too.
And I can squish a crystal anyway I want.
But you can only develop a strain if you deform a crystal.
And the symmetry of the crystal is going to determine how the crystal deforms.
So therefore, a cubic crystal should be able to deform only in a way that stays invariant to the transformations of a cubic symmetry, right?
So cubic crystals can only deform isotropically.
So if I wanted to design springs for an automobile so that elastic energy could be stored most efficiently in a solid, I would want to make these automobile springs out of a triclinic metal.
Because then the tensor could be one of general deformation.
OK, well, this, obviously, is a swindle like my assertion that stress could only be an isotropic compressional stress for a cubic crystal.
But the argument is a little more convoluted.
And strain, indeed, is also a field tensor.
Because, even though I can only create the deformation by application of a stress or perhaps an electric field or something of that sort in that the link between the deformation and what I do to the crystal is coupled by a property of the crystal, which has to conform to the symmetry of the crystal, nevertheless, given a tensor that describes deformation, I can always, independent of the symmetry of the crystal, devise a particular set of stresses that would produce any state of strain I want.
I would have to design, though, a particular state of stress to produce a desired state of strain.
And that coupling has to conform to the symmetry of the crystal.
But there's no reason why strain itself has to conform to the symmetry of the crystal.
I can create any state of strain I like by choosing the appropriate stress.
OK, so that is a swindle.
But everything now that I can say about the behavior of a second ranked tensor applies to the strain tensor.
I can take a strain tensor epsilon ij.
And I can perform the surface epsilon ij xi xj equals 1.
And that will be the strain quadric.
The strain quadric will have the property like any quadric constructed from a tensor that the value of the radius in a particular direction is going to be such that the strain in that direction is equal to 1 over the radius squared.
So what does that mean?
Well, we have to look at what the strain tensor is relating.
The direction, remember now that the definition of the strain tensor is that U sub i equals epsilon ij times x sub j.
So the quote "applied vector" is the direction in a solid.
And we will get, in general, a displacement, U, which is not parallel to the direction that we're considering.
And the epsilon in a particular direction is going to be equal to the part of U that's parallel to the direction of interest divided by distance in that direction.
And what this is going to give us is the tensile component of deformation in the direction that we're examining.
Now, the radius normal property works in this case.
And what the radius normal property says, if you look at a certain direction in the solid and then look at the normal to the surface at that particular point, this will be the direction of what happened.
So I've not drawn this correctly.
This should be the direction of U.
And this is the radio vector.
So the normal to the surface gives us the direction of the displacement that's going to occur.
And the reason I can say that is by definition that that property holds only for a symmetric tensor.
And by definition, we have defined the strained tensor such that it is symmetric.
So the radius normal property holds.
If we want to change from one coordinate system to another, we do that by the all for transformation of a second ranked tensor that, if we change coordinate system, the elements of strain change to new values epsilon ij prime that are given by cil cjm times epsilon lm, the same law for transformation of a second ranked tensor that we have seen earlier.
Finally, we can, because there is a quadric, we can change that by finding eigenvectors and eigenvalues, take a general form of the tensor epsilon 1, 1, epsilon 1, 2, epsilon 1, 3, and so on, and by solving the eigenvalue problem, convert this to a value epsilon 1, 1 prime 0, 0, 0, epsilon 2, 2 prime 0, 0, 0, epsilon 3, 3 prime to find out what the extreme values of tensile deformation are and the orientations within the original description of the body in which these extreme values occur.
OK, I'm just about out of time.
I'll save until next time another neat feature of the strain tensor.
And that is contained within these elements is information on the volume change that the body undergoes.
And this is something called the dilation.
And we'll see that this is related in a very simple way to the elements of the strain tensor.
OK, we've spent a lot of time developing the formalism of stress and strain.
And next thing we'll do is to move on to some interesting properties of crystals, which are defined in terms of tensors of rank higher than two.
So we'll look at some third ranked tensor properties.
That will include things like piezoelectricity, the stress optical effect, and things of that sort.
These are going to be really anisotropic.
Second ranked tensors were anisotropic enough.
But the variation of property with direction was a quasi-ellipsoidal variation When as 1 over the square of the radius of the quadric.
For higher ranked tensor properties, we will encounter some absolutely weird surfaces with dimples and lumps and lobes, not this smooth quasi-ellipsoidal variation, a property with direction, which was the case for second ranked tensor properties.
So we will get into some exotic anisotropies very shortly to finish up the semester.
So I will probably see some of you at the MRS meeting next week.
I won't see you on Thursday for reasons that I need not elaborate upon.
So I hope you have a happy holiday.
Relax, suck in air, and come back refreshed next Tuesday for some really wild anisotropy of physical properties.
I wanted to make a few comments on the third problem set, which was a funny puzzle.
And it was intended to be entertaining and get you thinking about things related to symmetry in an amusing context.
And there were two problems.
The first problem had a platter on which there were arranged cherries with either two or three berries to a spray, and apples and pears.
And they were either black or white, which didn't make then particularly appetizing.
But in any case, there were three platters and the fruits marched around on the platter.
And the puzzle was, what would be the fourth platter in the sequence?
This didn't involve symmetry in any sense of the word.
But yet, what was going on from picture to picture was a mapping of the motifs from one location to another.
Not just rotation or translation or anything nice in the crystallographic, but still they were being moved from one location to another.
And just about everybody got the right answer just by deduction.
But the reason I liked that problem, and the reason I gave it to you, is that it introduced another sort of transformation.
One of the pairs of fruits not only change locations, but they switched from black to white as you went from one position to the other.
And that is another sort of transformation we can introduce.
And this is something we could call color symmetry.
And I would observe that this is something that you are already very familiar with in patterns, in particular, in the pattern that is represented by a checkerboard where you have black squares and white squares.
This location here is a bona fide four-fold axis, as is this location here.
But there other locations, such as the corners of the black and white squares, where, indeed, there is a 90 degree rotation that takes this square and transforms it into this square.
But in doing that transformation, you change it from a black one to a white one, and then from a white one to a black one again, back to a white one.
And that is a color symmetry.
You could make it red and green if you like to be a little more attractive.
Similarly, there are mirror lines like this that are true mirror lines.
The pattern is left invariant.
There's another locus such as this one here where when you reflect you go from a black one to white one and then back to a black one when you do the operation twice.
So there are black-white mirror planes as well as regular mirror planes.
There are two-fold axes but all of the two-fold axes are black-white axes, white to black, black back to white.
So that's a new sort of transformation that we can develop if we wanted to, switching the color.
Does this is have any utility in the physical sciences?
Be nice for wallpaper designers, but does it have utility in physical sciences?
And the answer is yes.
Because what is going on here is that there is a binary character assigned to each of the motifs.
So any time the atom or whatever you have, the molecule, has some sort of binary characteristic that it can exist in two distinct states, the arrangement of that sort of motif requires a black-white symmetry.
And one example in crystals that requires this sort of symmetry is the case of a magnetic structure-- and I think I mentioned this last time-- where the magnetic moment can either point up or down.
Well, saying that you have an atom with spin up and an atom with spin down is the same thing as saying, in terms of transformations, having a black atom or a white atom.
It's a binary character that switches from atom to atom.
Could you have three color symmetries?
Yeah.
You can.
There is a Dutch artist named Maurits Escher who spent much of his, career deducing patterns, and did this in a purely intuitive way.
I mean you look at these patterns and say, wow, that guy is some crystallographer.
But I heard him, while he was still alive, speak twice.
And he describes it in metaphysical terms.
This is the twoness merging with the fourness, and just completely intuitive.
But his patterns are gorgeous.
And he has a number of very nice ones that are three-color symmetries.
And maybe later on in the term, I'll bring in a projector and show you examples of some of these.
They're very entertaining.
All right.
So the purpose of that first puzzle was to introduce you to black-white symmetries.
Then the second puzzle on the sheet was a pretty dumb thing.
There was a stack of blocks and you said how many-- I asked how many blocks are there in the stack?
And you could count them up, 1, 2, 3, 4.
And If you did it right, you came out with 32.
So that was pretty dumb.
Well, this was intended as an example of how one can use symmetry in the solution of a physical problem.
Let's suppose that you were a cowboy and you had to go out and buy horseshoes for a herd of horses.
Would you count the number of feet?
No.
You'd count the number of horses and multiply by four.
OK.
So there you'd be using symmetry.
Now The point of this problem really was-- You could have had this one.
We have just the exact number seats.
The point of this problem was that you have to, in assigning a symmetry to a system, usually make physical assumptions.
When you do a derivation, as we've been doing for the last several days, we say this is a four-fold access, and there's no ambiguity whatsoever.
Because it's my ball game, and I want it to be a four-fold axis, But when you're given a pattern or given a crystal and you have to assign a symmetry to that crystal, you invariably have to make assumptions.
The person buying the horseshoes, for example, has to assume that five- and six-legged horses are pretty rare articles and that a horse with only three or two legs is going to be of little utility to the rancher.
It Would probably have a limited existence at the ranchers expense.
So you're assuming that all horse have four legs.
In looking at a crystal, though, things are not that easy to decide.
And let me give you a few really practical examples.
A lot of mineral crystals grow in fissures in the rock.
And let's suppose we've got a crack, and bubbling through this crack is a solution that contains silica.
And as the solution cools off, it will deposit a crystal of quartz.
And this crystal of quartz would probably grow looking something like this.
And you would knock the crystal out of the rock.
And you would say, what is the symmetry of this crystal?
It doesn't appear to be anything hexagonal about it.
But what's happened is this crystal if it had developed in a uniform environment would have had a hexagonal shape.
But because the nutrients are flowing in from one direction, it tends to elongate in that direction.
So how could you assign the symmetry to that crystal that you dug out of the rock?
Unfortunately, crystals do not come with a legend on the bottom that says quartz SIO2 space group P3 subscript 121, made in USA.
You have to yourself assign the symmetry.
And what you would do is you would look at this face and this face and you say, gee, these angles, when I measure them, all turn out to be 120 degrees.
This face has a luster on it that looks like the luster on this face.
This face has little edge pits on it, and they point in the same way as the edge pits on this face.
I can measure the electrical conductivity and find that it varies in a fashion consistent with a hexagonal crystal.
And you would do any physical test that you decided you would need to make.
And then when you were all done, you would say everything that I can measure about this crystal other than it's shape is invariant to a 120 degree rotation.
Quartz has a hexagonal shape very often, but actually only this face, this face, and this face are symmetrically equivalent.
And there's only a three-fold axis in there.
And if you look very carefully at this crystal, you could see some difference there, too.
The neighboring faces very often have striations, very faint striations on one face, no striations on the neighboring face.
So you find out all that can.
You have to make some assumptions, and then on that basis decide the symmetry.
Let me give you another example.
This is a real one.
And one of the things that it's fun to do is to grow crystals from solution.
The Science Museum has little kits that you can buy that have some powder that you could heat up in water and then put it in a beaker and put a string in there, maybe with a little crumb of the stuff that you dissolved.
And one of the crystals that is nice for this purpose is alum.
It's a cubic crystal.
And if you let the stuff set there, it will form a nice, lovely equiaxed cubed.
Science Museum usually puts a pinch of another salt in there so the crystal is not only pretty and shiny, it's purple or it's the orange.
Now, unless you attach this little seed crystal carefully, there would be a good chance that it would fall off and land on the bottom of the beaker.
When that happened, your crystal would not have the shape of a cube.
It would have a shape in which these two top edges had length L and these faces on the side were exactly half that length.
And what's happening is that this crystal is growing in an unconstrained environment.
The nutrient is coming on and crystallizing from all directions.
In this case, nutrient can't get in from the directions that are below the crystal.
So the growth in that direction is eliminated, and this height is exactly half that at the edge of the crystal.
So that's sort of a very specialized case of a crystal growing in a constrained environment.
I'm digressing now, but let me make a few more remarks.
The shape that a crystal has-- and the shape is in crystallography designated by a special term.
It's called habit.
Crystals that undergo plastic deformation never die, they just develop bad habits.
Oh, come on.
This is a tough crowd.
I thought that was pretty funny.
Anyway.
The habit that a crystal has can depend very much on impurities in the solution or melt from which it grows.
And let me suppose I start with a little crystal that looks like this.
And there are two different kinds of faces on there.
And let's suppose one of them grows rapidly and one of them grows slowly.
Which face do you think would be the one that eventually forms the boundaries to the crystal volume?
The fast-growing face or the slow-growing face?
Fast-growing face
Fast, slow are both possibilities.
Let's do a time lapse experiment.
Let's suppose that this is the fast-growing face.
And, after a certain amount of time these faces would have all advanced to this location.
And this is the slow-growing face.
And after the same amount of time, it will have advanced to this location.
After the next time increment, this one would have grown up to here.
This would only have grown that far.
And you could see the only thing that's gonna be left is the slow-growing face.
So it's sort of counter intuitive.
The faces that survive to bound the crystal as rational planes are the slow-growing faces, not the rapidly growing faces.
This is of great use to the people who want to control crystal shape.
And this is very often in very every day, ordinary contexts.
The people who grow table salt, for example, want to grow a crystal that has nice pointy edges so when it comes down on your tomato, it doesn't bounce off.
It sticks to it.
So the shape of table salt makes a difference in how it behaves.
And the way you change the shape is to find something that will be absorbed preferentially on one set of faces that you would like to have be the faces that appear on the crystal.
And that will do the job for you.
I'll close with just one final example of this in real, every day industry or technology.
I once had a phone call from a company that made salt for sprinkling on roads in the winter time, particularly around New England when the roads tend to ice up.
And in an earlier era it, was politically correct to use sodium chloride.
But you wanted the sodium chloride, if you could, to have an acicular, a needle-like, shape to improve its flow properties so you didn't waste a lot when you shook it out of the back of the truck.
And the material that they use for that-- you think rock salt in the environment is bad-- they used a cyanide compound to modify the growth habit and make it needle-like.
And they wanted to get away from that before they were really dragged into some sort of trial by the EPA.
And did I know of something that could change the shape of the salt crystals for them?
And the answer to that question was quite straightforward.
No.
That was the end of the conversation.
But it was an interesting example of how surface chemistry can change crystal shape.
OK.
So all that long-winded explanation was a justification of the meaning of those two final two puzzles.
And the final thing I have to say about them is that these puzzles were lifted from a puzzle book published by MENSA.
Have any of you heard of MENSA?
MENSA is an association of self-declared geniuses.
So this was a puzzle book for geniuses.
And you all did horribly well on that third problem set.
Everybody at MIT is bright.
But you folks are not only bright, you are geniuses because you cracked the MENSA puzzles.
OK.
Enough enjoyment and fun and games.
Let's get serious, back down to what we were doing earlier.
We had established in earlier discussion that there are a very limited number of ways in which we can arrange a mirror plane and a rotation axis about one fixed point in space.
And these, accordingly, are called the point groups.
And in particular, they are the two-dimensional crystallographic point groups.
We had rotation axes by themselves, 1, 2, 3, 4, and 6.
And these are the only ones we need worry about because these are the only rotational symmetries that are compatible with translation.
Then we had a mirror plane.
And then there was no reason why we could not combine a mirror plane with these rotational symmetries.
And in doing so, we used a theorem that said if you take two reflection operations, sigma 1 and sigma 2-- and again, I use sigma to represent an individual operation of reflection-- and a mirror plane is the locus of operations of reflection, and then reflecting back again, there are two operations involved.
And if these are combined at some angle mu, sigma 1 followed by sigma 2, when they are separated by an angle mu is the same as the net transformation of a rotation about their point of intersection through twice the angle which the mirror planes are combined.
And that led us-- in addition to a one-fold axis, a two-fold access, a three-fold axis, a four-fold axis, and a six-fold axis-- let us combine reflection with these axes to get symmetry combinations that we named according to the symmetry elements which were present in the final combination.
We used two mirror planes if the mirror planes that arose where independent.
Independent in the sense that the two-fold axis never turned one into another.
And independent in the sense that if we draw a representative motif, the way those motifs were disposed about one mirror plane was different from the way they were arranged about the other one.
For a four-fold axis, we could add mirror planes.
And they would have to be at half the angle of the four-fold axis.
And as we saw, it was kind of a mouthful to call this 4MMMMMMMM.
There were just two kinds of mirror planes, so this one was called 4MM.
6MM, analogously, was a six-fold axis with mirror planes separated by 30 degrees, half the throw of the six-fold axis.
And the one that I left out is 3, not m m but 3M because a three-fold axis has mirror planes separated by 60 degrees.
And the same thing goes on at each of these mirror planes, pair of objects hanging at one end of the mirror plane, the other end is unadorned.
So they all behave the same.
There's only one kind of mirror plane.
The three-fold axis maps one mirror plane into another, and so they have to be the same.
So therefore, only one M appears in the symbol.
Independent of this, we had, when we combined reflection or rotation with a lattice, found that there were a very limited number of lattices in two dimensions, the oblique lattice which had T1 not equal to T2 and T1 inclined to T2 by some general angle.
The next degree of specialization was for a two-fold axis-- so either a one-fold or a two-fold axis could fit in here.
For a three-fold axis or a six-fold axis, we had to have a lattice in which the two translations were identical to one another, identical in magnitude and identical in the way the atoms were ranged relative to these translations.
And the angle between T1 over T2 had to be identically 120 degrees.
And by convention, we take the larger of two angles, 120 rather than 60, to define the inter axial angle.
A four-fold axis required a net that was exactly square, not approximately square but exactly square.
So T1 and T2 had to be identical in magnitude.
The angle between them had to be exactly 90 degrees.
And close is no cigar.
It has to be exactly 90 degrees because that angle is generated by symmetry.
And then at the place in which I'm gonna pick up today in just a moment is the case of a mirror plane and reflection, we saw depending on how we arranged the translations relative to the reflection plane, could give us either a lattice in the shape of a rectangle-- and here T1 and T2 made an angle of exactly 90 degrees.
But the magnitude of T1 could be anything it liked relative to the magnitude of T2, and then a double cell, which had the shape of a diamond.
But it was operationally much to our advantage to pick a double cell which had a right angle in it.
And the reason for that was that it is a nightmare to do calculations of inter-atomic distances and angles or angles between faces in an oblique system.
The right angle makes these calculations very simple.
And the fact that this cell tells you twice as much of the area as you really need to know about is a small price to pay.
So this is the rectangular net, as we'll call it in words.
And this is one that we'll call the centered rectangular net.
OK.
So let me pause here and suck in air and see if there are any questions.
This is where we left off.
And now having everything spread out on the board, we're going to start to make combinations and continue that process.
Yes, sir
So the centered rectangular has the same exact [?
intercept ?]
as the rectangular but the only thing you're doing is you're deriving it based on symmetry that you're using like a diamond kind of thing to get the center one?
Yeah.
So all these arose when we did it slowly and systematically.
We said let's let this be the location of the mirror plane.
What happens if we combine it with a translation?
We can take one lattice point on the mirror plane since there's no unique lattice point.
And then the mirror plane is gonna reflect this over to here.
So here's a first translation.
I've got a second non-colinear translation.
Wham-o.
Instant lattice.
So what I'm doing is taking this diamond shape and I am taking one translation here-- let's put some labels on this.
Let's call this T1 and this T2.
So this translation here, call it T1 prime, is my original translation T1 plus T2.
In this translation, T2 is the negative of the-- Yeah.
It's gonna be minus T1 plus T2.
So I've taken linear combinations of the two vectors of the diamond-shaped cell.
And again, this is contrary to the rules we have that say pick the shortest two translations.
Why?
The final larger area of volume then you need.
And the answer to that is occasionally the provision of a coordinate system, which is far easier to work in, is a small price to pay for that redundancy.
OK. And then the primitive rectangular net, just to remind you again, we said is there any case in which this is not true?
And that case was if you take T1 exactly perpendicular to the mirror plane and then you define only a lattice row.
So that gives you a second choice for T2.
You can't pick it anywhere you like, otherwise it gets reflected across just as in the first case, and you have two translations that are not compatible.
And the only way that they are compatible is that if you make the translation T1 straddle the mirror plane.
And that's what we've already got up here.
So that's how we got the rectangular net.
We could get a second distinct sort of lattice only if this translation was exactly in the plane of the mirror line.
Yes, sir
But the number and kind of symmetry elements are the same for both, right?
Exactly
Uh, so--
So there's a curious sort of duality here.
Here in the case of a hexagonal net, this is just a lattice of a very specialized shape.
And does that have a six-fold axis in it?
No, not unless I decide to put one in.
I could put in a three-fold axis, alternatively.
So here, I have one lattice that can accept two different kinds of symmetry.
Here I've got the reverse situation.
I have two distinct kind of lattices, both of which are happy and content with the same symmetry on them.
So there's going to be a far greater number of combinations of lattice with symmetry than simply a one-to-one correspondence between lattice types and point group types.
OK. Any other questions?
All right so let me now shift down to low gear and remind you of one thing that we did last time.
We asked what happens if I can combine a rotation operation, a alpha, with a translation?
Call this T1.
Sounds like something I already did in showing that the values of alpha are restricted to values.
But what I'm going to do now is use something that looks as though it's a similar starting point to arrive at a different result that we saw last time.
I'm going to deliberately look at a line that is alpha over 2 on one side of the perpendicular to T1.
And then the operation of the a alpha is going to take this translation with a lattice point of necessity at the end of it, and it's going to move it over to here, alpha over 2, on the other side of the translation of the perpendicular.
Then I'm going to pick this up with the translation T1 and move it and everything along it to a location like this.
And my question now is, what is a alpha followed by the translation T1?
The handedness of the objects have not changed.
So this has to be either translation or another rotation.
And what we can very quickly show is if upon dropping a perpendicular down to the original translation, these lines are parallel.
This line cuts across it.
So this angle is alpha over 2.
And this line is inclined to the perpendicular by alpha over 2.
So this angle in here is also alpha over 2.
So the answer is that if I rotate from one location to another by rotation alpha and then pick up that motif or the entire space and translate it by the original translation T1 to get a third one-- this is number 1, right-handed, say, number 2, right-handed.
This is number 3.
Stays right-handed.
Has to be related by a rotation.
And we next ask, what point is the point about which the rotation occurs?
Then we used-- what at the time seemed like a trivial observation-- that a rotation axis in space or a rotation point in two dimensions is the locus that is left unmoved by the rotation.
So if rotating from here and translating over to here is equivalent to a rotation and if the locus of the rotation has to be the point that's left on moved, bingo.
That's where the rotation occurs.
And it's through the same angle, so we can label this an operation b alpha.
And we know exactly where that point is gonna be.
It's gonna be along the perpendicular bisector of the original translation.
And it's gonna be up a distance x which is T over 2 times the cotangent of alpha over 2.
So now we've got another, what I call, combination theorems.
And again, this is nothing more than knowing how to complete the product of two operations in establishing the group multiplication table.
so let me write it down in the form of a combination theorem.
This says that a alpha followed by a translation is another rotation-- in the same sense, both counter clockwise in this example-- another rotation b alpha in the same sense about a point that is always T over 2 times the cotangent of alpha over 2 along the perpendicular bisector.
So that's how you would go about making an entry in the group multiplication table.
Sorry.
I put you off
Yeah.
Can you just, like, I just want to see what you did actually to the motif in that drawing
OK. Rather than taking an abstraction, an arbitrary line that is at alpha over 2, I put a motif in the space that doesn't necessarily hang on this particular locus that I drew for reference.
And the rotation through alpha would move it to here.
It's lagging behind this first line.
It lagged behind the second one by the same amount.
Then I pick it up and I move it by the same translation and that slides it over here, still canted over to the left.
And now what I claim is that I get from the first one to the third one-- and actually should have raised your hand on a different matter.
This is out in front of the translation.
So number 3 should sit here, also to the right to the translation.
And the way I get from 1 to 3 in one shot is by rotating alpha b. OK. Now, a cautionary note just so you do not use your new power recklessly.
Let me emphasize that this is an escalation in operations, not in symmetry elements.
And the reason for that is that a rotation axis, in general, contains a number of different rotations.
An n-fold axis has n different rotation operations implied in it, including the identity operation of rotating 360 degrees.
So if I say I'm gonna drop a four-fold access in a lattice, what I'm doing is dropping into the lattice a 90 degree rotation, a 180 degree rotation, and a 270 degree rotation.
And the location of the new operations depends on alpha.
So these new operations are not gonna pop up at the same location.
They're gonna be sprinkled over different locations.
So as an abstraction, that may be hard to appreciate.
But we're going to straight away derive a couple of more additions.
And then when we see what happens upon adding a rotation axis to a lattice in a couple of nontrivial cases, I'll just summarize the results and assume that you'll be able to derive them on your own.
The last time we had got around to doing just one of these combinations.
And we ask what happens when you combine with a translation a rotation operation a pi.
So we'll take some motif that sets up here, right-handed, rotate it by 180 degrees to get a second one, and then pick that up and translate it by T. So I'll get a third one sitting down like this of the same handedness.
And the question now is how do I get from 1 to number 3 directly?
Well, let's use our theorem.
Our theorem says we should go a distance x along the perpendicular bisector, which is half the magnitude of T times the cotangent of pi over 2.
The cotangent of pi over 2 is zero.
So we go up a distance x.
That's equal to 0, all in the perpendicular bisector, which means we stay put.
And-- in this class, there is always truth in advertising.
Around this locus here is a rotation operation b pi, which takes the first one into the third one.
OK. And now, lickety split, I will derive again-- and I apologize for going fast because I did do it last time.
Let's ask what happens when we combine a two-fold axis with a parallelogram net?
So we're taking a two-fold axis plus a parallelogram net.
In this two dimensional lattice there are two translations, T1 and T2, of arbitrary magnitude with some angle between them that's arbitrary.
And to this lattice point, I'll add a two-fold axis.
A two-fold axis involves the presence of two operations, the identity operation, and that we can forget about, and then there's the operation a pi.
By combine a pi with T1, I'll get an operation b pi that sits at the midpoint of T1.
If I combine a pi with T2, we get an operation c pi at the midpoint of T2.
If I combine a pi with T1 plus T2, I'll get another operation d pi at the center of the cell.
And we don't have to ask what happens when we go three translations over.
There's gonna be another operation of 180 degree rotation on that translation.
But we really only need bother about what goes on on the edges and in the interior of one unit cell, because whatever's there has to be repeated by translation.
So, in making these additions, we need consider only independent translations-- that is, not related to one another by the rotational symmetry-- independent translations that terminate within the unit cell.
And that means for the addition of a two-fold axis, I need do only what I have, in fact, just done.
I want to consider a pi with T1, T2, and T1 plus T2.
If I have the operation a pi or b pi or c pi or d pi at these four locations, that is all I need have present to say that I have, in addition to the two-fold axis that I added to the lattice point, a two-fold axis halfway along T1, a two-fold access halfway along T2, and another one smack in the middle of the cell.
These are lattice points.
So those axes have to be repeated by translations, this one from here to here, this one from here, this one from here to here.
So I'll have two-fold axes in the middle of all of the edges of the cell plus this fourth one right in the middle.
So this is an example of a two dimensional space group.
It's a symmetry that acts on all of space.
And the names that we will give to these combinations is a symbol for the symmetry element that we added, in this case a two-fold axis.
And then we'll specify the nature of the lattice to which we've added it.
And all that I have to say is that the lattice is primitive because you're all aware at this point that a two-fold axis requires only a primitive, a parallelogram net.
The only place we'll need a special symbol is when we add a mirror plane to either the rectangular net or the centered rectangular net.
And then we'll have to specify which type of lattice of that shape, primitive or centered, we're dealing with.
Yes, sir
So all you did with that derivation there is prove that you don't have to worry about any rotations outside of that?
That's correct
That's correct?
Yep.
Which is very fortunate 'cause there's an infinite number lurking outside the boundaries of the cell.
I'll show you another general truth.
What does a pattern look like?
The pattern looks exactly like the pattern of a two-fold axis with that pattern of motifs hung at every lattice point.
These new two-fold axes don't do any further repetition of the motifs.
They don't do any repetition of the motif.
They just express relation between things that you get when you take this pair and then repeat it by the translations.
So this two-fold access, for example, relates this to this.
This two-fold axis relates this one to this one, and this one to this one, and so on.
They're just relations that exist between the things that you get when you add the pattern of a two-fold axis to a lattice point.
So another generalization of what we're doing here that turns out to be valid, the pattern of motifs for a particular plane group is nothing more than the pattern of the symmetry element-- symmetry element or symmetry elements, sometimes we'll be adding more than one-- that we've added to a lattice point.
So in other words, if we take 2MM and drop that into a rectangular lattice, the pattern is simply gonna be these four atoms related by 2MM hung at every corner of the lattice.
If you add it to a centered lattice, these would be hung at every corner of the lattice, and also at the centered lattice point.
So it's that simple.
If you look at some of these high symmetry space groups or plane groups, wow, there's symmetry all over the place.
You say, how can I draw a pattern for that?
Easy.
Just do what the point group does and drop it in at the lattice points.
So that turns out to be a universal feature of all space groups and plane groups.
OK. Somebody had a question over here and I cut you off.
Yeah
Yeah.
On that two [INAUDIBLE] diagram, wouldn't you want to use four of those because you said it was-- you don't concern ones that are-- you're only concerned independent ones, the ones you can't get by--
You're absolutely right.
And let's look at this pattern again.
Here are a pair that are hanging close to this two-fold axis.
Here's a pair hanging close to the two-fold axis, very different arrangement, different arrangement relative to this two-fold axis, different arrangements relative to this two-fold axis.
So what we're saying is-- and this is the way, in fact, we derive them-- that these four two-fold axes are all distinct.
But the ones that are along the edges are just related by translation.
So why do I have to put them in?
You just put them in to just show the symmetry?
Yeah.
Exactly.
Exactly.
'Cause if I say this is a lattice and I have two-fold axes here, that's an incomplete representation because if these are translations, I jolly well better have the same thing at all the lattice points.
I'm just emphasizing the property of a lattice which is understood to be present in all of these combinations.
OK. Other questions?
And I would remind you that a reasonable question or request would be, could you do that all again, please, at half speed?
And I'd be happy to comply.
But really this is ground that we covered last time.
And I'm just reinforcing it.
So now that we've run out of time, we can go on to something new.
Let me do, in the couple minutes remaining, let me make one more addition just to show you how things change when you add a rotation axis that has more than one rotation operation.
Let me ask what happens when you add a four-fold axis to a lattice.
We've seen that a four-fold axis can coexist with no lattice other than 1, which is dimensionally square and which has the same thing hanging about, two translations that are exactly 90 degrees apart.
So what we're doing is taking a four-fold axis and drop it in at the corner lattice point and then step back before all hell breaks looks.
Now, first thing we're gonna do is to get the four-fold axis at every corner lattice point.
And then we're gonna use our theorem, taking into account that when I say I'm adding a four-fold axis, I am adding the operations a pi over 2, a pi, a3 pi over 2.
But let me call that, instead, a minus pi over 2, the same thing and a little easier to deal with.
And then the operation a2 pi, which is the same as the identity operation.
And that's dull and uninteresting, so we won't consider that addition.
We've already done this, a pi, in deriving P2.
And that's the nice thing about this derivation.
It's gonna snowball because as the symmetry gets higher, we've already done the work for the subgroup that contains the elements that are present in the higher symmetry.
So we'll get an operation a pi over a pi here.
We'll get an operation a pi here.
And if we take the diagonal translation, T1 plus T2, I'll get the operation a pi here.
Or I should probably call them a pi, b pi, and c pi to indicate that these are different two-fold axes.
OK. Let me emphasize again that this theorem that we have is a theorem in individual operations and not symmetry elements.
If I say a four-fold axis is present, it means that all of these operations exist about that locus.
I cannot say that a four-fold axis sits here.
I cannot say a four-fold axis sits here.
I'm gonna have to take each of these combinations in turn and then step back and see what operations exist about the various loci within the cell.
So the one that's really different is the combination of a translation T with the operation a pi over 2.
And my theorem says if I put in a motif and rotate it by 90 degrees to here-- so this is number 1 and this is number 2-- and then translate it to a location that's related to the first by symmetry-- by translation to get number 3-- they will be related by an operation b pi over 2 that sits up along the perpendicular bisector by an amount x, which is T over 2 times the cotangent of alpha over 2, T over 2 times the cotangent of 1/2 of pi over 2, 45 degrees.
The cotangent of 45 degrees is unity because the two edges of the triangle are equal in length.
So this says I should go up along the perpendicular bisector by an amount that is half the length of the cell.
And that puts me right in the middle of the cell.
So the operation b pi over 2 sits not at the edge of the cell, but at the center of the cell.
And that, I think you will agree, even in this hastily sketched diagram, is the way number 1 is related directly to number 3 in one shot.
Now I could do it on the cheap from here.
But let's show that we have first an operation a pi over 2.
We've already got the operation c pi.
Let's now combine the operation a minus pi over 2.
And that says that for c minus pi over 2, I would go up the perpendicular bisector a distance T over 2 times the cotangent of minus 1/2 of pi over 2.
And the cotangent of a negative angle is minus that.
So I go a distance minus T over 2 along the perpendicular bisector.
And what that is gonna do is to bring me to the center of the cell that's directly below.
And this will be the rotation c minus pi over 2.
But everything has to be translation equivalent at the interval T. So I can just move this operation c minus pi over 2 up to the center of my original cell.
So now I have at the center of the cell all the operations of a four-fold axis.
So a four-fold axis exists in the middle of the cell.
And let me write the results in here.
I've now got all of the operations of a four-fold axis sitting here.
I've got all the operations of a two-fold axis sitting here and here.
So this is the final result, if I translate the two-fold axes over to the other edges of the cell
What is that angle [?
by the d project?
?]
That should be 90 degrees?
That should be 90 degrees, if I drew it carefully.
Yeah.
So let me give the final result with a pattern in it.
These are the two translations, equal in length, four-fold here, four-fold here, four-fold here.
They are all the same four-fold axis.
Four-fold axis in the middle of the cell.
Two-fold axis in the middle of the cell edges.
And the pattern of objects is the set of things that are rotated by 90 degrees that form a square about the corner of the cell.
Another one here.
Another one that sits up here.
And another one that sits like this.
Four of them on the corners of the square.
And that's all we're going to get in this pattern.
Again, it's just a pattern of a four-fold axis that is hung at every lattice point of the square net.
And these other symmetry elements that arise simply are things that relate to the squares that are hanging at every corner of the square cell.
OK.
So the square in the center of the cell, for example, relates these four.
And you can pick any other four and they'd be related as well.
The two-fold axis relates this to this.
This two-fold axis relates this to this and this to this.
So the pattern is just the square produced by four stamped out at every corner of the square net.
And it's that simple.
The name of this thing-- what have we done?
We've taken a four-fold axis, we've put it in a primitive net.
I don't have to tell you it's squared because in your heart of hearts know that that is what a four-fold axis requires.
So this describes completely the combinations that's been made.
And this is a representative pattern.
A pattern of this sort is one that we've seen several times already in describing sample symmetries.
That is the plane group of square floor tiles.
It's the plane group of the square mesh that's in the overhead lighting fixtures.
It's the pattern that is present in the tiles up above the lighting fixtures.
It's a pattern that is very commonly present in square shirts, but not too often.
It's kind of dull and uninteresting.
Yes.
Fellow back there against the wall wearing P4, probably didn't know until now.
Thank you.
Actually, he's a shield.
I made him come in.
I called him up last night and said be sure you wear the P4 shirt.
And he said, OK, I will.
All right.
I'm getting silly.
So that's tells me it's time to quit.
So let's take our usual 10 minute break, And then we'll resume.
And I think I'll go lickety split through the remaining plane groups that consists of combinations of a rotation with a lattice.
OK.
So do come back.
Any questions before I obliterate all this lovely geometry?
No.
OK. We handled them during intermission I guess.
Let me do a couple more plane groups just very quickly to show you how they come out without going through all of the steps, because I think we've seen now what one has to do to derive these.
There are two that are left.
One is a threefold axis, and if that's all the symmetry we've put into the lattice, we're combining a threefold rotation axis with a primitive lattice, and we know that this has to be hexagonal net.
Because from our depth of experience, we know that is the shape of a lattice that is demanded by a threefold axis.
It has two translations, T1, that are identical in length.
And this angle between them is exactly 120 degrees.
That's what we saw a threefold access require.
So now what we are doing is putting in a threefold axis at one lattice point, and this means we are adding the operations A 2 pi over 3.
A minus 2 pi over 3.
I'll choose to define a 120 degree rotation that way, and the operation A 2 pi, which is the identity operations.
So I'm adding three operations.
Putting the threefold axis in at all these locations.
They will all be translationally equivalent.
The pattern of this particular plane group is going to be the pattern of a threefold axis.
And so there will be one motif here, 120 degrees away.
There will be a-- sorry to wipe out those operations-- 120 degrees.
There'll be another object located here.
And 120 degrees again, there will be another object located here.
So these guys will fall at the corner of an equilateral triangle.
And the same will be at the other corners of the net, and that, as I've said before, is the pattern of that plane group.
And that's a plane group that we would call P3.
Primitive lattice that has to be hexagonal, and what we've added to it is a threefold axis.
OK, let me down here just indicate the combinations that we would do.
And I could do it up on top, but I can work with this translation as well, which is easier to reach at this late hour in the afternoon.
Let me combine A 2 pi over 3 with this translation here.
And that says we have to get an operation B 2 pi over 3.
That is located at the original translation T times the cotangent of 1/2 of 2 pi over 3.
And that's at 1/2 of the cotangent of 60 degrees.
And this turns out not to be any nice neat number like 0 or 1/2, but if you evaluate what the distance up along the perpendicular bisector is by this amount, where you come out is right in the center of this triangle.
Trust me.
A little bit of trigonometry will let you convince yourself of that.
So if I rotate 120 degrees.
Bring this one to this one, and then translate over to here, the way the first one and the second one are related is by a 120 degree rotation about the center of this triangle.
If I combine the operation A minus 2 pi over 3, with this translation, I go down a distance x of minus this amount and that puts me in the middle of the triangle that is directly below.
And I can move that back up.
And now I have all the operations of a threefold axis about this location.
I can do the same thing with the threefold axis at another lattice point.
Another thing I could do is to just rotate this by 120 degrees.
And by either route, you find there must be a threefold axis here.
Notice that these threefold axes will once again, as advertised, simply take things that are at the different corners of the cell.
For example, this threefold axis will tell you how this motif is related to this one is related to this one.
And so it goes through the other threefold axes as well.
This threefold axis will tell you this one is related.
No, I don't want to draw it in.
So this is P3, a pair of threefold axes in the centers of the triangle, and another one at the corner of the cell.
And now I am going to do the remaining combination of a rotation axis with a lattice, so quickly, and it's going to take your breath away.
And that is seemingly the most difficult and complex one of all.
This would be P6.
Sixfold axis plus a primitive lattice.
And we know it also has to be hexagonal.
And that will be called P6.
What we're adding to the lattice is a sixfold axis.
And now what I'm going to do as a shortcut is to say a sixfold axis also contains all the operations of a threefold axis.
So I can take P3 and drop it right on top of P6, and that's going to give me threefold axes here.
Sixfold axis not only contains 2 pi over 6 and 2 pi over 3, it also contains the operation 2 pi over 2.
This is the operation of a twofold axis, and that says I have to, in addition to the two full rotation that sits here get twofold rotations in the middle of every one of the translations T2.
So actually, this is going to be P2 superimposed on P3.
And that's going to give me these twofold axis and this threefold axis.
And the only thing I have to do is to show you what A 2 pi over 6 combined with one of these translations is going to do.
What is A 2 pi over 6 combined with this translation?
And it's going to be new operation B 2 pi over 6, and it's going to be located at 1/2 of T times the cotangent of 1/2 of 60 degrees, cotangent of 30 degrees.
And the cotangent of 30 degrees is 2.
So this is going to be at 1/2-- no, what do I want to say?
This is 30 degrees.
This is one.
This is two.
Cotangent of 30 degrees is this over this, and that is-- no that's one
It's 2 pi [?
squared 3.
?]
Yeah.
OK, that's right.
So actually, what that does is to say the sixfold axis sits right up here.
So I don't get any new sixfold axis rotation of 60 degrees here, followed by translation is the same as 60 degrees about here.
So this is P6, and there is a lot of pure rotation axes combined with lattices.
We've got P1, P2, P3, P4, and P6.
Now, what I will do eventually, when we're all done here, the plane groups are not derived in any book or set of tables that I am aware of.
And next time, you will get some notes that do this in very slow motion fashion and give you all the individual steps.
But the international tables does give you diagrams of the resulting plane groups, very nice carefully done figures along with the representative arrangement of motifs that they generate.
But we're not done yet.
We have not let mirror planes enter the picture.
And so unless there is dissension or debate, I'd like to consider what happens when we take a mirror plane, and we can combine that with two different kinds of lattices, a primitive rectangular net or a centered rectangular net, which is called C. And in order to do that, we need yet another combination theorem.
Here are the lattice points, and let me first derive the plane group that is called PM.
And what I'll do is to put the-- let me use a squiggly line here, not because I'm excited or nervous about this, but just to distinguish it from the edges of the cell.
So here is the operation sigma.
The pattern that is going to be displayed by the plane group is once again just the pattern produced by a mirror plane hung at every lattice point.
But now we need a theorem.
What we have here is the operation of reflection, followed by a translation that is perpendicular to the locus of the reflection line.
And we will ask what is that?
Again, you get the answer by just looking at once and for all, and say, if here is a first one and it is right-handed, and I reflect it to one number to a second one, number two, which is left-handed, and then move that by translation here to get number three, which stays left-handed if I move by translation.
And ask now how was one related to three.
The chirality is changed.
Reflection is the only thing available to us, and lo and behold, if I say there is a new mirror plane here, that tells me how this is related to this, and this one is related to this one.
And this to this and this to this, so the answer to this question is that a reflection combined with a perpendicular translation is a new reflection operation sigma prime that is located at a distance removed from the first by 1/2 of that perpendicular translation.
And again, it's a plane old mirror plane just like the first one, but notice that the disposition of objects relative to the mirror plane in the center of the cell is quite distinct from the disposition of objects relative to the first mirror plane, so this is a second mirror plane.
It is an independent mirror plane from the first.
So that is plane group PM.
If there's symmetry in this business, you might ask is there a plane group AM?
The answer is yes in three dimensions.
There is a space group, PM, and there's a space group AM.
So there's AM and PM, and there is symmetry, and all is well in the universe.
OK, we're making great progress here.
And we'll be fairly well along before we have to bring things to a close.
Let's do the second addition that's possible with a mirror plane.
And that is to take a reflection operation and add it to the translations that are present in a centered rectangular net.
We've already done all the work for PM, so we can use that as a starting point.
The pattern of this plane group is going to look like a pair of objects related by reflection.
But now, we'll have an extra pair hung at the centered lattice point as well.
And the first thing we can note is that this mirror plane is no longer independent of the first one.
What goes on at this mirror plane is something that's related to what happens at the origin lattice point and mirror plane, and therefore, these two mirror planes are going to do the same thing.
The pattern of this plane group, we've taken a mirror plane and added it to a centered rectangular net.
This is called a C lattice, standing for centered.
And correspondingly, the symbol for this plane group is CM.
We know how this one is related to this one.
This one related to this one.
And now, we've got something of a problem.
All of these motifs are equivalent.
How is this one related to this one?
Or in more general terms, what we're asking is suppose I have a reflection operation sigma, and I add the translation not at right angles to it, as I did here, but place the translation at an angle with respect to the reflection plane.
So what we're going to do is to take a first motif that's right-handed, reflect it to a second one, which is left-handed.
And then slide it along so that it sits up in the same position relative to this centered lattice point.
So here's number three, translation leaves it left-handed.
So I've taken the operation of reflection combined it with a translation that has a perpendicular component plus a parallel component, perpendicular and parallel meaning the orientation of these two components of the translation relative to the reflection operation.
Anybody want to hazard a guess on how that first one is related to the third one?
Number one is right-handed.
Number two is left-handed, so it's got to be reflection, right?
If I put a reflection plane in here, this one ought to be tilted like this.
That's not going to do the job.
Anybody got any idea?
Yeah
What is that, T, T1 [INAUDIBLE]?
OK. T parallel plus T perpendicular means it's a component of this translation.
This is T. This has a part T perpendicular, and it has a part T parallel relative to the initial mirror plane.
My friends, we have just stumbled headlong over a new type of symmetry operation, which we have discovered upon making this combination of mirror plane with a centered lattice.
And it's come up to smack us rudely in the face even though we may not have been clever enough to think of it in advance.
This is a new type of operation, and it is an operation that cannot be reduced to one of the simple operations that we've defined so far.
You've got to take two steps to get from number one to number three.
The way you can do it is to reflect along a locus that is one half of the way along the perpendicular part of the translation, exactly the same location as we found the symmetry plane positioned when the translation was normal to the first mirror plane.
But yet we can't put the object down yet in the position that would be produced by translation, because our translation is inclined to the mirror plane.
So I've got to take a second step.
Reflect.
Don't yet put it down yet.
Before you put it down, slide it up parallel to the mirror plane by an amount that's equal to the part of the translation that is parallel to the mirror plane.
So to summarize this before all these words get too confusing, I'm saying that a reflection operation combined with a general translation that has a perpendicular component in the parallel component relative to the locus of reflection is going to be a new operation, which I'll write as sigma tau, a reflection part and a translation tau that is parallel to the mirror plane, and tau is equal to the part of the translation that is parallel to the mirror plane.
Astounding.
This is a two-step operation that cannot be described in terms simpler than saying do two steps to do it.
And we'll see as we go along, particularly into a three-dimensional space, that there are other two-step operations as well.
Now you're all familiar with a pattern like this, because in very short order when New England's winter descends upon us, as you go slogging along from your room into the Institute, your footprints will make a pattern like that in the snow.
Exactly what we've got here.
Reflect across and slide.
Reflect across and slide.
Reflect across and slide, and this is the glide component tau.
And this is an operation that is called a glide plane.
And it's a new type of symmetry operation.
It can only exist in a pattern, which has translational periodicity.
And if we were not clever enough to invent it, we would see it as soon as we combined a mirror plane with a lattice that was non-primitive and had a translation parallel to it.
So it's a very, very descriptive name, the glide plane.
Reflect and glide, reflect and glide, reflect and glide.
It sounds like something you'd be doing in the Arthur Murray dance studio, very melodious.
Reflect and glide, reflect and glide.
It's a nice operation.
All right, so what has happened then when we add a mirror plane to a centered rectangular net is that we get a new operation coming in, something completely new.
We've got a symbol to represent an individual operation, sigma the symbol for reflection with a subscript tau.
And the pattern we've already drawn, but let's do it again in a tidy fashion.
Exactly as advertised, the pair of objects related by reflection hung at every lattice point of-- holy mackerel, look at this.
That would give you the willies.
That's the nature of the motif.
Get that out of there.
Mirror planes going through the lattice points.
Glide planes halfway in between.
The mirror lines and the glide planes tell us how things on the right side, for example, of the lattice point are related to the motif of opposite chirality on the left-hand side of the centered lattice point.
And this is a plane group that is called
Question
Yes, sir
How can we just define a new operation that's a two-step.
Seems like we could do this forever, define two steps of any new operation, like you said, with two steps
That's a good question.
We found this, because we tripped headlong over it.
And we say OK, there it is.
We've got to deal with it.
But you're right.
How do you know that rotating once reflecting, and then turning end over end three times is not a new operation that cannot be decomposed, is the word that's used for it, into something simpler?
The answer is you've got to try it.
If they're there, when we make these combinations of a symmetry operation, and now the symmetry operation can be a two-step symmetry operation, now that we've discovered that, combine that with lattice and with rotation and with reflection, and ask what is the relation between the motif at the beginning and the motif at the end.
If you can't describe it any more simply other than a hop, skip and jump, you've got to introduce the hop, skip, and jump as an element that goes into the derivation of these groups.
Now before you get concerned and fill out an add/drop card, I have to reassure you that there are a couple of two-step operations that we have yet to discover, but there are no three-step operations that are necessary.
Whew.
Feel better now, don't you?
I was just going to say it's kind of arbitrary, because you have a and b, you're just applying the third [INAUDIBLE] T sub a and b, [INAUDIBLE] trivial multiplication tables
Well, actually, this is something that is distinct.
I mean here is the group, and you cannot describe the relation between everything that is in this pattern, which was obtained simply by taking the operation of reflection-- we know how to peacefully coexist with that-- and placing that pair of objects at lattice point of a centered rectangular net.
So that's nothing really freaky.
I mean it's a straightforward addition, but if it's a group, you have to know when you combine all the operations pairwise that it'd be able to show that these operations are members of a group.
And the answer in terms of the language of group theory, if you combine a reflection with a translation that has a component that is parallel to the reflection plane, then there is a new operation that comes up that has to be in the group multiplication table, and the operation has a reflection part and a translation part
I have a question
Yes
How do we go from the lower left to the upper right in one operation?
The lower left to the upper right
No, the upper right diagonally.
Across the diagonal, the upper right corner of the square
Upper right here?
Yeah.
And the left-handed guide below that
The left-handed guide below
No.
Down there
Where is down there?
The lowest edge of the--
Down here?
Yeah
OK. From this one to this one?
Yeah
OK. What I can say is-- may sound like I'm slipping off the hook too easily-- we said that we really only want to consider operations that terminate within the cell, and the way I get from here to here is to reflect and then translate up by the diagonal translation.
So that's something that lies outside the cell.
So I can always knock off an integral number of translations or add on an integral number of translations to any mapping transformation, modulo T. OK?
But surely, if you've done, one operation then it's [INAUDIBLE]?
From here to here?
No, not necessarily.
If I give you a very simple pattern and plane group P1, and you ask how do I get from this one here to this one that sits up here?
If it's outside the cell, I've got to go translation that's outside the cell.
But it's not any new translation or any new sort of operation.
This is sort of in the same category.
Yeah
You can make a glide plane in the center of the cell
Yeah, OK.
Thank you.
There is a glide operation that goes from here to here and then translates up by one full translation.
But a glide operation that is an integral number of translations says that you're dealing with-- there's another object that is removed by a translation that is the same thing.
So it is first and that simpler one that would be inside of the cell and would be only the unique sort of translation you need to consider.
Thank you.
That's a good question.
That was a good answer to his question.
In fact, anybody want to take over?
I didn't do well on that one.
Yeah
I was curious.
Does the glide plane exist maybe in this case because we're not using a primitive cell?
If you were to consider this [INAUDIBLE] as hexagonal?
OK, let me answer that question by saying that that's our first encounter with it.
But now, that it exists, that is a symmetry element, which we should consider adding to lattices in addition to pure reflection.
So let me proceed now to do another plane group since we had discovered the operation of glide.
And I will take a primitive rectangular lattice, and I will now add to the lattice point, not a mirror plane, but a glide plane.
OK.
The pattern is going to look like the pattern of glide, have things left and right on either side of the glide plane.
The same thing here.
And there is the pattern.
This is a pattern that's called PG.
And what I have to ask is what is the guide operation sigma tau combined with, let's say this translation T1.
And the answer is that if I reproduce number one to number two of opposite chirality by the glide operation sigma tau and then follow that by T1 the first and the third will be related by a new glide operation sigma tau prime that's located at half perpendicular part of the translation away from the first.
And that really is a generalization of this relation here.
If we make the pure reflection operation a glide operation, combine it with a translation that is perpendicular to it, it turns out that the net result is a new glide operation, sigma tau prime.
The two taus are equal, and this occurs at one half of the translation T perpendicular removed from the first.
And the complete generalization would be to say if I have a translation that has a parallel part and a perpendicular part relative to the glide plane, what I will get is a new glide operation sigma prime that's located at 1/2 of the perpendicular part of the translation from the first, and it will pick up a glide component that's equal to the original tau plus the part of the translation that's parallel to the glide plane.
So this now is this theorem involving translation and reflection type operations in its most general form.
OK.
So this is a new group called CM, and it consists of pairs of objects.
Sorry, we did that earlier.
This is PG, pairs of objects related by a glide plane, hung at every lattice point of the primitive rectangular lattice.
Is there a CG?
Actually, there is not, and let me show you why, and then I think we're just about quitting time.
If there is a centered lattice, and I hang a glide plane at the lattice points, we'll have glide planes in the locations of PG, the pattern will look like objects repeated by glide.
At this lattice point, and now we're going to have another object hung at the centered lattice point, and it's going to be in positions like this.
Now, you can see just by looking at the pattern that there is going to be-- whoops, what did I do here?
I want to move this one up to here, and I want to move this one up to here.
This one should be in this location.
If I look at that pattern, what I've done is to create halfway along the-- quarter of the way along the translation a pure reflection operation.
And I can find that using my general theorem.
I have a glide operation, sigma tau.
I combine with that the centered translation, which is 1/2 of T1 plus 1/2 of T2.
I've taken 1/2 of T1 plus 1/2 of T2, that's the centering translation, combined that with sigma tau.
I'll deftly jump to the side, so the people against the wall can see it.
I've taken a glide operation, combined it with this translation.
This should be a new reflection type of operation that will be located at 1/2 of the perpendicular part of the translation, and that's at 1/2 of 1/2 of T1.
And it will have a glide component tau prime, which is the original tau, which was 1/2 of T2.
And to that we add the parallel part of the translation, and that is 1/2 of T2.
So this is rewritten in slightly different form, is simply a new glide plane sigma prime, which has a glide component equal to the entire translation T2 at 1/4 quarter of T1.
And this is the same as a mirror plane at 1/4 T1.
And this, if we compare it with CM, is exactly the same thing.
Identical to CM with an origin shift of 1/4 T1.
So there is no CG in what we picked up in terms of rectangular nets and symmetry planes is PM, PG, and CN.
So there are three groups involving orthogonal nets in a single symmetry plane.
That is a good place to wrap things up.
And we'll next turn very, very quickly equipped with a set of notes, which summarizes the result, to what happens when we take two MM and put it into a rectangular net, and take two MM and put it into a centered rectangular net.
And then things get interesting, because we've got two planes.
We can make a both mirror planes.
We can make them both glides, or make one a mirror plane and one a glide plane.
So there are three possibilities with the addition of two MM to the net.
More than enough for one day.
It's Thursday.
Take the rest of the week and weekend off by doing no symmetry.
There's no assignment, and we'll have at it again next Tuesday.
Why don't we get started.
Good afternoon-- to stretch the literal meaning of the word considerably.
What a miserable day out there.
I get to and from the boonies via the commuter rail.
And this morning, we sat there for an hour part way in because a tree had fallen across the railroad track.
I don't know who went out to do it, but somebody had to get a saw and saw the tree up and take it off the tracks before we could proceed into Boston.
Even with that, I was sort of reluctant to get off the train.
I saw how it was pouring outside.
So I am highly flattered that you decided to trudge through the puddles and show up this afternoon.
All right, this is a momentous occasion because we are exactly at this point, halfway through the term.
And this is when we are about to change gears from our discussion of symmetry theory and switch over to physical properties of crystals and the way in which symmetry impacts on those properties.
What I wanted to do primarily today, though, is to say a little bit in a very fast and very simple way about the nature of crystal structures because that is one of the primary uses of symmetry theory.
It's to describe the periodic arrangements and symmetrical arrangements of atoms in crystals.
So it's the language that's necessary to describe such arrangements, particularly when they're beyond the red balls at the corner of the cube and black balls in the middle of the faces level of structure.
Let me first, for your amusement and edification, pass out another problem set.
This has only two problems on it.
You are not really equipped to do it yet because we haven't talked about space groups.
So don't worry about doing that now.
But I wanted to get it in your hand so that you could look it over.
It's a good problem set on which to end this part of the discussion-- but there's another one that I'll be passing out next week-- because this is an example of how you can use the notions of symmetry that we have established to say some fairly profound and non-obvious things about the nature of atomic arrangements.
So that's the purpose of this problem set.
It is to give you an opportunity to see the surprising things that can fall out of our material that we now have at our disposal.
Let me begin by telling you where we left off last time.
We had completed deriving the so-called Bravais lattices, the three-dimensional space lattices.
The process by which we did this was to take our two-dimensional space groups, the plane groups, and say that, if we add a third translation, we will have generated a space lattice.
But these different nets in the base of the cell are special and determined by the fact that there are symmetry in these nets.
So if we take, for example, a net that is exactly square-- it's square because there's a fourfold axis there.
And that fourfold axis, in a space lattice, pokes not only into the base of the cell but extends up through space.
So you must pick a third translation such that the symmetry elements line up.
We would take each of the crystallographic plane groups and pick a third translation.
You end up with almost all of the lattices that you're going to get except for the ones that involve cubic symmetries that are inherently three-dimensional.
And the reason for that is that the more complex three-dimensional point groups are, for the most part, obtained by adding inversion to the simple symmetries that exist in two dimensions.
But inversion doesn't require anything of a lattice.
So you can immediately say that 2M and 2/M are compatible with the same sort of space lattice because the symmetry 2/M is nothing more than what you get when you add inversion to either a twofold axis or a mirror plane.
And, finally, another way of classifying crystals, and a broader way of classifying them, is according to the coordinate systems that are necessary to describe the lattices.
And these are called the crystal systems.
And, as a mnemonic device for keeping straight what the crystal systems are as opposed to the crystal classes, which is another word that we haven't used but it's another word for the point groups, the crystal systems goes with coordinate systems.
And it's simply a word to describe the shape of the lattices, family of lattices, that we've described.
And we distributed a handout last time that gave the names of these systems and the relations that exist between the edges of the unit cell, which we use as the basis for our coordinate system and also the labels that we put on them.
I have another sheet, which I'll pass around, which is nothing new.
It just assembles-- I'll give you two sheets, which I'll pass around.
The first is an assembly of the crystallographic point groups in a somewhat neater and tidier way than we had on the sheet that we passed around last time.
So this is a picture of all 32 of the point groups.
Again, beware of the fact that mirror planes and guides to the eye that split things up into 90 degree segments or 60 degree segments come out deceptively similar on these sheets.
But there with the representative pattern are the arrangements of symmetry elements, a representative pattern of motifs, and the international and [INAUDIBLE] symbol for all 32 of the three-dimensional point groups.
And then the next sheet I'll pass around takes the point groups and groups them together with the lattices that can accommodate them and the crystal system that describes the arrangement of axes in the lattice.
So this is, in a nutshell, everything that we've covered up to this point.
And, at this point, I have to say that there is some disagreement on how many crystal systems exist.
And even though people who work with symmetry and mathematics are usually rational, almost to a fault logical individuals, there are several junctures as you build up a body of knowledge where you could go either way or adopt either convention.
You sort of got to pay your money and take your choice.
And this occurs in our arrangement of point groups and lattices among the crystal systems.
This occurs in the hexagonal crystal system.
Symmetry 3 and the other point groups that can be derived from it can be placed either in a hexagonal lattice, that is to say, a third translation C that is exactly at right angles to a pair of translations, A1 and A2, which are 120 degrees apart.
Or a three-fold symmetry can go into this funny double-body centered hexagonal lattice, which could be defined, if we were masochistic and enjoyed working in an oblique coordinate system, could be defined instead in terms of a rhombohedral lattice, which had the special characteristics that three translations are identical by symmetry, called therefore A1, A2, and A3; and the angles between the more equal alpha 1, alpha 2, alpha 3; and they could be anything we wished.
So that is a unique sort of crystal system and unique sort of coordinate system.
There are those who would say that if this is the case, we really have seven crystal systems-- triclinic; monoclinic; orthorhombic; tetragonal; hexagonal; and then this one that is compatible only with a three-fold axis and that is called the trigonal system or, by some folks, the rhombohedral system; and then, finally, the cubic.
Well isn't that a rational and logical thing to do?
Yeah, question at this point?
Do you have any extra copies of the sheet [INAUDIBLE]
Oh, yes.
There should be more than enough floating around.
This one here?
[?
The ?]
[?
chart ?]
Yeah, [?
not the ?]
[?
chart ?]
OK.
I'll give you-- I think there are more back there.
But why don't you take one for now?
Yeah, so it's a perfectly reasonable thing to do is to set up the trigonal system as a separate coordinate system to describe the rhombohedral lattice.
The problem that creates, however, is that if you do this, there is no longer any one to one correspondence between the point groups and the crystal systems.
A crystal with symmetry 3, for example, can fit into it either in a hexagonal lattice or a trigonal lattice.
US So sometimes a crystal with point group 3 would have to be called a hexagonal crystal, sometimes a trigonal system.
But which is it?
Well it can be either.
So admitting the trigonal system as a separate crystal system breaks down the one to one correspondence between point groups and crystal systems.
The very pragmatic reason for not admitting the trigonal system as a separate crystal system is that, although you can in this funny triple hexagonal cell define a primitive trigonal cell that has a special geometry, nobody in their right mind, as I indicated earlier, would want to work in such a coordinate system.
Better to take the triple cell, pay the price of three-fold redundancy, and yet have two interaxial angles that are 90 degrees and one that's specialized at 120.
That's a much more convenient reference system for our description of crystal properties.
One final thing I'll comment on but won't say anything about very much-- I always do that.
I say I won't say anything about it but, and then launch into a 10 minute digression.
Off in the right hand corner is something called the Laue group of which there are 11.
The Laue groups are those of the point groups that contain the operation of inversion.
And the reason they're called the Laue groups is that one of the primary ways, other than looking at crystal morphology to determine the symmetries of a crystal, is to look at the way it diffracts x-rays.
So if we had, for example a tetragonal crystal and there were lattice planes in here that were related by 90 degree rotations, then if we brought in an x-ray beam at such an angle to satisfy Bragg's law and saw some sort of intensity that came off, if we rotated the x-ray beam 90 degrees, we ought to see diffraction at the same angle and with exactly the same intensity.
So either by moving the x-ray beam, or simpler thing would be to leave the x-ray beam fixed, since it's attached to an x-ray generator that weighs a ton or more, instead move the crystal relative to the x-ray beam.
Now what would you do in this way of thinking to determine whether a crystal had the operation of inversion in it?
If this is the set of planes with indices HKL, you, again, would bring in an x-ray beam at an angle theta, in out.
Look at the intensity from the set of points HKL.
And then look at diffraction, both in terms of the Bragg angle theta and in terms of the intensity.
Look at that diffraction from the planes that were related to HKL by symmetry.
Well what plane is related to HKL by symmetry?
It's these planes down here which have indices minus H, minus K, minus L. So if we did the corresponding experiment to what I did to the left, we'd bring in an x-ray beam like this.
It would be diffracted at some angle theta.
And we'd look at the intensity for the set of planes minus H, minus K, minus L. Is that going to be any different?
No.
Right-minded person would say, why should this intensity be different?
All we're doing is bouncing the x-ray beam off the opposite side of one in the same set of lattice planes.
So why should the intensity be any different?
In this case, clearly if the set of lattice points is rotated by some finite angle, we are not going to see equivalent diffraction from planes that are 90 degrees apart.
So the upshot is that the diffraction symmetry of a crystal, the way in which planes with indices that are related by symmetry diffract in terms of position of the beam and intensity, the diffraction symmetry of a crystal always looks as though that crystal possessed an inversion center.
So, therefore, to-- and I qualify this by saying, to a good approximation-- the only point groups that you could determine and distinguish from one another by x-ray diffraction are the point groups that contain an inversion center.
Almost, but not quite.
If we look a little more closely at what is doing the scattering, it is not the Bragg planes.
These are imaginary, shimmering, absolutely planar things that are constructs that we use to interpret the process called diffraction.
Inside the crystal what is actually doing the diffraction is a collection of atoms, each of which are scattering the x-radiation.
And that physically is what causes the diffraction.
And the resultant intensity is the addition of all the little wavelets scattered by these separate atoms.
So in terms of the atomistics of what is involved, to look at diffraction from the set of planes HKL and contrast that with the diffraction for the set of planes minus H, minus K, minus L, we're coming in from opposite directions.
And here the beam would hit-- let's call the atoms, if I did not do so already, let's call them A and B.
A for a and b for big.
Diffraction from the set of points HKL would pass through the sheets of A atoms first and then impinge on the B atoms, whereas in the opposite direction, the x-ray beams would impinge upon the B atoms first and then hit the A atoms.
Does that make a difference?
Aha.
Believe it or not, it does.
And I can't explain why without going into a very detailed discussion of diffraction.
But if you compare the intensities that are scattered by these two sets of planes, which are related by inversion and therefore just opposite sides of one another, it does make a difference which atom the x-ray beam encounters first.
And the difference in intensity is on the order of 3% to 5%, although by picking a radiation appropriately for the particular chemical species in the crystal, you could increase this perhaps to 10% or 12%.
But that is something that you could not do until synchrotron radiation came along, which could actually tune the wavelength to a particular value that you wish to use and not use something that was spewed out by an x-ray tube that had a target such as molybdenum or copper or chromium, perhaps.
OK, so you can with this careful measurement of intensity distinguished between the opposite sides of the given set of Bragg planes, unless, of course, the crystal did have inversion in it.
And then there would be no difference at all.
It requires a very careful experiment, though, because there are many things that could make the intensities different.
For example, absorption by the crystal.
If the crystal has an irregular shape, the absorption, which can easily be factors 10, 20, or even more, the absorption would be different for x-ray beams coming in those two orientations.
So in any case, as I say, I won't say anything about it.
But 10 minutes later, that's the meaning of the Laue groups.
These are the point groups that have inversion in them.
That effect that I just described, by the way, is called anomalous scattering or anomalous dispersion.
There's really nothing anomalous about at all.
It has to do with the way electromagnetic radiation interacts with the clouds of electrons that are distributed about the nucleus of each atom.
Where we will go from here to finish up symmetry theory will be to now take the 32 crystallographic point groups and to drop those point groups into each of the distinct 14 Bravais lattices.
And then we'll need combination theorems to deduce where the additional symmetry elements arrive.
And when we go through that process, if we were to do this exhaustively, we would find that there are 230 distinct three-dimensional symmetries that involve point group symmetry operations and lattice type.
I saw several faces blanch.
Don't worry.
We're not going to do that.
What we will do next time is do a small handful and then conclude by showing you how this information is tabulated in the international tables.
And the quick, immediate answer to that question is that the problem set that I passed around has two pages for two different point groups that show the way that this information is tabulated in the international tables.
So that's where we'll end up.
And we'll come pretty close to completing that when we meet on Thursday.
I heard no sighs of relief.
I heard no cheers.
So you are presumably still engaged in this material somewhat.
I have two things I would like to do.
We talked about crystal systems and point groups.
And let me now ask, do you think there is a crystal known for every one of the point groups?
Think so?
I saw some skeptical shakes of the head.
The answer is yes, but just barely for some of the point groups.
One of the very rare point groups, in terms of structures that populate it, is our high symmetries but which lack inversion or mirror planes.
One very, very rare point group is 432-- just axes arranged in a cubic arrangement with no mirror planes or inversion centers.
And my rationalization of that is that if an arrangement of atoms has to conform to all of these rotational symmetries, it's pretty hard for it to do it and have that assemblage of atoms assume the lowest possible energy state if it doesn't pick up in the process inversion or mirror planes as additional symmetry element.
Sort of satisfying for me.
But that's not a rigorous argument.
OK, let me now take a poll.
Do you think, using the broader cut of the crystal systems, do you think there's a crystal known for every crystal system?
Are there triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal, cubic crystals known?
Yeah.
That's a pretty realistic expectation.
And that is indeed the case.
But now let me cut this a little finer.
Which do you think is the most popular crystal systems for real materials and which do you think is the rarest?
OK. Who thinks triclinic crystals are the most common?
OK, there are pessimists.
What is the definition of a pessimist?
The pessimist is somebody you should borrow money from because they'll never expect to be paid back.
Who would say cubic crystals?
There are naive [?
optimists.
?]
Now I will reveal all.
The answer is, it depends.
Depends upon whether you're talking about, interestingly, inorganic materials or whether you're talking about organic materials.
If you look at inorganic materials, the most popular crystal systems are-- and the results are, first of all, cubic.
Is this not a benign world in which we live.
Most inorganic materials-- if you're a material scientist who works with metals and with ceramics, you're going to work with cubic materials, thank God, most of the time.
The second most popular-- by a considerable margin but there are still a lot of them-- is orthorhombic.
And the third most popular, shortly behind orthorhombic and almost in a tie, monoclinic and tetragonal.
We'll launch into a quick overview of crystal symmetry.
And we'll see why inorganic materials have high symmetry.
It's true, in particular, when the material is held together by ionic bonding.
If the bonding is ionic, it's non-directional and atoms of opposite charge will therefore try to get as close as possible to the atoms of opposite charge.
And because atoms of like charge repel one another, the surrounding atoms will try to be as widely separated as possible.
And this grouping of atoms, positive around negative, tends to have a lot of symmetry.
And, therefore, the way in which these groups are linked together has a lot of symmetry.
OK, I said organic materials.
And I'll pass around some actual numbers that were assembled a long time ago but are still, in terms of proportion, still valid.
The organic compounds are broken down into whether these are compounds based on rings or extended, lopsided molecules.
And the numbers are very different depending on that.
But, for organic molecules [?
material, ?]
monoclinic is the most common, then orthorhombic.
And shortly after orthorhombic comes way, way behind is tetragonal and triclinic, which are almost tied.
And I think I can rationalize this observation, as well.
Organic materials contain discrete molecules.
And if we've got something-- and I'm going to, since I know I'm among friends, display my abysmal ignorance of organic chemistry.
I am unapologetically and unabashedly inorganic and even non-metallic in my research interests.
So let's suppose we have something that's composed of various strange rings with little side groups sticking off like this, another one off like this, and then maybe another ring up like this.
That, to me, looks like a reasonable molecule in terms of my ignorance of organic chemistry.
So if there are these finite groups and you're going to put them together in a crystal, what are you going to try to do?
They're going to be held together by very weak van der Waals forces, which are non-directional.
So think of the outline of this molecule as being some ugly thing like a ham hock.
And then your problem, if the forces between these molecules are non-directional, how do you close pack ham hocks?
The way you would do it would be to put down a row of them in a close-packed sequence.
So we'll put down things that look like this.
And then, maybe to make them fit together, dovetail another one in here.
And maybe something like that.
And then for the next row down, and you would maybe put the nose of one in here.
And then put the other one in here, the nose of one here, and something like that-- being very schematic.
The lattice that describes this packing is not going to be orthogonal.
It's not going to be square with high symmetry or even have two orthogonal translations.
It's going to be something that has this as translations.
There's going to be an A and a B that is not at right angles to it.
And the B translation will have different links, simply because the thing that you're putting together in close packing has a very irregular shape.
So I think that is the reason why molecules, discrete molecules that try to achieve something that is as close packed as possible, will have translations in them that are of unequal length and making arbitrary angles with respect to one another.
So having kept you guessing for so long, let me pass out another sheet.
This was a compilation that was done a long time ago.
And there were only 9,000 crystals for which complete crystallographic data had been obtained.
This was way back in 1967.
As I say, there's much more data now, easily in order of magnitude.
But I don't think the proportions will change very much.
They'll stay about the same.
Let me pass out these sheets.
So make sure you memorize them because the first question on the next quiz is, how many out of 8,759 crystals are actually triclinic?
I would never do that.
All right, any comments or questions?
Yes, sir
Why would organic molecules of such odd shapes even be crystalline to begin with?
Why wouldn't they just be amorphous?
OK. Clearly the interatomic forces are very directional and very strong.
And that's going to hold the molecule into a fixed configuration
Didn't you say there were van der Waals forces?
But they're these-- even for something that is not ionic or metallic, there still are weak attractive forces between these molecules.
Why?
Well, the charges are not uniformly distributed over these molecules.
So there are going to be some van der Waals forces that tend to hold them together.
And a van der Waals force is essentially non-directional.
So if you have this weak force, you're going to be able to squish the crystal into something amorphous very easily.
And organic materials are usually very subject to plastic deformation.
But if you assemble the molecules carefully from solution, or by vaporization, condensation, they're going to want the order if they are solidified with enough mobility to rearrange themselves in the lowest energy configuration.
I think what you're saying is these are not going to be very strongly bonded crystals.
They're going to melt at low temperatures.
But, nevertheless, prepare them properly, and the lowest energy state will be ordered
I'm sure [?
they're ordered.
?]
But would it be such that you have [?
translational ?]
symmetry?
Who's to say it's going to have a definite pattern that will repeat itself?
Yeah, if there are attractive forces-- let me again use a hand-waving argument.
If you've got a bunch of particles and they aggregate here in the lowest energy configuration, that is going to be the lowest energy configuration.
And a similar group of atoms will want to do the same thing here.
And if there are interactions between these groups, they're going to arrange themselves in an order configuration.
That is simply saying-- this is a very weak qualitative argument-- says if you have a configuration which is the lowest energy arrangement here, any other place in the system where those atoms get together and congeal are going to assume that same configuration.
OK?
Other comments or questions?
This is all very qualitative but actually concerns matters that we are describing a language to describe
[INAUDIBLE] Yes, there are.
If you look at this table that came around, if you look at cubic organic materials out of-- there-- what is the number?
117 out of what looks to be several thousand.
Yes, there are cubic organic crystals, but very few in number.
Hexagonal-- 56 out of a rather large number.
All right, when we took a poll a week and a half or so ago and said, how many of you had seen any material and crystal chemistry in solid state chemistry in a previous class on structure and bonding or crystal chemistry, and only a few people tentatively twitched a hand slightly.
So what I'd like to do for the rest of this session, since this is what symmetry theory is developed to describe, I'd like to say something about structures and why they form the way they do to give some qualitative insight.
This is a fat pack of notes.
And I'll go over it quickly with you and lead you by the hand through it.
But this is not material for which I'll hold you responsible.
In fact, I'll turn my back and those who want to get home early because of the rain can tiptoe out.
And I won't even notice who you are unless everybody goes.
And then I'll feel very hurt because I wrote these out.
And it took a lot of trouble.
All right.
I have to apologize and say that most of what is described in these notes applies to ionic bonding and, to a lesser extent but still in large measure, to metallic structures.
And what is the reason for that?
It's because the metallic bond, to a fair approximation, an ionic bond rigorously, if you believe Coulomb's law, is a non-directional sort of bonding.
So the structures that are assumed, the structures that are the lowest energy configuration turn out to be determined by interatomic distances and ionic or metallic sizes.
The Coulombic interaction that holds ionic compounds together is a central force.
And one of the features of electromagnetism is that, if you have some distribution of charge about center and another distribution of charge of a different sort about a another center, the cohesive force between, let's say, a positive ion and a negative ion is, along a line, joining their centers.
And, secondly, you can take, once you're outside the distribution of charge, you can clump all the charge together at the center of the distribution.
So that's what makes the ionic bond so simple.
And that's why everybody who talks about bonding and structure discusses ionic bonding because organic chemistry is just too complex in comparison.
So let's look at a given cation.
And cations generally tend to be smaller than anions because you've removed electrons from the outer configuration.
We'll look at some effective sizes of ions in just a bit.
But you've stripped electrons off to make a positively charged ion.
You've added electrons on to make a negatively charged ion.
So when you add electrons, the ion functions has [?
always ?]
had a larger radius.
So let's bring up a B minus ion as a neighbor.
And for those two species, there will be, if we put energy as a function of separation D, there's going to be a Coulombic attractive force that goes as 1/D.
But when the electron configurations get in to close proximity, at small separations, the electrons will start to push one another, beginning with the outer electrons first, into higher energy states.
And the energy of interaction, due to the interaction of orbitals, will increase the energy of the system.
And the result then is that the net potential, as a function of distance for an isolated ion pair-- I was looking for some colored chalk and it's gone-- is something that rises steeply at low separations and then levels out more gradually at larger separations.
So this then would be an equilibrium separation, D0, for that ionic pair.
If we got energy out, and for that isolated ionic pair it'll be just exactly this much energy, E0.
If it weren't for a pair of ions-- let's take another A and bring it into the system.
And there's a lot of room to put still another A around.
And we can add that to the system.
And every time we bring up an additional neighbor, we get approximately this much energy out of the system.
But not quite because there will also be repulsive interactions between the ions of like charge.
OK?
But to a good approximation to within about 10% to be quantitatively, every time we bring up another neighbor, we get out the same amount of energy again by forming this group.
And all that works splendidly until-- let me now look at the arrangement of anions about the little cation because they will have to have a shell as well.
So if we have a small cation and we bring up a big fat anion, there will be, first of all, two consequences.
These ions of like charge will interact repulsively.
So the lowest energy situation is going to be one in which these neighboring ions arrange themselves so that the repulsive energy is minimized.
And we're going to get the maximum separation between all of these anions and get the minimum energy when they arrange themselves on the corners of a regular polyhedron.
And that's where symmetry starts to come in to the importance of ionic structures.
That, for this particular configuration of anions, the B ions might be on the corners.
And I am not drawing them in contact now just so it'll be easier to draw.
The B ions will be equidistant and tend to arrange themselves on the corner of the regular polyhedron.
If we got energy up by bringing up extra neighbors, we will reach a point where the B ions will essentially be in contact.
And the A ion is rattling around in a hole.
It can touch perhaps two neighbors.
And we will get the full amount of energy on.
But for the other two neighbors, we're going to be out here on this curve.
And we won't get nearly as much energy out.
But at the same time, we're paying the price energetically of all of these repulsive interactions between the surrounding neighbors of like charge.
So the second consideration is that, as we said, first of all, this group of atoms, which is referred to as a coordination polyhedron, it first of all will be symmetric.
And then the number of neighbors, the number of ions of one charge that are coordinating the others-- and that's something called the coordination number, introduce some jargon-- has an upper limit that is going to be determined by the size of the ions.
So these are the two basic tenets of ionic crystal chemistry.
So having established those points, we can do some geometry and assume that the ions are hard spheres or at least elastic spheres with some stiffness because it's going to take energy to perturb the orbitals of the electrons and, just in terms of simple geometry, calculate the range of radius ratios RA/RB where A is the central ion in the polyhedron and B is the surrounding ion, and do this for different coordination numbers, picking polyhedra in which the surrounding ions of like charge are as widely separated as possible.
So starting at the beginning, coordination number one would be a pair of ions.
Obviously no constraints whatsoever on the relative sizes.
RA/RB can be anywhere it likes between zero and infinity.
For coordination number two, again the surrounding ions want to be as widely separated as possible.
So it'll be a linear chain.
But, again, the central ion can become as small or as large as it like.
So the range of radii is between zero and infinity.
For coordination number three, the polyhedron will be an equilateral triangle.
And if you calculate the geometry, when the B ions are exactly in contact, you have a little 30, 60, 90 triangle in which one edge of the triangle is RB and the other is RA plus RB.
And you can manipulate that into the form where you can show that there will be a lower limit RA/RB, which is 2 over the square root of 3 minus 1.
And that turns out to be a magic number, 0.155.
Going to coordination number four, you'd say, aha.
That's going to be square group.
No, a square group would not keep the surrounding ions of like charge as widely separated as possible.
And the geometry which would keep them as widely separated as possible but yet in contact with the central sphere would be a tetrahedron.
The Bs would be arranged on alternating vertices of a cube.
The magic number there turns out to be 0.225.
Coordination five-- very, very rare.
Five is a crystallographic abomination because five-fold symmetry does not conform to a lattice.
But there would be, in principle, one five-fold group.
And that would be a trigonal prism where you have three Bs arranged on an equilateral triangle about the central A.
And then you have one A up here, another B up here, and another B down here.
So this, then, is a bipyramid, a triangular bipyramid.
And you can derive a range of radius ratios for this.
And this is a dandy problem to give on a problem set.
But you find that the range of radius ratios RA/RB is the same as for six-fold coordination, which is the next one considered on the handout.
So we six coordination with the corners on a regular arrangement would be an octahedron.
And the permitted range there is 0.732 to infinity.
Five coordination, the trigonal prism, leads to exactly the same range of radii.
And the reason is, if you look at the geometries, this triangle here is the same for both an octahedral coordination and a trigonal pyramidal coordination.
Notice as we chug on that the lower limit to the radius ratio is gradually increasing.
And if we go to coordination number eight, cubic coordination, it goes up to 1 to infinity.
And for coordination number 12, RA/RB has to be exactly 1 period.
And that would be what you achieved in the close-packed number.
OK, turning the page.
There is also going to be a range of radius ratios that's determined by the number of As that can be packed around a central B.
And doing exactly the same calculation, if, for example, four-fold coordination permitted in terms of RA/RB, the range is 0.225 to infinity.
If we want to know the limits for packing of A around B with tetrahedral coordination, this would give us numbers RA/RB that were exactly the same range of numbers because we've just changed the labels.
If we want to put everything on the same basis, in terms of one particular radius ratio, if we do anion to cation radius ratio, the range would be 0.225 to infinity.
If we take the reciprocals of these numbers for four-fold coordination of B, this would be 1 over 0.225 to 1 over infinity.
So this will give us-- this is zero.
This is infinity.
So there'll be a range of radius ratios, 0.225 less than RA/RB, less than 1 over 0.225, which turns out to be 4.44.
So this would be the range of permitted radius ratios for CNA equals 4 and CNB equals the same thing, 4.
So there's going to be an upper limit that's imposed by the packing of As around B and a lower limit that's imposed by the packing of Bs around A.
But what are going to be the coordination numbers of A and B?
Let me finish this up.
And then we can take our break.
I'm going to now restrict things rather drastically.
I'm going to consider binary compounds in which all cations have the same coordination and all anions with the same coordination.
And the implication there is there's only one type of coordination number for both.
And that's not always true in compounds.
Sometimes you find a given species that has two different coordination numbers in a more complex compound.
If that's the case, if this is the A ion and this is the B ion, let me consider the chemical composition of a bond.
And that's going to translate into the same thing as the compound.
So let's suppose there's a number of Bs around the As, which is, by definition, the coordination number of A, and a number of As around a B and that is, by definition, the coordination number of B.
So let me look at a wedge that's associated with this bond.
And this is an amount of an A atom, which is 1 over the coordination number of A.
And if I do the same thing for the B ion the part of the B ion associated with one bond is going to be 1 over the coordination number of B.
So we can say then that the composition of a bond is going to be A subscript 1 over the coordination number of A, B, 1 over the coordination number of B.
The chemists abhor fractional subscripts on a chemical composition.
So let's multiply this by the product of the two coordination numbers.
And what we'll find is that the formula of the compound with integer subscripts is going to be A, C, and B; B coordination number of A.
And that is a subject to a very drastic set of assumptions that each species A and B has just one coordination number.
OK, so doing this in terms of composition, just turning this result around.
And then we'll quit and take a stretch.
This says that if we look at a particular composition, AnBm, something like TiO2 or Al2O3, then the coordination number of A in proportion to the coordination number of B is going to be equal to the ratio of M to N. And the more ions of one charge that we can pack around the other, the lower the energy will be.
And, therefore, the coordination numbers will be the maximum permitted values of CNA and CNB, subject to the constraints of radius ratio.
In other words, to conclude very quickly with one specific example, if we look at a compound like Al2O3, an important ceramic and a gemstone, this says that the coordination number of Al to the coordination number of the oxygen has to be equal to three to two.
So possible structures might be aluminum with coordination number three, oxygen with coordination number equal to two.
Or aluminum could have six coordination.
The oxygen could have four coordination.
Or the aluminum could have 12 coordination.
And the oxygen could have eight coordination.
Those three possibilities.
Next we have to use the ionic sizes to see what constraints there are that are going to limit the maximum coordination numbers because the greater the number of neighbors, the lower the energy.
And, in the case of aluminum oxide, it turns out that it's six to four.
And this is determined by the radius ratio.
And you know what?
If we used the ionic radii for Al3 plus and O2 minus, we were right one the nose.
That is the nature of the structure of aluminum oxide-- aluminum with octahedral coordination by oxygen and oxygen with a tetrahedral coordination of aluminum.
And that is the actual structure.
The three-dimensional linkage would have to be such that ions of like charge have the maximum possible separation.
And to keep you beyond your patience for just a little bit, what is that going to be?
We can use very simple considerations there.
If we have coordination polyhedra and we want to join them together, the worst possible situation is to have them share edges because the central ions then had minimum separation.
The better situation would be to have to the polyhedra share corners and have the ions in the center of these triangles form a straight line with the shared ions.
For a three-dimensional polyhedron, the sequence of increasing repulsive energy would be shared faces for something like cubes, shared edges-- that would be lower energy but not the best you could do.
Shared corners would be the best possible arrangement.
As you make each of those transitions, you take the central ion and keep the separation between the neighboring ion of like charge as low as possible.
The interaction energy would be as low as possible.
All right, so let me stop there.
I think we're probably about halfway through.
We can then determine ranges of radius ratio for various stoichiometries.
And all this is just a fantasy unless we can say, do ions have sizes that are the same from compound to compound?
And, if so, how do we determine them?
I hope that intriguing pair of questions will keep you here, dry and warm, rather than heading off early.
OK, so let's take our usual 10-minute break and resume to wrap this up.
All right, let's resume our discussion.
And we'll finish it for sure by the end of today.
And as you say, you can tuck this away for study on your own or for future reference or whatever you want to do.
So if we could assign sizes to atoms and ions, we could do a lot towards predicting what sort of structures they might form if we knew the composition.
The composition-- again, assuming there is only one type of anion and one type of cation, the composition is determined by the valence.
The composition fixes the ratio of coordination numbers.
And the structure will have lowest energy when we have the maximum number of anions around cation and vice-versa.
So we're in great shape.
And this is simple, idealized, but very powerful stuff if we knew the sizes that were assumed by ions.
And that, really, is an absurd notion, to talk about an atom as though it behaved like a sphere with finite dimensions.
If we plotted the density of the electrons as a function of distance R from an atom, it does something like this.
At very small distances, that defines a shell of a rather small volume, so there are not many electrons.
Rises up to a maximum, and then tails off gradually.
So what is the radius?
The maximum in the distribution of density.
So the notion of radius is absolutely nonsense in terms of the actual distribution of electrons on the atom.
However, one can get from your fraction measurements a determination of the lattice constant.
And even with routine methods, you can get the dimension of the lattice to extraordinary precision.
The typical lattice constant is about 10 Angstroms.
You can get this through routine methods to plus or minus 1 in the fourth decimal point.
So this is a precision of 1 part in 10 to the 5th.
You can get atomic positions when they are not fixed by symmetry to get a coordinate to plus or minus 0.yyyy plus or minus 1 in good cases.
So that is 1 part in 10 to the 5th.
Missing a factor or 10 somewhere.
This is plus or minus 1.
That's 1 in 10 to the 4th.
So you can get this information very accurately.
One thing I always like to point out here is that a thermal expansion coefficient for most inorganic materials is something that never gets much lower than about 1 or 2 times 10 to the minus 6.
So if you're going to push the measurement of a lattice constant to the ultimate accuracy, you'd better have a thermometer on the top of the X-ray generator, or your measurement is meaningless.
The thermal expansion can be more than the uncertainty in your measurement.
So you should report, for very precise measurements, the temperature at which you made your measurement.
OK, you can get interionic distances with very high precisions.
And if you look at a series of compounds-- and these are the ones that are discussed in the notes, which have the rock salt structure type.
So here is the cation, here is the anion.
The anion-cation distance is simply equal to 1/2 of A for the rock salt structure.
So that's easy.
You don't even have to know the atomic positions.
Just measure the lattice constants.
And then you would find, for example, that R, NA plus R, CL is equal to 1/2 of A for NaCl.
Very precisely know the sum.
How do we split this up?
Well, let's look at another compound.
Let's look at R, NA plus R for bromine.
And that would be 1/2 of the lattice constant of NaBr.
Hm.
More data, but we haven't solved anything.
We have three radii, only two measurements.
And you can see, if you try to push this further, you can never get enough information to split up the interionic distances into sums of individual radii in any sort of unique fashion.
So you have to start somewhere.
And there are different sets of ionic radii that were proposed early in the game.
And they are discussed in the notes.
I give you an example of some of the sums of radii for the oxides and sulfides and selenides of the transition metals.
You can see that as the anion atomic number increases, the size of the lattice constant goes up.
So you could get, for example, the difference in size between a sulfur and an oxygen and the difference in radius between a selenium and a sulfur.
But you could never get the absolute values.
There are four early sets of radii.
And they are tabulated for you part way through the notes.
And there's a cautionary lesson here.
One the early series of radii-- see, even the old workers in this field appreciated the role of ionic sizes in determining structure.
So they wanted to set up self-consistent schemes of radii.
And the earliest one is by, really, the founder of crystal chemistry, a man named Goldschmidt.
Linus Pauling got into the act very early on.
If you look at this table that I copied from these separate publications, four different people-- Goldschmidt, Pauling, Zacharias and Ahrens.
Some people left out the radii of certain species.
Other people left out different radii depending on the compounds that they looked at.
If people looked at the same ion, they very often got very, very different values, depending on their starting point.
Look at the size of barium, for example.
Zacharias said 1.29.
Ahrens said 1.34.
Pauling said 1.35.
Goldschmidt said 1.43.
Depends on the starting point that they used for the initial radius.
And once you've determined that, everything falls into place.
Another basis for determining radii, Bragg determined a number of the structure of important silicates.
Why silicates?
Because as you walk along the ground, you're walking on silicates.
These are part of the minerals that form the Earth's crust, so there was a lot of interest in those.
Bragg, from his early work way back in the late 1920s, found that the size and shape of the Si04 tetrahedron forms tetrahedral coordination because silicon is so small compared to oxygen.
The interionic distances were always about the same.
The tetrahedron was always regular.
So what Bragg said was, a-ha, this is a tough little nut in silicate structures.
Doesn't deform.
The size is always the same.
It's always regular.
The reason must be that the oxygens are just in contact.
So he used the dimensions of an Si04 tetrahedron, took the radius of oxygen as half the edge.
Pauling used a different starting point that has been followed.
There's still another way of determining radii.
Is there any physical property of a material which might also be shown to be proportional to the size of the ion?
And the answer to that, which I'll pass over quickly, is yes.
Quantity called the ionic electrical polarizability, which is the relation between an applied electric field and the dipole moment that is induced on the atom when you separate the electron cloud and the nucleus.
And a simple model shows-- and this is carried out for you-- that you can relate the ionic electronic polarizability to the volume of the atom, and it turns out to be beautifully simple, that the polarizability is simply 4 pi epsilon 0 times the radius of the ion cubed.
So therefore, you can split up ionic radii in proportion to the cube root of the polarizability of the ions.
Well, that trades one problem for another.
How do you get polarizabilities?
It turns out that the index of refraction of materials is given by the sum of the number of each species per unit volume times the ionic polarizability of that species.
And there is a famous equation called the Lorentz-Lorenz equation.
I can never resist calling it not the Lorenz-Lorentz equation but the Lorenz-Lorentz equation equation because it sounds like you're saying everything twice.
If you note that the dielectric constant of a material is related to the square of the index of refraction, instead of n squared, you can replace the n squared by epsilon, the dielectric constant, and then the equations become known as the Clausius-Mossotti equations.
And I always take that as a nice illustration of the fact that a very simple change of variables can be all it takes to become famous and immortal.
[INAUDIBLE] what happened to Clausius and Mossotti.
Where do you get polarizabilities from?
This was looked at by a number of people.
One of the most complete were three people, Tessman, Kahn, and Wild Bill Shockley of transistor fame.
So you can find these numbers, and you can get a self-consistent set of radii.
But note in the table that I give you here that the individual numbers are very different.
And two caveats that I have to issue is you can't mix ionic radii in different people's tables because they have different starting assumptions for the standard radius.
If you do that, you're mixing apples and oranges.
And I wish I had a cookie for every half-baked book on chemistry that said, hey, Ahrens has got something for americium and Pauling doesn't.
Let me put the value for americium in there.
Totally wrong that they are established on different bases.
So you can't mix apples and oranges.
You have to use a self-consistent set of radii.
The ones that were followed until very recently are Pauling's radii.
And Pauling's radii are based on a standard radius for oxygen, which is 1.40 Angstroms for the radius of oxygen.
The other thing that I have to issue as a caveat, which I hope you won't forget, is that radius ratios do a pretty good job of predicting the expected coordination number.
But if you are very, very close to a limiting critical radius ratio, all bets are off.
Because then you're really asking the question, if I have this polyhedron of ions, if I squish them a little bit so that I increase the coordination number, does that give me a lower energy in return for the price that I have to pay to deform the ions that I'm packing together?
And that is not possible to answer on the basis of a rigid sphere model.
So when you're very close to a critical radius ratio, you really can't say for sure whether the structure will go to the lower coordination number or the higher coordination number.
So you have to be careful of that.
Finally, people have continued on.
And stuck in the margin of these older radii is a reference to a paper by two people, Bob Shannon, who was a crystallographer at DuPont, and Charlie Prewitt.
I'm very proud that he became so notable, because he was my office mate in graduate school.
So I like to promote his radii whenever I can.
And in Acta Crystallographica in 1969, they published a set of radii, which is in the nice typeset version that's Xeroxed directly from their paper in Acta Crystallographica.
And the headings need explanation.
There's the ion and its valence, then the electron configuration.
And what's interesting is that some of the transition metals, for example iron, can be in a high-spin or a low-spin configuration depending on whether the moments on the electrons are all parallel or anti-parallel.
High-spin and low-spin configurations give ions with very, very distinct radii.
So the second column, EC, stands for Electron Configuration.
Then comes coordination number.
And here is another feature that we really should not overlook, that as even rigid spheres-- here's an A, and here's a B-- acquire progressively higher coordination numbers-- let's look at just coordination number three and coordination number four, because they're easy to draw.
In addition to the attractive force between anion and cation, there is also a repulsive force between the anions.
So here's the repulsive force between this B and its neighboring B and this B and its neighboring B.
And the resultant is a force that tends to push apart the A and the B ion.
For four-coordination, remember first of all, the Bs are closer, so the repulsive force is stronger.
But there are four of them, as well.
So this turns out to be a larger repulsive force.
So the conclusion is that the radius of A would be expected to go up slightly as the coordination number increased, simply because there is a repulsive interaction between the coordinating Bs.
And that will tend to push them apart more strongly as the coordination goes up.
So the third column in the Shannon and Prewitt tables is coordination number.
And if you look at several ions, they're-- for example, barium has 6-coordination, 7-coordination.
8, 9, 10, and even 12.
And as you go down that list of radii, it gets larger as the coordination goes up.
And it's a relatively small effect that's on the order of 5 or so percent.
But still, when trying to calculate an interionic distance, you should use the radius that is appropriate for the coordination numbers.
The fourth column is spin.
And the spin stands for high spin or low spin.
And you will find that for the transition metal atoms-- for example, CO2 plus low spin or high spin, and note the rather different radii for these two configurations of the electron moments.
Yeah
Under coordination number, what's the difference between 4SQ and just Q?
You would guess that the lowest energy configuration for four-coordination would be tetrahedral, because that keeps the anions most widely separated.
There are cases where the coordination is plainer.
So the "SQ" stands for square, and that's unusual.
And you'll notice that that appears for things like silver, which has a considerable covalent character to it.
For gold, 3-plus.
That's one of the noble metals, again, that has a considerable covalent character to it.
So you'll notice that as the coordination number goes up, you get a different radius.
Now, there are then a set of numbers, and these are numbers that are based on different starting points.
The furthest column to the right, quote "IR" quote, this is based on Pauling's starting point.
And Pauling's starting point, if you go way down the list onto the second page, is that for 02 minus for oxygen in six-coordination, the standard radius from which everything else hinges is 1.40.
The Pauling radii had been so commonly used that Shannon and Prewitt decided to use 1.40, Pauling's radii.
So all of their numbers are compatible with the early Pauling radii, except they're based on thousands more crystal structure determinations.
There is another column, second from the right, called CR.
And that stands for Crystal Radii.
And there were two people that proposed that Pauling's radii really didn't give a feel for the proper sizes of the ions.
And they went on at great length to explain what "proper sizes" is.
And the only difference for the sizes that are based on, quote, "crystal radii" is that if you go all the way down to oxygen, the size of the oxygen ion in six-coordination is given by 1.26.
So all of the anions in the crystal radius column are smaller by 1.40 minus 1.26.
I can do that in my head.
That's 0.14 Angstroms.
Run down the list.
They're 1.4 Angstroms smaller for the anions.
And the cations are 0.14 Angstroms larger.
So there are two alternative sets of radii, one based on the traditional Pauling radii, 1.40 for oxygen in six-coordination, and the Fumi-Tosi radii, 1.26, which maybe give a better picture of what the sphere sizes actually, quote, "look like," unquote.
But note the tendencies.
There is a progressive change-- whoever's starting point you use, there's a progressive change in size with coordination number.
And let me give you, if I may, at this point, give you-- again, a little knowledge is a dangerous thing.
One of the structures which I worked on many years ago was a puzzle, and this was the prototype fast ion conductor silver iodide.
And nobody had been able to determine the structure for several reasons.
Nobody had ever made a single crystal of it, and all the diffraction data you had available for the structure were five or six intensities.
And there were about seven parameters necessary to fully describe the structure, so you were out of luck.
We learned how to make single crystals.
But before that-- no, he will go nameless.
He said, OK, let's see what's going on here.
Oh, we'll get the radius of silver.
We know that in silver iodide, the iodines are in body-centered cubic packing.
And we know what the lattice constant is, because we got that from powder diffraction data.
Huh, the silver ion is too big to fit in either the tetrahedral interstitial site or the octahedral interstitial site in silver iodide.
So he said, therefore, the structure has to distort, and it's really tetragonal.
But there are going to be domains.
a tetragonal domain in one orientation here, a tetragonal domain in another orientation here.
So he gets himself a random number generator and takes numbers one, two, and three and puts it on a cubic lattice to determine what the distortion will be.
Said, hey, I got a bunch of ones here and a bunch of threes here.
This crystal is built of domains.
And the reason the silver ions migrate rapidly is that a domain can pop from one mode of distortion to another, and that's essentially going to transport the silver ion from one end of the domain to another.
This guy got two papers out of it.
What he had done was he looked up the radius of silver in Pauling's tables.
And he didn't realize that all of Pauling's original radii were normalized to octahedral six-hole coordination, which makes the radius larger.
And that's why the silvers didn't fit in any of the interstices.
If you take the tetrahedral radius for silver, which is where it actually sits in the structure, it fits as snugly and nicely as you please in the tetrahedral interstitial site in the body-centered cubic arrangement.
So this guy made an idiot of himself because he didn't realize the basis on which these tables had been established.
But he's nameless, so I can call him names.
So anyway, again, a little knowledge can be a dangerous thing.
Moving on, in the few minutes that we have left to devote to crystal chemistry.
Pauling's radii, and even better still now, the Shannon-Prewitt radii, let you predict nearest-neighbor configurations-- coordination numbers with fair degree of confidence.
Pauling-- and this was one of the things that he received the Nobel Prize for-- proposed a set of five rules that are called Pauling's Rules for ionic structure.
The first thing he said was something that we've already stated, a coordination polyhedra of anions is formed about each cation, and the cation-anion distance is determined by the sum of the radii, and the coordination number is determined by radius ratio.
So there are three principles in here.
The two most important ones is that expressing faith in a set of radii to determine interionic distances and the fact that if there are radii that can be assigned to the atoms, you can predict a coordination number.
So that's Pauling's first rule.
Actually, Pauling's first rule was first stated by Goldschmidt.
But to have a complete listing of guiding principles, he included that there and didn't claim it was his own.
But then he does something that is rather interesting.
This is the only one of the rules that is quantitative.
If you had an ionic structure, and you could determine its unit cell, the structure should be electrically neutral if it's composed of ions that are charged.
So therefore, you would not make a structure in which all the cations were up in one corner of the cell, and all the anions were down in the opposite end of the cell.
That would be something that might be electrically neutral, but it would be a high-energy configuration.
So Pauling's second rule is a very cute way of expressing the fact that a structure should be atomistically electrically neutral, as well as microscopically.
And he did that in the following way.
He said, let us look at the coordination number the cation, A, and look at the coordination polyhedron.
And let's define as the bond strength, S, as the ratio of the cation charge plus Q divided by the coordination number of A.
And then he said that in a stable structure, if we look at the anion, the sum of all the bonds donated to-- he visualize the charge being donated to the anion.
The sum of all of the bonds from the same cations or from different cations, if that's the nature of the structure, should be equal to the charge of the anion.
So let me say "cation" here.
So it's a cute way of expressing local electroneutrality.
Actually, that's been superseded by an extension of this general principle.
And I may not have time to discuss it today, but I'll give you a handout next time.
It's not in the notes, either.
This is a relatively recent tool that most people are not familiar with, so I'll return to it briefly next time.
The next rule says, OK, we know how to determine the coordination numbers.
To describe the structure, we should be able to say how these coordination polyhedra fit together.
So looking at the coordination numbers-- well, let me do a three-dimensional case, because that's the realistic situation.
Let's suppose we have coordination number eight, which we often do.
And besides, that's easy to draw.
So he says that in a stable structure, the coordination polyhedra tend to share corners.
And I might say also that they tend to have the line between cation, anion, cation to be a straight line, as opposed to sharing edges.
So for eight-coordination, this might be a pair of cubic coordination polyhedra sharing edges.
And especially corner sharing is more favorable than sharing faces.
It's a very specific statement on how the coordination groups whose geometry is determined by size should fit together in a structure.
And it has a very simple basis, an almost trivial basis.
As the polyhedra of the given size share corners and then edges and then faces, the cations inside of these polyhedra are progressively getting closer and closer together.
And clearly, that increases the repulsive energy.
So the cations are most widely separated if not faces, not edges, but corners tend to be shared.
Puckered a little bit.
Maybe this is not exactly a 180-degree bond angle.
But corner sharing is still favorable.
And rule D is essentially an appendix to that, and it says, and how, if the charge of the cation is high and the coordination number is low, the higher the charge of the cations, the stronger that repulsive interaction.
The smaller the coordination number of the cation, the closer together they become in, let's say, face sharing for tubes as opposed to face sharing for tetrahedra.
The lower the coordination number, the closer together the cations come, regardless of the precise shape of the coordination polyhedra.
So rule D says, and how for high charge and low coordination number.
The last rule, sometimes called the "rule of parsimony," the "rule of stinginess," says that the number of different kinds of atoms in a structure tends to be small.
You can read in two different interpretations of the term "different kinds of cations," and both of them are right.
You could mean different in terms of different chemical species.
In other words, if you take a pot and dump half of the elements in the periodic table into the pot, melt it up, and then cool it down to crystallize the phase, you're not going to incorporate all of those ions into one crystal structure.
They have different sizes.
They have different coordination polyhedra.
The different coordination polyhedra have different sizes.
So to try to fit all of those things together in one structure is just going to cost you a lot of energy.
It's going to be much more efficient to form two or more phases, which have much lower energy, and then pay the price of an interface between them.
So have a chemically complex melt or solution, you're going to get different phases.
You want to incorporate everything into one structure.
The other interpretation of different is in terms of their coordination number and their coordination polyhedra.
If the size is such that a particular cation wants to have tetrahedral coordination, you would expect them all to have tetrahedral coordination and not have a mixture of cubes and tetrahedra and octahedra.
If it works for one coordination through the chemical species, it'll work for them all.
That rule tends to have some, but not terribly many, exceptions.
But there are cases where you find two different coordination numbers.
But very often, that's because of a covalent character.
So there's some discussion and elaboration of each of those rules.
I'm just about out of time, and therefore out of crystal chemistry.
Another entirely different approach to structure is to say that the anions are the big things.
And if the energy of the structure is lowered by taking ions of like charge and getting them as close as possible to the larger ions, say the anions of opposite charge, there is a strategy for making the arrangement.
And I think it's the same strategy that I use when I'm packing a suitcase for a trip.
Take the big things-- the books, the shoes-- and arrange them in the suitcase as densely as possible.
Then I take the little things-- the toothbrush, the socks, and so on-- and I stick them into the interstices that are left.
And this other view of examining structures and interpreting says the same thing.
Take the big things-- not the shoes but the anions-- and arrange them as densely as possible.
And this gets into the question of close packing of spheres.
Then take the little guys, the tiny little cations, and stuff them into the holes that are left in the close-packed arrays.
So another way of interpreting structures and predicting structures is to look at the sort of interstices that exist in a close-packed array.
Regardless of the stacking sequence of spheres, there are always two kinds of interstitial sites, one tetrahedral, one octahedral.
And there are two common, but not exclusive, types of packing arrangements, which you probably know.
There is the cubic close packed array.
There's a little diagram of close-packed sheets here.
And there are two sorts of interstices shown in the figure on the first page of the notes on packing considerations.
There's one set of triangles that points in one orientation that forms a hollow in which you can drop down a second layer of spheres, another triangle that is the closest triangle to the first one that points in the opposite direction.
Each represents a potential location for the next close-packed sheet, but those two different locations are too close together to both be used.
So you have to pick one or another.
And if you label those three different locations A, B, and C, any arrangement of spheres that does not place two letters in succession will give you a close-packed arrangement with exactly the same density.
So ABC AC, AC, ABC, AC, AC is a packing of close-packed spheres that is exactly the same density as AB, AB, which is hexagonal close packed, or ABC, ABC, ABC, which could be shown to be the face-centered cubic structure.
So that is another way of looking at structures.
And specifying the space filling in simple ionic structures is another way of looking at structures.
And at that point, I'll quit.
There are some examples of close-packed structures and derivatives thereof that form the last pictures.
And I'm going to stop and suck in air and mention as one last thought that only five of Pauling's rules are stated here.
There was a sixth Pauling's rule, and that's not commonly listed.
And I'll tell you what it is.
The sixth rule says, massive doses of vitamin C can cure the common cold.
Actually, that was something that Pauling pushed.
And interestingly, about three years before he passed away, Pauling came to MIT and gave a lecture in 6120.
And as you might expect of MIT undergraduates, after he finished his lecture, there were a few questions.
But there were about a dozen students who came up with bottles of vitamin C and asked Pauling to autograph them for them.
And he did so with good humor and thought he had perhaps won a few converts in the process.
So that's it.
Getting silly, so it's time for all of us to go home.
My watch says five after 2:00, so why don't we get started?
This is going to be a strange lecture.
I'm expecting some wise guy in the back to say, all your lectures are rather peculiar!
But I have a phone call from Europe coming in sometime around 2:15, 2:30.
Somebody's coming to visit us next week, and we're going to set up a seminar.
So I don't know exactly when it's going to come in, but I have to be in my office.
So what I will do is adjourn at quarter after, 20 past the hour.
And then so I don't keep you here sitting restively wondering when I'm going to come back, we'll adjourn until 3 o'clock.
So you seem to have been taken longer and longer breaks during intermission.
So I'll give you a full 3/4 of an hour so you can get it out of your systems.
And then, we'll take shorter breaks from here on in.
OK this is going to be pretty much our last lecture on symmetry.
And we will begin, at least in half of the next class on Tuesday, to begin talking about physical properties, and in particular tensor properties.
But I'd like to say a little bit about the derivation of space groups and take a look at how this information is tabulated for you by the kind folks who prepare the international tables.
And there are really very strict parallels between what we did in two dimensions.
And that's the reason why we did it so thoroughly and systematically, except that there are three dimensions.
And because we are taking 32 point groups and putting them into 14 lattices, there are a lot more combinations to be considered.
And there are a lot more of them that are unique.
Also, they're a little bit more difficult to visualize and depict.
So there are some new conventions in preparing a representation of three dimensional symmetries on a sheet of paper or page of a book which, of necessity, has to be two dimensional.
So we'll go over some of those conventions.
But the main reason for going a little bit further is that there is another surprise lurking in there for us as soon as we begin to combine translation in three dimensions with a symmetry element.
But let me go through in some detail the monoclinic space groups.
So we have at our disposal to decorate with symmetry elements two lattices.
And they are a primitive monoclinic lattice with two translations, a and b, that define an oblique net, and a third translation, c, at right angles to that net.
And then, either two choices for a double cell-- and I'll draw these in projection because I can do both of them with one diagram.
Either this is a and b and the extra lattice point is in the center of the cell halfway up from the base.
This would be body centered.
And that's represented by I for innenzentriert.
In German, that's German for body centered.
Or the alternative would be to pick a cell like this, in which case the extra lattice point gets caught in the center of one of the faces.
So this depiction here would be a side centered lattice.
And it could have the extra lattice point in the middle of the face out of which b comes.
And in that case, the symbol for the lattice is B.
Or alternatively, without changing any of the nature of the specialness of the lattice, the extra lattice point could be in the middle of the face out of which a comes.
And that would be an A lattice.
With the first, setting with the c axis unique, there is no c lattice because the oblique base of the cell would not give us anything new if it were centered.
All right, so there are those two lattices.
And then, there are three point groups.
And they were 2, m, and then a combination of a twofold axis perpendicular to a mirror plane.
So it looks as though in principle there are going to be 6 combinations.
In point of fact, there are a lot more because of the surprise that we have in store for us.
Before proceeding further, though, let me ask rhetorically the question.
Which lattice type would you pick as the standard one?
And here, there's a major decision that has to be made.
Should the labels on an axes give a unique symbol for the lattice?
In other words, if you have the double cell for a monoclinic crystal, should you define the cell to make it body centered always, or to make it A centered or B centered always?
Or should the labels be defined by the relative lengths of the axes?
So those are the two possibilities.
And there's no right or wrong way.
You pays your money and you makes your choice.
And on pragmatic grounds, this is the convention that's followed.
And it's purely a pragmatic decision because lattice constants have been determined for literally tens of thousands of materials.
Sometimes, the lattice constants are known and the space group is not.
But it would be terrible if the label that you applied to the axis could be C attached to the longest one, the shortest one, or the intermediate one.
It would be impossible to calculate any database for lattice constants displayed by particular materials.
So there's all this-- this is what the decision was made.
This is the decision that was made many years ago.
And for monoclinic crystals, the labeling of the axes is determined by the c axis being the unique axis.
That is, the axis that is along the twofold axis or perpendicular to the mirror plane.
And then, the labels b and c are defined by magnitude of b greater than the magnitude of a.
So I inherently drew this is in the proper fashion.
So the labels are applied to monoclinic crystals in that fashion.
The price you pay for that is that the symbol for the lattice for a monoclinic crystal that has the double cell could either be A, B, or I depending on the relative lengths of the axes.
If these were the two shortest, it would be a body centered lattice.
On the other hand, if this and this were the shortest pair, then it would be a side centered lattice.
So the symbol for the lattice type will bounce around depending on the dimensions of the translations.
So that's a complication in the space groups.
In the two dimensional plane group, that issue never came up.
We just took B greater than A for the rectangular and the oblique nets.
OK, so let me do quickly a couple of the monoclinic space groups.
Let's take a twofold axis and put it into a primitive monoclinic net.
And this is exactly the same as our plane group, P2, with the twofold axes extended parallel to the c translation.
So we would have twofold axes in all of these orientations.
And that's simply P2, little p in two dimensions, with the twofold axes extended normal to the plane of the plane group.
So this is p2.
This is capital P2.
And that is way we distinguish plane groups from space groups.
So the symbol for the lattice is a capital symbol for the space groups.
So I didn't have to use any new combinations theorems.
I simply say there's a third translation perpendicular to the net.
So everything protrudes out of the plane group into a third dimension.
Let me now combine a twofold axis this a and this b and this c with a lattice that is not the primitive lattice but one of the double cells.
And I'll take it, for convenience, to be the body centered flavor because it's easier to draw.
And the operation that we've added to the lattice, if this is a combination of a twofold axis, is the symbol A pi.
And I know what happens if I combine A pi with a translation that is perpendicular to the rotation axis.
I get a new rotation axis, B pi, that's halfway along the perpendicular part of the translation.
And that's where all of these additional twofold axes came in.
But if a lattice is a body centered lattice, there is now a translation that goes up to the centered lattice point.
So what's that going to be?
What is T followed by A pi followed by T, where T is 1/2 A plus 1/2 of B?
And that is a part that is perpendicular to the axis, and then a third component, c, which is parallel to the axis.
As with all these theorems, what you do is you draw it out once and for all and see what it turns out to be.
So let's do that.
Let's say that this is the translation that goes up to the point 1/2, 1/2, 1/2.
Here is the parallel part-- the perpendicular part, rather.
This is 1/2 of A plus 1/2 of B.
And then, we have a part that's parallel to the rotation operation, A pi.
And that is 1/2 of c. So let's just do it.
Here's my first object number one.
It's right handed.
I'll rotate 180 degrees to get a second one that's right handed.
And then, I will translate it up to here.
And here sits the third one.
It's also right handed.
How do I get from one to three?
It's not really clear how I do that.
And again, I'll draw it in projection because this is looking pretty messy in three dimensions.
So here's my first one.
I rotate 180 degrees to get a second one.
The chirality is all the same.
And then, I translate up to this point.
So if this one is at z and this one is at z, this one here will sit at z plus 1/2.
And this is the third one.
Anybody got any idea?
Can't be reflection.
They're of the same chirality.
It can't be translation because the orientation is different.
Yes, sir?
Then it has to be some form of rotation
Right, has to be.
But how do we get the object up to a different elevation?
Well, we've stumbled headlong over a new type of operation, just as we discovered the glide plane when we started putting mirror planes into a lattice in combination with translation.
The only way I can get from the first to the third is to rotate 180 degrees about exactly the same point that I would have before, namely at 1/2 of the part of the translation that is perpendicular to the operation.
But then before putting it down, I've got to take a second step.
I've got to slide it up by a translation component that is parallel to the axis about which I've rotated.
So to complete my theorem, A alpha followed by a translation that has a perpendicular part plus a parallel part is a new type of operation that involves rotating through 180 degrees.
And this is located at 1/2 of T perpendicular.
But it has a translation component which I'll call generically tau.
And tau is equal to the parallel part of the translation.
So if I separate out this new sort of operation in all of its grandeur, what it does is it takes a first object, rotates us 180 degrees, does not yet put it down, first slides up by the amount tau.
Doing the operation, again rotates it 180 degrees, doesn't yet put it down, slides it up by tau.
Doing it again moves us to another one up here.
So what we've generated is a helical spiral of objects about the central axis about which we're rotating.
And this is a new two step operation that's called a screw axis.
There is a persistent rumor that it derives its name from the effect that it has on 360 quizzes.
But that is just an ugly rumor.
There's nothing to that at all.
So this is a new two step operation that we'll call, in general, A alpha tau.
And what it will do is to generate a screw like distribution of objects.
For those of you who have come in late, since you've been demonstrating a predilection towards longer and longer breaks, we're going to adjourn now and have a 40 minute break so you can get it out of your system.
And we will resume at exactly 3 o'clock sharp.
And we will launch into an elucidation of the nature of screw axes.
And again, for those of you who were not here at the very start of class, the reason is that I have a phone call that had to come in about this time of day and no other time.
So rather than keep you shifting, wondering whether I'm coming back, we'll start at three when I should be done with my business.
OK, so that's it.
You missed all the important discussion of what's going to be on the next quiz, and nobody will tell you because they want to keep class average down.
All right, so sorry for the longer than usual interlude, but we'll start again promptly at 3:00.
Let the minutes of the proceeding show that I re-entered the room at 3:00 and 12 seconds, true to my word.
OK, I wanted to give you one example of a screw axis that you're probably familiar with in everyday life.
That was a telephone pole that I was very familiar with when I was a little kid because we used to love to wait until it got dark and our parents couldn't see us.
And then, we'd climb up the thing.
And wow, you could see all over the whole neighborhood.
I'm surprised nobody said, what?
They had electricity when you were a little kid?
Yes, they did.
But now it's all underground.
So kids, I don't know what they do for recreation these days.
OK, let me pursue this question of screw axes.
We've seen what a 2 sub 1 screw access looks like.
I just erased it.
But it consists of objects of the same chirality extending left and right on either side of the principal axis.
The symbol for a twofold axis is this.
The symbol for a twofold screw axis is the symbol for a twofold access with alternate sides extended like a propeller, a very descriptive symbol.
The thing is rotating around.
And you can think of these as the little ribbons that you used to have on the handle bars of your bike when you were a little kid.
And as you pedaled along, these things fluttered out behind, and it was really cool.
OK, let's move on to the next family of screw axes.
And let me look at an operation A2 pi over 3 with a translation component, tau.
And these things are going to get awfully cumbersome to draw.
So let me use a device-- I wish I could claim credit for it.
But actually, a student in my class one year said, hey, I've got a really cool way to draw these things for you.
Let's imagine that this is a screw axis.
We put a cylinder of paper around the screw.
And then, let's divide up the surface of the paper in terms of end segments.
So if this were a threefold screw axis, these would be 120 degrees segments.
And then, let's draw horizontal lines on the sides of the cylinder.
Then, if we want to see how the screw axis reproduces a pattern from a given motif, you just fill in these boxes.
And then when you're all done, cut the cylinder and you can draw the pattern quite nicely on a two dimensional surface.
So this is what I call the unrolled cylinder device.
Somebody very imaginative invented it.
His name, unfortunately, is lost to history.
But I didn't want to claim credit for it myself.
So what is the pattern of a threefold screw axis going to be?
Let's let the difference between boxes be the translation component, tau.
And so what we will do is take an initial motif, rotate it 120 degrees, slide it up by tau relative to the axis about which we're rotating.
Performing the operation again would involve rotating 120 degrees, sliding up by tau.
Doing it yet get again brings us back down full circle.
So this would be the translational periodicity along the direction of the screw axis.
And this would be the value of tau equal to 1/3 of the translation.
So that is what a threefold screw access would look like.
And the pattern obviously, if we would keep repeating it, would do something like this-- so perfectly interpretable, easy to draw.
But things are actually more interesting and more complicated than this.
We are stating two things about this operation.
We're specifying a translation.
But then the other thing that we're specifying is the translation component tau.
And why should we be constrained to say that tau can only be one nth of the translation?
If we do the operation n times, then doing the operation n times would give us a total translation of n tau.
But there's no reason why that has to be one translation.
Why not two?
Why not three?
Why not four?
So the only real constraint is that n tau has to be some integer, m, times the translation that is parallel to the screw axis.
And this means that the value of tau is not just equal to 1 nth of-- restricted to 1 nth-- of t. Tau can be m/n times T. And that is perfectly compatible.
You do the operation n times.
You're going to be directly above where you started.
That's going to be a translation vector.
But why not two translations, or three translations, or four translations?
So there are infinitely more screw axes than just the n subscript something translations.
So let's look at some of the possibilities.
For a threefold screw axis, tau could be equal to 0T Tau could be equal to 1/3 of T. Tau could be equal to 2/3 of a translation.
Tau could be equal to 3/3 of the translation, 4/3, 5/3, and so on, on and on and on.
We could fill the whole board with possible screw axes having different values of tau.
Now, let me convince you, hopefully easily, that if tau is equal to 3/3 t, tau would be an integral number of translations-- in this case, one.
Down here, 6/3 t would be two translations.
We're already going to have those operations in the pattern when tau was equal to 0T.
So let me show you what I mean if it's not clear.
Here's a trio of objects on either side of the rotation part of the operation.
So what I'm saying is tau would be equal to two translations.
That would, for some perverse reason, defining this as the pitch of the screw.
Is it clear that that is going to be an operation that I already have when I say there's a threefold axis and a translation, T, parallel to that threefold axis?
That's going to give all sorts of screw operations.
But they are going to be integral multiples of the translation that's parallel to the axis.
So the rule is that we can always subtract off one translation from any of these definitions of tau and reduce it to something smaller.
So we can always define tau less than or equal to T without any change or artificial restrictedness on the nature of the pattern.
So that says that for a threefold screw, for a threefold screw axis, there are only three that we should consider.
Tau equal to 0, and that's a pure threefold rotation axis.
tau could be equal to 1/3 of a translation.
And now, we introduce the notation used to designate screws.
If n tau was equal to mT, the symbol for the screw axis is n, the rank of the axis, with m as a subscript.
So the pattern that we just drew here would be designated 3 subscript 1.
And that's says automatically that the value of tau was equal to 1/3 of the translation that's parallel to the axis.
So the only other ones we have to consider besides 3 and 3 sub 1 is 3 sub 2, where tau would be equal to 2/3 of T. Let's see what that looks like by quickly drawing it out again using this dandy little unrolled cylinder device.
Let's let this be my first motif.
And I'll take this as the value of tau, two boxes up.
So I rotate, slide up by tau, rotate, slide up by tau, rotate, slide up by tau, and I'm back up here.
And here is my translation.
tau is equal to 2/3 of a translation.
And my translation, then, should be equal to 3/2 times tau.
But this is supposed to be the translation.
And there's nothing in this box.
OK, this introduces another very important aspect of the patterns produced by screw axes.
And it's sufficiently important we must use first the basic screw operation, A alpha tau and the translation in terms of which we have defined tau.
So use the spiral that you have defined by stating A alpha tau is the basic operation.
But then, don't quit.
You're saying another thing, that tau is a certain fraction of a translation.
And you have to use that translation to generate additional objects in the pattern.
So in this pattern that we've drawn here, this is supposed to be a translation.
So I have to take this one and slide it up by T. I have to take this one and slide it up by T. I have to take this one and slide it down by T. So I'm using a different kind of shading to indicate the ones I produced by A alpha tau and the ones that I produced from that helix by using the translation, T. So this is the final pattern.
It has a basic screw operation that's equal to 2/3 of the translation.
And it has a translation that's 3/2 of tau.
So everything's fine.
Yeah?
For the very top one where you have it colored in, [?
should that stay ?]
outside the box?
Yeah, well, it's supposed to be another row
Shouldn't that be a box lower [INAUDIBLE]?
Yes, it should indeed.
Thank you.
When your nose is right in the middle of the thing, sometimes you don't notice that as readily as somebody in the audience.
OK, so that's the pattern of 3 sub 2.
And now, if we realize that we must use both the basic operation, A alpha tau, and translation to fill things in, it's clear that if we try to have tau greater than the translation, the translations, when we apply them, would give us a screw that had the possibility of being redefined in terms of a shorter tau.
OK, one thing that is apparent if you look at this pattern is that 3 sub 2 produces the same sort of helix but in an enantiomorphic sense.
This one is a spiral that goes this way.
And this one increases when we rotate in a clockwise fashion.
So are they distinct?
Yeah, they really are because you can have both the left handed spiral and the right handed spiral together in the same space group.
I don't remember whether that's true for the threefold screw axes or not.
I'm not sure.
I have to check that.
Yes?
So really [INAUDIBLE] enantiomorphic because they're all left handed.
So it's just minus 1, isn't it?
Good for you.
I was about to make that point.
The sense of the spiral is enantiomorphic in the sense that this one is a right handed spiral.
This one is a left handed spiral.
But they're not truly enantiomorphic patterns because if we start with a right handed motif, this one also has a right handed motif.
So I should qualify that statement, which I was about to do.
It's the sense of the spiral which is enantiomorphic.
This is not to say that one has motifs of one chirality and the other one has to have the opposite chirality
Couldn't you just [INAUDIBLE] being minus 1?
Just [INAUDIBLE] minus 1?
Yeah, I could do that, I could do that.
And remember that we can add or subtract a translation at will from tau.
And if I took a full translation and subtracted it from 3 sub 1, I would get-- I get tau equal to minus 2.
Nope, doesn't work that way
[INAUDIBLE]?
Yeah, it's the same tau but in the opposite sense, yeah.
OK, so what comes out of this is that there are three kinds of axes that involve a 120 degree rotation.
Three, which we can view as 3 sub 0, 3 sub 1, 3 sub 2.
The symbols that are used for them now, we know and love the triangle that represents the locus of a threefold axis.
Four 3 sub 1 and 3 sub 2, what one does is to extend the edges of the triangle.
There's the streamers on the bike handle again.
And if you look down on the pattern and use a right handed spiral, then you extend the streamers, the edges of the triangle, that goes down into the board this way.
And 3 sub 2 would-- if it was in the same pattern-- you would indicate by extending the opposite pair of the opposite set of edges
So would they be different space groups?
[INAUDIBLE]?
Yes, there is a P3 sub 1.
And there is a P3 sub 2.
And they are distinct space groups
So how can you distinguish [INAUDIBLE]?
You would find out where the atoms sit.
And in one case, the spiral would occur in a right handed fashion.
And in the other one, from a left handed fashion.
You can determine the position of the atoms unambiguously.
OK, let's do one that's more interesting where the rank of the rotation part of the operation is higher.
And let's look at fourfold screw axes.
So here, we would consider tau equal 0, tau equals 1/4 of a translation, tau equal to 2/4 of a translation, and tau equal to 3/4 of a translation.
So let's take our cylinder that was formerly wrapped around the rotation access, straighten it out.
And I won't bother to draw the pattern for a fourfold access.
But for a 4 sub1 1 screw axis, this would be tau.
And that would be one, two, three, one quarter of a translation that's parallel to tau.
Or conversely, so T is equal to 4 tau.
Or conversely, how is equal to one quarter of the translation.
So start with a first motif, rotate 90 degrees, slide up by one quarter of the translation.
Rotate and slide, rotate and slide.
Now I've come full circle.
Rotate and slide.
So this, then, is the pattern for the screw axis that I'll label as 4 sub 1 by analogy to what I've done with threefold screw axes.
Over here, let me immediately jump to 4 sub 3.
I'm going to have to be a little more stingy with the spacing of my boxes or I'm going to run out of room.
So if this is 4 sub 3, turns out that tau should be equal to 3/4 of a translation.
And conversely, the translation should equal to 4/3 of tau.
So that's the situation that I would obtain if one, two, three boxes gives me the length of tau and four boxes give me the length of the translation.
OK, so again I'll use different sorts of shading to indicate which way I have gotten this.
I would rotate 90 degrees.
I would jump up one, two, three boxes.
So here's the next one.
Rotate, one, two, three boxes up brings me to here.
Rotate, one, two, three boxes brings me up to here.
Rotating again, sliding up three boxes brings me to here.
OK, so there's the basic helix.
And how is indeed equal to 4/3 of the translation.
But I can't quit yet.
I've used the basic operation, A alpha tau.
Now, I've got to fill in with translation.
And according to my rule, the translation should be equal to 4/3 of-- I'm sorry, this should be the translation is 4/3 of tau.
So this should be the translation running up one, two, three, four boxes.
So that's-- I'll use a different shading here.
That would fill in one here.
One, two, three, four boxes up brings me to here.
One, two, three, four boxes up would bring me to here.
Go down by four-- one, two, three, four-- brings me to here.
And let me use still a different shading here.
One, two, three, four brings me down to here.
Filling in quickly, one, two, three, four.
Another one would sit here.
One, two, three, four, another one would sit here.
One, two, three, four, another one would sit here.
And you can see what is happening here, that 4 sub 3 looks exactly like 4 sub 1 but in the opposite sense.
So it's the same relation as between 3 sub 1 and 3 sub 2.
What do we use as symbols?
This is a 4 sub 1 screw.
And again, if we look down on the pattern from the top, use a right handed rule and extend every edge of the square, that would be the symbol for 4 sub 1.
4 sub 3 is the same sort of spiral, but in the opposite direction.
So we will, with a right handed rule and going between neighboring closest motifs, we would extend this, extend this, and extend this.
So that's the symbol for 4 sub 3.
With that, we come back over to here.
And that is everything except 4 sub 2.
And again, I'll mark up a cylinder that's been split into quadrants, draw reference horizontal lines.
This is 4 sub 2.
And we define this as tau.
Then, the translation, this should be equal to 1/4 of the translation parallel to the axis.
And conversely, the translation should be four halves.
So let's draw the pattern.
This was the first one.
Rotate, slide up by tau.
This is the next one.
Rotate and slide, rotate and slide.
Rotate and slide begin, rotate and slide.
So this turns out to be the nature of the helix that is produced by A pi over 2 tau.
But this is not yet a pattern that has the translational periodicity that I have claimed.
That's supposed to be the translation, twice tau, up to here.
So I would have two slide this one up to here.
I'd have to slide this one up to here.
I'd have to slide this one up to here.
And something's wrong here-- one, two, three, four.
I think I went too far up for one of them.
This one was one, two, three.
This one, I went up too far.
I skipped one.
What did I do wrong?
You did the rotation wrong
Hm?
The two [INAUDIBLE]
OK, let me do this all over again.
Again, I can't see standing right on top of it.
It's not a very good attempt to make it look easy.
Let's split this up into four quadrants.
If this one comes out quite differently, I'll make sure I have it right.
So this is my translation, one, two, three.
This is the translation.
And that should be equal to four halves of tau.
So this, then, is my value of tau.
It should go up two boxes to here.
And that should be equal to 1/2 of T. So I want to rotate, slide up two boxes, rotate, slide up two boxes, rotate, slide up two boxes, rotate, slide up two boxes, rotate, and slide up two boxes.
And now, I have to fill in with translation.
So I will take this one and go up one, two, three, four translations.
That's what I did wrong.
I'll take this one and go down one, two, three, four translations.
Take this one and go up one, two, three, four translations.
Take this one and go one, two, three, four translations.
Take this one and go one, two, three, four translations.
And now, lo and behold, what I have is one, two, three, four.
Curious sort of thing-- I have two motifs at every level.
And they are 180 degrees apart.
So a 4 sub 2 screw axis produces a pattern that looks like this.
Pair of motifs like this 180 degrees apart, pair of motifs skewed by 90 degrees to that pair, another pair translationally equivalent to the first.
But what is the chirality of this spiral?
3 sub 1 was right handed.
4 sub 1 was right handed.
4 sub 3 was left handed.
4 sub 2 was right in the middle.
Is that right handed or left handed?
The answer is it's both.
It's both.
There's one spiral that goes up this way.
And there's another spiral that goes up in the opposite sense.
So it's both.
It's a left handed spiral and a right handed spiral together in the same pattern.
And the pattern for this screw axis, the symbol for this screw axis if it occurs in a pattern, is every other edge of the square extended.
Which do you extend?
Well, it really doesn't matter because it's both left handed and right handed simultaneously.
So you can use either one.
Yes, a couple questions?
Can you explain why you explain you did that shading?
I just wanted to differentiate the ones that I used.
They're all the same.
And where they come from doesn't affect what the basic pattern looks like.
But I tried to make clear which ones I got by using the operation A alpha tau.
And for 4 sub 2, which I just did twice, once correctly and once incorrectly, these with shading were produced by the operation of rotating and sliding up by 1/2 of the translation.
The other ones, the ones for which I used this shading, I filled in by using what the translation was, namely twice tau.
So it's just a way of keeping track of what operations I used to produce everything that's in the pattern.
And for 4 sub 3, I had to use three kinds of shading.
Yes, sir?
Should I [INAUDIBLE] the fact that this was actually two superimposed twofold axis separated by 1/2 T?
Yes, you could read that into it.
And you would be very observant because a 4 sub 2 screw axis contains a twofold rotation as a subgroup
[INAUDIBLE]
And since you were shrewd enough to deduce this-- we ask the question with glide planes when we're dealing with two dimensional plane groups.
Is a glide plane a candidate location for a special position?
In other words, if I move the atom directly onto a glide plane, is there coalescence?
The answer was no.
Suppose I asked the same question now of a screw axis.
Is a screw axis a candidate location for a special position?
The answer is sometimes yes, sometimes no
Well, only if you have [INAUDIBLE]
Yeah, so actually if I move the atom, the first atom, onto the locus of the rotation part of the operation, I get pairwise coalescence.
And instead of getting all of these, I'll have just a string of single atoms separated by half a translation, half as many.
So that's a special position.
I don't get the full number out when I throw one into the space group.
Now, there are an infinite number of screw axes.
My old friend, the saguaro cactus, which has a 19 to maybe 23-fold rotational component to its symmetry, does not, if I look at it carefully, have the same thing on every one of these ribs because the tufts of spines spiral up like this.
So a saguaro cactus, if I take the spines into account, doesn't have 22-fold rotational symmetry.
It has some sort of 22-fold screw axis.
And I've never been determined enough to risk puncturing my finger by going down around a saguaro and seeing if there's some other tuft at the same level and whether it's a 22 subscript 3 or 22 subscript 2 screw axis.
But a saguaro cactus does have a screw axis as its symmetry element.
OK, let us generalize on the basis of what we've done so far.
For any screw axis whatsoever, crystallographic or not, we will have an n sub 1 screw, an n sub 2 screw, where tau is equal to 1 nth of a translation, tau equal to 2 nths of a translation.
And we will go all the way up n sub n minus 1, n sub n minus 2.
And always, regardless of n, the ones at either end tend to be spirals in an enantiomorphic sense.
And n sub 2 and an n sub n minus 2 are spirals.
If you end up with an odd one in the middle, this has no chirality to the spiral.
There's a left handed one and a right handed one.
On the other hand, for something like two and four and six, the ones at opposite ends of the series are enantiomorphs in the sense of the rotation.
For any six-fold screw axis-- and I will pass out the nature of the pattern without taking time to draw it-- 6 sub 1 and 6 sub 5 are enantiomorphous.
6 sub 2 and 6 sub 4 are enantiomorphous.
And 6 sub 3 stands alone.
6 sub 3 consists of a pattern of three objects on a triangle pointing in one sense, three on a triangle pointing in the opposite sense.
This is the value of tau.
And that is equal to 3/6 of the translation.
So no left handed or right handed sense.
6 sub 2 consist of pairs of objects separated by 60 degrees.
6 sub 4, exactly the same pairs, but they go in the opposite sense.
6 sub 1 is just a single spiral going up with atoms at intervals of 1/6 of the translation.
6 sub 5, a spiral in the opposite sense.
OK, with that, let me pass out a sheet that has patterns done in a decent fashion for all of the crystallographic screw axes.
Bear in mind that there are fascinating, interesting, and intriguing patterns for non-crystallographic screw axes.
And let me say a few words about the symbol used in the space group for glide planes.
Any questions on this, by the way?
I don't want to have beat it into the ground.
Yes?
I have a question on symbols.
How do you know which way the arrows point on the triangle or the squares?
Oh, here?
Yeah
OK.
Probably the best thing to do is to look at the patterns that I just passed around if you got one yet.
If you look at 3 sub 1 and look down on it from the top and indicate the sense of rotation that gets you from the one above to the one that's directly below, your hand will curl around in a clockwise fashion.
And the little tails on the symbol for the threefold axis trail out behind the way in which you're rotating.
If you look at 3 sub 2 and look down on it, you would have to, using your right hand, rotate in a counterclockwise sense.
And therefore, the streamers go out again in the opposite direction.
Same for 4 sub 1.
There, every edge of the square is enclosed.
If you look at how you have to rotate to get from the one on top to the one immediately below using your right hand, again, you have to rotate in a clockwise sense.
And therefore, the edges that are elongated trail out in the opposite direction.
And finally, drawn in a decent fashion are 6 sub 1, 6 sub 5, which are spirals in the opposite sense.
6 sub 2, 6 sub 4, pairs of things at each level.
And the level is 1/3 of the translation.
6 sub 3, triangles of objects at intervals that are equal to 1/2 of T, 3/6 of T. OK, does that answer?
That is really a convention of some higher order.
But nevertheless, it is a convention and it is consistent.
Other questions about screw axes?
Yeah?
So in nature, is there a difference between the right handed screw [INAUDIBLE]?
That's like asking is there something like a Coriolis effect in crystals.
You know the Coriolis effect that if you're in the southern hemisphere, the water gurgles down the drain one direction.
If you're in the northern hemisphere, the opposite direction?
There was actually a very famous scientist, what was his name?
I think it was Serret.
Serret, so famous I can't think of his name.
I remembers his name, Serret He developed a very famous theorem in kinetics, and that was thermal migration, migration driven not by a chemical potential gradient but by a temperature gradient.
And it's very difficult to measure.
And only recently have people been testing that theory and making measurements.
He asked this very same question.
And he said, I think it was potassium chlorate that he looked at which has no mirror plane or inversion in it.
So it occurs in left handed and right handed forms.
He asked, is there something like a Coriolis effect that would give you more right handed crystals in the northern hemisphere and more left handed crystals in the southern hemisphere?
So what he'd do, he drew a lot of crystals.
And he started counting.
And he counted, and he counted, and counted.
And he couldn't tell if there was a difference.
And counted, counted, counted.
And still, after lots and lots of counting, still could not get a result that was statistically significant.
And he went nuts.
Sorry to tell you this.
So I think it remains to be seen whether there is indeed a Coriolis effect in crystal shapes.
And that's true.
It's tragic.
We're all laughing at the poor guy.
But he actually just couldn't stand it.
He counted so many crystals, poof!
That's a true, unfortunate story.
No, I don't think there is any difference between left handed crystals or right handed crystals.
Remember my story about sugar and its chirality and the little bugs that can only eat one chirality.
Other questions of the less frivolous nature?
That was not frivolous.
That was a good question.
OK, well let's say a little bit about glide planes and how they appear in space groups.
Things were very easy in two dimensions because the glide plane was a glide line.
And everything had to be confined to a two dimensional space.
So we used a solid line to indicate a mirror plane, the locus of the operation, sigma.
And we use a dashed line to indicate the locus of an operation, sigma tau.
And tau was in the direction of the dashes.
Well, I've given it all away with the little diagram that I passed around.
Things are much more complicated in three dimensions.
If this is the locus of a glide plane and this is the direction of tau in a space group, we could be looking at that glide plane in a total of four different ways.
We could be looking at the glide plane from the side, edge on, and perpendicular to tau.
We could be looking at the glide plane edge on and along the direction of tau.
Or we could be between these two orientations and looking at the glide plane edge on but in between normal to tau and parallel to tau.
And finally, we could look at the glide plane down from the top.
If we view the glide plane from above, for example suppose we have a monoclinic crystal and we put a glide plane in the base of the cell which has tau in this direction, the way you indicate that is with a little chevron off to the side.
And if this is the direction of tau, you put a little barb on the chevron that indicates the direction of tau.
So the three possibilities for a monoclinic crystal is that how could be this way, in which case you use a chevron like this.
Or how could be in this direction.
And then, you use a chevron with a barb attached like this.
Or how could be a diagonal glide, in which case the atoms would be up and opposite chirality and down and back up again.
In that case, you add a little barb in between the two directions to indicate the direction of tau.
We put some labels on this, this a and this b.
This situation would be a tau that is equal to 1/2 of b.
In this case, how would be equal to 1/2 of a.
In this case, how would be equal to 1/2 of a plus b.
This is designated as a b glide.
And rather than having the symbol n appear in the symbol for the space group, a b would appear.
This is an a glide.
And the a would appear in the symbol.
And this is a diagonal glide.
And that, for reasons that I have never found anyone able to explain to me, is called an n glide.
I know of no language in which the word for diagonal begins with an n. But that's what it's called.
Now, those same glides can be viewed from directions that are parallel to the plane of the glide plane.
Looking normal to tau is exactly the same situation we had in plane groups.
So one uses the dash.
If you're looking along the glide plane end on but in the direction of tau, you use a dotted line.
It's easy to keep straight which is which.
If there's a little arrow in there and you look at it from the side, you see a line segment.
If you look at it from the end, you see a dot.
So it's easy to remember this convention.
If you're neither perpendicular to nor parallel to the glide plane, you just mix the two symbols.
So you use a dashed dot line.
So those are the symbols for glide.
Glide plane is something like a piece of wood.
It's got grain to it.
The direction of the grain is the direction of tau, if you'd like.
All right, in the hand out, I give you some examples of the way in which a pattern of objects that has glide in it would be interpreted in terms of a geometric symbol.
At the bottom right of the page, the left hand diagram has atoms related by a glide plane in the base of the cell where tau is running left to right.
And so you have a chevron with an arrow on the right hand branch.
If you have atoms alternately left handed, right handed at a level and 1/2 plus that elevation, then you're looking at a glide plane edge on and down along the direction of tau.
So you would indicate the locus of that glide plane by a series of dots.
What other type of glide plane is possible, not for monoclinic, but for lattices that have a centered lattice point in the middle of the cell?
So this takes a special type of Bravais lattice.
And this would be a lattice, for example, that had lattice points at the corners of the cell and another lattice point in the middle of the cell.
One face of the cell is centered.
Then, you have not only translations a, b, and a plus b, which can serve as directions for glide.
But you have another translation that is not present in the primitive lattice.
And this is 1/2 of a plus 1/2 of b.
And if that's a translation, you could have a glide plane parallel to the base of the cell and have tau equal to 1/2 of 1/2 of a plus b, namely have this little vector in here be tau.
And the symbol for that kind of glide is d. And I do know what that stands for.
That stands for a diamond glide because diamond, one of the very simple structures for an element or any compound and one of the early structures to be determined experimentally, does have a d glide in it.
So the name of the compound, diamond, gave its name to the type of glide plane.
OK, I've got still 10 more minutes.
What I would like to do next is to look at some of the other monoclinic space groups and also examine the way in which this information is presented for you in the international tables.
So I've got a big, fat pack of material that shows you all of the monoclinic space group, that is the space groups derived from point group two, the space groups with point group m, and the space groups with point group 2/m.
And we've done a few of these.
But what we haven't looked at is the Shcoenflies notation for the result or the way in which this information is presented in the international tables.
To introduce you to these gradually, this is the monoclinic space group-- these are the monoclinic space groups as presented in the old international tables for x-ray crystallography.
A little bit later, I will pass out to you a few higher symmetry space groups that are represented in both the old international tables and the new international tables.
And as I indicated somewhat disparagingly earlier on, there is so much information in the new version that they're very, very hard to use.
But OK, if you open the hand out, you find space group P2.
So that's a twofold axis added to a primitive monoclinic axis.
You see on the left hand page this situation, this ridiculous situation, that exists only for monoclinic crystals the first setting, where the axes are a on the left hand side running down, and b on the top running from left to right, and c coming straight up out of the page.
And the second setting, where the unique axis is b, and that means if you're going to draw the space group with c coming up, a down, and b to the right, b now is the direction of the twofold axis.
So the diagram looks completely different.
But it's the same space group except that the first setting is tilted on its side so that a comes down, b is from left to right, and c, which is now not a direction of the twofold axis, comes out.
So there's some jargon here, that geometric jargon.
The atoms occur at different elevations.
So in these two cases, the atoms occur only at elevations plus z or minus z.
And you see next to the little circles that represent the symmetry equivalent positions for the general position a plus.
And that means elevation plus z.
And for the first setting, all the atoms occur at an elevation plus z for the general position.
For the second setting in the right hand diagram, you're looking at the twofold axis from the side.
And therefore, one atom is at plus z.
The one that's related by symmetry gets rotated down to minus z.
So you see a little plus next to one atom, a little minus next to the next to it.
We need a symbol for a view down along the locus of a twofold axis.
And that's the little pointed oval that we became familiar with with the plane groups.
When you're looking at the twofold axis from the side, how are you going to indicate that?
Well, the convention is to use an arrow.
And that's what you see on the right hand side for the second setting of P2.
If this were not a twofold axis but a 2 sub 1 screw axis, it would be a one-sided barb.
So looking down at these two axes with rotations of 180 degrees, this is the symbol for two viewed from the side.
And a one-sided barb would be the symbol for a 2 sub 1 screw axis.
Mercifully, these are the only screw axes that need to be depicted in a view from the side.
So there's no standard symbol for a sixfold access or a fourfold axis looked at from the side.
In boldface, in the outer corner of these pages, you see the international symbol, which is what we used for the plane groups, the symbol for the lattice type, primitive, except now we use uppercase symbols for the lattice type to distinguish the space groups from the plane groups.
Underneath it, you see a C2 superscript 1.
This is the Schoenflies symbol.
You'll recognize C2 as the Schoenflies symbol for a twofold axis.
Schoenflies' symbol for the space group was to add a superscript, namely the first one that old [?
Artur ?]
got, the second one that [?
Artur ?]
got from this point group, the third one he got from the point group, and so on.
So it's pretty much an arbitrary order, except he starts with, as you'll see, a twofold access, then replaces the twofold axis with a screw access.
So if you turn to the next page, you see symbols for not twofold axes, but 2 sub 1 screw axes along the edges of the cell.
You see now a different sort of symbol alongside the atoms.
There's one at plus z.
That's the representative one at x, y, z.
If you repeat it by a screw rotation, you rotate it 180 degrees and slide it up by 1/2 of [?
c.
?]
So you see the symbol 1/2 plus.
So plus is plus z.
1/2 plus is 1/2 plus z.
Over to the second settings, trying not to sneer too vigorously, you'll see pointed barbs.
That's the symbol for a twofold axis viewed from the side.
And then, one of them is plus.
The one that's a little long parallel to the 2 sub 1 screw axis goes to the other side and goes from plus z down to the minus z.
Let me ask you a question which people usually don't think about.
If you look at the first setting and ask, what are the lengths of the translations?
That's a.
And that's so many angstroms.
And that's b.
That's so many angstroms.
If I ask you that for the second setting, the answer is a little trickier.
What is the length of this translation?
That's b.
What is the length of this translation?
I'm sorry, that's c. And this translation is a, right?
Wrong.
That is no longer a.
And the reason for that is that when you have a cell with non-orthogonal angles in it, let me indicate for the most general case for a triclinic crystal.
So this is a.
This is b.
And then the third translation normal to that plane is c. And if we have a structure with atoms in these locations and we want to project the structure along c, you do exactly that.
You don't plop the atoms down onto the base of the cell when you project it because if you did so, the atom that is up in the neighboring unit cell would come down to a different location.
And you would not have a pattern that was periodic based on a lattice.
You get that only if, when you're projecting in an oblique lattice, you project along the translation.
And then, and only then, can you end up with a pattern that looks like it has translational periodicity and really does.
So if a cell is oblique, you do not project just by plopping the atoms down on the base of the cell.
You project them parallel to the translation that extends up above the direction onto which you're projecting.
So if our cell in the second setting is monoclinic and this is a and this b and c comes up, if we look at that cell from the side as is done in the depiction on the right hand, then we let this be the direction of b.
So that is b.
But the thing that is at right angles to b is going to be a times the cosine of beta.
And it's not a itself.
So this is a right angle.
But this is not the lattice translation.
It is the lattice translation times a trigonometric function.
The triclinic case is a hairy beast.
To calculate the length of these axes and projection, it's clear what they are.
But to get these values involves alpha and beta and gamma.
And it's a very, very complicated function
So the angle between a and c in general is not 90 degrees in monoclinics?
The angle between depends on how you label your axes.
For monoclinic, if you do the first setting, then by definition this is a, and this is b, and this is c. So these are always by definition 90 degrees.
If you use the second setting, then this is the direction of b.
And this is the direction of a.
And this is the direction of c. That keeps the system right handed.
Then, this is a right angle and this is a general angle
But all you're doing is taking that top one and just [?
tilting it over ?]
No, I'm doing more than that because if there were a twofold axis, that would be the direction of the twofold axis.
Now, this is the direction of the twofold axis.
The other thing that happens is the symbols for the glides change.
So if you look at space group number seven, which is a case where the mirror plane has been replaced by a b glide in the first setting, and that means the glide is perpendicular to the twofold axis.
If the twofold axis is now the direction of b, the glide plane turns into a b glide.
But let's just thumb through them quickly.
And then, we're actually two minutes past my promised quitting hour.
First one, P2.
Then, replace the twofold access by a 2 sub 1 screw.
So that's Schoenflies symbol C2 superscript 2.
Then, do this with centered lattices.
The international table chooses to use a side centered cell.
So that's the third one you can get from a twofold axis.
So Schoenflies calls it C subscript 2 superscript 3.
And if you add that to a side centered lattice, you get screw axes interleaved between the twofold axes.
Put it on its side, it stays C2 superscript 3 in the Schoenflies notation.
But the side centered d become side centered c because it's b that comes out of the oblique net.
So the international symbol tells you exactly what you have.
The Schoenflies symbol is arbitrary.
It tells you the point group and it tells you the order in which Schoenflies derived them.
But that non-informative nature to the symbol has a blessing.
And that is it doesn't change with different labellings of a, b, and c. It doesn't change with a being shortest, b being shortest.
It just sits there and its dumb, uninformative fashion.
And that is nice.
So both of them have survived.
OK, so could we get another space group out of this by replacing the twofold axis by a 2 sub 1?
No, because we've already got 2 sub 1 screw axes in C2 superscript 3.
So then, Schoenflies moves on and begins to work with symmetry planes.
So number six puts a mirror plane in the primitive monoclinic lattice.
That becomes CS.
That's the Schoenflies symbol for point group m. And it's the first thing you can get.
Another piece of convention for displaying the representative pattern of atoms, in the first setting, the two atoms sit directly above one another.
So there are two new things here.
A vertical line through the atom indicates that there are two of them that are superimposed in projection.
And moreover, the little tadpole has appeared inside the frog eggs in the left hand part.
So this says on the right hand side of that split circle, the atom is at plus z.
On the left hand side with the comment to indicate an enantiomorph is 1 at minus z that sits directly below the first one.
And you'll notice in the diagram of the symmetry elements is a chevron working off the lower right hand corner of the cell.
And that's the symbol for a mirror plane that's being viewed directly from above.
And then I'll just go a couple more.
You could replace the mirror plane with a glide plane.
And that's a b glide.
Now, you can see the atoms going up at plus z, slide along by 1/2 z, and popped down to minus z, going down to an enantiomorph, so there's a comma inside the circle.
The chevron in the lower right hand corner of the diagram of symmetry elements now has acquired an arrow, a barb, that indicates the direction of tau.
The second setting, again, flops the thing on its side so that you're looking now down along the b glide.
And so you see a dotted line that indicates a glide plane viewed edge on in the direction of tau.
Then, you can put a mirror plane for number eight into a side centered b lattice.
And that gives you another space group.
And then, you can put a glide plane into the b lattice.
And that, interestingly, gives you a different space group.
Number eight has an alternating mirror plane with an axial glide.
Number nine has an alternating axial glide alternating with a diagonal glide-- two glide planes.
Not a mirror plane and a glide, but two different kinds of glide planes.
And then, you've used up all the tricks you can pull there.
So now, we take the last monoclinic point group, 2/m, drop it into a primitive lattice.
Replace the twofold axis with a 2 sub 1 screw axis.
That gives you P2 sub 1 over m. Then, leave the twofold axis alone and leave the plane alone and put 2/m in a side centered cell.
And then, replace the mirror plane by a glide.
That's number 13.
And then finally, the summa cum ultra, replace the twofold axis y a 2 sub 1 screw and replace the mirror plane with a glide, number 14.
And then, do the same thing in a centered lattice.
So we get a total of 13 different space groups from two lousy kinds of lattices and three possible point groups.
You get 13 different distinct space groups.
OK, that's a quick overview.
Notice that just as was the case for the plane groups, the tables go on to list the way in which atoms are related by symmetry and the number per cell and the site symmetry just as in the plane groups.
And the way a three dimensional structure is going to be described to you is exactly analogous to what we did with two dimensions.
Magnesium in position 3b, symmetry 6/m, oxygen in position 12d, symmetry 1.
And then, the tables give you the coordinates of all the symmetry S atoms.
OK, thank you for your patience.
That was a long stretch in one shot.
I'll say a little bit, very little, next Tuesday about the convention for orthorhombic space groups because there, no one direction is any more or less special than any other.
So the symbol for the space group changes all over the place when the axes take on different lengths, relative lengths.
But the symmetry is exactly the same.
So we'll see you--
All right, we've been slogging our way through derivation of the plane groups.
And I think I'll do a few more, because we'll stumble across some major tricks in deriving a subfamily of them.
But to not get lost in the forest because of all the trees, I have a set of notes.
They are handwritten because my secretary would resign if she had to fit in all these figures and subscripts and strange symbols.
So, they are as neat as I could make them.
Sorry to say that, in running them through the Xerox machine in an attempt to get everything on one sheet, some of the last lines got clipped.
So I'll run these through again and give you a copy that's minus those truncations.
All right.
What we've been doing so far, to have a brief reprise, was to take the symmetries, the 10 two-dimensional plane group symmetries.
And they were one-, two-, three-, four-, or sixfold axes, a mirror plane, 2mm, 3m, 4mm, 6mm.
So there are 10 of them.
And these are the so-called crystallographic point groups.
Crystallographic because we considered only those rotation axes that are compatible with a lattice.
And they are point groups because they leave at least one point in space, invariant-- stays there rigidly fixed.
And they are groups because the collection of operations that are present satisfies the postulates of the mathematical entity called a group.
We've then are in the process of taking these 10 symmetries and adding them to the 5 two-dimensional lattices.
The parallelogram net, the rectangular, the centered rectangular, and the square, and the hexagonal.
And clearly, we can't put each of those point groups in every one of the lattices.
For example, the lattice has to be square for either 4 or 4mm, so we would attempt to place only those two of the 10 point groups into a square net.
We've gone through quite a few of them.
I won't bother to draw the pattern of the symmetry elements.
But we put no symmetry at all in the general parallelogram net.
And that is plane group P1.
Maybe I will draw the figures, after all.
P2 was a twofold axis dropped into a net, which had to have no specialized shape, simply because a twofold [INAUDIBLE] axis requires nothing but a lattice row if one translation is combined with it.
The threefold axis could fit into an equilateral net.
And there's the threefold axis we added to a lattice point.
And we have two additional threefold axes in the centers of those 2 equilateral triangles, which each make up half the cell.
P4 was the square net.
We put a fourfold axis at the corner of the cell.
We've got another one in the middle.
Two folds in the middle of the edges because of the 180-degree rotation that is built into the fourfold axis.
And P6.
We put that into a hexagonal net.
We've got sixfold axes that we dropped in at the lattice point only, nothing else.
There's a 120-degree rotation at the lattice point, and that gives us the threefold axes that are present in P3, as well.
There's a 180-degree rotation contained in a sixfold axis, and that gives us the twofold axes in all of the locations of P2.
OK, I haven't drawn in any representative patterns, but let me remind you again.
The pattern that is characteristic of every one of the plane groups is just the pattern that the point group that you've placed at the lattice point would produce.
And that pattern is, in turn, hung at every lattice point of the two-dimensional cell.
So even though the huge number of symmetry elements that's present in the higher symmetry point groups is rather intimidating-- you say, wow, how would one draw a pattern for that-- the pattern is nothing more complicated than the pattern of the symmetry that you've placed at the lattice point.
And all of these other symmetry elements arise to express relations between the motifs that you've placed at the initial representative lattice point.
Deriving these, we used one theorem, which it's well to remind you of.
And that is that if I have a rotation operation A alpha, follow that by a translation that's perpendicular to the rotation axis, what I get is a new rotation operation is B alpha.
And it's located in a very specific location.
And that is at a location that's a distance x equal half the magnitude of the translation times the cotangent of alpha over 2.
So this combination term, if nothing else, reminds us that these combination theorems, as I call them, are not equations in symmetry elements.
They are equations in individual operations.
So, for example, if I combine the fourfold axis with the translation, the 180-degree operation that's present in a fourfold axis puts B pi here.
The 90-degree rotation puts B pi over 2 here.
And the 270-degree rotation-- which I can define just as well as A minus pi over 2-- puts the operation A minus 2 pi over 2 down here.
So it's not an equation in symmetry elements.
It's an equation in individual operations.
Then some peculiarities started to arise, which we perhaps might not think of.
If we put a mirror plane in a primitive rectangular net, that gave a group that we called Pm.
If we combine that with a translation, we need a theorem that says what happens if you combine a mirror reflection operation with a translation that's perpendicular to it.
We sketch that out, and once and for all could decide that it's going to be a new reflection operation that's located halfway along this perpendicular translation.
So this reflection operation sigma, combined with this translation, gave us a new reflection operation, sigma prime, halfway along the cell.
And, of course, we have this one hanging at a lattice point as well.
And then came the interesting one.
When we combined a mirror plane now with the centered rectangular net, we have all of the mirror planes that we have here, because this primitive rectangular lattice is a subgroup of the centered lattice.
Then the interesting thing happened when we combined the reflection operation sigma with this translation that went to the center of the cell.
And that gave us a transformation that was something we had not encountered before.
We took an object, reflected it in this plane, and then translated it down to a centered lattice point over here-- to give us one that sat here-- and then asked, how did I get from the first one, a right-handed one perhaps, to a second one that's a left-handed one, and then a third left-handed one that sits down here?
The answer is that we found there was no way we could specify getting from number 1 to number 3 in a single shot.
We had to take two steps to do it.
And there was nothing more simple than that.
We had to first translate down by the part of the translation-- let me call it tau-- which is equal to that part of the translation t, which is parallel to the reflection operation.
And then we had to reflect across, and that would get us from the first to the third.
That was a new sort of operation.
We'd indicate its locus by a dashed line to distinguish it from a mirror plane.
And it has a translation part and a reflection part.
Doesn't matter what order in which we do them.
We get to the same location if we reflect first and then slide or slide first and then reflect.
So this was a new operation that I'll represent by sigma tau-- looks sort of like a symbol for reflection, but the subscript reminds us you've got to translate by an amount tau that is parallel to the initial mirror plane.
And this gave us a new theorem that a general translation that had a part t perpendicular and a component t parallel, when it followed a reflection operation, was equal to a net effect of reproducing the object by a glide plane, sigma tau prime.
It had tau equal to the part of the translation that was parallel to the reflection part of the locus.
And it was located always at one-half the perpendicular part of the translation.
So using that theorem and completing the mirror planes that hang at the lattice points, we have a very interesting group that consists of a centered lattice, mirror plane hung at the corner lattice point-- this is also a lattice point, so we automatically get the mirror plane in the middle of the cell-- and then in between, we get this new 2-step operation, the glide plane.
And the pattern that's representative of this plane group is, as advertised, just what a mirror plane does.
And that is hung at every lattice point of the centered rectangular net.
So the glide plane, this new operation that popped up, does what the new operations did in all of the other preceding groups.
It tells you how do you get from the pair that you've hung at the lattice point to these that sit in the center of the cell.
They're related by translation, but the relation of one to another of these motifs is by the glide plane.
So that is an operation which has arisen in the group.
And this, then, is a group that we would call Cm.
C for a centered rectangular net, as opposed to the primitive net, which is the one we got immediately before.
Again, I sound like a cracked record sometimes, but let me emphasize the simplicity of these patterns.
Again, the pattern consists of what you would get when you hung a motif on one lattice point, and then that is repeated by m, which is the symmetry you've placed at a lattice point.
And that's hung at every lattice point of the centered rectangular net.
The glide planes just express relations between things that you already have when you've hung the motifs on the lattice point.
Any questions after this brief reprieve?
Comments?
Yes, sir
You called those [INAUDIBLE] glides, that's a sigma tau?
Yeah.
The relation between this one and this one, I've called sigma.
That's the reflection operation.
The relation between this one and this one down here would be the operation that has a reflection part and a translation part, tau.
Let me point out something that's worth observing when we start making some more complex additions.
We said early on that we only have to consider translations that terminate within the unit cell.
Because everything is translationally periodic-- not only the atoms in the motifs but the symmetry elements as well.
But the observation that I want to make is that if we put a glide plane in.
And let's do that for the direct addition of a glide plane to a lattice point.
And having done that, we have the potential of possibly having derived a group that we would call Cg.
OK.
This diagonal translation-- this is T1, and this is T2.
We might ask, what is the reflect [INAUDIBLE] glide operation that sits at the origin lattice point, followed by T1 plus T2.
Well, what does our theorem tell us?
It says that we should get a new glide plane.
It should be at one-half of the perpendicular part of the translation.
And the perpendicular part of the translation is T2.
So this is at one-half of T2, and it should have a glide component equal to the parallel part of the translation, and that is T1.
T1 plus the original glide component, tau.
What is this?
A glide component of half of T1 plus the entire T1.
If we ask, does that make sense?
Yeah, we would glide down to here and then we would translate down to here.
And that would give us sitting at half of a path of T1, a new object that sat here.
Is that a glide operation?
Yeah, it is.
But it is one that is not really distinct from the glide plane that sits here.
Because if we have a glide operation with tau equal to one-half of T1, and if this sits in a lattice, then there's going to be a glide plane has tau equal to 1 plus one-half of T1.
And there will have to be another glide operation that consists of 2 plus one-half of T1.
And the reason is that everything is periodic at an interval T1.
So the moral that I'm trying to draw here is that one can add or subtract to identify the actual nature of a symmetry.
One can add or subtract an integral number of translations.
And that permits one to reduce any tau to a sigma tau prime, such that tau prime is always less than the translation that's parallel to the tau.
In other words, lop off this translation of an entire T1, this translation of the entire T2, and you have identified the basic nature of the glide operation that sits here as something with a translation that's half of T1.
The translations that move motifs out of the cell may be related by a glide operation that involves an integral number of T1s plus half of T1.
But it doesn't change the basic nature of the simplest glide step that's in there.
That's a very obscure explanation of probably something that didn't puzzle you in the first place, but it's worth saying when we make some of these additions.
Let's finish this off.
We have another translation in here, and that's the translation T1 plus T2 over 2.
So, what would we get if we took the glide operation sigma one-half of T1 and followed that by a translation T1 plus T2 over 2.
And that operation, again, would be equal to a glide plane sigma prime with the tau equal to the original glide component plus the part of the translation that is parallel to the glide plane.
And it'd be located at one-half of the perpendicular part of the translation, which is T2.
So it'd be at one-half of one-half of T2.
So this says that the combination of the glide operation with the centered translation is a glide plane with a glide component T1.
And it's located at one-quarter of T2.
And that, indeed, is what you would do if you reflected and reproduced the object by a glide down to here and then translated by T2 over to here.
OK. No, I'm sorry.
We glided and then we added on the parallel part of the translation.
So we would end up down here.
And if one sits here, there has to be one repeated by T1 up here.
And this is exactly the same thing as a reflection plane that's been introduced.
So here is a case where we could subtract off the entire translation T1 and say this is identical to a mirror plane passing through the origin, a pure mirror plane sigma that is at one-quarter of T2.
So, let me clean this up and show you what we have.
Completing the operations, we would have a pair of objects related by glide, like this, that is hung at every lattice point.
It's also hung at the centered lattice point.
And what that is going to give us is a pair of objects that sit like this.
And what has come in as a result of those combinations is a mirror plane interleaved between the glide planes.
And this is exactly what we have in Cm, which was also interleaved mirror planes and glide planes with the origin shifted by one-half one-quarter of T2.
So this is not in a group.
Proceeding logically, we'd take Cm and replace the mirror plane by a glide.
When we do that, we have a consistent group.
But it turns out to be exactly the same arrangement of symmetry elements and exactly the same pattern as Cm, but with a little nudge over along T2 by one-quarter of that translation.
Yes
But in Cm, didn't we have mirror planes going through the lattice points?
Yeah.
And what I'm saying is here there are glide planes going through, but the lattice point is arbitrary.
I can put it anywhere I like.
And if I decide to put the lattice point here, that turns it into Cm.
OK?
So, let's put the two of them side by side.
Cm was this mirror plane, mirror plane, mirror plane glides, and the atoms motifs did this.
A lattice point here in the center.
And I'll deliberately, to make my case, offset the cell by one-quarter of T1.
Now I've got a glide plane here, same as this one.
Mirror plane at a quarter of T1.
Glide plane here at the centered lattice point.
Mirror plane here, glide point T2 away.
And here are the lattice points.
But the pattern of objects looks like this.
Armed in advance on what the thing has to look like.
Looks like this.
And this is exactly the same pattern of motifs.
OK.
So, coming out of this consideration, we have with the rectangular nets Pm, Cm, Pg.
But Cg, if we try to construct it, was the same as Cm.
And that exhausts the possibilities for a single symmetry plane and the rectangular nets.
OK. Let me move on then to the next step.
And I'm going to skip over a threefold axis.
And I'm going to look at the square net combined with the other symmetry that would require a net with this dimensionality.
And this would be a primitive square lattice plus 4 mm.
We've already done almost all the work.
So as we derive the symmetries that are subgroups of these higher symmetries, we've done P4.
And that says that if we put a fourfold access at the origin, that fourfold axis, we'll get another one in the center of the cell.
And we'll get twofold axes in the midpoints of all the edges of the cell.
And then 4 mm has one kind of mirror plane.
Two Ms because there are two kinds of mirror planes.
The mirror plane says, hey, if I'm in a lattice, I want to be at right angles to a rectangular or a centered rectangular net.
Well, OK.
This one is happy, because a square net is a special case of a rectangular net.
The two translations are merely equal.
So he is happy.
Combine that with T2, and we'll get another mirror plane like this, and another mirror plane like this.
Fourfold axis is going to rotate that mirror plane, so we'll have mirror planes running this way, and this way as well.
And now we have a different kind of mirror plane that we tried to put in in this location.
This mirror plane says, hey, I have to be parallel to the edge of a rectangular net or a centered rectangular net.
So if we look at the translations that are parallel to and perpendicular to this translation, lo and behold, this mirror plane is aligned along the edge of a centered rectangular net that has the additional specialization of being a centered square.
But that mirror plane now is perfectly happy.
And he says, OK, I'll hold my piece.
I have the arrangements of translations relative to my orientation that makes me happy.
So I can say now that there is a mirror plane running this way.
There has to be another mirror plane 90 degrees away from those orientations.
And this, then, is going to be the location of all the mirror planes in that net.
And at no time did the chalk ever leave my fingers.
Just doing what we said we're doing.
Putting in 4 mm, making sure the requirements imposed on the lattice are those that that symmetry element demands.
And here we go.
Except that there is now one other combination that we have not considered.
Here is a mirror plane.
And now this mirror plane is diagonal to our translation T2.
And here is a mirror plane.
And that mirror plane is diagonal to our translation T1.
So what is that going to require?
Here is the mirror plane, that's the operation sigma.
And here is our translation, T1, down to here.
We have a theorem that says that a reflection followed by a translation that has a general orientation is going to give me a new reflection operation, sigma prime, that's located at one-half the perpendicular part of the translation.
And it's going to have a glide component that is equal to one-half of the parallel part of the translation.
So what has that told us?
To get from this mirror plane down to here, we've gone this far in a sense that is perpendicular to the mirror plane, so this is T perpendicular.
And we have a part of the translation that is parallel to the mirror plane.
This is T parallel.
So the plane that results is going to be a glide plane in here.
It's located at one-half of the perpendicular part of the translation.
And that is one-half of one-half of T1 plus T2.
And it has a glide component equal to T parallel, which is equal to one-half of T1 plus T2.
So we get a new glide plane in here.
And that will require glide planes through similar arguments that go down the diagonals of the cell, like this.
And let me convince you that that, in fact, is an operation that must arise if I place, at the origin of the cell, a set of objects that have 4 mm symmetry.
We have one like this, one like this, one like this.
Another pair hanging here.
Another pair hanging here.
And if we do a reflection operation, let's say this pair up to this pair, and then slide it down by the diagonal translation-- and we have the same set, again, hanging down here at the diagonally opposed lattice point.
Reflect across, slide down to here.
The way in which you get that is, believe it or not, to reflect across this glide plane.
How do we get there?
We reflect across.
We translate down to here.
And the way I do that is by reflecting across this diagonal glide.
That probably has convinced no one.
But map it out with your own pattern, and I think you'll agree.
Let me observe, while you're considering that.
When we derive groups based on 3m or 6mm, we're going to have other cases where a translation along the edge of the cell is inclined to a mirror plane.
And the effect of combining a translation with a mirror plane that's inclined to that translation always has the effect of interposing a glide plane halfway in between the mirror planes that are related by translation.
So, let me say that in general, because that will be an observation that'll let us identify glide planes quickly.
So the general resolve is if we have a translation inclined to a mirror plane, which of necessity is repeated by translation, a quite general result is that a glide plane is interleaved always between the two.
And they'll be parallel because they're related by translation.
So we'll see some more cases where we can immediately state that without further thought.
OK.
There is something that we might consider doing.
I'd like to put it off, though.
Here we start with a mirror plane.
Can we replace the mirror plane with a glide plane?
The answer is yes, we can.
But it's not at all clear if I take a fourfold axis and put a glide plane through it, what this plane has to be.
So I'd like to leave this one for now and come back to that.
OK.
In the notes, I've tried to do all of these derivations thoroughly and logically.
So if this is a bit fast or me waving my hands, saying a glide plane is here, and this, that, and the other thing, when you can't really see what I'm pointing at, I think if you refer to the notes that'll be clear.
But yes, you had a question?
So you said that this glide plane [INAUDIBLE] one-half T perpendicular.
What did you write underneath that?
I guess that maybe is where your T and one-half T perpendicular is where the glide plane is
OK.
I was just demonstrating that, in fact, for this mirror plane with the diagonal translation, the glide plane would come in at one-half of the perpendicular part of the translation.
And the part of the translation that-- we were combining this fourfold access with this translation.
So the perpendicular part of the translation is half of the body diagonal.
And if a glide plane comes in one-half of the perpendicular part of the translation, that put it parallel to the mirror plane and at one-quarter of the way of the translation.
So this little scrawl here says it one-half of T perpendicular and one-half of one-half of the body diagonal.
So it's one-quarter.
All this equals one-quarter of T1 plus T2.
OK. Now something unexpected happens.
If we would move along, I skipped over 3m.
We know what P3 looks like.
And that is a hexagonal net, an equilateral net.
And we have a threefold axis that we've added to the lattice point.
And then we found additional threefold axes in the center of these triangles.
And, once again, the pattern that has this symmetry is just a triangle of objects hung at every lattice point.
And now we want to add a mirror plane to that threefold axis.
So we have a primitive equilateral net plus the two-dimensional plane point group 3m.
The question is, which way should we orient the mirror plane?
Well, a mirror plane says, I want to be along the edge of either a centered rectangular net or a primitive rectangular net.
There's nothing about this lattice that looks rectangular.
Yeah, there is.
I see a couple of people doing this, and they have the right idea.
Here's a lattice point.
If we go more than one unit cell in this diagram, here's a lattice point.
Here's a lattice point.
Here's a lattice point.
Here's a lattice point.
Go down to here and go over to here.
And lo and behold, here is a centered rectangular net, hiding incognito in a hexagonal net.
Well, that helps.
I can put in the mirror plane.
But let me point out-- this is getting very messy, so I'll redraw it on a smaller scale.
This will be worth doing so that it's perfectly obvious.
I'm going to draw a number of equilateral nets.
The reason I'm drawing more than one cell, I would like to point out that here is one centered rectangular net.
And it has its edge perpendicular to the translations in the hexagonal net.
But there is also a rectangular net that goes down like this.
That's a centered rectangular net that has its edge along one of edges of the unit cell.
So I have two possibilities.
One is to draw the rectangular net like this.
The other one is to draw the rectangular net like this.
And having confused you thoroughly, let me simply draw this larger cell again.
And say that I could, if I wanted to, put the mirror planes in in these directions, along the edges of the cell.
OK. And in this case, if you're convinced that there's a centered rectangular net sitting there.
Let me clean this up.
And here's one mirror plane.
60 degrees away is another mirror plane.
And there'll be another mirror plane at this lattice point doing this.
So now I've got 3m at every one of these threefold axes.
And I got all those simply by repeating these mirror planes by translation.
But in one case, the mirror planes are perpendicular to the cell edge.
This being T1, let's say, and this being T2.
In this case, the mirror planes are along the edges of the cell.
This being T1, and this being T2.
So, holy mackerel!
One point group.
Two different space groups, which differ in the way in which the symmetry is oriented relative to the lattice.
And if you think back a little bit, this particular net, with this dimensional specialization, was happy with a sixfold access as well, and will be compatible with 6mm.
So, really, these two different orientations for the mirror planes, even though we haven't got there yet, are the two different orientations which are both present simultaneously if we were to put 6mm into this hexagonal net.
So, same point group.
Same lattice.
Two different plane groups that depend on the orientation of the symmetry relative to the lattice.
So there isn't even a one-to-one correspondence between the point groups and the plane groups.
The pattern for either of these is, once again, going to be just what the pattern of the point group does.
If we start with a motif here and repeat that by the threefold axis and the mirror planes, we would have motifs like this.
And here the translation points out in between the mirror planes.
In this case, the mirror planes would repeat the objects in two locations, like this.
So, indeed, both patterns have 3mm symmetry.
And what's different is the way the translations come out in between those mirror planes.
OK. Now our notation has come up against an impasse.
We said that the notation for the plane group should be the symbol for the lattice type, which the initiated know has to be a hexagonal net.
The point group that we've added, which is 3m, but now how do we distinguish the two different orientations of the mirror planes?
And the way around that is to actually use three symbols.
And the second symbol will be what is perpendicular to the cell edge.
And the third symbol will be what's parallel to the cell edge.
So this one would be called P3M1.
And this one, just to distinguish it, would be called P31M.
So if this weren't bad enough, now you've got almost the same symbol with a permutation of two of the terms in it.
In this case, the mirror plane is along the cell edge along the translations.
In this case, it's perpendicular.
But we're not quite yet done, because we've got translations that are inclined to mirror planes.
And now you can see how prudent I was in saying quite generally, without trying to figure out what is the perpendicular part and what is the parallel part, here's a translation and it's at an angle to a mirror plane.
And therefore we get a glide that has to be halfway between these mirror planes.
And there'll have to be a glide halfway between these mirror planes, and a glide that's halfway between these mirror planes.
In this case, the mirror planes are this way.
Again, a translation that's inclined to the mirror plane.
We have to have a glide that is midway between these mirror planes.
In this case, they're a little easier to identify.
They sort of make a triangular box that surrounds the threefold axis.
So now we see another difference in the plane groups and the reason why they're so very distinct.
And that is the rearrangement of the glides relative to the threefold axes are quite different in these two cases.
So these are two quite different symmetries.
Notice that when we start identifying special locations in these two plane groups, here there is a location only of symmetry 3mm for a point group, 3m for a point group.
Here there are two different locations, one that's symmetry 3m, the other one is just symmetry 3.
So this is location of point group 3 and a location of point group 3m.
So the sort of special positions that exist in these two plane groups will be very, very different.
One more addition, and then I think we can leave with a light heart because we should be done.
What about a primitive hexagonal net combined with 6mm?
And if you think I'm going to try to draw that, you're crazy.
But I can say quite simply and hopefully convince you that this looks like P31M plus P3M1 right on top of one another, because we've got mirror planes that are 30 degrees apart.
So, as a schematic direction and an invitation to do this at home in your spare time, is to take what P6 looks like.
And that's sixfold axes at the lattice points.
Threefolds here.
Twofolds here.
And then, directly on top of this, place the mirrors and glides in this orientation, and the mirrors and glides in this orientation.
So we will have mirror planes coming out like that, and glide planes in between all the parallel mirror planes.
As I say, I'm heroic but not foolish.
Yes
Can you just explain that derivation again?
Your P3 [INAUDIBLE]
OK.
The object of this additional bit of confusion is to put in the symbol the point group, which is 3m, and then tell you how the mirror plane in a point group is oriented relative to the edges of the cell.
And we could call them anything we wanted to.
We could call them P3M perpendicular, subscript perpendicular.
Or P3M subscript parallel to indicate mirror plane parallel to the cell edge.
And mirror plane perpendicular to the cell edge.
Or we could call them some obscenity, which would be very descriptive, as well.
But this is just a way of keeping track of what's parallel to the cell edge and what's perpendicular to the cell edge.
So, this middle symbol is what's oriented perpendicular to the cell edge.
And in this case, this is the mirror plane and that is perpendicular.
All the mirror planes are perpendicular to the edges of the cell.
And then parallel to the cell edge, there's no symmetry at all.
So that's 1.
1 stands for no symmetry.
In this case, the mirror plane is along the cell edge, so there's an M in the third position.
And perpendicular to the cell edge, there's no symmetry at all-- no symmetry plane at all.
So it's purely arbitrary, but we need some mechanism for distinguishing which one we're talking about.
That pretty much works as well as anything.
And it's a minor perturbation of what we've done for the additional plane group symbols.
All right.
It's time for our midafternoon break.
It's a nice place to quit, because you're probably feeling good.
We've done this, and we can move on to something new.
Actually, there's one more trick.
And unless you were really clever, you wouldn't have thought of it.
But there's another small family of plane groups that we can get through one particular device.
I shouldn't have told you that, because you'll be less inclined to come back now.
But we're almost through.
And I think it'll be very clear what these additional possibilities are.
Examine every possible means for combining the symmetry at once, but there is a seemingly paradoxical trick that we can yet pull.
And let me indicate what is true here for 2mm.
OK, so this is a twofold axis.
Has mirror planes perpendicular to it.
If one of these is a mirror plane, the other one has to be a mirror plane as well.
So there's no way we could make one a mirror plane and one a glide plane.
OK, that requires a net that is exactly rectangular.
So let's put in the twofold axis.
And I add one to the corner of the cell.
As we well know we have to have twofold axes at all of these other locations.
We want to put a mirror plane in the cell.
We could pass it through the twofold axis, and that would be the same as P getting to P2mn back again.
But why do we have to put the mirror plane through the twofold axis?
We have to have the twofold axis left unchanged when we add the mirror plane, because if we created a new twofold axis we create a new lattice and we'd wreck the lattice that we've constructed.
But why, why, oh why couldn't we put the mirror plane in like this?
That's going to leave the twofold axis alone.
It's going to leave the translations invariant.
Why don't we do that?
Why not?
So here, trick number five, or wherever we are now.
You can add the symmetry elements of a point group to a lattice, but not necessarily at the same point.
You can interweave them.
But the constraint is that this addition must leave the axis invariant.
Let's leave the twofold axis invariant.
And vice versa.
That is to say the mirror planes can't create new twofold axes, the twofold axis can't create new mirror planes.
And let's make sure we understand the reason why.
If I've got a twofold axis here and a twofold axis here, I have to have a translation that's twice their separation.
If this distance is delta, then I have to have a translation that's automatically created at twice delta.
This is the same as saying twofold axis with a translation gives you a twofold axis halfway along.
If I take the twofold axis, I get the translation back as a consequence.
So what I'm saying is if we take the first twofold axis and repeat number one to number two, and then add another twofold axis which repeats number two to number three, you have a third one that is related by translation of twice delta.
So if you're going to interweave the symmetry elements, you have to leave the arrangement of symmetry elements at the same intervals, otherwise you would have created a translation or more that is incommensurate and incompatible with the other ones.
So this is a third trick.
And we are going to need a new theorem.
Let's first of all look at the possibilities.
We can have the twofold axis interweave with a mirror plane.
How about putting a glide plane in between the twofold axis.
In other words, replace the mirror plane by a glide plane.
This twofold axis hangs off here, reflect it across and slide it down to here, and you get this twofold axis here.
So that's also a self consistent compatible arrangement.
How about a centered net?
And that's more difficult.
A centered net has twofold axis and all the places of C2, and then twofold axes at locations like one quarter, one quarter as well.
We can't put a mirror plane in here.
That's not going to work, because it will generate a new twofold axis, and that changes the translation T1 to half its value.
We can't even put a glide plane here because this would slide down to here, reflect across, and that's not going to work.
So the number of possibilities is limited, but we'll see that there's one other case as well for a fourfold axis.
OK, we need a new combination theorem that says if you have an operation a, pi, and combine it with a reflection operation that's removed from it by some distance delta, what is the result?
That general theorem has worked out for you on the notes, but I think what I'll do it in the interest of time is simply look at how this arrangement of symmetry elements would move things around.
And here I've taken the motif and hung it at the lattice point, and repeated it by the twofold axis.
The mirror plane would take this and reflected it across to here, then take this one and reflect it across to here.
And do the same thing down at the bottom of the cell.
Let me now ask you what sort of plane has arisen that would be perpendicular to the mirror plane?
Could be interweaved or passed through the twofold axis.
Anybody want to hazard an answer?
OK. We have to have some sort of correspondence theorem that says if you've got a twofold axis, have one plane, you've got to have another plane 90 degrees away.
Do I know how everything is related?
I know how this is related to this.
I know this is related to this.
That's by a reflection plane.
I know how the twofold axes relate things.
How is this pair related to this pair?
And the answer is that the way they are related is by rights and lefts on here.
This is right, this is right.
Reflection changes handedness to left, so I've got to have some way of getting from this pair to this pair that involves a change of handedness.
And I see nobody is really jumping out of their seat, but there is a glide plane in here.
Take this pair, slide it along by half of T2, and flip it across by reflection.
And there is a glide plane right here.
And therefore of necessity there's a glide plane here.
And this glide combined with the perpendicular translation will put a glide plane in here as well.
And you look in your tables.
This is plane group number seven.
And this one has a twofold axis.
It has a primitive rectangular neck, so this is called P2m, and now the second plane at right angles is not an mirror plane, it's a glide plane.
So this is called P2mg, or there's a shorthand form that leaves out the two.
PMG is the shorthand symbol for it.
So there's a new plane group that is based on orthogonal symmetry planes.
One a mirror, one a glide, and twofold axes.
Let me ask you now what is the point group of a crystal that would have this relation between the atoms that are down in the guts of the crystal?
What would be the point group of the crystal?
We've got two different point groups.
We've got point M, we've got point group two.
What would be the point group of the crystal?
And this is a new problem, a new concept.
All of the plane groups that we looked at until this point, the symmetry elements on an atomic scale down inside the plane group were exactly the same as the point group that we had added to the lattice points.
But here we've got two different point groups that had been put into the lattice.
So do we say that this crystal has symmetry two or symmetry M?
Or just point group 2m.
That won't work because if you have one mirror plane in a point group, you have to have another.
What this introduces is a very subtle point.
The point group is inherently a macroscopic symmetry.
When I say a crystal has a particular point group, what I mean is that if I look at the exterior faces of the crystal, I would say there's a twofold axis here.
And that's about all, if these spaces are pair wise distinct.
I have a crystal that looks like this externally.
I would say that that crystal has symmetry 2mm based on the faces.
If I found that the etch pits on the surfaces were not quite the same on this face and this face, I would have to throw out the twofold axis, perhaps, which means this mirror plane would go out as well.
But also I would do things like look at the optical properties.
I would measure the conductivity as a function of direction.
I would look at the slip systems and yield stresses in different directions, and the assignment of a point group is an experimental observation that requires that you look at all possible properties.
Not only shape, but properties like conductivity and the magnetic susceptibility, and also perhaps the way the crystal diffracts X-Rays.
That's another physical property.
And then when you've done as many measurements as you choose to make, you say as far as I can tell, all behavior of this crystal is determined inconsistent with this particular point group.
But you're doing macroscopic things.
You're doing macroscopic things.
When you talk about the plane group or space group, you're talking what goes on on a local atomistic scale in the unit cell.
So what would be the point group of this crystal?
It would have a twofold axis that would be manifested externally.
It would have a mirror plane, which makes the left hand side want to be the same as the right hand side.
And if this were not a mirror plane here, but a glide plane, what you would say is that this face, if I reflect it and slide it along by one angstrom is going to become this face.
And this face here, if I reflect it and slide it back by one angstrom is going to become this face.
But what will that do to the external shape?
I just extend all these surfaces, and it's going to be exactly what I've drawn here.
How would you distinguish a mirror plane from a glide plane?
If a crystal has a mirror plane, a face that sits here passes through some atoms.
Those atoms are repeated by reflection.
And being slit up by an amount tau which is on the scale of atomic dimensions, and that would give rise to a face here.
But can you macroscopically assign any difference to the fact that two faces atomistically don't meet, but are separated by 3.2 angstroms?
No.
What you see is one face like this, and one face inclined to it with a slope that is the same for either a glide plane or a mirror plane.
So another truth about crystals is that macroscopically a glide line plane manifests itself as a mirror plane.
So paradoxically this crystal, which only has a twofold axis and one mirror plane down in its guts is going to look as though it has point group two in it, even though there is no sight, atomistically within this arrangement of atoms that has symmetry 2mm
I have a question
Yeah?
Sorry
[INAUDIBLE]
That's a very good question.
And actually, to answer it properly requires knowing something about diffraction.
The symmetry of the diffraction pattern that you observed would look as though it had a mirror plane in it.
Which is to say if we put a beam of white radiation down along this direction of a crystal, imagine these things all extending out in a third dimension, that what we would see among the arrangement of spots is a twofold axis in the center of [INAUDIBLE] photograph we'd see a mirror plane running this way, and a mirror plane running this way.
So we would see spots on the [INAUDIBLE] pattern, which would look like this.
So the glide plane would also behave it were a mirrored plane.
And, maybe, if you think of what goes on with the fraction.
If you ask yourself, if I brought in x-rays this way and diffracted them off a layer of atoms related by this glide plane, and then I can't do this actually to the crystal.
But, suppose I then inverted the direction of the incoming beam, and slitted along by five angstroms.
Doesn't make any sense.
The diffraction from these planes is going to look the same whether I bring in from one side or the other.
So a glide plane, in diffraction symmetry, manifests itself as a mirror plane.
So how can you determine the presence of glide planes using diffraction?
And you can.
And the answer is, that the glide plane causes the intensity diffracted it from planes with certain indices to be identically zero.
You've probably heard of these magical extinction rules they're called.
And it says that if h plus k, plus three times L is two pi plus four, and the crystal is green, then you've got such and such, the reflection being absent.
Actually it is-- I'm getting distracted-- but, it turns out that glide planes affect just a sheet of reflections in reciprocal space.
And the reason some reflections are absent is that that sheet of reflections corresponds to what the projection of the structure onto that plane would do.
And, if you project a structure that has a glide operation onto the plane of the glide plane, the structure looks like the lattice translation is half as big.
And, if that's the case, the diffraction spots are twice as widely separated.
And it turns out that all of the reflections that have an index corresponding to the direction of tau are absent if that index is odd.
So, if you knew a little bit about it to begin with, that perhaps answers the question.
So yes, you can determine the presence of glide planes.
And later on, we'll talk about screw axes.
You can determine their presence unambiguously, but you do that not the symmetry of the diffraction effects, but from systematic absences.
OK, another thing I suggested we could do is to put a glide plane in between the two-fold axes instead of a mirror plane.
And this says that I have a pair of atoms repeated by the two-fold axis, the glide plane would take that pair, shift them down by half of the cell, and reflect them over.
So this would be the pattern of objects.
This is not a lattice point, because if these are right handed, that's a right-handed pair, this is a left-handed pair.
Now let me try you again.
I know how these guys left and right to this glide plane are related.
How about these ones that are up, to the ones that are down here in the middle of the cell?
There's got to be, from our correspondence principle, some plane going in this direction.
Anybody want to hazard a guess at what it is?
The answer to that question resoundingly is, no, not at this hour the afternoon, thank you.
Yeah?
Left one
Yeah, very good.
Right ones have to go left ones, and they're going to go left and right about these planes here.
And, if I combine this glide plane with t2, there has to be another glide plane in here.
If I combine this glide plane with t1, there'll have to be another glide plane in here.
OK, and that turns out to be plane group number eight, P2gg, not P2mm, both mirror planes have become glide.
All right, we're almost there.
If you look at the hexagonal symmetries, there is just no way that you can interweave things, and an attempt is made in the notes that I handed out.
You just cannot do it.
But with P4M, P4, there is one final possibility.
And I'll just set that up and not carry them out.
OK, I'm going to put on just P4 with twofold axes in here.
Now we have planes that go in orientations like this.
And then there is another plane 45 degrees away that is a symmetry independent plane.
And the question now, I'll let you have a crack at it, is there any way in which I could take planes in the diagonal orientation or parallel to the cell edge and interweave them in between the rotation axes in such a fashion that the rotation axes were left invariant?
Again, conversely if they're not left invariant, we will have created new translations and wreck the lattice.
So how could we slip in planes, either reflection or glide, and leave the axes invariant?
One quarter and three quarters?
Yeah.
Do you want parallel to the cell edge?
Horizontal
Horizontal?
Yeah.
This [INAUDIBLE] here has a fourfold axis up, a fourfold axis down, a fourfold axis up.
Twofold axis down, twofold axis up, twofold axis down.
So this would be a lovely location for a glide plane.
So that's one.
From the fact that I am placing the same diagram on the board yet again suggests that there's another way.
That's a perfectly good suggestion, and that's one of the things we would have to examine.
Is there any other way I could put in a mirror plane or a glide plane and leave the axes invariant?
Yeah?
[INAUDIBLE] 45 degrees?
45 degrees.
Through here.
I don't want to do it through here because the fourfold axes sit on the same side, but you want to put it in like this?
I don't think you can do that.
Glide plane won't work here.
How about a mirror plane?
There's no mirror plane that goes through these twofold axes for P4mm.
What we had there were mirror planes here, and then going this way were glide planes.
We've already got that.
How about putting in a mirror plane like that?
This goes into this.
Halfway along there would have to be another mirror plane in that orientation.
Here's a twofold axis that demands that there has to be another one if it has a mirror plane passing through it.
So there's another way.
One final one, and I will not keep you guessing anymore, sneaking a peak at the answer.
There is another way you could do it, and that would be to put the glide plane down diagonally through the cell.
That is put the glide plane here.
That and take this twofold axis, reflect it across to here, and slide it down to here.
So this twofold axis would be related to this one by glide.
So those three possibilities are available for point group 4mm and a square net.
If you try them, and the simple way of doing it without any general theorem would be to draw in the way the symmetry elements repeat the motifs, and then ask what they require for the remaining symmetry elements.
All of these give the final result.
So they yield the same result.
You can develop a general theorem for what you obtain if you have an operation a, alpha combined with a plane, and make it a general plan, a glide plane that is removed from the axis by some distance delta.
And that theorem is written out for you in the notes on the very first page.
You take a glide plane, and this is a specific example, a, pi, if you have a glide plane that misses a twofold axis by some distance delta, then the effect of those two successive operations is a glide plane at 90 degrees to the first, and it's removed from the twofold axis by two delta.
It has a glide component, excuse me, has a glide component two delta.
So there's a general theorem for the twofold rotation.
The general theorem for a fourfold or a threefold rotation is considerably more complex to derive, but that's contained for you in the notes as well.
All right, we've come to the end of one other major part of our story of symmetry.
These are the periodic patterns in a two dimensional space.
And these are then the 17 crystallographic plane groups.
Crystallographic is rather redundant because they are groups that have [?
in ?]
translation and therefore these are the symmetries, the two dimensional symmetries that are suitable for the atomic arrangements in crystals.
As a little bonus, we also have, as a result, 17 of the three dimensional space groups.
Just imagine each of these plane groups with a translation perpendicular to the plane of the plane group, and that would leave all the symmetry elements coincidence.
So you could pick a third translation at right angles to the blackboard of any arbitrary length, and you would have a three dimensional lattice that contain the symmetries of a plane group.
I think I mentioned earlier that the way you distinguish a plane group with a twofold axis in it from a space group with a twofold axis is that a lowercase letter for the lattice type implies plane group, and an uppercase symbol, capital P stands for a primitive lattice in a space group.
So something like lowercase p4mm is a plane group, capital P4mm is a space group with a third translation of arbitrary length at right angles to the plane group.
We talked about plane groups and space groups, let's take a giant leap backwards.
How about the one dimensional space groups?
Bet you were wondering about them all along and were afraid to ask.
So what would be the story for one dimension?
It's nontrivial.
A one dimensional space group would be a lattice row.
You have just one dimension to play with, and you can make that space periodic by translation.
So there's one lattice type.
Just the lattice row.
And what sort of symmetries could you place in that lattice?
Now you have to define your ground rules.
In our two dimensional symmetries, we did not permit any operation which would pick the two dimensional space up and rotate it and flip it over and put it down again, because that is a transformation that takes you out of that space and pops you back into it.
And we will not in our discussion of three dimensional crystallography consider symmetries operations that take something, suck it up into a fourth dimension that we can't comprehend, and then all of a sudden pop it down in again, and suddenly it appears.
I mean, that sounds bizarre.
We wouldn't want to do that.
So for plane groups, we did not allow for any operation, say a twofold axis that would flip the object over and turn it upside down.
But why not?
I mean this, is mathematics.
If it's your ballgame, you can make up the rules.
So why couldn't we have a twofold axis in the plane or the plane group?
It would have to leave the net invariant.
Well actually such entities do exist, and they have been derived.
They're called the two sided plane groups.
And if you want to allow that when you make up the game, you can actually do that if you wish to allow it.
And also there could be a mirror plane in the plane of the space that would take the top side and relate it to the bottom side.
So you can do that, you can permit that.
That's a different beast entirely, but you can allow that to be one of the transformations.
So for our one dimensional space, I would submit to be consistent with what we've done and just finished in two dimensions.
And what we will do in three, we will not allow for any operation that will take the space and transform it into a second or third dimension, and then put it down again.
So that being the case, the only operation that is possible is a mirror point that would reflect things left to right.
And let me illustrate now with some patterns.
There's the lattice point.
So let me put in some motifs.
and I have to make a one dimensional motif, but to distinguish the ends I'll take a little artistic license and make one end of the motif a little fatter than the other.
Or alternatively, I could put a little one dimensional headlight on this thing that would shine just in one direction.
So here's a motif hung on every lattice point, and so this is no symmetry at all.
So the symmetry part of the symbol would be one, and the lattice is primitive, and what do you think people use?
We've used lowercase symbols, we used uppercase symbols, what's left?
Yes?
Greek?
Good try.
Gothic.
Actually, what you do is you use a Gothic symbol with all these nice little shaded angular shapes.
That's a Gothich P, isn't it?
Yeah, that's a Gothic P. It is.
It actually has thick and thin parts, curved parts [INAUDIBLE].
So that's P1.
What about another symmetry?
Well if we have a mirror point in there, another possibility would be to have these motifs point alternately left and right.
So I'll take some non one dimensional license and say that there would be mirror points at these locations.
Really nice symmetry, but actually it would be totally wasted on these poor denizens in a one dimensional world.
Life would be dull and uninteresting because they couldn't see anything except the motif on either side of them, because everything is constrained to one dimension.
There's an old saying about sled dogs that says if you are not the lead dog, the scenery is not very interesting.
And the same is true for these poor people in a one dimensional world.
They could never see the elegance and beauty of that.
But in any case, that would be Gothic P, no snickers, please, and that would be m. And those are the only two possibilities and a one dimensional space.
We have a couple minutes to go, but I think this is a good place to pause.
And I'll give just a brief indication of coming attractions.
We're now going to look at point groups in three dimensions, and a good way of entering this much more complex situation is to go back to something analogous to what we did for two dimensional symmetries.
And I'm going to first consider the arrangement of rotation axes in three dimensions.
We do not have the constraint that all the rotation axes be perpendicular to the plane of a two dimensional space, being therefore more properly rotation points.
We've got three dimensions we have to view a rotation axis as extending infinitely in a direction, and there's no reason why we can't have another rotation axis at an angle to it.
But you know that already.
Everybody's heard about cubic crystals.
Even if you think all crystals are either body centered cubic, primitive cubic, face centered cubic, or complex, four kinds of lattices.
But you've heard of cubic crystals, in the cubic crystals clearly the rotation axes are inclined to one another and arranged spacially.
So we're going to ask the nontrivial question, how can we combine more than one rotation axis at a time about a common point in space?
And the constraint is going to be that a rotation about the first axis followed by a rotation about the second axis, wherever it is, is going to have to turn out to be something that is crystallographically compatible with a lattice.
So that's the constraint.
It's not an easy question.
And just to make you feel proud of yourself when we get through this, this uses a geometry that is due to one of the great mathematicians of all time, Leonhard Euler.
Probably the most remarkable thing about Euler was that he's Swiss.
How many world class scientists have you heard of that come from Switzerland?
Not very many, and the reason is it's such a small country, and if genius occurs as a certain fraction of the population, that's not going to happen very often in a country like Lichtenstein or Switzerland.
Have you ever heard of anybody prominent in science who came from Lichtenstein?
No, probably not.
OK, this then is going to be a nontrivial piece of mathematics for us, largely because it involves spherical trigonometry.
Which I'm sure if you've ever heard of, you've forgotten, and probably don't see any utility in it.
OK, with that exciting prospect in hand, I look forward to seeing you on Thursday.
I think you will get going.
A lot of people are not here, but I think if the MRS meeting were going on across the river I would be there too if I didn't have a class to go to.
We had last time finished a discussion strain.
And we introduced strain by defining it as a displacement, each component of a vector displacement U as being given by a set of coefficients, eij times x of j, which says that the displacement in a deformed body varies linearly with the position of the particular point in the body.
We could also express the same thing in terms of change of length.
The change in the vector U would be given by-- we show the same set of coefficients-- eij times delta x of j.
So whether you want the displacement vector or you want the fractional change of length, you get this by an expression of the same sort.
We saw though that unless we define the coefficients correctly we could have a situation where a body is not only deformed but is rotated as well.
And we saw that in general unless we define these coefficients carefully we would include in this tensor a component of pure rotation, rigid body rotation.
And this would not measure deformation.
However, regardless of its suitability or unsuitability, the positional vector is a vector.
It transforms like a vector.
The displacement U is a vector, and it will transform like a vector; ergo, the 3 by 3 array eij is a second-rank tensor and we'll transform like a tensor.
We showed though that we could take eij and get rid of the pure rotation that might be contained in those coefficients by writing it as a tensor epsilon ij plus another tensor omega ij.
And that was a new wrinkle, the concept of adding two tensors together to get a third tensor.
We did run into that before.
And we define the elements of the two tensors epsilon and omega; epsilon ij was given by one half of eij plus eji, and omega ij is given by one half of omega ij.
And notice the order of the subscripts here in this difference.
That is a matter of definition, but it defines which term is omega ij and which is omega ji minus omega ji.
I'm sorry.
This should be eij minus eji.
And we showed that at a tensor of this form which would have all of the diagonal terms zero and the off-diagonal terms equal to the negative of one another that a tensor of this form does indeed correspond to pure body rotation.
We showed that by looking at a point at the end of a positional vector U and looked at a displacement which would be exactly at right angles to it.
And the U would be given by the tensor relation.
We showed that U dot r is identically zero when the displacement vector is given by this tensor with diagonal term zero and the off-diagonal terms equal to the negative of one another.
So we have then from a general tensor of the form eij constructed in the terms epsilon ij something that is a measure of true deformation.
It is a symmetric tensor by definition.
And therefore in addition to all of the other properties of second-rank tensor, namely having a law of transformation and staying symmetric for any arbitrary change of coordinate systems, the radius normal property also works.
Hence, the representation surface that we construct from the tensor epsilon ij xi xj equals 1, gives us the strain quadric.
And it has the property that the radius gives us the value of strain in a given direction.
And the value of the radius literally is going to be the vector that's on the left hand side, the magnitude of U times the component of U that is parallel to the radial vector per unit radial vector.
And this then is the tensile strain, the fractional change of length as we look in this direction.
The radius normal is going to give us the direction of the total displacement U.
And the value of the property in that direction is going to be the product of that displacement with a unit vector in the direction of interest.
And therefore that is the tensile strain as we said before from the general properties of the quadric.
So everything we said about second-rank tensor and in particular symmetric tensor holds for the strain tensor.
We can talk about diagonalizing the strain tensor into a form epsilon 1, 1, epsilon 22, zero zero, epsilon 33.
And I'll remind you that we know how to do this to set up the normal equations and solve for the eigenvalues and then look for the principal axes of a tensor.
There are special forms of the strain tensor.
Something that 1 diagonal goes into this form is called plane strain.
And I love the melodic nature of that, the plane strain.
If this were strain induced by a Modernistic transformation, we could call it the Bain plane strain.
If we diagonalized it, it would be the main Bain plane strains.
And if that deformation took place in an aircraft window, we could call it the plane-pane main Bain strains.
And we could continue on indefinitely, but I think you're finding this tiresome.
Another form of the strain tensor that doesn't look what it actually represents is epsilon 1 zero minus epsilon 1 zero zero zero zero.
That looks peculiar, very special.
But, again, this is analogous to what we found for stress.
If we rotate this tensor 45 degrees, we would find that treat tensor was diagonally, had all the diagonal elements zero.
And we would have the epsilon 11 prime-- let's just call it epsilon prime-- and this epsilon prime zero zero zero.
And this is pure shear Something that we'll show shortly, and in fact I left it to you as a problem on a problems set.
The sum of the diagonal elements, epsilon 11 plus epsilon 22 plus epsilon 33 is something that's defined as the trace of the tensor.
And for the strain tensor, the trace turns out to be equal to delta V over the fractional change in volume.
And we'll show that in just a little bit as well.
So a characteristic of pure shear is that it does not result in any change of volume of the solid.
There's deformation, but the volume stays exactly the same.
Let me then proceed to show that the trace of the strain tensor is indeed the change in volume.
And want I'll do is to look at an element of volume that is oriented along the principal elements of strain.
So if this as x1, this is x2, and this is x3, the strain tensor I'll assume is oriented.
So we have a term epsilon 11 zero zero zero zero; epsilon 22 zero zero zero; epsilon 33.
So let's suppose that the solid originally has edges L1, L2, and L3 along x1, x2, and x3, respectively.
So the initial volume before any deformation will be simply L1 times L2 times L3.
Let's suppose now we impose the strain, and the volume then we'll increase to some value of V plus delta V. And the length L1 will change to a value L1 times 1 plus epsilon 11.
That is to say it'll be L1 plus epsilon 11 times L1, which can be factored out in this fashion.
L2 changes to L2 times 1 plus epsilon 22.
And L3 changes to L3 plus epsilon 33 times L3, which I'll write in this fashion.
So expanding this product, it'll be L1, L2, L3.
And then if I simply multiply out these terms, I'll have 1, and then I'll have epsilon 11 plus epsilon 22 plus epsilon 11 times epsilon 22.
And this will be times 1 plus epsilon 33.
And if I carry out that multiplication, that will be 1 plus-- and I'll get exactly these terms again-- epsilon 11 plus epsilon 22 plus epsilon 11 times epsilon 22.
And then a term epsilon 33 plus epsilon 11 epsilon 33 plus epsilon 22 times epsilon 33.
And then a third-order term, epsilon 11 times epsilon 22 times epsilon 33.
Now having gone through the painful process of expanding that, I will proceed to say that the epsilons are always going to be very small.
Elastic strains are typically on the order of 10 to the minus 6.
So a product of two of those strains is 10 to the minus 12.
And a product of three of those terms is going to be on the order of 10 to the minus 18.
So this last term is going to be minuscule.
And now I use a higher order term intentionally rather than a higher rank term.
This is really a term of higher order.
The same is true of all the peer-wise projects of this [INAUDIBLE].
So I now have left if I turn out these higher order terms to pasture, I'll have 1 plus epsilon 11 plus epsilon 22 plus epsilon 33.
And the rest is negligible.
And this will be the original volume V times 1 plus epsilon 11 plus epsilon 22 plus epsilon 33.
So delta V over V, if I subtract off V on the left, delta V over V then is simply going to be equal to epsilon 11 plus epsilon 22 plus epsilon 33.
So this is indeed the fractional change of volume, and it's the trace of epsilon ij.
Now, physically, you wouldn't expect the volume to change if I change my reference axes because that's a scalar quantity.
So I've shown by a hand-waving, backdoor argument that in fact the trace of a tensor does stay invariant for at least this particular tensor.
But it's fairly straightforward to show that for any tensor aij the trace of the tensor is invariant when you change the coordinate system.
And that's not hard to do.
If you're clever, it takes about three lines.
So I'll invite you to the exhilaration of discovering that for yourself on a problem set.
Any questions or comments?
All right.
Having defined strain as a symmetric second-rank tensor when we factor out rigid body rotation, we can view strain as a physical property.
And I think I tried to hoodwink you last time by saying this is a property of material.
So therefore strain tensor has to also conform to all of the symmetry restrains that we have derived very exhaustively.
And you said, no, no, no, no, that can't be; I don't believe that I can only give a uniform uniaxial deformation to an orthorhombic crystal.
And I say, yeah, but to get the deformation, you have to squeeze the crystal.
And that is going to be linked to the deformation by the elastic constants, and they are physical properties.
And that maybe makes you think a little bit.
But actually even though operationally you impose a strain by imposing a stress or by doing something else to the crystal that results in a strain, you can nevertheless pick any stimulus you wish to achieve a desired state of strain.
So there is no symmetry constraint on the form of the strain tensor that can result.
So the stain tensor then is not a property tensor.
It is also a field tensor just like stress.
So it can have any form we choose to create in it.
So we now have the potential of having stress as a field tensor.
And we have strain as a field tensor.
And within limits, we can create strains of these two field tensors of any form that we wish.
And we can now regard these as things that we do to crystals and look at a whole range of properties that result when the generalized force is not a vector, a tensor of first rank, but the generalized force is a-- let's say-- a stress tensor.
It's a generalized force which itself is not a first rank tensor.
It's a second rank tensor.
So we'll see a whole collection of interesting properties that fall into this category.
But first I'd like to look at one remaining second-rank tensor property, and that is thermal expansion.
And this is a curious second-rank tensor property, and we couldn't discuss it until we had talked about strain.
And let me introduce it by one-dimensional example.
Let's suppose we have a rodlike specimen of length L. And let's suppose it is in equilibrium at some temperature T. And then we heat it up to some temperature T plus delta T. In response to that particular change in temperature, the length of the rod will increase to a length L plus delta L. It expands.
And we would find experimentally that delta L, first of all, is proportional to L. If you make the rod twice as long, the amount of expansion in an absolute sense is twice as large.
And secondly, provided you're well above very, very low temperatures, close to absolute zero, delta L will also be proportional to the change in temperature.
Thermal expansion is one of those properties which thermodynamically must go to zero as temperature goes to zero.
So the thermal expansion coefficient that we've yet to define will be a function of temperature as we get down to very low temperatures on the order of a few degrees Kelvin.
So we can define then the linear thermal expansion coefficient for this rod as 1 over L, delta L over delta T. So this is something that will give us the fractional change of lengths.
And what we can do now in general is to say that if we create a general strain, not a tensile strain, we have a general strain epsilon 11 plus epsilon 12, epsilon 13, and so on, has to be symmetric.
So epsilon 21 is equal to epsilon 12.
We'll say that this tensor epsilon ij will be written in general for the three-dimensional case as a second-rank tensor aij times a scalar quantity delta T. The strain is the field tensor.
That's the response.
The stimulus that creates this response is an incremental change in temperature.
But as this is a second-rank tensor and this is a scalar, an array of coefficients in which we multiply every element of a tensor by a scalar is also a tensor.
So this is called the linear thermal expansion coefficient tensor.
And it is a symmetric tensor by definition since a measure of true strain is by definition a symmetric tensor.
So this is the linear thermal expansion coefficient tensor
Quick question
Yes
[INAUDIBLE] the temperature gradient [INAUDIBLE] and talk about the directionality say between here and here because of the temperature [INAUDIBLE]?
Oh, yeah.
Sure.
Sure if you wanted to do that.
That assumes that the thermal conductivity would be very small so that the temperature inside the body would not attempt to [INAUDIBLE].
Yeah.
Yeah.
In the same way that you could create strains that are in homogeneous as well.
We'll now for the first time start talking not in terms of abstractions but in terms of some real physical properties.
What is the range of linear thermal expansion coefficient tensors?
And unlike many physical properties, that's an easy one to answer, 10 to the minus 6 degree C per degree C. Materials have a very, very small range linear thermal expansion coefficients.
And let me give you some examples for real materials.
They go up to maybe 10 to the minus 5 for a very soft weakly bonded materials.
But the for metals-- and I'm giving you a single number now because these are averages for polycrystalline materials-- the value of A times 10 to the 6.
For lead, a low- melting metal, is 28, 2.8 times 10 to the minus 5.
For copper, 18.
For iron, 12.
Four tungsten, a very refractory metal, 5.
And for diamond, a very strongly bonded materials, it's 0.89 times 10 minus 6.
If we look at compounds, again, the average linear thermal expansion coefficient times 10 to the 6 for polycrystalline material.
For aluminum bromide, hardly a technology material of commerce, but in light of this because this has one of the largest linear thermal expansion coefficients of any material, very weakly bonded to compound.
For a more typical ionic compound, NaCL, 40 times 10 to the minus 6.
For calcium fluoride, this has a bivalent cation in it, so the material is stronger, more strongly bonded, 20.
For MgO, both a divalent cation and the divalent anion, as you would expect, the thermal expansion coefficient drop still more.
For Al2O3, a trivalent cation that goes down to 8.8, and that's really an average because this is a hexagonal material.
And a second-rank property tensor for a hexagonal crystal has to have two equal diagonal elements and one independent diagonal element when you diagonalize the tensor.
Spinal, MgAl2O4 this is representative of all of the ferrites for example, 7.6.
Silicon carbide, a very refractory covalent compound, 4.7.
For glasses, here's a surprise.
You might think that a glass might have about the same linear expansion coefficient as the crystalline form of the same composition.
But a sodium calcium silicon oxide glass, so called soda-lime silicate glass, is 9, fairly low.
A borosilicate glass, like the Vicor that we love to make glass apparatus out of, 4.5.
And one of the near record holder is fused silica, silicio glass, and here a measly value of 0.5.
So glasses have very low linear thermal expansion coefficients compared to crystallize materials that are predominantly ionic.
And the reason lies in the nature of the structure.
Something like MgO has ordered ions, and when you heat it up, it expands isotropically.
A glass is this random network of edge-shared polyhedra.
So when you heat it up, yes, the dimensions of the polyhedra increase in proportion to the change of temperature.
But the framework because it's so meandering and open can buckle to accommodate the increased size of the tetrahedra that are in the linkage.
So the result is the network buckles, but the overall microscopic thermal expansion is very, very slight even though the tetrahedra are changing dimensions in the same degree that they are in these crystalline inorganic nonmetallic materials.
Let me give you a handout not that it's up to date, but there was a very nice, convenient one-page article in the Journal of the American Ceramic Society a number of years ago that lists thermal expansion coefficients for low expansion oxides.
So let me pass that around and let you take off for copy with that.
And that gives some examples of the value of this property for, again, for polycrystalline materials, which if you looked at a single crystal would show anisotropy.
And here you'll find a few materials in the list near the top they're arranged in order of increasing linear thermal expansion, you'll find some that are just about zero when you have a polycrystalline form of the material.
And you'll have some that have this very unusual behavior where the linear thermal expansion coefficient is actually negative.
So let me pass this around for reference.
Now this may seem to be a curiosity that you can fine for a polycrystalline material for which the individual grains expand anisotropically a bulk linear thermal expansion coefficient of zero.
Why should one care other than that being a curious result?
Can anybody guess?
Can you say that again?
That these are materials which in a polycrystalline form with random grain orientation have a linear thermal expansion coefficient this is essentially zero?
[INAUDIBLE]?
Yes.
Exactly.
The materials out of which you build furnace linings and tanks for melting glass or smelting steel have to be made out of polycrystalline materials.
You can't have a monolithic single crystal that's large enough to melt a significant amount of glass in for example.
And being polycrystalline, when the material expands, there is a great deal of stress between neighboring grains because they're randomly oriented.
If the bulk microscopic thermal expansion coefficient is zero, that is going to be refractory that is very, very resistant to thermal shock.
When you heat it up, you don't find intergranular cracking.
So the materials that have essentially zero spatially average thermal expansion coefficients are very attractive for refractories.
You can pick up any issue of the Journal of the American Ceramics-- not the journal, but the Bulletin of the American Ceramic Society and you will find refractory companies hustling materials like Cordierite, which is a silicate material that on average has one positive thermal expansion coefficient, one negative thermal expansion coefficient.
And the volume change of the individual grains is essentially zero.
So that makes a dandy material for refractories.
Cordierite has another interesting property that is a really nice example of anisotropy of a physical property.
If I ask you to rattle off some properties which you know to be very anisotropic, one of them that you wouldn't think of is color.
Your t-shirt is pink, so how can that be a function of direction?
Well your t-shirt is not a single crystal.
And their single crystals which if you hold them up to light and pass through a plain polarized beam of light the crystal has a very different color for one direction of polarization then another.
And these colors, interestingly, very often are strikingly different.
They're crystals that are green in light polarized in one direction and red for light polarized in the opposite direction.
And there is in the old Icelandic sagas-- few people have heard of them.
The Icelandic sagas which were written in about the year 500 are one of the early forms of Western literature that is really great, enduring world-class literature.
In the sagas, there's one episode where Thor is sleeping and Olaf comes along and steals his sunstone.
And Thor is upset he takes his battle as and cut Olaf's head in twain.
And people puzzled for hundreds of years, what is sunstone.
This doesn't compute in today's age.
And finally a Norwegian archaeologist came up with a hypothesis that in Iceland there are remarkably large and remarkably transparent and perfect single crystals of Cordierite.
And Cordierite is strikingly pleochroic.
That's color that is a function of direction.
And what this archaeologist theorized is that the Vikings would take a sunstone when they went sailing on the North Sea, which is notoriously damp and overcast and gray, and your principal means of navigation was the sun.
But when the sky was overcast, you couldn't see the sun.
But the sun as it comes through the cloud gives you light that's very strongly polarized.
So the theory is that the Vikings would have very perfect crystals of transparent Cordierite, would hold them up to the sky, and turn them around until the Cordierite crystal lit up a bright, golden sunny yellow.
What better name for the sunstone.
You found the sun from the direction of polarization of the light scattered from the cloud when you've got the crystal oriented just so.
In a different orientation, it would look murkier, look sort of a bluish purple for the other principal direction of the birefringence.
So there is putting crystal anisotropy hundreds of years ago to a very useful purpose to navigate on cloudy days.
Unfortunately, the Vikings used it to sail down the British coast and sacked the next village.
So it shows you that even the most simple of science can be put to application in war research.
Anyway, won't go there anymore.
So index of refraction and color can also be a strong function of direction The numbers that I put on the blackboard suggest that the magnitude of thermal expansion coefficients are influenced very strongly by the strength of the interatomic bonding.
And this shows up in a very, very striking way if you group together classes of comparable materials.
And if you plot the value of the linear thermal expansion coefficient in units of 10 to the minus 6 per degree C as a function of the melting point in degrees Kelvin, the numbers range from about 400 K for materials like sulfur up to about 3,700 for tungsten.
And the variation is so beautiful it could make you cry.
It goes as the inverse of the melting point.
And way up here are materials like for sulfur and lithium.
And the way out here are materials like tungsten.
And their relation is given by a equals 0.020 over the melting point in degrees Kelvin.
So very strong correlation between melting point and linear expansion for simple elements.
If you do the same thing for compounds, you find a similar parabolic relation except it is offset.
For oxides and halides, many of them with the rock salt structure, so they are isotropic.
And again, if you plot here the melting point in degrees Kelvin, the numbers here range up to 60 times 10 to the minus 6.
And again, a variation with 1 over T, but offset from the origin.
And the curve here is that a equals 0.038 over T, the melting point T sub m in Kelvin, minus 7.0 times 10 to the minus 6.
Finally, let me finish with some data for single crystals that provide examples of and anisotropy.
These are all hexagonal materials.
And I will give you the-- not all hexagonal.
I take that back.
But they're all uniaxial crystals.
The value of the thermal expansion coefficient along C and the thermal expansion coefficient that is parallel to C, so these are the elements that would be in the diagonal isothermal expansion tensor.
So for Al2O3, which is a hexagonal rhombohedral oxide, the two expansion coefficients are 8.39 and 9.0 times 10 to the minus 6, not terribly anisotropic.
But the structure of alumina is a close-packed arrangement of oxygen with aluminum filling 2/3 of the available octahedral holes.
So it's hexagonal only because of the sites that are filled in the array.
For TiO2, which is [INAUDIBLE] and therefore also uniaxial, 6.8 perpendicular to C; 8.3 parallel to C. For zirconium silicate, 3.7 and 6.2, almost a 2:1 difference.
For the quartz form of silica, which is hexagonal, 14 perpendicular to C; 9 parallel to C. For carbon, the graphite form, a layer structure-- you can almost guess how this is going to turn out.
It's going to be humongous perpendicular to the lawyers and very small within the plane of these tightly bonded hexagonal nets.
And that is indeed the case, 1 perpendicular to C; 27 parallel to C. So there is an anisotropy of a factor of 27.
A few more.
Aluminum titanate, perpendicular to C minus 2.6; parallel to C 11.5.
A very well known example of anisotropy, extreme anisotropy, where one principal coefficient is positive and two are negative is the calcite forms calcium carbonate, minus 6 and 25.
For zinc metal, a hexagonal close-packed metal, surprising degree of anisotropy, 14 and 64.
And for tellurium, which is a chain structure like selenium, 27 perpendicular to C and minus 1.6 in a direction parallel to C. Even simple hexagonal close-packed metals can do strange things at low temperatures.
For zinc, the expansion coefficients perpendicular to C and parallel to C as a function of temperature go at 300 degrees C from 13 and 64.
At 150, the values have dropped to 8 and 65.
At 60 degrees K perpendicular to C, the thermal expansion is negative and has not dropped terribly much of at all parallel to C. So this is a hexagonal close-packed structure lot.
A lot of action and strange things going on in the plane at the close-packed layers, but not very much change in a direction perpendicular to the layers.
Let me raise one more issue.
If we were to look at some of these strangely anisotropic materials which had negative thermal expansion coefficients in one direction and positive in another, what would the representation quadric look like?
Suppose we looked at calcite, for example, which has a large negative thermal expansion coefficient.
And if we look at the thermal expansion quadric when the tensor was referred to the principal axes, parallel to C-- well, the thermal expansion tensor aij along its principal axes would be along A1 in a direction perpendicular to C, it would be for the data given here, minus 6, zero, zero, zero, minus 6, zero, zero, zero.
And A33 is 25.
So that's very, very anisotropic.
But what is this going to be?
This is going to be a quadric that has the shape of an hyperboloid of two sheets.
So it's going to look like this.
And there'll be an asymptote like this and then asymptote like this.
Along the asymptote, the radius of the quadric is infinite.
So therefore along these directions the strain is equal to zero.
Strain goes as 1 over the radius squared.
In this range of directions here, the radius is imaginary.
But remember that the strain is given by 1 over the radius squared.
So this says that the strain is negative, so the material has contracted.
And finally, in this range of directions where the radius is finite and positive, if the value of strain is 1 over radius squared, this says that the maximum deformation is along the direction of the C axis, and it's positive.
The material expands along C; the minimum radius of the quadric corresponds to the maximum thermal expansion.
So let me close with a question.
Along the asymptote of the quadric, the strain is zero.
Does this mean that this direction in a crystal of calcite does not move, no deformation at all when you heat it up?
I see faint shakes of the head.
I don't know if that's just in awe of what's going on here or an opinion on the question.
Do you think it'll be no strain?
It would be [INAUDIBLE]
Good answer.
Let's remember a property of the representation quadric.
The direction of what happens is normal to the surface of the quadric.
The thing that is happening here is the displacement; ui is given by epsilon ij times U sub j.
So when we look in this direction, the displacement, which would be normal to the surface of the quadric, gives us this as a displacement.
So along the asymptotes of the quadric, you can't say that there's no deformation, but that the deformation corresponds to pure rotation and not any fractional change of length.
In these directions, the length changes, but it is a decrease in length within the asymptotes.
For those directions, the strain is an extension.
And again, if you want to know the direction of the displacement, you find the normal to the surface of the quadric in the direction of interest, and that's the direction in which the radius vector points.
All right.
That's I think a good place to quit.
I think I'll say a little bit more after the break about the atomistic reason for increases in interionic separations, which we can get some insight into from a simple model.
But before we disperse, I don't know if everybody got a copy of problem set 13.
If anybody didn't, there's some extra copies up here.
But I will with great pleasure pass out to you problems set number 14, which has served two purposes.
One is to have you show for yourself that the trace of a second-rank tensor-- I called it second-order tensor-- second-rank tensor the sum of the diagonal elements is invariant for a change of reference axes for a general direction cosine scheme for the change of axes.
And then the other two questions are to give you some practice in diagonalizing a second-rank property tensor which is not in diagonal form.
And I asked you to do this in two ways.
One is by direct solution of the secular equation and finding the eigenvectors.
And then the third example, which is for a completely general tensor, to do this by the method of successive approximations.
And you'll find after a couple of iterations you get fairly close to convergence.
So I shall pass this around.
And again, I think there's something to be learned from that.
Let's take our stretch then.
I'm sure that lecture involved a lot of stress and strain.
Ha, ha.
Let's resume in 10 minutes.
And we'll then move on to some additional higher rank tensor properties, which will be very, very interesting.
Go for it.
Any questions before we push bravely onward?
OK if there's no question on any of this I shall clean off the board.
And the last thing I wanted to do in connection with thermal expansion is to examine how the macroscopic linear thermal expansion coefficient relates to the nature of the potential minimum between-- in the attraction between-- neighboring atoms.
So let me suppose we have a pair of atoms separated by some distance, d, and there is a cohesive force between them or else the material would not be in a solid state.
And the nature of the potential-- and I'm going to call the distance between the atoms, x-- and this is going to be the energy.
And I'm measuring now relative to-- let me call this d as I did over here-- so the energy as a function of d is going to consist of a very short range, repulsive interaction, that rises very, very rapidly when you attempt to squeeze the atoms any closer than a fairly small value.
And then there'll be some sort of cohesive energy.
And everybody loves to model ionic compounds because in an ionic compound this cohesive force would be purely a Coulombic force that would be, in rationalized MKS units, 1 over 4 pi epsilon zero times the product of the charges and divided by the distance between them.
Taking those two potentials together, and you've all seen this many times before, there is a minimum in the energy.
And then absolute zero-- that is the separation at which this atom pair would sit-- and this would be the binding energy of that ionic pair.
If you carry this through to get the entire energy of the crystal, you find that the energy of the crystal goes as 1 minus 1 over-- haven't introduced this term yet, so let's hold off on that.
What I'm going to do now is to take a look at this potential and expand it around the equilibrium separation just with an empirical collection of terms.
But from the shape of this well, we can see why the bond length should increase as the energy of the atomic pair increases.
At absolute zero, the separation will be this equilibrium separation, d zero.
But as you increase the temperature and pump vibrational energy into this pair of atoms, they will get a certain increase in energy.
And as a result, the pair will be able to vibrate between these two limits.
Put in some more thermal energy by raising the temperature, and the pair will be able to vibrate between these limits-- larger limits.
And so the average position of the atom, the separation of the atoms, is something that is going to increase with increasing temperature.
And that is purely a consequence of the asymmetry in this potential well.
If this were a Coulombic bond, we know exactly how to model the attractive part.
The repulsive part is a little harder to model because that is something that's inherently quantum mechanical.
As the ion pair gets close together, the electrons start to perturb each other's orbitals.
And a very common thing to do is to model this as 1 over d to the sum power n. And n is typically on the order of 8 to 12.
So this is something that rises very, very steeply.
I was once at the Gordon conference where a fist fight almost broke out because somebody was giving a presentation and started from the beginning by introducing this model to the potential well.
And he says, "As usual, we'll write this as a power law, 1 over d to the n." And somebody in a front row jumped up and said, "what do you mean usually?
Sensible people use exponentials."
And the speaker said, "No, I'm going to use a power law."
That's dumb.
Exponentials work better.
Power laws, exponentials, power laws, exponentials.
But in point of fact they're both grossly empirical and one is not better than the other.
The algebra changes a little bit, but people get very, very emotionally attached to their physical models
[INAUDIBLE] law in the exponential
Good, you should've been there to break up the fight.
If you'd gotten between them, you'd have been likely punched in the nose.
[INAUDIBLE] OK so I'm going to use the power law because as we've seen they're the same thing as exponentials.
So now what I'm going to do is expand this in a purely empirical way-- measuring distance from the minimum of the potential well.
So what I'm going to say is that the potential as a function of x-- and now I won't call it d because I'm going to measure distance, x, from the minimum in this well.
I'm going to say V is a function of x.
There is a well in which the-- which keeps the atoms separate.
I'll assume that is a parabolic potential well, so it'll go as first some constant c times x squared.
Then the thing that causes thermal expansion is the asymmetry.
And I want the asymmetry to increase the energy more rapidly on the side of negative x than on the side of positive x.
So I'll put in a term, g x cubed.
And I put a negative sign in here because I want it to go up as x becomes negative.
That means the ion pairs are decreasing at separation.
And finally, a very fine touch, if x becomes very large, the potential well should soften.
The edges of this are starting to level out and get lower than the purely parabolic part for very large separations.
So that's what I'll use as a very simple model for my potential.
This is what gives the asymmetry, and this is what gives softening.
And now I'm going to ask, what is the average value of x?
The probability of finding a particular x is going to be proportional to e to the minus V of x over kt.
So I can say that the average value of x is going to be the integral from minus infinity to plus infinity of the value that I'm trying to find the average of, namely x times the probability of finding that x.
And that's going to e to the minus V of x over kt.
And I will integrate that over x.
And then I will normalize by a term which is the integral of the probability from minus infinity to plus infinity of e to the minus V of x over kt.
So that's a general expression for an average.
It's the thing you want to average times the probability of finding the particular value divided by the integral of the probabilities.
OK now let me do some very straightforward things and I'm just going to rattle these off very quickly.
We'll expand this in terms of a series of terms.
And I'm going to expand both the numerator and denominator using the approximation that e to the u is approximately equal to 1 plus u.
So the numerator-- x times e to the minus cx squared over kt.
I'm going to take the higher order terms and say that e to the gx cubed over kt times e to the minus-- this is a minus sign out in front-- times e to the fx fourth over kt, can be approximated by, since these are very, very small terms, 1 plus g x cubed over kt times 1 plus fx fourth over kt.
So that's going to be, if I put an x e to the minus cx squared, I'm going to have a term 1 plus g x cubed over kt, plus f x fourth over kt.
And then a very, very minuscule term, f times g times x to the 7th over kt.
And I'm going to throw that out.
If I look at these integrals one at a time, the integral from minus infinity to plus infinity of x e to the minus cx squared over k t, that's the first term in expansion, times dx.
That's going to be zero.
Because for positive x, the thing that we're integrating is plus.
For negative x, the thing that we're integrating has the same exponential but has a minus sign.
So that integral is zero.
So the next term I want to look at is x times gx cubed over kt.
And for this I turned to my handy book of integrals.
The integral from zero to infinity of x to the 2n, where that quantity is even, times e to the minus a x squared dx, is equal to, as you all know, 1 times 3 times 5 all the way up to 2n minus 1 divided by 2 to the n plus 1 times a to the n times the square root of pi over a.
And proof of that the integral is left as an exercise to the reckless student.
Yes?
The x cubed [INAUDIBLE]
That's why this is a 2n here.
So I can say that the x cubed term is--but we don't have an x cubed term.
The next term is going to be equal to x fourth.
And this I'm going to drop because that's going to be x fifth.
So the integral from minus infinity to plus infinity of x to the fourth power times g over kt times e to the minus cx squared over kt dx.
If I use this formula, turns out to be a very, very simple thing.
It turns out to be 3/4 times the square root of pi times g times kt over c to the 5/2.
And kt is to the power 3/2.
So that is-- if I throw away the term in x times x fourth which is odd, that will vanish and I've then thrown out only a really large term in x to the 7th.
For the denominator, the integral that is there before us is the integral from minus infinity to plus infinity of e to the minus cx squared over kt times 1 plus g x cubed over kt.
This thing is going to vanish because x is odd.
And then I'll have plus fx fourth over kt.
That will not vanish but it's going to be a relatively small term, so I'll go like that.
Which means I just have to integrate this Gaussian factor.
And that turns out to be equal to 2 times 1/2 times square root of pi kT pi times T over c. So this is the denominator.
And if I take the ratio of those two integrals, I have that the average value of x will be equal to 3/4 square root of pi times g times kt to the 3/2 over c to the 5/2, divided by the square root of pi times kt over c to the power 1/2.
And if I carry through the algebra correctly, this turns out to be 3/4 of g over c squared times kt.
So there is the average distance between this pair of atoms as a function of temperature in terms of characteristics of the potential.
The first thing that is very gratifying is that the bond length, according to the simple model, should increase linearly with temperature.
So it tells us we would expect things to expand linearly with temperature which corresponds to reality.
The next thing we'll see is that the amount of expansion goes inversely as the square.
That's a strong dependence-- inversely as the square of the parabolic part of our potential.
That was the c-- c was the coefficient in front of cx squared in our potential.
So the stronger the bond, the smaller the thermal expansion coefficient is in proportion to the reciprocal of the square of that term.
And then finally, the last important term to survive is g, the asymmetry.
And the thermal expansion coefficient should increase with the first power of the term that describes the asymmetry of the potential well.
So your intuition would tell you that that's the way the thermal-- linear thermal expansion coefficient should vary.
The interesting thing is that it depends more strongly on the parameter that gives the parabolic part of the potential well.
And secondly, that it does predict a thermal expansion coefficient that increases-- the separation that increases with-- in proportion to temperature.
And therefore there is a constant linear thermal expansion coefficient.
Now all this is true at elevated temperatures.
At low temperatures, thermodynamics kicks in.
And it turns out that the linear thermal expansion coefficient, as does energies of vibrating atoms, must go to zero as temperature approaches absolute zero.
So a typical variation of the thermal expansion coefficient with temperature is something that does this.
The same way as-- the same sort of variation as things like heat capacity.
And this temperature in here is on the order of the Debye temperature of the particular material.
All right, so much for linear thermal expansion.
And I would now like to turn to something much more interesting.
And that is physical properties that involve property tensors of rank higher than two.
One of the important and more interesting of these is a property called piezoelectricity.
Piezo means pressure, so the property literally means pressure electricity or pressure induced charge.
Now what is the charge?
First of all, this does not describe a property where-- if I were to pick up a chunk of crystal, and I drop it because I just got an electric charge.
These are not mobile charges, these are bound charges.
And the reason for the induced charge is the fact that we are squeezing or stretching an ionic material, and that's moving charged ions around.
So what we're going to do if that process involves motion of charged ions is to induce a dipole moment per unit volume.
I have a little primer on some quantities in electromagnetic theory and I-- I brought them with me upstairs to look them over and I forgot to bring them down.
But I don't think we'll need them in what's left of this hour.
So let me say that for matter and bulk, the quantity that is involved in piezoelectricity is the polarization.
And polarization is dipole moment per unit volume.
And a dipole moment, to refresh your memory, is that if I have a positive charge plus q and negative charge minus q separated by a distance, d. The dipole moment, which I'll indicate by a lowercase p, has a vector quality and it's equal to the magnitude of one of these charges times the distance between them.
And we'll define the distance as going from the positive charge to the negative charge.
That's a crazy way of defining a pair of charges of equal but opposite sign.
The reason for defining a quantity like the dipole moment is that this combination of charge and separation of the charges comes up in all sorts of problems in electrostatics and electrodynamics.
And this is the delight of people who teach elementary physics.
You are given, on the first quiz, a question that says-- calculate the electric field that is created by a set of dipoles on the Great Court which spell out MIT.
And we can come up with all sorts of crazy configurations of dipoles and then calculate the field that they produce.
And the field turns out to be something that always involves the product of charge and separation in a vector sense.
Probably the best illustration of this would be if I have a dipole charge plus q, charge minus q, separated by d. And I put it in the electric field.
The electric field, which is defined as the force per unit charge on a positive charge is going to pull the positive charge this way.
And it's going to pull the negative charge this way.
And what that's going to do is to place a torque on the dipole.
And that torque is going to be equal to a vector product of the electric field and a vector.
You can express it in terms of the separation of charges defined in a vector sense going from the positive charge to the negative charge.
So there's an example of how the torque on a dipole directly involves the product of charge and the separation between them and the angle between the field and the dipole.
So it's a useful quality and, without further apology, we will use it in our discussion.
Now polarization-- dipole moment per unit volume-- is just going to be the dipole moment of one dipole pair times the number of them per unit volume.
And that will be a macroscopic measure of polarization.
OK, if we subject a piece of material that contains ions to a stress- so here is a stress tensor, sigma ij.
Let me call it jk.
We will create reorientation of the dipoles.
We will have a polarization, a dipole moment per unit volume.
This will be a vector because there will be some net dipole moment per unit volume.
And that will just be the sum of all the little individual dipole moments on the ions-- ion pairs.
And if the stress is not too large, every component of polarization is going to be given by a linear combination of every one of the components of stress.
So we will have an array of coefficients, d ijk The i goes with the P and the j and k goes with the elements of stress.
This is something that's called the direct piezoelectric effect.
It's the basis of a lot of every day devices-- pick ups on the old records which nobody uses anymore, cigarette lighter which nobody uses anymore because everybody's quit smoking-- but worked the cigarette lighter, in some of the more modern incarnations of the lighter, was the piezoelectric effect.
It's important to note that the charges that are involved in the piezoelectric effect are bound charges.
Again you can't get a shock by picking up a piece of a strongly piezoelectric material like quartz.
Might say then, well if there's no spark, how could a piezoelectric material in a cigarette lighter cause the lighter fuel to ignite.
The charge doesn't move.
How could there be a spark?
How could the lighter fluid be ignited?
Anybody got an idea?
Charge can induce current flow in circuit elements that are removed from the charge.
So if you have a little capacitor in the lighter that is near the piezoelectric element that will develop the charge, you can induce current flow in a circuit as a result of that capacitor experiencing the electric charge of the piezoelectric induced charge.
And that's how the little piezoelectric cigarette lighters worked.
OK, this stress-- array of stress elements, nine of them, constitutes a tensor of second rank.
Polarization is a vector per unit volume.
This is a tensor of first rank.
This is a tensor of second rank.
Therefore it follows that this array of coefficients, d ijk will be three equations involving all nine elements of stress.
So it'll be a total of 27 coefficients in the array.
And this array constitutes a tensor of the third rank.
So this is clearly a property tensor which relates a generalized force-- the stress tensor to a generalized displacement, which is the dipole moment per unit volume.
Since this is a property of a crystal, that tensor is subject to symmetry restrictions.
In particular, if there is some direction, cosine scheme Cij, that corresponds to the operation of a symmetry element, then it follows that if we evaluate the new tensor elements, d ijk prime in terms of the old elements, that is going to be a transformation of the form Cil, Cjm, Ckn times all 27 of the original tensor elements, d lmn.
And the l, m, and n are dummy indices, just as with our second rank tensors.
And the ijk are real indices that go with the subscripts on the particular tensor element.
So there are 27 coefficients.
And each transformed element is going to be a sum therefore of 27 coefficients times-- each times three direction cosines.
So each transform tensor element is going to be given by a sum of 3 times 27 quantities.
And to do the complete transformation would involve doing it 27 times.
So 3 times 27 squared is the number of terms that we're going to have to write down to transform the entire tensor.
Not a terribly encouraging prospect unless you use the method of direct inspection, namely that the tensor transforms like the product of the corresponding coordinates.
And then it works fairly quickly, particularly for symmetry transformations, where there are a lot of zeros in the direction cosine scheme.
So we're going to have to do this for each of the crystal-- crystal point groups.
So we're going to have to do this exercise of 3 times 27 times 27 times the 32 point groups that exist, which becomes an even more daunting task.
But let's do one such transformation and let me show you that it, in fact, it's going to go fairly easily.
So let me follow the tracks of what we did with second rank tensors and examine the restrictions imposed by inversion.
OK so the transformation of the axes, as we've seen before, that's produced by 1 bar.
That's going to take x1 and change its direction to minus x1.
It's going to take x2 and change its sense to an x2 prime that points in this direction.
And take x3 and change its sense to something that points down here.
So that's x3 prime.
So the relation between the axes, before and after inversion, is that x1 prime equals x1-- minus x1-- x2 prime is equal to minus x2.
And x3 prime is equal to minus x3.
And we've seen this before.
It's deja vu all over again.
So Cij, the direction cosine scheme, is equal to minus 1, 0, 0.
0, minus 1, 0.
0, 0, minus 1.
So let's transform a representative tensor element for a change of axes that corresponds to that direction cosine scheme.
And that's the same as physically inverting the crystal and demanding that the property be unchanged if that is a symmetry transformation which leaves the crystal invariant.
So if we look at some d ijk prime, that's going to be Cil, Cjm, Ckn, d lmn, where this is a sum over the index l, the dummy index m, and the double index n, from 1 to 3.
If we sum over Ci something, the only term of the form Ci something, regardless of what i is, is going to be the diagonal term Cii, the diagonal direction cosine Cii.
In other words, if i were 1, the only term of the form C1 something that is non-zero, is C11.
If I sum over m, the only term that is non-zero is C, with m equal to j.
So this is going to be Cjj.
And if I sum over n, the only term that is going to be left is Ckk.
And all this will be times d ijk, the single tensor element.
Regardless of the values of i, j, and k, they are all equal to minus 1.
So the operation then of inversion, then says that for the transformed crystal, d ijk prime for any i, j, and k, is going to be equal to minus d ijk.
But if this is a symmetry transformation, the tensor element has to be the same before and after.
And the only number which can be equal to the negative of itself is zero.
So we have, with one swell foop, transformed 27th tensor elements-- each one of them a sum of 27 elements in the original tensor.
So the conclusion is, if a crystal has the operation of inversion, all d ijk vanish.
What does that mean?
Means the property just doesn't exist.
So no crystal which has an inversion center can display piezoelectricity.
Which means that none of the 11 Laue groups can be piezoelectric.
And the only point groups for crystals that can be piezoelectric are the 32 minus 11 equals 21-- and this isn't even one of my good days-- acentric point groups-- the noncentrosymmetric point groups.
And, these, in principle, can be all different.
So we can't get away with just looking at the restrictions of Laue groups as we did with second rank tensors.
We're going to have to look at every one of the point groups that have no inversion center.
And these can, in principle, be different, and they usually are.
Let me give you a physically-- physical feeling for why the piezoelectric effect cannot exist for a crystal that has an inversion center in it.
Let's suppose we cut a wafer from our single crystal, regardless of orientation.
And we subject it to a compressive stress.
We will see-- and I've got a primer on some simple electrostatics that may be useful-- that a polarization in the x1 direction manifests itself as a charge per unit area on the surface and an opposite charge on the other surface.
So P1 corresponds, numerically and physically, to a charge per unit area on a surface that's normal to x1.
OK now let's suppose that that crystal has an inversion center in it.
And that means we should be able to invert the crystal and measure exactly the same charge per unit area.
Well, if this is surface a, and this is surface b, we can imagine ourselves doing the thought experiment in which we invert the crystal.
So this surface becomes b and this surface becomes a.
And we apply the same sigma along x1-- a compressive stress.
It's really doing exactly the same thing that we were doing before-- inverting the crystal.
And now we say that the property has to stay the same.
So now we are demanding that surface b have a positive charge per unit area, and surface a develop a negative charge per unit area-- completely the reverse of what happened here.
But yet what we're doing is just squeezing the plate.
The crystal can't tell which side is up, which side is down.
We're applying the same compressive effect.
So these two distributions of charge are not the same and can be the same only if that charge is zero.
So physically we can see why the crystal can't be piezoelectric.
If you don't like to invert the crystal-- if you can't do that physically-- one thing we can do is leave the crystal alone and invert the compressive stress.
So this is sigma top and this is sigma bottom which is exactly equal to it but opposite in direction.
If we invert the stress relative-- relative to the crystal-- then what we've done is to take sigma top-- leave the crystal alone-- so this is now the sigma that was squishing the top.
Here's the sigma that was squishing the bottom and well, that's clearly doing the same thing.
So inverting the stress, relative to the crystal, is the same as inverting the crystal relative to the stress.
And nothing is supposed to happen but clearly the sign of the charge is going to be reversed.
So that just won't work.
The charge has to disappear.
OK this is the direct piezoelectric effect.
It can exist only for crystals that lack an inversion center.
Let me mention some other piezoelectric effects.
So we mentioned that a polarization, given by a third rank tensor times an applied stress, sigma ij, is the direct piezoelectric photoelectric effect.
However, if we have a stress, we also have a strain.
So another piezoelectric effect, which is not dignified with a name, is that we can apply a stress that creates a given state of strain.
And we again, we'll get a polarization per unit-- a charge per unit area of polarization.
And this is a set of coefficients which are represented by e. And we can write a tensor relation of this sort-- that components of polarization, charge per unit area, normal to axis xi, is given by a linear combination of every one of the components of strain.
This is something that is not dignified by any name.
And clearly, the piezoelectric coefficients, e, have to be related to the piezoelectric coefficients, d, by the elastic constant somehow-- unless we are doing the same thing.
We've got to create a stress, operationally, in order to create the strain.
So a stress creates a strain.
And that has to describe the same polarization if we've done the same thing.
So the e's and the d's have to be related.
Another piezoelectric effect involves a applying an electric field.
And if we apply an electric field, we will in general create a strain.
This direct piezoelectric effect is three equations that involve a linear combination of nine elements of stress.
This piezoelectric relation here is going to involve writing an equation for each of nine elements of strain.
And there would be three components of the electric field vector.
So it'll be three this way.
So both of these relations involve 27 elements.
So again we're going to have a third rank tensor because the electric field is a vector-- strain is a tensor.
And now the mind boggling thing that I will leave you with is exactly the same set of coefficients, d ij describe both effects.
And this effect is given a special name.
This is called the converse piezoelectric effect.
And it's not at all clear how strain, in terms of electric field, should be described in terms of the same array of 27 numbers as polarization per unit volume in terms of stress.
And that is an argument that is based on thermodynamics.
So that's a good place to quit.
We'll have some fun in looking at a few symmetry restrictions imposed on the acentric point groups, 21 of them.
We're not going to do all 21.
If you know how to do one or two, and marvel over the results, you can do them all.
But one thing that we'll be asking is how can we represent the variation of the piezoelectric effect with orientation of the crystal.
And the answer is we can't.
We can't.
Because what is the direction of what we're doing?
There are nine elements of stress, not just three components of a vector which defines an orientation.
There are nine parameters here.
And there are three components to the charge per unit area-- that's a vector.
So all that one can do for these higher ranked tensors is to look for a special, generalized force.
And then perhaps for a very special form of that generalized force, that tensor, look at how the polarization changes.
But you're going to have to look at six different representation surfaces-- each one which will describe a different piezoelectric effect.
So this will become less simple in many respects.
But correspondingly, more interesting because we'll have some really wild, mind boggling representation surfaces for piezoelectric effects.
OK so we'll stop there and resume on Thursday.
And I'll let you get back to the MRS if that's where you earlier today.
I think it's about five after the hour so we ought to get started.
Before we forge bravely ahead, I'd like to make sure that everybody has a copy of the things that we've been handing out.
There is a copy of the handwritten notes on the derivation of the 17 plane groups.
Need a copy of that?
There you go.
And then there is, in addition, a set of diagrams from the international tables that give very nice pictures of all of the plane groups in the arrangement of symmetry elements.
That's the only other thing that was handed out up to this point.
For your continued edification and amusement, I have another problem set.
And let me say something about this problem set.
The first problem gives you some angles between crystal faces and asks you to deduce a possible set of lattice translations which are consistent with those angles.
And in the days before diffraction, that was what crystallography was all about.
It really was mapping the geometry of crystals.
And the interesting thing is, you could measure these angles very precisely with a device called a reflecting goniometer, Provided you had a crystal with nice, shiny faces on it.
You had a two circle instrument, and you could adjust the crystal so that a light beam, a very finely collimated beam of light, was focused into an eyepiece.
And you could measure angles to within not one minute, but one second of arc.
And from this you could deduce if you could assign Miller Indices to the faces.
You could deduce not the absolute magnitudes of translations, but you could deduce the ratio of them.
And along came x-ray diffraction and shook everything up.
Turns out, sometimes you were right and other times you were wrong, because you had determined a self consistent set of indices, but not the ones that were correct.
So this is to give you a little look at the early days of crystallography, and to see if using the way Miller Indices are defined you can calculate the ratio of axes.
Not terribly demanding, but worth doing.
And we had talked earlier about these combinations of symmetry elements as constituting the elements of a group.
And in the second problem I ask you to take a not terrifically simple, but a fairly complex point group, 4mm, and show that, indeed, the operations that are present satisfy the group postulates.
And that is worth doing once so that you convince yourself these really are groups.
And then the third problem is easy to state and it is diabolically tricky.
So I'll let you have a go at it, but don't beat your brains against your desktop for an entire evening over it.
Try it.
If you don't see the solution to it, all will be revealed later on in class.
But it's based on the fact that the plane groups, which we've now derived, 17 of them, these can be viewed as the base level of a three dimensional space group.
And for each of them, you could take another translation, t3, that was perpendicular to the plane of the group, and just imagine all of those rotation axes and all of those mirror planes not being two dimensional operations, but three dimensional operations.
The only one where you could pick the translation, generally, would be the oblique lattice, p1, no symmetry at all.
And there you could pick the translation in any orientation you wished.
But for each one of the 17 two dimensional plane groups, there is a corresponding three dimensional space group.
So we got 17 three dimensional symmetries for free without really doing any additional work.
However, there's one thing that one should not gloss over.
We found that a rotation point in a two dimensional plane allowed only a very limited number of lattices.
But we were not thinking of things in terms of three dimensions.
So what I'm inviting you to do in the third problem is to consider, now, that construction that we did where we showed that rotation angles were restricted to values of cosine of alpha equals 1 minus p over 2.
Now generalize that to a three dimensional situation, where the translation does not have to be perpendicular to the rotation axis, but can be inclined to it.
Are there additional rotation axes allowed?
Are their fewer?
Something we ought to examine.
But there's a little bit of ingenuity that you have to use in that proof.
But I invite you to have a go at it on your own.
So with that tantalizing introduction that will want to send you running home and start it as soon as we finish our class this afternoon, let me hand out, without any further blather, problem set number five.
Pass that in the back corner.
Last time, without saying much about them, I handed out this extract from the international tables that summarized the 17 two dimensional plane groups and their properties.
And I'd like to spend a few minutes going over the considerable information that's contained on these pages.
Each one is treated in the same way, and let me start with one that's almost trivially simple, and that's the two-fold axis in the oblique net.
Across the top of the page you find, in boldface, the symbol for the plane group.
And then, just counting in order of increasing symmetry, the number of that particular plane group in the set.
And then a symbol that really has its full meaning only in three dimensions, but it gives the symbol for the lattice, again, 2 and then a 1 and a 1, which says that if this were a three dimensional space group, there would be a two-fold axis in one direction and no symmetry at all in the other two directions.
And then, next comes the symbol for the point group of the crystal.
And we derived this particular group by dropping point group 2 into a lattice.
And then the coordinate system that is necessary to describe the features of this plane group, if you take the edges of the unit cell as the basis of a coordinate system.
So this is something called a crystal system.
So for each of the plane groups, you have this information spread across the top of the page.
Then underneath that is a diagram that, by means of open circles, shows the way in which that particular symmetry moves a motif where atoms, in the case of a crystal, moves atoms around.
And for p2 we have the pair of motifs of the same chirality related by a two-fold axis.
And that, then, is hung at every lattice point of an oblique net.
And then, immediately to the right, is the arrangement of symmetry elements in the group.
Then, the way in which that particular plane group can move atoms around to generate a structure is summarized for you.
And the first thing they have to state is where you're going to choose the origin.
There is no unique lattice point.
It's convenient to take the origin at a location of high symmetry, and in this case the smart thing to do is to take the origin at one of the two-fold axes.
That's a non trivial question, because there are some groups that have different kinds of rotation axes.
P4 has a four-fold and the two-fold.
Taking the origin of the coordinate system at a location of high symmetry, then, gives you the scope of picking either a two-fold or a four-fold axes.
Yes,sir?
On the left hand figure, what's with the-- those aren't mirror planes [INAUDIBLE] are they?
On the--
[INAUDIBLE]
Here they have this.
That's just to split things up into quadrants, so you have a feel for how the atoms hung at the lattice points split up into different quadrants.
That is a very good point, though.
Mirror planes are shown as bold lines.
Sometimes the outlines of the cell look pretty bold themselves.
So if you turn the page to pm, if you have the notes with you, can see the lines that are reference lines for the cells are lighter in their weight than the symbols for the mirror planes.
But it's a very subtle difference.
And when you Xerox it a couple of times it becomes almost indistinguishable
Can you just say, one more time, [INAUDIBLE]?
That means that this is the location of 0,0.
And it is always assumed, but nowhere stated specifically, that the origin is in the upper left hand corner.
And what is also assumed, but never stated, is that the x-coordinate in the a-axis goes down and the y-coordinate in the b-axis goes from upper left to the right.
That is nowhere stated anywhere in the international tables, but that is the direction that is assumed for the reference axes.
So we'll take our origin of the two-fold axis.
That's a sensible thing to do.
And then there are two ways you can look at how the particular plane group generates a pattern.
You can do it with little circles or little commas, or something like that.
In other words, do it graphically.
But a way to communicate the atomic arrangement in a structure, which is going to be the most free of ambiguity and rigorous, is to do it analytically.
And that's what the tables do for you next.
They give an analytic description of the way in which the symmetry elements will move an atom around.
For this particular plane group, things are very simple.
This is the direction of x, this is the direction of y.
And if we plop one atom in here, it will be at a location x and y.
So that's what happens when you drop one atom in it.
That atom gets hung at every lattice point, and it gets rotated by the two-fold axis.
So you're going to get two of them per lattice point.
Somewhat arbitrary which pair you take.
The coordinates of this atom are x and y.
The coordinates of this atom inside of the box is 1, minus x and 1, minus y.
That would do it, but it makes sense, esthetically, and to see that the atoms are related by symmetry, if, instead, you specify as the two atoms per cell the two that are hanging at the lattice point.
And in that case, the coordinates of the first, if they are x and y, would be minus x and minus y.
If you have a pair of numbers, 0.283 and 0.456, then minus 0.233 and minus 0.456 leaves no doubt that these are positions that are related by symmetry.
If you have coordinates, like 0.2, 0.3 and then 0.8, 0.7, without doing some arithmetic in your head, it's not clear that those are going to be atoms related by symmetry.
So the coordinates that will be stated for you, then, will be the coordinate of the pair of atoms at the lattice point, and some of the coordinates would be negative.
This is something called the general position.
And that transformation of coordinates is different for every one of the plane groups.
That's what makes them different.
They're different in the way they move around an atom in a general location and fill space with it.
So that's the general position, and it is unique for each plane group.
And there are several ways in which this information is conveyed to you.
Move this over a little bit; x, y and minus x, y.
The first thing that's a characteristic of the position is the number per cell.
So you get the number 2, or this is sometimes referred to as the rank of the position.
Drop in 1 and you get a second one out related by symmetry.
The next is, and I'll go to the last next, this is the site symmetry.
And by definition, this is always 1, no symmetry at all, for a general position.
So why make a big deal about all of these characteristics of the general position?
The reason is that there are locations that are termed special positions.
And let me clean up this right hand part of my diagram.
And say that, for plane group p2 you will always get two atoms per cell if you place an atom in an unspecialized location.
What would be a specialized location?
Suppose we would drop the atom down right smack on top of that two-fold axis?
Then that two-fold axis is just going to twirl the atom around on its axis and it's not going to map it into a second location.
So following the general position is always provided to you a set of special positions.
And what's special about them?
They are on a symmetry element.
You can look at the characteristics of a special position in two ways, either geometrically or you can do it analytically, as I'll show you in just a moment.
Imagine that the atom is to sit here, x and y migrate progressively towards the location 0,0.
And then the other atom related to it by two-fold symmetry will move towards it until finally the two of them will merge into just one single motif.
So when that happens you get not two per cell, you get only one per cell.
And the reason for that is the site symmetry is 2.
It's on a two-fold axis.
But there are other two-fold axes as well.
What if the element migrated to the position 0, 1/2?
If that were the case, if the representative atom moves to here, this one up here will migrate down and the two will merge into one atom that sits at just 0, 1/2.
So there'll be another type of special position, one per cell, also, on a two-fold axis.
This one was at 0,0, and this one is at 0, 1/2.
You remember, we made a point of saying that there are four different kinds of two-fold axes within this plane group.
Different in how they're positioned relative to a pattern, different in that they are not mapped into one another by some other symmetry element which is present.
So each of these four locations is a location of another general position of rank 2, sitting on a two-fold axis.
x and y migrate down to 1/2, 0, and this one and this one will come together.
So there's another one, giving you one per cell, sits on a two-fold axis, and this would be at the location 1/2, 0.
And finally there's another one in the center of the cell, and if you let the atom migrate to that location, again, the atoms will coalesce pair wise into a single one.
So there'll be another one, one per cell, and also on a two-fold axis, and this would sit at the location 1/2, 1/2.
And those, then, are the characteristics of this particular simple plane group.
There's another symbol that's added, and this is something called the Wyckoff symbol.
Wyckoff was a crystal chemist who attempted valiantly, back in the early days of the century, when there were a couple of dozen structures determined per year because it was such a grand, new adventure, and few people knew how to do it.
But Wyckoff tried to summarize, between one pair of hard covers, all of the crystal structure determinations that had been performed in a single year.
And he went at this courageously for perhaps a dozen years, and then results began to accumulate so rapidly he just said, the heck with it, clapped covers on it, and that was the end of the series of books.
So they don't get very far, but it was a real useful contribution at the time.
This is just a shorthand way of referring to the position.
And he starts with a for the most specialized, runs the way up through the alphabet until you've assigned a letter to each of the positions.
This is a nicety.
All these special positions, for example, sit on a two-fold axis, and you don't really have to specify the coordinates, because they have to be either 0 or 1/2.
So this gives us a way of saying you have an osmium atom in position 1a and you have an oxygen atom in position 1c.
And it's a nice shorthand way of avoiding mentioning things which could be calculated once and for all and not stated explicitly.
So this, my friends, is the language in which you will see structural datas cited, when people have determined, using diffraction methods, the locations of all of the atoms within the units, some of a particular material.
If the coordinate is variable, with today's techniques and high speed computation, you can get x as a fraction of a cell to usually at least plus or minus 1 in the fourth place.
So it's data that can be determined extremely precisely.
Yes sir?
I didn't quite get the exact definition of the general position.
It's a set of points related by--
The general position, it's a set of equivalent positions that are related by symmetry.
Yeah.
And the general position, yeah-- Each of these is called a set of equivalent positions, equivalent in the sense of being related to one another by symmetry.
And the entire set is sometimes referred to as an, singular, equipoint.
Sort of a condensed jargon for a set of equivalent positions.
So when one speaks about an equipoint of rank 2, in this case, or rank 1.
Let me just, even for this trivial two dimensional symmetry, give you an example of some information that falls out of this immediately.
First of all, you can use these special positions only once.
If you put an atom in position 1c at 1/2, 0, it's used up.
You can't put another atom in there in a given structure.
Secondly, if you think more globally, not in terms of atoms, which, to a good approximation, are spherically symmetric.
But think in terms of a molecule, if you're a polymer scientist or an organic chemist.
and not place an individual atom, but place an entire molecule.
If there's one molecule per cell in this particular symmetry, that molecule has to sit on a two-fold axis.
And that means the configuration of that molecule is going to be limited to having to conform to a two-fold rotational symmetric.
So just from the density in the lattice constitute, if you find there's one molecule per cell, you know that molecule has to have two-fold rotational symmetry.
So you can say something about the structure of the molecule.
So there are lots of ways in which the symmetry information has something to say about the arrangement of atoms in the structure.
Any further questions?
Yes?
How did they decide, between b, c, and d, which was the most special form?
That is a very good question.
And the way it's decided is by you getting there first before anybody else did it.
So everybody follows, now, the conventions that are given in the international tables.
That's sort of the dictionary of all these terms.
But you're absolutely right.
What is more special about this one than this one?
Well, they're all locations of the same symmetry.
You can turn one into another by changing the origin of the cell, where you're going to define a lattice point.
So they are thoroughly arbitrary.
What is generally done is to take 0, 0 for your particular choice of origin as position a.
And then what should come next, 0, 1/2 or 1/2, 0?
The truly observant among you will notice I snuck a quick peek at the international tables before I wrote one of these down.
I can never remember which is which.
And it is arbitrary.
Usually you end up with the one that has both coordinates non-zero, but that is code that's embodied in the international tables.
But that's a good question.
Yes, sir?
How did you decide to make the motifs [INAUDIBLE] special decisions?
Oh, I just changed the value of x and y.
Those are numbers that can be, for the representative atom, each of them can be between 0 and 1.
So what I just said was, what will happen if we let x and y take on not general values, but very special values?
And the special values are going to be things like 0 and 1/2, if I have picked my origin at a two-fold axis.
It's another reason for picking your origin at a location of high symmetry
But how about the other non-origins of these special decisions [INAUDIBLE]?
There are two ways you can do it.
One of them we just did.
We just looked at the geometry and we said, if this representative atom, this is the one we define as being at location x, y.
If x and y both approach 0, this atom and the one over here are going to migrate towards the origin lattice point and eventually just coalesce.
If I let x and y and migrate, x go to 0 and y go to 1/2, then this one and this one will come together and coalesce.
So that's why I did that, because I knew that the number I would generate would be smaller than the number that I would get for a set of coordinates for an atom that was off the symmetry element.
I said there's another way of doing it.
For something this simple it is not really that profound, but it is a way of checking whether you've counted the same position twice.
I said you can imagine these special positions arising when x and y assume special values.
So let's let, for example, x be 1/2 and y be 0 and plug those coordinates into my equation for the general positions.
So I'll put in 1/2 for x, 0 for y, and then I'll put in minus 1/2 for x and minus 0 for y.
But what goes on at minus 1/2 has to be the same as what goes on at plus 1/2, if the structure is based on a lattice and periodic.
So what I'm really getting is 1/2, 0, 1/2, 0 twice, which is the same as saying, if x is 1/2 and y equals 0, I'm putting two atoms together right on top of one another.
So you can do it analytically.
And for the more complicated symmetries, where there might be for 20 or 48 atoms, that's a good way to see if you've looked at the same position twice.
So we're going to do a couple more before we move on, and this one is almost trivially simple.
Any other questions before we go forward?
Let's take one that is slightly more complex.
And I'll go to one of the rectangular groups.
And let me look at, let's take plane group p2mm, which has a fair amount of symmetry to it.
So this is the group that we got by putting 2mm into a lattice that had to be rectangular, then.
And across the top of the page, this is number six, you'll see that it is rectangular.
It has to have a lattice in a rectangular shape because of the symmetry.
Next comes the short hand symbol.
And I always feel a little bit apologetic when I have to point out the existence of this, and the reason is that people who work with symmetry are no more or less lazy than any other human being.
And some of the symbols become really ungainly when we go to three dimensional symmetry.
And something like a 2 sub 1 over a 2 over m 2 sub 1 over d is a mouthful.
And actually what's done is to just specify the minimum amount of information, which defines the symmetry.
So here, two m's that intersect have to give you point group 2mm.
Then comes the full symbol of the plane group, p2mm.
And for those who are counting, this is number six and the shorthand symbol is pmm, where the short symbol for the point group, along with the lattice, is sufficient to tell you that this is 2mm placed at the nodes of a rectangular lattice.
Then, underneath, comes the representative pattern.
Again, there's a cross in the middle of this just to split it up into quadrants.
The symmetry is one that we've already encountered and was fairly easy to come to terms with.
Two-fold axes at the corners of the cell, and in the center and in the midpoint of the edges.
That's just like p2, except the mirror planes that are present require that that parallelogram in p2 straighten out into a rectangle.
And then we put 2mm at the lattice points so the mirror planes run down through the cell like this, up and down and left to right.
So here is all the symmetry here.
Pattern, I'll say it again.
It's so easy to forget.
The pattern is nothing more than the pattern that 2mm produces hung at every lattice point of a rectangular net.
And all of the symmetry planes and symmetry axes that arise are just ways of defining all of the relations that exist between these things when you perform that process of addition.
So this is, then, the arrangement of all of the atoms in the cell and then some.
And there's a new symbol that's introduced, and somebody asked about that last time.
To me it looks like a little tadpole inside of a frog egg.
Has anybody seen a frog egg?
That's just what it looks like.
It's a little tiny tadpole in there, waiting to hatch out.
So here is our little tadpole.
This is used to indicate an enantiomorph.
If this were a motif with chirality, this one would be right handed, this one would be left handed.
So that indicates all of the atoms or molecules that are the same chirality.
The purists among you will notice that the dummy that produced this diagram did not show the tadpoles conforming to the two-fold symmetry.
Both of the commas point in the same direction.
Tsk, tsk.
Little oversight.
So this, then, is the arrangement of motifs in the pattern.
And now let's proceed to analyze that, according to the nature of the positions that are available of different sorts.
Again, coordinates have meaning only if you define the origin.
And nobody in their right mind would want to pick an origin that is off, at least, the symmetry plane.
But the thing to do is to take the origin at 2mm.
Again, I'll remind you that x goes down this way, y goes this way.
So if I have a representative atom at x, y, I'll have another one at minus x, plus x, another one at minus x, minus y, and one at plus x, minus y.
So there are the coordinates of all the symmetry of related atoms.
So putting down the general position, I get four of them if I place one in the cell.
By definition, the general position is always at a site of symmetry 1.
So this is the site symmetry.
Then we just rattled of the coordinates.
They're x,y, minus x, y, x, minus y, and minus x, minus y.
There are lots of special positions here, now, because we have not only two-fold axes with mirror planes intersecting at them, but they're also locations of just a mirror plane alone.
And there are a total of four distinct, independent mirror planes, this one, this one, this one, and this one.
So we're going to have four positions where the atom sits at a location of symmetry 2mm.
And those are going to be analogous to the positions that we found for the two-fold axes in p2.
And, not surprisingly, the coordinates look very much the same.
0,0 puts us on a location of symmetry 2mm.
0, 1/2 does the same, 1/2, 0 and 1/2, 1/2.
So there are four positions of symmetry to 2mm.
And then there are four positions just on a mirror plane, but not on a two-fold axis.
And we could do that for the mirror plane that runs along the x-axis, or the mirror plane that runs along the direction of x at y equals 2.
And the order in which you pick them is rather arbitrary, but there will be a pair of positions at x, 0 and minus x, 0 that would happen if the y-coordinate was exactly-- let me draw what the position is, just off here to the right in a small diagram.
This would be where you put the atom on the mirror plane running through the origin, then these things would give you only a pair.
The next one is at a position x, 1/2, and that's where the mirror plane would be the mirror plane that is along the position y equals 1/2.
And here we would get atoms coalescing like this.
x is general, but y is exactly 1/2.
So we'd have x, 1/2, minus x, 1/2, and I better put these off to the right, here.
The remaining two are 0, y and 0, minus y.
That is also a position of site symmetry m. And that would be a pair of objects that has x equal to 0. y is anything, so that's this mirror plane here.
And in that case we'd have a pair of atoms, left to right, reflected by the mirror plane that goes through the origin.
And then last one, thank goodness, is one at a location 1/2, y and 1/2, minus y.
And this would be the mirror plane exactly at 1/2.
y can be anything.
x is 1/2.
So that would be a pair that sits somewhere on either side of the mirror plane running through the center of the cell.
That's also a site of symmetry m. Now we assign the Wyckoff symbol by working our way up the alphabet.
So this is a, this is b, this is c, this is d, this is e, f, this is g, this is h, and finally we end up at position i.
For all of these, we get two atoms per cell.
These have rank 2.
For each of these we get just a single atom
Can you explain the two atom per cell thing?
How do you [INAUDIBLE]?
Which one would you like me to do, that last one?
1/2, y?
[INAUDIBLE]
So if x is 1/2 and y is anything, that's this locus here.
And that is a mirror plane.
And if we let this atom move from its location, x,y, down to this location, the one in the lower part of the cell is going to move up and these two will coalesce to a blob that sits there
That would be 2m?
So this would be site symmetry m. And instead of getting four per cell I would get two per cell; this one and the one that's reflected across.
Now the other way of doing it is to not think about it all.
Say here is x,y, minus x, y, x, minus y, minus x, minus y.
And then let's make one of the coordinates specialized.
Let's let x be exactly 1/2.
So I'll get 1/2, y, minus 1/2, y, and then I'll get 1/2, minus y and minus 1/2, minus y.
Looks like four atoms, except if I have one at plus 1/2, the one in the neighbor unit cell is at minus 1/2.
So actually I've had the atoms coalesce pairwise, because these coordinates are actually identical to saying 1/2, y plus 1/2, y, 1/2, minus y, and minus 1/2 is the same as plus 1/2, minus y.
So you can do it that way.
Just put in a special value for x or y, turn the crank on the general position, and you'll find you're going to get the same thing twice.
And if we look at position a, which is 0,0. put in 0 for x, 0 for y, then put 0,0 for all these other four positions, you're going to get 0,0 four different times.
I'm going to do just one more, and the results are here for all of the remaining 14.
So I don't think there's any need to do them all.
One of the things I would like to examine with you is a group that has a glide point in it, and see how that changes things.
Any other questions on this?
Can you just go over the rank of fours?
Four--
Four means four per cell
Four per cell
And if you count them around the origin as 1, 2, 3, 4.
If you count the number that's caught within the box it's 1, 2, 3, 4
And site symmetry of 1?
That's site symmetry 1 because 1 is no symmetry at all, and that is what's general about it.
It's not sitting on a symmetry element which would then fail to reproduce it.
And again, if the motif was not an atom but was a molecule, if you find only one molecule per cell from the density in the cell dimensions, that molecule has to have symmetry 2mm, because it has to sit.
If there's only one per cell, that set of atoms has to sit at a location that conforms to symmetry 2mm.
Let us go to one more.
And I'm going to take p2mg, because it's one that has a glide plane in it.
This is pmg for short.
This is number seven.
P2mg is the proper symbol, and that's point group 2mm.
And, again, the coordinate system is rectangular.
This is one that we derived last time.
The arrangements of elements in this is, once again, a rectangular lattice, two-fold axes.
If we take the origin at 2, would be at the corner of the cell and in the midpoint of the edges.
Then there's a glide plane passing through the two-fold axes.
And there is a mirror plane passing in between the two-fold axes.
And here's a good example of an arrangement of symmetry elements that has a mirror plane in it, and then a line that just indicates the edges of the unit cell.
And can you distinguish the bold line from the light line?
Barely.
But the bold line is the mirror plane.
And what does the arrangement of atoms look like?
This is a tricky one, because there are two different symmetry elements; a two-fold axis, and that's going to take an atom at x, y and repeat it to minus x, minus y.
But here, now, is a case where there's another symmetry element that does not intersect the first one.
So that mirror plane is going to reflect this atom down here to an enantiomorph and reflect this atom down here to another enantiomorph.
So if this is x then this is y, this one sits at minus x and minus y.
For this atom here, y is the same as before.
But if this distance is x, the distance up from the center line will be x, so this second coordinate is 1/2, minus x. I'll do that again since this is pretty small and tight.
This distance is x.
We reflect the atom across a mirror line at 1/4, and therefore the one towards the center of the cell sits up above the location 1/2 by the same number, x.
So this is 1/2 minus x, y.
And by the same argument, this one here is at minus y, and it sticks out beyond the position x equals 1/2 by the same amount, x.
So this one is 1/2 plus x minus y. Nontrivial.
So if I began to tabulate these-- We've picked the origin at 2.
Didn't have to do that.
If you wanted to, you could take the origin at m if you wanted to, and just switch everything by 1/4 of the coordinate x.
We found four per cell.
The site symmetry is 1, and the atom locations were at x, y, minus x, minus y.
And then we had 1/2, minus x, y and 1/2, plus x, minus y.
So the coordinates are getting permuted around in a very more complex fashion than just changing sign.
Special positions.
Just as before, any of these two-fold axes is going to be a site for a special position.
Let x and y go to 0.
This pair at the origin will coalesce and this pair will coalesce at a point that's halfway along the cell.
So there will be two of those, one at the origin, one in the middle of this edge, and their locations would be 0,0 and 1/2, 0.
And this is a site on a two-fold axis and we have only two per cell.
Would this be a location for a special position, this two-fold axis?
It was before.
Is it now?
Key point is independent sites of symmetry.
There's a mirror plane here.
What goes on here has to be exactly the same as what goes on here.
So this is not an independent special position.
In the same way, there's a mirror plane.
And if I let the atom migrate down to the mirror plane I will have the mirror plane at a location 1/4, y.
And these will be on a mirror plane.
1/4, y would be a first, 3/4, minus y would be the second one that I get, and they would coalesce pairwise.
Is this is another mirror plane that could be regarded as an independent position?
This mirror plane, again, is not independent, because it's related to the one we just considered by the two-fold rotation about the center of the cell.
So there's only one kind of mirror plane in the structure.
But, going back to the two-fold axes, there is a two-fold axis here.
There's another one that is halfway along y.
So there's another two-fold axis that is independent at 0, 1/2, 1/2, 1/2.
And this is the entire set.
So there are three special positions of rank 2, two atoms per cell.
We make the most specialized one be named a, work our way up to c, and the general position, then, is referred to as d. Again, so what I did was to say that this mirror plane is related to this one by symmetry, and if there's something on this mirror plane as a special position, I'm automatically going to get it on the lower mirror plane.
This two-fold axis is the same as this one.
So whatever is going on as a result of specializing the location on this two-fold axis is going to be provided for me automatically at x equals 1/2 by the action of this mirror plane.
There's nothing that throws this two-fold axis into this one, though, so they're both independent.
Now, one of the symmetry elements that I steadfastly ignored was the glide plane.
What happens if I place an atom right on the glide plane?
A glide plane, if you use a motif that has some handedness to it, we repeat the atom by translating half of a translation, not yet putting it down, first reflecting it across.
This is an enantiomorph.
Doing it again gives me an atom that's related to the first by a translation.
What happens if I let the atom migrate onto the glide plane?
Let this atom move down to here.
Now it sits here.
This one, repeated from it by glide, will migrate up to here.
Nothing happens.
It's just that the amount that the atom is displaced from the locus of the glide plane is different in the two cases.
But there is no coalescence of the atoms, because of this translation component to the glide plane.
So a conclusion, then, is that glide is never a candidate location for a special position.
This is a good example of a case where I can check what I've done, and to see if I counted the same position twice.
I said that both of these locations are not special position.
The mirror planes are related by symmetry.
Suppose I were not clever enough to notice that, and I said, OK, there's going to be a special position, 1/4, x, 3/4, minus y.
That's what happened if I use this mirror plane as a special position.
What would I get if I took 3/4, y and 1/4, minus y?
Looks like a different specialized position.
But let me put this number into my expression that is the formula for the general position.
If x is 1/4 and y is y, then this position here would be minus 1/4 and plus y, 1/2, minus x would be 1/4, and y is y.
Over here, I got the same thing back again.
And 1/2, plus x would be 3/4, and minus y is minus y, minus 1/2 and 3/4, plus 3/4 are the same thing.
So what I've gotten is the position 1/4, y, and 3/4, minus y twice.
So what this is telling me, analytically, if I just plug in the coordinates of what seem to be a distinct, special position is, I get the same atom on top of itself twice.
And it looks exactly like choosing the first mirror plane.
So that's a good way of checking, particularly in a very high symmetry where there are all sorts of sites of possible point groups that could cause coalescence, to just crank out the coordinates and see if, in fact, you have exactly the same pair with the same coordinates that correspond exactly to the x and y of some previous location that you identified.
I think, hopefully, you have a feeling for how it works.
If you look at some of the higher symmetries, the number of atoms that constitute the full equipoint set of equivalent positions.
For p6mm there are 12 locations in the general position, and because the coordinate system is oblique, x and y get permuted into linear combinations of one another.
So that's really, extremely complicated.
If you look at some of the square symmetries, the number of the special positions is very, very high.
There are two positions in p4mm, there are two positions of rank 1.
There is a position of rank 2 that sits on 2mm.
Three different mirror planes, and the general position with rank 8 requires working your way all the way through g in the alphabet.
Yes?
I had a question about the p2gg [INAUDIBLE] whether they're at the entrance of the [INAUDIBLE]?
For p2gg, the only site of point symmetry is a two-fold axis.
So the only thing that you could use for a special position in p2gg is a two-fold axis.
And there are two different two-fold axes.
They relate in a curious way because of a glide plane.
And this is a case where the site of a special position is not related by rotation or reflection, but by a glide.
P2gg looks like this.
Two-fold axes in the old familiar places, 0, 1/2.
The glide planes are here, here, here, and here.
So which two-fold axes are equivalent?
The glide would take this two-fold axis, reflect it down, and slide it over to here.
So whatever goes on here goes on at this two-fold axis
Sorry.
I guess my question was, can you put the glide planes at any edges of this sites?
[INAUDIBLE], or--
You could.
You could, but you would lose, then, the apparent similarity in coordinates in that, if you put the glide plane at the two-fold axis, if this is x, y, this one over here is at minus x, minus y.
And then if you repeat it by glide, that would slide over to here and then be reflected down here.
So there'd be one pair up here and one pair down here of opposite handedness, so you would not see the similarity.
No, that's right.
If you took the origin at the glide plane, you would not see the simple relation between the numbers that you do when there is that a rotation axis of a mirror plane.
But you can take the origin anywhere you want.
And the advantage of doing it at a symmetry element is then the numbers that describe where the atoms sit are simple permutations of sign, or adding 1/2 to the number of the preceding atom that was just mapped to a new location.
I think you're more than ready for a break.
Let's resume in 10 minutes, and we'll move on to something completely different.
OK. Before we get started, I'd like to deal with a small matter of some unpleasantness.
The class is sort of like a football game.
When there's two minutes to go, you shoot off a pistol.
But when there are two meetings to go until we have the quiz, we shoot off a pistol to wake you up.
We're scheduled to have a quiz on October 6.
And even though October seems far away when you're still in September, that is going to be a week from this coming Thursday.
So if you have problems that you're working on, try to get them to me on Thursday, or just come slide them under my office door, if I'm not in.
And I'll have them back for you on this coming Tuesday.
The way the material has played out is that we're really at a nice, convenient juncture between one chunk of interconnected material moving on to another.
So I think the first quiz we'll confine to two-dimensional symmetry.
And beginning in about two minutes flat, we'll begin to move into-- take the first small steps, at any rate, into three-dimensional symmetries, which will be much more complicated in which we will not deal with the exhaustiveness that we have been able to afford the luxury of in two dimensions.
Before I begin, let me-- does everybody remember their spherical trigonometry?
Has anybody had spherical trigonometry?
OK. What I will then do is give a short primer on some of the definitions and concepts in spherical trigonometry.
And then we shall immediately use this to combine rotation axes in space.
Pass those back
I had a quick question on this
Oh, sure, please
With the limiting possible--
Yep
--reflections and conditions, that does say they have to be figured out?
Actually, that was a good question and something that we are not going to use at all in the symmetry tables.
Somebody asked what about this notation in the far right-hand edge of all of the plane groups-- conditions limiting possible reflections.
One usually doesn't put a two-dimensional crystal in an x-ray beam fairly often, although I suppose a thin film actually is almost a two-dimensional crystal.
But you've probably all heard one way or another about the magic conditions relating the Miller indices of a plane that will require that the intensity diffracted from that set of planes is identically zero.
And they're linear combinations of H, K, and L. And one of the rules is that if H plus K plus L is even or not, the intensity may be zero.
These are rules for systematic absences.
And the corresponding information is given for you here for these not terribly realistic real two-dimensional crystals.
So, for example, if you turn to number seven, P2MG, it says conditions limiting possible reflections for the general position.
For H and K, there's no condition.
For H zero, H has to be even, if the reflection is to have non-zero intensity.
And for the last two, for the general reflections, H, K, if there's an atom occupying the position either zero, zero or the position zero, one half, then for the general planes with indices H, K, you will see intensity only if H is even, as well-- same as the condition above.
This is something that is not generally known that everybody knows-- that if the crystal is face-centered cubic, there is a pattern of absences.
But there are additional absences if the atom is in a special position.
And this can very often be used to advantage because you can single out certain classes of Miller indices for which one atom in the structure will not diffract or which another atom in the structure will not diffract.
And that can be of great utility in unraveling a structure.
We'll see some examples of this in useful form when we deal with three-dimensional space groups.
And the corresponding sheets for the three-dimensional symmetries will be handed out to you-- some of them, not all of them.
Let me take a little bit of time to remind you, if you've forgotten them, but to inform you of certain definitions and trigonometry in spherical geometry.
Spherical trigonometry differs from plane geometry in that all of the action takes place on the surface of a sphere.
And it'd be nice if I had a spherical blackboard.
Actually there is one in the x-ray laboratory that I can draw things right on that spherical surface.
But this is a sphere.
We'll see directly that the radius of the sphere is not important.
So we'll take that as a unity, which is a nice, even number.
And as my dichotomy of the afternoon, if we pass a plane through that sphere, if the plane hits the sphere, it will intersect it in a circle.
If the plane passes through the center of the sphere-- and we've assigned the radius of the sphere as unity, then this is a circle that's referred to a great circle.
Sounds like a value judgment, but it's simply saying that's as large as the circle is going to get is when it passes through the center of the sphere., it would have unit radius, just as the sphere does.
So if you take any other plane which intersects the sphere but doesn't pass through the center, it's going to have a smaller radius.
And this is something that's called a small circle.
OK, so if all of the action is going to take place on the surface of the sphere, and we have two points on the sphere, A and B, sitting on the surface of the sphere, how do we measure the separation of A and B?
Well, if you think in terms of a normal three-dimensional person, you say, zonk, connect them by a line.
And that's the distance between A and B.
Now, you can't do that because all the action has to take place on the surface of the sphere.
People who deal with spherical trigonometry all the time are airplane pilots.
And if your pilot is going to take you from New York-- where would you like to go?
Paris?
That sounds like a nice place.
But if you're going to go from New York to Paris, you don't plow your way through the intervening earth.
You follow something that is at a constant radius out from the center the Earth, at a height of 5,000 feet above the surface of the Earth, an additional radius.
So the way we'll define distance is to pass a great circle through A, B, pass a plane through A, B in the center of the sphere.
And then we will define distance between A and B as the smaller of the two angles subtended at the center of the sphere.
So this is a more reasonable looking great circle.
If this is point A and this is point B, pass a plane through the center of the sphere, O, and through A and through B.
And then we'll measure the length of the arc AB in terms of the angle alpha subtended at the center of the sphere.
So it's a crazy notion.
We're measuring distance in terms of an angle.
And if that's an angle, we can take a trigonometric function of that angle, like sine or cosine.
And that blows the mind that you can take trigonometric functions of a distance.
But we can.
We'll see it's going to be useful to us, too.
And then I emphasize again, we'll take this as the smaller distance.
It'll be 360 degrees minus alpha.
That would be the long way around from A to B.
All right.
We've defined now how we will draw distances between two points.
Suppose I have three points on the surface of the sphere-- A, B, and C. I can pass a great circle through A and B. I know how to do that.
I can pass a great circle through A and C. I know how to do that.
And I can pass a great circle through B and C. So now I have defined something that is referred to as a spherical triangle.
We know how to measure the length of the spherical triangle.
Let's call the arc opposite the point of intersection A as little a and the length of the arc opposite B as a distance and angles little b, and the distance from A to C as little c. But there's something in between these arcs that looks like an angle analogous to the angle in a planar triangle.
And how can we define that?
Well, the arc AB is defined by a plane, a great circle.
The arc AC is similarly defined by a plane that passes through a, c in the center of the sphere.
And what we will define as the spherical angle BAC is the angle between the great circles that define the two different arcs.
So if there's one great circle that defines the arc from A to B and another plane that defines the arc from A to C, we'll define as the spherical angle between those two arcs the angle between the planes that define the great circles.
So we're going to call this angle in here between these two planes as angle BAC.
Another construct that is a useful one-- suppose I look at the plane that I've used to define a great circle and at the center of the sphere construct a line that is perpendicular to the great circle.
And if I extend that line, sooner or later it's going to poke out through the surface of the sphere.
And I will refer to this point as the pole of arc AB, or the pole of the great circle that we've used to define arc AB.
So the North Pole is actually the pole of the great circle that defines the equator.
And clearly there are two poles.
There's one in either direction.
So there's a North Pole and a South Pole to this arc AB and to the great circle that defines it
I have a question.
Why couldn't you define the angle BAC as the angle [INAUDIBLE]?
You want to make a tangent here?
Yeah
I don't know if that's really defined.
In other words, if I'm saying I want a line that is tangent to the sphere, it doesn't fix its orientation
In the plane of the great circle
OK.
In the plane of the great circle.
Suppose you could if you wanted to.
There are trigonometric qualities to defining the angle in the way that we have.
And the construct really is not something confined to the surface of the sphere, and everything else that we are doing is.
OK?
So it's sort of the non sequitur because we started out by saying that everything has to take place on the surface of the sphere.
There's things that we would do in our three-dimensional world, like defining distances between points as the shortest straight line, that are ruled out in spherical trigonometry.
And I think something similar could be levied at your proposal.
And the answer is we just don't do it that way.
That's the real answer.
OK.
If I have not boggled your mind so far, let me go a bit further with another useful construct.
We can see how we can define a pole of a great circle or a pole on an arc that is a portion of a great circle.
Let me take a spherical triangle, A, B, and C. And I've got three great circles now, which have formed those arcs that make up the sides of my triangle.
Let me now find the pole of arc CB.
And that means we're going to go out 90 degrees to that plane through the center of the sphere.
And that's going to define some point that I'll call A prime, that is the pole of arc BC.
And I'm going to do the same thing for the other arcs that are sides of my great circle of my spherical triangle.
I'll find the pole of arc AC.
And I'm going to label that point as B prime.
And, finally, there'll be another pole that is the pole of arc AB.
And that's going to define a point C prime.
Now I've got three points, I can connect these together and make another spherical triangle
How do you know to determine where the pole is?
If you think of it in three dimensions, I got three different arcs.
And for each one of them I am drawing a perpendicular to the plane of that great circle and looking at the point where it emerges.
OK.
So now if there's another arc, there'll be another great circle coming around like this.
And I look for the pole of it.
And that would be another one of the corners.
This thing that I've constructed is bizarre.
But it's given a special name.
This is called the polar triangle.
And it has some useful properties.
A property of the polar triangle is that the two triangles, A, B, and C, and A prime, B prime, and C prime are mutually polar.
That is, if I use the spherical triangle ABC to define and locate the three points A prime, B prime, C prime-- now if I reverse the process and find the whole of arc A prime C prime, that turns out to be point B.
And if I take the arc of A prime B prime, that turns-- I'm sorry-- take the arc of B prime C prime, that turns out to be point A.
So the two triangles are mutually polar.
The polar triangle of the polar triangle is the triangle that we started with.
Yeah?
I guess I missed that.
So if you take B prime through C prime, then all of that is going--
Yeah, I'm saying that the pole of this arc, A prime C prime, is this point here.
And the way I can show that is to say that we got B by looking at the pole-- I'm sorry-- I got C by looking at the pole of the arc AB.
So B is 90 degrees away from C prime.
I found point A prime by finding the pole of arc CB.
So B is 90 degrees from any point on that arc.
So it's 90 degrees away from A.
So B is 90 degrees from A prime.
B is 90 degrees C prime.
And, therefore, it has to be the pole of that arc.
Now that was a little too quick.
That's written down in the notes.
And that's why I wrote them out.
One final thing and then we can put circle trigonometry to one side, and this is something that is not all obvious.
If we look at a spherical triangle and simultaneously the polar triangle-- so let's say this is ABC.
And here is the polar triangle A prime, C prime, B prime.
It turns out that the spherical angle in one circle triangle and the length of the arc opposite it, namely this arc B prime C prime, are complementary-- supplementary, not complementary.
And the way one would do that is to say that the measure of alpha is the length of this arc here.
And this total side, B prime C prime, is equal to this arc plus this arc minus this length.
And these two arcs are 90 degrees.
So let me do as I've done in the notes.
Let me call this point P prime.
And I'll call this point Q prime.
So my argument, it says that B prime is the pole of arc AC.
And, therefore, B prime Q prime equals 90 degrees in length.
And then I would say that C prime is the pole of arc AB.
And, therefore, the distance C prime P prime is also exactly 90 degrees.
And that says that B prime Q prime plus C prime P prime-- if I add those two together, it has to be 180 degrees.
But I can write B prime C prime as B prime P prime plus P prime Q prime plus Q prime C prime plus the side P prime Q prime.
And that's 180 degrees.
But these three things that I've lumped together here are exactly the same as the length of the spherical polar triangle A prime.
So what we've shown then is that A prime plus alpha is equal to 180 degrees-- QED.
So this angle plus the side of the polar triangle add up to 180 degrees.
And that is not obvious at all.
One final relation-- and this I will simply hand to you on a platter.
I'm not about to derive it.
Sides and angles in planar geometry are related.
And there's a particularly useful relation in plane geometry that's called the Law of Cosines.
So this is in plane geometry.
And if you have a triangle that has sides a, b, c-- a general oblique triangle-- and it has angles A, B, and C, the Law of Cosines says that the side A is determined by c and b and the angle between them.
And that's clear.
If I specify this length, specify this length, specify that angle, things set up like a bowl of supercooled jello.
And the triangle's completely specified.
So a squared in the Law of Cosines is b squared plus c squared minus 2bc times the cosine of angle A.
In a spherical triangle there is a similar sort of constraint.
If we have a spherical triangle with sides a, b, and c, and spherical angles capital A, capital B, capital C, in the same way as specifying the spherical angle A and the lengths of the two sides c and b, specifies and fixes the spherical triangle entirely.
This side must be determined by the length of side c, the length of side b, and the angle between them.
And that, since everything is in terms of angles, is something that doesn't involve squares.
It involves totally trigonometric expressions.
And it turns out the cosine of this missing side a is given by the product of the cosines of the two other sides.
So as I said, you can take a trigonometric function of a length, which sounds like an oxymoron.
And it's the product of the cosines of the two known sides and times the sine of b sine of c times the cosine of the spherical angle A.
And that is also called the Law of Cosines.
And this is the corresponding case in spherical geometry.
OK.
So there's some machinery-- yes, sir?
What's the difference between the sines of lowercase a and--
OK, the angles are the capital letters.
This would be the angle between the great circles that defines the--
Since the radius is one, there's no difference between the angles and the-- [INAUDIBLE]?
No.
This angle is something, for example, we can choose.
And depending on how long we want this arc to be, we can put the arc BC anywhere we like
Does that mean your radius [INAUDIBLE]?
No.
This would be, say, two points of the spherical triangle.
Now we can pick any third point on the surface of the sphere, connect that with great circles, and here is a spherical triangle.
OK?
So I see.
I think I see what your problem is.
Here are the two planes.
We define the angle of the spherical triangle as the angle subtended buy the great circle.
So this is the definition of A.
But now the other two points on the spherical triangle can be any point on these great circles.
So this can be point B, and this can be point C. And my spherical triangle can be something like this.
So the arc that defines the spherical angle A is a value that is independent from the length of the arc AC.
That would be what is subtended at the center of the sphere.
I'm going to have time just to set the stage for how we're going to use these relations.
And the problem that I would like to raise and then apply spherical trigonometry to is the question if I go into three-dimensional space, there is no longer any requirement that rotation axes be all parallel to the same direction.
In two dimensions the rotation points were really-- could be viewed as axes that were always perpendicular to the plane of the blackboard, the plane of the plane group.
But now when I'm dealing with three-dimensional spaces, this could be the operation A alpha.
And there is no reason whatsoever why we should not try to combine with this first operation a second rotation B beta-- a rotation through an angle beta about this axis B and a rotation of angle alpha through this axis A.
If we're going to come up with a crystallographic combination, the angles alpha and beta have to be restricted to the angular rotations of either a onefold, a twofold, a threefold, a fourfold, or a sixfold axis, if the combination is going to be crystallographic.
Yes, sir?
You're just rotating around those lines?
I'm rotating around those lines, yeah.
So I'm saying that we're going to rotate an angle alpha about axis A.
And now what I'm going to raise as a rhetorical question-- what is the rotation A alpha followed by B beta?
So I rotate through the angle alpha about A.
So here's my first motif, right-handed.
Then I'll rotate alpha degrees.
And this here's number two.
And that will stay right-handed.
Now, I will place on axis B the two constraints.
These have to be crystallographic rotation angles, namely 360, 180, 120, 90, or 60.
And I'll also, since I would like to obtain point group symmetries initially, I will require that axis A and axis B intersect at some point.
And one of the variables in the combination will be this angle between the two rotation axes.
So let's complete our combination of operations.
I'll rotate from one to two by A alpha.
If the first one is right-handed, the second motif is right-handed, as well.
And then I will rotate beta degrees about B.
And here will sit number three.
And it will stay right-handed.
So, again, the $64 question that we raise periodically-- what operation is the net effect of two successive rotations about a point of intersection?
[INTERPOSING VOICES]
Lots of opinions.
Let's sort them out.
I heard translation.
Well, let's put down what it could be.
We know that only translation and rotation leaves the chirality of the motif unchanged.
So it's got to be one or the other
Since there's no reason for the general orientation of the two to say it has to be rotation around the third axis, so the question is what angle and what--
OK. That is exactly the problem.
Now that we know the problem, we can go home early because we know what we're going to do next time.
Well, let me expand a little bit.
It can't be translation because clearly the separation of number one and number three depend on exactly where I place the first one.
If I place it a little further out from A, then it's going to rotate to here.
And then B is going to swing it off to some different location.
So it can't be translation.
And I don't think these guys, if I rotate this way and then I rotate this way, are going to be parallel to one another.
I doubt that very much.
So it's got to be a rotation.
So without knowing how to find it, let's say that we can get from number one to number three in one shot through rotation of an angle gamma about some third axis, C. So the answer, in general, without being specific, is A alpha followed B beta has got to be equal to a third rotation, C, about a direction that has the same point of intersection with the first two axes.
Now we've got some really, really tough constraints.
Alpha is restricted to one of five values.
Beta is restricted to one of five values.
The third rotation, gamma, jolly well better be a crystallographic rotation and not something that is not a sub-multiple of 2 pi.
And even if it is a sub-multiple of 2 pi, it has to be either 0 degrees, 120, so on.
It has to be one of the crystallographic rotation angles.
So what sort of relation can we get that would give us, first, the value of gamma in terms of alpha, beta, and the angle at which we combine them?
So taking A and B as the five crystallographic rotation axes two at a time, we want to put them together, if we can, such that the angle makes the third rotation axis also be crystallographic.
And then we would have to find its location.
It looks like an impossible constraint .
It looks absolutely impossible to do.
We've got to put this first in quantitative form and then simply put in the values for alpha, beta, and gamma and find the angle that they have to be combined on to make this, if possible, be one of the crystallographics sub-multiples.
That is not an easy problem to formulate.
And, as I said a couple of times ago, the geometry that is the basis of this derivation was originally proposed by Euler.
And it's known as Euler's Construction.
I will have for you next time my own set of notes on this.
We have finished with two-dimensional crystallography.
So we are back to Buerger's book again.
We had that little interlude.
Buerger deals with Euler's Construction.
But I don't think he's at his best in this particular section.
So we'll take it a little more slowly.
And next time we'll get around to deriving Euler's Construction.
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Hi I'm Jocelyn and today we're going to go over fall 2009 exam 1, problem 1.
So to begin with, we always want to read the whole problem.
So in this one it says uranium metal can be produced by the reaction of uranium tetrafluoride with magnesium in a sealed reactor heated to 700 degrees C. So we know that we're dealing with a chemical reaction here.
The byproduct is magnesium fluoride.
To ensure that all the magnesium is consumed in the reaction, the reactor is charged with excess uranium fluoride, specifically 10% more than the stoichiometric requirement of the reactor.
To produce 222 kilograms of uranium, how much uranium tetrafluoride must be introduced into the reactor?
So that's what the question is asking us.
Then it says we assume complete conversion of magnesium, express your answer in kilograms.
So we know that the question is asking us for the uranium tetrafluoride needed to make 222 kilograms of uranium.
Right?
That's the last sentence.
The sentence that's before that, however, also gives us very important information.
It tells us that we need the stoichiometric equivalent plus 10%.
And that's very important to keep in mind because, as you'll see later, that's where a lot of people got tripped up on this problem.
So then let's go to the first two sentences in this problem and extract the information given there.
We know that we're dealing with a chemical reaction.
So when we are dealing with one of those we always want to put down the chemical equation.
So we have uranium tetrafluoride reacting with magnesium to give uranium with a byproduct of magnesium fluoride.
And it tells us that it that it's at 700 degrees C. That's superfluous information but you can always put reaction conditions above the arrow just to keep track.
So now that we have this equation are we ready to move on?
No.
Right?
We need to have a mass conservation.
We need to balance this equation.
So to determine if this is balanced or not, we want to keep track of the amount of uranium, fluoride and magnesium on each side of the equation.
So we have uranium, fluoride and magnesium.
And then we have on the product side-- and that's not right.
This is the reactant side, right?
And over here is the product side.
So on the reactant side we have 1 uranium, 4 fluoride and 1 magnesium.
On the product side we have 1 uranium, 2 fluoride and 1 magnesium.
So we see that we have an imbalance in fluoride.
To fix that we can put a stoichiometric coefficient in front of the magnesium fluoride on the product side in order to have 4 fluorides.
We put a 2 there.
And we change that to 2.
But we also have to change the magnesium.
Right?
Now, although we fixed the problem in the fluoride, we have an imbalance in the magnesium.
So to fix that, we put a stoichiometric coefficient in front of the magnesium, switch our accounting down here, and we see that we now have conservation of mass.
The two sides are balanced.
So with our balanced equation we can now move on to what the question is actually asking us.
And we know that we have, we want to make 222 kilograms of uranium.
We then need to figure out how much uranium tetrafluoride is needed.
This chemical reaction gives us mole ratios.
But we are, right now, have kilograms.
So the first thing we need to do is convert kilograms to moles.
And we do that using stoichiometry or unit conversion.
So the other thing we need to know is the molar mas of uranium.
So this is something you can look up on your periodic table, online, whatever resources you have available.
So the molar mass of uranium is 238 grams per mole.
Now we just do a simple unit conversion to get the moles uranium that we want.
So first we need to convert to grams.
That's 1,000 grams.
And then using the molar mass, we convert to moles.
And because I have grams up top here I need to put the molar grams down here in order to cancel everything out.
And that gives us 933 moles uranium.
OK so we have the number of moles uranium we want to make.
Now we can use the chemical equation over here to determine how much uranium tetrafluoride we need.
So the chemical equation tells us that we need a 1:1 ratio of uranium tetrafluoride.
However this is where we need to go back to what the problem was actually asking us.
And remember that we need the stoichiometric equivalent, which is given by this mole ratio, plus 10%.
So the amount of uranium tetrafluoride we need is the stoichiometric equivalent plus 10% times that stoichiometric equivalent.
And that equals 1,026 moles of uranium tetrafluoride.
So we found the amount of uranium tetrafluoride we need.
Are we done?
No.
Right?
If we go back to the question we see that it asks for our answer expressed in kilograms.
So that's why it's always important to keep track of what the question is actually asking you and to make sure you've answered that.
So we have uranium tetrafluoride in moles.
Now we need to convert it to kilograms.
Again we need the molar mass.
This time of uranium tetrafluoride.
And it equals-- so we have 1 uranium per molecule and 4 fluoride.
And again we just look up that molar mass of fluoride on the periodic table.
And that equals 314 grams per mole.
So we do our unit conversion again.
And this is my favorite way of doing stoichiometry unit conversions.
You may have another way that you learned in high school or later on.
But this way I know that I'm keeping track of my units and everything's canceling out correctly.
So we have 1,026 moles uranium tetrafluoride.
We use the molar mass we just determined.
And remember that this is an industrial scale process, so we actually care about the amounts in kilograms.
And we get 322 kilograms uranium tetrafluoride.
Going back to what the question was asking us, we know that we now have the correct answer.
And we can put a box around it.
So that on the exam whoever's grading you knows that that is your answer.
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Hi, I'm Jocelyn, and we're going to go over Fall 2009, Exam 1, problem number 2.
So as with every problem, we want to first read the full question.
In box notation, give the complete electron configuration of each of the following gas-phase species: calcium 2 minus and magnesium 4 plus.
So the first thing to note here, is you need to know what electronic configuration is.
If you don't, you might want to go back and review that material, but it is basically assigning electrons to orbitals.
The next thing you need to know is what box notation is.
And Professor Sadoway introduced that in class, and you'll see, as I go through this, a refresher of that, if you forgot.
So to assign electrons to orbitals, we first need to know how many electrons we have.
So let's start with the first part of this problem, which is calcium 2 minus.
We will look at our periodic table, and see that neutral calcium has 20 electrons.
So calcium with no charge has 20 electrons.
The 2 minus tells us that it has 2 extra electrons.
Right?
Because electrons have a negative charge.
So we have a total of 22 electrons that we need to assign into orbitals.
In order to actually do the assigning, we need to follow the three rules that Professor Sadoway introduced in class.
So I will put that over here.
The first rule is that you need to fill them, at least for a ground state species, in order of increasing energy.
That is, you want to fill up the lowest energy orbitals first, and then move to the higher energy orbitals.
This makes sense because ground state species want to have the lowest total energy possible.
So the first is-- Now, how do we figure that out?
Well, there's a few ways.
One way is a nice trick that you can do on the side of your page or something.
And it goes like this.
So we have our s, p, d, and f orbitals.
Right?
We know that s can be in the first energy level up.
We know that p starts at the second energy level, d starts at the third, and f starts at the fourth.
The tricky part is that you don't just fill up 1s, 2s, 2p, 3s, 3p, 3d.
There's a little bit of mixing up of the energy levels versus the actual energy of the orbital.
So a nice side of the page trick is to draw diagonal lines down through these numbers.
And if you follow these arrows, you'll fill up in order of increasing energy.
For example, after you fill up the 2p, you're going to fill up the 3s then the 3p, the 4s, and then you'll go to the 3d.
If that doesn't make sense, you might want to go and review a little bit about the orbitals in your textbook, or other resources.
The other way you can think about the ordering of the energy of the orbitals, is to be familiar with the periodic table.
So we're just going to have an aside over here.
And a general, very rough sketch of the periodic table would look something like this.
OK?
You have your alkali, alkali earth metals over here.
You have aluminum, oxygen.
Your halides, your noble gases over there.
But the way I drew it here is in what we call blocks.
So this is your s block, and then we have our p block over here, d block, and the the f block is down here.
So once you get more familiar with electron configurations, you can actually looks towards the periodic table, and know, depending upon where your element is in the periodic table, what its electron configuration is.
Or at least the electron configuration of the valence electrons.
OK.
So moving back to our trick, which we'll use now, we know that we have 22 electrons, and we know the order that we need to fill the orbitals.
So let's start writing out our boxes, which is how Professor Sadoway wanted us to answer this problem.
So we have 1s, our 2s, 2p.
If you're unsure why I'm only drawing 1 orbital for s and 3 orbitals for p, again, you might want to go review that beginning orbital material.
So we filled, we have 1s, 2s, 2p, 3s, then we're going to do the 3s here, we're going to do the 3p.
And after the 3p, we do the 4s, and after the 4s, we do the 3d.
Which has 5 boxes, right?
So now that we know which orbitals we're going to fill, we need to know some more rules about filling them.
So the second rule, over here, is the Pauli exclusion principle, which says that no two electrons can have the same quantum numbers.
For us, that means that we basically can only have two electrons per orbital, and of opposite spin.
So that's something we need to remember for this.
The third is Hund's Rule.
And that is, we want to have as many unpaired electrons as possible.
So that will come into play when we're filling up our boxes in just a second.
If these are confusing, again, we want to look at the lecture material.
So I forgot my boxes here.
Now let's start the fun part of filling up the elections.
So we fill up 1s, spin up and down, the 2s, spin up and down.
So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, and we have 2 more.
So in the 3d, we're going to do 21, 22.
And I separate them out because of Hund's rule.
We want to maximize the number of unpaired spins.
They both spin down, or both spin up, but they will be the same spin, and they will be in different orbitals.
So this was the answer for this part of the problem.
Now let's move to the second part.
And maybe you can try by yourself, and then we can do it here.
All right.
Now we're going to do the second part of this first part of the problem number 2.
So it asks us for the electron configuration in box notation of magnesium 4 plus.
So again, we're going to determine how many electrons we have.
Which, neutral magnesium has 12 electrons.
But the 4 plus signifies that we have 4 less electrons than neutrality, and so we subtract 4, and we have 8.
With 8 electrons to work with, we again want to write down our boxes using the energy level tricks that we have.
And now we can fill them up.
The Pauli Exclusion Principle tells us 1 spin up and 1 spin down per orbital, and Hund's rule tells us that when we fill up this 3p, we're going to try to maximize the number of unpaired electrons, and then, when we have to, we pair up the spins.
So that's another main mistake that people make, is they pair up too many spins, violate the Pauli exclusion principle, putting 2 electrons of the same spin in one orbital, or something of that nature.
So this again is the correct answer, and would-- oh!
I'm sorry.
This is 2p.
I'm sure you guys caught that.
And would have awarded you full points in this part.
All right.
Now we're going to start the second part of this problem, part B.
So first we want to read what the question is asking.
Give the chemical identities of the species with these ground state electron configurations.
So now we're working in the opposite direction.
In part A, we were given a chemical species and asked for the electron configuration.
Now we are given the electron configuration, and asked for the chemical species.
So the first one has the electron configuration of xenon.
So this is a noble gas configuration, plus what turns out to be 27 electrons.
One way you can do this, is to go to xenon, which is in the fifth row of the Periodic Table, and count through 27 electrons, remembering that when you get to the d on the sixth and seventh row, you need to go to the f block, which is signified in this problem by having f block electrons.
Count through the f block, the d block, and end up at thallium.
Or if you are familiar with the Periodic Table enough that you can recognize the valence electrons are in 6p.
And the 6p only has one electron in it.
So that tells me, xenon is in row 5.
This element is going to be in row 6.
I go to the first column of the p block, which signifies that there's one electron in that 6p orbital, and I see that it's thallium.
That's the quicker way to do it.
On an exam, you might want to be that familiar with the periodic table to be able to do that, because you're certainly time-crunched, a lot of times.
So because that's a neutral atom, that method will work very easily.
The second part is a little trickier, because it has a charged species.
So we have a net charge of 4 plus, an electron configuration of argon and 3d5.
This may seem a little confusing at first glance, because we always fill the 4s before the 3d.
But when you're ionizing an atom, it turns out you take electrons from the 4s first, even though it's of lower energy.
So that's a little subtlety that's not too important to this problem, but it may have tripped you up when you were looking at this, and certainly caused some problems for the people in this class.
Again, we go to the periodic table and we see where argon is, and then we go to the next row.
The way I like to think about this, is this is the electron configuration, missing 4 electrons.
To find the element that has this charge, I'm just going to add back in the electrons, go to that position in the periodic table, and that is my elemental species.
So we want argon plus, instead of 5 electrons-- I'm sorry, the problem says this is a 3-- instead plus 3 electrons, we are going to go to argon plus 7 electrons.
And going through the periodic table, you'll see that that leaves you with manganese.
Adding the 4 plus, because that is part of this chemical identity, we get the answer of the problem.
So again, the charged species make this a little more complicated.
The easiest way is to go back and add in the missing electrons, because that is your chemical species, and then remember to signify that there actually is that charge for your answer.
So moving on to part C, it asks us, write the quantum numbers of one of the 3d and one of the 4s electrons in iron.
So this requires a knowledge of what quantum numbers are, and what values they can have.
So let's start with the 4s, because that's a little more straightforward.
He tells us that we have an n quantum number, l quantum number, m and s. And, hopefully, we all know that n is the principal quantum number.
It's basically your energy level.
l gives you what type of orbital you're in.
Right?
If you're in the s, the p, the d. So it just tells you your orbital shape, basically.
m gives you the angular identity of the orbital you're in.
We didn't really talk about that in depth.
We just need to know, it tells you which of the p orbitals you're in, or which of the d orbitals you're in.
So I'll say, which orbital.
And the s is your spin.
Right?
So for a 4s electron, it doesn't actually matter what the identity of our species is.
So that he tells us is iron is kind of extra information.
A 4s electron already has 3 of its quantum numbers set.
So we know that n equals 4, for s, l equals 0, and m, which orbital, s only has 1 orbital.
So m always equals 0 for s, and your spin.
And I'm going to make that a script, just so it's less confusing, is plus or minus 1/2.
So the answer to this problem could have been 4, 0, 0, 1/2, or 4, 0, 0, minus 1/2.
Three of them are set by the fact that this is the 4s orbital.
The spin can either be plus or minus 1/2.
If you're unsure of where I got these numbers, and why they have to be such, you may want to review the lecture on that.
So moving onto the 3d quantum numbers, we know that n is 3, l for d is 2, m is negative 2, negative 1, 0, 1, or 2, right?
Because there's 5 different orbitals in a d set.
And then again, s can be plus or minus 1/2.
So picking one set of those, you could have said 3, 2, minus 2, plus 1/2, or any combination of 3, 2, and the m and the s. You don't have to get the exact same answer as your neighbor or someone else you're-- well, you shouldn't be looking your neighbor's paper, or a classmate, but these do signify that you're in a 3d electron.
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Hi.
I'm Jocelyn, and we're going to go over fall 2009, exam 1, problem number 3.
With every question, we want to make sure we read the full problem first.
So answer the following questions about the difluoro-iodate ion.
Draw a three-dimensional representation of the molecular geometry around the central atom, not simply the Lewis structure.
Show all atoms and bonds between them.
Because this part seems pretty separate from the rest of them, we're going to start with that.
So we have a difluoro-iodate ion.
It has a positive charge.
And it asks us to draw the molecular geometry in a 3D representation, and not just Lewis structure.
However, to do the molecular geometry in 3D, we need to know the Lewis structure.
So we're going to start with that.
To draw the Lewis structure, we first need to figure out which atom is the central atom.
And it's always a good idea to do that by looking at the relative electronegativities.
As we know, fluorine is the most electronegative element in the periodic table, and so that's probably not going to be in the middle.
Therefore, iodine, being the less electronegative, is going to be in the middle, and it will have fluorines on either side.
This is our rough sketch of the molecular geometry.
Not too important right now.
Next we need to know how many electrons we have.
Iodine has seven valence electrons, fluoride has seven valence electrons, and we have a positive charge, so we know that we're missing one.
And that gives us 20 electrons.
Now that we know how many electrons we have to work with, we need start filling them in.
I always start with the outer electrons.
Those are usually, hopefully, the most electronegative, and therefore, the most likely to have the octet rule satisfied.
Also, if you remember, fluorine is in the second row, and therefore needs to have the octet rule satisfied.
Iodine is a little bit lower in the periodic table, so the octet rule isn't as important.
It can have an extended octet if we need it to.
So we will work with iodine second.
Let's start out with satisfying the octet rule for the fluorine, putting first the seven valence electrons that the fluorines brought themselves, and then noticing that each needs to share one from iodine, so that we have a full octet.
So each has one bond with iodine.
We've used 8 electrons, or 8 per fluorine.
So we have a total of 16.
We have 4 left, and we're going to put those on the iodine.
Counting up how many electronics are around iodine, 1, 2, 3, 4, 5, 6, 7, 8, we see that the octet rule is also satisfied for iodine.
This is looking like a pretty good Lewis structure.
Now, just to make sure it's really good, we're going to look at the formal charge.
So if you recall, the formal charge is the number of valence electrons that the element usually has, and minus the number of unshared electrons, and the number of dots.
So this sum subtracted from the valence electrons will give us the formal charge.
Looking at the fluorine, we have seven valence electrons.
We have 1, 2, 3, 4, 5, 6 unpaired, and one bond to the fluorine.
So that gives us a formal charge of 0.
The same goes for the other fluorine.
So I guess its formal charge is 0.
Now for the iodine, we have-- I'm going to do the formal charge down here-- it again has 7 valence electrons, usually.
We have 4 unshared electrons, and 2 bonds.
That gives us a formal charge of plus 1.
Because we have a charged species here, the full molecule has a plus 1 charge.
There's no way we cannot have a net formal charge.
Right?
We need to have a net formal charge of plus 1.
And it makes sense that that formal charge is on the least electronegative atom.
So this looks like the right Lewis structure for our purposes.
One more thing that you might want to write down.
I'm not sure if points were taken off from this.
But technically, if you have a charged species, put brackets around your Lewis structure, and put the net charge on the outside.
This is not, however, our final answer, right?
This is the Lewis structure.
But the question asks for the molecular geometry to be represented.
So if we remember how we determine molecular geometry, we need to look at the number of electron domains around the essential atom.
So we're going to move over here, and I'll redraw our Lewis structure.
And remember, an electron domain can be either an electron pair, or a bond.
So we see that iodine has 4 electron domains.
And if we remember our skeletal geometry, we know that that's tetrahedral.
Thus, to draw this in a more 3D fashion, we might want to say that one of the electron pairs is here, one of the electron pairs is over here, and then we have a fluorine coming out of the board, and a fluorine going into the board.
We didn't really go over how to do three-dimensional drawings.
So as long as you show something like, you know the fluorines will be an angle, not straight from the iodine, and that the electrons be on the other side, something to that effect.
A little bit more detailed than just a Lewis structure, was what Professor Sadoway was looking for in this problem.
And that would be your answer.
And for this one especially, I'd say it's important to box your final answer.
Because you hopefully would have figured out the Lewis structure, but this is not the correct answer.
So you always want to signify what answer you want to be graded.
Part B says, name the type of hybrid orbitals that the central atom forms.
So we are almost all the way to answering this question, so I'll put it down here.
Remember, hybrid orbitals are-- we have hybrid orbitals, because we can't really say what orbitals exactly are bonding to which atoms.
We want to have 4 equivalent orbitals for this system, right?
We have 4 electron domains, so we want 4 equivalent bonding orbitals.
And from our talks earlier in the lectures about hybridization, we know that if we want 4, we take 1S and 3P, and that gives us the hybridization of the SP3.
if we only had 3 electron domains, it would be SP2, because we only use 3 orbitals.
And the number of orbitals that go into the hybrid is the number of orbitals you will get out.
So we know that our hybridization is SP3.
And that goes along with the fact that our shape is tetrahedral.
Now we need to name the molecular geometry of this molecule.
so we already have the molecular shape and the kind of spacial arrangement is tetrahedral.
However, when we're naming the molecular geometry, electron pairs no longer matter.
So for part C, tetrahedral is our skeletal geometry, right?
It shows us our special arrangement.
The molecular geometry is ignoring those electron pairs.
So it is called bent when we only have 2 bonds.
And I think about that because in spectroscopy and ways that are used to study molecules, you can't see the electron clouds very well.
So they could only see the iodine and the two fluorines, thus looking like a bent structure.
So hopefully you're sticking with me here, because this problem has a lot of parts.
And we're going to move on to part D now.
Part D asks, is difluoro-iodate polar or nonpolar, and explain.
If you put down polar or nonpolar and did not explain, we assume you guessed, and you didn't get any points.
Explaining shows that you know why, and you understand the concepts behind this, and thus deserved points on this problem.
So there's two different parts of this problem.
First, to have a polar molecule, you need to have polar bonds.
So I would first ask myself, are the bonds polar?
We have a difference of electronegativity that you can look up on your periodic table and know is not negligible.
And so yes.
A difference in electronegativity signifies that our bonds are polar.
However, not all molecules that have polar bonds are themselves polar, right?
if we have spacial symmetry, those polar bonds could cancel out.
So now we need to make sure, is the molecule polar?
And for that, we need to have answered the first questions correctly.
Right?
We need to know that this is not a linear structure.
If this was a linear structure, those polar bonds would be geometrically symmetric, and cancel each other out.
But we all are very smart, and we got this part C right.
We know that we have a bent structure.
Thus, I'll rewrite this over here.
We know that if we draw in our polarity vectors, we can realize that we'll have a net dipole pointing down.
So the answer to this is yes, because of the difference in the electronegativity, and because of the spacial asymmetry.
If you said both of those things, you would get full credit.
If you said one or the other, you'd probably get part credit.
But knowing to look at these two things shows that you fully understand what it means to have molecular polarity.
All right.
So now we're going to move to part E. This part is changing gears a little bit, so if you didn't quite understand the first part, you can still try this one.
Part E, again, what is the question asking us?
Determine the maximum wavelength of electromagnetic radiation capable of breaking the IS bond.
So we need to find the maximum wavelengths which corresponds to the minimum energy.
Right?
Because we have e equals hc over over lambda for electromagnetic radiation.
We're trying to figure out the lowest energy of photon that will break the IF bond.
Now, what are we given to answer this question?
We're given the energy of the homogeneous fluorine-fluorine bond, is 160 kilojoules per mole.
And the energy of the homogeneous iodine-iodine bond is 150 kilojoules per mole.
And in order to find the wavelength that will break the IS bond, we need to know the energy of the IS bond.
Right?
That's what we're working towards in this problem.
So given the homogeneous energies, do we have a relationship that will help us determine the energy of the IS bond?
If you recall, in lecture 9, Professor Sadoway went over the Pauling formula for determining a heterogeneous bond energy from the homogeneous bond energies, and the electronegativity.
So I'll just write that up here.
The Pauling formula is, in a generic form, the energy of an AB bond equals the square root of the product of the homogeneous bond energies plus a constant times the difference between the electronegativities squared.
And I just said what all of these terms mean.
Because even if you had this on your equation sheet in the test or something, a lot of people wrote this down, and then used it incorrectly.
So it's really important to know not only what equations you have, but especially what each of the terms in the equations mean.
So we're given the homogeneous bond energies, and we have a periodic table, or some other resource, that has the elecronegativities.
Next step is to plug in the numbers and find the energy of the IF bond.
So we have the square root of 150 kilojoules plus.
And the exact values you get for the electronegativity may vary, depending on the source.
But they'll be close enough, and relatively, probably very-- the difference between them will be very similar.
So don't worry if your numbers were a little different than mine.
So we wrote all that out, and plugging it into the calculator, you get that the IF bond has an energy of 323 kilojoules per mole.
Now that we've found the energy of the IS bond, are we done with the problem?
You have an answer.
You might be ready to move on to the next question.
However, this isn't what the question is asking us, right?
We are asked for the maximum wavelength that can break a bond of that energy.
So we need to go back to our energy relation to our wavelengths.
And we know that the wavelength equals hc over the energy.
Now here's where some people tripped up a little bit.
You have a value for the energy.
h and c are constants.
And you find the lambda.
However, if you plugged that all in, your answer would be 6.15 times 10 to the negative 31 meters.
Some people, that was the answer that they found.
And it makes sense.
You have an energy, you have some constants, and you can get your lambda.
However, if you think about it, does this wavelength actually make sense?
The gamma ray radiation has a wavelength of about 10 to the negative 17.
That's very, very, very energetic electromagnetic radiation.
So this is 14 magnitudes smaller than gamma ray radiation, and it just doesn't make sense for a bond energy of this much.
Plus it doesn't-- it just-- you should look at the answer you get, and try to be comfortable with if it makes sense or not.
So where people got tripped up, is in the units.
We found Pauling's formula gives you the bond energy in kilojoules per mole.
However, if you look at Planck's constant, it has units of joules time seconds.
The speed of light is in meters per second.
And you want lambda in meters.
To make everything or cancel out, that means we want the energy to be in joules.
Not joules per mole, because we're looking at the wavelength of one photon, right?
We're looking at the energy of one photon to break this bond.
So we want the energy of the bond, not per mole of bonds, but just of one bond.
So instead of just having this equation, we want to have put in h times c over 323, change it to joules, and then multiply by-- sorry, moles are on the bottom, 1 mole and Avogadro's number.
And Planck's constant of the speed of light, you should have on your equation sheet, or somewhere else, some other resource that you have.
You don't need to memorize those in most cases.
If you remembered to divide by Avogadro's number, you got 3.7 times 10 to the negative 7 meters, which equals 370 nanometers.
And if we recall, the light we can see is in the 400 to 700 nanometer range.
So this puts us in ultraviolet radiation, which makes sense that it would take that much energy to break a bond.
The other problem some people had was forgetting to convert from kilojoules to joules.
That would still give you a physical answer.
But again, I can't stress unit cancel out enough.
So make sure you know that since your Planck's constant is in joules, you want to make sure you have joules on the bottom.
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Hi, I'm Sal.
Today we will be doing problem number 4 of fall 2009, exam 1.
Now before you attempt the problems there's certain background information that you should know before attempting it.
One is knowledge in ionic bonding.
Because the problem deals with ions.
Two, how to draw an energy-level diagram.
That's very important.
And three, what Madelung's Constant is.
Now before attempting the problem, it's a good thing to read the whole problem in detail and make sure you don't skip anything out of it.
So that you get all the information that's given to you.
Because that's very important.
So the problem reads as follows.
For a given cation, c and anion a, show that the following four energy states on the same energy-level diagram.
1, Ions in infinite separation.
2, 9-pair c and a, c stands for cation, a stands anion.
3, An ion line, c a c a c a, just a repetitive line of anions and cations.
And 4, crystalline solid of c a.
So a three-dimensional crystal structure with t and a.
And it says also that, assume that the comparison is based upon identical numbers of ions in all four states.
The diagram need not to be drawn to scale, however you must convey relative values of different energy states.
So that's the problem reads.
Now if you recall from lecture there's an important equation that actually describes the energy between two ions, between a cation and an anion.
And this energy is directly proportional to the product of the two charges.
And it's inversely proportional to the separation distance between them.
And it's multiplied by some constants.
So what does this mean?
Well the fact that it's directly proportional to the product of your two charges means that your energy for any ion or anything that ionically bonds, it's going to be negative.
Because your z minus will be negative and that's the whole value of your energy will be negative.
And that's what dictates stability.
How negative the energy is.
So with that in mind, we can go ahead and start the problem.
So we have a cation a and anion c. Now I like to draw diagrams, or draw little pictures because that helps me when I'm solving the problem.
So I'm just going to draw pictures of a cation and an anion.
So I have a cation, c and then I have an anion, a.
So c and a, and the value of r naught from our equation is pretty much the separation distance between the two.
So project this up, this is r naught.
And if you look from the picture, r naught is just simply the radius of your cation plus the radius of your anion.
Now this is already describing the energy between just a cation and an anion with these magnitude of charges.
So all that tells me is that, this is already answering number 2, pretty much from the problem that we're asked for, which is the ion pair.
So the problem asked you to draw an energy-level diagram.
That's what I'm going to do over here.
So I'm going to go ahead and start drawing the energy-level diagram.
Now I'm going to go ahead and label my energy axis to be positive to be up.
And I'm going to draw a baseline of 0 here.
So I know that to be able to incorporate my ion pair-- like I said the energy is negative, so if the energy goes up to here and let's say the value of the energy here is 0, then that ion pair should lie somewhere below it.
So I'm going to go ahead and draw this line.
And I'll label this, s 2 for our energy level diagram.
So this is the ion pair.
Now number 1 says ions at infinite separation.
What is the energy of two charged species that are separated by infinity?
Well if you look at your equation, your equation tells you that it's inversely proportional to the separation distance.
So all that tells you is that if you divide by infinity, your energy should be essentially 0.
Which makes physical sense because those two charged species can't feel each other when they're separated.
So this is actually already my number 1, which is 0.
Separated away.
So now we need to find what the energy is of an ion pair.
Which is the line of c a.
So we want to draw a little picture.
It's pretty much going to be a repetition of that.
And if I draw a line-- that's not very straight-- I'm going to draw a bunch of cations and anions.
Essentially I can just imagine that it extends to infinity in both ways.
So with my energy between these two ions right here is given by my equation, then I should be able to figure out what the energy is of the whole system.
Because we're talking about electrostatic interaction, which you can just add linearly.
So I can make one assumption.
But look at my cation and anion pair, I can go ahead and simplify my life by letting z plus be 1.
So I'll write that down.
I'll let z plus equal to plus 1, which is the charge on our cation.
And I'll let z minus equal minus 1, which is the charge on the anion.
So that's to help with the math.
So if I come back over here and if I look at my equation line then I know that if I draw-- let's see-- a reference access here.
Then the separation between the cation and anion is r naught.
And between the cation and the other cation is going to be 2 r naught.
And it's going to extend for that.
So if I go ahead and relabel my energy equation for a line.
I'll go ahead and call it e line of function of r naught.
It's simply going to be the first one, which is going to be negative e squared over 4 pi epsilon naught, r naught, multiply the 1 over minus 1 over n. So that's the energy between these two ions.
So now because the energy is electrostatic, this cation also feels the repulsion of this other cation.
So therefore with that you have got to add what that interaction is.
So the product of the two charges is plus 1, which gives it a plus in front of your equation.
And I end up getting e squared over 4 pi epsilon naught.
Now here's something that you should pay particular attention.
The separation between these two cation is not r naught.
It's actually 2 r naught.
Because of where exactly where it sits on the line.
So my separation distance now becomes 2 r naught and it's multiplied by the same 1 minus 1 over n factor.
Then if you repeat it again, you're analyzing the interaction energy between the cation and the other anion that's next on the line.
So as you can see you end up getting this nice-- 4 pi epsilon naught, 3 r naught, 1 minus 1 over n. And essentially it repeats until it ends.
Now I'm just analyzing on the right side.
Because it extends, it's an infinite line, because it extends from negative infinity to infinity, then you can also assume that there's an anion to the left.
So all you really have to do is, you take your overall energy for the right side and you multiply it by 2.
Because it's additive.
So this whole thing gets multiplied by 2.
And that's essentially what the energy is of a line.
Now in order to be able to know where it lies on the energy-level diagram, it's important to try to get it into the same basic form as the energy of an ion pair, just something simple like over there.
So just by looking at the equation I know there's a lot of factors here that are common.
Like that e squared, 4 pi epsilon naught, even r naught and this factor can all come out of the equation.
You can take it out.
So if I do that, my equation of my line then simply becomes negative e squared over 4 pi epsilon naught.
Now I'm going to go ahead and take out the r naught as well.
And I'm going to multiply it by the factor.
And essentially what stays from here is just your 1 over 1.
From this one it's your 1/2 and then your 1/3 and then we're happy.
And because I also took out the negative, I want to make sure that I take out the negative from all the other ones too.
So this product then gets multiplied by 1 minus 1 over 2, plus 1/3 minus 1/4 plus... and it goes on forever.
Now it so happens that this is actually a series that we have an answer or an answer to.
And this product that's being multiplied by your fundamental base of your equation is just simply-- oh sorry I forgot the times 2, don't forget that, that's very important because you're summing from both sides-- this essentially right here it's simply arised because of your geometry.
You're analyzing your systems.
And this value happens to be greater than 1, which is a good thing.
And the value of this is actually 1.386.
So the fact that the whole equation is still negative, because we're talking about coulombic interaction between a negative charged species and a positive charged species.
And we're multiplying that by a value that is greater than 1 this means that if this is 1 right here, then our energy level diagram should lie somewhere below it, like right here.
So this is 3.
And that's your ion line.
Now in real life we know that crystals don't exist in lines really.
Or they're not just an ion pair.
it's literally something that is three-dimensional and has an ordered structure.
Sometimes could be different geometries, but normally it's a cubic structure.
So because that forms in nature, then that tells me that the energy of a three-dimensional crystal, which is the fourth one that we need to put on the energy-level diagram, the value should be greater than 1.386.
And this value of 1.386 is actually Madelung's Constant, which you should know.
It was a required preliminary to the problem.
So I know that below this value of 1.386 I have my three-dimensional crystal.
Now an energy-level diagram is pretty much quantized by different steps.
Now what is quantizing our problem here?
Well this is 0 because our energy is 0.
So we can almost assume that we're multiplying-- or that the Madelung's Constant for ion pair with infinite separation is 0.
And this is 1, so we can assume that the Madelung Constant is 1 or we know that it's 1 for the ion pair.
So if I start writing this out on the side here-- give me some space-- I know that's one.
I'm going to go ahead and label this by Madelung's Constant, which is 1.386.
So I come over here.
Relabel this.
This value here's is 1.386.
And this value here has to be greater than 1.386 for a three-dimensional crystal.
And this little diagram here is the answer to your question from the exam.
So when you take the exam you don't really have to go through all this math to answer the problem.
If you know the material from lecture, then the solution should be very straight-forward very easily.
And also if you keep in mind that there's a lot of material that you need to know before you take the exam, and knowing how to absorb that material and apply it will help you in solving these type of problems, which can be pretty difficult.
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Hi I'm Sal.
Today we're going to be doing problem number 5 of fall 2009, test 1.
Make sure that you read the problem in full detail before attempting the problem.
And there's also some things that you should know, pretty much background material.
Before attempting the problem.
And that's what I need, which is W I N. You will win if you know how to manipulate these equations.
The first equation is the energy of two charged species as a function of their internuclear separation.
And the other equation is a function of just the separation between two ion pairs at equilibrium, which is given by r naught.
And r naught is just simply the sum of the radii of the cation and the radii of the anion.
So if you look at this equation there's two components to it.
The first component is actually the component that is responsible for the attraction of your charged species.
And the second component is the repulsion of it.
Because essentially ionic bonding happens because of your electrostatic interaction between an cation and an anion.
So with this in mind we can go ahead and read the problem and solve it.
So the problem reads as follows.
On the same graph below-- and this is the actual graph that we're going to be doing the problem on-- it says on the same graph below, for one, BeF2, so beryllium fluoride, and two, BeO, which is beryllium oxide, sketch the variation of potential energy with internuclear separation arc between a cation and an anion pair in each compound.
The diagram need not to be drawn to scale however you must convey the relative magnitudes of key features.
So the first thing that I would do is write down my compounds that I'm using.
So I'm focusing on beryllium fluoride, is the first one.
So if I look at beryllium fluoride, I'm going to look at the relative magnitudes and charges when they're in their cation and anion state.
So I know that for beryllium, beryllium forms a cation that is 2 plus.
So it has a positive charge of 2, which means that it lost two electrons.
So I can say that z plus for beryllium is plus 2 and for fluorine, when it's in the anion state, my charge is equal to minus 1.
So 2 plus minus 1.
So with this in mind, you can take this and it will help you solve the problem.
Because like I said, the first part of the equation is pretty much the attraction part, which involves the product of your charges.
And I want to go ahead into the same thing for-- so this is for BeF2.
And for BeO we have the oxygen 2 minus.
So my anion has a z minus of minus 2 and my cation just has z plus of plus 2.
Now if you look at your equation over here.
There's two things that will dominate at certain parameters.
Now this value of n is known as the Born Exponent, and that has a value between 6 and 12.
So it's always greater than 1.
And this is actually what's nominating your repulsion, so I'll do some colored chalk.
So for the attraction part I'll circle it.
And the fact that it's being attractive means that one of these charges has to be negative that will make the overall component negative.
Right here because you'll have q1 times q2.
And in our case if we look up beryllium fluoride, you have beryllium 2 plus you have fluorine minus, so if you make the product of those two you get minus 2.
And you have this other component, that is the repulsive part.
So if I was to draw these independently, not as a sum, I know that for my attractive part, which is the negative, it should look something like so, as a function of r. And my repulsive should look something like so.
And this makes sense because as your radius becomes smaller and smaller then essentially if you get below 1, your potential energy curve for your-- the component for your repulsion-- spikes up.
So if I was to superimpose these two, then I should expect that the curve to look something like so.
Now this is good because this is exactly how the curve should look.
And there's some characteristic features here that we want to go ahead and use BeF2 BeO2, draw on the same diagram to compare together.
And this point right here on your r-axis is simply just your r naught.
So it's just the sum of your cation radius and your anion radius.
And this valley here is your energy at r naught.
Which is the second equation that you should know how to apply.
So if I want to analyze my system, how do I know which one has a bigger radii than the other?
Well I know that my cation is common.
So in both cases I'm dealing with beryllium cations.
But I'm dealing with a different anion.
Now I know that both fluorine and oxygen anions have the same number of electrons, from looking at the Periodic Table of Elements, or their charges that they have.
But fluorine has one more proton.
So what that does is that it allows the fluorine to pull the electron cloud tighter together and that decreases your radius.
So just by looking at the Periodic Table of Elements, I can already assume that my radius for my fluorine should be less than the radius of my O2 minus anion.
So if I look at my system than I know that if I want to look at the internuclear separation, the r naughts of the two compounds, then I should assume-- or have an educated guess based on my data that I have here-- that r naught for beryllium fluoride should be less than the r naught for beryllium oxide.
So this allows you to put the position of your radius on your curve.
So I'm going to go ahead and just redraw the a new axis over here just to compare both of them so you won't confuse yourself.
So we know this is the energy of our system.
And we concluded that our, that the radius of your beryllium fluoride is less than the radius of beryllium oxide.
So I can go ahead and mark this as r naught of BeF2 and I'll mark that as r naught of BeO.
So this tells me exactly where the low point of my potential curve should lie.
But how do I know how deep it sits in the well?
Well that's going to be dictated by the equation that governs the energy at r naught.
And that's our simple equation of when it's the function of r naught.
And if you look at the equation, the equation is just the product of your two charges.
So if we come back and we look at our analysis here, we know that if I multiply these two charges together, you know the combination of these two, should be, q1 or q-- beryllium 2 plus, I'll call that q1 and I'll call this one q2.
The product of these two gives you minus 2.
But for this one, q1 q2 gives us minus 4.
So this tells me that the electrostatic interaction for beryllium oxide is a lot stronger than beryllium fluoride.
And it makes sense because you have plus 2 and minus 2 being attracted to each other instead of plus 2, minus 1.
So on my graph I can assume, I can guess, that my value of my beryllium oxide should sit lower, more negative, on the potential curve than beryllium fluoride.
So if I designate this as the low point for my beryllium fluoride then my curve for this system, for that particular compound should take a certain type of shape.
And so again, this is your energy, the bottom point of r naught.
And if this is the energy given to the fact that we have a higher magnitude when you multiply the two charges together, then it should sit lower on the potential well.
And this will be my low point.
So my curve should also look something like so.
So if you were able to do this on the exam you would have gotten full points, which is good.
And, again the key points about this problem is just knowing what equations to use by reading the problem.
The problem talked about internuclear separation, well we've learned in class how to treat those problems and how to use equations to be able to describe the way that the energy should look, which is a good visual.
And by looking at the Periodic Table of Elements, just thinking about how many protons, how many electrons certain cations have, you should have an estimated guess of what your ionic radius should be.
And with that you can go ahead and look at the math.
Look where that parameter plays a role in your equation.
And then decide whether or not it'll be greater or less or what have you.
So with that in mind, I dare you to try the problem now and give it a good shot.
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Hi I'm Sal.
Today we're going to solve problem number 6 of exam 1 of fall 2009.
Now before you attempt a problem, there are a lot of things that you should know, that is background material that's going to help you solve the problem.
Especially during an exam, given that this was the hardest on this exam.
So you should be able to finish this within 15 minutes or so.
So before you start the problem you should know what the energy is of a charged particle in an electric field, which is given by this equation.
The conversion from an electron volt to a joule, which is just a conversion of energy and this should be given to you on your table of constants.
Also the energy associated with an emission spectrum, which is given by this equation where k is actually the ionization energy of hydrogen, which is 13.6 electron volts and the z squared is the atomic number of your atom of interest.
And then the n f and the n sub i are your transition states.
The DeBroglie wavelength, which is given by lambda, is equal to a Planck's Constant divided by the momentum.
And always remember that you need to conserve energy.
This is what's going to get you the full points on the problem.
So the problem, it's best to solve if you draw a little image as you read the problem.
So the problem reads, atoms of ionized helium gas, He plus, are struck by electrons in a gas discharge tube operating with the potential difference between the electrodes set at 8.8 volts.
So I'm going to go ahead and start drawing an image.
So let's say I have two plates where the voltage between these two is set at 8.88 volts.
And what's happening here is that you're accelerating electrons through here.
So you can imagine that-- say this is positively charges and this is negatively charged-- that an electron starts from your negative plate and then it accelerates toward your positive plate and you can imagine your plate having hole or something small where your electron can then exit and then be in free space under the influence of no potential.
So then these electrons are actually being aimed towards your helium plus ions.
So that's your target.
Now if I continue reading the problem, it says that the emmision spectrum includes the line associated with the transition from n equals 3 to n equals 2.
Calculate the minimum value of the DeBroglie wavelength of scattered electrons that have collided with helium plus and generated this line in the emission spectrum.
So it's an energy conservation problem.
And all this tells you is that, you have an electron that is being accelerated through your potential.
The electron is going to strike a helium plus atom and upon that you detect an emission spectrum.
Well what does that tell you?
That tells you that that electron actually excited an electron in the atom.
The electron had to transition to a higher state to absorb energy and then decay and in the process of decaying it emits a photon, which is what you detect.
So this information is given to you.
But the problem asks to find the value of the DeBroglie wavelength.
So since this an energy conservation problem, and this is the very first step that happens and I know that my energy equation for a charged particle is simply q times the voltage.
So I know that q, which is a certain value of charge, the product of these two when I'm multiplying by 8.8 volts, this yields 8.88 electron volts.
So it's a new unit of energy where we just saw the conversion.
And the reason why this takes this shape is because one electron has the charge of 1.6 times 10 to the negative 19 coulombs and that's where the conversion from electrical volts to joules come from.
So I'm going to go ahead and call this e sub i initial.
Because this is our initial energy that we're given.
So we just start with this energy, nothing else.
Nothing else was given from the problem.
So we want to go ahead and conserve this.
So our final state, the combination of whatever we are solving, should not have a higher energy than what initially we started with.
So because the problem says that we have an emission spectrum, I'm going to go ahead and draw another image.
I'll label this one, initial and I'll label this one, final.
So my final image, you can imagine a helium plus atom.
And the electron strikes the helium plus atom and we detect a photon and you're getting a scattered electron.
That's what the problem says.
Scattered electron.
So what this tells me is that this energy now is going into creating a photon and scattering an electron from your atom.
So if I was to equate those two or what not I can already say that, well in my final state I detect an emission or an emission spectrum.
And I know what the transition says.
It goes from n equals 3 to n equals 2.
That's what the problem tells us.
So if I take my equation, my delta e of my emission spectrum, is simply going to be negative k, which is 13.6 electron volts.
And I'm going to multiply by the atomic number, squared now.
This is where people also make mistakes during the exam.
The atomic number is not 1.
It's actually 2 because we're talking about helium.
And the atomic number is not the number of electrons or values in terms of an ion, but it's actually the number of protons that you have.
So helium has 2 protons.
So I can multiply this by 2.
And I'm going to square it.
And I know that I'm going from n and equals 3 to n equals 2.
So if I plug this into my equation I have 1 over 3 squared minus 1 over 2 squared, which is 9 and 4.
And if you plug this in, and assuming that your calculator works and is correct, you get a value of 7.56 electron volts because that was the unit that I was using.
Everything else is unitless, so my energy is now 7.56 electron volts for my transition.
Which covers this part for my photon.
So now we're left with what the energy of this is.
So what is the energy of the scattered electron?
Well if I equate the initial to the final I know that the final is a composition of the transition in my scatter state.
So I know that e initial, which is 8.88 electron volts, this equals to the energy for the emission spectrum, which is 7.56 electron volts.
So I'll put the units in square brackets so you don't get confused.
And then plus the energy of the scattered electron.
So if I subtract the two, I get the energy of my scattered electron.
And that pretty much gives me a value of-- after I subtract that-- e of scattered electron is equal to-- well the distance between these two is actually 1.32 electron volts.
But don't stop here because the problem didn't ask you to figure out the energy of the scattered electron.
It actually told us to figure out what the DeBroglie wavelength is of this scattered electron.
Now the fact that the electron is scattered, to me that means that the only energy that this particle has is kinetic energy.
So it's a classical form of energy.
So I can go ahead and I can equate this to just 1/2 mass of the electrons times my velocity squared-- whatever velocity it has.
Now I'd don't need to figure out what the velocity is.
This is just important to figure out what the DeBroglie wavelength is.
So this is good.
Now if we look at what our DeBroglie wavelength is, we know that lambda is equal to Planck's Constant divided by p. And Planck's Constant has a value of 6.6 times 10 to the minus 34 with units of joules seconds.
So this is important too.
Always keep track of your units because dimensional analysis will help you a lot when solving these problems.
A lot of the times you're given energy values in electron volts but your solution will have a constant that doesn't have an electron volt, that will have a joule, so you're forced to make that conversion or else your problem is going to be wrong.
OK so with that in mind, we know what h is.
But what about p?
Well p is momentum.
And classical momentum is just mass times your velocity.
So this is just mass times velocity, but this is the mass times our velocity.
We know what the mass of the election is.
That's given on our table of constants.
But the velocity, we don't know what it is but we can grab it from our classical energy.
So if I look at my energy-- from the side so I'm going to look at my energy-- I know that my energy-- I'm going to call this scat, for scattered-- the energy of my scattered electron is simply 1/2 the mass of the electron times our velocity squared.
Now if I multiply both sides by 2m, I'm going to go ahead and get us to the point where I can get an equation that is just my mass times v. So I'm going to multiply both sides by 2m, So I get 2 mass of the electron, e scat of the electron, equals-- if I multiply this by 2m, the two cancel, so I get an m squared.
m squared times v squared is essentially just m times v squared.
And this becomes just mass of your electron v squared.
Now if I take the square root of both sides, I end up getting a nice relation for just m times v. So I know that m times v now is just simply equal to this-- canceled, the square root cancels the square.
The square root of 2 mass of the electron times the energy of the scattered electron.
So if I go back to my DeBroglie wavelength, I have m v. So I can take that value as a function of energy and just substitute it into my equation.
And that'll help us get the answer because we know what the energy of the scattered electron is.
We calculated that on the first part.
So with that in mind we'll come over here.
Again this is 6.6 times 10 to the minus 34.
And the units are joules seconds.
And down here I have the square root of 2 times my mass of the electron, which is on your table of constants.
And it's just 9.11 times 10 to the minus 31.
And this has units of kilograms.
And the energy of my scattered electron, which is 1.3 times 1.32 and this units of electron volts.
Now this has units of electron volts.
This has units of joules.
This problem's a little bit-- now you're not going to get a good answer with that unless you make the conversion.
I pointed it out, the bullet point in the beginning was that 1 electron volt is equal to 1.6 times 10 to minus 19 joules, so if I simply multiply this by that conversion, just 1.6 times 10 to the minus 19, this has units now of joules per electron volt.
So this has units of joules per electron volt.
The equation turned out to be pretty long.
But this cancels the electron volt and now has units of joules.
Which if you go through the math, you're going to go ahead and at the end get a unit of just meters.
Because joule is just kilogram, meters squared per second squared.
So all that factors out and you end up getting that.
If you go through the math and you get your math right, then this should yield a value of 1.06 times 10 to the minus 9 meters.
So this is the value that will get you the right answer on the exam.
Again this is lambda for your DeBroglie wavelength because that's what the problem is.
So the problem asked for the DeBroglie wavelength but it give you all this information to get to the point where you needed to solve.
And by conserving energy and knowing the right conversion factors between energies you're able to get an answer, which is good.
It's really easy to complicate things and get a wrong problem now.
I remember from grading the exams, the most common error that people faced was actually letting the energy of the electron be the energy of a photon.
Now that can't be because the energy of a photon is just your Planck's Constant times a frequency.
And that's the energy for massless particles, your electron has mass so if it's moving it's not going to be h nu.
The energy is going to be kinetic energy.
1/2 mb squared.
If it's not in an electric field, that is.
So with that in mind just be confident when you solve these problems and make sure that you know exactly what energy equations to use to get the right answer.
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Hi there.
Welcome.
I'm Brian Spatocco.
We're going to be going over exam 2 from the fall 2009 semester.
The logical place to start on this exam is problem 1.
And what I usually do for all the problems is discuss the things we need to know before we start it.
So I would recommend reviewing these concepts before attempting the problem.
And then attempt the problem and then identify your weak spots.
So here's what I personally found to be important information to know before attempting the problem.
So I call it what I need, (W I N), if you want to win.
There's four things.
The first are Miller Indices.
This is a crystallographic notation for figuring out and communicating planes and directions in crystals.
So review Miller Indices We have plane/direction notation.
I'll explain a little bit more what that is in a second.
Definition of planar density.
And for that matter, linear and three-dimensional density as well.
And 4, the concept of a crystal basis.
And this is actually the concept that most students had trouble with on this particular question.
So those are the things I would recommend you look at before starting.
And if you haven't, check it out now.
So let's start off with part a of problem 1.
And so the question asks us to draw the crystallographic feature indicated and label it clearly.
It doesn't actually say if we're drawing planes or directions and that's left to the exam taker to identify.
So the first thing I note on that problem, part a, is that we're asked to draw 0, 1 bar, 2 bar.
And 1 1 1 bar.
So the first thing I notice about these two particular crystallographic features is how they're actually encapsulated.
So the first one is in parentheses.
The next one is in square brackets.
That actually tells us some information about the problem.
The first part, a part one, is 0 1 bar 2 bar.
So you actually need to know what this implies.
Parentheses mean we're looking at a particular plane orientation.
Now I'm not saying a plane.
I'm saying a plane orientation.
Which means there are infinitely many planes in a crystal which have this orientation.
OK so that's distinguishing between just one plane and a crystal.
The second part, b, we have square brackets.
That implies we're either looking at a direction or a place in a crystal.
We also have things like this, which we're not seeing on the exam this time but may show up later.
We have these curly brackets and we have these angled brackets.
The curly brackets denote a family of planes.
And the angled brackets denote a family of directions.
And it's markedly different than a particular or plane orientation and a particular direction.
So let's do some visuals here to explain what we mean.
So let's start off with a 1.
Now we're asked for 0 1 bar 2 bar.
Now the bars indicate negative signs.
So you would know that from your Miller Indices review.
Now the way to do this problem-- what I find is the easiest way we reviewed in recitation-- was to not be afraid to move your origin.
A lot of people are afraid to move their origin around but we have to remember in a crystal, basically what a crystal is, it's a unit lattice repeated infinitely in three dimensions.
So it's OK to move your origin as long as you choose some reasonable point.
So in this problem we're looking in the back bottom left corner.
That's our defined origin on the paper.
But to do this problem, I would recommend choosing a new origin.
I chose this one.
And that's really going to facilitate doing this problem a lot more easily.
So let's look at what we have to dran now.
0 1 bar 2 bar.
The 0 indicates that we're never actually intersecting the x-axis.
OK, so at no point will this plane-- we know it's a plane because the parentheses-- intersect the x-axis.
The 1 bar indicates that we are intersecting the y-axis at 1 lattice parameter spacing.
Miller Indices are a little tricky because they're somewhat inverted.
1 implies you are going the full distance of the unit cell.
2 implies you're going half the distance.
3 implies you are going 1/3 the distance and so on and so forth.
So for 1, we're going to go the distance of 1.
We're going to move negative 1 in y.
So we're going to move negative 1.
We're going to go to here.
That's in our y.
And when we actually make this a little clearer.
I've chosen a new origin.
So here's my y prime.
Here's my z prime.
Here's my x prime.
So we're moving negative 1 and y.
We're going to move half, negative 1/2 and z.
And we're also told that this plane doesn't intersect the x-axis.
So it must run parallel to that asix.
Which tells me that it's going to run along the x-axis.
We're going to get a plane like this.
And here's our plane.
That's the answer to part a 1.
Pretty easy and most students got this correct.
For part a 2 we're asked to draw a direction or point to a particular place in space.
I also moved my origin on this problem.
And I found it easiest to redefine my origin up here.
Which means that my new set of axes are going to be x prime, y prime.
I'm going to keep z because z didn't move at all.
OK note how, when I'm drawing these things-- we're going to draw this one now-- that the way I choose to move my origin generally tracks with the negative signs that I have.
This problem I had a negative y and negative z, I moved in the negative y, negative z directions.
Here I had a negative z.
So I just moved up in the negative z.
That kind of told me how to move my origin.
It made life a little easier.
And now it's pretty easy from here.
Now what we're going to do is draw 1 in the x, we'll go 1 in the y, we'll go negative 1 in z.
We're pointing to this point.
There we go.
That's our direction.
So pretty easy.
Part a 1.
Part a 2.
So we're batting 100 right now.
And just don't be afraid to move your origin.
That's the take-away from this part.
The problem, of course, gets a little harder.
That's something that we pride ourselves on in 3.091.
If it was always this easy, everyone would pass the class and get 100s.
So the second part, part b, is asking us to calculate the density of atoms in a plane.
So this gets back to what I was saying before.
We need to understand the definition of a planar density.
OK, so I've sort of defined it here.
This is rho.
Not p. rho planar.
And rho planar, I define-- to help me, maybe it'll help you-- is the number of full atoms in a particular plane-- in this case we're looking at 0 0 1-- divided by some area.
Because the plane is denoted by an area not a volume.
Don't be afraid, I mean densities don't always have to be mass over volume.
It can be some unit over some specific distance or area or volume.
So that's general density.
So we're looking at the number of full items in a plane divided by the area.
We're told that we're looking at Dalium, of course named after Salvador Dali.
And so we're also told that we're looking at the 0 0 1 plane.
I've drawn a bcc crystal here.
As a scientist or an engineer, when you're given a bcc fcc, simple cubic, the first thing you put on that paper is the actual crystal structure.
So I drew the Dalium up here.
Not a real element, but a make-believe one.
And we're going to identify our 0 0 1 plane.
This is good practice.
So there's our x, y and our z is right here.
Our 0 0 1 plane, so we're not intersecting x.
We're not intersecting y.
We're only intersecting z.
So here's our 0 0 1 plane.
Let me sort of bring this down for you.
So we can look down on it from above.
It's a square, with sides a.
And here's our atoms.
So now the question is, how many full atoms do we have and what's the area?
The area's obviously a squared.
And the number of full atoms in this area is we have four quarters of an atom.
When you're in two-dimensional space an atom looks like a circle to you.
When you're in three-dimensional space it looks like a sphere.
So if you lived in the two-dimensional plane, you would see a quarter of an atom, a quarter of an atom, and so on.
You'd have one full atom in this plane.
So this is sort of what we're looking for.
And you say, oh, well I'm done.
A lot people left it at that.
But that's not the answer we're looking for.
We want an actual numerical answer.
So the question now becomes, what is a?
So we're going to digress and find a and then we'll know exactly the answer to the planar density.
So a-- this is where you have to be a little clever.
How do you find a?
The lattice parameter, the lattice spacing.
So the best way to do this is to think both small scale and large scale.
So first let's think about a unit cell.
Let me ask you, what's the three-dimensional density of a unit cell.
So how many atoms do we have per unit cell?
We have our bcc.
We know in bcc there are two atoms.
We have eight corner atoms and they're each 1/8.
We have one body atom.
So there's two atoms.
So we have 2 atoms per-- what's the volume-- well the volume is a cubed.
Now we got a cubed.
We didn't really make any progress here, we still have an unknown variable.
But the question is, how do we get solved for a cubed.
So let's zoom out from that unit cell.
Let's zoom out to a mole of material.
OK, so a mole of material, how many atoms are in a mole of material?
Well by definition, it's Avogadro's Number.
6.02 times 10 to the 23.
And then what's the volume of a mole of material?
Well looking back at the problem, we're given the molar volume.
Which is something that I would recommend you write down.
You know all the information you're given, see what you've got.
The volume of a mole is the molar volume, which we're given.
So now it's pretty easy.
We know Avogadro's, we know molar volume.
So we can solve for a.
So not too bad.
You'll find that a is about 2.8 times 10 to the negative 8 centimeters.
You could give me 2.85, 2.79.
We're worried about the concepts here, not the numbers.
So don't go back and waste a lot of time getting the exact number.
That's our a.
And now that we've got a, we've got our planar density.
Very easy.
I'm going to write rho plan, just to save some time.
And that equals 1 over this number squared.
That's one atom, by the way.
And that's all squared.
And we're going to find in the end that our planar density is going to be equal to 1.27 times 10 to the 15.
And the units are important.
That's atoms per centimeter, squared.
If you don't give us the units then we have no idea if you're right or wrong.
So always include your units at the end.
So hopefully that helped.
We're going to take a second and clean off the boards.
And we're going to start with part c. OK we're back.
We're going to finish problem 1 now from the 2009 exam 2.
We're on part c. So thus far in part a, we've talked about Miller Indices, we've talked about the plane and direction notation with the parentheses and the brackets.
We've also talked about planar density.
And I would invite you to review those things.
Now we're going to go to part c, which is really the concept of the crystal basis.
And this was the part of the problem that most students lost points on.
So it's the thing that I would emphasize the most on this problem.
The problem asks us to-- and I put a sad face because that's how people felt after this part-- the problem asks us to draw a particular plane in a fcc crystal.
So we're given Magnesia, MgO.
They want us to draw that plane and show the atoms.
And we also want you to show the relative sizes of the atoms.
You know we have anions and cations.
And we actually give you the sizes of those ions, so it shouldn't be too tricky on that part.
But I think the best way to start is to go over the answer that was given that was wrong.
And then we're going to talk about why it was wrong and how to get the right answer.
So to start off, we want you to draw this crystal.
The best place to start is by actually drawing the crystal in 3-D and then we're going to take the projection after that.
So people think, MgO, they see fcc, and they say, oh that's not so bad.
I know what fcc looks like.
Let me draw fcc.
I'm going to use for this problem, I'm going to use this red chalk to represent magnesium and the blue chalk to represent oxygen.
OK so magnesium, we're going to have it as a circle like this.
This is magnesium, 2 plus.
And we're going to have the oxygen-- it's going to be slightly bigger because it's an anion.
This is O 2 minus.
Maybe I'll color this in too.
People said, OK easy.
fcc.
This is a simple problem, easy points.
They drew what they thought to be an fcc crystal.
So they did-- let's say they put the Mgs on the corners.
And they said, OK now I need my faces.
So maybe the oxygen are the faces.
So I'm going to put in my 6 face atoms.
So here are my 1 side here, and I've got the bottom and the top.
There's my 6 faces.
And they said, OK, easy.
I'm done.
And then they took the projection of 0 1 1.
We also know, when you put the axes on here just to make it clear.
And here's our z, You know 0 1 1, we know it doesn't intersect the x but it intersects the y and the z at one lattice spacing.
So x doesn't intersect.
It intersects z here, intersects y here's.
We've got this kind of plane going on here.
We've got something like this.
And they just drew this.
It's a little bit of 3-D visualization But they drew that and they got 0 points.
The reason that that is not a correct answer is because that is not what MgO-- it's rock salt structure-- that's not what it looks like.
And the reason they got it wrong is because they didn't understand the concept of the basis.
So I'm going to make a statement right now.
This is a really important statements so I urge you to think about it.
A crystal equals a lattice plus a basis.
Here's the difference.
A lattice is a collection of points.
So you have an fcc lattice, you have a bcc lattice, you have a simple cubic lattice.
Those are points where atoms-- I use atoms plurally, perhaps-- can exist.
OK, a basis I like to think of as a stamp.
A basis represents the most fundamental pattern that you stamp at every lattice point.
So let's draw an fcc lattice.
So I'm going to drawn an fcc lattice on this cube right here.
I'm going to put points.
So I'm going to make them white.
They're just points.
I got all the corners.
They're hard to see.
But they're all the corners.
I got the faces.
Side, back, front and bottom.
You're saying to yourself, that's like the same thing you just drew.
It's really not.
What I drew the first time, I put atoms in.
That's a crystal I drew.
I just a lattice right now.
This isn't a crystal, this is a lattice.
So to actually answer the question, we want to figure out, we want to draw the 2-D projection of a crystal.
So we've got to draw the crystal eventually.
Now we're dealing with MgO.
We're told MgO is face-centered cubic.
Which means that our basis must have 1 Mg and 1 O atom.
So I'm going to do that right now.
I'm going to draw what I'm going to call to be my stamp.
You can use whatever method you think is best.
But I call this my stamp.
I'm going to draw it like this.
So here's an oxygen atom.
Here's a Magnesium atom.
And this little white dot is the point that's going to get stamped.
And here's my stamp.
And the idea is I'm going to take the stamp, this basis-- that's what it is, it's a basis-- and I'm going to stamp it at every lattice point.
OK so note I have 8 corner lattice points and I have 6 face lattice points so I'm going to be stamping 14 times.
So I'll do the first couple very slowly.
I'll take my stamp and I'll stamp here.
Here is my O.
Here's my Mg.
I'll stamp over at this corner.
Here's my O, here's my Mg.
I'll stamp in the face.
Here's the face, the front face one.
Here's my O, here's my Mg.
So you can see what's happening.
Let me just circle it for you.
I'll circle one of them.
Here's my stamp.
I'm going to go ahead, I'm going to finish drawing this in.
It's going to get a little messy for a board.
But I hope that you can go home and do it on some graph paper as well.
All four corners.
The face is here.
I have a tendency to leave off faces sometimes.
Let me make sure I've got them all.
I already did that one.
Yep that's right.
Is that right?
No.
OK, now let me finish with the magnesiums.
The magnesiums-- it's going to be hard to see-- exist at the center of every edge.
So it's a little tricky but let me explain what we're looking at.
What you should be looking at.
We have what looks to be oxygen in the face center cubic arrangement.
And we have magnesium at all the edges.
So you're thinking, this is weird this isn't face center cubic.
Look at the magnesium, this doesn't look face center cubic.
Well the thing I would challenge you to do at home is actually draw this unit cell out 4 more times.
And then you will actually be able to find face center cubic magnesium lattic cell in that larger unit cell.
So this is actually the answer.
This is what a rock salt structure looks like.
This is MgO.
You have oxygen, looks like your traditional face center cubic.
And you've got these magnesium stuck on all the edges.
So now we're just going to take our projection.
Let me use white chalk for that.
Or rather we're going to take our plane.
We're looking at the 0 1 1 plane.
Here's our axes again.
Remember it's always right-handed.
We're going to do 0 1 1.
Not intersecting the x and were going to intersect z and y.
So we're going to come down across the face.
We've got this.
This is pretty terrible looking.
But this is one of the tricks of the problem.
You have to be able to visualize these in 3-D space.
So let's slowly transcribe what this plane here looks like projected out on the board.
We know that at our corners and also at the faces of this cube, we've got oxygen.
So here's a corner.
These are our corners.
At the faces, two faces, we've got oxygen.
What else do we have?
We know we must be hitting some magnesium.
Well we know that at all the edges-- here's an edge, here's an edge-- we've got some magnesium.
What about the center?
Well the center we do too because we know that when we made this stamp, the bottom face with this oxygen, we also stamped halfway up a magnesium.
So there's actually a magnesium right here.
It's there.
It's right there.
Believe me.
There's our magnesium.
And this is what we've got.
This is the answer to the problem.
Notice how I've giving my oxygens larger than my magnesiums and I've got it looking like this.
In fact let me let me emphasize this even more.
The fact that you could actually have drawn this differently.
You could have drawn it like this.
And it would have been exactly the same thing.
You would have gotten full credit.
You could have drawn this.
This is the thing that I would recommend you try at home to prove that this is true.
These two are equivalent It's the same thing.
It all depends on what you choose as your stamp point.
So I could have chosen to put my red at every point, my magnesium.
So this is the answer to part c. This part, most people did not get correct.
Less than half the class got this correct.
So think about a little bit.
Identify crystal as lattice plus basis.
And then think about the full scope of the problem.
What do we go over?
when went over Miller Indices, went over designation and crystallographic notation, so plane and direction notation.
We went over the definition of planar density.
Think about what linear density means, think about what three-dimensional density means.
Don't be afraid to not use mass.
And the last thing we just covered was the crystal basis.
So review those and hopefully you did well.
Thanks.
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OK. We're still on Exam 2 of the fall 2009 class.
We're going to be doing problem number 2 now.
This is the x-ray problem, as I like to call it.
It's kind of exciting.
Let's talk about what we need to know before we actually reasonably try to attempt this problem.
So I would review these things again.
We want to know emission line nomenclature, how to name emission lines.
We want to know Moseley's Law.
We want to know Bragg's Law, and I would also emphasize, I'll get onto it later, you want to know how to derive Bragg's Law as well.
We want to know, this is pronounced Bremsstrahlung, it's a German word which means breaking radiation.
So Bremsstrahlung radiation.
And we also want to know some reflection rules, and I'll sort of elucidate that a bit more.
But let me start you off in that direction.
So getting on to part A, we are told that somebody has been horsing around with our x-ray machine.
We've all had this problem before.
And they've changed the target.
So let me first tell you what a target is, and how the x-ray machine looks.
Then we'll be able to understand what exactly it is we're trying to figure out.
An x-ray machine, basically the x-ray source involves electrons being accelerated into a target material, a metal.
And then that gives off a whole bunch of electromagnetic radiation, so photons.
Some of those photons we'll talk a little about later, in the Bremsstrahlung spectrum, are very characteristic of the material.
So I'm going to call, you know, we have long wavelength photons, short wavelength photons.
Some of the photons have a very characteristic wavelength or frequency that corresponds to a particular electron transition.
We're going to talk more about this in a little bit, but I'm going to call this the K-alpha photon, OK?
So that just happens to be in metals, x-ray.
We know that for particular materials, we have particular K-alpha wavelengths.
And what's basically happened in this problem is that, you know, maybe we knew what this material was before, and then all of a sudden we weren't around, somebody switched it on us, and now we have to figure out what it is.
Don't you hate when that happens?
So the best way to approach this problem is to understand Moseley's Law, and then have it on your equation sheet.
We allow all of our students to have an equation sheet, and also a periodic table.
So don't panic if you can't memorize this equation.
Our students are expected to, not to memorize it, but to understand it.
So this is Moseley's Law.
And Moseley's Law basically tells us, let me go through these variables.
We have the wave number.
We have a Rydberg constant.
This is a bunch of physical constants, sort of agglomerated into one big one, to save us some time.
We have the final energy level, where the electron ends, and we have the initial energy level where it begins.
And then we have z, which designates our element.
So it's the number of protons we have in the nucleus, for example.
And then we have this sigma.
OK?
So, you know, in order to solve this problem, we want to know what element.
So obviously, we're looking for z.
Now the question is, do we know everything else?
So yeah, we definitely do.
Let's just go through each individual thing that we know.
So we know Rydberg constant.
That's just, you know, have that on your equation sheet.
It's a bunch of constants agglomerated together.
We know that, for K-alpha we're told, this is K-alpha radiation.
Let me draw a cartoon for you here.
OK?
This is our nucleus, OK?
This is n equals 1, this is n equals 2.
I'm going to go kind off the board.
This is 3.
OK?
And K-alpha corresponds to an electron dropping down from n2 to n1 to here, and then giving off a photon.
So I'm going to draw a photon as a squiggly line like this.
OK?
So this is going to be h nu.
This will give you our K-alpha radiation.
So we know all of those things.
We know our wave number here.
So we know pretty much everything.
But what is sigma?
Well, sigma is the correction factor that Moseley found.
Now, if you're looking at this and you're thinking, wait, this looks a lot like the Rydberg equation, it's because it's sort of is.
The only difference is that the Rydberg equation is only for hydrogenic-type atoms.
OK?
So we're talking about hydrogen, we're talking about helium plus 1, lithium plus 2.
We can't assume that our target is hydrogenic, so we can't use the Rydberg equation.
So we're using Moseley's equation.
And what Moseley found, just to show you where the sigma sort of comes from, and this is actually reviewed in, I think, lecture 17 of the online postings.
What Moseley found was that if you plot your wave number versus your z, you've got points like this.
You've got some sort of line.
OK?
So this is what Moseley found.
And what Moseley basically did, was he fit an equation to it.
And the equation looks a lot like the Rydberg equation.
And because these don't intersect the origin, we've got this fitting factor right here.
We know that for a K-alpha transition, we're looking at a sigma equals 1.
OK?
So we're good.
We know our sigma.
We know our n's.
We know n final is going to be 1.
We know n initial is going to be 2.
We know the Rydberg constant, and we know our wave number, because we're given our wavelength here.
So we can go ahead and actually just calculate z.
And let's do that right now.
So let me rewrite this equation so that it makes more sense.
Rydberg constant.
1 over.
our n final.
That's 1 squared minus 1 over, our initial point is 2 squared.
We're looking for our z, and we're going to subtract off 1, because we're dealing with a K-alpha transition.
So you're expected to either remember that, or have it on your question sheet.
There's no way to derive that unless you've got all this data, which you won't have.
So we've got this squared.
We know everything, so we can basically reduce this.
This is why on the answer sheet, you see something like this.
You kind of skip all these steps, and we went right to this.
3/4 z minus 1 squared.
And now you can just solve for z.
No problem.
OK?
So that's 4 over 3 lambda Rydberg constant to the half plus 1.
And that's going to give you about 23.
You might get 22.9.
You might get 23.1.
We're talking about 23.
That's going to be the vanadium.
That's the answer to the first part of this problem.
OK. We're going to take a second and clean up, and come right back and finish the problem.
We're going to start part B now of problem 2 on the second exam.
So we basically have used the Moseley equation to find out an element.
And now we're going to switch the problem up a little bit more, and we're going to ask you to find the smallest defraction angle for a particular situation.
So what's the situation?
Let's go back to our x-ray machine.
The way you actually do x-ray defraction, you know, it's acronymized XRD.
The way you do it, is you have to generate x-rays.
So we accelerate an electron into some material.
That material gives off a whole bunch of photons.
You then filter all the photons, so you only pick up a certain wavelength.
And we're going to use the K-alpha wavelength.
And then you get these K-alpha radiation, which then probes your sample.
So here we go.
This is what we're looking at for this part.
We've got K-alpha radiation coming in, and it's hitting our structure, which has a crystal, hopefully.
You're going to do XRD.
And then you're getting off some defraction, anything that's reflecting off of the crystal in some way, like this.
So the thing you need to know to do part B, is you need to know Bragg's Law, Bragg's equation.
And that's number 3 on the win list.
So I drew this picture here just to define the variables, but one thing I would stress for you at home, is to please, go home and, you know, search online, Google image search even.
Look for a picture like this.
Search Bragg's Law and look for a picture like this.
And it's actually from this.
It's very easy, just using geometry, to derive Bragg's Law.
Bragg's Law is just the geometric interpretation of this picture.
OK?
So Bragg's Law, in its traditional format, is n lambda equals 2d sine theta.
And I've defined, I drew the picture to show the variables here.
We have n, which is-- we'll talk about n in a second.
We have lambda, which is our incoming radiation.
We have d, which is the spacing between planes.
Whatever plane orientation you're looking at.
2d is the spacing, is there because you need to come in, and then come out again.
So you're doing twice the distance between the planes.
Theta is the angle at which you're incident on the material.
And n is going to be your defraction index.
We're not going to worry about that in this class.
We're going to assume it's one, for this class and for this exam.
So the question is asking us to find the smallest possible angle of defraction.
What that means, is that we're getting all this radiation, it's hitting our material, we're moving our material around.
The question is, what's the smallest angle at which we're going to see a peak on our XRD graph?
I rearranged the equation here, so theta equals the inverse sign of lambda over 2d.
And we're looking for as small a theta as possible.
This question is really just mathematical manipulation.
I need to know a couple important details that hopefully you've read about in advance.
So we want to find the smallest theta possible.
That's the objective for this problem.
I've drawn the arc sine function here.
To minimize theta, we want to have the argument, which I call x here, we want the argument to be as small as possible.
The smaller it is, the closer to 0 you are for y, which is theta, in our case.
So we want to minimize the term in parentheses.
And the way you do that, small theta means small lambda over 2d, which means you want to have a large d. And the reason I'm not using lambda, the wavelength, as the knob, is because lambda is set.
Lambda is predetermined by the target material you've chosen.
So for our problem, we have titanium as our target material.
Which is, we can't change the lambda K-alpha.
It's set by that material.
So we have lambda K-alpha-- we have k-Alpha coming off, which is characteristic of titanium.
So we can't use lambda as a knob.
The only thing we can mess around with here, to figure out this problem, is we have to look at d. So we want to have a large d to minimize this, and to minimize your theta.
To get a large d, we need to know the equation for planar spacing.
OK?
This is your equation for planar spacing.
And as we sort of talked about in problem 1, in the prior video, d planar spacing is the distance between one plane of a particular orientation, and another plane of the same exact orientation.
OK?
So we're looking at maybe a 011 plane in this unit cell, and a 011 plane in this unit cell.
What's the distance between the two?
That's what this d is.
OK?
So we want to have a large d. And let's look at what we have in the equation for d. We have these hkl.
These are the coefficients the Miller indices of the plane.
And we have a, which is the lattice constant of the material.
We're looking at-- what are we looking at here?
We're looking at tantalum.
We can't change the lattice constant for constant temperature and pressure.
So this is nonnegotiable.
We can't change what a is.
The only things that we can toggle, the only switch we can toggle now, is h, k, and l. So the only thing we can do, is look at different plane orientations in the crystal.
So this is the logic I used to go through the problem.
So you want to have as small an h, k, and l as possible, because the smaller these are, the larger d is, et cetera.
So the final piece of information where we have to be clever is knowing how to get the smallest h, k, and l. We're told that we're looking at tantalum.
And the students in this class have access to a periodic table, which has a vast quantity of information about all the elements in it.
One of the things it has is the crystal structure of pure elements.
So for tantalum, we're looking at a bcc structure.
So we're looking at bcc.
bcc, from your reflection rules, which is the number 5 on the things we need to know to do this problem-- for bcc, you have to have your h plus your k plus your l-- let me write it over here-- h plus k plus l-- must be even.
That's a rule, OK?
Proving that is a little bit beyond the scope of this class, but that's a rule that we went over in class, and it's in the lectures as well.
So we have a couple things now.
We want to minimize h, k, and l, and they have to be even.
So basically, what that means is, we have to go through a couple permutations.
So let's think about it.
0 0 0.
That's even, but that doesn't really correspond to a plane.
That doesn't make, you know, that's not going to help us.
1 0 0, or 0 1 0.
So let me write some of those down.
Let's look at, you know, 1 0 0.
Well, that's not even, so we can't look at that.
So what's the next smallest thing we could go?
How about 1 1 0?
Those are the smallest coefficients we can have for a plane which has the rule that the h plus k plus l must be even.
So this could be 1 1 0, this could be 1 0 1, this could be 0 1 1.
But what we're basically talking about, if you remember from the first problem, is the family of 1 1 0 planes.
So I plugged in just 1 1 0 for h, k, and l. You get a d of 3.31 over the square root of 2.
And then with that d, you know that you've just maximized the size of your d, which means we've minimize the size of lambda over 2d, and that means we've minimized our theta.
And now for theta, we can easily calculate that we're looking at an angle, if we plug in for d, which is 3.31 over the square root of 2, here.
And if we plug in for lambda, which is given to us as 2.75 angstroms, we get an answer of 36 degrees.
Put that right in the middle.
And I put in the middle to emphasize that this is sort of the logic you have to go through, sort of a, you know, circular logic, to get to that answer.
So that's the answer to part B.
And remember, this is pretty much math.
Just manipulation until the very end, where we discussed the reflection rules for bcc.
I want to go on to the last part now, and I'm just going to erase this.
We only need a little bit of room.
This last question was actually my favorite question in the entire class.
And actually, we only had 3 students get it right, out of 480.
So this was a great question, and if you get it right, then you're doing really well in the class.
So let's look at C. It's really a chatty question.
We're not looking for a number.
The question basically asks us, we want you to talk about, to sketch the emission spectrum of an x-ray target that's bombarded by photons instead of electrons.
So let's go back over to what we said in the beginning.
When we generate x-rays, generally the way we do it, is we accelerate an electron into a target material.
In part B, the target material was titanium, but it could be anything.
A very common target material is copper.
copper K-alpha is a very standard target material, or wavelength to use.
You accelerate your electron, and you generate the spectrum of electromagnetic radiation.
OK?
So this is our spectrum.
We have large wavelength, we have small wavelength, we have some wavelengths in the middle.
And generally, what we would see, coming back over here, I'm going to draw it like this.
This is what you'd normally see, if you're using electrons.
Professor Sadoway refers to this as the whale-shaped curve, probably because we're in New England.
But this is what you would normally see.
Now, this y-axis is the intensity.
That's basically the number of photons you count at some specific wavelength.
And you see these spikes.
And these spikes correspond to K-alpha.
This is K-beta.
We'll talk about what they are in a second.
L-alpha and L-beta.
And et cetera.
You'd have M-alpha, N-beta.
You go all the way down.
But they have very low intensity.
So this is what we would normally see if we use electrons to hit our sample.
But we're not using electrons.
And this is actually the answer we got for probably 90% of the solutions from students.
They just drew the Bremsstrahlung radiation, Bremsstrahlung, breaking radiation.
And they walked away, and thought they had full credit.
But in reality, this is not the correct answer, because you think about what's actually causing this whale shape.
Once you understand where the whale shape comes from, then you understand what happens if we're using photons instead of electrons.
So let's actually understand where these characteristics come from.
There's two things to pick up from this plot.
We have a whale shape.
We've also got these spikes.
So first, let's talk about where the whale shape comes from.
Let me just do it with blue.
We'll talk about the spikes in a second, because they're actually a different phenomenon.
The whale shape comes from-- let me draw it for you.
Let's zoom in.
Let's zoom in on our target here.
So here's our target material.
We have an electron hitting it.
It's just a metal.
And we've got photons coming up.
Let's zoom in really close to the surface of that target material.
Let's go back over here.
So we've got-- looks something like this-- we've got an electron, which I'll draw in yellow, coming in.
So here's our electron.
It's about to hit our material.
Notice I've drawn it with some crystal structure, because that's what we're going to talk about, really zoomed in, the atomic level.
Now this electron, we've talked about this in class.
What can happen is that this electron could come in-- here's its path, normally, if it wasn't going to get deflected-- it can come in, and it can be deflected.
OK?
So it can get deflected off-- let me draw the following path in blue.
So it can go off like this, like this.
It could actually go straight through, without being deflected at all.
It could actually be reflected back, like this.
And these different reflections correspond to the electron being accelerated.
It's getting accelerated.
If we define our system like this, so here's our x and our y, the electron's getting accelerated in the y-direction.
This is just a simple location, but what happens, is when you accelerate a charge, you generate radiation.
You generate electromagnetic radiation.
Accelerating charge.
So, you know, here in the first path, where it comes through, and it gets deflected only by a little bit, you generate low energy radiation, low energy photons.
So what we're talking about here is a very large wavelength photons coming out.
So we have an electron.
It gets deflected.
It's accelerated.
And from that point, we're also generating a photon.
Same thing for all these other paths, but for the larger angle deflection.
So this one here.
And as you actually start deflecting back, you're generating very high energy electromagnetic radiation.
And this is all because the electron is basically, you know, is getting accelerated.
You can actually have, you know, the maximum energy photon you can give off here is a photon that corresponds to this electron coming in and getting stopped completely by the atom.
Think electrostatic repulsion.
So it comes in, and it gets stopped dead in it's tracks.
So you have some energy, it was moving some kinetic energy, and now has 0 kinetic energy.
The energy of your photon that gets given off is basically the difference in the two.
And that's what we call here, on this plot, the short wavelength limit.
We're going to write SWL.
Because this is lambda, which means that energy moves in the other direction.
Lambda goes up, energy goes down.
Lambda goes down, energy goes up.
So this is our short wavelength limit, which means that is the maximum, the highest energy we can generate from this setup here.
So all I'm saying is that you can get deflections in any angle across this way, you can get deflections in angle and generate any number of wavelengths of photons with the minimum wavelength being this one, that's generated from this situation.
So I'll just make it very, very small for you.
So you generate all these wavelengths, you create all this energy of photons, and that's what basically corresponds to the whale-shaped curve.
You're more likely to get deflections that correspond to something around here, because you have higher intensity, and over here, you're less likely to see those happening.
So that's our whale shape.
And now, let's return to the problem.
Let's ask, when we use photons, what's the situation?
Well, photon means we're no longer looking at an electron.
Our photon is coming in.
And the thing about photons, is that they're not going to get deflected like that.
They don't have a charge.
There's no electrostatic propulsion, for example.
So a photon will either only be absorbed and re-emitted, or it will go through the material.
So what that means is that there's no process now that will accelerate a charge to create Bremsstrahlung radiation.
And if there's no process to accelerate a charge, then there's no way to get Bremsstrahlung radiation.
So the first thing you need to do to get most the points on this problem is to say, look!
There's no Bremsstrahlung radiation anymore.
There's no charge being accelerated.
So that's sort of the first answer.
The second thing to realize is that these spikes still exist.
Because the reason for these spikes is a different mechanism than it is for the Bremsstrahlung.
These characteristic peaks correspond to the movement of an electron to a different energy level, and then a dropping back down.
So it's very characteristic of the material itself.
So a photon can still come in.
It could still liberate or move an electron.
It could just knock an electron out of its shell.
Perhaps n equals 1.
And then the n equals 2 electron will drop down and fill in the n equals 1 shell corresponding to K-alpha radiation.
If you had n equals 3 electron dropping down to n equals 1, you'd have your K-beta radiation.
And likewise from, you know, 3 to 2, 4 to 2, L-alpha, L-beta.
So that's the key take-home message here.
The message is that your Bremsstrahlung, the thing that you see in the notes in the class all the time, the whale shape with the spikes, that's two specific mechanisms.
The whale, the Bremsstrahlung, corresponds to the breaking of an electron.
These peaks correspond to the ejection of an electron, and an electron's moving around energy levels within the atom.
OK?
So our learning objectives for this problem, the things that we've taken home from the problem overall, is that we know, for example, Rydberg equation is only for hydrogenic-type atoms, and we know how to use the Moseley's equation now.
That was part 1.
Part 2, we know Bragg's Law, and how to use it, and what all the variables mean.
That's something you should learn and take home.
And I really highly recommend that you derive Bragg's Law from the geometric interpretation.
And part C, we really wanted to probe and understand who conceptually understood what was happening.
And part C tells us, the thing that we learned from it is that this Bremsstrahlung radiation with the peaks that we saw in class, that's two separate mechanisms occurring.
So that's problem number 2, and I hope you did well.
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OK we're going to do problem 3 now on exam 2 from the fall 2009 class.
Let's talk a little bit about what the problem concept is.
This is doping.
We're going to talk all about doping semiconductors.
As we say chemistry, where doping is legal.
So we're going to look at the things we should probably know before attempting the problem.
You want to review your doping principles.
How does doping works.
What are the mechanisms, your conduction mechanisms?
So why does the material conduct?
What's actually happening?
Your donor and acceptor levels, understanding which is which.
And your p versus your n type doping.
What those mean.
So look at those real quick and do the problem.
Let's go on.
I wrote down the information that we're given in the problem.
The problem part a asks us to find specifically how much gallium is needed to reach a certain carrier concentration in our germanium.
So we're doping gallium into germanium.
We're given the fact that the germanium band gap is 0.7 electron volts.
We know that we want to achieve a carrier concentration of 3.091 times 10 to the 17 carriers per centimeter cubed.
And we're working with 1 kilogram of germanium.
So we need to achieve this density or this carrier concentration in one kilogram of germanium.
I've drawn schematically-- now remember the germanium crystals are in 3-D so the bonds don't actually look 90 degrees like this-- but I've drawn just for visual a germanium crystal.
So this is what a pure germanium crystal would look like.
I'm going to draw the band structure for the pure germanium crystal now as well.
When you see a problem that gives you your band gap and it tells you this is about doping, the first thing you want to do is just get some points.
Put some stuff on the paper that you know to be true.
So let's draw the band structure.
I'm going to abbreviate conduction band with cb and valence band with vb.
So here's my conduction band, here's my valence band.
We're told that this is here.
We know our band gap energy is 0.7.
OK and I'm going to use this blue to show where electrons are.
This just shows that there's electrons in the valence band going all the way up to the top of the valence band.
That's how the valence band level is defined.
So this easy points.
You've got this on the paper, we're good to go.
Now what we're talking about is putting gallium into germanium.
So let's do that.
Let's take out this germanium.
And let's put a gallium in its place.
We'll put it in red.
OK so this isn't exactly correct yet.
Germanium makes 4 bonds.
If you look at the periodic table you'll see that it has 4 valence electrons to make bonds.
But gallium does not.
Gallium only has 3.
So what that means is that one of these bonds can't exist like this anymore.
So the way I like to draw it, is we still have an electron from this germanium but we're missing one for this gallium.
And this is what a hole basically is, schematically, in doping.
So we're going to write this as hole plus.
And the reason it has a positive charge associated with it is because in our neutral crystal with no charges you having an electron there.
And you've just removed an electron and now it's positively charged.
You have a positive charge in this region of the crystal, which corresponds to a missing electron.
So we just created a hole.
We have a hole here and then it's going to result in what we call-- I'll use red again-- an acceptor level.
And remember this band diagram is an analogy for, these are the energy levels where your electrons can exist.
So we have these energy levels and the valence band and we've got energy levels up in the conduction band.
What's just happened is that this dopant has created an energy level slightly above the valence band.
If you have a donor level from a different type of doping-- which we'll talk about in a second-- you'll create a level right below the conduction bands.
So we've got some points.
We understand the system now.
It's just going well.
So let's try to do a.
So a, as I said before is asking us to figure out the actual amount.
We're looking for grams.
This is what we're looking for.
Grams of gallium.
That we need to put into a kilogram of germanium to create a carrier concentration of 3.091 times 10 to the 17 carriers per centimeter cubed.
Not too bad of a problem.
This is basically just stoichiometry.
Dimensional analysis.
Here's how I did it and many people did it this way.
I start off.
times 10 to the 17.
And I always keep my units.
That's really important.
So we're going to write this as carriers per centimeter cubed.
Now we're also told in the problem, we have the additional information that the temperature is high enough that all of the sites are ionized.
What that means is that-- let's take a look at this band diagram-- we have electrons existing in the valence bands up to that level.
We've doped and now the temperature's high enough that these electrons can be excited, one of them in this case can be excited to this level here, the acceptor level, creating, of course, a hole in the valence band.
That's what it means that it can be fully ionized.
So basically every gallium atom that we put into our material creates a carrier.
And why do I say that?
Because conduction is either, the movement of electrons in the conduction bands, which we don't have.
Or it's the movement of holes in the valence band, which we now have.
So we're talking about conduction.
And so for every gallium atom we put in, we're creating an acceptor level, which takes an electron up, which creates some conduction in the valence band.
Now if we add lots and lots of gallium atoms-- there's a lot more gallium atoms in here-- we're going to actually see many more of these levels showing up.
And you're going to get what almost looks like very thin bands, an acceptor band or acceptor level there.
So we start off like this.
And I went into that's digression because now I'm just going to say that for every atom we put in we get 1 carrier.
And then we're going to say-- I want to be sure I get this right-- we have 6.02 times 10 to the 23 atoms per mole.
That's right.
And then we have 1 mole.
And notice how I'm being very careful about the dimensional analysis here.
Because if you put something in the wrong top or bottom you're off by 46 powers.
So don't make that mistake.
We saw that a lot on the exam.
So we're good here.
Let's do our dimensional analysis.
We start off, we're looking for this particular concentration carriers per centimeter cubed.
We have 1 atom creates 1 carrier.
1 mole has this many atoms.
And one mole of-- we're talking about in this case-- gallium.
I'll put Gallium here to be clear.
1 mole of gallium has this much mass.
And you're left with something like this.
You're left with 3.58 times 10 to the negative 5 grams per centimeter.
Grams gallium per centimeter cubed.
So we're looking good.
We're pretty close to the answer but we don't actually have the answer yet.
Remember we're looking for total grams of gallium.
Not how many grams per centimeter cubed.
That's pretty easy because we're told the we have 1 kilogram of germanium and we know the density of germanium from our periodic table.
So looking at our periodic table we can look up germanium's density and we can then back out how many centimeters cubed of germanium we have.
We'll do that right here.
So 1 kilogram of germanium is 1,000 grams, times-- we have the density, which is going to be times 1 over the density, rather-- grams per centimeter cubed.
And that's going to give us 187 centimeters cubed.
And now we're good.
Because we know here we have 3.58 times 10 to the negative 5 grams of gallium per centimeter cubed of germanium.
We have this many centimeters cubed of germanium.
Multiply them together and you get the answer, which is 6.69 times 10 to the negative 3 grams of gallium.
So that's part a.
Just dimensional analysis and stoichiometry.
But I stress, be careful about the way these things are.
Even I pause to make sure they're correct because if you get this wrong, the whole problem's wrong.
You're off orders of 46.
You're off by orders of 1,000 here, so just be careful.
Part b.
We've actually already answered when we drew it up, we threw down the things we knew.
We knew that we were creating a hole.
And a hole corresponds to p type doping.
How do I remember p and n?
It's easy.
p means positive.
We've created a hole, which has a positive charge.
n, I think of as negative.
So if you had put something else in there like arsenic, arsenic would've had an extra electron compared to germanium, which means you have a negative charge.
So n negative.
So we have p type doping.
We're done with part b.
Easy.
Part c. I'm just going to erase some of this stuff here and we're going to draw a couple more of these band diagrams.
So basically what we're asked to do now is to think about, what does this band diagram look like at different temperatures?
Because it actually looks slightly different.
I've drawn it here schematically.
But let's actually think about what it looks like in real life.
So let's do 2 more of these.
And we'll use that one as well for part c. OK so I'm drawing my conduction bands.
These are my conduction bands.
These are my valence bands.
I have the same band gap for all of them.
So you know band gap here, here's the band gap.
It's always worth putting down.
If you know it's true, put it down.
And we know that in both cases-- let me just get rid of this thing here; we're going to start from the beginning; put our electron back-- we've created all these acceptor levels.
Because we've doped with more than one gallium atom.
We have many, many, 10 to the 17 gallium atoms.
Which sounds like a lot, but in a mole we have 10 to the 23 spots in a mole of material.
So it's 6 orders of magnitude less.
So we have these systems and we're asked to do them at-- I'm going to keep going back and forth-- 300 k, 4.2 k and 1200 k. That's Kelvin.
Let's go with the extremes first.
And we'll do the middle one, 300 k, once we have an idea of what's happening.
So basically, in the very beginning when we to dope in a material, time t equals 0 and no electrons have had a chance to move around.
You have all your electrons in the valence band, you have these open holes in your acceptor levels.
And the reason that electrons would move between levels is if they have some energy associated with them.
Now why would something have energy in a crystal?
Well that's because of thermal vibrations.
It's a thermal energy they have.
So it's completely based on the temperature at which the crystal exists.
So at a very low temperature-- let me write that up there for you, this is our basic equation-- at a very low temperature we have a very low thermal energy.
k
this is the Boltzman's Constant.
At a very high temperature, we have a very high thermal energy, which means that these electrons have the ability to move between levels more easily.
So here's our very low temperature.
At a very low temperature we can actually calculate the energy.
We'll find that the energy here is about-- just so you have a scale-- it's about 0.00036 electron volts.
Now you say to yourself, wow that is a lot less than 0.7 electron volts, which is the energy of this gap.
So there's no way these electrons here can even make it up to this acceptor level.
That's not exactly true.
This is best thought of as an average thermal energy that an electron will have.
It's based off of a distribution, if you will.
So most electrons will have around this energy.
But some will have a little bit more and some will have a little bit less.
And very, very, very few at the tail of that distribution will have enough energy to actually make it up to one of these levels.
So the way to do this problem, to be actually fully correct, is to say that and explain it.
I actually, when I did this problem on the answer key, I have an electron, one electron going up into one of these spots and the rest are all empty.
I denote that with a little bit of blue here.
But everything else is empty.
We've got a hole here.
So there is a finite, very small, amount of conduction.
But it's extremely small.
And it's extremely unlikely to have electrons moving around but probabilistically it will happen.
That's 4.2 Kelvin.
Very, very low.
Let's go to very, very high.
Starting at the beginning again, we know all of our electrons are in the valence bands.
At 1,200 we have an energy of about 0.1 eV.
Still less than 0.7.
We have a significantly larger proportion electrons in that distribution, which are at 0.7.
So we actually do have electrons that will move up into the spots.
And even many that'll move up here as well.
Because they can overcome this 0.7 eV.
So we've got some electrons, they're going to move here, they're going to fill up most of these.
Because this delta here, the difference between our acceptor level and our valence band, is actually quite small.
It's extremely small.
That's another reason why we have an electron moving up at 4.2.
I should have mentioned that.
This is very small.
This gets pretty much completely filled.
And I'll denote that with a blue line here.
And we have some electrons, even going up into here.
What that means, that that has to correspond to electrons being depleted out of the valence band.
This is the way you would draw this picture.
So we lose these electrons from the valence band.
Some jump up to the acceptor levels and the rest will go up to the conduction band.
So this is 1,200.
We have more than enough energy to reach this acceptor level and we have probabilistically some electrons will make it up to the conduction bands.
That's the best way to think about it.
Now let's go back to our middle temperature.
I think the best way to do this one is to just go somewhere in the middle.
That's the safest play, right?
So we're at a temperature which corresponds to about 0.025 eV.
Now that is obviously smaller than the gap.
Probabilistically we'll still get some electrons in the gap.
Diminishingly small.
But we'll put one up there.
OK maybe one goes up there.
And it's still much bigger than the delta here between the acceptor level and the valence band top.
So these electrons will still pretty easily make their way into the acceptor level.
Now I'll put a little bit here.
We're going to lose electrons very slightly.
That's how I would have done that problem.
It sort of looks like that on the answer key.
Now I want to emphasize that, what is what's the importance of this?
Well when you dope, you're creating conduction.
Why is that?
Because the more we dope, the more holes we have that can be played around with.
So we have more and more of these acceptor level energy levels.
And what happens is these electrons at finite temperatures-- temperatures above 0 Kelvin-- will move into those levels.
And they will cause conduction in the valence band.
And you know at high temperatures, back over here, we have electrons moving to the conduction band, which is traditionally what we think of when we think of conduction, electrons moving around.
So we have in this case conduction in the conductor band as well as in the valence band.
So that's this problem.
This problem was generally pretty easy; many students got it right.
The learning goals and objectives we had from this problem were, number 1, to understand doping principles.
What it means to n and p type dope.
Holes versus electrons and how you get them.
So if you have germanium or silicon, those are very common examples and you dope things into them.
So gallium or arsenic or aluminum, for example.
We talked about conduction mechanisms-- just like here-- and we also talked about the donor and acceptor levels.
So the donor levels are just slightly below the conduction band and are full of electrons.
And the acceptor levels are just slightly above the valence band and initially holes.
So that's probablem 3 and I hope you had good luck at It.
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Hi, I'm Jocelyn, and today we're going to go over to the Exam 2 from Fall 2009, and Problem 4, Parts (a) and (b).
As always, we're going to start by reading the question.
Boron exists in the gas state as the dimer of boron 2.
Explain how the fact that B(2) is paramagnetic, two unpaired electrons, implies that, in this molecule, the pi 2p orbitals must lie at a lower energy than do the sigma 2p.
OK, so what is this question actually asking you to do?
Well, it's asking you why the energy of the pi 2p is less than the energy of the sigma 2p, given that boron 2 is paramagnetic.
So the first thing we need to know is what paramagnetic means, and he actually gives that to you in the problem, so it means that you have two unpaired electrons.
So this is unpaired electrons.
We are asked here then to give a reason why the fact that there are unpaired electrons implies a certain ordering of the sigma and pi 2p orbitals.
So the first thing we want to do, probably, is recognize that these are molecular orbitals, not atomic orbitals.
Therefore, we probably want to start by drawing the molecular orbital diagram for boron.
Now, how do we draw a molecular orbital diagram.
Well, the first thing to do is to start with the atomic orbitals.
So boron has five electrons, as we can find out from our periodic table, and they occupy the 1s, the 2s, and one electron in the 2p.
Again, if this seems unfamiliar to you, you might want to go back to earlier in the course, where we went over electron configuration.
However, when we're looking at molecular orbitals, we actually only care about the valence electrons, because those are what are involved in bonding.
The core electrons are much, much lower in energy, so they aren't really involved in the bonding.
So we're actually only going to look at the electrons in the second energy level.
So we have three valence electrons per boron, and they are in the 2s and 2p, which we remember, there are three p orbitals in each energy level.
So now let's draw those in.
We have 2s and 2p.
Remember to always have a scale here, and we're in an energy space, and this is not to scale, because we're just looking at relative energies, we don't care about the actual difference in energy.
And because we have two borons-- so this is boron one-- I'm going to draw the same thing over here.
Make sure they're at the same energy level because they are the exact same atom, and therefore, the same atomic orbitals.
So this is, again, boron, and we have three electrons for each.
We don't have to fill them in, but I'm going to do so, just to keep track.
And again, we're filling according to the orbital filling rules gone over earlier in the class.
So now that we have our atomic orbitals, we are going to combine them to, like, molecular orbitals.
And we want to remember that s orbitals combine to make a sigma and a sigma bonding and a sigma anti-bonding.
Remember that whenever you combine atomic orbitals, you create-- if you're combining two atomic orbitals, you create one of a higher energy and one of a lower energy, because the net energy stays the same.
Oh sorry, that's crooked.
All right, now we'll move on to the 2p, and because the problem told us that we need to explain why the pi bonding orbitals are lower in energy than the sigma bonding orbitals, we're going to start with that configuration.
The actual order depends on the molecule, and so without any extra information, we wouldn't really know if the sigma was above or below the pi.
So as the question, kind of, hints at: p orbitals make two pi bonds and one sigma bond.
This is due to the orientation of the pi bonds, or the p orbitals in space.
Then, as always, we need to put in the anti-bonding just to make sure we keep track of all of our orbitals.
OK, so we combined one, two, three, four, five, six, seven, eight atomic orbitals, and we made one, two, three, four, five, six, seven, eight molecular orbitals.
We want to make sure we always have the same number of orbitals before and after.
So now we can fill in the electrons into the molecular orbitals: that's these in the middle here.
And we use the same rules as we do when we're filling atomic orbitals.
So we start at the lowest energy, pairing spins, if we only have one orbital in the energy level; but as here, we have degenerate orbitals, so we can actually have unpaired spins.
So we have one, two, three, four, five, six electrons, and we have three from each boron atom, and so, we have the same number of electrons in our molecule.
Now that we have our molecular orbital diagram, we can go back to the actual question.
Some students stopped here and were like, I made the molecular orbital diagram.
I'm done.
But we need to remember that the question is asking us why the configuration we drew makes sense because of the unpaired electrons.
So if we go back to our molecular orbital diagram, we see that because the pi 2p have two degenerate orbitals, we can have unpaired electrons.
If we think about the sigma being below, we would have paired up those electrons, and therefore, gotten 0 unpaired electrons, and it would not be paramagnetic.
So let's just try that out.
Move over here, and we're going to draw just the molecular orbitals; since we already did the full process of making them from the atomic.
So this is the sigma 2s, sigma star 2s.
And now we're just looking at what if the sigma 2p was lower in energy than the pi 2p.
And again, we would fill up our orbitals, and we see that because this lowest occupied molecular orbital is singularly degenerate, we pair up our electrons, and so if this were the case, boron would not be paramagnetic.
However, because it is paramagnetic, we know that the pi 2p is lower in energy than the sigma 2p.
So that's what this question was asking, and if you said all those things, great job, and you answered the question correctly.
Now we're going to move on to Part (b), and it's very related to Part (a), so we're going to use the same diagrams here.
So it asks, is the gas molecule B(2) 2 minus more or less stable than the gas molecule to B(2).
Explain.
So here he's asking about the boron dimer that has a negative 2 charge-- that is, it has two extra electrons.
So we're still talking about boron; and thus the atomic orbitals that we're combining are exactly the same.
I'll write that down.
So we have we still have the 2s and the 2p from each of the boron.
That's not the same energy level, and they still make the same molecular orbitals, but we have two extra electrons.
So when we go to fill up our molecular orbitals, we now have eight instead of six.
So not only is the boron 2 minus no longer paramagnetic, because we don't have unpaired electrons, but we also have a difference in the bond order.
So if you don't remember what bonding order is, it is when-- I'll just put it here-- it's how we determine the strength of the interaction between the two atoms.
And because he's asking is the boron 2 minus more or less stable, we want to figure out the strength of the interaction, and that will tell us which is more stable.
So the bond order is the bonding electrons minus the anti-bonding divided by 2.
Sorry that's a little messy, but I hope you get the idea.
So for boron 2 minus, which we're talking about over here, the bonding order: we have two, four, six bonding electrons, electrons in bonding orbitals.
And we have two in anti-bonding orbitals, divided by 2 and that gives us a bonding order of 2.
So that's equivalent to saying there's a double bond between the boron 2 minus, the two borons in that molecule.
So let's move back to our diagram for the neutral boron dimer, and we have two, three, four bonding, and two anti-bonding, and so that equals 1.
So in the neutral dimer, we have a bonding order of 1, and in the charged 2 minus dimer, we have a bonding order of 2.
From this, we can see that the 2 minus has more bonding interaction, and so, it will be more stable.
So again, you can't just put that it's more stable, we need to have an explanation for why, and this is a way to explain that.
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Hi, I'm Jocelyn, and today we're going to go over Fall 2009 Exam 2, Problem number 5.
As always, we're going to read the problem first.
Which compound do you expect to have the higher boiling point: HF or ammonia?
Justify your choice with an explanation, using narrative or cartoons or both that makes reference to the operative chemical bonding.
So first, we are trying to figure out which has a greater temperature of vaporization, or boiling point; and we're looking at HF and ammonia.
Now when we're talking about boiling points, we're talking about breaking a certain type of bond, and that type of bond is intermolecular bonds.
We're not going to be breaking the actual molecular bonds bonding hydrogen to fluorine or nitrogen, we're breaking the bonds in between molecules of HF or ammonia, and so what types of bond do we have to look at?
And Professor Sadoway, in class, went over a couple of different types of intermolecular bonds.
And I'm going to write them down in order of increasing strengths.
So the weakest intermolecular bonds we have are Van der Waals.
The next is induced dipole-dipole.
Next we have dipole-dipole, and finally a special type of dipole-dipole is the hydrogen-bond, which is the strongest.
So here we're increasing strength.
Van der Waals is an induced dipole-dipole.
That means that although your atom or molecule does not have a net dipole, the electrons can shift, and at some point will cause a difference in polarity between different sides of the molecule.
Induced dipole-dipole is two different molecules interacting: one that has no net dipole and one that does.
Dipole-dipole: pretty self-explanatory.
You have two molecules that have dipoles, and they are interacting.
And then hydrogen-bond is when you have hydrogen bonded with x, where x is nitrogen, oxygen, and fluorine.
We call that a hydrogen-bond, because these are really, really strong dipole-dipole interactions and cause much different behaviors in molecules, and so we, kind of, separate it out at the subset.
So now that we've reviewed intermolecular bonds, let's go back to the problem.
So we were asked to figure out the difference, or which one has a higher vapor pressure, or temperature of boiling, so that's a relative thing.
And as we now know that we need to look at intermolecular bonds, a good place to start is with the Lewis structure, because that will tell us if we have dipoles or not, and then we can see what the strongest intermolecular bond is that we need to break to boil these molecules.
So starting with HF, we have hydrogen bonded to fluorine.
We know that fluorine has a much higher electronegativity than hydrogen, so we're going to have a net dipole.
Well, we have a polarized bond and a net dipole towards fluorine.
This causes there to be a partial positive charge on the hydrogen and a partial negative charge on the fluorine.
So what is our strongest intermolecular bond that we have for HF?
Going back to our list of bonds, we see that we have a dipole, so we might first think about dipole-dipole.
However, we need to remember about hydrogen bonds, and here in HF, we have hydrogen bonded to fluorine, and so that would be our strongest interaction.
So let's write that down to keep track.
Now for ammonia, we have nitrogen bonded to three hydrogens, and again we have a difference in electronegativity, and so each of these bonds is polar.
And because of the asymmetric geometry, we will have a net dipole, and each of these have a partially -- sorry, the hydrogens have a partially positive charge, right, because they are less electronegative, and the nitrogen has a partial negative charge.
So, which is our highest or strongest intermolecular bond that we have here.
Again, we have a net dipole, and so we have dipole-dipole interactions, but because hydrogen is bonded to one of the special three atoms, we have hydrogen bonding, and that is the strongest.
So, that is what we care about.
So now, we're, kind of, at a roadblock here; both have hydrogen bonding, which is the strongest intermolecular bond.
So how do we decide which one is actually going to be stronger than the other, and therefore, have the higher boiling point.
Well even though they're both hydrogen bonds, they will have difference in strength, and that's going to depend on the partial positive and partial negative charges, because all of these interactions are due to coulombic attraction.
In fluorine, we know that fluorine is the most electronegative element and has a higher electronegativity than nitrogen, so just to keep track of our thought process here: this will cause the partial positive on hydrogen to be larger in the HF than in the ammonia, and so from this we can say that the hydrogen bonding in HF will be stronger than the hydrogen bonding in ammonia, and therefore, the t evaporation of HF will be greater than the t-- the boiling point of ammonia.
So now we're going to move on to Part B.
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So now we're going to start Part (b).
First we're going to read the question.
To which does an atom of argon form a stronger bond: another argon atom or an atom of krypton.
Justify your choice with an explanation, using narrative or cartoons or both that makes reference to the operative chemical bonding.
So again, he's not just asking you for one or the other, but that you give an explanation to show that you understand the concepts that are being tested.
Here we're asked: does argon bond stronger with another argon or krypton.
And some students were confused by the usage of the term bond here.
But remember that bond doesn't necessarily mean a covalent or an ionic bond, it can also mean intermolecular bonds, which we have been using in Part (a), so it might seem logical that we would think about intermolecular bonds in Part (b).
Now we need to look at the specific species we are asked to consider here.
Argon and krypton are both Nobel gases, and what do we know about Nobel gases?
They have complete octets; they don't bond very often, and so, especially to each other.
Therefore.
we will probably be looking at intermolecular bonds, as I said before, but also the weakest form, which is Van der Waals.
We're looking at Van der Waals, because each of these are a mono-atomic species.
That is, they just are gases as one atom of argon and one atom of krypton.
And so, Van der Waals is the only type of interaction they can have.
As I said in Part (a), Van der Waals has to do with the movement of electrons in an induced dipole.
And what does that mean?
So, we're going to talk a little bit about polarizability.
And polarizability is basically how easy is it to induce a dipole into the atom or molecule.
And because these are non-polar atoms, obviously, we want to think about what can cause a difference in polarizability between argon and krypton.
So if you think about it, we're going to do a very simplistic model of the atom.
We have a cloud of electrons, and to get intermolecular interactions, we want to have a partial negative and a partial positive induced in the atom.
And this can be due to just spontaneous fluctuations in the electron density or to an external charge or another molecule or something.
But here we have two non-polar atoms, so we want to see which one will inherently be easier to polarize.
And so if we go back to our two species here, we have argon, which has 18 electrons and krypton, which has 36 electrons.
So based on our discussion of polarizability, we can see that krypton is going to have a greater amount of polarizability, because it has a larger cloud of electrons.
Those electrons are bound less strongly.
As we know from our periodic trends, krypton is a larger atom, and so we are going to say that the polarizability of krypton is greater than argon.
As we talked about in Part (a), even though these had the same type of intermolecular bonds, if one of them exhibits a higher difference in partial charges as krypton will, then it results in a stronger bond.
And so the answer here would be that krypton has a stronger interaction because it is more polarizable.
And that because is important.
Remember, when you're answering a question, you don't just want to put down the answer, you want to show that you know why that's the answer, and that you understand the concepts.
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Hi.
I'm Sal.
Today we're going to be solving problem one of exam three of fall 2009.
Now before you attempt the problem, there's a couple of things that you should know or have background knowledge of these materials before starting.
First thing is understanding the properties of crystal defects and for this problem particularly, Frenkel defects, which we'll talk about.
The formula, which is an Arrhenius relationship, as a function of temperature and energy enthalpy of the fraction of vacancies that are formed in a crystal and the conversion again between one electron volt and to the joule.
This will help you solve the problem.
So the problem reads as follows.
Silver bromide, which is AgBr, has rock salt crystal structure.
So it's an FCC Bravais lattice with the ion pair Ag plus and the Br as the basis-- Br minus the basis.
The dominant defect in AgBr or silver bromide is the Frenkel disorder.
So it's telling you what the dominant defective is.
So the question asks, for part A, so I'll put A here.
Part A for the question asks, does the Frenkel disorder in silver bromide create vacancies of silver plus, vacancies or Br minus or both?
Explain.
Dionic radii are 0.67 Angstroms for silver plus and 1.96 Angstroms for bromine minus.
So that's data that's given to us.
I'm going to go ahead and write that down.
So I know from reading about point defects and crystals that a Frenkel defect is pretty much formed when you have an ion pair of dissimilar sizes.
So if one of the radii-- like, say, your anion is a lot bigger than the radius of your cation, then one should expect that you would have a Frenkel defect to form.
And what a Frankel defect is-- which you should know from reading the material-- is that it's when one of the ions in your crystal leaves and leaves back a vacancy-- so just migrates-- just hops out of its lattice site and leaves back empty spaces, which is called a vacancy.
So we're given two things.
If I draw a little picture-- I can call this my silver and I'll call this my bromine.
And I'm given the fact that the radius of Ag plus is equal to 0.67 Angstroms-- and an Angstrom is just 10 to the minus 10 meters-- very small number.
And I'm also given the radius of bromine, the anion, is 1.96 Angstroms.
Now I would argue that these two have very dissimilar radii.
Obviously, the bromine looks to be about three times bigger than your silver.
So by understanding the definition of a Frenkel defect, I can claim that I would expect the smaller ion of the two to be the one that leaves the vacancy behind, hence forms the Frenkel defect.
So for part A, do I expect it to create silver plus vacancies or Br minus vacancies or both?
I would expect just to be silver plus, given the size of your cation.
So if you were to write on this problem, just expect only Ag plus, which is your silver cation, or the smaller one.
Expect Ag plus smaller ion to create a vacancy and hence, this leads to the Frenkel defect.
So yes, I should expect it to form a defect now.
If I was given a cation that had a similar radius that I-- I wouldn't know if I could argue the fact that this will form a Frenkel defect or not because the conditions are that you have to have the similar radius between your cation and your atom-- and it makes sense because the fact that your silver is small-- it has more freedom to hop around.
So it requires less energy for this smaller atom to then start hopping around your lattice site without penalty or without major penalty compared to your bromine ion.
So therefore, this is the one that should be expected to form that.
OK.
So that's part A.
And part B reads-- calculate the temperature at which the fraction of Frenkel defects in a crystal with silver bromide exceeds one part per billion.
The enthalpy of Frankel defects and formation, delta h sub f, has a value of 1.16 electron volts per defect.
And the entropic pre-factor A has a value of 3.091.
So it's giving you data.
Now the way I would solve this problem is that the first thing I would do is that I would write down what my data is.
So I'll call this data and the first thing is that our fraction of vacancies that are formed in silver bromide is one part per billion-- so 10 to the -9.
That's a fraction so no units.
The second thing that we're given in the problem is that our energy of formation-- our enthalpy energy-- is 1.16 electron volts per defect.
This is the energy penalty for every time one of your silver cations jumps out of sight to create that.
Nothing is free.
Everything requires energy.
We're also given the fact that A, the entropic pre-factor, is 3.091-- with no units-- and given our equation that I showed you on the bullet point.
That you should know-- you can go ahead and see what's missing.
So if I write down my equation-- this is all-- I'll box this off as my data because I'm going to refer to this to solve the problem.
And the problem talks about temperature.
So I know that f sub v is going to equal to the pre-factor times the exponent of negative delta h, of formation over kb t. Now you notice that kb wasn't given to you and kb is both in its constant.
Now a lot of student forget that they have a table of contents in front of them and that value is in there.
So if you don't know it by heart, then you want to make sure that you reference to that table because a lot of information will be given to you, because you're expected to look at the table of contents or your periodic table.
But if I include this as my data-- I'm going to go ahead and write it over here-- that kb-- half the value of 1.38 times 10 to the -23 and this has units of joules for degree Kelvin.
So this is something that you should pay particular attention to because now kb is in joules for Kelvin, but our energy was given in electron volts.
Now this is the number one thing that will take off points.
You'll get points taken off if you don't notice this-- that you need to go ahead and do the conversion from electron volts to joules or joules to electron volts.
Either which way, you're going to get the answer.
But the problem asks, calculate the temperature.
The very first sentence.
So what does that mean?
Well, I'm going to go ahead and look at my equation and I'm going to look at the data that I have and the constants and see if I'm missing anything else because obviously you can't solve an equation that has two unknowns and just one equation.
Solve the equation, you've got to only have one unknown for the equation.
So f sub v-- no.
Check-- no.
That's given.
Is A given?
It's right here.
That's given as well.
What about delta A sub F?
Well, that's the energy-- the energy penalty to create a defect and kb, which we got from our table of contents, and t. So this concludes that the only thing we're not given is temperature because that's what we're asked to solve.
So I need to do some math on here to go ahead and solve for temperature.
And the first thing I want to do is take the natural log of both sides.
So by taking the natural log of both sides-- so natural log of f sub v-- this equals to natural log of A plus the natural log of the exponent part and we know from math that the natural log of an exponent cancels each other out because they're inverses.
So the natural log of exponent, of negative delta h sub f, over kb t-- this cancels that and we just get the natural log of A plus negative because you have a negative up here-- delta h of formation divided by kb t. So the only thing we don't know here is temperature.
So if I can rearrange this equation and solve for temperature, I can get an answer.
So if I do that-- what's the best way of doing that?
Well.
I can add-- I can move this to the other side and move this to the other side and that gives me delta h sub f divided by kb t equals natural log of A minus natural log of your vacancy fraction and I can then multiply both sides by t so it cancels this one and it arrives over here and then divide both sides by ln of this.
So I'll go ahead and do that.
So I'll multiply this by t. Multiply that by t. So that cancel that and I end up having delta hf over kb t-- the t got canceled-- this equals to the natural log of A minus the natural log of your vacancy fraction times t. So now if I divide both sides by this factor, I solve for temperature.
So this gives me an isolation that t ends up being delta hf over kb times the natural log of A minus the natural log of the vacancy fraction.
Now all we have to do is plug in the numbers, but again, if you look at the units of Boltzmann's constant, which are joules per Kelvin and the value that we're given up here is electron volts for defect, we want to go ahead and convert the electron volt to the joule because it'll just be easier to do the math that way.
So I can go ahead and write it out and I do that by multiplying the top by that conversion factor and if I do the math, t ends up being-- we have 1.16 of this-- has units of electron volts for defect and then I want to get joules because that's what Boltzmann's constant has.
So in order to get joules, if I multiply this by the conversion factor which is 1.6 times 10 to the -19-- this has units of joules per electron volt.
Now the electron volts cancel and I get-- the numerator has units of joules per defect.
That's good because that's going to cancel with the units in Boltzmann's constant.
And then I put in 1.38 times 10 to the -23-- this is joules per Kelvin-- and this factor is multiplied by the natural log of 3.091 minus the natural log of 10 to the -9.
So unitless-- so if you do the math out, you end up getting a value that at the end of the day, your temperature-- this t right here-- will be equal to-- let's go ahead and write it over here.
So now that we go ahead and do the math-- that way it's nice and clean.
We know that the t, the temperature that we got from plugging all that math-- ends up being just 615 Kelvin.
And where did the Kelvin come from?
Well, it came from your joules per Kelvin, from your Boltzmann's constant, because that's what the dimensional analysis does.
So this is the answer.
This is good.
The problem doesn't say show the answer in degrees Celsius, but since everybody knows degrees Celsius-- everybody in science relies on degrees Celsius-- you can just convert this over to degrees Celsius, which happens to be 342 degrees Celsius.
So this right here is your answer to part B.
And again, I want to express again that the crucial part in getting this is knowing the fact that you're given an energy in electron volts, but the constant that you use has units of joules.
So you need to convert to get the right unit out or else you will get not Kelvin-- you would get something different here, which wouldn't make any sense given the question that was asked.
So with that, again I advise that-- always read the problem in detail.
Look at the units that you're working with and make the appropriate conversions because everything should be on your table of contents.
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Hi.
I'm Sal.
Today we're going to solve problem number two of test three of fall 2009.
Now before you attempt the problem, there are certain things that you need to know to solve the problem.
One is mechanical properties of metal and also the properties of amorphous metals, which are known as metallic glasses, and how to draw crystal structures given to you.
The problem reads as follows-- for a given alloy composition, explain why the yield strength of the amorphous form, the metallic glass, is greater than that of the crystal form.
Now in order to answer this question, you want to start thinking about the mechanical properties that these two raw materials have.
So if I look at a metal, I know that the way a metal deforms before it reaches classic deformation-- it deforms the planes sliding past one another-- so it's known as slip.
So I know that metals-- they form via slip and slip is facilitated by dislocations and you should know what a dislocation is by now, which is just a line imperfection in your crystal.
So slip is facilitated by dislocations.
So I know that my metal has planes that are arranged because it's a nice crystallographic structure-- and that the slip-- when it undergoes deformation, that this is actually assisted by dislocation.
So having dislocations helps to form our metal.
So now I want to start thinking about what a metallic glass has-- so the amorphous form.
So if I look at the metallic glass-- and all I'm doing here is comparing the basic properties between the two-- so this is going to get me metallic glass.
I know that for metallic glass, there's no long range order.
So the fact that there's no long range order means that these metallic glasses don't form a perfect crystal array.
And that lack leads to metallic glasses having no dislocations.
So if a metal slip is facilitated by dislocations and if a metallic glass doesn't have dislocations, then I would assume, just by comparison between the two, that the threshold for one of the materials to yield would be due to the fact that-- whether or not I have dislocations.
So with this argument, I would say that the metallic glass has a higher yield strength than my metal because of this property.
So that-- if you did that-- if you went ahead and wrote all that out for your answer in part A, you should be able to get most of the points for this problem.
Now part B-- it asks, on each of the following separate drawings of one face of an FCC unit cell-- FCC stands for face centered cubic-- indicate one of each of the following.
So now we're going to move on to part B-- and part B wants-- there's three scenarios.
One-- it wants us to indicate on a face of a crystal, a substitutional impurity-- so substitutional impurity.
So if I look at-- if I draw a face-- here.
I can-- so to find out what the definition of a substitutional impurity is, I know that it's not going to be anything that can be just embedded in between atoms-- like an interstitial.
I know that it's going to be a substitutional impurity.
So if this is one of the faces of your crystal-- and it's FCC, which is face centered, from the face-- the face is centered on the center of an atom on your face.
Then I would say that if I substitute this out for a different atom of type B compared to these atoms, then that's a substitutional impurity because that atom is not the same as the other.
It's like having silver and gold, for example, in composition.
And then it also asks for the second part.
So that's number one.
For number two, it wants us to demonstrate a vacancy.
Now a vacancy is simply an atom missing from a normal lattice site on your crystal array.
So if I draw again another face, then I can leave that blank and I can go ahead and I can point and I can say that the vacancy is actually-- sits right here.
So this is my vacancy.
There's no atom that's there.
It moved out of its place.
So that's another defect that we put in there.
And for three, it asks, an interstitial impurity.
So what's interstitial?
Well, an atom that can fit in between two atoms that are different than the one that fits.
So if I draw again an FCC face, I notice that there's certain spots where an atom of a different composition can fit and I can go ahead and I can just put a-- that's an atom of different composition and that can very simply be your interstitial.
So that's your interstitial impurity.
So knowing the definitions, you're able to answer this problem very simply.
There was no math involved in this and all you needed to know was knowledge of our mechanical properties of metals versus amorphous metals, which are known as metallic glasses.
And knowing that information, you're able to answer a question like this, which is pretty simple after you finish and you think about it.
And if you didn't know what FCC was-- which stands for face centered cubic, you could have probably lost a lot of points on this problem.
But if you know that, if you know how to draw it, then you're good to go.
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Hi.
Jocelyn here and we're going to go over fall 2009 exam three problem number three.
As always, let's read the question first.
Calcium ammonium phosphate dissolves in water according to the dissolution equation, for which the value of the solubility product, ksp, has been determined to be 4.4 times 10 to the -14.
Calculate the solubility of the compound in water.
Express your answers in units of molartiy, ie moles of the compound per liter of solution.
So the first thing we want to do is write down what the question's asking.
Part A-- how much calcium ammonium phosphate can dissolve in water?
And we're going to call this cs-- or the saturation concentration.
Next we need to figure out how to find this out, right?
So it gives us that the ksp-- 4.4 times 10 to the -14.
So we need to know something about the ksp and although it's called the solubility product, it's the same as other equilibrium constants.
So equilibrium constants tells us about the concentrations at equilibrium and therefore the maximum solubility.
Looking at the equation, we have calcium ammonium phosphate dissolving-- which is a solid-- dissolving into calcium ions, ammonium ions and phosphate ions.
As with every equilibrium constant, we can write down an equation relating to the concentration of the species.
So our ksp is the product of the concentration.
And a common mistake made on this problem was that people included-- students included the calcium ammonium phosphate solid on the bottom as you normally would-- where you would normally put the reactant.
However, remember that for equilibrium constants, we don't put solids in there because the activities don't change-- or the concentration doesn't change.
It doesn't really make sense to add it in.
So here we just look at the solvated ions and we see that we have a 1:1:1 molar ratio here.
That's going to make our life a little bit easier.
So now we need to figure out what the solubility of the compound is and to do that, we want to look at this chemical equation.
We see that for one mole-- so one mole of calcium ammonium phosphate dissolved gives you one mole of the calcium ion, the sodium ion and the phosphate ion.
Thus, we can say the amount of calcium ammonium phosphate dissolved, which we called Cs, is going to be equivalent to each of these concentrations.
Because we're asked for how much of the compound is dissolved, but we're given the ksp, which is in terms of the concentrations of these solvated ions.
I'm sorry.
This doesn't equal the product.
These are all equal.
Now we can plug the Cs into our ksp equation that we had before.
So instead of the concentration of calcium ions, we have-- it's equal to the concentration of the compound dissolved.
The same goes for the ammonium and the phosphate, because again, we have 1:1:1:1 stoichiometric coefficients.
And we can be a little more concise.
Now it's just a matter of solving for the saturated concentration.
So doing some algebra here.
Then we plug in the numbers.
And we get that the solubility of calcium ammonium phosphate is 3.53 times 10 to the -5 moles-- molar, sorry.
So that means 3.53 times -5 moles per liter of water.
That's not very much.
So we can say generally or relatively this calcium ammonium phosphate is not that soluble in water.
So let's move on to part B.
Part B says, calculate the solubility of calcium ammonium phosphate in 2.2 molar calcium bromide.
Express your answer in units molarity.
Assume that in water, calcium bromide completely disassociates into calcium 2 plus cations and bromide anions.
This is basically asking us for the same value.
It's asking us for the solubility of calcium ammonium phosphate, but under slightly different conditions.
So we want to write those conditions down.
Now we have to ask ourselves why would the fact that we have a concentration of calcium bromide affect the solubility of calcium ammonium phosphate?
And the thing to remember here is the common ion effect, right?
2 molar calcium bromide will-- when dissolved in water, the concentration of calcium from just the calcium bromide will be 2.2 molar, right?
Because we're told that it completely disassociates.
So if we look at the disassociation reaction, we see that for every mole of calcium bromide, we get one mole of calcium and two moles of bromide.
So having calcium bromide will alter our answer from the previous problem because the ksp will always be the same.
It's an equilibrium constant, right?
So even though it has to do with calcium ammonium phosphate, we have to take into account that we already have calcium in the system.
So before we start plugging in any numbers, we should think about this problem.
Do we think that already having calcium in the system will increase or decrease the solubility of the calcium ammonium phosphate?
Hopefully we can agree that it would probably decrease the solubility because you already have those calcium ions.
So putting more calcium ions into the water is going to be harder and therefore less calcium ammonium phosphate will dissolve.
Now that we know what kind of number we're looking for, we can start doing the actual calculation.
So again, we'll start with our equation for the ksp, which for calcium ammonium phosphate has not changed.
I'm just going to rewrite it on this board here.
But now instead of having each of these be equal concentrations, we already have some calcium in the system.
So we need to take that into account.
And I'm going to call the saturation concentration Cs star because we're under different conditions, right?
The calcium bromide does not contribute any ammonium or phosphate ions so those concentrations will just be determined by how much of the calcium ammonium phosphate dissolves.
From before, we can use what we found in part A to simplify this a little bit.
So in part A, we know that without anything else, without a common ion effect, that we have decided will decrease the solubility.
We have 3.35 times 10 to the -5th molar solubility.
Therefore, we can say with good certainty that our saturation concentration with the common ion effect will be much, much less than 2.2 molar because 2.2 molar is much greater than our pure solubility.
Going back to our equation over here, that means we can have-- and now we have a fairly simple algebraic problem.
Dividing by 2.2 and then taking the square root, we get that the saturation concentration under these conditions-- 2.2 molar calcium bromide-- will be 1.41 times 10 to the -7 molar.
And going back to our answer from part A, we see that this is indeed a lower solubility.
There's less calcium ammonium phosphate that can be dissolved and that makes sense from our previous thought about this problem.
And so now that you've found the answer, box it and we're done.
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Hi.
Jocelyn here and today we're going to go over fall 2009 exam three problem number four.
Starting with part A-- let's read the problem.
In the 1920s, Jack Breitbart of Revlon laboratories found that acne could be treated by the use of benzoyl peroxide.
The oxygen-oxygen bond in peroxide is weak and under the influence of modest heating, benzoyl peroxide readily decompose to form free radicals according to the reaction given.
The rate of decomposition is measured at 92 degrees C at various concentrations and found to be-- and you're given two different concentrations and two different rates.
So part A asks, determine the order of reaction for the decomposition of benzoyl peroxide.
The first thing to do is, when asked for the order of the decomposition reaction, we want to think about the general rate loss, right?
We're given concentrations.
We're given rates.
We're asked for the order.
Something you might want to have written down on your equation sheet or look up is the general rate law equation.
Remember that students taking these exams did have an equation sheet so they could have this ready.
The important part is that you know when to use this equation, what each of the terms mean.
So r is the rate.
k is a rate constant.
It is a function of temperature, but if you're at the same temperature, that rate constant will not change.
It's constant.
c is your concentration.
And n is your reaction order.
So this would be the place to start for this question.
And now we know that we're solving for n. So what is n in this case?
Looking at what we're given, we can write two different equations.
The first line gives us a specific rate for a specific concentration.
The second line gives us another rate for a second concentration.
So now we have two equations and two unknowns.
That means we can solve for one of the unknowns and then could solve for the other one if we wanted to, but here we only care about n. So how I'm going to choose to solve this and you may choose to do it differently-- I'm going to divide the first line by the second line and I get rate one over rate two equals c1 divided by c2 all to the n power.
So that made the k drop out, which is one of the unknowns that I don't care about because the problem isn't asking for that.
Now all we need to do is solve for n. So if we remember algebra, we're going to take the log of both sides.
Because of the properties of the log, we can bring the n out front.
So n equals-- and plugging in the numbers given in the problem-- they're given in the same units so everything will cancel out and that's good-- we get that the n equals one.
So this is a first order reaction.
Now you could've probably done that by inspection if you're familiar with the rate law and how to determine that by just seeing that as the concentration went down by a fourth, the rate also went down by a fourth.
However, the question asks for you to determine the order of reaction and so we were looking for more of a derivation of the problem instead of just an inspection because that shows that you have a familiarity with the general rate law equation.
So now moving to part B-- we are asked, on the plot below, sketch the variation in energy with extent of reaction for the decomposition of benzoyl peroxide.
Assume that the ratio of activation energy to-- delta e, which is the energy of the reaction-- equals -2.5.
Label the energy states and label the delta e of reaction, the activation energy for the forward reaction and the activation energy for the backward reaction.
So that's asking us to do a lot.
So first we're going to write down, what are we actually asked to do?
So the main thing is that we're asked for the variation in energy with the extent of reaction.
So as the reaction progresses, what's the energy of the molecule?
The energy state that it's going through?
In that plot, we're asked to label the energy states of the reactant and product and we're asked to label the delta e of reaction.
So what's the net change in energy as well as the activation, both forward and backward?
I would always write this to the side or scrap piece of paper or something because this is a lot of things to put on one plot and this will be a nice reference to help us to make sure we answered all the things the question is asking.
So on the plot given, which I'll reproduce here, it has axes of extent of reaction and energy.
The first thing to do would be to label the beginning and start energy states.
So because we know at 92 degrees C, this-- the benzoyl peroxide-- decomposes readily, we can assume that there's a decrease in energy.
So we'll have our products start at a higher energy than our-- sorry-- our reactant start at a higher energy than our products.
And because the question asks us to label that, we're going to put in the actual labels.
And so that's your reactant and then we have the radical product.
Now that we have our beginning and end energy states, we need to think about what's happening in the middle.
A clue is that he talks about the activation energy, right?
And we know that most processes that we talk about have a certain activation energy.
You can't just go straight from the molecule to the radical.
You have to have a little bit of energy cuffed.
And so we know we're going to have some type of hill that we have to go over.
Furthermore, in the problem he states that the activation energy divided by the delta e is -2.5.
So we know that because this is a negative delta e-- so this is our delta e here and it's negative-- our activation energy is two and a half times as large as that net change in energy.
So we want to kind of just eyeball that in here.
You didn't have to be exact, but relatively-- it's going to be larger than twice the distance between here.
And remember that activation energy is the hill, the energy costs that you have to go through to get to your lower energy state in the product size.
So now that we have all of our energy states, we want to draw a smooth curve to show the variation of energy with the reaction.
So connecting all these points, we get two valleys and a hill, right?
We have two stable energy states and we have an activated complex corresponding to the activation energy there.
So we have our energy diagram, but we need to go back to what the question is asking and see that we've labeled the energy states, we've labeled the delta e of reaction, but we need to label both the forward and backwards activation energy.
And this is one of the things people had issues with-- the backward activation.
So forward activation is pretty self explanatory, right?
It's the energy we need to overcome the activation of this process and so this is the Ea of the forward.
Now the activation energy of the backward reaction is just the same concept, but going backwards.
So if we started down here, what is the energy we need to get over this activation barrier?
And so that's this full energy, this full amount of energy here.
And so we can see that it's bigger than the activation energy for the forward reaction by the delta e of the reaction.
So going back to our checklist here, we've answered all over the questions and so we can move to part C. So part C asks us, on the same plot above-- which we have over here to the right-- sketch the variation in energy with extent of reaction for the decomposition of benzoyl peroxide under the influence of a catalyst.
So with a catalyst-- first to answer this question, we need to know what a catalyst does.
And in class, we learned that a catalyst will lower the activation barrier, thus allowing for the reaction to proceed faster.
You don't necessarily get more product because that's governed by a different set of laws, but you do get the product faster.
And so the way we can change this is that instead of having our activation barrier up there, we can say it's going to be somewhere here.
It's going to be lowered by the presence of the catalyst.
So we want to make a smooth curve again and connect this lower activation barrier.
And you'd want to label this-- I'll label it in white-- this is the catalyzed curve there.
As long as you've showed that the activation barrier was lowered by the presence of a catalyst, that's fine.
That was what the question was asking for.
So we've labeled everything, answered all the questions and we're done with this problem.
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Hi, I'm Sal.
Today, we're going to solve Problem #5 of Exam 3 of Fall 2009.
Now before you start this problem, there are a couple of things that you need to know in order for you to fully understand the science behind it.
And so what you need to know is Fick's first law.
Not just knowledge of it, but how it works.
Also, Fick's second law.
And not necessarily the equation form, but the solution to Fick's second law, which is a function of concentrations, and which we will refer to to solve the problem.
And also know popular dislocations, how they help your material and whatnot.
So the problem reads as follows.
Specimens of steel are being surface hardened by the introduction of carbon.
The concentration of carbon at the free surface of the steel is kept constant at c sub s. Chemical analysis of a number of specimens indicate that after a time-- t sub 1-- the carbon concentration has the value of c star at a depth from the surface-- x1-- shown below, which is shown on your graph.
So this is given to you.
So this pretty much is your concentration profile as a function of position.
And it's telling you what's happening.
So a good thing to do is draw a picture of the science that's happening here.
And they tell you, first of all, I'm going to draw it underneath here that you have a rod of steel.
And so this is steel.
And I have a certain fixed concentration here at the surface of carbon.
So I have some type of a carbon flux in my surface.
And this is a visual image of how it looks.
And what the problem says is that you have a certain concentration on your surface, which is given by c sub s. And then it starts telling you that chemical analysis was able to go ahead and measure this concentration at two different values inside your position at different times.
So if I continue reading the problem, it says that chemical analysis of a number of specimens indicate that after a time-- t1-- the carbon concentration has a value of c star.
So this one right here-- this is x1-- so this has a value of x1 at t1.
And it also tells you that after a certain time-- t2, which is equal to 2 times t1-- the carbon concentration has a value of c star, so the same value at a depth of x2.
So now, this thing says that at a depth of x2, at a time of t2, we measure the same concentration, c sub star in the specimen.
And it tells you that x2 equals 2x1, and it also tells us that t2 equals 2t1.
Now they tell us that for a reason, because it's probably expected for us to use this to solve the problem, so don't forget that.
So the question is, under these conditions, would you describe the rate of carburization as one, steady state diffusion.
Two, transient state diffusion.
Or three, not governed by any diffusion, i.e., rate limited by some other process.
Now you have to justify your choice.
So this is pretty much a problem of process of elimination.
And for part a, which is-- I'll label this part a-- it wants us to do three things.
So the first thing, it asks us if it's steady state.
So the first thing to ask yourself is, well, what's steady state?
And the definition of steady state is that whatever phenomenon is happening doesn't vary with time.
But the problem already tells you that it varies with time.
It tells you that at a certain time you measure this and at another time you measure this, and you have this flux coming in, and this is what's happening.
So from just reading the problem, I can already say that this process can't be steady state, because your concentration profile varies as a function of time.
So for one, I would write, can't be steady state due to the concentration being a function of time.
So one is out, so we can scratch that.
It's not a steady state.
Now two.
This is where it gets a little bit more complicated.
Two is a transient state diffusion.
Well, how do we know whether or not this is governed with time?
Well, you would argue, because this is not steady state, then it has to be time dependent.
But, there are certain rules that you need to know to be able to make that justification.
First of all, if it's time dependent, then whatever your concentration and whatever your position in your time is, has to fit into your solution to Fick's law.
So if I look at my solution to Fick's law right here, I know that on the left side of my equation, I have the concentration at any point in x-- that's what c stands for-- minus the concentration at my surface, which is given as cs-- sorry, this is supposed to be sub s-- divided by the initial concentration.
So c sub naught is if there was already some carbon in my steel, minus, again, the surface concentration.
This equals to an error function-- which is the solution to Fick's second law-- and this error function has the values of position, diffusion coefficient, and time.
So in order to verify if it's a transient state, we're going to go ahead and take that equation and break it down.
So if I assume that there's no carbon, I can go ahead and write this down.
Assume c0 is 0, just to make the equation easy.
I can, because there was no information given about that.
I can go ahead and-- because we're trying to figure out if it's a transient state-- write this down.
So c minus c sub s over negative c sub s. This equals to my error function, which is x divided by 2 squared of Dt.
And I know that at x1 and time-- t1-- I have this value of my concentration, c sub star.
So I'm going to go ahead and plug that in here into this equation.
So I know that my concentration is c sub star.
So that's what we measured, c sub star minus c sub s over negative c sub s. This equals to my error function, x1 over 2 square root of Dt1.
So this is for the first time point, x1, t1.
Now I want to go ahead and do the same thing for the second point.
And if I do it for the second point, I know that just by looking at the equation, I have the same concentration at x2, and I have the same surface concentration.
So what that tells me is that this value is going to be the same for x1, t1, and x2, t2.
So if I go ahead and plug this into my equation.
I can go ahead and look at this and say, this is for t equals 1 for that, and this is going to be-- or, t equals t sub 1, sorry-- and this is t sub 2.
I have c star minus cs all over negative cs.
This equals to my error function of x2 over 2 square root of Dt2.
So I know that this value is the same as this value.
So therefore, this should be the same as this.
The argument in both your error functions.
So we have to go ahead and prove that.
Well, how are we going to do that?
Do I know x2 as a function of x1?
I do, because that's given in the data.
And I know again, that x2 equals 2x1 and t2 equals 2t1.
So what I want to know, does x1 over 2 root Dt1, does this equal to x2 over 2 root Dt2?
Well, I'm going to go ahead and plug in my values that I know, and this becomes 2x1 divided by 2 times the square root of D-- and t2 happens to be 2t1.
So those two cancel each other out, and I end up getting that this is actually equal to x1 over square root of 2Dt1.
Now this and this are not the same.
Because this has a square root of 2, where the other one is just a 2.
So because it did not satisfy the solution to Fick's second law, we can say that it's not case two either.
This is not transient state diffusion.
So therefore, by the process of elimination, it has to be governed by some other phenomena.
And that is the answer to this problem.
The fact that it's governed by something that we don't have enough information to answer.
We don't have enough data points to answer.
So that's part a, which is good.
Now part b asks, if the steel specimens in part a were replaced with steel specimens of identical composition but with a dislocation density 10 times greater than that of the steel in part a, how would the rate of uptake of carbon change?
Explain.
Again, this requires you to know properties of dislocations, and what dislocations do to your materials.
Your metals, for example.
What is a dislocation?
Well, a dislocation is-- if you have a bunch of a plane of atoms-- what a line dislocation is, or an edge dislocation is, this line defect such that it just terminates.
So you have a plane of atoms that just comes to a point and then it stops.
And what do you create here?
Well it creates void, and void facilitates diffusion.
It increases-- in our case, we're talking about carbon diffusion into steel-- so I would argue that the more dislocations that you have, the greater the diffusion should be.
Because you have more void-- more spaces-- for your carbon atom to jump into and from.
So for part b, just by arguing the fact that dislocations represent defects in your crystal-- specifically a defect that can create void-- then with that argument, the fact that more void means higher diffusion, I would conclude that-- for this part-- that my rate of diffusion of carbon should increase when using steel of a greater dislocation density.
So the rate of carbon uptake or diffusion of carbon into your steel increases as you increase your dislocation density.
And with that, you're to be able to solve a lot of these problems.
And these are common in 3.091.
And again, knowledge of Fick's first and second law, how to use the solutions to Fick's first, second law, and knowing what a dislocation is, that will help you a lot in solving these problems.
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Hi, I'm Sal.
Today, we're going to solve Problem 8 of the Final Exam of 2009 for 3.091.
The problem reads as follows.
The phospholipid-- the chemical formula is right here-- is added to hexane-- which is C6H14, which is a hydrocarbon-- in sufficient quantities to enable interaction between phospholipid molecules.
Draw a cartoon depicting one possible molecular arrangement that results.
You may stylize your parts of the phospholipid molecule with the squiggly line-- which is shown in the exam-- for a hydrocarbon chain.
So if we look at our molecule here, the first thing that we want to understand is what the structure is.
And as you can see, the molecule has a nonpolar component and it has a polar component.
So for the problem, it will be good to isolate the two with different colors, and then we'll stylize them such that the nonpolar part is the zigzag line and the polar component is the red circle.
So we want to go ahead and draw the arrangement together when it's added to hexane.
So the easiest thing to do is to draw a picture.
So we come over here to the right.
And this is part a. I'm going to draw a beaker with hexane.
And I know that as soon as I look at my chemical formula for hexane-- which is C6H14 and is a hydrocarbon-- I know that it's nonpolar.
So I know from my knowledge of 3.091, when chemical bonding, that like dissolves like.
So the molecules that are polar are going to want to stick together, and molecules that are nonpolar are going to want to stick together.
So by looking at my phospholipid, I already know that-- the problem tells me that-- I'm adding a sufficient amount such that the molecules can interact with each other.
So the only thing that comes to mind for me that will be a possible arrangement, would be that-- in the hexane, which is nonpolar, and if we have a distribution of these phospholipid molecules-- then one possible arrangement will be that you have a bunch of your polar heads interacting together with the hydrocarbon components out here interacting with the hexane.
Because the hydrocarbon component again is nonpolar.
The hexane is nonpolar.
So those want to interact with each other.
But the polar component, which are the heads for the molecule-- the phospholipid as a molecule-- are going to go ahead and interact with each other.
And pretty much if you were to write this down on the exam, you would get full points for part a.
Now we'll move on to part b.
Part b says, to the beaker described in part a, an equal volume of liquid ammonia-- NH3-- is added.
The mixture of hexane-NH3-- or ammonia containing the phospholipid-- is now subjected to vigorous agitation for several minutes and then allowed to equilibrate.
Draw a schematic representation depicting one possible molecular arrangement that results.
The values of density are 0.65 grams per centimeter cubed for hexane and 0.6 grams per centimeter cubed for liquid ammonia.
Assume that hexane and liquid ammonia are mutually immiscible.
So there's a lot of information within the paragraph.
And the first thing that it gives us is the densities between the two components.
It tells us they're immiscible, so immiscibility means that they're not going to mix.
The question that you want to answer now is, which layer is going to be at the top?
Which layer is going to be at the bottom?
Well, the more dense layer should be at the bottom, and the more dense happens to be hexane.
So if I go ahead and draw my beaker again for part b, and then I mix my components.
And now I have two components that are separated by an immiscible boundary.
And I know that the density of hexane is given by 0.65 grams per centimeter cubed.
And the density of liquid ammonia-- which is NH3-- is given as 0.60 grams per centimeter cubed.
So I know that the hexane is going to be down here, so I'll go ahead and label this C6H14.
And I'll label this as NH3.
So one important thing to know is, what is the structure of ammonia?
Well, we know that ammonia happens to be a polar molecule.
And its structure is simply given by-- you have a central nitrogen atom, and you have three hydrogens and a lone pair of electrons-- and the lone pair of electrons is what gives the ammonia the structure.
Hence, you have a polar dipole moment on your molecule.
The fact that this is polar means that it's going to interact with the polar head of your phospholipid molecule.
And we already know we have hexane plus your phospholipid.
Well, if in our arrangement with the absence of ammonia the heads group together because that was the only polar component in the mixture, then over here I should expect the heads to rearrange themselves such that they are in contact with the ammonia.
So one possible arrangement for the system would be that I would have my polar part of my molecule interacting with the ammonia, and the hydrocarbon chain interacting-- which is the nonpolar component-- interacting with the nonpolar mixture.
So if you were to draw this schematic, then you would get full points on the exam as well.
So that's again, like dissolves like.
And if you fully understand the nature of secondary bonding, then this problem shouldn't be that difficult to go through.
Now there's a final part to this problem, which is part c. And part c says, to the beaker described in part b, a small amount of iodine-- I2-- is added to the mixture, and then subjected to vigorous agitation for several minutes, and then allowed to equilibrate.
Explain what you expect to observe.
Again, we're going on this whole issue of like dissolves like, or that polar molecules want to be with polar molecules because they should be miscible together.
And we are given that now we're going to add I2-- iodine.
Well, iodine is a linear molecule that's nonpolar.
There are no other atoms that are bonded to it that can give any polarity.
So if I was to go ahead and draw this diagram, the first thing that comes to my mind is that the iodine should mix with the hexane, because essentially it's a nonpolar component mixing with another nonpolar component.
So if I draw my diagram, I'm going to go ahead and draw the two immiscible regions.
And one question that a lot of students have is, how do you know that the iodine is not going to be on top of the ammonia?
Well, iodine is pretty dense.
It's actually more dense than these two molecules over here.
So the fact that it's more dense, I know it's going to want to sit at the bottom where the other nonpolar molecule is at.
So I would argue that down here we would have hexane-- C6H14-- mixed with iodine.
Over here, we will still have our ammonia.
And then our polar heads would still be-- the polar part of the phospholipid molecule will be in contact or bonding with the ammonia.
And the nonpolar tail would be with the nonpolar components which are hexane and iodine.
So if you did this on the exam, you get full points.
And being able to fully understand secondary bonding will help you solve this problem in no less than 10 minutes.
So that's it.
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Hi.
I'm Sal.
Today, we're going to solve problem nine of fall 2009, the final for 3091.
The problem reads, the phase diagram of the binary system neodymium, praseodymium and EPR is given below.
There are two allotropes-- alpha, which is hexagonal close pack-- HCP-- and beta, which is body center cubic-- BCC.
Now this problem obviously is about phase diagrams.
So before you attempt to solve the problem, you need to make sure that you understand how to read phase diagrams and also how to apply the lever rule and know where in the phase diagram to apply the lever rule.
So if we look at our phase diagram, which I have drawn over here for part A-- the part A asks, explain why a lenticular phase diagram is to be expected for the binary system.
So this is a lenticular phase diagram and a lot of students when answering this problem get confused and say, oh, because the components are soluble together, that's why you get a lenticular.
That's correct, but that information you can get from the phase diagram.
It's obvious by looking that you have an alpha phase that you have from zero to a 100% that can be missable together, but it doesn't really explain the answer.
Now the best thing to do is to look at the periodic table of elements and look at the physical and chemical properties of the elements.
So here we're comparing to-- we're comparing neodymium and praseodymium, which are two components that are making the alloy.
So here this is atomic percent praseodymium on your x axis and on your y axis, you have temperature in degrees Celsius.
So at zero percent praseodymium, we have a 100% neodymium and at 100% praseodymium, we have zero percent neodymium.
That pretty much comes from the phase diagram.
So why would these two have formed a lenticular phase diagram?
Well, if you look at the periodic table, you'll notice that these two elements lie right next to each other.
So what does that mean?
That means that chemically, they're pretty much almost the same.
To answer this question for part A, I would look at it and I would notice that actually both these elements have the same crystal structure.
So I would write down, they have the same crystal structure.
They also have pretty much the same electronegativity or very similar electronegativity.
So very similar electronegativity.
Also, they have pretty much the same atomic radii--atomic radius.
And also very similar, very close to each other-- the same density.
So the density is almost the same.
So because these two components have the same properties together-- the same size, same crystal structure, then you should expect that they should form a missable lenticular phase diagram, which is what we have here.
So if you were to answer that for part A, that would give you pretty much all the points and that hints that you understand how to look at the chemical properties of the elements and the physical properties of the elements and compare them together and rationalize why the phase diagram looks the way it does.
Part B asks-- I'll come over here to do part B.
At each point, I identify all the phases present in equilibrium.
II-- state the composition of each phase and III-- calculate the relative amount of all phase presence.
And it asks us to analyze the three points that we have on our phase diagram.
So we have one for I, II, III-- So number one-- if we look at our phase diagram, we lie exactly where your alpha phase is.
So to answer the first part of the problem is identify all the phases present-- we only have the alpha phase here.
So if you were to write down alpha phase, that's correct.
The part two says, state the composition of each phase.
So what's the composition of the alpha phase in here?
Well, if I trace down my line down to my x axis, I notice that for the atomic percent of praseodymium, pretty close to about 18%.
So it's 18 atomic percent here.
So to answer the question for part I, I would notice that-- I would write down that it's alpha phase.
For the second part, we have 18%-- 18 atomic percent praseodymium, which leaves behind about 82% atomic percent of neodymium.
And for part three-- part three asks about the relative amounts of the phases present-- we only have the alpha phase in that part so I would say that we have 100% pure alpha, which is-- if you write this, that's correct.
Now for the second point that we're going to analyze in our phase diagram number two-- if we come over here and look at it, you'll notice that now we lie in between or between two phases so where the lenticular region is at.
So here, I don't have pure beta or pure liquid.
I have a combination of both beta and liquid and this is exactly where you need to apply the lever rule.
You can't apply the lever rule here because it's just 100% pure phase.
You can only apply when you're in a region where you have two phases that coexist in equilibrium.
So if I go ahead and analyze part two-- part I-- the phases that are present are beta and the liquid phase.
For part II, it asks, what are the compositions?
Well, if I blow up my lenticular phase diagram-- so I'm sitting right here and this is sitting at around 62 two atomic percent of Pr-- praseodymium and it asks about the phases or the relative composition of each phase.
Well, over here, I have beta, over here I have the liquid phase and in between I have the beta plus the liquid that are coexisting in equilibrium.
So if I want to know how much-- what's the composition of my beta phase that I have here-- well, I'm going to have to go to the line where both intersect with the beta and the beta plus liquid intersects.
And if I analyze this point-- if I look down what the composition of that is for that line, I see that I have about 56 atomic percent of praseodymium.
So to start answering part two, I would say that if I look at beta-- in the beta phase, I have 56% or atomic percent of Pr and if I have 56% of Pr, then this leaves just 44 atomic percent of neodymium and if I look at my liquid-- so I'm going to just move this over here.
I look at the liquid, which is going to be where it intersects in the liquid region and if I trace it down, I see that this is about 70% or so.
So here I have around roughly 70%.
If you trace it down to your x axis, this is 70% or 70 atomic percent Pr, which leaves 30 atomic percent of neodymium.
So if you were to write this down for the liquid, just write an arrow here so I won't have to rewrite that-- that would be the answer for part two because again, we're analyzing this point, which is point number two and we're in between two phases.
So that covers the second part of part B.
And for the third part-- I'll go ahead and I'll put up here.
That one now is where we have to apply the lever rule because now we have to figure out exactly how much of each phase is present at equilibrium.
So if I look at my diagram, it's going to be very simple.
If I want to know how much beta I have in this mixture, you're simply going to apply the lever rule and for beta, it's going to be whatever this magnitude is or whatever the difference between these two points are over the length of the whole existence at that temperature of the length of the whole composition that you have.
So if I do the simple calculation for beta-- beta is going to be simply 70 minus 62 over 70 minus 56 and this essentially just equals eight over 14, which is about 57 atomic percent praseodymium.
So-- sorry-- not percent Pr-- percent beta phase.
So in beta, I have 57%.
So that means that for my liquid-- I can do two things for this problem, for the liquid.
I know that everything is normalized to 100%.
So the difference between 157 will be the liquid composition where I can go ahead and apply the lever rule again and do the math.
So if I apply the lever rule again for my liquid phase, it's going to simply be 62 minus 56.
We're coming back on our-- looking at the liquid phase that is coexisting in this equilibrium at this temperature and I want to know how much liquid-- it's going to be the magnitude of this over the total length of my composition there-- that's what I'm doing here.
Again, this is 70 minus 56.
So this would end up being just six over 14, which is essentially just-- if you have 57% here, if you do the math, you're going to get 43% of liquid.
So part three-- when you apply the lever rule, you see that at this point you have more beta than you have liquid-- and it kind of makes sense because if you look at your diagram, your point is sitting closer to the beta phase.
So I would expect intuitively that I would have more beta present then the liquid present and that's exactly what the lever rule tells us.
So that's point number two for part B.
And if I look at point number three now, which I'll write right here-- this is essentially the same thing as point number one.
But now point number three-- we're no longer lying in alpha.
We're lying in the beta phase.
So if I trace down what my composition is here, I'm about 78% of Pr-- so to answer the first part I-- beta-- the second part II, which is dealing with the compositions of that beta phase is going to be 78% or 78 atomic percent Pr and that leaves just 22 atomic percent neodymium-- and for the third part III-- what is the relative amounts of the phases present?
Not the composition, the phases.
I'm exactly where only pure beta exists so I can say that I have 100% beta.
So that pretty much does it for this problem.
Again, knowing how to read phase diagrams, knowing how to apply the lever rule and where to apply the lever rule will allow you to very easily solve the problem.
A lot of students make the mistake of trying to apply the lever rule when you have a pure phase.
I mean, that makes no sense because said lever rule pretty much calculates what fraction of the phases are present in that equilibrium mixture.
So keep in mind that you can only do that when you have a region in your phase diagram, when you have two phases that are coexistant in equilibrium.
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Hi.
Jocelyn here.
And today we're going to go over fall 2009's final exam, problem 10.
As always we're going to start by reading the question.
You have 333 milliliters of alkaline solution at a pH equal to 9.9.
You wish to neutralize this by reacting it with 222 milliliters of acid.
What must be the value of the pH of the acid?
So first off let's write down the information that was given and the information that he's asking for.
So we have 333 milliliters of a solution with pH equal to 9.9.
And he's asking you what must the pH be if you want to put in 222 milliliters in order to completely neutralize that basic solution.
And the key here is to recognize what neutralize means.
And a neutralization is when you take acid, or protons, and base and it reacts to form water, thus creating a neutral solution.
So the next step would probably be to figure out how much excess hydroxide we have in this solution.
And then we can figure out how much hydrogen we need to react with that hydroxide and to form water in order to completely neutralize this solution.
So first step is to find out what our concentration of OH minus is.
And we do that by remembering a few things about acid/base solutions.
Right?
We know that if you have the pH plus the pOH it equals 14.
And this is from the equilibrium constant of this neutralization reaction.
And then we can figure out what the pOH is because we're given the pH of our initial solution-- 9.9-- and that equals 4.01.
Now using the definition of pOH, which is that the pOH equals the negative log of the concentration of the OH minus, we can rearrange that and get that our OH minus concentration equals 10 to the negative pOH.
Right?
Just using our log rules there.
So this gives us that our current concentration of the hydroxide ion is 10 to the negative 4.1.
That gives us the concentration of the hydroxide, but what we really want to know, if we're going to use our neutralization reaction here, is we want to know how many moles of hydroxide we have in that solution.
And we can just do that with stoichiometry.
Right?
The concentration is in moles per liter.
And we know-- so our moles hydroxide equals our concentration, which is 10 to the negative 4.1 moles per liter.
Times-- if we want to get rid of that liter unit we multiply it by our volume, which is 333 milliliters, converting it from milliliters to liters.
And this gives us 2.65 times 10 to the negative 5 moles of OH minus in solution.
So that's just the first step of this problem.
Now that we have the number of moles of OH minus we need to find the number of moles of H plus necessary to react with this excess hydroxide, and then we will be able to find the concentration of hydrogen, or protons, in that second solution.
So first off we're going to look at our neutralization reaction over here and realize that it's a one-to-one stoichiometric ratio.
That means in order to react with 2.65 times 10 to the negative 5 moles of hydroxide, we need the same number of moles of hydrogen.
So over here I'm going to say that this also equals the moles of H plus needed.
All right.
Now that we have our moles H plus needed, we need to do the same process here but in reverse.
So first we need to find the concentration of protons that we have.
So given that we want to only have 222 milliliters of this solution, right?
So we have-- and we just divide that by our volume, 0.222 liters.
Right?
Because we divide by 1,000 to convert from milliliters.
And that equals 1.19 times 10 to the negative 4 moles per liter.
So this is another problem where you could get the answer but not actually be answering the question.
Right?
Because this is the concentration of protons that we need, but now we need to convert that to pH.
So moving over here we'll just write the final steps.
We know the definition of pH is the negative log of the concentration.
And so because we know the concentration of protons we know that the-- we just need to plug this into our calculator and we get our answer, which is a pH of 3.92.
Put a box around that one because it's our correct final answer, and you'll be good to go for this part of the problem.
Now moving on to part b we are asked, name the conjugate base of each of the following.
So we'll just start part b right here.
And the first thing to recognize is what does a conjugate base mean.
And in an acid/base pair, or when you have a kind of dissociation of an acid, in the most simple terms we can think about it just losing a proton.
In the leftover molecule, that's the conjugate base right there.
So in this problem we're asked if the molecule loses a proton, what is the resulting molecule?
Ion, basically.
So part i we have HPO4, and that will dissociate to hydrogen phosphate ion.
Simple enough.
This is your conjugate base.
So that's your answer for part i.
Moving to part ii.
We have CH3NH3 plus.
So it's kind of like an ammonia ion but not quite-- sorry, ammonium.
And we know that this dissociates by losing a hydrogen off of the nitrogen, because that's where the positive charge is.
It makes more sense there.
And we get-- sorry, keeping the same order as above-- a proton plus.
And this would be your conjugate base.
If you were a little bit confused on this one it might help to draw the Lewis structure, and you would see that the formal charge does actually reside on nitrogen and, therefore, it makes sense that the nitrogen would lose a proton.
In addition, if you think about the ammonia/ammonium ion relationship you know that you have NH3 going to NH4.
And in this case we just have one of the hydrogens replaced with a methyl group, and so you can make the analogy there that, oh, the hydrogen would come from the nitrogen in that case.
All right.
So hopefully this was pretty straightforward.
As long as you're comfortable with what a conjugate base is and what the dissociation reactions look like for acids.
So now on part c we are asked to classify each of the following as a Lewis acid or a Lewis base.
Now first we need to remember what a Lewis acid and a Lewis base are.
So if you remember from lecture or in your reading, we know that a Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor.
These definitions are just a little more general than the classic proton acceptor or proton donor.
The first part is asking us about the cyanide ion.
And to kind of clarify it for yourself and because we're asked about Lewis acids and bases, which are to deal with electron pairs, it might be helpful to draw the Lewis structure.
So for this molecule a good Lewis structure looks like this, and it has a net negative charge.
And you can see that we have a couple of lone electron pairs there that aren't involved in bonding.
And because it has a net negative charge you can imagine that these electron pairs would want to kind of grab onto something else that's a little bit more electron positive.
And because of that we know that this is actually a Lewis base.
It acts as a base and would grab something like a proton and thus become more, I don't know, stable in some cases.
All right.
Moving to part 2.
We are asked about water, which is a molecule that we deal a lot with in acid/base chemistry.
Again, you might be able to answer this right off the bat, but we're going to look at the Lewis structure here.
Water has two lone pairs.
So that means it could very well act as a Lewis base.
Hopefully that makes a little bit of sense.
Right?
Because these electrons can go and grab something like a proton and form the hydronium ion with a positive charge, thus acting as a Lewis base.
It's important to remember that even though we're creating an acid in the hydronium ion, this water molecule in converting to the acid acts as a Lewis base.
However, we also know that water can lose a hydrogen and accepts those electrons to form the hydroxide ion, and thus acts as a Lewis acid.
So in the case of water either Lewis acid or Lewis base was a correct answer in this case.
Now moving on to part d. We will go over here, I think.
I'm sorry.
d, not b.
Again, it's going to be about acids and bases.
Consider the effect each of the following substances has on the ionization of the weak base ammonia.
For each state whether the substance, one, suppresses ionization, two, enhances ionization, or, three, has no effect on the ionization of ammonia.
In each instance give a reason for your choice.
Going back to the first part of this question we see that it's asking us about the ionization of ammonia.
A good place to start from here would be to write down the reaction, just so we have that on our paper, in our mind, ready to think about it.
So we have ammonia-- and aqueous means that it's in water, right?
And that is reacting with water to form the ammonium ion-- sorry-- by adding a proton and a hydroxide ion.
OK.
So this is the reaction we're asked about.
Now we need to look at each of the three substances listed and decide how it affects this ionization.
So number i, we have KOH.
Again, a good place to start is to write what happens to potassium hydroxide when it goes into water.
And we know that KOH is a strong base and will dissociate into potassium plus hydroxide.
So we see that it produces hydroxide.
And this should indicate to you that, well, it has the same product as one of the products of the ionization of ammonia.
And if we remember back a little bit before this section, we know that if you have the same ion in solution it will definitely affect the solubility, or ionization.
So that's one way to think about it, is that I now have a common ion effect that will, because there's more hydroxide in solution now, it will push this reaction, this equilibrium, back to the left.
Another way to think about that is Le Chatelier's Principle.
Kind of the same thing, although it's a little bit more versatile, I think, so pretty easy to think about.
Right?
If you add a product your reaction will shift to the left.
If you add a reactant your reaction will shift to the right.
So here we're adding a product, our reaction will shift to the left.
Same as a common ion effect, these two combine to decrease, or both suggest that we will decrease, ionization.
Moving to the second part of this problem.
We're asked about hydrogen chloride.
Again, let's look at what that does in water.
We know that it's a strong acid, so it will dissociate into hydrogen, or protons, and chloride.
Now this one's a little bit trickier because we don't have either of these ions in our initial ionization reaction, but if we think about the fact that ammonia is a weak base we can write an additional ionization reaction.
So that additional would be ammonia reacting with a proton to form ammonium.
Right?
So this is another way that the ammonia molecule would be ionized.
And we can see that if we add protons from this acidic molecule there we will increase the ionization, because it will allow the ammonia to work more as a base because of this reaction here.
So we are increasing ionization.
Another way to think about this is completely valid and kind of the same thing, but if we look at this ionization reaction here where the hydrogen chloride is dissociating into protons and chlorides we can recognize that these protons will then react with the hydroxide that is produced in the initial ionization reaction.
This will result in a decrease of hydroxide in solution, right?
And so by Le Chatelier's Principle we know that a decrease in a product causes the reaction to shift towards the product, to make more hydroxide.
So that's just another way of thinking about the same process, and it results in the same conclusion that addition of hydrogen chloride increases the ionization of ammonia.
OK. Moving to part iii.
We have ammonium chloride, and this will dissociate into ammonium and chloride.
And even if you're not sure how much it will dissociate, it will still dissociate to some amount.
So it will have at least a little bit of an effect on our initial ionization reaction.
So here we see the dissociation produces ammonium ions, and just like the first case that is one of the products in our ionization reaction.
And so Le Chatelier's Principle, common ion effect, whatever you want to call it, tells us that it will shift the equilibrium to the left and decrease ionization.
So with that we're done with problem 10 from the final.
It had a lot to do with acid/bases, so hopefully you feel more comfortable with them after this.
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Hi.
Jocelyn here.
Today, we're going to go over fall 2009 final exam problem number five.
As always, we're going to start by reading the question.
The skeletal structure of the amino acid alanine is given below as it exists in its neutral zwitterion.
To the right is shown its titration curve in aqueous solution.
The abscissa expresses concentration in terms of degree of protonation so that a value of 0.5-- that the neutral ion is the only species present.
At a value of zero, alanine is totally deprotonated and at a value of one, alanine is totally protonated.
So let's move to part A.
What is the hybridization of each of the three carbons in alanine?
So to begin with this one, I would start by rewriting the structure of alanine because as it's written now, it's not as clear what the hybridization of the alpha, beta and gamma carbons are.
So starting with the alpha carbon, connecting the beta carbon-- and here is the gamma carbon-- and of course, we don't want to forget the amine group.
So this looks fairly similar to what is on the page, but I've kind of expanded the structure so that it's more straightforward for us to determine the hybridization.
The alpha carbon is the central carbon here and we see that it has four bonds and four electron domains.
If we go back to the beginning of the class, we learned that when you have four electron domains around a central carbon, we have Sp3 hybridization.
Going to the beta carbon here, we see that it also has four bonds, but one of them is a double bond to oxygen.
So we need to remember from our earlier lessons that the double bond only counts as one electron domain.
Therefore, this carbon has three electron domains.
And going back to our hybridization rules, we know that this gives us an Sp2 hybridization.
Lastly, we have our gamma carbon and it gives us one, two, three, four bonds and four electron domains.
Therefore, our gamma is also Sp3 hybridized.
This problem on the final was dealing with first beginning material, but then in part B and C, we're going to move on to the material we've learned more recently.
So moving to part B-- we are now are dealing with the titration curve that is given to us.
So this looks hopefully fairly similar to what is on your page and the question-- it's asking us to indicate on the titration curve-- one, the PKA for protonation of the zwitterion.
Two, the PKA for deprotonation of the zwitterion.
And three, the isoelectric point.
So before we start that, let's talk a little bit about this titration curve here.
What is it showing us?
What is it depicting?
What I like to do is-- we see that we have two buffered regions.
That right off the bat should tell you that there are two different equilibriums going on.
There are two different protonation/ deprotonation reactions.
So I'm going to split this right down the middle here.
And that way we can think about the two protonations/ deprotonations separately.
So on the left side here, we have the more acidic portion of our titration curve and even if you didn't quite understand his description of the abscissa, we know that because this side-- the solution has a lower Ph, it's going to be dealing with the more protonated species of the alanine.
This-- on the right side, we have the higher Ph, so we're going to have a more deprotonated alanine on this side.
And he already told us that we had the zwitterion in the middle.
So next to 0.5, we're going to write a generic term for a neutral acid.
If we look back to our structure, we see that we have-- the neutral zwitterion has one acidic proton here on the amine group and it has a neutral charge.
So that's why I'm going to depict that zwitterion as HA Moving to the left, more acidic portion of the titration curve, we have-- we know that there's a protonation going on.
So I'm going to write that as H2A plus because whenever we add a proton, we're adding a positive charge.
And then, at the more basic side, we have doubly deprotonated alanine and so we have just our backbone, which I'm going to call a and a negative charge.
So from the question, we should know that this is true, correct?
Because he told us that we have the totally protonated alanine on the left side.
We have the zwitterion in the middle at 0.5 and we have the totally deprotonated on the right side.
OK. What else can we know from this titration curve or what can we add to it so that we can understand it a little bit better?
Well, let's write down the equilibrium reactions that are happening that make this buffer region here.
On left side, we have the doubly protonated alanine going to proton plus the zwitterion.
And that's what's creating this buffered Ph region.
On the right side here, we have the zwitterion in equilibrium with-- again, a proton and the doubly deprotonated alanine.
These are things that are true for any titration curve you see.
Depending upon the number of protonations-- the number of times a buffer region might change, but for each buffered region, you can always determine an equilibrium constant that is causing that stability in Ph.
OK. Now let's go to the question.
We need to indicate on the curve the PKA of both this equilibrium equation and this equilibrium reaction.
So to do that, we hopefully will remember the Henderson-Hasselbalch equation and that gives us a relationship between the Ph of the solution and the PKA of the acid that's in solution.
So writing down the Henderson-Hasselbalch equation-- we have-- not the Ph-- and this is the same equation that Professor Sadoway derived in lecture.
And here, I've written it in the most generic form, but no matter what our equilibrium species are, we just know that we put the more basic species on top and the acid that got deprotonated on the bottom.
So for the first point of this, we are looking for the PKA of the protonation of the zwitterion.
So we know that it's an equilibrium between the doubly deprotonated and the zwitterion.
That is the equation we're looking at.
So let's write down our Henderson-Hasselbalch for that and I'm going to call this PK1 just to keep them separate.
Remember here, my Ha is on top because we have-- our doubly protonated species is the acid in this case.
Now in order to put on the titration curve where the PKA is, we look at the Henderson-Hasselbalch equation and see that where the Ph equals the PKA is when the concentration of the acid and its conjugate base are equivalent.
because if Ha equals H2a plus, the log term goes to zero and we have our Ph equals PK1.
Going back to the titration curve, we see that the concentration between H2a plus and Ha is equal right in the middle.
We can assume that this is a to-scale concentration abscissa here.
And so there we go up and we can say, OK, this point on the titration curve here corresponds to our PK1.
So from the Henderson-Hasselbalch equation, we can see that given the titration curve, this is where our PK1 is and it's about two.
For our purposes, that's-- we weren't asked to estimate it, but if I had to, I'd say it's around two.
Now we have to do the same thing for the other side of the titration curve.
So if you haven't done that already, maybe you can stop now and try that on your own because it's the same process.
OK.
So it's do of the second one-- which again, is just the Ph is equal to the PKA when the two species are equal, except for now we have different species.
So basically we're looking for-- our new Henderson-Hasselbalch equation is-- with using the new equilibrium reaction, our doubly deprotonated alanine is now on top and our zwitterion is on the bottom.
So this might be a little confusing because we have the same species going from the top to bottom in our Henderson-Hasselbalch equation, but remember that we're always talking about the acid is in the denominator and the conjugate base is in the numerator here.
But either way, we again want to look for where the Ph equals the PK2 and so our zwitterion concentration will equal our doubly deprotonated alanine.
And that occurs right here.
So the last part of this question is to indicate where the isoelectric point is and the important thing for this part of the problem, of course, is to know what the isoelectric point meets-- and it's where the dominant species in the solution is the neutral zwitterion.
And Professor Sadoway already told us where that is and it is where we're at 0.5 here, which we've already indicated as where the zwitterion is dominant.
And so we just can go up here and circle that and we can call that number three.
So you didn't have to go through all of the work I did and if you were familiar with the titration curve, you could have simply put these down, but understanding where these points come from is quite important for understanding acid base chemistry and how amino acids work.
All right.
So let's move to part C. In part C, we're asked to draw the skeletal structure of alanine when it is solvated in an aqueous solution at extreme acidity-- ie, Ph less than one.
So I'm going to write that down and then we can go back to our titration curve and see where that is because we've already spend a lot of time understanding our titration curve and that tells us a lot about the acid base chemistry of alanine.
So at the titration curve, we see that at a Ph less than one, we're going to have mostly our doubly protonated zwitterion.
That said, now we need to go back and look at the structure of alanine and we can put in the protonation.
So we're given the zwitterion form of alanine, but at very low acidity-- sorry-- high acidity, low Ph.
This carboxy group is going to be protonated.
And so we have both of our acidic protons are on the alanine and this would be a structure of alanine at a Ph less than one.
Now moving to part D-- we are asked for an aqueous solution of alanine-- calculate the ratio of the concentration of neutral alanine zwitterion to the concentration of deprotonated anion when the Ph is 8.091.
So a couple things I would write down.
Ph equals 8.091 and we are asked for the ratio of the concentration of the zwitterion to the doubly deprotonated anion.
When we're asked for this, hopefully you think first, Henderson-Hasselbalch-- I can use that.
We're given a Ph.
We can find the PKA and we're asked for the ratio and concentrations.
So let's write down the Henderson-Hasselbalch equation again.
And because we're asked for the ratio, we don't need any more information except for the PKA.
Luckily, from part B, we know what the PKA is of this equilibrium between the zwitterion and the doubly deprotonated anion.
So moving back to your titration curve, we see that the PKA is roughly 10-- doesn't need to be our-- our estimation doesn't need to be exact, but if it's in the ballpark range, then you have the right idea.
So I'm going to write that.
We can plug that in just down here.
Log of our concentration.
That's a log term.
And from here, it's just algebra.
So simplify a little bit.
If we bring the 10 over to this side, we get -1.909 equals the log of a minus over-- I'm sorry.
I'm getting a little bit crooked here.
And then if you take both sides of the power of 10, we'll get-- the ratio equal to this.
And if we take the reciprocal of this, we'll get a nicer answer.
We'll get HA over A minus and it won't be a fraction over here.
So I suggest that you do that and you'll see that our final answer is-- let's put it right here.
81.1:1-- the ratio of our zwitterion to the doubly deprotonated anion.
And then you've answered all parts of the problem.
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Hi I'm Jocelyn and we're going to go over fall 2009, the final exam, problem number 6.
It says, the dipeptide shown below is derived from alanine and serine.
And I'll write up that structure here.
And we know it's a dipeptide because we see that it has two-- what look like-- amine groups and two kind of carboxyl groups.
So that's an indicator that we're dealing with a dipeptide not just one amino acid.
Something to notice right off the bat is that we're given the zwitterion form.
We have the amine group at the end here is protonated and the carboxylic group on the other end is deprotonated.
So moving to part 1.
It asked us to indicate the position of the peptide bond.
Hopefully this is a pretty straightforward task.
We realize that the dipeptide bond is what makes the dipeptide.
It's the bond between the amine group of one amino acid and the carboxyl group of another.
Right there.
So circling that carbon nitrogen bond would get you full credit on that part.
Moving to part 2.
We are asked to draw the skeletal structure of each constituent amino acid as it would be present in an aqueous solution of an extreme basicity, i.e.
pH greater than 12.
So before even starting to write we want to think about what extreme basicity means.
Well it means that both our amine group and our carboxylic acid group are going to be deprotonated.
So let's start with alanine first.
We have the central carbon.
The amine group.
And the carboxylic acid group.
Now I just drew that totally protonated.
However we want to draw it totally deprotonated because it's in basic solution.
So I'm just going to erase the extra hydrogens.
And we have a deprotonated amine group and a deprotonated carboxyl group.
And that is what it would look like in basic solution.
Now serine is a little different because you have a side group with functionality.
It has not just carbon and hydrogen.
First drawing up the backbone here, we have a central carbon, which has a methyl alcohol group off of it.
And attach the carboxylic acid.
And amine group on the left here.
So this one I drew as the zwitterion, which would be in moderately or kind of neutral.
It depends upon the amino acid.
But again we want to deprotonate our amine group here because pH greater than 12 will not have a protonated amine group.
Now the question is, what do we do with the side chain here?
It has an H attached to an oxygen.
Does that come off in super basic solution or not?
A lot of students got tripped up on this.
An alcohol actually has a pKa of about 16.
And so it won't be deprotonated in an aqueous solution.
Therefore we want to keep this hydrogen on.
It will not be-- this will not be a charged oxygen in the basic solution that we were asked about.
So now we are going to move on to the second part of this problem.
The second part asks us about the tertiary structure of the length of protein that is shown.
I am not going to draw the whole length of protein but we're going to-- at the four specified locations-- we're going to indicate what we would call the tertiary structure; or what functionality is causing the tertiary structure.
So looking at that protein chain.
Professor Sadoway has indicated four different distinct parts of the chain.
And the question is asking us, at each of the four numbered positions name one chemical change to the environment of the protein that would destabilize the associated feature in the tertiary structure.
Explain the relevant chemistry so it's asking us to say what can we do to disrupt that tertiary structure?
And the first thing we need to do is figure out what the tertiary structure is.
So if we look at number 1, we see that it has two sulphur atoms connected together.
So we call that a disulfide bond.
The second part, we see that we have a phenyl group, so it has a very electron-deficient hydrogen interacting with a deprotonated carboxyl oxygen anion.
And so we have a hydrogen bond there and that's what's causing those amino acids to be pulled together.
So let's just say hydrogen bond.
At the third number we see that we had a deprotonate carboxyl group and a protonated amine group interacting.
And that's not quite a hydrogen bond but it's a negatively charged part of the molecule interacting with the positively charged part of a different molecule.
And so we call that electrostatic interaction.
It's similar to a hydrogen bond in that it's a intermolecular interaction that's causing these molecules to move closer, preference to be closer together.
And now we come to the fourth part.
Which is hydrophobic interactions.
So we see that there are a lot of just carbon hydrogen chains.
No oxygen is around, no amine groups around.
Purely carbon and hydrogen.
And because this is an aqueous solution, those hydrophobic parts of the molecules will preferentially go together and kind of make a little-- almost like an-- oil droplet in water.
They're trying to decrease the amount that they experience the polarity of the water.
And so the interactions are hydrophobic.
So now that we know what the functionality at each of these points in the DNA structure are, we can go back to what the question is actually asking us, which is, how can we actually disrupt these features.
So the first one is the disulfide bond.
And in class Professor Sadoway talked about how to disrupt the disulfide bond.
And one of the ways is to put extreme heat on to the protein.
And that breaks the disulfide bond and you can straighten out that portion of the DNA.
This is how you straighten your hair if it's curly or other types of things with the disulfide bond.
So heating the disulfide bond to extreme heat will cause it to break and thus the protein will lose that, that folding or whatever the disulfide bond imparted to that portion of the protein.
Now going on to number 2.
We have hydrogen bonding.
And we see that it's between a alcohol group on a ring there and a carboxyl group.
So one way to disrupt that hydrogen bond is to make the carboxyl group no longer deprotonated.
So if we put back on the hydrogen to that oxygen it will become less partially negative.
It will have a lower interaction with that hydrogen.
And so we can say, to protonate that carboxyl we want to make the solution more acidic.
So we can add acid.
So the disulfide we're going to heat, the hydrogen bond we're going to add acid.
Looking at number 3 we see that we have a electrostatic reaction.
And so we want to not have that electrostatic reaction.
And one way to do that is to make one of those entities not have a charge.
So one way to do that would be to reprotonate the carboxylic acid, just as we did in the previous step.
Or we could deprotonate the amine group.
And that would cause it to lose its charge and thus the electrostatic interaction would be weaker or gone.
So to deprotonate the amine group we should know that we want to add base.
And the final part is, the portion with hydrophilic interactions.
And here we want to figure out a way to make those hydrophobic portions of the amino acids not hold as tightly together.
And one thing we learned about in class is something like soap.
Something that has a hydrophobic side and a hydrophilic side.
And so if we put soap in with this protein it can kind of stabilize the hydrophobic part and make it stretch out a little bit more, not ball up so tightly.
So soap or detergent would be a good answer for this one.
And so for this question, as long as you indicated the relevant chemistry, which is specified what tertiary structure you were looking at and how your suggested procedure would disrupt that tertiary structure, that's what Professor Sadoway was looking for and that would mean that you answered the question completely.
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Hi there.
I'm Brian.
We're going to be doing problem number 7 of the 2009 final examination.
So the way I like to approach these problems is to identify what I think is important to know before attempting them.
Things that you should review that will help you successfully navigate the problem.
I call this the what I need, W I N. So for this particular problem, which I think of as the phase diagram problem, you need three things critically in order to successfully complete it.
Number one you need to have a general understanding of what a phase diagram is, how to read it, how to write it, how to interpret the information it gives you.
You can find a lot of this in chapter 9 of Shackleford.
Along the same lines you should probably know how to go in between binary and unary phase diagrams.
How they are related at least.
In this problem it's not absolutely necessary to know this.
But in similar problems we might ask, this is critical to understand how these two are interrelated.
The second thing you should know is some language like critical point and triple point.
Where these are and what they mean on the phase diagram.
And the third thing is the phase coexistence.
So what does that actually mean?
How do you find it on the phase diagram?
So with these three particular points, I'd review them and then I would recommend attempting the problem afterwards.
So we're going to start the problem.
We're going to start over here.
I'm actually going to start this from a reverse direction.
And review really quick what a binary phase diagram is.
because this is actually the phase diagram which you will see most often as you progress in your career as a scientist or an engineer.
This is something you'll see a lot of the time.
It'll have either one element and another element.
Or one compound and another compound along the axis.
And it shows the different phases which exist at certain temperatures and compositions.
So in general it's important to sort of understand what a phase is.
So a phase, is a chemically and a structurally homogeneous material given certain state conditions.
So a phase diagram depicts picks all these phases.
It's the master plot of phases that are in equilibrium given a particular set of state conditions.
When I say state conditions I'm actually referring to things like temperature, pressure, I even sometimes consider composition to be one of those state conditions.
So in a binary phase diagram, what we're looking at is the temperature versus the composition.
So if you have some composition of silicon and germanium at some temperature, you know exactly what phase it should be at thermodynamically.
It doesn't actually mean it will be, but it should be.
So that's what a binary phase diagram looks like.
So this is something you'll see a lot and this is something you probably will become used to seeing a lot as you go on in your years.
We're going to come back to this in a second.
But first, for now, we're going to it actually attempt the problem that's given.
And then I'll tie both of those together for you.
So let's go over and let's actually look at the problem now.
We'll go back to that in a second.
We're asked to draw a unary phase diagram for silicon.
And we're given some critical information.
Specifically on the problem we're given the triple point, we're given the critical point, and we're asked to draw this.
Not to scale, of course, but what it should generally look like.
So the first thing I do is label my axes.
Pressure, temperature.
And then we're given certain data.
So I'm just going to throw it down right in the beginning.
We're given pressure and temperature for triple point.
We're given pressure and temperature for the critical point.
So I'm going to put those points down right now.
Because I know that's part of the answer.
So let me first put down the triple point.
We're told the triple point occurs at 0.15 atm.
And we're told that that is at a temperature of 1415 celsius.
So temperatures are going to be in celsius and our pressures are going to be in atmospheres.
So we're going to go up to find this point.
In fact I'm going to give it a different color.
That is our triple point.
I'm next going to plot the critical point here.
We're told that that occurs at a pressure of 6,600.
So not to scale of course.
But that's going to be somewhere all the way up here.
Now we're told that that is at a temperature of 4880.
So here's a point here.
So we've just put two points down.
But that's not enough to complete the problem.
It doesn't actually answer all the questions.
The problem actually also asks us to provide the normal boiling point and the normal melting point of silicon.
So you're not actually given those temperatures because you've got to be clever.
And all the students in our class are given a periodic table during the exam.
So I would encourage you to have a periodic table with this type of data.
So if you look at the periodic table for silicon you can then find your melting point and you can find your boiling point.
So this data, a normal melting point and a normal boiling point will occur at 1 atmosphere.
Let me just put 1 atmosphere on the pressure here.
Again not to scale.
So anything that occurs at this particular pressure, this isobaric line, is going to be normal.
So our normal melting point occurs at 1414, which is right next to 1415.
There we go.
And our normal boiling point occurs at 3265.
These are approximate numbers.
So we'll put that here.
And now if you've seen a unary phase diagram before, which I hope you have during your studying, you'll know that it looks somewhat like a y shape.
So one thing we're going to do now is we're going to connect these dots.
We know that at very high temperatures we're going to have what's a gas, basically.
And at very low temperatures we're going to have something like a solid.
And we have some liquid in the middle.
And the way we're going to delineate these phases is to connect these dots.
And this is what our unary phase diagram is going to look like.
So in order to complete this problem-- it was actually not too bad of a problem-- all you have to do is label some particular aspects of this graph.
So number 1, we have a solid region, we have a gas region.
That implies this must be the liquid region.
I'll put an L there.
You're asked to identify 1, 2 and 3 component regions.
Or three 1-phase regions, 2-phase regions.
So number 1, you could have chosen for a 1-phase region, you can chosen a point here in the gas, a point here in the liquid, or a point here in the solid.
I'll just choose one in the gas.
So something here, let's call that i, for answering part i.
You could have chosen a point along any of these lines for a 2-phase equilibrium.
Which implies that at that point, so at that specific pressure and at that specific temperature, you have both phases of equilibrium.
Let me just choose one and explain it.
So let's choose something here.
Between 0.15 and 1.
So let's choose this point.
There we are.
So at this point, at whatever pressure we're at and whatever temperature we're at here, we have both liquid and gas in equilibrium.
They both coexist.
You're not tending towards one or the other.
They're both there.
So that's part B, the 2-phase coexistence area.
So we put that here.
i i.
And the third part is the 3-phase coexistence, which is-- almost you can pull it out of the name-- the triple point.
At the triple point-- and there's only one of these-- you have all three, gas, liquid and solid coexisting together.
So we've already sort of made that.
That was our blue dot.
1, 2, 3.
That's that point right there.
So not to scale.
But this is the answer.
This is full credit.
One thing that's really interesting to note is that on this particular answer we have a negative slope on our solid-liquid coexistence line.
Now you've probably seen this before in your studying and it occurred with water.
Most materials actually will have a positive slope.
If you think about it it's sort of counterintuitive.
What this implies is that if I'm at a constant temperature and as I raise my pressure I'm going to go from the solid to the liquid.
Intuitively, we would think that the solid is more densely packed and therefore as we increase the pressure we're going to tend towards a solid.
That's not the case in water and in this problem is not the case with the data we're given for silicon.
In this case as we raise the pressure we're actually turning into a liquid.
It's kind of counterintuitive.
Another way of actually saying that is the liquid phase at this particular temperature is more dense than the solid phase.
And you see that with water, which is why ice floats on top of water.
So that is our unary diagram.
That is the full credit on this problem.
I do want to tie it in really quick though, with what a binary diagram is.
Because this is a very important concept to get and a potential problem.
So we're going to go back over to binary phase diagram over here.
And so what we're looking at on this binary phase diagram is actually, it's the temperature versus the composition at a specific pressure.
Now for a binary phase diagram you're going to assume-- unless you're told otherwise-- that it occurs at 1 atm, 1 atmosphere.
If I want to go between the binary and the unary that we just looked at, you need to add your third axis.
So if we had a third axis-- let me draw that here.
This is going to be our z-axis.
There's the z.
We know that if we put a pressure along this axis, we're going to be able to extract what we had for our unary diagram.
So let me do that for you.
I'm going to call this the pressure axis.
Going back in and out of the board.
And actually this is at 1 atm.
Which implies at some point farther back we have the origin.
You can just do this lightly so it doesn't get too confusing.
We have an origin for our pressure.
So at this point we have 0 pressure.
We don't work with negative pressures.
0 pressure.
Which implies-- let me just draw the Cartesian form of this-- so we have composition, we have temperature, we have pressure.
That's our x, y, z.
Now what happens is that as we increase pressure what happens to, for example, this point?
Now what is this point?
This point is the point where we go from the solid to the liquid.
That's the melting point.
What happens to the melting point as we increase or decrease our pressure?
Well that information is given on our unary phase diagram.
Let's go back really quickly.
As we increase our pressure we can see that what's going to happen with our point is it's going to go up.
So what basically we have to do is we need to take-- you have to think three-dimensionally-- and take this graph and you're going to take it an spin it and put it on to one of those planes there.
So we're going to get an idea.
This is something that I would challenge you, the listener, to do and think about and do on your paper after you've completed this problem successfully.
I'm going to draw it right now.
And I'm going to let you look at it and think about it, what it means.
So let me do that.
Let's take a look now.
This is our temperature axis.
Here's t. Here's p. Remember unary was p versus t, so you've got to do a little flipping.
But let me just draw it now.
It kind of goes up to the words.
But you get the idea.
This is-- if you do a little flipping and I challenge you to do that, some three-dimensional thinking-- this is the unary phase plot that we gave for our answer.
And the point I'm trying to make is that you can go in between your unary and your binary diagrams by understanding that.
For example, binary diagram is at a constant pressure and your unary diagram, unary means that it has one material, one composition.
So in this particular problem I've created germanium to be our b element.
And we're operating here with pure silicon.
So you can go in between if you have the grasp of that concept.
So I would challenge you to think about that.
If you've gotten the problem right and you understand this concept, I think you're good to go on phase diagrams.
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OK, welcome back.
We're going to be looking at problem number 11 in the fall 2009 final exam.
This is one of those problems on the exam that kind of tests a whole bunch of other stuff we couldn't get into one big question.
So it's looking at a bunch of different areas.
But the main themes are going to be bonding and reaction rates and half lives.
So the things I think you need to know, the W I N list as I usually call it, you need to understand reaction rate orders.
What they mean, how they apply to reactions.
What a half life is and how you use that.
And the third thing is melting points and how they are related to intermolecular forces and how intermolecular forces affect melting points.
So review these three things and once you've got them under your belt then let's attempt the problem.
So let's start the problem.
The first part is asking us to determine the half life for a particular reaction.
So there's two places you could start.
We saw both answers.
You could start with the definition of half life.
If you've got it down on your equation sheet, you could just say, well I know this is the equation which is the natural log of the starting concentration-- c naught-- divided by the current concentration that you're looking at-- c-- equals kt, where t is time and k is your constant for k. Or your reaction rate constant.
Or-- and this is the place that I would recommend starting because it's universally useful to use for doing kinetics problems and also for doing half life decays-- you start with your rate equation.
This is a very general rate equation, which I'm going to show you how it can be transformed to solve this very simple problem.
We're told here that we're looking at the half life of a particular reagent, it's not listed.
And we're also told that it falls by factor of 11 in 11 minutes.
Because we're looking at one material, we don't need two different items.
So we're just going to look at a.
Let's forget about b.
There's no b. OK, we're just looking at how a changes over time.
We're also told that it's first order.
So we're given this information in the problem.
First order means that you have an n equal to 1.
So n equals 1, which turns this equation into k times the concentration of species a.
We also know that rate is nothing more than the changing concentration with respect to the change in time.
So in calculus that would be da / dt.
So looking at this now, we've already got a differential equation, which we can solve and given certain boundary conditions we can apply as a naught as our starting concentration, we can create a new equation.
So the way we're going to do this is we're going to do the calculus-- which I'm not going to do for you here-- it's very simply you're going to find this equation.
This is all coming from the very general rate equation that we started with.
So there we are.
That's where we're at and you'll notice if I bring this a to the other side, I take the natural log, I put a negative sign in there, I return back to my original equation that I sort of luckily-- or didn't luckily-- have on my equation sheet.
So now we've got this equation.
And now we can actually go ahead and solve the first part of the problem very easily.
So I'm going to move over here.
We're going to start off.
We're going to say we're told that in 11 minutes we have now 1/11 of the concentration we originally had.
So let me write that mathematically for you.
I'm going to write a naught as my starting concentration of a.
So in 11 minutes we're going to have 1/11 a naught a naught e to the negative k. And that occurs in 11 minutes.
Just express the statement mathematically.
We've can cancel out these a naughts so they don't matter.
Which is good because we don't know what it is in the first place.
And now we can rearrange this equation pretty simply.
We can just do it as, solve for k, our reaction rate constant.
That's going to be negative 1/11, natural log of 1/11.
That's k. And now we have a value for k. Now that we know the value for k we can go back and find the half life.
And half life is simply defined very, very similar to how we approached the problem to begin with.
Half life is how long it takes for the concentration to drop from its original value to half of that original value.
So let me write that now in mathematical terms.
In this problem we know k now but we don't know the half life.
We don't know t 1/2.
I'm going to call it t 1/2 to designate it and to mark it differently than before.
Again we can cancel these off.
And now we can solve.
Because we know k. So let's plug in everything.
We can reverse everything.
So let me do that for you.
We're going to have this.
Natural log of 1/2 being equal to negative k, t to the 1/2.
And now we can solve for our t 1/2 because we know that k-- let's bring k to the other side, and the negative sign as well-- k is going to be negative 1/11 natural log 1/11.
We can then change natural log 1/2 into natural log 2.
That actually looks like our original half life equation if you had that on your equation sheet.
So when you solve this out, you're solving for t 1/2, t 1/2 the long and the short of it is 3.18 minutes.
The main takeaway from this problem is that you don't need to have memorized any number of equations to do it.
You can start from a very general equation, which is the rate equation, which we showed in the beginning, and go from there using logic and then just mathematically represent what the problem is telling you to do in the first place.
So that's part A and that's the majority of the points on this problem.
Moving over, part B and C are very quick, very short.
Just to check if you know the facts.
Let's do them right now.
We're asking you-- we're going to do B 1 and B 2 together-- to tell us which one of the compounds has a higher boiling point.
And we give you two different cases.
In each case there's two compounds.
So number 1, you're given silene and phosphene.
So the first thing you should do is actually draw those molecules out.
So that's what I did.
Silene looks like this.
And then phosphene-- put the bonds in there-- phosphene looks something like this.
So now we have to sort of identify which one of these has a higher melting point.
Now in order to do that you need to have reviewed and know that melting point is related to the intermolecular forces between the between the molecules.
The forces between the molecules are dependent on both their geometry and their chemistry.
So knowing that, if something has stronger intermolecular bonds, it'll have a higher melting point.
So let's take a look at these two molecules.
Which one has stronger intermolecular bonds?
We have in silene case, we have van der Waals forces, they're generally very weak.
We've got those tpo for the phosphene.
But in phosphene we additionally have a dipole.
We have partial-- we put maybe delta minus here and delta plus here and here on this side.
So our molecule is actually polarized.
There's a plus region and a minus region.
And because of that you have much stronger intermolecular forces between the phosphenes then you do the silenes.
And because these have stronger intermolecular forces, this one has a higher melting point.
The last part of the problem, part 2.
We try to throw a little trick in there with you.
We look at ammonia and phosphene.
So now we're doing NH3 and PH3.
And they actually both look identical in terms of their geometry.
So let me do that now.
It actually looks tetragonal if you were to draw the structure out.
There's ammonia and here's our phosphene again.
You can tell they look the same.
So I think you have to sort of reason out which one has the higher melting point, which is the same thing as saying which one has stronger intermolecular forces.
A very common answer we got was that the nitrogen is a smaller atom, therefore everything is held more closely, therefore the forces are stronger.
That's not really the answer that is most dominant and it's also not the answer we were looking for.
The correct answer here is that even though both of these have the same-- probably very similar-- van der Waals forces, they're both polarized molecules, nitrogen has hydrogen bonding.
So remember the hydrogen bonding elements you should probably know are fluorine, oxygen, chlorine, nitrogen, four of them.
So nitrogen will hydrogen bond, which means that this nitrogen will have a hydrogen bond with some other ammonia molecule over here.
This H might be bonded to an N somewhere else.
And because there are additional hydrogen bonding force in the system, the intermolecular forces are stronger and more prevalent in ammonia, which means that ammonia will have a higher boiling point, which is true.
The answer here is ammonia.
So all you have to take way from this is that the mechanical properties as well as the melting and boiling properties of a material are very much governed by the intermolecular forces that exist in that particular material, that the liquid or the solid state.
And so in this case we're looking at these compounds and we've identified this one as being a higher melting point because it has a polarity whereas this one doesn't.
And in this case where we try to trick you up, we identified that nitrogen has this additional hydrogen bonding force.
And that's all there is to it.
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Hi.
I'm Brian.
We're going to be going over problem number 13 of the fall 2009 final exam.
I like to think of the theme of this problem as polymers.
So that's the overarching concept.
But I'm going to give you some things I think are important to review and to know before actually meaningfully attempting this problem.
So there's three things that I that I sort of thought were very important.
The first one is polymerization processes.
So we learned in class and in the book as well that there's two ways that we've covered that you can create a polymer.
And that is through either addition polymerization or condensation polymerization.
And I'll talk about that more in a second.
The second thing you want to know is basically molecular weight and how it applies to polymers.
And you want to know about elastomers for this problem.
If you don't know about elastomers then you'll have trouble with the last part of the problem.
So that's kind of what I would review.
And then once you've got that under your belt give the problem a try.
So we're going to go and we're going to start the problem now.
We're given in the problem the 6-aminohexanoic acid, which I've drawn here.
And we're also given at the beginning of the problem the structure of nylon 6.
This is the actual reaction that's used to create nylon.
So what I'm going to do is I'm going to show you-- or ask the question of how do we create nylon from 6-aminohexanoic acid.
And that's actually going to answer part A and part B to this problem.
So to create a polymer-- in this case we're talking about the polymer nylon 6-- you actually have to add together many mers.
So merge is the term we used to denote the single unit which is repeated n times to create the poly mer-- polymer.
So in this case the mer, the individual unit of nylon 6, is 6-aminohexanoic acid.
And that's this.
That's the same thing right there.
So we're going to add these together and we're going to start creating our long chain of nylon 6.
Remember, nylon 6 a polymer is a really, really long chain.
Think of it as spaghetti, very long spaghetti.
It isn't just two molecules that react.
So let's just start somewhere.
It's got to begin somewhere.
You have to have two molecules at some point in the beginning reacting with each other.
So let's look at our two molecules of 6-aminohexanoic acid-- I have to keep looking back at how to say that-- so let's take a look and we have to think about now our two possible processes or routes towards polymerization.
There's addition and there's condensation.
Now addition, as you probably read and heard, requires there to be a double bond present between two carbon atoms.
The reason for this is because usually have a free radical coming in, which will then initiate.
That's why oftentimes the free radical molecule or free radical provider is called the initiator.
You need an initiator to come in to break that double bond and then create another free radical.
Which then lets the process continue along in a chain like that.
If we look at our molecule here, we don't actually have any double bonds between two carbon atoms.
We have a double bond with an oxygen but that's not sufficient for an addition-type reaction.
For a condensation reaction, what that basically implies is two mer units, or two things coming together to react and they give off a byproduct.
So that's how I remember condensation, very often the byproduct is water.
You can have other byproducts, like HCl or any other sort of small molecules can be given off.
But most times what we'll see in 3091 will be water given off in condensation, so it's sort of intuitive to call it condensation.
In this problem, if we look at this, we have a molecule that has the Hs on one side and has Os on the other side, we can actually almost see this immediately as being a condensation reaction.
What's going to happen is we're going to have this O reacting with the positive end of this molecule, which has Hs.
And we're going to have an H2O liberated and given off.
We're going to have the new molecule formed.
Let me draw the molecule for you.
Let me write it out.
So we've just begun forming our polymer and this is a polymer which has an n of 2.
It has 2 mer units building it.
Notice I've kept the ends here.
The positive and the negative here.
In reality what's going to happen is we have a big vat of this 6-aminohexanoic acid and they're going to keep reacting.
So we might have another one of these molecules float over and react with this O.
Or conversely, we could have another one of these molecules come over and react with the H. So this thing is going to keep building.
And the way these polymerization processes begin is the control of some parameter of the system.
It could be temperature, pressure you can add something in to sort of initiate it.
But that's how this is going to start.
So we've actually sort of answered part a and b.
We've come to the conclusion that this is obviously a condensation reaction.
Because we're going to give off-- I left that out there, see if you guys caught it-- H2O.
So we're losing this O and two of these Hs and we're forming water.
And that's the byproduct.
And a byproduct immediately should tell you we've got condensation going on.
So condensation polymerization, part A.
Part B, we've actually already done to sort of logic out the answer to part A.
This is the answer to part B.
Part B is just the reaction.
I'll read it very specifically.
Write the reaction that converts two molecules of 6-aminohexanoic acid into a dimer of nylon 6.
So we have two molecules forming the dimer.
You are going to have trimer, and it goes on and on.
The most common mistake on this problem was to give us the full repeating units.
Let me just show you what that would look like.
And then we'll move on with the problem.
So the most common mistake on this problem was to not have these end units.
And instead say, oh it's created this huge, long chain.
So I'm going to write it as follows.
And this is the symbol to imply that it repeats.
Likewise over here.
So people would give us this.
And that's not correct.
And the reason that's a problem is because not only do you lose points on this part, you also lose points on the next part.
So part C is now asking us to calculate the molecular weight of a molecule given a certain n. n is the degree of polymerization.
So in this case, n equals 2, we've got a dimer.
In part C, we have n equals-- surprisingly-- 3,091 You'll see that's a recurring theme in this class.
So we have n equals 3,091 And the question is, how do we approach the molecular weight problem?
Well we're told that we have 3,091 one of these mer units making up this polymer chain.
So let me just write this down.
We're going to form an equation.
If we take the mass of the full chain and divide it by the mass of a single mer unit we should be able to extract the number of units in the chain.
So let me write that out for you.
So the total molecular weight of the molecule, divided by the weight of a particular mer unit.
OK so we're going to put that as mer.
And so this is easy.
Let's first identify what our mer unit is.
And this is why it was easy get tripped up on this problem.
Because if you made a mistake on the part B, you have trouble on part C. What is our mer unit?
So in our mer unit, what we're talking about is a chain which has 3,091 of these linked pieces here.
So let's look at this.
What do we have?
Here's our mer unit.
Let me circle it.
We have 6 carbons.
We have 11 hydrogens, 1 nitrogen and 1 oxygen in this mer unit.
We're not going to take the full, complete end unit.
We're not counting this O.
We're not counting those 2 Hs.
We're dropping off 18 grams per reaction into water.
So this is our mere unit.
Let's calculate the mass of our mer unit.
So it's going to be-- let me write this here-- we're looking for molecular weight.
We're going to divide that by the mass of the mer unit.
Let me just write out the full line of what it would be.
Approximately 6 times 12 plus 11 times 1.
So this is 6 carbons, times the mass of carbon.
11 hydrogens times the mass of hydrogen.
We're going to have 1 oxygen times the mass of oxygen.
And we're going to have the nitrogen as well.
So let me add that down here.
1 nitrogen times the mass of nitrogen.
I rounded.
That's OK. As long as you get the idea right.
So now we can easily just solve for the molecular weight.
And we'll find that molecular weight in this problem will be equal to 3.5 times 10 to the fifth grams per mole.
There's another way to do this problem, actually.
And what you can do is you can calculate n times the number of this-- this particular molecule-- and you can subtract off 3,090 orders.
18 grams.
3,090 times eighteen.
You subtract that off.
Think about a little bit why it's 3,090 and not 3,091.
So that's just maybe a brain teaser for you.
That is the answer to part C. And now part D asks us if we're able to convert this into an elastomer.
So this is where it's really required that you understand what an elastomer is and what that implies and what the structure looks like.
So an elastomer-- and I sort of paraphrased this from the book and added my own thing to it-- is basically a randomly oriented amorphous polymer with some cross-linking such that you have the ability to move the chains but not slide them completely over each other.
So in order to have an elastomer you must have cross-linking.
In order to have cross-linking, what must you have?
Well from lecture and from the book we know that to cross-link you must have double bonds in your carbons.
So you must have 2 carbons-- going back over here-- which are double-bonded together.
And why is that?
Because what happens is those double bonds will be broken by some initiator element-- many times we'll have sulfur-- will come in, break the double bond and then you'll have links forming between things.
So let me actually give you real-life example of this so you can see what I'm talking about.
We're going to show disulfide bridges bounding and creating cross-links in a molecule.
Something we're very familiar with is rubber.
And we're also very familiar with car tires.
The big difference between those two things is that rubber has vulcanized Which means we're actually putting sulphur into it to create a different type of polymer.
We're looking at a cross-linked, much stronger rubber which we use on our car tires.
So polyisoprene looks something like this.
You should be familiar with this notation by now.
We have this denotes it repeats.
We have our double bonds, we have our carbons and hydrogens.
This is polyisoprene.
And this is what we're going to add sulphur to vulcanize rubber.
We're going to create vulcanized rubber.
sulphur generally actually looks, it's a ring of 8 sulphur atoms.
We'll then add it to rubber, maybe heat it up mixed.
And you'll make what you're used to on car tires.
So what's actually happening here is that you can see we have this double bond between carbons.
The sulphur's going to come in, it's going to react with this double bond, and is actually going to pull it off.
And it might form a chain like that.
And then this sulfur will then connect to another one of these polyisoprene molecules that was originally double bonded.
And I'm not going to draw the rest of it out.
And it's going to react, kill this double bond, and it's going to connect it.
So basically what's happened in this process of forming sulfide bridges is you've got this polymer, polyisoprene, connecting to this polymer, polyisoprene, and you have cross-linking going on.
And that's what we would need to have in our case for the nylon 6 in order to get an elastomer, And because nylon 6, because our mer unit, 6-aminohexanoic acid doesn't have double bonds, we can't break them.
We can't network.
We can't cross-link.
And therefore we can't make an elastomer.
So the answer to this problem-- in a very long-winded manner-- is no.
You cannot create an elastomer given that mer unit.
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We, we are solid state devices.
This is soft matter.
My eyes are photo detectors.
Band gap of about two electron volts.
They're just not made of a compound semiconductor inorganic.
And this motion is just conformational changes in a polymer.
See it's all chemistry.
The rest is stamp collecting.
So the principles that apply to inanimate objects apply to animate objects.
And the solid-state chemistry umbrella is very broad.
If I say solid-state chemistry, people are apt to think of electronic devices, transistors, what have you.
But that extends to the life sciences.
I mean we, as human beings, are chemical machines.
So this is the classical view.
Electronic structuring informs bonding but then in 3.091 bonding informs atomic arrangement, which tells you everything.
The shape of all things, it's all electronic structure.
There's nothing more.
The difference is that the focus in the chemistry department, their focus is the molecule.
Whereas in 3.091 I like to think of the focus being aggregates of molecules, namely solids.
Because engineering systems are made of solids.
If you're steeped in chemistry from high school, you've had the AP curriculum, again 3.091 might be something that is a fresh look at chemistry from a slightly different perspective.
What I wanted to play for you as you're leaving today is an old Beatles song, Lucy in the Sky with Diamonds.
Being able to bring in subject matter, music, art, literature, into a chemistry class and make the connections to the world around us, I think is most important in the 21st century, when problem solving even in the technical domain is going to require skills that build bridges to the humanities, arts and social sciences.
And so I like to think of 3.091 not as a chemistry class, but rather a chemistry-centered class.
The majority of students taking 3.091 at MIT may never take another chemistry class.
So I ask myself what topics are critical for someone who might major in electrical engineering?
Or for someone who might major in physics?
Or what's the chemistry that's critical to an MIT student who might major in political science or economics?
And that informs my choice of topics.
And it turns out that that doesn't diminish from the education of the student who may choose to major in a chemically intensive discipline such as material science, chemical engineering, or even chemistry itself.
I don't view the presentation of the subject matter as a linear track, where this is the beginning and this is the end.
I view it more along the lines of a spiral.
And so the first time you encounter a certain concept it's at a rather superficial introductory level.
And then at some point later in the semester you spiral back and see it again.
And then maybe there's a little more context and a little more intensity.
Maybe there's a little more mathematics associated with it.
And then you'll see it applied later again.
And through that revisiting-- repetition by design-- I think that key concepts get driven home.
Some people are shocked that I stand in front of a chalk board and I write on it and I believe that that's still a valid means of teaching.
For the first-time learner I think it's important to activate all of the learning channels.
The act of drawing and inspiring them to draw is important.
And maybe the person at home ought to put themselves in a chair in the auditorium and be taking notes as though they are attending the lecture.
Watching those videos, especially for someone that has had a fair bit of high school chemistry, you're going to hear a lot of the same things but maybe coming at you at a slightly elevated level.
But not so elevated that you won't follow it.
And I think that's good for a high school student to be talked to at a higher level.
I think the videos also have utility for people that are interested in specific topics.
I mean for example, if somebody needs to know something about x-rays, or the use of x-rays, dropping in to that three-lecture sequence on x-rays could be quite useful.
I think it's possible to sample with a goal of mastering a particular concept.
So clearly for the learner out in the World Wide Web, going to recitation and interacting with a recitation instructor is not practical.
So in lieu of the recitation section OCW has created these help session videos.
We actually hired some recitation instructors from the fall of 2009 and asked them to work through some sample problems and what you see in the help session video is essentially what would have happened in a recitation section.
I issue the homework questions with model solutions at the beginning of the unit.
So the homework becomes a study aid.
And I recommend to students that they try the homework problems without referring to the answer key initially.
And to turn to the answer key when they reach an impediment.
What's the message of 3.091 for the man on the street?
An understanding of those relationships.
Electronic structure to bonding.
Bonding to atomic arrangement.
Atomic arrangement to properties.
Is the key to understanding the world around us.
And in particular it's the key to fueling invention.
And I believe in practice very much and in the application of science in service for humanity.
Why do we worry about designing a car that's a zero-emission vehicle, as if to do no harm to the environment?
Why don't we go even further and say, what about a car that purifies the air as it drives down the road?
So that the air in the wake of the car is cleaner than the air in front of the car?
And I'm finding that increasingly that's a message that resonates with the students.
They want to take the tools of technical analytically centered education and put them to work to solve societal problems.
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Let's get started.
Settle down.
Settle down.
Welcome to 3.091.
I'm Donald Sadoway and I'm going to be your instructor this semester.
So what I want to do today is to introduce my plans.
I've got plans for you.
Plans for learning.
I want to talk a little bit today about those plans, introduce myself, introduce my class curriculum-- the path forward.
So let me begin by saying that 3.091 is the most important class you will take at MIT.
It's true.
But, you know, anybody who stands in front of you to lecture should say the same thing about his or her class.
If they don't believe that they shouldn't be standing in front of you lecturing.
The difference is when I say it, I'm right.
[LAUGHTER]
And it's not just an opinion, it's because of the content of 3.091.
3.091 is about chemistry.
It's about the central science, but it's not just about any old chemistry.
This isn't a class that you could take anywhere else in this country.
This is the only place that you can take 3.091.
3.091 is solid state chemistry.
Now why do we talk about solid state chemistry?
Because engineering systems are made of solids.
Now I know what you're thinking-- oh, he said solid state, he's going to talk about the chemistry that constitutes his laptop computer or the chemistry that constitutes this laser pointer.
And that's true.
We will talk about that chemistry.
But we will also talk about soft matter.
We as human beings are chemical machines.
When this hand changes shape, it is a polymer that is changing conformality.
These eyes are photodetectors, band gap of about two electron volts.
They're not made of gallium nitride.
They're made of organic compounds.
Inside, what supports us, it's a ceramic skeleton.
So solid state chemistry describes life science as well.
Well, what we're going to learn in 3.091 is the rudiments.
We're going to learn the rudiments.
So what I'm going to do today is now go through some of the basic organization.
So tomorrow you're going to meet your recitation instructors and get to know each other and get to know the recitation instructor.
Today I'll say a few words about myself so you know who's standing in front of you.
I graduated from the University of Toronto in chemical metallurgy, came down here in 1977 to spend a year as a postdoctoral fellow and, I guess, I lost track of time.
Now what's my research?
My research is in electrochemistry.
Electrochemistry is the most important branch of chemistry.
Do you notice some theme in my professional life?
See, I have tenure.
So what does that mean?
It means you find your passion and pursue it.
You don't waste time on trivia, all right?
And that's what I urge you to do: Find your passion and pursue it.
So what's my passion?
My passion is electrochemistry in nonaqueous media.
Anything but water.
Let the rest of the world work on water.
I work on molten salts, ionic liquids, and polymers.
And what's the reason for this?
There's an application.
I'm in the School of Engineering.
So I'm interested in environmentally sound technologies for metal production, all right?
Right now looking at titanium, iron recycling as well.
I'm also interested in electrochemistry as it applies to energy storage, energy storage for mobile power.
See this gal here?
She's got the cell phone.
She's not even looking where she's driving.
[LAUGHTER]
So making safe batteries out of earth-abundant, accessible materials for portable power, ultimately to drive the car with electrons, electric fuel, to eliminate this country's dependence on imported petroleum.
We can do it.
How?
By inventing.
By inventing.
And the way we're going to invent is to learn the lessons in 3.091 that will give us the chemistry we need to invent a battery that can send that car 250 miles on a single charge, and put it in a show room for the same price as a car with an internal combustion engine.
The only thing that stands between that image and where we are today is invention, and the requisite material is right here in this class.
We're also looking at colossal batteries.
Store the grid.
Enable renewables, wind, solar.
And then, lastly, let's never forget about dreaming.
So if we want to dream and imagine that man will return to the moon or maybe even go to Mars, we're going to have to be able to produce oxygen.
We'll live like the pioneers, produce oxygen from local resources, produce structural metals, and even produce photovoltaic silicon so they can generate their own energy from local resources.
And so electrochemistry is the key enabler for that as well.
So now let's turn to the whole underpinning of 3.091.
If we take a look at the performance of any engineering system it's a combination of the design and the construction.
Now the construction is both the workmanship and the choice of materials.
Now how do we choose materials?
We choose them on the basis of their properties.
So, for example, here's an application: a beverage container.
This one I'm holding is made out of an aluminum alloy.
It's a metal.
When I was your age, they had steel beverage cans.
You can take this same beverage and contain it in a glass bottle.
You can contain it in a polymer bottle.
Why do we make those various choices?
Because they have the right mix of properties.
Now how do you determine the properties?
Properties obviously are determined by the composition.
We wouldn't make this thing out of sodium chloride.
It would dissolve.
So composition is important, but atomic arrangement is important as well.
Example is 1/2-inch-thick pine board compared to a 1/2-inch-thick piece of plywood.
Both pine, but the 1/2-inch-thick pine board is one solid pine board.
I can take that pine board and I can fracture it if I strike it along the grain.
But the 1/2-inch-thick plywood, I can't do that because it's 1/8-inch sheet pine cross-laminated.
1/8 inch north-south, 1/8 inch east-west.
If I try to cut through that board, I can't advance the fracture.
Same composition, different atomic arrangement.
So the thesis of 3.091 is that the electronic structure of the elements holds the key to understanding that relationship between composition, atomic arrangement, and properties.
And once you have properties that's how you make your selection.
And away we go.
And that's how we got the syllabus of 3.091.
So the syllabus has two major blocks.
The first block is general principles of chemistry-- it's going to be about the first five weeks-- and that's the same stuff that you get here, 5.111, 5.112, the basics of electronic structure and bonding.
And then we part company and in 3.091 we dig down into solid state chemistry.
You don't have to take notes.
You can relax.
All this is going to be burned to PDF and posted at the website.
So everything that you see in this class that goes up on the screen is archived for you.
This lecture is being video recorded right now and within an hour will be posted at the website.
So in the unlikely event that you can't quite come to class, it is available.
[LAUGHTER]
So now what I want to do is introduce some of the operational aspects of the class.
And we had some handouts going around.
If you didn't get them we can get you extras.
But again, this is all posted at the website.
So you know who I am.
The key person in this operation is my administrative assistant, Hilary Sheldon, and she's just down the hall.
You can roll a penny down the hall from here and get to my office.
So if you tell me you tried looking for me, you couldn't find me, I know you couldn't have been trying very hard.
The text is a two-volume set consisting of the text Chemistry: Principles, Patterns, and Applications by Averill and Eldredge, and then a second volume that consists of miscellaneous readings taken from other Prentice Hall texts.
And it's identical in content to the blue book that was used last year and the year before.
So if you get a copy of that blue book, you don't have to buy these.
I don't work for Prentice Hall, I'm not trying to sell books, but we will take the homeworks out of the books to a great extent.
And the page numbering is the same, so anything this year or the last two years will be fine.
If you have to buy something, you'll have this.
This one will come with a CD that will get you into a mastering chemistry, which is a sort of a tutorial, computerized system to help you with certain homework problems.
The lectures, Monday, Wednesday, Friday, here in this room at 11 o'clock.
The lecture starts at five minutes after the hour-- gives you five minutes to get in-- and then the lecture stops at five minutes to the next hour-- time for you to leave and then for the next class to arrive.
And what I'm going to try to do is establish some plug flow here.
So what I'm going to try is to have everybody leave by the exit to the north here to my right, and there's also an exit over the top back.
So leave to your left and that way it'll be easier for the next class to come in.
And maybe we can persuade the people here before us to do the same and then it'll be easy to change.
I don't know why but there's this sort of-- everybody has to charge to this door.
And I just stand here and watch people collide.
And I don't know.
Just interesting social experiment.
And there are two doors.
I don't understand why people do this, but I'm not in the social sciences.
Recitations.
Recitations will meet Tuesdays and Thursdays.
So you'll go to the same hour on Tuesday and Thursday.
The section should be roughly 20 students.
And that's where the question and answer occurs.
Here it's largely I talk, you listen.
I've got time for a question like, hey, shouldn't that be a minus sign?
I can take something really quick.
But where you get to really interact with the instructor is in recitation.
And I've given a direction to my recitation instructors not to give you a fourth and fifth lecture.
You control the content of the recitation.
So you have to come to the recitation prepared with questions.
So they're supposed to walk in and say, good morning, good afternoon, what are your questions?
And you can say, I didn't understand the last five minutes of the lecture yesterday.
Would you do number five on the homework?
Would you go over secondary bonding?
That's the sort of thing that's supposed to happen in section.
You've been assigned by the Registrar.
You are forbidden to change your section on your own.
You can't just squat in another section.
We do this with the intention of trying to keep the enrollment roughly uniform.
We don't want sections growing to 35 because then that limits your ability to interact.
If you've changed classes or you've picked up a UROP, or what have you, and your situation has changed, please go down the hall and meet with my assistant Hilary, who will then arrange to move you.
And that way we can keep some control over the section size.
Again, you may not change simply on your own.
And it has to be with cause, academic cause.
You can't go in and say, I was assigned to the 9:00 am section and I don't do mornings.
That's not going to work.
Homework.
Homework is very important here.
Homework is a little bit different, though, in 3.091.
You're not going to be asked to turn in homework for grading.
Instead, at the beginning of each unit, you will be given homework with the model solutions right at the beginning.
This is a study aid, and you can use the model solutions to help you understand the homework material.
Now since we're not asking you to turn anything in and we do want to stimulate interest in the homework, we found a way to stimulate interest.
And that is that once a week in section you will have a 10-minute quiz based on the homework.
That will ensure that you've at least looked at the homework.
And those weekly quizzes will be graded, and the aggregate of all of those quiz scores will constitute your homework grade in the subject.
And you must take those quizzes.
If you don't take the quiz-- you got sick or some personal emergency came up-- contact your recitation instructor and within a week of the date of the original quiz arrange to take a makeup.
If you don't take the makeup, we're going to give you a zero.
And there's no dropping of lowest score.
Somewhere there's this in the lore here that, oh, you can drop the lowest score, the lowest two scores.
People come to me with this proposal and I say no.
[LAUGHTER]
So let's get started.
This is homework number one.
It's assigned today.
We're going to save paper-- just go to the website-- and you'll be tested on Tuesday, next Tuesday, September 15, at the beginning of section.
At the beginning of section there'll be a 10-minute quiz based on that homework, and you can get the homework and so on.
By the way, here's what the homework looks like.
Chapter 1, Chapter 3 of Averill.
Averill is the major text.
We just refer to it by the author's last name.
So taken from Averill.
By the way, I don't like to use this unpleasant term "test."
I like to refer to it as a celebration of learning.
[LAUGHTER]
So it's a celebration.
We're going to start celebrating on Tuesday.
There's the website by the way.
You probably want to bookmark that.
You're probably going to go to that a lot.
We're going to keep celebrating, and so we will have some monthly celebrations, monthly tests, and they've already been scheduled.
And you will write those tests during the normal class time.
But on those days we will spread you out a bit.
I like to see some vacancies.
So you will not be sitting one next to another.
There'll be at least one human vacancy next to every human, and that way you've got room to spread out, keep your eyes on your own work.
So those are the dates and there'll be more of that in time.
Now when you take the monthly test, I allow you to use aids.
So everyone will be given in recitation tomorrow a Periodic Table of the Elements, very nice one, laminated one.
So you take that with you to the test.
In fact, you should take this with you everywhere.
[LAUGHTER]
If I run into you at Harvard Square, I want to see the Periodic Table, because every educated person has the Periodic Table.
You get a table of constants.
So you'll have one of these.
You'll get that tomorrow as well.
Paper copy, and on there are all the constants so you don't have to remember that the permittivity of vacuum is 8.85 times 10 to the minus 12 farads per meter.
It's on there.
I urge you to use that when you do your homework.
So it's always amusing to me during the time of the first monthly celebration, somebody's looking at this as though it's today's newspaper and they're looking, where is that?
Sort of revealing about the intensity with which the homework has been embraced.
Calculator, something to calculate with, and I don't care if you bring in a graphing calculator or a mainframe.
I don't care.
[LAUGHTER]
And you're allowed an aid sheet, an 8 1/2-by-11 sheet of paper.
And you can write on the front, you can write on the back.
You can photoreduce previous exams down to micro-dot size.
I don't care what you put on it, but with this, you have then no excuse to say I really understood this stuff, but I couldn't remember a formula.
Actually, many students tell me that the act of preparing the aid sheet organizes the subject matter in their minds.
They bring this with them to the exam, and they never consult it, but it just soothes the nerves and makes sure that everybody does well on the monthly test.
The weekly test, no.
The weekly test you bring your Periodic Table and table of constants, but no aid sheet.
For the weekly test, it's a very concentrated amount of material.
And I'm not going to test your memories, because that doesn't prove anything to me.
And then, of course, the celebration of celebrations is the final exam.
[LAUGHTER]
That's huge.
That's a huge celebration.
It's three hours.
See, these are really 50 minutes.
This is three full hours.
The time and location will be set by the Registrar.
We should know by October 1.
The final exam period is December 14 to 18.
And so what I urge you to do is as soon as that schedule comes out and we know when all the final exams are going to be held, then you book your passage for the holidays.
Do not get that order reversed.
You cannot come to me and say, I got a really good price on a ticket.
I'm going Acapulco on the 15th of December, and your exam is on the 16th of December.
I'll say, you just got a zero on the final.
You have to be here for that.
You know why?
I have to be here for that, you can be here for that, too.
There are about a quarter of a million students in Boston.
It's a great college town, but at Christmas time, it's pandemonium at Logan Airport.
So you want to book your passage early, but you can't do it until you know what your final exam obligations are.
Grading.
Freshman, you know, it's pass/no record.
Pass/no record.
And so that means that if you struggle and things don't go well, you don't have any blemishes on your record.
Unfortunately, some of the upperclassmen will tell you as pass/no record, you now, barest pass imaginable/ no record.
Well, I hate to let you know this, but increasingly I am being asked by medical schools, law schools, scholarship providers to reveal the scores on the freshman year.
So think about that before you call Hilary in late November saying, what do I need to get a 50?
I need a 32 on the final?
OK.
But Lord help you, you get a 31 you go down.
[LAUGHTER]
Upperclassmen get the luxury of the entire alphabet: A, B, C, D, F. The final grade composition.
1/6 for homework-- that's the aggregates of the weekly test scores-- and 1/6 for each of the three tests.
And then the final is 2/6, or 1/3.
I didn't want to get into transcendental numbers, so I made it 33 exact, 16.75 exact.
I could have made it 16.77, I could have made it 16.81, I could have made it anything I want.
I'm the professor.
But I chose 16.75.
Bottom line here is it's really dumb to fail the final.
If you passed the final, you're pretty much assured that we're going to be favorably disposed to passing you.
But you fail the final that says two things.
It says, number one, you don't have a grasp of the overall year, but it also sort of indicates that you ran out of steam partway through the semester and stopped working.
It's very bad.
You get a lot of good advice from upperclassmen, but sometimes-- I'd ignore that one that says, hey, you don't have to do that well on the final.
Anyway, so you have to get a C level as a freshmen or greater.
And, by the way, we do not grade on a curve.
I've seen it in magazines this last year that MIT grades on a curve.
I don't where they get that from.
I don't grade on a curve.
Your success does not come at the expense of your neighbor.
As far as I'm concerned everybody in this class can get an A.
Again, I'm the professor, all right?
So you say, well, how do you know that 50 is the right number?
Why isn't it 55?
Why isn't it 75?
Well, I know.
I know.
How do I know?
Because when we grade, we set up the point scheme so that if the student has the mastery of the barest level of competency of the key concept the point scheme must reflect a passing grade-- 5 out of 10, 5 out of 9-- and you propagate that through.
I don't care how much is written, if it doesn't demonstrate basic mastery of the key concept the point scheme must give 4, 3, 2, 0.
Maybe a 1.
And so that, if you propagate it through the whole semester, means 50 is a pass.
There's some wiggle room there.
How do I know 49.7 is a fail and 50 is a pass?
That's when I call your recitation instructor.
[CELL PHONE RINGING]
Let's kill that.
We're going to get to that in a second.
Let's call the recitation instructor.
And what happens?
I ask the recitation instructor, well, what can you tell me about this young lady?
And, oh, she came to all my classes, she tried really hard, she came to office hours.
I don't think the exam is a proper reflection of her understanding.
I'll listen to that and maybe we'll promote.
If, on the other hand, I get the response, I never saw her, didn't come to class.
I repeatedly reached out to her, ignored my entreaties, she's going down.
[LAUGHTER]
OK. Website.
This is what the website looks like.
There's a number of tabs here.
The readings are there, the videos are there archived.
The schedule-- what's going to be coming up.
So, for example, this is what's going on today.
It says what the topic is, roughly what the readings are.
I know today you didn't come to class having read.
And it's OK, we'll get through it.
But from now on I urge you to do the readings.
A couple of topics I'm required to talk about.
My management requires me to do so.
Academic honesty.
There's a lot of texts here, but in plain English, this says don't cheat!
You know what this means!
Now I don't want to hear, well, in my country, the custom is-- I don't care what they do in your country!
You're in my class, and if you cheat in my class you will pay for it.
Very simple.
Accept information of any kind from others-- wrong.
Represent somebody else's work as your own-- wrong.
You know this.
It was Juvenal, the Roman politician, said men need not so much be instructed as reminded.
I'm not telling you anything you don't already know, but I'm going to say it so no one can say, well, nobody told us it wasn't OK to erase our answers and hand them back in for more points.
I just did.
All right?
So if in the unlikely-- but it happens every so often.
Maybe every two, three years.
And people get caught.
You know why they get caught?
Because for the first time in your life you're being taught by people who are as smart as you are.
[LAUGHTER]
My TAs are really smart.
And it never ceases to amaze me-- somebody succumbs and does something dishonest, and they get caught!
They get busted!
And what happens then?
Then I get angry.
You know why?
Because we can't settle that in my office.
You can't come and cry in my office.
I have to take this episode to a committee here at MIT called the Committee on Discipline.
It is staffed by faculty, by administrators, and by students.
And the case is brought before the COD at a hearing and a punishment is decided.
And it can be anything from suspension to expulsion.
And of the three categories, faculty, administrators, students, which category do you think is most severe on infringers?
Students
Your peers.
You got it.
You know, because people my age, old faculty, they'll be like, oh, they're just kids, whatever.
The students say, no!
Throw them out!
[LAUGHTER]
So it's not going to be me that's going to expel you, it's going to be your peers.
So don't do it.
And if you're ever in an exam and somebody's pressuring you, just raise your hand and ask to be reseated.
For all I know, the guy next to you has got a bad case of B.O., you just want to move.
We will not ask any questions, so take yourself out of the situation.
Conduct yourself appropriately.
Classroom behavior.
Now this is the first lecture, there's a whole bunch of violations in here right now.
So I'm going to say it now and we'll fix it for next time.
We've got 425 seats here, we've probably got 475 people, and there's only one way we can make this system work and we have to observe certain rules of decorum.
And I make the rules.
So if I want to maintain a fertile learning environment I'm going to ask for these rules to be observed.
No talking at all.
No talking.
Little conversation here, little conversation here-- it disturbs people.
I don't want any food or drink.
No food, no drink.
One exception is the professor.
[LAUGHTER]
Because I do not want to have my throat get so dry that I can't finish talking my way through the end of the class.
Otherwise, no food or drink and no disruptive behavior.
No horseplay or anything like that.
And wireless communication devices must be silenced.
Cell phone goes off, you get up, you leave the room.
That's it
Is water OK?
If you need water because it's some kind of a health thing, fine.
But I do not want to see a whole bunch of people drawing on water bottles.
Don't need it.
Don't need it.
You can go 50 minutes without your little-- whatever.
[LAUGHTER]
You know why?
I'll tell you why.
It's not because I'm trying to be a control freak.
This is a chemistry class, but it's chemistry-centered.
You didn't just come to MIT to learn some geeky techno stuff.
You're preparing for a professional career, and part of that is how you behave, how to act as a professional.
And you cannot learn behavior by doing problem sets.
How do you learn behavior?
You learn behavior by observation.
And how does one teach behavior?
By modeling.
If you're in some high-level committee meeting and all of a sudden your cell phone goes off and you're scrambling and you can't find the damn thing, it's in the bottom of your briefcase, right?
I can't tell you how vulgar that is.
You're in a professional setting, you commit professional suicide.
And I need to tell you that.
If you think it's OK you're wrong.
So let's get in the habit.
Disable the damn thing.
What do you need?
You're going to call your stockbroker?
What's so important right now?
And on an exam, that thing goes off I take the exam in front of everybody and-- [MAKING RIPPING NOISE]
--we introduce a defect, it's called a tear.
[LAUGHTER]
And then I put a zero on it like this, with a circle around it.
It's called a donut.
[LAUGHTER]
All right.
Now let's talk about some lighter things, more upbeat things.
Look, I want you to succeed in this class.
Now how are we going to succeed?
You go to the listing in the bulletin you'll see this, right?
And you zoom in here it says 5-0-7.
What's this mean?
Well the 5 is 5 contact hours.
3 here and 2 with your recitation.
The 0 is lab.
There's no lab with this class.
So what's the 7?
The 7 is the reading, the homework, preparation, et cetera.
I pledge to you, you give me 7 hours-- you go to the 5 hours contact and 7 hours-- you will not just pass this class, you will flourish in this class.
How do I know this?
Because I used to chair the committee on admissions and I've read applications.
I know the quality of individual in this room.
You should have seen, there were 11,000 applications piled up like this, each dossier like this, and they just get copied down on a four-by-six card.
We get here on a Presidents' Day weekend and we got stacks and stacks of the cards.
We're looking at your whole academic life is on a four-by-six card.
And I pick that up and I go, yes.
Literally at this speed.
About 45 seconds.
No.
[LAUGHTER]
Hell no!
[LAUGHTER]
That's how I spent Presidents' Day weekend.
And so you got through a very grueling selection process.
What's the corollary of that?
Listen carefully to this: Everybody in this room has the intellectual apparatus to pass 3.091.
The only people who fail 3.091 are people who choose to fail 3.091.
They choose not to come to class, they chose not to go to recitation, they chose not to work the homework.
I don't know why they make those choices.
But I guarantee you if you give me this amount of time, you'll do well.
It's straightforward.
Number one problem you are going to face this fall is not secondary bonding, it's time management.
So I used to call this strategies for-- what did I used to call it?
I forgot.
I used to have it something like, survival strategies.
But I don't want you to survive, I want you to flourish.
So I call it recipe for success.
There are different venues for learning.
OK?
So lecture, that's here.
That's my responsibility.
Recitation, that's Tuesdays and Thursdays.
That's my staff.
Now reading, that's you.
Homework, that'd be you.
Weekly quizzes is you, monthly tests is you, final exam is you.
[LAUGHTER]
What we have here is a partnership.
See?
[LAUGHTER]
So I'll do my part, you do your part.
You know, one of the things beyond the basic learning of the chemistry that we're going to attempt here is a transition.
You are going to change in ways you can't imagine.
I want you to think about the way you are right now, and I'm going to ask you to think about how you feel about yourself on the last day of class.
You will be amazed, and it's not because you know a few more chemical equations.
And what one of the things that I want to see happen and to help facilitate is the transition from student, which you are as a high school graduate, to scholar.
And the difference between a student and a scholar is a scholar takes ownership of his or her learning.
So you're going to take ownership of it.
You know, is this going to be on?
Are we responsible for this?
Go back to high school.
Here you're a scholar.
You say, how can I learn more about this?
That's the difference.
So I think we've covered enough.
I think we don't want to just have the whole day, welcome to MIT, welcome to MIT.
So what we're going to do is we'll get into the lecture, and in the very brief amount of time we have left I'm going to talk about the beginnings of chemistry.
We're going to talk about taxonomy, classification, nomenclature.
And to help introduce this I'm going to refer to the writings of William Shakespeare.
We're going to integrate some humanities here.
We're going to read from Romeo and Juliet.
Maybe there are people here who were admitted on the strength of the performance in Romeo and Juliet.
Maybe one of you was Romeo, one of you was Juliet.
Maybe one of you did the lighting.
Maybe one of you did the set design.
Maybe one of you took the tickets.
Somehow you were involved, right?
Because we're all involved.
So let's take a look.
Act 2, Scene 2.
Romeo: "But soft, what light through yonder window breaks?
It is the east"-- that's the east.
East Campus is that way, right?
[LAUGHTER]
"It is the east, and Juliet is the sun."
That has nothing to do with nomenclature.
Maybe photon emission, but-- that's a joke.
We'll get to it.
Now Juliet: "O Romeo, Romeo!
Wherefore art thou Romeo?
Deny thy father and refuse thy name.
Or, if thou wilt not, be but sworn my love, and I'll no longer be a Capulet."
See she's a Capulet, he's a Montague.
You know, the Hatfields and the McCoys.
And this is a metaphor.
It's as old as literature.
It's about warring factions, two communities that can't stand each other based on prejudice.
And then these two youngsters fall in love, and how love triumphs over hatred, and so on.
It's powerful stuff and it's written here.
Romeo: "Shall I hear more, or shall I speak at this?"
Remember, she's up high, he's doing "Shall I hear more, shall I speak at this?"
Fellows, the answer to that question is don't speak.
Don't interrupt her.
All right.
So now she goes, "'Tis but thy name that is my enemy.
Thou art thyself, though not a Montague.
What's Montague?
It is nor hand, nor foot, nor arm, nor face, nor any other part belonging to a man.
O, be some other name!
What's in a name?
That which we call a rose by any other word would smell as sweet."
That's properties, right?
"So Romeo would, were he not Romeo called, retain that dear perfection which he owes without that title.
Romeo, doff thy name, and for that name which is no part of thee take all myself."
Beautiful writing.
400 years ago and it's just fantastic.
Well, that's no good.
That's not the way it works in science.
In science we have to agree-- [LAUGHTER]
We have to agree on the name.
And so this is taken from Chapter 1 of the text, and this is the classification of matter.
This is about stuff and the different forms of stuff.
Over here we have the simplest form of stuff, which is the element.
And we're going to start here in 3.091 and we're going to work our way all the way through this table, starting with electronic structure and how electronic structure governs stuff.
So let's start with a history lesson.
We'll start with a history lesson.
And the history lesson goes like this.
What are the origins of chemistry?
The ancient Egyptian hieroglyphs refer to khemeia, which was a chemical process for embalming the dead.
You know the Egyptians were very fixated on the afterlife.
And the chemists, the chemists were revered in that society-- not like here.
They were revered in that society because they knew how to prepare the body for the afterlife.
Embalming is a chemical process.
A few years ago I was in London.
I toured the British Museum and came upon this.
This is 18 inches tall.
It's the mummified cat.
It's a fantastic example of mummification.
It's a beautiful-- most people go zooming right past it.
They're looking at the big dinosaurs and all that other nonsense.
But this, this is beautiful.
Mummified cat.
And then khemeia expanded to other chemical processes: dying of cloth, glassmaking, and metals extraction.
The chemist could take dirt and turn it into metal.
Sorcery.
Some things haven't changed.
I'm in that tradition: producing metal from dirt.
And they were always looking for an overarching theme, to unite the heavens with the simple elements.
So the seven known, naked-to-the-eye, astronomical bodies were associated with the seven known chemical elements.
And you could even talk in this priestly language.
So if you wanted to make a bronze, which is an alloy of tin and copper, you could say, well let's have the confluence of Jupiter and Venus.
It's all there.
Mercury.
Why do we call it Mercury?
Mercury's a metal that's liquid at room temperature.
It's fast.
Because Mercury is the planet that moves quickest around the sun.
Very nice.
So this is what we knew 2,400 years ago.
We had the seven metals, carbon, and silicon.
And the beginning of the shift from practice to theory, or from craft to science, is with the work of Democritus.
Democritus, who lived around 400 BCE.
Democritus, who was Greek.
He described the physical world as consisting of a combination of void plus being.
Void plus being.
These are lofty words.
Listen to this.
Void, being not some little equation.
It's big, big ideas.
And how did he describe void?
Void is something that we would recognize, in his language, as akin to what we recognize to be vacuum.
And being, he said, was comprised of an infinity of atoms.
He coined the term "atom."
"Atom" comes from the Greek tomoi, which is to slice.
And then if you put "a" in front of it as in apolitical or amoral, cannot be sliced: indivisible.
The atom cannot be sliced, and so to these atoms he attributed these properties.
They're indivisible and eternal.
I mean, this takes you all the way to E equals m c squared.
From 400 BCE, and that's all he had to work with.
This is brilliant.
Absolutely brilliant.
But to show you that things don't always go in a linear fashion for the better, along comes Aristotle.
Aristotle, another Greek, and he decided, nah, we don't need all this carbon, sulfur, and so on.
He said, we will have four essences that will describe the earthly world.
Four essences.
And here are the four essences.
Earth, water, air, fire.
So those are the Aristotelian essences.
And there's actually a fifth essence that describes the heavens.
That's why we say something is quintessential.
It's heavenly, it's et cetera.
So these are the four essences.
All right?
And then it gets even worse because he has compounds, a combination, so you can take fire plus earth and make dry, earth plus water make cold, air plus water make wet, air plus fire make hot.
This is nuts.
[LAUGHTER]
And it dominated science for a long, long time.
Because we know really earth is an aggregate.
It's an aggregate of different minerals which are compounds.
Water, as you know, is a compound H2O.
Fire is the product of combustion.
It doesn't even belong in this set.
And air is a solution.
It's a homogeneous mixture of nitrogen, oxygen, argon, rising levels of carbon dioxide, sulfur dioxide.
And if you're next to an aluminum smelter, tetrafluoromethane.
And so on and so forth.
And finally this thing was knocked down.
So we can look at this chart and go back to this one and say, now we can make the connections.
All right.
Here's the way the elements looked at the time of the American Revolution.
The alchemists gave us arsenic, antimony, and bismuth in the 12th, 13th, 14th centuries.
In 13th century India, there was zinc isolated.
There's a tall zinc pillar still standing there.
Platinum is an American metal.
It was unknown to the Europeans.
It was discovered when the Spanish came to the Americas.
Actually, it's kind of ironic because plata is silver, so platina is sort of like a diminutive of silver, which in point of fact is backwards because silver melts at 962 degrees Celsius, platinum melts at 1768.
It is nobler than silver.
It has catalytic properties.
It is even fashionable in jewelry now if you know how to work with it, because gold melts at 1063 and platinum melts at 1762.
A lot of jewelers can't work with platinum.
It's too high-melting.
It's a fantastic metal.
It's an American metal.
[LAUGHTER]
It really is.
I mean it's an American metal.
I didn't make this up.
It's true.
And then we see these other elements.
Discovered, discovered, discovered, discovered.
Now we see modern science.
So what does it mean, discovered 1766?
There was no hydrogen before 1766?
They found it?
No.
It means that Cavendish isolated it and documented its properties.
Hence, it was discovered.
Now I've been cheating here.
This is actually lined up in a way that we already know, where the story is going to end with this, with the Periodic Table.
I'm just lying them on the table in a way that makes it possible to anticipate.
But this was the first table of elements that is of record, and this was by John Dalton, the English chemist, who put these in order of atomic mass.
They're organized in ascending order of atomic mass.
And he also started a system of chemical symbols.
And you can see that iron has an "I" with a circle around it, and zinc has a "Z" with a circle around it.
And the Swedish chemist Berzelius said, you know I don't think the French are going to like to use "I" with a circle around it for iron, or Germans aren't going to like that.
We better choose something a little bit neutral.
So they chose Latin.
And that's why iron is "Fe" for ferrum, and gold is "Au" for arum, and so on.
But this was the first attempt.
John Dalton actually was a polymath.
He was also working on vision, human vision, and he suffered from an affliction that is present in about 10% of men, which is red-green color blindness.
And he did the original work on red-green color blindness.
In fact, in some circles it is known as Daltonism.
It's the same man, John Dalton.
Other classifications.
Dobereiner in Jena talked about triads.
And if you took the atomic mass of chlorine, added it to the atomic massive of iodine divided by two, you get something that's not too far off the atomic mass of bromine.
Newlands was a musician and he talked about octaves.
So if you start here, if this is a diatonic scale, so this is C, D, E, F, G, A, B, C. So potassium lies an octave away from sodium.
He was ridiculed.
They said, have you considered perhaps putting the elements in alphabetical order?
They were cruel.
Scientists can be very cruel to new ideas.
And, in fact, in your book this is some of Newlands work, and you can see to what extent it helps understand things.
But the first proper organization came in 1869 with Mendeleyev, and also 1870 with Lothar Meyer in Tuebingen, the Periodic table as we know it.
And this is the set of elements that were known at the time.
And this is a page from the paper in which Mendeleyev published the Periodic Table.
And here's the smoking gun right here.
It's right here where he says, in this new system, in this proposed system, there are very many missing elements.
Very many missing elements-- that was the key.
Because he knew that even though arsenic has the atomic mass next highest to that of zinc not to put arsenic under aluminum, not to put it under silicon.
It had more in common with phosphorus.
So he put it under phosphorus, and so did Meyer.
But what Mendeleyev did, which was a first, he said, there are missing elements here.
There is an element that lies between silicon and tin, and I predict what it's mass will be.
I predict what it's chemical formulation will be, how it will react with oxygen.
The predictions.
And how did he get this?
Because he used to travel by train and he would sit in the train station on his trunk playing solitaire.
And he'd be going down one suit, and then there'd be a gap.
And he'd go down the other suit and he could go farther.
And so the concept of, I know there should be an eight of spades here, but I have to stop at the nine.
I've got a seven but there's something in there-- triggered his imagination and led him to have the courage to say, there are elements there and I will predict their properties.
And when they were discovered how close he was is shocking.
Absolutely shocking.
So we'll get to that next day.
But before you go I want to leave you with this.
This is the portrait of Mendeleyev that you typically see.
This man on in years, disheveled, sort of mad scientist look.
Don't remember that.
Look at this.
This is Mendeleyev, age 35, when he proposed the Periodic Table of the Elements.
OK. We'll adjourn.
We'll see you on Friday.
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Tuesday will be the first weekly quiz, celebration that is.
That will be at the beginning of recitation.
You'll have 10 minutes.
It'll be a short one-pager.
You just write on the page.
All you bring is your periodic table, which you should have gotten in recitation yesterday.
Periodic Table, table of constants, calculator, something to write with, but no aid sheet on the weeklies.
Readings: Readings, I urge you to read before you come to class.
And so if you go to the website, you can go to Schedule, and in the schedule, you'll see stuff like this that tells you what the readings are for the day.
As I mentioned last day, the lectures are being videographed and posted probably within an hour on the website and any of the images that I show are also recorded, burned as PDFs and uploaded.
So you don't have to be put in high-speed stenographic mode in order to attend class.
What's the other thing I wanted to tell you?
If you're new to the class or if you need to change your recitation section because your conditions have changed, do not simply go to the other class.
We're trying to regulate enrollment, particularly on Tuesdays.
If the TA shows up expecting 20 students and has 20 copies of the quiz and 25 people show up, that's not a recipe for success.
So you must go to my administrative assistant, Hilary Sheldon in order to change recitation.
And if you need any of the handouts and so on, it's just down the hall here in Building 8, Room 201.
I think that's all that I had to say.
If you go here the videos are all listed.
So last day we started talking about taxonomy and that led us to the beginnings of atomic theory.
We visited with Democritus, 400 BC.
We had that detour with the idiocy of Aristotle, and then eventually got back to our senses, and we saw John Dalton with his table of the elements, and then ultimately onto Mendeleyev.
And I wanted to pick up the thread there.
But before doing so, draw attention the fact that John Dalton did more than simply develop a set of fonts for us.
So he proposed the model of the atom and this goes back little over 200 years ago, and these are the features of the model.
First of all, that matter is composed of atoms that are indivisible and indestructible.
So that goes all the way back to Democritus, nothing new there.
All atoms of an element are identical.
Atoms of different elements have different weights and different chemical properties.
This is the emergence of modern material science, the connection between properties and elements.
So the weight arguably is one of the properties, but the only way they could distinguish elements at that time was by their atomic mass.
Atoms of different elements combine in simple whole-number ratios to form compounds.
Well, that makes sense because they're the elements.
They are the elemental building blocks.
If I told you you could build a structure made of blocks and part-way through your construction I say, why don't you cut that block in half, you'd say, well, then the block isn't the building block.
It's the half-block that's the building block.
So axiomatically if these are the elements they must combine in simple whole-number ratios to form compounds.
And lastly, atoms cannot be created or destroyed.
Well, he wasn't foretelling E equals mc squared.
What he was saying was that if you take elements and you combine them to form a compound, if you subsequently decompose the compound you get the elements back as they were.
So those are the features of John Dalton's model.
And we fast forward to 1869.
And this is the knowledge that was available at the time in terms of the elements that had been isolated and characterized.
And it was with this set of elements that Mendeleyev operated.
On file cards, in his breast pocket, he carried with him everywhere.
And he wrote down the names of the elements and their atomic masses and their properties and whatever else he could use the way of characterizing them.
And during the course of writing a textbook-- he had just finished a chapter on the alkaline metals and he was sitting in the railway station playing solitaire, and boom!
The flash came to him that you don't put arsenic underneath aluminum even though it's next in mass to zinc.
You move it over and you don't even put it under silicon.
You put it under phosphorus.
And furthermore, what he said was there's going to be an element here discovered under silicon and it will have these properties.
And let's look a little bit more deeply at the properties.
But before doing so I want to say that by announcing this prediction of what the element should be that's missing is that we start to see the evolution of principles of modern chemistry.
First of all, he recognized the pattern.
So did Lothar Meyer in Tuebingen.
So they both proposed a periodic table of the elements, but where Mendeleyev pulled away from the pack and distinguished himself was that he developed a quantitative model.
And I haven't shown you the quantitative aspect yet.
That's coming next.
That explains the observations, and that's good.
You might say, well, that's just curve fitting if you're a cynic.
But it makes predictions that can be tested by experiment, tested by experiment.
So let's take a look.
He said that under silicon, but above tin, there would be an element.
He called it eka-silicon.
Eka is a Sanskrit word, which means one after.
So this is the element one after silicon.
It was eventually isolated and given the name germanium.
Mendeleyev said it would have an atomic mass of 72 grams per mole.
In fact, it's 72.59.
He said it would have a density of 5.5 grams per cubic centimeter.
It's 5.36.
This is 1869.
He said that it would have a high melting point, whatever that means, and it melts at 958 Celsius.
It's compounds.
he said it would form a dioxide with a high melting point and a density of 4.7.
It forms a dioxide and its density is 4.70.
In fact, there's a story about a French mineralogist who came upon some of the stuff that ultimately became germanium dioxide, measured its density and reported it to Mendeleyev in a letter, saying, you know I measured the stuff and it's 5.3 grams per cubic centimeter.
Mendeleyev wrote him back and he said, make the measurement again.
You're wrong.
He wrote back three months later and said, I measured it again.
It's 4.7.
That was the genius of Mendeleyev.
To go way out on a limb and make those predictions.
And so I've made the case for the table of the elements.
Why do we call it the Periodic Table?
What's the periodic about?
Well, the periodic-- take a look here, if you go to the website there's a tab called Courseware and there's a tab called Periodic Table.
And you can go to the Periodic Table and ask the software to plot property.
So for example this is boiling point versus proton number or atomic number.
And so you see the boiling point varies as you move from low atomic number to high atomic number.
But it's not totally random.
It's not a Gaussian distribution.
There are features.
It goes up and down, up, down, up and down.
If you train your eye a little bit you'll actually see some regularity, a pattern there.
Maybe that's not so good.
Let's look at this one.
This is electrical conductivity.
And again, up, down, up, down and look at those red lines.
There, there, there, there.
Don't you see something?
That's a pattern.
And they're almost equally spaced.
So that was where Mendeleyev announced his Periodic Law, where he said the properties are related to the identity of the atoms.
And furthermore, he announced, that the properties are a periodic variation in atomic mass.
So let's get that now Mendeleyev's Periodic Law, and the properties of the elements vary periodically with atomic mass.
That was Mendeleyev.
So now that we know that we can go forward, and here's now the full-blown Periodic Table according to the framework that Mendeleyev established.
Now if you look at this carefully, you'll see down here things get whited out and there's these strange notations, uu, m, and all that stuff.
What's that all about?
This is where the super heavies lie.
These are all synthetic elements.
Transuranic, they're made by high-energy reactions, so-called high-energy physics in what you might call accelerators, atom smashers, what have you.
And there's only three places on the planet where you can conduct such reactions.
One of them is in Darmstadt in Germany.
One is in Dubna, just outside of Moscow.
And if you want to stay home-- and eschew the frequent flyer miles-- you can go to Berkeley, California.
These are the three places where we have the accelerators capable of making such compounds.
And so, take a look carefully at what the nomenclature is.
The way you name them is by using these Latin ordinals.
So un, bi, tri, quad and so on.
So if you wanted to name element 115, it's ununpentium.
You want the ium ending.
And you can make these up.
You could make up element 205 if you want to or whatever.
My favorite is 111 because that's unununium.
But there they are, so you can have fun with those.
But with time, the elements are being named and these have been synthesized since your version of the table was printed.
And so number 110 is named Darmstadtium in honor of the team at Darmstadt that first isolated it.
And number 111 was just named two years ago and the name is roentgenium after Wilhelm Roentgen, who discovered x-rays.
Now what is it about discovery?
Well, here's an example of one such reaction that would give you an element.
So if we had access to one of these devices we could take, for example, lead and nickel and accelerate them to very, very high energies.
And then we could make 110, ununilium.
Or now we'll call it Darmstadtium plus neutron.
And in doing so we've generated the new element.
But we can't just say we've made the element and publish.
We have to be able to characterize it.
Remember the reason that we gave Cavendish the credit for discovering hydrogen wasn't that he's the first to know that hydrogen exists, but he isolated it and gave it value.
So if you look at the rest of the periodic table, you get things like boiling point, melting point, density, electronegativity, first ionization energy.
There's a lot of information there.
If you go down here there's nothing.
It's all blanks.
These things have very, very short lifetimes.
Fractions of a second.
But you have to isolate them.
There's certain criteria before you can publish.
And all this is regulated by this governing body called the International Union of Pure and Applied Chemistry, So UPAC, the organization that finally rules on the legitimacy of any of these.
And actually there have been some retractions in recent years, where people published claiming-- I think there was a report out of Berkeley claiming that they'd synthesized 115 and then subsequently that was retracted because they couldn't support the property measurements Last thing is, if you're interested, want to do some extra reading, there's a fantastic book about Mendeleyev.
He was the youngest of 14 children, came out of a very poor family in Siberia and rose to be a giant of his day.
He was a polymath.
He, among some of the other things he did, he worked for the Ministry of Weights and Measures under the czar.
The czar was interested in taxation of alcohol.
And if you mix equal volumes of water and vodka you don't get additivity.
So 100 mL of water plus 100 mL of vodka doesn't give 200 mL.
It gives less.
And so Mendeleyev did a study to determine what the optimum ratio is so that people couldn't misrepresent the amount of alcohol in the beverage.
And set the standard at 40% alcohol by volume, which is used the world over to this day.
He also came to the United States in 1876 to go to Titusville, Pennsylvania, where the first oil well was drilled.
And did an exhaustive study of what was the American petroleum industry at the time.
And then went back to Imperial Russia and did the same survey for the Czar in Imperial Russia, including a report that recommended how to develop the natural resources of the time.
He was really an amazing man.
He wrote text books and so on.
And nobody in science-- I would venture to say-- has not heard of the periodic table.
Mendeleyev died in 1906.
The Nobel Prizes were first offered in 1901.
So there were five years where he was close to the top for winning the Noble Prize but was eked out by somebody else.
When you look back at those other Nobel Prizes, they were deserved.
But none more so than that for Mendeleyev.
So ironically the man who gave us seminal knowledge of all chemistry was never awarded the Nobel Prize.
And there's probably a lot of politics in there.
And as I said last day, here's the typical picture of him.
In this he sort of looks like a street person, disheveled and so on.
But this was the man that gave us the periodic table.
That's him at age 35 when he annunciated the Periodic Law.
So good for him.
Alright.
So now, let's take a look a little deeper about the properties of the elements.
How do we understand the properties of the elements?
For the properties elements we're going to have to look inside the atom.
If you did your reading you undoubtedly came across this table.
Which at first pass, deconstructs the atom into three simple particles: the electron, the proton and the neutron.
Here are their symbols, e, p and n. And they're distinguished by charge and mass.
So the electron has charge, minus 1.6 times 10 to the minus 19 Coulombs.
And a very low mass: 9.11 times 10 to the minus 31 kilograms.
The electronic charge is balanced by the protonic charge.
The atom is net neutral.
So the proton has a charge of plus 1.6 times 10 to the minus 19 Coulombs.
The neutron, as the name implies, has 0 charge.
The proton and the neutron have very nearly equal masses, however.
Right.
And just a word about the units.
The units here are given in terms of the Systeme Internationale.
So when we use the term, C, capital C is for the Coulomb.
And that's the unit of charge.
And it has an uppercase letter because it's named after a scientist.
In this case, the French scientist, Coulomb, whereas the gram is not named after a scientist and so it's lowercase.
And then we can amplify by powers of three.
So if I want 1,000 of these, I put a lowercase k here.
If I put an uppercase K, I end up with the unit Kelvin, which is the unit of temperature named after Lord Kelvin.
And all of this is known as SI units, which is the International System.
And it's not because the scientists don't know how to develop an abbreviation.
This was originally developed when French was the international language of science.
So this is known as the Systeme Internationale and all of these units were defined at that time.
And the term SI sticks that's the legacy.
All right, so now if we go to the Periodic Table.
When we start looking at the elements, we can look at any entry on the Periodic Table, and we have the chemical symbol that I'm designating here as uppercase X, and this was originally John Dalton with the I and the circle around it.
And about 30 years later the Swedish scientist Berzelius suggested that we use neutral units and so therefore we have the Latin coming in for many of the elements, such as iron, Fe, ferrum, and gold Au, aurum.
In the upper-left corner, we have the quality I'm representing here, A And A is the mass number.
Some people call it the atomic weight.
And it is the sum of the masses of all the constituents.
So it's the sum of the mass of the protons, so it's the proton number plus the neutron number plus the electron number.
But since the electron weighs 1/1800 of what these others weigh, you normally don't consider this.
It doesn't matter.
So just adding protons plus neutrons gets you to what we call the atomic weight.
And then down in the lower left corner we have Z and Z is the proton number.
And as the name implies, it's equal to the number of protons in the nucleus, which then equals the number of electrons outside the nucleus in the neutral atom.
Now I'm specifying neutral atom, because it's not necessary for atoms to be neutral and we'll take a look at those in a moment.
A point about redundancy here.
We don't really need the proton number and the chemical symbol because the proton number really defines.
The proton number is like the Social Security number.
This is the identity number of the atom.
If we change the the proton number, we change its identity.
So for example, I could write sodium.
Sodium 23 and 11.
I don't need the 11.
11 means it's sodium or sodium means it's 11.
So I could just write this as 23 sodium.
So I know it's sodium, that means it's got 11 protons and 23 minus 11 must be neutrons.
Or if I wanted to be a smart aleck, I could write this : 23 11.
That's sodium.
I don't need to put anything here.
But there's no smart alecks here, of course.
So for example, we could then show this reaction as-- this is what?
208.
This is lead, 208.
And nickel, 62 gives us Darmstadtium with a value of 269 and the neutron is 1.
You can see how these reactions can be made to go.
Now atoms don't necessarily have to be net neutral.
We can have something that is net non-zero charge.
Net non-zero charge on the atom gives it the term, ion.
Ion is an atom with net non-zero charge.
And we have two cases where the atom is net positive.
If the atom is net positive that's the result of electron deficiency.
The atom is electron deficient.
And we term such an atom the cation.
There's two types of ions.
The cation.
And then we have something that is net negative.
If it's net negative, it means it's electron-rich.
That is to say, there are more electrons than protons and the net negative ion is called the anion.
And you can try to figure out ways.
I sometimes think that cation has a t, which looks a little bit like a plus sign.
Anion has five letters, minus has five letters.
And they both end in n, but this has an n, which is negative or something.
You'll figure something out.
Now we've talked about varying charge at constant proton number.
But the other thing we can do is we can look at-- you can vary the neutron number.
Since the neutron has no charge you can vary the neutron number and not harm the identity and still have a neutral atom.
So vary neutron number at constant proton number.
And let's see what that is.
That gives you something that looks like this.
So for example, if you if you look at carbon, the atomic mass that's shown here is 12.011.
And you'd say, well, gee, if it's got 6 neutrons and 6 protons, why isn't that 12 exactly?
Well, this is the answer here.
You can vary the neutron number at constant proton number.
So let's take a look at how that plays out.
The way that plays out is as following.
Let's see I'm going to make a little table here.
So we'll start with carbon 12.
Carbon 12, so that means-- now I know what I'm going to do.
I'm going to bring this down and make some headings for me.
This will be my proton number and this will be my neutron number.
And finally I'm going to talk about abundance, natural abundance.
So carbon 12, since it's carbon, axiomatically it must have 6 protons.
And 12 minus 6 is 6, so it's got 6 neutrons.
And this is the dominant form of carbon.
If you took a chemical analysis of the carbon you'd find that over 98%, 98.892% of the carbon atoms that you examined would be of this form, carbon 12.
Now there's also carbon 13.
Has to be 6, otherwise it's not carbon.
That means it's got 7 neutrons.
And it's a minority species.
1.108%.
And then there's a third form of carbon and that's carbon 14.
Again, has to be 6 and it's got 8 neutrons.
And it's found in vanishingly small quantities, one part in 10 to the 12.
Or we could call it ppt, parts per trillion.
So this is same atomic number, same proton number, same Z but different mass numbers.
Different A's.
So all of these variants of carbon are found on the same place, the same spot on the Periodic Table.
The Greek word for same is iso, and the word for place is topo, so these are called isotopes.
The isotopes of carbon are species that have identical proton number but different neutron number.
How about the units?
What are the units here?
Well, we have to give some kind of unit.
I've been sort of freely going around and counting protons as one and so on.
And here's the standard.
The standard for mass is defined, and the definition goes like this.
If you take carbon 12, which we just introduced to you, and we say that we're going to specify a mass of 12.000 grams exactly for a specified quantity, in other words, a specified number of these atoms.
We have to say we'll take a certain number of these carbon atoms and specify the mass of that number is 12 exactly.
And specified number of atoms being the mole.
The mole.
And it turns out that the mole has a value of 6.02 times 10 to the 23rd.
How do they get that number?
A little bit more in the way of definitions.
It was a concept put forth by a professor.
So we're going to take some time on it because we respect professors, in this class at least.
And so this was a concept put forth by Professor Amadeo Avogadro.
Professor Avogadro, who was a professor of physics at the University of Turin, Torino.
And he was a contemporary of John Dalton's and they were both studying gases.
And it was Avogadro who taught us that, when you keep the pressure constant equal volumes of different gases contain equal numbers of molecules.
It doesn't matter if you have argon, which is by itself atomic, or we have oxygen, which is diatomic, or you have methane, which is CH4, five atoms making a compound.
Equal pressure, equal volume, equal numbers of those species.
So that was Avogadro's Law.
So let's put that down.
At constant pressure equal volumes of different gases, contain identical numbers of atoms.
Equal volumes of different gases contain equal numbers of molecules.
And here I'm using the term molecule as a counting unit.
So it could be, strictly speaking, an atom or it could be diatomic and so on.
That's what it was.
And out of honor for Avogadro, we name the number of atoms in the mole the Avogadro number.
Which I've written 6.02 times 10 to the 23rd.
Now how do we determine Avogadro's number?
That's an interesting story.
So first of all, we need two pieces of information.
Because we're going to do this by the noblest form of chemistry, electrochemistry.
So the first thing we're going to do is we're going to look at the work Michael Faraday in England.
And what Michael Faraday did is he studied the electrodeposition of metal.
And specifically he passed current through a cell and he electrodeposited silver.
So he starts with silver plus, that's silver a cation, and by the action of electric current attaches an electron to silver and renders it neutral.
Silver, which now plates out onto an electrode and they measured the mass.
They measured the mass of silver-plated and they compare it to the amount of charge that was passed.
They measured the charge.
And you can get charge, because you know current.
So charge is simply equal to the integral of the current times the time.
You know the current, that's easy.
And what Faraday found was that to make what we now know to be 108 grams of silver, 108 grams of silver, which we're going to subsequently recognize as the mole, which is identical to the amount, the number of particles in 108 grams of silver, is equal to the number of particles in 12 grams of carbon.
Sort of an Avogadro-type harkening.
He found that that is-- the equivalent requires 96,485 Coulombs.
So you can say 1 mole of electrons gives me 1 mole of silver, so that's the charge on 1 mole of electrons, where Coulomb is the elementary charge, because we know 1 electron per 1 silver atom deposited.
So now if I know that's a mole of electrons, I need to find a charge on one electron, divide through and I get the Avogadro number.
And to finish the story we have to wait about 50 years and come to the United States, where it's Robert Millikan, Robert Millikan at the University of Chicago doing the oil drop experiment through which we learn the elementary charge.
And here's the cartoon of the oil drop experiment.
I took this from a different text.
It's not shown in your text.
So I actually did this experiment as a sophomore at the University of Toronto.
They had us repeat some of the great experiments of physics, the ones that were accessible, obviously.
I couldn't do high-energy physics in an afternoon.
That would have taken me a little bit longer.
But we did this one.
And so it consists of an atomizer, sort of a perfume atomizer, in which there's oil.
And by the action of atomization we form a shower here, a very, very fine dispersion of tiny droplets of oil.
And then-- this cartoon is hard to make sense of so I fixed this-- we shine high-energy radiation on this.
And by the action of high energy radiation we take these neutral droplets and we turn them into ions.
We eject electrons.
And so now these are charged.
And then we charge the plates.
So if we have neutral species and they simply come out of the atomizer, they'll settle under gravity.
But now if they're charged and I put a charge on the plates-- let's say as here the upper plate is positive-- if any of these particles is charged positive, the action of the electric field will accelerate the descent, because the bottom plate is negative attracting and a positive plate at the top is repelling.
And vice versa.
If I have a particle that's negative, the upper positive plate will actually cause it to slow down, and in the extreme, it may actually start to rise.
And so what Millikan did is a set of experiments in which he studied all the different particle sizes.
See this telescope?
Right over here is Millikan.
And Millikan's sitting there and he's squirting and he's watching.
He's measuring the settling velocity.
And he changes the magnitude of the electric field.
He changes the intensity of radiation.
He changes the nozzle.
He changes everything he can.
And what does he find?
He finds that the distribution of velocities is not continuous.
It's not continuous.
You think, well, gee if you just keep dialing you should get every variation of velocity.
Well, he doesn't.
He finds that he gets variation down to a single value, below which he can't go.
He determines that electric charge is quantized.
That is to say there's a base unit.
It's an element.
I just talked to you about the elemental building block.
That's an element in mass space.
Now I'm going to go conceptually into charge space.
There is an elemental building block of electric charge.
Electric charge is quantized.
And he found that the elementary charge, which we gave the symbol, e. e is not the symbol for electron.
e is the symbol for elementary charge.
It has a value, if you convert it to modern SI units, of 1.6 times 10 to the minus 19 Coulombs.
So now I can take these two pieces of information, Faraday which is up here.
This is known as the Faraday Constant.
Script f, Faraday constant.
So if I divide the Faraday constant, which is the charge on a mole of electrons, by the elementary charge, which is the charge on one electron, presumably I should end up with the Avogadro number.
It should be the ratio of the Faraday to the elementary charge.
And it gives us-- for the third time this morning-- 6.02 times 10 to the 23rd.
If you like per mole, yes or no, doesn't matter.
So now, what's the atomic mass unit?
Now we can say the atomic mass unit is, 1 atomic mass unit then must equal what?
It's going to equal 1/12 of the mass of carbon 12.
1/12 of the mass of carbon 12 divided by the Avogadro number, which gives us 1.661 times 10 to the minus 27 kilograms.
Now be careful because the system is just a little bit rickety.
You know we went SI, but look, this is still defined as 12 grams.
And so sometimes if you look depending on where this is, 10 to minus 27 kilograms or 10 to the minus 24 grams.
Just be careful.
If you ignore this you'll be off only by factor of 1,000.
That's a joke.
But it's lost here.
People are too serious.
We'll lighten you up.
All right, so enough of the history.
Let's now do something dynamic.
So far we've been studying static elements.
But chemistry is really the action of elements in motion.
So how do we describe a chemical reaction?
Let's look at that.
What are the rules to describe a chemical reaction?
Write an equation.
Write the equation of the chemical reaction subject to these rules.
There are two simple rules.
One is conservation of mass.
We've been told the repeatedly since Democritus, conservation of mass.
And the second thing, we use Dalton's Law of Molar Proportions.
That is to say, the building blocks in integer ratios.
And so I thought I'd do this in context.
So I've got a specific example here.
So this is something that I'm interested in.
Some of my research is in metallurgical extraction by benign processes.
What you're looking at is a billet of titanium.
To give you a sense, you can see the stairwell back here.
So this is about 4 feet, a little over a meter here.
So you can see this is one honking big piece of titanium.
This came out of the primary reactor, the Kroll reactor and this is subsequently swaged and hot worked and so on to form these billets.
So this is the first step of turning dirt into metal.
That's called titanium sponge.
And titanium sponge occurs inside a Kroll reactor.
It occurs inside a Kroll reactor, which was invented by a man of the surname Kroll in Luxembourg in the 1930s.
And then with the advent of World War II, he decided to be smart, to get out.
And he ended up in Oregon where he became a professor.
So he's known as Professor Kroll, although the truth be told he really made his discovery before he became a professor.
But he's still a professor and so we'll honor him.
And so the Kroll process for making titanium centers around this reaction.
Here's the reaction written according to the rules above.
We take titanium dioxide, which is found in the Earth and by some prior chemistry convert it to titanium tetrachloride, and in a reactor that I'm going to show you in a moment, we react titanium tetrachloride with magnesium.
And magnesium has a higher affinity for chlorine than does titanium and steals the chlorine from titanium to form magnesium chloride, leaving behind titanium metal.
Now we have to have conservation of mass.
So you can see, I've got 4 chlorines on the left but only 2 chlorines on the right.
So I'm going to put a 2 here and double the magnesium chloride.
But now I've got 2 magnesiums on the right and only 1 on the left.
So I'll put a 2 in front of the magnesium and now we have a balanced equation.
And here's what the reactor looks like.
You can imagine a giant vessel with a pressure seal on the top and a couple of valves, big enough to make this.
So this is about 15 feet by 30 feet.
And so we introduce titanium tetrachloride, which is a gas, and magnesium as a solid and heat to 900 degrees C. And at 900 degrees C, if you look on your Periodic Table you'll know that magnesium melts at 650 degrees C. So we have a liquid sitting here, titanium tetrachloride here, and this thing is sealed.
It's called a bomb reactor.
Nothing can get in, nothing can get out.
The pressure builds up here.
And right at this interface the titanium tetrachloride reacts with the magnesium according to this reaction.
Now this is very interesting.
It's beautiful reaction because the titanium tetrachloride is a gas; magnesium is a liquid.
Magnesium chloride is a liquid, but it is of different density, and it is insoluble in magnesium, and titanium melts at 1670 and it's a solid.
So what happens over time is this.
The magnesium chloride that forms pools underneath the magnesium liquid, gets out of the way so that we can continue to keep this interface clean and have the reaction proceed.
You don't want to reaction where reactant A reacts with reactant B, makes a product that covers the interface and now the product is in the way of future reaction.
So this is very elegant because I don't need any fans, I don't need any nose propellers, nothing.
By density the magnesium chloride settles and the titanium settles.
And it's sitting here at the bottom.
And you can imagine if we do this long enough, this titanium at the bottom will continue to build until it looks like this.
As long as you keep feeding TiCl and Mg. See I'm talking metallurgy now.
TiCl and Mg, that's what you make.
So that's how we make titanium, first step.
And so suppose you get hired and it's your first day on the job and you're working at Cambridge Titanium and the boss says let's put in 200 kilograms of TiCl and we'll put in 25 kilograms of Mg. And the question is, what is the yield?
What is the yield?
How much titanium are we going to make?
Well, you say, just multiply it out.
But first you have to see if things are in balance.
We have to study the stoichiometry of the reaction.
Stoichiometry, what does this mean?
It's from the Greek, stoicheia, which has to do with measurement proportions.
So if these are not put in to the reactor in proportion to what they are in the equation we're not going to get the yield here.
So first thing I gotta do, this is in moles, this is in kilograms.
So I have to convert the kilograms to moles and then maybe I can make some sense of this.
So if I divide by the atomic mass of titanium, four times the atomic mass of chlorine and convert; I will discover that I have 1,054 moles of TiCl.
And I've got about 1,029 moles of magnesium.
Well, this equation says I need 2 times the amount of titanium tetrachloride.
Well, it's obvious to the naked eye, 1,029 isn't two times 1,054.
So I've got a problem here.
I'm not going to get as much titanium as I put in.
Titanium chloride.
This yield is going to be restricted.
It's going to be restricted by-- this is sort of a chain is as strong as its weakest link-- the yield is restricted by the amount of limiting reagent.
And in this case, magnesium is-- this is less than 2 times the mole number of titanium chloride.
So this means this is the limiting reagent.
Alright so now if we use that principle then I'm only going to get as much titanium as I had magnesium and you can see from the stoichiometry here, if I've got 1,029 moles of magnesium I'm going to have half of that number of titanium.
So therefore the amount of titanium is equal to 515 moles of titanium.
And you notice I'm not obsessed about a significant figures and so on.
It's a metallurgical plant.
Half of 1,029 is 515.
Is it 514.5?
If you wish.
I don't care.
So 515 moles and then I convert that, which gives me 24.7 kilograms of titanium when I use that amount of magnesium.
And if you go to the text, Section 2.7, you'll see the nuts and bolts of how to run these reactions.
For those of you who had a lot of chemistry in high school, I know this is review, but I want to bring everybody up to the same page.
So we're starting with this.
All right.
I think that's a pretty good place to stop with the delivery.
But I don't want you moving.
You don't move yet.
Because the last 5 minutes I'm going to still continue to talk.
But on a slightly different topic.
And so I don't want to hear the binders snapping and so on.
We're here; you paid your money.
Five more minutes.
Five more minutes and then you're out there.
Out there, then begins le weekend.
But not until then.
So a couple of things.
First is, the music today.
I try to link the music thematically.
So the music playing today was Polovstian Dance number 17 from Prince Igor, by Borodin, Aleksandr Borodin.
Why were we listening to this music?
Well, because I insisted that we listen to it.
Well, what about Borodin?
Borodin lived in Saint Petersburg.
He was a friend of Mendeleyev.
OK, that's cute but more importantly Borodin wrote his music in his leisure time.
He had a day job.
His day job was professor of chemistry.
And he worked at the Medical Surgical Academy in Saint Petersburg.
He was an exceptional human being.
In those days, women were forbidden to attend institutions of higher education.
He set up an entire curriculum for women in a night school at the Medical Surgical Academy.
He cavorted with artists and therefore obviously his politics were radical.
And they were trying to reform the political scene in Czarist Russia at the time.
And he was also quite a bon vivant.
And he died on his feet dancing at a ball.
So that's the way to go.
Having a great time.
That was Borodin.
One other thing before you go.
You were very very dour, so I thought I'd try to put you in a good mood to the extent this is possible with this group.
And I wanted to share with you some news.
There's been a new element discovered.
You know these atoms smashers, they're always working.
And so the discovery of the heaviest element known to science has been reported.
The element, tentatively named administratium.
I don't know if UPEC is going to go for this, but you can suggest names.
So they're going to name is administratium, the discoverers.
It has no protons or electrons.
So that means its atomic number is 0.
It does have one neutron, 125 assistants to the neutron.
75 vice-neutrons and a 111 assistants to the vice-neutrons.
This gives it a mass number of 312.
The 312 particles are held together in the nucleus by a force that involves the continuous exchange of meson-like particles called memo-ons.
There's no electronic mail, because there's no electrons.
There may be neutronic mail but we don't know yet.
Now you've already learned something today.
You know something.
Since it has no electrons, what do we know about its chemical reactivity?
It's inert.
It has no electrons.
It can't exchange.
So this is chemically inert.
So you say, how did they detect it?
Because it seems to impede every reaction in which it is a present.
According to the discoverers a few nanograms rendered a reaction that normally takes a fraction of a second, it took now four business days to conduct that same.
There are a few other properties.
We know so far that it's radioactive.
And we're going to study radioactivity later, so there's a little bit of foreshadowing.
It has a half-life of about three years, at which time it stops decaying and instead it undergoes a reorganization, in which the vice-neutrons, assistants to the neutrons and assistants to the vice-neutrons, exchange places.
Some studies indicate that the mass actually increases after each reorganization.
So you can imagine now we'll have something like this.
See how this increased?
So if they occupy the same place, they have the same proton number, but a different neutron number, in the case of administratium, they're called isodopes.
So with that I will say, have a good weekend.
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Time to learn.
Time to get back to business here in 3.091.
So let's do the announcements first.
Tomorrow quiz one, based on homework one.
Ten minutes.
Bring your Periodic Table, your table of constants, something to write with and a calculator.
There's about a dozen or so copies of the text on reserve at the Hayden Library.
I want you to be aware of that.
Now some of you approached me last day and said is it true that we have to memorize the Periodic Table?
Yes.
But not the whole Periodic Table.
I know some of you are going to say, well that's so 19th century, rote learning and so on.
To which I say, you're wrong.
Every educated person should know that potassium lies under sodium, which lies under lithium.
And you are going to be educated whether you like it or not.
But this is a lot to learn and so we're just going to look at the main block.
We're not going to ask you to memorize the lanthanides, actinides and the super heavies.
We'll leave that out.
So it's quite straightforward.
We've been doing this for many years.
This is from 2004.
And it's not that hard.
I give you the blank here.
And I even put the numbers on for you.
And all you have to do is add the two-letter or one-letter symbol.
You don't have to give me the full name of the element.
You don't have to get me its density or its electron affinity.
You've just got to put argon, in there.
Ar.
That's all you've got to put in there.
OK?
And every year people grouse when I announce this.
But you'll find out that you'll breeze through it.
You'll have 10 minutes.
This will be on September 24.
So next week there will be two minor celebrations.
One on Tuesday and one on Thursday.
So it's going to be a very festive week.
And people blast through it.
It's one mark off for every error down to zero.
We don't give negative scores.
And people just go right through it.
And then they urge me to make sure that in future years I victimi-- I mean I ask students to continue this.
People are very proud of the fact they know the S, P, and D blocks of the Periodic Table by memory.
But I don't want to neglect, disrespect the lanthanides and actinides.
So we're going to have two contests.
Mnemonics to help remember the names of the lanthanides and actinides.
That's tricky to remember.
So these are mnemonics contests.
So here are some examples.
Here's one, see lanthanum, cerium, praseodymium, neodymium and so on, all right?
Obviously this one came from industry because of the slur against the Academy.
Obviously, somebody in industry has no idea how hard we work here.
And plus, look at this: "to dramatically help".
This gigantic split infinitive.
Clearly, clearly someone from industry.
Here's another one that would help you memorize, I think.
Now, this is obviously not referring to 3.091.
It must be referring to some other chemistry professor here.
I'll let that go unchallenged.
Now, occasionally people ratchet up.
And this was about six, seven years ago, a young man by the name of Blake Stacey, who was aware of the fact that there are 14 lanthanide elements because there are fourteen F electrons.
And an Elizabethan sonnet has 14 lines.
And so he took Sonnet 57 by Shakespeare in reference to the fact that lanthanum is element 57.
And he wrote this in reference to Sonnet 57.
And, of course, he knew he was going to win when he made samarium the word smelting, and being the chemical metallurgist I am, he had me at the fifth line.
Actually, I showed this to Professor Sonenberg in Humanities, and she said, actually, this is pretty good.
That's one of those times you look at and say, you wouldn't see this at Harvard.
So now I'm going to show you one.
Sometimes people ratchet up.
A couple of years ago, there was a video.
So let's take a look at that one.
[VIDEO PLAYBACK]
And she goes through various poses as a superhero of different elemental value.
[END OF VIDEO PLAYBACK]
But we don't have that much time.
So anyways, September 25, 5:00 p.m. Eastern Daylight Time.
Submit your entry, whether it's the big canisters of the Hollywood footage or whether it's just a 14-line mnemonic device.
And the prizes-- it's a contest, there are prizes-- I have ties and scarves from the American Chemical Society.
[LAUGHTER]
What are you laughing at?
These things!
Now look at me!
This is going to be stylish.
They're black.
They've got elements from the Periodic Table on them.
Very hot.
Very hot.
They're black.
They look great.
And we're going to present them on the Friday of the parents' weekend because the parents eat this stuff up.
So now I think it's time to get back to learning.
So last day we ended with the stoichiometry, and I showed you the Kroll process, how to make titanium.
So the question that we never got to was how do we tell that magnesium has a higher affinity for chlorine than titanium does?
How do we know that magnesium is going to reduce titanium tetrachloride to a sponge?
So for that we have to look inside the atom.
We have to look inside the atom.
So that's what we're going to start doing today.
And we're going to begin, as often we will, with a history lesson.
So let's go back for a status report around the end of the 19th century.
So what did we know towards the end of the 19th century?
The atom is electrically neutral.
The negative charge is carried by electrons.
We knew that there was some carrier of negative charge.
We knew further that this electron is very, very tiny.
Small mass.
And so if it has a very small mass, then by difference, then the bulk of the mass of atom must be contained in the positive charge.
So now the question is, what is the spatial distribution of charge inside an atom?
You might say, well why do we want to know that?
Well, we're trying to get a physical model of the atom so that then we can start to address the question of chemical reactivity.
So we have got to know where all the players are before we can attribute action to them.
So let's now take a look at models of the atom.
For the first modern model of the atom we go to J. J. Thompson.
J. J. Thompson, he was a professor of physics and director of the Cavendish lab at Cambridge University.
The other Cambridge.
Cambridge University.
And this Cavendish Lab was named after Henry Cavendish.
The same one who is credited with the discovery of hydrogen in 1766.
He never married, and when he died he bequeathed this fortune to Cambridge University, and they took that money and established a physics laboratory, which is still in operation to this day.
So what did J. J. Thompson tell us?
He said the electrons were distributed throughout a uniformly charged positive sphere of atomic dimensions.
So that's the text.
Now let's make this cartoon, because that makes more sense.
Visualize it a little bit better.
So it looks a little bit like this.
So here's a sphere of atomic dimension.
Electrons are these tiny little things distributed throughout the positive sphere.
And this is massive because the electrons have very, very low mass.
And just to be clear, there are no protons.
No protons.
So this isn't positive.
It's a positive chargeness.
It's just a big positive blur.
And then the little electrons inside.
So the electrons are negative, and they're tiny and low-mass, and they're mobile.
He didn't say much about how they move, but you can imagine that they move, and maybe they even spin.
And so on.
And this was known as the Plum Pudding Model.
There's a big cultural bias here.
So even though I grew up in British Canada, I never had this stuff.
But my understanding is plum pudding is this dish that's served at holidays and Christmas-time.
So the custard is the charge, the positive charge.
And these little bits of fruit are the electrons.
And they're throughout.
At Christmas-time, they might even put little charms in here.
Like a little wishbone or something for good luck.
And you pull it out and all that kind of stuff.
Yeah that's merry, old England.
So this is the Plum Pudding Model.
And that's what was in place.
By the way, we've been using the term electron, but J. J. did not like the term electron.
Actually, he won his Nobel Prize for the-- quote, unquote-- the discovery of the electron.
Just as Cavendish didn't discover hydrogen, he characterized it.
J. J. Thompson characterized the electron.
What did he in particular characterize?
He characterized the charge-to-mass ratio.
He made the first quantitative measure of the charge-to-mass ratio of the electron.
But he never used the term, electron.
He called it the corpuscle.
He called it the corpuscle of electrical charge, whereas this is really sort of an elementary charge.
The corpuscle.
And fortunately there was another British chemist.
In fact, he was an electrochemist.
His name was John Stoney, who, while I think that Thompson was a brilliant man, I wasn't crazy about his literary choices.
And it was Stoney who said, let's call the term electron, coming from the Greek term, elektra, which is the Greek word for amber.
And you know if you rub amber, you'll build up static charge, and so on.
That's where we got the electron from.
So that looks pretty good.
So that's our theory.
And what happens next?
Well, what happens next is that we've got to put this theory to the test.
And the person who puts the theory to the test is a man by the name of Ernest Rutherford.
Rutherford was an interesting character.
He came from New Zealand and he was born on a farm.
It was a family with 12 children.
You notice Mendeleyev came from a family of 14, this guy comes from a family of 12.
So you learn how to survive.
He worked on a farm so he was very handy, and that came in to play as his career progressed.
He was a brilliant experimentalist.
He was able to do experiments where others failed.
And he did this experiment where he was a professor of physics at Victoria University in Manchester in the UK.
Now, he came out of New Zealand, got a scholarship to study at Cambridge University.
And he was actually a research student for J. J. Thompson.
And he conducted his thesis research under J. J. and during that time his thesis research was about the properties of charged particles.
And they were very interested in gases and radioactive elements, ionizing radiation, the whole thing.
The Cavendish lab was a beehive of activity.
And what he did under his PhD was he understood both the alpha particle and the beta particle.
First, he identified what they are.
The alpha particle was determined to be the helium nucleus.
So this is helium that's lost both of its electrons.
So it's just protons and neutrons in the nucleus.
Completely naked.
No electrons.
And then the beta particle, which other people in physics had been referring to this mysterious beta particle, he made the connection between the beta particle and the electron.
So the same electron that's over here and attributed to the negative charge within the nucleus is also a free particle that can go through space.
And these are taken from your reading.
And you can see that if you send the particle beam through a charged plate, the negative plate here being the lower plate, the positive particle, which in this case is the alpha particle, will bend down towards the negative plate.
And the negative particle, being the free electron, is bent in the opposite direction.
So these are the kinds of experiments they would do.
And they also looked at their ability to penetrate different media.
So when it comes to their ability to penetrate solids, they differentiate themselves.
So they alpha particle is quite poor at penetrating solids.
So you see in the cartoon there, the alpha particle is stopped even by that little sheet of paper.
So it doesn't do so well here, whereas the beta particle, the electron, can go zooming right through the paper and is stopped by the lead.
And then over here they were also studying their ability to ionize gases.
So if you take a gas and you bombard it with alpha particles, will you get a lot of ionization?
Yes or no, et cetera.
And what they found was that it was a complementary, that the alpha particle is pretty good at ionizing gases, whereas the beta particle is poor at ionizing gases, and in fact, is stopped.
This is why, if you have an electron beam you need a vacuum.
Otherwise, the electron will be dissipated in very, very short order.
So after doing this work with J. J., he got a teaching job in Canada at McGill University up north.
And while he was there he continued to do work on the origin of alpha particles and won the Nobel Prize for his work while he was at McGill.
And so when he determined that the alpha particle origin in the decomposition, the disintegration of elements.
And believe it or not, in this work back in 1906, 1907, he was already anticipating nuclear fission because he was saying that if you put thorium or polonium into a tube, you can expect that it will emit alpha particles.
Well, conservation of mass says they have got to be coming from somewhere.
And if they're not coming from out here, they have got to be coming from inside.
This was brilliant work.
This is Rutherford, young man in Montreal, gets the Nobel Prize in 1908.
And then he goes to Victoria University, and he sets up this experiment.
Critical experiment.
I mean, imagine this guy.
I mean, he's already got the Nobel Prize.
He could just put his feet up.
No, he keeps going.
So this is the experiment.
It's called the Rutherford-Geiger-Marsden experiment.
Because Rutherford was the brains behind it and Geiger and Marsden were the guys in the lab.
And so here's the experiment.
So what he's going to do is he's going to put this model to the test.
We're going to see if this model makes sense or not.
So he's got a gold foil-- now they used different elements, but ultimately they loved gold because gold, first of all, is noble and it doesn't oxide.
So it's pure, even in atmospheric oxygen.
And secondly, we've known since antiquity we can physically deform gold and make it into something sub-paper thin.
So that's what they did.
They beat this thing down to 600 nanometers thick, 0.6 micron.
And they used alpha particles here.
Radium, polonium, thorium as the source in the lead container and so on.
And this stream of alpha particles comes zooming out at energies of 7.68 million electron volts per particle.
So this is like a cannon with a pumpkin in front of it.
And they're going to shoot these at the foil.
So what's the purpose here?
They want to see what's going on.
And if you imagine that we are shooting alpha particles, which are positive, and I've got a wall here of all of these nuclei that are essentially positive chargeness, uniformly distributed with these tiny, tiny little electrons of almost no mass, you'd expect that most of these particles are just going to go zooming on through.
Beyond a certain thickness, I might start seeing some attenuation.
But if I get the gold film really thin, I should just go blasting these all through.
Well, what they find is, two things.
First of all, most of the particles do go through.
When they go through, they're deflected a little bit.
They don't just go through, they are deflected.
And most of them are deflected through low angles.
But a small number of them are deflected through high angles.
They almost go straight back at the source.
So if you've got that set of data, that doesn't support this model.
This model cannot be sustained in the light of those data.
It's a very important experiment.
By the way, how do they count these things?
Well, Geiger is the same guy who give us the Geiger counter.
Geiger was a technician that worked for Rutherford.
And he developed this screen, they called it a scintillation screen.
It was coated with zinc sulfide.
And so when alpha particles bounced and they hit the screen, they would excite electrons in the screen and then cause a scintilla, a spark.
And then what you do is you just get a graduate student or some other source of abundant, cheap labor-- and you have a whole bunch of graduate students sitting here-- and they count.
That's how it was done in good old days, 1909.
And so they counted what the distribution was.
So that's the experiment.
So this is taken from your book.
So this what you'd expect.
And this is what they saw.
So this is small angles and this is large angles.
Theaters of the angle here.
Now Marsden, who's the third?
There's a third person.
Marsden.
Marsden's not too different from you.
Marsden was about 20 years old at the time.
He dropped out of school and then he came to Rutherford, and he said, I'm looking for a job.
In those days, you couldn't return to college.
If you dropped out, finished.
You had to stay all the way through.
Once you dropped out, they slam the door behind you.
But Rutherford, he was different.
He looked at people on the basis of their intrinsic value.
He didn't care about title.
And so he said to Marsden, OK, you want to do something for me?
Here's your question, I like you to imagine an alternative model of the atom.
And here's the alternative model of the atom.
What Rutherford said was, suppose instead, if this is the dimension of the atom, instead of having positive charge uniformly distributed, he said, what if we have positive charge concentrated at one point and then out here we have the negative charge?
So here's the positive charge.
And here's a whole bunch of negative charge.
And now here's the next one.
And here's the next one.
And so on.
And now we're going to shoot these helium nuclei.
See what's in between here?
This is a void, taking from Democritus.
And so now if I shoot the positive helium nuclei here, if they get close to this positive charge they'll be deflected.
If they don't get close to the positive charge, they just go right through.
And once in awhile they get almost on axis, in which case they fling back.
This model would be consistent with the data set I just showed you.
And then to go further, he says to Marsden, I'm going to tell you what the distribution of high-angle scattering was, you tell me what is the ratio between the radius of the nucleus, where the positive charge is to the-- excuse me, I'm showing you the diameter, so this is the diameter of the nucleus versus the diameter of the atom.
What would be the ratio of the two?
I mean, it's gold, so I know I got 79 plus here.
And I've got 2 plus.
And I've got a 79 plus.
What would be the size of the sphere of 79 plus that would deflect something of 2 plus through this angle?
So that's Marsden.
It's not bad for a kid that's been out of school for a couple of years.
That's his problem.
So he did.
He did the calculation.
He did the calculation and Marsden figured out that the radius of the nucleus to the radius of the atom is on the order of about 1 to 10,000.
That means most of the gold foil is void.
Most of the gold foil is nothing.
And you've got all this mass, highly dense, concentrated here.
So this is the Rutherford model, which was called the nuclear model.
This was proposed by Rutherford because it has such a thing is a nucleus.
He coined the term, nucleus.
So that's a big step forward from this thing here.
So you'd think that the physics community would be ecstatic.
That now we've got a model that squares with the data.
Well, what do you think the reaction to Rutherford's announcement was?
Condemnation.
Derision.
People didn't like it.
They laughed at him.
They said, well, this is crazy.
There's a couple of problems with this.
So strong negative reaction to Rutherford's model.
And here the two major objections that people posed.
First of all, they talked about nuclear collapse.
They said if you've got a system of negative particles revolving around a positive nucleus, electrostatic forces are going to cause them to be drawn to one another.
What's going to keep the electrons from collapsing into the nucleus?
That was the first objection.
Coulombic attraction.
Coulombic, electrostatic, same thing.
If you want to use a man's name, Coulombic.
If you don't want to use a man's name, electrostatic.
Coulombic attraction between plus and minus.
And the second one was, OK, suppose for some reason we don't understand that the electron doesn't collapse, it stays in orbit, it's still not going to work.
Because you run out of energy.
If you take a charged particle, a charged particle that accelerates consumes energy.
Acceleration, as you know, can mean either change in speed or change in direction.
So if the electron is revolving at a constant speed, it has to change direction; otherwise, it will fly out of the room.
So changing direction means acceleration, which means what keeps the energy in the electron?
Why doesn't the electron just run out of energy and just stop?
Even if you can prevent it from collapsing, it will just stop.
So this is the problem of energy deficit.
Energy deficit.
Well, what powers the electron?
What powers the accelerating electron?
So that's a problem.
So now the story continues.
The next chapter is 1912.
Because this is 1911.
In 1912, a young Danish businessman by the name of Niels Bohr.
Niels Bohr just finished his PhD in Copenhagen and he got a fellowship from the Carlsburg Brewery Foundation.
Carlsburg Brewery put up money for science.
So he won a Carlsburg fellowship.
And he was a young physicist.
And he knew that in the UK there was a controversy brewing between the model of Rutherford and the model of J. J. Thompson.
So he says, you know what I'm going to do?
I've got a year, I got money.
I'm going to spend six months with J. J. Thompson, I'm going to spend six months with Ernest Rutherford.
And that's what he did.
So he goes and he spends six months at the Cavendish lab with J. J. and then he goes up to Manchester and he spends three months at Manchester with Rutherford.
Why didn't he spend six months?
Because he got a job offer and he went back to Copenhagen to start his teaching duties.
And what he did is he developed a quantitative model for us.
He developed a quantitative model.
And that's all part of what I showed you the other day.
We recognize patterns.
That's what the Rutherford, Geiger, Marsden experiment gave us.
The pattern of the distribution of alpha particle deflections.
So now we need to develop a quantitative model that will explain the observations, and if it's a really good model, it will make predictions that, again, can be tested by experiment.
So we ratchet back and forth between experiment, model.
Experiment, model.
So he proposes the model, not with J. J., because J. J., I don't know what he did there.
This is where he wants to propose the model.
Quantitative model.
To Rutherford's atom.
Let's take a look at what we've got there.
Now, big lesson here.
How do you announce a model?
You don't put an ad in the newspaper.
This is the scientific literature.
This is not Google.
This is not Wikipedia.
This is a primary source.
This is how science is communicated.
So this is a journal.
It's called Philosophical Magazine and Journal of Science.
It goes all the way back to the 1600s.
In fact, when I was a freshman at the University of Toronto we had copies of Phil Mag going back all the way to the 1600s.
And these were not facsimiles, these were real copies.
And I was in the stacks and I was reading something by-- I don't know-- Lord Kelvin, you know late 1800s, and I said gee, well, if I've got Lord Kelvin, then maybe I can go over here to the early 1800s and read Humphry Davy and, gee, if can keep going, and going and I pull out a leather-bound edition of Phil Mag from the 1660s, and I open the book and there's a paper about gravity by Isaac Newton.
And that's how science is communicated.
So I urge you to go to the library and read primary sources.
Most of this stuff is available online.
You can just sit at your desk and you can zoom in to history.
So this is Phil Mag and this is London, Edinburgh and dublin.
See this is 1913, so Ireland was not a free state yet.
So Dublin was a British city.
London, Edinburgh, Dublin.
There's a close-up of it.
July 1913, On the Constitution of Atoms and Molecules by Niels Bohr, Doctor of Philosophy, Copenhagen.
And it was communicated by Professor Ernest Rutherford, FRS, Fellow of the Royal Society.
So, let's read.
It's in English.
In order to explain results of experiments on scattering of alpha rays by matter, Professor Rutherford has given a theory of the structure of atoms.
You see the dagger here?
Down here, Ernest Rutherford, Phil Mag, duh-duh-duh-duh.
There's a citation.
See, attribution.
According to this theory, the atoms consist of a positively charged nucleus surrounded by a system of electrons kept together by attractive forces from the nucleus.
That's what keeps them from just fleeing anywhere, and they don't go in, but they vwomp.
The total negative charge of the electrons is equal to the positive charge of the nucleus.
Further, the nucleus is assumed to be the seat of the essential part of the mass of the atom and have linear dimensions exceedingly small compared with the linear dimensions of the whole atom.
This is beautifully written.
Clear.
It's textbook quality and it's written by a man whose native language isn't even English.
Read the literature.
The number of electrons in an atom is deduced to be approximately equal to half the atomic weight.
Great interest is to be attributed to this atom model.
You have to know a little bit about Bohr.
Bohr used this phrase, great interest.
That was his way of saying, embroiled in controversy.
If someone said something that he didn't agree with, instead of saying, I think that's nonsense, he say, very interesting.
Very interesting.
So great interest is to be attributed to this atom model.
That means people are in pitched battle on both sides.
For as Rutherford has shown, the assumption of the existence of nuclei as those in question seems to be necessary in order to account for the results of experiments on large-angle scattering of the alpha rays.
Another footnote: Geiger and Marsden.
Geiger and Marsden published their results in this journal.
In an attempt to explain some of the properties of matter on the basis of this atom model, we meet, however, with difficulties of a serious nature arising from the apparent instability of the system of electrons.
Difficulties purposely avoided in atom models previously considered.
For instance, in the one proposed by Sir J. J. Thompson, which is to say, mmm, you know, it's curious.
You condemn the Rutherford model because he's got electrons in motion, which means they're accelerating in energy deficit.
But in the Plum Pudding Model, the electrons are in motion, and J. J. Thompson is silent about it.
So this is a very aimed barb right at J. J. Thompson done elegantly.
With this elegant slap in the face.
You know this is how scientists do it.
They don't go, oh you're ugly , and your mother dresses you funny.
Instead they write things like this, difficulties purposely avoided in atom models previously considered for instance.
So this is sort of yeah, how about you, huh?
But anyway, we can go on and on.
But down here, the result of the discussion of these questions seems to be a general acknowledgment of the inadequacy of classical electrodynamics in describing the behavior of systems of atomic size.
So he's saying that you can't use classical electrodynamics down to subatomic dimensions.
He's saying that those models, any more than you can use planetary models down to human dimensions, this, in many respects, is foretelling nanotechnology.
He's saying the properties we know, of how things behave in the Newtonian world are not necessarily valid at atomic dimensions.
This is very important stuff.
Anyway, so what I did is I went through the paper and I've reduced the content of the paper to these things called postulates.
So Rutherford's atom is correct.
That's the first thing.
Classical electromagnetic theory not applicable to the orbiting electron.
And you're going to get-- the PDF and all this is going to be posted, so don't feel like you have to be a super stenographer here.
Newtonian mechanics is applicable to the orbiting electron.
So you can see, wait a minute it's sort of like cafeteria physics here.
Electrodynamics doesn't apply, Newtonian mechanics does apply.
And the answer is, yeah.
That's what we're going to do here.
We'll build a model and we're going to try these ideas out.
And if they're crazy, how are we going to know they're crazy?
Because there's going to be data that refute.
But if we get data that supports all this, then there's only one conclusion.
This thing makes sense up to a point.
The energy of the electron is conserved in the system.
It's bond is kinetic plus potential.
The quantization through angular momentum.
And lastly, the Planck-Einstein relation applying to electron transition.
So now I'm going to take all this and I'm going to write out the model for you.
Now, I don't want you to say, oh, he's going to derive something.
Do we have to know derivations?
This is the only time I'm going to derive something for you.
The reason is, it teaches the model.
It teaches how the model is based on assumptions and so on.
But I don't expect you to memorize things because that's not the way things work.
So let's go starting with this point here.
This point here is Bohr model atom.
So it's Bohr's representation of the Rutherford model.
We can call it a nuclear model or a planetary model.
Because you've got orbiting electrons.
Now, first important point.
The Bohr model is the simplest model.
It works for a one electron system.
So he has a single electron.
Planetary model, single electron.
And this orbits the positive nucleus.
So you might say, one electron, well, what's that mean?
Well, obviously it means atomic hydrogen.
But it could mean helium plus.
That's a one electron system.
It could mean lithium 2 plus.
It could mean roentgenium 110 plus.
That's a one electron system.
And this is all gas.
Gas phase.
No solids here.
Gas phase.
So it's an isolated nucleus with one electron around it.
That's how we're going to begin.
And here's what it looks like.
I'm going to put the nucleus, first of all not to scale, because we're not going to draw 10,000 to 1.
Over here is the nucleus with z positive charge.
And then out here is the electron in orbit.
This is the lone electron in orbit.
And this distance is r. The distance from the nucleus.
You're going to say, well, is that from the outside or the inside?
It's 10,000 to 1; it doesn't matter.
Forget about it.
So now let's use this concept.
We're going to say it's a conservative system and so the energy of the system is simply going to be the energy of the electron.
You might say, well, why don't you consider the energy of the nucleus?
Because the nucleus is, relatively speaking, stationary.
To use Rutherford's colorful language, he says when an elephant has fleas, it's the fleas that do the jumping.
So you don't care about that energy of the nucleus in this case.
So the energy of the electron is the kinetic plus the potential.
So the kinetic energy, we're saying it's 1/2 mv squared because it's Newtonian mechanics apply.
See number 3.
So that's 1/2 mv squared, where v is the velocity.
So that's the kinetic, and then what's the potential energy?
The potential energy here is electrostatic.
Because that's the energy that's stored here is due to the Coulombic forces between them.
So that's going to be q1, q2 over 4 pi epsilon 0 divided by r. So what I'm going to do here, we have to just for convention I'm going to always choose that q1 in my derivation, q1 will be the charge on the nucleus.
so q1 will equal z, which is the proton number here, times the elementary charge.
Remember e is the elementary charge.
It's not the charge on the electron.
So sometimes I use the word e with a minus sign, meaning the electron.
But this is e meaning the elementary charge.
z times e and then q2, which is the charge on the electron, is minus the elementary charge e. So we're always going to have minus e out here.
What varies is z.
And so we're going to make some substitutions.
So this is going to be what?
It's going to be 1/2 mass of the electron times velocity of the electron squared.
And now q1 is ze and q2 is minus z, so that's minus ze squared.
And then that'll be over 4 pi epsilon 0 r. And I'm going to call this thing equation 1.
And by the way you can find all of this information by looking at your table of constants.
So see, epsilon 0 sitting here.
Permittivity of vacuum.
There it is, 8.85 times 10 to the minus 12 farads per meter.
Elementary charge 1.6 times 10 to the minus 19 Coulombs.
Electron mass, 9.1 times 10 to the minus 31 kilograms.
And what's this 4 pi epsilon 0 doing here?
Well we know this is going to give us joules.
And this is going to give us joules.
And the epsilon 0 is the factor that renders electrostatic units on to the same plane as mechanical units.
If I plug in-- this is going to be Coulombs, it's Coulombs squared, and this is going to be meters, and now this things is farads per meter.
And I got farads per meter times meters divided by Coulombs squared.
I don't care.
This, with impunity is joules.
Because this is an energy unit.
And this is Systeme Internationale.
And this is kilograms times meters per second, meters per second and how many apples in a bushel and-- forget it!
Put this in kilograms.
Put this in in meters per second.
With impunity, it's in joules.
So that's what that thing does there.
Alright.
Good.
So let's keep going.
Postulate 3.
Newtonian mechanics applies to orbiting electrons.
So orbiting electron, it's in a stationary orbit.
So that's the thing.
So postulate 3 implies that the sum of the forces acting on the electron must be 0.
This is a force balance, right?
If the forces are out of balance, the electrons are either going to move closer to the nucleus or move farther from the nucleus.
So that's equal to, again-- it's a sum of a dynamic force and a Coulombic force, or electrostatic.
Here I just noticed a nice parity here.
So these are technical terms, but we can give them human terms.
So the dynamic force, there's a Newtonian force, this is a Newtonian force and this is a Coulombic force, after the two people that are associated with the ideas.
Charges interaction are Coulombic, blah, blah, blah, blah.
So now lets substitute in.
So we know that if you've got something orbiting, something tethered, then the dynamic force on that, that pulls the object away on a tether is mv squared over r. And the electrostatic is q1 q2 over 4 pi epsilon 0 r squared.
Force goes as 1 over r squared.
Energy goes as 1 over r. How do you know?
Well, one way to think about it is, you know energy is the result of a force moving through a distance.
You say, I'm working really hard, and I say I don't see any force moving through a distance.
That's a joke.
So here's energy.
So force.
So if this is 1 over r squared, the integral of 1 over r squared is minus 1 over r. If you get these backwards, you integrate 1 over r, you're going to get natural log of r, which makes no sense.
That's one way.
Or the other way is, you can just remember.
So anyways we're going to plug into those values.
And what do we get?
What we get there is this.
We get mv squared over r. q1 is z times e. q2 is minus z.
So it's minus ze squared over 4 pi epsilon 0 r squared.
And I'm going to call this equation 2.
So now we keep going.
Next one, energy quantized through its angular momentum.
This is really critical.
This is the major breakthrough by Bohr.
Now what do we mean by quantization?
I told you that in 1909, Millikan had figured out that electrical charge is quantized.
Now quantization had already been annunciated by Max Planck in 1909.
You see, the classical theory of radiation gave this prediction for intensity as a function of wavelength.
They called this the explosion, the collapse of the theory.
As you go to low wavelength, the intensity goes without abatement.
But these are the spectra that were measured.
Depending on the temperature of the gas, you get this distribution.
So how to get the distribution to turn around at lower values of wavelength?
And in order to do that, Planck suggested that light is composed of energy packets, or quanta.
That light is a stream of individual energy bits.
And the elementary unit of electromagnetic radiation is the photon.
And you can say, well I don't understand, where goes he get that from?
That's not the way to think about it.
Say, if you make this assumption you get these curves.
And then figure out what it means.
You can't make everything anthropomorphic.
If this were the case, and furthermore if the energy of one of these quanta was related to its frequency-- and wavelength and frequency are inversely related-- then the proportionality constant is h, which we now have designated the Planck constant in honor of Planck, then you get this.
People grew to accept it.
1900, he gets a Nobel Prize for it.
La de da!
But who knows what light is anyways.
It has no mass, it's kind of mysterious.
1900, the whole concept of action at a distance was strange to people.
The notion that something could affect you, from me, without physical contact, this was a new idea.
We take it for granted, but this was very new.
So it's kind of mysterious, they said, OK, he's a physicist, who knows?
And it worked!
But it's light.
Now what Bohr is going to say is the electron moving in orbit, he says, I'm going to apply the same concept of quantization to the electron moving in orbit.
Well, that's different.
And using this value he gives us mvr, which is the angular momentum is quantized n h over 2 pi.
h is the Planck Constant and n is an integer.
1, 2 3 4, et cetera.
So this is the quantum condition and this will be equation 3.
And I have three equations and three unknowns, and what I'm going to do next day is come back and show you the solution of the equations and how they give us a set of equations that allow us to compare with data.
So let's jump to the thing.
And by the way the Planck constant is there.
Now Niels Bohr eventually won the Nobel Prize.
He was widely revered.
He achieved the stature on a par with Einstein.
If you go to Denmark, it's still not part of the European Union, you can still get the 500 Kroner note if you change a 100-dollar bill there.
And there's Niels Bohr with this pipe.
And you can see the concept of things rotating around it an so on.
There's a young Niels Bohr.
This is Bohr with Werner Heisenberg.
Heisenberg was a post-doc who worked with Bohr in 1925 and annunciated the Uncertainty Principle, which we will study a few lectures from now.
This is a chemical conference sponsored by chemical company Solvay.
Brussels 1930, there's Bohr and Einstein.
They loved hats.
This is the Hamburg, this is Borsalino.
Here's Bohr mixing it up with royalty.
This is a young Queen Elizabeth of England.
There's Prince Phillip.
This is the royalty Danish family.
Bohr loved music and here is with Louis Armstrong undoubtedly talking about the resonant structures within the tube here and how to get the various harmonics and overtones.
Now last thing, a little bit about hydrogen.
We've been studying hydrogen here, the one electron atom.
But there are isotopes of hydrogen.
The original was discovered by Henry Cavendish.
There is one form of hydrogen that has a neutron in addition to the proton.
That's the deuterium, from which we can get heavy water, D2O.
This was discovered here in New York at Columbia University by Harold Urey.
And then lastly, there's a second isotope of hydrogen that consists of two neutrons and a proton.
That's tritium and it was discovered at the Cavendish lab by Ernest Rutherford.
He was active right up to his death.
In fact, it is argued that Rutherford did his best work after he got the Nobel Prize.
Your book is showing this.
It's nice to follow.
We started with Dalton and now we're here and then later, in another week or so, we'll be down here.
And here's the sort of the intellectual road map that takes you to modern atomic theory.
So with that I'll dismiss the class and we'll see you on Wednesday.
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OK, settle down.
Let's get started.
One announcement, yesterday we had our first weekly I would say minor celebration.
And by and large, it went well.
Please make sure that you go to your assigned recitation.
If you miss your recitation, you need to get down to see Hillary so that we make sure that we have enough copies of the weekly quizzes on hand.
If you've joined the class, you have to check in with her.
She's down the hall here in room 8-201.
And you'll be assigned to a section.
What else do I have by way of announcements?
Oh yes.
Just reminding you, this was from 2003.
See it doesn't change.
It's the same s-block, p-block, and d-block elements.
So that's coming up a week from tomorrow, two celebrations next.
And of course the contests, the contests with hot prizes.
All right.
Let's get down to business.
Last day we looked at the Rutherford-Geiger-Marsden experiment.
Oh there's one other one.
If you look at the readings, there's this one section called the archives.
My predecessor, Professor Wit, wrote a set of lecture notes.
And they look something like this.
And some students have said that they find these a little more expository in certain sections on certain topics than the book to be.
And I have no preference.
But if you take a look at this it'll say LN1, lecture notes 1.
If you go to this, read this.
If you find that's helpful, good.
If you don't find it helpful, than stick with the book.
Just letting you know what that is.
All right, so last day we looked at the Rutherford-Geiger-Marsden experiment in which a high energy beam of alpha particles bombarded a thin, gold foil.
And on the basis of the scattering results, namely most of the particles went through with minor scattering.
And a tiny fraction of them were scattered through large angles.
The Thomson plum pudding model was rejected in favor of Rutherford's nuclear model of the atom.
And then subsequently, Bohr came up with the quantitative representation off the Rutherford nuclear model.
And we were partway through the treatment of Bohr last day when we adjourned.
So let's right pick up the thread from where we left off.
And so just to remind you, the Bohr model is for a 1-electron atom gas phase.
So this is either atomic hydrogen, it could be helium plus lithium 2 plus roentgenium 110 plus Its doesn't matter how many protons.
there's always only want electron.
And it's a planetary model.
The positive charge concentrated in the nucleus, Z is the proton number.
And at a distance r from the nucleus is a circular orbit in which resides 1 electron.
It has a charge of minus E. And I just designated them q1 and q2.
You could have done it the other way.
But I had to choose something.
So we went through and looked at the constitutive equations here.
So first of all, the energy of the system-- and this is only going to be the energy the electron.
Because we assume that the nucleus is far more massive.
And so we don't have to get into things like reduced mass or anything like that.
So just measure the energy of the electron.
1/2 mv squared is the newtonian component.
And then coulombic energy that's stored is z times e squared over 4 pi epsilon zero r, where epsilon zero is the permittivity of vacuum.
And it's the factor, the 4 pi epsilon zero, is the factor that allows us to take electrostatic energies and put them on the same plane as mechanical energies.
When we run through this, we always end up in joules.
Then there's a force balanced to make sure that the electron neither falls into the nucleus nor flees and breaks free of the atom.
And the force balance is if you put a ball at the end of a string, and you whip it around on a tether, you have a centrifugal force that's trying to get the ball to break away.
And then the string is pulling in.
So the pull in, in this case, is the coulombic force.
And the force that makes the ball want to flee is this mv squared over r. And that must be net zero.
Otherwise we're going to have a shift in orbit.
And then lastly we have the quantum condition.
And this was the breakthrough of Bohr where he enunciated that the quantum condition is going to give us this energy level quantization.
And this was a big departure from what had been in the past.
The only antecedent idea of this nature was the work by Planck who said that light is quantized.
But as I told you last day, who knows what light really is.
The Newtonian notion of a ball orbiting was very compelling here.
And the notion that the movement of the electron could in some way be discontinuous was quite a major departure.
So I left you last day with three equations and three unknowns.
And what I'm not going to do right now is solve the equations.
Because I've been lecturing long enough to know that's the way to kill interest, quench a lecture.
And so if you really want to go through the algebra, be my guest.
You're smart enough to do that.
Instead I'm going to show you the results.
But those are the three equations that you need.
So we have an equation in r. We have an equation at v, and an equation in e. So let's go after them.
If you first look at the solution for r. This is the radius of the orbit of the electron, the orbit of the electron.
If you go through and solve, you'll end up with this, Epsilon zero times the square of the Planck constant divided by pi times m. It's always the electron.
So this is the radius of the electron orbit.
This is the mass of the electron times the square of the elementary charge-- that whole thing I'm going to group-- times the square of the quantum number divided by z, the proton number.
So what we see here?
Well, everything inside the parentheses is constant.
These are all constant.
Pi obviously is geometric constant.
And the rest of these are constants you could look up in your table of constants.
And we noticed that there is a set of solutions to this.
The radius of the electron can occupy various discrete values defined by n. So we say that the radius takes on a plurality of values a function of n. And furthermore, the functionality goes as the square of n. It's n squared times a constant, where that constant is inside those parentheses.
And we notice that because the r goes as n squared it's nonlinear.
It's nonlinear.
This is so important I'm going to write it down one more time.
So the radius of the electron orbit takes multiple values.
It takes multiple values.
And they're discreet.
The physicists like to use a different term.
When something is discretized, the physicists say it is quantized.
So these values are quantized.
You cannot continuously vary the radius and nonlinear values, multiple values.
So let's plug in.
Because I want to get a sense of scale.
So let's look at the simplest one.
The most primitive 1-electron atom would be hydrogen.
In which case, Z equals 1.
So I've just got a proton orbited by an electron.
So look at atomic hydrogen.
So in that case, Z equals 1.
And I'm going to look at n equals 1, which is the lowest number here, right?
R scales as n squared.
So the lowest value or r is obtained when n equals 1.
And this is termed the ground state.
The ground state.
So I want to ask what is the radius of the electron orbit ground state in atomic hydrogen?
And if I plug in these values, I'll call this r sub 1.
It turns out to be 5.29 times 10 to the minus 11 meters, or 0.529 angstroms.
I love the angstrom.
It's a great unit.
It's a great unit.
It's not an SI unit.
But I like the angstrom.
I'll show you why.
If you try to express this in SI units, well there's 10 to the minus 11 meters.
The Si units go in units of clusters of 1,000.
So for example, you've got the meter.
You've got the kilometer.
You've got the micrometer.
You've got 10 to the minus 9 meters, which is the nanometer.
This is a Goldilocks problem.
This one's too big.
And then the next one down here is 10 to the minus 12 meters, which is the picometer.
So this is either 52.9 picometers, or a 0.0529 nanometers.
And that's no good.
I want numbers like 3, 7, simple to remember.
So 0.529, this is about 1/2 angstrom.
It's good to know.
But you try to publish, you know what happens in a scientific literature today?
The Literary Lions that control the journals, they'll circle that and say you have to convert to SI units.
And so they have to right some goofy nanometer thing or something.
I know you think I'm crazy, but I love the angstrom.
So there.
Anyway, so here it is.
Once you know that this is 0.529, this is on your table of constants.
It's right on your table of constants.
So you don't have to go and calculate all this stuff.
Which means if you do your homework with your table of constants, you will know where those numbers lie, as opposed to opening this thing up for the first time on the first celebration of learning on October the 7th, and with 47 entries and they're tiny, tiny font.
And you're wondering where is that thing.
Just a word to wise.
So now we know what this is.
We know this is 0.529 angstroms.
So now I can write an equation for the radius of a 1-electron atom anywhere, anytime.
r of n is going to be equal to a naught which is this value here.
And it is termed the Bohr radius.
So you can write it as a function of the Bohr radius, times the square of n divided by Z.
So that's for all 1-electron atoms, gas phase.
And you can see that as Z goes up, the r goes down, which makes sense.
So suppose instead of hydrogen, we talk helium plus What's the only difference?
Helium plus has 2 protons in the nucleus.
Which means that the coulombic force of attraction between the same 1-electron and now a doubly charged nucleus is going to be stronger.
So the first orbit is going to get pulled in.
And all the other orbits are going to get pulled in.
By how much are they going to get pulled in?
By that much.
So this is the functional representation of all of that physics.
All right, there's three equations, three unknowns.
Let's look at energy.
So if you go through and solve for energy, you get this one.
Minus this big monstrosity, mass of the electron to the fourth power of the elementary charge times 8 times the square of the permittivity of vacuum times the square of the Planck constant, all times the square of the Proton number divided by the square of the quantum number.
And I just to make sure everybody is with me here.
I always want to write n here.
n equals 1, 2, 3, takes on integer values.
And we'll say it again here.
N equals 1, 2, 3, et cetera, et cetera.
So again we say we see that e is a function of n. It's discretized.
It's quantized.
Why?
Because once you impose the quantum condition here on angular momentum, it propagates through the entire model.
So radius this quantized, energy is quantized, you're going to see velocity is quantized.
Because the quantum condition is pervasive.
So e to the n, and I'm going to take this whole quantity in parentheses and just call it giant K. These are all positive quantities.
Mass is positive.
And squares and fourth powers of numbers must be positive.
So this is K times Z squared over n squared That's good.
And we can go and evaluate K. And when we evaluate L in SI units, we get 2.18 times 10 to the minus 18 joules.
This is joules per atom.
Or you can multiply this by Avogadro's number.
If you multiply it by Avogadro's number, then that will give you 1.312 megajoules per mol.
So that's the energy of the electron in the ground state of atomic hydrogen.
And then we can mediate that with Z and n, and go to electrons that are outside the ground state, above the ground state, or ground state electrons in atoms that have more than 1 proton, or both.
And so let's take a look at the graphical representation of that.
So instead of Cartesian coordinates, because this is spatial distribution, I'm going to go to energy coordinates and give you an energy level diagram.
So again, not to scale.
Because this thing goes is 1 over the square.
So that's going to be messy.
So let's start here.
And on the left side I'm going to designate the energy.
And on the right side I'm going to designate the quantum number.
I'm going to start down here.
That's the lowest energy.
You see these are all negative values, first of all.
They're all negative values.
Because Z is a square, n is a square, and K is a positive quantity.
So n equals 1 is the lowest state.
It's the ground state.
It has a value of minus K. So I'm going to do this one just for atomic hydrogen.
So I'm going to write atomic H. So now Z equals 1.
You can do it later for Z equals 2, 3, whatever.
So this is atomic hydrogen.
So ground state energy is minus K. What happens if we go to n equals 2?
n equals 2 it becomes K divided by 2 squared 4.
So it should be 3/4 of the way up.
I'm not going to go quite 3/4 of the way.
Because I want to leave room for some fine structure.
That's why it's not to scale.
All right so this is minus L over 4.
What if we go to n equals 3?
Well it's not symmetric here.
It's nonlinear.
But this should be really what?
Minus K over 3 squared is 9.
You get the picture.
3, you can go 4, 5, and so on until n equals infinity.
What happens when n equals infinity?
I've got minus K over infinity, which is vanishingly small, zero.
Where is the electron when n equals infinity?
r is n squared times the Bohr radius.
That's a great, great distance away.
What does it mean?
Physically it means that the electron is so far away that it is no longer bound.
It's no longer part of the atom.
And when it's no longer part of the atom, and the potential energy that's stored is a result of the charges coming together from infinity.
I'm starting to talk like someone out of 802.
What's the energy if I take 2 charged particles at infinite separation?
I bring them into some finite separation.
Voila.
There it is.
So when they're in infinite separation there's no energy stored.
Hence, you are at that point.
So n equals infinity means r equals infinity, which means E equal zero.
There's no stored energy.
So this means the electron is no longer bound.
And therefore, if it's no longer bound, we have a term for it.
It's called free.
It's a free electron.
And if the electron is free, then the atom is electron deficient.
So if the electron is free, that means the atom is now an ion.
Because it's lost an electron.
It's no longer net neutral.
Or we say an atom hasn't turned into an ion.
Or the electron has been ionized.
What's the energy for that?
We can calculate what that energy is.
It would be called the ionization energy.
So if I started with an electron down in here, and I sent it all the way to infinity.
See that's an energy space.
Which is the equivalent in Cartesian space to go from here to infinity, same idea.
Do you see the models?
This is Cartesian.
It's like a. Map This is energy coordinates.
It's different.
And you're going to be able to think from one model to another.
What's the energy consequences?
What are the Cartesian consequences?
Until we get to the point where there is no Cartesian representation.
Because the abstraction level is too high, we'll have to content ourselves with this.
So get comfortable moving from there to there.
And then some day we're going to say it's too complicated.
There's no Cartesian thing.
We'll be comfortable by then with this.
OK so now we're taking an electron from here up to here.
While we ask, what is the ionization energy?
The ionization energy must equal the delta E of the transition.
So what's that?
The delta E of the transition is going to equal always E final minus E initial.
E final minus E initial.
Which is equal to E at infinity minus E1, the ground state.
Well E infinity, we just said, is zero.
And the ground state energy is equal to minus K. So minus minus K is K. So you also get the energy.
From infinity down to ground state is the ionization energy.
So we can define the ionization energy in terms of this transition.
Define ionization energy as the minimum energy to remove an electron from the ground state of an atom in a gas phase.
So that means there's no solids, no liquids.
There's no work function here.
There's no lattice energy, and so on.
So there's a definition of the ionization energy.
And we can be a little bit more elaborate.
Even though right now I'm just going to do a little break here.
I don't want to mislead people.
But just an aside.
I'm nonlinear.
I can have multiple conversations at once.
And you are capable of stacking.
So we're going to break now.
We're not going to talk about 1-electron atom.
We're going to follow the thread of ionization energy.
I'm going to take lithium.
Lithium in its normal state has 3 protons, 3 electrons.
So I'm going to take lithium gas.
And I'm going to ionize it and make lithium plus.
So this is a lithium plus ion in the gas phase.
It's still got 2 electrons.
So this isn't Bohr model stuff.
But anyway, here we are.
So the energy for this action would be called the ionization energy.
Because I took a neutral atom, and I pulled an electron out of the ground state, and so on.
So this is an ionization energy.
But now I can continue this process.
And I can take Lithium plus.
And I can lose an electron from that, which will than give me lithium 2 plus.
And this is called also an ionization energy.
This is called the second ionization energy.
So this is the first ionization energy.
But just as when you write an equation, when the coefficient is 1, you don't write the 1.
I don't write 1 lithium here.
I know it's 1.
This is the second ionization energy.
And then I can keep stripping away electrons.
And I can take lithium 2 plus in the gas phase, and take away that electron leaving me with just the lithium nucleus, lithium 3 plus, plus electron.
And that's called the third ionization energy.
OK, now what can we say?
What's the relationship here between any of this except the definition and the Bohr model?
well?
The Bohr model applies only to 1-electron atoms.
Are there any 1-electron atoms on this board?
So we can calculate the energy, the third ionization energy.
We can get that from the Bohr model.
You can do it in your head.
You can do it in your head right?
It's just K times Z squared right here.
It's going from 1.
So if this is 2.18, it's going to be 9 times that trivially.
OK, so this is just 3 squared times K. And this when you have to get from the literature.
So you have to go to primary sources.
Which is why you're going to learn how to use the proper database.
And this one here also you get from the literature.
But this is on your periodic table.
Your periodic table, one of the data points it gives is the first ionization energy of all of the elements.
And so even though lithium normally is a solid at room temperature, the ionization energy for lithium as given on your periodic table is for this reaction.
It's for the gas.
Anyway.
So that's little aside.
All right, the last quantity that we could get from the Bohr model is v, the velocity.
And I solve for the velocity.
We don't talk about this very much.
But we're going to do so once today.
And here it is.
So I went through the algebra.
And you get nh over 2 pi mr, where r is the radius.
There's the quantum number.
This is quantized as well.
So I regrouped this.
And I already have a nice, cool expression for r in terms of the Bohr radius.
So I use that because that's on the table.
So this is 2 pi times the mass of the electron times the Bohr radius, 1/2 angstrom, times Z proton number divided by n, where n equals 1, 2, 3, and so on.
So let's again get a sense of scale.
So let's try for sense of scale.
Let's do velocity of the ground state electron in atomic hydrogen.
So that means Z equals 1, n equals 1.
So I plug in the numbers.
And I get v1 for hydrogen, atomic hydrogen, gives me 2.18 times 10 to the 6 meters for second.
I don't know.
Is that fast?
Is that slow?
I don't know.
But I do know this much.
I know that the speed of light is equal to 3 times 10 to the 8 meters per second.
So 10 to the 8 divided by 10 to the 6.
2 and 3, that's roughly 1, speaking as an engineer.
Who cares?
So this is about 1% of the speed of light.
That's pretty good.
That gives me something I can hang on to.
I would say that if this thing is zipping around at 1% of the speed of light, I would say that's relatively fast.
One more time, 1% of the speed of light.
That's relatively fast.
Remember last day I told you that Bohr simply dismissed the concept of the use of classical electrodynamics down to atomic dimensions.
Well here's another example of why a lot of these assumptions aren't going to work so well.
We're talking about the ground state electron velocity in this putative planetary model of a 1-electron atom.
You're already getting into relativistic effects.
So, just another example.
By the way, why do we use the letter c for speed of light?
It comes from the latin word celeritas, which means swiftness.
And we get the modern word acceleration, deceleration from that.
OK. All right.
So the Bohr model, we've now rolled it all out.
We have the energy portrait.
We have the radii, discrete.
We have quantization.
And we have velocities if we ever want to look at those again.
Now what's the next thing we do in science?
We compare the predictions of the Bohr model with data.
Are there any data to support this?
Because remember, all Rutherford said was plum pudding doesn't make sense.
Instead I'm going to concentrate the positive mass in the center.
And then Bohr came along and said, not only is it going to be a planetary model, I'm going to have circular orbits.
So now we've gone a long way from Geiger-Marsden.
So is there any data?
Well there were data in 1853.
Remember, Bohr published this in 1913.
In 1853 there was a spectroscopist by the name-- I'm going to tell you his name-- in Uppsala, Sweden.
And his name was Angstrom.
Angstrom was doing experiments on hydrogen in gas discharge tubes.
So he measured emissions from gas discharge tube.
And it was filled with various gases including atomic hydrogen.
And in order to take his data, what he used was this device here-- there's the Bohr radius, just an example.
He used the prism spectrograph.
So here's a gas discharge tube.
And I'm going to show you the physics of that in a second.
Basically you've got a pair of electrodes.
You've got gas in the tube.
And as this cartoon shows, you apply a potential across the electrodes.
And beyond a certain threshold potential, the tube begins to glow.
And the glow goes in all directions.
And a blinds you when you're in the lab.
So what you do, is you cover this up a bit.
You have a narrow slit.
And then you force the light to come through in a thin ribbon, and then expose it to a prism.
What the prism does, is it takes the light and breaks it into its components, sort of rainbow-like, and magnifies the difference.
As refraction goes, different wavelengths will refract different amounts.
And then you shoot this across the room.
There's two counters.
One is a scintillation screen and an army of graduate students who sit there in the dark.
But they're no good because you can't stick them into the publication.
You need to have data that people will rely upon.
So instead you use a photographic plate.
And if you put even a tiny, tiny angle of separation across a great enough distance, you start to get enough line splitting that you can see.
And then you go backwards.
And you know the geometry here.
And you can figure out what the wavelength must have been to go this distance, et cetera, et cetera.
And these are all color coded, not because they had color film in those days, but just to let you know that 656 nanometers, if you were the graduate student sitting there, you'd see a red line, a green line, a blue line, and a violent line.
So that's how he made the measurements.
And he published those measurements.
And so they lay.
And then the story gets a little thicker.
In 1885, there's a Swiss high school math teacher.
I mean, I can't make this stuff.
This is true story.
There's a Swiss high school math teacher by the name of J. J. Balmer.
We had J. J. Thomson.
Now we've got J. J. Balmer.
And J. J. Balmer, he loved to play with numbers.
And he was studying this set of lines.
And he was trying to come up with a pattern.
Can you see a pattern there?
410 434 486, 656, do they go squares?
Are they primes?
What's the pattern there?
So Balmer puzzled over this for awhile.
And he finally came up with the equation to represent those lines.
So he studied Angstrom's data found the pattern.
And here's the pattern that he found.
He said that those are wavelengths.
If I take instead wave number, nu bar is called wave number, which is the reciprocal of the wavelength.
So if I take the reciprocal of the wavelength, I end up with those 4 lines conforming to a series that goes like this.
1 over 2 squared minus 1 over n squared, where n equals 3, 4, 5, 6.
And there's a constant here which we're going to designate r. And the value of R-- I'm going to put that on the next board-- the value of R as expressed in SI units today, would be 1.1 times 10 to the 7 reciprocal meters.
Wavelength is a meter.
Reciprocal wavelength or wave number must be reciprocal meters.
So how this all of this support the Bohr model?
Well in order to explain it, I've got a first tell you what the physics of the gas discharge tube are.
So let's go inside the gas discharge tube and understand those physics.
So here's the gas discharge tube.
It's made are borosilicate glass.
And we fill it with gas.
And in this case, the gas is going to contain among other things hydrogen.
So this is a hydrogen gas phase atom.
And this is probably at low pressure.
And then I said I need electrode.
I'll get the electrodes inside.
So I've got to have a really good glass blower who can make a glass to metal seal, and have a feedthrough to an electrode.
So this is still a vacuum seal.
How do they get the gas in the first place?
We don't show you this in the books.
I'll tell you because I did this in my Ph.D. What you do is you have a little side tube here.
You evacuate.
This goes to a vacuum pump.
And then over here you have a gas source.
You evacuate.
Put in the gas to whatever pressure you want.
And then the glass blower disconnects.
This pulls this down to a reduced pressure, and seals it.
But the books don't show you that.
That's a secret.
So we've got a gas at low pressure.
And I've got an electrode over here and an electrode over here.
And they're connected to a variable voltage power supply.
So I'm going to put an arrow with the V meaning it's variable voltage.
I can change the voltage.
And this convention, this is the negative side.
So this means the electrons leave the power supply and go like this.
Which means this electrode will be negative.
And this electrode will be positive.
And thanks to Michael Faraday, we will call this one the cathode.
And this one we will call the anode.
Now what happens?
We start turning up the pressure, turning up the voltage rather.
Low pressure here, but electrical pressure is voltage.
The voltage gets high enough, eventually the electrons will boil off the cathode.
And this is a gas at low pressure.
And they will accelerate from rest and go all the way across the tube and crash into the anode.
And we complete the circuit.
But if there's a gas in here, some of these electrons are going to hit the gas molecules.
And when they hit the gas molecules, if they have enough energy to do so, they will cause electrons inside the gas molecules to be excited.
And if they're excited enough, the electrons will jump up to a higher energy level.
But they can't be sustained.
Because this is a ballistic collision.
It's a one of.
It's like a bowling alley.
One ball, one pin, one impact.
Now the pin is in the air.
What happens to the pin?
It falls back down.
Why?
Because gravity pulls it down.
In this case, you've got the energetics pulling the electron back down.
Now the electron goes from high energy to low energy.
And when that happens, the energy difference is given off in the form of a photon.
So I get photon emission when this falls back down.
And this photon has a wavelength.
What I'm going to show you is that the set of lines that you get from exactly this configuration using this equation give you the Balmer series.
So now you've got a model that Bohr postulated for atomic hydrogen on 1-electron atom that exactly predicts that set of 4 lines.
Which were measured 50 years before.
So let's go.
So first of all let's get the energy here.
And I'm going to get the energy of this electron.
This electron I'm going to call a ballistic electron.
Why do I call it a ballistic electron?
Because it's not bound.
It's free.
It boils off the cathode, flies through free space, and crashes into an anode.
Clearly it's not part of an atom.
But there's a second electron in this story.
And it's the ground state electron in hydrogen.
And it lives here.
So what's the energy of the ballistic electron?
Well that's just 1/2 mv squared And where did it get its energy from?
It got its energy from the power supply.
And what's the electrostatic energy?
It's the product of the charge on the species times the voltage through which it was accelerated.
So away we go.
I know the charge on the electron is minus E. Whatever the voltage is there, 1 volt, 10 volts, 100 volts, whatever.
Away we go.
By the way, I'm going to show you just one other thing in terms of order of magnitude.
The kinds of voltages you see along here are 1 volt, 10 volts, that sort of thing.
So suppose, to get an order of magnitude, suppose we had a species of charge E. So in other words, it's only 1 times the elementary charge.
A species of charge E influenced by voltage of 1 volt.
So this is 1 in the voltage units, and 1 in the elementary charge units.
How much energy would that be?
That's equivalent to making 1 volt and accelerate an electron from rest across this gap.
And the result would be, the energy then would simply equal 1.6 times 10 to the minus 19 coulombs times 1 volt.
And what's the energy going to be?
Well I've got coulombs times volts.
And I don't know how I convert one to the other.
I don't have to.
Why not?
Because that's an SI unit.
And that's an SI unit.
This is an energy.
So with impunity, I write 1.6 times 10 to the minus 19 joules.
That's the beauty of SI units.
So that's a good news.
I know it's joules.
The bad news is I hate this number.
It's a stupid number, 1.6 times 10 to the minus 19. it's crazy.
Why don't I come up with a number like 3, 7?
So what I could do, is I could define.
I could define a unit such that when the elementary charge is accelerated across the unit voltage, I would call that unit 1 electron volt.
And so somebody thought of this before me.
And hence, this is the unit of the electron volt.
It takes these crazy things that we've been spewing here up until now, and rationalizes them into numbers that people can carry around in their heads.
So what's K now?
K is 2.18 times 10 to the minus 18 joules.
Yuck!
Let's convert that to electron volts.
So I divide by 1.6 times 10 to the minus 19.
And I got 13.6 electron volts.
You'll remember that on your death bed, ionization energy of atomic hydrogen.
Maybe we don't have to give out tables of constants.
You just know this stuff.
It's OK. All right.
Now one last thing about this.
So this has got gas in it.
This is the cathode.
And this beam of electrons, back in the 1800s, there was a popular term, it was called the ray.
So instead of a beam of light, people refer to the ray of light.
So then when they got to particle beams, they talk to them as rays.
So this is now not an electron beam, it's an electron ray.
And it comes off the cathode.
And it's in a vacuum tube.
So this could be called a cathode ray tube, a CRT.
Now see, I could flatten this.
And I could spray it with phosphors.
And then I could put some charge plates here.
The electrons have a negative charge.
So if I charge these plates, and I was clever about how I charge them and varied the charge, I could raster the electron beam like this about 30 times a second all up and down the screen.
And then I could put some program signal in there.
And I could sit here.
And I could watch TV.
It all started with the gas discharge tube.
It says nothing about the content unfortunately.
Very nice physics, but no content.
All right.
So now we've got the electron.
The electron is moving, the ballistic electron.
Now I want to look at what happens when the ballistic electron smashes into one of those hydrogen atoms.
So let's go back over here.
So here's the incident particle.
And it's going to be an electron in this case.
So I'm going to designate this electron.
This is my ballistic electron.
So here's the incident electron.
And this is ballistic, just to be clear.
It's the ballistic incident electron.
Now this is a mixed metaphor here.
Because I'm representing this in Cartesian space.
But I've moved into energy space.
So some people are going to get really upset.
Because they're going to say, well this is Cartesian, but this isn't.
It doesn't matter.
It's my lecture.
It's my model.
It works.
I'm the professor.
So we've got this mixed metaphor here.
But anyways, it helps a lot.
So what happens when this thing comes in?
It depends on how much energy it has.
Now if the incident energy, if E incident is tiny, nothing happens.
This thing just zooms right on through.
But if the incident energy, if E of the incident ballistic electron is greater than delta E for any transition that's feasible-- and in this case I'm going to assume I don't have thermal distribution of electrons.
If I gave you Avogadro's number of hydrogen atoms because of the thermal distribution of energies-- and we'll come back to this later-- there might actually be, at any moment, some electrons that are thermally excited above the ground state.
We're going to forget about that for now.
We're going to spiral up the learning curve here.
So first time we're going to assume all the electrons are in the ground state.
If I don't enough energy to go from n equals 1 to n equals 2, nothing happens.
If I have more than enough energy to go for n equals 1 to n equals 2, I will take that energy.
And the electron will jump, steal that amount of energy, and then this thing moves on-- and I'm purposely making this vector shorter than the incident vector-- with that amount of energy raw.
And this is called the scattered electron.
Go back to the bowling ball analogy.
The bowling ball comes in with a certain energy, hits the pin, continues to roll.
But you know that there's a loss of kinetic energy in the bowling ball.
That's what we're seeing here.
It's purely ballistic.
Now let's say we do have more than enough energy.
Suppose I have enough energy to go from n equals 1 halfway between n equals 2 and n equals 3.
There's only n equals 1, n equals 2.
I can't take the electron up to n equals 2.3.
It's unallowed.
These are the only allowed states.
So that differential amount of energy then resides with the electron that's ballistic.
And it moves on here.
So we've got conservation of energy here.
We can say that E incident will then equal the energy that's lost in the transition plus the energy that's still left with the scattered electron.
And we can calculate what that transitional energy is.
That transitional energy is going to equal 1/2 mv squared incident.
This is the velocity, the incident electron.
What's the energy to go from n equals 1 to n equals n?
Whatever it is, its minus K times Z squared If it's hydrogen, Z is 1.
It's going to be 1 over nf.
The final quantum number squared minus 1 over the square of the initial quantum number.
And then, what's left over after this has been robbed from the incident ballistic energy is 1/2 mv squared of the scattered ballistic electron.
And the quantization dictates that only if E incident is greater than delta E going 1 up to n-- here I'm assuming everything is in its ground state.
Later on we're going to be more sophisticated.
But for first time through, all are ground state electrons.
I have to have enough energy to go at least to n equals 2.
I can go to n equals 3, m equals 4.
In principle, if this thing had more than 13.6 electron volts, what would happen?
It would kick this electron out.
Gone!
And you'd have 2 free electrons.
So if it's greater than this, then the consequence is electron promotion.
So we're moving along.
But this excited state is unstable.
This excited state is unstable because it's like the bowling pin that got thrown up.
So what happens?
The electron standing up there on n equals 2, for example, looking down to n equals 1.
And it falls.
When it falls it gives off radiation.
And that radiation is conservation of energy there.
And what we know is when it gives off an energy of the emitted photon, the energy of the emitted photon must equal delta E of the transition falling from 2 to 1.
And we know how to calculate that.
That's just that thing flipped around.
And this thing is equal to what?
This is equal to h nu.
According to Planck it's hc over lambda is equal to hc nu bar.
You know what this one is.
This is minus KZ squared 1 over-- in this case it's going to be-- 1 over nf squared 1 over 1 squared minus 1 over, in this case, 1 over 2 squared And you can generalize this 1 over nf.
So where am I going with this?
Well I'm going to flip all of this around.
And when I flip it all around, what I'm going to end up with is this equation here.
And better than that, I'm going to end up with this equation.
And I'm going to end up with this as the constant.
And when I get that, we're going to say Bohr has done it.
The data support the theory.
So that's what we're going to do.
But I think we're going to stop at this point today.
So let me just jump here.
I mentioned to you that if you go here on your periodic table, there's the 13.6 electron volts.
In the case of lithium, this is 5.4 electron volts.
So you can see all the various values.
This by the way, people no noise.
No noise.
It's 11:52.
I'm holding court until 11:55.
I'm simply changing topics.
It's not, oh this is the part where I can talk to my neighbor.
What are the rules?
No talking.
No food.
No horseplay until 11:55.
Then, still no horseplay, gentle talking, no food, no drink.
This is a sacred space.
I'm not kidding you.
Do you know why?
In this secular America, this is sacred space because this is where people learn.
The lecture hall is sacred space.
Now, here's the 13.6 electron volts.
And there's the 1.6 times 10 to the minus 19 joules.
And if you multiply those 2, you'll get the 2.18 over here.
All right.
Here's a cartoon showing the photon, higher energy orbit, lower energy orbit, electron emission transition, right out of your book.
And there's a postulate 6.
And we're going to finish this up at the beginning of the Friday lecture.
So here's the whole series from n equals 3 to 2, n equals 4 to 2, and so on.
See how that works.
OK. One of the things that we've learned here, is that it doesn't matter what the incident energy is here.
The emission is characteristic of the energy levels inside the gas.
Instead of using an incident electron, I could use an incident proton.
I could use an incident alpha particle.
I could use an incident neutron.
Anything that has enough energy to kick this up from n equals 1 to n equals 2 will result in photon emission of this frequency.
That means the set of those lines is unique to the target.
And this is the beginning of chemical analysis.
I can use this to characterize species, to it, stars.
How do we analyze the composition a stars?
Well, do we send a NASA spaceship out 25 light-years and grab some gas and bring it back to the lab?
No.
All we've got is the spectrograph.
The star is hot.
That means thermal excitation and cascading down with photon emission.
And the lines we get are related to the energy levels within the stars.
So from a distance of a 100 light-years, I can tell you what the composition is.
And if there's two gases there, what if there's hydrogen and helium?
the helium lines will be there.
And they'll be superimposed on the hydrogen lines unless they lie directly on top of one another.
I'm going to be able to figure out what's there.
That's how it works.
Now here's a story about an astronomer, Cecilia Payne.
She's the first woman graduate student in astronomy at Harvard.
She went on to chair the Faculty of Arts and Sciences, awarded tenure, but denied a professorship for 18 years because she was a woman.
And in her thesis, she was the first person to figure out to the sun is dominantly hydrogen, not iron, which is what most astronomers thought.
Why?
Well, because the earth is made of iron.
Meteorites are made of iron.
The whole universe must be made of iron.
Never mind the fact that the sun is glowing.
So here's how spectroscopy works.
Look at this.
See, what does this mean?
This is an analogy.
What could that mean?
Well, they said iron again.
I know this is misspelled.
But maybe it's just a glitch in the instrumentation.
So most people would look at that and say, the message is iron.
But it's not about the word.
It's about the pattern.
It's about the pattern.
And that's what's spectroscopy is.
By the way, if you somehow didn't catch this lecture.
And you walked in, and all you saw was those four lines.
You know those four lines.
That set of lines is characteristic of atomic hydrogen, and nothing else.
By the way, those lines are very faint.
This is not to scale.
And they're so faint, they're ghostlike.
And what is the latin word for ghost?
Specter.
So what is a spectrum?
It is a set of ghostlike lines.
To this day, the term spectroscopy refers to the ability to study data that are so faint they're ghostlike.
All right, we'll see you on Friday.
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All right, a couple of announcements.
Next week two minor celebrations, maybe receptions we would call them.
Quiz 2 Tuesday based on homework 2, and periodic table quiz Thursday.
I provide the numbers, you provide the letters.
I guess that's how it works.
And the contest ends Friday five PM.
Last day we looked at the Bohr Model and we developed equations for the radius of the electronic and the orbit of the one electron atom.
The energy of the electron and the velocity of the electron.
And we found that for all of these they were a function of n. Quantum number.
n takes on discrete values.
One, two, three, and so on.
We say that these energies radii velocities are quantized.
They take discrete values.
And then later in the lecture we started looking for evidence.
And we found ourselves in an exercise of reconciliation with data taken by Angstrom about 50 years earlier and fit to an equation by J.J. Balmer.
And we were part way through that and adjourned.
So I'd like to pick up the discussion at that point.
I've done a different drawing of what's going on inside the gas discharge tube.
Last day I had the ballistic electronic here, and this is boiling off the cathodes.
The cathode is inside the gas discharge tube and this electron, if the voltage is high enough, will leave the cathode and shoot across this low pressure gas, which contains, among other things, atomic hydrogen.
And I'm trying to depict the atomic hydrogen atom here.
Here's the proton, which is the nucleus-- that's the sum total of the contents of the nucleus-- and here's the lone electron that is orbiting the nucleus at some initial value.
n sub i.
Could be ground state.
Doesn't necessarily have to be.
With some thermal energy this could be n greater than one.
Then we reason that if the electron, ballistic electron, and it's trajectory across the gas discharge tube over to the anode, which is charged positively.
If it collided with this electron it could impart some of its energy, thereby promoting the electron from ni up to nf, the final level.
And the electron would be up here.
And for this transition there would be an energy cost.
That energy cost is delta e. Delta e is the energy to go from ni to nf.
And so the kinetic energy-- half mv squared of the incident electron-- is diminished by this amount.
And the electronic continues on it's merry way at a slower speed.
We assume it's mass doesn't change.
The only way we can change its energy is to slow it down.
And there's a conservation of energy, so the sum of the energy of the scattered electron and the transition energy of the electron within hydrogen must equal the incident kinetic energy of the ballistic electron.
But there's more.
This is a-- no pun intended-- a one shot deal.
This is ballistics, and so the electron is not sustainably promoted.
It falls back down.
And when it falls back down we have the transition energy now given off.
Here, to promote we had to call for energy.
When the electron falls down it gives off that energy.
And that energy is given off in the form of an emitted photon.
And it's that emitted photon and it's that emitted photon that ultimately gives rise to the lines.
The lines in the spectrum are generated by the emitted photon here.
Everything else is preamble to this event, and this event gives rise to the emitted photons.
And I think that's about where we got last day with the reconciliation.
So let's look carefully here.
We recognize that there needs to be conservation of energy again.
In other words, the energy of the emitted photon, the energy of the photon, which we know from Planck is h times nu.
I'm trying to distinguish.
This, I'm making nu, the Greek symbol nu.
And I put a little descender on it.
It looks like a v, but I put a little ascender here to distinguish it.
This is lowercase v, as in mv squared.
This is nu.
So h nu, or it could be hc over lambda.
Or it could be hc nu bar.
Three ways of writing the energy of the photon.
And that must equal delta e of the transition.
So let's keep going.
We know that the delta e, the transition is given by the Bohr Model.
Delta e transition will equal e final minus e initial, which will be minus kz squared-- I'm writing this generally, in this case with atomic hydrogen z as 1-- with minus k times 1 over n final squared minus one over n initial squared.
What we can do then is equate these and roll them around to isolate nu bar.
nu bar, then, equals minus kz squared over the product of the Planck constant, the speed of light, 1 over nf squared minus 1 over ni squared.
Now for the Balmer series, that is to say the Balmer series of lines, that turns out to be a series where all of the transitions end up on n equals 2.
We said nf equals 2. z equals 1 because we're talking about atomic hydrogen.
Then what we have is a set of translations that go 1 over 2 squared minus 1 over ni squared.
ni must be greater than nf, so ni must come from the set 3, 4, 5, et cetera.
Furthermore, I'm going to put z equals 1.
Let's evaluate k. We know that's 2.18 times 10 to the minus 18 joules, or 13.6 electron volts.
And we know the Planck constant, 6.6 times 10 to the minus 36.
And this is 3 times 18 of the 8th all in SI units.
So this gives me 1.1 times 10 to the 7th in reciprocal meters.
And if I put all this together I end up with exactly they equation that was published by Balmer.
Exactly Balmer's equation in 1885, rewritten to express it in SI units.
1 over 2 squared minus 1 over-- I'm just going to put ni squared-- or ni equals 3, 4, 5, 6.
This is Balmer exactly.
Balmer exactly.
So the assumption of this planetary model, with all of the restrictions that Bohr placed on it in order to get this set of equations, reconciled with laboratory data.
Very, very significant.
So here we are.
Those four lines all can be derived from the Bohr model.
And here's another cartoon from your book and showing what I'm trying to depict here, namely it's the falling down, the return to the state from which the electron was promoted that generates the photon.
And the set of those lines is what gives you this.
There so there's the validation of the sixth piece.
So Bohr Model agrees with Angstrom's data, but it also suggests other experiments.
Let's think about this for a second.
OK, here's another cartoons from your thing.
You know, I told you that this thing is unstable.
And in the Balmer series it goes from n equals 2 up.
But there's a ground state, n equals 1.
What was wrong with those electronics in Sweden in 1853 than Angstrom could never find any electron that would fall all the way down to the ground state?
What's wrong with it?
Well, here's the answer.
It has to do with instrumentation.
So this is an example where science goes further thanks for the advent of new instrumentation that allows this to make measurements that previous people couldn't make, even though they were very competent experimentalists.
Angstrom could have found n equals 1 series, but couldn't see them because he was using a photographic plate.
This shows you the range of sensitivity for photographic plates.
Here's the electromagnetic spectrum.
Out here you have low energy radio waves and up here you have x-rays and gamma rays and so on.
And the visible spectrum is parked right here in the middle, and here it is unpacked for you.
And it roughly runs from 400 to 700 nanometers.
That's the invisible spectrum.
So wavelength increasing from left to right, which means energy frequency in wave number increase from right to left.
They're complimentary, right?
e nu, nu bar on the top, lambda is on the bottom.
Some spectra are plotted in lambda, some are plotted in wave number, whatever.
And by the way, I want to show you the power of knowing a few things.
I don't expect you to know a lot of facts, but I expect you to know a few things.
Every educated person ought to know that the visible spectrum runs round numbers 400 to 700 nanometers.
But look, I can take 700 nanometers and use this formula and convert it to energy.
And I'm going to get something like 3 times 10 to the minus 19 joules.
Yuck.
Instead, I go in electron volts.
1.8 ev.
Over here, 400 nanometers is 3.1 ev.
So round numbers, the visible spectrum spans 2 to 3 electron volts.
Our eyes are photo detectors that operate on the band width 2 to 3 electron volts.
that's easy to remember.
Those are good numbers.
So where does that leave us?
It leaves us here.
We go back and we see these numbers, 656, 486, 434, they're all in the visible spectrum.
So I went and I did a little calculation.
I said, well what would I have if we'd gotten the wave length for the transition from 2 down to 1?
This is n equals 2 down to ground state.
If you plug in the numbers to the Bohr model, you'd find that that would give you 122 nanometers.
122 nanometers?
Well, 122 nanometers is going to put you way over.
It's too high energy, right?
122 nanometers is going to put you off to the left there into the ultraviolet, where the photographic film was not sensitive.
So he couldn't measure those lines.
So now I'm going to end by putting the master equation that captures all of this.
And the master equation that captures of all of this is here.
It's that nu bar goes as r times z squared, 1 over nf squared minus 1 over ni squared.
So this is the most general form for all 1 electron atoms.
That's why I've got z squared in there for all 1 electron atoms.
And this is called the Rydberg equation.
Named after another Swedish spectroscopist at the University of Lund.
I think a Swede would probably pronounce this something Rydberg, but you don't have to say that.
You can just say Rydberg and it'll be fine.
And in honor of Rydberg, the constant here is given the symbol, capital R. The capital R as the Rydberg constant and it has a value of 1.1 times 10 to the 7th reciprocal meters in good ST units.
Well, there was more evidence for the support of Bohr's Model.
More evidence for the support of Bohr's Model.
By the way, as the detectors got better and better we could get more and more lines.
You see these, as you get higher and higher series ending on higher and higher end numbers, you move off into the infrared.
Because this is not to scale.
These n equal 4, n equal 5 are closer and closer to closer together in terms of energy.
They're farther and farther apart in terms of spacing, but they're closer and closer together in terms of energy.
Because they're farther from the nucleus.
You say, gee, shouldn't it cost more energy to go farther?
Uh-uh.
Because you're farther from the positive nucleus.
Be careful.
Don't let your intuition send you in the wrong direction.
It's all about Coulombics.
Anyway, so the Lyman series ends at n equals 1.
And these are different scientists.
Paschen, Bracket, Pfund, Humphreys, and so on.
So maybe, I don't know, if somebody hasn't claimed n equals 214, all the lines that end there, you know, maybe that could be your name on the series.
As if anybody cares.
Looks like this quantum condition is validated.
See this is really important because this was the big break away from classical theory.
That the motion of a body, something with mass, could be quantized in its behavior shook this physics community.
But this reconciliation of the data says that assumption is valid.
There's more that happens.
So in 1913 in Berlin-- remember 1913 is when Bohr published the paper-- 1913 in Berlin there was James Franck and Gustav Hertz.
James Franck and Gustav Hertz.
And they were conducting experiments on gas discharge tubes.
Only they filled a gas discharge tube, instead of with hydrogen, they filled it with mercury vapor.
So gas discharge tube-- GDT-- gas discharge tube containing mercury vapor.
The same thing.
Put the electrodes, connect on a power supply, and started varying the potential.
So I'm going to show you what they found.
This is the Planck voltage, and this is the current between electrodes, or if you like, across the tube.
Between the electrodes, or through the tube.
Or if you like, tube current.
Meaning from one electrode to the other.
The tube current.
Well, low voltage, low current.
High voltage, high currently.
They get up to a certain value of voltage, all of sudden the tube starts glowing blindly and the current falls to 0.
Then they continued to raise the voltage.
More voltage, more current, up, up, up, up, up.
And then they get to another critical value of voltage, even more intensity.
And then the current falls to 0.
So you look at those data and say, well, what's that got to do with the Bohr Model?
Because mercury is not a 1 electron atom.
It's got a boat load of electrons.
This is not a 1 electron atom.
So you say, I know what it is.
It's ionization energy.
Must be ionizing the mercury.
So you go to the periodic table and you look up the ionization energy of mercury and you discover that that's 10.4 volts.
10.4 electron volts is the ionization energy.
And this first null is at 4.9 volts.
Well, 4.9 is a long way from 10.4.
And this second null occurs at 6.7 volts.
6.7 volts.
So what's this telling us?
What this is telling us is that when you get to a value of 4.9 volts, you've hit a certain value that allows you to promote electrons within mercury between one level and the next level.
And those electrons are being promoted and then cascading down.
And they're cascading down and they're emitting in the visible and it's blinding you.
Say, OK, so what does that mean?
Well, it means that the Bohr Model, which is for a 1 electron atom, assumes that energy levels within it are quantized.
These data indicate that on the basis the behavior of this gas discharge tube, there must be quantized energy levels inside of mercury, which means all atoms have quantized energy levels.
You understand?
Everything is quantized.
That's really powerful.
It starts off with this nerdy little 1 electron atom, and now he's applying it across matter.
And this is gas, this is more elaborate gas.
Heaven forbid, it might exist in liquids and solids.
So that's the Franck, Hertz experiment.
So his stock goes way, way up as a result of that.
And they win a Nobel Prize.
Here's James Franck.
Here's Gustav Hertz.
You know what the Hertz is.
200 kilohertz, so on.
That's Hertz.
James Franck was at Gottingen when he won this, but he ultimately came to the United States when the political changes started occurring in the thirties in Germany.
Franck decided to seek safer surroundings and ended up at the University of Chicago, where there is to this day the James Franck Institute of Physics.
Very, very high-end physics institution.
So this is good.
But all good things come to an end.
So 1913 was a bittersweet year for Bohr.
Because he got some good news, but he also got some bad news.
So now I want to move over to limitations of the Bohr Model.
Limitations of the Bohr Model.
So I know what you're going to say-- well, it only talks about 1 electron atoms, so that's a limitation.
No, there's more to it than that.
Even the 1 electron atom model doesn't capture everything.
I'm going to summarize the limitations.
I'm going to show you three, and they all fall under the general umbrella of fine structure.
Fine structure.
In other words, the Bohr Model is good, give us the big lines, but when you start looking more carefully it fails to capture some of the physics.
So first of all, let's go back to some earlier data.
1887.
1887, there were already data out there that were going to give heartburn to the Bohr Model.
And those data were taken by Michelson and Morley.
Michelson and Morley.
Everything I've taught you so far, with one exception, has been European science.
Americans were not active in science because this was a young country.
We were really good engineers because we were blockaded by the rest of the world.
We had to live by our wits-- that's where you get the term "yankee ingenuity."
Science was hifalutin stuff.
We didn't have time for it.
But towards the latter half of the 19th century, we started moving into fundamental science.
The first American to win the Nobel Prize was Michelson.
Michelson was doing work at Case in Cleveland, which eventually became Case Western Reserve University.
So he was at Case in Cleveland and he was studying optics.
And he was a brilliant experimentalist.
In fact, he made the first reliable measure of the speed of light.
Back before 1900.
What they were doing is they were looking at Angstrom's lines and they noticed something peculiar.
If you take a look at even this drawing, you notice the red line is a little bit fatter than the others.
Now you might just say, well, that's just the artist taking liberties and somebody didn't catch it in proof reading.
But in point of fact, what he found was that if you look at that line, which is really the line for the 3-2 transition-- the 3-2 transition in the Balmer series-- what you find is that if you look at the photographic plate more carefully, you find that this thing in fact is a pair of lines, but very, very closely spaced.
This is known as a doublet.
Two lines very closely spaced, centered at 656 nanometers.
And with his interferometer he gets super, super good data.
And he could split the doublet.
Well, what's that mean for Bohr?
Bohr has no way of explaining this.
If you look at the Bohr Model, you've got n equals 2, you've got n equals 3, alright?
So this is energy 2, energy 3, right?
And so when the electron falls from 3 to 2, we get a photon of a certain value.
It's going to be nu 3 to 2.
That's the frequency or wave number, what have you.
Now, the fact that you've got a doublet here means that there must be two transitions, but darn close.
There's either a 3 and a 3 primed, or there's a 2 and a 2 primed, but it's not simply 3 and 2.
So that piece of information runs counter to the Bohr Model.
Bohr Model is silent about it.
It gets the big picture, but if you look more carefully it can't capture the doublet.
And Michelson ultimately gets the Nobel Price.
And I think I've got him here.
There he is.
The Nobel Prize.
By the time he got the Nobel Prize he was at the University of Chicago, but he did the work that won the Nobel Prize for him at Case.
So sometimes when you see even Millikan, Millikan did his work at University of Chicago, but eventually took a position at Caltech.
So the Nobel Prize says, Robert Millikan, Caltech.
But he didn't do that work at Caltech.
He did it at Chicago.
Anyways, you can go to the Nobel website.
You can read about these people.
And what's really cool is when you win the Nobel Prize-- you notice I didn't say if-- I say, when you win the Nobel Prize, what you do is you get on an airplane, you go to Stockholm, and then you go and you have dinner in this beautiful hall.
I've been there and it's gorgeous, gilded and so on.
Very nice kitchen, excellent wine list.
And-- yes-- and you can go there and they serve meals.
the menu is taken from previous Nobel Prize dinners.
So you can sit and-- whatever it is, it could be the Nobel Prizes of 1927 and that's what's going to be on the menu today-- and after the dinner they have a presentation ceremony with the King of Sweden.
You get your Nobel Prize, and then people listen to your lecture.
And those Nobel lectures are really, really expository.
So if you want to go and read the Nobel lecture that Michelson gave on the occasion of winning the Nobel Prize, you'll probably learn all of about this and more.
It's really, really good, so go there and read.
Now back to the story.
Second problem with the Bohr Model.
1896-- see, all this data had been accumulating-- 1896, there was a postdoc by the name of Zeeman.
Piet Zeeman.
He was a postdoc at Leiden.
Leiden in Holland under Lorentz.
You'll learn about the Lorentz force when you study 802.
What he was doing-- again, gas discharge tube.
So this was gas discharge tube, and what Zeeman was doing on his postdoc was in a magnetic field.
These people were doing all sorts of experiments.
They were trying to block out the whole experimental space.
So one guy, his specialty is high energy.
One guy's specialty is low pressure.
These people are taking a gas discharge tube and putting in the jaws of a powerful, permanent magnet and then measuring the spectrum.
And what he found was that for certain lines, this was the rest-- b, I'm going to use as magnetic field-- in the absence of applied magnetic field you have a line.
And this is not a doublet, triplet-- it's just a plain old line.
Well behaved line.
But when they take that gas discharge tube and put it into a magnetic field, they see a plurality of lines.
And furthermore, the spacing-- I'm going to use c here-- the spacing in the lines is proportional to the intensity of the magnetic field.
No magnetic field, single line.
Modest magnetic field, modest amount of what is called line splitting.
So a modest amount of applied magnetic field, modest splitting.
Intense magnetic field, intense splitting.
Bohr Model is silent about that.
Because you know, if you've got different lines, it means you must have different energy levels.
It's as though the energy level diagrams opens up in a magnetic field.
The Bohr Model can't account for that.
And parenthetically, they got the Nobel Prize, too.
So there's Piet Zeeman.
Got his PhD in 1896.
He's got his Noble Prize, 1902.
He's off to a good start, I'd say.
And there's Lorentz.
Two of them.
We'll get to him in a second.
So third piece of bad news for the Bohr Model.
And that comes, again, in 1913 in November.
In November of 1913 there was a man by the name of Stark in Germany.
And Stark was doing analogous experiments.
He was studying gas discharge tube in electric fields.
Obviously, you've got an electric field across the electrodes to excite the electrons.
But he's taking a whole gas discharge tube and putting it between flights and then applying an electric field.
And what did he find?
He found the same sort of thing.
He got line splitting in an E field, and furthermore that extent of splitting, extent dependent upon the intensity.
E intensity.
So no field, no splitting.
Modest field, modest splitting.
Intense field, intense splitting.
Well, again, that's a headache for the Bohr Model.
So this is all three problems, and it's all under aegis of fine structure.
So we know the Bohr Model has its limitations.
OK, Stark.
I know he's got his Nobel Prize.
There he is.
So 1913 ends on a sour note.
But people don't give up.
1916, Arnold Sommerfeld in Munich.
He was a professor of physics and he proposed modifications.
Modifications to Bohr Model.
It's a patch, we would call it a patch.
going?
To put a patch on the Bohr Model.
And what's he going to do?
What's the gist of his idea?
Well, he retains the planetary structure.
He liked that idea-- nice orbits, so on.
But he took a page out of Kepler's book.
The planets in the Kepler model, when they revolve around the sun their orbit is not circular.
It's elliptical.
So Sommerfeld said, why don't we give that a try?
What if we said the electronic orbit can be elliptical or circular?
And he was quite specific.
He said, suppose-- and this, again, is not to scale, but to emphasize this is going to be elliptical or circular, but very, very mild eccentricity.
What I'm going to draw for you is extreme eccentricity to make a point.
But suppose we had the circular orbit as I'm drawing it now, and then we had an elliptical orbit that is centered on that circle.
So it's mild eccentricity.
We might have another one-- let's do one more.
This is good enough.
The gist here is that we have a circular orbit and an elliptical orbit, but the bandwidth here is very, very narrow.
So this is very, very thin.
And it's sort of like an egg shell.
So if I asked you, what's the dimension of an egg?
You'd say, well, it's dimension of the surface of the egg.
Then I'd say, but the egg shell has some thickness, right?
But that thickness is relatively small in comparison to the total dimension of the egg.
So an analogy.
He said that the range of distance from the nucleus, whether it's circular or elliptical, is very, very narrow.
So we can say the set of circular and elliptical orbits lie within a shell, as in egg shell.
So this is a shell model.
It's a shell model.
So now how do you designate the different orbits?
You've got some that are circular, some that are elliptical.
He needs to distinguish them and he needs to be able to label them.
So he introduces new quantum numbers to allow us to name them.
So let's go and take a look at the quantum numbers that Sommerfeld gave us.
So he starts off with n. He retains that from the Bohr Model and he calls that the principal quantum number.
And it's primary attribute is size.
It captures the distance, the principal r from the nucleus.
And it takes values 1, 2, 3, all the way up to infinity.
So n equals 1, small radius.
n equals 10, large radius.
Oh, by the way, there's another numbering system.
This is what we use, but the spectroscopists use letters.
The spectroscopists use letters-- why?
Because remember the Balmer series?
Everybody was hooked on the Balmer series and it ended up being n equals 2.
And then later with better detectors we find there's an n equals 1.
So the spectroscopists said, we're going to get fooled again.
So we're going use letters.
And we're going to start with the letter k. It's in the middle of the alphabet.
That way if we discover even lower energies, we've got some head room here, we can label those.
But we never found any.
So if you go over to Building Thirteen and you do some x-ray refraction and you use the line that emanates from a copper target, n equals 1-- it's called the k alpha line of copper.
To this day.
So k, l, m, and so on.
You can't get to infinity, obviously.
You know, I didn't think this thing through.
Now the l. l is, what's his name?
Sommerfeld.
And it's called the orbital quantum number.
Why?
Because he said that the electron is in an orbital instead of an orbit.
Orbit is Bohr, orbital is Bohr-Sommerfeld.
And it speaks to the shape.
Somehow, I've got to distinguish between elliptical and circular.
And it takes values 0, 1, up to n minus 1.
So the n number controls the range of l. And again, the spectroscopists, they're real number weenies, they're afraid, so they use s, lowercase.
See this is uppercase, this is lowercase.
s, p, d, f. For sharp, this is the sharpest line from the l equals 0-- then the principal because as you go to z they all seem to converge and look like hydrogen-- d is diffuse, f is fine, and then after that they ran out of ideas so g and h. So you'll talk about the one s-orbital, meaning n equals 1, l equals 0.
And there are some values here for shapes.
I'm going to put that right above it.
When l equals 0, l equals 0 means you have a circular orbit.
And when l equals 1 it's elliptical.
And when l equals 2 it's much more complex, and we'll just leave it at that.
1, 2, and 3.
So there's l values.
And then m is the magnetic quantum number.
And it talks about orientation.
I'll show you what I mean by that in a second.
The values are governed by l, which is governed by n. Starts at l, l minus 1, goes down through 0, goes to minus values and ends at minus l. So for example, we could do something like this-- when n equals 1, then l most equal 0, so therefore m must equals 0.
So this means for n equals 1, it's only a circular orbit and this thing is going to be immune to line splitting in a magnetic field.
When n equals 2, l can equals 0 or l can equal 1.
When l equals 0, m equals 0.
That's boring, that's circular.
But here's another possibility.
And that is, when l equals 1 then m can equal 1, 0, and minus 1.
Now I said it has something to do with orientation.
Most of quantum mechanics doesn't translate into the Cartesian world, but this one does, mercifully, and I think it's a cute analogy.
If I were to tell you that I've got three different quantum numbers and I've got an elliptical thing-- and one way to think, see, the number 0 looks like a circle and the number 1 has some asperity associated with it, so you can think of that as the ellipse-- so I know that I can, with no prior knowledge of where the true origin of the universe it is, I can arbitrarily define a set of rectangular coordinates, orthogonal coordinates, x, y, and z.
And that means I could put one orbital here, one orbital here, and one orbital here.
So those are three orthogonal orientations, which I think is consistent with the fact that m takes on three values.
OK, that's cute.
So that's as far as Sommerfeld went.
I'm going to go and do something as a retronym.
I want to get the fourth quantum number up here now, but we're going to pause the story.
We're going to fast forward to 1925 so I can get the last quantum number up here.
And that's called the spin quantum number.
And it takes values plus or minus a half.
Where did that come from?
Well, in 1922-- oh, you know everybody's getting Nobel Prizes and I didn't give Niels Bohr his proper recognition.
He gets the Noble Prize, as well.
Oh, when Sommerfeld turned 80, they had a symposium in his honor and they published a book.
And the book had papers and well-wishes, papers that were given at the symposium.
And in the front they had Sommerfeld's picture and they also had this twin picture, this diptych.
So on the right is Sommerfeld, and on the left is the same picture but they've morphed it.
Now remember, there's no Photoshop.
Horrors, there's no Photoshop.
Can you imagine?
So how could they do this?
They had to take the negative, which was a photographic plate, and when they were printing the negative using a light box they had to hold the negative on an angle to get the distortion.
And in holding it on an angle to get the distortion, they turned this image into something that was a little more spread out.
And the caption that went with this, "To Arnold Sommerfeld, who taught us that the circle is the degenerate form of the ellipse."
Now that's geek humor.
I mean, they laughed.
They thought that was so funny, hahaha.
You know.
They were having a great time.
It was Germany, and nineteen twenties.
And there he is.
OK, so now let's go to 1922.
This is the Stern-Gerlach experiment.
Very interesting experiment.
This is really physical vapor deposition.
Over here I've got a crucible and it's full of molten silver.
So Stern and Gerlach were studying the magnetic behavior of liquid metals.
So what they were doing is they had this-- over here you see it's red even though it's silver, because this is that about 1,000 centigrade.
Silver melts at about 960.
Everything, I don't care what it's color is at room temperature, at 1,000 degrees it's red.
It's called red hot.
All right.
So this is red hot silver and there's a vapor here and there's a slit and the silver atoms come out of the slit.
And they go across over here to a substrate and then they pause it on the substrate.
So making little band of silver on the substrate.
And furthermore, he sometimes put them through a magnetic field.
So he's got a slit here that narrows the beam, and then he sends it through a magnetic field that is asymmetric.
It's divergent.
Can you see here?
Look at the end.
The south pole as a tip and the north pole is this arc, this cup.
So the field lines don't go just directly from tip to tip, they go from tip off to the side.
So you can see the divergence of the magnetic field.
And so he looked at what kind of deposits he got as a function of the magnetic field.
Here's what the observed.
Very puzzling.
The whole thing was about Maxwell's equation.
So he's got a silver beam.
And when it went directly from the furnace to the substrate he just got the shadow of the slit.
And they have the split crosswise with respect to the divergent magnetic field.
So if you look at the substrate you just see a band.
So this is a band of silver and you can imagine there was a slit out here and it just cast a shadow and that's the band of silver.
This is PVD, physical vapor deposition of silver.
Now when b is not equal 0, what would you expect?
You'd think the beam would bend, right?
So what do you think happens?
The beam bends up?
The beam bends down?
Or beam bends to the right or to the left?
Think about it.
I don't want to hear your answer.
Think about it.
What do they observe?
What they observe is, if this is where the original one is.
Two.
The beam splits in two.
And it gets two deposits, one above, one below, of equal intensity.
That's a problem.
Beam splitting.
But now it's a beam of matter.
Beam splitting.
Boy, they had them scratching their heads on that one.
No way to explain that.
So along come a couple of graduate students.
1925, couple of graduates.
So this is 1992 in Frankfurt.
1925, two graduate students in Leiden, again.
Gaudsmit and Uhlenbeck.
They're just like my TA's.
Grad students.
And they looked at this thing, I don't know, maybe sitting around over a beer one night, and they said, you know, so far what we've been saying is the electron revolves around the nucleus.
And sometimes it revolves in a circular orbit, and sometimes it revolves in an elliptical orbit, but here's the electron revolving.
And they said, what if in addition to revolve, the electron rotated so that it's going like this?
[GESTURES] But there's two choices.
It can be going like this, or can be going like this.
[GESTURES] Now, it's a charged species and it's rotating, which means that it's going to have a magnetic moment depending on rotation.
And now I'm going to send it through a divergent magnetic field.
Doesn't it follow to reason that if I put it through a magnetic field and I've got some of them doing this and some of them doing that, they're going to go in different directions?
Opposite directions?
And what do you think the numbers are?
If I give you Avogadro's number of silvers, you think I'm going to get a dominant clockwise and a minority anti-clockwise?
No.
We're going to get equal numbers.
Some are going to spin like this, some are going to spin like that.
And you're going say, but electrons don't spin, they're not doing this.
But if you model them as though they are doing this, you get those results.
Those results make sense.
And so they introduced the spin quantum numbers.
And I think these are the ones that have been erased.
Such is education.
OK.
So you know it.
S plus or minus a half.
By the way, Gaudsmit and Uhlenbeck were here during World War Two.
They worked in Building Four.
You go down the corridor, Building Four just off the Infinite Corridor, there's a plaque there for the Radiation Laboratory.
That's where they worked, in the Rad Lab.
That's where radar was first engineered.
There was work in the UK, there was work in other places.
But this is the Radiation Laboratory, started here and they were both here at the time.
OK, well, I think that's a-- So this is a plate from the paper.
No magnetic field with the magnetic field.
And by the way, why did they choose silver?
They chose silver because it's atomic number is 47-- it has an odd number of electrons.
You're going to learn later that you get two electrons in an orbital, and if you have two electrons, one will be spin up, one will be spin down.
There's no magnetic moment.
So they were clever about choosing an element that had an odd number of electrons so that there would, at the end, be an unpaired electron.
And there's Otto Stern with his Nobel Prize.
And he came to the United States, as well.
And you're going to see the ascendancy of American Science as people flee Europe up in the nineteen thirties.
And America is the beneficiary, and then you see American science rise.
But for now it's European science.
OK, so I'm going to talk a little bit about hydrogen and transportation.
And we're going to talk about the Hindenburg because it was full of hydrogen.
And to give you a sense of scale, this is what a 747 would look like and is what the Titanic would look like.
It was almost as long as the Titanic.
It was built in Germany by the Zeppelin company.
And the Titanic, the Hindenburg rather, was LZ129 serial number.
That's Luftschiff Zeppelin.
Airship Zeppelin.
135 feet in diameter.
804 feet long.
How long is a football field?
So that's a big boat.
Seven million cubic feet of gas, giving you 112 tons of useful lift.
You ever have to lift something very heavy, there's your sky crane.
So why are they using hydrogen?
Well, when the Nazis came to power in Germany, Congress passed the Helium Control Act.
The dominant supplier of helium to the world was the United States.
Helium comes from helium wells in the earth.
And so as of 1933 the United States refused to sell Helium to Germany, so the engineers were forced to use hydrogen.
Next best thing.
Here are some posters.
"Only 2 1/2 half days to Europe."
And here one, a German one.
"And now over the North Atlantic."
That's Manhattan.
That's the lower tip of Manhattan.
There's the Chrysler Building, look at that.
Now look at that picture, isn't that magnificent?
10 transatlantic flights, 1936.
1002 passengers.
Cruising speed, 78 miles an hour.
Took two and a half days.
By the way, 100 feet in diameter-- when people traveled, they traveled in style.
They had a ballroom there and a grand piano.
People didn't sit like this.
[CROUCHES] With a plastic knife and fork.
That's progress, right?
Two and a half dancing, tux, tails, champagne.
Now, like this.
[CROUCHES] May sixth, 1937.
Arrival of first flight to the U.S. while docking in Lakehurst, New Jersey.
Why were they docking in Lakehurst, New Jersey?
If you go to the top of the Empire State Building, look and you will see at the corners moorings.
Moorings sticking out.
The plan was to dock airships at the Empire State Building.
So you'd come in from Europe, you'd dock at Fifth Avenue, get on the elevator, and there you were.
When they tried to dock the air currents were so violent that they couldn't safely dock the ship.
So then they moved across to the fair grounds at Lakehurst, New Jersey, where obviously this mooring is much closer to the ground than the top of the Empire State Building.
The wild currents, they're bad, but they're manageably bad.
At the top of the Empire State Building, impossible.
There's another image.
So what happened?
It did not explode.
It did not explode.
It couldn't explode.
Seven million cubic feet of hydrogen to explode requires seven million cubic feet of oxygen instantaneously.
And air is 20% oxygen.
So it was a very violent fire, roman candle from the point of egress of the hydrogen.
Most of the people on board walked off the Hindenburg.
Most of the people walked off the Hindenburg uninjured.
They think it was electrical discharge in the vicinity of a hydrogen leak.
Recent research has indicated the skin was made of resin finished with a lacquer dope.
And then to make it shiny they put aluminum powder.
And why they put iron oxide on the inside I don't know, but this is what NASA uses for solid rocket motor grains.
So when this thing catches fire, this is a thermite reaction and could be very violent.
And this spelled the end of rigid error ships in commercial air transportation.
Now this a U.S. Navy airship filled with helium.
And there was a small gasoline fire, look what happened.
Again, it was the skin.
That's a blow up of that one.
So I looked at that and I thought, geez, that looks Lichtenstein, doesn't it?
You know this one?
This one.
Look at that.
Look at that.
So you know, I can be an artist, too, right?
OK, I'm going to tell you one more story.
Another Niels Bohr story.
So in 1896 there was a guy, and astronomer at Harvard called Pickering.
Pickering at Harvard, 1896, and he was studying the lines in star light.
And he attributed to some of the spectra that he was getting, he said he was seeing atomic hydrogen in star light.
And then there was a fellow in London called Fowler.
And Fowler, in 1912, reproduced the experiments in the laboratory.
He put gas in a tube and got the same thing in the lab.
So this guy's at Harvard and the other guy is at London.
Well, Bohr looks at this stuff and he says, you guys are wrong.
You guys are wrong-- your lines are off by a factor of 4x.
You've got the right series, but you got the wrong element.
What you guys are looking at is helium plus.
You're not looking at hydrogen, and you know, from-- it goes to z squared.
So the lines are going to be shifted by factor of four because this z is equal to 2.
So Fowler was a pompous ass and he didn't like being called on his bad science.
So he does a calculation and he looks more carefully and he says, Bohr, you're wrong.
In fact, our lines are off by 4.0016.
Now don't laugh.
The reason is the spectroscopy was so precise that they could go to five significant figures.
So Bohr says, hmm.
And he goes back and he says, you know, we've been doing all these calculations with a one electron atom just neglecting the center.
So he redoes the calculations for the entire Bohr model, including considerations of the mass of the nucleus and the mass of the electron in the form of the reduced mass.
The reduced mass is-- you're learning this in-- the reciprocal of the sum of the reciprocals.
And when he does that, he gets that the value of the line shift should be 4.00163.
So he says, you guys are wrong.
It should be 4.0016.
You got 4.0016, you idiots.
You're looking at helium plus.
That was Bohr.
Did not want to get into an argument with Bohr.
All right.
Have a nice weekend.
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So, a couple of announcements.
Tomorrow we've got Quiz Two.
Based on Homework Two And then, Thursday, we'll have the periodic table quiz.
I give you the numbers, you give me the letters.
And Friday is the last day for the contest.
I'll have office hours.
I know when others have office hours.
I'll be available 4:30 to 5:30 down the hall in my office.
Where is Grant's, who's buried in Grant's Tomb?
Grant.
Where are my office hours held?
In my office.
Good.
Last day.
Last day.
It was a rough weekend.
I can't get a smile, can't get a laugh.
So, last day we looked at the validation of the Bohr model.
And we had two pieces of experimental data.
First were the hydrogen spectrum lines, measured in 1853 by Angstrom and fit to an equation by J.J. Balmer in 1885.
And, secondly, we saw the Franck-Hertz experiment, in which we were able to get the sense that energy levels within a multi-electron atom, like, mercury, those energy levels are also quantized, which gave credence to the quantum condition.
Which said that the movement of an electron, the movement of an electron, is quantized.
And then we started looking at the limitations of the Bohr model, problems with the Bohr model.
And I said, in two words, fine structure.
And we started looking at fine structure.
We see that the 656 nanometer line in hydrogen, in point of fact, is a doublet.
It's actually two lines very closely spaced.
Bohr model is incapable of explaining this.
Zeeman looked at gas discharge tube, measured hydrogen spectrum in a magnetic field, and found line splitting proportional to the intensity of the field.
Stark did similar experiments only in an electric field, and also found line splitting, the degree of which was proportional to the intensity of the field.
And the Bohr model is incapable of explaining this.
Sommerfeld, in 1916, put a patch on the Bohr model and proposed that the electron moves not only in a circular orbit but also in elliptical orbits.
And there's a plurality of these orbits.
But overall, the distance from the nucleus to the electron orbits is more or less given by the principal quantum number n, and the degree of eccentricity is tiny.
But, more or less, described by n. And then to further depict what's going on, he introduced the orbital quantum number or, as your book calls it, the azimuthal quantum number.
And the magnetic quantum number.
And further said that the energy of the electron is given by the specifications according to all three quantum numbers.
And then, lastly, in order to get completeness, we jumped ahead in time to 1922 with the Stern-Gerlach experiment, which was the beam of silver.
Atoms through a divergent magnetic field, the beam split in two symmetrically about the point at which the beam would land on the slide in the absence of a magnetic field.
And that can led to some deeper thinking by two graduate students in Leiden, Goudsmit and Uhlenbeck, who proposed the notion of electron spin.
And from that came the fourth quantum number, s, and we're going to throw that into the mix even though Sommerfeld didn't give it to us back in 1916.
But I just want to move forward with all of them.
And we recognize that s could take on two values, plus or minus 1/2, or we could call it spin up or spin down.
And, as you're going to learn in 802, the convention in electromagnetism is the right-hand rule.
That is to say, the thumb indicates the vector and the curl of the fingers indicate the rotation.
So we would argue that this is an electron spin from looking top-down anticlockwise and then vice versa.
So that's as far as we got.
And so now what I'd like to do is to take a look at the periodic table and get a sense of electron filling.
And whether that explains the trends in the periodic table.
And for that I'm going to go to table 6-3 in your reading.
And all we're going to do, basically, is say, well can we can we use this idea of n, l and m, and explain the order of filling in a periodic table.
So when n equals 1, l must be equal to 0.
And the m must be equal to 0.
So that means there's only one orbital.
And, in that orbital, we can have two electrons.
So we've got the possibility of two different electrons going into the 1s orbital.
Then we go, n equals 2. l can take a value of 0.
Which is the same as what we had before.
Just one orbital in that subshell.
Or we can have l equals 1, in which case m can take three different values; minus 1, 0, and plus 1.
So there's three orbitals there.
3 plus 1 is 4.
And that gives us the possibility of putting 8 electrons in.
Because this is n, l, and m. And then we've got the choice of s plus or minus 1/2.
And we'll do one more.
We'll go to n equals 3.
So at l equals 0, we just have 1.
When l equals 1, we have the same thing as we had with the 2p.
And then when l equals 2, we have what's known as the 3d.
Remember the spectroscopists there.
They don't like numbers.
So they use s, p, d, f. But we're smart.
We can go 0, 1, 2.
It didn't hurt us.
And we go minus 2, minus 1, 0, 1, 2.
So there's 5.
So 5 plus 3 is 8, and 9.
9 times 2 is 18.
Now let's go to the periodic table and see if this reconciles.
So, when we have 1s, there's hydrogen, helium.
And then the next is n equals 2.
So that should give us 4 times 2 is 8.
So, 3, 4, 5, 6, 7, 8, 9, 10.
So there's the 8 different electrons that get us all the way over to neon.
And now let's go to n equals 3.
So, I have 1, 2, 3, 4, 5, 6, 7, 8.
And then I'm over to 4.
4s.
But look, this is saying I should have 18.
18 electrons in n equals 3.
So what I'm pointing out here is that there's a disconnect, there's a disconnect between the populating of electrons just in ascending quantum number, and the way the elements are arranged in the periodic table.
There's some other factor at work here.
So we want to take a look at what that could possibly be.
And, so what we need is to go to a modified energy level diagram.
A modified energy level diagram, that can explain what the filling sequence is in the periodic table.
And the modified energy level diagram is drawn on the basis of the Aufbau principle.
The Aufbau principle.
Aufbau, German, meaning construction.
I think it actually means out-build.
But, anyways, it generates the filling sequence.
So, there are three parts to the Aufbau principle.
And the Aufbau principle is going to govern, it's going to govern or direct the electron filling sequence.
Directs the electron filling sequence.
So, the first component of the Aufbau principle is the Pauli exclusion principle.
Pauli exclusion principle.
Wolfgang Pauli was an Austrian.
He did his Ph.D under Sommerfeld in Munich, and then he post-docced with Born in Gottingen and on to Niels Bohr Copenhagen.
These people worked together.
They traveled from lab to lab, and it was a very vibrant conversation going on.
Eventually, he became a professor physics in Hamburg.
And the Pauli exclusion principle, simply stated, is that in any electron system, each electron has a unique set of four quantum numbers.
A unique set of four quantum numbers.
Any atom set, four quantum numbers are unique for each electron.
For each electron.
So you can think of this as the set of n, l, m and s as sort of the social security number, if you like, for each of the electrons in the set.
And he eventually gets the Nobel Prize for this, in 1945.
Virtually everybody I'm going to talk about today, with the exception of Sommerfeld, gets the Nobel Prize.
Sommerfeld is a very interesting character, though.
While he himself never won the Nobel Prize, many, many of his students and proteges won Nobel Prizes, to which you must include that there was something very, very special about the quality of the mentoring that he gave, that so many of his proteges went on to win the Nobel Prize.
So this is the first part of the Aufbau principle.
The second part is that the electrons fill from lowest to highest energy.
So electrons fill orbitals.
You can think of the orbitals as placeholders.
They're really energy concepts.
But we populate orbitals from lowest to highest energy.
From lowest to highest energy.
Or if I wanted to be a wise guy, I would say, I'm going to define energy in such a way as I get that as the filling sequence.
One's got to fit the other.
And the energy is itself a function of the four quantum numbers.
So energy, once I specify n, l, m, and s, you can give me the energy and away we go.
And the thing is that you need to realize that the energy levels are actually a function of electron occupancy.
So you can look carefully, if you get down into the d orbitals and the f orbitals, as you add one more electron moving one element to the right on the periodic table, sometimes you see the population not adding by one, because the addition of an electron changes the relative energies of the various orbitals.
So let's remember that, that energy is a function of electron occupancy.
Energy is a function of electron occupancy.
And then the third part of the Aufbau principle is called Hund's rule.
Hund's rule.
Now, Hund is German for dog, but this isn't named after a dog.
This is named after Friedrich Hund, a professor of physics at Frankfurt.
He taught for many years.
He died at the tender age of 101.
And he spoke about degeneracy.
Not societal degeneracy, but degeneracy in an atom.
And this degeneracy is the condition where you have a plurality of orbitals at the same energy level.
So, in orbitals of equivalent energy, in orbitals of equivalent energy, we strive for unpaired electrons.
This is filling, now.
This is how to direct the filling sequence.
Strive for unpaired electrons.
Let's say unpaired electron spins.
Let's make it a little clearer.
Unpaired electron spins.
So I'm going to give you an example, work through an example.
So, what have we got here?
Here's carbon.
And if you look at any element on the periodic table, you'll see this green.
And this is the electronic configuration.
This tells you what the sequence looks like for that particular element in the neutral form, et cetera, et cetera, subject to many, many considerations.
So let's look at carbon.
So if we look at carbon, it tells us 1 s2, 2 s2, 2 p2.
So what's all this mean?
Well, it's all written in code for us.
So the first number here is the n number.
So this means n equals 1. s is the spectrocopist's notation, so this means that l equals 0.
And then the 2 here means 2 electron occupancy.
So the 1s orbital is filled to the tune of two electrons.
And the 2s orbital has two electrons, and the 2p orbital likewise has two electrons.
So now we want to do is put these electronics into their orbitals.
And we can use, there's a variety of notations.
This is one I'll use today.
This is a box notation.
So this is 1s.
This is 2s.
And then this is 2p.
If you recall, 2p there's going to be three of them.
If you go back here, 2p has three orbitals in that sub-shell.
And we said that this is the one time that m makes some sense with respect to Cartesian coordinates.
m is -1, 0, +1 so we can call this 2p x, 2p y, and 2p z.
And I don't know which is which.
I don't know what the arbitrary standards of x, y, and z, I don't know where the origin of the universe is, so I can't tell you all this.
But, arbitrarily, if I choose one as x, I know where the other two are according to right-hand rule.
So let's start filling.
This says 2, so I'll put in one spin up, and one spin down.
Now it says 2s 2.
One spin up, one spin down.
And now here's where the Hund rule comes in.
I've got to put two electrons into these three boxes.
Now, I could be a librarian and start left to right, and make them nice and neat.
Or, I could just put them in wherever I want.
What the Hund rule says is, strive for unpaired electron spin.
So for the Hund rule, you'd put them in both same spin and in different orbitals.
And then when you get to nitrogen, nitrogen will have a third one here.
And when you get to oxygen you've got three plus the fourth one's going to be one of these three and I don't care.
I mean, some people are really anal about it, and they want you to go from left to right.
As long as you've got two of them unpaired and one of them paired for oxygen I'll be happy.
So now we can use this concept and look at the energy level diagram for multi-electron atoms.
And this is taken from the book.
So this is different from the energy level diagram for hydrogen, because you can see some compression here.
And I want you to know first of all, all of these values are negative.
It's true that energy increases vertically, but the zero is way up here.
So these are all negative values.
And I want to zoom in here.
Because the energy difference between 1s and 2s is large.
And we know in hydrogen it's 3/4 of the total energy difference.
Because there's some compression in a multi-electron atom.
But nevertheless, n equals 1 to n equals 2 is a huge energy difference.
So let's zoom in on here.
And what do we see?
Well, 2s, 2p, 3s, 3p and look.
4s lies below 3d.
4s lies below 3d.
So that tells us, there's my 3s.
3s 1, 3s 2.
There's 3p x1, 3p y1, 3p z1, 2, 2, 2.
And now what's next?
It says go to 4s.
So potassium is 4s 1.
Calcium's 4s 2.
And scandium is 3d 1.
So this energy level diagram is the map.
It tells you how to put electronics in sequence.
So, this now gives us the rational basis for e equals n, l, m and s. All right.
Well this is nice.
It's been graphical, and so on.
Now I want to go to the same position.
I want to get back to this, I want to get back to this point.
But I want to go by different route.
I want to go by different route, and for that we're going to go by wave mechanics.
So, same destination.
Same destination, via wave mechanics.
Now, you don't have partial differential equations as a prerequisite for 3.091, so I'm not going to go through the math.
I'm going to give you the features of the wave mechanics.
So that later on you're going to spiral around and study this again.
You'll have seen it before.
And again, there's going to be people involved, and they're all giants in modern physics.
The first one is de Broglie.
The first one is Louis-Victor de Broglie.
Let's get his name on the board.
Louis-Victor de Broglie, he was an aristocrat from Normandy, who had gone to the Sorbonne.
He was studying humanities.
Political science, literature, and around about the time of his senior year he decided to switch horses for graduate school and forget about a career in the diplomatic corps, do a Ph.D in physics.
So, he did a Ph.D in physics.
And in 1924 he published this Ph.D thesis, beautiful piece of elegant writings.
Less than 30 pages long.
And I'll give you sort of the summary, the one-liner that summarizes his Ph.D thesis.
Most of you are going to have to write a thesis.
The word thesis comes from the Greek, and it means, sort of, my position.
My statement.
Before I have a thesis, I have something that is not a thesis.
It's my trial balloon.
That's a hypo-thesis, a hypothesis.
Now, the key to writing a good thesis is to ask a really good question.
If you ask a pedestrian question, you're probably going to get some pedestrian answers and ho hum.
If you ask a really interesting question, you give rise to the possibility of interesting answers.
And what de Broglie did is, he asked a really interesting question.
So here's his question: He says, if a photon, which has no mass, photon's just an energy packet.
If a photon which has no mass can behave as a particle, and we've seen.
We model ray optics as particle beams.
And a photon, and Max Planck said equals h nu got no mass, but we can think of it in our little anthropomorphic brains as, photon photon photon.
So if a photon, which has no mass, can behave as a particle, does it follow that an electron, which has mass, can behave as a wave?
It's beautiful.
Let's do it one more time.
If a photon, which has no mass, can behave as a particle, does it follow that an electron which has mass, can behave as a wave?
So he asked the question.
And he answers it.
In less than 30 pages.
So, let's get this on the board.
Because this is beautiful.
It's like, ask not what you can do for your country.
If a photon, which has no mass, can behave as a particle, or can be modeled as a particle, behave really means, so that the theoreticians can model it as a particle, does it follow that an electron, which has mass, can behave as a wave?
See, if you understand the question then the impact of the answer, and the answer is, if it does, this is what its wavelength is going to be.
So de Broglie said, the wavelength of an electron, if it were to behave as a wave, would be given by the ratio of the Planck constant to the Newtonian momentum, which you know from 8.01 is simply h over the product of electron mass and its velocity.
So that's de Broglie's thesis.
So let's take a look what we can do with this.
Now, you remember in the Bohr model, recall Bohr.
Well, Bohr taught us that mvr, that's the quantum, condition, mvr equals the ratio of h over 2 pi times n, where n takes on the discrete values.
1, 2, 3, et cetera.
That's the quantum condition.
Now let's take this idea of de Broglie.
And first of all we have to put the electron in its orbit.
So we'll put the electron in its orbit.
And now I'm going to have it behave as a wave.
So if it behaves as a wave, I'm going to draw it as a wave.
Now, why did I draw it this way?
There's two kinds of waves in this world.
There's standing waves and there's traveling waves.
Now, this orbit's station, it's in this orbit.
So it better be a standing wave.
I think there's a cartoon in the book.
There you go, standing wave.
In order for this to be a standing wave, there's a geometric constraint on this.
Listen carefully geometric constraint.
I'm not saying anything about quantum mechanics.
Geometric constraint, the geometric constraint for a standing wave is what?
You know what this distance is, right?
I mean, this is not to scale.
This should be 10,000:1, in which case these ripples are barely visible.
But here it looks kind of exaggerated.
This is a hyper wave.
This is emphasis added in proof.
So that means that the circumference here, 2 pi r, the circumference must be an integral number of wavelengths for a standing wave.
But we know that from de Broglie, I can write n lambda as n h over m v. I've just put in de Broglie's definition of the wavelength of an electron.
And now I can cross-multiply and I get mvr equals h over 2 pi times n. Which is Bohr's quantum condition.
So we've got validation of the Bohr model, so that's a pretty compelling case that maybe the electron really does behave as a wave, and that explains why we have the quantum condition that we do.
So de Broglie, that's his Ph.D in 1924.
Einstein read the thesis, loved the thesis.
But we don't care what Einstein says, because he's a theoretician.
So one theoretician praising another theoretician.
That's not how science works.
How does science work?
Data.
We need data.
And the data come in 1927.
1927, at Bell Labs in New Jersey.
At Bell Labs in New Jersey come the critical data.
And they were taken by Davisson and Germer.
Davisson and Germer.
Davisson and Germer were studying crystals.
They were studying crystals of various elements, and in particular metal crystals.
Metal crystals, by X-ray analysis.
And in order for you to appreciate what I'm going to show you of Davisson and Germer's work, I'm going to take you back to high school to those thrilling days with the wave tanks.
Remember the wave tank?
This is the top view, this is the side view of the wave tank.
And you might have some kind of a mechanical device here that has a paddle.
And it starts vibrating up and down, and it starts sending waves into the tank.
So the waves come like this, from the edge it looks like this.
Remember that?
Sure you do.
You're toying with me.
Oh, I don't remember anything, we never did that.
Sure you did.
OK, so you can send waves down.
Now, what we can do is, we can put a dam here.
I'm going to put a dam.
And depending on, if this is the wavelength, this is the wavelength, it's the distance between two successive crests.
And, if this spacing here, the gap between the wall and the edge of the dam, d, if d is greater than lambda, the waves just propagate but for the place where they're blocked by the dam.
So you get, you essentially cast a shadow.
You've seen that.
And so I could model this system as a beam.
This is a beam.
This is a water beam.
This is a water beam shadow.
And this is equivalent to ray optics.
Straight lines, if something gets in the way it's opaque.
Blocks transmission.
End of story.
So you've seen all that.
But you also did this other experiment, I'm willing to bet.
So let's do the other experiment.
We're going to do the same thing.
I'm going to send waves down here.
Only, this time we're going to make the dam a little bit different.
This time we're going to bring the dam in from the wall.
And we're going to put a tiny opening.
And I'm going to go some more into the tank.
And then another tiny opening.
It could be the same.
It's probably best to keep it the same dimension.
Now, in this case, the spacing, d, is much less than lambda.
d is much less than lambda.
And what happens in this case?
When d is much less than lambda, you don't get the shadow.
You don't get something like this instead, remember.
You've got the rings.
This is called diffraction.
Diffraction.
And there is no way to explain diffraction modeling water as a beam.
You must implore the wave-like behavior of water in order to explain diffraction.
Explain only by invoking wave-like properties.
So with wave-like properties, we get something that makes sense in terms of the data.
Now, let's do this same experiment, let's do this same experiment on a metal crystal.
So if you go back to the gas discharge tube.
Remember the gas discharge tube that we were looking at for lecture after lecture?
If you take a look and go through the energetics of it.
If you put one volt across the plates, you know the energy is going to be a product of charge times voltage.
And that's equal to 1/2 mv squared.
Which you know from 801 is p squared over 2m.
And p is equal to h over lambda.
Pretty soon you can come up with the wavelength.
And that will give you the wavelength of the electron.
This is a ballistic electron now.
See what I'm doing?
See, once I said that this indicates that the electronic in stationary orbit can be modeled as a wave of a certain wavelength, so now the free electron.
It's got m, it's got v, there's Planck constant.
I can go ahead and compute its wavelength.
I can compute the wavelength of a baseball.
So, you go through it.
And you get a value of about 12 Angstroms.
12 Angstroms.
Now, if I want to see whether there's wave-like properties, I need to have a condition that gives me diffraction.
So I'm going to have to find something that gives me a dam with an opening that's less than 12 Angstroms.
If I'm going to use 1 volt.
So what can I do?
Well, turns out you're going to learn this in greater detail later, but if this is a crystal of nickel.
Crystal of nickel, the atoms are arranged in regular arrays.
And this is what the face of nickel looks like, 4 atoms each at the corner.
And one in the center of that face.
This distance is 3.53 Angstroms.
Perfect.
Perfect.
So what Davisson and Germer did is, they irradiated this.
They irradiated this first with X-rays on the order of lambda, on the order of, say, 10 Angstroms.
And what did they get?
This is the output.
This is the output.
You get a diffraction pattern.
It's a set of rings, concentric rings.
So this is the X-ray diffraction.
This is the X-ray diffractogram, if you like.
And then, what did they do next?
They irradiate the same crystal with an electron beam.
lambda of the electron beam, 10 Angstroms.
And what did they get?
Are you ready?
Drum roll.
Now, there is no way that you can get a ring pattern from a beam of electrons acting as a particle beam.
The only explanation for this is that the electrons were behaving as waves of this value to give us the same spacing as we got with X-rays.
And you're comfortable if I say that X-rays are a type of light.
So, therefore, it's got wave-like properties.
And it's got particle-like properties.
Well, now I've just made the point that this is the electron diffractogram.
So this is evidence of electron diffraction.
And this shook the world.
Because now it's real.
There's no way you can get this otherwise.
This is electron diffraction.
This was 1927.
1929, de Broglie gets the Nobel Prize.
1937, Davisson gets the Nobel Prize.
So this means the wave-particle duality is complete.
It applies not only to light, but it applies to matter.
So, wave-particle duality, they call it.
Wave-particle duality is complete.
Matter can act as waves.
Electromagnetic radiation can act as particles.
So, sometimes people refer to de Broglie's accomplishment as matter waves.
Matter waves.
And what do you call the behavior of billiard balls banging around and so on?
We call that mechanics.
So now we're going to use what might seem as an oxymoron, contradiction, wave mechanics.
Wave mechanics.
That means matter behaving as a wave.
But still behaving as a matter.
So this is the dynamic.
So that's pretty good for de Broglie.
So let's go to number two.
I said there were going to be three people here.
Number two is Werner Heisenberg.
Werner Heisenberg.
Heisenberg studied with Pauli.
Sommerfeld.
Did his Ph.D in Munich in 1923.
Got his Ph.D with Sommerfeld at the age of 22.
And then he decided to take a postdoc with Bohr.
And he was working with Bohr for a couple of years, he was feeling a little bit burned-out.
And decided to take three weeks off.
Went up to a deserted island off the coast of Norway.
Came back three weeks later with the mathematical formulation of quantum mechanics.
I'm not kidding you, that's what he did in his time off, to kind of unwind.
And so one of the things that he used as a critical piece of this derivation is that the position and velocity of an electron cannot be fully specified.
They cannot be fully specified below certain limits.
There's a threshold below which we can't go.
This is sort of like, if I asked you to time a 100-meter sprint, which typically takes less than 10 seconds.
But I give you a clock, and the nearest unit on the clock is the minute.
So you wouldn't be able to distinguish.
So, he says-- and the reason for this is, it's a consequence of quantization.
Light itself is quantized.
So, at some point you're asking for a continuous splitting and splitting and splitting into finer and finer time segments.
And you can't get there.
So we already knew this from Planck.
And so one of the ways that he expressed the inability to go below a certain threshold is the uncertainty principle.
The uncertainty principle.
And that's unfortunate that, see, it was originally published in German.
And the idea really is the indeterminacy principle.
But, English says uncertainty.
So it's a limit to determination, but there it is.
And so one expression of this is, the product of the velocity.
Only, he wrote it in terms of momentum.
So the mass isn't going to change.
So, think of this as the uncertainty and the velocity.
And the uncertainty in the position.
So this is the x-coordinate of particle.
You can break its trajectory into three orthogonal components.
So the uncertainty in the x direction of the momentum times the uncertainty in the position is greater than or equal to the Planck constant divided by 2 pi.
The Planck constant divided by 2 pi.
And so what this means is that we're going to see a transition in models from individual atoms.
If we want to describe what happens with individual atoms, we need what is known as a deterministic model.
Sort of, Newtonian mechanics.
You tell me the initial position and velocity.
You tell me the forces, and I can predict where it's going and where it's going to be.
So that's deterministic models.
So, deterministic models describing individual atoms are going to give rise to probabilistic models.
Probabilistic models.
And probabilistic models obviously can't be talking about individual atoms.
Must be talking about ensembles of atoms.
So I can't say where any individual atom will be, because I don't have the ability to do so.
But I can tell you, if you give me a large number of them, I'll tell you roughly what the expected outcome could be in terms of energy.
And ultimately predict the spectrum, and so on.
So instead of chicken and egg we have now chickenality and egg-ness.
Everything is just sort of getting a little bit murky.
Little bit murky.
You can do a calculation on this.
You can do a calculation on this, a very simple take.
Take the Bohr model and take the ground state electron in hydrogen.
n equals 1 in atomic hydrogen.
And you know this is about half an Angstrom.
So the distance across here is 1 Angstrom.
So make 1 Angstrom your uncertainty, and you'll find that the uncertainty in the momentum is on the order of 15%.
But what this is saying is, when you get down to atomic dimensions, you can't just shine light on it and reveal what's going on.
Because you're going to disturb the very thing you're trying to measure.
Some people say that, every time you try to work at the atomic level, it's as though you're trying to take a picture with the sun at your back and your shadow is in the picture.
So you can't get there without disturbing the very thing.
Another way to think about it is, the photons that are capable of this resolution are going to have such high energies they'll knock the very thing you're trying to measure.
All right.
He gets Nobel Prize in 1932.
And then the third one is Erwin Schroedinger.
Let's get him up here.
Erwin Schroedinger.
Also an Austrian, University of Zurich.
He too was burned out.
They get burned out, these guys.
So at Christmastime, 1925, took a vacation.
At Villa Herwig, in Aurora.
And comes back two weeks later with the wave mechanics formulation, of quantum mechanics.
See, sometimes going away on a vacation.
So he took de Broglie's notion of the electron as a wave and wrote equations to model wave-like behavior.
So, let's look at how to get there.
And here's what he did.
So, you know, for example, that we could start with a violin string.
And it has a geometric constraint.
It must be fixed at both ends.
And if I pluck that string, it can vibrate as long as it conforms to the geometric constraint of a standing wave.
So here's one possibility.
Wherein we would call this n equals 1.
I have simply the entire string vibrating in the matter that's shown.
But here's a second possibility.
I could have it vibrating as is shown here, n equals 2 with a node in the middle where that node doesn't move at all.
The string is stationary at its mid-point.
And what's the characteristic here?
This is operating at a certain frequency.
Let's say it's middle C. And this is the overtone.
This is the first harmonic.
And it's going to be an octave higher, because it's as though we have two strings, each fixed.
See, from a physics standpoint I could literally cut this string in half and fix it there, and this is now n equals 1 for the half-length.
So it's going to have the same pitch as the half-length.
Which means that this is an octave higher, and this is going to be two octaves higher, and so on.
And all of these conform.
So you get a plurality of solutions.
You get a plurality of solutions.
And the solutions look something like this.
They'll eventually teach you the string as a simple harmonic oscillator.
Simple harmonic oscillator.
And it has equations that look like this.
If you want to plot its position, this is x going from 0 to l. And this is the y-coordinate.
So you can, for example, write something like this.
So, the function will look like this.
Some pre-multiplier times cosine of kx plus another pre-multiplier b times the sine of kx.
And the geometry will dictate that the value of k is n pi over l. So, pi over l is the geometry.
And n takes multiple values.
Just as you see here, there's not a unique solution.
So listen carefully.
Wave equation, plurality of solutions.
But subject to some constraints.
Subject to some constraints.
So what Schroedinger did is, he wrote a wave equation to describe the motion of electron in its orbit.
And guess what he gets?
He gets a plurality of solutions.
And when you look at the plurality of solutions, the plurality of solutions ultimately map into what we know as the distinct values of n, l, m and s. See, this is the one-dimensional.
So this is giving us n numbers.
n equals 1, this is now n equals 2.
So I'm getting quantum numbers here.
Now, if I did this in three dimensions, I'd have a plurality of quantum numbers.
And Schroedinger gets us all the way to n, l, m and s. And so here's what it looks like.
This is the equation, it's a wave equation, so there's a double derivative in space.
There's a forcing function.
And this is i, square root of minus 1 in a time base here.
So it's a harmonic kind of equation.
Psi is the wave function, it's an abstract concept but we'll show you how to make sense of it.
And these are the various solutions, the plurality of solutions.
And we can now map those into what we know as 1s, 2s, 2px, 2py, 2pz, et cetera, et cetera.
And you see this number a sub 0.
That's our Bohr radius.
Comes right out of the equations.
0.529 Angstroms.
So, this is quite good.
But, as I said, the psi is the wave function.
Or psi, however you want to call this.
This is called the wave function.
Wave function.
And we have plurality of solutions.
We call these Eigenfunctions.
Eigenfunctions.
And the closest we can get to something physical is the product of psi and its complex conjugate.
And that is related to the probability of finding the electron.
Probability of finding the electron.
Which, in essence, gives the boundaries of the orbitals.
So now I'm going to put, we're going to get a Cartesian shape.
I told you, 1s, 2s.
What do they look like?
Here's what they look like.
So these are the square of the wave function plotted.
So this is in a radial distribution function.
It's only in one direction, out from the radius.
Now, if you whip this around in 3-D, you'll generate the surface.
But already you can see, here's 1s, and it peaks at about 1/2 an Angstrom.
And here's 2s, and it peaks at 4 times the Bohr radius.
And psi squared 3s about 9 times.
This is from your book.
So it's a maximum.
But there's some uncertainty.
See, it's not a simple line fixed at 0.529 Angstroms.
This is another way of plotting.
So these are spherical.
What we were calling circular now becomes namely spherical.
And there's this node here.
This is the p orbitals.
They're dumbbell-shaped.
With two lobes.
And if you have a single electron, it doesn't reside in one lobe.
It can jump from one side to the other.
You might say, well, how does it get from one lobe to the other when halfway between, there's a nodal plane.
It has zero probability.
Well, it's behaving as a wave.
Behaves as a particle, you can't get through a wall that says zero permission.
That's how, you can transfer energy from here to here and have that node's perfectly stationary.
Anybody skip rope?
You know how this works.
Now, this is where I quarrel with the book.
This is another drawing.
But I'm uncomfortable with the fact that they chose different colors.
Because I think to the first time learner, you might be tempted to think, well, one electron lives here and the other electron lives here.
No, the electron, if there's only one, it can go from one to the other.
If there are two, they can go from one to the other.
And, see, they do this all the way through.
So please don't start rationalizing in your mind that one electron goes in the yellow and the other electron goes in the grey.
There's the d. Aren't they pretty?
If you find the f orbitals, that's wild.
So I think this is probably a good place to stop.
We've got a few minutes here.
If you want to read more about uncertainty, this is a very nice book by David Peat that goes into the meanings, including this indeterminacy, and so on.
Good book here on hydrogen.
Please, I don't want noise.
We've got, still, a few more minutes.
Still got a few more minutes.
This book here talks about hydrogen.
Goes right back to Democritus.
One chapter is a beautiful thing on the use of hydrogen as a potential fuel.
All the Bohr and whatnot.
This is a play that won the Tony Award in the year 2000.
Written by Michael Frayn.
And it's about the fact that Niels Bohr was the mentor to Werner Heisenberg.
And now it's 1941, the Nazis have invaded Denmark.
And Bohr is essentially waiting to get out of Denmark before the war overtakes the rest of Europe.
Meanwhile, Heisenberg is now the head of the Nazi equivalent of the Manhattan Project.
And he goes to Copenhagen to visit his old mentor.
That's a fact.
They have dinner.
That's a fact.
They go for a walk.
That's a fact.
They never speak to each other after that night.
That's a fact.
So the question is, what went on that night.
And that's what Michael Frayn uses as the dramatic point of departure.
So, did Heisenberg go to Bohr to get Bohr's opinion about nuclear weaponry?
Did he try to find out whether the Allies were working on a bomb?
Did he go to say, look, we should on both sides not develop nuclear weaponry?
What went on in that conversation?
And so you see Bohr at the center.
There's Heisenberg who's the one electron.
And there's Margaret, who is Bohr's wife, who was the observer.
And so there's the play between the uncertainty in quantum mechanics and the uncertainty in human relations.
It's a really nice play.
Here's a rendition of it.
This is Stephen Rea playing Bohr, Francesca Annis playing Margaret Bohr, and playing Werner Heisenberg is, do you recognize him, it's Daniel Craig.
This is a book that came out not too long ago about Heisenberg.
A lot of controversy about him.
Some people accused him of being a collaborator.
Other people say that he was absolutely brilliant in the right amount of foot-dragging.
He did not want to give Hitler nuclear weapons.
If he'd been a total disaster he would have been replaced by somebody who might have been more zealous.
And if he went too fast, he might have figured out how to make nuclear weapons.
So, very interesting book about him.
And here's a nice photo.
This is Bohr.
This is Werner Heisenberg.
And this is Wolfgang Pauli.
Undoubtedly talking about what goes on when you pluck that string.
So we'll see you on Wednesday.
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OK, a couple of announcements.
Tomorrow will be the Periodic Table test.
I provide the numbers, you provide the letters.
Friday, the contest ends, 5:00 pm.
Get the submissions to me, either over the internet or if you have something majestic, you have to cart it in with the riggers.
Get it to my office.
Last day, we had a lesson on the Aufbau principle.
And the Aufbau principle gave us the-- There's still too much talking.
Still too much talking.
I will tolerate zero, zero talking.
When you walk through that door, I'm assuming it's an act of free will.
And when you walk through that door, we enter into a contract.
The contract goes something like this: You have certain expectations of me.
You expect me to come to class prepared.
You expect me to treat you with respect, not to use vulgar language, not to say things that are insulting, insensitive.
And I have certain expectations of you.
I expect you to come to class prepared.
I expect that you've done the reading.
And I expect that you're going to sit quietly.
Absolutely quietly.
Because it's my duty to preserve a fertile learning environment.
If you don't want to learn, I don't care.
I don't care.
But if you impair the ability of your neighbor to learn, I will take action.
It's very simple.
Very simple.
Where was I?
I think I was talking about the Aufbau principle.
And the Aufbau principle tells us what the filling sequence of electrons is in a multielectron atom, which ultimately we rationalized in terms of the four quantum numbers and the filling in ascending order of energy.
And then we tried to understand what's behind the Aufbau principle.
And lastly, we took a journey through central Europe in the 1920s and 1930s.
And we met de Broglie, who gave us the concept of matter waves.
If matter has wave-like properties, the wavelength would be given by the ratio of the Planck constant to the Newtonian momentum.
Heisenberg taught us that we can't deal with precision down to atomic dimensions, and there's a certain degree of uncertainty or indeterminacy.
And it's given by this relationship here.
And then, lastly, we saw Schroedinger, who wrote the wave equation saying that if we have wave-like properties then let's model the atom as a wave, and he gave us this equation here, which is essentially the equation of the simple harmonic oscillator.
The simple harmonic oscillator, which allows you to tell us what's going on with the plucked string.
And it's essentially just this, the double derivative x double dot plus kx goes to some function.
And the solution of that is a bunch of sines and cosines.
And that's this equation, but dressed up for night-time.
It's a little bit more sophisticated and gives us the three dimensions, eigenfunctions that ultimately, we can take and generate the pretty pictures of the orbitals.
So, today what I want to do is go a little more deeply into properties.
And if you learn nothing else from me in 3.091, this is what I want you to retain.
It's that electronic structure dictates properties.
Electronic structure dictates properties.
You know this on your death bed.
It's that simple.
Electronic structure dictates properties.
Then we're going to take 14 weeks and describe electronic structure, right?
But that's it.
That's it.
So I want to go back to this.
We got into this energy, Aufbau principle because we saw on the Periodic Table there's some problems here.
If we just go in terms of quantum numbers, we see there's two elements in n equals 1 shell.
There's 8 elements in n equals 2 shell.
And, if we just kept going, we would expect to find 18 elements in n equals 3 shell, but instead we found only 8.
And that's where the journey began.
Well, there's a way to sort of help remember what the filling sequence is.
And that rule is encapsulated by the n plus l relationship.
So for equivalent, n plus l values, for equivalent n plus l values, you fill in ascending n. Fill in ascending n. So let's take a look.
So, here's Aufbau.
n plus l rule.
Here's an example where we have 3d.
d means l equals 2. p means l equals 1. s means l equals 0.
So 5 plus 0 is 5.
4 plus 1 is 5.
3 plus 2 is 5.
So what do I do?
Well, it's saying go in order of ascending n plus l. And so, you can snake your way through this n plus l, and get the filling sequence for the whole Periodic Table.
So let's take a look.
We'll start with 1s.
Well, that's easy.
That's hydrogen and helium.
Then we go next to 2s.
There's lithium and beryllium.
And then next goes 2p.
I mean, nobody's going to try to stick 3s ahead of 2p.
That's pretty straightforward.
2p, and that gets us all the way over to neon.
And then we wrap around over here to 3s.
That gets us from sodium to magnesium.
And then we jump to 3p, gets us from aluminum over to argon.
And then, we don't go to 3d, because 3p is 3 plus 1 is 4.
4s is 4 plus 0 is 4.
So it says we fill in ascending n. So we go to 4s next.
And that gets us potassium and calcium.
Then comes 3d.
So that gets us scandium over to zinc.
Then comes 4p, gallium to krypton.
5s.
I'm having fun here, I'm going to keep going.
5s gets us over to strontium.
Next comes 4d.
Gets us all the way over to cadmium.
5p gets us to xenon.
Now it's turn 6s cesium, barium.
4f.
We've got to jump down here.
Finally we're going to get to the f shell.
See?
The 4f is in here.
Now, this is an item of convenience.
Strictly speaking, lanthanum is 57, cerium is 58.
So all of these elements belong in here.
But if we were to put this to scale, the Periodic Table would be way over here.
It wouldn't fit on a page.
So people, over time, have gotten used to just putting it down here.
But these elements are in the middle here.
So you go 4d, 5p, 6s, then 4f.
Finally we get over to the edge here, to erbium, and then we jump over here to hafnium, 5d, over to mercury, 6p, et cetera.
So the n plus l rule gives you the filling sequence in ascending order.
That's good.
So we've got a nice compact way of grabbing this e goes to n plus l. Now let's look at some properties.
We said it's a table of the elements, but it's a periodic table.
So let's see what this periodicity looks like.
Now, here's a variation of first ionization energy.
So here it is in kilojoules per mole, and here's my pet unit, the electron volt per atom.
Because here's hydrogen, of 1.6 megajoules per mole, but I like the 13.6 electron volts because I can remember 13.6.
I can't remember 1312 megajoules per mole, kilojoules per mole, whatever it is.
13.6 electronic volts.
So here's hydrogen, there's helium.
And then we drop down to lithium, and then we move across up to neon, and so on.
So you can see 1s, we filled the 1s shell.
Now, we're going to go to shell n equals 2.
Here's 2s.
And then we go to 2p.
And you can even see, look, boron, carbon, oxygen, the ascending value of the first ionization energy.
And then there's a little jog here at oxygen.
Because here we've got the three unpaired electrons for nitrogen.
And then there's the oxygen starts to pair, and we continue to neon, and so on.
3s, 3p, 4s.
So you can see the relationship between that property, and the place, the element, in the Periodic Table.
So how do we measure these values?
Well, I thought it might be a good opportunity to look at, revisit the whole question of gas dynamics, and also understand the measurement.
Measurement of ionization energies.
And the technique that's used is called photoelectron spectroscopy.
Photoelectron spectroscopy.
All right.
And it's got a three-letter initialization, PES.
I want to show you a cartoon of how this works.
Actually, I took one -- this is taken from the text we used to use.
I like the text that we're using now better, but there are a few things in the other text that were good.
So, we've scanned these few pages and posted them at the website if you want to go through and read this stuff.
So let's jump over to that.
So here's an example of a terrible drawing.
I look at this and I haven't got the faintest idea how this thing works.
It's not the artist's fault.
Somebody should have proofread this thing.
What they should be doing is showing something like this.
So, let's see what's going on.
So what we're going to do is, we're going to bombard a specimen right here.
Which they never show.
I suppose this atom beam is supposed to be colliding with the photons.
That would be quite an apparatus to build, let me tell you.
Instead, what you have is, you have the apparatus sitting so that you've got the material in the center of the chamber.
And you irradiate with photons.
And the photons have very, very high energy, and they dislodge electrons.
When they dislodge electrons that's the ionization event.
And then what we do is, we measure the velocity of those ejected electrons.
And if we know the velocity, we know their energy.
We know the energy of the incident energy and by difference, we calculate the binding energy of the electron and the atom.
So let's take a look.
I'm going to draw a cartoon here that gives you a sense of what's going on.
So, here's the n equals 1 shell, or the k shell.
And then I'm going to show, say, a second shell here.
One, two, three.
So this is the l shell, or n equals 2.
And you might say, gee, he's got that wrong.
It's kind of simple, and so on.
It's as complex as it needs to be for the explanation.
Don't let accuracy trump clarity.
The real accurate picture here is too detailed to convey what I'm trying to convey.
So, this is the specimen, right?
Let's call this the specimen.
And I'm going to irradiate the specimen with some radiation of high energy.
And so, I generally indicate the photon by a squiggly line with an arrow.
That's to indicate it's got wave-like properties.
And I'll usually write h nu next to it, to indicate that it's a photon of energy h nu.
So the photon comes in, and if the energy is high enough, it will dislodge an electron.
So this electron is dislodged.
It's now ballistic.
It's rendered ballistic.
It's free.
It's ionized.
It's gone.
It's gone.
And what we're going to do is, we're going to take an energy balance here.
And at some point you may come back and learn some quantum mechanics.
This photon is annihilated at this collision.
So the photon doesn't act the way an incident electron does where it loses some of its energy.
All of the energy is lost.
So this photon's gone.
All of the energy is given to this electron.
So let's do an energy balance here.
So I can say that the energy of the incident photon, the energy of the incident photon, is now going to be given to the electron.
It's now got some kind of kinetic energy.
E kinetic plus the energy it took to pull it out of the atom.
Plus E, let's call it binding.
E binding And we know the incident energy of the photon because we're running the experiment.
So we set the value of the incident energy of the photon beam.
And then we send this to a detector.
This goes to a detector.
And the detector measures the velocity and ultimately gives us the energy of the scattered electron.
This is called the dislodged electron, or photoelectron.
It's the electron that was kicked out by the photon.
So it's known as the photoelectron.
Photoelectron.
Pardon me, let's try that again.
Photoelectron energy is measured at the detector.
And then by difference, we get the binding energy.
We get the binding energy.
So this is h nu.
This one here is going to be a 1/2 m b squared, and then by difference we get the binding energy.
And what kind of photon energies do we need?
We need fairly high energies.
And so typically, we might use wavelengths down around 1 angstrom, which then makes it an X-ray.
And so, over here in Building 13, we have such instrumentation, and the material scientists would call this XPS.
X-ray photoelectron spectroscopy.
And if you don't want to blast all of the electrons out of the specimen and just get the most weakly bound, then you want a lower incident photon energy, which means a longer wavelength.
You might come in at around 100 angstroms, which is ultraviolet.
You know, people, the general public, is so afraid, they don't know science.
If they hear X-ray they get all panicked.
You know, it's radiation.
Their children are going to die and all this kind of stuff.
So, what people will do is, they'll say instead of soft X-ray they'll say hard ultraviolet.
And then the public thinks, oh, as long I wear sunscreen I'm going to be OK.
So if you use hard ultraviolet, it's called UPS.
That's nothing to do with brown trucks.
It's ultraviolet photoelectron spectroscopy.
And collectively, this is known as PES-- photoelectron spectroscopy.
All right.
So now, let's go back to this schematic.
Now you can see how bad this schematic was that these photons are coming in as so.
And striking the specimen here, ejecting the photoelectrons which then are focused and ultimately measured here at the detector.
So there's the energy balance.
The energy of the incident photon is distributed across the kinetic energy of the photoelectron plus the binding energy, and this is how we measure all of these quantities.
Since you're using very, very high intensity radiation, there's nothing saying that you're going to eject only one electron.
You can eject a plurality of electrons.
And the different electrons have different binding energies.
So you're going to get a set of binding energies, which means you'll get a set of lines.
You'll get a spectrum of energies.
So this is what the spectrum looks like, and this is, by the way, just a general way of looking at any spectrum so that you train you eye to know what to look for.
If I look at any spectrum, I'm going to have a plot of some kind of intensity, some kind of intensity unit versus some kind of energy unit, versus some kind of energy unit.
And sometimes, energy increases from left to right.
But sometimes they might put energy increasing from right to left as you see in these spectrum.
But energy is along here.
And intensity is related to population.
All other things being equal, if I have a specimen that has two times the concentration of active species, I'm going to get twice the strength of the line.
So the intensity measures population.
The energy is related to the characteristic binding energy of a particular electrons, so the energy is related to the identity of the species.
This is how we can identify certain species.
I've already shown you if you walk in and you see those four Balmer series lines on a spectrum, you know that's characteristic of atomic hydrogen.
Nothing else.
So those are energies.
And then, where do we go?
So we have a line here, but we know that real data have a little bit of a spread to them.
So you don't see discrete lines; you'll see a spread centered at about the value.
So let's look at these.
These are highly stylized, but here you can see-- here's boron for example.
So these are the 1s electrons of boron, and they're held at a binding energy of about 19.3 megajoules per mole.
And then this is 2s, and here's 2p.
Now there's a difference between 2s and 2p, but they're of the same order.
They're roughly about a megajoule per mole.
There's subtle differences.
And this goes back to Sommerfeld, who said that even though there are s and p orbitals, there's a circular orbit and there's an elliptical orbit.
But they're roughly in the same shell and that's what you see.
But there's a huge difference in going from shell n equals 1 to n equals 2.
This is not to scale.
You see they got a little break there.
And now look, what's the electronic structure of boron?
1s2 2s2 2s1.
There are two 2s electrons, but only one 2p electron.
And this is trying to show that the height here is twice the height of the 2s peak.
So you see that we've got two electrons here, one electron here, and it's easiest to pull out the 2p electrons first.
And you go on, there's beryllium and so on.
So we see that we can build data in this fashion here.
We keep going.
Here's carbon, which is 2, 2, 2.
And then oxygen has 2s2-- pardon me.
1s2, 2s2, 2p4.
So this is two electrons, this is four electrons.
Showing that the 2p height is twice the 2s height.
And then finally, neon is 2s2, 2p6.
So this is roughly three times the height of the 2p.
But all of the 2p numbers are roughly of the same order of magnitude and decidedly smaller than what's going on at 1s.
And what else do you see?
Well, the carbon has 6 protons, so the inner electrons of carbon are held not as tightly as the inner electrons of neon, which has 10 protons.
All of this comes out.
The data are all good.
So now what do we want to talk about?
What's all this about?
I said electronic structure leads to properties.
The first thing we want to talk about is reactivity, chemical reactivity.
Well it's pretty clear from looking at these XPS data that those inner shell electrons aren't going anywhere.
They're too tightly held.
So first thing is the only electrons that we can even think about involving in chemical reactivity must be the electrons in the outermost shell.
These are called valence electrons.
The outermost shell is called the valence shell.
So let's get that down because that's probably important.
So chemical reactivity determined by only electrons in outermost shell.
And we're going to term that the valence shell.
And those electrons are called the valence electrons.
And, these are the only ones that we expect to see involved in chemical reactivity.
And we want to have a measure.
You can see that there are subtle differences between the energies that hold the valence electrons of carbon as opposed the valence electrons of neon.
And so what we can do is have a measure of the ability for valence electrons to react.
That is to say, to participate in chemical activity.
And that measure is called the average valence electron energy.
And there's no special symbol for that, so we just call it AVEE.
And it's just the values of the valence electron energies averaged.
So if I want to take the average valence electron energy for say, oxygen.
So I can take it right off of there.
I can see I'm going to need two times the ionization energy of 2s plus four times the ionization energy of 2p.
And those data are up there.
All divided by 6.
2 plus 4, the total number of valence electrons.
And if I go through the math up there I get 1.91 megajoules per mole, which I prefer to expresses as 19.8 electron volts per atom.
And when we go through and calculate the values of average valence electron energies, we find trends.
And here's what the trends look like.
So you have, first of all, following along with the ionization energies, very similar values.
Pardon me.
So here's the plot of average valence electron energy in bar heights arranged as the elements are found on the Periodic Table.
So you have hydrogen here, helium on this scale would be way up.
And then here's lithium, beryllium, boron.
And so you see a monotonic increase as you move from left to right.
And you see a monotonic decrease if you stay in the same column from low mass to high mass.
So what's going on there?
Why do we have that change?
We have similar electronics structures.
We have similar electronics structures in a given column and lithium is s1, sodium is s1, potassium is s1, rubidium is s1.
And let's put some values on here.
Let's put some values.
For the majority of elements in the Periodic Table, we get values of average valence electron energy less than 11 electron volts.
And when we have such values, we have the following properties: the valence electrons are weakly held.
The valence electrons are weakly held.
How do we know this?
Because the binding energy isn't so strong.
That means the electronic are weakly held.
So we say that the element is a good electron donor.
It's a good electron donor and we term this a metal.
The property that makes an element a metal is that it has a low value of average valence electron energy and therefore, it is a good electron donor.
And it turns out about 75% of the Periodic Table is made of metals.
At the other extreme we have high values of average valence electron energy.
High values.
So when the average valence electron energy is high, we have the complementary set of properties.
So that means the valence electrons are tightly held.
Valence electrons tightly held.
Tightly held.
And so the element is a poor electron donor.
But complementary fashion, it is a good electron acceptor because when it gets near electrons it tends to grab them and hold on to them.
So this is a good electron acceptor.
And we term such a element a nonmetal.
And so roughly, about 25% of the Periodic Table is non-metallic.
And I think I've got that shown here.
And you see just a few elements off to the upper right.
And then in the middle, we've got this-- about a half a dozen elements and they have values of average valence electron energy intermediate between 11 and 13 electron volts.
And so, they can behave either as electron donors or electron acceptors depending on who else is in the room.
So if you put an element like silicon with a very strong metal, silicon will act as an electron acceptor.
If you put silicon in the presence of something that's a very strong nonmetal, silicon can act as an electron donor.
It has that dual property.
And these elements are called semimetals or metalloids.
And I think I've got some images to portray that.
There's the semimetals.
And if you look carefully at the Periodic Table that you have, you'll see that there's this red staircase here.
So there's silicon, arsenic, tellurium, antimony and so on that act as semimetals.
They can behave in both fashions.
All right, so now with this classification scheme, let's start thinking about chemical reactivity.
And seeing if we can, on the basis of where the elements are found on the Periodic Table, start to make some judgments.
Well I know one thing right off the bat that along the right-hand column, the noble gases are chemically inert.
That much we know.
Noble gases are chemically inert.
I know there's a Nobel Prize out there for getting xenon to combine with fluorine, and the chemistry textbooks love to point out these exceptions.
But by and large, the group 8 elements are chemically inert.
So what do we know about their electronic structure?
They all have the same electronic structure ns 2, np 6.
With the exception of helium.
Helium, because it's 1s 2.
There's no p. So beside helium, starting with neon, we've got 2s2, 2p6.
Argon is 3s2, 3p6, and so on.
n2 p6, it has 8 p electrons.
It has a full valence shell.
And because the last electrons are p electrons, this is termed octet stability.
There's something about having a full shell that renders the system satisfied energetically.
And it doesn't want to react anymore.
So we say octet stability, and if you want to be a wise guy, parenthetically you whisper helium is due at stability.
But you know, whatever.
All right, so let's make this octet stability a hypothesis and see how far we can go with it.
So the next one I want to look at is sodium.
Now sodium, it's got the electronic structure 1s 2, 2s2, 2p6, 3s1.
Now, I really don't need to talk about anything up to n equals 3 because this is not valence electrons.
I just put it up there for completeness, but this is all we have to worry about.
Now we know that sodium is a metal.
It's a metal.
It's got an average valence electron energy of about 5.2 electron volts, which happens to be the ionization energy.
Because that's trivial; it's only got one electron, so the average valence electron energy is equal to the ionization energy.
So it's a metal.
So it's a good electron donor.
But let's think about that.
If we could get rid of that electron in sodium, we could then turn it into something that has the same electronic structure as neon.
And why do we want to do that?
Well because there seems to be an energy well there.
If we can make something neon-like, then it's going to be happy.
So let's do it.
Let's write a reaction sodium to lose an electron and become sodium plus.
And sodium plus now is this minus 3s1.
So it's isoelectronic with neon, and neon has the octet stability.
But charge neutrality forbids me to say, OK, lose an electron.
It won't.
The only way for sodium to lose an electron is to find an acceptor.
So if you put a good electron donor in the presence of a good electron acceptor, then the donor can give an electron to the acceptor and both profit from the transaction.
So, where are we going to find a good electron acceptor?
Well I just told you, the nonmetals are the really good electron acceptors.
So let's go zooming over to the right-hand edge of that row and look at chlorine.
We need an electron acceptor.
So choose just for argument's sake, chlorine.
It's got an average valence electron energy way, way up there.
So let's go.
Chlorine, it has the electronic structure of neon plus 3s2, 3p5.
So chlorine, if it could acquire an electron, would then become isoelectronic with argon.
So imagine chlorine plus electron becomes Cl minus and Cl minus is isoelectronic with argon.
So now I've got two elements.
One that's one electron richer than inert gas structure, and one that's one electron poorer than inert gas structure.
If I put them together, put them both together in the same reactor, they will react and electron transfer occurs via electron transfer.
Chemical reaction occurs.
Chemical reaction occurs through electron transfer.
And what's the purpose of electron transfer?
It's to achieve full valence shell.
Achieve full valence shell occupancy, if you like.
Achieve full valence shell occupancy.
So, now we've got it.
But there's more.
There's more.
What happens after the electron transfer step?
We don't have neutral species anymore.
The sodiums have given up their electrons to become sodium ions.
And the chlorines have acquired electrons to become chloride ions.
And these are free.
Let's make this simple.
Let's make it a gas phase reaction.
So sodium vapor and chlorine gas have become sodium ion vapor and chloride ion vapor.
What do you know happens when you have charges of opposite value, opposite polarity?
There's a coulombic force of attraction.
So the sodium will attract the chlorine.
So let's put the two together.
So sodium attracts chlorine, but it's not over.
There's more sodiums, there's more chlorines.
So chlorine attracts sodium, which attracts sodium.
Which could attract more chlorines.
Which could attract more sodiums.
And what do you see happening here?
Well we're now starting with a gas phase and we're forming some giant atomic aggregate.
This is a giant ion aggregate.
What's the state of matter going to be if this thing's honking big?
It's going to be a solid.
We're going to form a solid starting with those two gases: sodium vapor and chlorine vapor.
And what about the solid?
What do we see about the atomic arrangement here?
All the sodiums are the same size and all the chlorides are the same size.
They have the same coulombic forces of attraction in between them.
Can you see that they're going to form not just a solid, but they're going to form a solid consisting of atoms in a regular array?
This is going to be an ordered solid.
And there's a plain Anglo-Saxon word for an ordered solid, it's called a crystal.
Now look at how far we've come.
Look at how far we've come.
10 minutes ago we hadn't done anything about chemical reactivity.
We made one observation.
We had the tools of course.
We were ready.
We're ready for discovery.
We had the tools.
We had the average valence electron energy.
We knew that helium, neon, argon, krypton, xenon, radon, et cetera, all have this inertness.
We postulated that the octet stability was some-- as a sweet spot.
And then we went with it.
And operating with that one postulate, we get to electron transfer and we're concluding that if I take a really, really good metal-- see, I can generalize this now.
There's nothing peculiar about sodium that makes it react this way and no other element will react this way.
I could replace sodium with any other metal.
I could say any metal, any metal will react with any nonmetal in order to achieve octet stability.
So now you can take big pieces of the Periodic Table and conclude that entire sets of elements will act in this way to form ionic compounds, crystals, atoms of regular periodicity.
That's a nice pun there, right?
We started with the Periodic Table and now we have periodic spacing of atoms.
I think that's beautiful.
Obviously this view is not shared by the rest of the audience.
This crystal consists of ions.
It's held together by what chemists like to refer to as bonds.
This is called an ionic bond.
And what you're witnessing here is ironic bonding via electron transfer.
So we've now categorized the first form of primary bonding.
Sodium chloride, magnesium oxide-- ionic bonding.
And ionic bonding must give us solids at room temperature.
If you've got ionic bonds and you don't have a solid, clearly you are at elevated temperature where at elevated temperature, the thermal energy is disruptive enough to break those bonds.
Those are very, very strong bonds.
And what we're going to do next day is look at the nature of those bonds.
So I think I'm going to stop the formal lesson at this point.
And by the way, perhaps I failed to mention on the first day of the class.
What I'm doing is I'm lecturing for, out of the 50 minutes, about 45 on hard core topics.
And then, about the last 5, 7 minutes, I want to go to something related to these-- chemistry in the world around us.
That's why you see this little break.
But it doesn't mean that class is dismissed.
There's a difference between changing topics and dismissing class.
I don't know how people confuse the two.
But somehow I've inadvertently communicated to you evidently.
So we're not dismissing yet.
We're going to try to apply this knowledge somehow.
So what I wanted to do was to show you what happens with these ionic solids in commerce.
And in particular, I wanted to talk about metallurgy and the best form of metallurgy, which is of course, electrometallurgy.
Which is the kind of research that I'm involved in.
And electrometallurgy is quite pervasive.
The aluminum beverage can is made by electrometallurgy, as is magnesium.
So I'm going to talk to you a little bit about magnesium.
This is a bar of magnesium.
Obviously, it's magnesium because if it were steel, it would be so massive that I wouldn't be able to fling it around like this.
The density of iron is 7.87, the density of magnesium is about 1.76.
It's less than 2.
It's 2/3 of the density of aluminum.
It's lighter than aluminum.
Now perhaps some of you've been told that magnesium burns.
Well, let's see what happens.
There's a billet of magnesium and this is the key made of steel.
We're OK. We're OK.
It's not going to burn.
That's a myth.
Magnesium in powder form, magnesium in ribbon form will burn.
The surface to volume ratio is so high that the oxidation generates a lot of heat.
But here the surface to volume ratio is so low that it's fine.
In fact, I went to school at the University of Toronto.
And at the time the department head was a man by the name of Pigeon who invented a process during World War II to make magnesium cheaply.
And when he came into the classroom to lecture on the day that he would talk about magnesium, in those days they had lab benches at the front of the class with sinks and gas outlets.
And he would put the magnesium billet on a stand and light a Bunsen burner under it.
And of course, people in the class sort of leaned back, and he would lecture for 45 five minutes with the Bunsen flame burning the whole time.
And then towards the end he would extinguish the flame, let the billet cool.
And then pick it up and go back to his office with it to make the point.
All right, so how do we make these metals?
Well, what we do is we reverse the electron transfer step.
And we start with the cations and anions, and we force the electrons back onto the cations anions.
Thereby, creating the neutrals again.
This is called electrolysis.
So cation plus electron gives us neutral.
Anion minus electron gives us neutral.
So in the case of magnesium, we start with magnesium chloride, which dissolves and forms magnesium cations and chloride anions.
So, at elevated temperature-- in this case, at about 700 degrees Celsius, the ionic solid becomes an ionic liquid.
And it's clear, colorless, and has the fluidity of water.
And this is a very simple schematic, so we've made the cathode of steel negative and on the negative electrode, magnesium ions are reduced to magnesium, which is a liquid at this temperature.
And these are little bubbles of magnesium and they pool here.
And we collect them and syphon them off.
And on the other electrode, we're making bubbles of chlorine.
And they rise and we collect them.
So we reverse nature.
If you put chlorine in the presence of magnesium it wants to form magnesium chloride.
Here we do the reverse.
So electrolysis undoes the spontaneous electron transfer of ion formations.
So I decided I better get something more authoritative than my little cartoon.
So first I looked on the library website and I found that there was an article in this volume of Advances in Molten Salt Chemistry.
And there's this article here, an authoritative article about chemistry and electrochemistry of magnesium production.
And so I decided to turn to the article and let me read the first paragraph of this authoritative article.
A cubic kilometer-- what is a cubic kilometer?
It's a cube, 1 kilometer on edge.
A cubic kilometer of sea water contains approximately 1 million tons of magnesium.
1 million tons of magnesium as the salt magnesium chloride decahydrate dissolved in sea water.
More than has ever been produced in one year by all the magnesium plants in the world.
The world production of magnesium right now is about 600 million tons.
No, furthermore, sea water contains only 3.7% of the total magnesium present in the earth's crust.
Clearly magnesium resources are ubiquitous and virtually inexhaustible.
So when people tell you we've got to recycle because we're running out of resources, I read this-- I don't know.
You might want to recycle for other reasons, but you can't make the case it's because of scarcity.
This is magnesium that has been made by this electrolytic root.
And what's the value of this?
It can substitute for steel in automobiles.
It's density is less than 2.
Steel is about 8.
Factor of 4 you lightweight the vehicle, which means per unit distance traveled, less fuel consumed.
Fewer emissions.
And less fuel, which means less dependence on imported petroleum.
So why aren't we doing this?
Why aren't we lightweighting our vehicles with magnesium.
Because the bonds are so strong the energy required to make this metal is so high that it's very costly.
If I give you a light bulb that costs three times what the other light bulb does and burns only twice as long, you won't buy it.
If it costs three times as much and burns 10 times as long, then you're willing to pay the premium.
So what we do in my research group, among other things, is study the electrochemical processes by which we make these metals in order to make the process more efficient.
Thereby reducing the cost, making these lighter materials more competitive.
Thereby, making the world a better place by electron transfer in service of humanity.
And I invite you to do the same.
But in order to do so, you have to learn the lessons of 3091.
Because after all, this is the most important subject you will take at MIT.
All right, we'll see you on Friday.
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OK. OK. Settle down.
Settle down.
Settle down.
So the weekend doesn't begin until after the 3.091 lecture.
OK. A couple of announcements.
Tuesday, Quiz 3 based on Homework 3.
The final exam has been scheduled.
In fact, all the finals are scheduled now, so please consult the Registrar's listing.
The celebration of celebrations, the 3.091 final exam, will be Tuesday, 15th of December in the morning, 9:00 am to 12:00 noon over in the Johnson Athletic Center.
So I urge you to go through the final exam schedule and then make your travel plans when you know what your last obligation is.
And book them soon, because we've got about a quarter of a million students in the Boston area and the semesters all come to a close within a very narrow time window, and everybody's trying to get through a wormhole called Logan Airport.
So you want to be ready.
Do not try to leave town before you've met all of your obligations.
You can't leave before you've finished all of your finals.
But you know what they are now, so call your travel agent or make your internet booking.
And an announcement for Professor Paul's section.
Owing to the holiday on Monday, he's going to have office hours from noon to 1:30 today.
So if you're in Professor Paul's section, you'd like to catch up with him, noon till 1:30 today.
All right.
Let's get to the lesson.
The last day we started looking at octet stability, and we looked at octet stability and what it means in terms of shell filling and some sweet spot in energy with respect to reactivity.
And a filled shell leaves us with this electron configuration, ns2np6 for n greater than 1.
In the case of n equals 1, there's just ns2 for helium.
Otherwise, ns2np6.
2 plus 6 is 8.
There's the octet stability.
It's observed trivially in noble gases.
And then we saw that it can also be observed in certain ions.
And how do we get to octet stability?
Electron transfer.
And here's the prototypical reaction.
I like this.
This is very iconic.
It puts it in broad terms.
The marriage of an electron donor with an electron acceptor leads to the formation of a cation and an anion, thanks to electron transfer from the donor to the acceptor.
It doesn't end there, though.
You have cations and anions in the gas phase and they're attracted to one another by coulombic forces, or electrostatic forces, and that leads to ionic bonding.
And today I want to go deeper into it and study the energetics of ion-pair formation.
So let's do it.
And what I'm going to do is study this with reference to an example of sodium chloride.
So we will go through the energetics of sodium chloride, which we know is going to exist as Na plus and Cl minus-- sodium being the electron donor and chlorine being the electron acceptor.
So what I'm going to do is show you a plot of energy as a function of separation.
So you're going to have to be pluralistic here in your ideas.
So, for example, if I wrote this word by itself you don't know if I'm saying lead or I'm talking about the metal lead.
The only way you know is in context.
And so you have to know in context here.
So r is not the radius of the atom.
In this case it's the symbol.
And I'm using this symbol not to confuse you.
This is what professionals use.
So if you go to the text and the literature you'll see r. This is the interionic separation, and it's determined nucleus to nucleus.
So I'm going to put a sodium ion at the origin.
This is the origin.
Energy is on the ordinate.
So I'm going to model this as a hard sphere.
So this is sodium ion, Na plus, and it's got some finite radius.
And its radius, I'm going to call it r-plus.
That's the radius of the cation.
And next to sodium I'm going to put the chloride anion.
And it's bigger.
It's bigger.
It's not to scale, but it's bigger.
So this is chloride, the Cl minus, and it has its radius, rCL minus.
And then the interionic separation is measured from the sodium nucleus to the chloride nucleus, and it's given the symbol r0.
That's the interionic separation.
And we could we write that r0, in fact, is strictly r-plus plus r-minus because we're using a hard sphere model.
So there's no shrinkage when the two touch.
So that makes sense.
Now what I want to do is calculate the energy here.
The only energy that we have here is electrostatic.
So we're going to start and say, imagine if we had these two ions separated by infinite distance and we brought them to a separation of r0.
How much electrostatic energy would be stored there?
So I can write that as E, and I'm going to call it attractive.
There's a force that attracts these two ions together, and that's given by Coulomb's Law.
So that's the product q1 q2 over 4 pi epsilon0r, in general.
I'm making this as a function of r and then we're going to figure out how to get to r0.
Let's put a little ledge in here.
So q1, I'm going to make that the sodium just for grins and chuckles.
So q1 is equal to the z, the valence on the sodium, times the charge, the elementary charge.
So the charge on the sodium is plus 1-- so it's plus 1e-- and q2 is going to equal the charge on the chloride times e. And in this case the chloride ion is negative 1, so q2 is minus e, q1 is plus e. Now if instead I were doing magnesium oxide, magnesium would be plus 2, oxide would be minus 2.
So that's where the charge on the ion comes into play.
So we can put that in here and will make this z-plus times e, z-minus times e. So there's q1 q2 over 4 pi epsilon0 r. And in this case, for sodium chloride, this is plus 1 times minus 1 is minus 1 e squared over 4 pi epsilon0 r. So this is a function of r. This e is 1/r.
That's the hyperbole, and I can draw that here.
So it's going to look something like this.
All right?
It's going to be a right like this.
Good.
So that's the attractive force.
Now how do we avoid the ions just blending together?
Why do they stop at r0?
What puts the brakes on the coulombic force?
Ah!
For that we have to look at the fine structure of sodium.
So let's look at the fine structure of sodium.
So sodium is net charge plus 1.
But sodium more properly, if I go to the next level of structure, is really 11 protons in the nucleus and it's got electrons around it-- in total, 10 electrons.
So it's net charge is plus 1, all right?
Now if I'm way over here and I look at sodium I just see something of charge plus 1.
And, in fact, I could model sodium as just a point charge, a plus 1, and do all the electrostatics and I would be perfectly accurate in my estimation.
But as I get closer I see there's fine structure.
It's like so many other things in life.
You know, from a distance it looks really good, and you get up close you go, eww.
[LAUGHTER]
I'm not going to mention any names, all right?
But it's Friday.
Be careful.
So as I get up closer I realize that it's plus 1, but the plus 1 is plus 11 mediated by minus 10.
So when things start getting close together this exterior of negative charge becomes manifest, palpable.
So let's look at the chlorine.
The chlorine's coming, sees plus 1, plus 1.
It starts getting closer and closer.
And now what happens?
The negative electronic configuration on the outside of sodium is interacting with the negative electronic configuration on chlorine, and the result is you have electron-electron repulsion.
And we've got an equation for that, and that's given by E repulsion is equal to some constant b over r to the n. And this n is called the Born exponent.
It's not the quantum number.
So r today means interionic separation, n means Born exponent.
And you have to determine it by experiment and the value lies between 6 and 12.
And we have to determine b by experiment.
So let's say this thing's got a-- pick a number in the middle.
Say it's 8.
So let's plot r to the eighth.
So that's going to hug the abscissa, and it's going to come in really, really close.
And then somewhere around this distance here, where the electron-electron repulsion starts to be felt, this thing takes off.
But 1 over r to the 8 goes way, way up fast.
So instead of being a gentle curve it's more like a hockey stick shape here.
And so now the net energy is the sum of the negative attractive energy and the positive repulsive energy.
And if we sum the two what are we going to get?
Well, way out here 1 over r to the 8 is negligible.
And the net value-- so E net, in red-- is essentially equal to E attractive.
And at very, very low values of r, r to the 8 dominates r. So we have this as the net value.
And the two sum-- and I can't quite do it because of the way this is drawn, but I think you can see.
These two eventually equilibrate and go through a minimum.
So I'm going to cheat a little bit here and I'm going to add these two in such a way as to go through a minimum value of energy at r equals r0, which is the sum of E attractive and E repulsive.
So we can sum those two and let's see what we get.
At r equals r0, E net is equal to its minimum value.
So how do you find a minimum in a function?
You take the derivative.
I'm going to put some math to work here.
So we'll take dE by dr and set it equal to 0.
And the value of r at which it equals 0 is termed r0.
OK?
So we'll just go through and take the derivative of that thing.
And I'm not going to go through all the math but just show you the set-up here.
What's the other thing that we know?
dE by dr represents what?
That's force.
So I could say that at r equals r0 the net force is 0, which is what you'd expect.
Because if the net force isn't 0 it's going to either push the ions farther apart or draw them closer together.
So this is mathematics imitating reality.
What a concept.
Math working for you instead of you working for math.
So what do I have?
I have everything in these equations except the value of b. I'm assuming we know the value of n from experiment.
We don't know the value of b.
And so you can solve to get the value of b, and once you've got that you can put everything together and give an expression for the energy of the system at r equals r0.
When you plug everything in you get this: z-plus, which is the net charge on the cation, times z-minus, which is the net charge on anion, divided by 4 pi epsilon0 r0.
1 minus 1/n, where n is the Born exponent.
And this is valid at r equals r0 only.
If you're not at r equals r0 then you can get the value of b and then put it into this expression, where E net will equal E attractive plus E repulsive.
So there it is.
And so this represents-- this is the energy of a single ionic bond, because that's all the energy that's there.
It's the single ionic bond.
And the second thing that we realize is plus times minus is net minus, so this means that it's negative quantity, as it should be.
It has to be a negative quantity.
All right.
So what do we have here?
What we've seen by going through this derivation is the recognition that the ionic bond is electrostatic attraction mediated by electronic repulsion.
It's the balance of the two.
And those words sound so good to me that I'm going to write them down.
Electrostatic attraction mediated-- another lovely word-- mediated by electronic repulsion.
So that's how you get to the final setting here of the interionic separation.
So what are the characteristics?
What does this lead to in terms of characteristics of this bond?
Characteristics of the ionic bond.
First of all, it's omnidirectional.
This is a concept based on the fact that the electric field radiates in all directions uniformly.
So the negative field coming from the chloride ion is uniform in all directions.
There's no preferred direction.
Omnidirectional.
OK?
E field-- oh, I better not say that.
Electric field, not the energy field, radiates in all directions uniformly.
And that's going to have consequences.
I'm not just telling you this because we like cataloguing things.
This isn't a bookkeeping class.
So we're going to come back, use this fact.
And the second thing is that the bond is unsaturated, which is a chemical way of saying that a given ion can bond to more than one other ion.
In other types of bonds that's not the case.
A given atom can only bond once and then it's done.
Whereas in this case the ion can bond to a plurality of other ions.
So ions bond to more than one.
OK?
Plurality of bonds is formed.
They're polygamous, if you like.
So what does that mean?
That means that here is what happens.
We've got the blues as the sodiums, and for any given sodium it forms bonds without limit until the number of bonds is stopped by physical limitations-- not because the E field was saturated.
It's unsaturated.
You just can't jam any more chlorides physically around the sodium.
That's why the sodium is only bonding to the number of chlorides that it bonds to.
There's no intrinsic limitation.
So what happens when you get to this situation where you have omnidirectional forces, unsaturated bonds, and ions that you can model as hard spheres of constant radius?
All the sodiums have the same radius, all the chlorides have the same radius.
You make a 3-dimensional ordered array.
So you can make an infinite atomic ordered array, which we use the simple Anglo-Saxon word last day to describe: crystal.
You form a crystal.
And as a result ionics have to be solid at room temperature, because if you've got thousands and thousands of atoms together in one aggregate they're not going to float.
They're going to settle.
Put another way, the strength of the bond, the amount of energy in here, is so great that the thermal energy of the room isn't great enough to disrupt this bond.
It's a combination of unsaturated, omnidirectional, and high energy.
So we form solids at room temperature.
OK. Now I want to show the energetics of that one because this is good.
You know, I promised you I wouldn't do derivations, so I'm not going in detail on this.
I'm giving you just enough so that I can introduce the characters here.
You know, how else am I going to introduce the Born exponent?
Am I'm just going to say, there's this exponent n, the Born exponent.
We're going to introduce it in context.
So now you know what the energetics are.
So now I want to prove to you energetically along this line that crystals will form.
So let's imagine-- we're going to do this thought experiment.
We're going to take three ion-pairs of sodium chloride.
So here's three ion-pairs of sodium chloride.
And I want to compare these three ion-pairs.
So this is an ion gas.
And the distance between ion-pairs is great enough that one pair doesn't affect the other.
The electrostatics are only strong within the pair.
So we'll just label this infinity with quotation marks around it.
They're very far apart.
When a physicist says they're very far apart, very is code for infinity.
So this one doesn't interact with this one, which doesn't interact with this one.
And I want to compare the energetic state of the ion dispersion to what would happen if I were to put all of those in a single line.
Plus, minus, plus, minus, plus, minus.
So then this has a certain energy state, it's the energy of the ion line.
And I want to show you that there's an energy decrease in collecting all of these and ordering them into a line.
So there's more energy in a line dance than in ballroom dancing.
That's what we're going to say ultimately.
OK?
So let's compare the energies.
That's all we're going to do.
So what's the energy of this?
Well, we're not going to do 3 versus 3 Let's think big.
It's Friday.
So let's take Avogadro's number of pairs, shall we?
So the energy of the ion dispersion would then equal-- that's the energy of one pair, and they're infinite distance apart, so there's nothing to be gained by putting them in the same chamber.
So it's just going to be N Avogadro, if that's the number, times the energy evaluated at r equals r0.
So we're done.
And we know what that is.
I don't have to rewrite it for you.
OK.
So now what I have to do is get an estimate of the energy of the line and show you that the line is at a lower energy state.
Well, let's see.
Jocelyn, take the top one and the middle one, please, but not the bottom one.
Thanks.
OK.
So now let's look at the energy of a line.
So we're going to do-- here's my line.
I'm going to get the colored chalk.
Green and blue.
You know, some people think being a professor is so cool because you get to travel, you get to research, and so on.
It's the colored chalk.
[LAUGHTER]
It's the colored chalk.
All right.
So there's a sodium.
And on each side of the sodium we'll put a chloride.
All right?
And I'm going to just keep going this way.
Here's a sodium and here's a chloride.
All right.
And now what I'm going to do, I'm going to start here at this sodium and I'm going to count the energy, electrostatic energy, that's in this system.
So we're going to count to the left, we're going to count to the right, and then we're going to multiply it by the total number of ion-pairs.
And I know that the ones on the end aren't the same, but the number of ends over N Avogadro, that's peanuts.
The edge effects are negligible because there's such a giant middle.
That's how you model this stuff.
You don't obsess over the fact that the last ion doesn't see anything on that side.
You just do it and forget about it, because you know you wouldn't do it for 5.
This would be a big problem.
But if you have Avogadro's number, who cares?
It's called risk assessment.
All right.
So let's look at the energetics here.
So what we've got is the energy of the ion line.
OK?
So let's start with this central one.
And separated by distance r0 is the chloride, and there's an attractive energy here.
So that's going to equal minus e squared over 4 pi epsilon0 r0.
Now I'm going to keep going, because the field is unsaturated and goes in all directions.
So this sodium, that's a distance 2r0 away.
It's got a repulsive force exerted on this sodium.
So let's add that.
So that's going to be plus.
A repulsive force raises the energy of a system.
That's e squared over 4pi epsilon0 times 2r0.
And let's keep going.
So now let's go 3r0 away.
So 3 times r0, that gets me out to the chloride over here.
Let's put 3r0.
Now that will take me from the center of the sodium to the center of the next chloride.
And that's going to be attractive.
So that'll be minus e squared over 4pi epsilon0 times 3r0 plus, et cetera.
So you see how this goes.
So you go all the way out, you add them all up, and you go the other way.
And so on and so forth.
So that's how the derivation goes.
You might say, hey wait a minute.
What happened to the Born exponent and the repulsive energy term?
Well, where's the repulsive energy term going to be felt?
It's called electron-electron repulsion.
Axiomatically, the electrons here are nowhere near these electrons.
See, you only have to count it for the nearest neighbor.
So we can patch that in at the end.
And we do.
I haven't forgotten.
But I'm not going to spend a whole day trying to derive this thing.
I'll show you how it starts to evolve.
At some point you end up with something that looks like this.
e squared over-- in fact, I'll put the minus sign up here.
Minus e squared over 4 pi epsilon0 r0.
And you're going to double it, because you're going to go one side and the other side, and you're going to get a series that looks like this: 1 minus 1/2 plus 1/3 minus 1/4 plus blah, blah, blah.
Yeah.
OK.
So what does this look like?
I've really broken this into two pieces.
So this coefficient out in front here, you should now be able to repeat this in your sleep: e squared over 4 pi epsilon0 r. This is electrostatics, isn't it?
Electrostatics.
This is the consequence of Coulomb's law.
What's this second term here?
What's this all about?
Geometry.
This is dictated by atomic arrangement.
So I could calculate this if I took, instead of a line, what if I put them in a sheet subject to the constraints of those sizes and plus 1 and minus 1?
So what would be?
I'd start at the sodium and count how many chlorides?
If I'm on a plane there'd be one, two, three, four.
And then how many sodiums?
Well, they'd be on the backside of each of the chlorides.
And I'd add them all up in 2-space and I'd end up with another coefficient here.
All right?
And we compress all of this into a coefficient which we call the Madelung constant.
And it's a function of the atomic arrangements.
So different crystal structures have different Madelung constants.
It's named after a German professor, Madelung.
In 1910, he published calculations for the energy of a system of point charges-- just abstract theoretical paper.
And then about 10 years later another German professor by the name of Paul Ewald-- he did his PhD for Sommerfeld-- he published a paper in which he actually made the calculation for ion crystals, and he came up with this constant.
And to show you the class of the guy, instead of naming the constant after himself he named it after Madelung.
Now that's class.
So Madelung did the first calculation so he gets named.
So now what we're going to do is we're going to multiply by the N Avogadro, because I've got N Avogadro of these things, and we're going to put in the Born exponent patch and so on.
And here's what the final expression looks like for the line.
There are a few algebraic tricks that I'm not willing to do in class because I don't think that's a profitable use of our time in a chemistry class.
But if you want to try the derivation I have it in full and we can compare notes.
So once we get the patch in it's going to look like this.
It'll be minus.
There'll be the Madelung constant times N Avogadro e squared-- and this is already assuming it's plus 1, minus 1.
If this were magnesium oxide there'd be a 4 in here.
4 pi epsilon0 r0 times 1 minus 1/n, where n is the Born exponent.
So compare this to this one here.
What's the only difference?
The only difference is the Madelung constant, right?
It's the only difference.
So E of the pair dispersion is really equal to E of the line divided by the Madelung constant.
So what I'm trying to prove to you is that E line is more negative than E of the dispersion.
So it all hinges on the magnitude of M. If M is greater than 1 we win.
If M is less than 1 I've just proved to you that water runs uphill, so that's a bad day.
All right?
So let's calculate the value of M. And you can go to your algebra books.
And you've got this series natural log-- and engineers write natural log "ln."
I know the mathematicians write "log."
Uh-uh.
Engineers-- uh.
That's 1 plus x. OK?
Natural log, 1 plus x.
You can expand this as x minus x squared over 2 plus x cubed over 3, dah, dah, dah, dah.
You know, look that one up.
Set x equal to 1, which is what we've got, right?
Because we've got 1 minus 1/2, 1/3, dah, dah, dah, dah.
Go through it and you'll get the value that M, according to this, will give you 2 times the natural logarithm of 2, which is 1.386-- which is greater than 1.
And so we're golden.
That means that the energy of the line is lower, more negative, than the energy of the dispersion.
So I'm going to do this pictorially.
Let's make an energy level diagram.
And the energy level diagram will look like this.
So up here the energy is 0 and everything is negative.
So if I put this as minus 1 unit.
All right?
These are all negative values increasing in this direction.
So this is the dispersion of ion-pairs.
So all we've done is take a cation and put it to the anion.
We've just seen that if we do the calculation for the line it's 1.386 times whatever this is.
So that gives us-- this is for the ion line, which I'm going to take the liberty of calling a 1-dimensional crystal.
It's a 1-dimensional ordered array.
And what I can do is go to the 3-dimensional array, start at the lower right-hand corner with that chloride.
Calculate the distance to each of the nearest neighbor sodiums.
Go through the geometry.
The next nearest neighbor chlorides, the next nearest neighbor sodiums.
And you'll build an infinite series and you'll evaluate it.
And in three dimensions it's even lower.
It's 1.7476.
So this is for the 3-dimensional crystal.
This is for the ionic crystal.
All right?
3-D crystal.
So what this is showing is that the system keeps making more and more bonds.
Why does it make bonds?
Because the more nearest neighbors it has the lower the energy goes.
So making a 3-dimensional crystal is energetically favored.
Now there are different Madelung constants for different crystal structures.
You say, well wait a minute.
How do you get different crystal structures?
Suppose instead of sodium it's potassium.
What's the only difference?
Potassium is plus 1.
What's the difference?
Size.
Potassium is larger.
They're not going to pack quite the same.
And so depending on the relative ion sizes-- I mean what if I have something like silver iodide?
Iodide is huge.
Silver is so small it'll fit into the interstices between touching iodines.
So it's going to have a different crystal structure, and the different crystal structure will give us a different Madelung constant.
And there it is.
So we've come a long way with that little assumption of octet stability.
So now let's take a look at what the properties are of these things.
They're solid at room temperature because we've got strong bonds.
High melting points.
Bonding is related to melting point.
Now think point is dictated by bonding, because now you're comparing thermal energy versus the cohesive energy of the crystal.
So tightly bonded substances melt at high temperatures, weakly bonded substances melt at low temperatures.
Transparent to visible light.
How do I know that?
Because I'm the professor.
No.
How do I know that?
How do we think about it when someone says to you is something transparent, in this case to visible light.
Well, what I do is I say, here's the solid.
This is the ionic solid and here is visible light, h nu.
And I'm going to write 2 to 3 electron volts per photon.
And what happens when I want to decide whether this is transparent to visible light?
Now look at the modeling here.
Photon is a squiggle.
I don't know if that's Cartesian space.
This is like a crystal, right?
I'm speaking California.
It's like a crystal.
So now if I go inside I want to make the energy diagram.
So what's the energy diagram look like?
All right?
So now the question is how does-- if this is the energy diagram of the crystal and I make this the energy of the photon of visible light, I'm going to compare how much energy the photon has versus how much energy it takes to excite electrons all the way to a new state.
Because if you don't excite them all the way to a new state-- they can't go part way, so nothing happens.
And if nothing happens the photon goes through and that's transparent.
Now what do I know about the binding energy and the energy level diagram of Na plus?
Well, it's isoelectronic with neon.
And I know that neon has an average valence electron energy of about 20 electron volts.
So my guess is the visible light is not going to do anything.
And so it just passes right on through.
See, we got all that.
Electrical insulator.
How do I know that?
Well, all that glitters is not gold, but it must have free electrons.
And these electrons are all bound, and they're tightly bound.
Hard and brittle.
If it's going to be ductile the atoms need to be able to slide over one another.
Well, there's no way these can slide over one another because to slide over one another requires that at some point the two sodiums are going to be nearest neighbors, and the repulsive forces are so high the crystal fractures.
So if you try to deform an ionic solid you will get it moving in accordance to the elasticity.
So force will be proportional to the extension-- Hooke's Law-- but if you try to plastically deform it, you shear it.
Soluble in water.
We'll come back to that later.
Melt to form ionic liquids.
And good for electrolytic extraction of metals.
I showed you magnesium last day.
Today I'll talk a little bit about aluminum.
But that comes later.
OK.
So where do we find elements that are going to form ionic solids?
Well, you go back to the beginning of the lecture.
What do you look for?
You've got to find the box of really good electron donors and the box of really good electron acceptors.
That's where you go.
So the good electron donors are at the left side and the good electron acceptors are at the right side.
So if you take sodium plus chlorine you get sodium chloride.
If you get calcium plus fluorine, calcium fluoride.
Magnesium plus oxygen, and so on.
OK?
Yeah, this shows.
Yeah.
Aluminum plus oxygen, yeah.
There's Max Born.
Max Born, he got a Nobel Prize.
And so did Fritz Haber, but we're going to come to him in a minute.
All right.
So now I want to do this energetic calculation one more way, because right now we've been operating with ion gas but sodium isn't found normally in the form of an ion gas.
So let's do something that starts with elements found in nature.
So we want to form an ionic crystal from elements in their natural state.
And what's going to happen is en route we're going to be able to define a few more terms.
So that's going to make everybody happy because we get more definitions.
And this is called the Born-Haber Cycle, named after Born and Haber.
And what we're going to do is we're going to-- pardon me, there's a C in there-- we're going to invoke Hess's Law.
And Hess's Law is sort of like Kirchhoff's law for electric circuits.
Hess's law says that the energy of a chemical change is path independent.
So energy change in a chemical reaction is path independent.
It's sort of like potential energy in Newtonian mechanics.
It doesn't matter if you take the elevator to the top of the Hancock Tower or if you walk up the stairs, the change in potential energy is the same when you express it from the top of the Hancock Tower, although you might be somewhat more exhausted having walked the steps instead of taking the elevator.
But you don't get any credit in terms of the potential energy.
So let's use Hess's Law in order to describe the formation of sodium chloride.
So I'm going to start with sodium as it's found in nature, and I'm going to talk about room temperature.
So sodium is a solid at room temperature, and I'm going to react it with chlorine gas.
Chlorine's a gas at room temperature and it's a diatomic molecule.
And we're going to react it to form sodium chloride, which is a solid and a crystal.
Now you might say, well isn't that kind of redundant?
No, because later on I'm going to teach you about a form of solid matter that does not consist of atoms in a regular array-- disordered solids.
So we're specifying, I want to form crystal and solid sodium chloride, because that's the reaction that would occur if we were to do it in the lab.
And what have we calculated so far?
What we've calculated so far is this.
We've calculated chloride ion in the gas phase plus sodium ion in the gas phase reacting to form the crystal.
And we've called this the energy of crystallization.
That's this Madelung stuff.
This is the Madelung energy here.
All right?
I'll even put M here.
Madelung energy.
And why am I using H?
Because that's what the books use.
H is enthalpy, and for condensed matter the difference between enthalpy and energy doesn't amount to much.
So H, just for the record, is enthalpy.
And we've been operating with E as energy.
It's almost equal to E, which is energy for condensed matter.
For the gas phase it gets hairy.
OK.
So I want to get us from sodium solid, chloride solid over to here.
So how am I going to do that?
Well, first of all, I know how to make sodium ion gas.
I start with sodium gas and then by ionization I make the electron plus sodium ion.
So this is called the ionization energy, isn't it?
This is the ionization energy.
Sodium gas goes to that.
And now how do I get sodium gas from sodium?
Well that's just called sublimation.
So this I'm going to need delta H of sublimation.
Sublimation is the conversion of solid to vapor.
And you can look that up on the Periodic Table.
I'll show you how to get that in a second.
And now how do I get chloride gas?
Well chloride gas is going to start with atomic chlorine gas, but instead of losing an electron I've got to acquire an electron.
And this action of adding an electron is sort of an inverse ionization, and this is called electron affinity.
And there are tables of electron affinity.
So each element has the ability to lose an electron, it has the ability to gain an electron.
Losing an electron is ionization energy, acquiring an electron is electron affinity.
And just as with ionization energies, if you have multiple electrons you have a first electron affinity, second electron affinity, and so on.
And now how do I get to atomic chlorine?
I've got to dissociate diatomic chlorine.
So this is called dissociation.
So that's the whole thing.
And I'm going to now put some numbers on here.
Let's see.
I'm going to call sublimation step one, dissociation step two, ionization I've got here is step three, electron affinity is step four, and crystallization or Madelung is step five.
And so working off of Hess's law we can say that the total energy required for the formation of the crystal-- delta H for the reaction.
What's the reaction?
The reaction of sodium plus chlorine to make sodium chloride-- is going to be the sum of all of the constituent components.
The sum of all the delta's H. Not delta H's.
delta's H, like attorneys general.
All right?
So now let's add these up.
So we go number one.
Number one I can look up on the Periodic Table.
Where is it?
There's Fritz Haber.
So that's given here.
If you look on the Periodic Table that'll give you the number.
And for sodium it's 108 kilojoules per mole.
Number two, get that from tables.
It's 122.
Number three is just the first ionization energy.
You look it up on the Periodic Table.
First ionization energy of sodium is about 5.3 electron volts, which turns out to be 496 kilojoules per mole.
But look, these are all positive energies and we're trying to make a net negative energy.
So these three steps are all raising the energy of the system.
Finally, acquiring an electron by chlorine is going to decrease the energy of the system, because chlorine is a good electron acceptor.
So that's minus 349.
But watch this people.
The energy in forming the crystal from the discrete ion-pairs is 787 kilojoules per mole, which gives us a net value of minus 410 kilojoules per mole.
So what this is showing you is what the relative contributions are of the different components of that thing.
So here it is in graphical form.
There's the vaporization of sodium.
This is the dissociation of chlorine.
This is the ionization of sodium.
All positive energies.
And now electron affinity.
And look at this contribution from the Madelung energy.
So when things crystallize a lot of heat's given off.
In fact, we can use that in cooling and moderating climate if we're clever about it.
All right.
Now just to show you what the different values are here that you're going to go in and get the lattice energies, you need to know the various r values, you see.
The r0 is simply going to be the r-plus and the r-minus.
So lithium fluoride, there it is.
There's the lattice energy and it's based on the combination of lithium cation and fluoride anion.
So they've gone through and calculated these values.
So there's sodium chloride is 787.
And, you know, you even get things like the boiling points, melting points.
So sodium chloride, for example, melts at about 800 degrees Celsius.
Now if you take magnesium oxide, magnesium oxide is, look, 3,700 versus 700.
And the melting point of magnesium oxide is 2,800 degrees C. Look at aluminum.
Aluminum plus oxygen, look at the end binding energy there.
It's phenomenal, which means it might be useful for-- I'm standing underneath the shuttle here.
This is the tiles underneath the shuttle, and they're made of aluminum oxide because the Madelung energy is so high so it's got the thermal shock resistance.
So now you know how to go and design things for thermal ablation resistance.
All you need is this table.
That's all.
No, you need a little more than that, but this is a good place to start.
If you don't understand this table I don't want you working on the project.
OK?
What else?
All this shows you-- yeah, we're going skip that.
All right.
So now we've got a few minutes here.
What I want to do is last day I talked about magnesium, today I want to talk about aluminum.
And it is also made in an electrochemical process.
In this case the electrodes are horizontal.
We feed aluminum oxide in and we pass current-- and huge currents.
This thing typically runs at 300,000, 400,000 amperes and about 4 volts.
So with the cathode we are running-- remember, we're running nature in reverse.
Instead of the electron donor that aluminum is, we're shoving electrons onto aluminum ion and converting it back to aluminum.
Unfortunately, on the anode side we have to use carbon, and the carbon itself is consumed.
So we consume about a half a ton of carbon to make a ton of aluminum.
So aluminum smelters generate a lot of greenhouse gases.
So you can see this is like a drafting pencil: it's constantly being fed.
And to give you a sense of scale this is probably about 10 feet across, and this gap is about 2 inches, and this is about, I don't know, a foot and a half deep.
It's going to get about 1,000 degrees Centigrade-- liquid aluminum and liquid salt.
So this is what a smelter looks like.
What's the sound of electric current?
Yeah.
The only sound you hear is the fans on the top that keep the place clean.
There's the busbars that are bringing in the current.
And all of these various posts are these things.
So all of the magic is occurring below the floor here.
It was invented simultaneously in the United States by Charles Martin Hall and in France by Paul Heroult.
In the same year they filed patents independently and eventually crossed license when they collided at the World Court.
So this is what happens.
We dissolve aluminum oxide in a molten fluoride called cryolite, which originally came from Greenland, make liquid aluminum carbon dioxide.
Now if we wanted to make this truly green, we want to eliminate greenhouse gas emissions, you need to find an inert anode.
So on an inert anode aluminum oxide would be converted into aluminum and oxygen.
So not only would you not produce greenhouse gases but you'd produce tonnage oxygen, which is marketable.
So some of the work that goes on in my lab is in advanced materials with a view to trying to find an inert anode that would then make this process very, very clean and justify substituting aluminum for steel in cars.
By the way, when you have the field, even though it's a DC current it's a divergent field.
And this is me in a magnesium smelter just in Utah.
There is the busbar for the magnesium cell.
So I'm about, oh, two meters away from the edge of the busbars.
The magnetic field is so high I've got one, two, three, four, five paper clips standing against gravity.
And I asked them for paper clips and there was no office there.
They managed to find a few.
I wanted to see how many would go.
I'm willing to bet I could probably have about seven or eight paper clips up before the gravity would cause them to collapse.
That's the intensity of magnetic fields in these smelters.
So when you drive up to the smelter for the tour you park a fair distance away, you leave your wallet with the credit cards in it, because this is the biggest bulk demagnetizer you can imagine.
If you've got a watch that's got hands that move they're going to be going like this.
[GESTURING]
Yeah.
This is the shuttle.
This is forged aluminum wheels.
The shuttle lands on Centerline racing wheels.
They're forged at a place in California.
They're made of aluminum alloy.
And I don't know if you can read on the side here, it's a little bit light, but these are special tires.
They're made by Michelin So, anyways, the whole message here is learn the lessons here in 3.091 and then we can work together to make metal in an environmentally acceptable way.
And just before I send you on your way for the weekend I thought I'd tell a little joke related to the uncertainty principle.
So the joke goes like this.
Heisenberg is racing down the Autobahn and he gets pulled over by the state trooper who comes to the car and says, where's the fire buddy?
Do you know how fast you were going?
And Heisenberg looks at him he says, no, but I know where I am.
[LAUGHTER]
All right.
Get out of here.
Have a good weekend.
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This Wednesday will be the first celebration of learning.
Test 1 on Wednesday, October 7, you will write during the normal class time.
So you'll have 50 minutes.
And we want to have a little bit of comfort here, so you won't be sitting cheek to jowl.
So before long I'll have the room assignment.
So some of you will write in here.
We'll have fewer people than seats so that there'll be vacancies next to each person.
And then some will write in a few of the other locations, probably 26-100 and the gym above the Walker Memorial.
And we'll get that out to you.
And next week on the 6th we will have no weekly quiz, because enough celebrating.
No point in testing you on the 6th and then on the 7th.
There will be, of course, the mini-celebration tomorrow, quiz 3.
And I'll be available for office hours later today.
Oh, and the coverage, just to remove the mystery, will be right up to the 7th of October.
I've been doing this for over 30 years and I've learned that in order to inspire interest on the part of the student it really pays to have the coverage of the celebration extend up to the lecture before the celebration.
Now obviously I'm not going to drill deep on something I taught you on October 5, but it would be a good idea to stay awake during all of the lectures between now and then.
So last day we talked about ionic bonding.
And ionic bonding occurs with electrostatic attraction between ions that have formed through electron transfer.
And we saw the energy of the ion pair given by this formula, where we have Coulomb's law with the Born exponent.
And then this is plotted.
This is E at r equals r0, and we learned that there were two terms.
The attractive term, which is the Coulombic force here shown.
And then there's a repulsive term, which results from electron-electron interaction when the two lines get very, very close together.
And this is taken from your text.
And I think they did a very nice job here of illustrating as you go to high values of r they're depicting that you have the ions separated by considerable distance.
And there's a certain amount of stored energy, but not a lot.
And then if you go much, much closer than the hard sphere sum of the ionic radii, I think they're depicting here that there's some squashing of the electron clouds.
And you can see that the energy has gone way, way up.
So this is an unfavorable situation, meaning that the energy here is greater than 0.
And there's a sweet spot here at 236 picometers, which represents the ideal location.
And that is the sum of the radius of the sodium ion and the radius of the chloride ion.
And so you can see how energy tracks.
And if you go far, far, far away to the point where they're at infinite separation, there's no energy stored.
So everything makes sense.
And then we said, well what happens if we keep packing these things?
We rationalized that they would continue to do so and ultimately form a 3-dimensional crystal.
And so you can see there's a lot of similarity between what's above and what's down here.
This has been written for a 1-1 system.
In other words, a cation plus 1 and an anion minus 1.
But it could be mediated by the valences.
And what we have here is the structure factor.
This Madelung constant tells us how much we get decrease in the energy of the system by going to a 3-dimensional array.
So depending on the atomic arrangement we'll have a different value of Madelung constant.
We saw that for sodium chloride the value is 1.7476.
And different crystals have different things.
And what determines the crystal structure?
It's a combination of the size of the two ions and their valence.
So what we saw for sodium chloride, this is a structure type.
So obviously sodium chloride is sodium chloride crystal structure, but there is an entire suite of compounds that have radius ratios and charges that end up with a sodium chloride type crystal structure.
And then towards the end we started looking at the Born-Haber Cycle.
And the purpose of the Born-Haber Cycle was to give us a sense of scale of the various constituents in the formation of a crystal.
And what we noted, the takeaway message from the Born-Haber Cycle, is that this enthalpy of crystallization, which is basically this term here, is huge.
It's huge.
It was the big component of the energy required to form the crystal.
It's large and it is negative.
So what I want to do today is start by talking about shortcomings of the business of ionic bonding.
See, how did we get to ionic bonding?
We started with this idea of octet stability.
Octet stability was the driving idea behind all of this.
Octet stability, and in the case of ionic bonding this was via electron transfer.
And so that got us a long way, but it has its limitations.
So let's put up some new data.
So suppose I look at compounds like H2, N2, O2.
Do these things form ionic bonds?
How does octet stability play here?
And so let's start by looking at hydrogen.
So if we took hydrogen and started with atomic hydrogen and added an electron to it, then we would form an anion known as H minus.
And H minus looks pretty good because it's isoelectronic with helium.
So maybe this isn't going to be so bad a day.
But if we're going to have a bond then we need to form an H plus.
So let's do that.
So that would be, then, H goes to H plus, plus an electron.
And that's really nothing more than a proton.
So that doesn't look too appealing.
That's probably a high energy state.
And besides, in the same location at the same time-- in other words, same temperature, same conditions-- half of the hydrogens have to acquire electrons and half of the hydrogens have to lose electrons.
And that's not going to happen.
They're either going to have a propensity for electron gain or a propensity for electron loss.
So it looks like ionic bonding is not going to help us explain the formation of molecules such as H2, N2, and so on.
So who came to the rescue in this case to get us out of the conundrum?
G.N.
Lewis.
G.N.
Lewis was actually born in Weymouth, Massachusetts and he finished his PhD at Harvard in 1899.
And then, like so many Americans of the day, went off to Europe and he postdoc'd in Europe for a while.
And then he came back and got a job at MIT.
And he taught at MIT from 1905 to 1912.
And then in 1912 he was lured to the West Coast where they were starting to establish the chemistry department, the University of California at Berkeley, and he went out to Berkeley and that's where he spent the rest of his career.
And we can speculate why he went.
Maybe he was fed up with the weather here.
Actually, today is one of those few days-- write it down, because one of the few lovely days in Massachusetts.
So G.N.
Lewis, what did he say?
He said, well I've got an idea here.
He said, what if hydrogen achieved shell filling not by electron transfer but by electron sharing.
So he posited the idea of shell filling by electron sharing.
This is in contrast to electron transfer.
So let's see.
Oh.
There's an image of G.N.
Lewis.
He died, actually, on the job.
He came back to his lab one day after lunch and hit the floor.
So he worked right to the very end.
Here's some data taken from a lab notebook and memo, actually.
1902.
And what do you see here?
Well, he developed a notation for us, and we still use this notation to this day: Lewis notation.
So here's lithium and he's got one electron.
But we know lithium has three electrons but only one valence electron.
And then there's beryllium and magnesium-- two electrons.
Aluminum with three.
Here's fluorine chlorine, and when they ionize he puts the eighth electron right here.
And look at this one for silicon.
He's got probably some kernel inside the atom, thus.
So he's even starting to think about concentric shells.
This is 1902.
Remember the Bohr model isn't until 1913.
So you can see people struggling.
And notice that we have eight electrons in a shell-- that's where we're getting the octet stability-- and he's using cubes.
Now we know that the cube isn't the shape of the shell, but it's a pretty good device to help you keep track of electron number-- because there's eight corners on a cube.
So it's another example of how that's not what it is but it's a really good model and it keeps you out of trouble and allows you to go forward.
So this is going back-- way, way back-- for G.N.
Lewis.
Now let's use this idea and account for the formation of H2.
So here's hydrogen, and using the Lewis notation we'll put a dot here for its one electron.
And we'll bring in a second hydrogen and we'll use a cross, or an x, to indicate the electron from the second hydrogen.
And now we're going to double count-- in other words, double attribute.
These are shared electrons so they count for both atoms.
Double count the shared electrons.
And when you do so what do you come up with?
Well, the element on the left has two electrons and, therefore, is isoelectronic with helium.
OK?
Maybe it was a California thing.
They were sharing.
And then there was sort of another California concept, like.
So it was like helium.
And then on this side this is also sharing.
And it's kind of like helium.
So now we've achieved the stability of the filled shell by sharing the electrons.
OK. And I think I even have another slide of how-- this is the more modern version of it.
Electron dot.
So the nucleus and the inner-electrons are contained inside the chemical symbol.
And, actually, this goes all the way to modern quantum mechanics.
Density functional theory does the same thing: lumps all of the inner-shell electrons plus the nucleus into one piece, and then the valence electrons are outside.
And so starting in 1902 with some little dots and crosses we go all the way to DFT today.
All right.
So let's do another one.
How about nitrogen?
Let's try nitrogen.
So when we going to nitrogen we know the valence electron's 2s2 2p3.
So put nitrogen here.
One, two, three, four, five.
Now these three electrons here are according to the Hund rule.
So it's px, py, pz, and this is the 2s2 sitting over here.
And I'll bring in a second nitrogen, and there's its 2s2.
2px, 2py, 2pz.
And now what do I have?
Look at the nitrogen on the left.
Two, four, six, eight.
So the nitrogen on the left feels as though it has access to eight electrons.
The nitrogen on the right-- two, four, six, eight-- it feels as though it has access to eight electrons.
So both nitrogens are isoelectronic with neon if we push on this concept of electron sharing.
Now there's a second thing I want to do.
It's to draw attention to two types of orbitals.
So these three orbitals in the center consist of electrons that are shared.
So these are going to be called bonding orbitals.
And "bonding" and "blue" both begin with a "b", so I'm going to denote the bonding orbitals, or bonding domains, as distinct from the nonbonding domains in red.
Red are nonbonding domains.
Always two electrons per orbital.
They like to live in pairs.
That's the way it works.
OK?
And so each one of these pairs is a bond.
So I can then write nitrogen with three lines through it indicating I have a triple bond.
Three pairs of electrons, three bonding domains, triple bond.
This is all in formation according to the concept of electron sharing.
And Lewis coined a name for the type of bond that is formed in this way.
He said we get bond formation involves cooperative use-- sharing, cooperative-- of valence electrons.
So now we can take the "co" symbol here and the "valence" here and come up with the term "covalent bond."
Covalent bond, thanks to G.N.
Lewis.
So, again, to make sure we're very clear, ionic bond results from electron transfer, covalent bond results from electron sharing.
Now we can do this-- so let's go to heteronuclear molecules.
These are homonuclear.
So let's go to heteronuclear molecules.
And so let's see.
I've got some rules up here, I think.
Yeah.
Drawing Lewis structures.
So let's go to a heteronuclear molecule.
And I'm going to choose as an example sulfuryl chloride.
And I don't expect you to be able to name these things on site.
I will always give you the name.
I'll say sulfuryl chloride, parenthesis, SO2Cl2, blah, blah, blah.
OK?
So sulfuryl chloride.
I want to put up the Lewis structure of sulfuryl chloride.
So center the element with the lowest average valence electron energy.
So it turns out that the average valence electron energy stack like this.
Sulfur is the lowest, then chlorine, and then oxygen.
This is this ranking of average valence electron energies.
And you'd be given those data.
So it says put sulfur in the center.
So I'll put sulfur in the center.
And then what does it say?
We're going to count all the valence electrons.
So sulfur over here is 3s2 3p4.
So that gives me six valence electrons.
And there's two oxygens.
And oxygen lies above sulfur, so that's 2s2 2p4.
So that's 2 times 6.
So that's 12.
All right?
Let's put the 6 over here.
And then there's chlorines in this compound, so that's 3s2 3p5.
So that's 5 plus 2 is 7, 2 times 7 is 14.
And we add this whole thing up, we get there's 32 valence electrons.
And draw a single bond from each surrounding atom to the central atom.
All right.
Again, this is a model.
I'm not saying that this is the shape of the molecule, but it's a way to count.
All I'm doing is trying to keep track of bonds and paired electrons.
So I can put chlorine on either side.
And I'll put an oxygen below and an oxygen above.
All right.
So that's already two, four, six, eight.
So I'm losing eight.
So 32 minus 8 is 24.
And so with the 24 that means I've got 12 pairs of electrons to place.
So let's start putting the Lewis structures up.
So chlorine consists of one, two, three, four, five, six, seven.
And I'll do the same thing on the other side.
Two, four, six, seven.
And oxygen has six, so that's two, four, six.
And then the lower one, same thing.
Two, four, six.
And sulfur has six.
I'm going to use x's for sulfur.
So I'll put one x with the chlorine, another x with the chlorine.
Two with the oxygen, two with the oxygen.
And so now we're in pretty good shape, right?
We can identify bonding and nonbonding domains.
Here's the bonding.
One, two, three, four.
And then the nonbonding.
Looks like there's 12.
And sure enough, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
The 12 nonbonding domains.
4 bonds.
And we have the Lewis structure for this particular compound.
And one last little piece worth pointing out.
Notice that in the bonds to the chlorine you have two electrons, as you need.
One electron comes from the chlorine and one electron comes from sulfur.
But in the bonds between the sulfur and the oxygen the sulfur's so desperate to form a bond that it actually donates both electrons to the bond.
And oxygen's happy because it's isoelectronic with neon, and sulfur's happy because it's going to be isoelectronic with argon.
But, you know, it has to go to some lengths.
So this is called a dative bond, when both electrons come from the one element.
OK. Well, this is great.
I'm going to do one more.
How about methane?
CH4.
So I'm going to start with carbon.
Carbon's going to go with this on there.
And a carbon is 2s2 2p2.
And so I'm going to use the box notation now.
See, this is the Lewis structure, this is chemical equation, now we're going to a box structure.
We can move fluidly from one model to another.
We had cubes up there.
It's all good.
So this is 2s.
And now this is 2px, 2py, and 2pz.
Now according to this, 2s2, that gives me an electron pair.
And now I've got 2p2, which according to the Hund rule goes in like this.
Well I've got a problem here.
How many unpaired electrons?
Two.
Now what's my maximum number of bonds I can form by electron sharing?
It's two according to this.
So the best I can do, best possible here, is CH2.
And that's no good.
We know from mass measurements it's CH4.
The stoichiometry's CH4.
And besides, what are these orbitals going to look like?
These are the p orbitals, so they're dumbbell-shaped and they're orthogonal, right?
They're 90 degrees, which means if I formed this thing-- which is called methylene-- if I form methylene I'd end up with CH2 looking like this, which has a dipole moment.
And we know from spectral measurements and electrical properties measurements that this thing is symmetric.
So this thing-- electron sharing isn't working.
It's not working.
So we need another patch here, and that patch comes from none other than Linus Pauling.
Another American.
You see, it's American science today and it's in the 20s.
That's why we have Gershwin playing at the beginning.
Celebration of American science.
So Pauling was born in Portland, Oregon.
He was the son of a pharmacist.
And he went to Caltech, got his PhD in 1925.
So the Rhapsody in Blue came out in 1924 when he was just hunkering down to his thesis.
Probably listened to it, got some pleasure out of it, as most people did.
So then he finishes, 1925, at Caltech in Pasadena and he goes to Europe.
Now he chose wisely.
He chose four postdoctoral positions.
These are people he postdoc'd with.
First with Sommerfeld, then with Bohr, then with Schrodinger, and then finally with Bragg.
You'll learn about Bragg.
Bragg got the Nobel Prize for X-ray diffraction.
So that's not a bad preparatory start.
So he comes back and teaches at Caltech.
In fact, I have a picture of Linus Pauling.
There he is.
That's a middle-aged Linus Pauling, probably around the time he got the first of two Nobel Prizes.
So what did Pauling do?
Pauling said, why don't we mix the orbitals?
They're all in the shell n equals 2, and what we're trying to do is to fill the n equals 2 shell.
So how about mix?
Let's mix 2s and 2p states in order to maximize the number of bonds.
Remember, when you form a bond you decrease the energy of the system.
Four bonds is a greater decrease in energy than two bonds.
So the system, if it could-- and they're all within the same shell.
You notice he didn't say, gee, if mixing 2s with 2p is good let's go get some 1s.
Well 1s is down in n equals 1, and there's no way you can mix.
They had to be in the same shell.
Mix states in order to maximize number of bonds that can form.
It's all about maximize the number of bonds that can be formed.
And this, of course, is by electron sharing.
We're talking about covalent bonds here.
OK?
And he termed these mixed orbitals "hybrids."
Termed the mixed orbitals as "hybrid orbitals."
They're cross-breed-- part s, part p. So now let's look at the energy level diagram-- or the box notation.
Forgive me.
So I'm going to mix s and p. So I've got a single s and I've got three p's, so this is called sp3.
Each one of these is a mixed sp3 hybrid orbital.
And I've got four of them.
And how many electrons do I have?
Four.
So now I use the Hund rule and in go the electrons.
One, two, three, four.
And now I have the ability to form four bonds.
But it gets better.
Here's the next thing.
These are degenerate.
They're all in the same state.
That's why we're using the Hund rule.
And so degeneracy in energy implies degeneracy in spatial orientation.
So what does that mean?
It means that if these are four bonds equivalent, then the way those bonds will arrange themselves in space is to be equivalent.
So if I've got a central carbon here and I'm going to put four sticks from the central carbon so as to make the four sticks symmetrically disposed in space, that dictates the architecture of the molecule.
And how do I put four sticks off of a central point symmetrically disposed in space?
One, two, three, and four.
This is meant to be the corners of a tetrahedron.
Each one of these is 109 degrees apart.
And this describes a tetrahedron.
So now I've got carbon in the center, and now I've got the hydrogens at the four corners of a tetrahedron.
There is the structure of methane.
And each of the hydrogens has a shared electron with the carbon, making it isoelectronic with helium.
And the carbon has four of its own electrons, four shared with the four hydrogens to make it isoelectronic with neon.
So everybody's happy.
Shell filling, and it's all good.
So now it's symmetric and it has no net dipole moment.
Everything squares with the data.
Well, good for Pauling.
But he went further.
He went much further.
What Pauling wanted to do was to make it quantitative.
And so he wanted to have something analogous in covalent bonding to what we have in ionic bonding.
So what Leslie is now rubbing off the board there is-- no, keep going.
It's good.
It's OK.
This is a rule of academics: You always erase that which you will refer back to.
We need more boards in here.
How many boards do I fill in a period?
9, 18?
Maybe what?
We need about 24 boards.
That's a good lecture.
All right.
So here's what Pauling was thinking about.
He was thinking about the analogy, for example, if I want to get the energy of magnesium oxide I can use the formula that Leslie has just erased, and it looks like this.
So if all I need to know is the radius of the anion, the radius of the cation, it's charge, and the Madelung constant and then I just plug in, I get the crystallization energy.
But then suppose instead of magnesium oxide I want to go to magnesium chloride.
I can use the same formula only I need the Madelung constant.
This is the Madelung constant for magnesium oxide.
If I have the Madelung constant-- forgive me, script M-- for magnesium chloride, and I know the ionic radius of magnesium cation and chloride anion, away I go again.
I need this.
I need, of course, the Born exponent.
This Born exponent and away we go.
The same formula applies.
So I can build with a library of basic physical data.
So what did Pauling do?
Pauling said, what if we can do the same thing for covalent bonds?
Is there some kind of an analogy?
So he said, let's take a look at an arbitrary heteronuclear compound.
So I'm going to do this with HF, hydrogen fluoride.
So, first of all, let's build a hydrogen fluoride molecule.
H with its one electron, and fluorine with its seven.
So now hydrogen sharing an electron with fluorine is isoelectronic with helium.
It's happy.
And fluorine sharing the electron hydrogen is isoelectronic with neon.
It's happy.
So again we see shell filling by electron sharing.
So what Pauling wanted to ask is, can I get a measure of the HF bond energy knowing only the bond energies of H-H and F-F?
So then if I knew all the homonuclear bond energies and then I mixed these to make heteronuclear bonds, is there a path from homonuclear bond energy to heteronuclear bond energy?
So let's look and see what the numbers are.
So hydrogen.
The hydrogen bond's fairly strong.
It's 435 kilojoules per mole.
That's mole of bonds.
435.
Fluorine-flourine is 160.
And so what do you think the value of the H-F bond should be?
Well when I first look at this I say, well it's part H and it's part F, so it's somewhere between 435 and 160.
I don't know if it's the arithmetic mean-- you know, add these two and divide by two-- or maybe it's the geometric mean-- multiply them together and take the square root-- but it's got to be somewhere in between.
What do the data show?
The number's 569, which is greater than 435.
So I take a bond of 435 and a bond of 160, I put them together I get 569.
That's very, very strange.
But Pauling was smart.
Pauling said, I have an explanation.
He says, suppose when these electrons are shared in between the two atoms, suppose they're not shared equally.
Suppose there is a displacement of the electrons.
So instead of putting them dead center, as I've been doing up until now, suppose the electrons are actually drawn closer to the fluorine.
So we still have octet stability, or in this case duet stability, but the sharing of the electrons is not equal.
So this is charge displacement.
And what does charge displacement constitute?
Well, charge displacement means stored energy.
And Pauling quantified that stored energy.
And so what he did is he said that you increase the bond strength by thinking of it as a two-step reaction.
So in the heteronuclear bond that is a bond between two different atoms.
So in a heteronuclear bond we form by what-- and this is my coinage, you don't see this anywhere in the book-- two-step, share and then pull.
So share is, as the name implies, we share electrons to achieve octet stability.
But then because we have unequal atoms we pull towards one of the atoms.
And which one do we pull towards?
Well, we pull towards the one that's got a greater appetite for electrons.
And we've already gone through this concept.
Which atoms on the periodic table have the highest appetite for electrons?
The nonmetals.
The weakest appetite is the metals.
The metals are good donors, the nonmetals are good acceptors.
And fluorine's up in the top right corner, so fluorine has a very, very high appetite for electrons.
And, indeed, in this bond the electrons are pulled to the right.
And why Pauling got the Nobel Prize and Lewis didn't-- it's my theory-- is that Pauling was quantitative.
So he came up with a quantitative measure.
He devised a quantitative measure for the degree of unequal sharing, thereby allowing us to make these calculations with some accuracy.
And he called that quantity electronegativity, and it's denoted by the Greek symbol chi.
OK?
And he devised a whole scale.
How did he get the scale?
He looked at bond energies for all sorts of pairs of elements across the periodic table and went through an exercise with pencil and paper that today we would call multivariable regression analysis.
And came with a set of-- [SLIDE APPEARING]
Oh, I'll come back to this.
This is the structure of methane.
This is the s and p. Oh, let's take a break.
You can stack.
All right.
So this is what methane looks like.
There's the s, there's the p. And the sp hybrid looks like this.
It's sort of an asymmetric dumbbell.
And these four things stick out.
And then you bond the hydrogens and there's the methane.
OK.
So here's what the electronegativity scale looks like.
It looks a lot like the scale for average valence electron energies.
The nonmetals have the highest appetite for electrons period, which means in a bond they're going to hog the electrons.
And the nonmetals have the weakest appetite, and so they're going to end up having the electrons in a covalent bond pulled away from them.
So nonmetals have high electronegativity, metals have low electronegativity.
And now here's taken from the text.
And you see that the electronegativity is periodic.
If you go across a period the metal has the lowest value and the nonmetal has the highest.
And there's fluorine, number nine, at a value of about 4.
It's got the most intense appetite for electrons.
And then you jump down here to sodium, et cetera, et cetera.
Here we are going across the lanthanides and whatnot.
And this is taken from your text.
There's fluorine, 3.984.
That's the thing.
And down here we have very low values of electronegativity.
So with electronegativity we are now able to make calculations.
And this is the Pauling formula for calculating the bond energy in a heteronuclear bond starting from homonuclear bond energy.
So let's continue with the HF.
So if I want to get the bond energy of HF I'm going to take-- and this is the Pauling formula-- the geometric mean.
So I take the bond energy of hydrogen-hydrogen times the bond energy of fluorine-fluorine, and square root.
So that's the geometric mean of the two.
And then comes the Pauling piece that gets him the Nobel Prize.
You take the difference in the electronegativity between the two elements squared, and then the factor 96.3 gives us the unit consistency with kilojoules per mole.
So the greater the difference in electronegativity the greater the contribution here in terms of the deviation from just the geometric mean of the two homonuclear bond energies.
Or put another way, if you have a homonuclear atom such as H2, if it's chi H minus chi H is 0, so this second term goes to 0.
And obviously when fluorine is one of the members you're going to get a very, very high number, because this has the most-- and it doesn't matter which order you put them in because you're taking the square, so it's always going to come out positive.
And I want you to appreciate the sense of scale here, so if we go in here we'll multiply.
This is going to be 435 times 160.
And I'm going to take the square root of this.
And then this is 96.3.
And you look on your periodic table this is 2.2 for hydrogen.
Fluorine is 3.98.
And I know there are different tables of electronegativity.
I don't care.
Just whatever you've got on your periodic table.
The one in the book is a little bit different but it all comes out in the wash.
So you multiply all this out, and we find that the first term is 264 kilojoules per mole and the second term is 344.
So this second term is even greater than the first term.
So the amount of energy in that electron displacement is substantial.
And if you sum the two of these you get 608.
Now you might say, well wait a minute.
The real number is 569.
But 608 takes you in the right direction and accounts for the contribution of electron displacement.
264 is just plain wrong.
So this was an important start for Pauling.
And he has labels on these two contributions.
This first term, which is just the combination of the homonuclear bond energies, is called purely covalent.
It's the purely covalent contribution.
And it's what I've been referring to as the sharing.
This is what you get from sharing.
OK?
And then this second term here with the difference in electronegativity is what you get from what I've been calling the pull on the electron pair.
And Pauling called this the partial ionic character.
He's not saying that there's electron transfer, but it's a move in that direction.
Partial electronic character.
So what I've done here is I've decided I'll make a sort of a panorama of what we've seen up until now.
And so I'm going to make something called the electron sharing meter.
All right.
So if I look at a homonuclear system like hydrogen.
So my meter reads neutral.
So the arrow's at 12 o'clock.
The electrons are shared equally.
And then if I go to HF what do I have?
Well, I know that the fluorine is pulling the electrons.
And so we can designate that by writing delta minus, delta plus.
delta the physicists use.
The lowercase Greek delta means little bit of.
All right?
So delta minus means it's a little bit negative.
And we've got charge neutrality, so if the fluorine end is a little bit negative then the hydrogen end has to be a little bit positive, which means this thing has a net dipole moment.
It's a dipole.
And the arrow points to the negative end.
One way to think about it is I put a little slash there and that starts to look a little bit like a plus sign.
You can come up with your own way to remember it.
So it's got a little bit of a dipole moment.
And people depict dipoles usually as ovals, and they'll put a minus end and a plus end.
So it's net neutral but the charge is not uniformly distributed.
OK?
So our sharing meter in this case is going to show something to the right.
We've got electrons that are unequally shared, and that moves over to the right.
And, you know, the dipoles have interesting properties.
Oh, there's a plot of electronegativity 3-bar in the bar plot.
And actually this is an interesting one.
Just parenthetically, you see hydrogen here?
Hydrogen's weird.
They put it in the periodic table above lithium but it's not an alkaline metal.
And you can see it just doesn't belong there.
And there's a lot of conversation about putting it maybe somewhere centered above the p block elements, because it certainly doesn't belong next to helium.
But it probably doesn't belong above lithium either.
Anyway, I thought that was very interesting.
I can tell from the response of the class, why does he care?
[LAUGHTER]
All right.
Now this is really-- I'm going to use an adverb here-- this is really important.
All right?
So here's HCl, is a cousin of HF, and you see in the upper frame it's just a bunch of HCl molecules just bopping around any which way.
So there's the delta plus and the delta minus.
Now if you take these dipoles and you put them in an electric field they will align themselves, and the positive ends will face the negative plate and the negative ends will face the positive plate.
And there's energy stored when the random orientation goes into an ordered orientation.
This is the principle behind a capacitor.
A capacitor is nothing more than a whole bunch of aligned dipoles.
So if you want to invent a supercapacitor that we can use on a car to extend the range of the automobile so we can reduce our dependence on imported petroleum, you're going to look for molecules that have a honking big dipole moment.
That way you get more energy per unit electric field.
So, again, a simple idea that tells me how to go and invent.
I can go back to my office and go and invent something right now just based on this lecture 9.
[LAUGHTER]
See, you go and invent.
You start the company, you make the billion.
Remember good old Professor Sadoway at MIT, and established the fellowship for students, and so.
All right.
But you have to know what a dipole moment is.
Got to know what a dipole moment is.
OK.
So there's the dipole moment.
And then lastly I'm going to put sodium chloride.
So what's sodium chloride look like?
Well it's Na plus and Cl minus.
So the electron has transferred completely.
So this isn't even sharing at all.
So this is really bury the needle.
This is not sharing.
In this instance the sodium doesn't even get visitation rights to the electron.
The electron's gone.
Whereas here hydrogen gets to see the electron on Saturdays kind of thing.
Depends what kind of lawyer fluorine had.
That's what it all boils down to.
All right.
This is the same thing that I just showed you.
But you see, the textbook gives you, as the name implies, dense text.
I gave you the sharing meter.
The sharing meter is far more expositive.
All right.
And then, finally, the percent ionic character is given by this formula here.
So this is 1 minus the exponential.
So the exp term, this exponential of-- what is it-- minus 1/4 times the difference in electronegativities squared.
This notation means e base natural logarithms, minus 1/4, blah, blah, blah.
That's what this thing is.
So if you plug in, multiply by 100% you get something that goes from 0 to 100.
So obviously when delta chi is 0 you get 0%.
e to the 0 is 1, 1 minus 1 is 0, and so you have no ionic character.
And so if you plug in the numbers for HF-- so you're going to take this difference here, square it-- it ends up giving you 1.8, which gives you a value of about 56% ionic character.
So it's as though the electron is sort of half transferred.
But you might also look at it from this perspective.
If you take 344-- because this is the partial ionic character, which is the energy of electron displacement over the total energy in the calculation-- that turns out to be 57%.
So this stuff makes sense.
There's a sensible metric here at work.
And so this is what Linus Pauling got his Nobel Prize for, and it's the description of polar covalency.
And polar covalency is operative when you have heteronuclear bonds, because the two different elements don't share the electron equally.
And the Pauling formula allows you to calculate that.
And his formative book was written in 1937, called The Nature of the Chemical Bond.
OK.
So turning to the last five minutes, I want to bring to your attention some covalent molecules.
Today we're going to talk about Freon.
Freon was an invention, it was a designer chemical, invented by Thomas Midgley.
This is me.
I named him "sp3."
That's his nickname.
Thomas sp3, for the hybridized orbital.
So he was working at the Dayton engineering laboratories in Dayton, which was owned by General Motors, and he was working in the 20s at a time when there were no refrigerators in American kitchens.
The only refrigerants that were used were either toxic or flammable, things like ammonia, methyl chloride, sulfur dioxide.
And you read about horrible accidents.
People making ice cream at some plant and the compressor blows up and two or three people are killed.
So it was deemed unsafe in the American kitchen.
In the 20s Midgley discovered this molecule, which looks just like methane only we've replace the hydrogens with two chlorines and two fluorines.
So this is called dichlorodifluoromethane and it's a chlorofluorocarbon, a CFC.
And this was fantastic stuff.
It it was colorless, odorless, tasteless, non-toxic.
It was not just used as a refrigerant, it was used in propellant.
When I was your age all of the sprays-- whether it was hair spray, shaving cream, any aerosol-- was propelled by Freon-12.
It was fantastic stuff.
Well, it turns out that in the upper atmosphere-- you know, you go pss pss pss.
You got people all over the world doing this, eventually this stuff starts floating away.
And what turns out in the upper atmosphere where we don't have shielding from ultraviolet-- you know how to do this calculation, because you can look up the energy.
And, in fact, it's part of your homework, where you look at the energy differences and the electronegativity differences, you can compute the wavelength of light that's capable of breaking the carbon-chlorine bond.
And it turns out to be in the ultraviolet.
Once the chlorine is broken you have a chlorine radical, and that chlorine radical goes over here and attacks ozone.
[CELL PHONE RINGING]
Cell phone-- out.
Just get up and leave out of courtesy.
[CELL PHONE STILL RINGING] [LAUGHTER]
Hello?
The first year I was teaching 3.091 there was a Nobel Prize awarded to Mario Molina, who was a faculty member here in Earth and Planetary Sciences who had worked years earlier at University of California, Irvine, and had speculated on the mechanism by which ozone depletion occurs and linked it to rising levels of CFCs.
Initially-- that's why it says a vindication-- people pooh poohed it, said it was crazy.
There wasn't enough of this pss pss to cause any trouble.
But then later with the NASA program they started taking a lot of images and they could track ozone levels in the atmosphere and start seeing that not only was ozone changing but there were actually pockets where ozone was being depleted at an accelerating rate-- because obviously the atmosphere isn't constant composition and constant temperature.
Duh.
So anyways, yeah.
There he is.
And this was the paper that was published in 1974 in Nature.
And this was done before computers.
The PC wasn't invented and commercialized until the early 80s.
So this was typeset, and the person who typeset it obviously didn't take 3.091 because instead of "atom" hyphen "catalysed" we have "atomc-atalysed."
But even ignoring the spelling error in a Nobel Prize winning paper-- [LAUGHTER]
--the Nobel committee overlooked this.
Yeah.
So there it is.
And then they went to HFCs and so on.
There's a lot of activity in this.
And what happened is when we changed from CFCs to HFCs we had to change the design of the compressors.
And what happened was everything got much, much more efficient.
So this was an example of necessity for a change that was driven by concern for the environment.
Instead of putting people out of work and killing an industry, gave us much more efficient refrigeration.
And the last thing I'll show you is this to draw your attention.
This was in your textbook.
This is the cap at the top of the Washington Monument.
The Washington Monument was built to celebrate the American centennial, 1876.
They finished it in 1884.
And this is 100 ounces of aluminum, because aluminum was a precious metal.
It was priced higher than silver.
1884.
Two years later Charles Martin Hall and Paul Heroult invent an electrochemical process that drives the price of aluminum down to the point that we make beer cans-- I mean soda cans-- out of it today.
[LAUGHTER]
And a good example of how chemical innovation can lead to superior products.
I'll see you on Wednesday.
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Settle down.
Settle down.
Settle down.
All right, first announcement is the celebration of learning.
Reminding you that the celebration of learning is a week from today.
Go to your assigned rooms.
A through Ha will write in here.
This group will go into 26-100.
And the last group over to 4-270.
And you will take with you your periodic table, table of constants, aid sheet, something to write with.
And we'll give you paper.
You'll write on the exam paper itself.
And I'll say more about test taking strategies on Monday.
Coverage will be right up through Monday.
With emphasis obviously on the material that you've had some time to digest.
And there will be no weekly quiz next week.
So let's get right into the lesson.
Last day we looked at Louis, who gave us the notion of achieving octet stability by electron sharing.
And that led to the concept of covalent bonds.
And then Pauling helped us understand the energetics of covalent bonding by putting forth the idea in a heteronuclear molecule there's unequal sharing of the electrons.
And out of that emerged the concept of polar covalency and the definition of electronegativity.
And on the slide you see how electronegativity varies across the periodic table.
Electronegativity being a measure of the pull an atom has for electrons within a covalent bond.
And as you would expect, the non metals which are good electronic receptors are also the ones that have the highest electronegativity.
And the metals, over here, which are good electron donors are those that have the the lowest value of electronegativity.
And Pauling was quantitative in his formulation and gave us this equation that tells us how to measure the energy of the x-y covalent bond.
If x is not equal to y, then axiomatically this is going to be a polar covalent bond.
And you take the geometric mean of the homonuclear bond energies.
This is x-x bond energy in x two.
This is the y-y bond energy in y two.
You take the product square root of which.
And then the partial ionic character, this is the contribution to unequal sharing of the electrons.
You take the square of the difference in the electonegativities of the two elements.
And the 96.3 is a factor that allows you to get the overall quantity in kilojoules per mole.
So you need these numbers in kilojoules per mole.
And we further had a formula for the percent ionic character, which you can get by looking at the difference in electronegativity.
And then through this formula, you end up with a scale that runs from 0 to 100%.
And just by way of example, we had a look at H-F. We spent a fair bit of time on that.
Obviously, it's a heteronuclear molecule.
We calculated its bond energy and so on.
And then to indicate polar covalency, we use the dipole notation.
The dipole shown here.
It's just an oval.
It's net neutral.
But the charge is not uniformly distributed.
You see one end is a little more negative.
The other end is a little more positive.
Sometimes people write lowercase Greek delta, indicating little bit negative here, a little bit positive here, net neutral.
There's the arrow indicating the dipole.
And furthermore, we argue that this is a polar bond.
And since this is just a molecule with the two atoms, then this is also a polar molecule.
Which means it has a net dipole moment.
And I made some observations about dipole moments and the ability to store energy capacitively.
And how you'd go about finding a really good capacitor.
And so that all came out of the desire to find something that has a net dipole moment.
And I want to continue that conversation.
And so I want to look at some other elements.
And what I want to look at in particular is the compound methane.
Let's look at methane from this new found appreciation of polar covalency.
So first thing I want to do is to put it's structure up.
We've gone through the Louis notation and the structure.
It forms sp3 hybrids.
And you end up with a structure that looks like this.
These are all at a 109 degrees, symmetrically disposed in space.
And we know since we have a heteronuclear system here, we're going to have some polar covalency.
You can look up that the electronegativity of carbon is 2.55 electronegativity of hydrogen is less than that from its position in the periodic table.
But to be quantitative it's 2.2.
So that means if I look at the carbon hydrogen bond, carbon has the higher electronegativity.
So it's going to pull the electrons.
So that means that the carbon end is going to be a little bit negative.
And the hydrogen end is going to be a little bit positive.
So that means I've got a dipole moment here.
Polar bond.
Everything is the same as above here.
Polar bond.
But, I want to address the question, is the molecule polar?
So the way I interpret the question, is it a polar molecule, I ask is there a net charge displacement?
Well, we're off to a good start here.
We see that we have charge displacement on the bonds.
But, is there a net dipole for the molecule.
So the way I think about that is to say, where is the center of positive charge for the molecule?
Where's the center of negative charge?
So I know all the hydrogens are a little bit positive.
And they're all the same distance from the nucleus.
And I can draw a circle that will capture all four hydrogens.
Or more appropriately, it's a sphere, right?
These three on the bottom don't lie in the plane.
So the corners of the tetrahedron lie on a sphere.
So where is the center of positive charge?
The center positive charge is right here at the center of the molecule.
And where's the center of local negative charge?
It's on the carbon because the carbon is the negative end of all the bonds.
So the centers of positive and negative charge for the entire molecule are colocated at the center of the molecule.
So this means no net dipole moments.
No net dipole moment for the molecule.
So this is a non polar molecule.
It's a non polar molecule consisting of polar bonds.
Non polar molecule.
So there's two ways that I can get a non polar molecule.
One way is to have a homonuclear molecule, right?
So, homonuclear molecule.
Homonuclear molecules axiomatically must be non polar.
Because they have equal sharing.
So if I give you anything that's homonuclear trivially, it's non polar.
So you can look at things like H2, P4, S8.
These things are all non polar.
I don't care what their structures are, it doesn't matter.
This one here is definitely polar.
And then the last one here that we're looking at, the the methane, is non polar because we have spatially symmetric disposition of identical polar bonds.
So spatially symmetric, that means there's three dimensional symmetry.
Spatially symmetric disposition of identical polar bonds leads to non polar molecule.
Because the centers of positive and negative charge are colocated.
So, now it's time to move on.
Now I want to look at covalent bonding from an energetic standpoint.
I want to go back to energy level diagrams.
We've looked at energy level diagrams in the past for atoms, single atoms.
But now, what I want to do is build energy level diagrams for molecules.
So, for that we're going to go back to the Schrodinger equation.
I want energy level diagrams for molecules.
And for that I'm going to call upon the Schrodinger equation.
And we're not going to go through the quantum mechanics in mathematical detail.
We're going to do quantum mechanics pictorally.
We've got a long way.
And in particular, what I want to do with regard to the Schrodinger equation is to recognize that we can move from atomic orbitals to molecular orbitals by exploiting the fact that the Schrodinger equation is a linear equation.
Using the linearity property of Schrodinger equation.
All right, what do I mean by linearity?
Well, I'm going to do just a little bit of math.
Just enough to make you sit up in your math classes and find out that there's some utility there.
Here's what I mean.
Let me talk about what the linearity principal is.
So if I have some equation, f of x, y, and z.
It's a three variable equation.
And the equation is f of x, y, z is k1.
k1 could be a constant, it could be a function, it could be anything.
And let's say that it has as its solution, the solution is s1.
And then I've got a second variant of this x, y, z.
And it equals k2.
So it's the same function but it equals k2 here.
It could be a constant, it could be something else.
And it has as its solution, s2.
If this f of x, y, z is a linear equation, then if I'd give you f of x, y, z equals k1 plus k2, you don't have to go and solve the equation with impunity.
You can write that the solution is equal to s1 plus s2.
And that doesn't hold if it's a non linear equation.
Just put y equals x squared.
And if I tell you y equals one, y equals two, and then I give you y equals 1 plus 2, it doesn't equal the sum of the solution to that.
You can prove it to yourself.
So this is the fact that superposition holds.
You can superpose solutions and build a library.
So superposition holds when you have a linear equation.
And that's all we're going to use in order to make equations for the molecular orbital.
So that way I can write that the wave function of a molecular orbital then is a linear combination of the atomic orbitals.
That's all this is doing.
This is setting the stage for, if you take quantum mechanics later, you'll go through this in gory detail.
But I'm saying that you know enough now to appreciate that we can do what we're about to do so.
That means I'm just going to sum the wave functions of the atomic orbitals.
In this case, i goes from 1 to 2 because there's only two orbitals in a bond.
And there will be some pre factor here, a sub i times ci.
And this whole business is called linear combination of atomic orbitals into molecular orbitals.
So it's an SLI, it's a six letter initialization.
You know like FBI is a TLI, it's a three letter initialization.
This is an SLI, in 3L91 we go big.
This is an SLI, six letter initialization.
So we're going to do some examples here.
And all you need to do in order to run the examples is use two ideas in LCAO-MO.
First of all, conservation of states.
Conservation of orbital states.
And the second one is, we're going to fill the newly created molecular orbitals according to the Aufbau principle.
Fill MOs by Aufbau.
If we do that, we're in good shape.
So let's take a look.
So first think I want to do is just rationalize this one here.
H plus H goes to H2.
I want to demonstrate that there's a rational basis for this.
So what I'm going to do is make energy level diagrams.
So there's an energy coordinate here.
Energy goes up in the vertical direction.
And out here is zero.
And what I'm going to draw for you is energy level diagrams for two atomic hydrogen gas atoms.
So if this is the zero, we know that somewhere down here is the 1s.
And I'm going to be complete in my label.
This is the 1s atomic orbital of hydrogen.
And over here there's a 1s atomic orbital of atomic hydrogen.
And furthermore, I've got an electron sitting in 1s.
Now these are at infinite separation.
These are far apart.
Far, you know what that means.
Far with quotation marks means that they're separate quantum systems.
So I'm not violating the poly exclusion principal by having both of these electrons sitting in the ground state.
So they both have the same set of quantum numbers.
1, 0, 0, 1/2.
Both of them.
Same thing.
Let's put it over here, 1, 0, 0, 1/2.
So they're very, very far apart.
Now if I bring them close enough together to make the molecule H2, what happens is I'm going to violate the poly exclusion principal if I have both of these atomic orbitals at the same level.
Because I start filling them according to Aufbau principal I'm going to end up with more electrons than two at the same state.
So what happens is this splits.
One orbitals ends up at a lower energy and one orbital ends up at a higher energy than the ground state energy in the atom itself.
And so now this is called the sigma 1s molecular orbital.
And the one above it is called sigma star 1s molecular orbital.
And sigma is at a lower energy.
So if electrons populate this orbital, the system's energy will decrease and a bond will form.
So this is called a bonding orbital.
If electrons populate the upper orbital, they will raise the energy of the system, destabilizing it.
And so this orbital denoted with the star is called antibonding.
It's an antibonding orbital.
So now I've got the energy level diagram for molecular hydrogen.
So now let's go and populate according to the Aufbau principal.
I've got two electrons and they go in spin up spin down.
And now you can see that this occupancy put the electrons at lower energy than they were by being in the energy state of the atoms.
And so I can argue that by this diagram, I haven't predicted anything, but I can use this diagram to rationalize that for this reaction delta E is negative.
And we know how negative it is.
It's minus 435 kilojoules per mole.
It's hugely negative.
Minus 435 kilojoules per mole.
All right.
So, that's good.
In fact I think I've got some pictures.
You can go through the whole quantum mechanics.
And just as is the case for a single atoms, you can use the product of the wave function and it's complex conjugate and so on and make plots.
Oh by the way, this was just making the point that you can pull out the electronegativity off of the periodic table.
It's given here.
And the periodic table is pretty good.
But obviously somebody got a little bit ahead of himself.
They called this the first ionization potential, which is the potential you'd put across the plates in a gas discharge tube.
But the unit is not the electronvolt.
If it's a potential, it's a volt.
Somebody here that put this together seemed to think the electronvolt is a unit of potential.
And I want to make sure that nobody in this class believes that.
You got to get a little bit of nail polish or something and cover up the little e there.
Or take the nail polish, paint over potential and write first ionization energy.
One or the other, but not this.
This was a plot if percent ionic characters.
You can see that the strongly covalent compounds down here have very very low electronegativity differences and therefore very low ionic character.
And way up here are the ionics.
And in between is HF.
It's almost at the cusp.
But, can you see that there must be a mistake on this diagram?
Because this thing has got a greater electronegativity difference than lithium iodide and yet lithium iodide has a higher percent ionic character.
How can this function zigzag like that?
Something's wrong.
So when you look at something, you go wait a minute, that doesn't make sense.
So I go back and I say, do I trust anything here?
Trust, but verify.
Read critically.
It's a good book.
But, it's a big book and there's going to be some mistakes.
All right, so here's some pictoral stuff of what we were just doing over here.
So here are the two spherical 1s orbitals and now they come closer and closer together and they overlap.
And this is what the shape of the sigma 1s orbital looks like.
It's like an oval with the two nuclei inside.
Here's taken from a different text book.
The overlap of atomic 1s atomic and now here's the molecular orbital.
Now you can also look at what the shape of the antibonding orbital would be.
This is what the shape of the antibonding orbital would be.
It would have two lobes with a nodal plane in between.
I think this is taken from yet another book.
Oh this is our book.
There we go.
1s, 1s, as there we go.
Hydrogen molecular orbitals.
Now look, suppose we do the same thing for helium.
If we do the same thing for helium, helium starts with two electrons in 1s.
And now it's going to have four electrons.
Two will go in the bonding and two will go in the antibonding.
And the two that go in the antibonding raise the energy of the system more than the two that go into bonding decrease the energy of the system.
There's a net increase in energy.
And this is the way you could rationalize that helium exists as the atom in the gas phase.
You don't see He2 gas molecules.
So using this energy level diagram you can go through an rationalize.
I would never ask you to predict.
I would say, fact, helium is found as the atomic species in the gas phase.
With the use of energy level diagrams, rationalize.
And that's what I would expect you to do.
Take this and show that the two are of different stability.
One other point to to make, the level of 1s.
If I wanted to put helium on this board here.
Not the scale, but just roughly.
Where would helium 1s be relative the hydrogen 1s.
We got three choices.
Same level, closer to zero energy, or more negative.
How do you think about the problem?
What determines what this energy is?
It's the electrostatic force of attraction.
Is the electrostatic force of attraction on the 1s electron in helium greater than or less than it is in hydrogen?
It's greater.
So that means that the energy is going to be more negative.
And so if I were to put over here helium, I'd put down here this would be 1s atomic orbital for helium and then we go through the analysis.
And why does that come into play?
Well you might want to do a heteronuclear molecule.
Suppose you wanted to do the bonding diagram for hydrogen fluoride.
So we'd have hydrogen here and the fluorine and would be here.
And all the fluorine orbitals would be much lower and then they'd combine to make the molecular orbitals.
And I think there's something opportunities to practice that in the homework.
OK let's do one more.
Let's do one more.
How about lithium.
Let's do lithium.
I want to ask, is dilithium stable?
Li2.
And this is all gas phase.
Is dilithium stable?
So I'll start off with here's the zeroes.
The zero of energy for infinite separation.
So this'll be a lithium gas atom.
And this is the putative dilithium gas atom.
We're going to figure out if it's stable or not.
So it's going to have bonding and antibonding orbitals.
And then this is the 1s, 2s.
And this will be sigma 1s, sigma star 1s, and so on.
And then we'll have 2s atomic orbital splitting into sigma and sigma star of 2s.
And now lithium is 1s2, 2s1.
So there's one, two, three.
And one, two, three.
And so now let's use the Hund rule.
So I've got four electrons in the n equals one shell, and they populate in this manner.
And then I've got two electrons that go only into the bonding orbital.
And according to this it appears that lithium 2 is favored over atomic lithium.
And that in fact is the case.
That in fact is the case and so that applies to all of the all of the alkalide metals.
Because they all have the ns1 configuration.
So when you're driving down the highway and you see those orangey yellow low pressure sodium vapor lamps.
What you're looking at is emission not from atomic sodium but from Na2 vapor.
And you've got the energy level diagram to convince yourself of that.
So next time you see those, just smile, knowing that you're looking at disodium, not sodium.
OK, well so far we've only looked at single bonds.
Now I want to look at multiple bonds.
Double and triple bonds.
So let's look at nitrogen now.
N2.
Remember that gave us the the triple bond?
We had nitrogen, one, two, three, four, five.
Second nitrogen, one, two, three, four, five.
So in order to get octet stability, we had three pairs of electronic sharing.
Which then give us a triple bond here.
And so what's that going to look like?
what's that going to look like pictorally?
And the way to think about that is first of all these are all p orbitals.
If we look at, this is 2s2, 2p3.
So 2s and these are the 2d orbitals.
So I've got s is filled first and then I've got the lone electrons in each of the p orbitals.
And these things are shaped like figure eights.
And the p orbital, this is the p atomic orbital.
It has two lobes.
Each of these zones are just called lobes.
Same word is earlobe.
And this zone in the middle, this one point in the middle, is called a node.
And that's a sight of zero electron density.
Zero electronic density.
And remember the electron, even if it only has one electron here, the electron can be in either a lobe.
It can move from lobe to lobe even though it can never be at the nucleus.
And how does it get from one lobe to the other while never crossing the nucleus because it's never supposed to be in the nucleus?
By behaving as a wave.
The same way that you can have a jump rope and you can have a fixed point of zero motion, yet you can transmit energy down the rope passed the node.
So, let's draw these things.
A word about how you draw, you have to use the right hand rule.
Have to use the right hand rule.
So when I put the coordinate system up, it's going to have to conform so that the thumb is x and they go y and z.
And if you don't use the right hand rule later on if you get into electromagnetics, you start looking at forces and vectors, you're going to end up with things moving in the wrong direction.
So we conform to the right hand rule.
And the other thing, and it's kind of nice, it's not mandatory.
But chemists generally like to have atoms bond along the z-axis.
So that means we start with the z-axis here, if I'm going to put my second nitrogen, have them bond on the z-axis.
And so that means if z is in the plane of the board pointing to the right, then pointing up must be y, and then pointing into the board must be x.
So I'll have, here's my px orbital, then the py orbital, and the pz orbital.
And the three of these are all symmetrically disposed around the nucleus.
And next to it at the same kind of floral arrangement.
I'll start with px, py, and pz.
And now these are going to come close together and overlap.
So I want to figure out what those orbitals are going to look like.
So let's start along the z-axis.
The z-axis is the easy one.
So I've got two of these things lying on their sides like infinity signs.
We're going to do quantum mechanics pictorally.
Why?
Because it's a linear equation.
So I can add pictural.
So this is 2pz of one of them.
And a 2pz of the other.
And this is the nitrogen atomic orbital.
And I'm going to smear these things and what are they going to look like?
The nucleus is here, the nucleus is where I'm indicating the dot.
These two combine, very simply, it's going to look like this.
So there's one nitrogen.
Here's the other nitrogen.
And what's this thing?
This is electronic density.
And so this is our sigma bond.
Looks a little bit different from the case of of hydrogen, because hydrogen was the blending of two s orbitals and s orbitals are spherically symmetric.
In this case, we've got two lobes.
But what's characteristic about this one and hydrogen, hydrogen looked like this, remember?
This was hydrogen.
This was H2.
And this was a sigma bond.
The characteristic of a sigma bond is that when you start from one nucleus and you go to the other nucleus, you move through unbroken electron density.
So there are no nodes, no holidays, between the nitrogen nucleus on the left and the nitrogen nucleus on the right.
There is a node here, but that's different.
I can go from one nitrogen to the other with unbroken electron density.
That's what makes this a sigma bond.
So that's good.
So this thing here is going to be called stigma 2p molecular orbital.
Sigma 2p molecular orbital And I think I've got the slide that shows this.
There's some art work.
People really get excited about this.
OK there's dilithium.
Or dipotasium, disodium.
OK so here we are.
This is the smearing of two pz atomic orbitals to make the sigma 2p bonding orbital.
And there just for grins and chuckles is what the antibonding would look like.
But we don't care, because it doesn't form.
Well this is the book.
And you know what I'm going to say.
I've got my little hobby horse here.
I don't know why they change color on the lobes.
Because when I look at that, it starts conjuring up to me the image that one electron stays in the blue lobe and one electron stays in the yellow lobe.
Besides, I've seen them and they're not different colors.
They're the same color.
So now let's look at what happens when we blend off of the z-axis.
So let's blend the two py orbitals.
See what that goes like.
OK so let's do that one.
And that one is going to look like this.
We're to start with, again figure eights.
But now they're their side by side, they're lateral.
So this is 2py atomic orbital.
2py atomic orbital.
And then we're going to blend them along the z-axis to give us, and I'm going to do this stylized, OK?
So there's the the two nitrogen nuclei.
So I put the nuclei up.
And I'm going to smear the upper lobes.
So these two lobes are going to smear.
And I'm going to get really stylized.
I feel like it's France and it's the late 1800s.
So there it is.
And I'm going to smear the two bottom ones.
And it's going to look like this.
So what do I have here?
Now I have two lobes as before.
But if I look at the second nucleus from the first nucleus, not only do I fail to have unbroken electron density, I have zero electron density.
See, this was a nodal point in the atom.
With the two atoms together, the plane orthogonal to the board is a nodal plane.
There's no electron density in the plane orthogonal to the board.
Nodal plane.
So this is definitely not a sigma bond, this is a pi bond.
This is a pi bond.
And it's characterized by smearing of atomic orbitals, just as the sigma bond is.
But it has a nodal plane that separates the two lobes.
I think I've got some cartoon illustrations from other books Here OK this is from one book.
This is good.
They call the px, I call it py.
There it is.
And I'd go further and I'd say that if I were to slice this and look at it from angle, if we were to cut this and look from here, I'd venture that you'd see something that's sort of figure eightish.
Oh here's our book, bless them with two colors.
But anyway, there's what it looks like.
That's the pie.
And then the same thing happens with the x.
So this is this is going to be pi 2py.
This is pi 2py.
And it's a molecular orbital.
And there's going to be a pi 2px.
And it's going to blend front and back.
So we can make a catalog.
So we're going to do quantum mechanics in pictures here.
So I know that s plus s must always make a sigma bond.
There's no other way.
Because I've got electron density all around.
Let's do it pictorally.
That's easy.
And this is sigma.
We know this has to be sigma.
What about something like HF where the H is an s and the F is going to have the one last electron missing in the p orbital.
S plus p must give sigma always.
Because that's this cartoon.
See there's no way that when this smears with this, there's going to be zero electron holidays from the hydrogen nucleus to the fluorine nucleus.
So you're going to end up with something that looks like this.
It starts around and gives you something that's going to be a little bit asymmetric.
So this will also be a sigma.
So I'll just put HF here as sort of prototypical of that.
And then if I take p plus p axially, on axis, that also gives us a sigma.
Because that's this one, the infinity signs.
Two infinity signs give us this.
And then finally, if we get p plus p longitudinally.
So that will give us a pi bond and that's the 88.
8 plus 8 gives me, and this is pi.
So that's quantum mechanics.
The math will follow.
So now what I want to do is go back to this energy level diagram and show how these energy level diagrams can work.
So here's the energy level diagram for nitrogen, N2.
There's two f's, two p, and the scaffolding is in place.
There's the energy levels.
Now here's the molecular orbitals and here are the atomic orbitals.
And now here they are occupied.
So nitrogen has one, two, three, four, five according to the Hund rule.
And here is the set up for the N2 molecules.
So the 2s's and the 2s's go bonding antibonding.
Now I've got three plus three is six.
Two, two, two.
Everything's paired.
So we get the triple bond, 946 kilojoules per mole.
Enormous energy in nitrogen.
Enormous energy in nitrogen.
Now, let's keep going.
Now let's look at oxygen.
This is the scaffolding for oxygen and fluorine.
And there's a little change here.
A little change, I'm going to draw your attention to it.
And you can't predict this.
We would give you this.
I would tell you what the energy sequence is of the energy levels.
But look here carefully, you see in the case all nitrogen, the pi's lie below the sigma.
In the case of oxygen and fluorine the sigma lies below the pi's.
These are tiny, tiny differences.
But, they are measurable.
That's the little difference.
Now let's see what happens.
Actually here's from out text book and it shows the stigma 2pix slowly meandering down, down, down, down, down.
And somewhere between nitrogen and oxygen it crisscrosses.
All right, so now let's fill oxygen.
So oxygen is two, four, five, six.
So two plus two is four.
There's the two, two, two.
And then these last ones go up into the antibonding.
But look at the antibonding.
We have, according to the Hund rule, not two electrons in the first orbital, but one and one.
So we end up with unpaired electrons in the antibonding orbital.
So these offset the three pairs here, and so we end up with a double bond.
And its energy us 498 kilojoules per mole.
Substantially less then nitrogen.
And this energy level diagram rationalizes that.
And then here's for fluorine.
If you go to fluorine it's the same scaffolding only there's two more electrons.
These are paired and you have a single bond.
160 kilojoules per mole.
Now there's a property that we get from the fact that we have these unpaired electrons.
We're going back oxygen now.
We have unpaired electrons in the antibonding orbitals.
We saw unpaired electrons.
What did they do in the Stern-Gerlach experiment?
Changed the magnetics substantially, right?
We ended up with the splitting of the silver bean.
So this will also have an impact.
An impact known as a paramagnetism.
What's paramagnetism?
It's shown in this little cartoon.
If you have a substance here that's balanced and it is paramagnetic, if you engage the magnetic field, the magnetic field will pull on a substance that's paramagnetic.
And oxygen is paramagnetic.
Not only in a gas state, but paramagnetic as liquid.
And here's an illustration from your text book.
You may have seen this and said, yeah a guy is pouring liquid oxygen, wow.
Look carefully now.
The boiling point of oxygen at one atmosphere pressure is 90 Kelvin.
So room temperature is substantially higher.
This is the equivalent of taking water and putting it in an oven at about 300 degrees celsius and watching it pour.
It would still be liquid, but it would be liquid trying to boil.
Why doesn't it all turned to gas?
Because there's a time to heat everything.
So this is sitting at 90 Kelvin, he pouring it down, or she's pouring it down, and these are the jaw of a permanent magnet.
And it's in a gravity field, it doesn't keep falling, it stops.
And it continues to boil away and as fast as you can pour it, it sits between the jaws of the magnet.
Because it's paramagnetic.
You can think of this as the liquid equivalent of iron filings.
If I told you that were these were iron filings you'd say, yeah the iron filings go and they stick to the magnet, that's what magnets do to iron filings.
They do the same thing to liquid oxygen.
And why?
Because of this.
This explains this.
So we can do a lot with these primitive little diagrams.
Let's do one more.
I want to do one more of these things.
I want to go to hybridized systems.
I want to go to a hybridized system.
And I want to look at ethylene.
C2H4 is ethylene.
OK, so if I told you give me the Louis structure of ethylene, you just start going according to the rule.
So I'm going to have carbon, carbon, hydrogen, hydrogen.
And carbon has one, two, three, four electrons.
The hydrogen has one electron.
I'll put the one electron from hydrogen.
And then the other carbon over here, I'm going to give it dots.
One, two, three, four.
And then these hydrogen each have one.
And now let's see what I have here.
These hydrogens are all isoelectronic with helium, so they're happy.
And the carbon now has two, let's see, what does he got?
Yeah, carbon is happy.
Two, four, six, OK good.
So now carbon is.
Happy as well.
So what do we have here?
This is the equivalent to a carbon carbon double bond.
So what have I learned with the nitrogen example?
What I learned with the nitrogen example was, that if I want to make more than one bond, I have to have a combination of a stigma and a pi.
You can only make one sigma bond because there's no room.
Once you've got zero electron density between the two nuclei you're going to violate the Pauli exclusion principle if you have another orbital cutting through that same domain.
So this means that multiple bonds require a mix of sigma and pi.
Multiple bonds require a mix of sigma and pi.
If you have a single bond, all you need is the sigma.
So let's go back to how we got the original hybridization.
Remember, we started with carbon, with the box notation looking like this.
Here's the 2s and the 2p's and we started with just native carbon off the periodic table.
And we have this.
Just if we go according to the 2s2, 2p2.
And then in order to get hybridization to rationalize methane, where we've got four equivalent bonds.
In that case, we hybridize the s and all of the p's.
We took the s and all three of the p's to make the sp3 hybrid orbitals.
And then we were able to take these electrons and put them in one at a time.
And then bring in the four hydrogens and we end up with something that is symmetrically disposed in space.
The carbon, hydrogen and so.
So now I give to you ethylene I say, well how do I make ethylene.
Well, if I started with this sp3 thing maybe I could bring another carbon over here.
And I'd have three hydrogen sticking out.
So there's my sigma bond.
So I'm off to the races.
This is good.
But now I need to build a second bond.
I'm going to throw away a couple of hydrogen.
So i'll throw this hydrogen away and this hydrogen away.
So I'm going to have now C2H4.
I'm almost there.
The only problem is, this orbital is sticking out this way, this orbital is sticking out this way.
And I don't have the license to bend these orbitals.
And build a double bond.
Because they're 109 degrees apart and they're inflexible and they won't bend.
So this hybridization technique will not work as such to give me what I need to build ethylene.
What do I need?
If I'm going to build a sigma and a pi bond, I'm going to need to preserve one of the p orbitals so that it can still be available for lateral smearing Because how do I make a pi orbital?
I make a pi orbital middle with an 88.
So I need to preserve a p orbital so that in both of the carbons I still have this pi bonding capability.
So what I'm going do is instead of taking the s and all three of the p's I'm going to take the s and only two of the p' s and reserve one of the p' s to be sitting there for lateral smearing.
So instead what I'm going to do is this.
So this is going to be an unmixed p and this is going to take the s and not three p' s but only two p 's.
And this would be called sp2.
And now what do I have?
I've one, two, three.
And how do I put three of these in space?
They lie symmetrically in a plane at 120 degrees.
And then this thing is normal to the plane of the board.
It's sticking out.
Actually, I should've maybe drawn it in perspective like this.
And so now I've got the ability to put two of these together.
This'll giving me my sigma.
And then these two things lying on their side will smear with the lobes to give me the pi.
Now that's what gives me the double bond.
And here are the cartoons that show this.
Oh here, I took this from an old text book, it's text book I had when I was your age.
The wrote great books.
And then I flipped this around see.
I did all this just for you.
So I took this one, I flipped the image around.
So there are the sp2's these are the unmixed p's.
And you bring them close together and bingo.
There's the sigma.
That's this one, p plus p axially, that gives me the sigma bond.
And then I smear those two and that gives me the pi.
And there's ethylene.
You put the hydrogens here, one, two, three, four.
Two carbons.
Tada.
Isn't that cool?
Here's from another book.
It's looks like a Boston traffic map, doesn't it?
Just crazy.
This is from our book.
So they're showing you there's the sigmas and the pi's.
There we go.
More pictures.
OK, I'm going to take three minutes at the end here and show you an example of electronegativity at the extreme.
So if you take a look at the periodic table and look at a compound like sodium iodide.
If I just told you sodium iodide, covalent or ionic?
You'd say, ionic.
Because you got something from the the most metallic of the metals and something from the most non metallic of the non metals.
And you're right.
And the delta chi is 1.73.
And the covalent character is high.
And you end up with something that's very polar and so on.
Now here's a very interesting one, if you compare cesium and gold, you get 1.75.
Which is greater than what it was for sodium and iodine.
Now no one would argue that sodium and iodine is covalent, you'd argue that's ionic.
Well by the same metrics, cesium and gold is as ionic.
The plot thickens.
Cesium, if you melt it it's a metal, liquid metal.
If you melt gold, it's a liquid metal.
But if you mix them in equal number.
So you make a alloy of 50 mole percent cesium and 50 mole percent gold and you've got a delta chi 1.75, you've essentially made a cocktail with equal numbers of really good electron donors and really good electron acceptors.
And guess what happens, electron transfer.
And the melt is not metallic, it turns clear and colorless just as molten sodium iodide would be.
It turns into a molten salt.
So cesium gives its electron to gold.
And gold becomes the negative gold ion.
And what color is every ion?
It's got stable octet configurations.
It's got be the same color as neon, argon, and helium.
They are clear and colorless.
Big drop in electrical conductivity and a shift from electronic to ionic conduction.
Metals have electronic conductivity, ions, what do ionic liquids have?
They have like ionic conductivity.
Here's some data from the literature.
This is the log on the connductivity as a function of concentration.
So here's pure cesium over here, here's pure gold over here.
This is 600 degrees C. So the line stops here because gold melts at about 1060.
So gold is a solid beyond this alloy concentration.
But you can see this is roughly what you'd get.
So electronic conductivity up here at about 10 to the 4 siemens per, this is reciprocal ohms, but siemens per centimeter.
And down here, this is very low value ionic conductivity.
So this is a liquid metal.
And this is a molten solid.
And it all happens just when you get very very close to 50/50.
So you end up with something called cesium oride.
And it's sorcery.
You have one vial of liquid metal.
You have a second vial of liquid metal.
You pour then and it turns clear and colorless.
And the conductivity drops three orders of magnitude all because of electron transfer due to this electronegativity difference.
That's so cool.
That is so cool.
OK, with that I'll let you go.
We'll see you on Friday.
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OK. Settle down.
Settle down.
The weekend doesn't start until after the lecture.
First we learn.
All right.
So couple of announcements, the big one being the reminder about that little celebration next Wednesday, coverage through the end of the lecture on Monday, there'll be no weekly quiz next Tuesday, and Hilary wanted me to announce that there's a job opening for someone who's in class anyways, and wants to-- if you take decent notes and your handwriting is legible, you want to make $9 an hour, go see Hilary.
All right.
Let's get into the lesson.
Got a lot to cover today.
Last day, we were looking at the formation of molecular orbitals by linear combination of atomic orbitals into molecular orbitals.
So we had our SLI six-letter initialization, and we saw the formation of sigma and pi molecular orbitals.
And the way you distinguish between sigma and pi is based on electron distribution.
Specifically, in a sigma molecular orbital, you have continuous electron density between the nuclei.
That is to say, no holiday.
So I've drawn a sigma orbital here where I've taken two P orbitals lying on their sides, so they're bonding longitudinally and when the bond is formed, the one nitrogen nucleus is here and the other nitrogen nucleus is here and you can see that there's continuous electron density, no holidays between the two nuclei.
So this would be called a sigma bond.
The pi molecular orbital, on the other hand, has a nodal plane containing both nuclei.
So since the nuclei are in a nodal plane, that means that you've got to have two lobes, one above the plane and one below the plane, and that's shown here.
This is the example of nitrogen, where it forms both sigma and pi bonds and to form the pi bonds, you take the lateral smearing.
And so this is what I call the 88.
If you take the 88, you're going to make a pi bond.
So that's lateral smearing of the p atomic orbitals.
And the last thing I want to make sure that you appreciate is that when you form multiple bonds, you have to have sigma and pi.
You can only form one sigma bond because you have no holidays between the two nuclei so now you can't go and invade that same space with a second bond.
So that means you're going to have to get beyond that zone and that means you have to go into pi.
So if we've got only bond, it's a sigma.
If you've got two, it's a sigma and a pi.
If it's three, it's a sigma and two pis.
That's what we saw in the case of nitrogen.
I didn't complete it here, but there's two pi bonds because this is a triple bond.
So there's a sigma and there's two pi's.
One pi is shown here, and that's vertical to the board, and the other pi is the sigma x, which would be orthogonal coming in and out of the plane of the board.
So those smear give us two of these and then we've got the sigma bond.
And we even extended this to hybrid systems, and towards the end of the lecture, we looked at ethylene, and I just wanted to come back to that for a second because I raced through it there.
And so we wanted to look at C2H4, which is ethylene.
And in the case of C2H4, we recognized that when we went through the Lewis structure, we had a molecule that looked like this, and the thing to note here is that we have a double bond.
So once you see this double bond, then that means I'm going to need a mix of sigma and pi.
Now earlier we had studied methane-- methane, which was CH4, and to get methane, we hybridize sp3.
sp3, which gave us four identical struts coming off the carbon.
Well, we recognized that wouldn't work here because we needed to make a sigma and a pi so the gambit here was to reserve a p orbital.
Reserve a p orbital.
Don't combine all of the three p orbitals.
Combine only two of the p's and that gave us sp2 plus a p. And the box notation for that looked like this.
Here's the p orbital that's not been hybridized.
These are the sp2 orbitals that are a mix of S, and I've taken two of the p's and I'm leaving one of the p's unmixed.
And now I've got four electrons to put in here and that's where I ran out of gas last day.
I have to put the fourth electron in here.
These are still degenerate.
So now I've got still the capability of forming four bonds.
You can see-- whenever you see carbon, there's an interesting-- just something to train your eye, if you see carbon in a structure, you need four sticks coming out of the carbon, or else there's a spelling error.
So look here.
This is carbon with four hydrogens.
One, two, three, four.
This is carbon one, two, three, four.
Here's a carbon one, two, three, four.
It's always got four struts.
That's not what's different here.
It's the arrangement of the struts.
These are all four going at 109 degrees in three space.
This one is going to have three of them identical, because we studied last day the principle that equal energies, equal bond energies, imply-- I want to get the spelling here.
Equal bond energies imply equal spatial disposition.
Equal spatial disposition of those bonds.
So if we have four bonds of equal energy, we have the tetrahedron.
If we have three bonds of equal energy, how do I take three sticks coming out of a central carbon and make them uniformly distributed in space?
They're going to lie in a plane and they're going to be 120 degrees apart.
That's the consequence of having this structure.
All of this follow and why did we get to this?
Octet.
So here we are.
Here's the three of them and here the fourth bond is p, and it's normal to the plane of the board, and now we can take one of these and bond it to another one, bring them in along here.
This is an sp2 and an sp2 and they bond to make a sigma bond.
How do I know?
Look, no holidays.
Continuous electron density.
And now in the plane normal to the board, I have lateral smearing, and I make a pi.
So I've got a sigma and a pi.
That gives me the double bond and let me just show you the cartoons here because we've got some nice artistic renderings.
So here we are.
This is the top view of the three sp2's and then the p's are vertical.
so we're going to move along the z axis, bring two of these together-- there's the smearing of the sp2's on axis.
So this is a sigma bond, No electron density holidays from one nucleus to the other, and now we're going to smear the lobes of the p orbitals of the unmixed carbon.
That gives us the double bond and there we are.
This is the same thing taken from your book.
This is the sigma orbital and then chemists like to do this.
You see this sort of slashed line?
It sort of indicates it's receding.
It's going into the plane and then these are coming out.
This hydrogen is coming out at us and that other one's going way, way in the back and this is in the plane and there's the lobes colored is they irritate me always.
By the way, they've chosen to make this the z-axis, which I don't like.
The z-axis should be the axis along which they bond.
And then here's acetylene.
Acetylene is a compound that C2H2.
C2H2, and I'm not going to go through this in class.
You can try it when you have nothing else to do.
and so it's got a triple bond.
it's got a triple bond between the carbons and then hydrogens on either side.
Again, look at the carbon: one, two, three, four.
But now we're going to have to have two lateral smears, so that means I'm only going to take one of the p orbitals and leave two in reserve.
so when I do that, I'm going to end up with this, starting with the box notation.
Here's carbon as it would exist without any hybridization.
An S and three p orbitals.
And what I want to do-- I want to have two pi's and a sigma, so I'm going to just capture one of these and reserve two p's as unmixed, so that will give me this.
This is one S and one p, so this is called sp hybridization.
And then I've got the two p's unmixed, so now I've got one, two, three, four.
are both orthogonal to one another and these two are what?
I've got two struts coming off the carbon.
How do I get two struts coming off a carbon symmetrically disposed in space?
That's it.
So look at the shape of the molecule.
You see it here.
There's the-- the sigma orbital is here and then we have vertical and then out on the plane.
And so you first would say, wow, I understand that.
That looks a lot like nitrogen.
And then, of course, I looked at that, and I said, what about the right-hand rule?
And according to the right-hand rule, the z is going in this direction.
It's a little more subtle, but I pick up on stuff like that.
I can't help it.
OK.
So this is good.
So what I've done here is gone through a number of instances where we start with electronic structure.
sometimes we hybridize, sometimes we don't.
But what follows here?
When we start with the octet rule and we start playing with the energetics and forming the orbitals, can you see that there's a connection between electronic structure, which then dictates the bond disposition in space, which ultimately dictates the architecture of the molecule?
Molecular architecture.
So there's a way of doing this There's a way of doing this systematically.
I just got ahead of myself.
Before we get to that, I wanted to just pause here for a second and reflect upon the properties of the covalent bond as contrasted to the properties of the ionic bond.
So let me get to that first.
Properties of covalent bond.
OK.
The first property I want to highlight is the fact that the bond is saturated.
Remember when we said that the ionic bond was saturated?
So what does it mean?
Saturated?
It means that there's two atoms only.
Once you've got the electron pair between the two atoms, the bond is formed and that's the end of the story, whereas with ionic bonding, you can keep by electrostatic attraction piling on and piling on.
So this is the concept of saturation.
So we have the shared electron pair.
Shared electron pair and that's all there is to the story.
The other thing about the covalent bond is that it is directional.
And so the bond spatial arrangement-- these are the positioners.
You can't put atoms where there are no bonds and since the bonds have a direct spatial arrangement, then that's going to inform molecular architecture.
Dictates molecular architecture.
So what I'd like to do next-- so for example, here's one that we know.
You've seen this one.
This is methane.
This is sp3 hybridization.
So we learn that methane is tetrahedronal.
We say that the four corners of the hydrogen positions form a tetrahedron.
So we call this molecule tetrahedral.
So I could look at other molecules that have sp3 hybridization.
So if I look at something like titanium tetrachloride, which we met earlier to make titanium, this is titanium in the center and the four chlorines.
It's the same molecular shape.
If you see leaded gasoline, the compound in unleaded gasoline is tetraethyl lead.
Lead at the center, ethyl's at the four corners of the tetrahedron.
So all of these-- and same over here.
This one here, sp2, this means the molecule is planar.
The carbons and the hydrogens lie in a plane.
The whole molecule lies in the plane of the board.
If I give you any molecule that's got sp3 hybridization and has one, two, three, four, five, six atoms, in other words, all the bonds have atoms, the molecular is going to be a planar molecule with this, the 120 degrees.
So we've got a grand scheme here that allows us to codify this behavior and the connection between molecular structure and the electronics arrangement and it's written up here under the term Valence-Shell Electron-Pair-Repulsion Model.
Let's write that down.
I'm going to go through some examples.
Valence-Shell-- hyphenated-- Electron-Pair-Repulsion.
That's the model, and it goes by the acronym VSEPR, which is a five-letter acronym.
This is not an initialization because chemists reverse these, and they call it VSEPR, because the V and the S is kind of hard to pronounce unless you have Slavic blood.
So it's called VSEPR, and that is the scheme that's up here.
So I'm going to go through some examples and show you how you can start with electronic structure and figure out what the molecular architecture is.
So I'm going to follow this.
So first thing we're going to do is write the Lewis structure.
And the example I want to start with is sulfur hexafluoride.
I want to-- so the question is, what is the molecular architecture of sulfur hexafluoride?
Can we predict the shape?
And this is an interesting molecule.
It's very dense.
It's about five times-- it's a gas at room temperature.
It's about five times the density of air.
It's got an atomic mass of about 150, and it's used in the light metal industry.
It's used as a blanket gas for casting of aluminum and magnesium to keep the air away from the liquid metal.
Otherwise, the metal would oxidize.
But unfortunately, the sulfur fluorine bond is a really good infrared absorber, and when this stuff leaks out, it's a fantastic greenhouse gas trapper.
It's got a greenhouse gas coefficient that's, I don't know, something like 10,000 times that of CO2.
So there's a lot of conversation about banning this.
If we ban this, we're going to have to find a substitute.
There is no readily available substitute.
And what that'll mean is the cost of aluminum castings and magnesium castings will go up, which means that they won't be used in automobiles.
Instead, cheaper steel will be used in automobiles, which means the mass of the automobile will go up, fuel economy will go down, tail pipe emissions will go up and greenhouse gas emissions will go up.
So how do we get ourselves out of this situation?
Do we ban SF6 or not?
How do you go about thinking about that problem?
Well, if you're in Washington today, it's the lobbyists with the biggest bag of money that comes in.
That's what's going to dictate policy.
But if we lived in an ideal world, we'd have people like you who understand the chemistry and what would be the analysis?
You'd look at all the SF6 emitted from the cast shops in a given year and compare that to the increase in tail pipe CO2 emissions if you replaced the aluminum with steel, and then you figure out what the trade off is.
And maybe what you'd do is set up a policy where you would gradually phase this out with-- heaven forbid-- solid research support to give us what we need.
That way we get clean air and everything else.
OK.
So that's what we would like.
Now let's look at the Lewis structure of this thing because that's the first thing we're supposed to do according to VSEPR.
So I'm going to put sulfur here and I know sulfur's got-- it's 3s2 3p4.
So it's got six valence electrons.
I'm going to put them symmetrically around the sulfur, and then I'm going to bring in the fluorine.
I'll bring in one fluorine, and I know it's got seven valence electrons: one, two, three, four, five, six, seven.
And the same thing with the other fluorines.
So I'm going to put one, and then for the electron pair, I'm just going to draw a line.
OK. You're smart.
You can figure that out.
That's just shorthand.
OK.
Here's another fluorine.
One, two, three.
So there's two, four, six, seven.
There's one, two, three, four, five, six, seven.
One, two, three, four, five, six, seven.
One, two, three, four, five, six, seven.
Beautiful.
OK.
So now what I want to do is identify all of the bonding orbitals.
So I see around the sulfur, I've got one, two, three, four, five, six, six bonding orbitals, six bonding domains, if you like.
Well, that's 12 electrons.
I said octet stability.
Looks like sulfur took that idea and ran with it.
And it turns out that this does happen on occasion in elements that are very electronegative.
Very, very strong non-metals will form what is known as an expanded octet.
And here we've got-- here's a list of elements.
[MICROPHONE ADJUSTMENT]
We need a high-tech tie clip.
Pull out a little more, maybe it's constrained.
It's tethered.
the bond is too tight.
That's what it is.
It's not flexible.
It's covalent, you see.
All right.
So here we are with the elements, and the ones shown in the in the grayish-blue here are the ones that on occasion might form an expanded octet.
Not always, but if you do see 10 or 12 electrons around one of those elements, don't be alarmed.
If you see 10 or 12 electrons around the potassium, that's probably a mistake.
OK.
So we have six bonding domains and what else do we have?
We have six bonding domains, and we have zero non-bonding domains, and in total, we have six electron domains, in other words, six electron pair systems around the central sulfur.
And on the basis of the total of bonding and non-bonding, this is what dictates the skeletal structure.
In this case, the skeletal structure is trivial.
It's the same as the atomic structure because there are no non-bonding domains.
And so I've got six of them.
They're equivalent.
So what's that going to mean in terms of shape?
Well, I'll put the sulfur in the center and I've got one, two, three, four, five, six.
So two in one plane crossways and then two vertical.
And then I put the fluorines-- one, two, three, four, five, six.
So there's the molecular architecture and it's termed octahedral.
I know at first this throws you because you see six struts.
Where do you get the octahedral?
Well, if we connect the four fluorines along with the sulfur, they lie in a plane, and if we were to pitch a tent here, we could pitch a tent to the upper fluorine, and we'd end up with four faces.
So this is Greek.
Eight faces.
So there's four above and four below.
I know it's sort of like a Jasper Johns painting.
You say, octahedral.
You hear octa, you're thinking eight and you're seeing six fluorines or seven.
Anything but eight.
So this is anything but eight, but it's eight faces.
Last thing-- let's look at the energetics.
To hybridize like this, we needed to do something different because we know that the base structure of sulfur would be 3s2 3p4.
So this is 3s and this is 3p.
So that's 3s2, and then 3p4 would be one, two, three-- whoops!
I violated the polyexclusion there.
OK.
There we go.
So I can't form six bonds with this.
I only have two bonds.
So you say, well, okay, let's make sp3.
Well, if I make sp3, sp3 will give me this.
So now I got one, two, three, four, and then I'm going to put five, six.
I'm still down with two bonds.
I need to have six unpaired electrons.
So I've already consumed my s's and my p's.
Where else can I go shopping for orbitals?
I go to the orbital store and the orbital store is over here.
Here's 3d.
One, two, three, four.
So I've got one, two, three, four, five d orbitals.
So what I'm going to do is, I'm going to take the s, I'm going to take all three of the p and I'm going to lop off two of the d's.
So now I'm going to have one, two, three, four, five, six and I'll leave three d's by themselves and here's now sp3d2.
And now I've got six electrons-- one, two, three, four, five, six and now I have six equal energy states, which means I have to have six struts equally disposed in space and everything makes sense.
All right.
Is this polar or nonpolar?
Is this molecule polar or nonpolar?
Nonpolar, good.
Let's look at it carefully.
What do we have?
I've got sulfur fluorine.
Whenever you see fluorine, that's the biggest electron hog on the Periodic Table so you know you're going to have a dipole moment on the bond.
But I have six such bonds and they're symmetrically disposed in space.
All of the fluorines lie on the surface of a sphere so the center of local negative charge is right on top of the sulfur nucleus, which is where the center of locally positive charge is.
No electron displacement, no net dipole moment.
OK.
This is working.
Let's try another one.
Let's see.
What do I have here?
I got bromide pentafluoride.
So start-- let's go back to the VSPER rules.
There we go.
OK. VSPER rules.
So I put bromine in the center and it's got seven valence electrons-- one, two, three, four, five, six-- and I'll put the seventh up here.
And I'm going to bring in the fluorines-- one, two, three, four, five and just as before, there's one, two, three, four, five, six, seven.
So you get a bond with each fluorine.
A bond with each fluorine.
All right.
So what do we have now?
How many electron domains?
One, two, three, four, five, six.
Six electron domains as before, but one of them's nonbonding.
I get to use the colored chalk again.
All right.
So this is one nonbonding domain and five bonding domains.
One, two, three, four, five.
One bond to each of the fluorines, five bonding, and the total is six electron domains, so that means octahedral structure.
The total number of electron domains dictates skeletal structure.
So man said put six.
So I put the bromine here, and I go, one, two, three, four, five, six.
That's the skeletal structure.
Those are the bonds and space.
Now I know one of those is a nonbonding domain.
So where do I put it?
Well, they're all equivalent.
It doesn't make any difference.
Nobody knows where the real x is in this world.
So just for grins and chuckles, I'll make-- I'll put the nonbonding pair down here.
And then I put the fluorines at the other struts.
And so now what's the shape of the molecule?
You don't see these electrons.
And in fact, this is only done by chemistry professors, the other ones, not me.
I only do this to let you know what the books are doing because we all know that those nonbonding electrons are in here.
They're right up tucked against the bromine.
So if you walked in the room, what do you see?
The bromine and the four fluorines lie in a plane, and then you've got this thing sticking up.
So the molecule is a pyramid with a square base.
So it's called square pyramidal.
And how about polar or nonpolar?
Is there a net dipole moment?
Yes or no?
Yes, because we got the-- in the plane with the bromine, everything is centered, right?
The fluorines pull out, but they all pull out symmetrically, but then there's the fluorine pulling up.
So this is polar.
This is polar and it's got a dipole setup like this.
Good.
All right.
Let's do one more.
I'm having so much fun.
I'm going to do as many as I can until I run out of time.
This is the best part because colored chalk.
That's the end.
All right.
Let's do this one.
Tetrafluoroiodate.
IF4 minus.
This is cool because it's got a net charge, but it's a covalent anion.
So same thing.
Lewis structure.
So we're going to put iodine first.
Iodine and it's just like bromine.
It's got one, two, three, four, five, six-- I'm going to put a seventh electron here because I'm anticipating.
Just to mix things up.
See, last time I put the seventh electron up.
I could put the seventh electron down, but it's got a net charge.
How'd it get the net charge?
By accepting an electron.
So that electron's got to be here, too.
So I'm going to indicates that that extra electron's here as well.
So iodine's got eight already.
So we're off to a roaring start.
It's got eight and it hasn't even started bonding yet.
So we know where we're going here.
We're going to expanded octet land.
All right.
So we need four more fluorine.
I'm going to put four fluorines-- one, two, three, four-- and as before, so we've got bonds.
One, two, three, four.
And we've still got some other orbitals lying next to the iodine and they bond to nothing.
So we have two nonbonding, and we have four bonding for a total of six electron domains.
Again, octahedral structure.
Octahedral skeleton.
So I'm going to put iodine here.
One, two, three, four, five, six.
All right.
Now comes the part-- see, we said-- this VSPER consists of valence-shell.
I've been doing a lot of valence-shell thought here, but I haven't said anything about electron-pair repulsion.
Now I'm going to show you how electron-pair repulsion comes into play.
So I can put four fluorines around here in two different ways.
One way is to put the four fluorines into play and then I have the-- these are called lone pairs.
Why are they lone?
Because they don't bond to a second.
So sometimes chemists call these lone pairs.
They're all alone.
It's Friday and all alone.
So here they are.
So that's one way to put them, but there's another way to put them, and that's this one here.
I'm going to go way over to the other side, and here's the other way.
The other way to put them is to put the pairs in the same plane next to each other, because if I put the pairs opposite each other, that's equivalent.
It doesn't matter.
It just-- whether on the x-axis or y-axis.
So if you think about this as a globe, you think about the molecule as a globe, so in this case over there, I put all the fluorines on the equatorial position.
You can think of this plane at the waistline as the equator.
So these are all at the equatorial position, and the electron pairs are at the poles.
So the lone pairs are at polar positions.
In this case, the lone pairs are lying in equatorial positions.
Now what do you know about electrons when they get close together?
They repel.
And it's a really strong repulsion.
That's the Bourne exponent.
Electron-pair repulsion is really strong and it'll jack the energy way up.
So if you've got a choice of putting electrons here close together or here farther apart, which one is going to be the lower energy condition?
This, because this minimizes electron-pair repulsion, and in fact, this is the one that's favored.
So by using valence-shell thought, bonding, nonbonding skeletal structure, and then finally electron-pair repulsion, you conclude that this is the correct form, and so you don't see these electron pairs because we know they're really tucked in tight here.
What do you see if you do a spectroscopic analysis of this thing?
All five atoms lie in a plane.
So this molecule is planar.
It's planar, just as the sp2 ethylene was planar, but this is planar for a different reason.
Is this polar or nonpolar?
See, it's negative.
It's not negative.
If I were a cation, I'd be drawn in like this.
So is it polar or nonpolar?
Where's the net displacement?
The fluorines are pulling away from the iodine, but they're pulling away symmetrically.
So there's no net dipole moment.
It's cool, huh?
No net dipole moment, but net negative charge.
I love it.
It just-- it's the best.
OK.
So now let's look at some others.
We've looked at CF4.
We know CF4.
CF4, you could do this one in your head.
This would be just like methane, right?
This would be tetrahedral.
I'm not even going to write that out.
This tetrahedral and it would be sp3 hybridization and we just did IF4 minus.
See the stoichiometry.
It's four fluorines.
Four fluorines with something and it's not the same architecture.
So stoichiometry doesn't dictate molecular architecture.
Very important.
You can say, well, it's a net charge.
A net charge doesn't do-- it's this.
This is square planar.
Yeah, so I'm going to add that.
It's not only planar, it's planar and it's a square, whereas the other one is trigonal planar.
So this is square planar.
So I'm going to do another F4, just to show you, make the point that this thing doesn't-- so I'm going to do sulfur hexafluoride.
You say, gee, sulfur hexafluoride.
I bet it's just like carbon because carbon, sulfur-- no.
Let's look carefully.
So I'm going to put the sulfur here and we're going to put-- how many electrons?
One, two, three, four, five, six.
OK. And then we're going to bring in the fluorines, one, two, three, four.
So each of the fluorines is going to form a bond.
Do I have to do this?
Can I just do this?
Yeah, that's enough.
I'm not-- I'm raising-- we're accelerating now.
Now I'm going to stop talking and I'll just think about it and you'll get it.
All right.
So here we go.
Bonding domains.
One with the fluorine, two with the fluorines, three with the fluorine, four with the fluorine and there's two electrons, and they'll just sit together in a nonbonding domain.
So how many domains?
One, two, three, four, five.
Five electron domains.
Four bonding, one nonbonding It all adds up.
Good.
But now I got to put five struts.
This gives me the skeletal structure.
As before, skeletal structure.
So I've got to-- how do I put five struts around something uniformly in space?
One, two and I'll put three in the plane.
Three in the plane.
So three at the equator and two at the poles.
That's the skeletal structure.
This is called a trigonal bipyramid.
Because I can make a pyramid above, right?
I got a tent with three corners.
So that's a pyramid with three faces.
So that's called trigonal.
It's trigonal-- three-- and then I've got it above and below.
So that's trigonal bipyramid.
See, you're going to learn all these new words.
When you go out tonight, you're going to be able to impress people to the point they'll walk away from you.
All right.
So now let's start putting the fluorines-- again, we've got choices here.
They're not equivalent.
So one possibility is to put the lone pair at the equator and then the fluorines go here, and the other possibility is to put the-- let's see.
I've got room up there.
The other possibility is-- let's make the trigonal bipyramid-- is to put the lone pair at a polar position and put all the fluorines in the equatorial plane.
And so now we have to think about this for a second and try to convince ourselves which situation has greater electron-pair repulsion.
So the way to think about it is, in which position does that lone pair come into more contact than otherwise?
The other thing to know is-- I didn't say this earlier, but it just occurred to me-- is that if you've got a choice of bonding electrons versus nonbonding electrons, nonbonding electrons want more room.
Why?
Because they're not tightly confined.
When you get something in a bond, it's on axis between the two atoms, and so things are much more tightly confined.
So these things want a lot of room.
So if I'm sitting here, I interact with this fluorine and this fluorine-- these two are far away.
If I'm here, I interact with three fluorines.
So it turns out the equatorial position is favored.
Give more space nonbonding-- place nonbonding domains at equatorial positions in trigonal bipyramid.
That's what number five means.
If somebody asked you to test the microphone, instead of saying number nine, number nine or one, two, three, say, place nonbonding domains at equatorial positions and a trigonal bipyramid.
And they'll ask you to step away from the microphone.
OK.
So this structure here is, in fact, the shape of the molecule.
You don't see this.
And this molecule is called seesaw, like a teeter totter.
You remember, when you were kids.
You don't remember that.
OK.
So how about polar versus nonpolar?
It's obvious.
If it's asymmetric and you've got the fluorines and the non-pair, this is definitely polar.
And where's the net dipole?
Where's the negative side?
Well, along this axis, it's pulled right to the center.
So along this axis, clearly the fluorines are pulling so you have something that looks like this.
And how about the energetics?
How do we get five of these?
If we've got five struts, it means we must have five hybridized orbitals of equal energy.
Well, we got six before and before that, we had four.
Before that, we had three.
Well, now we're going to get five.
So how do we make five orbitals?
So in this case I start-- I want s, p, d, and I want five.
So one plus three is four plus one is five and that gives me sp3d.
Gives me five and then, what do we got?
One, two, three.
It's good.
OK. Now-- so we've covered a lot.
All right.
So some parting comments.
We've been sticking other things-- what we could do is we could do this one.
We could take carbon and I'm going to go now sp3.
only instead of putting hydrogens, I'm going to put carbons here.
So what happens now?
One, two, three, four.
I'll put another carbon here.
One, two, three, four, et cetera.
So this is sp3, all carbon, and this is called diamond.
This is called diamond sp3.
So these are all sigma bonds, all sigma bonds everywhere, sp3.
And these are very, very strong bonds and these are very tight bonds-- and one of the properties of diamond is that it has a very high refractive index because high energy-- and these are tight, tight bonds.
So the electrons aren't easily excited.
It's transparent, divisible light, which means that when light shines on this, there's no expectation, but there is bending.
So it's very, very high refractive index and a high refractive index is what makes this such a troublesome compound because it concentrates light.
So you can imagine there's somebody in a dimly lit cafe, and she's sitting over in the corner with a candle in front, and she's wearing a diamond stud, and it takes those few candellas of energy.
It concentrates, it shoots it in the room, and some unsuspecting fellow gets hit in the eye, is attracted to the person wearing the earrings, and that's when the trouble begins.
So this is very, very dangerous, very dangerous, and it's a consequence of these strong sigma bonds.
But there's another way.
There's another way that we can bond carbon.
We can make sp2 hybrids.
So if we make sp2 hybrids, that now looks like this.
The carbon lies in the plane, 120 degrees, which means we're going to end up with something that looks like this, et cetera, et cetera.
And this is graphite.
This is graphite.
And now we have a mix of sigma bonds in the plane that are very strong.
In fact, they're every bit as strong as these bonds, but there's a fourth.
I said, you always have to find four struts off the carbons.
So where's the fourth strut?
It's the pi bond that comes out of the board.
So there's a pi bond here and a pi bond here and a pi bond here and what we're going to learn later is that these things are delocalized.
Because of the spacing here, the electrons can form a bond here that then forms a bond here, but it can actually resonate in such a way that the electrons can move everywhere throughout the crystal of graphite and this is what gives graphite its electrical conductivity, whereas diamond is a strict insulator.
So these are very strong bonds in the plane, but not between planes, which gives graphite its lubricity.
It's a dry-lock lubricant and so on.
OK.
So this is very good.
So we have a mix of sigma and pi bonds.
And it's an absorber and it's dark.
It's an absorber so it appears black, whereas this diamond is transparent to visible light.
And if you look in your book, this is all categorized for you, everything that we've done here with VSPER.
All right.
Look at this.
All nice big system.
OK.
So here's diamond, which I was trying to draw over here, four struts sp3 hybridized.
this is graphite sp2, and these are pi bonds in between.
OK. Now suppose this fellow asks this girl out, and things get serious and they get-- I'm going to use an adverb here-- really serious and he decides maybe he wants to make a commitment and he wants to buy her a ring, an engagement ring, and the custom is that the engagement ring has a stone in it.
Now being an MIT student, he knows that he wants to have symbolism in that stone.
Now the custom is to get diamond, but it turns out that if you look carefully, the stable form of carbon at room temperature and atmospheric pressure is not diamond, it's graphite.
So if he wants to give his love this stone that symbolizes eternal commitment, wouldn't he choose the stable form and not the one that's metastable and doesn't represent the natural state at room temperature?
That's the truth.
That's the truth.
Graphite is the stable form.
I leave it up to the young man to persuade the young lady that the gemstone is going to have graphite in it and not diamond.
OK. Now something very interesting happened.
What I wanted to play for you as you're leaving today is an old Beatles song, Lucy in the Sky With Diamonds.
And what happened-- before I get to it, what happened was-- and this is even in today's Le Tech if you look on the second page, "New Fossil Skeleton from africa Predates Lucy."
It was just discovered.
I'm not kidding you.
This isn't a joke.
I heard this on NPR last night.
So it's a female skeleton of a pre-hominid, 4.4 million years old, which is a million years older than Lucy.
Lucy was discovered in Ethiopia in 1974 and considered to be 3.18 million years old.
This new one is 4.4 million years old, about four feet tall, and it just turns the clock even farther back.
The interesting thing is the name.
How did the skeleton discovered in 1974 get the name Lucy?
Interesting story.
There were two American archaeologists that were head of the team that did the dig, and they were so excited the night they found this pre-hominid skeleton.
They knew it was a female, and they hadn't dated it yet, but they knew that it was roughly way, way back, but it was definitely female.
So they were back at the campfire, kicking back, drinking, probably heavily, and singing songs, and one of the songs they were singing was Lucy in the Sky With Diamonds and that night, they named the skeleton Lucy.
But the story doesn't end there.
Where did this Lucy in the Sky With Diamonds come from?
John Lennon, one of the-- you may have heard of this group or not, but there was this group called The Beatles.
So one of the lead composers was John Lennon.
His six-year-old son came home from school one day with a piece of artwork, and the father asks, what is this?
He says, it's a painting of my classmate Lucy, and it was Lucy and he had diamonds in the sky.
So it was Lucy in the Sky with Diamonds, and then they wrote the song about it.
The real Lucy just died this week.
She had Lupus.
She died at the age of 46.
And all of these things have come together just because we had this lesson today about covalent bonding, which took us to this, which then took us to Ethiopia, which then takes this up here.
and so with that, I think we'll go for an early dismissal, and I'll wish you a pleasant weekend.
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Let's go to the lesson.
So the last day we studied VSEPR, which allowed us to infer the shapes of molecules, covalent molecules.
Today I want to talk about the state of aggregation.
What do I mean by state of aggregation?
Is something a solid, a liquid or a gas at a particular temperature?
So 3.091 is introduction to solid state chemistry.
One of the things we need to know is, under what conditions is the solid state stable?
And the state of aggregation, when it applies to covalent molecules, is going to lead us to a discussion of secondary bonding.
So today's lecture is state of aggregation or secondary bonding.
Now let's just reflect upon what we've learned up until now.
We studied ironic bonding and we knew that ionic bonding necessarily leads to crystal formation because we have unsaturated bonds.
And that leads to an ion array of unlimited size until you run out of reagent.
And something honking big made of ions is going to be a solid at room temperature.
Last day we appreciated with covalent bonding we have two options.
We can either make discrete molecules such as HCl.
And HCl as a discrete molecule, depending on a number of factors that I'm going to show you today, could be a solid, liquid or gas.
And we're going to understand more deeply how to sort out between the three of them.
Or, I showed you at the end of the last lecture, you can make a three-dimensional network and diamond was one example.
If it's a three-dimensional network that is a large array of solid then that means that you're going to end up with a formation of a crystal, which is going to give you a solid.
Diamond is solid at room temperature, graphite is solid at room temperature.
So let's now go to the one case that we haven't dealt with, and that's the formation of discrete molecules.
So let's look at discrete molecules.
And what we want to understand is whether they're going to be solid, liquid or gas at room temperature and what's the relevant physics here in order to make that determination?
The relevant physics is the following.
We're going to compare two forces.
We're going to compare the cohesive force between molecules versus the disruptive force.
And the disruptive force in 3.091 is always going to be thermal.
It's thermal energy.
We know this.
As we heat things they go from solid to liquid to vapor as temperature increases.
So thermal energy plays the role of the destructive force whereas the bonding is something that is the cohesive force.
So let's go back to HCL.
Last day we looked at HCl.
So here's one HCl molecule.
We have a covalent bond.
It's a covalent bond within the molecule.
We know this is a polar covalent molecule, with the chlorine having greater electronegativity and pulling the electrons towards its end.
And furthermore, this bond I want to categorize as within the molecule so I'm going to call it intramolecular.
Intramolecular.
Now if I want to answer the question, is hydrogen chloride going to be a solid, liquid or gas at room temperature, I don't have the information on the board.
The only way I can answer that question is to put another hydrogen chloride.
This simply forms the bond between hydrogen and chlorine.
To determine whether hydrogen chloride is a solid, liquid or gas, I need to look at how one hydrogen chloride molecule bonds to another hydrogen chloride molecule.
So now this end is the delta plus this end is delta minus.
And we have again an intramolecular bond.
So the question is, how does one hydrogen chloride bond to another hydrogen chloride?
This is the intermolecular bond.
Intermolecular bond.
So intermolecular bonds govern the state of aggregation.
And I'm going to show you the different types of intermolecular bonds.
This intermolecular bond isn't the primary bond.
The primary bond is here inside the molecule.
So the intermolecular bond is known as a secondary bond.
And there three types of secondary bonds.
We're going to look at them in turn.
So the first one is depicted here on the board.
And it's called dipole-dipole interactions.
And these obviously occur only between polar molecules.
Operative and polar molecules.
Not in.
Between.
Take out in.
Operative between polar molecules.
So we've made something covalent and now how does one stick to the other?
Alright and I think I've got a cartoon here that shows this.
There we go.
That's taken from an old book that I had.
So you see the H and the Cl and we have a dipole moment and the negative end of one dipole attracted to the positive end of another dipole.
So we have a net dipole moment here and we can measure the distance between the center of the dipole-- center to center spacing and call it r. Center to center dipole spacing.
And say that the energy in dipole-dipole interaction-- I don't expect you to know this by heart and do anything with it, but I just want you to realize that it's a weak force.
This is very weak.
It's proportional to the magnitude of the dipole moment.
You'd expect that.
Weak dipole moment, weak energy.
Strong dipole moment, strong energy.
Turns out it goes as the fourth order of the dipole moment.
And it's inversely proportional to the separation.
Only it's not coulombic, it goes as 1 over r cubed.
And there's a temperature factor.
1 over t. As temperature goes up, this energy goes down.
And these are very weak forces on the order of about 5 kilojoules per mole.
You remember what the crystallization energy of sodium chloride was?
787 kilojoules per mole.
So this is very, very weak.
It's operative at, between, at low temperatures.
At very low temperatures.
So for example, hydrogen chloride, in the case of hydrogen chloride, the melting point of hydrogen chloride is minus 115 degrees C and the boiling point is minus 85 degrees C. You can see at very low temperatures we already have enough thermal energy to disrupt.
So if we have solid hydrogen chloride, the forces between the hydrogen chloride molecules in the solid are these very weak dipole-dipole interactions.
And I think there's a couple more cartoons here.
This is taken from your text.
So there we go.
There it is.
That's the soup that might be HCl liquid.
OK so that's the first type of secondary bond.
Dipole-dipole interaction.
Let's look at the second one.
The second one is called induced dipole-induced dipole.
And it's operative in non-polar species.
Dominant in non-polar species.
Why am I using the word, species?
I'm trying to be a pedant and use fancy words?
No.
Because I'm going to make it generic.
I'm going to show it works in atoms and in compounds.
So as examples, what are some non-polar species?
Well how about something like argon?
Argon if you look on the periodic table it'll show you that it has a melting point of 84 kelvin and a boiling point of 87 kelvin.
So if I cool argon to below 84 kelvin, I get argon ice.
So what are the bonds between one argon atom and another argon atom?
It can't be this.
There's no net charge.
It can't be ionic.
It doesn't form a covalent network the way diamond does.
How do you justify the existence of solid argon?
It's just so cold that it just sits there and freezes?
I mean, how does it bond?
There needs to be some kind of bonding.
And other non-polar species.
So for example, the molecule iodine.
Iodine melting point is above room temperature.
It's a solid crystal at room temperature.
Well there's a strong covalent bond inside iodine.
But how does one iodine bond to another and to another and to another?
And we can even go to polyatomic species, such as methane.
You know there was data from the Cassini Probe.
Look at this.
This is an image from the Cassini Probe.
This is an island group.
The yellow is an island group.
The blue is a methane sea.
A sea of liquid methane on the moon of Saturn, Titan.
So there's liquid methane.
How does liquid methane form?
What causes one methane to bond with another methane?
That's the question that we're wrestling with.
So let's look at it first in a simpler context.
Let's look at it with argon.
So I put argon here.
It's a spherical atom.
And the question we had before says, how does one argon bond to another argon, as has to be the case in the solid or liquid argon.
So we know that this has a lot of electrons and the atom is net neutral.
But the electrons are in motion.
If I had an attosecond camera and I went in here and I went, snap, I could catch a freeze frame where the electrons aren't symmetrically distributed around the nucleus.
This nucleus, the atom is in constant fluctuation.
But net neutral.
Time average it's symmetrical.
So at some moment I might have a preponderance of electrons over here at around three o'clock.
So this end of the argon is a little bit delta minus.
Which means the other end is delta plus.
Now what happens if this is delta minus, this is delta plus?
Can you see that the positive end of this argon atom will then pull on the electron distribution of the adjacent argon atom rendering it delta minus at three o'clock and delta plus?
So this is the dipole in one, induces a dipole in another.
And why does it occur in the first place?
It occurs because the electrons are in motion.
Electrons in motion lead to time fluctuating dipoles.
They're time fluctuating.
Time average, there's no net dipole moment.
I'm not going back on what I said before.
There's no net dipole moment here, but this is time fluctuating dipole.
Time fluctuating dipole.
So who was the first to explain this?
The first to explain this was Fritz London in 1930.
In 1930 Fritz London gave us the explanation because this troubled people for a long time.
And he did so in a quantitative manner.
And the mathematics of the treatment are identical to the mathematics for the dispersion of light under certain conditions.
This has nothing to do with the dispersion of light.
The mathematical formulation imitates the formulation of the dispersion of light and hence the force here is known as the London Dispersion Force.
So people will say that solid argon is held together by London Dispersion Forces.
Or the bond is known by the name van der Waals.
We either call them van der Waals bonds or London Dispersion Forces.
Van der Waals was Dutch.
I know how to spell the word wall.
This is the dutch spelling.
W A A L S. Van der Waals.
So in this system here, the London Dispersion Force or the van der Waals bond is in fact not the secondary bond.
It's the primary bond, isn't it?
It's the only bond.
So by definition it must be the primary bond.
But can you see that in every compound, including diamond, we have time fluctuating dipoles in all of the atoms?
It's not just argon that has time fluctuating dipoles.
Every atom in you and me has time fluctuating dipoles.
But we're held together by much greater forces.
So that's why I say that in this case this is the dominant force.
But it's operative in everything.
Because wherever we have electrons, they're in motion.
So this is the dominant one.
Alright so we can look at it in a few other instances.
So for example we can look at it in the case of iodine.
We can look at it in the case of methane.
All the same.
Time fluctuating dipoles.
And this is a very, very weak force.
So the energy in London Dispersion Forces, or van der Waals bonds, is proportional to a quantity called the polarizability.
The polarizability is-- and this was defined by London-- polarizability is the measure of how easy it is to induce this dipole.
A measure of the ease of electron displacement within an atom.
And it depends on, it's influenced by the size and it's influenced by the number of electrons in the atom.
So what do I mean by that?
Well let's take two systems.
Suppose I've got argon and I'll go up two members in the same series and I've got helium.
So what are the forces that hold helium together?
Same thing.
But the boiling point here is 87 kelvin.
The boiling point here is 4.2 kelvin.
Why does this have such a low boiling point?
Well, polarizability.
Size-- helium is smaller.
So the degree, the corral in which the electrons can roam is smaller.
So the extent of electron displacement is smaller.
And secondly there are only two electrons.
And they're in 1s, so they're tightly held.
And so the delta minus delta plus capable of being created in helium is tiny compared to the delta minus delta plus that can be created in argon.
So the energy goes as polarizability squared and divided by r to the sixth, where this is the separation.
Not the radius but this is separation.
Dipole separation.
And this is a font issue.
This is proportional to alpha.
So you can tell the difference.
This is a proportional that's to alpha.
Of course you can't tell the difference.
This is contextual.
If I just wrote this by itself, you don't know what it is.
My goodness you're nervous.
So nervous.
All right let's take a look at-- here's the cartoons.
Induced dipole.
And here's London's paper.
Here's London's paper as it first appeared in 1930.
Theory and System of Molecular Forces.
And here's some cartoons from your book.
This is helium, showing helium.
It's kind of funny how the artist shows.
Like there's these two helium atoms and they're absolutely immobile and all of a sudden one of the helium atom starts jiggling and then it induces a dipole moment in the other one.
That's not how it happens, but anyway.
Alright this is hydrogen.
Same thing as iodine.
All right so there's delta plus, delta minus.
Because the electrons are moving even though there's a strong covalent bond inside.
So let me just make the point, I want to make sure people are very clear about primary versus secondary bonding.
So if I look at iodine.
I2.
This is a covalent bond.
It's homonuclear.
This covalent bond, there's no net dipole moment.
Here's another iodine.
But then we have time fluctuating dipoles.
So this delta minus isn't because this is the bond here.
This is the primary bond.
This is the primary bond and it's covalent.
And this is the secondary bond.
And the secondary bond is London Dispersion Force or van der Waals bond because this is induced dipole.
So now let's look at-- oh here's another one.
This one gets on my nerves.
Oh actually this is good.
This is this the bad one.
All right so I'm going to show you polarizability.
So here's a series of sp3 hybridized chains.
Propane, octane, icosane.
They're all the same.
They look like this.
Ch is sp3 hybridized.
And so you just have-- so what is that, C 3 H H. So it's just-- so you have 1, 2, 3 4. so there's hydrogen, hydrogen, hydrogen, and on the fourth one I'm going to make it flat instead of trying to make a tetrahedron.
So 1, 2, 3 hydrogens.
The fourth one is carbon.
1, 2 hydrogens, carbon carbon.
1, 2 3 hydrogens.
So this is C3H8.
And compare that to the longer one.
Which is octane, which is C8H18.
So it's 1, 2, 3, 4, 5, 6, 7, 8.
And all you have got to do is, 1, 2, 3, 4, 5, 6, 7, 8.
You put four sticks off of every carbon.
You can't go wrong.
You count them up.
You got C8H18.
Look at the structures.
They are the same.
So how come propane is a gas at room temperature?
Whereas octane, which is the principal constituent of gasoline, is a liquid at room temperature?
It all comes down to polarizability.
This one's got a longer corral.
So if you like the delta minus versus the delta plus, it's basically the same here except that separation is bigger.
All right, so that's pretty good.
This is the one that gets on my nerves.
You see this?
It's exactly what I just showed you.
So here's methane.
And there's the propane.
Here's butane and somewhere between butane and pentane we cross the the line at room temperature.
So pentane is liquid, butane is gas.
And you keep going up, up, up, and finally if you get to C20 you go from liquid to solid because now the van der Waals forces are strong enough that even at room temperature you make a solid.
It looks like paraffin.
That's why you can melt paraffin, because it's got weak van der Waals forces.
So temperature disrupts and it reforms.
If you try to break a covalent bond, you pyrolyze the thing and you don't get it back.
Secondary bonds allow for ready processing.
Here's what bothers me about this.
Instead of talking about polarizability, which is a physical quantity that means something.
They say molecular weight.
Well it's true that these scales, these are heavier and it's monotonic.
But to me what's the relevant physics between the ability to form van der Waals bonds and the mass.
It's just a dumb thing.
It's not a gravitational effect.
So that's why this thing is stupid.
And you see it all over in chemistry books.
And it's just dumb.
And if you put that on my exam you're not going to get points for it.
Because that's dumb.
OK let's go to the next one.
There's a third type of bonding.
There are certain things I feel strongly about and that's one of them.
OK so let's look at the third type of secondary bonding.
The third type of secondary bonding is called hydrogen bonding.
Hydrogen bonding.
It's a type of secondary bonding.
And it occurs between hydrogen and the most electronegative elements, fluorine, oxygen, nitrogen.
Why these?
Because they have very, very high average valance electron energy.
So the average valance electron energy I'm quoting in that second column, not megajoules per mole, but in sensible units like electron volts.
And so you can see, when you get up around 18, 19 electron volts you cross a threshold.
And that's the electronegativity as represented by average valance electron energy.
So you can see that as the electronegativity gets beyond some threshold value, roughly 3, then you can form hydrogen bonds.
So this is owing to high average valance electron energy, or if you like electronegativity.
Which means very strong polarity in the covalent bond.
So you say, well there's polar and there's even more polar.
So let's see what happens.
So I'm going to use a prototypical value here.
I'm going to look at HF.
So if I look at HF let's go through the Lewis structure.
There's H with it's 1 electron.
And F with the 7.
1, 2, 3, 4, 5, 6, 7.
And so we have a covalent bond here.
We know fluorine is the most electronegative so we have a dipole moment here.
Now the dipole moment is very strong here.
This is an accounting procedure to put the two electrons, but it no way represents the physical position of these electrons.
They're brought in very close to the fluorine.
So it's not some symmetrically disposed between H and the F. So I can't tell anything about whether HF is a solid, liquid or gas.
Why?
Because there's only one sitting here.
I have got to put at least one more.
Because this is a primary bond.
And it tells me how H bonds to F. It doesn't tell me how one HF bonds to another HF.
So I'm going to put another HF over here.
So here's another HF.
And it's also dipole.
But there's something special about that.
Already there's a dipole-dipole interaction.
But the hydrogen bond is much stronger.
It's much stronger.
And why is it stronger?
The electron in this hydrogen on the right is pulled towards the fluorine to such an extent that this hydrogen is so denuded of its electrons that it's acting as proton-like.
It's proton-like.
Now don't tell people that Professor Sadoway said that hydrogen inside an HF is a proton.
It's not a proton but it's starting to get more nearly like a proton.
Now what do we know about a proton?
Positive charge.
Tiny high-charge density.
So this hydrogen's looking forlornly over at its electron that's being hogged by the fluorine to a right.
And what do we know about these?
Oh it's time for colored chalk.
It's time for colored chalk!
What color are those?
They're red because they're non-bonding.
And the bonding are the blue in-between.
And what do we know about the volume occupied by a non-bonding pair versus a bonding pair?
It's larger.
Because they're not constrained.
So not only do we have a non-bonding pair, they're not just here.
They're sort of flopping around.
Hanging way out and there's this thing here that's almost denuded of its electrons, so it starts looking over here saying, if I can't get any action over here, what about here?
So that proton starts establishing contact with the non-bonding pair of electrons on the adjacent fluorine And this is the hydrogen bond.
The hydrogen bond is here.
The hydrogen bond is not here.
If you write this I will give you a 0.
I'll give you a 0 with a circle around it.
It's called the doughnut.
That's what you get when you write something so stupid.
This is not the hydrogen bond.
This is the hydrogen bond.
OK?
So it works.
It works.
Now let's see the effect of it.
Let's see the effect.
Alright so here's a cartoon showing that in water the hydrogen-oxygen spacing is about 1 angstrom.
And the hydrogen-oxygen spacing in adjacent water molecules can be less than 2 angstroms.
So there's certainly a difference.
I mean it can't be the same.
If it were the same it would be a covalent primary bond.
You'd have a network.
Now this is interesting here because this shows the values that are given on your periodic table, which were obtained without the use the average valance electron energies.
These trends are correct.
But look at this one.
They've got chlorine up at 3.16.
And all of these have been revised.
Now on a test, just use what you've been given on the periodic table.
But I want you to understand how this is rationalized with better data coming from photoelectron spectroscopy.
So now I want to show you the implications of hydrogen bonding.
The implications.
So I've got 4 homologous series.
So they're all element plus hydrogen.
So let's start with the group 14.
That's shown here in purple.
And what do we have?
All of these, we'll start with methane.
CH4, so that's a central atom.
1, 2, 3 4.
They're all tetrahedral.
Hydrogens at the corners.
Non-polar.
And no hydrogen bonding capability.
So one methane bonds to another methane by weak van der Waals forces.
That's how it does it.
It doesn't matter if I substitute the carbon with silicon or germanium or tin.
I can put SnH4.
And how does SnH4 bond to another SnH4?
It's just by London Dispersion Forces.
That's all that's operative here.
And so we have, what makes sense here, is that the heavier, more massive-- no, the ones that have greater polarizability have a higher boiling point.
Because their van der Waals forces are stronger.
That means you have to go to a higher temperature to achieve disruption.
Same temperature, weak van der Waals force: gas.
Same temperature, strong van der Waals force: liquid or solid.
And you see the boiling point here.
Monotonic from the lightest to the heaviest.
Now let's go to the next one.
Let's go to the green line, group 15.
Well group 15, what's that look like?
Let's look at the structure of group 15.
Group 15 is-- ammonia is one of them.
So we can look at the structure of ammonia.
And if we use VSEPR we'll end up with something that looks like this.
I'll go through the whole thing.
You're going to end up with three bonds like this.
It's a tetrahedral skeleton with a lone pair.
Ah, colored chalk.
So now, what happens?
I can't say anything about this.
Why can't I say anything about whether this is a solid, liquid or a gas?
It's the only one there.
I have got to put another one.
So put another one up here.
N. H. H. H. Same gambit with the hydrogen fluoride.
This hydrogen sees this lone pair and establishes a hydrogen bond.
And that adds to what otherwise would have been a dipole-dipole.
This has a net dipole along moment, agreed?
There's a dipole-dipole moment here.
But this bond is even stronger.
The hydrogen bond is even stronger than dipole-dipole interactions.
So now let's look at the graph up here.
So what about in phosphene, PH3?
No phosphorous isn't electronegative enough.
So in phosphene, in arsine and in stibine we don't have hydrogen bonding.
So you see this series in the group 4 here?
It goes monotonic from the heaviest element down to the lightest element.
But here, heaviest, less heavy, less heavy, and the lightest element that should be down here is up here.
Why is the lightest compound not down here?
Because of the addition of hydrogen bond.
So ammonia is off the line.
This line shows the trend based on dipole-dipole interactions only.
And you can see the difference.
See if you have van der Waals forces alone, versus dipole-dipole, dipole-dipole are stronger.
So SnH4 has a lower boiling point than SbH3.
I can't predict this but I could ask you, if I gave you this data, I'd say, can you explain this to me?
Let's keep going.
Let's go to group 16.
So, telluride, selenide, suphide, the oxide, H2O, water.
It should have a boiling point of minus 100 centigrade were it not for hydrogen bonding.
How does water work?
Again, SP3 hybridization.
Oxygen, 1, 2, 3, 4.
Two lone pairs.
1, 2, H, H. And now what happens?
I bring another water molecule over here and hydrogen bonding.
And that hydrogen bond raises the temperature, raises the requirement for thermal disruption and moves the boiling point of water up to 100 degrees celsius.
It it weren't for this we wouldn't have this conversation.
Because we wouldn't have evolved as a species capable of conducting business at room temperature if water boiled at minus 100 celsius.
Hydrogen bonding is critical.
It's absolutely critical.
So now you know.
And this isn't just some little bit of pedantry for a professor.
This is very important.
Because we're going to learn later that, when it comes to biochemistry, we will appreciate that most biochemicals are made of carbon, oxygen, nitrogen and hydrogen.
And that means you can have hydrogen bonds here.
And you can have hydrogen bonds here.
Hydrogen bonding is critical to life.
So this is a very important thing to know about.
OK we have a minute or two.
So you can see polarizability increases here but hydrogen bonding operative here.
That's explains the mystery.
All right we're going to jump over this because we are out of time.
And so I will simply show you a few pictures.
Pictures!
So this is a conference in Copenhagen in June of 1936.
And there's Fritz London.
But look at who else is at the conference.
Niels Bohr, Wolfgang Pauli, Werner Heisenberg, Max Born.
Remember the Born Exponent?
You better remember it for Wednesday.
This is Lise Meitner, you haven't met her yet.
We'll get to her.
This is Walter Stern.
Stern-Gerlach.
And this is James Franck.
That's just the first two rows!
That's quite a conference.
Here's a picture of Fritz London sitting on a bench in Berlin with Erwin Schrodinger.
Schrodinger is brooding; he's thinking.
Fritz London is smiling.
You know why?
Because he's figured it out.
He's figured out.
It's time-fluctuating dipole, but he's not going to tell Schrodinger.
He's going to say, you have to read about it.
But you have to go to the library.
Because if you don't read the primary sources, if you go to Wikipedia, you won't find this.
Because this is not going to be in Wikipedia in 1913.
OK, last thing-- hold on, hold on, hey wait a minute!
Where are you going?
We're not done yet.
Did the professor say class is dismissed?
No.
So here's a biography of Fritz London, which I would recommend if you have a few minutes and you'd like to unwind.
Go out, sit in the sun and read something like this.
It tells the story of how he came up with these ideas.
The rise of fascism in the '30s.
He comes to the United States, takes a teaching position at Duke University.
All of the people that he met along the way.
All these people that we study, all this stuff, it's all there from his perspective.
And the perspective of his biographer.
Plus the pictures.
So anyway, primary sources.
Go read the primary sources.
All right, class dismissed.
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OK. Folks, I may not look like Don Sadoway, but for today, I am Don Sadoway.
So let's-- I guess you can hear me pretty well, right?
OK.
Professor Sadoway is in a faraway, terrible place.
It's called Vienna.
Yeah, I know.
We're here.
What can I say?
And he'll be back next Tuesday.
And so my name is Ron Ballinger, and I'm taking his place.
Put a muffler on that.
OK, great.
I'm sure you're all anxious, before we get started, to see that.
Right.
The average-- it was about 66, which is about 10 points lower than the last five or 10 quiz ones.
So it's a little bit lower.
It was a little bit harder test.
I'm sure Professor Sadoway will mention this when he comes back, but for those of you who are below the 50 mark, it's time to do something about it, and when I say do something about it, I'm sure your recitation instructors will be very happy to help out in any way possible, but there are other options, not the least of which is a tutor.
And so if you think you need one and you need to be kind of ruthless about examining yourself-- if you think you need one, go down and see Hilary and there are tutors which are available for individual instruction.
So I think it's a great time to take it take advantage of that because I can guarantee you that test number two will not be any easier.
In fact, it'll be considerably harder because it'll be pretty much new material.
OK.
So that's enough for that.
Last day, we were-- I wasn't here.
I was in Washington.
They're actually trying to build what's called an exoflop computer.
Does anybody know what exo means?
10 to the 18th.
10 to the 18th floating point operations per second.
Why do they need that?
Because when you go beyond 3091 then starting modeling these atoms-- especially F-block atoms-- you need a supercomputer that large and to do one run on a petaflop machine takes 100,000 CPUs 10 hours for one atom.
So solving the Schrodinger equation's a bit tricky.
OK.
Remember last time, we talked about metallic bonding and we're sort of sneaking up on it.
Remember, Paul Drude-- well, let's look up here.
These are the characteristics of metallic solids.
They have high electrical conductivity.
They have high thermal conductivity.
They shine and they have ductility.
So we're going to deal with the first three today, and later on in the course, we'll talk about the ductility part.
But recall that Paul Drude modeled the solid as a series of cations, which amounted to the nucleus plus the inner core electrons and then allowed the electrons-- the remaining, the outer shell electrons, the valence electrons, to float around, called it a free electron gas.
And that model explained the temperature dependence of heat capacity, but it was not so good when it came to electrical conductivity.
It explained the fact that you get electrical conductivity, but not the fact that some materials are conductors and some materials are insulators.
So we need to deal with that and for that, we need to talk to two sets of folks.
The first one is Felix Bloch and he was a 1928-- he was a-- he got his PhD under Heisenberg.
Boy, it would've been nice back in those days, working for those great folks.
And he applied quantum mechanics to solids.
He said, OK, let's consider that in a solid, the atoms are arranged in an array.
What's an array?
Well, it's actually called a crystal.
It's an ordered array of atoms.
We're going to say a lot more about that as we go along, but-- and then he said, well, then let's apply the Schrodinger equation to this system.
Now it's a big difference between the Schrodinger equation for a single atom in a gas and multiple atoms in a solid.
It's a different set of boundary conditions, different set of conditions all together.
And when you do that, you get a set of solutions for the valence electrons which is, as you might expect, it's periodic.
It's periodic and it invokes wave-like properties of the electron and you end up with a set of values of the wavelengths for the electron that are such that it allows mobility, which is, after all, what we're after.
These electrons got to move through the solid if we're going to have conductivity.
And this is an example where classical physics wouldn't allow that.
So you can't get this kind of behavior with classical physics.
So that's one piece.
And the second group, two guys.
Walter Heitler and Fritz London.
We know Fritz London from London dispersion force-- the same guy.
These guys were very, very productive.
And he did a post-doc-- that's another word for slave labor, by the way.
Folks will know that here.
For guess who?
Well, he did it for Schrodinger.
Interesting story, I guess.
These guys showed up for their slave labor at Schrodinger's lab, only to discover that Schrodinger had taken a position at a different university.
So these guys showed up and he said, goodbye.
Anyway, that's a bummer.
OK. And they sat down and they said, well, OK. Let's see if we can go at this from an energy point of view as opposed from the quantum mechanical point of view and let's see if we can apply LCAOMO-- now don't break out in hives because of the test-- to a solid, to large aggregates.
Large what?
Of atoms.
How big is large?
Well, I don't know.
Dream up a number-- say, 10 to the 23rd.
Lots of atoms.
And let's see what happens.
Well, we've already done a little of this in the past.
We looked at the energetics of the stability of hydrogen or the stability of helium.
So we can take that and we can kind of add up and we'll see what happens.
Well, remember, we had atom A and now let's only deal with the valence electrons.
And so this would be 0 and there's another atom A-- 0-- and this would be A2 and we've been through this before.
Let's just take-- we know that we get splitting and we get a sigma bonding orbital and a sigma star anti-bonding orbital.
Well, let's start adding some atoms here, additional atoms.
What do we get?
Well, let's just try for A sub N-- lots of atoms.
What do we get?
We end up with lots of states.
Remember, we have to have conservation of states.
So for every atom we add-- let's say this is copper, for example, which has an s1, one s. It's an odd number of electrons.
So every time I add a copper to this, I add extra states.
And then what do I do?
I start filling them using Aufbau, just like we've done in the past.
Well, if we keep going, and by the way, not to scale, it starts looking like a whole bunch of states.
And this energy level here, these energies, what?
We know for hydrogen, this is what?
Minus 13.6 electron volts.
So what are we dealing with here in terms of state differences?
Well, let's take 10 to the 23rd atoms and let's see if we can calculate energy.
Well, let's take for copper-- the molar volume of copper is what?
7.11 centimeter cube for mole.
All right.
And that's N to the 23rd-- actually, 6.02 times 10 to the 23rd atoms.
And let's just ask ourselves, what's the sort of range here?
Well, we know it's 13.6 here.
We know that's 0.
We're all friends.
So let's call it 10 ev, just for grins.
Not a bad number.
And let's ask ourselves, well, if we've got 10 to the 23rd atoms and 10 ev-- let's take 10 ev over 10 to the 23rd and you get what?
Well, 10 to the minus 22 ev per state.
That's small.
That's really, really, really, really small and if you convert that to joules, you end up with 10 to the minus 41 joules.
That's really small.
So what are we saying?
We're saying that this organization here-- when we and put a lot of atoms in there, we end up with what amounts to a band.
A band of states.
And what do we do?
We populate this band just like we did using the Aufbau principle, but what does it really look like?
Well, in the case of copper, you remember, copper has a 1s electron.
Copper is 3d10s1.
If we start filling start this band, we're going to get-- and so we start filling and we get up here and we find that we end up with a half-filled band, because there are states that are not occupied.
But remember, the distance between this guy and this guy, an occupied and unoccupied state, is only 10 to the minus 41 joules.
So that's pretty small.
10 to the minus 22 electron volts.
To give you an example, 0.025 electron volts, which is, if you want to convert that to temperature, that's about 300 degrees Kelvin.
So very, very small energy.
So that means if I were to take and if I were to apply a potential here, then what happens?
Well, I add a little energy-- and I don't have to add much energy-- and it's very easy for me to promote one or more of these electrons up into the above here and then I can get migration.
So the endpoint is that if we apply a potential, we end up with conductivity, which is what we were after.
Moreover, we can say a few more things.
If I shine-- now this is a sort of mixed metaphor here.
This is an energy diagram and this some plates on an electrode so be a little bit careful.
We shine photons on this thing.
What happens?
What's the energy of the photons, of light?
Between 200 or 400 and 700 nanometers.
It's about one electron volt.
So there's plenty of electron volts here.
I'm going to-- with a metal, not only will I have conductivity, but there'll be enough energy here to promote electrons and those electrons will move around and we'll end up with re-emission and we end up with opaqueness.
In other words, we absorb light and readmit it and so we end up with, in the case of a metal, luster.
So there's a couple of things.
So if we recall back here-- we're talking about high electrical conductivity.
We had-- we need to say a lot more about that, but we're OK here.
But Drude, it was OK as well.
High thermal conductivity, Drude did that.
That was fine.
Luster.
OK, So it works for copper, seems to be OK.
But what about another one?
Like, say, beryllium.
What's with beryllium?
Well, beryllium is 2s2s2.
So now let's draw the thing for beryllium.
We have beryllium.
We have-- must be a 2s and now it's 2p and we populate this.
And now we want to add a beryllium-- n beryllium atoms, all right?
What's happening?
well, it'll be a little bit.
We got this band that we've talked about earlier.
Now we go to fill it.
So we're filling these guys and we fill it, and lo and behold, we find out that we fill it all the way to the top and so we're screwed.
We have no conductivity.
What I can imagine now is there's 100 computers in here.
50 of them have Skype.
Guess what's going to happen?
By the end of the day, there'll be 50 things up there.
So we know beryllium's metal and it has conductivity.
So what's the deal?
Well, it turns out that while beryllium has the 2s band full, the 2p orbitals still are there.
They're still there and so there's going to be a band unfilled.
There's going to be a band for the 2p orbitals.
Well, guess what?
It turns out that the way I've drawn it, the 2p band overlaps the 2s band.
And so what that means is that I can promote into the 2p band and I can achieve my conductivity.
So that's another-- so we solved the problem of the filled s-orbitals, and we've got conductivity in both cases.
So we're OK so far, but Drude wasn't far behind.
What can we say about insulators?
It looks like we have a problem.
Well, let's take a look at another example.
And let's try carbon.
Well, we know carbon is what?
2s2p2.
And so we can go-- and we know, by the way, that most of the time carbon will hybridize and so we'll end up with 2sp3.
So n equals 2sp3 hybridized.
So we see that happen.
And we also know that carbon's not a metal and that we have strong bonds.
In the structure that we're going to talk about, they're covalent bonds and so they're very strong bonds.
So let's do this.
Carbon in the gas phase-- we have 2s and 2p.
We know that what happens is we end up with hybridization and so we ends up and we fill these guys.
So that's what we get.
Now this would be for diamond in the gas phase.
So with this hybridization, we get bands.
We start adding carbon atoms to this and what do we get?
Well, we get a band that's the 2p band.
The sp3 band is different looking than the bands for magnesium or copper or beryllium in that there's a separation between the sigma and the sigma-star anti-bonding orbitals.
So we start doing these guys up using Aufbau, and we discover that there's an energy gap e sub g between the sigma bonding orbital part of the band and the sigma-star band.
So what's happening?
Well, that's actually pretty good sized.
It's about 5.4 electron volts.
Now compare that with what?
Visible light is around one electron volt.
So now we have a very, very different situation and there's some terminology we need to have here.
This is the so-called conduction band and this is the so-called valence.
band.
Valence band and conduction band.
So in order for us to get electrical conductivity, we have to somehow promote an electron across the 5.4 electron-volt band gap.
This is the so-called band gap.
Now we know that visible light is about 1 electron volt so we know that's not going to do it.
And in fact, we might expect that even a diamond is what?
Transparent to visible light.
So if the photons come in, if there's no promotion, there's no mission, transparency to visible light.
So that's a big number.
Let's try another one.
Let's try silicon.
Now I'm going down group four, where silicon is going to be also sp3 hybridization.
So we can do that, only this time over here, carbon is what?
n equals 2.
For silicon, n equals 3 and so we can do the same thing.
Here's the 3s.
Here's the 3p.
And we hybridize and we end up with before and then we end up with a band, same kind of band structure where this is a sigma-star, star, this is e sub g, this is the valence band, conduction band.
Now we have states up here, but in this case, what do you figure?
n equals 3.
So those valence electrons are hanging out further away from the nucleus.
And we know generally that the energy drops off, the valence electron energy drops off as we get further away from the nucleus.
So you might expect that the energy of this gap would be a little bit smaller and indeed.
it is.
Very fortunate for us, it's 1.1 electron volts.
Now we're getting close.
And so this is 1/4 or even 1/5 of that for carbon and remember, visible light is on the order of one electron volt.
So one, one and half electron volts.
So what happens?
This is close enough so that we get semi-conduction.
It's called a semiconductor and we'll make a definition.
If the band gap is greater than 3 electron volts, we call it an insulator.
If the band gap is less than-- well, let's give it a range.
1 to 3 electron volts, we call it a semiconductor and of course, if the band gap is equal to 0, we call it a metal.
So that's a distinction, which is a little bit arbitrary, but pretty good.
So now, if I take a look at silicon, what do I see?
Well, if I had a piece of silicon here as opposed to diamond and I looked at it, it would be gray.
We would have color and the reason it would have color is because I get some promotion here and I get re-emission and so now I get color.
So it's not transparent to visible light.
OK. Let's take a look in general at what we-- what we're talking about is photoexcitation.
Now this is a diagram which is-- you should have in your aid sheet.
Sooner or later, you should have it in your aid sheet.
OK.
So let's draw a general band.
Here's our two bands.
This is the valence band.
This is the conduction band and we have a band gap.
And now what happens if I were a photon here?
Well, I guess it depends.
If the photon-- if e photon greater than e band gap, then what happens?
I get promotion of an electron up from the valence band to the conduction band.
OK.
So then what happens?
Well, the photon is quantized.
It's one shot.
The photon's gone.
Now I guess we need to settle one little thing.
What happens if the photon is really a lot greater than the band gap?
In another words, let's say it's a million electron volts or something like that.
Well, for purposes of 3.091, what we're going to assume is that any excess energy goes to heat.
Let's not worry about what happens for other things.
In fact, that's not far off.
Some of these LEDs that you see, they get warm.
So there is some heat.
So the photon-- once it goes away, I get-- the electron falls back down.
Well, if the electron falls back down, then what do I get?
I get another photon out, but now I've basically built myself a diode or some kind of device.
That photon-- the energy of the photon is equal to what?
It's equal to the band gap, which is equal to HC over lambda.
So I can adjust the wavelength here if I can adjust the band gap.
See where we're going with this?
Well, what happens if I-- let's say I hook this up to a resistor and I draw a current.
Well, as long as I keep the light shining on here, photons, then I can keep promoting these guys, and I can keep drawing current.
So what do I have?
I have something that I can generate electricity with light.
And so that's sort of solar-powered something, isn't it?
We can generate current.
What happens if I take-- and now instead of doing that, I hook up a battery to this thing and now I pump current in here?
I pump current in here.
In that case, I can force the electrons up here.
So I can force electrons to go in the reverse and I could make a photosensor or I could force the electrons up here and let them come back down and I know this wavelength here, I can make a light-emitting diode.
So just this one little concept here, which is very simplified, gives us the basis for photovoltaics.
And that's exactly the way it works.
OK. Let's put this in another way.
Let's plot the percent absorption.
In other words, if the energy's high enough, I absorb-- versus wavelength this way and since energy is inversely proportional to wavelength, we have energy going this way.
Let's put some numbers in here.
Let's say this is 400, this is 700 and this is Professor Sadoway's dreaded nanometers.
So this is visible light and let's put carbon on there.
Well, if you do the calculation, convert iy, it turns out that for carbon, with this band gap, you end up with behavior where if the energy is above the band gap, I get-- well, let's call this 100.
That's 0, right?
I get 100% absorption.
When I get to the band gap, below the band gap, what happens?
Well, I drop off and it goes like that.
In the case of carbon, this wavelength, right, which is, by the way, called the absorption edge-- that number comes out to be about 229.
If I try it with silicon, that number-- so this would be carbon, this would be silicon and carbon-- 1125.
All right.
So the absorption edge for silicon is 1125, which means it absorbs in the visible range and so that's the way we get the luster.
And this would be in the what?
Infrared.
And this would be in the UV region, if you wanted to-- which would be far hard UV and this would be far infrared.
OK.
This is the paper-- Heitler and London's paper.
It's in German and you can see in Zurich.
And this is their original paper.
You can go down to the library and you can get this original paper and if you know how to read German, it's-- we have it.
This is some of the original paper.
You notice the Schrodinger equation up there.
Nasty-- really nasty, but this is the calculations.
This is radius.
This is energy and you can see a point here where you get an energy minimum and that's where the bands operate.
This is another way to look at it.
This would be the bottom of the bonding orbital or the top of the anti-bonding orbitals.
This is another way to look at what we've drawn so far.
Somebody is talking.
Something you guys ought to know, this room is acoustically perfect.
If you pass gas in the back row, I'll hear it, OK?
So if you're talking up there, I'll hear it.
OK.
So you can see what's going on here.
We've drawn that and this another way to look at it.
By the way, there's a mistake in your book that doesn't make any sense.
2s3p-- it's 2p in the description for beryllium.
OK, this is another way to look at it.
This is in the archival notes.
It's one of the few pieces of the archival notes which I don't understand.
So don't worry about that.
OK.
Here's what happens when you-- where we illustrated this where you apply a potential.
What happens is you depopulate the bonding orbitals, and you populate the antibonding orbitals rules, and you get conduction.
This is another-- this is the right way to look at it.
This sort of illustrates the things we've gone through the whole time.
A metal has no band gap.
That no band gap is achieved either by half-filled set of orbitals or overlap between one or between different levels.
An insulator has a large band gap and a semiconductor has a sort of intermediate band gap.
Let's get a little-- sort of wet your whistle for your reading for next Tuesday.
I think we have Monday lecture on Tuesday.
If you go over to the reactor-- they have a reactor here at MIT-- once a week, a truck backs up and it's full of silicon logs.
These things are about 12 inches in diameter and they're about this tall and they bring them into the reactor and they're irradiated with neutrons.
Well, what do you think happens?
Since everybody in here knows the Periodic Table by heart, what's the next atom-- what's the next element over from silicon?
Phosphorus.
OK.
So guess what?
You take and you irradiate the silicon.
You absorb a neutron.
It adds one to z, doesn't it?
And it becomes phosphorus.
Well, phosphorus has one more electron than silicon.
So I have implanted phosphorus into silicon by transmutation.
And there's one more electron that's in there.
Now where's it come from?
I don't know.
The electron bank, right?
But what do you figure the energy of that guy is?
Well, it's a lot higher than the electrons in silicon.
And so guess what?
That electron, we'll find out, doesn't reside down here.
It resides up here or very close to that.
So that's what's called-- and we'll talk about it next time-- it's called doping, only now we're doping it in a special kind of way.
OK. We've got-- let's keep going.
I want to get this one up there.
There we go.
OK. Now-- so far we've talked about photons.
They're a one shot deal.
What about thermal excitation?
Well, we can make our-- the same drawing we have in the past.
We got our material here where we have a valence band, conduction band, and now we put-- we add thermal energy.
By the way, what's the one electron volt in temperature?
Just to give you a feel-- one electron volt is 11,600 degrees Kelvin if you convert that to temperature.
So one electron volt seems like a small number, but in temperature, is pretty warm.
OK.
So now the thermal energy is what?
It's constant.
Unlike the photons, which are one off, right?
So it's constant.
So what happens?
Now I take an electron from the valence band and I promote it up here.
So it looks sort of like-- so far up there with photons.
But there's one critical difference.
It stays up there.
It stays up there because the energy is constant.
So what does that mean?
Well, it leaves behind a broken bond.
And that broken bond has a lot of energy, and in fact, it doesn't like to stick around.
So given a chance, it'll move.
It's like a hot potato and so that broken bond is called a hole.
We're not too imaginative.
It's a hole in the lattice.
But it has high energy.
So what's the deal with this hole?
Well, this is-- a hole is a 0 in a land of minus 1.
So what does it mean?
It's actually net positive.
So now in the case of thermal excitation, I end up with an electron in the conduction band and a hole in the valence band.
So I end up with an electron in the conduction band and a hole and the electrical engineers call this p. They're not very imaginative either-- positive in the valence band.
So we get 241.
We get two charge carriers for every event.
That, we get one.
This, we get two charge carriers for every event.
So now we have what?
Let's ask ourselves, what's the-- can we do a little math on this?
And the broken bond, by the way, is a hole and so the number of electrons is also equal to the number of holes because it's a one for one, and in the electrical engineering world, they say n equals p. And so we can now go back and say something about this conductivity.
We can do some calculations here and we can calculate the conductivity due to the electrical connectivity, and it's really the sum over i of n of i, which would be the number of the population of carriers times-- that's the value of the charge on the carrier-- times something called the mobility.
OK, so what's the mobility?
Well, you could imagine that these electrons have to move back and forth in the lattice and in a metal versus a covalence solid or something like that, it might be a little bit different.
The resistance might be different and so-- in fact, it is.
mu i is equal to the velocity of the charge carrier divided by the electric field.
OK.
So we're almost there.
Now we need to-- and we can simplify this because of the n equals p business, and we can say that the electrical conductivity is equal to what?
It's equal to n sub e times e, which is the electronic charge times mu sub e plus n sub h, which would be the holes, the electronic charge, times mu sub h. But since n is equal to p, we end up with n sub e times the electronic charge times mu sub e plus u sub h. OK. And so that-- what we really need to get is this.
Well, we don't have time to go through that, but from a quantum mechanical calculation, we can get that.
n sub e is equal to what?
It's equal to some constant times T to 3/2 times the exponent of minus E sub g over 2k sub B times T. So what is all this stuff?
Well, this is a constant.
This is the temperature.
This is the band gap.
So it's a representation of the binding.
What's down here?
Well, k sub B is Boltzmann's constant.
This is absolute temperature and so this represents the disruptive force.
So it's a balance between how tightly they're bound and how much energy I've got to do this.
Well, let's-- we've got one minute.
Let's do a quick calculations for silicon, just to close the loop here.
For silicon at room temperature, that number comes out 1.3 times 10 to the 10th per centimeter cubed.
Say well, that's billions.
That's a big deal.
What about copper at room temperature?
Well, that number comes out about 8.5 times 10 to the 22.
Now we're talking big numbers.
So there's a 10 to the 12th difference between these two Well, let's take a quick look before we finish up.
OK.
This is the band gap as a function of position in the Periodic Table.
Now you guys ought to be able to rationalize this.
Lead is a metal, bigger atoms you can see.
10 actually comes in two flavors.
Gray and white and one of them-- I think it's white-- actually forms covalent bonds and they use tin for night vision.
Think about where that band gap is.
But here's the punch line.
We're trying to rationalize this large difference in conductivity.
Well, there's copper up there at 10 to the 7th, there silicon at 10 to the minus 4 and ten to the minus 4, 10 to the minus-- about 10 to the 12th or thereabouts-- and so, lo and behold, this-- we were able to rationalize with these fairly simple models, the range of conductivities that go all the way from metals to semiconductors and even diamond.
Look at diamond-- 10 to the minus 11.
You can rationalize that the band gap-- where's the band gap?
5.4 electron volts.
OK, we're out of time.
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OK, so this takes us into the paper by Heitler and London, where, by using the concept of linear combination of atomic orbitals into molecular orbitals for a large number of atoms, instead of splitting one plus one atoms, making a bond into two energy levels of bonding and AN antibonding, we take a large number of atoms, bring them together, and we get a large number of bonding orbitals and a large number of antibonding orbitals.
And they have to lie within a zone of about 5 to 10 electron volts, which means if you put Avogadro's number of these things together, you're going to end up with an energy difference between successive energy levels on the order of about 10 to the minus 20 electron volts, or 10 to the minus 40 joules.
And for all intents and purposes, that is a continuum, and hence, the term band.
And now we start populating this with electrons, and each one of these is a separate state.
And electrons go in two by two, just like Noah's Ark, all the way up-- either halfway up or all the way up.
But now the energy difference is very, very tiny.
And a second instance, we looked at insulators where-- for example, in diamond-- where the bonds are really strong.
So that means if the atomic orbital is roughly at the middle here, the depression to make the bonding orbital is very, very great, and very negative.
And therefore, the elevation to the antibonding orbital is great and relatively positive.
And when we start putting Avogadro's number of these things together, we still end up with the 10 to the minus 40 joules.
But when we get to the top of the bonding orbitals, we have still got 3 or 5 or 5 electron volts to get to the bottom of the antibonding orbitals, hence the formation of the band gap with an energy of-- as I mentioned.
So up here, we have the conduction band where electrons are free to move about.
And this is derived from the antibonding orbitals.
And down here, we have the valence band, which consists of the bonding orbitals derived from sigma and pi.
And then the semiconductor is sort of like an insulator-- as they say in California, it's like an insulator-- but the band gap isn't quite so severe.
It's down on the order of 1 to 3 electron volts, and that gives it unique properties.
Why?
What's magical about 1 to 3 electron volts?
Number one, visible light.
Visible light is 2 to 3 electron volts.
So visible light shines right through an insulator, because visible light has an energy of about 2 to 3 electron volts.
If that's the energy of 2 to 3 electron volts, and this is 6, it zooms right through.
But if it's 2 electron volts and this is 1 electron volt, we're going to get thermal excitation.
I know Professor Ballinger talked about thermal excitation the last day and what happens, and it's very reminiscent of excitation in the case of the Bohr model.
Photoexcitation across the band gap.
Incident photon with energy greater than the energy of the band gap.
And it's the same drill as you've seen before.
That's why those early lessons on-- photon comes in or electron comes in, strikes the Bohr atom, electron rises up, momentarily sits there, falls back down, photon emitted.
It's the same set of ideas.
Only instead of moving from n equals 1 to n equals 2 in the Bohr atom, you're moving from the valence band to the conduction band, sitting in the conduction band momentarily, you fall back down, and you get a photoemission.
So that was covered last day.
And I know towards the end of the lesson, he started talking about thermal excitation.
So let's look at that in a little more detail.
Thermal excitation.
So thermal excitation of what?
Thermal excitation of electrons across the band gap.
So this is matter energy interaction, only in this case, it's thermally inspired.
So let's draw our cartoon again.
Up here, we have the conduction band, and below we have the valence band.
So in the conduction band we have free electrons.
In the valence band is where all the bonds live.
And we've got an energy gap here of some value.
Could be semiconductor, could be insulator.
In this case, we're going to use visible light.
And we've gone through the analysis with incident photon and the emission.
In this case, I want to look at thermal excitation.
So with thermal excitation, I start with an electron down here in a bond.
This electron lives in a bond.
And with thermal excitation, the electron jumps out of the bond and up into the conduction band.
So I'm going to designate the electron in the conduction band with an e minus and a circle around it.
And it leaves behind a broken bond.
As you know, every bond has two electrons in it.
Well, one of the electrons left the bond, leaving a hole.
Little h with a plus sign.
This is a broken bond, and a broken bond is highly unstable and very mobile.
If I put a potential across here.
If I put this material-- and I'm doing a mixed metaphor-- this is an energy level diagram.
But now I'm going to put a couple of plates across an energy level diagram.
But you're smart.
You can understand this.
So this is not energy.
Now I'm going to make the right electrode negative and I'm going to make the left electrode positive.
What would happen?
Well, the electrons in the conduction band are going to be attracted to the positive electrode or repelled by the negative electrode, but the complementary situation occurs in the valence band.
This broken bond is like a hot potato, and it has an effective charge.
This is a zero in a land of minus one.
A broken bond is a zero in a land of minus one.
So therefore, if I make the right side negative, this hole is going to be drawn.
So I have two forms of electronic conduction.
I have electrons moving to the left in the conduction band and I have holes moving to the right in the valence band.
So for every thermal excitation I get two carriers.
Every excitation of electron from valence band to conduction band generates two carriers.
Carriers of what?
Water?
Carriers of electrical charge.
Two carriers of electrical charge.
And what are they?
The electron in the conduction band and the hole-- little h with a plus sign-- in the valence band.
So that's how thermal excitation works.
And you can see that the numbers have to be equal.
So we write, therefore, that the populations are equal.
The number of electrons in the conduction band equals the number of holes in the valence band.
Now electrical engineers have a similar expression, but they don't care about numbers and electrons and holes, they care about the functionality.
And the electron is a carrier of negative charge and the hole is a carrier of positive charge.
So the electrical engineering books will write, thermal excitation gives rise to pair generation.
Pair of what?
Pair of charge carriers.
And the population of negative charge carriers equals the population of positive charge carriers, which in chemistry is the number of electrons equals the number of holes, or the number of broken bonds.
So that's the way it works, but there's a conundrum here.
Thermal excitation is based upon the thermal energy, and we know that the average thermal energy in an environment is roughly given-- I mean there are fancier calculations-- but to a first order, I can approximate it as the product of the Boltzmann constant and absolute temperature.
So at room temperature-- room temperature, I like to say, about 300 Kelvin.
300 Kelvin is 27 Celsius.
That's kind of a warm room, but 300 is a nice number.
All right?
So if you take 300 Kelvin, you get on the order of about 1/40 of an electron volt.
1/40 of an electron volt, which is a lot less than one electron volt, which is roughly the band gap of a semiconductor.
So how is it that we get any thermal excitation of data indicate that there is some thermal excitation even at room temperature?
So the question is, if the average thermal energy is a fortieth of electron volt according to this-- and this is correct-- and the band gap is one eV, then it should be no thermal excitation.
But we know what happens.
So, one eV, which I'm going to say is Eg.
All right?
So how do we get any thermal excitation under these conditions?
So there's an answer to this.
And for this we go to merry old England and James Clerk Maxwell, that you know or will come to know in 8.02 as the annunciator of Maxwell's equations of electricity and magnetism.
He also thought about physical chemistry, and in particular he thought about the physical chemistry of gases.
And what did he teach us about gases?
He said, if you look here in this room, you see nitrogen and oxygen and all the other constituents of air.
But to a first approximation, we have four parts nitrogen, one part oxygen.
The room is roughly 300 Kelvin.
What Maxwell taught us is that velocities of the gas species zooming around in this room are not all equal.
Some of the gas species are moving around as though they're at 500 Kelvin.
And some of them are moving around very slowly as though they're at 100 Kelvin.
On average, it's 300 Kelvin, because we can feel it as 300 Kelvin.
But some of them are zooming around.
Some are really lazy and they're just moving along really slowly.
And furthermore, he was quantitative about it.
He said, I'm going to tell you what the distribution of energies is.
So let's get that on the board for starters.
That in any ensemble of gas species, could be gas atoms or it could be gas molecules or gas compounds, in any ensemble of gas species, that all molecules-- because you can always say the atom is a degenerate form-- all molecules do not have the same velocity.
And furthermore, he gave us the distribution.
He calculated the distribution of velocities.
So that was about 1859.
Forgive me here.
Velocities.
OK. And then Boltzmann came along in 1872.
And Boltzmann generalized this idea.
He made the link between gas velocity and temperature.
So, he took the idea and he said, if the gas molecules move at different velocities, that means they have different energies.
And why do we call this the Boltzmann constant?
Because Boltzmann made the link between particle energy and temperature.
So he said there's a variation of instant temperature.
All right?
So Boltzmann generalized Maxwell's idea to energy and then ultimately temperature.
And so I'm going to show you, instead of the Maxwell distribution, I'm going to give you the Maxwell-Boltzmann distribution.
But before I do, I want to make sure you understand the math and appreciate what I'm going to plot for you.
So we just had a test, right?
In fact, we can go back to the slide.
All right.
So what do we have here?
I could plot on this axis, the abscissa, the grade-- g-- and on the ordinate, I'll plot the number of people who have any given grade-- n of g. Now the class av was 66, all right?
This is going to go from 0 to 100.
Now what we see here is-- and by the way, I'm going to normalize it.
Instead of dealing with 463 or whatever, I'm going to divide by the total class.
So the number is going to go from 0 to 1.
So this a normalized distribution.
What I'm teaching you applies all through science.
As I told you, this is the most important class.
Not because I'm teaching, but because the subject matter is general, just as Boltzmann did.
So let's look at the normalized distribution.
Now the class av was 66.
Now we didn't have everybody getting 66.
That was not the distribution.
That's one solution to the problem, isn't it?
Instead, what we got was we got that.
Now one possibility would be this one, the bell-shaped curve.
Goes like this.
That's bell-shaped.
This is Gaussian.
This is a Gaussian distribution, and its functionality is y equals e to the minus x minus x average, quantity squared.
Now, to have a normal distribution, you have to have a class of normal people, so that didn't happen on this test.
That's a joke.
God, you're so serious.
Have you no fun in your lives?
Have you ever laughed?
[LAUGHTER]
Overachievers, you.
OK.
So we're not going to have a Gaussian distribution.
We're going to have that one there.
It's kind of a skewed Gaussian with a long tail.
All right, so that's the one thing.
But what Boltzmann gave us, instead of the number of atoms of a given grade, he plotted it as a function of energy.
So instead, the Maxwell-Boltzmann distribution has energy as the ordinate instead of the grade, and up here, the number of atoms-- or more importantly-- the atom fraction.
The fraction of atoms in the distribution that have that energy.
And it doesn't look like a normal distribution.
It looks sort of off-normal.
Skewed, like this.
It rises to a peak and it has a long tail that moves off to the right.
So this is the Maxwell-Boltzmann distribution.
It's not y equals x minus x bar squared.
So this is Maxwell-Boltzmann distribution of energies.
And furthermore, what we see here is that the area under this line has to equal 1.
If you integrate the fraction from 0 to infinity, because this asymptotically goes out to infinity, the integral under this line is 1.
OK. Area equals 1, which is the integral of ndE.
Now, because of the asymmetry here, clearly this is the average energy.
The average energy isn't the maximum energy.
Whereas in the Gaussian-- let's keep the Gaussian over here, just for grins and chuckles-- so the Gaussian looks like this.
It's symmetric.
And this is the n of g max and also a g average, whereas here, the average is a little bit to the right.
A little bit higher.
OK.
So now, if this is the distribution, and this at room temperature is 1/40 of an electron volt, and way, way up here is the band gap energy, there's a tiny but non-zero fraction of this distribution that has energy in excess of the band gap energy.
And so this tiny fraction of the total distribution has the thermal energy to allow this excitation and pair generation to occur.
And furthermore, this is a function of temperature.
This is a function of temperature.
So E average is related to temperature.
So, I'll say E average 1 at T1.
And then what I'm going to do is I'm going to heat this to a higher temperature, so I'll use red chalk to indicate it's hotter.
And what happens to the distribution under an increase in temperature?
Well, the average had better move to the right, and that's what happens.
But it doesn't just move to the right, it skews a little bit and broadens.
So in point of fact, at a higher temperature, it looks like this.
OK.
So this is now E average 2 at T2 where T2 is greater than T1.
T2 is greater than T1.
OK, so that's the Maxwell-Boltzmann distribution.
But here's where the interesting stuff occurs.
So I'm going to blow up this high energy end of the distribution.
So again, this is going to be n of E normalized, and this is E. And what we're going to find here is that at T1-- this is the curve at T1, and the area under the line is shown here-- and then-- not to scale-- we're going to put a break in this.
And I'll show you how much of a break is needed.
And then, at T2 I'm going to have this come down.
This is T2.
And the area under the line T2 captures the T1 plus all of this.
And so what we see, is that for a modest increase in temperature, this maximum shifts modestly, but the area under the line in excess of this critical energy increases radically.
And that's the powerful leveraging factor of increase in temperature.
And I've done some calculations just to show you what this does.
The functionality of the Maxwell-Boltzmann distribution is e to the minus some critical energy divided by the product of Boltzmann constant temperature.
So we can put in a value here-- eg-- what have you.
All right.
So I chose for silicon-- for silicon, I know the band gap energy is 1.1 eV-- and I have it at room temperature-- when I go through the calculation-- the fraction underneath that line is 10 to the minus 19.
One part in 10 to the minus 19.
This is the area.
Area under the curve is greater than the band gap energy of 1.1 eV.
So then I said, let's go up to the melting point.
I'm not going to melt the silicon, I'm just going to bring it up to its melting point but not melt it.
So it's like ice cubes at 0 Celsius.
It's still solid.
OK?
The melting point of silicon, you can look in your Periodic Table, it's over 1400 degrees C. And this is 10 to the minus 4.
10 to the minus 4.
Area under curve: da da da da.
So that means that the red area-- A2, the red.
Actually, let's use red chalk so that everybody is in tune.
A2 to A1 is equal to 10 to the 15, whereas T2 over T1-- well, red chalk.
It's my chance to use red chalk and I'm not going to squander it.
T2 divided by T1-- it's about 1700 Kelvin divided by 300 Kelvin-- is about 6.
So for a change in temperature by a factor of 6, I get a change in population of 10 to the 15.
So you can see how modest increases in temperature have a huge impact on thermal excitation.
So now you say, if we wanted to make a device, we can't heat our computers up to 1600 degrees, 1500 degrees C in order to get an adequate number of charge carriers.
There has to be another way to get charge carriers.
I'm going to show you a third way that's called chemical promotion.
So now I want to talk about charge carrier generation.
Charge carrier generation by change of chemistry, and it's going to be specifically by the introduction of impurities.
Charge carrier generation by introduction of impurities.
You might say, gee, aren't impurities bad for a system?
Well, I want to tell you, there are two kinds of impurities in this world.
There are bad impurities which we call contaminants, and that's what you're commonly taught to think about.
That's bad.
But there are good impurities, and these are called dopants.
Dopants are impurities that we are happy to have in our system.
And so by the introduction of impurities of value, we can lead to engineered materials.
So I'm giving you the introduction to high technology, where we're going to control the engineering properties of a material through control of chemistry.
So that's chemical composition tailored-- for specific-- for targeted values of properties.
And that's the essence of material science.
We control the chemistry.
We control the properties by controlling the chemistry.
So what I want to do is give a vivid example of this, and it's not going to be an isolated example.
It's going to be the example that pertains to all of the microelectronic devices in our world.
So that's silicon.
So let's look at how doping works in silicon.
It also applies to germanium, but we don't use much germanium in devices.
It's dominantly silicon.
So what we're going to do is dope.
The gambit here is to dope-- that's introduce an impurity-- with aliovalent impurity.
What do I mean by aliovalent?
You know the word alias?
It's another name.
Alio in Latin means other.
So we're going to dope silicon with something that has a different valence from that of silicon.
Can't before.
So let's look at the simplest example.
And I'm going to put phosphorus into silicon.
I'm going to dope phosphorus into silicon, and phosphorus is Group 5, or more properly, IUPAC calls it Group 15.
So we're going to put Group 15 element into silicon, which is Group 14 element.
So we call phosphorus a supervalent dopant, meaning it has a valence higher than that of the silicon.
So it's a supervalent dopant.
Supervalent dopant.
So now let's look at how that supervalent dopant works.
I'm going to show you some silicon here, and I'm going to give you the silicon.
This is sp3 hybridized.
it's sitting here as a single crystal.
Here are three silicons.
So it's a single crystal.
All these silicons on the bottom row line up.
All the silicons in the center line up.
And there are silicons at the top and they all line up.
So that if I look at it on the edge, it's a perfectly, atomically smooth system.
And I have one of these.
I know Professor Ballinger made reference to it last day.
This is last-generation technology.
It's only eight-inch diameter.
But this is an eight-inch diameter silicon wafer.
So this started as a giant salami, eight inches in diameter-- probably two meters long.
Single crystal.
Every atom here is adjacent to the next atom according to the rules of atomic arrangement.
And when this thing is cut and polished, it's flat to one atom.
It is atomically flat.
And it makes a fantastic shaving mirror.
Actually Dave, can you cut to this?
Yeah, here we go.
Here's the penny.
And this is silicon here.
It's going to drive this thing nuts because it can't focus on this thing, because it's so-- there we are.
So this was a part of a giant salami, and now it's been cut.
And obviously you can't cut it one-atom thick.
It's some submillimeter thickness, because you need to have some mechanical strength to it.
But the surface is absolutely flat to one atom.
All right.
And that's what we're going to dope into.
We're going to dope into this thing, and I'll show you how we're going to dope it later.
Actually, why don't we cut to that.
Here's something that's already been doped and made into devices.
So there's the plain old silicon, which you're having a hard time seeing anything except a little bit of reflection.
And now here's the thing that's been processed, so let's zoom.
Zoom.
OK.
So now you can see all the features.
So these features involved gas phase reaction to introduce dopant atoms into the silicon lattice.
And then we're going to chop these little things up and put them in your cell phones and in your laptops and whatever other devices are out there.
OK.
So this is what we're talking about, and if we live long enough, we'll see one of these.
This is a Pentium chip.
OK.
So where's the chip?
The chip is underneath this piece of gold.
Where did I put my pointer?
Ah, it's over here.
All right.
So the actual silicon device is underneath this.
And all of these are this spider's web of leads.
It's trying to access all the tiny devices.
It doesn't do you any good to have a device density that's greater than your access density.
So the technology that goes into this is phenomenal.
Plus, these things all generate heat.
When you put all these tens of thousands of devices into something the size of your fingernail, you've got a toaster oven working here, and you have to get that heat out.
And this is several generations old.
What you have in your machines today is even denser than this.
And supercomputers?
Forget it.
You have to have those things so aggressively cooled or else they'll heat up and it'll be just like a light bulb, just bursting.
So that's what's going on inside.
So let's see how that happens.
Let's cut back to-- well, you can leave that up.
It's a pretty image and will probably soothe the nerves of the students.
So now I'm going to put phosphorus in here.
I'll put phosphorus in here, but where does phosphorus go?
Phosphorus goes onto a silicon site.
So it actually comes in and sits here.
Phosphorus comes and sits at a silicon site.
Now what do we know about the number of bonds that phosphorus likes to form?
It has five valence electrons.
So it forms one, two, three, four bonds, and it has a fifth electron.
And that fifth electron is sitting there somewhere.
It's sitting there somewhere in the lattice.
All right.
It's sitting somewhere in the lattice because it has no place to go.
Now in energy space, I know where it goes.
So here's the conduction band of silicon, and this is the valence band.
And this is all of the silicon host crystal, because the dopant is in a tiny, tiny amount.
We're talking parts per million.
So for all intents and purposes, it's still a silicon crystal.
It has tiny amounts of phosphorus in it.
So this is the band gap of silicon, where this is about 1.1 electron volts.
And where does this fifth electron go?
This fifth electron sits up in here in the conduction band, doesn't it?
And every time I introduce a phosphorus, I have a fifth electron that goes in the conduction band.
But I don't have to generate a hole.
I don't need to break bonds in order to put electrons in the conduction band.
Because the fifth electron is like the fifth wheel.
You know?
It goes in the conduction band.
So there's no generation of holes here.
All right.
And furthermore, I can control the conductivity.
At the end of last lecture, Professor Ballinger showed you that the conductivity of a substance is related to the number of charge carriers.
So if I want to double the number of phosphoruses, I will double the number of electrons in the conduction band, independent of temperature.
All right.
So for every phosphorus dopant atom, we get one electron in the conduction band.
We get one electron in the conduction band and no holes.
No holes in the valence band.
It's direct.
It's direct.
All right.
Now, electrical engineers would look at this and they'd say, I don't care if it's phosphorus.
They don't care about the chemistry, they care about the functionality.
What kind of carriers am I throwing into this crystal?
For an electrical engineer, there are only two kinds of carriers.
Positive carriers and negative carriers.
For every phosphorus, what's the outcome?
A negative carrier.
So this is now doped silicon.
It generated negative carriers, so this is called n-type.
This is now n-type silicon.
And its properties are governed by the population of these-- because I'm going to show you in a second that the number of these far exceeds the number that are thermally generated-- so the properties here are called extrinsic.
So now we have a crystal that is exhibiting it's extrinsic behavior, thanks to the doping with the supervalent impurity.
One more thing.
One more thing.
Something very, very cool here.
You see this electron?
It's sitting here somewhere in the-- now this is Cartesian map.
I'm looking through a scanning electron microscope.
I see all the silicons.
I get the phosphorus.
And there's an electron in here somewhere.
And this has 15, right?
This has 15 protons.
This has 14 protons.
So in a land of 14 plus, 15 plus is relatively positive, isn't it?
The phosphorus is locally positive in comparison with the rest of the silicon lattice.
And it's pinned, because it's covalently bonded.
And there's an electron sitting here with no place to go.
Do you know of a model that describes the behavior of a one-electron system around a pinned positive nucleus?
The Bohr model
Bohr model?
Just for grins and chuckles, what if we were to use the Bohr model to describe the behavior of that electron?
You can say it's crazy.
That doesn't make any sense, because that's not a one-electron system.
But what if we modeled it as though it were?
I'm going to-- just for grins and chuckles-- get a sense of what that ground state value is.
And recall that in hydrogen, the ground state-- E1, ground state of the electron-- is equal to minus k. Which is minus me to the fourth over 8 epsilon naught squared times Planck constant squared.
Which is minus 13.6 electron volts So what is the energy of that electron there if we use the Bohr model?
Well, we have to modify it a little bit, because the electron isn't moving in a vacuum.
Depending on how far this electron is from the center, it could be moving around, and there are all kinds of silicon atoms and bonds and whatnot.
So there are all kinds of stuff between the electron and the center here.
So I have to use a modified form.
I don't use the permittivity of vacuum.
I map the permittivity of vacuum into the dielectric constant.
This is the dielectric constant of silicon.
Remember there's so little phosphorus there that it doesn't alter the properties.
And then for reasons I can't go in to here, I have to change the mass of the electron into the effective mass-- we saw this when Bohr did the calculation about the lines of helium plus-- effective mass of electron in the conduction band.
So if you put both of those in-- and I've got numbers for this-- this value is 11.7 and the second one is about 1/5 of the rest mass of the electron.
So if you put all of that in, you end up with the energy of this ground state electron equal to minus k times 0.2 divided by 11.7 squared, which equals minus 0.02 electron volts.
0.02 electron volts.
So this is 2/100 of an electron volt below something.
What's it below?
It's below the conduction band.
It's right here.
This is the ground state of the electron from phosphorus.
It's the ground state of the electron from phosphorus.
E1 of electron from phosphorus.
And phosphorus, the electrical engineers call a donor.
A donor of what?
It's a donor of electrons.
So this thing here is called the donor level.
And that's the ground state.
So the electron should be sitting.
Isn't the ground state here lower than the bottom of the conduction band?
Sure it is.
So the electron really should be down in here.
There's the electron.
Now you're saying, how are you going to get conductivity?
Well, we just figured out that the energy level of the ground state of the donor-- so I'm going to call this the energy of the donor level, ground state-- is 0.02.
How does that compare with thermal energy at room temperature?
It's comparable.
So at room temperature with 1/40 of an electron volt, there's enough energy to promote this electron up into the conduction band by thermal excitation from the donor level.
Thermal excitation from the donor level.
And in fact, if we want to do something really, really-- I want to use an adverb here-- really, really cool.
If that's the ground state-- I'm going to blow this up.
So this is the bottom of the conduction band.
And now this is the ground state.
This is the donor level.
This is the donor level, which I was calling E1.
Is it just the donor level here and the conduction band, or can you see that there's an E2 and an E3 and an E4?
And I have a whole set of quantum states from the donor level right up to the conduction band.
And if I have enough energy-- if E thermal is on the order of E donor-- I get promotion of these up in here.
Now what would we say if this were the Bohr model?
You'd have an electron loose in a ground state up in here where it's free to move.
What's that process called?
Ionization.
And you know what the electrical engineers call these electrons?
They call them ionized electrons.
But they're referring to the ionization of the electron out of the donor level into the conduction band, not out of the silicon crystal.
So these are ionized electrons out of donor level.
So we've come full circle.
By the way, this is a calculation.
You know what the measured value is?
It was measured as minus 0.045 electron volts.
You might look at that and say, whoa, that's about two times this.
Hey, wait a minute.
We started at 13.6.
We went from 13.6 to this, and the real value is this, and so we are in good shape.
Now, what happens if I put another phosphorus in here?
Another phosphorus, it's also going to sit at this-- I'm going to do it over here.
If I put another phosphorus, it also has the same donor level.
And another phosphorus, it has the same donor level.
How can I justify this?
Am I not violating the Pauli exclusion principle?
How come all the phosphorus donor states are at the same level?
Or put another way, under what circumstances is this diagram accurate?
If the number of phosphorus atoms that I introduce is tiny-- parts per million-- the separation, the physical separation of this phosphorus from its nearest phosphorus neighbor is so great, that for all intents and purposes, they are at infinite separation.
And therefore, all of the phosphorus donor atoms establish a donor level at the same value.
And if you dope to very, very high levels, you see this break down.
If you go to very, very high concentrations of phosphorus dopant, you will discover what happens if these two energy levels get close enough together in real space.
They have to split.
And so you'll see all of that happening down here at the miniature level.
The last thing I wanted to do was to push the Bohr model just a little bit harder.
And so I've given you an energy, and the energy seems to make sense.
Even though you might say, this is the crazy.
It's not a one-electron system, but it gave us a reasonable number.
Let's just for interest's sake determine: what's this radius?
I wonder how far this electron is away from the donor phosphorus?
So I plug into the r1.
That's the radius of the ground state electron.
And that's the Bohr radius, right?
0.529 angstroms in elemental hydrogen, but now this has to be modified by adding the dielectric constant and the effective mass.
And the number is 30 angstroms, or 3 nanometers, if you must.
All right?
And compare that to silicon-silicon bond distance.
The silicon-silicon bond distance is 2.35 angstroms.
So clearly this electron is very far removed from the central phosphorus.
It's not as I depicted.
It has to be way over here, going way, way around.
This is really, really something.
So now you understand how you can get extrinsic properties and the behavior of the system governed by the introduction of impurities.
What I'm going to do next day, is to show you that at typical doping levels, when you dope, dope at about 10 to the minus 6-- 1 part per million-- 10 to the minus 6 phosphorus per silicon.
It turns out that the number of electrons from this operation is much greater than the number of electrons from thermal excitation, which is tiny.
And therefore, we argue that the properties are governed by the dopant atom, and so we have extrinsic behavior that we see.
OK, well that's a day's work.
David, may we go to the slides, please?
OK, so you saw all that last day, and this, this, this.
All right, there's the insulator, and that.
So I'm just doing the lecture again and fast.
You can read.
You can parse visual images.
See?
There's the n-type semiconductor.
We're going to look at p-type next day, quickly.
There's a silicon crystal.
There's the salamis-- there's the little wafers that are cut.
And there's the spectrum, the x-ray spectrum indicating you have a single crystal.
There's the first transistor-- this was a single crystal of germanium lying here on its side-- the point transistor in the fall of 1947.
Three gentleman at the Bell Labs that demonstrated the rectification.
And they won the Nobel Prize for this in 1956.
Now, you can also make semiconductors that have compounds.
So these are compound semiconductors-- Professor Ballinger showed you last day-- tin, germanium, silicon, carbon.
And look at all of these.
This is a Group 3 element with a Group 5 element.
5 plus 3 is 8.
Group 3 and 5.
This is Group 2 and 6.
2 plus 6 is 8.
It's always about octet stability.
But they make strong covalent bonds that give band gaps all over the map.
So suppose I wanted to have something like a stoplight.
Suppose I wanted a stoplight, and I wanted to make a red light.
Well, I need-- I can go backwards from 660 nanometers, and I can go e equals hc over lambda and calculate-- I need a material with 1.97 electron volts band gap, and I look on here and there's nothing that's exactly 1.97.
Well, I can mix these things.
I can mix them.
I can mix something that has 1.52 with 2.3 and get 1.97.
That's called band gap engineering.
And there's more than one mix here, so why would I choose one mix over another?
It's cheaper, or it's easier to process, or it's stable in the atmosphere.
You don't want a semiconductor that can't stand rain and snow.
And so I'm showing you here a variety of compound semiconductors, and this is the basis of things like-- the LED, the CD reader, the DVD reader is based on light that's generated from these.
What was the big deal about Blu-ray?
Why is Blu-ray so cool?
Because the original reader was red.
And what's the wavelength of red light?
Red light is around 600 nanometers.
And what's the wavelength of blue light?
It's down around 300 nanometers.
So all other things being equal, my stylus is half the size, which means for a given area, I can get double the density of information without changing the size of the disc.
But it wasn't trivial to find a blue laser.
And the person who found the right mix of compounds that could give blue in an efficient manner became a very wealthy individual.
So this is our stoplight now.
These are the materials that go into the stoplights.
No more do you have that giant lens with a 150-watt incandescent bulb.
Now you have the array of these LEDs that are designed to be in this band gap.
So I think at this point, we've probably run out of time.
So we'll resume the discussion tomorrow.
Same time, same place.
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OK, OK, settle down.
It's time to start learning.
One announcement.
Tomorrow we'll have the weekly quiz.
In lieu of the Tuesday celebration, we're going to move to Thursday, since the holiday moved everything, compressed the week.
So last day, we were talking about doping of semiconductors, and I want to finish up that unit.
Just to refresh your memory, we looked at how we could change the behavior of the semiconductor by introducing impurity atoms.
And when the behavior of the semiconductor is dominated by the impurity atoms, we term that behavior extrinsic.
In other words, it's not the behavior of silicon itself, but it's the behavior of silicon as determined by the presence of some dopant atom.
And we looked at the special case of doping with a supervalent impurity.
And here's the energy level diagram for that situation, where we've got the valence band down here, we've got conduction band up here, separated by an energy gap.
That's all characteristic of plain vanilla silicon.
But then when we dope with the supervalent impurity, there is an extra electron.
And that is put into a donor level that lies just a tiny bit below the bottom of the conduction band.
And then, thanks to thermal excitation, which gives us an average energy of about 1/40 of an electron bolt, we get, for all intents and purposes, 100% excitation of the electronics that sit in the donor level, excitation up into the conduction band.
And this is what electrical engineers term ionization.
And I also reminded you last day that this is the donor level, and for each donor atom, there sits a donor level at the same value of energy, but these are so far apart point, owing to the dilution, the dilute concentration of the impurity, that we don't violate the Pauli Exclusion Principle.
So these are all sitting at the same level and that's why they've given temperature promotes all of them up into the conduction band, where they are mobile.
And then, just to complete the review, we call this n-type because, thanks to doping, we generate electrons in the conduction band, and electrons are negative.
Supervalent doping gives us n-type.
Then the total number of electrons in the conduction band is the sum of those that would have been generated by plain old thermal excitation.
Thermal excitation operates all the time.
So with thermal excitation, recall, we promote from the valence band up into the conduction band.
So we break one of these bonds, shoot an electron up here, and leave a hole behind.
So this generates the pair.
Hole in the valence band, and electron in the conduction band.
That's always operative, but owing to the energies involved, the band gap is on the order of one electron volt, whereas the donor level sits only about 1/50 of an electron volt below the conduction band.
So we get very, very little promotion at room temperature.
And in fact, it's something like 10 to the minus 19 is the fraction of promotion, and if you say roughly 10 to the twenty-third per cubic centimeter, that means you've got about 10 to the fourth electrons per cubic centimeter due to this thermal excitation from the valence band to the conduction band.
Normally, when you dope a semiconductor, you dope it around parts per million level.
And parts per million, that's 10 to the minus 6.
If you take 10 to the minus 6 times 10 to the 23, you get about 10 of the 17.
Now you can see, 10 of the 17 is vastly larger than 10 of the fourth.
So for all intents and purposes, once you've doped something, this contribution is essentially 0, and so the electron population, the conduction band, is that dictated by the impurity atoms, which is why we say that this is exhibiting extrinsic behavior.
The intrinsic properties are not visible, they're not palpable.
They're immeasurable.
So intrinsic is substantially less than extrinsic.
Now, the last thing I wanted to do for you before we say goodbye to this fascinating topic, which lays the groundwork for everything that we know in the modern electronic era, parenthetically, is to look at the other type of doping.
I've showed you how to make an n-type semiconductor.
How do we make a p-type semiconductor?
So in that case, what we can do is dope with a subvalent impurity.
What do I mean by subvalent?
Well, the valency of silicon is 4, so I want something less than 4.
So a good example is boron into silicon.
So boron is group 13, or 3, all right, if you look on your periodic table.
And silicon is group 14, according to the IUPAC notation.
So let's go back and take a look at how that might appear.
So I'm going to draw the silicon crystal.
Remember, this is going into a single crystal of silicon.
So silicon is sp 3 hybridized, and we have silicons everywhere in the structure.
So we're going to put silicons.
And I'm trying to show sp 3 hybridization.
So this is 3 legs each in the plane, then you've got 109 degrees here.
So that's silicon, and so on.
I get the three of them just to complete enough of the picture.
And now what we're going to do, is introduced boron.
And boron goes into the silicon lattice and covalently bonds.
It doesn't go sit in some void space in between the silicons.
It actually sits on a silicon site and substitutes for the silicon.
Now, boron is group 13.
It's got 3 valence electrons.
And so it forms bonds with 3 silicons.
And now there's a fourth silicon here, and that silicon has an electron, but the boron doesn't have a fourth electron.
Here's where it gets interesting.
The driving force to complete the picture here is so great that the system will actually pull an electron out of a silicon-silicon bond-- and so I'm going to use a different color chalk, and I'm going to indicate, just for argument's sake, that we're going to pull an electron out of this bond here, break this bond, and shoot that electron over to here.
That way, we get r months around the boron.
It says p3 hybridized.
It just doesn't have that extra electron.
But now it rips the electron out of this silicon-silicon bond.
And what's the consequence of breaking this bod?
What's the electrical feature that we've created here?
A hole.
So now we've created a hole somewhere else in the crystal in order to satisfy the desire of boron to get that fourth bond.
And now can you see that for every boron that I introduced into the crystal, I'm going to make a hole somewhere in the crystal.
And on the energy level diagram, where do those holes live?
Those holes live in the valence band.
So let's go make the energy level diagram.
So up here we have the conduction band, and downstairs, we have the valence band, as before.
And I'm going to assume that we got the thermal promotion.
But it's so tiny in comparison to the amount of boron we're going to put in.
I'm not going to muddy the water here and show that you've got pair formation.
Because the extent of it is so small, it doesn't make any difference.
So now what happens?
Now, I know I'm going to generate holes here, and I've got to show this energy level.
And this bonding level is different from this bonding level.
Agree?
Silicon-silicon bond has different bond energy from boron-silicon bond.
So that means the top of the valence band is different from-- because this valence band is silicon-silicon-silicon.
So it turns out that, and this is not to scale, that the energy level of this bond is just a little bit higher.
And this is an energy level that involves the accepting of electrons, whereas in the other case, we were donating electrons.
Hence, that's called the donor level.
This is called the acceptor level.
And it's about 1/50 of an electron volt above the top of the valence band.
And so what happens is that in order to make this bond, we move something out of the valence band up into the acceptor level and generate a hole.
And so for every boron that we put in, we have another broken bond.
And all of these lie at the same level, but the dilution is so high, that they're so far apart, that they don't violate the polyexplosion principle.
They're thermally promoted up, and away we go.
So now, under these circumstances, the number of negative charge carriers no longer equals the number of positive charge carriers.
Because I've got all these positive charge carriers.
Thanks to the introduction of boron, p is greater than n. So we've made a p-type semiconductor, by the introduction of a subvalent impurity.
And we can redo the entire central part of last day's lesson, which was the Bohr model.
Now, this is the sexiest part of the Bohr model I've ever seen.
This is really, really cool.
The hole is mobile!
The hole is mobile!
And the boron is a little bit shy of protons, right?
It's 3 in a land of 4.
So it's negative.
So I've got a stationary negative center, and I've got something positive revolving around a negative center.
This is an inverse Bohr model!
Because the Bohr model has a honking big positive center with a dinky little electron revolving around.
We've got the immobile negative center, and the mobile hole revolving around.
Go through, you get the same set of quantum-- I mean, there's a whole bunch of quantum levels in here.
Fantastic!
How far that Bohr model can take us.
So now you know how to make a p-type semiconductor, you know how to make an n-type semiconductor.
And then what you can do is put them together.
And you can put, say, boron-doped silicon opposite phosphorus-doped silicon.
Be careful here.
The fonts are really critical.
This is uppercase P, for phosphorus.
This is lowercase p, for positive doping.
So don't go, oh, P, phosphorus, that means it's p-type.
No, P gives you extra electrons.
It's n-type.
So that's what uppercase P. This is a lowercase p. This is p-type.
So now I've got p-type, n-type, and so what do I have here?
I've got a p-n junction.
And this is the beginning of solid state devices, rectification of A/C current, diodes, you name it.
And it all starts with this, the chemistry.
This is the birth of the transistor, and all the modern electronics that we have based on that.
Later on, we'll show you how you get the boron in, and how you get the phosphorus in.
You don't just come in and sprinkle it on top, and hope that it goes to the lattice site.
There's a lot of processing involved.
OK.
So this is where I want to hold for the unit on semiconductors, and how it fits into the grand scheme of things.
So in the books, you'll see-- this is a very crude drawing, because this is showing n-type semiconductor with grossly exaggerated numbers of donor electronics up here.
There's no mention of the donor level.
So this is sort of semiconductivity, sort of pre-3091.
This is extrinsic semiconduction for idiots, I guess you'd call it.
And then, same thing here.
You see all of the holes in the valence band.
We know better, because we know there's an acceptor level and there's a donor level.
OK.
So now we're going to switch gears.
Now here's the big transition, right now.
If the last two lectures didn't leave high school behind, today we leave high school behind.
So those of you who have been coasting up until now, thanks to your very good high school background, start paying attention.
Because it's high school no more.
So just to remind you where we've come.
We started over here, at the beginning of the semester.
And the big theme is going to be, electronic structure informs bonding, which informs state of aggregation.
State of aggregation is solid, liquid, or gas.
So I want to show you how we get to where we are.
So we did all of these topics, right?
Bohr, or Sommerfeld quantum numbers, multielectron.
And then we came to octet stability.
And with octet stability, we went a long way.
We got into ionic bonding.
We got into completing the valence shell.
And in fact, all of this is a consequence of the octet stability.
And that led us into the various types of primary bonding.
Ionic, covalent, metallic.
And in some instances, van der Waals.
Solid argon is van der Waals bond.
It's the only kind of bonding, so it's the primary bonding.
But in some cases, like HCl, how do you get liquid HCl?
How do you get solid methane?
So we had invoke secondary bonding for some of these covalent molecules, and that invoked dipole-dipole interaction, the London dispersion force, which is the same as the van der Walls bond, but I decided to mix things up, so that both men get a little bit of attention, none of them feel slighted.
And finally, hydrogen bonding, in those special instances.
And that allowed us to then determine whether something was a gas, a liquid, or a solid.
And when it's a solid, we're happy.
Because this is 3091, and we're interested in the solid state.
By the way, what is a solid?
Well, here's an operative definition.
That which is dimensionally stable, has a volume of its own.
It doesn't distort in a gravity field.
All right?
Liquids will distort.
Fluids, gases will distort.
Now, there's two ways of classifying solids.
One is bonding type.
That's what we did here.
But the second way of classifying solids is based on atomic arrangements.
All right?
So let's look at that.
There's only two types of atomic arrangement.
Ordered and disordered.
I love these taxonomies that split everything into a choice of 2.
That way, you just sort of go through life and decide, is it left or is it right?
Is it up or is it down?
Is it ordered or is it disordered?
So what are the characteristics or an ordered solid?
It has a regular atomic arrangement.
That means it's got a long-range order.
And I'll show you that in an ordered solid, I know not only where my next nearest neighbor is, where my tenth nearest neighbor is.
I know where my hundred and fifty second nearest neighbor is, because it's an ordered solid.
And we have a simple Anglo-Saxon word for an ordered solid.
It's called crystal.
So when I say something is a metallic crystal, I don't mean it's something that's got magical properties, and if I put it over my head, I get powerful or something.
It just means I've got atoms in an atomic regular arrangement.
Now contrasting to that is a disordered solid.
And in that case, we have a random arrangement.
But I put an asterisk here because, as they say in California, it's not totally random.
It's only random up to a point.
No long-range order.
There may be short-range order.
In other words, I know what my next nearest neighbor is, but I probably don't know where my tenth nearest neighbor is Because the coordination shell is established, but the next coordination shell, and the next coordination shell, they're not established.
And I'm going to show you, with examples, what that means So we call those kinds of disordered solids amorphous solids, as opposed to the regular solids, which are crystalline solids.
And there's a plain Anglo-Saxon word for a disordered solid.
It's glass.
Now you might think, well, gee, doesn't glass mean that it's transparent in visible light?
No, no.
Maybe up until now, but I want everybody who's ever taken 3091 from me to be disabused of the notion that glass means transparent.
How do we think about the question of transparency?
And here I mean transparency vis-a-vis visible light.
I'm not going to say, what's transparent to x-rays?
That's pedantry.
So how do we think about whether something is transparent to visible light?
We just ask ourselves, is the band gap greater than 3 electron volts?
The band gap is greater than 3 electron volts, visible light zooms right through.
It doesn't have enough energy to excite the electrons, be absorbed, and re-emit.
So that's what transparency means.
So I can have the kind of material that's used in things like eyeglasses.
It's transparent to visible light because of its high band gap.
Diamond, diamond is transparent to visible light, and you wouldn't dare insult diamond by saying that diamond is glass!
In fact, watch the old gangster movies from the 1930s, that was the derogatory term for fake diamonds.
You call them glass.
So how is it that diamond, which is crystalline, it's transparent to visible light, and yet the window glass that we have is transparent to visible light?
It has nothing to do with state of order.
Has everything to do with this.
In fact, David, if we could shoot to the document camera, I'll show you a piece of glass that's not transparent to visible light.
How do we zoom in on this thing?
OK. Auto focus.
So what I'm showing you is a piece of obsidian.
It's a naturally occurring mineral.
It's rich in silica, and it's turned dark because it's got some iron impurity, and the iron absorbs in the visible.
And this is clearly not transparent to visible light, and yet, it is a disordered solid.
This is a glassy solid.
In fact, the white speckles there are devitrified obsidian.
It's actually started to crystallize.
So the crystalline form is the part that's transparent.
If you could peel off this crystalline form, you'd actually end up with something that's transparent to visible light.
So here's an example of where the crystalline form is transparent to visible light, and the glassy form is opaque.
So glass has nothing to do with transparency.
This is how you think about the question of transparency.
OK.
So let's cut back to the slides, David, please.
So we're now going to start by looking at ordered solids.
So we're going to take next several lessons and talk about ordered solids, and then after that, we're going to take a few more lessons and talk about distorted solids.
Now first of all, what's the term crystal?
Where does it come from?
It comes from the Greek word, crystallus, which is the Greek word for ice.
That's where we get the term crystal.
So we're going to start with a history lesson.
That's how we start everything in 3091.
A history lesson.
So we go way, way back to Isaac Newton's time.
The Reformation, Charles, Oliver Cromwell, all those exciting times.
In merry old England, there was Robert Hook.
You know him from Hook's Law, in mechanics.
So he was doing military research in 1660.
He was trying to understand what was the optimal way to stack cannon balls.
So when you're on a battle front, you want to figure out how to keep your material compactly arranged.
You don't lie them on the ground.
So from that, he concluded that crystal that is a regular array must owe its shape to the packing of spherical particles.
So that's what he was thinking, or trying to imagine.
Remember, Democritus said that we have these indivisible particles, but no one had ever seen them before.
Then in 1669, a Dane by the name of Steenson who was working in Italy was looking at quartz crystals.
And he was studying various quartz crystals.
By that I mean, quartz crystals of different sizes.
And what he observed was, didn't matter what the size of the crystal was.
They always had the same angles between corresponding faces.
So if I give you a big crystal and a smaller crystal of the same material, and a smaller crystal, and they all have the same shape, Can you imagine that you might conclude that that is reflective of something down at the atomic level, and if you could get down to the atomic level, that spacial arrangement would hold right now?
Because that's how it grew, right?
It started from a small number of atoms and grew, grew, grew.
Why would it start with one shape, and all of a sudden change to another shape?
It doesn't make any sense.
At least, not to me.
And certainly not to Steenson.
And then we go to Holland, 1690.
Christiaan Huygens.
And he was studying calcite crystals.
And he drew drawings of atomic packing and bulk shape.
This is 1960.
I'm going ot show you an image from his book.
This was drawn in 1690.
See, this is stacking of cannonballs, but he was looking at a calcite crystal.
Actually, Dave, could we cut back to the document camera?
We're going to do a fair bit of this.
I'll show you calcite crystal.
This is what he had to work with.
It looks like this.
To give you a sense of scale-- what do I have?
Do I have anything you'd recognize?
OK, here's a soda can to give you a sense of scale, OK?
This is one honking big calcite crystal.
I got it as a gift from someone who watched 3091 lectures on OpenCourseWare.
And he said, if I ever come to town, can I attend your lecture?
I said, sure.
And he showed up, he attended the lecture.
He was a man probably in his early 40s, and as a pastime he collects gemstones.
And he gave me this giant calcite crystal, because the one I was using before was so pathetic.
He said, I've got to give you one.
And then, I thought, jeez, if I could ever come upon somebody who's in the gold business-- and you know, he might give me a big honking crystal of face-centered cubic metal called gold-- which broke $1,000 an ounce yesterday, but-- anyways, if you're out there, and you're watching these lectures, I'm here.
It's 8201, and you can see any anytime.
My assistant's name is Hillary, she'll take the call.
All right.
So let's go back to the slides.
Dave, could we cut to the slides, please?
Thanks.
All right.
So now-- oh, I'm going to show you a few others.
Remember, here's the idea, that we're looking down to the elemental level.
So if you look on your periodic table, it says 10, you see, 10, it's supposed to be tetragonal.
So here's a crystal of 10.
Dave, cut back, please.
We're going to go back and forth today a lot.
There's a piece of ten 10.
So you might look at it and say, gee, that looks kind of cubic.
But if you look carefully, the vertical dimension is greater than the horizontal.
In fact, it's got a square base.
And then, this is the tall one.
Which, if we cut back to the slides, let's see.
And that's what tetragonal looks like.
And we'll study all of these in a few minutes.
These are some giant crystals I didn't dare drag in.
This is sort of human-sized scale.
This is balsalt, which is a magnesium iron silicate.
It's on a coast of Iceland.
Dave, let's show them a few more.
Here's one more, if we go to the document camera.
This was beryllium aluminum silicate.
This is a barrel.
And you can see, this has got a hexagonal habit.
So something's going on at the atomic level that's different as we go from one crystal structure to the other.
What else have I got here?
Here's some calcite, here's some fluorite.
Look at this one.
This is square pyramidal but it's ionic.
So don't give me any of that sp3 d2 stuff.
Because it's not going to work here.
All right.
Let's go back to the slides, please.
So then we jump to 1781, to France.
Rene Juste-Hauy was at the Sorbonne, and he was studying the cleavage of calcite crystals.
In other words, he was taking these things and breaking them.
So he'd go to work every day and break things.
And he studied the shards, and he found that the shards were all rhombahedral.
the same shape as the parent crystal.
No matter how small the shards got, they always looked like this.
He didn't end up with cubic shards.
He didn't end up with long slivers.
He always ended up with rhombahedral slices.
So the other thing is that he didn't end up with any voids.
No matter how he cut this thing, he never got voids.
So he reasoned that this must be the way things are packing in three space perfectly.
So then he said, gee.
If they pack perfectly rhombahedrally-- we all know that we can stack boxes together perfectly, cubes together.
So he, being a Frenchman, said, what's the mathematical formulation?
They were steeped in mathematics there.
That wasn't a slur against the French.
It's a compliment.
He said, what are the mathematically distinct shapes, that if we stack them together in three space they will fill three space with no void?
And the answer is seven.
Seven space-filling volume elements.
By the way, the gabled milk carton isn't one of them.
And so he called the crystal system, so I call seven distinct shades of milk cartons.
And these are the seven crystal systems, and they're described geometrically, and you'll get a chance to go through them.
The cube is one of them.
Tetragonal.
Here's the calcite, rhombahedral.
There's barrel, hexagonal, and so on.
The seven different ways you can feel three space.
And this is from the archival readings of my predecessor, Professor Witt.
Actually, just to give your a sense, here's the two-dimensional analogy.
If I said to you, we're going to open a company, and we're going to make bathroom tile, and we want to make it sort of an upscale company, you know, not everybody wants boring old square tiles.
But definitely, I can cover two space with square tiles by putting them side by side, one on top of the other.
What's another way I can fill?
I can go with rectangular tiles.
So this dimension is different from that dimension.
What's another way?
I can use a rhombus, right, at 60 degrees.
Or I can use just a plain old parallelepipede at some arbitrary angle.
And if you go over to the Stata Center, to the Gehry building, take a look at the shape that they chose as the unit cell.
This is the repeat group, if you like, or the unit cell, because that's the one that's going to be cookie cutter replicated, side by side.
When you're putting the skin on a building, you have to make everything fit with no holidays, right?
Because you don't want gaps.
what they chose was this area element.
And all the pieces of stainless steel are cut to this shape.
And that's how you make things fit.
How you take something and make it fit around a three dimensional object is tricky.
It's the same thing as the clothing problem, right?
How do you take a flat piece of cloth and make it fit the contour of the human body?
At least, you know, in men's clothes, you know, we've got dimensions.
We've got sleeve length, got waist, chest.
We've got all these numbers.
For a woman, 8.
One number?
One number?
Fashion designers don't know crystallograpy!
If they knew crystallography, they'd know how to specify the shapes!
If you understand this, you can start your own business and make clothes that fit.
And I guarantee you, there's a market out there for clothes that fit.
What about this one?
Pentagon.
We're going to make pentagonal bathroom tile.
Well, when we install these, we're going to have to use two different colors.
You can see this up here, right?
This is pentagonal bathroom tile.
It doesn't fit.
You've got all these white spots.
So this is not a unit cell.
This is not one of the crystal systems for two dimensional.
So this one's no good.
We better put an x through it, in case someone fell asleep, and wake up, and then tell me that that's OK. All right.
So then we move about half a century forward, to the turbulent year 1848, also in France.
Auguste Bravais.
And what Bravais did, is he said, all right.
This is just filling space.
But where are the atoms?
I haven't told you where the atoms are, right?
You know, as far as you're concerned, maybe there's one atom in each of these boxes.
Or maybe there's atoms at the corners of the box.
Or maybe there's atoms at the corner and the center of the box.
Where are the atoms?
I've simply told you that I can fill three space with boxes, or with these rectangular boxes, and so on.
So he set out mathematically to prove how many different arrangements of points there are in space.
It's sort of like saying, if I want to tell you where all the apples are in the orchard, I'm going to tell you where the trees are, and I'm going to tell you where the apples are on a tree.
So where do the atoms go inside these boxes?
Turns out there's 14 different arrangements.
And what do I mean by that?
This is a good example.
So we've already established that the cube is a space-filling volume element.
It's one of the seven.
But look.
If I'm one of the atoms at the corners here, I've got six nearest neighbors.
This is also a cube, but it's got atoms at the corners and an atom in the center.
You might say, well, what's the big deal?
If you're in the atom in the center, and you look to the nearest neighbors, you've got eight nearest neighbors.
Here you only have six nearest neighbors.
So this is spatially differentiated from this.
So what why don't we keep going?
And by the way, the same environment of that central atom that's coded green is the same environment of any of the atoms on the corners.
They're all symmetric.
So both are cubes, but they have different space point environments.
And here's a third one, where we put atoms at the corners and atoms on each of the faces.
So this atom on the face, you can see, it's got one, two, three, four atoms in the plane, three behind, three in front.
It's got 12 nearest neighbors.
This has got eight nearest neighbors, this has got six nearest neighbors.
So Bravais went through and he said, let's take those seven crystal systems.
Let's put atoms in distinguishable ways.
How many distinguishable ways can we come up with atom arrangements?
And the answer is 14.
Now, what we're going to do in 3091, is we're just going to confine our conversation to the cubics.
Because you use orthogonal vectors, they're all the same length, the mathematics are simple, and the principles are the same.
And by the way, boatloads of materials are in the cubic system.
So it's not as though we just chose something that's convenient but irrelevant.
So here are the different Bravais lattices.
So you can see the cubic, there's simple cubic, body-centered cubic, face-centered cubic.
With tetragonal, there is simple tetragonal, body-centered tetragonal.
There's no face-center tetragonal.
If you try to make face-center tetragonal it's degenerate, reverts to one of the other systems.
So these are the distinguishable systems.
And everything, including crystalline proteins-- if the protein crystallizes, it forms atomic arrangements in one of these Bravais lattices.
Everything conforms to one of these Bravais lattices.
And if it doesn't, it means it lacks long-range order, in which case it is a glass.
That's it.
So here's face-centered cubic.
You can see the placement.
But there's something that's been simplified here.
Right now what you see is a single atom, designated by a hard sphere.
But at each of these lattice points, I can put more than one atom.
So let me show you what I mean by that.
What I want to do, is put different numbers of atom arrangements at the lattice point.
So those positions, the distinguishable positions, are called lattice points.
And now I want to define the crystal structure-- the crystal structure is the complete accounting of atomic arrangement.
It's the complete description of atomic arrangement.
And we don't have to go too far, because we just have to get the base unit, the unit cell, and then just repeat the unit cell.
Complete description of atomic arrangement.
And so how we going to go about this?
We're going to start with a Bravais lattice.
And what is the Bravais lattice?
The Bravais lattice is simply a point environment.
And this will mean something to you in about three minutes, when I give you the rest of the description.
So it's a set of points in space and the basis.
And what is the basis?
The basis is the atom group at each lattice point.
So what do we mean by that?
Well, let's look at what's up on the board.
What's up on the board is face-centered cubic.
So let's do all of these.
So we're going to say that the Bravais lattice, in this case, in face-centered cubic, FCC.
That's what this one is, up on the board.
And we've got dots indicate the points.
Because really, the points are nothing.
It's a concept.
See, it's French.
It's a concept, right?
What's this?
See what I'm holding?
I'm holding a Bravais lattice.
It's just a set of points.
It's like a John Cage piece, you know, a bunch of rests.
So if I go to the piano and I play rests in five four time, listen.
Isn't that great?
Man, it's fantastic.
All the different rests, and different-- So this is a set of lattice points.
Now I'm going to put something on them.
You think I'm kidding.
I'm not.
All right.
Now what's the basis?
If I put a single atom, we have what's up on the slide.
And so that's the representation of metal.
So for example, gold is FCC, with a single gold atom at each lattice point.
Aluminum, copper, platinum.
These are all FCC metals.
So there's a single atom at Bravais lattice point, and so the crystal structure is known as FCC.
It's the same as the Bravais lattice.
It's called FCC.
OK. Now, we can also put a molecule at each lattice site.
So I'm going to put up the FCC here.
Here's the FCC.
There's the cube, the set of lattice points, and the lattice points are-- I'm going to just illustrate on the front face.
One, two, three, four, and a fifth atom at the front face.
Now these are the points.
I could either put simple atoms there, and the central image there shows what happens when you have the close-packed arrangement, the atoms are so big that they touch.
Or what I could do, at each lattice point, I could put a molecule.
So for example, I could put methane.
I could put CH4.
A methane molecule.
Because when one of you gets to Jupiter, and you see the methane ice, and you go, oh.
That's FCC.
Because everybody else on the mission hasn't taken 3091.
They don't have a clue what they're looking at.
But you go, that's face-centered cubic.
And they go, huh?
And this is at each lattice site.
You see, my point is, now look, here's the difference.
All five of these, the carbon and the four hydrogens, sit at this lattice site.
All five atoms sit at this lattice site.
It's not that the carbon goes here, one hydrogen goes over at one lattice site, one hydrogen over here.
You don't have the methane straddling the unit cell.
You have all five atoms here, five atoms here.
At every lattice point, I have all five atoms at each lattice point.
So in that case, we also have, it's FCC.
So methane ice is face-centered cubic.
Because it's spherical.
They pack just like gold atoms.
You know, from a distance, we can model this as a sphere, as a point.
Or we could put an ion pair.
An example of that is sodium and chloride, and this is called rock salt crystal structure.
Rock salt.
Let's look at that one.
So at each last lattice point, I'm going to put two ions.
Not one, two.
This is taken from your text.
So you might look at this and say, well, here's the front.
You might look at this and say, well, isn't this kind of like simple cubic?
Because I've got one, two, three, four, but I've got different atoms if I try it that way.
See, this is taken from a different book.
I did some highlighting on it.
So this is sodium chloride again.
One, two, three, four, five.
With this lattice site, I put the chloride atom and the sodium atom.
The pair of atoms are associated with this lattice site.
You might say, yeah, but this is way over here.
Doesn't matter.
I'm going to anchor all of those to this lattice site.
To this lattice site, I'm going to anchor this and the sodium ion.
So this is not simple cubic.
Because if I try to call this simple cubic, I have different atoms at different lattice sites, and that's a no-no.
Has to be the same.
So if I associate the pair, chloride ion and sodium ion, chloride ion and sodium ion, chloride ion and a sodium ion not depicted.
It's off the diagram here.
And I put these dots at the center of each chloride ion.
Can you see that I'm at the face of FCC?
So that's an FCC Bravais lattice, but it's got two atoms, in this case, both ions, associated with each lattice point.
I will come to the dogs in a second.
And now there's a last one I can do.
And that's got two atoms, but not an ion pair.
We're going to put an atom pair.
Got an atom pair, and that looks like this.
I'm going to put a carbon atom, one carbon atom, and a second carbon atom.
And this angle here is 109 degrees.
And I'm going to put this pair of carbon atoms together, and when I do that, and I replicate that through FCC, I end up with diamond cubic.
And I'll show you a three-dimensional example of that next day.
So just bear in mind that if you have an ion pair, you end up with rock salt.
If I show you this pair, this is due to the sp3 hybridization.
And so examples of this one are diamond, not graphite, just diamond, silicon-- so everything that I've shown you about single crystals, and intrinsic and extrinsic semiconduction, is diamond cubic crystal structure.
But it's not one of the Bravais lattices.
It's an FCC Bravais lattice with two atoms per lattice site.
That's diamond cubic crystal structure.
But it extends to two dimensions.
So here's an Escher print.
It's got all these dogs.
In two dimensions, what's the crystal structure here?
Well, there's no such thing as body-centered cubic.
There's no third dimension in two dimensions.
So you've only got simple cubic or face-centered cubic.
So what do we have here?
Well, you see, you've got a problem.
Because the black dogs are facing the opposite direction of the white dogs.
So you can't go one, two, three, four, and make anything that way.
So instead, I put dots on the, I don't know what you call this part of the dog.
I don't know.
This part of the dog.
The leg?
Whatever you call this thing.
Yeah.
You put it on the same spot of the dog-- maybe I should have used his eye, I know how to call it.
Anyway.
So I put these things up, right?
And that means that if I associate the four dogs with this point, the two dogs facing the right, and these two dogs facing the left.
So these four dogs are associated with this lattice site.
Then I move up here, and I've got the same thing.
One, two, three four.
Two dogs facing right, two dogs facing left.
So that means I've got a simple cubic bravais lattice, and the basis is four dogs.
Four dogs as my basis.
And with that combination of a simple cubic lattice and four dogs is the basis, I can make the entire structure.
That's what this is all about.
That's my unit cell, that's my repeat unit.
So there's simple cubic, right there.
So instead of a single atom at each corner, or these are spheres that are touching, we got the four dogs.
Here's another one.
Once I got the Escher, I kept flipping through the book.
I don't know what this thing is, but it's creepy.
Anyway.
It's wallpaper.
If it's wallpaper, it's ordered.
If it's ordered, it's a two-dimensional crystal.
If it's a two-dimensional crystal, it has to conform to one of the 14 Bravais lattices.
So which one is it?
So I looked at this for a while, and I took this happy entity here, and I put a dot in its belly.
So wherever I found one of these, I put a dot.
and then I connected the dots, and what do I get?
I got the rhombus.
Now, what's the basis?
That's the Bravais lattice.
What is the basis?
So I went over here and I said, well, if I start from this thing and I go all the way up, I got some tail, blue tail here.
And this thing way down here looks like it's missing a tail.
So if I start from this, go all the way down to here, and capture all of this material in between and associate it with that lattice point, that can be my basis.
And then I take that set of imagery, and I put it on top of this, where it's centered at about this thing's navel, and put it around here.
You can see, the translation is obvious there, and the translation is obvious here.
There's the wallpaper.
So it's part of a Bravais lattice, only in two dimensions, you see?
This is crystallography.
That's all it is.
Crystallography.
All right.
So the next thing that we have to do, it's kind of a little bit boring, but you need to know this stuff, because it's formative for the next unit.
It's the characteristics of cubic lattices.
Now what do I mean, the characteristics?
I mean the geometric characteristics.
So there's a table here that I've taken out of your archival reading.
If you go to the website, there's a set of notes that were written by my predecessor, Professor Witt.
And if you go and look this up, this unit on crystallography, there's this table.
And it shows such things as the unit cell volume.
Well, what's the unit cell?
The unit cell is this cookie cutter cube that's the repeat unit, and it has a dimension here called a, which is the lattice constant.
It's the size of the box.
So what's the volume of the unit cell Well, duh, it's a cubed.
That's obvious, all right?
Doesn't matter whether it's simple cubic, face-centered cubic, body centered cubic.
How many lattice points per unit cell?
Well, that's-- how many atoms do I get per unit cell?
If you look at this, you say, well gee, it's-- you know, if you've got the simple cubic, you've got them at the eight corners, so it's eight, right?
Well, no, because each atom on the corner is inside this cell, but it's also inside the neighboring cell, and it's inside each of the eight neighboring cells.
So I have eight times one eighth.
So the total amount of atom within the cube is only one.
That's why you've got to go through this derivation.
So here's an example for face-centered cubic.
You see the corners, one, two, three, four?
There's eight of those.
Eight times one eighth is one.
And then you've got these atoms on the face, and there's one, two, three, four, five, six, right?
Six faces.
But each of these atoms on the face is only half inside the cube.
The other half is in the other cube.
So I got six times one half is three, and that gives me four.
So that's how you go through, and figure out what's going on.
And then you can look at things like nearest neighbor distance, and what's the relationship between the radius, if you're looking at hard sphere packing.
So here's a good example.
Here's the hard sphere packing model.
So I've got atoms all the same size, and they're packed to touch.
And they're packed to touch so that in the end, I end up with a face centered cubic arrangement.
And so what's the relationship between the radius and the lattice constant?
And it's just simple trigonometry, right?
4r equals 2a squared, and you go through it, and that's how you end up with all these formulas.
So that's what going on.
All right.
So I think we're going to move to the last five minutes here.
So just a couple of comments.
Today we played music when you came in.
It was Burning Down the House.
An old piece by Talking Heads.
Why was I playing Burning Down the House?
Well, I want you to study this painting.
This is a painting by Georges Braques, from 1908, And it's called The Houses at L'estac, or if I speak Canadian, The Houses at L'estac.
And you see the thing about these houses?
You see the ones in the background have better definitions than the ones in the foreground?
And the ones in the foreground are all distorted, and kind of falling over?
Well, when this was exhibited at the Salon, on the critics lambasted it.
They said, this thing is absolutely childish.
And one of the critics said, this isn't a painting of houses.
This is nothing but a stack of cubes.
And from this painting, and that negative comment, came the term cubism.
And so the entire movement of cubism comes from this.
And since we're studying the properties of cubic crystals, I thought the connection was pretty tight.
And that's why I chose that piece.
Don't go anywhere yet.
We still have a couple of minutes.
It's not better out there than it is in here, I guarantee you.
All right.
So now I want to show you one of the properties.
Why do we care about atomic arrangement Suppose some of you are interested in fiberoptics.
One of the things you have to do in fiberoptics is make junctures, because you can't run a cable 100 miles without making some junctures.
Well, atomic arrangement reflects microscopic properties.
And one of the properties that you can have in a crystal is something called birefringence, depending on the atomic arrangement.
So what happens is if you have a ray that comes in from the left here, it splits into two.
And that's no good, in certain instances.
In other instances, if you want a beam splitter, you want something birefringent.
So look at all these crystals.
Quartz is birefringent.
Ice is.
You've seen it.
Calcite is birefringent.
What do I mean by that?
Well, Dave, can we cut to the document camera?
Right.
So here's a here's a crystal of-- well, this is my handwriting.
It's beautiful.
Let's write-- here's more.
Isn't that beautiful?
OK, enough about me.
Now let's-- All right.
So this is birefringent.
But what happens if we rotate it?
We can change the degree.
And if we're really clever, and we get it onto the optical axis, we can actually get these things to superimpose.
Here's a piece of cryolite.
This is the stuff they make aluminum out of.
They dissolve aluminum in it.
And you can see that, see, I loaned it to somebody, and they dropped it.
And when they dropped it, they damaged it.
That's what happens.
Look, it's all crunched up there.
But anyway, I don't want to get into that.
But so you can see, if we turn this and we get it right on the optical axis, we can actually make these things line up sometimes.
Anyway.
So you can see the property of birefringence.
OK. Let's go back to documents, Dave.
One more thing.
Last thing I want to show you.
You know gold.
Gold looks like this.
It's yellow, right?
Well, depending on how you, you know, those of you have seen certain wedding bands know that if you want to make a white gold, you add nickel to it.
And once the nickel content gets above a certain level, the gold looks white.
In other words, just metallic reflective.
In Eastern Europe, it was common to add high levels of copper, in which case, you get a beautiful rose-colored gold, or pink gold.
I'm going to show you some blue gold.
This isn't a joke.
This is 14 karat.
It's 40 plus percent by weight gold, and the balance is indium.
And when you add indium to gold, you get blue.
Document camera, please.
Just so that you remember.
OK. That's gold.
This is gold foil.
Rutherford, et cetera, et cetera.
You know, this is yellow gold.
Ho-hum.
This is white gold.
Same deal, but with some nickel.
Now it's reflecting off of the yellow gold.
Boy, there's a lot of yellow in here.
Where's that coming from?
It's hard to tell.
All right.
Now Let's look at some blue gold.
This is blue.
So, but it's brittle.
So you know, what are you going to do with something that's brittle?
You can't shape it.
So I have an idea.
So here's my idea.
See, what we do is-- see this?
We do, instead of this, we put this.
We drill a hole in it, we put hands on it, we sell it for a lot of money.
It's blue gold.
It's blue karat gold.
What you can do if you understand crystallography.
All right, get out of here.
I'll see you on Friday.
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OK, Let's get down to business.
Last day we started study crystallography.
And we were looking at ordered solids.
And we recognized the crystal structure is the sum of the combination of a Bravais lattice, which is the way that points are identified in space, plus the basis.
And we saw that we could put a variety of elements at the corners of the Bravais lattice.
And this is taken from the archival notes where we see the simple, single atoms at the various lattice points.
And we recognized last day that we could put clusters of atoms and so on, and even proteins will crystallize in this form.
And actually, to make a point, about five years ago, a group of students came to my office after the final exam, and recognizing that I drink Fresca during the lectures, they came to me with this body-centered Fresca unit cell.
So this just makes the point that you can put anything at the lattice point.
So there it is.
Oh, here's the point.
This is me at a conference in Copenhagen.
This is outside the Admiral Hotel.
And just to show you I'm always thinking about 3.091, I came outside-- and this is a colleague of mine from France, Marcelle Gaune-Escard.
How do we know she's French?
She's got a cigarette.
She's a scientist.
She's a woman.
And she smoking.
And it's fantastic.
And she's a great gal.
Anyways, so there we are out in front of the Admiral Hotel.
And I notice these cannon balls.
And what do you notice about the cannon balls?
They are sitting in one of the Bravais lattices.
So I couldn't resist having my wife take this picture.
Because I knew that this would come back to 3.091.
So here we are with the individual cubic lattices where we're going to focus.
And we started looking at the properties of cubic lattices.
And I started working through this table.
And we got down to nearest neighbors, and so on, and this relationship between the lattice constant and the radius.
And I just wanted to show this last point here about packing density.
Because it comes up considerations of properties of elements.
And so I wanted to go through the packing density calculation.
And so the way we handle the packing density, remember we're looking at a unit cell here.
This is the cube that's the repeat unit.
And it has a lattice constant of dimension a.
And then depending on where we put atoms, I want to show that we've got three different options here, simple cubic, body-centered cubic, and face-centered cubic.
And the last line of that slide shows that the densest packing of atoms-- this is hard spheres, the cannon balls-- the densest packing is face-centered cubic.
And I wanted to show the derivation of that where if we face-centered cubic where the atoms are big enough to touch.
If we have 4 atoms at the corner touching on the face diagonal and then continuing through the cell, we want to get packing density as the ratio of the volume occupied by the atoms-- and here we're modeling the atoms as hard spheres-- divided by the volume of the cell itself.
And so that's pretty straightforward calculation.
We're modeling the atoms as hard spheres.
And we saw last day that if we take the fraction of the atom, 1/8 times 1/8, and 6 times 1/2, we have 4 atom equivalents per unit cell.
And each of these has a volume of 4/3 pi r cubed.
And then we divide that by a cubed.
And then we know the relationship between a and r. This is 4r and that's root 2 a.
So we can say root 2 a equals 4r.
Either eliminate a or eliminate r. We eventually end up with this dimensionless group, which is pi divided by 3 roots of 2.
And that is 74%.
And that's the densest package.
If you go through the same calculation for body-centered packing, you'll find it's 68% packing.
And for simple cubic it's 52% packing.
That's the highest density that you can have.
So I urge you to spend some time.
If you work the homework, you'll work your way through that chart.
So now I wanted to take a few minutes and talk about crystallographic notation.
Because we need to know this in order to describe various features of a cell.
So I'm going to start with a Cartesian coordinate system that obeys the right-hand rule.
So we'll put down x, y and z.
And then we'll put up a unit cell here with edge a.
So we have a on the x, y and z coordinate.
And we know that a equals b equals c. That's the unit vectors here.
There's a little a here, a little b here, and a little c here.
These are all equal to the lattice constant value a.
And all of the angles are 90 degrees.
This is what defines the cubic system.
So I'm going to put several positions here.
So here is the origin.
And the origin we designate as 0, 0, 0.
I'm going to put a position up here, a.
And you just use the same notation as you'd use mathematics.
Only crystallographers don't put parentheses around it.
So what I'm really teaching is notation.
You already know the math.
So a up here is what?
It's 0 units on the x-axis.
It's 1 unit on the y-axis, and 1 unit on a z-axis.
So that would make it 0, 1, 1.
And I'm going to put a b out here.
B is out 1 unit on the x-axis.
It's 0 units on the y-axis.
And it looks like it's up at about 1/2 unit on the z-axis.
So that makes 1, 0, and 1/2.
And so now what I want to do is show you how we designate directions and planes.
You use a slightly different notation from what you used to in math.
So this chart will be posted at the website.
Move the coordinate axes so the line passes through the origin, define a vector from the origin to the point on the line, and choose the smaller set of integers.
So let's do a simple one here.
Let's do from origin to b.
So o to b.
That's this vector here.
So how would we do that one?
Well we're already at the origin.
So that's trivial.
o to b, so we're going from here out to here.
And then what does it say?
It says choose the smallest set of integers, no commas, clear fractions.
So that's going to take us out to 1 0 1/2.
But the crystallographers don't like fractions.
So you're supposed to clear the fractions.
You clear the fractions by multiplying through.
So make that 2 0 1.
It's the same direction.
See if I came out two units here, I'd end up being one unit here.
So it's the same direction.
It really is the same direction.
Don't put any commas.
And put brackets around it.
So that's the 2, 0, 1 direction, which is o b.
And I want to do a different one.
I'm going to go this way, a to o.
So let's look at that one, a o.
And what happens on a o?
Well now I'm going to put the origin at a. I'm going to go to 0 in the x-axis.
I'm going to go minus 1 in the z-axis, and minus 1 in the y-axis.
So that gives me 0 minus 1 minus 1.
But see they don't use commas.
And so this looks kind of goofy.
Because you get these minus signs in here.
So what the crystallographers do is they put the minus sign on top of the number.
So this becomes 0 1 with a macron it.
If you'd studied Latin, you'd know this thing is a macron, a line over a vowel.
So 0 1 bar 1 bar.
And that's the direction a o.
So that's the macron acting as the minus sign.
So that's the a o direction And then furthermore, we can designate families.
This is just the 0 1 bar 1 bar.
If you want to be a hipster with crystallography, you call this 0.
You say it's 0 1 bar 1 bar.
But if you're a nice chemistry student, you'll say zero, minus 1 minus 1.
No we don't do that in 3.091, 0 1 bar 1 bar.
Now suppose I wanted to describe all of the faces, every face.
So what I could do, is I could say I want the set of all face diagonals.
Well, that's a face diagonal.
But what I can do is in a compact notation, I could write 0 1 1 and use the carat.
This is called the carat.
So if I enclose in carats, this means the set up all 0 1 1 type planes.
And so what that means is I can unpack that, and write 0 1 1, 0 1 1 bar, 0 1 bar 1, et cetera.
And what I'll end up doing is describing the set up all face diagonals.
So if I want to talk to you later when we're studying x-ray diffraction, I want to talk about face diagonals, I don't use that word.
We're quantitative here at MIT.
I say look at the 0 1 1 planes.
So you go face diagonals.
That's the compact notation.
If we want to look at all cell edges, what's the cell edge?
Here's the cell edge.
This is an edge here, isn't it?
It's right on the edge of the cell.
So it's 0 on the y, 0 in the z.
It's 1 in the x.
So the set of all cell edges would be 0 0 1.
And you could write 1 0 0, that's OK too.
Or you could write 0 1 0.
That's OK.
But if you want to be a hipster in crystallography, put the zeroes out in front.
So the set of 0 0 1 directions is all the cube edges.
And then lastly, if we wanted to get all the diagonals that starts from here and goes up, we'll have all cell diagonals, all, if you want to call it, body diagonals, not face diagonals, body diagonals.
Body diagonals is 1 1 1.
And the mathematics imitates reality.
How many combinations are there here?
I've got 0 1, and I've got minus 1.
So I have 0 0 1, 0 0 1 bar, 0 1 0, 0 1 bar 0, 1 0 0, 1 bar 0.
There's 6 And how many edges are there?
6.
Mathematics imitating reality, what a concept.
That's great.
OK.
So that's good.
Now we want to look at planes.
Let's look at planes.
Now planes are a little bit different.
This is basically what you know from math with a little bit of notation It's basically the same thing.
Planes are quite different.
So let's go to the planes.
So you know the equation of a plane.
It's x over a plus y over b plus z over c equals 1.
And a b and c are the intercepts at the x, y z axes.
Well, in order to make this notation compatible to what we have here for lines, a British meteorologist by the name of William Hallowes Miller in 1839 said let's define h, k, and l as the reciprocals of these intercepts.
And now we'll write the equation of a plane as hx plus ky plus lz equals 1.
And we'll use those values of h, k, and l as the indicators of a plane.
So here's an example.
So again, I've got the right-handed rule coordinates.
And here's the plane.
And you can see that it cuts the x-axis and a.
It cuts the y-axis at b.
And it's parallel to the z-axis.
So if we were just to use the old mathematics notation, and put the a, b and c, you'd have 1/2 here.
You'd have 1 here.
And you'd have infinity here.
And that looks goofy.
So instead what we want to do is use the Miller indices, which are the reciprocals.
So now the reciprocal of 1/2 is 2.
The reciprocal of 1 trivially is 1.
And the reciprocal of infinity is 0.
And so now you have something that makes sense.
And you put no commas in between.
And you put the whole thing in parentheses.
So that's the 2 1 0 plane.
And there's a really cool property here.
And that is, using this same formulation I just showed you for directions, the 2 1 0 directions-- you see 2 1 0 with square brackets, that's a direction.
2 1 0 with parentheses is a plane.
The 2 1 0 direction is perpendicular to the 2 1 0 plane.
And that's one of the advantages of using Miller indices notation with the classical notation for direction.
Here's another one.
This is 0 1 0.
This is the x, y, and the z.
So this cuts at 1, 0 1 0.
This is 0 2 0.
An 0 1 0 directions are perpendicular to 0 1 0 plane.
0 2 0 directions are perpendicular to 0 2 0 plane.
Oh here's 1 1 1.
That's cool.
See it cuts it at 1 1 and 1.
And 1 1 1 direction cuts through the 1 1 1 plane perpendicular.
This is an interesting one.
See the origin.
The plane cuts through the origin.
Now what do you do?
What do you do?
You move the origin.
Because we don't know where the origin of the universe is.
It's just a unit cell.
It doesn't matter.
So we move the origin up here.
And now what do we have?
We have 1 1 and minus 1.
And just as before, we put 1 with a macron over it.
That's 1 1 1 bar plane.
And there is the 1 1 1 bar direction.
It doesn't matter.
It'll always give you perpendicular.
It's really cool.
All right.
And we can also specify a set of planes.
So instead of using parentheses, we use brace brackets.
And if we use brae brackets, we get families of planes.
So for example, we could say if we wanted a particular face so that the face in the xy plane, facing the xy plane would be 0 0 1.
Right, because it cuts the z.
So this is 0 0 1.
But what if I wanted to say in a compact way, all faces of the unit cell, all of the faces.
So this is the particular plane.
If I wanted to say all faces I would write 0 0 1.
But I can't use any of these containers.
So I'll use brace brackets.
So 0 0 1 in brace brackets means the whole setup, 0 0 1, 0 0 1 bar, 0 1 0, 0 1 bar 0, 1 0 0, and 1 bar 0 0.
So there's only 6.
And there's 6 faces.
Again mathematics imitating reality.
What a concept.
Make the math work for you.
You don't work for the math.
All right and just to make the point about the h k l plane, is perpendicular to the h k l direction.
So that's good.
And the last thing I want to show you that comes out of the use of the Miller indices for planes, is that another property that we'd like to know is the interplanar spacing.
What's the spacing between like planes.
So interplanar spacing, that is to say the distance between like planes, adjacent planes identical index.
I'm going to write it to be specific to say Miller index, h k l in parentheses.
And the property that comes out of this-- and you don't have to derive it.
Somebody worked this out for you about a 150 years ago-- is that the distance between adjacent planes with the index h k l is given by the ratio of a, which is the lattice constant, this is the cubed edge, divided by the sum of the squares of h, k, and l taken to the power of 1/2, So the square root of h squared plus k squared plus l squared divided into the lattice constant, gives you the value of the spacing in between.
And this is the a just to be clear about it.
So for example, I think I've got some cartoons up here that show this.
So here's 0 1 0.
It's trivial.
What's the distance here?
The d 0 1 0 is just day.
How about here?
This is 0 squared, 2 squared.
0 squared, this is d over 2.
So this is a 0 2 0 plane.
But if the distance between 0 2 0 planes is 1/2 a, well, then that's an 0 2 0 plane.
Then this is an 0 2 0 plane.
So point of fact, every 0 1 0 plane is an 0 2 0 plane.
But every 0 2 0 plane isn't an 0 1 0 plane.
You can see that some planes have multiple names.
Here's another one.
Look, this is the distance between adjacent 1 1 1 planes.
It's a over root 3.
So as the index goes up, as h k l goes up, the distance between successive planes goes down.
And the angle between the planes goes up.
The easiest one is orthogonal.
OK, that's good.
That's good.
All right.
I think that's a good place to stop.
I dread these kinds of lessons.
And I'm sorry that the parents ended up here on a day like today.
But there's a little bit of this notation and taxonomy that they have to know.
And so we suffer through it.
And then we get on to the meat.
Now we're ready for some meat.
Now that we can describe crystal structure, we're ready for physical measurement of crystal structure.
So what I want to start today is the characterization of crystal structure by x-rays.
How do we measure?
How do we know that gold is fcc.
We use x-rays.
So today what I want to do is start as I always do on new unit with history lesson.
We'll start with the discovery of x-rays.
And we'll get into the underlying physics.
And the next day we'll look at applications including x-ray diffraction.
So first of all, what are x-rays?
They're a form of electromagnetic radiation.
And they have the property of that.
They have very short wavelength.
How short?
It's on the order of about angstroms.
So strictly speaking, it goes from about 1/100 to 100 angstroms.
So centered at about 1 angstrom is the center of the x region of the spectrum.
And because they're a form of electromagnetic radiation, their energy is given by these equations, hu equals hc over lambda or hc u bar.
Those equations that you've seen over and over again apply for the energy of an x-ray, because an x-ray is identical to a photon of wavelength somewhere down at this level.
Why do we care about this?
Because this is atomic dimensions.
If I want to measure the dimension of the human hair, I don't use a yard stick.
I use something that is of the human hair dimension.
If we want to get into the characterization of atomic structure, we need a measurement tool that is on the order of that blank scale.
And that's why we're going to use x-rays.
Now other property, I said very, very short wavelength.
But because of this inverse relationship between wavelength and energy, I want to draw attention to these superbly high value of energy.
So let's plug in the energy for lambda equals 1 angstrom.
When lambda equals 1 angstrom, we end up with 12,400 electron volts per photon.
Now what's that mean, just to put it in perspective?
Look up here at the table I've shown.
We know the ionization energy, the lone electron in hydrogen is 13.6 electron volts, which is 14 to 2 significant figures.
And then we go to helium.
You pull these off the periodic table.
This is the first ionization energy.
And they're down around somewhere about 10 electron volts.
Look at.
This is 12,400.
So let's go down to the 1 electron atom.
So this is helium plus.
This is lithium 2 plus.
And we see a set of numbers here that follow this sequence 1 4 9 16, which is nothing more than the sequence you would expect for a series of 1 electron atoms.
It's K, which is 13.6 times the square of z/ So z here is 2, z here is 3, z here is 4, and so on.
But even so, this is the last electron of nitrogen.
The most tightly bound electron of nitrogen, 668 electron volts.
That's a huge amount of energy.
But that pales in comparison to 12,400 electron volts.
So this is clearly ionizing radiation writ large.
What do I mean by ionizing radiation?
It has the capacity to ionize everything.
Every electron will be blown away.
This is called knocking the stuffing out of the electron structure.
So that's what we're dealing with.
So now let's go to the origin of x-rays.
So how did we discover x-rays?
Well, it was a dark and stormy night, not unlike the weather today, apropos of the lesson.
It was November 8, 1895.
And it was at the home and laboratory of a professor.
Of course it's a professor, Professor Wilhelm Roentgen.
And he was a professor of physics at the University of Wurzburg in Bavaria, now part of Germany.
And he was doing research, as many people were in the 1890s, on the properties of gas discharge tubes.
He was studying gas discharge tubes.
And he had a special interest in gas discharge tubes.
Remember Lorentz?
He like to put the gas discharge tubes in a magnetic field.
Stark liked the put them in an electric field.
And what was Roentgen's pet project?
Roentgen loved high voltage.
He loved high voltage an low pressure.
He was focused high voltage, low pressure, near vacuum.
And so let's have a drawing of what he used.
And here's an image of the museum that is literally in the house that was occupied by him.
So the apparatus is sitting right there on the table.
And here's the heart of the apparatus.
I'm going to go to a fresh board because I need a little extra room here.
So we know what the gas discharge tube looks like.
It's a glass tube, sealed.
And we've got feedthroughs for electrodes at both ends.
And this goes down to a power supply.
And the power supply that he used was batteries of chromic acid.
And they were open beakers.
And they were down in the basement two floors below.
And he had wires going from the basement up to the second floor.
And he was able to generate about 20 volts.
So he had a number of these in series.
Because this is roughly a 1-volt system.
So round numbers, you got about two dozen of these things in series.
And I'm going to write here, in the cellar.
Now how did he get the high voltage?
He put in here something that we wouldn't recognize today as an inductor.
He called it a choke coil.
And what the choke coil did, is it would charge up until it got to 35,000 volts.
When it got to 35,000 volts, it would discharge and send a current at 35,000 volts to the cathode here.
And with 35,000 volts, the cathode would send electrons boiling off the cathode, heading towards the anode with the kinetic energy associated with 35,000 volts.
And this discharged at about eight times a second.
So I'm going to write 8 hertz here.
So eight times a second this thing, bam, bam, bam, bam.
And the electrons go firing off here.
And this to make sure he had really low pressure, he had the cell open actually.
And this went to a vacuum pump.
He had almost no gas molecules in here.
Now, the way he did the experiment is, he'd set this thing up, started it going, and this was after dinner.
So on a Friday night after dinner, he was upstairs.
And he wanted to measure emissions here.
And so what he had, you either get graduate students to sit there and you exploit their labor, or what you do is you put up a detector here which was a screen.
So this is either a sheet of paper or a sheet of cloth.
So we call this a screen.
And so you're going to look here.
All right you're looking at this, seeing what's coming out of this gas discharge tube.
And it was a screen up paper or fabric.
And it had been painted with a chemical that would glow if it were struck by a particle, or by a photon of a given energy.
And they used in this case, coated with the chemical barium platinocyanide, so cyanide four times.
So you make an aqueous paste of this stuff.
And you gob it on to the screen.
And when the screen is hit by something, it sparks.
And the latin word for spark is scintilla.
Have you ever heard of someone being described as a scintillating conversationalist?
It means they are a sparkling conversationalist.
So this is called a scintillation screen, a scintillation screen.
And so presumably, every time there's an event, the screen glows.
And so Roentgen starts the experiment.
And it's raining outside.
There's thunder.
There's lightning.
There's a street light as you can see right there.
There's windows.
And the thing is glowing.
And he really can't detect what's going on here.
So he decides to cover the tube.
So he puts a wooden box around the gas discharge tube.
So now the gas discharge tube is enclosed in a box.
And he continues the experiment.
And what he sees is glowing on this scintillation screen.
The scintillation screen is glowing even though the gas discharge tube is in a box.
There was another observation he made.
The student earlier in the day, who had painted the screen with the barium platinocyanide.
I guess he was practicing his calligraphy.
And he left a paper towel on the edge of the bench.
And he painted the letter A on the paper towel.
And the paper towel was glowing with the letter A.
And this thing is enclosed in a box.
So he's saying what's going on here.
So he decides that there's some kind of an interaction between the screen and something going on in here.
So the first thing he does is he says, well maybe it's electromagnetic.
Maybe it's charged particles.
So he takes a magnet.
And he moves a magnet around.
And nothing changes.
He kills the power to the gas discharge tube.
Everything stops.
So it's definitely inspired by 35,000 volts of energy going into the tube.
So then he says, OK I'll take a piece of black paper.
And he puts the black paper in between.
And the screen continues to glow.
So then he takes a book.
And he puts the book in front.
And he sees the shadow of the book.
And he's standing here like this.
He sees the shadow of the book.
And he can see the shadow of the bones of his hand.
So he says, you know I don't think it's particles.
I think it's radiation.
I think it's radiation.
And maybe it's something like that radiation that Hertz has been talking about.
It's some kind of radiation.
But I don't know exactly what it is.
It's mysterious.
And x is the letter of mystery.
I'm going to call it x-radiation.
So he's discovered this mysterious form of radiation.
Does he rush to publish?
No.
He's not going to risk his scientific career by announcing that he's discovered a form of radiation that can see inside the human body.
That would be professional suicide.
So he continues to repeat the experiments through November and December of 1895.
And in January of 1896, he finally publishers.
And it takes the world by storm.
By the way, how did he record this stuff?
He used photographic plates.
And he stored the photographic plates in his lab in a wooden file cabinet.
And so many of the plates were fogged even before he got them to the experiment.
Because of the x-ray, they were going everywhere in the lab.
And he was working there the whole time.
So what happens?
What's the world's reaction?
January 16, 1896, New York Times reports radiation that can see through matter and peer inside the human body.
Peer inside the human body, you say.
Well, why is that?
What's the big deal?
Well, in London, immediately there was a market response.
A manufacturer claimed to be a purveyor of x-ray proof lady's undergarments.
I mean if you can see inside the human body, it's Victorian England.
How modest can you be?
This is a calamity.
In France, of course, they don't care about such things.
So they had a different response.
Remember this is where Daguerre invented halide photography.
So there it's very deep.
It's very existential.
Somebody claimed that with x-rays he could irradiate and get a picture of the human soul.
How French.
But I'm sorry to say it, the wackiest comes from guess where?
The United States.
No lesser an institution than the New York City College of Physicians and Surgeons.
They proposed to use x-rays to project drawings from text books of anatomy onto the brains of students thereby creating, and I quote, an enduring impression.
In Iowa, we plumb new depths.
There was a group out there claiming that they could turn copper pennies into gold.
So this was the response.
But the thing that makes it so fantastic is we take it for granted that if I want to exercise power and exert a force on that can, I don't necessarily have to make physical contact with it.
We accept this as given that you an either have direct mechanical contact, or indirect contact, magnetics, what have you.
But in 1896, this wasn't given.
So the notion that I could exert a force on something from a distance was very fresh.
It wasn't so easy.
But there were benefits of the x-radiation already in middle of 1896.
There was a seamstress who had a broken needle in an industrial accident in Scotland buried into her hand.
This is trivial.
Look, all you need is low pressure in a gas tube, and you can set this up in high school.
It's nothing.
People had this set up in dentist offices.
It's nothing.
So this surgeon was able to find where the needle was, and then use precision to go in and take the needle out of the injured person's hand.
In eighteen 1899 they were already using radiation in treatment of cancer.
So what is the relevant physics?
what's the relevant physics here?
Let's look at the relevant physics.
Well, when we're talking about 35,000 volts, and we're talking about energies in the vicinity of tens of thousands of electron volts, you know that this can't come from emission from an excited electron in a gas molecule.
The gas binding energies are way lower than that.
This must come from the solid.
The electron leaves the cathode, and it crashes into the anode.
Here's the smoking gun.
There's something going on here when the electron crashes into the anode and then gives rise to the radiation.
So let's look inside the anode.
So what's the energy level diagram?
Let's say the anode is made of copper.
So what's the energy level diagram look like?
So we know up here we're going to have energy is 0.
And the energy quantum state is infinity.
Down here is the ground state.
And the energy is the ground state energy.
And we know this is up in the tens of thousands of electron volts.
And then up here there's an E2 for state 2.
And up here is an E3, energy 3.
Then maybe an E4 for energy 4, et cetera.
And I'm going to write not to scale.
This is just to give you a sense.
And so the ballistic electron with, in this case, 35,000 volts of energy, comes crashing in to the anode.
And this is one of these mixed metaphors.
So this is a Cartesian drawing of an electron.
All right?
This is the incident electron.
And its energy is given by the charge on the electron, which is E, and the plate voltage, which is 35,000.
So it's got 35,000 electron volts, which turns into 1/2 mv squared.
It goes crashing into this thing.
And it's got enough energy it can kick out top level electrons, mid-level electrons, and in the extreme, it an eject ground state electrons in 1s.
By the way, the spectroscopists, they don't like numbers.
Remember, this is K. This is L. This is M. This is N. So we can eject K shell electrons.
What happens if we create a vacancy down here?
all?
Of these electrons above are unstable.
If its copper, there's 29 of them.
Now there's 28 with a vacancy.
And they fall down.
And when they fall down from N equals 2 to N equals 1, there's a photon emitted.
And that photon emission is in the x region of the spectrum.
But we don't just get one wavelength.
We get a plurality of wavelengths.
n equals 2 to n equals 1, we get a photon.
There's a possibility to go for n equals 3 to n equals 1.
We'll get a photon of even higher energy or a lower wavelength.
Well, heck, if you've got enough energy to kick out an electron from n equals 1, you've certainly got enough energy to kick out an electron from n equals 2.
So you can get a cascade from 3 to 2, which gives us a photon, or from 4 to 2, which gives us a photon.
Can you see that you're going to get an entire set of lines?
We're going to label them because we don't have a single line.
So what do we label them?
We label them on the basis of the destination shell and how far they traveled.
So this line here is known as a K alpha line.
Why is it a K?
Because that's the nf.
That's the n number of the final state.
And why is it alpha?
Because the delta n equals 1.
It came from 1 shell away.
What's this one?
It's gotta be a K line, because we ended at n equals 1.
But the delta is 2.
It went from n equals 3 to n equals 1.
So this is called K beta.
So there's a K beta line and a K alpha line.
This ends at n equals 2, which the spectroscopists call L. So we call this L. Delta is 1 L alpha.
This is L, but delta is 2.
So this is L beta.
And what I can do, is I can plot all of this on a curve showing the spectrum of this particular element.
And what do I expect to see?
I expect to see a spectrum that plots intensity versus some energy coordinate.
It could be wavelength in this direction, which means that energy increases in that direction.
And which has the highest energy?
The highest energy up here is K beta, n equals 3 to n equals 1.
So there's K beta.
And then close to it is K alpha.
And why do I put the K alpha line higher than the K beta line?
I don't know what the relative amounts should be.
But I suspect that if there's a vacancy, it's like musical chairs.
If there's a vacancy in n equals 1, and an electron only has to go from n equals 2 versus n equals 3, the chances are the journey from n equals 2 to N equals 1 is easier, and therefore more frequent than the journey from n equals 3, even though n equals 3 to 1 is greater energy.
But 2 to 1 is more likely.
And intensity reflects frequency.
The wavelength reflects the energy.
And then over here is L alpha L beta.
Now there's one more piece here.
The precise value here is determined by the identity of the element in the anode.
If it's copper-- and I had to do this as a junior at the University of Toronto.
I spent a whole year doing x-ray crystallography.
And I will know on my death bed the wavelength of copper K alpha radiation is to five significant figures, 1.5418 angstroms.
Now if someone had fallen asleep in this lecture and open their eyes right now, and looked at the wavelength of this line as 1.5418 angstroms.
And I erase the letters Cu.
You take one look at that, and you say copper.
Because only copper has a K alpha wavelength of 1.5418, which means I can use x-ray analysis to determine the identity of a unknown specimen.
And furthermore it's additive.
So if I had a brass, which is copper and zinc, the zinc K alpha line is over here.
And I will have both lines.
And the relative intensities of the lines are related to the relative concentrations of the copper in zinc.
So now I have a tool for chemical analysis.
All of this from going upstairs after dinner and horsing around with the gas discharge tube.
OK, so we're going to end the formal lesson here.
But we have a few more minutes.
So everybody is instructed to not make a lot of noise and just sit quietly.
So let's take a look.
This is an image.
This is an image of Bertha.
This is Roentgen's wife.
This is her hand.
He invited her into the lab.
And she stood for a full 15 minutes in front of this tube.
And in that 15 minutes, got a lifetime's worth of radiation.
But what we're looking at here, and what I'm going to point out to you in the last several minutes here, Roentgen was a prissy physicist, a really prissy guy.
And he was German, which meant that the German academics had no contact with industry whatsoever.
On the night of November 8, 1895, he invented three modern technologies.
This is Bertha's hand.
It represents the birth of medical radiography.
Is there anybody in this auditorium who has never had an x-ray, a dental or body x-ray?
It's very rare.
We see them all the time.
We use them all the time.
This morning I had one taken about 9:10, right here.
And I'm OK.
So that's radiography, medical radiography.
Roentgen had a balance in his laboratory.
The justice, and she's got the torsion balance.
So you put the unknown on one side, and then you put weights that are calibrated.
These are made of brass.
And they're in a wooden box.
And so Roentgen has already covered the tube with a wooden box.
And so he took the emission from this, and he radiating his box of weights.
And what he see?
He saw dark regions where the weights were.
Because metals have a different electron density, because they've got a greater atom density than wood.
Wood is a more open structure.
And so by contrast, he could see inside the box.
This is the basis for airport security.
Roentgen had a gun.
He was not holding up convenience stores.
He was a hunter.
He went into the woods.
And he x-rayed the gun.
Now why did he x-ray the gun?
Now the gun is made of steel.
But deep inside the steel, there could be a flaw.
There could be a crack.
There could be a blowhole.
The surface of the steel looks beautiful.
But inside is a crack.
And that crack could lead to failure under pressure.
So Roentgen irradiates.
And we already know from this that different electron densities give different shading, which means if there were a crack right here, you might see it in the radiograph.
And this we use for certification of pressure vessels today.
Certain pressure vessels that have to be welded, have a specification that requires that every inch of the weld be inspected by radiation, to be certain that there are no cracks deep inside the weld that under high pressure could open up, have an explosion, and cause injury, or in the worst case, death.
So what's the outcome of it all?
In 1901 when the Nobel Prizes were offered for the first time, Roentgen by unanimous vote of the selection committee was the first recipient of the Nobel Prize in Physics.
So on that occasion, he was invited to come from Wurzburg and to deliver a lecture in Stockholm.
I've been to the place, it's a spectacular hall, gilded, beautiful champagne reception, elegant dinner.
And he gives a lecture.
And what's his lecture on?
The discovery of x-rays.
He received a huge amount of money at the time in prize money.
And he returned to Wurzburg What did he do with the prize money?
he?
Donated it to the University to be used for scholarships for students studying physics.
And to this day, students who study physics at the University of Wurzburg are beneficiaries of the generosity of the first winner of the Nobel Prize in Physics.
Such was his humanity.
I hope you have a great weekend.
We'll see the students back here on Monday.
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Let's get right into it.
So last day, we started looking at Roentgen again and the generation of x-rays, which we saw occurred when we operate the gas discharge tube at high voltage and low pressure and this is the image that we saw of his wife's hand.
Birth of medical radiography.
And we started looking at the origin of x-rays and we started looking at the energy level diagram of the target anode in the gas discharge tube and this is the energy level diagram of the target.
So the ballistic electron-- I'm calling it incident-- this is the electron that's making its journey across the x-ray generating tube.
It left the cathode and it's moving across and crashing into the anode.
So this is the target or the anode.
So let's label it as such.
So we're looking at one of these mixed metaphors, where we've got both a Cartesian image here of the electron and we've got an energy image.
So this is the target or it's the anode and it's charged positively and so the electrons are crashing into it.
And we reason that what would happen is, if the electrons, the ballistic electrons, had high enough energy, they could actually dislodge 1s electrons and when they do so, they make the conditions for a cascade.
So you can see electrons falling from n equals 2 to n equals 1 and when they do so, they emit radiation.
They emit photons and these photons are called K alpha.
K because the final shell number was n equals 1.
And remember, the spectroscopists prefer to use letters.
So when the electron ends at n equals 1, that's n final, we call it a K photon.
And furthermore, the subscript alpha means that the delta, the n initial to n final is only one.
So when you go from 2 to 1, you get the K alpha.
And less likely but still possible is the transition from 3 down to 1.
So it's still called a K photon because the electron that generated the photon ended in the K shell, but it traveled from a delta n of 2, went from 3 to 1 so that's a K beta.
Over here, obviously if we have enough energy to kick out K shell electrons, we have enough energy to kick out L shell electrons.
And so if we lose an L shell electron and we have a cascade, then anything that ends in n equals 2 will be called an L photon.
If it comes from 3 to 2, that's a delta of 1.
That's the L alpha.
It it falls 4 to 2, that's a delta of 2.
That's the L beta.
And so you can see you get a set of lines.
And that set of lines looks like this.
We can plot a spectrum of this entire set of lines, which will be characteristic of the identity of the target.
And I think I showed you last day that it's going to look like this, where we will plot intensity and the intensity on a spectrum is related to the frequency of occurrence and on the abscissa, we're going to have some kind of energy coordinate.
We can put lambda increasing from left to right, which means energy increases from right to left.
And we saw that we have this family of lines where we have a discrete line at the energy of K alpha and a second line at the energy associated with K beta and K beta has a higher energy, therefore, a lower wavelength, and the relative heights related to the relative frequency, again, not to scale, but still the general relationship is correct.
The likelihood of falling from 2 to 1 is higher than falling from 3 to 1 and so you'd expect to have the intensity of the K alpha line greater than the intensity of the K beta line and the L lines have to be of lower energy because transitions 3 to 2 are less energy than 2 to 1 so we expect those lines to be out here and again, a relative frequency where the L alpha has a slightly lower energy and L beta has a slightly lower frequency of a current.
So that's the spectrum and this we call characteristic.
Characteristic of what?
Of the identity of the anode or the target.
This is generating the x-rays.
So if I want x-rays of this value of wavelength, I choose the target appropriately.
That's why we have the relationship here.
And then the other thing is that-- it's obvious, but I just want to make sure that we put it up here-- it's quantized because we're looking at the photons coming at discrete values of energy.
So that's what we speculated on.
Are there any data?
The answer is yes and the data come from a young man by the name of Henry Moseley.
He was doing his PhD up in Manchester.
He was is working for Rutherford.
Rutherford was his PhD thesis supervisor and 1913, 1914, he was making systematic measurements.
He was conducting a systematic study of the characteristic spectra of no fewer than 38 elements of the Periodic Table.
And what did we learn from his measurements?
Learned from his measurements that there's was a pattern here and here's what the pattern is that Moseley found.
He found that if he took the value of-- I'm going to just take one line.
If he took all of the K alpha lines from 1K alpha per element and he plotted the value of the wave number of a particular line, he found that the value of the wave number of a particular line scaled with the identity of the element by the square of the proton number.
So, for example, we would have, say here we started with aluminum.
So aluminum proton numbers 13 so 13 squared.
And he went all the way up to gold.
Didn't do all of them but he did, as I say, 38 from one to the other.
And these are discrete values.
You have different elements here and gold is up here and he found that he could take the set of data for, say, all the K alphas and they lie on a line-- new bar-- as if proportional to the square of the proton number and he did the same with other lines-- say, L alpha and so on-- found that they lie on a line.
So what does all of this mean?
Well, let's take a look.
Here's the image of the paper.
High-frequency spectra elements.
Henry Moseley, Master's degree, so he's working on his PhD, and the data were taken as follows in photographic plates.
And so by trigonometry, you can figure out what the wavelength would be of the line that would go to this degree of distortion, et cetera, et cetera.
It's a beautiful set of lines.
Look at how these date are posed.
So there's calcium.
Scandium is frightfully expensive so you don't see it here.
Then there's titanium, vanadium, chromium, manganese, and as you change the element, the wave number, the wavelength, the energy of all of these lines changes systematically in accordance with the square of the proton number, and we got over here to copper.
Next one is zinc.
Zinc melts at 420 degrees Celsius, which is relatively low melting.
And under the bombardment of electrons-- the bombardment of zinc would cause it to melt and so rather than work with zinc, he instead worked with brass because brass is an alloy of copper and zinc-- and look carefully.
You can see that brass has four lines, two lines identical to the lines of copper and these two new lines are the lines associated with zinc.
And I mentioned last day that you could actually deconvolve the complex spectra and identify the constituent elements.
So this was really fantastic, really fantastic.
Si what did we learn from all of this?
What did we learn by these data?
well, first of all, corrected Mendeleyev What do I mean by that?
Not here to say bad things about Mendeleyev, but Mendeleyev had told us that periodicity is a function of the atomic mass.
That's what Mendeleyev said.
Periodicity is a function of atomic mass.
And now Moseley says no.
Moseley says that periodicity is a function of proton number.
And you can see here-- I wanted to show you the-- this is an image taken from his paper.
He worked for Rutherford.
These people were brilliant experimentalists and so here he's got-- the cathode is up here at the top.
You see cathode of x-ray tube and the feed through and so on, so the anode is down here and it's connected and so on, and rather than take the x-ray tube apart in order to change the target, he went to the toy store and he bought-- this is the train from the toy store and he's got different elements seated next to each other on the flat car of the train and he's got feedthroughs with silk fishing line so that after he's done the experiment with one element, he can pull the train car over and change the anode without having to take the apparatus apart.
These guys were very, very good experimentalists.
So here's from his paper.
The author intends first to make a general survey of the principal types of high-frequency radiation and then to examine the spectra of a few elements in greater detail with greater accuracy.
The results already obtained show that such data have an important bearing on the question of the internal structure of the atom and strongly support the views of Rutherford and Bohr.
It's 1913.
Remember, that's when Bohr paper came out.
All these people were working in the lab at the same time, supporting each other.
You see, this doesn't make any sense with a Plum Pudding Model, does it?
See, he continues: "We have here a proof that there isn't the atom of fundamental quantity, which increases by regular steps as we pass from one element to the next.
This quantity can only be the charge on the central positive nucleus on the existence of which we already have definite proof."
See, not only-- remember, the Plum Pudding Model has this big blob of positive charge nests.
There are no protons.
In the nuclear model, we have a nucleus that has positive charge and he's saying, I know there's positive charge, and it increases as you go from one element to the next.
We are therefore led by experiment to view that N-- we use the letter Z today.
He's using capital N. N is the same as the number of the place occupied by the element in the periodic system.
This atomic number-- for the first time, the term atomic number is used.
That's why I was being coy here and I kept saying proton number, because that's the way they knew it.
Now he's say, this is the social security number of the element.
Proton number is then for hydrogen 1, helium 2, lithium 3, calcium 20, zinc 30, et cetera.
We can confidently predict-- look at-- this is the masters student and he's going on a limb.
He says: "We can confidently predict that in the few cases in which the order of the atomic weights A clashes with the chemical order of the periodic system, the chemical properties are governed by N, while A itself is probably a complicated function of N." He's right.
You look at the Periodic Table.
You say, yeah, yeah, they just go in descending mass or ascending proton number.
Look carefully.
Potassium has a lower mass than argon.
And nobody's to-- well, if you were Mendeleyev, you'd say, go back and measure it again.
Well, they have measured it.
These are the accurate values and no one's going to put potassium underneath neon, but it comes next in the order of ascending atomic mass.
Cobalt and nickel-- nickel actually weighs less than cobalt.
But they're transition elements so who cares if you get those two mixed up?
You find them in stainless steel.
It doesn't matter.
This is a good one.
Iodine is lower mass than tellurium.
Iodine is a halogen.
It belongs under fluorine, chlorine and bromine.
You're not going to put it under oxygen, sulfur, selenium.
But look, there's the data.
And then last one that Moseley couldn't have known about is transuranic synthetic element, but that's just the fourth case in the periodic table.
So that gets you the last wedge in Trivial Pursuit.
What are the four pairs of elements that are out of sequence?
OK.
So now we can say that proton number is the atomic number.
It's the identity.
So that comes out of Moseley, comes out of his work.
The second thing that he did as a result of his study of 38 elements is he figured out what to do with the lanthanides.
Remember our friends, the lanthanides.
Place the lanthanides correctly in the Periodic Table.
Many of the lanthanides have a valence 3 so people were struggling.
They were putting them underneath aluminum.
They didn't know what to do with them.
He placed the lanthanides correctly in the Periodic Table.
And it's not as though these were brand new.
Lanthanum itself had been first isolated in 1839, and all through the 1800s, they were picking them up and finally, lutetium was the last one discovered, categorized in 1907.
But people didn't know what to do with them and remember, they were obsessed with atomic mass measurements.
So I'm going to give you this.
I'll give you this piece of information.
There's the atomic mass of lanthanum, and the atomic mass of lutetium is 174.97.
So what?
I don't know what to do with that information.
That doesn't help me at all.
But now comes Moseley and he says, we're not talking about atomic mass.
We're talking about atomic number.
So now I tell you this atomic number is 57.
So where do you put it?
Duh, you put it right next to barium.
There's no debate.
And furthermore, this one is 71-- atomic number.
Well, if this is 71 and this is 57, I can tell you with impunity-- how many lanthanides are there?
There's 14 of them from here to there.
OK. 14 elements, which makes sense because in s we've got one orbital, in p we've got three orbitals, and d we've got 5 orbitals and these are f and there's 7 orbitals.
7 times 2 is 14.
Everything makes sense.
And the last thing that's a corollary to the item two-- we were able to give uranium its proper atomic number is 92.
So now I ask you, how many elements are there up to uranium starting with hydrogen?
92.
All of this comes as the result of Moseley's experiments.
But it's not over yet because he worked for Rutherford, and Rutherford pushed his people really hard.
He wasn't abusive, but he brought out the best in them.
So who else was in the building?
Bohr, and what were they doing?
They were doing theory and so he said, well, that's nice, but he says, I want those things fit to an equation.
So Moseley said, all right.
I'm going to use-- there's already an equation in the building for a new bar.
It's the Rydberg equation.
So I'm going to use a Rydberg-like equation.
So this is what he does.
This is Moseley's fit.
He goes the wave number of whatever the line is-- whether it's K alpha, L alpha, what have you-- is going to go as the product of the Rydberg constant-- 1/NF squared minus 1/NI squared.
So far that's just the Rydberg equation.
But now comes Moseley's contribution.
We're going to put z squared, but one more piece.
Notice the way I've drawn those lines.
They don't go through the origin, do they?
There's an offset.
So he said, let's allow for the offset-- z minus sigma quantity square.
And this is known as Moseley's Law.
And you've got values for sigma.
sigma for K alpha equals 1 and for L alpha equals 7.4.
So now I can-- with impunity, you give me the wavelength.
I can get the wave number-- it's 1 over the wavelength and I can plug into this equation and identify the element or turn it around if I want to get wavelength radiation of, say, 1.25 angstroms, I can plug into this equation and come up with the z, which will tell me what the target element choice should be.
So let's go and plug this in because we're only going to look at these two.
So we can say that the wave number for K alpha-- it's always going to be 1/2 squared minus 1/1 squared, which is always going to come up 3/4.
So it's 3/4 times Rydberg z minus 1 squared.
So this is for the 2:1 or, if you like, L to K transition and then the other one of interest here is nu bar of L alpha.
nu bar of L alpha's going to be 1/3 squared minus 1/2 squared, which becomes 536 times the Rydberg constant z minus 7.4 squared and this is for the transition 3:2 or KLM, M to L. OK.
So there it is.
And now, what's the significance of all of this?
I try to give a physical significance of this sigma, which is known as the screening factor.
We're going to call sigma here the screening factor.
You'll see why in a second.
Why are we calling it the screening factor?
So let's consider what's happening in the case of the K alpha lines.
Let's look at K alpha generation.
So not to scale.
Let's draw-- here's the nucleus with all of its z positive charges and then I'm going to draw the K shell, and its got two electrons in it, and I'm going to show there's a hole here.
Because without this hole, there's no reason for the cascade.
And then I'm going to draw the L shell and it's got-- what?
There's 2 from the s, 2s, and then 6 from the 2p.
At most.
I know this is a terrible model.
It violates all kinds of things, but it's as complex as it needs to be for the explanation I'm about to give and I don't want to load you down with a whole bunch of extraneous information.
So consider the electrons in the L shell.
They see the electron vacancy in the K shell and they're going to fall down.
What's the Coulombic pull of the nucleus on the L shell electrons?
Can you see that it's not z plus, but at z plus screened by 1 minus.
So the nuclear charge is mediated-- in other words, it's reduced by 1 thanks to the presence of the one negative charge here.
So these electrons in L shell feel z less 1 and hence, we get z minus 1 as screening factor.
It's plausible.
It's at least physically consistent.
Now let's look at the next one.
That's the transition from M to L. OK.
So we get to M. We've got 18 electrons, up to maximum of 18, and if they're going to fall down to n equals 2, there needs to be at least one vacancy here.
So now let's use the same logic.
So an electron in the m shell sees the positive charge of the nucleus mediated by the electrons between the shell n equals 3 and the nucleus.
Now, I don't know how many electrons there are.
What's the extreme case here?
Two, four, six, seven, maybe eight, nine.
Because this would still give me the conditions to generate the transition 3 to 2.
I need a vacancy in two.
I don't need a vacancy in one.
So this is the maximum.
so that would be two, four, six, eight, nine.
So it could be z minus nine, but that would mean that all of these electrons are on the same side and I don't have any vacancies lower.
I only have one vacancy here.
That's an extreme.
It's not observed.
And the other extreme is, we blow away all the electrons.
So somewhere in between 9 and 0, it turns out it's 7.4.
I can't predict that it's 7.4, but 7.4 makes sense.
If the number were greater than 9, I would be distressed.
So it makes sense, rationalizes.
OK. And by the way, if you use this formula, Moseley's Law-- let's bring that back down.
Remember, I told you I can still wake up in the middle of the night and quote you the wavelength of copper K alpha radiation of five significant figures.
I had it drilled into me in my junior year.
It's 1.5418 angstroms.
That's the lambda.
So you can use this formula for new bar and you know that new bar is equal to 1 over lambda.
So I can use Moseley's Law.
But true value, lambda-- I love this one.
Watch.
I'm going to do a triple subscript.
Lambda of copper K alpha.
Isn't that cool?
Lambda of copper K alpha is equal to 1.5418 angstroms.
And if you use Moseley's law, you get 1.546 and the delta here is 1/3 of 1%.
This man was a positive genius.
He was heading pell-mell for the Nobel Prize, but he never got it.
Why not?
World War I broke out in 1914 and Moseley was passionately concerned about World War I.
He wanted to fight.
He wanted to fight for the Allied cause and he enlisted in the Army.
Rutherford was furious.
Rutherford called the Minister of War, which is analogous to the Secretary of Defense and said, give him a desk job.
Put him up in Oxford at a military laboratory.
And Moseley said, no, I refuse.
I'm going to fight.
And so he was sent to Gallipoli.
Gallipoli, as you may know, was one of the bloodiest sites of World War I.
A quarter of a million Allied troops and 1/3 sort of a million Turkish troops were killed at Gallipoli.
Almost 2/3 of a million people died for that little piece of land.
And on August 10th, 1915, at the age of 27, Henry Moseley was killed in the battle of Suvla Bay, and they don't give Nobel Prizes posthumously, and so that was the way it ended.
There's a shot of Henry Moseley with one of his books.
The tributes poured in from all over the world.
The physics community was devastated because they knew all of this stuff.
This is the one that really, I think, says it best.
This was written by Robert Milliken, an American.
He's the one that gave us the elementary charge of the electron, from the University of Chicago at the time.
This is what Milliken wrote-- wrote this to Rutherford to read.
"In a research which is destined to rank as one of the dozen most brilliant in conception, skillful in execution, and illuminating in results, a young man 26 years old threw open the windows through which we can glimpse the subatomic world with a definiteness and a certainty never dreamed before.
Had the European War no other result than the snuffing out of this young life, that alone would make it one of the most hideous and most irreparable crimes in history."
That's when American scientists knew to write.
It's beautiful writing.
What a tribute.
OK.
So let's move on.
Now that we've got Moseley's Law, we've straightened out the Periodic Table.
So we keep moving.
And I say, well, I gave you the drawing of the spectrum.
Remember the spectrum?
The spectrum was here.
You see it?
I'll put it up real quick again.
This was the spectrum.
OK.
This is intensity and this is wavelength and this is K alpha and this is K beta and this is L alpha and this is L beta.
These are the data coming from the x-ray tube.
This happens to be a molybdenum target.
Well, it doesn't quite look like what I've drawn, does it?
It looks a little bit like it, but not quite.
This is definitely there.
You can see some of these lines so I'm going to call this spectrum A and it looks like spectrum A has been added on top of something else.
I'm going to call the something else spectrum B and the spectrum B looks like this.
And in New England-- this curve has a shape because it's New England.
This is called whale-shaped.
I don't know what they call it in the rest of the country, but it even looks like a whale.
So you can go on a whale watch.
So it starts-- it's a sharp front, goes straight up, hits the maximum and then-- are you ready for this?
It tails off, all right?
Whales have tails, yes.
So what's the difference here?
Well, the spectrum A, we already observed is quantized, whereas this one isn't.
This one's continuous.
It has continuous values up to this minimum value or, if you like, maximum value of energy, minimum value of wavelengths.
So we got that.
It's not quantized.
And the other thing that's interesting is-- spectrum A, we said, is a function of the identity of the target.
z of the target, whereas this one, it's a function of the plate voltage.
So you can see in the diagram I've shown you, as the plate voltage changes, we get a series of enveloping curves.
So this is V1 and this is V2-- greater than V1.
So this whale-shaped thing is somehow related to plate voltage.
And then beyond a certain critical value, can you see when you're down here at 15 or 20,000 volts, all you got is the whale-shaped curve?
But somewhere between 20 and 25,000 volts, you hit a threshold and now you switch on the characteristic lines.
So with low voltage, you don't get the characteristic lines.
You only get the whale-shaped curve.
And then beyond a certain critical value of V, it's as though all of a sudden these lines appear.
Low voltage, no lines.
High voltage, you get the characteristic lines.
And what can we do here?
Well, we can compute K alpha, L alpha by Moseley's Law.
The continuous spectrum can't do anything with that, with one exception-- here.
So let's look inside and figure out what's going on.
We have a proposal here of what's going on inside that target atom.
So I'm going to make-- this could be the molybdenum target up there.
So these are molybdenum atoms.
And just as in Moseley's figure-- so this is body centered cubic-- I can look that up on the Periodic Table.
I know this is BCC crystal structure, molybdenum atom sitting here.
This is the anode.
It's charged positive and way up top, I get the cathode.
So the cathode is shooting off ballistic electrons and they go zooming across the gap and crash into anode, but up until now, we've just said the anode is some material.
Now we're going to take one more peel off the onion and say that these are discrete atoms.
It's not continuous molybdenum.
It's molybdenum atoms.
And what do we know the molybdenum atom looks like?
It's got a dense nucleus where all the positive charge resides and then there's this almost vacuum-like zone with the negative charge.
So when the electron comes in, this electron sees the negative charge around the molybdenum atom.
What happens?
It's deflected.
Maybe it comes in on a closer angle and it's scattered through a higher angle.
Maybe it comes in almost in between and it hardly moves at all.
Now can you see that when you have a charged species that changes direction, that's called an acceleration, and an acceleration gives rise to an emission of radiation?
So because the angle of deflection-- so this is low-angle deflection, this is high-angle deflection.
So low-angle deflection means low energy emission.
High angle means high-energy emission, and the result is this continue spectrum that I've shown you here.
So this is the result of low-angle scattering of the ballistic electrons.
This is the result of high-angle scattering of the ballistic electrons, and somewhere in the middle here is the dominant angle.
I can't calculate this curve with one exception.
Imagine the electron comes from the cathode, and with all of its energy is dead on and stops, gives up all of its kinetic energy to a photon.
That's the maximum amount of energy possible.
Let's look at that.
That's the case where an electron comes in, stops dead here and then emits the photon.
Sp the kinetic energy is translated into the photon energy.
So we can do that one.
That's a straightforward calculation.
So the energy of the incident electron-- E of the incident electron is equal to product of the charge on the electron and the plate voltage.
Well, the charge on the electron is the elementary charge and the plate voltage is whatever it is, and I'm going to equate that with the energy of the emitted photon, and that's equal to hc over lambda.
So now I can cross multiply and I can call this the lambda of the shortest wavelength.
That's the shortest wavelength on the whale-shaped curve.
So by algebra, you're going to get the product of the plane constant times the speed of light divided by the elementary charge times the plate voltage, which turns out to be 12,400 divided by plate voltage where the wavelength's given in angstroms.
Just try it.
If you put 10,000 volts, you're going to get 1.24 angstroms, which is smack dab in the middle of the x-region of the spectrum, and this is called the Duane-Hunt Law.
That's the only thing we can compute in the continuous spectrum.
So I can get this one.
Lambda shortest wavelength, because shortest wavelength is maximum energy.
Now there's a fancier name for this whale-shaped curve, and it's the scientific community's term, and it's a German word.
It's called bremsstrahlung.
I love it.
You need to know this.
You can impress your friends at parties.
What does bremsstrahlung mean?
Brems is the German word for brake, as when you put on the brakes of a car.
And strahl, strahl is the word for ray.
And ung is like ing.
So this is raying radiation and this is the radiation of braking.
So the electrons are coming in and when they come up against the negative charge of the outer shell the target atom, they are slamming on the brakes and skidding everywhere.
So this is what bremsstrahlung means: braking radiation.
So we're going to call-- I'm going to put a B here-- see that?
I was thinking ahead.
B is bremsstrahlung.
OK.
So we've got a lot going here.
for us.
I think we've explained a fair bit.
Now I want to talk about modern x-ray tubes.
What do modern x-ray tubes have that these primitive ones that Moseley worked with and Roentgen worked with didn't have?
So I want to show you that the modern x-ray tube is the result of improvements made by an MIT alum.
His name was William Coolidge, and he's the class of '96-- 1896.
And he made a number of improvements.
He actually taught for awhile then eventually spent a good part of his professional career working as a research scientist at the General Electric Labs out in Schenectady.
If you go down to the lobby of Building 6, on the south side of the lobby, there's a showcase.
Look inside the showcase.
You'll see there's a little display in honor of Coolidge.
So what's the first thing Coolidge did?
He was an engineer so he was thinking about making things more efficient.
First thing he did is he turned the discharge tube into a vacuum tube.
Remember, Roentgen worked at low pressure, but there was still gas, and he was blinded by the light and so on.
This means you don't get any visible light, and secondly, it's more efficient because if you've got the tube like this with the two electrodes, the feedthroughs and you've got gas inside, some of the electrons are crashing into the gas molecules, and we don't care about the gas molecules.
We want to get the electrons crashing into the target.
So by going to vacuum, this improves the efficiency.
No glow in the visible and higher energy efficiency.
More x-rays out per unit power put in.
What's the second thing he did?
Second think he did was hot cathode.
Remember, you're trying to rip the electrons out of the cathode and send them on their journey.
So Coolidge reasoned that if you heated the cathode, you'd weaken the bonds, and the electrons would come off.
They'd boil off much more readily, OK?
Raise temperature to reduce bond energy, to reduce binding energy of the electrons.
The electrons in the cathode makes them easier to boil off.
Think I've got a cartoon of that.
Yeah, here it is.
So here's the tube lying on its side.
So the cathode is over here to the left.
It's negative.
Here's the anode to the right.
That's this purple thing here and the electrons are moving from left to right and so he's got-- see, you can have multiple electrical signals going through the same conductor.
It's not the same.
You can't have an AC waveform and a DC waveform in the same conductor.
So you've got a big DC voltage between the cathode and the anode and you'd got a separate little circuit going through the cathode, running almost like a toaster to make it super hot.
So the potential between here and here is 35,000 volts.
The potential along this stretch of real estate might be several hundred volts, and furthermore, this could be an AC signal.
It's Coolidge.
He was smart.
So this makes this hot and now for per given voltage, you get much more yield of the electrons.
So that was pretty good.
I like that.
Smart.
Third thing he did was he heated the cathode and he cooled the anode.
Water-cooled anode.
Why?
You got to dissipate the heat.
All these electrons crashing into the anode.
You raise the temperature of that anode so high you'll melt it and so they had to run the tubes intermittently.
They just get a decent signal.
You have to take these x-ray measurements over a long period of time because the amount of x-rays you get is small, but you have to keep shutting the thing down.
Otherwise, you melt through the anode.
So he put that on a water-cooled copper hearth and was able to run continuous.
Continuous operation.
No more pulse current.
But sometimes when nature hands you a lemon, you make lemonade.
So I'm going to turn this thing around.
I'll say, hey, wait a minute.
I don't want any water-cooled copper anode.
Suppose I want to weld some titanium.
Titanium melts at 1675 degrees Centigrade and it's got a voracious appetite for oxygen.
Well, this thing's a vacuum tube.
So what if I were to take my part and I put the two pieces of titanium in the path of an electron beam and I don't cool the titanium?
Eventually, I get the temperature so high that I can weld titanium.
This is the birth of electron beam welding.
And that's how you weld refracting metals.
You get one of those titanium bicycle frames that cost about $4,000.
It might be tungsten inert gas.
It might be electron beam, depending.
So that's the flip side of the technology.
By the way, if you were in attendance observing electronic beam welding, what do you think is being generated in the electron beam welding apparatus?
X-rays, by the boatload So there's a fourth thing-- maybe the most important thing that Coolidge came up with-- shielding.
You see the yellow here?
That's lead shielding.
Why did he choose lead?
Well, it's got a very high z.
That means it's got many energy levels.
So that if you start looking at the energy level diagram of lead, you've got lots of action up in here.
So what can happen?
When the x-rays come with their high energy, the x-rays from the tube on their way out to get you, and everybody standing observing this marvel.
They excite electrons inside.
So these x-rays are absorbed.
The electrons inside the lead rise, cascade down and now they come out.
And what's the difference in energy between the x-ray and these?
The ones up here are much lower energy.
So this is, if you like, a frequency shifter or an energy shifter.
So now we're using photons to excite electrons to generate photons.
And that's how the shielding works.
So you want to absorb and re-emit.
And while we're on the topic, he gave us lead shielding and he gave us beryllium windows because they were using silicate glass, just as you use in your home, but the silicate glass was absorbing some of the x-rays.
So to get higher efficiency, he chose beryllium.
Why did he choose beryllium?
Well, it's got a low z.
So that means it has few energy levels.
So therefore, there's less absorption and re-emission-- again, higher efficiency.
I think this was taken from one of the other readings.
So they flipped it around.
See, here you can see the cathode here, the anode.
That's one of these things.
All right.
So the window is up here.
You don't use lithium because lithium is unstable in the air unless you had a giant room of humidity set at a dew point of minus 38 degrees, see, and then you can use a lithium window.
Otherwise, you use beryllium and lead down here.
Why'd he choose lead?
Well, you're pretty much at the bottom of the naturally occurring elements in the Periodic Table.
So it's the cheapest of all of these heavies.
All right.
So now let's take a few minutes and talk about the use of x-rays in characterizing art.
So this painting here at once time was arguably one of the most recognizable paintings in the western world.
It's The Angelus by Jean-Francois Millet.
It was painted at 1857 to 1859 on commission for an insurance company here in Boston.
The Boston Brahmins loved BA's paintings of rural life in 19th century France.
And this is couple of peasants who are giving thanks to God.
The Angelus is a prayer, and they're thanking God for this pitiful bounty of potatoes evident.
Salvador Dali, as an art student, was required to paint this as part of training.
You know how when you learn to play the piano, you reproduce the works of the Grand Masters.
So when you go to art school, you have to paint.
He hated this painting.
For many years, it was here in Boston, and then it was repatriated around World War I, it hung in the Louvre, and then ultimately, now it's in the Musee d'Orsay.
In 1963 when Dali was at the peak of his career, he asked the curator of the Louvre if she would have this painting x-rayed.
He said this painting spooks him.
He doesn't like this painting.
And they x-rayed the painting.
What did they find?
They found that it had been painted over down here.
Wasn't an art forgery, Millet himself had painted it over.
You may have heard two years ago there was this art find.
It was a van Gogh painting.
They found that van Gogh had painted over one of his own paintings that had never been seen before even though the pencil sketches survived, so they assumed that maybe the painting was destroyed in World War II or some such thing.
It turns out that van Gogh was so poor at one point that he valued the canvas over the art, and he took his previous painting, and he painted over it.
OK.
So this was going on here.
This is the truth.
Now my speculation begins.
So what was underneath here initially?
It was the casket of a baby.
Look at the pose.
These people don't look like they're happy with their bounty of harvest.
They're grieving.
This was this futility of peasant life.
It was a hard life and they lost their baby.
So imagine Millet paints this, puts it on a boat, it comes over to Boston, they unveil it, and the directors of the insurance company go, we can't hang this in the lobby of an insurance company.
That's my speculation.
It didn't take him three years to paint this thing.
I think it spent most of its time on the boat.
They sent it back and said fix it.
So he put the basket of potatoes.
That's my theory.
All right, look at the pose.
This is from the Galleria Borghese in Rome.
I was there about two years ago and saw this and went, wow, referential.
You see, everything's been said.
All of art has been said.
Now we're just recasting it.
So now here's Dali's revenge.
Now this is what he paints.
You see the man is shorter than the woman.
His hat is down a little bit below the waistline.
There's all sorts of psychosexual things going on here, but we're running out of time.
So here's Dali and his father and his dad is saying, see, this is life, et cetera, et cetera.
So you can see the reference.
See, man taller here, woman taller, et cetera, et cetera.
Have you seen this one?
The Hallucinogenic Toreador.
OK, he went outside one day in Manhattan to buy some pencils.
There was a company called the Venus Pencil Company, you see.
And so he came back and made this trompe l'oeil.
So this is the toreador.
Do you see the breast here is the nose?
There is the face.
There are the flies of the Thames and here is the cape and so on.
Symmetry plane-- all of those Venuses facing back, these Venuses facing forward.
There's the bull.
The life force of the bull is a crystal.
What is the crystal structure?
It's one of the 14 Bravais lattices.
If you ignore color, it's simple cubic, isn't it?
And if you don't believe me, check.
And if you don't appreciate it, study.
Here's the symmetry plane.
Atoms, atoms, fly, fly et cetera, et cetera.
What else do we have here?
Oh, this is the one he painted on the occasion of the revelation of the structure of DNA.
It's hard to see, but over here is the double helix of DNA.
This is his wife Gaia and here are people standing in cubic arrays with guns pointing at each other.
And the point that he's making is that now that we've discovered the instruction set for reproduction of life, not just human life, but life, we are still at a point where we can't resist killing each other.
So this was the point and he chose to use cubic arrays, and I would venture to say this is simple cubic.
If you put a fifth person here, it would be-- are you ready for this?
Body-centered cubic.
Oh, yes!
And there's more, but I think we're running out of time so I think at this point we will dismiss the class.
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OK, OK, Settle down.
Settle down.
It's 11:05.
It's time to learn.
All right.
Last day we were looking at the emissions spectrum of the target in the x-ray tube.
The target is the anode, and it's being a bombarded by electrons that have been accelerated across a potential difference of some tens of thousands of volts, and this was what we saw as the output, intensity versus wavelength.
We see this whale shaped curve which we have attributed to bremsstrahlung, which is the breaking radiation.
And then if the voltage is high enough, high enough to do what?
High enough to eject inner shell electrons, we will get the cascade, and the cascade will give rise to specific values of wavelength associated with transitions within the target atom.
There's too much talking.
I want it absolutely silent or else somebody is going to leave.
That's the only way it works here.
I talk.
You listen.
And if you have a problem with that, there's a door here.
There's a door there.
And there's a door at the back.
So let's get it straight right now.
It's the only way it works here.
Now what happens is that these characteristic lines are calculable in part.
The K alpha line, the L alpha line we saw from Moseley's law, which is up here on the slide.
and the leading edge of the whale shape curve, the bremsstrahlung is calculable by the Duane-Hunt Law.
The rest of this is not calculable.
So there it is.
What I want to do today is harness this radiation.
The reason we were studying x-rays in first place was that we said they had a length scale that was comparable to that of atomic dimensions.
And so now we want to close the circle here and probe atomic arrangement by x-ray diffraction.
And it goes by this three-letter initialization XRD.
And in order to use x-ray diffraction to probe atomic structure, we're going to make another model.
And this model is going to be simple.
And it's going to be inaccurate, but it's going to be good enough to explain the phenomena that we're going to study.
So the first thing I want to do in constructing a model that will allow to use x-ray diffraction, is to model the atoms as mirrors.
That's the first assumption that we make.
We model the atoms as mirrors.
And when we model the atoms as mirrors, this means that we will invoke the laws of specular reflection.
So if this table top is a plane of atoms, and I have a beam coming down at an angle, the laws of specular reflection say that the angle of incidence will equal the angle of reflection.
So that's all we're doing by modeling the atoms as mirrors.
Laws of speculative reflection apply.
So that simply means that the angle of incidence, theta incidence equals theta reflection.
And you'll see how that comes into play in a minute.
And then the second thing we're going to do in order to get intensities is to use the concept of constructive and destructive interference.
OK.
So we apply interference criteria.
And I will explain what that means in a second with a diagram.
But in dense text, it simply means that in phase rays-- it sounds like this rain in Spain falls mainly in the plain-- in-phase rays amplify.
And out-of-phase rays dampen.
So at some point, you'll study this in physics, but we need it right now, so I'll give you a little bit of foreshadowing, and then you'll be ahead of the game when you meet this in physics.
So what I'm going to do is I'm going to show you two rays, and they both have the same wavelength.
And I want to show you in phase.
And so I want to illustrate this concept here, in-phase rays amplify.
So let's give one ray here.
I'm going to depict it as a wave.
All right, so that's ray number 1.
And beneath it, is ray number 2.
It's got the same wavelength.
So that means crest-to-crest distance is the same, and if it's in phase, in phase means that crest lines up with crest, trough lines up with trough.
So I'm going to make another one seemingly identical.
It's supposed to be identical to the capacity of my ability to draw.
And so what this means is that these will amplify.
So 1 plus 2, 1 plus 2 will give me something that looks like this.
I'm going to line it up again.
Only now the amplitude is double the amplitude of a single ray.
That's what I'm trying to depict here.
OK.
So this is amplification.
Now over here, I want to do the same thing.
Only I'm going to have two rays, same wavelength, but out of phase.
OK, so in phase is here, out of phase is to the right.
So we'll have ray number 3, which does this.
And ray number 4, I'm going to line up with the crest of ray number 3, the trough of ray number 4, but it's going to be the same wavelength.
So the distance between successive crests is the same.
Only they're going to be offset by exactly half wavelength.
So the crest here of wave 3 lines up with the trough of wave 4.
The trough of wave 3 lines up with the crest of wave 4, and so on, and so on, and so on.
And if I drew this thing accurately, it would be obvious to you.
But since I can't draw very well, then we're going to take 3 plus 4.
And the sum of the crest and the trough is 0.
The sum of the trough and the crest is 0.
And so the sum here is 0.
All right this is total destruction.
This is cancellation.
So in the extreme, you can have destructive interference that dampens to the point of obliteration.
OK, so that's the two extremes, full amplification and complete cancellation.
So now, I want to take this concept of planes as mirrors and rules of interference criteria, and now what I want to do is illuminate.
And I want to use the full board here.
So here's what we're going to do.
We're going to take in phase, coherent-- that's what coherent means.
Coherent means that the radiation is in phase and monochromatic.
What's that mean?
Monochromatic means one color, which means single wavelength.
So I have something of a single wavelength in phase.
Coherent, monochromatic, incident radiation, so I've got a plurality of rays.
I have this thing flooded by a beam.
So what I'm going to do, is I'm going to draw some atoms here.
And I'll put a second batch on top.
And now I'm going to model these as mirrors.
So I'm going to have the incident beam come in like so.
The incident beam comes in like so at an angle theta.
So this is theta incident.
This is the angle of incidence.
And according to our model, the beam is reflected at the same angle.
So this is theta, reflection of theta.
Reflection equals theta incidence.
So that's the first thing that we do.
And then the second thing we do, is we use the laws of interference.
So I'm going to take a second beam, and it's going to go down to this atom here.
So I'm going to bring it down like so.
And it's going to come in at the same angle.
And it's going to leave at the same angle.
And if you'll excuse the drawing, these lines should be parallel.
They should be coming in parallel, and they should be leaving parallel where these angles theta i and theta r are the same.
Now how do we invoke the interference criteria?
Well what I'm going to do is put a marker here.
I'll put a marker.
And I'll call the first one ray number 1, and the second one ray number 2.
And what I'm going to show is if I start from ray number 1-- maybe it's time for colored chalk-- so if I start from ray number 1, to show that it's in phase with ray number 2, I'm going to draw this little waveform, and they're both lined up.
So you can see crest lines up with crest.
Trough lines up with trough, and they both have the same wavelength.
So that makes the point coherent, monochromatic.
But now things get interesting, because you can see that ray number 1 travels a shorter distance than ray number 2, and we can put on some coordinates here and mark the geometry, so we can say that the point at which the ray number 2 has to start traveling a longer distance, I'll drop a normal down, and I'm going to call this point A.
This is point A.
The bottom here is point B.
And then up here where it catches up with ray number 1, the reflected ray number 1, I'm going to call this point C. So you can you can see that ray number 2 has to travel this extra distance, AB plus BC.
And now, if I want to get out to here where I'm now in the reflected zone, and I want ray number 1 reflected and ray number 2 reflected to continue to be in phase, there's a geometric constraint on the dimension of AB plus BC, isn't there?
That length, AB plus BC must be a whole number of wavelengths.
It must be an integer number of wavelengths of whatever this thing is, otherwise, these two will not being in phase.
So that's the interference criterion.
And n can equal 1, 2, 3, and so on, so this is called the order of reflection.
And then we can go through the trigonometry.
I'm not going to do it in class.
It's just a waste of time.
It'll bore everybody to tears.
But if you want to do that at home at some point when you've got nothing to do, you can go through.
You've got all the numbers here.
You know the value of lambda.
You know the distance AB, and you know what this distance is.
This distance between successive planes is your dhkl, isn't it?
So if you go through all of this analysis, you will show that n lambda is equal to 2 times the d of the hkl spacing times the sign of theta.
And this is called Bragg's Law.
And Bragg's Law governs the reflection of incident radiation by a crystal.
Now you'll notice I didn't put an apostrophe here.
Some people put the apostrophe here, because the man's name was Bragg.
Some people put the apostrophe here because, in fact, it was father and son.
And they both together won the Nobel Prize in 1915.
Now there have been people who lived to see, as Nobel Prize winners, one of their children win the Nobel Prize.
but this is the only time in history a father and son team together won the Nobel Prize for the same work.
And there are undoubtedly people sitting in this room who think the fact that the father and son could work together in physics for an extended period of time alone is deserving of the Nobel Prize.
Now in 3.091, I'm going to keep it simple, always choose first order reflection, always n equals 1 in Bragg's Law.
So therefore, we will write Bragg's Law as lambda equals 2d sin theta.
And to make the point, the d is specific to a particular set of planes.
So it's a d spacing of the hkl planes.
And it's the theta associated with the correct reflection of the hkl planes.
Now how does destructive interference come into play?
Destructive interference comes into play should there be a situation where I have a third ray.
I'm going to bring a third ray in also at the same angle of incidence.
And that ray is halfway between these two rays.
And what situation could that be?
That could be a situation in which the crystal structure has in a plane out from the board, an atom that sits halfway between these other two atoms.
And if that happens, you can see.
We can go through the derivation.
But there's going to be a path length difference that's exactly 1/2 wavelength different.
That's bad.
It's not just there's going to be some measure of reduction.
It's going to be total cancellation.
And so by going through the set of crystal structures and recognizing that when you have 1/2 wavelength difference, you get destructive interference.
And therefore, you will see no line.
The incident radiation will come in at this angle, and at the reflection angle for that particular plane, there will be nothing detected.
So you can say that you have a combination of selection rules that involve the integral of both the rules of interference and interaction with the crystal structure.
So let's generalize this and say if we take interference criteria plus the crystal structure, that is to say, the instant relationship of atom positions for a particular specimen, the combination of that will give rise to the set of expected reflections.
So only when you satisfy the Bragg criterion do you get reflection.
So you move the specimen.
And only at that special angle will you get the reflections, will you get constructive interference.
So this, in fact, is a fingerprint.
This is a fingerprint of the crystal.
In this case, we're not getting the chemical identity.
We're getting the structural identity.
So we can determine if something is BCC, FCC, and so on.
So for example in BCC, that's exactly the case that I just illustrated.
So in BCC we have atoms at the eight corners.
And we have an atom dead center.
There's a central atom.
And the central atom lies in 0 0 2 plane, doesn't it?
The base of this is 0 0 1.
And the top is an 0 0 1.
But the central atom is an 0 0 2.
And 0 0 2 is halfway between successive 0 0 1.
So can you see that light that comes in?
And it's going to be constructively reinforced.
Off of 0 0 1 is going to be destructively reinforced off of 0 0 2, and the result is that in BCC, no 0 0 1 reflection observed.
The same thing happens in face centered cubic, right?
What's the face of face centered cubic look like?
It looks like this, doesn't it?
So there's an 0 0 1.
Here's an 0 0 1.
What about this space centered atom?
Where is it?
That's an 0 0 2.
It's halfway between 0 0 1.
So at the angle where you would've expected, if you use the Bragg Law and calculate the angle at which for your particular crystal, because you know the d spacing.
You fix the wavelength of your x-radiation coming out of the generator.
At the angle where you would expect to see reflection off of 0 0 1, you will see nothing, because 0 0 2 is canceling 0 0 1.
So you don't have to worry about all this.
This has all been tabulated for you.
Somebody has gone through and done all of this.
Well, this is just making the point.
See, there's a simple cubic.
All the planes are reflecting.
There's body-centered cubic.
That's a/2.
There's face-centered cubic a/2.
0 0 2 is going to cancel 0 0 1.
You're not going to see anything there.
So people have gone through.
And they've made this set of rules for reflection.
So simple cubic, you get reflection from all the planes.
Body-centered cubic, you just get these planes here.
And it turns out that there's a simple rule that compactly represents which planes are going to reflect in body-centered cubic.
And in BCC, the hkl, it's the planes for which h plus k plus l is an even number.
h plus k plus l is even.
And 0 was counted as even.
So for example, 0 0 1 h plus k plus l is 1.
It doesn't reflect.
0 0 2, 0 plus 0 plus 2 is 2.
It does reflect.
And so on.
You just go down the whole line.
And so these are in ascending order of h squared plus k squared plus l squared, because that's a nice way of deciding how to add them up.
And now in FCC the selection rule is a little bit different.
In FCC you get reflection from planes when you write the HKL such that h, k, and l must be either all even numbers, or all odd numbers, or some people like to say unmixed.
Some crystallographers say unmixed, meaning you can't have a combination of even numbers and odd number.
So for example, 0 0 1 won't work because 0 is a zero.
That's even.
1 is odd.
But 0 0 2, h plus k plus l all even or all odd, there you go all even or all odd unmixed.
So this will work.
And then so on.
So you can go through and see which ones work.
All right.
So the next one here is in the sequence.
This sums to 1.
And what would sum to 2?
It would be 0 1 1.
And 0 1 1 doesn't work either because zero is even.
And 1 is odd.
The next one of the sequence is 1 1 1.
These are all odd.
So that one reflects.
So in FCC, you don't see 0 0 1, 0 1 1.
But 1 1 1 does reflect.
And then 0 0 2, which is the next one, because 0 squared plus 0 squared plus 2 squared is 4.
This is even even.
So that one works.
So there's the sequence.
And so now what we can do is use this technique in order to make measurements.
But before we do so, I want to show you the experimental measurement, one way.
There's several ways of conducting the measurement.
And so the first way I'm going to show you is called diffractometry.
And you can do this over in Building 13.
If you get yourself a UROP, you might be assigned to make some measurements.
So diffractometry is a form of x-ray diffraction.
It's one of the techniques.
And the way it works is you fix the wavelength and vary the angle of incidence, you don't rotate the x-ray generator.
You rotate specimen and present continually varying angles to the thing.
So this shows the technique in operation.
So you have a specimen sitting here.
The specimen can either be a thin film, or in other instances, we have a finely ground powder.
So each of the powder grains presents a different angle.
And coming out of the collimator is the monochromatic.
We want a single wavelength.
So this collimator means it's monochromatic and coherent.
So that beam comes and strikes the specimen.
And for historical reasons, instead of calling this the angle of incidence, and this the angle of reflection, they call the angle that goes to the detector 2 theta.
In other words, the projection of the beam, this is theta incident equals theta r. So this is really 2 theta incident, or it's 2 theta reflected.
It's the same thing.
So you'll see x-ray data usually reported in units of 2 theta.
And here's the detector.
And then all you do is you rotate the specimen.
And by rotating the specimen with a detector, you're able to get the entire set of reflections.
And whenever you move through an angle that satisfies the Bragg Law, you get a peak in intensity.
And here's what the output would look like.
So you're plotting intensity as a function of 2 theta.
So somewhere along here at around, it looks like about 38 degrees, you have satisfied the Bragg angle, and you get constructive interference.
When you move off of 38 degrees, destructive interference reigns supreme, and you get almost no reflection.
Then you get to the right angle for 2 0 0, you get reflection.
Well, all you get is these lines.
You don't get these numbers.
These numbers you don't get for free.
So here's the experiment that we're going to do.
We're going to run our x-ray generator with a copper target.
Why do I tell you the copper target?
Because you're going to fix the wavelength.
And you fix the wavelength by choosing the target.
So we've got copper target in our x-ray generator, OK?
So remember, the sample is not the target.
The sample is what is being irradiated.
This is being bombed by the electrons in the tube, copper target in x-ray tube, and that means that the lambda of the radiation is lambda copper.
And I'm going to use lambda copper K alpha because I know it's wavelength to five significant figures.
It's 1.5418 angstroms.
And here's the data set.
I looked this up in the literature.
Here's the data set from the experiment.
And that's all you get, a set of 2 thetas.
So here's my challenge to you.
I'm telling you you've got an unknown sample that's cubic, and there's your data set.
The task for you is determine two parts to the question.
And we're going to do it together.
First part, determine the crystal structure.
So it's either FCC, BCC, or simple cubic.
And the second part is determine the lattice constant, the value of the lattice constant.
So you can get quantitative measurements.
So I'm going to show you how to do it.
So how are we going to do it?
We're going to use this technique.
This is my self-help book for you.
And here's the key.
Here's how I came up with this.
There's a way to unravel this.
And the way you unravel this is to take these two relationships.
You have lambda equals 2d sin theta is 1.
And you also know that d-- in fact, I'm going to keep writing hkl subscripts here-- you know the dhkl is equal to a.
That's this lattice constant.
Be sure this is lattice constant.
This is the cube edge measurement lattice constant divided by the square root of h squared plus k squared plus l squared.
So what I do is I combine the two.
If you combine these two, you can end up with this relationship.
If you combine the two, you get lambda squared over 4a squared equals sin squared theta over h squared plus k squared plus l squared.
And this is the key.
Why is it the key?
The value of lambda is set by you.
You chose the target.
The value of a is set by the sample.
That's the lattice constant.
So the ratio of two constants is a constant.
Agreed?
So this is a constant.
So if the left side of the equation is a constant, the right side of the equation must be a constant.
But h, k, and l vary.
And theta varies.
But the ratio of the variation, when mapped into sin square theta over h squared plus k squared plus l squared, this must be a constant.
And that's going to be the way I work through this delightful problem.
So let's see.
Start with 2 theta values, and generate a set of sin squared theta values.
That's what Sadoway says first.
So I took 2 theta.
And all I did was make 1 theta, and then took the sin of it, and then took the square the sin.
So this is the data set transmogrified into sin squared theta.
All right, what's the next thing he says?
Normalize by dividing through with the first value.
So I'm just going to take this whole series and divide it by 0.143.
And now I end up, instead of 0.143, 0.191.
I have 1, 1.34, and so on.
That's what I mean by normalize.
Normalize to the first entry.
OK, What's next thing he says?
Clear fractions.
Clear fractions from the normalized column.
So I'm going to multiply this by a common number.
Because 1.34, remember this is experimental data.
It's noisy.
1.34 looks like 4/3.
Doesn't it?
It's fuzzy logic.
And 2.67 looks like 2 and 2/3, 3 and 2/3.
That's roughly 4.
5 and 1/3.
The 3 looks like a magic number.
So if a multiply this column by 3, I get 3, 4, 8, 11, 12.
What does it say next?
Speculate on the h, k, l values, that if expressed as h squared plus k squared plus l squared would generate the sequence of the clear fractions column.
And then that's going to take me to the selection rule.
So I say, well how do I get 3?
It's 1 plus 1 plus 1.
4 is 2 squared plus 0 squared plus zero squared.
8 is 2 squared plus 2 squared plus 0 squared.
So I'm generating this thing.
And now it's pretty obvious, right?
Because now I've got to use these.
And what do I see here?
Well, 1 1 1 are all odd.
2 0 0, all even.
2 2 0, all even.
Gee, this looks like it conforms to the selection rules for Bragg reflection from an FCC crystal And then to make sure that I'm on to something, what I do is this.
Compute for each theta the value of sin square theta over h squared plus k squared plus l squared on the basis of those assumed values.
What I'm doing is I'm saying if my hunch is right, whatever I choose for the theta and the assumed value h squared plus k squared plus l squared, that should be a constant that doesn't change.
So let's test it.
And sure enough, look.
0.0477, 0.0478, so it looks like I'm on to something.
And furthermore, I know what this is.
That 0.0477 is this.
This is 0.0477.
And I know my lambda.
That's 1.5418.
So now I can calculate my a.
So now I've determined that it's FCC.
Plus if I go ahead and I calculate the a value, I get 3.53 angstroms.
And if it's FCC and it's 3.53 angstroms, I bet it's nickel.
So that's how you index this stuff.
So there we are.
There are the selection rules.
Now here's a little trick.
Let me show you.
Because when I go to index this, the first thing I do is I go for low hanging fruit.
If you start looking for whether it's simple cubic, face-centered cubic, or body-centered cubic, let's look at the sum.
So I've got simple cubic, body-centered cubic, and face-centered cubic.
So the first plane on simple cubic, it's going to give me everything.
It's going to give me 1, 2, 3, 4, 5, 6.
And FCC, what does it give us?
If you start looking down there, it gives you 3, 4, 8, 11, 12.
Well, this is so different.
3, 4, 8, 11 is so different from 1, 2, 3, 4.
What does BCC give you?
It gives you 2, 4, 6, 8, 10, 12.
Can you see you have a problem here?
It's pretty easy to pull out FCC.
But look at these two.
Since you don't know, you're just normalizing.
You can't tell.
If I give you a number sequence that is in the order 2, 4, 6, 8, 10, how do you know that that couldn't be 1, 2, 3, 4, 5?
There's a hook here, though.
Look at the sequence of h plus h squared plus k squared plus l squared.
Do you notice that there's no combination of h squared plus k squared plus l squared that gives you 7?
You get 6.
And then the next one is 8.
But there is a sequence that gives you 14.
So if you have a seventh line, if the seventh line to the first line, the ratio of h squared plus k squared plus l squared for angle number 7 to h squared plus k squared plus l squared.
For angle number 1, if it's equal to 7-- and point of fact it's not 7:1, it's 14:2-- and if it's equal to 8 for the seventh angle-- then it must be simple cubic.
And now you've sorted it all out.
So you're an expert now.
Now you can do it.
OK, so you're going to get some practice on homework.
So you can index crystals.
You can determine a crystal structure.
All right, a couple of other things we can do.
There is a second technique.
And it's called Laue diffraction after von Laue, who got the Nobel Prize in 1914.
He beat the Braggs by one year for this technique.
So he shines.
In this case for Laue, it's a slightly different technique.
What Laue does, is he uses light x-rays.
That means variable lambda.
So how would you get a fixed value of lambda?
Well, you would pick off one of those lines, like maybe the K alpha line because it's a nice, singular line.
And repress the bremsstrahlung and so on.
But if I want a variable lambda, where do I go for variable lambda?
I repress the lines.
And I go for the bremsstrahlung.
So now I've got lines varying all across the x region of the spectrum.
So I use variable lambda.
And I fixed the angle of incidence, whereas with diffractometry, I use a single value of lambda and vary.
So here's the cartoon which shows what's going on.
I'll do it one more time.
So this is called camera obscura, which is darkened room.
That's all this means is dark room.
So let's say I've got the specimen on the back wall of this thing, and what I've got is white x-rays coming in.
So the white x-ray enters through the face.
I've got the plate here, and I've got the specimen sitting somewhere in between.
And what happens is I get a spot pattern.
I get a spot pattern.
And the spot pattern is imitative of the symmetry of atomic arrangement.
So I have to say a little bit about symmetry so we know what that means.
So let's take a look at symmetry.
So I'm going to talk a little bit about rotational symmetries.
So let's look at that.
So first one I'm going to look at is an 0 0 1 plane.
So if I just take a projection of an 0 0 1 plane, it looks like this.
Doesn't it?
It's just the cube edge, or the cube bottom.
So it's got a and a as the edge.
And now, the rotational axis, it's about a normal rotational axis.
A normal rotational axis.
So what I'm going to ask is how many degrees do I have to go through before I get this same image back.
You can see that you go 90 degrees, it'll come back.
If I stop at 45 degrees, it's going to look like this.
You're going to say I know he moved the specimen.
If I go 90 degrees, you can't tell it apart.
So we define fold.
The fold is equal to 360 divided by the basic angle of rotation.
So we would say that this plane here exhibits 4-fold symmetry, OK?
Let's do a second one.
The second one is 0 1 1.
So if I look at 0 1 1, 0 1 1 at this plane.
0 1 1 goes across the diagonal.
So I'm going to take the diagonal here and plot it like so.
So it's going to have an edge of a.
And it's going to have the diagonal, which is root 2 times a.
And if I put a rotational axis in the center of that, clearly if I go only 90 degrees, I'm going to tell that the thing is rotated.
I have to go 180 degrees before I can't tell that there's been any disturbance.
So this one here has 2-fold symmetry.
We'll do one more.
Because there's only three major ones that we have to deal with.
And that is going to be the 1 1 1 plane.
And the 1 1 1 plane.
And the 1 1 1 plane, I think I've got an image of it here.
You can see the 1 1 1 plane drawn in this cube.
So what's the face of 1 1 1 look like?
The face of 1 1 1 looks like this.
That's face of 1 1 1.
And if you go through this analysis, this is root 3a because it's a diagonal of a difference persuasion.
So this is 3-fold symmetry.
This has 3-fold rotational symmetry.
And that means if I take a crystal such as this, for example.
I've told you before, this is my silicon wafer, and this silicon wafer has been cut from a single crystal.
So I'm looking at the edge of an atomic plane.
The question is, which one is it?
The crystal didn't come with a label on the side.
I don't know what the crystal growth axis was.
I started with a single crystal, and I dipped it into molten silicon.
And just like rock mountain candy, I drop the heating coils and cause the liquid to solidify around the seed crystal.
And I grow this salami that's about 8 inches in diameter, about 2 meters long.
So now I go and I cut these things with the diamond wheel.
So I'm cutting them normal to the growth axis, but I still don't know what plane I'm looking at.
So if I use the Laue technique and put the crystal like so, bombard it with white x-rays, and I look at the spot pattern.
The spot pattern is going to give me one of those symmetries.
And if I get a 3-fold symmetry, I know I'm looking at 1 1 1.
I mean, it's not going to be some wacko plane.
It's going to be either face, edge, or diagonal.
And on the basis of the symmetry in the Laue pattern, I can tell which plane I'm looking at.
They use this stuff.
They use it to make the devices that are in your cell phone and your computer.
I'm not just telling you a story.
All right.
So let's look at some others.
We'll go back to the Escher prints.
Everything has symmetry.
OK, so what's this one?
See this?
What's the symmetry?
4-fold.
How about this one?
What do you think?
3-fold.
I even went into Photoshop just to show you how dedicated I am.
So I took this image.
And I told Photoshop to give me 120 degrees rotation.
And I got this.
You can even see the fold in the book there.
So I'm really doing it.
It's coming back.
It's 120 degrees rotation.
How about this one?
That's 3.
Now these are real Laue patterns.
This is 4-fold.
This is obviously 3-fold.
This one is hard to see.
But it turns out it's 2-fold.
The one axis is just a little bit longer than the other axis.
What about this one?
That's our simple cubic puppy.
Yes, 1-fold.
Translational symmetry without rotational symmetry.
You can make this by just moving them sideways.
But you have to go all the way around.
So that's 1-fold symmetry translation.
Look at this one?
What's going on here?
Is that one of the Bravais lattices?
This is called Penrose tile, again rotational symmetry without translational symmetry.
You can take a patch of this and move it rotationally.
But you can't take and cover a wall with it.
So that's rotational without translational.
See?
You can go around this way.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and reproduce this.
But you can't use this as a unit cell and keep it hopping laterally and tile a wall with it.
Here's another one, not a Bravais lattice.
All right.
So I told you at the beginning of this unit that there are ordered solids and disordered solids.
We've been focusing on ordered solids.
They have a unit cell.
They're periodic.
And we call them crystals.
And next week we're going to start looking at disordered solid.
So they have no building block, no long range order.
And we call them glasses.
For a long time, people thought that solis have to fall into one box or the other box.
And then the best thing that can happen in science is not, oh, yeah.
That's exactly what I expected.
It's, I wonder what that means.
And there was one of these moments in 1982.
In 1982, a man by the name of Danny Shechtman from Technion in Israel, was over here in the United States working at the National Institute of Standards and Technology, which used to be National Bureau of Standards down in Gaithersburg, Maryland.
And he was looking at a set of aluminum-manganese alloys.
So aluminum is a metal.
Manganese is a metal.
You can make a solid solution which we call an alloy.
And these are highly ordered.
And what he found in his Laue measurements, were rotational symmetries that are impossible in a crystal.
He was getting 5-fold symmetry.
You can't get 5-fold symmetry.
There is no set of planes here that will give you a 5-fold symmetry.
They lack translational symmetry.
They were called aperiodic.
And the popular name for them was quasicrystals.
So here's the Laue pattern of one of Danny Shechtman's aluminum-manganese specimens.
I think this is 25% manganese and aluminum if I'm not mistaken.
So let's count, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
You can't have 10 or 5.
Atoms don't work that way.
But he got it.
He got it, 5- fold, 72 degrees.
Now this is really, really exciting.
This is a shot of his lab book.
You're looking over his shoulder.
These are the specimen numbers.
And this is aluminum 25% manganese, April 8, 1982.
So it's just another day at work.
You could be doing this.
And he can't believe what he's getting.
He's wondering what's going on.
You can't contribute this to a calibration error on the instrument.
This thing is on unassailably 5-fold symmetry.
The joke is that this is not far from the Pentagon.
Only within shouting distance of the Pentagon would you discover 5-fold symmetry.
So there it is.
So where have we come with all of this?
We've come with the ability to start with some very simple ideas about x-ray generation.
And we've come to the point where we can characterize crystals, get quantitative measurements of their lattice constants, and by using Laue techniques, investigate symmetry.
OK. Let's adjourn.
We'll see you on Friday.
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I know I'm not Dr. Sadoway.
I'm one of Dr. Sadoway's younger colleagues so-- but class is beginning, so let's get started.
Dr. Sadoway, of course, would like you to know that he would very much love to be with you here today and he would probably like that very much more than what he's doing now, which is sitting in a secured area secured by the Secret Service about to meet the President.
So you can ask him all about it when he comes back.
Associated with that, I have one request for you.
Today, when you're leaving, some of you may be used to leaving that way.
Don't leave that way.
Leave any other way.
This way's fine.
In the back is fine.
Just don't leave that way.
OK.
So last time, we talked about x-rays defraction and Bragg's Law.
And x-ray defraction and Bragg's Law has a lot to do with perfect crystals.
So perfection, right?
We've been dealing with perfect crystal so far and today, we're going to be dealing with imperfections.
So defects.
And in the field of material science, the field that I work in along with Dr. Sadoway.
there's a saying due to one of the most famous material scientists that says that crystals are like people.
It's the defects that make them interesting.
So now you've heard that saying and you can roll your eyes from now on every time you hear it because it's repeated it very often.
So let's see.
What do we mean by defects?
Well, there's two types of defects, broadly speaking, that we're going to be talking about, and we can classify those into two categories.
One of them is chemical.
So far, we've been dealing with chemically perfect materials.
So ones that are made up of one element or ones that are made up in stoichiometric quantities of two elements, but you can also have chemical imperfections.
You can have impurities.
You can have alloying elements and we're also going to be talking about atomic arrangement.
And in the case of atomic arrangement, we're dealing with structure, right?
Here we're dealing with chemistry.
Here we're dealing with structure.
You know about crystal structure so you know what perfect crystals look like, but real materials are not perfect, neither in the chemical sense nor in the structural sense.
So perfect crystals don't really exist.
You have always some forms of imperfections and we'll go over those today.
So in the case of chemical imperfections, we have, broadly speaking, two types.
Good ones-- Dr. Sadoway labels them with a smiley face so I will do that, too.
And bad ones are a sad face.
So both are defects and whether they're good or bad depends on their utility.
Are they useful or are they detrimental?
If they are detrimental, we call them impurities.
And if they are good, we call them other things.
For example, dopants.
We studied dopants, right?
Dopants are a type of impurity, they're a kind of imperfection, but they're good.
We use dopants.
We use them in semiconductors.
Also, alloying elements, you can put in the elements yourself.
Those are examples of good chemical imperfections.
When it comes to atomic arrangements, we're going to spend a little bit of time talking about that in more detail in just a minute.
But broadly speaking, they are situations where you don't have perfect crystal in order.
You have disruptions and that perfect crystal in order-- either in the form of missing atoms or extra atoms or differently oriented unit cells of the crystal and so forth.
So we'll talk about that a little bit more in just a minute.
So one thing that I forgot to mention maybe is that we have a test coming up-- That is, I'm sorry.
Celebration of learning, second celebration of learning.
And those are your room assignments.
You don't have to necessarily write them down now, but you'll see them again.
So I just wanted you to see them.
A through Ha in 10-250, He through Sm, 26-100, So through Z, 4-270.
OK, so let's go into the taxonomy of defects.
So I mentioned that there are a number of different kinds of defects, and we can do better than to say simply, defects exist.
We can actually start to classify them.
So there are, broadly speaking, four types of defects, which we classify based on dimensionality.
So there's 0-dimensional defects and 0-dimensional defects are point defects or point defect clusters.
It's just like in math.
A point has 0 dimension, right?
1-dimensional defects or line defects, things that thread through a material like a shoestring or something, and we'll have an example of that in the form of dislocations.
Then we have 2-dimensional defects.
Those are interfacial defects.
So if you have interfaces between two different kinds of materials or if you have two crystals that are misoriented with respect to each other, those are examples of interfacial defects.
And we also have bulk defects-- 3-dimensional defects like inclusions.
We'll go over some of those.
So let's start out with point defects.
Point defects are those 0-dimensional defects.
They're points, just like in math.
And there are a number of different point defects that we can look at.
We have our substitutional impurities, interstitial impurities.
The general overall recurring theme in these point defects, regardless what type they are, is that they are localized disruptions.
So a lattice, a crystalline lattice, is a regular arrangement of atoms, and a point defect is a very local disruption in that regular crystalline arrangement.
And those disruptions can occur on lattice sites.
So basically, on the positions where you would see atoms normally, they can also occur between lattice sites.
So they're localized rifts, you can say, in the periodicity of a crystalline material.
And here's just an illustration of the way that these different point defects play out.
So let's start out by looking at this one.
This is a substitutional impurity atom.
So it's an impurity.
So it's a type of chemical imperfection.
You can imagine that this is, for example, some kind of FCC material like copper, for instance, and suppose that you have an impurity of some sort like iron, and it sits there and it replace one of the copper atoms.
So this is called a substitutional impurity.
It substitutes for the regular atom that would have sat at that location.
Substitutional impurities.
There's another picture.
There's your substitutional impurity up there.
We'll get to the other ones in just a moment.
We already talked about some substitutional impurities.
So dopants.
If you have boron or phosphorus or silicon, that boron or that phosphorus replaces the silicon that would have sat at a certain lattice site.
So the dopants we've studied so far are types of substitutional impurities and we call them good impurities because they give us desirable properties.
They are dopants.
They're good impurities.
Then we have alloying elements.
I am not a very big fan of sodas, but if you were somebody else who is a big fan of sodas, then you might, for example, drink a lot of sodas out of aluminum cans.
Those aluminum cans are really not pure aluminum.
They're alloys.
They're alloys with other metals to give them the good properties that they need in order for the forming process to occur.
So if you ask yourself, how do you actually make aluminum cans?
They're stamped out from sheets of aluminum, but it's not just aluminum by itself.
Aluminum by itself would just tear if you do that.
So if you alloy it, you give it better properties so it's ductile.
You can deform it to very large strains and that's a good thing in the case of alloying elements.
Another one that he showed you was nickel and gold.
That's if you want to change the color of gold, if you want to be very creative when you're proposing.
So that's a good kind of substitutional impurity.
There are also contaminants.
Contaminants are bad kinds of impurities, but before we get to the contaminants, let me just show you what good impurities can do.
So this is the Hope Diamond.
It's in the American gem collection.
You can actually find out more about it.
I'm not a big gem expert, but anybody who looks at this Hope Diamond can immediately see that it's a really pretty diamond.
But you guys are ahead of everybody, because in addition to knowing that it's a pretty diamond, the fact that it has boron impurities in it already tells you what the majority charge carrier is.
So when you go to this gem collection, you can educate everybody else.
So those are all good impurities.
There are also bad impurities-- contaminants.
Lithium in sodium chloride is an example.
So if you were using saline solution, for instance, in an IV and you had lithium in there, that is very detrimental to the person who's receiving that.
That's definitely a contaminant.
You don't want those.
That could make you die.
So let's move on to some other types of point defects.
Here's another type of point defect.
So just now we talked about the substitutional impurity atom.
Now we're going to talk about the interstitial.
The interstitial's very different from the substitutional.
In the case of the substitutional, we have a lattice site, which instead of being occupied by the regular atom that would have occupied it in a perfect crystal, is occupied by a chemically distinct atom.
In the case of an interstitial impurity, that impurity can sit in the space in between lattice sites, so-called interstitial sites.
That's why we call them interstitial impurities.
And interstitial impurities or interstitial atoms can be both chemical impurities, and obviously, they are rifts in the atomic arrangement so they are structural impurities as well.
Why do I say chemical?
The reason is because if this is, for example, iron, if this is an iron matrix and you have a carbon atom sitting in the interstitial site which makes steel, then that's an example of a structural defect, but it's also a chemical defect.
If, for example, you have, on the other hand, iron sitting in a nuclear reactor and it's getting bombarded all the time by energetic neutrons, then interstitials are being created by atoms getting knocked out of their lattice sites and they're getting put into interstitial sites.
So that's creating defects that are chemically the same as the surrounding matrix material, but which are structurally distinct.
So those are interstitial atoms.
So here's another picture of interstitial atoms.
There it is.
It's not sitting on a regular lattice site.
It's sitting in between lattice sites.
It's sitting in the space, in the interstitial space between atoms.
And I already mentioned to you the fact that if you put carbon into iron, those atoms go into interstitial sites and there would give iron some of its beneficial properties, which we look for in steel.
So some of the good mechanical properties.
Here's another example of a situation where an atom goes into a lattice and create an interstitial impurity.
So lanthanum-nickel 5 is a prototype hydrogen storage material.
It takes up a huge amount of hydrogen.
It takes up a greater density of hydrogen than liquid hydrogen, actually, so if you expose this to hydrogen, the hydrogen just goes right in, and it sits in the lanthanum-nickel 5 lattice as an interstitial atom.
So this is considered to be as a prototype hydrogen storage material.
Unfortunately, it's extremely expensive so it's not being used very much these days, but on the other hand, it goes well with our gem theme in this particular lecture in terms of expensive things.
Here's another example of an interstitial impurity.
This one is not an alloying element.
It's a contaminant.
So hydrogen, but this time in iron.
So hydrogen in lanthanum-nickel 5 is good.
We want to store it.
Hydrogen in iron is bad because it actually degrades the mechanical properties of the iron, unlike carbon, which gives us steel, which is good.
You put hydrogen in, it embrittles it.
So hydrogen embrittlement in steels is a big problem.
And it's actually one of the challenges to a hydrogen economy.
If you have steel pipelines or valves or various pieces of machinery, structural components that are made out of iron that are constantly exposed to hydrogen, over time, they're going to brittle.
They're going to become very difficult to use.
It's one of the challenges in that whole undertaking.
OK.
So we're going fairly at a clip here through these taxonomy of point defects.
So in the taxonomy of point defects, perhaps the easiest defect to understand is the vacancy.
So when you think of the vacancy, think of the hole that we talked about in the case of semiconductors.
Vacancy is nothing.
It's a missing atom.
So if you have a crystalline lattice and it's FCC or it's BCC or it's simple cubic, whatever you like, and you know that there's supposed to be an atom at a certain lattice site and it's not there, that's a vacancy.
That's a situation where you have an unoccupied lattice site and there are different ways to form these vacancies.
They can be formed during crystallization.
If you heat up a material, the number of vacancies decreases.
So if you quench it really quickly, you can actually trap the vacancies before they can leave in service under extreme conditions.
I mentioned just a moment ago that interstitials can be created if you irradiate a material, if you bombard it with energetic particles, like neutrons for instance, you'll create interstitials, sure, by knocking atoms out of their atomic sites, but what's left behind after you knock that atom out?
Vacancies.
So actually, you create two defects at the same time: vacancies and interstitials.
So I think we probably have a picture here of a vacancy.
A vacancy is nothing.
It's just empty space.
That's a vacancy.
There you go again.
Nothing.
So we've gone through a bunch of taxonomy, right?
So we know now that there are a number of different kinds of point defects in crystals.
We've talked about interstitials.
We talked about vacancies.
We've talked about substitutionals.
What can we say about these defects quantitatively?
So let's take the vacancy and derive or write down what is the number of vacancies that you can expect to find in a given material at a certain temperatures?
So to do that, let's be a little bit more quantitative.
Suppose that you have a crystal.
I'm showing you a plane, for example, a 1 1 1 plane in an FCC material and how do you create a vacancy?
You simply remove an atom.
So you had an atom.
Now you have no atom.
You've created a vacancy.
When you created that vacancy, you broke all the bonds between the atom that used to be there and the neighboring atoms.
Breaking bonds costs energy.
So it costs energy to create a vacancy.
It costs energy to remove an atom from the place where it would have been because you're breaking bonds.
So in this case, I have six bonds that I broke.
If I were looking at some material, for example, placing our cubic material in 3D, I would find that I would be breaking 12 bonds to the 12 nearest neighbors, and so on for all the different crystal structures.
So we actually then take that information, the fact that we're breaking bonds, and encapsulate it in a single descriptions of how much energy it costs to create a vacancy.
And we call that the vacancy formation energy.
So this is an energy and so it's expressed in eV.
Its units are electron volts.
You can convert them to joules, anything you like.
And if you wanted to then compute how many vacancies there are in a given crystal, well, first of all, it costs energy to make them, so why would you ever even have a vacancy in the material?
Well, no material is perfect.
We know that from studying materials, but what causes it is the fact that at finite temperature because of the Boltzmann distribution that Dr. Sadoway told you about a few lectures ago, just like in the case of intrinsic charge carrier promotion in semiconductors, you can get thermal formation of vacancies.
So that Maxwell Boltzmann distribution can actually cause there to be vacancies despite the fact that there are-- that it costs energy to do that.
So how can we actually use that to express how many vacancies we have in a given material?
So I'm going to write down an expression for how many vacancies you can expect to find at a given temperature on the basis of their formation energy, OK?
So let's do that.
This is going to be the fraction of vacant sites.
So if you have a given material-- FCC, BCC, whatever you like-- you know that there's a certain number of lattice sites per unit volume, and you learn how to calculate those things.
And to quantify the number of vacancies, you have to basically say, what fraction of those sites does not contain an atom?
Is that 1/100 of a percent or is it 1% or how many?
So this is actually going to be expressed as a ratio, which I'm going to call this.
This is the number of vacancies per unit volume and this is the number of atomic sites, also per unit volume, OK.
So this is the definition of the fraction of vacant sites.
And how do we express it?
Well, we express it using a very simple formula.
This formula contains a factor here which is experimentally determined.
This is an empirical factor and then an exponential.
So the exponent we take here, the vacancy formation energy, and we divide it by the thermal energy at the given temperature of interest.
So this is actually telling you that to form vacancies, you actually have two competing factors.
On the one hand, you have the bonding energy that makes the difficult-to-form vacancies because you're breaking bonds, you're taking it an atom out.
On the other hand, you have this thermal energy, Boltzmann constant times the temperature, and the thermal energy is competing with that bonding energy.
And when that thermal energy is high enough, you can actually start knocking atoms out despite the fact that it costs you some energy, and obviously when this ratio is very large, that means that the bonding predominates, and when it gets smaller, that means the thermal energy is more and more sufficient to actually knock atoms out of their positions and cause there to be vacancies.
So when you do these calculations, make sure to use the absolute temperature in Kelvin, and dimensional analysis will tell you the units of the Boltzmann constant have to be energy units per degrees.
So eV's per Kelvin, for instance.
So this is the absolute temperature.
This is the vacancy formation energy and this is the Boltzmann constant.
So what that means is that at any given temperature, you'll actually expect to see some fraction of vacancies.
So let's actually try to do a calculation with an actual vacancy formation energy and see how many vacancies we get at a given temperature.
So to do that, we've been provided with what is the common currency in science, which is published experimental data, which we find from published journals which we find online through, for example, Web of Science.
And from this journal, we find that for copper, the vacancy formation energy is 1.03 electron volts.
Furthermore, in the same journal in the abstract there, you'll find the magnitude of this constant A, which as I told you is experimentally determined.
This quantity A we call the entropy factor.
And even though there's no way to very easily derive it-- I can't tell you what's the entropy factor for iron and you can't really tell me just by thinking about it-- nevertheless, it turns out that this factor usually fall within a certain range.
It's usually between 0.1 and 10.
That's usually the range and what you find As.
And in the case of copper, it's very much in that range.
It's just 1.1.
So let's use this information to actually figure out how many vacancies we expect to see in copper at the given temperature.
OK.
So here's my fraction of vacancies and we know that it is going to be expressed as 1.1 times exponent minus the formation energy, which is 1.03 eV, and then we'll put in Boltzmann's constant, and then let's pick a temperature.
For the moment, let's just take room temperature.
So T, room temperature, and room temperature's about 300 K. So when we put in all these numbers, it's just a matter of calculating.
This is on your sheet of constants.
This is the temperature which we choose at room temperature.
Let's actually write down T, room temperature, is about 300 K, and we find a certain number of vacancies.
And the number of vacancies that we find is 2.19 times 10 to the minus 18 vacancies.
So this is the fraction of vacant sites.
Well, is that a lot?
Is that a little?
How can we determine whether this is a lot or a little?
We compare it to the number of atoms that there actually are, and in the case of solid materials, we expect there to be something on the order-- this has to be compared to something on the order of Avogadro's number, but we can actually calculate more explicitly that this turns out to be something like 10 to the 5th vacancies per centimeter cubed.
And if in a real material, a solid material like silicon, for instance, or boron or whatever you're interested in, your number of atoms is something like 10 to the 23rd, Avogadro's number, right, then this is actually a very tiny number.
Compare 10 to the 5th to 10 to the 23rd.
The number of atoms that are missing at room temperature is very low in copper.
We can compute that using this expression.
However, even though it's low, it's not zero.
And if we continue going down in temperature, we'll find that the number is lower and lower and lower, but it never really goes to zero because we have this expression which gives us the total number.
So let's do another one and this time, let's take a different temperature.
Let's take the melting temperature of copper.
The melting temperature of copper is considerably higher.
It's about 1085 degrees Celsius and we can go through exactly the same calculation.
1.1 times all these quantities here times melting temperature, OK?
And when we do this, we get a vacant site fraction which is 1.67 times 10 to the minus 4 and that corresponds to 1.41 times 10 to the 19th vacancies per centimeter cubed.
So what do we know by-- what can we see now by comparing these two quantities?
Number of vacancies, add room temperature, number of vacancies at melting temperature.
How many orders of magnitude do they differ by?
Huge difference in the number of vacancies that we find at two different temperatures.
And why do we see such a huge difference?
Let's actually write down what that difference is.
Let's write down the ratio.
The fraction of vacant sites at melting temperature divided by fraction of vacant sites at room temperature is something like 7.6 times 10 to the 13th.
That's the difference in number of vacancies you get just by increasing the temperature from room temperature to melting temperature.
Why is that?
Well, one way to look at that is just from the expression that we have to calculate the number.
Here's where the temperatures go.
So any difference in temperature if it's a factor of two or if it's a factor of three or if it's a factor of four, it's going to go into the exponential.
So that exponential is actually making a huge difference as a function of temperature in terms of number of defects that you get.
The higher you go up in temperature, you don't get just-- you go up a factor of four in temperature, you don't just get a factor of four increase in the number of defects.
You get that in the exponential.
So the factor of four is hugely magnified by the exponential.
So that means that at any given temperature, even the lowest temperatures, you expect to see some defects, but if you increase the temperature, you see a hugely larger number of defects.
And you can use this sort of expression for any kind of defect.
So I talked about vacancies right now and vacancies have a specific formation energy, but interstitials also have formation energies, substitutionals also have formation energies.
So you can use this expression to determine the fraction of defects per lattice site for any kind of defect so long as you have the formation energy of that defect.
So just to show you how difficult it is actually to remove defects, if you have a crystalline material, defects want to stay even at the lowest temperatures.
I have this interesting little piece of art to show you.
This particular piece of art was actually discovered, I guess, by some scientist at the University of Toronto where Dr. Sadoway went to school, and this particular-- what do they call it?
They call it The Atomix.
Let me write down the name of it in case you want to look it up because I think they sold a lot more of these to material scientists than they sold to anybody else.
And all this is two plastic plates, two PMMA plates, polymethyl methacrylates, or plexiglass.
And in between them, there's a little hole that's cut, a gap, and in that gap are a number of ball bearings.
And the ball bearings are kind of like atom, right, so they move around.
And here we're going to the document camera right now.
So if you shake these things around and the ball bearings are moving around, that's like introducing temperature.
That's like taking a crystal, melting it, all the atoms are bouncing around.
You will even see some vapor atoms when these sort of leave the surface and fly through the air, but then if you stop, that's like quenching.
That's like suddenly I've taken this crystal, which was originally molten, and I've dropped the temperature.
Here's what I see.
So can you see that?
How to use this thing-- there we go.
So you can actually see a lot of the defects that we were talking about just a moment ago.
Here you can see areas of perfect crystalline order.
Here's another area of perfect crystalline order.
So you have many, many crystals that are adjacent to each other.
They're oriented differently.
So they're forming misorientation defects between themselves.
So for example, here's a grain boundary.
This is a crystalline grain.
And inside this crystal, you see a vacancy.
It's right there, So here's another instance of the vacancy.
There's another vacancy.
And what happens if I try to remove some of the disorder?
By the way, interesting thing is that some of the vapor is also left here.
So there's the vapor.
If I then try to remove some of these defects, I can just tap on this.
If I just continue to tap on it, I'm removing defects.
The whole thing is crystallizing more and more and more so now I put it down again.
OK. You still see the vapor phase.
Now you see a big huge crystal right here with some grain boundaries around it.
Here's another big huge crystal.
Here's another crystal.
There's a smaller crystal.
Each one of them is bounded by grain boundaries, but what do you see in each of these crystals?
There's a vacancy.
There's a vacancy.
There's a vacancy.
And you can actually sit here-- well, not here.
Maybe later, but if you want to take a look at one of these things, you can probably go to Dr. Sadoway's office and pick it up and he'll let you play with it for a little while.
You can try to get all these vacancies out.
You can try to get all the defects out.
No matter how hard you try, if you spent hours, you'll get things to be more and more and more perfect by tapping on it, but there's always going to be a vacancy.
Always.
And even if you are extremely patient, you get things down to just one vacancy and you think that all you have to do now is just give it a little bit more of a tap to remove that one vacancy, well, more often than not, you'll find that with that tap you'll remove the vacancy, but create another vacancy somewhere else.
So you'll always find these vacancies in these crystalline materials.
It's just a consequence of the fact that when you're agitating a material like you're doing here-- you're shaking it-- that's the same thing that happens if you have some finite temperatures, some non-zero temperature in the material, you're always going to be creating some defect.
So that's exactly what this expression is giving you.
It's telling you that there's going to be defects at any temperature, but their number goes up dramatically as you increase the temperature.
So let's go back to the slides real quick now.
OK.
So everything I've told you so far is concerned with defects in crystals that are of one type, so that are made up of a single element.
But in the case of, for example, things like ionic crystals, and ionic crystals are made up of multiple elements with multiple charge states, you can get new kinds of point defects that you can't see in a single element material.
So we'll talk a little bit about those and when it comes to those defects in ionic materials, we have to expand the taxonomy a little bit.
The taxonomy is now going to include defects called Schottky imperfections, Frenkel imperfections, and F-centers.
So let's go to a visualization of what these are.
Here you see an ionic crystal.
So you have alternating types of atoms.
You have these big ones, which are presumably the negatively charge ones, and then you have the small positively charged ones, and here you have an example of a Schottky imperfection.
So a Schottky imperfection, it's kind of like a vacancy.
It's missing atoms.
The main difference between a Schottky imperfection and a vacancy by itself is the fact that in a material that involves charged atoms-- ions, right-- in an ionic crystal, when you take atoms out, you have to make sure to maintain charge neutrality.
So these materials are charge neutral.
They have equal numbers of positive and negative ions, but you want to make sure that you take these out in equal numbers.
So in the Schottky defect, you take out one negative and one positive ion, or basically one unit, one stoichiometric unit.
If you add zirconium oxide, which is ZrO2, you would have to take out three atoms to maintain charge neutrality.
That would be the Schottky defect.
So here is another example, another visualization of a Schottky defect.
So when you take these two atoms out, nothing says that they have to be taken out from right next to each other.
You can take out one here.
You can take out one there.
It's good both ways because in the end, it's just about charge neutrality.
It's about maintaining a charge neutral material.
So we can actually write down reactions to describe the formation of these Schottky defects.
So let's do that.
When we write down reactions to describe the formation of Schottky defects, we first recognize the fact that we're dealing with empty sites.
We're dealing with void.
We want to see how this void, or in scientific parlance, null, is decomposed into two vacant sites in the particular ionic crystal that we're dealing with.
So let's-- to make things specific, deal with sodium chloride.
So sodium chloride, we have two type of atoms.
Sodium and chloride.
And to maintain charge neutrality, we have to remove one of each.
So this is actually-- this null space is actually going to be composed of two vacancies: a vacant site on a sodium lattice where a sodium atom would have been sitting and a vacant site where a chlorine would have been sitting.
And because this entire process takes place in the presence of materials that are composed of ions-- so ionic solids-- these two vacancies, in fact, have some charge them.
So when we think about the charge of these vacancies, what we actually have to think of, it's a little bit counterintuitive.
The sodium in a sodium chloride crystal is charge positive, but what does that make the vacancy?
So if you have a lattice of atoms that are charge positive like the sodium atoms and you remove one of them, what is the effective charge of that vacancy?
You have nothing, as Dr. Sadoway would say, in the land of plus.
So the charge of this vacancy, the effective charge of this vacancy, is going to be negative.
And we mark that with this little thing right there.
So void or nothing in land of plus is negative.
We mark that with one of these apostrophes.
So what about this vacancy?
That vacancy is created on a lattice of sites which are usually charge negative.
So if you have these chlorine atoms that are charged negative and you remove one of them, now you have void in a land of plus.
So this is going to be effectively charge positive and so we have this void in land of minus.
It is positive.
And we do it-- we annotate it with one of these dots.
We can do the same thing for other kinds of crystals.
So I mentioned that in this case, a Schottky defect would just be composed of these two atoms, because these two atoms constitute the structural unit for this crystal.
But if we had some other, more complicated crystal, like, for examples, zirconium oxide, then we would still write down a similar reaction.
So we could put a Schottky defect into zirconium oxide.
OK. And so now we have to figure out how this reaction plays out.
We have to create vacancies.
Here's our vacancy on the zirconium.
And because the structural unit contains two oxygens, we have to create two oxygen vacancies.
And now we have to think about how do we maintain charge neutralities?
What is the effective charge on all of these atoms?
So the zirconium sites are usually the positively charged atoms, so when we remove one of them, that's removing something, that's putting void basically into the land of plus.
So we have now a negatively charged vacancy, and this negatively charged vacancy has four negative charges, an effective minus 4 negative charge.
And the oxygens are the negatively charged ions, and so when we create voids there, these two vacancies have an effective positive charge, and because the charges have to balance, we know that this has to be effectively two negative charges and there are two vacancies.
So we have charge neutrality.
So that's how the Schottky defects work.
OK.
So let's move on to Frenkel defects.
In the case of Frenkel defects, it's really not so different from Schottky defects in some sense, except now instead of dealing with a situation where we maintain charge neutrality by removing two atoms, we actually displace one atom from its initial location to a different part of the crystal.
So we're creating actually a vacancy right there, and this atom, which was originally sitting on a lattice site, is now displaced to an interstitial location.
So that's an interstitial.
So a Frenkel defect is a vacancy and an interstitial.
And here's another view of a Frenkel defect.
Here, I've removed an atom from one of the lattice sites and moved it over to here.
So that's my Frenkel defect.
And in the case of Frenkel defects I can also write down reactions for Frenkel defects.
Let's do that.
OK.
So Frenkel defects usually occur in crystals with widely differing atomic radii, and in ionic crystals, the radius of whatever is positively charged would have to be much, much less than whatever the radius of whatever's negatively charged occasionally, or very rarely, but I suppose it's still possible, you would have to have basically a very big difference in radius between these two cases.
An example of one of these situations is, for example, silver bromide, where here the silver's positively charged and the bromide ion is negatively charged.
And we can write down a reaction to describe the Frenkel defect formation just like we did in the case of the Schottky defect formation.
So in the case of a Frenkel defect formation, how do we actually go about this?
We have, for example, can start out with a silver atom site and we're going to displace that silver atom away from its original site so it now sits in an interstitial position and we're going to leave behind a vacancy.
So let's do that.
OK.
So we're going to have silver interstitial site and let's just be clear-- say that this is on a regular silver site.
And then we have a vacancy left over on the silver site and this is the reaction for the formation of our Frenkel defect.
Frenkel defects, just like Schottky defects, preserve charge neutrality.
So how do we actually mark down charge neutrality in this case?
Well, there's no change in charges in the original state so we'll just mark that with an X.
And the vacancy is sitting in a place that used to be positive so it's a hole in the land of positive so we know that this one's going to be negative.
And this silver ion, which was displaced from its original lattice site, is now going to be sitting in an interstitial site which was originally a land of zero.
So this one's going to be charged positive.
And now we've balanced, once again, this reaction.
We have charge neutrality.
We've created a defect and this is defect, just like in the case of vacancies, costs us energy to make.
So we can actually write down a formation energy for Frenkel defects.
And this formation energy is usually a little bit higher than the formation energy of vacancies so it's going to be something more perhaps on the order of 2 eV.
OK.
So we have gone through-- before I go to this one, let me just mention one last defect that you might run into if you're working in ionically bonded materials.
An F-center is a situation where you have a vacancy with a bound electron inside of it.
And F-centers have the property that they give ceramics a tint, a hue.
They actually scatter light in the visible range so you can see them.
So with a very short amount of time left, I wanted to tell you about a couple of the other defects that you will encounter in crystals and one of the most interesting ones to me is the dislocation and dislocations are line defects.
They're 1D defects.
So if you look in, for example, a corn ear right here, you can see dislocations as the termination of a single atomic plane.
So here's a picture of what a dislocation looks like in a crystal.
You have these atomic planes and here you have one that just ends.
That's a dislocation.
Some of you probably have dislocations in your fingerprints.
So after the class is over, you can look for dislocations in your fingerprints.
They're line defects.
So here's another example of that.
You see that there's an extraatomic plane which ends.
That's a dislocation.
We mark it with one of these Ts.
Here's a picture of a dislocation.
Dislocations actually help materials to deform easily without having to shear off.
Here's a little picture of mine which I really like.
This is the only way that people really had to study dislocations before computers existed.
So now we can study, for example, dislocation by simulating them in real crystals.
And this is a mid-20th century computer.
It's a raft of bubbles.
It's just a bunch of bubbles that somebody blew on the surface, soap bubbles, and you can deform them.
You can deform this raft of bubbles by shearing it and here you see-- what's wrong?
Why isn't it going?
There it is.
There's your dislocation running through and you see lots of dislocations running through this crystal.
This was, by the way, done by Lawrence Bragg, who you heard about in connection with x-ray scattering.
You'll hear about Bragg again when we get to DNA.
So keep him in mind.
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Today, big announcement.
Wednesday, no lecture.
Instead, Celebration Part 2.
And I'm going to say a few words about it.
There is the room assignments.
I think they're the same as last time.
If you miss the exam because of illness, there will be a make-up test on November 4th during class.
There's no weekly quiz tomorrow.
I'll be available for office hours 4:00 to 5:30 this afternoon over here at 8-205.
So a couple of things.
I'm going to say few words about the quiz, the test, rather.
The celebration.
So the coverage is going to go starting from the time of the last celebration up through now, but I think in view of the fact that we started defects on Friday, and my thanks go to Professor Demkowicz, who jumped in the breach and lectured in my place so that I could get drawn into some of the activities associated with the visit of the President.
You will not have had the time to go through your recitation cycle on this new material.
So I'm not going to put anything on defects on the exam, all right?
So we'll go up through the end of the unit on x-rays right up through Bragg's Law, indexing crystals.
So that starting Friday, defects.
I'm not going to teach you Friday, teach you Monday, examine you Wednesday and Tuesday.
Tomorrow I expect you're going to have all sorts of review-oriented activities in your recitation.
So I don't want you torn between trying to master a lot of this new material, which is quite different from what you had in high school, and at the same time trying to have a catch up on capturing all that material since the second test.
So that having been said, I'm going to remind you.
I'm going to basically say the same thing I said before the first celebration, but we need to be reminded.
So what's the purpose of the test?
The purpose of the test is to give you a status report and send you a message.
And there's really only two answers here that we're looking for.
You're either going to get a smiley face or you're going to get a frowny face.
That's all it's about.
I don't care whether it's a 78 or a 75, but hard 50 is a pass.
And if you're not mastering the material, I intend to tell you so and then invite you in so that we can figure out how to get you through.
Because everybody in this room has the intellectual apparatus to pass this class.
Some of you just are not managing your time well and you need to be told that.
So we will tell you that.
With one of these.
That's going to be the message.
Go to the right room.
Bring five things with you: Periodic Table, table of constants, your age sheet, calculator, and a pen, something to write with.
You'll write on the question paper.
No wireless devices.
Shut them off.
Don't even bring them.
What do you need your cell phone for?
Don't need it during a test.
We're going to start at 11:05 promptly and stop at 11:55.
What's the strategy?
Read the entire paper.
Try every question.
You know this.
You get part marks if you write something.
If you don't write anything for the question, you get a zero.
So try every question.
And start with the easy one.
The easy one for you may not be the easy one for your neighbor, but how do you know what the easy one is?
I just put them down as they came out of my mind.
I didn't put them down in ascending order of difficulty, chronological order, topical order.
They're just-- they're there.
Leave tracks, tell us how you got to the answer, Work parametrically.
Don't start plugging in numbers on the first line, because if you make a computational error on the first line and drag it through the entire derivation, you'll make a mess of things.
You'll have wavelengths, size of distance here to Chicago, light moving at ten times its normal speed and stuff like that.
It's no good.
Focus on the question.
Don't just say, oh, it's something to do with x-rays, just give me some stuff that you memorized about x-rays, some stuff you wrote down in your age sheet about x-rays.
If it's not about the question that we asked, I'll give you a zero.
I don't care how much you write.
Write on the topic.
And refer to electronic structure.
When you don't know what to say, start talking to me about electronic structure.
That's the theme of chemistry.
electronic structure dictates behavior.
So talk to me about electronic structure.
Do your own work.
And lastly, it says something here, oh, good luck.
That's what it says.
Good luck.
So last day, Professor Demkowicz got to the line defects.
He was talking to us about line defects and how they have an impact on behavior.
So I wanted to pick up the treatment there.
So what's the whole issue behind line defects?
Well, what line defects do is they explain this anomalous behavior where the measured value of yield strength-- so I'm going to write sigma y; sigma is the symbol for strength.
the yield strength as measured is on the order of only about 1/10 of the theoretical value.
It's only about 1/10 of the theoretical value for the yield strength.
You might, say, wait a minute.
How do you get that theoretical value?
You can calculate this.
Not too badly.
If I take a look at a system of atoms, and this could be a metallic crystal, you could argue that there are metallic bonds between all of these atoms.
And if what I wanted to do to shear something-- so this is the yield strength for shear-- if I wanted to shear the top plane of atoms versus the bottom plane of atoms, can you see that to a first approximation, the energy required would be the energy to break each one of these bonds?
Because I have to break the bond to release the plane and then move it.
So if you make an estimation on the basis of bond strength-- so you take individual bond strength times bond density or bond concentration.
When I say density, I don't necessarily mean mass per unit volume.
I could mean bonds per unit area.
You saw your homework questions.
How many atoms per unit length?
And we said what's the linear density?
And some people were saying, but density is mass per unit volume.
No, I'm counting atoms per unit length.
Open your mind, OK?
So bond strength, what's bond density?
That's atoms per unit area.
So we can calculate this and we're way off.
We're way off by about a factor of 10.
So it was the dislocation that allowed us to explain why this happens.
And it was the work of two scientists, one by the name of Orowan in Hungary and Taylor in the UK.
Orowan, after World War II, went to the UK and then ultimately came to MIT.
And he spent most of this later part of his career here in the Mechanical Engineering Department, from which he retired.
And actually, he's too modest to say it, but Professor Demkowicz, his PhD was for Professor Argon, and Professor Argon was one of Orowan's students.
So this is direct lineage.
This is all very clear, so he was the appropriate person to introduce dislocations.
So what did Orowan do?
Orowan came up with this idea of an interruption.
And how did he get it?
It shows how to look around and, in the world, see analogies.
Now, we didn't have the powerful electron microscopes then that we have today.
This was in the 1930s, so people didn't even agree on an atomic view of things.
So people were thinking, thinking, thinking.
So this is the dislocation.
You can see.
Here, we have three rows of atoms, now there's two rows of atoms.
The middle row stops there.
And I'll show you what the consequence of that is in terms of being able to shear the crystal of lower values of applied stress.
So that's the dislocation.
You can see here.
If you break this bond and then propagate that through, you don't have to shear them all.
You break them one at a time.
So Orowan came home one day in Budapest, and the cleaning woman was moving the runner on a hallway, the story goes, and she was about to move the runner.
And Orowan came up the stairs and offered to help her, and she told him, get out of the way.
Academic.
And what she did, instead of pulling the rug, she did what we see in the cartoon here.
She put a little kink in the rug, snapped her foot and propagated the kink and moved the rug a half a foot down the hallway with minimal energy.
Instead of having to pull the rug with it's coefficient of fraction over three, four meters, all she had to do was come up with enough energy to make the kink and then stamp the foot.
This person's really doing it the hard way.
I don't know.
This guy must have taken classes in chemistry or something.
But see, all you've got to do-- once you get that kind you just push and it'll go by itself.
So Orowan said could this be a metaphor for what's going on at the atomic level?
And that's where it came up with this idea of the dislocation.
So there we have it.
So where do we find the dislocations?
Where do we find them?
So what is it?
The dislocation-- I keep talking about it, what is it?
The dislocation, this is, if you like, it's a line defect.
It's a line defect formed by misregistry of atoms.
And with the misregistry, that leads to a reduction in bond density.
And that broken bond is like the hole, if you like, in the valence band.
It can propagate very easily.
We're going to learn a little bit later in the lecture where we find dislocations and how they contribute.
But I want you to hold that in near-term memory.
OK.
So we're going down the list of the different types of defects.
Oh here, I found this off a coffee table book that's indicating-- I think this is a crummy analogy.
It shows this caterpillar moving by.
Instead of moving all of its legs, it causes this thing-- I think this is a bad analogy.
Just look down here.
You see, you breaking this.
And once you break this, you just keep moving it along.
Propagate the single broken bond instead of breaking them all at once.
I don't think the caterpillar moves that way, anyway.
It's cute, but it's wrong.
This is a dislocation.
I don't think it has anything to do with mechanical behavior, but it's a misregistry.
So you can-- these guys, they write the books and they come up with this stuff and you study it, and you're going, I don't get it.
You know what?
You shouldn't get it.
Because this is crummy.
It doesn't make any sense.
How does that impact the yield stress of the corn?
Someday you could write a book.
Better one than this.
All right, so now we're going to go to two-dimensional defects.
Let's go to two-dimensional defects.
So two-dimensional defects, there's a couple types.
The dislocation is one dimensional.
Two-dimensional defects.
There's two types.
One is the free surface.
We're going to call the free surface a defect, and I'll explain why in a second.
And then internal surfaces.
Internal-- I'm going to use surface in quotation marks-- they're interfaces.
And there's different types of them, OK?
An interface.
There's an interface inside the thing.
So what's the physics here?
Surfaces are high-energy regions.
Why?
Because when you're inside the crystal, you have the prescribed number of nearest neighbors.
Let's say you're in an FCC crystal.
You're an aluminum atom.
So you're an aluminum atom and you've got 12 nearest neighbors because you're FCC, except if you're at the free surface.
If you're an aluminum atom at the free surface you don't have 12 nearest neighbors.
Because if I look down I've got aluminum atoms all over.
If I look up, I've got nothing.
So axiomatically, I've got a different number of bonds.
Why do we form bonds?
We form bonds in order to lower the energy of the system.
But axiomatically, the ones at the top can't form the same number of bonds, so they are not in the same low energy state.
So once you get the connection between lower number of bonds, higher energy state, then any place, free surface, internal surface, is going to be relatively high energy.
Relatively higher energy than atoms in the matrix.
Matrix.
What do I mean by matrix?
In the middle of a crystal.
In the matrix, OK?
So that means that it's not quite as tightly bound, so the consequence of this is if they're high-energy regions, so high-energy implies higher reactivity.
And that can be good or it can be bad.
It can be good if you're trying to design a catalyst.
And it can be bad in the case of corrosion.
Where do you think corrosion occurs?
Not deep inside the crystal.
It occurs at the surface along the grain boundary.
So here's a cartoon that shows different regions within the crystal.
And Professor Demkowicz showed you on Friday-- and I'll cut to the thing in a second, the Atomix.
but here we see atoms.
So this is a simple cubic, this is simple cubic, this is simple cubic.
But when they meet they don't line up.
They don't line up because this formed from solidification from the melt.
So there might have been a seed here and it grows outward.
A seed here and it grows outward.
A seed here and it grows outward.
And when they grow outward they impinge, but they're not all growing with the same orientation.
You already know what the planes are.
And they could even be the same plane, but they're starting from a different point.
So there's this misregistry here.
But it's not a misregistry like the dislocation.
It's a two dimensional misregistry.
Actually, let's cut to the document camera.
Dave, could we do that please?
OK, so here's Atomix.
This was, as explained, an object of art made by Francois Dallegret in Quebec in the '60s.
So it's kinetic.
It's supposed to be after the artist, see?
You know, you, too.
You pick it up on a coffee table, you put it down, now you have become the artist.
You become the artist.
Postmodernism.
After the death of the artist.
See?
I'm an artist.
Look.
I'm an artist.
One of my best works.
So a couple of my professors at the University of Toronto were at a cocktail party one night and they saw this thing on a table.
Must've been a wild party because they kept looking at this thing.
Finally decided to write a paper about it.
And as Professor Demkowicz said, I bet Francois Dallegret sold more of these to professors of material science than he sold to art collectors.
So it's 200 ball bearings strategically placed between two plates of plexiglass.
And as he pointed out on Friday, you can't get everything to go away in terms of dislocations, but now you can see misregistry-- pardon me, vacancies here.
Here's a stacking fault and so on.
And now here you see grain boundaries.
There's the free surface.
Here's the vapor phase.
Everything.
We're having a lot of fun, here.
So we can look at it again.
We can change.
We'll try to anneal it.
So how do we anneal it?
You put it in an oven and you eat it.
This is the mechanical equivalent of annealing.
I'm tapping it, for those of you in the back.
Now, I'm going to put it down and you see I have a much more refined structure.
So now you can see a large section here of a single crystal.
Each one of these is called a grain.
A grain is a zone that is a single crystal with some defects.
You can see the vacancies.
And then look at the grain boundary here.
See this is a nice crystal, this is a nice crystal.
So one grain, two grains, and there's the grain boundary where these don't line up with those.
Because they started from a different angle so they can't impinge and blend.
There's grain boundaries here, here, here, and there's beautiful stuff.
You can see dislocations.
Everything.
And I know he showed you at the end the Bragg bubble wrap.
And you could see-- [SOUND EFFECT]
--dislocations zipping.
When people were trying to compress the bubble wrap, the way the bubble wrap was compressing the bubbles didn't all get flat.
The bubbled glided over one another, and the dislocations relieved the stress.
See what I'm telling you here?
What I'm telling you here is something that you don't get in a Gen Chem class anywhere else.
If I told you about the chemical origins of reactivity, you'd go, yeah, that's chemistry.
But what are we talking about here?
We're talking about the chemical origins of mechanical behavior.
Now that's different.
The chemical origins of mechanical behavior: only in 3.091.
And why shouldn't we?
Because at the end, everything is chemistry.
Everything is chemistry.
The rest is stamp collecting.
So the grain-- make sure we know where the grain is.
The grain is like, I want to call single crystal.
The grain is like a single crystal.
And the grains are separated by grain boundaries.
So by managing grain boundaries, control of grain boundaries, or if you like, management a grain boundaries, this has an impact on mechanical properties.
This is called grain boundary engineering.
How we control mechanical properties.
Grain boundary engineering.
So for example, if we want high strength-- and I just told you that the dislocations will allow the atoms to glide.
Can you see that if the dislocation is present in this large grain on the bottom here and the dislocation, just as in the bubble wrap movie, is zipping across, what happens when it gets to the grain boundary?
The misregistry of atoms stops that dislocation dead in its tracks.
So if I want high strength, do I want a large number of grain boundaries per unit area?
Or a small number of grain boundaries per unit area if my goal is to stop the dislocations?
I want a high number.
High strength means fine grain structure.
Fine grained means small grain size.
Fine grain structure.
You can think of it as like a river.
You know, you're running, running, running-- all of a sudden, you stop because you've got to ford the river.
On the other hand, if you want high ductility-- why would you want ductility?
For example, you want to make a beverage container.
How do you make a beverage container?
Not by casting.
You take a sheet of aluminum and you punch it and you cause it to deep draw.
That's atoms sliding over atoms.
So do I want to facilitate that process or do I want to impede the process?
Well, to facilitate the process, I want to make it as easy as possible for those atoms to glide over one another.
So high ductility means coarse grain structure.
Coarse grain means large, if you like, large grain diameter.
Fine grain structure means small diameter.
So there we go.
That's the beginning.
And there's a lot more to it, but I just wanted you to see that there is something.
And David, may we cut back to the slides, please?
OK, I think the next one shows.
All right, so this is a-- oh, we don't want to hear that.
Whoops!
Well, I've got music embedded in this one.
This is Philip Glass.
[MUSIC PLAYING]
Because we're going to talk about glasses later.
Koyaanisqatsi.
It's beautiful.
Can you feel the bass?
There's the subwoofer right here.
You stand right here.
It's like a message.
Low frequency, of course.
All right, so here's this.
But this is a piece of copper.
Now, this was the advent of modern materials characterization when Sorby, in the late 1800s, figured out that if you polish the piece of metal supersmooth-- that's good, huh?
Maybe we better turn the volume down.
I can't compete with these.
There's more of them than me.
I can't compete.
That's called the mute button.
[MUSIC ENDS]
So what did he do?
He polished the copper and then he etched it.
And what happens with an acid etch?
It will etch at different rates because the different grains have different atom densities.
You see, if I take a piece of metal and I polish it super, super bright and I shine a light on it and look at it in a microscope, what am I going to see?
I'm going to see my eyeball, right?
I'll blind myself.
So the key was to etch.
And then when you etch, you differentiate the different grains.
So that's what you're looking at-- the grain boundaries.
And this is now the same thing, only under a field ion microscope.
This is a bicrystal of tungsten.
It's hard to see, but this is all one.
This looks like a lolly pattern, doesn't it?
You can see the symmetry here.
I don't know if you can see right here.
There's a line, sort of v-shaped here.
And that's the second crystal.
So there's one crystal here, a second crystal here.
So this two-dimensional defect is now characterized.
And we'll get to that later.
OK.
So now we want to do three-dimensional defects.
What are the three-dimensional defects?
Three-dimensional defects are important.
And they have beneficial and they have harmful consequences.
Three-dimensional defects.
The dominant mechanism here is coalescence.
So three-dimensional defects formed by coalescence.
And you can coalesce two things.
You can coalesce voids, or you can coalesce-- actually, you can coalesce void, and you can coalesce-- I'm going to use Democritus talk here.
You can coalesce void, or you can coalesce being.
What do I mean by that?
Well, void, this could be a vacancy cluster.
So if you have a large number of vacancies coalescing, you'll actually end up with a hole, a blow hole inside the material.
And as you can imagine, that has the mechanical strength of nothing.
That's bad.
And being?
This could be impurity clusters.
And there's two types of impurities, as we've learned.
There are good impurities, such as dopants, and there are bad impurities such as contaminants.
And what can they do to the mechanical properties if they are bad?
So what I like to say is impurity clusters-- so we've got, for example, there are good ones that are managed.
The ones that are managed can be used to our advantage.
So I'm going to say the managed ones get a smiley face, and the ones that are uncontrolled-- I'll give you a vivid example of that at the end of the lecture.
Uncontrolled, they get a frowny face, and they can lead to major disasters.
So for example, you're really talking about clusters that exsolve as a separate phase.
So these things exist as a separate phase.
So they're no longer dissolved.
And these precipitates, some of these impurity clusters, if they're managed, we can call them precipitates.
i'll give you an example of precipitates.
Some of you might be going home at some point in the not too distant future, maybe for Thanksgiving.
And if you're flying on an airplane, the airplane is made of an aluminum alloy.
Aluminum copper.
And one of the precipitates that you can form is various copper-aluminum compounds.
And these form precipitates that the volume of copper-aluminum 2 is greater than the volume of the constituents.
So if you take a copper atom and two aluminum atoms in the lattice and they react to form copper-aluminum 2, there's a volume increase.
And so that works to give you constituents.
Isn't that great?
I left you in suspense there as to rest of the word.
So what happens?
This works a little bit the way carbon does in iron, where carbon atom is bigger than the interstitial void space in iron, and that force fit leads to a strengthening.
This leads to a strengthening, oK?
So that's good.
So this is called precipitation hardening.
I'm just going write precipitation pptn, precipitation hardening.
Very important in metallurgy.
Whereas carbon in iron is called solution hardening.
Because it doesn't exist as a separate phase.
The carbon atom is sitting on an interstitial site so it's a force fit, but it's still within the lattice, whereas the copper-aluminum actually exists as a precipitate outside the lattice.
So there we go.
In order to give you a little bit more, I said we're going to talk about the chemical origins of mechanical behavior.
And I want to show you how this all leads to strengthening, strengthening in terms of the yield.
So let's talk about deformation and the relationship to chemistry.
There's two types of deformation.
And this really accounts for the importance of metals in the history of man.
Why?
Because they have this unique property.
Until the modern era, when we discovered plastics.
But until modern era, metals are deformable.
They can deform without breaking.
That's what made them so desirable as materials of construction.
So I'm going to talk about deformation, how we deform them.
So really, metallurgy consists of-- really, there's two major divisions of metallurgy.
The first part is chemical, chemical metallurgy, and that's called dirt to metal.
That's the conversion.
This is the intensive chemistry.
And then the second part is called physical metallurgy, which is what I'm going to talk about next.
Physical, which is change of shape.
That relies on the ductility.
So deformation.
There's two types of deformation.
The first one is called elastic deformation.
And it is reversible.
So it is strain only under stress.
Or if you like, displacement only under applied force.
So this is Hooke's Law.
Hooke's Law applies.
So let's take a look.
I'm going to draw a little curve here.
You've seen this before.
I'm going to draw one of these energy curves.
So this is interatomic separation, which we might use the lower case r. And this is the potential energy that's stored between two atoms.
Remember, when they're far apart, they want to attract.
When they get too close together, their electronic shells repel.
So we've got-- time for colored chalk-- we have the attractive component, which looks like this.
And then we have the repulsive component, which looks something like this.
And then we put the two of them together, and I think we did this for ions, but there's an analogous one for other materials, and so you add the two of these, and you eventually end up like this, where this is r naught.
r naught, the optimum separation.
Here's attraction, this is repulsion, et cetera, et cetera.
So this is e minimum.
We've seen all of that before.
Now what's this have to do with mechanical behavior?
Well, I'll take it just a little bit farther.
And so I'm going to take the derivative of that and look at the force.
All right, so now I'm going to take the first derivative of that.
So this is going to be force, which is equal to de dr. Force is the derivative of energy.
And what do I find?
Well, look.
If I take this derivative, below r equals r naught.
I've got a negative slope.
Above r equals r naught-- I have a positive slope.
And at r equals r naught, zero slope.
So I'm want to put that down here.
So it's zero right at r equals r naught, and above r equals r naught it's positive, and below r equals r naught it's negative.
Not to scale.
But one thing is certain.
It is positive above r equals r naught, negative below r equals r naught, and to a first approximation, I can linearize about r equals r naught, Or Emin.
So what's the real shape of this thing?
You could use a Taylor series if you want.
And just take the first-- what's the first term of a Taylor series?
Constant.
Second term, linear.
Third term is squared, da, da, da, da.
So neglect higher order terms.
You get a straight line.
And what does this mean?
It means that if I try to disturb the atoms from one another and move them away from their rest position, there's a force that pulls them back, and the force is proportional to the displacement.
If I move the atoms farther apart by a distance r1, I get a certain force.
If I move them apart by r2-- 2 times r1-- I get twice the force.
It's linear.
Now let's go over here.
Now let's draw another analogy.
So here I've got something that's pegged.
And now I'm going to put a spring onto this secure wall.
So here's my spring, and I'm going to hang a mass on the spring.
Now I'm going to bring this into Hooke's Law.
I'm going to show you that what you know to be Hooke's Law is a direct consequence of this.
The atomic origin.
So what's the force on this?
There's a force on this, and that's equal to product of the mass times the gravitational constant.
And if there were no mass, this spring would be compressed and it might be sitting here.
I will call this x naught.
And now that I put the spring here I'll measures, say, this point.
And this point here is now at some value x.
So this is the displacement, isn't it?
This is the delta x.
So Hooke's Law is simply f equals k delta x, isn't it?
If I double the delta x, requires double the force.
Well, that's this.
This analogy goes right down.
And if I remove the mass it springs back, which is what I meant when I said it's reversible.
This is not permanent.
No stress, no strain.
Strain is the deformation.
No deformation, no force.
If I put a force, non-zero force, non-zero deformation.
By the way, Hooke-- little bit of culture-- Hooke announced Hooke's Law in 1676.
Here's how Hooke's Law was first published.
It's an anagram and it's in Latin.
And it was in a book entitled A Decimate of the Centesms of the Inventions I Intend to Reveal, which is English for, This is a Tenth of the Hundreds of the Inventions I Intend to Reveal.
He did not suffer from an overabundance of modesty.
Actually, Hooke and Newton sparred.
They hated each other's guts.
They fought bitterly.
Two giant egos.
Anyway, so here's the thing.
So in 1679 he published a book called De Restitutiva.
On The Spring.
Very presumptuous.
On The Spring.
And here it is in its unraveled, ut tensio, sic vis.
As the extension, so the force.
Nice, huh?
This is u.
That's why the u, but it's really v. Vis.
We get the English word vim from this.
This is the accusative form of vis.
The Romans didn't care between the u and v. In fact, if you go on the steps of 77 and you come up, it says Massachusetts Institute of Technology, you ever notice this?
Yeah.
I always thought that that's because the Romans, they couldn't-- no.
And I thought, well, maybe it's because it's hard to make a u, but then it says Institute of-- I figure if you can make one of these, you should be able to make one of these.
So if you can tell me why they do it that way, I might be able to find another Periodic Table beach towel.
So that's another thing.
All right.
So that's the elastic deformation.
And now we want to look at the other type of deformation, and that's called plastic deformation.
And that's permanent.
It's a permanent shape change.
So let's get that up here.
Permanent shape change.
And this requires ductility.
Because otherwise you'll have fracture.
If you take something that has no ductility and you apply a super critical force, you just crack everything.
So ductility allows for the atoms to glide over one another.
So what is the mechanism?
The mechanism for plastic deformation?
Plastic deformation in crystals, we're talking about here, not liquids.
It's called slip.
Atom's slip over one another.
So time for one more cartoon.
So I'm going to say, well, what happens if I start again with this?
I'm going to put a thin wire, a thin wire of metal.
Make it an FCC metal, let's say it's copper, and I'm going to put a mass here-- a honking big mass-- it's going to give me a force, and the force is great enough that it is actually going to cause the wire to elongate.
That is permanent shape change.
So what's going on?
Well, let's take this and blow it up.
When we blow it up, we have what's on the screen.
We have grains of different size.
And I'm just going to show atomic planes.
And that's why we have grains, right?
Because the atomic planes don't line up.
There's a misregistry along here.
So now I want to go down to the atomic level.
So now I'm going to take this and blow it up.
And what do I see here?
I see atoms like so.
I'm going to use that hard-sphere, close-packed model.
So here's the force.
The force is vertical, agreed?
We've got the mass and it's pulling straight down according to gravity.
So what's going on at the atomic dimension?
Because ultimately, this is what we're working with.
So what happens?
One possibility, and don't laugh-- this was actually suggested-- when you cause extension in this manner, you actually take some of these and you can turn them into elongated atoms.
No, no.
That's no good.
What in fact happens is slip.
So the force-- in fact, let me be more specific-- the applied force is resolved at atomic level.
So what happens is that the atoms slip over one another because they can slip along the planes that have the least bonding between them, right?
The chain is as strong as its weakest link.
So what will happen here is that in order to affect the elongation, these will move relative to one another along this slip plane.
This is resolved at the atomic level along slip planes.
I'll write that larger for the people way, way up in the back.
I haven't forgotten about you.
And what's the slip plane?
How do I determine which is the slip plane?
The slip plane it's going to be the plane that's strongest because it retains its integrity.
So which plane will be the strongest?
The one with the most bonds.
And which plane has the most bonds?
The plane with the highest number of atoms per unit area.
And that's why we study this stuff.
Because now we know, in an FCC crystal, which plane will slide.
So in FCC, the high-intensity plane is 1 1 1.
So the 1 1 1 planes do the sliding one over another.
And they'll zigzag the way you saw in the bubble wrap movie.
They'll zigzag down, across, across, across, But from a distance, it appears as the thing is elongating.
But at the atomic level, it's zigzagging along the close-packed planes.
Now I'm in a close-packed plane.
By the way, if I have the highest density in the plane of the floor, can you understand that orthogonal to the plane on the floor, I must have the weakest density?
All right, now I'm in the plane.
Which direction in the plane?
Any direction?
I'm going to push, I'm going to push some pasta.
I'm going to push uncooked pasta or cooked pasta?
Do you want to push on a rope?
Do you want to push on a broom handle?
The stiff.
So how do I determine, within the plane, which direction is the stiffest direction?
The one with the most atoms, which is the close-packed direction when I'm in the plane.
So if I'm in the 1 1 1 plane, which direction?
The 0 1 1 direction.
So this is direct line of sight from the original crystallography.
So that's how we get the slip.
And in BCC the highest density plane is 0 1 1 and so on.
And the close-packed plane simply means 1 1 1, because FCC is the closest-packed structure, so that's a close-packed plane.
Otherwise, they're planes of highest density.
It's a trivial-- forget the last column-- it's pedantry.
Forget it.
I'm sorry I have it up there.
So this is the slip system.
So the slip system is the combination of close-packed plane-- or closest-packed plane-- close-packed plane and the close-packed direction.
They all have close-packed directions because we've got atoms touching.
Close-packed direction.
That's how it deforms.
And then if I get a dislocation in a close-packed plane, voom!
Away they go.
1/10 of the applied sheer stress theoretical.
So if you want to get it down to a simple phrase, you can say, what causes slip?
Atoms slip, dislocations glide.
But please don't think for one minute-- listen to me carefully-- a lot of students come away from this and they say, oh, I know why we have deformation.
Because of dislocations.
No, no, no, no.
Atoms will slip no matter what.
It's just that you're going to pay a lot more to cause the deformation.
The dislocations allow you to have slip at lower cost.
Yeah, this is good.
And I think we're probably at the point where I can show you the enormous consequences of a three-dimensional defect.
Why did the Titanic sink?
It's a story of three-dimensional defects and human greed.
So this Tim Fecke at the National Institutes of Standards and Technology with an optical microscope and a rivet from the Titanic.
And what I'm going to show you in the rivet is a three-dimensional defect that took the boat down.
So this is the microstructure and what they're doing is they're showing you a blow up of this section of the rivet.
Now, the rivet is made of steel.
Before I can explain to you what's going on there, I have to show you-- [MUSIC STARTS]
Oh, shut up.
[MUSIC ENDS]
All right, so this is how they made steel.
Imagine this area the front of 10-250 as an open-hearth furnace, and I'm up to about my knees in liquid iron.
And on top of the liquid iron is a slag made of silica, calcia, and alumina.
It's the stuff when it solidifies it makes something not too different from window glass.
But we use it for refining to get some of the impurities out of the metal.
And those impurities go up into the slag phase.
And to keep this thing it about 1,300 degrees Centigrade, you've got burners.
Two.
One at each end like dragons-- [SOUND EFFECT]
--blowing flame into here for about 12 hours.
12 hours.
And that's what it took to make a heat of steel back in 1910.
This was made in Ireland, which was, at that time, part of the United Kingdom.
So on that day that they made the heat of steel that ultimately was shaped into rivets, evidently somebody became greedy, and instead of waiting long enough for the slag to phase separate and form a film on the surface of the metals, much like oil and water don't mix, but you can shake them up in your salad dressing application tool device and then pour it on quickly.
But if you let it sit they phase separate.
Well, somebody didn't allow this to phase separate.
They were told, pour the heat anyways.
So now you have globs of what is the precursor to silicate glass in the metal.
And now we take the metal and we cut it into little billets and the billets are extruded to form rivets.
And then the rivets look like this, where you have glass stringers.
Three-dimensional defects, impurity clusters, uncontrolled.
What are the mechanical properties of steel?
Strong.
Ductile.
Tough.
What does toughness mean?
Impactability.
When you say someone is tough, it means they know how to take a hit, whereas hardness means it takes enormous strength to cause deformation.
They're very different.
You know?
You say someone is tough, you mean one thing.
You say they're strong, it's different.
You can be strong and yet not tough.
You can be strong and brittle, so you can't take a hit.
I know a lot of people like that.
That can't take a hit.
Anyway, so enough about the people.
So here are the stringers.
So what happened?
What do you think the mechanical strength of this is?
If you have an impact like so, instead of having the toughness to absorb the impact, these things broke.
This is like perforations on a sheet of postage stamps.
And so these rivets were on the hull.
And when the ship hit the iceberg, what should have happened is there should have an a small hole, the water would've taken, who knows, a day or two to fill up, and there was plenty of time to get rescue vessels.
Instead, these rivets were brittle, thanks to this.
And so they just unzipped.
Just pop, pop, pop, pop, pop, pop, pop.
Opened up a giant hole and then took on the water.
Now, did nobody check?
Along the way, no inspection of the steel?
Nobody looked?
The guys that we're doing the riveting, they must have known the rivets were acting strange.
They must have broken a lot of them just putting them in.
But hey, it's not my job.
Not my job.
Not my job.
No testing anywhere along the way.
What's shocking is not human greed.
That's as old as humanity.
What's shocking is there was no checks.
So we have a different way to make vessels like this today.
But that's the story, right there.
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So these are the scores from test number two.
The celebration.
And as you can see, the average has gone down.
So as I've said in the past, I think everybody in this room has the intellectual capacity to be here.
Some people choose to be there.
I don't know why they choose to be there, why they choose not to go to recitation, why they choose not to try the homework, why they choose not to go to office hours, why they choose not to read.
But I guarantee you, when you make that choice you will be here.
And I have no requirements whatsoever to pass a certain number of people.
I can give everybody A's if I want.
And I can take a large number of people-- you know, look at this.
The way this is right now, the failure rate will be abnormally high.
So I would warmly recommend that if you're down here, that you look in the mirror and ask yourself a few questions.
OK. Let's get to the lesson.
We're going to start a new unit today.
The new unit is going to be the second half of the classification of solids.
We looked at solids and we reasoned that there were ordered solids and we've looked at crystals and their arrangements and so on.
And today, we're going to start talking about disordered solids.
And they're characterized by no long-range order.
They have short-range order.
You might know who your next nearest neighbor is, or your second next nearest neighbor, but you certainly don't know who your 10th nearest neighbor is or who your 100th nearest neighbor is.
And these solids are called amorphous solids.
And there's a simple one salable Angle-Saxon word for this.
It's called glass.
So we're going to talk about glasses.
What kind of materials can form glasses?
Well, obviously, materials that have trouble crystallizing.
And they come from a variety of walks of life.
So we have in organic compounds.
Some inorganic compounds can form disordered solids.
And a good example is silicates.
And this is what you know as window glass, glass that's used in bottles, cookware, and so on.
There are organic compounds that can form disordered solids and a variety of polymers.
Things like food wrap and so on, are also prone to forming disordered solids.
Some elements.
Some simple elements can form disordered solids.
A good example is sulfur.
Sulfur can form solid in crystalline form, but more often than not, it can form disordered solids.
And what I'm going to show you towards the end of the lecture is metal alloys.
Metal alloys can form disordered solids, and you're going to see metallic glasses.
And they're typically 80% metal and 20% metalloid.
That is somewhere along that red staircase on your Periodic Table that divides the metals from the nonmetals.
So a good example here is iron 80-- this is on mole basis-- and boron 20, boron being the metalloid.
And some, some do both.
Some compounds can form both crystal and amorphous.
So one example of that is SiO2, silica.
When it forms crystalline solid, we call it quartz.
Crystalline SiO2 is quartz, and amorphous SiO2 is the silicate glass that we know for windows and bottles.
And so as I've said many times, the term "glass" is related to atomic arrangement.
It has nothing to do with the ability to see through something.
Transparency has no place here.
So what are the conditions?
I mean, if you have silica and it can form crystalline or amorphous forms, how does it decide which to do?
And so I'm going to look at conditions promoting glass formation.
And I'm going to form glasses on the basis of solidification.
So I'm going to start from the liquid and go to the solid.
So I'm asking the question, why, on some occasions, does the liquid go to a solid that's amorphous and in other occasions it will go to a solid that is crystalline?
As in the case of quartz.
So there are primarily three factors that it boils down to.
And I like to make the analogy to the game of musical chairs.
So imagine I've got chairs placed around this central table.
You know how the game goes.
There are more people than chairs.
And the music starts and you walk around, and then at some point the music stops and people race for the chairs.
And some people are going to get a seat and they get booted out of the game.
And then one of the chairs is removed and then so it goes.
You know this game.
I know you don't want to admit to playing it, but your little siblings play it.
But of course, you don't play it.
But imagine if you were to play it.
The same situation here.
I've got silicate floating around in the liquid state, and the chairs are the crystal lattice sites.
So the question is, what promotes the ability to get to the lattice sites or not?
So obviously, one of the first things we think about is atom mobility.
Atom or compound mobility.
This is in the liquid phase.
And the way I want to write this, I want to talk about promoting glass formation.
So obviously, high mobility is going to enhanced crystal formation, isn't it?
If you've got high mobility, you're going to be able to find the right lattice site.
So I'm going to talk about if atom mobility promotes crystallization, the reciprocal of atom mobility will promote glass formation.
The second thing is the arrangement of the chairs.
If the chairs are in a simple line, it's easy to get to the chairs.
All other things being equal.
But if the chairs are arranged in some complex formation, it's harder to get to the chairs.
So that has an analogy and that's the complexity of the crystal structure.
The more complex, the more likely you are to form glasses.
And then the last thing, an analogy to the musical chairs, is the rate at which the music stops.
If the music stops suddenly you could get trapped in the liquid state.
If I gradually turn down the volume, you know, OK, wait a minute.
The music's going to stop.
You start moving towards the seat.
So that's the analogy.
Here is the cooling rate.
So a high cooling rate is going to make it more difficult for the system to find the crystal structure because as the cooling starts, the atoms start thinking, gee, it's time to form the solid.
But the thermal energy is removed from the system before the atoms have had a chance to find a crystal structure.
So the reciprocal of atom mobility, we write in a positive term and call that the viscosity.
Something, a fluid, a liquid, that has low atom mobility has high viscosity.
So I'll take this out and instead represent it this way.
This is the way to think about the formation of the amorphous state.
And why are we studying glasses?
Well, in the old days it was bottles and food and cook ware and so on.
Today, fiber optics.
Fiber optics is based on silicate chemistry.
Very important.
So understanding silicate chemistry has high-tech implications.
So let's look at, for the first study, the silicates.
The silicates for their value in fiberoptic technology.
So these are based on SiO2.
And some nomenclature.
This is called silica.
So the oxide of an atom is named by adding the term "a."
So silicon gives us silica as the oxide.
So here's the atom silicon, the basic element.
And if we fully oxidize-- see, here I've taken silicon, I put two oxygens, so I made a neutral compound-- that's silica.
And then if I fully oxidize-- the maximum number of oxygens I can put around is four-- and this is going to have a net charge of 4 minus.
And this fully oxidized anion is called the silicate.
And that's where we get the name of the family of glasses.
So this is fully oxidized.
So now let's look at the structure.
Silicate glasses.
sp3 hybridized.
Just like carbon above it.
sp3 hybridized.
So that gives us the four struts off of the central silicon at 109 degree angles to one another.
And let's put some flesh on the bones here.
So I'll put a silicon here in the center.
One, two, three, four.
But instead of putting silicons everywhere, I'll put oxygens at the end of the struts.
And then I'll put the second silicon, and you can see the oxygen is acting as a bridge between the two silicons.
One, two, three, four.
Let's do one more.
Another silicon.
One, two, three, four.
So again, you can see the oxygen acts as a bridge between the two silicons.
So you see four oxygens per silicon, but I've got two silicons per oxygen.
Hence, the structure looks like SiO4, but the stoichiometry of the compound is SiO2 because the oxygens bridge.
So this doesn't give you any clue as to what the structure should be, does it?
You really have to write this thing out.
Now, here's where the thing gets interesting.
This is a three-dimensional network.
All of these oxygens bridge to silicons.
And I was talking here about viscosity.
You can see that this thing in the liquid state is huge.
It's like a giant battleship, and this is just one.
So now we can have another silicate.
It can form chains.
It can form meshes.
And these meshes entangle.
So viscosity very high.
So when it goes to solidify-- David, if we can go to the document camera, please-- so here you see the silicate.
So the yellows represent silicon.
And you can see the oxygens, here, are at 109 degree angles.
So this is the SiO4.
We'll do some nanotechnology here.
So here's the SiO4.
One, two, three, four.
And now I'm going to make the bridge.
But look.
The bond between the oxygens and the silicon is defined in three dimensions.
It's a 109 degree angle.
But the bond between the two oxygens is not defined.
It's only defined in two dimensions.
So you can see that if I put all of these oxygens on the same plane-- so somebody borrowed this and look, they lost an oxygen for me.
Got a missing oxygen, here.
If you find this thing, please bring it to my office.
So now all five oxygens are on the same plane.
One, two, three, four, and five.
And then up here are the two silicons.
They're in the same plane.
The oxygens are in the same plane.
What I'm showing you here is the beginning of crystalline quartz.
Crystalline quartz, this stuff.
But now what happens if we cool quickly?
This bond is only specified in two dimensions, which means I can hold this bond at the proper value and this is free to rotate.
This is free to rotate.
Look at the autofocus on this thing.
Who designed this?
Boy, what a stupid machine.
So anyway, so here we are.
Here is the point.
This bond is free to rotate, and when it rotates, look-- now this oxygen is no longer in the plane with this oxygen.
And as a result of that freedom to rotate, we can end up running out of thermal energy.
And this is nowhere near its regular crystalline array.
So this gives you the indications of the high viscosity, moderate cooling rate will end up giving you the amorphous silica.
But in all cases, this bond is the linkage through.
So we end up with this three-dimensional network.
So what do we do with to show that we have some evidence for this?
David, may we cut back to the slides, please?
So how do you characterize?
I'm going to use x-rays.
So here's an x-ray diffraction pattern of the-- the upper one is cristobalite, which is one of the polymorphs of crystalline SiO2.
And you can see you have distinct peaks indicative of satisfying Bragg's Law.
There's some width to them, but that's because it's a real crystal.
Down here, this is the amorphous silica.
You don't see all of these features.
You see only one very broad peak.
Now if I said it's crystal-- pardon me-- if it's glass, you'd say, well, it has no long-range order.
So why do we have even this one feature?
What's this one feature indicative of?
Doesn't matter how much disorder there is, I still know that no matter what, I'm always going to have these four oxygens as my nearest neighbors.
So these four oxygens minus any long-range order gives us this one line here.
So we have evidence for it.
Now let's look at the energetics, because clearly, these are two different states.
One of these is lower energy than the other.
Which one is it?
How to think about the problem?
There's a simple way to do it with the tools we have in 3.091.
Energetics can be given by bond formation.
Energetics via bond density.
When things form bonds the energy of the system decreases.
So which one is going to have higher bond density?
Well, the higher bond density clearly is going to be exhibited by the one that has the tighter packing.
And which one has tighter packing?
The disordered solid or the ordered solid?
The ordered solid has much more dense packing.
So the ordered crystal exhibits tighter packing, therefore, this means more bonds per unit volume.
So that means, to me, that the energy of the crystalline state must be more negative than the energy of the glassy state.
That makes sense so far.
But now I want to show you one other thing that we've just inferred from this little exercise.
If we get more bonds per unit volume, then can you see that we've made an association between the binding energy of any arbitrary ensemble and it's molar volume?
So volume now, is a very easy thing to measure, isn't it?
You can see it with the naked eye.
So the volume of something with a given composition-- and we're talking about a mole, we're talking about equal numbers of atoms-- so the molar volume is indicative of binding energy.
It's a one-to-one correspondence.
If you like, the Vmolar is a measure of disorder, meaning the higher the molar volume, the greater the disorder.
Or put the other way, the smallest molar volume is that exhibited by the crystal.
So we have some traces here that come from the reading.
So this comes from the archival lecture notes that were written by my predecessor, Professor Witt.
There's a little typo here that obviously heating means increasing temperature, not decreasing temperature.
Make a little correction here.
So what we're plotting is the volume, the molar volume of silicate glass as a function of temperature.
And imagine we start with some blob of glass-- of some known mass-- so we can divide and figure out what the molar volume is, and we started cooling it.
We cool down until we get to the crystallization temperature of quartz.
And when we get to that temperature crystalline quartz forms and along with it, a tremendous decrease in the volume.
Unlike water ice, which is a rare exception, where the ice occupies a larger volume than the liquid, for most systems, the solid is more compact than the liquid.
And that's what you see here.
There's an abrupt drop here at the melting point.
And then when we continue to cool, there's some thermal contraction.
And as you can imagine a hot solid occupies a greater volume than a cold solid.
And so this is the cooling curve and you can retrace it, heat up, and when we get to the melting point, there's a tremendous expansion and then off we go.
So that's the classical form of the crystallization of a material that is undergoing the normal process of liquid to solid transformation.
I'm going to use some tea colors today.
So we went down the red line.
Now let's go down the green line.
So we're going to go down the green line.
We go down the green line, we're going to use the same stuff, this silicate network, only we're going to cool a little bit faster.
And we cool a little bit faster, we can zoom right on past the normal melting point and create a supercooled liquid.
And that supercooled liquid gets lower and lower in temperature and all the while the volume is shrinking.
Again, a hot liquid what occupies a smaller volume than cold liquid.
You know this from a mercury-- the old days, you remember these old thermometers?
You probably have never seen one of these things.
It's all done digitally.
But we used to have these liquid and bulb thermometers.
You'd have numbers on here, and what happens is the temperature goes up, the liquid in here rises to a higher temperature.
And at a lower temperature this contracts.
Isn't the solid expanding, too?
Uh-huh.
So what's the other thing you learned from this?
It must mean that coefficient of thermal expansion, which we affectionately will call CTE, the coefficient of thermal expansion of the liquid, must be much greater than the coefficient of the thermal expansion of a solid.
Otherwise, the two would expand and you wouldn't get any sensible measurement out of this, right?
Well, we see that on a curve.
We see that on a curve, because when we plot volume versus temperature, when we're up here in the liquid regime we have a steep slope.
This slope here is dv by dt.
And when we get down on the solid regime, it's a gradual slope.
It's still a slope, but it's a gradual because the coefficient of thermal expansion down here is very low.
And that's what you're seeing here.
So at some temperature the system changes from a supercooled liquid to a solid.
But it's a disordered solid because we've quenched in all of that remaining liquid disorder.
And take a look at this-- you've changed from the slope that's characteristic of the coefficient of thermal expansion of a liquid down here to the gentle slope coefficient of thermal expansion of solid.
You know what this proves?
This proves that glass is a solid.
There are many people out there, even in the popular press, who will say, glass is just a very, very viscous liquid.
Nonsense.
Look at this.
It has this coefficient of thermal expansion.
And don't fall for any of that nonsense they tell you when you go to the cathedrals in Europe and the glass is thicker at the bottom because it's been dripping for 400 years, and that proves-- you know why the glass is thicker at the bottom?
Because they made it by spinning.
And when they spun it, it was graded in thickness from the center out.
And now, if you're the glazier and you're putting the glass in a window, which way would you put a pane of glass that had variable thickness?
Would you put the thickest part up or the thickest part down?
It's thicker on the bottom because that's the way it was made.
Lord help the tour guide when there's an MIT student taking 3.091 on that tour.
All right, so here we are.
Look at what you have here?
You see this?
At this temperature-- let's say down here where the lines end, it's room temperature.
At room temperature the volume of the crystalline solid is low.
The volume of the amorphous solid is higher.
That's proving this.
Vmolar is a measure of the disorder.
And sure enough, the glass that was cooled quickly ends up quenching in more of the liquid state disorder, and you see that in terms of what I call the excess volume.
The excess volume is a measure disorder because the crystalline solid non-zero volume.
So we can define, we can go from this directly over and say that v, that's the excess-- that's a pun.
You know, instead of writing this?
This is the way they do in high school.
This is the excess volume.
Bologna.
We don't write like that.
Excess volume is equal to v of the glass minus v of the crystal.
And the greater the excess volume, the greater the degree of disorder.
So let's do one more.
Let's go down the orange line.
The projector isn't giving us a good yellow component, here.
OK, so this is the orange line.
And what's the difference between the orange line and the green line?
The difference between the orange line and the green line is that in the orange line we're going to cool more slowly than we did on the green line.
Because green means go.
So that's the fast one, right?
So here we are.
We still get supercooled liquid, but we get down to a lower temperature before we have ceased to have higher and higher viscous liquid and have formed the solid.
The knee in that curve occurs at a lower temperature.
And that value is called the glass transition temperature.
Why do we call it the glass transition temperature instead of just saying it's a solidification temperature?
Because when I say solidification, before today you would say, OK, liquid became a solid.
But after today, if someone says to you, solidification, you say, do mean ordered solid or disordered solid?
So we distinguish.
So every crystallization is a solidification, but every solidification is not a crystallization.
So here we are.
Look at this-- this has a smaller excess volume.
Because if it was a slower cooling, that meant that's the equivalent of giving more time, more thermal energy, for things to find their crystalline position, which means there's less excess volume quenched in.
OK.
So let's just get that down, define these things.
So this is a measure of disorder.
Or glassiness, if you like.
OK.
So now let's define these two different temperatures so that we have the distinction.
So we have, first of all, the classical one, which is crystallization.
And crystallization represents the reactions of liquid.
In both cases, we're going to convert a liquid to a solid, but a liquid goes to a crystalline solid.
And that occurs at a unique temperature called the melting point.
Imagine if I talked to you about the freezing point of water.
Freezing point of water at atmospheric pressure is 0 degrees centigrade.
If I asked you, well, if I cool the water really quickly or if I cool it slowly, does it make any difference to the freezing point of water?
No.
Why not?
Why isn't all this happening?
Because water is a tiny molecule and so cooling rate has no impact on the temperature of conversion.
Those water molecules will always find their lattice sites.
So this is a function only of composition.
If I put something in the water and I make it impure, I know I can change its freezing point.
If I put salt in water, I will depress its freezing point.
But if I put pure water, it will free at 0 degrees centigrade.
Later on, we'll learn that there are some pressure effects, but today that's not going to elucidate anything.
It'll just be a distraction.
So pure water, always the same thing.
But now if you go to something like silica, which has the capability of forming a glass, the glass formation is a different reaction.
Glass formation is written in this manner.
We will start with supercooled liquid.
It's liquid, but I'm already going to stipulate that it's liquid that's been cooled below the melting point.
Supercooled means it's cooled below the normal melting point.
So a supercooled liquid is going to form a glassy solid, and this occurs at tg.
tg, which is the glass transition temperature.
And that is very much a function of cooling rate.
And, of course, the function of composition.
Obviously, if we change the composition from SiO2, we're no longer comparing apples to apples.
So composition is important in both instances, but only in the case of supercooled liquid do we form the glassy solid.
And the degree of disorder is a function of the cooling rate.
And how is it a function of the cooling rate?
As the dt by dt, right?
Change in temperature with time as dt by dt goes up, the degree of disorder goes up.
So if I want to quench in the liquid state, can you imagine if I had liquid and I could instantaneously cool it?
I would quench in all of the liquid disorder.
If I quench it less rapidly, there will be some time for the atoms to strive for a degree of crystallinity.
So all of this we've said with reference to silicate glasses.
Now, there are other glasses.
What are other glass forming oxides?
Well, what do I have to look for?
I should look for other compounds that form bridging oxygens, right?
That's the key here.
What's the unifying feature?
It's bridging oxygens.
Bridging oxygen leads to glass formation.
So what are other compounds that could give us such bridging oxygens?
Well, you could be lazy and say, well, if silica does it, then why don't I look up and down the column on a Periodic Table.
If you go up to carbon that's no good because it forms gas, but if you go underneath you will form germania glasses.
So this is germania, and the glasses are germinate glasses.
You can also go to group three.
B2O3.
B2O3 forms sp2 hybrids.
And the sp2 hybrids off of each boron, we have three oxygens.
And that oxygen can bond to another boron.
One, two, three.
And this is all going to lie flat in a plane, isn't it?
But this oxygen bridge is only specified in two dimensions, whereas the boron struts are specified in three dimensions, which means this oxygen bond between the two borons doesn't have to lie in the plane.
It could tilt this up and if it does, this is going to have a greater volume.
You can see this with the naked eye.
I mean, I could have just taught the lessen by putting this slide up.
If those are equivalent numbers of borons and oxygens, it's plainly obvious that on the right side you've got excess volume.
This occupies a much larger volume than the image to the left.
The image to the left is crystalline B2O3.
The image to the right is the same stuff, except in many instances this boron-oxygen-boron bond doesn't lie in the plane.
It tilts out of the plane and it causes all sorts of distortions and leads to excess volume.
So this is called a borate glass.
All right, so if we can do it with the borates, we can do it with anything that will form covalent bonds.
So we can do it with P2O5, phosphate glasses.
V2O5, vanadate glasses.
As2O5, arsenate glasses.
And Sb2O5, stibnite glasses.
And these are used, actually, as additives to some of the glass that's put in computer screens so that they will gobble up excess oxygen on cooling and avoid bubble formation, which then makes the glass foggy.
And a foggy computer screen is no fun to try to look through.
So what are the properties of these oxide glasses?
Well, first of all, they're chemically inert.
Why are they chemically inert?
Because they've got strong covalent bonds.
So if you try to react something with them, it's going to take a very special compound that can trigger reactions.
Which is why they're used for bottling beverages and packaging foods.
You put vegetables and fruits in glass jars going back to ancient times.
Until recent times, with the advent of the soda can, there were glass bottles and so.
They're electrically insulating.
Why are they electrically insulating?
Electronic structure.
These are all covalent bonds, high band gap, which is why the amorphous version is used in window glass.
High band gap, which means light goes through.
How do I think about whether something is transparent?
Do a little finger demonstration.
This is the band gap of the glass, and this is the band energy, the photon energy.
If the photon energy is small it goes right through.
That's called transparency, see.
You can study quantum mechanics, but you know what?
This is it.
That's all you have to know.
Now what happens if the band gap is small-- like around 2eV-- and here's the photon energy?
That's called absorption and re-emission.
Transparency, absorption.
That's all it takes.
You think I'm kidding.
That's all you need to know.
So now, the next thing.
Mechanically brittle.
Why are they mechanically brittle?
Strong bonds.
No opportunity for slip.
Please don't tell me-- this is what students tell me every year and they get zero for the stupid answer-- that the reason glass is brittle is it's a distorted solid and therefore has no dislocations.
Uh-uh.
Dislocations take the stress required to cause slip and reduce it to a lower value.
But you cannot cause this thing to slip once it is solidified.
The only way to get this silicon to move relative to that silicon in a vertical shear-- let's say I want to make this silicon move up and this silicon move down-- there's only one way.
It's called break that bond.
When you break that covalent bond, that's called fracture.
So there is no slip allowed because we have strong covalent bonds.
If you want to reshape glass, what do you have to do?
You have to heat it back up above its glass transition temperature.
But you do not shape glass once it is solidified.
You may have tried in the home, and then you end up with something called shards.
And that's the reason.
Optically transparent, we know that one already.
OK. Last thing is the very high melting.
Melting point of silica is over 1,500 degrees Celsius.
You're saying, well, wait a minute.
He's telling me we're going to make beverage containers, we're going to make food containers, we're going to make all sorts of useful objects, but that's going to cost a lot of energy to go away up there to melt this glass.
But it does have desirable properties-- chemically inert, and so on.
So what can we do?
Back here.
Glass transition temperate is a function of cooling rate and a function of composition.
So next step is let's modify the composition of the silicate network in order to drop it's processing temperature so that we can make beverage containers at acceptable energy.
So what's it going to take?
I'm going to have to do something about those bonds.
Because those bonds are what caused me to have to go to such high temperatures.
So in order to reduce the processing temperature of silicate glasses via change in composition-- so this is material science-- change the composition in order to get desirable properties.
And specifically, I want to lower the processing temperature.
Even when I get these things liquid, they don't deform very well, because all these networks are entangled.
So I'll give you another analogy.
Suppose you're in the kitchen and you're going to cook some pasta.
So you take some linguine and it's about a foot long.
So you got this pound of linguine and you cook it in the boiling water and you pour in into a colander.
And you may or may not rinse it with cold water.
Whatever.
Just leave it in the colander for a little while and what you'll find is the whole thing turns into one big mass.
I'm not talking about the stuff got all gooey.
I'm just saying it just hangs together.
And if you're clever you can just gently slide it out.
And it'll slide out of this one big mass.
What's holding those strands of pasta together, by the way?
Well, are they covalent bonds?
Are they ionic bonds?
Are they metallic bonds?
I don't know, if you've got metallic pasta, you've got digestive problems.
So what's the bonds?
It's van der Waals bonds, right?
Now, I can cause them to slip again, can't I?
I can put a little water in there.
And what does the water do?
It goes in between and then it's got hydrogen bonds and they slide.
I can put a little oil in there and then that slides.
or.
The other thing I could do is-- you know, have you ever seen some people, they break the pasta before they boil it?
I did this experiment.
So you take the pound, divide it two halves.
By my math, that's two half pounds.
So you cook the one half pound one foot in length and you cook the other half pound-- break them in three.
So they're about four-inchers.
And then you put them each in a separate colander.
Which one do you think is going to be much more difficult to move around?
It's the long stuff, right?
The long stuff entangles.
So we're going to do the same thing here with the processing.
And basically, you can study and learn pretty much everything you need to know about polymeric networks in the kitchen with a pound of pasta.
All you need to know.
All right, so let's go and look at it.
I want to reduce the amount, the length, of those chains.
If I reduce the length of those chains, I can process at much, much lower temperatures.
So what's my weapon, here?
My weapon here is oxygen.
Oxygen is going to go in there in the form of the oxide ion.
And the oxide ion, O double minus, has a very, very high affinity for the bonding.
And so what'll happen is oxide ion attacks the oxygen-silicon bond.
It attacks the oxygen-silicon bond and breaks it.
It breaks it in two and then incorporates itself into the network in the following way-- you see I have to silicons joined across an oxygen?
This free oxide ion will come in here, interrupt that network in the following manner.
So it's now broken the silicon chain.
Now, I've got conservation of charge.
This was 2 minus.
I don't see any exposed charge here.
I need 2 minus.
This'll be minus 1, this'll be minus 1.
I have conservation of charge, I have conservation of mass.
And what I've done is I've broken the chain.
This is called chain scission.
Shorter chains, higher fluidity.
Higher fluidity, which means lower processing temperature.
Now, where am I going to get my-- I can't go to the lab and get a bottle of oxide anions.
I have to have charged neutral species.
And so what I'm going to look for is an oxide ion donor.
I need an oxide ion donor.
And where do I find an oxide ion donor?
Well, better be something that's going to be a cation, isn't it?
Because if I've got an oxide anion, I need a cation.
And what kind of a cation?
How do I make an oxide anion?
I have to have electrons to give to the oxygen.
So I need a good electron donor.
Or to use a simple Anglo-Saxon word, a good metal.
So a good metal oxide like, say, calcium oxide.
So if I take calcium oxide and I dissolve calcium oxide in silica, it dissociates to give calcium cation plus oxide anion.
And then the oxide anion migrates over here to our pal the oxygen bridge and results in two of these broken pieces.
So we call this one a bridging oxygen, as I've been saying up until now, BO.
And this one is called a terminal oxygen.
This is bridging oxygen, this is terminal oxygen.
Or some people, I don't know why-- they have no literary skills-- they call it non-bridging oxygen.
I hate that because it tells you what it's not.
So you might see non-bridging oxygen.
So all of these up here, the silicates and so on, these compounds that have the ability to form oxygen bridges, these are called network formers.
These are all network formers.
Because they have the capacity to make oxygen bridges.
And then compounds like calcium oxide that have the ability to donate oxide ions that break bridges, these are called network modifiers.
So good examples-- any good metal.
So I can look at lithium oxide, sodium oxide, these will all disassociate to give oxide anion.
If calcium will work and you believe Mendeleyev, then you should probably think about magnesium oxide, calcium oxide, and anything in that series.
A good ionic oxide like lanthanum oxide, yttrium oxide, these will be oxide ion donors.
If you want to get yourself fired, use scandium oxide because it's frightfully expensive.
And you can go to Group four.
You can even use something like lead oxide or tin oxide.
Even they will donate oxide ions.
In fact, you can put a lot of lead oxide in, you'll modify this thing so much that it'll start to allow you to cut crystalline facets and this'll be called lead crystal.
Why do we use lead?
Well, number one, it modifies so I can cut it and I get a straight line instead of that conchoidal glass edge that if you've ever cut yourself you'll never do it again.
The other thing is lead, because it's got so many electrons, has a very, very high index of refraction.
So when you make your billion and you've got your crystal chandelier hanging, of course you want the cut crystal with a high index of refraction because it's going to make that candlelight look really elegant and romantic and so on.
And if you make a super boatload of money, then you're not going to go with lead crystal.
You're going to make diamond pendants, right?
Because they've got a higher index.
And we all want the best index, don't we?
OK, so I think this is probably a good place to stay.
So let's jump to the end, because we've got a few things here.
So I said I was going to show you about metallic glass.
Here's metallic glass.
1959, Pol Duwez, who was a professor at Caltech, reasoned that if he were able to cool liquid metal very quickly, at, say, a million degrees per second, he could freeze the random orientations of atoms in the liquid state and prevent them from finding even something, either simple cubic, body-centered cubic, face-centered cubic lattice.
So he worked with gold silicon.
Gold silicon has a very, very deep eutectic.
Even though gold melts to over 1,000, mixed with silicon, it gets down to about 400 degrees Celsius.
So he dropped this through a little orifice here.
This is molten and it drops onto a water-cooled copper wheel spinning at a very high speed.
And it makes a ribbon some tens of microns wide.
And what came out here was disordered solid.
It was metallic glass.
This was the birth of rapid solidification, and this is metallic glass.
This is metallic glass.
This has no long-range order, no grain boundaries, no dislocations, but it's not transparent.
Why?
Because it's a metal.
And what do you know about band gaps and metals?
And it's different.
It was a different Q-factor.
It has different mechanical properties.
It has no ductility in a classical sense.
This is just for your reference.
This is aluminum foil.
And you know what aluminum is.
It's got ductility.
It's got no grain boundaries so the corrosion properties are different.
It's got no magnetic wall boundaries.
So that a transformer made of this stuff is half the mass of a transformer made of crystalline glass.
It has some other uses.
So it's used in magnetoelastic resonators for theft prevention-- oh, I'm sorry.
I'm in Cambridge.
I can't say theft prevention-- for inventory control.
And so you see that it's 20% boron and all of this metal-- iron, chrome, and moly-- and they put this inside the object in the store.
And then there's a permanent magnet there, iron-cobalt-chromium, that sets the metglass to a field.
And then when you walk through the uprights, there's a 58-kilohertz signal that causes this thing to ring.
So it excites and listens for the ring.
And if you still have that little piece of metglass in there that hasn't been demagnetized-- Bing!
And then, oh, jeez, I'm sorry.
I really am.
I meant to pay for this.
Explain it to the police officer.
All right.
So that's inventory control.
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OK, OK, OK. Let's settle down.
Weekend is over.
Tomorrow, weekly quiz.
Today I'll have office hours 3:00 to 4:00.
I have to go down to Washington, so I've got to leave a little bit earlier than normal.
So I will be available from 3:00 to 4:00 The lecture has started and there's still way too much talking in here.
Way too much.
You know how much is too much?
Any.
Any.
So last day, we started talking about oxide glasses, and we reasoned that we could have control of the properties by control of the composition.
We started with a network former, which is some oxide that has the capacity for forming covalent bonds through a bridging oxygen.
And then we wanted to drop the processing temperature, and we did so by adding modifiers.
Intermediates, we haven't talked about and we're going to do that in just a moment.
So if you look in the readings, this is from archival notes that were written by my predecessor, Professor Witt, these are compositions of some typical glasses.
I don't expect you know these from memory, but I would expect you, if I gave you the composition, explain to me why the various constituents are there.
So let's try few examples.
The first one is soda-lime glass.
And you see it contains silica, which is the network former.
It contains sodium oxide, calcium oxide, and magnesium oxide.
And these are alkaline earth oxides that are ionic, and so these are acting as network modifiers because they're donating oxide anions that go in and break the silicate chains.
And then there's this Al203 and that's sort of halfway in between, isn't it?
Silica is Group 4, or 14, if you want to use the modern notation.
Sodium is Group 1.
Calcium, magnesium, Group 2.
Alumina is Group 3, and it's sort of halfway in between.
It's amphoteric.
It can either be a former or a modifier.
And in these instances, depending on how much modifier is present, alumina can act as an intermediate.
And what's an intermediate do?
An intermediate is a covalent oxide.
It's a covalent oxide, or an oxide that can act as both covalent and ionic.
But in this instance, it's acting covalent oxide with a different coordination number.
And what does that mean?
Different coordination number?
It coordinates-- remember last day I showed you B2O3?
Borate glasses?
So they have a coordination of three, whereas silicates have four.
And what that means is that there's going to be a mismatch.
There's still strong covalent bonds, but they don't fit quite right.
And that's going to give even more free volume.
And that excess free volume, through covalent bonds, gives you the ability to endure thermal shock.
So if you want to impart thermal shock resistance in a glass, you give it lots of free volume so it can take the rapid change in temperature.
So it's a covalent oxide with different coordination number.
That is, nearest neighbors, in a covalent sense, from that of the network former.
So you can add a borate or aluminate, what have you.
And that'll give you thermal shock resistance.
And so alumina here is acting as an intermediate.
Let's go down here.
There's borosilicate.
borosilicate is the generic term for Pyrex.
So Pyrex is a trade name.
Pyrex was invented by Corning, and it contains silica as the network former.
There's some sodium oxide, potassium oxide-- in rather small amounts.
You can see this isn't a heavily modified network, but look at this.
13% B2O3 and 2% alumina.
That's the modifier.
And what was the hallmark of Pyrex?
It had thermal shock resistance.
So you could take it out of the oven and put it under cold water and it didn't shatter.
And we have the same analogous behavior for glass ware in the laboratory.
For all room temperature and low temperature work in the laboratory, we use Pyrex that has this resistance to chemicals and resistance to heat.
This is the big one, here.
And that gave birth to functional crockery in the kitchen, where you could work in glass instead of in metal.
And down here we see glass-- let's see, well, here's one.
Light flint optical.
See, that's 54%.
It's down to 54% silica.
And look at this-- boatloads of lead oxide.
And the lead oxide is acting as a modifier and also changes the index of refraction.
It modifies so much that we have almost down to the orthosilicate.
So that the chains are modified to the point where they're almost all terminals.
There's very little of this, and most of it is just terminal oxygens.
And that means that it's more nearly crystalline, and therefore, you can cut it.
And this is the lead crystal.
Lead crystal has that high value.
So you can see how that comes out.
And this is also taken from the reading.
It's a plot of viscosity versus temperature.
Only it's a logarithmic plot.
And what do you see?
Here's pure silica.
That's SiO2.
And if I want to take it down to the point where, if the viscosity goes down, you have the ability to work with the glass.
So if I want to melt silica, I've got to way, way up here.
Over 2,000 Centigrade.
So if I want to make bottles, if I want to make cookware, I don't want to run a lehr at this temperature.
Lehr.
L E H R. It's where you melt glass.
Lehr.
I don't want to run a lehr at this temperature.
But you can see if I add modifier, the more modifier that I add, the more I break the network.
So increasing modifier decreases network connectivity, and that means I can go to a lower temperature and get the same level of fluidity.
So there's a whole bunch of definitions here that I'm not going to go over.
You'll have this slide.
But basically, it's just different points in processing.
So if you go up here.
Strain and annealing.
I mean, these are very, very-- you're especially working with solid, whereas softening point, you start getting the glass to flow.
And the working point, that's the viscosity that you have to get below.
Otherwise, the glass is going to be to resistant to flow.
You want to get the glass tacky so that you can put it into an injection mold, shape it, as they do with bottles.
They take a blob of glass and boom, they just throw it into a mold and it sprays out.
And if you've got the right mass and the right spinning, it makes the wall thickness proper.
But you can imagine, if the glass is really, really viscous, it's not going to flow well enough.
If it's too fluid, it'll drip all over.
So there's an optimum in there.
And this is telling you how to figure out what that optimum is.
And you can see that as you change the composition, here's the working value.
This is the viscosity that you have to get below which in order to work.
And you can see that as you add more and more modifier, you can take the temperature and get it way, way down.
So to work with silica you have to be about around 2,000.
With soda-lime you can be down around 800.
So that's going to cut your energy costs, isn't it?
And it's going to make it easier to recycle.
What's the point of recycling if you consume as much energy to recycle as if you started with virgin material?
At least with virgin material you can guarantee the quality of the feeds stock.
So there's got to be some big saving.
All right.
So that gives you a some sense as to what we can do technologically with glasses.
I want to show you one last thing with glasses, and then we're going to move on to another topic.
You know, new week, new day, new topic.
So I want to go back to this curve.
So what am I showing you here?
I'm showing you that as we change the cooling rate, we change the amount of quenched-in excess volume.
So fast cooling quenches in more of the liquid free volume than slow cooling does.
That's why this V excess is small here, whereas V excess is large here for the other one.
I'm going to use that in glass strengthening.
So I want to strengthen.
Strengthening glasses.
We're talking about silicates or borates.
Strengthening oxide glasses.
Glass is a fantastic material.
It's really good in compression, but it's no good in tensions.
You know this.
If you try to bend glass, it'll break.
Why?
Not because it doesn't have dislocations.
It's got strong covalent bonds.
But you know, suppose you want to strengthen the windshield of your car so that when a stone hits it, it doesn't shatter.
What can we do to give it added strength?
So I'm going to show you two ways, and both of them operate under this principle that the yield stress-- this is the stress that will break the glass, and I'm going to say the effective yield stress, what you experience in life-- is equal to the sum of what I'm going to call the natural yield stress, which is the basic property of the glass.
Plus, I'm going to increase the surface stress.
What I'm going to do is I'm going to modify the surface, and I'm going to put an additive stress at the surface, and that's going to be a compressive stress.
So that means now the effective stress that I need to break the glass is going to be greater than it otherwise would be.
So the whole gambit here is surface strengthening.
Service strengthening, which means surface modification.
And I'm going to show you two ways to modify the surface to bring the strength of the glass up.
The first way is thermal.
Thermal treatment.
And the thermal treatment is to strengthen the glass.
Well, let's see.
Let's keep it in the context of the windshield.
The technical term for this, the technological term is tempering.
So what I'm going to show you was how we can look at that volume versus temperature curve and understand how we make tempered glass for windshields and so.
So I'm going to show you a slab of glass.
Here's a slab of glass.
And we just got below the softening point-- so T just less than T softening.
So now this thing's going to start, it's continuing to become more and more viscous and getting closer and closer to glass transition temperature.
And what we're going to do is we're going to introduce air jets at the surface of the glass.
What that's going to do is it's going to cause accelerated cooling at the surface.
And the same thing happens on both surfaces, so I'm going to cut this piece of glass in half, and we're just going to look at the upper surface.
The same thing happens in a lower surface.
So let's now blow this up.
I'm just going to look at the upper-half surface and I'm going to divide it into two zones.
This is the first level, most primitive finite element analysis.
Finite element analysis.
I'm going to divide it in two, and I'm going to say, this has got two zones.
Two cooling zones.
Here is the zone of the center, OK?
This is the center.
And I'm going to call this the inner portion.
And then there's the outer portion.
Well, take a look at this curve here.
Which is going to have slower cooling?
In the center or near the free surface?
The slower cooling is in the center and according to this graph, the slow cooling has a smaller residual volume.
I've written, V-interior.
That's the green line.
And then the upper one is a yellow line.
So the upper one where it's high cooling, it's going to have a higher volume.
So this is the second piece of finite element.
So I'm going to model this one like so.
So this is outer.
And the same thing happens on the other side, OK?
So it's happening on the bottom as well.
But we're just looking at the top because there's symmetry here.
So you see what I've done?
This is longer because that graph says it wants to be occupying a larger volume.
Problem is the glass can't do this.
I can do this with a piece of chalk, but the glass isn't going to look like that.
The glass is going to have a flat edge.
And how can it have a flat edge?
Because the bottom, here-- this is not to scale, let's make it more to scale-- the top is a narrow zone and the bottom is a big, thick zone, isn't it?
So this big, thick zone, which has a small volume, is going to pull on the thin upper zone, which has a large volume, and pull it in.
Can you see that?
And it's going to cause the introduction of compressive stresses.
So we got to compressive stresses simply using that graph and a little bit of air.
So you take that graph, put differential cooling, and now you've introduced compressive stress.
And that's all tempered glass is.
So the V-excess of the outer layer is greater than V-excess of the inner layer, or the interior, if you like.
And why?
Because the cooling rate, the change of temperature with time of the outer zone, is greater than the cooling rate of the inner or interior.
And there it is.
That's the beginning.
And so now to fracture you have to apply a greater stress then you would have otherwise.
So that's good, and that saves a lot of lives.
There's another way.
There's another way to surface strengthen, and that's a chemical treatment.
And again, what am I trying to do?
I'm trying to introduce a compressive stress, but I'm going to use a chemical means.
And this one is called ion exchange.
And this is used in technology, too.
So now I'm going to take a piece of glass here.
This is solid glass and let's put some components in here.
So I'm going to put some silica as my network former.
I'm going to put some sodium oxide.
And I'm going to put some modifier, B2O3.
So I got all three here.
Former, modifier, intermediate.
And just to put a little skin on the bones here, I want to show what the sodium oxide actually looks like.
Sodium oxide goes in as sodium cations, and oxide anions.
The oxide anions go in and they break some of the silica chains.
But the sodiums don't get involved in that.
They just sit around as spectators.
Now what I'm going to do is I'm going to put this, I'm going to soak this in molten salt.
Soak in molten salt.
This is huge area of my own research.
And the molten salt, one example might be I'm going to take potassium chloride.
Remember, we talked about making aluminum or magnesium?
This is one of the constituents of the melt in which we make electrolytic magnesium.
Potassium chloride.
And it exists as potassium cations and chloride anions.
Chlorine is green, except we know-- you know, this is the way chemistry books write it, but that's stupid because this is isoelectronic with argon.
And it's not green.
I know.
I've looked at this stuff.
It's clear, colorless, and transparent.
The chloride ion has to be clear and colorless.
It's got a complete shell, but the chemistry books will make it green.
Anyway, here's the potassium ion here.
And there's a lot of potassium ion here.
There's no potassium ion inside the glass.
There's sodium ion in the glass, there's no sodium ion in the molten salt.
But these are both media in which ions live.
So can you see this is sort of like the perfume bottle?
Things move from high concentration to low concentration.
Some sodium wants to leave and enter the melt.
And some potassium wants to leave the melt and enter the glass.
And where's the potassium going to go?
It has to go where there used to be sodium.
What's the relative size of potassium ion versus sodium ion?
Potassium ion is bigger.
So potassium ion goes and occupies a site formerly occupied by sodium and causes compressive stress, because you get all these big ions jamming in there and that's going to lead to compressive stress through ion exchange.
So through ion exchange we can raised the temp.
Since r, the radius of potassium is greater than the radius of the sodium, we get a force fit, and this is how we got solution hardening.
Hardening is the metallurgical term for raising the strength.
Hardening is equal to strength going up.
So we get hardening by solution.
So this is solution hardening.
Solution hardening because the potassium ion is in there.
So we have surface.
And this gives a lot of strength.
We do the same thing metallurgically if you carburize the surface of a piece of steel.
Tool steels, for example, surface carburized.
So we get hard surfaces that can cut.
You say, well, why don't you just put the carbon in all the way through?
Well, then the tool bit will be brittle.
So you want something that's got toughness in the center so it can take the impact, but it's got surface hardness so it'll cut.
And that's an engineered material.
We've got one set of properties at the surface, and we've got another set of properties in the bulk.
And it's all brought to you by control of chemistry.
And then that's also control of chemistry.
OK.
So I mean, I could talk more and more and more.
Glasses are just fascinating things.
But we've got to get moving.
So we're going to move to a new topic.
We're going to move to a new topic today.
We're going to start talking about kinetics.
Kinetics.
And what is kinetics?
Kinetics is the topic that we put into 3.091, because it's important, of course, but it's all about the study of reaction rates.
It's the study of reaction rates and mechanism.
So why do we study?
Well, we study it because I think it belongs here.
But first of all, kinetics is related to productivity and resource utilization.
You've got two factories.
One produces 200 units per time, the other produces 100 units per time.
The one that's producing 100 units per time probably isn't going to be in business that much longer.
And it's all about understanding how to get more throughput per unit time, and that leads to competitiveness.
So if you're interested in international competitiveness, you've got to know something about kinetics.
The two go hand in glove.
Second thing is energy and the environment.
A number of you have come to talk to me after class, sent emails because you're interested in energy and the environment.
Efficient use of energy involves understanding the kinetics so you can force chemical change with the least amount of energy utilization and do so in a way because its mechanism-- if there's two mechanisms that allow you to get to the same end product, choose the one that has the least toxic impact on the environment.
Otherwise, you're going to have to spend money to avoid the effluents, which then gets you back up to number one.
Your cost of doing business is higher than the other guy.
Guess what?
You're creamed in the marketplace.
And lastly, societal.
And this is one that excites me, is kinetics leads to productivity, which keeps factories open, which keeps people working.
So through understanding kinetics.
You go and you look at places in the United States where factories have closed.
In many, many instances it was because the technology lagged.
If you could just keep things going faster.
You've got to make things-- you know how fast you've got to make them go?
You've got to make them go so fast that even if the foreign workers are paid zero, you can still beat them in the marketplace.
That's your standard.
So that's how you design your processes.
You design your processes so that you can produce at very, very low cost.
And there's a huge range in reaction rates.
You know, you can start the range of rates, rates of reaction.
At the one that you have very, very slow.
And way over here you have very, very fast, all right?
So on the slow end, you have stuff that's sort of cosmic scale.
Geological.
And then over here, what do we have?
We have explosions.
You might say, well, explosions are usually bad, aren't they?
Well, no.
I can give you an example of where explosion is used to your advantage.
The airbag in the automobile.
There is no pump, no mechanical pump fast enough to inflate an airbag on demand.
So how do we make those airbags work?
And I hope none of us ever has to be present at the deployment of those airbags.
But you know, just academically let's talk about how they work.
There's like an explosion that occurs when you're driving.
There's an accelerometer and this is a key piece.
The accelerometer has to make a decision whether you're applying the brakes just because you've lost focus and you're, you know, whoops!
You've had that jerky brake.
Or maybe this is one of those holy-mackerel moments, and your jamming on the brakes, and we'd better deploy the airbags.
Once the accelerometer decides that this is one of those holy you-know-what moments, it sends an electric current through a wire, which then raises the temperature to 300 degrees C and causes this reaction to take place.
When this reaction takes place, this is a solid, nitrogen is a gas.
So the gas has a much, much higher volume than the solid and within milliseconds you get inflation.
Well, that's good and that inflates the bag and prevents you from hitting the hard parts of the car.
But you see the other byproduct?
That's sodium.
And if you look on your Periodic Table, elemental sodium is liquid above 98 degrees Celsius.
So now you've got liquid sodium.
There's no point preserving the safety of the occupants of the car just to cover them with liquid sodium.
So we better do something about that.
So being chemists, what we do is we add potassium nitrate inside the bag, as well.
so.
That that sodium is mopped up with potassium nitrate, which converts it to sodium oxide, potassium oxide, and a little more nitrogen.
[BLOWING SOUND EFFECT] Keep the bag nice and firm.
But now what happens when the Fires show up and they [SOUND EFFECT], you know, they start pouring water on it.
This turns into caustic.
So now you're going to get covered and wet lye.
So that's probably not too good, either.
Just remember, you've been physically saved from smashing your skull on the tempered windshield.
So it's so far, so good.
So now what?
You want a mechanical, you want a Newtonian death?
Or do you want a Coulombic death?
I guess that's the question, here.
And so we're going to keep doing some chemistry here and we're going to add silica.
And what happens if we add silica to sodium oxide potassium oxide?
Well that's a network former, these are network modifiers.
We'll make an alkaline silicate glass and that's OK because we put food and beverages in alkaline silicate glasses.
So everybody is happy.
So that's all the chemistry that goes on inside an airbag.
And we've got to understand the kinetics.
So I hope I've whet your appetite.
OK, that's enough of that.
Now let's get to the real stuff.
So let's write a general chemical reaction.
The general reaction.
What's the formalism if we want to set up the metrics?
So I'm going to write just a plain, old equation.
This is the classical P-Chem stuff.
Little a moles of A, plus little b moles of B go to little c moles of C plus little d moles of D. So this could have been lecture two.
Just a straight chemical reaction.
Remember we were studying stoichiometry?
So the constituents of the left side of the equation are called the reactants, whereas on the right side of the equation we have the products.
And what kinetics gives us, kinetics tells us the rate of conversion.
Or the rate of reaction.
I'm using all these terms so you understand their equivalent.
So we can write something like this where the conservation of mass kicks in.
So I can't make products any faster than I consume reactants.
There's no sources or sinks here.
When I lose reactants, I make products.
So I'm going to say, the rate of change of the total mass of the reactants-- this is summation sign-- I'm just saying the rate of change of the sum of all of the reactions must equal the rate of change.
So the loss rate of reactants must equal the gain rate of all of the products.
That's just simple stoichiometry.
Or some people call this conservation of mass or Law of Mass Action, what have you.
And then we count by concentration, usually.
Just reminder.
This is lowercase c. So this is the concentration of species i goes as the mole number.
n is mole number divided by the volume of the reactor.
So this would be in moles per meter cubed if we were in SI units.
So therefore the rate of change, the ci by dt, is really the rate of change of mole number, isn't it?
It's moles of i disappearing, but the volume doesn't change, typically.
And then the last thing is we can use this idea up here that concentration of products can only equal the loss rate of the reactant.
So I can write, term by term, the normalized rate loss of the concentration of a divided by its stoichiometric coefficient, for example, would equal the normalized weight gain, or pardon me, the concentration gain of the concentration of d. So this is just saying a disappears no faster than d appears, mediated by the stoichiometric coefficients.
So this is just Law of Mass Action, which is stoichiometry in motion, isn't it?
That's all it is.
Now here comes the cool thing kinetic theory.
We look at any reaction, I can tell you right now what it looks like.
If I applied concentration of i as a function of time, I start off with some value c naught, some initial value, and it just attenuates.
We know that's going to happen as we consume the-- it falls and there's some curvature here.
What we're trying to do next is to give some mathematical representation to this.
Can we come up with a mathematical formulation of the shape of that curve?
And we can, and here's the central tenet kinetic of theory.
Kinetic theory says that the reaction rate can be expressed in terms of a driving force.
And that driving force is the instant concentration.
OK.
These are lofty words.
I'll show you what it means with illustration.
So that's the overall idea.
So now let's say that the rate of change, the instant rate of change of concentration ci, is proportional to the instant value of the concentration.
And you know, the concentration is changing, right?
As the concentration decreases, the rate of change decreases.
So just this alone, this idea alone would rationalize a curve like that.
As the concentration falls, the rate of change falls, which means the concentration falls, which means-- and raised to some power-- it's not linear, necessarily-- raised to some power.
There's a power law at work here.
So now let's do this one more time.
I'm going to write it mathematically.
But I want you to first see the word, concepts.
So let's write it mathematically.
That means minus dci by dt-- so this is time rate a change of concentration i, the minus means it's falling-- is equal to the concentration of i raised to some power n. This is order reaction.
It's not mole number over there.
You've got to be pluralistic today.
n is going to be used in different ways.
So this is called order of reaction.
And there's a constant.
See up there?
It's a proportionality.
Here it's an equal sign, thanks to the constant.
And this is called the rate constant.
And this what all I need for plant design.
Suppose I'm running a chemical plant and I'm taking a plus b and making c plus d and the management says, we want to double the rate of productivity.
So you go, I know what to do.
We'll increase the concentration.
They say, by how much?
Well, this allows you, if you know the value of k and n, you want to double this?
Then you know what to do here.
It's not necessarily double the concentration, Because this is a 1.5 power.
By the way, this doesn't have to be an integer.
Could be anything.
Could be not necessarily integer and must be determined by experiment.
You can't look at an equation and say, oh, that's going to be second order because there's a two in front of something.
So here we are, plant designers.
And the management wants to know, how do we make things happen?
So I'm going to go back to that equation over there and I'll say that most generally, the rate of change of concentration of a will then go as rate constant times the concentration of a.
See, the rate of change goes as something to the power alpha.
But it could also be influenced by b.
So I'm going to put-- this is the most general form.
Then once you make your measurements, a lot of these fall out.
I'm even going to put c I'm going to put d. Everything was in that equation.
You might say, well, wait a minute.
How can the rate of consumption of a be influenced by the concentration of d?
d is a product.
I'll give you two examples.
One is that d, when I'm making d, d happens to have some catalytic value.
So when I start making a convert to d, d catalyzes the reaction, which makes it go faster, which makes more d, which makes the reaction go faster, in which case this is going to have a profound effect.
And unfortunately, there are some other situations where a converts to d and d retards the reaction.
And so I start off with pretty decent conversion efficiency, but as I make more and more d, the d chokes the reaction, in which case the value here is going to have negative implications.
So that's why you write it most generally, and then you go and you make some measurements in an experiment, and you figure out what these are.
Some of them might be zeroes, and away you go.
So I'll give you an example of one.
Here's one.
It's Monday after Halloween, so we'll get something kind of toxic.
This was the manufacture of phosgene that was banned by international convention.
It was used as one of the toxic gases for trench warfare in World War I.
So it's made by the reaction of carbon monoxide and chlorine.
COCL2.
This is called phosgene.
It's very bad stuff.
I used it, actually, in my research.
It has really good dehydration properties for salts.
If you've got any moisture in potassium chloride, this'll go after it and turn the water into carbon dioxide and HCl.
But you've got to be really, really careful with this.
Gas leaks lead to a bad day at the lab.
All right, this was banned.
It's still use by tyrants throughout the world, and that's why we have international war tribunals.
So here's the kinetics of it.
Here's the kinetics of it.
dc by dt.
So this is the rate of consumption of carbon monoxide determined by experiment goes as the concentration of carbon monoxide and the concentration of chlorine raised to the power 3/2.
So what I've done is mimic here.
So in this case, gamma and delta are 0.
Gamma equals delta equals 0, and it looks like beta equals 1.5 and alpha equals 1.
So I would say that this reaction is first order in carbon monoxide of order 1.5 in chlorine, or overall, of order 2.5.
Because, right?
This is the order of reaction.
This is c to the power 1, isn't it?
We don't write the 1.
This is great.
So we know, to increase the rate of reaction we increase the concentration.
There's another way to increase the rate of reaction.
What else can we do to increase the rate of reaction?
Increase the temperature.
Things go faster at higher temperatures.
Again, how?
How much?
Suppose I said, I want to double the rate of reaction.
Do I double the temperature?
Do I raise the temperature by 10 degrees?
How do I do that?
So to increase rate of reaction, we can increase temperature.
But how?
And by the way, where is temperature here?
I don't see temperature.
I don't see temperature in this rate equation.
You could say, well, if you increase the temperature you're going to increase the gas volume.
But that's piddling effect, right?
If I double the temperature, I double the gas volume.
It doesn't change the mole number.
The temperatures in here.
That's where the temperature.
Increase temperature, which then operates on the rate constant.
That's the only place left in that formalism.
But what's the quantitative value?
Quantitative value was annunciated to us in 1889 by the Swedish chemist Arrhenius.
And what Arrhenius found, he found that the rate constant varied with temperature in the following manner.
Rate constant goes as the exponential of the ratio of minus some quantity call the activation energy e sub a-- I'm going to tell you what that means in a minute-- divided by the ratio of the product of the Boltzmann constant and temperature.
And there's a constant of proportionality which we denote A in honor of Arrhenius.
Or another way to write this is A times exponential-- are you familiar with this?
Instead of writing e to the something, if there's a messy argument up here you can write it exp.
This is the same as this minus ea over the ratio of Boltzmann constant and temperature.
So what's in here?
What's in here?
Boltzmann constant temperature, we've seen that already.
What's the Boltzmann constant temperature product?
That's the environment.
That's the energy of the environment.
So this is the final environment.
And this thing up on top, ea, is a barrier.
It's a barrier.
Remember we saw the band gap?
Same idea.
Those ideas go all through physical chemistry.
You got thermal energy that allows you to do this 10 trillion times a second.
Everybody in this room is doing this 10 trillion times a second.
That's what kT is.
And then there's a barrier energy.
Every once in a while something happens.
Sometimes one in a million, sometimes one in a billion.
Whatever.
What Arrhenius found was this relationship, which means we could plot the variation of the rate constant.
This is a mess, right?
This is an exponential-- look, the eye can only see straight lines.
What if I show you this?
What's that mean?
I don't know.
Is that first order?
Second order?
Is that Lagrangian?
Is that exponential?
I don't know.
I know it's a curve.
But what's this?
I know what that is.
And you know what that is.
And you can tell goodness of fit.
This is what science is all about.
Forget all the other stuff.
You know, all that UROP stuff.
You're going to work in a lab and discover stuff.
Forget that.
This is what it's about.
You got data.
Data are all over the map.
Here's the thing-- I need to come up with some f function here versus a g function here that linearize my data.
That's all science is about.
And how do I get to f and g?
Well, there's two ways.
One is trial and error.
The other way is physical understanding.
And this is what Arrhenius taught us.
Arrhenius taught us that the f function for how k varies with temperature, right?
The problem, I'm trying to say, k versus t. That's what I'm trying to do.
And if I plot k versus t I get a curve.
No good.
Instead, if I plot some f of k versus g of t, I get a straight line.
And this is all science.
So what's f of k?
The natural logarithm of k, right?
If I've got k equals a exponential, I take the natural log of this, and the natural log of an exponential is just the argument.
And what's in the argument?
What's the T function?
1 over T. So if I plot the natural log of k versus the reciprocal of the absolute temperature, I get a straight line and its slope is minus ea, forgive me, minus ea over k-Boltzmann.
Minus ea over k-Boltzmann.
So now it's linearized and I can look at that and within an instant, anybody in this room can look at and say, the data conform or they don't.
They don't.
So now let's think about this.
We say well, what's the value of this barrier?
It's called activation energy.
This barrier energy is called activation energy.
ea is called the activation energy.
And it typically, it varies, but it has values-- just to give you a sense-- about 1 electron volt.
1 electron volt, which you know now is about 100 kilojoules per mole, isn't it?
I hear the chorus of yesses.
1 electron volt.
What is thermal energy?
What's kT?
kT at room temperature, k-Boltzmann at room temperature is about 1/40 of an electron volt.
Now, we are chemical machines.
This conversation is occurring because all sorts of chemical processes are at work in me and you.
And how's that happening?
When all we've got to suck out of the environment is 1/40 of an electron volt and we need on the order one electron volt to drive certain processes.
I guess we're all dead.
Any ideas?
Ay yay, yay.
You guys need-- what's that thing?
You've got to call your lifeline or something?
Come on.
How does anything happen?
How do we get-- Why does anything happen in this world?
Why should anything happen in a world with an environment of 1/40 of an electron volt?
Catalysts
Oh, catalysts.
Catalysts.
Yeah, yeah.
I just went to the catalyst store and I got catalysts.
Come on
Potential energy
Oh, potential energy.
Jeez, this is good.
Who can give me the wackiest answer?
Can anybody give me the right answer?
How come anything is happening?
Over here.
Again?
Distribution of temperature
Thank you.
Thank you.
For a moment I thought we were all dead.
At some level we were.
So, yeah, it's Maxwell-Boltzmann.
It's this, isn't it?
And here's room temperature.
And here is 1 electron volt.
And this is it.
And if you use this and you put it back to there, pretty soon you derive the Arrhenius equation.
And if you increase the temperature you know what happens.
After you increase the temperature, this.
Time to use a hot color or chalk.
So now this is T2 greater than T1.
That's what's going on.
That's what's going on.
So now what is this activation energy?
What is the meaning of it?
You know, I've told you there's a barrier energy.
What does it mean?
Well, you go to the textbook, you see goofy stuff like this.
And this is not a slam against the textbook.
This is a slam against all chemistry textbooks, because they all write this stupid stuff.
You see?
a plus b goes c plus d, the energy falls and you have to go over this activated complex.
You memorize it, and I ask you to repeat, it and we leave the room and we think, wow, we know physical chemistry.
There's nothing here.
This is nothing.
This is stupid.
Now this is a little bit better.
Can you see this?
This is the Maxwell-Boltzmann.
This is actually a beautiful graph up to a point.
See the Maxwell-Boltzmann?
Low temperature, high temperature.
Only they've turned this thing on its side.
See?
And there's the energy you need to get over the activated complex.
You see what's wrong with this graph?
What's wrong with it?
The product
Yeah, the product is at a higher level than the reactants.
I looked at that and I thought, whoa!
I guess you'll activate them, but they won't go anywhere.
Anyway, so this is all stupid.
What can we do?
What can we do?
So I decided to give you a mechanical analogy.
So imagine this is the loudspeaker.
Can you say the x?
People in the back, can you see the x?
Give me a thumbs up.
Great eyes.
All right.
So what do we see here?
The center of mass, I've indicated here.
All right.
So this is in a certain energy state.
Let's get the right graph up there.
This is awful.
All right.
So this is a plus b.
Now this is c plus d. You see, the center of mass fell.
It's lower.
It's closer to the table than it is here.
Can you see when I go like this, the photon goes?
Watch this.
Yeah, only I can see it.
Your eyes don't go to that end of the spectrum.
All right.
But what's wrong with it?
So why does the box not fall over?
Why does it not fall over?
We agreed that this is a lower energy state, and it is.
You're correct.
So why does the box manage to stay here and not fall over?
[INAUDIBLE PHRASE]
Exactly.
This is the box at 0 Kelvin.
Now we start raising the energy of the box, it starts doing this.
Because above 0 Kelvin, it vibrates.
The higher the temperature, the greater the vibration.
But we're still in trouble here, you see, because you've only got 1/40 of an electron volt.
So now we invoke distribution.
I don't know what's happening with any box.
Heisenberg tells me that.
I don't know what's happening to any box.
But if I take Avogadro's number of boxes I'll get this distribution.
And some boxes will vibrate very little.
And some boxes will vibrate a lot.
And what's the critical level of vibration?
It's to get to here.
Because once it gets to here it's downhill all the way.
And that would be the activation energy.
And the fraction of boxes that get to the value of activation energy tip over and then I restore the distribution, because now this one's out of the game and now I distribute that energy over the remaining ones.
And as I increase the temperature, the amplitude of the vibration increases, the average increases, the fraction in that red zone increases, and at some temperature it's so, so hot that they actually do this, in which case I have equilibrium.
The two states are in equilibrium.
So that's it.
This is what's going on.
So in three space, you can't go from here to here without going to here.
You can try going this way so you-- no, you have to do this.
It still has to be mechanically activated so we can show that this is what activation means.
So we can plot-- this deserves it's own board-- so we will plot something like this.
And now you'll see this.
Sometimes they write this in the books.
It's kind of goofy.
But they write stuff like this.
This is some kind of an energy coordinate.
This is an energy coordinate.
And this is called a reaction coordinate.
OR sometimes they write extent of reaction.
Again, some P-Chem term.
They usually use Greek C just to make it lofty, but it's meaningless.
And then so you write like this-- reactants here, products here, and this is exactly what we're seeing.
What do we see?
We see this.
Here we have the box sitting like so.
And then over here we have the box sitting like so.
And right here we have the box up like so.
And so now you have, here's the initial energy.
This is the energy of the reactants.
E of the reactants.
This is E of the products.
So then this distance here must be delta E of the reaction.
And what's this?
That's the activation energy.
I have to come up with this; otherwise, I can't make the box fall over.
So this is ea to go from here up onto the corner.
And now you understand what all this stuff means.
So that's here.
And this is the activated complex.
What's an activated complex?
It's a box on its edge ready to fall over.
That's what it is.
All right.
I think we're out of time, so let's cut to the end here.
What've we got.
Oh, just some plots.
We'll get to that next day.
All right.
So I want to show you a little bit about first-order reactions.
This is radiocarbon dating.
n equals 1.
And strictly speaking, this is not a chemical reaction, but it is first order.
So I'll lump it in here.
So this is a nuclear reaction.
And what happens in the upper atmosphere?
Radioactive carbon produced by cosmic rays, which generate neutrons in the upper atmosphere and then those neutrons attack nitrogen to make carbon 14.
That's the radioactive form of carbon.
Remember?
It's present in one part per trillion, if you look on your Periodic Table for the isotopes.
Now carbon 14 enters the carbon cycle.
And so in all of us and in all living organisms the ratio of carbon 14 to carbon 12 is one part per trillion.
And carbon 13 is also present.
I don't know.
What is it?
1 point something percent?
The thing is, the carbon 14 is radioactive.
And it decays.
Like this.
But what happens after somebody dies?
Well, they stop exchanging nutrients with the surroundings.
They stop breathing, so their carbon level is pegged as it was at the time of death.
And now it's a one0way street.
It's just carbon decay.
And it turns out you can measure the ratio of carbon 14 to carbon 12 to determine the age, which is given by 5,730 years, which is the inverse of the rate constant.
Now, you can't use this for crime scene investigations, but certainly you can start dating things hundreds of years old, not hundreds of minutes old.
So there's a whole bunch of other things that can be used, not just carbon.
So this is radiochemical dating.
And these are used in trying to nail down art forgeries and all sorts of things.
All right.
So here, you can see these.
These have all been radiocarbon dated.
You know, the Dead Sea Scrolls were radiocarbon dated and they look like they're from about 2,000 years ago.
Makes sense.
You go out over here.
They they found these Indian sandals from Oregon, and then they looked on an electron microscope, and they found something on the sandals that looks a little bit like that.
That's a Nike swoosh.
I don't know.
You people are so-- All right.
So now I want to show you the Shroud of Turin.
The Shroud of Turin, as you may know, for a long time was reputed to be the burial shroud of Jesus Christ.
And this is taken from the National Geographic back in the late '80s.
Back in the late '80s, they did a major study on it.
So this is the artist rendition of how the shroud might have been wrapped around a body.
The question is, what's the date of this?
So we can use radiocarbon dating.
And so here's the formation of carbon 14.
Carbon 14 exchanges with carbon 12 in flax.
So it's got one part per trillion.
When the flax is harvested the carbon 14 starts to decay.
And then we look at the weaving of the shroud and so on.
And they had three different labs, they took samples from the shroud.
And they concluded that the shroud dates from this period-- 1260 to 1390.
And the first time it was mentioned in the literature was around 1354.
So everything seems to make sense that this is not the burial shroud of Jesus Christ.
It's something that has importance to certain members of the Roman Church, but it's scientifically not.
You know, this stirs up a lot of passion.
So people started saying, wait a minute.
Wait a minute.
There was a fire in 1532 and people repaired the shroud and they were using candlelight and maybe the paraffin and mold and so on covered the fibers.
So now, are we looking at the surface effect?
How do we know what really happened?
The only way to know is to take the whole shroud, put in a big Cuisinart and break open the interior.
Well, in point of fact, what they did was take fibers and look inside.
And I think as of the late '90s, the Roman Church says, it's not the burial shroud of Jesus, but it is an object of reverence.
And I think it's a good example of how science sometimes comes up against spirituality, and we have to be careful how we handle it.
It's a delicate matter.
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OK, OK. Let's get started.
Going to go right into the lesson.
Last day we started looking at kinetics, which was the description of rate and mechanism of chemical reactions.
We looked at the general rate law, which shows us that the instant change in concentration of species i goes as the instant concentration.
So rate is proportional to concentration.
And the order of reaction is shown here as n and proportionality constant is the rate constant k. So we saw that we can make the rate of reaction increase by increasing the concentration in the reactor.
We saw there was a second way to increase the rate of reaction, and that was to raise the temperature.
And this is the Arrhenius relationship that shows that the rate constant changes with temperature according to this exponential relationship.
In the denominator, we have the product of the Boltzmann constant and the absolute temperature, which is a measure of the ambient thermal energy that's available to promote reaction.
And in the numerator we have a barrier energy, which we call the activation energy.
We'll say a little bit more about that today.
And the constant of proportionality out here is A, which is in honor of Arrhenius, who first annunciated the law.
So what I want to do today is to look at some common orders of reactions.
So we're going to look at the values of n that are encountered in not insignificant number of reactions.
So let's start with the simplest one.
The simplest one is first order.
First order reactions.
That means n equals 1 in this rate law expression.
And here's an example of something that we could use.
The decomposition of N2O5 into NO2.
And gives us half a mole of oxygen.
And we can write this rate law in terms of the rate of disappearance of N2O5.
So I can write minus d by dt of-- I'm going to use square brackets to indicate the concentration-- of N2O5.
And this has been measured.
You can't predict this from the stoichiometry of the equation.
You have to measure this, and on the basis of measurements it's found that the rate of change of N2O5 goes as the concentration of N2O5.
And there's a constant out here.
The rate constant.
So just to make the point, I'm going to put a 1 here.
Meaning that n equals 1 and that's why this is a first-order reaction.
But this is messy.
There's too many characters here.
So I'm going to compress this and give myself a nicer font, so I'm going to let the concentration of N2O5, just in this expression, just be c. Lowercase c. That way I can speed this thing up.
So I can rewrite this expression here as minus dc by dt equals k times c. All right.
And so we know what this is going to look like.
They all look like this.
If we plot concentration versus temperature we expect to have the maximum value of this at time zero.
So at time zero I start off with-- if you want c-star or c max-- and you know, this thing attenuates.
But I can't look at that and decide whether it's first order or not.
The only thing the eye can really tell is a straight line.
So what I want to do is I want to modify this thing.
I want to map c into some f of c. And I want to map t into some g of t. So that if I plot f versus g, I get a straight line.
And then I can take one look at that and go, bingo, this is first order.
Or not.
Or maybe it's first order and the data are very noisy.
So what we'll do is we're going to integrate that equation.
So I'm going to play with this a little bit.
So I'm going to transpose.
I'll get minus dc over c equals kdt.
And then I can integrate this because I know at time 0, t equals 0, c equals some value-- I can call it c-star or c naught-- and at any arbitrary time later, I can go down that curve and put some value of time and then I'll end up with some arbitrary value of concentration c. And I know there's some math people in the audience who are probably horrified that the limits and the integrand are both the same.
So I'll put a little tilde here that way I don't go to math jail.
So now I'm not integrating c. If I wanted to be a pedant and really make the math instead of me enslaving the math, the math enslaving me, watch this.
What I could do is I could make this a carrier variable, s. You know what I'm talking about now, right?
But let's get really crazy.
This is how bad the math can get.
You can do this in a chemistry book.
Let's use some Greek.
So I'll make this d xi over xi.
Now I've totally left the class behind, but it's mathematically correct.
What we will do is we'll put the t back and put a tilde over it.
And now everybody is happy, and heaven forbid, the student might even understand the lesson.
All right.
So we do that and what do we get?
The integral of dc over c is a natural log of c. This'll be natural log of c equals natural log of c naught minus kt.
OK, so integrating this thing gives us this expression.
And that means if I plot the natural log of c versus time I should get a straight line.
And furthermore, the slope is equal to minus k. So here's an example that's taken from your book.
It happens to be another first-order reaction and it shows the decomposition of cisplatin.
This is a compound that's used in certain cancer treatments.
And you can cisplatin hydrolyzes to form this monster cation plus a chloride.
And it turns out that the rate of decomposition of cisplatin is first order.
Will you look at that curve?
That purple line?
I don't know what that means.
But watch.
If I instead plot the natural logarithm of the concentration of cisplatin as a function of time, you get a nice straight line, and furthermore, the value of k is obvious from the slope.
So this is an indication of the power of taking the proper order of reaction and then using it in-- if you'll pardon the pun-- straightening things out.
And a couple other things worth noting.
I mentioned at the end of last day, radioactive decay.
Even though radioactive decay is not a chemical reaction-- it has to do with nuclear phenomena-- and we know chemistry is restricted to the life and times of the electron, we don't go into the nucleus.
We simply know there's something in there, there's neutrons, protons, and a whole bunch of other particles.
But chemistry is about the life and times of the electron.
Radioactive decay turns out to be first order.
It's good to know that because then you can appreciate what goes on.
So for example, you could take something like the decomposition of uranium-238.
One of its decompositions gives you thorium-234 plus helium.
And it turns out that this is first order, which means that if you were to take a plot of the natural log of the concentration of uranium versus time, you get a straight line.
And this slope would be the k, the rate constant.
But it turns out that the nuclear chemists, nuclear engineers, prefer not to talk about the rate of reaction in terms of k. But instead, they like to use the inverse of k, which is the half-life.
So instead, express k through half-life.
And the half-life, as the name implies, is the time it takes to decrease the concentration of the species by half.
So if I start off here at c1 and down here I have c2 equals 1/2 c1, then this is the time it takes to cut the concentration in half.
So through half-life-- and we can go through this equation here, just put c naught over 2 and do a little bit of manipulation, you'll end up with the natural log of 2 divided by the rate constant, which is 0.693 divided by k. So you can see that there's a direct relationship.
It turns out that the half-life for this reaction is 4 1/2 billion years.
4 1/2 billion years.
So that means after 4 1/2 billion years the concentration has gone down to a half, but even then it's probably not safe for the kids to go out and play.
So you really need to think about it in this way.
So here it is shown in your textbook.
Concentration as a function of time.
They could have plotted it log linear, then you would have ended up with this.
But anyways, they're chemists, so what can you expect.
So here it is.
Here it goes down to a half, and then this is a half of a half as a quarter.
It's the same time interval.
And then half of a quarter is an eighth, and it's the same time interval.
So you get the sense.
And the most important thing is the half-life is independent of concentration.
It doesn't matter what the starting concentration is-- if you know what the half-life is, you know how much time it takes to get to a half.
And you know enough math.
What if I said, well, what if I didn't want to go to a half?
What if I wanted to go to 10%?
Well, you can figure out how to modulate half-life into tenth-life.
That's trivial.
But the book, being a chemistry book, will show you a table in which they give half-life values for orders of reactions other than 1.
And it's stupid because except in the case of first order reactions, the value the half-life is a function of the concentration.
So what good is a constant that's variable?
I don't know, but I didn't write the book.
The only time I care about half-life is in the case of a first order reaction, which is when you have nuclear decay for sure.
All right.
So that's enough about first order.
Let's go look at second-order reactions.
So here's an example also taken from the book.
So second-order reaction, which means that n equals 2 in the rate law expression.
n equals 2 in the rate law expression.
And the example that they gave was NO2.
See, look.
We went from N2O5 to NO2.
And then we can keep the decomposition going.
NO2 goes to NO plus O2.
And again, I'm going to let, in this case, c will equal the concentration of NO2.
And what you end up with in this case is, if you go through the similar analysis, you get minus dc by dt.
It's been measured that the rate of loss of NO2 goes as the square of its concentration.
So this is based on experimental evidence.
It's not because this is a 2.
This is not.
Please pay attention here.
This 2 is not the consequences of the stoichiometry.
This 2 was determined experimentally.
I could have just written this reaction NO2 goes to NO plus 1/2 O2.
That doesn't make this first order, does it?
Experimentally in the laboratory.
All right.
So now what we can do is we can integrate this thing.
And so we'll get minus the-- let's do it this way.
Boards are cheap, so let's jump down to the next one.
So now I can then put minus the integral of dc tilde-- cross multiply-- so dc tilde over c tilde squared equals k, the integral dt tilde.
And we can integrate this from zero to some time, and this will go from c naught to some concentration.
And that will give us 1 over c equals 1 over c naught plus kt.
So this is now the new, improved way of mapping c versus t into f of c versus g of t. So f of c as 1 over c, which means now if I plot 1 over c versus time, I'll get a straight line.
In this case, it's got positive slope.
It's got positive slope, so that's what it's going to look like.
And this is k. So if I take our data set and I plot it like this, then I can make sense of it.
And I think we've got some data here.
So here's the data from this decomposition.
And again, what do you see?
A line that attenuates.
Can you look at that line?
Can you tell that line is second order, whereas the other one that looks like this was first order?
Of course not.
The only way you can tell his to fit it into some linearizing function.
So look at that.
What they've done, that's the least squares fit.
They plugged it into a least squares fitting equation.
You can fit a straight line through anything.
It doesn't matter.
But what's that mean?
I don't know.
So here's the log.
This is really important-- look carefully.
So somebody took these data and then plotted them as the natural log, which is what you do if it's first order.
And look at what happens-- it gets a lot better but it's still not a straight line.
But if you were in a hurry, you'd just say, force the fit.
You'll get a straight line and you'll assigned the value of k to it and party on.
And it's wrong.
So now let's try this.
Beta.
1 over c versus time.
So what do we get from that?
I get two things.
First of all, the fact that it's a straight line means that it confirms n equals 2 and the slope of that line is the value of k. So in point of fact, what I've been doing here is teaching you how to determine the rate of reaction.
So let's do it.
Let's codify this.
So determination of order of reaction.
So in other words, the way the story goes is I give you a data set that looks like this-- pardon me.
I give you a data set that looks like this and I say, determine the order of reaction.
So there's two general ways to do this.
The first one is called the integral method.
And that's what we've been doing up until now.
The integral method, it's called the integral method because it's using the integrated form of the radio question.
See?
This is the integrative form.
This is the integrative form.
Hence, the integral method, and it's trial and error.
That's what we did.
Trial and error.
But we don't say trial and error because we're at MIT.
So what do we say?
We have a lofty way.
What's the lofty way of saying trial and error?
By inspection.
By inspection.
Never tell anybody you do things by trial and error.
We don't do trial and error.
By inspection.
It's $50,000 a year.
It's what you get.
OK, so here's what I do.
I'm impatient, so I try n equals 1.
If I get a straight line on the log plot, good.
If not, I try n equals 2.
If I got a straight line on 1 over c, good.
If not, go to the second way of determining.
I give up.
Won't waste my time.
It's a two strikes and you're out thing.
So what's the second method?
The second method is called the differential method.
The differential method.
And this is fun because it's data fitting and it gets into some nice detail.
So what I want to do is to define the rate.
What's the rate?
The rate is over here.
Minus dc by dt, right?
You see up in the top left corner the rate law.
Minus dc by dt.
I'm going to call that positive r. I'm just compress-- this is like French cooking.
I started with this.
Look at all those fonts.
Then I compressed it down to c, and now I compressed it even more.
It's just so much information compressed in that little r there.
Poor little r. All right.
So now, let's rewrite the rate law expression.
So that's r equals kc to the n. There's the rate law expression written very compact.
And now I'm going to take the logarithm of both sides.
So taking the log of both sides, I can write natural log of r, the log of a product is the sum of the log.
So that's going to be natural log of k plus the log of, da, da, da.
n log c. n log c. And I don't care what kind of logarithms you use.
You can make this log base c. You can make it log base 10.
If you're feeling like you want solidarity with the Mayans, you can make it log base 5.
I don't care.
But make them the same.
Make them the same.
And then what happens?
Well, let me show you.
So let's say we've got our data, start off with this data set like so.
So this is concentration versus time and we end up with this.
It comes down, down, down, attenuates.
Remember, what are we trying to do here?
We're trying to take this data set and figure out, what is the value of n?
We're trying to determine n. So what this thing says is.
if I make a plot of r versus c so that it's on a log scale, so if I plot log r versus log c, the slope is going to be n. So how do I take c versus t and construct log r verses log c?
That's what we're going to do.
So here we are and I can start here that at time zero.
This is time zero.
That's the slope, right?
The slope here is the rate.
So slope, I'm going to call it r naught.
Because that's the time-- it's dc by dt r naught at t equals t naught.
But there's no t on here.
So what am I going to put?
It's when t equals t naught and c equals c naught.
So now I'm going to take some colored chalk and we're going to show you how to construct this.
So I've got the slope by taking minus dc by dt here.
So I'm going to take this value, r naught, and this value, c naught-- I want to gang them together, I'll bring over here and put them right out here-- so now I've got r naught.
Well, actually I should put the abscissa first.
So this is going to be c naught r naught.
And now let's do it again.
So we'll come down here to some later time.
Let's call this t1.
And at t1 the concentration has fallen to c1, and I take the slope of this.
Why am I taking the slope?
Because it's dc by dt.
That's what the rate is.
So here now I've got slope r1 at c equals c1.
r versus c. Forget time.
Time doesn't belong here.
So take those.
It's a lower concentration, so it's kind of like a lower energy.
So we'll take a lower wavelength here.
So r1 and c1 come over here.
c1, r1.
We'll do one more.
Three's a charm.
Come up to here.
This is t2.
Concentration has fallen to c2, and I've got a slope.
You see how the slope is getting less and less steep?
Because the rate loss says the concentration falls, the rate falls.
So this is slope r2 at c equals c2.
That's lower energy.
So I'll get the orange.
Or maybe this is fuchsia.
So we'll take this one.
And this will put these two together and bring them in like so.
And this is now c2, r2.
too And what's this thing say?
Take the slope.
Look, it's a log, log plot.
If you've got physical data and you don't get a straight line on a log, log plot, you're incompetent.
You're a complete idiot.
All right?
There it is.
And the slope here is n. And where it intersection, obviously I can't get a negative number, so just allow me to fix this just a wee bit.
So there's the axis, and where it hits the axis, this is now log k isn't it?
So I determine the order of reaction.
Good.
So now you know how to do it and you'll get some practice if you work the homework.
All right.
A couple more things to say and then we're done.
So last thing, we talked about increasing the rate of reaction by increasing the temperature and increasing the concentration.
There's one more thing we can do.
And what we can do is use a catalyst.
Increase rate by use of catalyst.
OK. How does the catalyst work?
How does the catalyst work?
Well, let's go back to this diagram here.
Where we've got, this is called the extent of reaction.
Some kind of reaction coordinate.
And over here we had some kind of energy coordinate.
We had this energy coordinate.
Remember, we had this thing?
This silly thing here?
So this is the reactions.
And this is the product.
And we reasoned that the activation energy ea is sitting here.
And we saw with the box, you had to get a box up onto its ear, otherwise it wouldn't fall over and so on.
And there's a finite amount, finite fraction of the Maxwell-Boltzmann distribution.
So if we plot energy and the number of particles in the distribution have of given energy, we get something that looks like this.
Tails off.
And we said that out here we have the fracture under this line.
This is the fraction that has the energy to jump over the activation barrier.
So fa is the-- f is the activated, is the fraction activated.
And those are the only ones that can participate in the reaction.
These other ones can't surmount the activation barrier.
Now, what the catalyst does is it lowers the activation barrier.
Catalyst lowers the activation barrier.
It doesn't change the n points.
The n points represent the energy state of the reactants and the energy state of the products.
What the catalyst does is it finds a low-energy barrier pathway.
A low-energy barrier pathway, which on this curve would mean that if this energy is lower, then the fraction of any distribution that has the requisite amount of energy is much higher.
Can you see the area?
The blue area is much, much greater than the white area.
So that means that this fa catalyzed-- I'm just going to write cat here-- fa catalyzed is much greater than fa in the absence of a catalyst.
And that promotes.
What's an example of it?
Well, in an automobile catalyst one of their reactions we're trying to promote is unburned hydrocarbons.
There's some unburned gasoline that comes out in the form of carbon monoxide.
What we want to do is convert the carbon monoxide to carbon dioxide.
And this occurs on the catalyst and I'll show you more of this at the end of the lecture.
But these are both gas molecules and they're zooming through space.
And if they're going to react with one another, they really need to somehow move almost on parallel lines so they get close enough to each other for a long enough period of time that they can start thinking about lower energy state.
And say, well, gee, if we break this bond and form a new bond, it's going to go to-- you've got to get to know each other, go have a few drinks, and da, da, da.
So it's not happening.
In the gas phase it doesn't happen.
So what we do instead is we provide a service.
This is the catalyst.
The catalyst is a solid surface and the CO sits down next to another CO.
They sit down and now they're in place long enough in the right orientation to then form CO2 and then desorb.
So this is how the catalyst assists.
We call this-- especially, imagine if it's the other way-- what if this carbon monoxide molecule is turn around so that the two carbons are side by side.
That's not going to promote the reaction at all because this thing's got to whip around and so on.
This is a.
Bad orientation And what's the Greek idea for orientation?
It's stereos.
That's what stereophonic sound is.
If I close my eyes, the two speakers give me spatial rendition.
So this concept here, where things are backwards and they're not the right place.
It's called steric hindrance.
This is steric hindrance.
One of the functions of a catalyst is to combat steric hindrance and put things in their most favorable orientation to facilitate reaction.
So what are characteristics of the catalyst?
The catalyst needs to have the following characteristics-- first of all, it has to have an affinity for the reactants.
In this case, the CO.
It has to have an affinity for carbon monoxide.
Affinity for reactants, so they'll be attracted to the catalyst.
But there's a second piece here.
If the catalyst doesn't get rid of the CO2, can you see that after a very short period of time, the surface of the catalyst will be covered with carbon dioxide and then the reaction starves.
It's choked.
So it has to have a disaffinity for the products.
Disaffinity for the products.
And in the real world, such as in the gas exhaust stream of a car, it's not just carbon monoxide.
There's all kinds of other stuff going on in there, and so the third thing that the catalyst needs to have is selectivity.
So it doesn't get distracted by other reactions.
And selectivity, most important when you have a mixture of all sorts of other species.
So that's what the catalyst does.
Now there's one more thing.
Just one second.
Perfect.
Gotcha just in time.
So one more thing here.
One more thing.
We've shown what happens when you catalyze.
So I'm going to put this.
This is ea under the influence of catalysis.
OK.
This is the e of the catalyst.
There are other substances that can retard the reaction.
These are called inhibitors.
And what's an inhibitor do?
It does the complimentary action.
So it raises the activation barrier.
So this is ea under the influence of inhibitor.
And what's the value of that?
Well, if you buy some antifreeze for use in the automobile cooling system.
If you read carefully the label, it's not just ethylene glycol, and water, 50/50 volume percent.
There are other additives.
What are those additives?
Some of them are called inhibitors.
Oh, I didn't spell this right.
No wonder you looked so puzzled.
I thought you didn't understand the concept.
Inhibitor.
So it inhibits the reaction.
What's going on?
You've got the iron or aluminum block of the engine, you got the water pump.
It's taking heat away from the engine, you're heating everything up, stirring it around.
It's a perfect environment for enhancing corrosion.
And we don't want the cooling system to act as the corrosion system for the engine.
And so we add components to the antifreeze too reduce the rate of corrosion, to lengthen the service lifetime of the vehicle.
And so what does the inhibitor do?
It raises the activation barrier for the corrosion reaction and thereby reduces its rate of reaction substantially.
So this is inhibited.
OK, so this is catalyzed, this is inhibited, and you can see that the fraction of active species over the activation barrier, under the influence of inhibition, is much less than the fraction otherwise.
There's a time and a place for all of these wonderful things.
So that's the story of kinetics.
That's the story of kinetics, but it's 11:35 and it's time to start a new topic.
And now, we can erase.
Thanks.
OK.
So let's talk about something else.
Let's talk about something else.
What I want to talk about today is another topic that comes from the general area of rate phenomena.
I want to talk about something else in rate phenomena.
We're going to talk about diffusion.
And that's from the general class of rate phenomena.
That's why it appears with chemical kinetics.
And what do I mean by rate phenomena?
Rate phenomena means d by dt is in the room.
All right?
We're looking at time, domain processes.
You know, chemistry, before you came into this class, you probably thought, I know what chemistry is.
Chemistry is chemicals reacting.
That's what people think.
Isn't that what the ancient Egyptians said?
khemia That's where the guys knew how to embalm the dead and so on.
It was about chemical reactions.
We've been talking about structure, properties, all sorts of things.
Well, now we're talking about rates of reaction.
If we want to run a reaction, there's two things that we need to move around to sustain a reaction.
So transport of two important quantities.
One is matter and the other is energy.
We star something for energy.
As we just saw, the thing doesn't go.
And if we starve it for matter it doesn't go.
So energy in motion is under the general rubric of heat transfer.
But we won't talk about heat transfer because we've only got 14 weeks.
I'd love to talk about it, but we're out of time.
So instead, we will talk about mass transport.
And mass transport in solids occurs by diffusion.
And we know what's critical.
For example, we know in the case of the Hindenburg, the Hindenburg didn't explode.
Why didn't the Hindenburg explode?
Because there were 7 million cubic feet of hydrogen, but we couldn't get 7 million cubic feet of oxygen to the site.
Because of the limitations of mass transport, 2/3 of the people walked off of that vessel.
It was on fire, and many of them died because they panicked and they jumped to their deaths.
Very few people were actually affected by the fire.
So it's a good example of mass transport, negating the rate of what otherwise could have been an explosive reaction.
So I'm going to look at a simple example today.
And what I want to do is look at the doping of semiconductors.
You know, we've studied semiconductors.
Dave, could we cut to the document camera, please?
So what I've got here is I'm going to show you a silicon wafer.
And the silicon wafer has got some features on it.
So it's already been doped.
Oh, that's pretty, isn't it?
So cute.
All right.
So what we've got here, remember this is the 8-inch diameter single crystal.
So this was cut from a single piece of silicon 8 inches in diameter, about 2 meters long.
How did they make a single crystal?
They immersed the tiny single crystal into the melt and gradually pulled the single crystal out.
Sort of the way you can grow rock mountain candy.
And if you're really careful, the entire giant, 2 meter length, 8 inches in diameter, is one single crystal.
And then we take this and we cut it with a diamond saw into wafers, Each of them about a millimeter in thickness.
And, of course, the width of the saw turns into dust So your yield [SNAPS] right off the bat is about 50%.
Because [SAW SOUND] it's all gone.
[SAW SOUND] it's all gone.
And now what we're going to do, is we want to dope it.
So how do I get the boron in here?
We say, why don't we add the boron to the melt, but what if I want to make pn-junctions?
So I want to get some boron and some phosphorous.
If I add the boron and the phosphorous, it then goes nuts.
I need to have p-type against n-type.
So how do I get it in here?
That's we're going to talk about.
We're going to talk about how we make those features.
See those features?
Because if you're really good, you start with those features and you cut them up and then you end up with this, which is this.
This is a Pentium chip.
The amount of material science, physical chemistry in that device is absolutely phenomenal.
What lies underneath this piece of gold is absolutely phenomenal.
It's what powers the information age.
And it all starts with this stuff here.
It's that simple.
So how do we get the boron in?
Dave, let's leave this up.
It's a pretty image.
It might soothe the natives, so let's just leave it up.
By the way, that spider network is all the gold wires that are addressing.
It doesn't do you any good to have a gazillion features there but you can't get information in and out.
So if you've got a gazillion features, you need a gazillion leads to take information in and out.
That's very, very thin gold.
How do you do that?
Not with a paint brush.
It's all with this stuff.
That's why I teach you this instead some of the other stuff.
OK.
So now we're going to dope.
We're going to dope the silicon with boron.
Number one point to remember-- we're going to dope the silicon in the solid state.
We do not add boron to molten silicon.
We're going to add boron to the silicon in the solid state.
So I'm going to draw this wafer.
Only it's not to scale.
It's going to look more like a hockey puck than a thin wafer.
Anyways, imagine it's this.
OK?
So this is the silicon wafer, the silicon wafer on edge.
And this is where all the features were that I showed you.
So we're going to do is we're going to try to introduce boron.
We want to get boron into the silicon, we're going to bring it in from the surface.
So this is a silicon wafer, which means it's a single crystal.
It's a single crystal.
And we could use Laue methods to figure out which is the face.
Is that cut on the 0 1 1?
Is it cut on a 1 1 1?
And so on.
So now we're going to dope with boron.
How do we do it?
We decompose diborane.
We introduce diborane, it's a gas.
B2H6.
And we flow diborane over the surface at some constant concentration.
And what happens is that the diborane decomposes.
B2H6 on silicon.
And I'm going to write silicon with an x.
What's the x mean?
Crystalline.
Single crystal.
Make sure no one is thinking this is being performed in the liquid phase.
And this diborane decomposes to elemental boron plus hydrogen gas.
So hydrogen is a gas, and the boron goes into the silicon.
Boron goes into the silicon where it goes on to the silicon sites.
I could just write a rate constant for this.
We just studied kinetics.
I don't know.
What is it?
Second order?
You know, it doesn't have to be integral order, by the way.
It could be of order 1.5.
Nothing is saying it has to be integral order.
OK.
So now what I want to do is I want to ask, what happens as the boron goes in?
So that's the next question.
As it diffuses in.
So what I'm going to do is I'm going to draw this little cartoon here.
So I want to look at here's the surface of the silicon.
So I've got a gas phase here and I've got boron going into the silicon crystal.
So I'm going to now get a little bit quantitative, if you'll permit.
And so it's going to look like this.
And put it right underneath here.
So this is x. this is depth from the surface of the silicon.
So this is the free surface of the silicon.
And I start off with some concentration surface value, c sub s. Meaning concentration at the surface.
And that's fixed.
So cs fixed by B2H6 concentration in the gas phase.
So that's fixed.
Now, question is, what does it look like?
The advance of the boron?
So this is now the concentration of boron in silicon.
This is the silicon crystal here.
So what does it look like?
Well, one possibility is it goes in like this as a front.
See if I wait, goes in farther and farther and farther.
That doesn't happen.
Sorry.
Take that off.
You know what happens?
Same thing we've been looking at all day.
So now what's the question?
Question is, what's the functionality of this?
Is this 1 over c?
Is this log c?
Is this some other function of c?
I want to map this into some f of c versus g of x so I can get a straight line.
Why do I want a straight line?
Because I want to be quantitative.
Because you are in charge of designing a fab line and they want to know what the resonance time of a silicon wafer is supposed to be at billion dollars multiples.
You know, the units are in billions of dollars.
So you don't want to make the fab line any bigger than it needs to be, the resident's time any longer.
This is only way you're going to be able to design it.
You have to understand this because the design specification is going to be dope to a certain concentration of a certain depth.
So how long does it take to get boron in?
You can tell me.
Set the concentration here and wait 5 minutes, 10 minutes, 30 minutes.
Two companies.
One takes 30 minutes to process the wafer, the other takes 10 minutes to process the wafer.
Guess which company is going to be in business two years from now?
It's not going to be the one that takes a long time.
Hint.
So we need to know what the rules are here.
And the mathematical formulation was set by Adolf Fick.
Adolf Fick, 1855, answered the question, what is the form of that line?
He's an interesting guy.
He was at the University of Zurich.
And it's sort of like the Balmer story.
Remember J.J. Balmer was the school teacher in Switzerland.
This guy was in Switzerland, too.
Something going on there in Switzerland in the late eighteen hundreds.
So remember J.J. Balmer?
He fit not his own data, he fit the data of Angstrom.
Angstrom made the measurements, Balmer fit the data.
Well, same thing here.
Fick did not make the measurements.
He looked at some data that had been taken back in 1833 by Thomas Graham.
Thomas Graham had been making measurements back in the 1830s of gases diffusing through porous membranes.
Or just say, through membranes.
Obviously, if they weren't porous they wouldn't be diffusing.
Redundancy.
Department of Redundancy.
This is diffusing through membranes.
So what happened was that he modeled it.
And here's what he came up with.
He came up with-- oh, by the way.
I didn't mention what he was.
He was a physiologist.
He was a medical doctor.
He just poured through the scientific literature for amusement, evidently.
OK.
So this is what Fick gave us as the model for the data.
He gave us what we know as Fick's First Law, Fick's First Law of Diffusion.
And we'll just call it FFL, Fick's First Law of Diffusion.
And we'll write Ficks' First Law J. J, which is the flux.
This is the unit of measure of mass transport, the flux, which obviously is the mass flow rate.
We said this is mass transport, so this is the mass flow right of species i, OK?
And we're going to say in the x-direction because things diffuse in all directions, right?
If I open a perfume bottle here, the diffusion will take place in all directions.
So I'm going to make this one dimension.
So the flux is a mass flow rate of i in the x-direction.
J sub i in the x-direction.
Remember we said the Chemical Rate Law?
We said the Chemical Rate Law was that the rate of progress of a chemical reaction goes as the concentration.
Well, Fick studied Graham's data and he said, the flux doesn't go as the concentration.
It goes as the gradient in the concentration.
That's how you linearize those data.
So you take the concentration gradient, which is dc by dx.
OK.
So that's the concentration gradient.
Gradient in concentration of i in the x-direction.
You could write this as-- make this vector-- but we're not going to do that.
It would make it one-dimensional here.
And the constant of proportionality is d sub i, which is called the diffusivity or the diffusion coefficient.
And puts the minus sign.
Why?
Well, you know things move from high concentration to low concentration, right?
So I open the perfume bottle, the odor gets-- the odor, pardon me-- the fragrance gets stronger as you get closer to the bottle.
Things move from high concentration the low concentration.
So if I gave you this concentration profile-- what's profile mean?
Profile means-- this is a concentration profile-- if I draw someone like this, this is head on, right?
And if I do the right view like this, this is the profile.
Profile.
Side view.
This is profile.
Concentration profile, obviously, the concentration is higher to the left.
So things are moving from left to right.
But look at the slope.
The slope is a negative number.
So if the slope is a negative number, but I want to get a positive flux, I need to have the minus sign in here.
Diffusivity is a physical constant.
It has to be a positive number.
So this is Fick's First Law.
I'll just give you the units and then we'll stop.
So what are the units?
Well, the units in flux are in SI units kilograms per second.
And we're going in one direction, and so as we're trying to diffuse into this piece of silicon, we have to normalize it per unit area.
So it's kilograms per meter squared per second.
You know that concentration is kilograms per cubic meter.
d by dx is 1 over meter, which means then that the diffuse coefficient or the diffusivity has units of length squared per unit time.
This is the units of the diffusion coefficient.
And we'll get back to business next day.
David, may we cut to the slides again, please?
OK.
So I talked to you about-- oh, yeah.
There's all this.
You see they give you a half-life?
Crazy.
Forget it.
OK.
There's the book talking about activated catalog, da, da, da.
This is Fick's First Law.
This is the paper from 1855.
Go ahead.
"On the Influence of Diffusion" by Dr. Adolph Fick.
So the number one catalyst in the world, number one application is automotive catalysts for exhaust.
So gasoline is octane.
C8H18, which we burn in air.
We burn it in air, we make hydrocarbons.
Pardon me, we burn hydrocarbons, which obviously means the carbon goes to CO2, ideally, and the hydrogen goes to H2O.
But sometimes you get incomplete combustion.
I already showed you that.
Maybe we stop it CO.
But remember, air is 4/5 nitrogen.
So when you burn hydrocarbons you also, at these high temperatures, mix nitrogen and oxygen and make NOx.
And is was actually the precursor to smog.
So what we want to do is to eliminate these imperfect combustions.
So we want to convert the unburned hydrocarbons.
There's actually some of the CH compounds, volatile, then don't get burned.
We want to burn those to H2O in water-- pardon me, H2O in CO2.
The unburned CO has to go to CO2.
You might say, but isn't CO2 a greenhouse gas?
Yeah, it is.
But you know, given the choice of carbon monoxide and carbon dioxide, my advice is go with carbon dioxide.
Carbon monoxide will kill you at about 200 parts per million.
And then NOx, we want to convert that to carbon dioxide.
So GM alone tried over 1,500 catalysts.
Why GM?
Why not the auto industry?
Because the U.S. Government, in its wisdom back in the 1970s, invoked the Sherman Antitrust Act and forbid the automobile industry to collaborate in the search for a catalyst.
And since they didn't understand the science of catalysis it was trial and error.
By inspection.
And it's a huge space, trying things.
But they eventually found, after 1,500 tests, platinum-palladium works for hydrocarbons and CO, and rhodium works for NOx.
It's almost as though mother nature said, I'll give you a catalyst and it's going to be the most expensive part of the Periodic Table.
You buy this stuff at the jewelry store, you see.
Furthermore, this is an oxidation reaction.
This is an oxidation reaction.
This is a reduction reaction.
So if I told you I want you in the same place at the same time on a vehicle moving 80 miles an hour to conduct simultaneous oxidation and reduction reactions with no intervention, you'd say impossible.
They did it.
They did it.
This is the power of selectivity of the catalyst.
So this is an old diagram from GM literature.
Now we have fuel injectors, here, but fuel goes into the engine, out to the exhaust.
There's an oxygen sensor, which I'll show you next day.
Oxygen sensor measures what composition of the output gas is, sends it to an electronic control module, which then goes back into the carburetor.
And the catalytic convert is right here.
And Dave, if you show the document camera.
This is the monolith.
This is a piece of ceramic because platinum and rhodium and all of these other elements are so expensive.
See, this thing's having a hard time.
So we'll cheat a little bit.
We'll put this here and then it'll focus.
All right.
So what you're looking at, these are long channels.
And just to give you a scale, this is about a half of a millimeter.
This is about a half a millimeter square and they go all the way down for a length of about eight inches.
And then the walls, the interior walls, are coated with a very, very thin layer of platinum, palladium, and rhodium.
Because these are expensive metals.
And it's a surface effect.
So you get no value if you have an inch deep of platinum.
The only stuff that works is the free surface.
What you'd really like is a monolayer.
But a monolayer of platinum has no mechanical strength.
So you put the platinum, palladium, and rhodium down on a catalyst support.
and this stuff is made of a ceramic, it's magnesia, alumina silica that's been extruded like pasta, and then fired.
The technology is absolutely phenomenal.
Goes into this.
And then to get the platinum-palladium-rhodium they have a very thin acid wash. chloroplatinic, chlororhodic acid, wash, little bit of surface tension, holds just a little bit of water, fire, out comes the metal.
Just thick enough.
If it's too thick, dilute the concentration of the acid.
But use that precious metal sparingly.
Platinum hit $2,000 an ounce last year.
$2,000 an ounce.
You steal a car, throw the car away, and get the catalytic converter.
So anyway, that's underneath.
That's what's going on underneath the car.
And actually, I'm holding for you the elements of clean air, the catalytic converter.
But the catalytic converter can't work unless we control the air to fuel ratio, which we have to do with an oxygen sensor.
I'll show you next day.
And you've got to feed it into a CPU.
If you don't have a CPU, it can't make rapid feedback to keep adjusting the air-to-fuel ratio.
All right, Dave.
Let's cut to slides and we'll get out of here.
So I'll show you what else is in here.
OK, there's that.
So there's the old thing.
There's the engine, da, da, da.
Exhaust, the control module.
And look, in the old days they actually had-- they look like little ball bearings-- cost of these things was phenomenal.
So last thing, we had to go to unleaded gas.
Unleaded gas was brought in because the lead would be broken down into elemental lead because lead is volatile as an oxide.
The lead alloys with the platinum.
Turns it into a lead platinum alloy.
Lead has no catalytic value.
Because if it did, we would use lead and not platinum.
OK. We'll see you on Friday.
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OK. OK, settle down.
Settle down.
Let's get started.
So last day we started talking about diffusion.
Diffusion is a solid-state mass transport by random atomic motion.
I made the first point last day and today I'll make the second point.
We'll see why it's random atomic motion.
We reasoned that we could describe the ingress of material.
Here I'm showing a concentration profile.
A profile is nothing more than a plot of something as a function of position.
So, this is looking at, for example, the doping of a wafer.
This is the free surface at x equals 0.
As we go to the right, it represents depth.
And we fix the concentration at the surface and the material tails off to zero deep inside the specimen.
Fick in 1855 annunciated the law of diffusion, shown here, that says that the rate of ingress expressed by the flux is proportional to the instant gradient in concentration.
And the constant proportionality is the diffusion coefficient.
This is where all the material science lies.
It's inside the D. That's what we're going to talk about more today.
And so we see things tail off and what we want to do is figure out what the mathematics are that'll allow us to render this thing into a functional representation.
Because all we know right now is that we've got something tailing off.
Ok?
And so this is the gradient multiplied by the diffusion coefficient and you can see that the red line represents the flux.
The gradient is maximum at the surface, so the flux is maximum at the surface.
And as we move farther and farther into the piece, the gradient in concentration becomes less steep, which means the absolute magnitude of the flux becomes less steep, and we go deep inside where there is no doping yet.
The variation of concentration is constantly 0, so the slope is zero, which means the flux is zero.
So the two sort of track, one and the other.
And we recognize these are negative slopes since the minus sign is here.
Even though we have a negative slope, we have a positive ingress.
And then the other thing that we realized was that the diffusion coefficient has a temperature dependence.
And the temperature dependence looks like this.
It's an exponential of some kind of a characteristic energy divided by the product of -- this is the gas constant, this is not the Rydberg constant.
The gas constant is simply the Boltzmann constant per mole.
So if you take the Boltzmann constant, multiply it by the Avogadro number, you get this.
But if you see something times T, you know it's got to be either gas constant or Boltzmann constant.
This two-letter character string has to have something akin to a Boltzmann constant, otherwise you don't end up with an energy term.
And this in the top is a barrier energy.
It's a barrier energy to diffusion and out here we have the pre-exponential.
Unfortunately we didn't give it the letter F in honor of Fick, it's just D naught.
And so there it is.
So if we take this equation and we plot it, natural log of d versus 1 over T, that'll linearize the equation, and we end up with something that gives us a straight line.
And the slope is minus Q over R, where this Q is some kind of a barrier energy.
And the R is the gas constant or the Boltzmann constant.
So this is the gas constant.
It's on your table of constants.
It's got a value in SI units of 8.314 joules per mole Kelvin, OK?
Just the Boltzmann times Avogadro number.
So now I want to invite you to join me in some pattern recognition.
If I give you the temperature dependence of a quantity that looks like this , in other words, the logarithm of that physical parameter versus 1 over T, gives you a straight line.
You know from the work that we did on chemical kinetics, this is the representation of something that is an activated process.
Remember the box that would fall on its side, and it had to go up on its corner?
So we had some physical sense of what activation is.
Now this has a log, something versus 1 over T dependence.
This represents some kind of an activation energy.
An activation energy.
But what's the activated process that's going on here?
So that's where I want to take you into the atomistics.
So here's a cartoon taken from one of the readings.
And it shows a set of atoms here and the atom next to the vacancy wants to move into the vacancy.
And you can see the artist's rendition is showing that in order to get from the current position into the vacant position, it's got to squeeze through this narrow channel.
And it takes energy to move through that channel.
And there's an activation barrier associated with that motion.
Can you see when the atom is sitting right at the saddle point?
The system is highly activated because it's pushing out against those atoms.
And then finally it falls into the new slot and the energy drops.
So the activation energy is associated with moving through this saddle point.
So what really happens here, I'll tell you right now.
First of all, we have to recognize that the system is depicted there, because it's a freeze frame, but we know that above 0 Kelvin, everything is in motion.
So the only way we can rationalize what's going on is to, first of all, recognize that we have a pulsating lattice.
We have a pulsating lattice.
Everything's vibrating.
And what's the heartbeat?
What's the idle speed?
I've told you this before.
Everything that's going on in this room has an idle speed, a heartbeat, of 10 trillion times a second.
That's the Debye frequency, OK?
The idle speed or the heartbeat.
The heartbeat is called the Debye frequency.
Here's frequency, lowercase nu, in honor of Peter Debye.
And it's on the order of about 10 to the thirteenth per second.
10 to the thirteenth Hertz.
That's 10 trillion.
So 10 trillion times a second atoms are going like this.
Pulsating.
All right?
So what has to happen for that operation to occur?
Well, I've got an atom here, vacancy next door, but because it's close packed the atoms on either side close in.
There's no way this can squeeze through.
But, imagine, 10 trillion times a second.
If I'm lucky enough to see the moment when the atom above pulsates up, the atom below pulsates down, a channel can open up and then this thing can squirt through.
Now, that doesn't happen every time.
If it did, history would have ended.
At 10 trillion times a second, it's all over for everybody.
Right?
So what happens?
What's the frequency at which we will get this unique combination?
It turns out that frequency, noted gamma, is called the jump frequency.
That's the jump frequency.
And the jump frequency is on the order of about 10 to the eighth Hertz.
10 to the eighth Hertz, which is about 100 million.
So if you take the ratio of the Debye frequency to the jump frequency, you see you get a value of about one try in about 10 to the fifth is successful.
One try in about 10 to the fifth is successful.
And that explains what's going on here.
So, what now does that mean in terms of Q?
What's the breakdown of Q?
Well, clearly, Q, the activation energy for diffusion, must involve both the energy to form the vacancy -- I'm going to call it delta Hv, and that's the energy for vacancy formation -- if you don't have vacancies, you cannot have motion -- plus the delta H sub m which is the energy for atom migration.
So you have to form the vacancies and then you've got to squeeze through those saddle points.
So that's how we get a sense of it.
And I think I've got another slide.
OK, so they're I've simply put the markings on for you so you can see the atoms above and below jumping in the right direction.
That allows the atom on the left to squirt to the right.
Now, this is also related to energetics.
It's related to energetics because the energy to form a vacancy and the energy to squeeze through those saddle points must be related to the binding energy.
And what else is related to binding energy?
Melting point.
So, this is an elegant plot that simply shows that the energy -- these are all FCC metals: lead, aluminum, silver, gold, copper, and iron.
They're all FCC on this chart here.
Iron has both BCC and FCC, but this is comparing apples to apples.
And you can see that there's lead melts at 327, aluminum at 660, silver 980, all the way up.
Iron 1535.
And you can see that the activation energy for -- that's this Q-value -- the activation energy for diffusion scales with melting point.
Again, an indication that this analysis makes sense.
And then, here's the other one, the interstitial.
If you have interstitial diffusion you've got plenty of interstitials, so you don't have to pay the penalty to create the interstitial, but it's much, much more difficult to squeeze through the saddle point.
So, it turns out that activation energy for diffusion by interstitial means is still fairly highly activated.
So this is in the case of substitutional systems, all right?
This is for substitutional.
Substitutional atoms.
What do I mean by that?
An atom that sits on a lattice site.
If an atom sits on a lattice site, it has to jump to another lattice site, which means there needs to be a vacant lattice site.
On the other hand, if we're talking about interstitial special atoms -- not substitutional atoms, interstitial atoms -- the assumption is that there's plenty of interstitial sites.
There are very few systems that I can think of where the interstitial sites are so heavily occupied that this assumption is invalid.
So in that case, you don't have to pay for vacancy formation.
It's simply equal to the enthalpy, which we know is analogous to the energy for a condensed system of atom migration.
Atom migration gives you the whole story and it goes back to the same picture, and so on.
Now, I said it's random walks, so let's look at another one.
So, this is an interesting example.
This is self-diffusion in cobalt.
Cobalt has a number of isotopes.
One of them is 59 and one of them is 60.
60 is a radioisotope.
Cobalt 60 is a radioisotope and it's used in a variety of technical endeavors.
And so what we've got here is a sandwich of cobalt 59 -- cobalt 60 and cobalt 59.
So even though cobalt 60 has radioactivity and it's got a different number of neutrons in the nucleus, chemically it is identical to cobalt 59, just as carbon 14 is chemically identical to carbon 12.
So there's no concentration gradient.
Fick's Law says it's supposed to go by concentration gradient.
The concentration gradient is zero across that piece.
And yet if you wait for long enough, you will start to see -- the dark atoms here are meant to represent cobalt 60 in the sandwich -- after some period of time, the cobalt 60's will move outward.
And if you wait long enough, the cobalt 60 concentration will be uniform throughout the specimen.
So what's going on?
There's no chemical concentration gradient, and yet over time, the cobalt 60 spreads out.
And the answer is this is happening by random walk.
We've got the pulsating lattice.
Pulsating lattice.
10 trillion times a second those atoms are vibrating and there's going to be some vacancy.
All of a sudden an atom falls into a vacancy.
And the mathematics of this -- look at this, this is no accident -- this curve here represents something that looks like a Gaussian distribution.
At any given time, the amount of cobalt 60 that's veered from the center band can be portrayed by the Gaussian distribution, the bell curve.
And what's the physical model for the bell curve?
It's the drunken sailor.
Or I guess I have to be PC.
I'm not going to just pick on sailors.
Maybe it's the drunken cougar, the soccer mom that went out drinking.
OK, so what happens with the drunken cougar, she comes to the door, and she starts to walk, all right?
And then she walks one way or the other way, until she finally falls.
And if you go through the number of steps that it'll take before the drunken sailor falls, you get this distribution.
It's an old physics problem.
The drunken sailor problem.
And it's a normal distribution.
And that normal distribution tells you that everything's going about by random jumping.
Random jumping, according to this, all right?
So that's why we say that diffusion is a random walk problem.
Before I turn this off, I want you to take a look carefully.
What's wrong with this textbook cartoon?
Any ideas?
It's a substitutional system, right?
It's cobalt and cobalt.
This is absolutely impossible.
What's wrong with it?
There are no vacancies.
There are no vacancies.
It's impossible.
There's no vacancies to jump into.
This thing is jam-packed, so it's physically unrealistic from a thermodynamic standpoint.
There has to be some finite population of vacancies.
Given this system, there's no way that the black atom could jump into the neighboring site, because you'd have to have the entire system open up, and the chances of that happening are vanishingly small.
So it just shows you, just because something costs 150 bucks and is pressed between two hard covers doesn't mean it's full of truth, OK?
Be careful here when you look at this.
So, again, the artist needed to be guided.
We need to put a few of these things out, just to give symbolic recognition to the fact that we need to have some vacancies.
OK. All right, let's keep going.
Let's keep going.
Here is some data.
Here is some data.
Various data.
This is diffusion.
This is the logarithm of the diffusion coefficient versus 1 over T. So let's look at the top.
This is hydrogen in iron.
So temperature increases from right to left.
So, go up, up, up, up, up.
And roughly 900 degrees Centigrade, iron changes from BCC to FCC.
And you know from our previous unit, the number of nearest neighbors in FCC is greater, so there's less void fraction, right?
So that means that it's much more difficult to move.
And you can see, first of all, the order of magnitude drop in the diffusion coefficient, and also, it's subtle, but the slope is steeper, because the activation energy for diffusion is steeper in a closer-packed system.
This is carbon in iron.
Same thing.
Relatively gentle slope at the same temperature, there's a phase change to FCC and it takes much, much more energy per unit increase in temperature to get the comparable increase in the value of diffusion coefficient.
This is iron, self-diffusion in iron.
So this would be radio tracer iron in iron.
Very much lower.
You see these values are up around 10 to the minus 5, 10 to the minus 6.
This is down to 10 to the minus 11, 10 to the minus 12 centimeters squared per second.
Iron diffusion changes at the transition temperature.
Now iron going through FCC iron.
Here's the last one.
Look at this one.
This is carbon in graphite.
So how does carbon -- suppose I had some carbon 14 in graphite.
What's the bonding in graphite?
Silence.
It's covalent.
It's sp2 hybridized.
It's really strong bonds.
The activation energy -- you know, forget just jumping through that saddle point -- you've got to break those covalent bonds.
It's very, very high energy to get carbon to move through graphite.
And so you see very, very low values and very, very steep temperature dependence.
Now, there's another way to think about this.
Again, this is now looking at the atomistics.
So this is the same material trying to indicate that if you use Fick's Law, moving in from the surface into the bulk, we have a certain front.
So this is some isoconcentration line.
So we could take this value off of the graph arbitrarily.
So they're saying c greater than c i.
Because it trails off to zero, right?
It's asymptotic.
There's no sharp cut-off here.
It trails off.
So what they're doing in that graph is arbitrarily saying, let's call this ci.
And where's the front at concentration ci, given a constant value of cs at the surface?
And you can see that there's a near-constant rate of advance through the bulk.
But along the grain boundary you go much, much farther in the same amount of time.
Why?
Because the atoms in the grain boundary are not bounded the same way as the atoms in the bulk.
There's this gap, this misalignment.
And so, it's easier to make that jump, and you can see you get much, much more advancement along the grain boundary.
And this is a crack.
And this material covers the surface.
You get surface diffusion.
Surface diffusion is very important in many processes, and it's very fast, because on a surface, you've got no atoms on the top.
It's almost liquid-like, isn't it?
You've got atoms to the bottom, but you've got no atoms to the top, so you can move unconstrained.
Go back to this one.
Imagine here.
Imagine if you didn't have any atoms on the top, how easy it would be to make that jump.
All right, let's go and see what the data are.
Nothing like data.
And I got data for you.
Let's look.
Click, click, click.
All right.
Here's the data.
This is all for the diffusion of silver.
So I'm going to use this value, this schematic, and I'm going to give you the value for silver diffusion in the bulk, silver diffusion in the grain boundary, and silver diffusion along the surface.
So these are the data.
So this is log diffusion coefficient versus 1 over T. Only, I hate this graph, because somebody got really wimpy.
You know, it bothers them.
You see, when you plot something like this, you end up with, if 1 over T increases from left to right, this really means that high temperature is over here, right?
1 over T increases this way, which means temperature is increasing this way.
So somebody, some weenie, got all nervous here and put the numbers backwards, you see?
That way, the high temperature is over here, but it looks stupid, because when I see an activation plot, I want a negative slope.
It's non-physical to have a positive slope.
So I condemned this and I turned it around.
So now let's look at it the right way.
So this is log d versus 1 over T, and this is the bulk, lattice or bulk.
This is grain boundary and this is surface diffusion, all in solid silver.
OK, look at this.
First of all -- I'm going to jump from here -- one, two, three, four.
Four orders of magnitude faster diffusion along a grain boundary, which makes sense, because it's less constrained.
And now let's go from grain boundary at constant temperature up to surface.
One, two, three.
Three orders of magnitude along the surface.
So you can see the effects.
And there's the melting point of silver, right here.
So at the melting point of silver, the diffusion coefficient is about 10 to the minus 8 centimeters squared per second, which is the diffusion coefficient of virtually every metal at its melting point.
So I happen to know that number.
10 to the minus 8 centimeters squared per second.
Not that I know it as a function of temperature, but I know it can't be any faster than that.
So I said, this is near liquid-like behavior, what's the next thing to look for?
Let's get some data from real liquids and see if this hunch is correct.
What's the number here?
10 to the minus 5 to 10 to the minus 4 centimeters squared per second.
This is diffusion in molten ferrous alloys.
This is manganese in molten iron.
10 to the minus 5, 10 to the minus 4.
Bingo.
It really does behave like a liquid, in terms of diffusion.
Surface diffusion is very fast, very fast.
And it's all explained by this simple model of atom jumping.
And I continued it.
Here's data from glasses.
We studied glasses.
So here's a suite of different glasses.
Log k. This is permeability, which is related to diffusion coefficient.
There's a gas constant in there, but forget about it.
So this is essentially log of the diffusion coefficient versus 1 over T. We've got a family of lines.
And what's the difference here?
This is pure fused silica up here.
SiO2.
There's borosilicate, there's soda lime, and there's lead borate.
So what's happening?
We're adding more and more modifier as we go from top to bottom.
And as we add modifier, we break the length of the silicate network, which means it packs tighter and tighter, and as you get tighter and tighter packing, the diffusion coefficient gets lower and lower.
So train your eye.
Let's pick 200 degrees Centigrade, right here.
So at constant temperature, the more modifier you put in, the lower the diffusion coefficient because things are more tightly packed.
It's like walking down the infinite corridor in the middle of the hour, versus walking down on the hour.
On the hour, there's too many people.
It's tighter packing.
And if you're the least bit civilized, you're going to have to move a little bit slower, whereas, if you walk down the infinite at around now, no problem.
It's liquid-like behavior.
And now this is the isotherm at 300 hundred degrees C. This is from some old Corning literature.
Corning used to make glass, at one time.
And this is the logarithm of diffusivity.
And this is now the set of silica B203 and P205, and these are all network formers, right?
They all form covalent bonds.
So the more silicate, borate, phosphate you put in, the more stretched out becomes the network.
And sure enough, look up here.
See they didn't have names for their glasses like the Malibu or the Corvette or something.
They called their glasses by four-digit numbers.
So someone would say, oh, this customer needs, oh, I'd sell them a 7040.
All right?
Because it has a certain mix of properties.
So forget about what those numbers are, just know that this axis indicates that as you move up, up, up, up, up, the higher diffusivity is associated with a much more open structure.
So, again, a link between atomistic behavior and atomistic structure.
It's all there.
OK.
So let's leave that up there while we go a little bit forward.
OK, so now, so we've been down buried at the atomistic level.
Now let's jump up to the continuum level.
We'll go back to Fick's first law.
And I want to do something a little more quantitative.
So I want to look at diffusion across a permeable membrane.
So this is gas through a membrane.
Gas through a membrane.
And the membrane is shown here.
This is the walls of the membrane.
Membrane has thickness L. Membrane of thickness L. In fact, we'll give it a name.
Let's call it membrane.
All right, so here's the membrane of thickness L, and I'm going to put gas on the left at P1, and it's a constant P1.
So we keep -- we've got a ballast here, so even though something starts diffusing, the amount that we lose doesn't affect the constant pressure over here.
And on the right side, I'm going to put a second pressure, P2.
And just for argument's sake, I'm going to say that P2 is less than P1.
So that means we're going to end up with mass transport from left to right.
And so, I want to ask, what's this going to look like if we use Fick's Law?
And so beneath I'll make a plot, a concentration profile, from x equals 0 to x equals L. And I can morph this into concentration.
We know the gas law is PV equals nRT.
This is the gas law.
P is pressure.
P is pressure.
V you know is volume.
But I'm going to write all these out, because some of these letters are used so many times that we don't know, is this the Rydberg constant?
No, it's the gas constant that you've seen on the first board.
It's the Boltzmann constant times the Avogadro number.
T is absolute temperature, temperature in Kelvins.
And n is not the quantum number, it's mole number.
Mole number.
PV equals nRT.
And we know that concentration of i is mole number of i over V. Both of those appear in here.
So if you take what's underneath there, and I'll expose it in a second, you'll convince yourself that Ci equals Pi divided by RT.
And that's how you get that permeability thing to turn around as well.
OK, so, I started with gas pressure, but I'm now going to write this in terms of concentration.
So I'm going to morph P1 into C1 and over here, P2 into C2.
And I want to plot the profile.
So what's the profile look like?
I'm going to plot the profile at steady state.
At steady state.
At steady state, after I've waited a while, I have a straight-line profile across here, at steady state.
It's pinned at the two ends.
It's pinned at the two ends and varies continuously across here.
And you can see from Fick's Law, that if you have a situation in which the gradient is invariant across the piece, the gradient is -- dc by dx is the same here, as it is here, as it is here, as it is here.
dc by dx is invariant, right?
It doesn't change.
This is a straight line.
Well, if dc by dx is invariant, and d is a constant, then J is invariant, which just means no sources are sinks.
All of the material that goes in, must come out.
This means that J is not a function of time.
J is invariant.
J is not a function of time.
I come back here ten minutes later, I have the same profile.
That's what steady state means.
That's what study state means.
OK, and then, of course, if I wanted to be a little bit pedantic, I could say, in certain systems, the diffusion coefficient is a function of concentration.
So, when the diffusion coefficient is a function of concentration, then even at steady state, you might have some bowing of the profile, but that's a little bit more advanced.
OK.
So, I can describe everything that's going on here.
If I wanted to know how much material leaves from inside over a certain period of time, then I can simply say that total material lost is given by Fick's first law when you have steady state.
It's simply going to equal the product of the flux, which is mass per unit area per unit time times the area times the time.
And that's sort of analogous to Faraday's Law, right?
If I wanted to look at how much electrical charge I get for a certain period of time with current passing, I could say that the total charge is the product of the current times the time.
You know this relationship.
Well, can you see that J times A is analogous to the current?
It's the mass current.
It's mass flow rate.
Mass flow rate per unit area times area gives me something that's analogous to the current, and the time is the time, this is the total material lost, this is the total electrical charge.
These two are the comparable equations.
OK. Now, I want to go to the more sophisticated question, and that is, what happens before steady state?
Before steady state is achieved?
Suppose at time zero, the membrane has no gas in it.
And at time zero, I inflate the left side.
And just to keep things simple, I'm going to set c2 equals 0.
It's a linear equation.
You can always add it, but I just want to keep it simpler for the analysis.
So, suppose at t equals 0, we inflate to c1 on the left side of the empty membrane.
There's nothing in the membrane.
So, let's see what happens.
First, I'm going to solve the problem graphically.
Because we know what's going to happen.
I'm going to have to do the mathematics.
So, here's what we expect.
This is going from 0 to L. And we've got nothing in the membrane on both sides, so that's good.
And then, at time zero, I inflate to c1.
There's nothing in here.
So, what do I have?
Well, I know if I wait long enough, it's eventually going to look like this.
Because that's the steady state solution.
I've put here c2 equals 0, for simplicity, but we can change it.
It's a linear equation.
We can just add.
So, what happens at time 0 plus?
I start getting diffusion in from left to right, following Fick's Law.
So it's going to look like this, isn't it?
This is the approach to steady state.
Some books will call it nonsteady state or unsteady.
Unsteady to me means unsteady, so I refuse to use those terms.
I call this the transient.
That's the literary term.
The transient.
We're talking about the approach to steady state.
And in some systems, you never get to steady state.
But that's OK.
So here's what it's going to look like at some very, very short time afterwards.
It's going to look like that, only somewhat modified for this geometry.
So it's going to look like this, isn't it?
It's pinned at c1, and it's pinned at 0, and it varies continuously.
So this is at time t 1.
What happens at time t 2?
At time t 2, it advances a little bit farther.
But in all instances, the slope and the shape is given by this, Fick's first law.
So this is at time t 2.
And let's do one more.
Three's a charm.
So, let's take a green one.
All right, so here's at t 3, still pinned at c 1 and still pinned at 0.
So this is at t3.
And then finally, this is steady state.
So if we park at any place -- let's park at x1.
Watch here.
Can you see that the slope at t1 is gentle?
The slope at t2 is steeper than t1.
The slope at t3 is steeper than t2, and this is the highest.
So, we go from nothing, to gradual, gradual, gradual, until we finally reach steady state, and we stop.
And then at that point, we just continue to have to mass flowing through.
And look, the concentration is zero, but the flux is not zero, because the flux isn't related to the concentration.
It's related to the concentration gradient.
Zero concentration, but non-zero concentration gradient.
In fact, just for grins and chuckles, I could plot the ascent of this.
I could plot time at x equals x1.
I could plot J.
And what's J?
At time zero, it's zero.
Then it's small, then it's larger, larger, larger asymptotically.
And this is J of steady state.
And this is t1, t2, t3.
That's the approach to steady state.
And you might say, well, this is kind of a professor's pedantry.
It's not!
This is doping of semiconductors.
You don't run processes at equilibrium.
You run them far from equilibrium.
So in the transient mode, doping, outgassing, drying, what have you.
So, now I want to give you the shape of the concentration profile.
So how do I do that?
OK, so the shape of the transient profile is not given by Fick's Law.
The Fick's Law just gives me the gradient.
In other words, here's what I want.
I want C as a function of x at any given time.
That's what I want.
I want to be able to plot the profile.
So, profile is C of x, but C of x varies with time.
So, I need C of x at different times.
What gives me that?
What gives me that is Fick's second law.
Go to FSL, Fick's second law.
Now, Fick's second law, I'm just going to put it up here.
I'm going to show you the solution.
I don't expect you to solve it.
I would, you were all required to have differential equations as a pre-req, but you don't.
So, I'm just going to put it up here so that you see it.
It's a partial differential equation, and it looks like this.
And it's really beautiful.
It's got a beautiful symmetry and the fonts look great, so that's why we put it up here.
So, this is the partial in time goes as the double derivative in space, because we need a two-variable function, right?
We want x and t. And this assumes that the diffusion coefficient is not a function of concentration.
If it is, the equation is messier.
This assumes a constant diffusion coefficient.
So, this, as you can see, is a partial differential equation, which is a bummer, because it's hard to solve, all right?
But, it's linear.
It's a linear partial differential equation, which means the solutions are additive.
We know this already from LCAOMO.
So, I can handle, because I can solve it once and then I can give you ways to patch it, to make it useful.
So, the solution.
First of all, how do I specify this fully?
Every time you differentiate -- I'm going to give you some math here, real math that you can use.
When you differentiate, you throw away information, right?
The derivative of a constant is zero.
So, what was the value of the constant?
It's gone.
You lost it.
So you have to bring it back.
So when I double derivative this, I need two pieces of spacial information.
And when I take a derivative with respect to time, I need one piece of temporal information.
That's math.
All the other lemmas and postulates, forget them.
This is how you use it, all right?
So, I need two pieces of spatial information and one piece of temporal information.
So these specify our boundary conditions, right?
So, what are our boundary conditions.
So, I will need the concentration of all x at time 0.
That's this.
The answer to this question.
And what we're going to do is say that it's a constant.
I don't know what the constant is, it could be zero, it could be any arbitrary number, but I'm going to give you the solution for the situation in which the initial concentration is a constant.
So, you could look at this and say, well, what happens if it's a non-zero constant?
You ready for linearity?
Watch this.
This is how cool linearity is.
You see this set of solutions?
This is the set of solutions for C equals 0.
What if C equals some non-zero value?
I just shift the origin down to here to the non-zero value and everything sits there.
That's what superposition gives you.
That whole set of solutions just gets jacked up by the value of C naught.
If C naught is 0, this slams down onto the x-axis.
That's real math.
You're the master.
The math is the slave.
Never let math enslave you.
I won't.
I refuse.
So that's how you make C equals C naught work.
And then you have to peg the concentration at the surface.
I can give you examples where that's not the situation, where you have a variable concentration at the surface.
The solution I'm going to give you won't work for those.
So, x equals 0 is the surface.
So, at x equals 0 at all time, I have some fixed value, which I'm going to just call C surface.
And I need a third boundary condition.
And the third boundary condition is a mathematical trick.
We're going to say that this only works at short times.
It's called the short time.
And at short times, I'm going to say that -- from the perspective of the diffusion experiment -- at very, very short times, this sample appears to be infinitely long.
Because the amount of material that goes in doesn't hit the other side.
So, at very, very short times, we can pretend that it's semi-infinite.
And so then we can write that the concentration -- this is the short time, which means infinite size, and you'll see that the two are related -- that the concentration at infinity, then, doesn't change from the initial value.
The concentration at infinity for all time must equal the initial value C naught.
And so if you put these three boundary conditions into that equation, you end up with this.
This is the thing.
You get C minus Cs over C naught minus Cs.
This is surface concentration.
This is initial concentration.
This is C. It varies by this amount.
Error function of x over 2 times the square root of Dt.
So, you tell me the time and I can tell you the relationship between concentration and position.
Now I've got to tell you what the error function is.
Here's the error function.
This is this problem before.
This is the solution I'm going to show you.
Here's Cs.
Here's C naught.
The most general case.
C naught could be zero.
And this curve here is given by this equation, OK?
And this is an example of doping, isn't it?
This is how doping would work.
And this is the case where C s is greater than C naught.
If we turn it around and we make the surface concentration less than the initial concentration, we'll get the complementary situation.
The complementary situation looks like this.
So in this case, C naught is up here, Cs is down here, and we end up with this.
So now matter is going out, because Cs is less than C naught.
So this is -- what do you want to call it -- this is effusion, this is some kind of a drying process, outgassing.
And this is doping.
And what's the shape of this curve?
The shape of this curve.
ERF.
That's what math is.
You find a mathematical function that templates for the curve.
And ERF, I'm going to show you, is related to random walk statistics.
So, it's physically validated.
It's the best fit.
I mean, heck, if you find anything that is pinned at two ends and has curvature, you'll get a reasonable fit, but what's the right fit?
The right fit is this.
The error function looks like this.
Here's the error function.
Error function.
I'm going to plot ERF of z as a function of z.
And it goes from 0 to 1.
It looks like this.
All right?
And here's an error function table.
Those are exact values.
And it turns out that ERF is pretty much linear up to about point 6.
Up to about point 6.
ERF of point 6 not to less than about 1% is point 6.
And then as you go farther and farther out, and eventually, ERF of infinity equals 1.
But you don't even need tables for 0 up to point 6.
And you can push it, go to point 6, 5, if you want.
Yeah, let's do it.
Let's push harder.
6, 5.
It's linear.
You don't need tables.
Right?
And so ERF of 0 equals 0.
Oh, let me give you the function.
You know how you can define sine?
Sine is that integral of something over 1 minus blah, blah, blah?
We've got a definition for this one, too.
It's cool.
It's cool.
Here's what the error function looks like.
ERF of z equals the integral from 0 to z of e to the minus u squared du.
You might say, why is he showing us all of this?
Because he wants to torment you, that's why.
No, it's because I'm trying to teach you something.
What's e to the minus u squared?
Actually, I think I can squeeze it in right there.
What's e to the minus u squared du?
This function here, that's the bell curve, isn't it?
e to the minus u squared.
That's this thing.
It's symmetric, right?
It's a minus u squared, so it doesn't matter if u is positive or negative.
And e to the 0 was 1, and then back and forth.
And so what this is doing, is it's integrating from zero out to some value.
That's all.
That's the random walk.
This is the drunken sailor.
How far does the drunken sailor go?
That's the shape of this curve, the integral.
And we'd like it to be so that if you integrate from zero to infinity, the area will be one.
It turns out that this area here is root pi over 2.
So, therefore, to normalize, we put 2 over root pi.
That's how you get the error function.
All right?
So, the integral from zero to zero is zero, and the integral from 0 to 1 is going to be, whatever it is, point 8, 4, et cetera, et cetera.
So, that's it.
That's it.
Now we have the functional relationship to describe any of these curves.
And you'll get some practice in doing that.
And it's a lot of fun.
Because now you can talk about how long it's going to take to dope something, and so on.
Outgassing.
And now I'm going to show you something really cool, and I'm not kidding you.
This has served me well.
I've been here 30 some odd years, and I've been in consulting situations where I can, in my head, figure out order of magnitude of what it's going to take to run a diffusion problem.
Because you've got, up here, C minus Cs over C naught minus Cs equals ERF error function of x over 2 roots of Dt.
So, this runs from 0 to 1.
The right side.
When I went to school, the left side of the equation had to do the same thing as the right side of the equation.
So if the right side of the equation runs from 0 to 1, the left side of the equation runs from 0 to 1.
So, let's pick a -- are you ready -- average value from 0 to 1.
What would you pick?
I'd pick point 5.
Now, I don't have to go to the error function tables because point 5 is less than point 6.
So ERF of x over 2 root Dt is essentially equal to x over 2 roots of Dt.
So, now I've got one half equals one half x over root Dt, from which I can say that x is approximately equal to the square root of Dt.
And I can be sitting in a meeting and someone says, gee, I don't know how long it's going to take.
This thing's going to go down about 100 microns.
And how long is it going to take?
So, I pick a number, like, OK, if this is 100 microns, and this thing here is 10 to the minus 8, now you just take the square root of that, and tell them, oh, it's going to take so many minutes, and they go, huh?
How did you get that?
It's right there.
And I'm not trying to tell them a number for three significant figures.
They just want to know is it going to take a minute, an hour, a day, or are we out of business?
And you can make that calculation with this.
This is so powerful.
And everything I told you about diffusion applies to conductive heat transport.
You can do the same thing for heat transfer, figure out how long it's going to take for a wave to go in.
This is really powerful stuff.
Anyway.
Let's move on.
Well, there's our pal.
OK, so I was going to give you the last element today.
The last element.
Remember, we talked about clean air.
So, I'm showing you the catalytic converter.
That's this thing here, coated with platinum, rhodium, and so on.
And I'm showing you the electronic control module.
That's the CPU.
It's a little bit toned down from a Pentium, but it certainly started from one of these units.
So, we've studied all of this.
Right?
OK. And, so now, I want to talk about this piece here, the oxygen sensor.
why do we need an oxygen sensor?
Remember, the last day I told you that to get rid of NO, the NOx, you have to reduce, And to get rid of CO and unburned hydrocarbons, you have to oxidize.
Well, you have to pick.
You can't have two different atmospheres in the same place at the same time.
Turns out there's a lucky sweet spot.
Can you see that the conversion efficiency is high for both reduction and oxidation reactions if you peg the air-to-fuel ratio at 14.6.
So you can't just set the carburetor or the fuel injectors at some arbitrary ratio, because as you change humidity, as you change temperature -- what happens when the temperature goes down?
How much oxygen is there in the air?
It's still 20%, but PV equals nRT.
You ever try barbecuing when it's -- I mean, I don't know, maybe you're not from the Northeast.
But if you want to barbecue when it's 0 Fahrenheit?
Do you think you get a hotter fire or a colder fire?
How much oxygen is there per unit volume in cold air versus warm air?
Think about it.
It'll be on the next exam.
I'm just kidding.
OK.
So, anyway, here's the sweet spot that gives you both oxidation and reduction.
And so we need a feedback loop.
That feedback loop is provided by the oxygen sensor.
Dave, would you mind cutting to the document camera?
So this is an oxygen sensor that's used on automobiles.
Here's the plug.
This goes into the pin set.
And the oxygen sensor is inside this housing.
This housing, you can see -- OK, what I'm going to show you is what's going on inside here.
This has got all kinds of protective stuff on it so that it doesn't get dented during installation and so on.
But it sits inside the exhaust train.
It sits inside the exhaust train.
Now, may we go back to the slides, please?
Every car that's running a catalytic converter has to have one of those in order to control, right here, to make sure the air-to-fuel ratio is optimum.
Otherwise, the catalytic converter is either not doing a good job on NOx, or it's not doing a good job on hydrocarbons and whatever the other thing is.
Anyway, so here's what it looks like.
So, this is what's inside.
It's a closed, one-end tube made of zirconia.
Zirconium oxide.
And zirconium oxide conducts oxide ions.
And there are platinum electrodes on the front and the back.
And what I showed you was from this side.
And this is the exhaust gas going along here, and this is the edge of your exhaust train.
And two wires coming off of that sensor go to the CPU.
The CPU measures voltage and then from there, it regulates the air-to-fuel mix to the engine, to the exhaust gas, and around and around and around we go, so that when exhaust gas gets to the catalytic convert, it is optimally being converted.
So, we want to get fast response.
We don't want slow response.
And it's a solid-state sensor and solid-state diffusion is slow, but that's we have to rely upon.
So we add a dopant.
We add a dopant to zirconia to stabilize its crystal structure and create more oxygen vacancies to get faster diffusion and a shorter response time.
So what we add is calcia, one of the candidates, and it creates oxygen vacancies, and these oxygen vacancies allow us to get more rapid response time.
So, these oxygen vacancies compensate for the charge imbalance, thanks to the addition of calcia.
So, it's an engineered material.
Without this it doesn't work, all right?
And so, now what I've shown you is that on the basis of this oxygen sensor, we can monitor the exhaust gas flow so as to optimize the conversion here in the catalytic converter and do so by sending information to the CPU on the car.
And the number one consumer of CPU's is the automobile industry, because every car has at least one, if not multiple, CPU's.
So this is a good example of how understanding the lessons of solid state chemistry can allow us to build fuel efficient automobiles that have minimum toxic emissions to the environment.
All right, we'll see you on Monday.
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So we will have weekly quiz tomorrow.
There's been a lot of coverage, and so to focus you a bit, I'm going to confine the weekly quiz to glasses and chemical kinetics.
So you don't have to worry about diffusion.
We'll catch up on that, but I know there's so much material there.
Let's keep it confined to glasses and chemical kinetics.
And I'll be available today 4:30 to 5:30.
If you can't be at that time, send me a note, and I'm sure we can figure out a time to get together.
So what I want to do today is to start a new unit.
We're going to start talking today about solutions, and do some solution chemistry, OK?
Today we talk about solutions.
And you might initially say, why are we talking about solutions?
This is solid-state chemistry.
I think up until now, you've seen that it's pretty rare that we use solids in their pure form.
We usually have mixtures.
So for example, when we studied glasses, we modified the glasses with an alkaline earth oxide.
Well we, in fact, were making a solution of more than one component.
So it's to get certain properties that we study solutions.
Secondly, coming out of liquid phase is a way to make solids.
So in terms of processing, we need to understand something about solutions.
And then lastly, we're going to be studying, towards the end of the semester, a big unit on biochemistry.
And biochemistry, why, you say, biochemistry?
Why is he doing biochemistry?
I thought this was solid-state chemistry.
We are solid-state devices, but we're made of soft matter.
At least the exoskeleton is soft matter.
The endoskeleton is ceramic, right?
Our bone structure is ceramic.
Hydroxyapatite, calcium hydroxyapatite, and the outside is a polymer.
So look at this.
Confirmational changes in polymer.
But the chemistry, the biochemistry, so much of it takes place in aqueous solution.
So hence, we better know something about aqueous solution.
So to this, we go almost right back to the first day.
You remember, was the second lecture, and we showed this figure, and all the different categories of matter.
And we started over here with elements, and we moved into some pure substance compounds, et cetera, et cetera.
Now we're going to move over to here.
So homogeneous mixture containing uniform composition and properties, as opposed to a heterogeneous mix.
So we're now over here.
We're working our way through the diagram.
All right.
So let's get a couple of basic definitions up.
So the solution is really a mix of at least two constituents.
One, the majority constituent is called the solvent, and then we can have one or more solutes.
So the solvent, this is the majority constituent, and then the solutes are the minority constituents.
And in some instances, it's pretty hard to tell which is which.
And as far as types of solutions, I want to broaden this.
You know, when I say solution, you're probably thinking about aqueous solution.
But I want to take a minute and work through this chart.
And this is all posted, so you don't have to write it all down.
Just follow with me.
So most of the general chemistry subjects will just stop at the end of the first line.
They'll treat solutions as aqueous solutions.
But in material science, we think of things broadly.
So a good example of a simple aqueous solution is sodium chloride and water.
Sodium chloride is the solute and water is the solvent.
But you can have a liquid solute.
So wine-- in case you've ever encountered this beverage, it's primarily water, but it can contain up to about 14% ethyl alcohol, and there are other constituents that give the color and the flavor and so on.
And you can have a gas in a liquid, and that would be seltzer, where CO2 is dissolved.
It's actually dissolved.
If you take a look at a bottle of bubbly water on the shelf, you don't see the bubbles.
The carbon dioxide is actually dissolved.
You can have a gas as a solvent, and air would be an example of that, where the solvent is nitrogen.
And then we have as solutes oxygen, argon, carbon dioxide, if you live next to a power plant, it's sulfur dioxide, if you live next to an aluminum smelter, it's tetrafluoromethane, and so on.
So we have all kinds of solutes in the air.
And then we can have solid solutions.
And OK, as soon as you see the word solid, you know, that means you're 3.091.
So what do we see for solids?
Metal alloys.
We saw carbon dissolved in iron.
Well, that's a solution.
It's homogeneous, single phase, just as the definition on that chart 1.11 said.
Semiconductor, boron doping into silicon, the boron sits on a silicon lattice site.
So this is a true solution.
The boron is the solute, and silicon is the solvent.
We can have the ceramic.
We talked about the oxygen sensor.
The oxygen sensor is zirconia, which has been stabilized with the addition of calcium oxide.
As one example.
In contemporary work, they might use another oxide.
But what's the purpose of the calcium oxide?
Well, I told you that it's to increase the vacancy population, give you a rapid response on your oxygen sensor.
And as we're going to learn later, that's true, and there's a second value in putting in the calcium oxide.
And it stabilizes, that's why they use the term stabilize, it stabilizes the cubic form of zirconia.
Zirconia has a different crystallographic modifications.
The cubic one is the one that gives us the best ability to transfer oxygen, and the addition of calcium oxide as a solute, the calcium ions actually sit on the zirconium lattice, that's a true solid solution.
It also makes the cubic zirconia stable, and with Christmas coming, you know, it's the poor man's diamond, and that's the same material there.
And to modify a glass, as I mentioned earlier, adding a alkaline earth oxide breaks the silicate network.
That's a solution.
Now here's an inverse one, where the solid is the solvent, and the liquid is the solute.
So this was dentist's office practice going back to the time when I was your age.
If you had a feeling, the dentist had silver and mercury in the dentist's office, and would add liquid mercury to silver and make up this amalgam, and then jam that into the tooth.
Actually, that was the B part of the operation.
There was this dentist here in the United States, and if I could find him, I would like to have a word with him.
But he had this theory.
You see, the amalgam is a metallic alloy.
And so obviously, it has metallic bonding.
And the tooth is-- it's a ceramic.
So what kind of bonds are going to form between a metal and a ceramic?
They're not very good.
So this dentist had the theory that the way to increase the bonding capability of the amalgam to the tooth was to maximize contact areas.
So when you went in with a tiny, tiny little cavity, the dentist would drill the tooth out, removing about 75% of the volume of the tooth, and then they would make this amalgam and shove that in.
And you'd wear that for about 20 years, until one day you bite into a muffin that's got a little piece of walnut shell in it, and the amalgam flexes, and your thin-walled tooth goes bang!
Like that.
And now you get to go back to the dentist's office for yet some more medieval treatment.
Actually, things have improved somewhat.
Thanks to material science.
But they're not using this amalgam anymore.
But there are a number of us who are still walking around with silver and mercury in our mouths, and-- yeah.
Enough about my dental problems.
Now let's go on to intercalation.
You can put a gas into a solid.
And I told you about this one, where if you want to store hydrogen, and this is a big problem, if we're going to talk about hydrogen-powered vehicles.
A problem as big as, how to get the fuel cell cheap enough, is where are you going to store the hydrogen on the car, and one of the materials that's been proposed is this alloy of lanthanum and nickel that can intercalate huge amounts of hydrogen.
So that forms a solid solution.
So those are examples of solutions, and we're going to go back and make sure that we cover the basic Gen Chem at the top here.
But I want you know that solution chemistry is very broad.
Now when you dissolve something, you actually have things down at the atomic level.
So for example, in brine, you actually have below two nanometers as particle size.
You actually have sodium ions and chloride ions dissolved completely, which means that the solution is clear, colorless, and transparent to visible light.
Water is clear, colorless, transparent to visible light.
If you had sodium chloride, it remains clear, colorless, transparent to visible light.
You can't filter, and you can't wait for the sodium chloride to settle out, because it's bonded within the structure of the liquid water.
And this has implications.
There's so much water on the planet, but there's not so much fresh water on the planet.
And desalination can't be accomplished by filtration.
A filter that had a pore size small enough to trap sodium ions, the pore size would be so small, water molecules couldn't go through.
So this is the concept.
At the other end, you can get something called a suspension, where the particle size is greater than about 1,000 nanometers, and blood is a good example of that, where it's opaque.
Visible light doesn't go through, but you can filter out some of the matter in blood.
And if you let it sit for a while, it'll settle in it gravitational field.
And in between, you have this whole zone of colloids.
And so between colloids and suspensions, the only difference is particle size.
And this whole thing we would just call a dispersion of another phase, either another solid phase, or another liquid phase.
And it's kind of an interesting physical chemistry about the dispersion, what makes it work.
For example, in milk, milk is a good example of a dispersion.
You have a fatty phase, and you have an aqueous phase.
The aqueous phase, where all the minerals are.
Why?
Because the minerals are-- what kind of compound?
They're not metallic, they're not covalent.
They're ionic.
The ionic compounds dissolve in the aqueous phase, and then in the fatty phase, that's where you have the protein and so on.
But the fatty phase is clear and colorless, and the aqueous phase is clear and colorless.
And yet milk is white.
It's, as the term implies, it's milky.
Why?
What's going on?
You have a second phase here.
This could be the fatty phase, and this is the aqueous phase.
They're both clear and colorless, but they have different indices of refraction.
And since this has n fatty phase, and this has n aqueous phase, this interface scatters the light.
So if you want to start a business, if you want to try to tailor the index so that the index of the aqueous phase matches the index of refraction of the fatty phase, you'd have the two dissolve, one and the other, and it would be transparent, divisible light.
So you would have milk that isn't milky.
I mean, the public would be very confused.
But anyways.
So that's what you do.
So why does this thing not settle?
Because they do have a density difference, and again, going back to the days when I was not a college student but a youngster, there was still this form of milk called pasteurized milk.
Well, all milk is pasteurized, but this milk was not homogenized.
And what would happen is, there was this person that would deliver milk.
This was a borosilicate glass bottle.
And here would be the cream.
The cream would rise to the top.
We have all these expressions in our language.
And then this would be the milk here, and you could skim this off for coffee or sugar, and then this would be a low-fat milk.
But people wanted this all mixed, so then they went to homogenized milk.
So what is homogenized milk?
This is your red cap, now.
All the milk is homogenized.
This was a big thing.
This was simply called pasteurized.
I'm going to get to the physical chemistry here.
It's very interesting.
Because this is lower density, and yet in homogenized milk it doesn't rise.
So let's take a look at what goes on in the physical chemistry of homogenized milk, because it's all about these various systems.
So I'm going to take this particle here, and its sum insoluble cluster.
And I'm not specifying the cluster size.
It's probably greater than about two nanometers.
So there are two forces acting on this.
There's a settling force and there's a buoyancy force.
Obviously, otherwise it wouldn't float.
So there's some kind of a buoyancy force.
Settling force and a buoyancy force.
Well, the settling force, this is just the gravity.
Right?
This is the force of gravity.
And we know the force of gravity.
That goes with the mass.
Come on, get that cell phone out of here.
This force of gravity goes as the mass, and the mass, we know, goes as the volume.
And the volume goes as the cube of the radius.
I'm assuming this is a spherical particle, all right?
Now, the buoyancy force.
The buoyancy force is the interfacial force between the two.
There's some binding across here.
Maybe weak van der Waals, or if this fatty phase has molecules in it that are polar, then there could be dipole-dipole interaction.
But in any case, there's some kind of an interfacial force here.
And this is all chemical bonding between solute and solvent.
But you see, the force is a weak force.
If it were a really strong force, it would dissolve this thing.
It won't quite dissolve it, but there is some kind of dipole-dipole weak force.
And this one here is operating across the surface area.
That's the contact.
So this force goes as the area, and area goes as the square of the radius, whereas mass goes as the cube of the radius.
So you know, from your math, that r cubed dominates r squared, but only at large r. It's not always the case, is it?
Maybe before you got here, you thought that.
But now that you've been at MIT a few months, you know that r squared can dominate r cubed at small r. Interfacial forces dominate.
And that's exactly what happens in these dispersions, and that's why they don't settle out.
Now, homogenized milk is simply milk that has been agitated in such a way as to reduce the fat globule size below a critical value so that these interfacial forces hold the fat globules in suspension.
If you waited long enough, they would agglomerate and settle, but that time is probably longer than the shelf life of the milk.
So the milk probably spoils before stuff settles out.
So this is all very important to understand.
The range that exists between insolubility and this sort of clustering, and suspension, and so on.
By the way, a lot of pharmaceuticals are like this.
A lot of pharmaceuticals.
So when it says, shake well before using, they're not kidding.
Because the active ingredient will settle.
And you're drinking just the solvent, just the vehicle.
And all of the potency is on the bottom of the bottle.
Shake that thing up!
I can't say what it does to the taste, but that's another problem.
Actually, I threw in this slide here.
We're not going to spend any time on it, but you can look at it at some point.
This is a whole taxonomy of colloids.
Solid-liquid emulsions, aerosols, they're all part of this magic zone between solubility and just brick, all right?
There's this whole fine, pardon the pun, the whole fine structure.
OK, So let's get to the chemistry.
Obviously there's something to do with bonding here, right?
So here's a simple experiment, and this is taken right from the reading.
So I've just taken this episode that is written up in the reading.
So we've got two beakers here, and in each beaker, we have a bilayer.
We've poured in some carbon tetrachloride, liquid, and we've poured in some water.
And these two are immiscible, because carbon tetrachloride is obviously a non-polar liquid, and water is a polar liquid with hydrogen bonding capability.
And in one beaker, we introduce crystals of iodine.
In the other beaker, we introduce crystals of potassium permanganate.
And then we shake them up, and we wait.
And eventually we see that on the left, the iodine dissolves in the carbon tetrachloride.
And we're using the purple color as an indicator.
And this is kind of cute, because both potassium permanganate and iodine will render things purple.
So you're comparing purple to purple.
They could have chosen something else, but this is kind of cute.
All right.
So here you end up with a solution of iodine and carbon tetrachloride, whereas on the right side, you end up with a solution of potassium permanganate and water, and nothing in the carbon tetrachloride.
So what can we infer from this?
Well, let's take a look at the possible interactions.
So first of all let's categorize H2O.
This is polar, it's a polar liquid, with hydrogen bonding capability.
Carbon tetrachloride is non-polar.
It's very toxic.
When I was a child, we had this in the medicine cabinet.
It's a non-polar solvent.
It's fantastic for getting grease stains off.
Every man had this.
With a little handkerchief, you'd take a little grease stain off your tie.
You can't buy this stuff.
You can't even guy it for research anymore.
It's too bad.
It's great stuff.
Highly toxic, though.
But you know, there's a time and a place.
All right.
So then here's iodine.
Iodine, we know, is non-polar.
It's a homonuclear molecule, it has to be non-polar.
And so what holds iodine together in the crystal?
There's only one bond, and that's van der Waals, right?
It's a van der Waals solid, whereas potassium permanganate is an ionic solid.
Potassium permanganate is ionic, and it consists of potassium cations and permanganate anions.
And what we find is that the non-polar solute dissolves in the non-polar solvent, and the ionic solute dissolves in the polar solvent with hydrogen bonding capability.
So from this, we can infer the general rule, which is encapsulated in the language used in chemistry textbooks.
Like dissolves like.
And what they're really saying here, is that like solutes dissolve in like solvents.
Solute-solvent is like dissolves like.
OK.
It's a good place to start, but it's an incomplete picture.
So I want to show you that there's some sophistication here.
This is taken from one of the other books.
And it shows just the rules for ionic compounds in water.
And I just showed you, from this example, that the ionic compound dissolved in water.
And what you see here is that some ionic compounds dissolve in water, but there's a whole set or insoluble ionic compounds.
So it's not straightforward.
But we know from 3.091, we know that we can make sense of this on the basis of competition.
Competition between what holds the compound in the solid state versus what will pull it into the liquid state.
So for example, we can compare sodium chloride, which we know dissolves in water.
So sodium chloride, as a crystalline solid, will dissolve to form sodium chloride aqueous solution.
Whereas if you look at magnesium oxide, which is also an ionic crystal, it does not, to any standard imaginable, dissolve to form an aqueous solution of magnesium oxide.
And what's the difference here?
The difference here is, compare solvation energy, in other words, the energy that you got by pulling this into solution, and forming bonds between sodium and chlorine in water, with the crystallization energy.
And what's that all about?
Well, that's the Madelung constant, remember?
Madelung constant.
q1 q2 over 4 pi epsilon 0 r, where this is the cation anion.
And you can see here that sodium chloride has sodium cations and chloride anions, and there's a certain binding energy in the crystal.
Magnesium has 2 plus.
Oxide is 2 minus.
The binding energy between magnesium and oxygen is so great that there's no driving force to dissolve.
Now, I don't expect you to be able to look at this and tell me whether something's going to dissolve or not.
But if I were to say to you, explain why sodium chloride forms solutions with water, and magnesium oxide doesn't, that you could go through this rationalization.
And here's a cartoon from the textbook that tries to illustrate this solvation.
Here you can see a crystal of sodium chloride, in the inimitable fashion, as drawn by chemistry books, where the chloride ion is green, and the sodium ion is blue.
And that's OK.
I can live with the color-coding, as long as we agree amongst ourselves, these ions are clear and colorless, because they have octet stability in their electronic shell.
And here's water, with the hydrogen shown in white, and the oxygen shown in red.
And you can see that the oxide end, the oxygen end, the delta negative end of the water, is trying to wrest sodium cation out of the lattice, and ultimately surround it by a cage.
And the same thing here.
You see the hydrogen ends of the water trying to pull chloride out, and ultimately surround it with a water-like cage.
So this is the competition I was talking about.
In the case of sodium chloride, water wins.
In the case of magnesium oxide, water loses.
The binding energy between magnesium and oxygen and the crystal dominate.
So you don't see that solvation.
OK. Let's talk about metrics now.
Let's look at some metrics of solubility.
It's quantifiable.
So we can express a measure of solubility in terms of a quantity known as molarity.
So we can express moles of solute per liter of solvent.
And this is called molarity, and the symbol is capital M. So we can say, for example, a 1 molar solution of sodium chloride in water, we'll write 1 molar NaCl, and then we'll write subscript aq, meaning aqueous.
So it's an aqueous solution at a concentration of 1 mole of NaCl per liter.
And the liter is named after a person, so that's capital L, per liter of solution.
And remember, the solution is the sum of the water plus solute.
For dilute solutions, there's very little difference between the total amount of water and the total amount of solution.
But in certain instances, the presence of the solute actually has a volumetric change on here.
So strictly speaking, it's per liter of solution And as I've shown you, there are degrees of solubility.
So people represent the threshold of solubility as c of the solute.
When c of the solute is less than about a million molar, 0.001 molar, we call this insoluble.
So something is vanishingly soluble, we say that's the value.
So we'll call this the threshold.
And then, something that's quite soluble, we'd say that the concentration of the solute starts to exceed at about 0.1 molar.
And then we would say, that qualifies as soluble.
So now let's look at two extremes in solubility, operating off of this.
So in the one case, we can have complete solubility.
So examples of that, where things are completely miscible in one another.
Complete solubility.
By the way, some people will use the term miscibility.
When something is miscible it means it's soluble.
Same idea.
If something is insoluble, some people might say it's immiscible.
Same thing.
OK?
So complete solubility is ethyl alcohol and water.
You can mix them in all proportions.
Continuously variable.
On the solid alloy is silver and gold.
Silver and gold, you can make alloys of any composition between 100% gold and 100% silver.
Now, that's the exception.
Most cases are situations of limited solubility.
So they go up to a maximum, which we can denote C-star or C saturation.
This is the maximum solubility.
So an example of something that's sparingly soluble in water is silver chloride.
Let's look at silver chloride.
So silver chloride, I'm going to start here, silver chloride as a crystal, and I'm going to dissolve that in water.
So that gives me AgCl aq.
So that's just simply the formation of an aqueous solution of silver chloride.
And when the reaction moves from left to right, we call that dissolution.
The silver chloride is dissolving, and when the system moves from right to left, we call that-- now here I'm going to nitpick with the book.
The book calls the left reaction crystallization.
And that's correct.
It is crystallization in this case, because silver chloride is a crystal.
But it is possible, in other systems, to have the solute come out of solution, and make a solid that is noncrystalline.
And you know that all crystals are solids, but not all solids are crystals.
You can have an amorphous solid.
So what would happen if you were to bring out of solution an amorphous solid?
It would be silly to call it crystallization.
So I prefer to use the term precipitation.
And there's another term that you can use, and I learned this one from reading the literature of geochemistry.
What the geochemists call the reaction going from solution to make a solid, they say the system exsolves.
This is dissolve, this is exsolve.
So this is called exsolution.
It's amazing what you can do when you know a little bit of Latin.
So this exsolves.
So that's the exsolution, or the crystallization reaction.
Now I'm going to show you what won Arrhenius his Nobel prize.
Arrhenius did not get the Nobel prize for his brilliant work on activation energy.
He got his Nobel prize on the theory of electrolytic dissociation, which was, people knew that you could dissolve salts and water, but they didn't know how.
And it was Arrhenius who said that the salts go into solution by dissociating and forming ions.
So goes in as Ag plus Ag plus aq plus chloride ion-- thank you.
And that, ladies and gentleman, was a Nobel prize for Arrhenius.
And you can see that there's a relationship between the amount of silver chloride and the amount of ions.
So there's a mass balance there, that the concentration of silver chloride dissolved, in fact, equals, in this case, by stoichiometry, the concentration of Ag plus, because of the nature of the dissociation reaction on a mass balance basis.
And that also equals the concentration of the chloride ion, OK?
So that's the way we can look at the system.
And how do we know that this thing has limited solubility?
Well, there's various ways of measuring it, and one of them involves conductivity.
Here's the conductivity of pure water.
And you know that water has very, very poor conductivity, and in fact, what's happening here, when we add silver chloride is, we're adding charge carriers, because the audience are charged species.
So they can carry charge.
And you can see that this is a measure of conductivity as a function of silver chloride concentration.
And as you add silver chloride to higher and higher values, the conductivity goes up.
And look at even the tiniest amount of silver chloride has a conductivity that's about, what, half an order of magnitude higher than the conductivity of pure water itself.
So I would say that this is akin to doping, isn't it?
So up here, when you've got 10 to the minus 6 Siemens per centimeter conductivity, that aqueous solution is demonstrating the extrinsic behavior.
This is very similar to doping.
And at some point, we get to this value here, around 10 to the minus 5 moles of silver chloride per liter, and then adding more silver chloride has no impact on conductivity.
Which tells you that you've hit saturation.
This is akin to adding more and more sugar to the cup of tea, until finally the sugar just falls to the bottom.
You can stir all you want, but you won't get it to dissolve, because you've hit saturation.
So that indicates the presence of a saturated solution.
And so we can talk about that value, and we can say that for silver chloride, the concentration at saturation is equal to 1.3 times 10 to the minus 5 moles per liter.
Moles of silver chloride per liter.
And obviously, that's equal to, according to that, it's equal to the silver ion concentration.
I'm going to use square brackets to indicate moles of silver, ion per liter of solution, which is also equal to-- pardon me-- it's also equal to the chloride ion concentration at saturation.
Now I'm going to ask you a simple question.
Suppose I've got a beaker here, and I know the maximum I can get here, the maximum is 1.3 times 10 to the minus 5.
Now this is a really simple question.
Suppose I am about to add silver chloride-- let's say this is 1 liter already.
All right?
I've got one liter.
And you'd tell me, well, you can put in 1.3 times 10 to the minus 5 moles to get the saturation.
Suppose instead of 1 liter of pure water, I had 1 liter of water already containing, say, 1 times 10 to the minus 5 molar silver chloride.
Well, that's kind of obvious, isn't it?
I'm only going to be able to put 0.3 moles in, because 0.3 times 10 to the minus 5, because I've already got silver chloride in there.
That's easy.
Now let's make the question a little more interesting.
Suppose instead of a certain amount of silver chloride in there, I have no silver chloride in there.
But I've got, say, 0.1 molar sodium chloride.
It's a salt.
It's a difference salt.
So the question is, does the presence of a different salt have an impact on how much silver chloride I can put into this solution?
And the answer is, yes.
The answer is yes.
So what we find is that the presence of the other salt, in this case, has an impact, because sodium chloride goes in as sodium plus, and chloride minus.
So there's already a boatload of chloride ion in there, and that has an impact on this relationship.
So how do we answer the question, what is the solubility of silver chloride in the presence of other chloride ions?
And for that, we define something called the solubility product.
And you need it in order to answer the question, how do you determine the solubility of a solute when there are other solutes present already?
And it's denoted capital K, lowercase sp, subscript.
Solubility product.
And it's equal to simply the ion products of the constituents.
So the solubility product of silver chloride is the product of the silver ion concentration, and the chloride ion concentration.
And you know that in a solution of silver chloride alone, if nothing else, that the concentration of silver ion equals the concentration of chloride ion.
That's the whole business of dissolving by itself.
And so I can then just say that Ksp.
will then equal the concentration of silver ion squared, which we also know is equal to the concentration of silver chloride aqueous.
See, all of these are the same.
So this solubility is the same.
I can just put that in, which is just concentration of silver chloride.
Square that.
So if I square 1.3 times 10 to the minus 5, I end up with a solubility product of 1.8 times 10 to the minus 10.
So now I can use this in order to determine how much solubility I get in the presence of another salt.
So in this case, I'm going to put 0.1 molar sodium chloride.
And these are strong salts, so we're going to get complete dissociation.
Gives me 0.1 molar sodium ion, and 0.1 molar chloride ion, when it dissociates.
Now you see the difference.
Because the silver chloride, by itself, gives me 10 to the minus 5 molar chloride ions.
When I add sodium chloride, I get 4 as a magnitude more chloride.
So let's go back to the Ksp.
So Ksp, this is for silver chloride.
Ksp for silver chloride is going to be equal to the silver concentration and the chloride concentration.
And in this case, the silver concentration is just equal to whatever that solubility is.
Because there's only source of silver ion, and it's silver chloride.
So I can write that as concentration of silver chloride.
That's good.
And now this one here is what?
I've got two sources.
I've got silver chloride, I I've got sodium chloride.
So it's going to equal this thing here.
0.1 plus whatever I get from silver chloride.
And it's vanishingly small, isn't it?
The concentration of silver chloride, whatever it is.
This is nothing, so I'm just going to neglect it.
It's dominated now.
The chlorine is flooded by the silver chloride.
And this product is a constant.
That's still equal to 10 to the minus 10.
So I can turn this around and solve for the concentration of silver chloride, which is equal to the concentration of silver io.
And that's equal to, what is it, 10 to of the minus 10 divided by 0.1, which then gives me-- what's the number here?
Plug that in.
And I end up with 10 to the 1.8 times 10 to the minus 9 molar.
Right?
So look.
Look at what's happened by having the chloride present from sodium chloride in such a large amount.
It's repressed.
It has repressed the dissolution.
The presence of chloride then has a negative impact on solubility, and instead of having 10 to the minus 5 molar, it's down to 10 to the the minus 9 molar.
And this effect of repressing solubility by adding a second solute is called the common ion effect, OK?
Solubility repression by second solute is known as the common ion effect.
And this is used in processing.
If I want to trigger the precipitation-- see, if I started with a solution containing 10 to the minus 5 molar of silver chloride, and I throw in some sodium chloride, it'll start precipitating out silver chloride.
So if I wanted to make a fine precipitate of silver chloride, I make a pregnant solution.
And I could drop the temperature, because you can imagine that solubility is a function of temperature, or I could keep it isothermal, throw in some sodium chloride, and out comes silver chloride.
So since the common ion effect on it and its value in processing.
OK. Good.
Well, I think I'm going to hold it there.
I"ll show you just one more thing.
If you've got stoichiometry like this-- this is a difluoride of magnesium.
If it goes into solution, you get magnesium cation plus 2 fluoride anions.
And so if I wrote the Ksp for this reaction, it would be the product of the magnesium concentration and the fluoride ion concentration.
Because there's the two here, this is squared.
So this, too, will transfer up there, and then throw in some sodium fluoride, and cause the other thing to exsolve, and away we go.
All right.
I've got a couple of things to show you.
We're talking a lot about Arrhenius.
This is a book I have.
It's an English translation of a book written by Arrhenius in the late 1800s, and it was printed in English around 1908.
And Arrhenius was a genius.
He wrote on all sorts of topics here.
Biology, physics, you name it.
And one of the things and he was interested in was the origins of the earth.
So this chapter is called, Celestial Bodies as Abodes of Organisms.
Already speculating on whether you could have life as we know it exist elsewhere in the universe.
And one of the things he talks about is global warming.
So this is Arrhenius on global warming.
I'll read you the last paragraph of this chapter.
There had been some major volcanic eruptions that had caused cooling when Krakatoa in 1883 and Martinique in 1902.
Major plumes of soot that caused dramatic decreases in temperature.
So now here comes the last paragraph.
We often hear lamentations that the coal stored up in the earth is wasted by the present generation-- remember, this is written 100 years ago-- without any thought of the future.
And we are terrified by the awful destruction of life and property which is followed the volcanic eruptions of our day.
We may find a kind of consolation in the consideration that adheres in every other case.
There is good mixed with the evil.
By the influence of the increasing percentage of carbonic acid in the atmosphere-- that's CO2-- we may hope to enjoy ages with more equitable and better climates.-- Remember, he's a Swede; it's cold-- especially as regards the colder regions of the earth, ages when the earth will bring forth much more abundant crops than at present for the benefit of rapidly propagating mankind.
So it's interesting to see the world-- it's a great book to read.
And you can see people in the 1830s were already calculating heat transfer coefficients to how much the earth was changing.
We're going to post these to the website.
This was, last year, in the New York Times.
Every Tuesday they have a science section.
And this was about glass.
And very, actually, with your 3.091 knowledge, you'll read this like a newspaper, and it'll be very easy.
And they go through the structure.
Here's the structure of a window glass.
You can see the network, former network modifier, and so on.
And they talk about how difficult it is from a first principle's standpoint to model the structure of glass.
These oxide glasses are complex and not easy to model.
So when you're trying to engineer the glasses, instead of trial and error, it's hard to do so by theory.
And it talks about some efforts at theory, and so on.
And the last thing I want to talk about is bulk metallic glasses.
You recall that I showed you metallic glass that was made by melting gold silicon, and dripping it onto a water-cooled copper wheel that was zooming around to give us a cooling rate of about a million degrees a second.
And those strips had to be very, very thin, the metal strips.
Because you've got liquid dripping down, and we're pulling the solid away, and there's a finite thickness here.
Let's use a Greek letter, xi.
There's a finite thickness xi.
Because what's happening is that this is the water-cooled copper wheel, and you're extracting heat here.
But eventually, the thermal conductivity of the metal is the limiter.
In other words, I don't care how cold.
You can put this in liquid helium if you want.
You can't get the heat through the metal fast enough.
And what happens is, when you look down here, the lower part is amorphous and the upper part is crystalline.
So you don't end up with metallic glass.
You end up with some metallic glass, and the upper part is crystalline.
So what happens when you end up with the limitation being the thermal conductivity of the metal?
At that point, you're finished.
And this was typically on the order of microns.
And then they got up to, sort of, submillimeter.
And that was it.
So the glass that I showed you was foil.
Now what happened was, with more research-- so here we are.
This is Pol Duwez at Caltech, gold silicon.
And this is the thickness in centimeters.
So you were down here at some tens of microns, all right?
Now, in the '60s, research at Harvard uncovered a set of palladium alloys that had better thermal conductivity and, remember the first day, about the analogy to the musical chairs?
These things have slightly more complicated crystal structures.
So for a given cooling rate, the metal has more difficulty finding the proper lattice site.
And they were able to make metallic glasses that were on the order of 0.1 centimeters.
That's still fairly thin.
And then back to Caltech in the late '80s, early '90s, and a man by name of Bill Johnson was able to develop a set of alloys that can be made in bulk form.
Totally amorphous metal.
And these are known as bulk metallic glasses.
And look at the complexity of the alloy designation.
So now you see, well, why are they doing this?
Because look, this is strength versus elastic limit.
So you can either have things like polymers, that you can stretch very, very far, but they don't have much strength.
Or you can have things like certain steels, that are very, very strong, but you can't bend them very far.
And bulk metallic glass has put you right up here.
You have strong alloys that have very, very high elastic limits.
So they make great golf clubs and tennis rackets and so on, because they can flex way back, store enormous energy, and then spring.
So here are some bulk metallic glasses, and here's a classic one.
This is zirconium, titanium, copper, nickel, beryllium.
All right, now how do we get that?
Well, zirconium and titanium are body-centered cubic, we've got copper and nickel are face-centered cubic, and beryllium is hexagonal close-packed.
So the idea here is the principle of confusion.
So the alloy is fighting with itself.
You know, am I face-centered cubic, am I body-centered cubic, am I hexagonal?
And this confusion about what the crystal structure is to be leads to quenching and the disorder of a liquid state, and preventing the formation of grain boundaries, dislocations and so on.
So I want to show you one example besides golf clubs.
Dave, could we cut to the document camera here?
Get this thing down.
So this is a-- I'm not endorsing the product at all, but you know, this is one of the companies that makes, this is SanDisk, I think.
And they make these flash memories.
This just looks like a another piece of metal, and in fact, it's very disarming, because they call this model the Titanium.
Well, it has about 13% titanium.
The interesting thing here, what's so cool about this, this is bulk metallic glass.
And what's attractive about the fact that it's bulk metallic glass is, that it can be shaped by casting, from the liquid state to very, very fine precision.
So you see all of these slots and everything, and on the side, all of this kind of stuff, and on the end.
All of this is done by casting from the liquid state, in one operation.
And this is a clam shell.
There's two pieces here.
I don't know if you can see, but there's two pieces that have been sandwiched together.
You can see on the edge there.
And the impact that has on manufacturing costs is phenomenal, because normally you make the basic shells, then you have to drill, and you've got to auger out, and so on.
This thing, one step operation, including the labeling.
There's no subsequent processing.
And this is done by a company out in Michigan called Liquid Metal.
And they've licensed the technology and so on.
And again, I'm not trying to make a commercial thing, but I'm just trying to show you that these ideas of changing properties for engineered materials, you know, are around us everywhere.
And this is relatively recent.
Bulk metallic glass.
Fantastic example of structure property relations.
OK.
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So a couple of announcements.
Weekly quiz Tuesday.
And also on Tuesday, I want to draw your attention to the event in the slide.
We have a poster here for a lecture that will occur right in this room, at 4:00 on Tuesday.
And it's entitled the Wolf Lecture.
The Wolf Lecture was established here about 30 years ago in honor of Professor John Wolf.
John Wolf was an antecedent of mine.
He was the person that invented 3091.
And in 1961, he dared, as a metallurgist, to launch a variant of freshman chemistry.
And you're the beneficiaries of John Wolf's initiative.
And he also was a spectacular teacher, quite a showman.
And in his honor, they instituted in the Department of Material Science and Engineering, the Wolf Lecture, which, as the poster says, is the entire community is invited to attend, but it's geared towards freshman.
So this is a lecture that the rules are.
Talk about whatever you want in terms of material science and engineering, but make it entertaining, and make it engaging and accessible to freshmen.
You should go to lectures.
If you can't go to this one, then by all means, go to other seminars.
If all you do when you come to MIT is go to class, you're missing out on some of the richness here.
We have all sorts of people coming through here every day.
And you can go to these seminars, you can learn a lot.
You might say, well, I'm just a freshman, I'm not going to understand everything.
I go to these things.
I don't understand everything.
But I learn something, because the first few minutes of the lecture, if the speaker is any good, he or she is going to set up the topic for you.
If you follow the first five minutes, you'll have something that, you know, gets you oriented to the topic.
And then the other thing, the other reason to go-- you see this on the poster-- room 10-250, reception immediately following.
That means there will be refreshments!
Food, for that, you know, pick-me-up in the late afternoon.
So you go to the seminar, you know, sometimes they have refreshments before the speaker.
I'm not going to tell you have to go into the seminar, having eaten the food and partaken of the refreshments.
You didn't hear me say that.
But anyways, there's always food around here.
Go to this one.
I think you'll enjoy it.
And the speaker is Professor Michael Rubner.
He's a faculty member in course 3.
An excellent teacher.
As you see, MacVicar Faculty Fellow, which means he's been acknowledged as a good teacher.
I think you'll learn a lot.
And he's talking about nature-inspired, so biomimetics, how we see things in nature and then mimic that design in advanced material science.
OK.
Enough said.
Let's get on with the lesson.
Last day we started looking at solutions, and we recognized that bonding is the key to understanding, and it's encapsulated in that catch-all phrase, like dissolves like.
And towards the end, we talked about Ksp as a metric that helps us understand common ion effect.
Common ion effect, again, if I tell you you have, say, 100 units of sodium chloride that can go into solution, you would say, that's the solubility limit.
But then if I tell you, I've got a solution that already contains 25 units of sodium chloride, that's a no-brainer.
You've only got another 75 units to go.
Now I say I've got potassium chloride in there.
So now you stop and think, wait a minute.
It's not sodium chloride, but it is a chloride.
How do I think about this?
Common ion effect through Ksp helps you reason through it.
It's when you have more than one solute, and one of the constituents of the solute is already present.
So today I want to talk about a subset of solutions, and in particular, I want to talk about acids and bases.
And this is important, not only in materials processing, but we're going to have to understand this if we're going to go forward later and talk about biochemistry.
So, you know, they're around us everywhere.
You know, you probably got up this morning, you washed your hair with a pH-balanced shampoo, maybe had orange juice or grapefruit juice with citric and ascorbic acid.
Your radio is powered by zinc alkaline batteries.
I started my car, it's got the lead acid battery in it, and electricity for all my appliances came from coal-fired power plants, spewing out SO2, turning out acid rain.
So we're off to a good start!
It's Friday.
So now we want to go back and understand this from the beginning.
So we're going to start with a history lesson.
And the history lesson starts in ancient times.
Acids were known all the way back in early times for processing of food and materials.
So who among us hasn't eaten something that has been pickled?
Pickle means to have been processed in acid solution.
In fact, the word acid comes from the word acidus in Latin, and acidus means either sour or tart.
But the modern chemistry of acids and bases starts with Lavoisier in France in 1779.
And I'm going to be naming many scientists from different countries today, so I'm going to use the international symbol.
You know, these ovals that you put on the back of the car, so now if you see a car with this on it, you know it's a French registry.
So he's probably got a Peugeot, and he's driving around with his Peugeot.
And what Lavoisier said, in trying to understand acids and bases, is-- it's a really interesting story.
He said that oxygen is present in all acids.
And why?
Because he spent most of his career studying combustion, so he was very concerned about oxygen.
And the interesting thing is that oxygen, the name of the gas, oxygen, actually comes from the greek oxy, which means sharp, and, you know, the particle gen, as in to generate or to be born.
OK?
So it's interesting that this gas is named incorrectly.
It's named for an attribute that is ascribed to acid, and it's wrong.
And it's in other languages, too.
The word oxygen translates into other languages as meaning something sharp, and it has nothing to do with it.
But while we're on the topic of Lavoisier, Lavoisier did study oxygen.
And there was an intense rivalry between Lavoisier in France, Joseph Priestley in Britain, and Scheele in Sweden.
And all three of them were working simultaneously to study combustion and understand oxygen.
And here I've got a very nice image.
This is a portrait of Lavoisier with his wife.
And it's hard to see.
If the lights were a little bit dimmer, you'd see-- notice here, you see all of this chemical apparatus.
There is bell jars, and various glass apparatus.
His wife was his partner in the laboratory.
She was unique among French women in that she read English.
And in his rivalry with Priestley, he relied on his wife in order to read Priestley's writings.
So they worked together.
She even helped him in the laboratory.
They married at the time he was 27 and she was 13, and they were-- what are you so shocked about?
It's France, and it's 1779!
Get over it!
Anyways, she helped him a lot, and we'll say a little bit more about that later.
OK.
So now let's go on to the next part of the history lesson.
So so far, we've got one explanation, and it's wrong.
So let's go to the next one, and we'll go to Britain.
Sir Humphry Davy in London in 1810.
So we'll put GB here, and he's driving around in his little Jaguar, I suspect.
And he said something that was correct.
He said that it's hydrogen present in all acids.
And that was pretty much it.
He was a great scientist, did some very good work, but really didn't do anything quantitative here.
So for the quantitative stuff, we have to wait for almost the end of the century.
And we go up to Sweden again, to Arrhenius.
And Arrhenius, in 1887-- so I'll give him a Volvo.
And he said that the acid is a substance that disassociates to produce protons.
So the acid is defined as a solution that dissociates to give protons.
I'm going to say H plus, meaning the hydrogen ion, or P plus, and we're going to keep using the term proton.
I'm not going to say hydrogen ion, I'm going to say proton.
So proton in solution.
So this is what he defines.
And he further defines the base.
He defines the base as the complement to the acid.
And he says that the base is something that dissociates to give us hydroxyl.
The base dissociates to give OH minus the hydroxyl.
And so now we've got this whole concept of electrolytic disassociation, which is what wins him the Nobel prize in 1903 for this thing.
And then we saw last day how the addition of ions to water gives charge carriers.
And so the presence of protons and hydroxyl is, in fact, the way we have any charge carried through water.
So up until now, when I said electrical conductivity, we were pretty much referring to electronic conductivity.
But there's a second kind of conductivity, and we can have ionic conductivity.
Ionic conductivity, as the name implies, is not by electrons, but by ions.
And this is typically ions in solution.
Ionic conductivity in solution, and a solution that is an ionic conductor is called an electrolyte.
So we have electrolyte in our bodies, saline, about 5% chloride.
We have electrolyte in batteries that are ionically conductive.
And the term simply means, we have conduction by ions.
And so it was the theory of electrolytic dissociation that won the Nobel prize.
So how does this work?
So we can start with something like HCl gas.
I'm going to dissolve it in water, and this will give me proton, H plus.
And I'm going to write aq, meaning that it's dissolved in water, and the chloride ion, also dissolved in water.
And so this gives me an acid.
This is a proton donor.
And now let's look at something like sodium hydroxide.
Sodium hydroxide at room temperature is an ionic compound, so it is a solid, but it is soluble in a polar, hydrogen-bonded liquid, so H20, to give hydroxyl aqueous plus sodium ion aqueous.
So again, we see the dissociation.
And then we can have, from here, a neutralization reaction.
And the neutralization reaction is simply reconstitution of the solvent.
So neutralization, another way to think about it, because we're not going to be confined to water by the end of the lecture, neutralization is simply a reaction that results in reconstitution of the solvents.
So the solvent, in this case, is water.
So how would we reconstitute water?
We combine proton with hydroxyl.
So let's do that and see the result of that reaction.
So if I take, and run it in the vertical direction, proton plus hydroxyl, we'll give water again-- let's give H20 liquid-- and now you can see Na plus Cl will give me NaCl, aqueous dissolved.
So this is what you probably learned in your high school chemistry, that acid plus base ca give you salt plus water.
So you see all of this resulting from just the simple Arrhenius definition.
So this is good.
We've got off to a decent start here with Arrhenius.
But then the theory has its limitations, as they all do.
And how do we discover the limitations?
With some new data.
So let's look at some data.
So it had been known for a long time that ammonia, when it's dissolved in water, can act as a solution capable of neutralizing an acid.
So if ammonia dissolved in water neutralizes an acid, then this must be a base.
But look, there's no hydroxyl here.
So the Arrhenius definition of base is inadequate to account for this.
So let's just get that down.
Ammonia, which you know is a gas at room temperature, will neutralize acids.
But no hydroxyl present.
So something's going on here that we can't account for by the simple Arrhenius definition.
So to get us out of this conundrum, we had to wait until 1923.
And two scientists simultaneously enunciated the same ideas.
So I'm going to put them both down.
1923.
Bronsted, who was in Denmark, and Lowry, who was working in the UK.
Also 1923.
And I'll give him a Jaguar as well.
And so the two of them proposed a broader definition to account for what's going on here.
And what did they say?
They said that acid-- they'll keep the same definition as Arrhenius.
So an acid is going to be a proton donor.
So that's good.
Let's even put here, same as Arrhenius.
But now here's the difference.
The base is no longer confined to hydroxyl chemistry.
They call it a proton acceptor.
And that's different.
So hydroxyl-free.
Now, let's be careful here.
If I propose this new definition, I can't throw out hydroxyl.
So you've got to watch to make sure that by broadening the definition, we don't exclude hydroxyl.
So the theory has to encompass what we already know, and then continue to encompass things that tend to contradict what we already know.
So let's take a look at what this gives us.
So I'm going to write a broader equation here.
I'm going to write this as the acid.
So what do I have?
I have every acid has to have a proton in it, plus some residual.
So I can rewrite the acid in this form.
I'm going to react it with a base.
So according to this definition, this has to be a proton donor.
And this, you're going to watch me on this one.
This is going to be a proton acceptor.
And we'll put some identities here.
Eventually we're going to write this with ammonia in it, but let's just do the broad definition.
So what happens is, if this is a proton acceptor, on this side of the equation, it takes the proton from here.
This is the proton donor.
It gives it away, and we end up with a BH plus.
So this proton acceptor has accepted the proton, leaving behind the deprotonated A minus.
That's a nice little reaction.
But now, if you'd nodded off for the last 90 seconds, and then opened your eyes and said, well, he said acid is a proton donor, this thing can give up a proton.
So OK, I'm going to call this thing a proton donor.
And this is most certainly a proton acceptor.
Look, it has accepted the proton.
So this is going to be a proton acceptor.
So now I've got in this equation two proton donors and two proton acceptors.
And they're linked.
This proton donor, BH plus, is linked to B.
So I'm going to put a little yolk over this one.
And this HA, the original acid, is linked to this base, A minus.
So I'm going to put a yoke over that.
And the Latin word for yoke, to yoke, is jugere, from which we get the word conjugate.
So I'm going to call these things conjugate acid-base pairs.
And every equation has two.
Because you've got a proton donor proton acceptor.
So now this is, follow the bouncing ball.
If you want to understand acid-base chemistry, just follow the proton.
Here's proton here, goes over here.
Here's something deprotonated, now it's protonated.
That's the rhythm here.
So the whole thing is, acid-base reactions can be defined as a proton transfer reactions.
Remember we talked about ionicity and electron transfer in order to achieve octet stability?
So we saw the whole life and times of electron transfer reactions.
Now life and times of proton transfer reactions.
OK. And last thing I want to do, is to just make sure that we see the definitions.
You know, all queens are female, but not all females are queens.
So this one here, what's this?
This is definitely, it's a proton donor.
So it's an acid.
So it's definitely an Arrhenius acid, because it's got a proton, and it's also a Bronsted-Lowry acid.
Now, what about this one?
This is a base.
It may not have hydroxyl in it.
May or may not.
So this is definitely a Bronsted-Lowry base, but I can't say for sure that it's an Arrhenius base, because Arrhenius has to be O minus.
This one here, what about this?
Unless it's OH minus, this one here is, what?
It's Bronsted-Lowry base.
And this one here, this can be both Bronsted-Lowry acid and it can be an Arrhenius acid.
So this is how we can differentiate them.
OK, good.
Now let's go to the ammonia, see if we've got ourselves out of the conundrum so we can write now NH3.
And I'm going to say that this is ammonia that's already dissolved.
So ammonia is already in aqueous solution, and it reacts with water, H20 liquid, to do what?
This is supposed to be a-- if this is a base, remember, this is the thing that we're trying to demonstrate.
If it's a base, according to this, it's a proton acceptor.
So I'm going to stick a proton on this, and I'm going to get NH4 plus the ammonium ion.
And how did I get the ammonium ion?
I took a proton from water, leaving behind OH.
So can you see that how this thing works?
By gobbling up protons, by becoming a proton vacuum cleaner, it takes protons out of water, leaving an excess of hydroxyl.
And in effect, now we've got something that's tantamount to an Arrhenius base.
But it all started off with this.
So now we can link these two.
I'm going to yoke ammonium and ammonia as conjugate acid-base, and there is water and hydroxyl as acid-base pairs.
So again, let's get the colors going.
Here we've got Bronsted-Lowry-- the bases are always going to be in blue.
Blue and base both begin with the letter B.
So this is definitely Brosted-Lowry base, but it's not an Arrhenius base.
This is a Bronsted-Lowry base, and it's also an Arrhenius base.
This one here, we can call this one-- in this case, it's a proton donor, so it's both an Arrhenius acid and a Bronsted-Lowry acid.
But look at this.
You'd say, OK, I know what he's doing.
Arrheius, Bronsted-Lowry, Arrhenius, Bronsted-Lowry.
This is Bronsted-Lowry only, this must be Bronsted-Lowry only.
But look!
This is a proton donor.
It dissociates to give protons.
So this is both Bronsted-Lowry and Arrhenius.
Most people will miss that.
We'll get to see, maybe, on a subsequent celebration, how many of you miss that.
OK.
So.
So now, let's go into the chemistry here.
Because I said up here at the beginning of the lecture-- what is it-- bonding is the key to understanding.
So what is it about bonding here that defines the Bronsted-Lowry base?
So I'm going to write this reaction in this way, now.
I'm going to put-- here's the proton, which is coming from water.
And I'm going to write that with NH3.
I'm going to use the Lewis structure.
So, you know, Lewis structure looks like this, Nitrate has got five valence electrons, three of them are bound, and now I've got this long pair.
So what do we know about the electronic structure of proton?
What does it look like?
What's the Lewis structure of proton?
How many electrons on proton?
None.
This is nothing.
This is very needy.
It's like the neediest friend you have.
You know, the one that's always taking stuff from you, taking your emotional energy, giving back nothing?
That's proton.
That's the human equivalent of proton.
So if proton wants to make a bond to form NH4, proton is going to come over here and exploit both electrons.
So if this is going to be a proton acceptor, the only way it can be a proton acceptor is to have two unused electrons available.
This is going to be a dative bond, agreed?
Dative, because both electrons from the bond come from the nitrogen.
The proton doesn't contribute anything.
So that's the hallmark.
The proton acceptor axiomatically must have an available non-bonding pair.
So wherever you see non-bonding pairs, wherever you see a compound that has this capability, it could serve as a Bronsted-Lowry base.
We know that water, because it has hydroxyl in it, has to satisfy this.
So let's take a look.
H20 we can do similarly.
Oxygen.
It's got 6 electrons, 2 are in the, bonds and we have 2 non-bonding pairs.
So then when we attach proton here, we make something called hydronium.
We make hydronium ion.
So it's a little fancier than just saying, I've got this naked proton with no electrons swimming around in water.
In fact, it's more coordinated like this.
H30 plus, and while we're in the neighborhood, we might as well show.
This is sp3 hybridized.
We've got one, two, three, four orbitals.
Hydrogen's on three.
It's got a net charge of plus 1, and lone pair on the fourth side.
So there's the structure of hydronium.
And we can now write the equation.
We can say H20 liquid plus H20 liquid gives me H3O plus, plus OH minus.
So this is now the self-ionization or self-dissociation reaction.
So now I can pair these as well.
I can yoke these.
I have acid-base pair.
So this is acid here, and the conjugate base is water, H20.
Hydroxyl, we know, has to be a base, and its conjugate acid is water.
So you see, in the same equation, same place, same time, water is acting as both acid and base relative to these two species.
And so we call such a such a compound that acts as both acid and base, we call it amphipathic.
Has two moods.
You know, amphi, like in amphitheater.
Have you ever seen a Roman theater?
It looks like this from the top down.
It's got all this, you know, like this, and all the people are here, and there's the stage up here.
But if I make a theater in the round, that is amphitheater.
Or as it's commonly mispronounced, ampitheater.
3091ers do not say ampitheater, they say amphitheater.
So this reaction here is self-dissociation.
Right?
This reaction is self-dissociation of water.
Or because we're forming ions, it's also called self-ionization.
Self-dissociation or self-ionization of water.
Now, turns out that the chemical's power of H3O plus and OH minus is very high.
So this reaction doesn't go very far to the right.
The amount of the dissociation is tiny, and to be specific, at 25 degrees C, the amount of, if you take deionized, deaerated water, absolutely pure and pristine.
The amount of native H30 plus and OH minus is on the order of one part in 10 million.
So H3O plus concentration would be 10 to the minus 7 molar.
And obviously, from the stoichiometry of the equation, you get the identical amount of OH minus.
Well, that's in the self-dissociation.
And so you have very, very few charge carriers, which is why high-purity water is a very poor conductor of electricity, even as an electrolyte.
And that's, I've told you in the past, that could be one indicator of certain toxins.
If you measure the conductivity of water and you discover it's abnormally high, and it's putatively supposed to be pure water, that's an indication that it's not pure water.
There's some other charge carriers present.
So we can write a Ksp we analogy.
And it's called the water ionization constant, which is the product H3O plus and OH minus.
And you can do the math.
That's 10 to the minus 14 is the product.
And you can use the common ion effect here.
If I introduce some other acid, look at how this equation helps you the same way that Ksp did.
If I have acid, in other words, protons donated from some other source, that means this number is going to be higher than 10 to the minus 7.
If this number is higher than 10 to the minus 7, and the product must be 10 to the minus 14, this must be lower.
So when I have, H3O plus goes up, then OH minus must go down, and under these circumstances, we have something that is called proton-rich.
And we call this acid, simply because the proton concentration exceeds the hydroxyl.
And likewise for the other.
And so we can make a plot of this, and that's shown here.
And all we've got is OH versus H3O plus.
And you can see that it's a right hyperbola.
But it goes back to that comment that I made in the past about how you graph data.
I don't know what to do with that, because it's curved.
I can't tell whether the data are good or bad.
So instead of looking at something like this, y versus x, I want to transform so that can get a straight line.
Because then I can make a judgment about goodness of fit and so on.
So I need some f of x versus g of y to, quote unquote, straighten this out.
And we have that, thanks to another Dane, by the name of Sorensen.
Sorensen was the one that gave us another way to think about it.
Sorensen, who in 1901, and he's got the DK on his-- he's probably got a Volvo, too.
Or maybe he's driving a Saab.
So he was a biochemist at the Carlsberg brewery.
Yes, they had biochemists, because they wanted understand the chemistry of beer production.
What a concept.
Having people that understand the technology of the business.
So he was working at the Carlsberg brewery, and he decided to take this acid-base business and turn it into something that's much more readily recognizable.
And so he defined a concept called the chemical potential, which, if you like, is the reactivity, the chemical potential of hydronium.
And he gave it the symbol.
Lowercase p for potential, and uppercase H as, obviously, for the hydrogen ion.
And he said, I'm going to make this a logarithm.
So it's logarithm, and in those days, everybody was using slide rules, so it's log base 10 of the concentration of hydronium.
And being a good engineer, he recognized since this starts off at 10 to the minus 7, the log of a number less than one is going to be negative.
And who wants to deal with negative numbers?
So you put a minus sign here.
That way, pH normally is a positive number.
And now you can see the results of Sorensen's straightening things out for us.
So now you go over a wider range, and you can see, you have nice straight line relationship, and then you can get data on there, and so on.
And this is taken from your book, and it shows the pH values of some common substances.
If you start here at neutral pH 7, you have milk, you have human blood, normal things that you would expect, fluids and so on, are hovering around 7.
If you take some coffee, you're going to go into the acidic region.
You see tomatoes down here at about 4.
Wine, you'll see in the reviews of wines, they'll say it has balanced acidity and so on.
Yeah, it's down around pH 3.
And carbonated soft drinks, some of them get down to 2.
Vinegar is wine that has spoiled.
Vin aigre, which means eager.
The Middle French eager meant impetuous, or tart, or something.
So this is spoiled wine.
And the pH changes.
So by measuring the pH of wine, you can tell if it's changing or not.
There's lemon juice.
Gastric juices in the stomach can get down to 1, pH of 1.
But you want that happening only at one point in the digestive cycle.
If you run around for long periods of time at pH 1, you will ulcerate, in other words, you'll puncture the walls.
And if that's about to happen, you need to neutralize.
So how do you neutralize?
You go up with something that has a high pH.
So you can start with something like Alka-Seltzer, sodium bicarbonate at about 8.4, or milk of magnesia, 10.5.
Why is it milk of magnesia?
From last day, because it's not a solution, it is a suspension.
The magnesia is in suspension.
The magnesium hydroxide.
And that's one of those mandatory shake well before using, because all the magnesia is on the bottom, and you've got this almost clear, colorless liquid on the top.
So you're just drinking water.
Not going to help you if you've got stomach pain, but gastric juice is down here.
If you're really desperate, don't reach for ammonia.
You have to be patient.
They have to work with milk of magnesia.
OK.
So that's the range.
So what else do we have here?
Let's take a look.
So far we've been assuming that when we add acid to the system, we get 1 to 1 dissociation.
So the next message that I want to give you here, is that not all acids are of equal strength.
See, I could look up here and say, well, maybe the reason that the lemon juice is down it at 2 and the gastric juices are at 1 is simply a concentration effect.
How much acid was introduced.
No, there's another explanation for it.
And that is that certain acids don't fully disassociate.
So for example, if we look at 1 molar HCl, hydrochloric acid, you look at 1 molar hydrochloric acid, that goes 100% into solution and gives us 1 molar H30 plus.
So 1 to 1 correspondence between how much hydrochloric acid we introduce, and how much hydroxyl that we generate.
So this is total dissociation of the HCl.
And as a result, we call this a strong acid.
In contrast, let me show you a weak acid.
So a weak acid is going to dissolve, but it doesn't dissociate.
So there's really two steps here.
First you've got to get the stuff in solution, and then once you get it in solution, it has to break apart.
It's possible to go into solution and not break apart.
In which case, you don't have the protons, but you've got the solution.
That's not an acid.
So a weak acid would be, let's, in contrast, look at a weak acid.
A weak acid would be something like acetic acid, which is CH3COOH.
And this is for historical reasons.
Normally, the proton that's going to dissociate is at the front of the formula, but this was written long ago, before people understood this theory.
And that's the way you'll see acetic acid written.
So really, this is the proton, and the CH3COO is really the acetate anion, Ac, acetate anion.
So we're going to put that into solution, like this.
I'm going to put it into water, H20 liquid.
So this is the salt out of solution, and all we're going to do is make it into the aqueous solution of same COOH, and I'm just going to write aq.
Now, if this were strong acid with impunity, I would write whatever the concentration of acetic acid is, I break it apart, and I get the full amount of the proton.
What happens is that this, now plus H20, gives me H3O plus plus the remaining acetate, which is CH3COO minus.
And now, here's where the weakness is manifested.
The degree to which the acetic acid grabs protons from here, and then ends up being protonated, is very, very small.
It turns out that in the case of 1 molar-- I'm going to write acetic acid this way, now.
HAc.
So this thing here is the CH3COO minus, all right?
It turns out that in 1 molar acetic acid, we get only 0.4% reaction.
0.4% reacts and dissociates.
So to give protons, it hangs onto most of its protons.
So we can then represent this in the form of an acid dissociation constant.
And that it looks like this.
K sub a for acid.
So on the right side, what we're going to do is make a mass balance here.
We're going to take the product of the proton, the acetate divided by the undisassociated acetic acid.
So we'll put H3O plus concentration times the acetate concentration divided by the undissociated HAc aqueous.
So I'm trying to represent what the ratio is here.
And instead of being one to one, it's 10 to the minus 5.
Very, very weak.
You get a little bit, but not much.
Now, you could say, well, 10 to the minus 5, this is, just for reference, it's 100 times 10 to the minus 7, which is the Kw.
So the point here is, if you put in acetic acid, you'll end up with 100 times the proton population that you would have had by just self-dissociation.
So it isn't acid.
It is donating protons.
But it's not doing it a lot.
So this is called a weak acid, because it's a poor proton donor.
And so we can then look at a comparison to, say, over here, HCl.
Let's go back here, and we can write the acid dissociation constant for HCl, for the strong acid.
And in that case, Ka over here, which is going to be the H3O plus Cl minus, H3O plus.
In the case of HCl, we're going to get proton and chloride over undissociated HCl dissolved in water.
And that's 10 to the plus 6, which for all intents and purposes is infinity.
You put in 1 molar HCl, you get 1 molar H plus.
And look at the ratio.
10 to the sixth, 10 to the minus 5.
There's a 10 to the eleventh relationship between the weak acid and the strong acid.
And here's the cartoon that shows this strong acid, put in this amount, 100% dissociation, weak acid gives you small amounts of proton, and a very weak acid is imperceptible amount of disassociation.
And here's a table that quantifies it.
So the strong acids up here with the Ka.
This value of Ka has to be greater than 1, because it's saying that we're getting a lot of dissociation, the reactions moving to the right.
And you have these numbers here.
10 to the ninth, 10 to the eighth, 10 to the sixth.
Those are differences without distinction.
if I say the ratio is a million to one, or a billion to one, for all intents and purposes, you have dissociation.
And then here are the week acids.
Look.
10 to the minus 3.
There's phosphoric, hydrofluoric, and so on, all the way down.
There's carbonic acid, 10 to the minus 7.
So there's the list.
And just another cute demonstration of the relationship between bonding and acidity, the acid strength increases from left to right, and the bond strength increases from right to left.
You know, HF is very strong.
It's a hydrogen-bonded.
Very tight HF bond.
But what determines if it's going to be a strong acid or not?
It's how willing HF is to give up its H. But its H is tightly bound.
So oddly enough, in HF, you don't give up much of the H. HF is over here.
It's less than one.
It's a weak acid.
HCl is a pretty strong acid.
If you want to go heavy-duty, HBr and HI are even stronger acids, because they have a weaker hold on the hydrogen within them, so they are much, much more generous proton donors.
So you can see that from there.
OK.
So what's the takeaway message here?
The takeaway message is that equal acid concentration does not mean equal acid strength.
You have to be mediated by the Ka.
Equal acid concentration, which means, how much did you dissolve?
Does not equal acid strength.
And what's the reason?
Because equal acid concentration, this is a function of solubility, whereas this is a function of dissociation.
So many dissolve, few dissociate.
How about that?
There's a nice tagline you can use.
Use that at a party this weekend.
Many dissolve, but few dissociate.
All right.
Now I want to go very, very high.
Big concept, all right?
Last definition of acid-base comes from the United States.
G. N. Lewis.
Same G. N. Lewis who gave us the Lewis cross and dot structures, covalent bonding, didn't stop inventing.
So we're going to put U.S.A., with a Chevrolet out in California.
And what did G. N. Lewis tell us?
He said, I want to extend the Bronsted-Lowry concept.
But you know, let me give you an analogy.
Have you ever stood on a bridge and watched the traffic go?
You see it starts from a stop sign or a traffic light.
Light turns green, the first car pulls away, the second car pulls away, the third car pulls away.
The other way you can look at it is, when the first car pulls away, there's a car vacancy.
And instead of watching the cars go from right to left, watch the car vacancy move from left to right.
And the two are linked.
And point of fact, if you're the fourth car from the traffic light, light turns green and you hit the horn, why?
You can't drive.
You have to drive only into a car vacancy, and it takes time for the car vacancy to get to you.
When the rate of vacancy flux is not matched by the rate of car flux, then we have a collision, and then we exchange papers.
So it's a mass transport problem.
So now what G. N. Lewis said was, instead of looking at this reaction as a proton transfer reaction, and looking at proton attachment, or proton acquisition, he said, why don't we look at this from the other side of the relationship?
And so he said, let's look at something like, for example, we've got NH3 here.
So let's put the N with the 2 electrons.
And we've been talking about this from the perspective of the bond being formed here through the acquisition of the proton.
Lewis said, I can view this from the other perspective, and say, I'm going to view it from the perspective of the nitrogen.
From the perspective of the nitrogen, this reaction represents the donation of the electron pair.
A very high concept.
So this base is not about proton attachment.
It's about donating an electronic pair.
So let's get that down, because that's really good.
So I'm going to say something that's a base looks like this.
This is what a base is.
It's a lone pair.
This is electron pair, very high level base.
And what's the proton give us?
What we're doing, is we're talking about electrons and donation.
So what's this other thing in that same category?
An electron pair can match up with an empty orbital.
That's the only place electron pairs can go.
They go to the empty orbitals.
So electron pair is a base, and this is just a circle around nothing.
OK?
This is nothing.
But it's a special kind of nothing.
This is called a vacant orbital.
And I'm going to call a vacant orbital an acid, in the most general sense.
So now I'm going to take an electron pair plus vacant orbital, and what did they react to give me?
Covalent bond.
You'd expect that from G. N. Lewis, because G. N. Lewis enunciated covalency.
That's why you'd expect it from G. N. Lewis.
So this is, vacant orbital plus electron pair gives covalent bond.
Very high level.
Very high level.
So what is a base?
A base is now not just a proton acceptor.
It's an electron pair donor, and the acid is an electron pair acceptor.
So now there's no chemistry here.
No chemical identities.
Which means, I can even go from-- I don't even have to be in solution.
I don't have to be in a liquid.
This could be a gas and a solid.
Could be a gas plus a gas.
So this is, like, Darwinian.
You know, we've come out of the primordial ooze, and now we can fly, because we can talk about any chemical reaction we want.
Any state of matter.
It's fantastic.
Absolutely fantastic.
So here, let me do one last color drawing.
We want to do this one here.
This is Lewis and the biggest concept.
So we're going to go like this.
Base like this.
Base with its lone pair hanging out plus acid gives us-- three colors.
This is great.
We've got all this colored chalk, it's Friday, what could be better.
All right.
So now we're going to end up with A, B, there it is.
This is the Lewis concept.
Lewis acid-base concept.
In the broadest thing.
Now I'm going to show you an example of how you can use this.
I'm not going to stop here.
OK, let's go.
All right.
So we're going to talk about acid rain from burning of coal.
You know, about half of the electric power in this country comes from burning coal, and coal contains about 1% sulfur.
You can purify it of sulfur, but it takes money to do so.
A ton of coal will give you about 25 million British thermal units, and that's the number in SI units and joules.
So 3 tons of coal will give you 1 megawatt per day.
By the way, it takes about one gram of uranium to do the same thing.
So someday, when you're sitting there as a policymaker, you've got a choice of 3 tons of coal or one gram of uranium, keep this fact in mind.
A 10 megawatt plant burns 30 tons of coal per day, containing a third of a ton of sulphur, which makes 2/3 of a ton of SO2.
And SO2 is a precursor to acid rain.
Now we can reduce that SO2 emissions by reacting SO2 with a lime, CaO, according to this reaction, to make calcium sulfite.
And now I'm going to bring in this.
Lewis.
Oh, here's from your textbook, there's the scrubbers with the calcium oxide, and water missed, and so on, and eventually they trap the SO2.
They do nothing to the CO2.
So all the CO2 is going up.
So here's calcium oxide, which is a solid.
It's lime.
You know, stuff you sprinkle on football fields.
And this is SO2, and it's got this structure with the resident bond double single.
The oxygen here acts the way it did in the glasses, as a modifier.
It goes in and breaks this double bond, and now we have three single bonds.
So we can look at this from Lewis acid-base.
You know, calcium oxides, electron pair donor.
If it's an electron pair donor, it's a base.
So it's a base.
And you know SO2 better be an acid, or your theory is nuts, because SO2 is a precursor to acid rain.
So it has to be an acid.
It's an electronic pair acceptor, which is what we said.
It's an electron pair acceptor.
So this is a Lewis acid-base reaction that is a gas-solid reaction.
So we started with aqueous solutions, and we end up generalizing all of these types of reactions.
So that's pretty good.
Here's a plot of acid concentration up in the sky, here.
You're Here.
Yeah.
Last thing I'll do, I talked a lot about Lavoisier and Scheele and Priestley.
This is a play, if you've got the little bit of time this weekend.
You might try reading it.
This is Carl Djerassi and Roald Hoffman, both Nobel Prize winners in Chemistry.
Not the same year.
They won independent Nobel Prizes, and actually speak to each other, and they collaborated on this play.
Here's the setup, the premise.
The premise of the play is that the Nobel committee decides to give Nobel Prizes retrograde.
Before 1901.
You see, you can't get a Nobel Prize posthumously.
You've got to do something great, and you've got to keep on living until you get the prize.
So they decided they're going to go backwards, you know, maybe give prizes to people like, I don't know, Maxwell or Faraday or whatever.
And they decide, it should be simple in the old days.
Because we didn't have all this dog-eat-dog competition in science, and big grants, and corporate interests.
Should be simple.
Until they hit the Nobel Prize for the discovery of oxygen.
And then they hit the wall with these three different competitors.
And so the story goes that the three competitors and their wives end up in Stockholm, and the interactions the take place.
And it's really fascinating, because you have Lavoisier, who was the political conservative but the chemical radical, Priestley, who was the chemical radical but the political conservative-- by the way, Lavoisier.
You know how he died?
He was guillotined during the French revolution.
And people think it's because he was a tax collector.
Not true.
He was a tax collector.
The real reason is Marat, who was one of the chief justices of the Jacobin terror, was also a scientist.
And around 1740, Marat had submitted an article for publication, and it was rejected.
And Lavoisier was on the editorial board.
And so sometimes when I get an article to review, and I have to give it a negative review, I think about this story.
And sometimes, I will just say, you know, I'm really busy teaching 3091 and I won't get to this on a timely basis.
OK, people.
Have a nice weekend.
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OK.
Couple of announcements.
First of all, the Wolf Lecture is tomorrow, right here in 10-250 at 4:00.
As also, there will be the weekly quiz, the mini celebration.
But there's going to be, a week from today, celebration part trois.
Yes, third celebration of learning.
And remember, what we're shooting for is the smiley face.
None of this.
And those of you who are over here on the grade point distribution curve, we want to move you to the right.
So I'll say a little bit more later in the week about what the coverage will be.
I'll be available for office hours later, and I thought to try to cheer you up, and remind you that studying is not something you do only the night before the exam.
You have to be studying on a regular basis.
All right?
So from the iconic film Ghostbusters, there's this one scene.
[FILM PLAYBACK] For a moment, pretend that I don't know anything about metallurgy, engineering, or physics, and just tell me what the hell is going on.
You never studied.
[END FILM PLAYBACK] You don't want to be a character in that scene.
So what are you going to do about it?
You're going to study.
All right.
Let's get down to work.
Today I'm going to take one day and I'm going to talk about organic chemistry.
Now, I'm not going to pretend that after one lecture, you're going to walk out of here and know much organic chemistry.
I'm going to give you, if you want to know organic chemistry, you're going to have to take 512, and many if you will.
But I'm going to give you enough that it will set up for the subsequent units on polymers and on biochemistry.
So we'll know enough that we can move forward.
So let's get into the definition of organic chemistry.
It's the chemistry of compounds, containing both carbon and hydrogen.
There can be, of course, other elements present, but you've got to have carbon and hydrogen to get yourself into Orgo.
And what makes carbon and hydrogen so special?
We've been studying these elements through the semester.
But just to be reminded.
First of all, hydrogen is peculiar because it's got the lowest atomic number.
It's got the lone proton, and a single electron.
And when hydrogen ionizes, it forms the hydrogen ion, or the proton.
That's all that's left.
And this thing's very tiny, and it can form covalent bonds.
It forms covalent bonds as primary bonds, and it can form hydrogen bonds as secondary bonds.
So it's peculiar for that reason.
It does not form ions.
You cannot imagine something is tiny as a proton occupying a lattice site in an ionic crystal.
OK. Now when it comes to carbon, carbon is also peculiar.
It's smack dab in the middle of the periodic table, in the second row.
Four valence electrons, which means that it's very difficult to ionize.
If it wants to achieve octet stability, it's going to have to form covalent compounds.
It's not going to acquire four electrons or lose four electronics.
In rare exceptional circumstances it, might, but commonly, it does not.
And with that intermediate average valence electron energy, it's going to be capable of all sorts of bonding.
And in particular, because of its small size, it's capable, with its small size and the intermediate value of average valence electron energy-- and this is critical-- is that it's capable of forming multiple bonds.
And this is key, as we'll see later when things try to polymerize and form multiple bonds.
And this is not common, for example, not the case if you look underneath carbon.
Silicon, germanium, for example.
Same thing.
Group 4, same kind of mid-stream average valence electron energy.
But they're too big.
Silicon and Germanium do not form double and triple bonds.
So the other thing is that it's capable of self-linking.
It can link to itself, form carbon chains, and also with nitrogen, oxygen, sulfur.
So this gives it-- and let's put up phosphorus as well.
Phosphorus and sulfur.
So with these kinds of links-- we've already seen oxygen links.
We'll see, pretty soon, nitrogen links and so on.
There's millions of organic compounds, and today I'm just going to look at a few.
I'm just going to give you some taxonomy and nomenclature, and we're going to walk through this.
And here's the way to keep it straight.
You know all this stuff.
I'm just going to organize it in your minds for you.
We've already studied the three different types of hybridization.
sp, sp2, and sp3.
And we've already seen that if you're going to have a double bond, you need sp2 hybridization to reserve 1p orbital, to make that pi second bond.
And if you want to make triple bonds, you make sp hybridization, thereby reserving two p orbitals to make two pi bonds.
So this is sigma, sigma plus pi, sigma plus pi plus pi.
That's it.
Now what we're going to do is to is go back and look at this in the context of hydrocarbons.
So hydrocarbons is going to be now the focus here, and the hydrocarbons are the simplest of the organic compounds.
So let's look at hydrocarbons.
I'm going to look at their nomenclature and their properties.
But the structure behind everything, the treatment, is the type of hybridization.
So these consist only of hydrogen and carbon.
Compounds containing only hydrogen and carbon.
So I'm just going to go down that chart.
And we'll go through.
So the first one on the left is the alkanes We'll look at the alkanes.
And the alkanes are characterized by sp3 hybridization.
And so this gives the maximum number of carbon linkages.
Why?
Because you can only form single bonds.
So if I can only form single bonds, every bond goes to a different atom.
Whereas if I form double bonds, then I'm burning up some of the capability.
Because if I used two bonds to get to one neighboring element, instead of two single bonds each to two different neighboring elements, I don't have as many linkages.
So the maximum number of carbon linkages-- and because they have the maximum number, we're going to borrow terminology from the last unit on solutions.
What's the maximum amount of solute you can get into a solution called?
It's called the saturation value.
And they use that term here.
These are called saturated hydrocarbons, because they've got the maximum number of linkages, bonds.
And they're all sigma bonds.
All of this is a consequence of the choice of sp3.
I could have led you through all of this, but we're putting it up quickly here.
All right.
And so they have a common formula.
CNH2N plus 2.
I think I've got some-- yeah, this is taking from one of the other books.
I know you've got a list like this in your own book.
So here they are.
So this is just CNH2N plus 2, and you go down, and there's the nomenclature.
The meth, for historical reasons represents n equals 1.
So here's what the hidden message is.
You name them by the carbon number plus the suffix A N E. So the -anes, the alk-anes, are all these sp3.
So 1 is meth, 2 is eth, 3 is pro, and 4 is but.
And they all have historical derivations.
But- comes from butyric acid, which is the acid that's formed when butter turns rancid, and so on.
After you get up to N equals 4, it's just your Latin.
So you just take your high school Latin, pentane, hexane, heptane, and so on.
And if you didn't take high school Latin, this is a great chance to learn the ordinals in Latin.
So there they are.
Now, I'm not going to expect you to-- if I say decane, you're supposed to slam down C10H22.
But I would expect that I could say decane, and I'd give you that formula and you'd be comfortable with it.
What's the other thing to note here, while we've got this up on the chart.
You can see that all of these are symmetric molecules, and they're non-polar.
They're perfectly symmetric non-polar.
And what happens as the molecular mass increases, the melting point increases, the boiling point increases, and the state of matter transitions from gas at room temperature to liquid at room temperature to solid at room temperature.
So here's an excellent example of polarizability in action.
Because the only way one methane can bond to another methane is by induced dipole-induced dipole interaction.
So the only difference between methane and icosane is that icosane is a honking big molecule, non-polar, but very polarizable.
And so as the polarizability increases, you can see the effect here.
OK, so that's good.
So you get to learn your Latin ordinals.
Now these are also termed straight chains.
They can be termed straight chains, but I want you to look carefully at what's going on here.
So here's a diagram used, the first one on the left, you've seen methane many times.
There's ethane and then propane.
And what do you notice here?
Because of the sp3 hybridization, all of these angles are 109 degrees, including the carbon-carbon angles.
And when the thing is really short, you get this zig-zaggy nature.
So because of the sp3 hybridization, you see the carbon doing this.
This is carbon-carbon, and then the rest, forget it.
I mean, this is the backbone.
But if I get up to icosane and I have about 20 of these things, forget the zig-zag.
From here, yeah, I know there's a little bit of fine structure, but for all intents and purposes, this as a chain.
Straight chain.
And that's where you get the terminology from.
But keep in mind that there is this zig-zagging back and forth, due to the sp3 hybridization.
OK?
So that's the first point.
The second point is that the bonds between the carbons are not fully specified.
So for example, take the carbon here on the left.
The sticks to go to the 3 hydrogens here depicted in white, and then the fourth stick goes to the carbon.
And then off of that carbon comes 3 more sticks, and all of these are 109 degree angles.
All of that specified, and nothing more.
There's nothing saying that these 2 hydrogens in the back have to line up.
These two hydrogens out front have to the line up.
That's all free to rotate.
And so we have that degree of freedom there.
The result is that you can have something that zig-zags back and forth.
So for example, this is called eclipse, because if I were to stand here, these two hydrogens are on a direct line.
So the front hydrogen covers the back hydrogen.
It eclipses.
This hydrogen out front to your and my right is on the same line is this hydrogen to your and my left.
And so if I were to look here from the end, this hydrogen eclipses, and so on.
So if I get something eclipsed, I'm going to have a perfect straight line.
If I have something that's staggered, it's free to zig-zag, but it's free to zig-zag and meander, and this is what happens.
So here's 2 C17H36 molecules, and the top confirmation is more or less straight.
These are more or less eclipsed configurations.
Whereas the lower one, you can see, there's a fair bit of stagger, and as a result, the backbone can actually loop back on itself.
But in both cases, we still call these straight chains.
They're still straight chain.
Because as you start at one end and go from carbon to carbon, you linearly go to the other end.
It doesn't mean it's straight like straight as an arrow.
It just means there's no branches.
So now I'm going to distinguish this from the other one by branching.
So let's just get this up here.
So we have straight chains, is one case.
Straight chains.
And the straight chains can either be staggered or eclipsed.
And the closer you get to eclipsed, the closer you get to-- the more eclipsed, the straighter.
I think you'll understand what this means.
Staggered allows it to make large loops-- I'll just put here, straight but looped.
You can use that to describe some of your friends, too.
All right.
Now-- it's a Monday, it's a joke!
Wake up!
Wake up!
All right.
Now we can look at another one.
Branched chains.
And now, this will be a major departure, whether the chain is super-straight or does a lot of looping back and forth, the branched chain is in marked contrast.
Now let me give you an example of the branch chain.
Best to just do it.
So here's the branch chain.
I'm going to use a butane.
So you know B U T is 4.
So that means 4 carbons, and A means 4 carbons, sp3 hybridized.
So that's it.
So now I'm going to write this.
Here's butane.
C, C, C, C. And I'm going to write this as just a straight chain, instead of doing this.
I'm going to contrast.
I'm going to do 1, 2, 3, 4.
1, 2, 3, 4.
1, 2, 3, 4.
1, 2, 3, 4.
OK?
So there we are.
So this is, instead of writing all this stuff, this is butane, and this is butane.
But now here's the thing.
Four stick rule.
Whenever you're doing Orgo, four stick rule.
That means, four sticks out of every carbon.
So here's the carbon-carbon bond, so I need three more sticks.
1, 2, 3.
And I know these are sp3 hybridized, but just, in compressed notation, you can do this.
And I'm not even going to write all the hydrogens.
Here I'm going to write the hydrogens because it's our first time through, and it's Monday, and everybody needs a little bit of TLC.
But now the TLC is gone.
This is it, all right?
I'm not even going to put the H's on the end.
The stick coming off of carbon means there's a hydrogen here.
All right?
So how about this one?
1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4.
And now let's count the hydrogens.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
C4H10.
Butane.
All right.
Now there's another way to write this.
I could go like so.
1, 2, 3.
And I could put a carbon up here.
This is a branch, you see?
It's not in a straight line.
And I'm not talking about eclipsed.
Because if I start at 1n and go down the backbone, I get to the other end, I only hit three.
I never got to this one.
But let's see.
You know, maybe this is different.
So four stick rule.
1, 2, 3, 4.
1, 2, 3, 4.
1, 2, 3, 4.
1, 2, 3, 4.
Now let's count the hydrogens.
1, 2, 3, 4, 4, 6, 7, 8, 9, 10.
This is C4H10.
It's a different molecule, you can see.
With what you've already learned in 3091, this has a different dipole moment, doesn't it?
It's got a different dipole moment.
It's going to have a different polarizability.
But it's a butane, it's a C4H10.
So this is called butane, right?
And this one, over here, historically, was called isobutane.
Because it's an isomer.
It's got the same chemical composition, but it's got a different makeup.
It's got a different structural makeup.
It's got a different-- I'm going to use the material science-y word.
Different structure.
It's got a different molecular structure.
But the chemists going back 50 years ago didn't use these terms.
They called it different constituents.
So I'm going to use that as well.
Different molecular structure or different constituents.
And what do we mean by constituents?
Now comes the color chalk.
So we're going to do, is I want to circle the makeup of this.
So at this end, I've got-- this is a CH3 here, correct?
There's a CH3 over here.
And then in between, I've got CH2 and a CH2.
Now let's go over here and diagram this one.
The isobutane, I've got a CH3 at the end, and a CH3 at the opposite end.
So far, same as the butane.
But now look.
There's a third CH3 here, and then there's this thing, which is a CH.
You could argue it's a CH2 that got methylated, but, you know.
So I'm going to put a CH2 here that's been methylated.
And I'll tell you what that means.
OK?
So it's really a CH2 group that got methylated.
So it's called isobutane, but you can see, it's got different-- all I'm doing here is making the justification for the different constitution, different dielectric constants, and so on and so forth.
So this is where the chemists must come in.
They would say, different constituents.
And so these are isomers that have the same chemical composition, and are different on the basis of their constitution.
So they call these constitutional isomers.
This is a constitutional isomer.
Isobutane is a constitutional isomer.
Now the last piece is nomenclature.
They don't use this today.
Because isobutane was a term that came out of chemical industry.
Because it looks like butane, has got the same molecular weight, but it's very different.
Today we use the IUPAC notation.
Remember this?
International Union of Pure and Applied Chemistry.
They're the ones, when one of you discovers and stabilizes element 113, And you want to name it after dadadadada, you have to get past the IUPAC committee.
OK?
So their nomenclature committee would say, this is an alkane, but it's only three carbon units long.
So it is not a butane.
It is a propane.
It is a type of propane.
But it's not simple propane, because there's a methyl hanging off the side.
So in point of fact, this, in IUPAC nomenclature, is a methyl propane.
And furthermore, if you want to get really pedantic, and IUPAC loves to get pedantic, is what they'll do, is they'll number these, starting from left to right.
So this is the number 1 carbon, this is the number 2 carbon, and this is number 3 carbon.
And the methyl group is pendent on the number 2 carbon.
So if we want to get really, really, really pedantic, and who wouldn't, on a Monday morning, what we would do, is we would call this 2-methylpropane.
So it's a propane, 3 carbon units, there's a methyl group hanging off of number 2.
Now, I don't expect you to be able to do these on sight, but I would like you to be appreciative of what this is.
So I'll give it to you on an exam.
All right.
So we've got the isomers.
And I think I've got some-- I think I've got this.
No.
OK, this is good.
Oh.
There's one other thing I want to talk about with alkanes.
There's another thing that we need to know, and that's about the radicals.
The radicals that form.
The radical is a species with one or more unpaired electrons.
Right?
This is broken bond.
One pair electrons.
That's equivalent to broken bond.
OK.
So this is highly reactive.
Because this unpaired electron is going to go look for a mate.
So if you take a look at CH4-- now I'm going to put the hydrogens in.
So this is methane, CH4.
I'm going to break this bond in the lower right corner, and just have an unpaired electron.
And this is called the methyl radical.
Very reactive.
So you can see that this methyl radical could go up and stick on to the fourth strut of that number 2 carbon, and thereby form the carbon-carbon bond.
So that's why they call this methylpropane.
So the methyl radical.
And you can write it this way, H3C with dot indicating unpaired electrons.
But I want you to be literate with the way chemists will write this in various compact notations.
So you might even see it written this way.
Now, that doesn't mean that the unpaired electron is on a hydrogen.
This is just a way of saying, you know, CH4 is methane, CH3 is methyl, and who cares.
There's only one possible meaning to this, so it's all good.
All right?
Let's do two unpaired electrons.
So I'm going to take two unpaired electrons.
Again, this is out of the alkane, so I'll put a hydrogen above and below.
So that would be like this one here.
You see the CH2?
So it's got carbon-carbon bonds on either side, but in point of fact, this is part of a zig-zag.
So I'm going to break bonds here and here.
All right?
So now I've got the possibility of putting a bond on either side.
And this one here is called methylene.
This is methylene.
This is the methylene radical, and it's got two unpaired electrons.
So we could write that as H2C with 2 unpaired electrons.
And, you know, these are electrons.
They're not just anywhere.
They're sitting in an orbital, and that orbital is as at 109 degrees from the other bond.
So it has to be sitting there in reserve, in waiting.
So if I bring another carbon here, I get the carbon-carbon bond, and I start forming the zig-zag.
And there's one more, a common one that we'll need to know.
And that's from ethane, which is C2H6, right?
Let's start with ethane, C2H6, and I'm going to make a radical out of that.
It will be C2H5, like this.
To which I can add a hydroxyl, and now I'll make ethyl alcohol, and then it's party time.
But so this is how you start.
All right?
This is the ethyl radical.
Starting from methane.
And that's pretty much it.
That's about all I want you to really take away with the alkanes.
C4H10.
Yeah, there you go.
So there you can see, to the left, the zig-zagging of the linear and the butane, and then the methyl propane.
That's good.
OK. Now let's go to sp2.
Now we're going to look at the alkenes.
So alkenes represent-- the alkenes-- this is sp2 hybridization.
And they contain-- this means there's the possibility of a carbon-carbon double bond.
And they're also known as unsaturated.
Because they don't optimize.
The fact that you spent two of your bonds going between two atoms instead of taking one bond to one carbon and another bond somewhere else, means that you're undermaximized.
So unsaturated hydrocarbons.
OK?
And the formula for these is CNH2N, where obviously N is greater than 1.
We can't form a carbon-carton double bond with only one carbon.
So you know, the prototypical one we looked at was this one here, just to C2.
That's the first one.
So it's going to be C2H4.
So carbon-carbon double bond.
There's a sigma bond here, and a pi bond.
And they lie in a plane.
The molecule is planar.
120 degrees here.
So we'll put the hydrogens on it-- you know already, I don't have to put the hydrogens there.
That's good enough.
All right.
So what's this one called?
This is called ethylene, historically.
Ethylene.
But the IUPAC notation is similar to this one.
It's carbon number plus -ene, E N E. So strictly speaking, this is what everybody knows.
But IUPAC insists-- and I don't know.
As you publish, some editors are really fastidious, and they'll force you to change if you use this kind of terminology.
So you're going to have to call it ethene.
Or in the UK they call it ethene.
But everybody knows it is ethylene.
OK.
So there it is.
And you only need 1 carbon-carbon bond.
So you can have long chains where the position isn't fixed.
So let's look at a couple of examples.
So let's look at butene.
OK, let me just write that down.
That's good.
The position of carbon-carbon is not fixed.
It just has to be somewhere in the molecule, which consists only of hydrogen and carbon So here's a couple of examples.
I'll start with this one.
This is going to be-- so this is 1, 2, 3, 4, so that's a but- something.
And it's got a double bond, so that's a butene.
It's all hydrocarbon.
So that's butene here.
But I could also-- well, let's finish this.
How many?
This is 1, 2, 3, 4.
See, I even tried to indicate the 120 degrees here.
This is going to be 1, 2, 3, 4.
1, 2, 3, 4, 1, 2, 3, 4.
All right.
So there's the butene.
But there's another place I could put the carbon-carbon double bond.
I could do it this way.
Make a carbon-carbon single bond, move the carbon-carbon double bond over to the number 2 position.
this is 1, 2, 3, 4.
So this is going to be 1-butene, and this one here is going to be 1, 2.
So we'll make this 2-butene.
And let's keep counting here.
1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4.
And I've proven to you, I hope, by now that I can count to four.
So this is 2-butene, right?
2-butene.
And if you go through this using the same kind of diagrammatic analysis as over here, you're going to come to the conclusion that even though these both have the same composition, which is going to be C4H8, they're both C4H8, but as you diagram them, you'll find that they have different constituents.
So these, in fact, are constitutional isomers of one another.
Constitutional isomers.
Yeah, that's pretty good.
There's even more.
It gets even better.
I want to focus on the bottom one, the 2-butene, and I want to write it out with a little bit more detail.
So let's start with that carbon-carbon double bond.
And we know the carbon-carbon double bond requires planar structures at 120 degrees from one another.
So let's look at that one.
And what we've got here, is on the carbon, I've got on the one side, I've got a hydrogen strut, right?
And on the other side, I've got a methyl.
I've put a methyl here, like that.
OK?
Or actually, let's get-- OK, I'll do that.
I was-- what I wanted to do, to get really-- OK, watch this.
All right.
So what I'm going to do on this side, this side, same thing.
The carbon has a hydrogen strut down here, and on the other side, you've got the methyl.
OK?
So that looks good.
But there's another way to do this, and still keep the carbon-carbon bond in the center.
So I'm going to do that over here.
So 1, 2, 1, 2.
And in this instance, I'll put the same thing I have on the left side, H3C.
So so far, so good.
I've got the same molecule.
But now what I'm going to do, just for grins and chuckles, I'm going to put the hydrogen in the upper position, and the methyl in the lower position.
Now, these are not constitutional isomers.
They have the same constitution, right?
Carbon-carbon double bond, two methyl groups, two hydrogens.
Carbon-carbon double bond, two methyl groups, two hydrogens.
So they're constitutionally identical, yet these have very different properties.
Can you see?
I mean, the dipole moment is going to be different.
Look.
Here's the charge, all up on one side, and here the charge is more symmetrically straddled.
Which means, they're going to have different melting points.
They're going to have different boiling points.
And they have the same chemical formula, and they have the same chemical constitution.
So they're different kinds.
There's some kind of isomers-- what's the difference here?
It's not the constitution, it's the way things are arranged in space!
This is great material science.
Spatial orientation is everything.
And what do you call it when you're listening to some loudspeakers and a really good audio system, and you can close your eyes, and the violins are over here, and the cello is here, and you can hear the percussion way in the back.
What is that called?
Stereophonic sound.
Stereo phonic, meaning you can render spacial resolution.
So these are called stereoisomers.
One's a stereoisomer of the other.
So they're spacially distinguishable.
So they have-- for example, I went and looked this up.
Oh.
Let's give some names to this.
In this case, all the methyls are on the same side, and the hydrogens are on the same side.
So I can a dividing plane here, if you like, parallel to the floor.
And I'll just cut right through the center of the carbons, parallel to the floor.
And so in one instance, all the methyls are on the same side, and here the methyls are on opposite sides of the double bond.
So when they're on the same side, this is called cis-butene.
And let's be super pedantic.
It was a 2-butene, so now it's a cis-2-butene.
This is terrific.
And this is a 2-butene, but it's on opposite side.
So this is called a trans-2-butene.
Cis-2-butene and trans-2-butene.
And I looked up-- so let's see, which one do you think has tighter, which one's going to be more tightly packed?
Which one's going to have the greater density?
Which one's going to pack better?
Cis or trans?
The density here is 0.627, and the density here is 0.61.
OK?
Now, which one's going to have the higher boiling point?
Which one's going to have the higher boiling point?
Gee, that should be a no-brainer.
Once you know which one's got the higher density.
Why does it have the higher density?
Because it's more nearest neighbors is more tightly packed.
So I would bet on the one with the higher density.
And this one boils at, boiling point is 3.7 degrees Celsius, and this one, the boiling point is-- hang on, I've got this backwards.
This is 3.7.
Boiling point is 1 degree Celsius.
So those are the data as they come up.
So stereoisomer, spacially distinguishable.
By the way, we can form multiple bonds.
We can form multiple double bonds.
So they're -enes as well.
So if we have two double bonds in the molecule, so we'll call that a diene.
And we can have-- you can keep going with the Latin, but common ones you might find is the triene.
And this will come back later when we look at polymers.
So if we have two bonds, we will have-- You know, some of that hard luggage that you're seeing that's coming back, they got the roller boards, but instead of the fabric, now they're coming back with the really hard polymer?
That's ABS, and the b is butadiene.
So let's look at a-- here's a pentadiene.
I'll give you a pentadiene.
So that's going to be 1, 2, 3, 4, 5.
It's five carbons, for starters.
And I'm going to put the double bonds on the 1 and the 3 carbon.
It'll be a penta-, it's five carbons.
And it's an -ene, because there's going to be some double bonds here.
And it's a di-, because I'm going to put double bonds at the 1 and the 3.
So 1 gets a double bond, and 3 gets a double bond.
So this is a pentadiene.
OK?
And then the last thing is-- pardon me.
Last thing is radicals.
So here's the common radical.
We'll start with ethylene, and I'm going to make an ethylene radical here.
I'll break off the hydrogen at this point, and I can write that as CH2CH.
And we can't call this ethyl radical, because ethyl was already used for C2H5.
So we need to indicate that there's a double bond, and so this radical is known as vinyl.
This is the vinyl radical.
And so you could react this with chlorine, for example, in which case you'd make vinyl chloride, and then you'd polymerize that, and you make PVC.
And we're going to see that in the next coming days.
Or you could you can put alcohol on there.
I've already told you that.
I don't need to remind you twice about that, I bet.
OK.
So now-- oh, there's one other one.
There's one other one that will come up if you take a lot of biochemistry, and that's the one that comes off of propene.
So if you start with propene, propene's going to look like this.
1, 2, 3.
So there's propene.
3.
There's a saying double bond here, and I'm going to put 1, 2, 3 in here.
1, 2, 3, 1, 2, 3.
1, 2, 3, 4.
OK.
So this is the radical that comes from propene.
So this one is derived from ethylene, and this one here is derived from propylene.
So again, as in this case, we couldn't call this ethyl.
We can't call this propyl, because that's the one that you get from the alkane.
So instead, this one is called allyl.
A L L Y L. And if you take 7012 at some point, Professor Weinberg will refer to these pendant groups.
He likes to pronounce this all-YL.
You'll hear him talk about, and the all-YLS are over here and here.
That's what we're talking about.
OK, good.
So this is C3H7 dot.
Good.
All right.
Moving right along.
Now I want to turn to the alkynes.
That's the sp hybridization.
sp.
And that means there's going to be at least one carbon-carbon triple bond.
One carbon-carbon triple bond.
So the formula there is CNH2N minus 2, and obviously N has to be greater than 1, otherwise you can't form a bond.
And there's very little that you need to remember about this one.
The dominant one that you'll come upon is C2H2.
So according IUPAC nomenclature, the two carbons means it has to be F, and these are all alkynes, so this should be ethyne.
And if you ask for ethyne, nobody knows what you're talking about, although it's very correct.
This is known as acetylene, which is a fuel gas for welding, cutting, torches and so on.
You've got two lines, you've got a big tank of acetylene, you've got a big tank of oxygen.
And this is used because there is enormous energy stored in this carbon-carbon double bond.
And so when you combust this with the stoichiometric amount of oxygen, you can cut through steel, you can use this to make very elaborate glassware with pure silica, making fused quartz glassware.
Very, very high energy contained in here.
And just to give one more example, let's look at something that's got four carbons in it.
So it's going to have four carbons and a triple bond.
So if it's 4, it's going to have the B U T but-, and it's going to need an -yne.
So this will be a butyne.
And depending on where we want to put the triple bond, I don't know, if you want to put a triple bond here, there's 1, 2, 3, 4.
So this we could call 2-butyne, and it's C4H6, If you go through the stoichiometry.
Let's try it, just for practice.
So this carbon on the end is 1, 2, 3, 4.
Look at the second carbon.
It's got one strut bonding to the neighboring carbon, and three struts bonding to the neighboring carbon on the right.
It's saturated now.
Nothing comes of without carbon.
Look at this one.
1, 2, 3, 4.
Nothing comes off of this one.
And then finally 1, 2, 3.
So there's three hydrogens here, there's three hydrogens here, there's your sixth hydrogen.
C4H6.
OK. That's good.
All right.
There's one other set that I want to look at, and these are called the arynes.
These are aromatic hydrocarbons.
Some people refer to them as arynes.
And the most notable one is C6H6, which is benzine.
And benzine's molecular weight was determined long ago, in the early 1800s.
And it was Kekule who, in 1865, dared to suggest that the chemical structure of benzine was a ring structure.
So Kekule suggested a carbon ring.
Up until this time, people didn't have the common understanding that carbons could link to themselves, let alone form a ring.
And so now, if we go through this, we will put, from each of these junctions there's the carbon, so we have to have four struts.
We need hydrogen, so let's put hydrogens 1, 2, 3, 4, 5, 6, So there's the 6 hydrogens.
And now, each carbon needs 4 struts.
So 1, 2, 3, and I'll put a double bond.
That will make 4.
1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, and now this one here.
So you have an alternating structure of double bond, single bond, double bond, single bond on the carbons.
OK?
So that's the way things stood until the twentieth century, when on the basis of new data, it was discovered that all carbon-carbon bonds-- and I don't want to make this a single bond, I'll just say all carbon-carbon bonds in benzine, C6H6, were measured to be the same length.
Well, that's a problem, isn't it?
That's a problem, because we would expect that, you know, we know that carbon-carbon single bond is about 1.47 angstroms, and the carbon-carbon double bond has to be shorter, because it's double-bond, it's tighter, so it pulls things in.
So that's 1.33 angstroms.
And the measurement here was found to be 1.39 angstroms.
And 1.39 angstroms is midway between 1.47 and 1.33.
So people were frustrated with this new information.
And it was Linus Pauling who made sense of this for us.
Pauling, who in 1931 came along and said, what we have a mix?
What if we have a mix of two structures?
And he said, suppose you had something that looked like so.
1, 1, 1.
So here's the set of double bonds off of the ring.
And what if there is a resonance with an opposite hybrid, where this double bond, instead of being here, goes over to here?
So now we have this?
And he said, if the system resonated between the structure on the left and the structure on the right, the time average value of the carbon-carbon bond length would be around 1.39 angstroms.
So this is a resonant hybrid structure, to account for the mix of single and double bonds, which gives rise to the current representation of benzine as simply this.
So we're not saying this is single or double.
It's a resonant between one and two.
OK.
The other thing is, if all of these bonds are the same length, and that means that between the 12:00 position here and the 2:00 position, there is pi bonding, and then along this double bond, there's pi bonding, and then along this double bond, there's pi bonding.
But if all of the bonds are the same length, I can't say that the pi bonds are confined to where the double bonds are indicated, because resonance indicates that it could be between adjacents.
So that means that the electrons, in fact, can go all around.
So resonance, then, means that the pi electrons are delocalized.
And now here's some better drawings.
It's hard to draw that.
So here they are.
OK.
So this is a nicer drawing of benzine.
So you can see, when all of these bonds are resonant, it doesn't matter whether I choose this front one as the double bond, or the one next to it.
They're all the same length, and so the electrons can go all around.
And you can imagine, if I did this with graphite, and I kept going, well, if the electron can go here, then it means it can go here, which means it can go here, which means it can go everywhere.
Which is why graphite is an electronic conductor, because of delocalized pi bonds.
OK.
This is 1-3-butadiene.
Double bond, single bond, double bond.
Again, all the same length, and you can hybridize, and-- This is how you start making electronically conducted polymers.
You just go double bond, single bond.
Double bond, single bond.
Smear.
That's it.
OK. Let's jump over that.
Let's talk about Kekule So he studied architecture, switched to chemistry.
And then he got a job in London, and he used to fall asleep on the bus to his apartment.
And one day he was sleeping on the bus.
He woke up, and he'd been dreaming about carbon forming chains in 1855.
And he proposed carbon chains.
And then he got a job as a professor in Belgium, at the University of Ghent.
And one night he fell asleep at the fireplace, and he dreamt of a benzine molecule as a snake biting its tail while spinning.
And that's where he got the idea of the ring molecule.
All he knew was that this thing had a formula, C6H6.
And for this, he's dubbed the founder of structural chemistry.
So what's this formula for success?
Well, he moved into chemistry from another field.
But to dream, you've got to get some sleep.
And I'm probably talking to one of the most sleep-deprived populations on this campus.
So I urge you to get some sleep, and then maybe you'll be able to dream.
Hey, don't make noise.
We've got two minutes left.
nobody's going anywhere.
All right.
So the next thing is, I want to talk a little bit about octane ratings and automobiles, and how we go from straight chains to chains of different lengths.
If you're looking at combustion, the idea is, you admit gasoline, which vaporizes, and then the piston comes up, and under high compression, the spark plug fires.
When the spark plug fires, this causes an explosion which then pushes the piston down, and then you've got the camshaft that takes the vertical motion and converts it to rotary motion.
Now, if you're running an automobile after a certain period of time, the combustion chamber becomes so hot that just admitting the fuel, having it vaporized, before the piston has risen fully to get maximum compression, this thing could just explode.
And when it explodes, it'll send the piston down, and it's out of sequence with the other pistons.
And that's when you get the knocking.
Sometimes you're driving up hill, you hear this [GRINDING SOUND].
That's the knocking going on.
You need to get a tune-up, or switch to higher octane gasoline.
So what's going on in here?
What you're doing, is you're changing the chemical composition, the constitutional isomerization.
And you do that in the process by which you synthesize the gasoline in the first place.
So the figure of merit is called octane number, which was first instituted here in the United States.
And they started with 2-2-4-trimethyl-pentane.
It's an octane, but it's an octane that's 2-2-4.
Three methyl branches.
So that means there's only five left.
So it's a pentane with three methyl branches.
And that's called 100.
And then is 0 is heptane.
If you put heptane into the combustion chamber, you'll knock yourself silly.
And the engine will just shake and shake and shake.
So gasolines have to be somewhere on that interval.
So there's the-- you know, if you tested and you had something that was trimetylpentane, 90%, versus 10% heptane, you'd call it a 90% octane.
And you can have octane numbers higher than 100, by the way.
It just depends on what the repression is.
And oddly enough, when you when you increase the octane, you make the fuel more difficult to burn.
Because you want it to burn only on demand.
Fuel that just burns whenever it wants leads to random events in the engine, and that's not good, if you want to get good thrust.
So oddly enough, high octane gasoline requires more ignition than low octane.
And the additives are tetraethyl lead or ethyl alcohol.
And from that, you know now, we get E10, which is gasohol, 10% alcohol.
E85 is 85% ethyl alcohol.
And all of this you get by synthesis.
You know, playing with the catalysts, temperatures, and so on, to get the right mix to get good fuel, and get good emissions.
And you can decide amongst yourselves whether going to E85 high levels of ethanol, ethanol derived from corn, is that smart, is that food for fuel?
You know, what's the sensible use of agriculture, algae, switchgrass?
All of that stuff.
I mean, it's all loaded into here.
All loaded into here.
So yeah.
With that, I think we'll adjourn.
I'll see you on Wednesday.
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We had a staff meeting yesterday, and we decided that what we're going to do is, we'll start with defects and solids.
We didn't examine you on it with the last celebration.
So we'll pick it up with defects, and we'll go through to the end of solutions, acid-bases.
So the stuff that we covered on Monday, the nomenclature of organics, we're going to leave out.
Because you don't have enough time with the recitations to really dig into that, and I think it's-- with the test being on a Monday this time, and not on a Wednesday, the schedule is little bit of out of balance from the way I'd like it to be.
But I thought having the third celebration the Wednesday before Thanksgiving was not wise.
I know that a number of you are going to be in transit that on Wednesday.
But let me remind you that Wednesday, we will have a full-blown lecture.
I'm going without any slow down.
So next Wednesday, we will start biochemistry.
So those of you who are not here, please make sure that you have viewed the lecture.
Because when you return on Monday, you will have no idea what's going on.
And it'd be a good idea not to blow biochemistry, because on the final exam-- if you fail the final, it's not good.
It's really not good.
So just to be clear, we're going to start with defects and solids, and go through, and end with acids, bases, and solution chemistry.
OK.
So today I want to start to get some mileage out of what we learned on Monday.
I want to talk today about one of the applications of organic chemistry, and that's polymers.
So we're going to take two lectures on polymers, and then we'll feed into biochemistry, and you'll see a very natural progression from what we're doing now into the biochemistry.
So let's begin with a little bit of reflection on what we've done so far.
Up until now, we've looked at various solid structures.
We've started with single atoms.
We looked at crystals, disordered crystals, and so on.
We've looked at compounds, and we've even looked at small change.
We looked at alkanes, straight chain alkanes, branched alkanes.
And we've looked at network solids.
Here is a regular network.
This is the structure of graphite, for example.
Nice, ordered crystalline structure.
Diamond grows in three dimensions without abatement.
We've looked at disordered networks, such as silicate glasses.
So that's what we've looked at up until now.
What I want to talk about today is polymers.
Polymers are macromolecules.
These are long chain molecules.
By long, we're talking about thousands and thousands of repeat units.
And this is the distinguishing feature between macromolecules that are found in the body, versus macromolecules that are man-made.
Mother Nature is a polymer engineer.
We are a macromolecular organism.
This is all polymer.
It's all polymer.
The changes in shape is all elasticity in a polymer.
And inside, of course, I've got a ceramic skeleton, which keeps the frame in place.
But this is all polymers.
But the difference here is, this is not a repeat unit.
Whereas when we have a man-made structure, it's the same repeating chemical structure.
And the other thing I want to say at the beginning is their importance in commerce.
Polymers are found everywhere today.
Trash bags, auto parts.
And in culture, they're absolutely essential to culture.
Let's think about it.
When you play that DVD, that DVD is information embedded in a polymer.
The magnetic drags.
What are magnetic drags?
These are polymers that have been coated with gammaferic oxide or some other magnetic material.
Without polymers we couldn't have the modern era.
And if you go back before the DVD, with the CD, before that was magnetic tape and cassette, and before that was this.
You might see some of these around.
This is an old way of presenting musical information.
This is the phonograph record, photograph recording.
And this is information that is scribed into a platter, and it's made of polyvinyl chloride.
That's why you hear the term vinyl.
Why is Virgin, the big Virgin enterprise, called Virgin?
Because Virgin Records, when they made their platters, they started with virgin vinyl, not recycled polymer.
And so their platters would lie flat on the turntable.
Whereas some of them you buy, you put them down, and they warp.
It's OK, because the needle would still track, but it looked kind of-- playing like this.
And it's too bad that we've lost this.
Because this gives rise to the real estate possibility for artwork, and liner notes, and so on, that you don't get when you download something for $0.99.
OK?
So an electromechanical device moves along here, and jiggles in the grooves, and those mechanical wigglings get converted into energy that ultimately gets fed through the loudspeakers.
And now let's talk about visual information.
Daguerre invented halide photography in the early 1800s, but we didn't get motion pictures.
Cinematography had to wait for the advent of polymers, because only when we could make material that would come out miles long on which we could coat with halide photosensitive emulsion, could we imagine images flashing past us faster than the persistence of vision.
You need-- what's the persistence of vision?
Around a twentieth of a second.
Motion pictures are 24 frames a second.
So you couldn't have glass plates on which you've got halide film zipping past at 24 frames a second for 90 minutes.
So only with the advent of polymers did we have the birth of cinematography.
So what I'm talking about here changed the world.
It changed the world, and gives birth to the fantastic access that we have to information.
And now, of course, as things move digitally, and they dematerialize, you're going to see a shift away.
But all this stuff was enabled by the advent of polymer chemistry.
So with that as a motivator, let's dive in.
This is good stuff.
It's really, really good.
So now the payback comes.
You know, all those little lessons about, where's the electron, where's the orbital, you know, who wants to learn that stuff.
You have to learn it!
If you're going to write the great novel, you've got to learn how to spell.
All right.
So let's go.
So the polymer, it comes from the Greek poly, meaning may, right?
Poly is many And the mer, the mer is this repeat unit, if you like.
Many units, many units.
So let's take a simple example.
Last day we studied polyethylene.
So polyethylene looks like this.
Double bond, one, two, three, four, four hydrogens.
So this is just ethylene.
And what I can do with ethylene, is I can react it.
Ethylene's a gas, room temperature.
Put it into a reactor and expose it to an initiator.
And the initiator is a radical.
You know what the radical.
It's something that's got this unpaired electron.
It's very active.
And the radical sees a unit of ethylene, which I'm going to write like this.
We don't need to put the 120 degrees now.
And what the ethylene does, is this radical attacks this double bond and figures, well, it's better to have two single bonds than one double bond.
Reacts the double bond, breaks it, and then forms the following.
The radical now bonds to the ethylene, breaks that double bond, and transfers the unpaired electron to the end.
But now this thing is itself a radical, and it's swimming in a gas of ethylene.
So now a second C2H4 comes up against this, and it's turned into the following, are now C C C C C, one, two, three, four, one, two, three, four.
And so on.
And this can go without limit, until we shut down the reactor, or we introduce something that terminates, that will cap this.
All right?
So this is the beginning of it.
And what you see is attachment, attachment, attachment, breaking of double bonds.
And so we have a repeat unit here.
This repeat unit, the mer unit, is this ethylene unit, and it can go to a value of n that is very, very high.
What kind of numbers can we expect to get?
We can have numbers here where n takes values-- and these aren't hard and fast, but just to give you some idea-- you could have a, you know the oxymoron jumbo shrimp.
You could have a short chain polymer.
A short chain, long chain.
You know?
Boy, you're really dead today.
What's the matter with you?
Sleep deprived?
You've-- All right.
So n can go up to about 10,000.
Now what's the atomic mass of this, just as representative?
This is 2 times 12 is 24, 25, or 26, 27, 28.
28's roughly 30, so 30 times 10,000, you've got 300,000.
That's 300,000 grams per mole.
This means you can have molecular weights on the order of a million grams per mole for one molecule.
And by the way, the polymer people, they come from a different branch of science.
So they don't use this grams per mole.
Instead of grams per mole, they like to use the word, the dalton, as a unit of measure.
So you might see in the pollymer literature, you'll see the polymer with so many kDa, kilodaltons, thousands of grams per mole.
That's the nomenclature you will see.
OK. Now, what are the properties of these things going to be?
Oh wait, I'm going to show you something.
To give you a sense.
You need to be awakened.
So let me give you a sense of just how long these molecules are.
So I'm going to give you a little mechanical model here.
I'm going to put up here so we get some-- So what I've got here is just a pull chain.
If you go in the basement, you might see these pull chains.
They're just made of brass, little brass balls, and then they've got links between them.
And when you cut them-- you buy them by the foot at the hardware store.
You put it on a light, you can turn the light on and off by toggling up and down.
So I'm going to say, let's say this distance here represents the distance between mer units.
This distance could be between 2 mer units.
And I figured, let's see, I've got 40 feet here, and I did the math.
So this thing here, with the 40 foot pull chain, has a polymerization index-- let's call this the polymerization index.
So my polymerization index, I calculated it for this thing.
It's on the order of about 3,000.
So that's pretty good.
It's right in the middle of this.
It's actually-- this qualifies as a bona fide polymer.
So let's take a look at what happens here.
This is one molecule.
So what do you know about interactions between molecules?
Now, in the liquid state, what's this going to be?
Is it going to be fluid, or is it going to be viscous?
It's going to be viscous.
Look, this is one molecule, and it entangles with itself.
Now, to make a liquid, I have thousands of these things swimming around, above their crystallization-- oh, crystallization temperature.
Freudian slip.
Above their solidification temperature.
When they solidify, are they going to form an ordered solid?
One of these at each lattice site?
Probably not.
They're probably going to form disordered solids.
So most of the polymers that we encounter are probably going to be disordered.
How is this held together when it forms a solid?
It's polyethylene.
Well, what's your menu?
Is it ionic bonding?
Is it metallic bonding?
Is it hydrogen bonding?
How is it held together?
Weak van der Waals.
Weak van der Waals bonds.
But look at the surface area.
So this is what it looks like.
See, and it entangles, is even entangles with my clothing.
It just grabs onto everything.
All right.
Actually, why don't we-- David, let's cut to the screen.
So here we are.
This is it.
This is the polymer.
And every once in a while, a polymer will do something like this.
It'll say, you know, I still, even though I'm in this big long chain molecule I still want to try to order.
Because I read in 3091 that ordering lowers the free energy of the system...
So what it does, it starts doing this.
And can you see when it starts doing this, that it starts to take on-- if you walked into the room just now, and this were blowing up really high, you might see just this little snippet and say, wow.
This thing is starting to look like a cubic array isn't it?
This is a really good model.
It's a really good model, because this is what really happens.
And so that decreases the energy of the system.
The bonds are greater there.
So what's that going to do to the mechanical properties?
Going to strengthen it, yeah.
It's going to make things stiffer.
Instead of being this soft, squishy polymer, it's going to have some stiffness to it.
And you've come up across that stuff.
The most notable one being the CD case.
The people that make those, I'd like to bring them here and sit them down and teach them some polymer science.
Those things are so brittle.
They crack, right?
There's only two classes of jewel cases.
Those that are cracked, and those that will crack.
Because these people don't know what they're doing.
OK.
So now let's have some time here to codify what we've just see.
So first of all, we suspect that they're going to be solid at room temperature.
Dominantly.
We can engineer them to be liquid, but dominantly solid at room temperature.
And van der Waals bonds abundant.
OK. And the liquid, as a liquid, we expect viscous liquid.
Good.
Got that down.
What else to have to know?
Let's do calculation here.
So I did one here where I said, what's this distance?
We saw last day that this distance, carbon-carbon, remember, we're looking at single bond, double bond.
A single bond carbon-carbon is about 1.5 angstroms.
So this is 1.5 angstroms.
The distance between successive mer units is about 3 angstroms.
And so if you take something 3 angstroms, and you make it this number of mer units, I came up with a polymerization index of 3571 when this thing weighs 10,010 kilodaltons.
And so then that means that you'd end up with a length, molecular length is on the order of 10,714 angstroms, which is on the order of 1 micrometer, which then makes it greater than the wavelength of visible light.
So this is really a different type of matter.
And we can go there.
We've talked about viscous, so I'm going to remind you of two observations here.
Viscous liquids, amorphous solids.
If you put those two ideas together, what else can we call upon?
We can go back to this.
Yeah.
Remember this?
What's this?
This is super-cool liquid, super-cool viscous liquid.
This is the solidification temperature, which in the case of liquid to amorphous solid is the glass transition temperature.
And here we have the excess volume, right?
This is the volume, and then there's some crystalline volume, and so on.
Right?
I can cool at a second rate, and I get this.
So this is slow cool, this is fast cool.
We have a Tg up here.
Tgf I'll call Tg fast, and Tgs, Tg slow.
So that'll give me a different volume.
So this is volume fast and volume slow.
All right?
And since I know that density is equal to mass over volume, so therefore v fast is greater than-- I observe v fast is greater than v slow, so therefore density of the fast cool must be less than the density of the slow cool.
Now I'm going to put some names on here.
I'm going to call this the cooling curves for polyethylene.
These are the cooling curves for polyethylene.
So this, down here, gives me-- this is now high-density polyethylene, and up here is low-density polyethylene.
Same thing, just different processing.
So faster cooling quenches in more free volume.
And since volume is a measure of disorder, right?
What's the difference between great disorder and not-so-great disorder?
David, again cut to the projector, please?
Forgive me, the document camera?
So the difference between high disorder and low disorder is this.
This is the order, here.
So what that tells me is that in high-density polyethylene, there is a greater percentage of zones like this.
And we just reasoned that this is going to give stiffness and so on, and sure enough, for the low-density polyethylene, low-density polyethylene is used in things like food wrap, things like stretch and seal, where you can pull because you can move those macromolecules relative to one another without fracturing the material to stretch it over.
Whereas the high-density polyethylene is used in such things as milk jugs, where you want a little bit of stiffness.
Polyethylene milk jugs are still kind of floppy, but there's a little bit of stiffness to it.
The other thing is, because the low-density polyethylene is dominantly amorphous, with almost none of this second zone here, it's transparent to visible light.
But now, let's think about what's going on here.
Where I've got this chain, and then all of a sudden I get to this.
All right?
So this is clear and colorless.
Right?
It's a high band gap material.
There's no free electrons.
It's going to be transparent to invisible light.
So this is transparent to visible light all along.
But can you see that because I have this zone of ordering, the density of matter here is different, and so therefore, when a photon comes in, the index of refraction-- now, you know, this is one of those days where n is going to come up.
So here n is the polymerization index.
Here it's index of refraction.
So maybe we'll use a different color.
How about that.
All right?
So the green n is index of refraction.
So the index of refraction in the amorphous zone, is different from the index of refraction in the partially crystalline zone.
So index of refraction varies from zone to zone.
Even though each zone is clear and colorless, can you see that this boundary between the ordered and in the disordered zone acts like an interface?
And what happens when you have an interface with a different index of refraction?
It scatters light.
And as a result, the milk jugs, they appear white.
You can't see through them, even though they're constituted of the same continuous material.
But there's these density fluctuations, because this is a higher density than this.
So you can rationalize all of this stuff.
OK.
So this is what we would call partial crystallization in this zone.
And that gives rise to the changes.
Now, how would we distinguish these?
What technique would I use to see if I've got order in this polymer, that I don't know anything about?
How do I interrogate atomic order?
What technique would I use?
X-ray defraction.
Thank you.
Back to the slides, please, David.
I'll find you the piece.
This is called polyethylene, but remember last day, the IUPAC notation for this is ethyne, so this becomes polythene.
And in the UK, it's known as polythene, and there's a Beatles song, Polythene Pam.
The only reason-- I put up here for two reasons.
One is, it has something to do with polymers.
At least they knew something about polymers.
It's apparently, they must have been in a drug haze at this period in their careers.
This is absolutely terrible.
Some of your parents probably adore the Beatles.
If you want to provoke a conversation at Thanksgiving, pull out this one and ask them to talk about the lyrics here.
But this is just garbage.
This is garbage!
This should be trash bag, is what it should be.
But evidently, this woman used to show up at parties dressed only in a polythene bag, a see-through polythene bag, I might add.
So this is a paean to her.
Anyways, it's bad.
Yeah, yeah, yeah.
Anyway.
So here we are.
This indicates chrystalline polyethylene.
Here you can see the C2H4 unit.
And it's attempting to try to occupy, as these beads line up here, they're trying to set up a faux lattice.
And this is what you might see.
And towards the end of the lecture, I'll show you a transmission electron micrograph where you can see this by dying it with different colors.
OK.
So now here's some x-ray defraction.
So this is crystalline, where they zoomed in on one of these zones, and they've just gotten the defraction pattern from that zone.
And you see Bragg peaks.
And then this is in the amorphous region.
it's not featureless.
There's one broad peak.
Why one broad peak?
Because even though there's no long-range order, you have short-range order.
You know that you've got a carbon on either side, you've got hydrogens and so on.
So that short range order gives you the broad peak.
But look at C here.
Isn't this interesting?
That if you have a polymer that has both some order and a lot of disorder, if you take the x-ray defraction pattern more broadly, more globally, you get the additive spectrum.
So you can see that this is amorphous, but there are some zones of crystallinity.
And that's the additivity power of x-ray defraction that allows us to interrogate.
So now let's talk in a little more fine structure about molecular architecture.
So tailoring molecular architecture.
And why are we doing this?
Because we want to engineer these materials for desirable properties.
Like CD cases that break after about one or two uses, OK?
Of polymers.
Actually, there's probably a business opportunity there.
You start a company where you make jewel cases that actually work.
People might be willing to pay, you know, a penny more for something that works.
All right.
So how do we change this?
What I'm going to show is a whole bunch of cartoons of architecture.
But how do we control this?
It's processing.
It's processing.
Processing and in the synthesis algorithm.
I'm assuming I already have polyethylene, so how am I going to tailor it?
It's in the processing.
And so what I can do, is I've got a number of levers I can move.
One is the composition of the polymer, and the second one is, I can use catalysis.
I already hinted at that last day in talking about gasoline.
How do you start with petroleum and get octane and not heptane and so on?
By playing with cataylsis, we can direct certain forms.
You can actually preferably synthesize a certain architecture by using catalysts.
So let's look at these variables.
First one is composition, obviously.
If you change the composition of something, you can very much expect to change its properties.
So what do we have here?
We can-- here's a simple example I can start here with.
This is ethylene.
Or I can put a chlorine here, so this now becomes chloride, right?
This will become, what's the radical, is vinyl, so this is vinyl chloride.
So if I polymerize this, this becomes polyethylene.
If I polymerize this, this becomes polyvinyl chloride, PVC.
Which is why you heard the fellows from Blue Man Group at the beginning banging on PVC tubing to make their music.
All right.
So I use a pure, in other words, only one mer.
It's pure polyethylene.
This it's called a homopolymer.
Only one mer type in use.
If I want to make the polymer analogy of an alloy-- in ally metals, alloy has more than one metal mixed in, or a metal can even have nonmetals mixed in.
Then the polymer term is called copolymer.
I'll call this a copolymer.
And a copolymer has greater than one mer type.
So for example here, this could just be polyethylene.
Whereas here, an example would be polyethylene, and then the notation is hyphen lowercase c hyphen, which is an indication of the fact that you're making a copolymer.
And this copolymer has mer units of ethylene, and mer units of vinyl chloride.
So this is a copolymer of polyethylene and polyvinyl chloride.
And then we can start looking at the various ways of arranging these.
So we can start with, for example, we can have a sequence of random mer types.
So as you're going down the chain, you get either an ethylene, or a vinyl chloride.
So this is called a random copolymer, and it's designated polyethylene-r-polyvinyl chloride.
And actually, I think I've got a cartoon showing this.
Yeah.
So A and B are different mer types.
And so a random copolymer just has A, B, B, A, A, A, B, A, whatever.
And this is controlled in the synthesis process.
So I can take the same mix of A and B, in other words, the same two mer types, but mix them in an alternating sequence.
So they're very regular.
It's 1 mer of ethylene, and then one mer of vinyl chloride, alternating all the way down the backbone.
So this gives you a regular copolymer, and it's designated A for alternating, because we've already used R for random.
So instead of a sequence of random, we have a sequence of alternating mer types.
So that would be, in this case, polyethylene alternating PVC.
So that's two of them.
And then you see the block copolymer shown.
And in that case, the mers grouped into-- they call them blocks, even though it's a line.
You know, maybe it's like walking down the street.
I've walked down one block.
So this is one block, and then there's the next block.
They're grouped into blocks.
So in that case, we have the block copolymer, lowercase b, and then PVC.
And what you see there is a run.
A run of-- So all of the A's are grouped, and then we have all of the B's.
And this is actually an artist's misconception.
In point of fact, most block copolymers that you find, this is the stuff they use for the soles of your sneakers and so on.
They're block copolymers.
They're usually just a diblock.
There's usually just two different mer types.
One half of the macromolecule is one mer type, the other half of the macromolecule is the other mer type.
Sometimes you might find a triblock, where you might find block A, block B, and then another block A.
But this is a pentablock and he's even got ellipses here as though this thing keeps going.
There's no commercial product that looks like that.
It's usually a diblock, occasionally a triblock.
OK?
And then the last thing you can do, as is shown up here, is what is known as the graft.
And in the graft, the thing that distinguishes the graft is mers grouped into blocks.
In this case, a long side chain of other mer.
So that's different from side group that we saw last day.
These are long macromolecular chains.
So you see here the primary backbone is A, and then you have this very, very long macromolecular chain of B.
And this is called the graft copolymer.
So that would be polyethylene, could be the backbone, and then long side chains of polyvinyl chloride.
And all of these have their different architectures.
And I think I have an example here of one.
This one, I think I mentioned last day, when we were looking at the butadiene.
So this is hard-sided luggage.
And if you go to grandma's house, and she hasn't gotten into the digital age, and isn't an old hipster with a cell phone, she's still got the old hard plastic telephone, it's made of this stuff.
ABS, which is acrylonitrile butadiene styrene.
And the backbone is butadiene, and then you have-- OK, so here's the backbone.
It's long butadiene, which we saw last day.
And then you've got side chains of two types.
And this is just indicating A, A, but this is a macromolecule.
This might be 3,000 units long, whereas this is 10,000 units long.
And this might be 2,000 units long.
So you have side chains of acrylonitrile and side chains of polystyrene.
And that's the hard plastic.
It's coming back, because people want luggage that doesn't destroy their contents anymore.
So that's coming back.
OK.
So this is what we can do in terms of composition.
Let's look at, also, the side group arrangement.
And this is called tacticity.
Tacticity, which is the equivalent of what we saw last day as stereo isomerism.
Remember, I showed you the cis and trans on the butadiene.
On the cis, you've got, in one case, you've got whatever the functional group is.
I'm going to put A, A, B, B.
And then so this is the cis version, or I can do a trans version, which is instead, I'll put in A up here and an A down here.
I'll put a B up here and a B down here.
So this is a trans version.
So imagine the analogy for polymers.
So let's look at that.
It's easier to see it, and then we can just document it.
So here's three different polymers, all right?
So I'm going to start with the lowest one.
This is called isotactic, because this is vinylchloride.
There's vinyl chloride by itself.
There's the vinyl radical from ethylene.
And then we tack on chlorine.
And now when we polymerize, we're going to break this double bond in order to link this carbon to the neighbor.
So now the backbone only has single bonds, right?
You start with the double bonded precursor, and now you've got this chain of single bonds.
Right?
But look at the chlorine.
The chlorine could either be below, or it could be above.
Well, in this case, the chlorine is always below the chain.
So this is called isotactic polyvinyl chloride.
Now this one, you know, they're trying to mix two things at once.
I would have shown this with vinyl chloride in all three, but OK.
So imagine now, in this case it's syndiotactic.
This happens to be polystyrene, because this is vinyl benzine, or what's the other way to call it, phenyl ethylene.
You can have either way.
All right?
So you can say you're putting a phenyl group onto vinyl, or you're putting onto the ethylene, or vice versa.
But the radical, this thing here is called styrene, and we're going to break that double bond and away we go.
So this is the benzine ring, and it alternates from above the chain to below the chain.
Above the chain to below the chain.
So in this, this is called syndiotactic.
And the top one is a polypropylene, so you start with propylene, that has the double bond here.
It's got three carbons and the methyl group coming out, and so you break that double bond and make the chain.
And it seems to be on a random basis where that methyl group appears.
Sometimes above the chain, sometimes below the chain.
So those are three different ways of arranging, so three different tacticities.
So we've got isotactic, we've got syndiotactic, and we've got atactic.
So the way to remember them, isotactic obviously means, everything's on the same side.
And when I'm talking about the same side, what is it?
Same side of the backbone.
And we're talking about the position of functional groups, right?
That's what this is all about.
The positioning of functional groups.
In other words, a methyl, or a benzine, or what have you.
So same side of backbone.
Atactic is random, and then by elimination, this one must mean alternating, above and below.
On opposite sides of the chain.
And these have different propensities for crystallization.
Which of these three-- imagine that all three of them were polyvinyl chloride.
Which of the three, atactic, syndiotactic, isotactic, would be most favorable from the standpoint of partial crystallization, to loop back and forth?
The one that's most regular, so the isotactic one has.
OK. Third one.
Third one is the backbone configuration.
This is the configuration of the main chain, all right?
And this is called conformality.
You see, they have different words for everything that came from a different heritage.
So to me, it answers the question, how distended is the chain?
And we saw this last day, when we looked at what happens when we start with something like this, where we have staggered or eclipsed, remember, orientation of the hydrogens here.
In the case of staggered, we get the lower.
Both of these are straight chains, because if you start at one end, you move monotonically all the way down the chain to the other end.
There's no branching here.
But one of them, the lower one, has staggering of that carbon-carbon bond, with the result, things form this giant loop.
So in polymers, imagine this, instead of being 36 units long, imagine it being 3,600 units long.
So now you can have things to do this.
You can even take a little break here and crystallize and so on.
So how do I distinguish that from something that does this?
So how distended is the chain?
This one here is as distended as it can be, and others can be much more coiled.
So this one is called linear chain.
This is branch chain.
Now, this is not graft.
Don't confuse this with the graft copolymer.
Graft copolymer means, I have one mer down the backbone, and a second mer off to the side.
That's a graft copolymer.
This is a homopolymer.
A homopolymer that has different branches.
So all of these, let's put this here to remind us.
These are homopolymer cartoons.
Homopolymers that haven't changed anything.
So this is something that actually has different branches.
So this is branched chain, branch chain architecture.
Good.
So which one of these is going to be harder to crystallize.
Which one is-- well, obviously, I've given it away, I put that little piece of crystallinity in there.
Can you see that when these two solidify, that the one on the right, because it's got these branched chains with the covalent bonds sticking out.
It's not going to pack as well.
And if it doesn't pack as well, it's going to have a higher free volume.
If it's got a higher free volume, it's got a higher degree of disorder.
So the branched chains have a greater propensity for disorder.
So let's put that down.
Branched chain, harder to crystallize.
And crystallize, it's not a giant crystal or polycrystal.
We're talking about the degree to which we can get that amount of ordering.
And then the last one I want to show you is, here's three different chains.
1, 2, and 3.
And what we're going to do is we're going to cross-link.
We're going to form bridges.
These are covalent bridges.
These are not hydrogen bonds.
These are not weak van der Walls bonds.
These are strong covalent bonds, all the way along here, linking backbone to backbone.
So this is cross-linked.
And what do you think the mechanical properties of this are going to be?
If I want extend this, like stretch and seal, I can pull chain number three relative to chain number one quite easily, until this covalent bond has bent over as far as it can, and then what happens?
I can't pull it anymore.
And what happens when I let it go?
It springs back.
So this imparts elasticity.
This makes the polymer rubbery.
Cross-linked polymers are rubbery.
But you know, the polymer people, they want elevated words.
If you say to a polymer person, oh, so you've made a rubber, they cringe.
They want to have a fancy word.
They call this an elastimer.
It's got the word mer in it, so they're happy, and it's elastic, so it's an elastimer.
Rubbery.
So how are we going to make these cross-links?
What are we going to have to look for as an architectural feature to make cross-links?
We look at the backbone.
If we've got a backbone going like this, all these carbon-carbon bonds, and down here I've got carbon-carbon bonds.
If I break one of these bonds in order to go up in this direction, I've broken the chain.
So how am I going to have the bonding capability to form a covalent bridge between chains without breaking the very chain I'm trying to link?
What feature am I going to have to have in the chain?
I need, after polymerization, to still have some double bonds.
And if I've got still double bonds in the backbone after polymerization-- and now would I do that?
I have to have a double bond to break in the first place.
So what would happen if I started with a unit that had two double bonds in it?
That way, I give up one double bond in order to make the chain, and I still have a second double bond.
That's why the rubber has butadiene.
The diene has two double bonds, after polymerization, still has one double bond, and now what I can do is break this double bond and this double bond, and link them.
But if I link them, they're going to be too close together.
They're going to be scrunched in.
I want to have some play, here.
I want to make this rubbery.
So what do I do?
I put a spacer in here.
What do I use as a spacer?
An atom.
And what kind of an atom do I need?
I need an atom that's capable of making one, two covalent bonds.
What am I going to choose?
What atom do you know likes to make linkages?
How did we make silicate?
What's the linker in silicate?
Oxygen.
You could use oxygen, but I want to make this thing even farther apart, and a little bit, you know, oxygen's small and it's got too much strength.
If I want to make it weaker, but something that behaved like oxygen, sulfur, I'd go down one row in the periodic table.
So I'd put a sulfur in here.
And if I wanted to do a really good job, I'll put a sulfur here, since I got a double bond from both sides, I'll put two sulfurs, and make a disulfide linkage, and now I've got the making of rubber.
Disulfide linkage.
And again, the feature, I have to start with the backbone that has a covalent bond at the end.
OK. Well, let me tell you a little bit about the birth of rubber.
It started here in Massachusetts.
It started just across the river, in Roxbury.
Nathaniel Hayward discovered that rubber treated with sulfur was not sticky.
If you take natural rubber-- you ever work with natural rubber?
It's very sticky.
You can't do anything with it.
And Hayward reasoned that by playing with the sulfur, he could-- he didn't understand the molecular architecture.
But he learned that by playing with sulfur, and introducing sulfur to the rubber, he lost the stickiness and he got an enhanced elasticity.
So Charles Goodyear came to Boston from Ohio to meet with Hayward, and he learned about disulfide linkage and so on.
Took the idea back to Ohio.
And one day in the laboratory, he was trying to prepare a batch of sulfonated rubber.
And it was a laboratory accident.
He knocked over the vessel containing this sulfonated rubber, and it landed on a hot stove.
Things were powered by fire.
Landed on a hot stove, and then by heating it, he gave birth to the process of vulcanization.
And hence was born the American rubber tire industry, by that accident, starting with the trip to Roxbury to learn about this.
And you could have done it all with what you learned today.
You're just about 150 years too late.
OK.
I will see you on Friday.
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OK.
Perfect timing.
Let's turn to the announcements.
You know the big announcement is Monday, Celebration Part three.
I think Hilary sent all of you the coverage.
Just to be clear, start with defects, amorphous solids, kinetics, diffusions, solutions, acids, and bases.
No orgo, no polymers.
We'll catch the orgo and the polymers on the celebration of celebrations, the final exam.
OK, so let's get moving.
Last day we started talking about polymers.
These are macromolecules.
Here we've got an example of a simple compound with a double bond, vinyl chloride.
And by the action of addition polymerization, we can turn this into a macromolecule where you see a repeat unit.
The double bond has been broken to allow but carbon to bond to another vinyl chloride, and so on.
These are very, very large molecules, can be on the order of microns in length, mass measured in kilodaltons, or thousands of grams per mole.
So today I want to go back and look a little bit at polymer synthesis.
Then we're going to talk about properties of polymers, and end with some cultural implications that I alluded to the very beginning.
So we've already seen how this reaction begins, but let's see how it proceeds.
So here's an example of starting with an initiation reaction that generates this hydroxyl radical that starts from a peroxide.
And then you can see, in this first instance, the unpaired electron attacks the double bond, breaks the double bond links, and then moves that unpaired electron to the end, and so on, and so on.
So here I'm indicating that we're going to keep attaching, in this case, ethylenes to make polyethylene.
And then at some point, we decide that we've got the molecule long enough, and then we have to quench the reaction.
So then we add something that will cap this unpaired electron, and this step is called termination.
So this is how we make the long-chain molecules by this particular reaction.
And the thing to know, trivially, is that the composition of the polymer equals the composition of the mer.
You can see it right here, that I start with vinyl chloride.
Yeah, we've changed the double bond to a single bond, but it's still C2H3Cl, C2H3Cl taken many times.
That's it.
OK.
There's another way to make polymers.
And this is really important, not only in commerce, but it's most important in biology.
We do not have the biological apparatus to create radicals.
So during the course of this lecture, we have cell death proceeding in all of us, and cell generation and polymerization reactions occurring, amino acids being linked to form proteins and whatnot.
And that process goes by the second mechanism, condensation polymerization.
And I'll give you a simple example of that where we start with more than one mer, OK?
Involves more than one mer type.
Typically two, all right?
So let's look at a reaction.
So this is going to be a first kind of a mer, and R is just a general placeholder the people use in organic chemistry.
It comes from the word residual, but it's some giant piece of molecular mass, and there's a hydrogen here.
And this one then reacts with a second mer, which I'm going to designate R2.
And the second mer has a hydroxyl.
And what happens in condensation polymerization, is that we link R1 to R2.
R1 is now linked to R2 in a covalent manner.
And then, we take the hydrogen and the hydroxyl, and we send them over here.
So the hydrogen goes here, the hydroxyl goes here, to form water.
And this is typically conducted, in industry, not in our bodies, but in industry, it's typically conducted at elevated temperatures where this eventually goes into some kind of a condenser.
And then the water condenses and makes liquid.
And for this step, the condensation step of the byproduct of the reaction that we name the process.
The process isn't named for what happens here.
It's named for what happens over here.
So this is condensation polymerization, and you can see, we have covalent linkages at both ends, and we can add another R1H here, and then another R2OH here, and so on, and so on, and so on.
So clearly, we're losing some of the mass of the reactants here.
So the mass of the final polymer is less than the sum of the masses of all of the reagents.
So let's make a note of that.
That the mass of the polymer formed by condensation polymerization is less than the sum of the masses of the constituents.
Or let's say reactants, if we like.
Reactants.
Because we lose the condensate.
And so here's an example from commerce.
Oh, before I get to that, I wanted to show you some of the common addition polymers, which ones that you're familiar with that are made by this first process.
So at the top, we have polyethylene.
That's we started with.
And you know that's used in food wrap, it's used in milk jugs, and so on.
There's the polytetrafluoroethylene.
You remove the hydrogens, you put fluorines, and then you make Teflon.
So that's also made by addition.
Polypropylene, also used as separator.
In fact, polypropylene is found in every lithium ion battery.
It's the separator between the two electrodes, and it's engineered so that its glass transition temperature is about 60, 70 degrees Celsius.
So that if there's some event, and you start getting thermal runaway in your battery, and the temperature rises to whatever, let's say the number is 70 degrees C, the pores in the polypropylene collapse.
And when they collapse, they form a barrier, and it kills the battery, and prevents the battery from exploding.
And that's polypropylene.
There's butyl rubber.
What else do we have?
Polystyrene, you know, which is used in these cups and food containers.
Orlon, here.
Polyvinyl chloride, which is used in everything from old vinyl phonograph records to plumbing supplies.
That white PVC tubing that's used for piping.
Polymethyl methacrylate, this is plexiglass, which many of you are wearing in your eyeglasses.
That's this material here.
Butadiene, OK. You see there's the after-polymerization.
There's still one unused double bond, so then we can mix this later on for cross-linking.
And the various rubbers, and so on.
You can see, they all have residual double bonds, which allows them to then participate in disulfide linkages.
OK.
So now here's the example I wanted to show you that you will be familiar with in commerce.
This is the PET, the polyethylene terephthalate, the plastic bottles.
The two-liter soda bottles are made of this stuff.
So you start with this thing, which is known industrially as terephthalic acid, which is really 1-4-benzene.
You see the benzene ring here, so one, two, three, four.
So off the one position and the four position, you have a carboxylic acid, and this is critical in proteins.
You're going to see the COOH structure over and over again.
And that H is detachable.
So this, then, it becomes a proton donor.
If it's a proton donor, it's an acid.
So this is the 1-4-benzenedicarboxylic acid.
And this is ethylene glycol, which you know as the base constituent of antifreeze for automobiles.
It's really ethane with two alcohols.
So it's a diol.
Ethanediol, 1-2.
And you mix the ethanediol with the terephthalic acid, spit off the proton and the hydroxyl, leaving behind this, which then forms the bridge here, and there's the basis for the plastic soda bottle.
This is the same material from which you can make Dacron, which is a fiber used sometimes in textiles, and Mylar, which is used in sheet films for polymer films.
So this is the PET.
And note here, you see this oxygen is acting as a bridge.
It's acting as a bridge between the terephthalic acid residual and the ethylene glycol residual.
So again, we see nature relying on the fact that oxygen forms those two bonds and acts as a bridge for us.
OK. A couple of other things to note.
Here are the typical polymers that are made by condensation polymerization.
All of the nylons.
The nylons are made by condensation polymerization.
And the bond here that exists between the nitrogen and the carbon is called the amide bond.
Even though it's spelled A M I D E, it's pronounced amid.
So these are all nylons.
The generic term is polyamides.
And you see the different types of nylon here.
Kevlar, used as a substitute for steel belting in tires, derives its strength from this, the benzene ring here.
And polyesters, OK?
The Dacron, the Mylar, I just showed you that.
Lexan.
This is used in everything from sports equipment, if you wear eyeglasses in sports, they're probably made of Lexan.
Again, look at these twin benzene rings, and then you've got the linkage on the end here.
And also used when you're sitting seven miles above the surface of the earth.
The only thing that prevents you from being asphyxiated or frozen to death in that airplane is a window made of Lexan.
And lastly, here are the silicones, which are also made by condensation polymerization.
And just a little note here for you.
The person who invented the silicones didn't know his chemistry.
And he thought that the silicone, in fact, had a double bond here.
See, this is acetone, and it from comes from the family of ketones.
So you have methyl above, methyl below, carbon, and a double bond off to the side.
Well, it turns out that silicone doesn't form double bonds.
Everybody in 3.091 knows this.
It's too big.
So it only has a single bond.
If it's a single bond, it's not an -one.
It's just like a pentane, methane, so on.
So this should be a siloxane.
So what people refer to as silicones today properly are called siloxanes.
OK. Another thing to note.
Let's go back to the-- you see here?
We go down the backbone of the nylon, there's the carboxylic acid.
And there's a C with a double bond to the O, and it spit off its H, right?
That was the proton donor.
Look at over here.
There's an amino group here, a nitrogen with a hydrogen.
Well, if I line those two up just right, look at what can happen.
Here's a carbon-oxygen, and here's a nitrogen-hydrogen.
What do you know if you've got hydrogen sitting next to an oxygen?
It can form hydrogen bonds.
And so nylons derive their higher strength from the hydrogen bonds that form between the chains.
Doesn't this remind you of cross-linking?
It's sort of same church, different pew.
It's not as strong as a disulfite covalent linkage, but it does form linkages.
That's good.
It's good because it gives you strength.
But it doesn't come with the penalty that cross-linking comes with.
And what's the penalty?
You can't recycle cross- linked polymers very easily, because you've got these strong covalent bonds in the chains.
If you heat the polymer up, those bonds don't break.
If you heat this up, the hydrogen bonds will let go.
And you can remelt nylons, you can recycle those polymers much more easily.
So I want to make another taxonomy here that distinguishes polymers on the basis of their ability to be recycled.
Very important these days.
One of the first questions people ask if you're using a polymer in a process, is it recyclable?
And what do you do?
You start thinking about the molecular architecture in order to answer that question.
OK.
So let's talk about the ones that can be recycled.
So recyclable ones are called thermoplastics.
Weak van der Waals and hydrogen bonds only.
And this is between the chains.
You don't break the bonds up and down the backbone.
If you break the blinds up and down the backbone, you'll pyrolyze the material.
If you go to a high enough temperature to break covalent bonds in the backbone of a polymer, you're on fire.
So don't go there, all right?
So hydrogen bonds only, so this can reheat and reprocess.
So now we're talking about something that's almost like recycling metal.
You remelt it and reuse it.
The ones that have the strong covalent linkages are called thermal sets.
And the thermal set is difficult to recycle.
This is cross-linked-- difficult.
Notice I'm not writing impossible.
It's difficult to recycle, which is a polite way of saying, you can do it, and it's really expensive to do so.
You might be better off to just start with virgin material.
By the way, where do we get the material?
How do we make polymers?
Where do we get this carbon from?
From petroleum.
You say, well, why don't we use coal?
Well, you need carbon and hydrogen in these systems.
And coal is carbon.
Petroleum is hydrocarbon.
So these are petroleum-derived.
So all these questions about resource utilization and so on come right up in center stage.
OK. Now I want to take a few minutes to talk about use of crystallization to strengthen polymers, OK?
So crystallization for mechanical performance.
We want to improve the mechanical performance.
And I alluded to this last day.
We;ll do this really quickly.
So we've seen that if we want to, we know that if we-- let me give you the little icon here.
So you're going along like this, and some nice, long chain, and all of a sudden, you get a zone of crystallinity.
So I'm talking about adding crystallinity in order to strengthen the material.
And so we know that this has more bond density than the straight chain here, so this is going to be stronger.
So if I want to strengthen the polymer, introduce more crystallization, so what favors the onset of crystallization?
Or we'll say, partial crystallization.
The whole thing doesn't turn into a crystal.
To favor crystallization, what do we do with the various set points that we have in the material?
Well, first of all, we know with composition, composition is one of our levers.
And we want to make it uniform.
So that means homopolymer is favored over the copolymer.
Or maybe there are other instances where you don't want crystallinity.
In the soles of your running shoe, you don't want that turning into something that's semibrittle.
You want it to be soft.
So that's why we use copolymers in the soles of running shoes, in order to disfavor.
And if we want to do something like those jewel cases that I keep railing against, the jewel cases are a homopolymer that have a fair bit of crystallinity.
The second one we talked about last day, tacticity, this is a second lever that you have.
So we know that isotactic over atactic.
The more uniform, the more likely the material will be to crystallize.
So the isotactic polystyrene makes those jewel boxes, whereas the atactic polystyrene makes the food containers that you get when you buy food from the trucks out here.
And number three, we talked about conformation.
And we know that conformation-- linear over branched, right?
Linear over branched.
If we have a branched polymer, it's not going to pack very well, so it's less likely to crystallize.
The linear one, as I've demonstrated in this cartoon, has a much, much higher propensity for crystallization.
All right.
How about properties?
Let's talk about properties.
Properties of polymers.
And I'm talking here about the bulk commodity polymers.
Sure, you can find specialty polymers that have remarkable properties that aren't characteristic of polymers, but I'm not talking about those.
So first of all.
Electrical properties.
They're insulators, electrical insulators.
Why are they electrical insulators?
We're talking about mainly covalent bonds here.
Covalent bonds, which means tightly bound electrons, tightly bound and localized, because you saw that in graphite, they're tightly bound, but they're delocalized.
So tightly bound electrons.
And no ions, because we've now recognized that we can conduct electricity not only by electrons, but by ions.
Tightly bound electrons, no ions.
That means no carriers, that means no connectivity.
Number two.
What's the second property?
Transparent to visible light.
Again, same thing.
Tightly bound electrons, not possible to excite, et cetera.
So these are clear, and furthermore, are generally colorless.
And this is especially in the case of amorphous materials.
Amorphous materials are clear, colorless.
What happens if we get semicrystallization?
Those crystalline zones, as I showed you last day, those crystalline zones are clear and colorless as well, but the index of refraction of the crystalline zone is not equal to the index of refraction of the amorphous zone, and so this boundary, if you will, scatters light.
So we have two clear and colorless zones of different index, and so if we put them together, we end up with something that is opaque, and that's in the case of the crystallinity.
When you have some crystallinity, you'll give up on that.
And actually, I wanted to show you a few things from the popular culture.
So I think what I've got here-- oh, these are the-- yeah, OK.
This is a little scene from the musical Chicago.
It's Mr. Cellophane.
Cellophane is polymer, viscose.
All right?
Listen to the properties.
They get it right.
[MUSIC PLAYBACK] -I tell you Cellophane, Mr. Cellophane, should I bend my name, Mr. Cellophane.
Because you can look right through me, walk right by me, and never know I'm there.
-Oh, I didn't see you.
[END MUSIC PLAYBACK]
OK.
So sometimes the popular culture gets it right.
All right.
So now, what's the third property of polymers?
Chemically inert.
And this is why we use them in food packaging, and also in other forms of packaging.
Again, strong bonds, strong covalent bonds, means that it's not very reactive with the contents of many containers, including foods and beverages.
A fourth property that's exploited in polymers is very low density.
Why?
Well, what are the constituents of the polymers?
Contains things like carbon, nitrogen-- well, let's put carbon and hydrogen first.
Those are the two biggies.
So we've got carbon and hydrogen.
You saw some nitrogen, oxygen, and then trace levels of other substances.
So these are all low Z elements.
So for example, in a beverage container.
I can make a beverage container.
You see a beverage container.
I see a bundle of properties.
What is the bundle of properties?
It has to be some physical rigidity, but not much, because I can squeeze this.
It's OK. And it has to be chemically inert with respect to its contents.
So I could build this out of a borosilicate glass, a glass bottle.
And the density of borosilicate glass is about 3.54.
Or I can make it out of aluminum, and the density of the aluminum is about 2.7.
Or I can make it out of polymer, and the density is down close to 1.
So now if I load up the beverage truck, instead of transporting the contents plus the mass of the container, the higher fraction now is just the contents, and the container weighs less.
When I was your age, we had still soda cans made out of steel.
Steel soda cans.
Yeah.
They didn't have pull tops.
You had to have an opener to open them.
And the density of steel is 7.87.
Iron is 7.87.
So from steel, down to this, down to this, and so on.
So you see the low density has an advantage.
But, you know, we're going to have to go back over here and start thinking about recyclability.
So you see the balancing act here in choosing materials.
You can't just seize upon one attribute and say, oh, this is less dense than that, let's go with it.
But it's made from petroleum, and maybe it's not so recyclable.
So now what do you do?
Maybe you do make it out of steel, because you know, what you can recycle steel.
You can take the door off a 1928 Model-T Ford, melt that door down, and you can put it on a 2010 Taurus.
That's how recyclable iron is.
I'm just telling you what the properties are.
You're going to have to make the tough decisions.
And five, solid at room temperature.
And why are they solid at room temperature?
Because the interchain bonds, even though they're weak, they have a long, long, long, long surface contact area.
So entanglement of van der Waals bonds.
So you have the best of both worlds.
It's solid at room temperature, but in the case of a thermal plastic with modest heating, you can restore back to the original case.
Oh.
Here I have, I found the Polythene Pam.
I'll play you just a little segment of it, the first verse.
The interesting thing to listen to, in many of the Beatles songs, you can't tell this.
In the song, you can really tell the accent from Liverpool.
Listen carefully.
You'll hear the pronunciation.
That Merseyside accent.
And the way they're playing is the old skiffle.
This is how they started, in the clubs in Hamburg and so on.
This is this is not advanced Beatles.
They were playing here, it's sort of little joke in the middle of the Abbey Road album.
They go retro and show the way they were when they first formed the band.
[AUDIO CLIP REMOVED ]
Did you catch that?
The Liverpudlian accent?
It comes complete with the glottal stop?
[IMITATING ACCENT] You know how you eat the tea?
Give me a bottle?
Yeah.
That's all from Liverpool.
Merseyside.
OK.
So now what do I have here?
Oh, here's recyclability codes.
Throw this in, you know, it's a part of the part of the pack.
So there's the polyethylene terephthalate, that I've shown you.
This is high-density polyethylene, PVC, low-density polyethylene, and so on.
And look here.
Invented by, invented by, invented by, invented by.
These are all man-made materials.
They didn't exist before.
Keep going.
Invented by, invented by.
And they're all invented in the 20th century, with one exception.
A pharmacist in Germany by the name of Eduard Simon was playing with the vinyl benzene and managed to make this goo that became what we know as polystyrene.
And nobody could explain.
Remember, in 18-- he was doing this in 1839, before Kekule had even speculated that you can get carbon-carbon linkages!
No benzene rings!
So that he was linking benzene, that they didn't even understand existed as a ring, and he was making polymers that they didn't appreciate that you could even make C17H36, let alone C10,017.
So in 1922, Hermann Staudinger gave the explanation that what we're seeing here is the polymerization into macromolecules.
And for that, he got the Nobel Prize in 1953.
Note, it took 30 years.
You don't invent something that's really, really prize-winning, and on the way home, turn on the radio and discover, oh, I won the Nobel Prize.
It takes time.
And especially, the more remarkable the discovery, it takes time before people accept it and say, this is not a hoax.
This is real.
They didn't have the characterization.
How were they going to verify in 1922?
So you keep waiting, waiting, waiting.
Look at what happened to de Broglie.
De Broglie said, lambda equals h over p, and they went, uh-huh.
And then, as soon as the fellows did the experiment down at Bell Labs and they got electron defraction-- boom!
He gets the Nobel Prize.
They said, you know what?
You're right!
That's how it works.
OK.
This was one of the giants.
This was Wallace Carothers.
Carothers, who for a short time was a lecturer at Harvard.
He left Harvard and joined DuPont, and he gave birth to an incredible array of synthetic materials, as they say here, similar to cellulose and silicon.
Invented neoprene, the first synthetic rubber.
This was very important for the United States and the Allied Powers during World War II, because we did not have access to natural rubber.
And the fact that they were making synthetic rubber here from materials that we had here allowed us to continue building for the war effort.
Invented nylon, in 1937.
An amazing man.
Unfortunately deeply troubled, very depressive.
And two days after his 41st birthday, he checked in a hotel in Philadelphia, and took his own life.
This is Wallace Carothers.
So I'm going to show you now some synthesis.
About ten years ago, I had a live demonstration here.
There was a gal who was doing her PhD over here in Building 13, and she'd come in and do these demos.
But, you know, it's been ten years now.
We let her graduate, so she's not around anymore.
So we took some videos of it.
So the first thing I want to show you is the synthesis of nylon.
So you're going to see this reaction.
So in this case, it's hexamethylene diamine and adipic acid.
And that's going to give you nylon-6,6 plus water, just as this reaction indicates here, and these are immiscible.
See, the adipic acid is in aqueous phase, and the hexamethylene diamine is a non-aqueous phase, so they form two phases right here, two liquid layers, and this reaction occurs at the interface between the two layers.
So you're going to see a close up of the beaker, and she's going to now dip an instrument in, and start pulling the solid.
She's making the solid from two liquids, and she's going to pull it.
And as she pulls it, she clears the interface.
Because once the interface is covered, new reactant can't get there.
So ironically, the product separates the two halves of the reaction from each other.
So you have to get the product out.
This is really cool.
So there's the chemistry.
So let's look.
This is Heidi Burch.
David, could we dim the lights, maybe?
[FILM PLAYBACK]
The derivative of a carboxylic acid then is mercuric chloride.
I have it dissolved in hexane.
My base is hexamethylene diamine, and I've got it dissolved in water.
So to do this right, I have to make them float.
So we're going to pour, if I can get it open
This is the old 10-250, too.
It's got a carpet here
--because it's dissolved in water, which has a higher density than hexane.
And then I have my acid dissolved in the hexane.
So what's going to happen when I pour that in?
It's not going to mix, right?
It'll separate
It's going to phase separate, exactly.
And we're going to have strata.
Is that going to help or hurt our reaction?
It's going to hurt our reaction, right?
To have a good reaction, you need intimate mixing.
Well--
This is like the Uzo water problem
So this is pretty, actually pretty unexciting.
Nothing really interesting has happened.
It's just kind of sitting there.
Nothing's exploding out of the beaker, nothing's coming out.
It's kind of tame.
But if you look carefully, you can kind of see-- where did my pointer go?
You can kind of see this film right there.
See that big bubble?
Well, there's a scum that's floating at the interface between the two layers.
That's actually the nylon that we formed.
What happens is, I mentioned that nylon has that excellent chemical resistance.
Well, when we formed it, it made a barrier film between the two layers so that no more reactants could get through to each other to form more nylons.
So what I have to do is remove that barrier film to allow the reactants to come into contact.
And I'm doing that by drawing off the nylon as it's formed.
Because I'm drawing off the film, are very haphazard.
And what's happening is, the polymer I'm forming has a very broad molecular weight distribution.
It's not all 240 grams per mole or whatever.
And that greatly affects the mechanical properties, and, in fact, it affects them detrimentally.
So the mechanical properties of this nylon that I'm forming aren't very good.
[END FILM PLAYBACK]
OK.
So that's the first one.
Now, the second one I'm going to show you talks about glass transition temperature, and how you engineer a rubber.
So it might shock you to know that a rubber is, in fact, a liquid.
It's a polymer that's above its glass transition temperature.
It might be supercooled liquid, but it's liquid.
And those disulfide linkages give you the flexibility so that the chains can move against one another.
If you drop the temperature below the glass transition temperature, the whole thing turns into an amorphous solid.
So what are the mechanical properties?
What I'm indicating here is the ball, all right?
So the bouncing ball is up here, above the glass transition temperature it's cross-linked and behaves as an elastomer.
If you get the temperature too low, you'll end up solidifying everything, and then it turns into something with the mechanical properties of a billiard ball, all right?
So we call that tough.
Tough means it has impact resistance.
It will absorb a lot of energy.
So if I drop the ball, it won't bounce, it takes the energy of falling and absorbs it and holds it.
So if you want a ball to bounce, you want to be above here.
So what we're going to do, is we're going to take two different polymers.
One that at room temperature is here, and one that at room temperature is here.
The one that's down here, we're going to put in the boiling water and get it over here.
The one that is naturally a bouncer here, we're going to put into liquid nitrogen and send it over here.
So we're going to make a bouncing ball not bounce, and we're going to make a flat unbouncy ball bounce.
We're going to teach these balls.
And what are we going to do?
We're going to teach these balls how to bounce and not bounce on temperature cue, by moving on opposite sides of the glass transition temperature.
So this is a glass, and this over here is a rubber, OK?
All right.
So here they are.
Those are the two rubbers.
One is norbornene, and the other is isoprene.
And both of these, if you go back to the earlier slides, have a double bond in the polymer, And that double bond has now been broken for disulfide linkages.
This is Heidi again.
[FILM PLAYBACK]
OK.
The thing I want to talk about now, is the effect of temperature on polymer molecules.
Basically, when do I get rubbery behavior and when do I get glassy behavior?
Because I can get the same polymer to exhibit rubbery behavior or glassy behavior, depending on the temperature at which I use that material.
As Dr. Sadoway told you, polymers have what's called a glass transition temperature, and the type of behavior that I get is determined by my use temperature relative to my glass transition temperature.
So I have here six little balls with faces painted on them.
The ones with the happy faces are polyisoprene.
The ones with the sad faces are polynorbornene.
Now, polynorbornene is a synthetic rubber that was developed during World War II when we didn't have access to all the rubber tree groves in Southeast Asia.
So the happy balls are happy because their glass transition temperature is minus 67.
So our use temperature at room temperature is much greater that their glass transition temperatures, so they behave as rubber would.
The polynorbornene has a glass transition temperature of 40.
So we're actually below its glass transition temperature at room temperature, and it just dissipates all the energy.
We get a much more glassy-like behavior from this material, and it doesn't bounce.
So what we can do here to demonstrate the effect is we're going raise the use temperature of both materials, and we're going to lower the use temperature of both materials.
So I'm going to drop them in boiling water, and hope that the little faces stay on, because I just painted them on with white-out.
I don't know how that's going to work.
And then I'm going to drop two of them in liquid nitrogen.
Now, these will be our controls.
Remember, polynorbornene, polyisoprene.
Now something to think about, which I know Dr. Sadoway always likes to ask, is why does one have a high glass transition temperature, and why does one have a low glass transition temperature?
And I'll leave you to ponder that little conundrum, while I try to fish these bad boys out.
OK. Oh!
White-out worked.
So this is the polyisoprene.
Can you hear the sound that it makes?
When I drop the one that's at room temperature, I get a nice rubbery sound.
When I drop that one, I get a plastic, hard glassy sound.
I've cooled it below its glass transition temperature, and now I get a totally glassy impact, like if I were knocking pool balls together.
Now let's go in for the polynorbornene.
Oh!
His eye came off.
Same thing.
I pulled the polynorbornene below its glass transition temperature as well, so it, too, exhibits a glassy collision, like a pool ball.
Now, let's go in here.
This is the polynorbornene, so it should still continue to bounce, because we've raised its use temperature further above its glass transition temperature.
That's not the exciting one.
The exciting one is-- if I can get it out.
What happens to the polynorbornene?
This is the sad ball, the one that wouldn't bounce at room temperature.
By raising its use temperature above its glass transition temperature by dunking it in boiling water, I get a rubbery response.
[END FILM PLAYBACK]
Super.
OK. House lights, please.
I don't read well in the dark.
OK.
So I hope you've seen some examples of how these work.
So now I want to talk a little bit about the broader societal issues.
So this, I think, is sort of a pattern of adoption of new materials.
You start off with this wonder, you know, this cool property that you get at the lab bench.
You know, when nylon was invented, it led to democratization.
Do you realize stockings, womens' stockings, were only owned by the wealthy?
Because they were made of silk.
When they were made out of nylon, they became cheap and abundant.
So that changed the way people dressed.
It changed the way people made clothing in a way that was unimaginable before.
So you could, for example, make a billiard ball, or piano keys, without killing an elephant.
You didn't have to use ivory.
You could use polymers.
Huge, huge implications.
So this is one of the first commercial polymers.
It's phenolic resin, and it's made by a condensation polymerization between carbolic acid and formaldehyde.
It was called bakelite.
and its corporate symbol was B with the infinity sign.
Material of a thousand uses.
And this, normally, is-- look at the elaborate art deco coffeepot.
And this is all-- it's thermal insulated, and so on.
These are all plumbing parts made out of bakelite, that previously would be made out of many pieces of metal that had to be machined and fashioned together.
You could take bakelite powder, put it into a die, heat it, and out comes the part, finished, in one step.
So this has a huge implication on productivity.
But then, so now we start talking about substitution, making the cheap things, the piano keys, the fork here, and on and on and on.
But then things that couldn't exist before.
You know, the movie film.
Look at the steering wheel here.
There's the, that's indicating the lady's stocking, and so on.
This is from Look magazine, 1940.
People thought these things were just godsends.
This is Synthetica.
This is a country, and it's got the province of Cellulose, Petrolia, Castphenolic and so on.
This is the 1960s.
This is plastics, the future has arrived.
It's at a museum in New York.
You're looking at the interior of a home with furniture built in, and it's all plastic walls, and formed, and so on.
I think maybe these guys were in a drug haze or something.
Those are the windows.
Maybe they thought those windows were square, rectilinear, I don't know.
So this is going way overboard.
And then let me show you a piece of iconic film from The Graduate.
You'll see a young Dustin Hoffman.
He's just come back from getting a bachelor's degree.
So this could be you someday in the not too distant future.
And he's at a party that his parents are throwing in his honor.
[FILM PLAYBACK] -We're all so proud of you!
Proud, proud, proud, proud, proud, proud.
What are you going to do now?
-I was going to go upstairs for a minute.
-I mean with your future?
Your life?
'Well, that's a little hard to say.
-Ben.
-Excuse me.
-Mr. McGuire?
-Ben.
-Mr. McGuire.
-Come with me for a minute
See the lipstick?
-I want to talk to you.
Excuse us, Joanne
See, it's the '60s.
They blow off the ladies.
Now the men are going to talk.
-I just want to say one word to you.
Just one word.
-Yes, sir?
-Are you listening?
-Yes, I am.
-Plastics.
-Exactly how do you mean?
-There's a great future in plastics.
Think about it.
Will you think about it?
-Yes, I will.
-Enough said.
That's a deal.
[END FILM PLAYBACK]
All right.
So this was 1967, '68.
So what's happened since then in plastics?
Well, all of these other considerations about recyclability, environmental health and safety issues.
In order to facilitate molding, when you want to mold the plastic, to get those bonds to loosen up a little bit, we use a chemical known as a plasticizer.
It's a low molecular weight polymer that helps the-- it's sort of the equivalent of putting a little bit of olive oil in with the pasta, to give it that freedom to move.
And so this, then, when you sit in a new car, the new car smell is the smell of the plasticizer, which is still evaporating.
And some of those plasticizers have molecular architectures that are similar to hormones in our bodies.
And so it's not healthy for us to be breathing this stuff.
So maybe the bright future that was talked about is not quite as bright as some people might have thought it to be.
The other thing is-- you know how I showed you the plumbing parts?
Well, we started making all kinds of things out of polymers.
As a child, I remember my father bringing me, one day, a shoe.
The shoe was 100% man-made.
It was all polymer.
And we thought this was the coolest thing, because of all man-made.
Of course, there was a complete reversal.
People said, you know, your foot can't breathe, and this is a silly thing to have.
But at the time, we started making-- and they were ugly.
And you know, for me, if it's ugly, I don't want to wear something that's ugly.
So what happened was, we started making all kinds of things out of polymers, or as the public knows them, as plastics.
And by the mid-'70s, to call something plastic was to make a derogatory comment.
It's plastic, it's cheap, it's shoddy.
So you know, making silk stockings cheaper as nylon stockings was good.
But if you get them too cheap, they're cheap.
See?
So there's an optimum there.
And all of that went into play.
By the way, how do we get the word plastic?
It comes from the Greek plastikos.
And same word as potter.
So when you're working with clay, you're working with a plastic medium.
It's deformable, it's malleable.
There's Samuel Johnson "let Thy plastick hand."
So all plastics are polymers, but not all polymers are plastics, because some polymers can't be deformed.
If they're below their glass transition temperature, they're brittle solids.
All right.
So I've told you enough about concern.
Let's look at recycling.
In the United States right now, 100 billion pounds annually of polymer, 50 million tons of which 3 to 4 million are recycled.
It's a very low recycle rate.
So this ends up in the waste.
Contrast that to steel.
Steel, we still make 80 million tons a year of virgin metal in the United States.
It's still a world-class steel industry.
You can bet it's world-class.
With our wages and our high standards, if you're still in business making steel in the United States, you better be damn good.
Very good.
So 80 million tons virgin metal, 60 million tons recycled scrap.
We recycle more steel per year from automobiles than we consume building new automobiles.
It's a complete ecology.
And compare that to aluminum.
Aluminum, 4 million tons a year, 1 million tons recycled.
Very high rate on used beverage containers.
But that's the exception.
Because this was DFE, Designed For Environment.
The alloy choices were made in order to facilitate recycling.
So there were certain metals that were forbidden to go into the alloy because they would pose problems later in recycling.
So there are many ways to get the same deep-drawing characteristics.
You know, you make a beverage can out of a sheet, and you snap.
Punch down, punch up, there's no seam on the bottom.
This is made from a single sheet of metal.
And then a second piece comes down on the top.
And so this has only aluminum, silicon, and magnesium.
You could choose other metals, but with aluminum, silicon, magnesium, it is recyclable.
But aluminum ladders-- if you're building an airplane, let's say you're the chief purchasing agent for Boeing.
And you've got a choice of aluminum scrap, which, you know, people might have thrown some old fishing weights with lead in there, and that's now contaminant in the aluminum.
Or you're going to buy high-purity virgin metal, and you're going to bet the company's future.
What are you going to buy?
You're going to put recycled metal into an airplane.
Have the thing fall out of the sky.
No way.
So you have to be careful.
Those alloys, if you take the, I told you, the door off the 1928 Model-T, and you put it on a 2010 Taurus, you can't take the wing off a DC-3 and melt it down and put it on a 777.
Because the top has an aluminum alloy with zinc, and in the bottom one has an aluminum alloy with copper.
And if you co-melt those, you'll end up with intermetallics, and the alloy is unusable.
So you have to separate them.
Now it's not so simple.
So before you go make pronouncements saying, yeah, recycling is great, think about it.
Follow through.
OK.
If you find this interesting, and the impact that polymer had on American culture, this is a lovely book by Jeffrey Meikle on polymers and their use.
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All right.
So here's the box score from the celebration on Monday.
Class average improved considerably.
A number of people moved from the sad face to the smiley face.
I would still tell anybody who is down over here that the final exam is coming, and if you perform well on the final exam, we'll take a look at the overall performance of the semester.
So get in and talk to us, if you want to, to see what we can do to help you succeed here.
But it takes a little bit of effort on your part as well.
I think that's all I want to say.
There will be a weekly test, a weekly minor celebration on Tuesday, as normal.
So today we're going to start the last of two units.
We have two units left before the end of the semester.
And this one is biochemistry.
And it's one of these integrative topics that's going to bring together a lot of the material that we've looked at in the past.
And biochemistry overview, it's the chemistry of living organisms.
And the first thing to point out is that biochemistry is governed by the same laws that apply to inanimate matter.
And apropos of that, I'd like to read something that was written by the Nobel Prize winner, Richard Feynman.
"If in some cataclysm, all of scientific knowledge were to be destroyed, and only one sentence passed on to the next generations of creatures, what statement would contain the most information in the fewest words?
I believe that it is the atomic hypothesis, or the atomic factor, whatever you wish to call it, that all things are made of atoms.
Little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another."
So that's the first thing to point out.
That it's the same chemistry that we've been learning up until now.
And in fact, I think I've got a a slide that captures the-- there's the Feynman quote, the last part of it.
Second thing is, by way of introduction, you see a boom around us today in biotech.
And what is the biotech?
It's the molecular biology, it's the commercialization of molecular biology.
Biology, when I was your age, was classification, taxonomy, naming things and figuring out what bin they fit into.
Today it's not that.
It's molecular biology.
Well, what's the molecular science?
Molecular science is chemistry, and it's solid state chemistry.
Biology is solid state chemistry.
We are solid state devices made of soft matter with an endoskeleton that's ceramic, and so on.
But we are solid state devices.
So this is solid state chemistry and material science, and that's why we talk about it in 3.091.
And why now?
Why is the biochem revolution only now?
Well, it has to do with complexity.
You can think about complexity from the standpoint of materials, and in fact, lay out the ages of man.
The Stone Age.
The Stone Age precedes the Bronze Age.
Why?
Because stone is found in nature naturally.
To use stone requires only to cut it and change its shape.
But you make bronze, you have to engage in chemical processing.
There's a chemical conversion that requires a higher level of societal sophistication.
And then iron follows bronze.
Why?
Because iron melts at a higher temperature, so you have to have a higher level of technology in order to make the furnaces and to conduct the reduction reaction to extract iron.
And then comes polymers.
Polymers, the birth of synthetics, come when?
First polymers were around, what, 1829, when the pharmacist was playing with the styrene.
And then it explodes in the 20th century.
And then mid-century we get silicon.
And silicon is even more sophisticated, because we had to understand quantum mechanics to even understand that we needed silicon.
We could convert beach sand into silicon back in 1500, but who knew what to do with silicon in 1500?
And then finally comes the end of the 20th century, and now we're dealing with this stuff.
So that sort of puts things in perspective.
So what we're going to do today is start the treatment of biomolecules.
And the first thing I want to say, is we're going to be talking about macromolecules but not polymers.
So all polymers are macromolecules, but all macromolecules are not polymers.
What's the difference?
In polymers, we have a common repeat unit, whereas in the life sciences, Mother Nature is a polymer engineer gone wild.
She takes different functional groups and puts them at every stage along the way, as opposed to polyethylene, where it is the same functional group at every stop.
So it is exploded polymer chemistry, macromolecular chemistry.
So there's four classifications of biomolecules.
The proteins, carbohydrates, lipids, nucleic acids.
And we're going to look at three of the four.
We'll start today with proteins.
We will study lipids and nucleic acids, but in the time allowed, we're going to lose the sugars.
So you'll take 7.012 or something to satisfy the Institute biology requirement, you'll go much more deeply into it.
But I think you're going to have very good preparation after three lectures here, and you'll be able to put together a lot of the material that we've looked at prior.
So the first thing we're going to look at is proteins.
Proteins are macromolecules, and they're formed by polymerization of monomers known as amino acids.
So before we can talk about proteins, we're going to first look at amino acids.
So let's do that.
And amino acids are substances that have the following skeletal structure.
I'm going to start a new board for this.
So this is the basic skeletal structure of the amino acid.
It's got a central carbon, sp3 hybridized with four struts on it.
Down in the lower left corner, we have the amino group, which is nitrogen and two hydrogens.
So there's one, two, three bonds off the nitrogen, and I want to emphasize but it's still got this unpaired set of electrons, which we know makes this thing a Bronsted base.
And it's going to have a site here for proton attachment.
And that's a big feature here in amino acids.
So this is the amino group.
And it acts as a Bronsted base.
So it's a proton attachment site.
Second strut is hydrogen in every amino acid.
The third site, over here, is the carboxylic acid end, and that's carbon with a double bond, and one, two, three, four, all right?
Every carbon needs four struts, and the fourth one has the OH.
And this is the H that can fall off.
So this is the proton donor here.
This is carboxylic acid end.
So this is three of the four ends of the carbon bonds are fixed in all amino acids.
And then the fourth one is free to be specified.
This is all R. R is a placeholder.
It stands for the substituent.
This is the substituent.
And this is nature's choice.
So you can put anything here.
There are hundreds of amino acids.
There are only 20 seen in protein.
But anything that goes up here, with this structure, is an amino acid.
I don't care.
You could put a cyanide up here, for all I care.
It'll still qualify as a amino acid if these other three sites are filled in the following matter.
So 20 seen in amino acids.
20 of those in amino acids.
And now I want to look at the substituents and break those into categories.
So there's four categories of substituents.
In other words, the 20 here can be broken into four categories.
Four categories of R. And we can group those by their chemical properties.
The first group, I took this from one of the other texts.
Yeah, this is from one of the other texts.
So there's the 20 amino acids.
So I'm now highlighting some of the ones that are nonpolar.
So what do you see there?
The nonpolar ones, you see for R. The R group is above here.
so?
This is just plain hydrogen, which is nonpolar There's a methyl group, CH3, so that's alanine.
Here's a methylene with two methyls, so this is alanine, and so on and so forth.
So the different groups hang off that R position at the top.
So these are nonpolar, and this also is characterized by being hydrophobic.
So that end becomes hydrophobic when we use nonpolar group.
The second option is to have polar substituents.
And some of them even have hydrogen bonding capability.
That's the second category.
And of course those are going to be hydrophilic.
The third group is charged.
The third group actually has a charge end, which becomes a negative ion in aqueous solution.
And these are going to be strongly hydrophilic.
And also-- [BREAK IN VIDEO]
Number nine, number nine, number nine.
Can you hear?
Yes?
Yeah, that's good.
Hydrophilic and acidic.
The acoustics here today are very different, huh?
We're missing a few bodies.
We need to put the crash test dummies in here, and then we'll have the same acoustic sound.
All right.
And then the fourth category gives you a cation, R plus, in aqueous solution, which then, strongly hydrophilic and basic.
So we'll go through.
Now, our bodies can synthesize only 10 of the 20 amino acids, and then the other 10 we have to get from diets.
So when you look at some of these fad diets, you might want to check and make sure you're getting all of the proper nutrients.
Now another of the structural features of amino acids that's important is chirality.
Chirality is a feature of amino acids.
And what is chirality?
I'll show you by way of example.
Chirality talks about handedness.
Comes from the Greek word for glove.
So let me show you molecules that have the chirality that distinguishes them.
So I'm going to give an example with alanine.
With alanine, the R-substituent group is equal to the methyl, CH3.
So I'm going to put CH3 up here for the substituent, and then the other three stations are as we know them to be.
So there's the amino group, there's the hydrogen, and here's the carboxylic acid.
And note, look at the compact notation here.
We write COOH.
This O is not bonded to the O next to it.
Both O's are bonded to the carbon, but this is compressed notation.
COOH, and then this.
So this is alanine, and I could just as easily write the molecule in the following manner.
I would still satisfy the requirement that I have an amino group, a hydrogen group, a carboxylic acid group.
Actually, if I want to get really fancy in chemistry, I could write it this way.
COOH, and everybody knows H can't bond to two species.
In fact, it's the carbon here.
But it's OK.
It's compressed and in the methyl group up here.
And what you notice is that the molecule on the left is distinguishable from the molecule on the right.
These are not the same molecule, even though the same substituents are here, and the other features are preserved.
This cannot be superimposed.
So you could distinguish these two.
If I just threw them on the floor, you'd be able to pick them up and say, one has the amino group in this orientation, and if I put the other one, with the methyl above and the hydrogen below, I will have the amino group on the other side.
And these things also are optically active.
And through their optical activity, we give them their names.
And so let's take a look here.
All right.
This is just a little bit more of the amino acids.
And actually, one of our former 3.091 TAs, Andrew Magyar, who just defended his PhD thesis.
I was on his committee, and he did a splendid job, and used his 3.091 principles to get through his advanced biology.
This is a list which is going to be part of the PDF that I'll post.
But there are 20 amino acids, and as you know, there are 26 letters in the alphabet we use in English.
And so if you want to write them in compressed form, you just use one of these letters.
There's the three-letter notation as well, so ALA is alanine.
And in the past, I used to have contests in 3.091 to ask students to give me the airports, if these were three-letter airport codes.
You see, it always comes around Thanksgiving, and people are busy heading to-- and it turns out that 19 of the 20 amino acids have airport codes that correspond to the amino acid three-letter abbreviations, with the exception of glutamine, GLN.
There's no airport that has GLN as its thing.
That's a piece of trivia that will get you the last wedge in Trivial Pursuit.
And then this is the one-letter abbreviation.
So Andrew has put them in this form.
So there again, I have the basic, the acidic, the aliphatic are the ones that are nonpolar.
And this is another way to look at it.
And here's the base structure, and here's the whole thing in monochrome.
All right.
So that's part of the thing.
OK.
So here's what I'm talking about.
The mirror plane.
So this box can be mirrored as shown.
But as soon as you put a distinguishing mark in the lower right-hand corner here, the box on the left is not superimposable to the box on the right.
They are distinguishable.
And so these things will have different chemical reactivities, not melting point boiling point, but how they react with other molecules.
So here's the chirologics.
Your gloves are chiral.
If you try to put your right-handed glove on your left hand, it's a poor fit, whereas the flask here is achiral.
And now here are some molecules, just to show you how this translates into chemistry.
So this molecule is chiral, this molecule is not chiral.
And it has to do with the number of positions that are distinguishable.
And how does this fit into the grand scheme of things?
It's sort of a form of stereoisomerism, isn't it?
Remember, we looked at the geometric isomers.
Here's an example.
This is dichloroethylene.
We looked at butene.
But here we've got cis-isomer, here we've got the trans-isomer, where the chlorines are on opposite sides of the double bond here.
They're on the same side of the double bond here.
These are types of optical isomers.
These are the mirror images, the same functional groups, but in this case, in one orientation, the plus enantiomer, here's the minus enantiomer.
They're called enantiomers because it comes from the Greek word for mirror.
So we're using the Greek word for glove and the Greek word for mirror.
Now we'll talk about the optical activity, and then we'll take a few notes.
So what's the optical activity?
It turns out that if you put these in aqueous solution, they will cause the polarization vector of light to rotate.
So here's a cartoon that shows a light source with unpolarized light, and then the light goes through a polarizer, which then forces only vertically polarized light.
In other words, the electric factor is in the vertical orientation.
And so now this electric vector enters this tube containing an aqueous solution of something that has optical activity.
And the degree of rotation is proportional, obviously, to the concentration of the chiral species, and also the path length.
So if you have a dilute solution on a long path, or a concentrated solution on a short path, it's the number of collisions.
And ultimately, you have a degree of rotation, et cetera.
And so based on how the electric vector changes is how we got the names for the various enantiomers, because we've got to name them somehow.
Otherwise we can't distinguish one from the other.
So it turns out that the way I've drawn this one, with the amino group in the lower right, and the carboxylic acid group in the lower left, causes the electric vector to rotate clockwise.
So I'm going to write h nu, just so that you know, or if you want to put the electric vector or something.
It rotates page to the right, and so the Latin word for right is dexter.
So this is called dextrorotatory.
This is the dextrorotatory enantiomer.
So now you've got really nice terminology to use at the table over the Thanksgiving turkey, all right?
So this is the D-form, D-circle, or, as every right-handed person knows when speaking to a left-handed person, this is the positive direction, OK?
And now this one causes an anti-clockwise, or counterclockwise rotation of light in an aqueous solution of this.
And rather than use sinister, which is the Latin word for left, and it has all these negative connotations, they went to the Slavic, and shows levorotatory.
So this is levorotatory, L-form, or minus N, with apologies to the left-handed people in the audience.
That's what it's called.
OK.
So these are called enantiomers, and hence, you understand the choice of Man in the Mirror for the opening music.
So 19 of the 20 amino acids are chiral.
The only one that's non-chiral, the only non-chiral amino acid, is glycine.
Glycine has hydrogen as a substituent.
So now you have hydrogens above and below, an amino acid here, and carboxylic acid here.
And this one is achiral.
This is achiral.
So the other 19 so that's 95% of the amino acids, are chiral, and they exhibit this interesting behavior.
Now the other interesting piece is that only the L-enantiomer-- by and large, there a few exceptions that you can find.
There's some research on this, trying to find, always the exception.
But by and large, only the L-enantiomer of amino acids are found in proteins.
So there's no preference in nature for synthesizing one over the other, but somehow, somewhere, a long time ago, there was a preference given to the L-enantiomer.
And so all of us have the L-enantiomer present in the proteins in our bodies.
So I'm going to say only, but you know that there's always going to be some exception here.
But by and large, only the L-enantiomer of amino acids found in proteins.
And you're going to see the protein synthesis in a moment, and try to figure out why that happened.
By the way, carbohydrates, which we're not going to talk about-- carbohydrates are chiral as well.
And in sugars, only the D-enantiomer of amino acids found in sugars.
And you'll see later, this is all important in terms of a lock and key mechanism.
And there are some L-sugars.
L-sugars you know as invert sugars.
And our bodies don't recognize invert sugars.
Our body apparatus requires that there be D-sugars.
There's some science fiction novella I read many years ago about this group of people that's marooned on the proverbial desert island, and it's lush with vegetation, and they're eating all of these fruits and vegetables and wasting away.
And at the end of the novel, it comes to be known that this island, because it was isolated, somehow, back in time, the plant life there had the opposite enantiomer, and so their biological apparatus didn't recognize anything.
They were eating all this stuff, and it was just going right through them with no nutrient value.
So if you're ever marooned, check!
Find out which enantiomer you're eating!
Otherwise you're wasting your time.
Honey has the L-sugar, because bees have a different biological apparatus.
Now, when we synthesize these things chemically, we can get both enantiomers coming out of the synthetic apparatus unless we take pains not to.
And so when both enantiomers are present, we call that specimen, that's term racemic.
So you have both the dextrorotatory and levorotatory present.
Now, normally we take the five minutes at the end to talk about chemistry in the world around us, but this is, you know, day before Thanksgiving holiday, and we've got a natural entry point here.
I want to talk about what the consequence here is for human health that drives directly from chirality.
About 40 years ago, there was a drug developed in Europe called thalidomide.
And it was developed by a German firm called Chemie Gruenenthal.
If I can just spell it.
And it was developed as an anticonvulsant.
And it was a sedative, and it had fantastic properties.
Namely, that it was something that you could use in treatment of people who are very depressive, but they couldn't kill themselves with this.
If you tried to overdose with thalidomide, you would simply go into a sleep, and then you could be rescued Well, that was its original intention.
Just by accident, it was discovered by various people who were using it that it was a very fine palliative for morning sickness.
And so women started using it in the first trimester of their pregnancies in order to calm their morning sickness.
And so there were calls for the importation of this drug to the United States.
And the FDA said no.
And of course there was an outcry from women's groups that said, oh, the FDA is just a bunch of men that are insensitive to women's health issues.
Let's speed this stuff up, let's get it over here.
They're using it in Europe.
What are we doing?
It turns out that this was during the time of President John Kennedy, and he had appointed the first woman director of the FDA.
And she said, it hasn't been tested thoroughly.
We're not importing it to the United States.
People got on airplanes, they used post office, what have you.
They brought it into the United States, and it was used widely in the United States, And then subsequently was discovered that this is a teratogen, which causes severe mutation of the DNA, and children being born minus limbs.
There were grotesque deformities.
And so the position of the FDA was right.
So you say, well, but didn't they test this stuff?
And they did.
Well, didn't they test it on animal models?
And they did.
Turns out the animal models that they used lacked a certain enzyme that's present in our bodies, that converts thalidomide downstream into something that's a teratogen.
It's a really, really sad story.
It turns out now, knowing what we know, only one of the enantiomers of thalidomide causes birth defects.
Not both.
So racemic thalidomide was loaded with both palliative and toxic components.
And so this is a powerful lesson of getting the-- avoiding the toxic enantiomer.
And also, there may be people here who are opposed to animal testing.
And I understand your concerns.
But we're far from being able to sit with the Shroedinger equation and a supercomputer and predict the metabolic activity of these drugs.
And so this is the dilemma for people.
Do you want to test it or not?
Do you want to get these things--?
It's actually coming back now, in the treatment of HIV-AIDS.
It's being used as part of a cocktail with other medications.
But now we know not to give the racemic form, but only to give a particular enantiomer.
Another one that is a chiral molecule is Ritalin, which is used, among other things, for the treatment of ADHD.
And there are some negative side effects of the use of Ritalin.
And again, it's being discovered that it's only one of the enantiomers that has the negative side effects.
And so the whole question of chiral molecules is a very, very hot topic in chemical research.
And so, how do you how direct the synthesis to get to one enantiomer and not the other?
Catalysis.
Again, which is something that is very important and not very well understood.
Highly empirical, still.
So all of those lessons that we've learned are very important.
I think there's a cartoon here that actually shows-- all right, so here's the example.
This is a molecule that's chiral, and it's very primitive.
You've got a square and a triangle and a circle and so on.
And you see, both of these have the same chemical properties.
But when you're looking at this in terms of biological activity in the human body, you can think about a lot of these interactions as lock and key.
Now, there aren't these little holes here.
It's not putting a square peg in a round hole versus what are they really trying to model here?
They're trying to model the types of secondary reactions.
So for example, there could be a polar center here, and this could be charged negatively, and this could be charged positively, and so they will be attracted coulombically.
Maybe this has a polar end, and this has a dipole, and they'll be attracted.
Maybe this is nonpolar, and this is nonpolar, so they all fit, whereas if you turn it around, you've got something that's positively charged here, and this thing's a dipole, there's no adhesion.
So that's where this lock-and-key idea comes into play.
And so what we're going to do later on is to figure out how to match things up in terms of dipole-dipole interactions, coulombic interactions, and so forth.
OK.
So what are the properties now?
Let's look at the properties of these things as just straight chemicals.
So we had amino acid, I had the solid neutral form of the amino acid.
What are its properties?
Not the protein.
Properties of the amino acid.
OK.
So first of all, they are solids at room temperature.
And why would we do that?
Well, because look.
You've got hydrogen bonding capability here, and that hydrogen bonding capability allows it to form bonds great enough to overcome thermal energy.
Molecules like this, do you think they form crystalline solids or amorphous solids?
They're small enough that-- this one looks a lot like methane, doesn't it?
Looks like methane with a couple of extra things.
But from a distance, it's a sphere.
maybe A little bit oblate, but it's a sphere.
So these things form crystalline solids as opposed to amorphous solids.
Their optical properties, I've already shown you.
They're colorless.
They're not good absorbers.
Where's the covalent bonds with tight electrons?
So there's no reason to have excitation and color.
And then, they're moderately soluble in water.
And that water chemistry is what we're going to turn to next.
So now I want to look at how they behave in water.
And where I'm going with this is to show you the first stages of how we animate matter.
Because I talk about inanimate matter.
We are animate matter.
Where does the animation come from?
What's the chemical origin of our animation?
How is it that I can do this?
See, you're looking at changes in the conformation of a polymer.
That's all you're doing.
You're saying, well, how-- he's moving fast.
What's the Debye frequency?
It's 10 trillion times a second.
10 trillion times a second the atoms vibrate in my body, and the speed at which I'm rotating my hand, compared to 10 trillion times a second, very slow, right?
Which actually explains why we talk at the speed we do.
And why is it that a human being can only run-- or not only run, but runs at the speed that he or she does?
It has to do with the chemistry.
With the Debye frequency, and what the skeleton can do-- I mean, why aren't human beings 40 feet tall?
I mean, stand back here and-- all right-- we're going to create this universe.
Why are we this tall, and not 40 feet tall?
Because make a skeleton 40 feet tall would be chemically very difficult.
It's all chemistry.
The rest, as we know, is stamp collecting.
OK, so now let's look at the properties of the amino acids and water.
So I'm going to write them again in a different form.
So here I'm going to put the H2N.
Here's the amino acid N. Here's the central carbon.
And I'm going to put the hydrogen, instead of writing it down here, this is the carbon with its little hydrogen.
This is the carboxylic acid N, and then the substituent group is above.
So this is again the thing.
By the way, there are two carbons here, agreed?
And the central carbon is also known as the alpha-carbon.
This is the alpha-carbon.
That's the beta-carbon, I bet.
All right.
So now what I'm going to do is, I'm going to dissolve this in water at neutral pH.
And what happens?
What happens is that once this thing solvates, the proton falls off the carboxylic end and comes over and attaches to that electron pair at the amino acid end.
And we end up with this, at neutral pH.
H3NCHCOO, and the substituent is a spectator here.
Well, I lost a proton, so this is locally negative.
And I gained a proton with its charge over here.
So this is locally positive.
I conserve charge.
This is not neutral and this is not neutral, but it has a negative end and a positive end, and it's not a dipole.
I mean, this is negative ion-like and this is positive ion-like.
So let's say, net neutral locally positive and negative.
And just as we've fallen in love with some German terminology for words like bremsstrahlung, we have a German word for this one.
This is like a twin ion, or it's like a hermaphrodite dual gender, so this is called zwitterion.
It's got both genders, if you like.
It's positive and negative.
OK.
So there's zwitterion.
And this is the way that we start to introduce the dynamic behaviors.
So now I'm going to show you how to make something come to life.
So what we're going to do, is we're going to show how this zwitterion can respond to changes in its environment.
And we're going to invoke in order to do is to say that to impart animation, we invoke the le Chatelier principle.
It's sort of like Newton's third law for chemistry.
You know Newton's third law.
Every action, there's a reaction.
So le Chatelier was a French engineer, but his name is associated with an important chemical principle.
And it basically says that if you disturb a chemical system, the chemical system will respond in such a way as to minimize the disturbance.
So the example I'm going to give is, I'm going to start with a neutral pH, and I'm going to change the pH.
And then the zwitterion is going to respond to minimize the disturbance.
In other words, if I drop the pH, zwitterion is going to try to raise the pH back to what it was before.
And in doing so, I think you're going to start to see how we can get animation.
So it's sort of like that a chemical disturbance-- this is the le Chatelier principle.
Chemical disturbance is mitigated by a chemical response.
So here's the example that we're going to use.
We're going to say, let's take something and radically change the pH.
So let's drop the pH.
So drop a pH means, the proton concentration is going to go up.
So how would zwitterion respond if it wants to reduce the proton concentration?
Well, it's going to try to gobble up the protons.
Well, how's it going to do that?
Well, we know that there's a proton attachment site.
Look, there's a proton attachment site at the carboxylic acid end.
How do we know?
Because it used to have a proton there.
So it's going to go and vacuum up protons to try to bring the pH back to what it was before.
So let's look at that reaction.
So here's H3NCHCOO minus plus.
So this is neutral zwitterion.
And what it's going to do in response to the increase in proton concentration is try to gobble these up.
And the following is the result.
That proton is now going to attach here, and cap that negative end.
So now we've got a proton on this end, and we've added a proton at the other end.
So what's happened to the neutrality of zwitterion?
It's lost!
Now zwitterion is net positive, but we've started gobbling up protons.
So this reaction is called protonation.
And the reverse reaction is deprotonation.
And this is messy, all these darn characters.
So I'm going to write the neutral zwitterion as just HA.
There's the proton here, and the rest of this stuff is just details.
So that's A.
So HA plus H plus gives me this thing here, which is HAH plus there.
Isn't that cleaner?
That's a lot easier to read.
And we can write an equilibrium constant for that reaction.
I'm going to call this reaction 1.
In reaction 1, I can write K1.
And it's written, the convention is that you always write the reaction constant for the deprotonation reaction.
So I'm going to write the reaction constant for this reaction from right to left.
So that's going to be this.
H plus over HA divided by concentration of HA H plus.
And then we're going to bring Sorensen to the party, and take logarithms, and make it nice and clean.
As we know, when Sorensen enters the room, he straightens things out.
And so this will then become PK, and if I take the log base 10 of this, I'm going to end up with minus log base 10 of H, which then will be PH.
And then this flips over with the minus sign, and becomes log base 10 of the ratio of protonated zwitterion over deprotonated zwitterion.
And this is known as the Henderson-Hasselbalch equation.
And what does it answer?
It answers the question, you tell me what the pH of the system is, and I'll tell you what the ratio of protonated to deprotonated is, because the PK is the function of the R group.
So different R groups have different PKAs, and they'll respond differently.
Some are very aggressive, and some are not so aggressive.
And look at what happens.
At one end, when we have neutral pH, we have all of the zwitterions sitting here.
And as we go to higher and higher pH, more and more of the neutral zwitterion converts to the protonated zwitterion.
And in the extreme, when I've got very, very high proton concentration, I will protonate all the zwitterion.
I'll protonate it out of existence.
And the only amino acid that exists is fully protonated.
So I've got a sliding scale from 100% deprotonated to a 100% protonated.
Well, somewhere along this scale is 50-50.
And what happens when I've got equal amounts of protonated to deprotonated?
That means the concentration of protonated equals the concentration of deprotonated.
I have the log of 1, which is 0, and that's the pH at PK1.
So now you see physically what PK1 means.
It's the pH at which you're halfway in between, which makes sense.
If I wanted to characterize something, it has values between 0 and 100, I'd give you the value at 50, and you know, it varies around 50.
Either 50 plus 50, 50 minus 50.
So that's what the PK is.
Good.
Now let's do the same thing for the opposite side.
Let's, instead of a sudden drop in pH, let's look at a sudden rise in pH.
Now you see, just to finish the point.
Look at what happens here.
What I'm trying to show you is that this change in pH caused this molecule to change.
you I mean, this could be in your stomach.
And now you drink some coffee, or you drink some cola, and the pH goes way, way down, because you've drunk something that's acidic.
Well, there are functional groups in the lining of your stomach that are going to now start protonating.
And when they protonate, they're going to change.
This could have been bonding to something, right?
COO minus could've been bonding to some other molecule that's locally plus.
When the proton comes in here and caps it, that bond is broken!
And that's the beginning of animation.
So that when I drink something, and all of a sudden, my stomach goes into action.
Well, we're going to have to process this stuff.
Have to turn on some switches here.
It's not my brain that's doing it!
It's not that the brain gets all these sensor signals.
We couldn't function that way.
We're functioning just by these things acting right there at the site.
This is the origin of animation.
I'm telling you that this is the secret of life.
I can't tell you the meaning of life.
You have to go somewhere else to answer that question.
But I can tell you where life comes from.
There it is.
So now, sudden rise in pH, this means that the proton concentration goes way down.
See, zwitterion is like one of these friends you might have.
Do you have any friends that are, what do you call them, rescuers?
You know, if they see anything wrong, they're the ones that get involved?
They've got to help everything?
In the extreme, they're quite meddlesome.
Well, this is what zwitterion is like.
See, zwitterion sees a sudden drop in pH and goes, I'm zwitterion!
I can help!
I know what to do, because I'm zwitterion!
I fix things!
And I'm conflict-averse, I don't like this!
OK.
Imagine you are zwitterion.
So here's zwitterion, sitting there, minding its own business [BREAK IN AUDIO]
--for a minute.
Why?
Because it's in a neutral pH solution.
Minding its own business.
That's a chemical joke.
All right.
So there's zwitterion, just sitting there, minding his own business.
And all of a sudden, the pH rises, and the proton concentration drops.
So now we've got an abundance of hydroxyls.
We're in an aqueous solution.
So the zwitterion looks around and says, I've got to fix this, but there's no hydroxyl attachment site!
So the zwitterion can't gobble up the hydroxyls.
So what's the zwitterion do instead?
Instead, it starts shedding protons.
Because if it sheds protons, then what can happen is, if it forms this plus proton, then the proton can react with hydroxyl to form water, and then shoot the pH back to near neutrality.
So let's dump the proton here, and we'll just end up with H2 and CHCOO minus with the R group.
So this is deprotonation in order to address the sudden rise in pH.
And now we can write a K2.
Let's call this reaction 2, and with K2, since it's a deprotonation reaction right away, we can just go ahead and write, this is going to be H plus.
And what's left?
This is just a denuded, deprotonated zwitterion, just the A minus, all over neutral zwitterion.
And it looks like this.
And we can invoke Sorensen and get PK2 equals pH plus log base 10 of the neutral zwitterion over the deprotonated zwitterion.
And there you go.
And then, of course, that 50-50 concentration of deprotonation, the pH gives you the value of PK2.
And now, at one end, I get deprotonated.
The other end, I get fully protonated, right in the middle.
I get the 100% zwitterion.
I said at neutral pH, but that doesn't mean 7.
It means the pH at which this particular species is stable.
And that's different from 7, and it's somewhere in between, isn't it?
Because if I go to very acidic solutions, I will start to protonate.
If I go to alkaline solutions, I'll deprotonate.
So somewhere in between.
And that somewhere in between is called the isoelectric point.
The isoelectric point is the place at which we get the neutral.
So that's HAmax.
The concentration of this is maximal at the isoelectric point, and that's called P sub I, and that's given by PK1 plus PK2, divided by 2.
And that's it.
You don't have to run it.
We're done.
Thank you.
OK.
So this is the simple story.
It gets more complicated if, for example, the R group can protonate or deprotonate.
So now we got two or three-ring circus.
It gets more interesting.
But this is the start, and there's some homework problems that will help you muddle through.
OK.
So I think this is a good place to switch into the final five minutes.
So here's the titration curve.
I'll show that at the beginning of the next lecture.
OK.
So now let's talk about kinetics.
And the extreme in kinetics, I said, is an explosion.
And we're coming upon December 6, which is the anniversary of the largest explosion in the history of the world before the nuclear age.
And it occurred in 1917, and Halifax Harbor, Canada, at that time was a British colony, and therefore Canada's foreign policy was dictated by Britain, which was involved in a war.
And there was a French supply ship and a relief ship, and they collided in Halifax Harbor.
This is what the French supply ship had on it.
Notice 20,000 tons of TNT.
At 8:45 AM, the Mont-Blanc is hit by the Imo.
But the TNT was not ignited, and instead we ended with sparks and a fire.
By the way, this is TNT.
If you look, this is toluene.
Toluene is methylbenzene.
It's just the methyl group.
And this is trinitrotoluene.
And if you go to Wikipedia, Wikipedia gives this formula, which is wrong.
The methyl group is missing here.
Just a word to the wise.
Be careful when you use that tool, I put in quotation marks.
Because it's worth about as much as you pay for it, which is why we teach you to use the bibliographic things.
OK, so there it is, right here.
I didn't make this up.
Look at this.
Completely wrong.
Anyway, so what happens?
"The crew of the Mont-Blanc, aware of their cargo, immediately took the lifeboats, screaming warnings that no one heeded.
They rode for Dartmouth"-- this is across the channel from Halifax-- "leaving the now furiously burning ship to drift towards Halifax, propelled in that direction by the Imo's impact.
The Mont-Blanc drifted by a Halifax Pier, brushing it and setting it ablaze.
Members of the Halifax Department responded quickly, and were positioning their engine up to the nearest hydrant when the Mont-Blanc disintegrated in a blinding white flash, creating the biggest man-made explosion before the nuclear age."
It was 9:05 a.m. on a Thursday morning.
"Over 1,900 people were killed immediately.
Within a year, the figure climbed to well over 2,000.
Around 9,000 were injured, many permanently.
325 acres, almost all of North End Halifax was destroyed.
Much of what was not immediately leveled burned to the ground, aided by what are stockpiles of coal in cellars.
As for the Mont-Blanc, all 3,000 tons of her were shattered into little pieces that were blasted far and wide.
The barrel of one of her cannons landed three and a half miles away.
Part of her anchor shank, weighing half a ton, flew two miles in the opposite direction.
Windows shattered 50 miles away, and the shock wave was felt even in Sydney, on Cape Breton Island, 270 miles to the northeast.
"There was about 20 minutes between the collision and the explosion.
It was enough time for spectators, including many children, to run to the waterfront to watch the ship burning, thus coming into close range."
I mean, they were little kids, and it's a fire, and the fire engines are coming!
It's fun!
It's more interesting than going to school, right?
"It was enough time for others to gather at windows.
Office workers came to the windows, and with the explosion, the glass shattered and blinded them.
"Not surprisingly, the hospitals weren't able to cope with so many wounded.
There was also a desperate need for housing, and the misery was compounded"-- are you ready for this?-- "by the blizzard that struck the city the following day, dumping 16 inches of snow over the ruins.
"With astounding speed, relief efforts were set in motion.
Money poured in from as far away as China and New Zealand, Canadian Government gave--" blah, blah, blah.
"But most Haligonians"-- which is what you call a denizen of Halifax-- "most Haligonians remember the generosity of the state of Massachusetts, which donated $750,000 of money and goods, and gave unstintingly in volunteer assistance through the Massachusetts Halifax Relief Committee."
People from the hospital district here got on trains, and went up there to assist, especially from Mass Eye and Ear, to assist the people that had been blinded.
"Starting in 1971, to this day, Halifax sends an annual Christmas tree to the city of Boston in gratitude."
And there are specifications on what the tree should look like, and how big it should be, and so on, and they bring it over to Boston Commons.
So if, in the days that come, when you get back from Thanksgiving, if you happen to be across the river and you see something that's about 45 feet high and it looks like it came from Canada, this is what it is.
And it's all about the Halifax explosion.
OK.
There you go.
All right.
Well, have a nice weekend.
We'll see you on Monday.
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OK, OK.
The weekend is over.
One announcement.
Last weekly test will be tomorrow, and it'll be based on homework 11, focused on polymers.
And then the following week is the very last week of semester, and there's no testing, no work required of you, when the subject has a final exam.
And this one has a big final exam, big celebration, so there will be no weekly test next Tuesday.
So this is the last one, so get out the Kleenex.
All right.
So today we want to continue the treatment of biochemistry.
Last day we started looking at proteins, and the building blocks of proteins are amino acids.
And so we looked at the basic structure of amino acids, shown here, which is an sp3 hybridized central carbon, which I've designated alpha carbon.
And three of the four bonds are the same in every amino acid.
Every amino acid has, off the central carbon, an amino group.
This is the nitrogen with two hydrogens.
It has a hydrogen, a lone hydrogen, and it has a carboxylic acid, which is this COOH.
Double bond here, I'm looking at four sticks off the carbon.
One, two, three, four.
And this proton here is the one that can fall off and go over to the side.
And when it does so, then we end up with the acid.
It's a proton donor.
And then the fourth bond is this designated R. R is the substituent, and it could be anything.
We could put a cyanide up there, if we want.
Anything that conforms to this architecture with whatever choice of R is still qualified as amino acid.
There are only twenty choices of R that are found in protein.
We also studied chirality, which is handedness.
Some of these amino acid molecules are chiral.
That is to say, if you put this carboxylic acid in the lower left, and the amino group in the lower right, you have molecules that are seemingly identical, but not superimposable.
We call them chiral.
And the two forms are called enantiomers.
And only the L-enantiomer is found in proteins.
And we saw that the L is determined on the basis of polarization rotation of light in an aqueous solution of this.
Except for lycene.
In the case of lycene, we have H up here.
So we have an H above and an H below.
That molecule is not chiral.
So nineteen of the twenty amino acids that we find in proteins exhibit chirality.
And later in the lecture we started looking at aqueous solution behavior of the amino acids.
And what happens is, when they dissolve in aqueous solution, they form zwitterions.
Zwitterions by proton transfer.
So the proton goes off the carboxylic acid and joins the amino end, so you have a plus end and a minus end.
And then we looked at what happens with the behavior in aqueous solution, and we came up with the Henderson-Hasselbalch equation.
And here it is, plotted.
This is the case for alanine.
In the case of alanine, we have the methyl group here.
So the R is CH3.
Everything else is the same.
So you can see that here's the neutral zwitterion, where the H off of the carboxylic acid, has transferred over to the amino end.
We have charge neutrality.
The molecule started off net neutral, and now the plus is over here, conferring a positive charge to this end, and leaving this end less a positive charge, and therefore minus.
So it's globally neutral, but locally positive and negative.
Hence the term zwitterion, because it's essentially dual gender, both positive and negative.
And we looked at the aqueous solution behavior of the zwitterion, and we saw that zwitterion is responsive to its environment.
And we reasoned that this is how you start to impart animation to something.
How do you get something animated?
You get something animated by having it respond to its environment.
A rock sits there.
You can talk to the rock, nothing happens to the rock.
But animated objects, if they're endowed with hearing, if you talk to them, sometimes they respond.
So what's going on there?
It to do with the ability to respond chemically.
And so what we saw here is, when we went into low pH regime-- low pH means very, very high proton concentration.
The zwitterion responds by the le Chatelier principle to try to minimize that acidity.
So it gobbles up protons.
And over here we see that it gobbles up protons by attaching to this carboxylic acid end.
That's a proton attachment site.
And in the extreme, all of the zwitterions have been capped by protons.
So over here we have complete protonation, and that's what we're showing in this graph here, which is just basically the one that's up on the screen, but I'm using different-- see, this is taken from one of the biology readings.
They say equivalence of hydroxyl.
There's no hydroxyl here.
I don't know how these-- you know, the language of some of these branches of science puzzle me.
That's equivalence of hydroxyl, but we don't have hydroxyl!
It's all about protons.
So here it is, in terms of the equivalence of H plus.
So here's the neutral zwitterion.
That's this thing here.
And this H, if it falls off, it can then become the neutral zwitterion.
And then what happens is, as we go to very, very low pH, what happens is that the zwitterion starts attaching protons.
So this is a neutral zwitterion to which we've attached the H. So it's much-- to me this is transparent.
What this reads-- I don't know.
Equivalence of OH?
There's no OH.
That's a way of saying it's fully protonated.
BUt to me that's sort of a backwards way.
Don't tell me what it isn't.
Tell me what it is.
So this is fully protonated over here.
And as the pH rises, the zwitterion feels less compelled to gobble up the protons, so it sheds protons.
And finally the pH gets to a point that's high enough that the zwitterion plays no role, and it exists as the neutral zwitterion.
That's the isoelectric point.
We have a 100% concentration of zwitterion.
Then we go to the other extreme.
The other extreme is the high pH regime, where it's proton deficient.
And so zwitterion responds by-- what?
Throwing out hydroxyls?
No!
In spite of what's written here, there are no hydroxyls.
What zwitterion does, is it tries to shed protons to compensate for the proton deficiency.
And so some of the HAs become just A minus, because the H's are being liberated.
And over here at the extreme, all of the zwitterion has lost its H here.
And so now you have something that's net negative.
Over here you have something that's net positive.
In the center it's net neutral.
So this is the high pH regime, low pH regime, and this is the Henderson-Hasselbalch equation, which answers the question, what is the ratio of neutral zwitterion to either the deprotonated form or the protonated form as a function of pH?
You tell me the pH, and I'll tell you what the ratio is.
Last thing that's really cool about this, and shows you the power of this mechanism of attachment or detachment, is as follows.
Look at here.
This is not exactly to scale, but it's reasonable.
You see the pK?
This is pK1.
This is for the protonating reaction.
So over here I have neutral zwitterion, over here I have no neutral zwitterion, all the zwitterions are fully protonated.
And in between, I've got-- this is supposed to be sort of halfway across here.
So this is 50-50.
But look at what happens.
For very, very large changes in ratio of zwitterion to protonated zwitterion, the pH doesn't change that much.
The pH changes a lot here, and it changes a lot here.
But over a vast expanse of protonation or deprotonation, the pH kind of hovers around pK1.
So the zwitterion is really doing its job.
It's trying to hold that solution from shifting pH dramatically.
We say that zwitterion, by doing so, acts as a buffer.
A buffer holds pH near constant.
And the near constant value it's going to hold it to is its pK.
So in the acidic regime, in the very low pH regime, over a broad range of protonation deprotonation, it's going to be roughly a pK1.
And the same thing happens over here.
You see?
Over a broad range of A minus to HA, the pH kind of holds around pK2.
And then finally, over here, the system gives up and then it shoots up to the 100% deprotonated.
So that's the story there, and you see the isoelectric point in the middle.
Now, so far all we've done is we've said what happens when we have a simple amino acid that has just the one proton attachment point on the carboxylic acid and the amino acid.
But it can get more complicated.
And there is an example in the homework that you might want to look at before the final exam.
And that's the case where the R group itself has the ability to protonate or deprotonate.
Suppose this R group had a COOH ending here.
Can you see that it could also participate?
And so then the isoelectric point isn't simply going to be the average of the pK1 and the pK2.
Because now there's a pK for this thing, as well.
So this is the example of the tritratable-- that's the term they use.
If I could just spell it, I could convey the information.
Titratable R group.
That's what they're talking about.
Otherwise, if the R group is a neutral, it's just a spectator.
Doesn't doesn't participate any more than this.
See, here's a hydrogen.
Look at this one, people.
You see this hydrogen?
It doesn't participate at all.
Please don't tell me at some point that the zwitterion says, I'm out of hydrogens from here, I'm out of hydrogens for here, I'm going to start shedding this.
This is a really strong covalent bond.
It's not going anywhere.
If you want titration, you've got to put the titratable R group up here.
OK?
This is so cool.
So now, I'm going to show you an application of this stuff that's used in biology.
And I know, because one year they gave a crash course for engineering faculty in biology, and we did we actually did this experiment in the lab.
So I'm going to show you how you can use the fact that zwitterion changes in response to its environment in the laboratory.
So what we're going to do is I'm going to make a column here.
This is a column, and it's an aqueous solution, a column of varying pH.
So it starts at the top, and just for argument's sake, I'm going to put high pH here at the top, and low pH here.
So it's an aqueous solution.
You can say, wait a minute.
He just taught us fixed laws.
This is proton deficient this is proton rich.
Aren't the protons going to go north and the hydroxyls are going to go south?
Yes.
So what are we going to do?
We are going to slow down diffusion.
We're going to change the diffusion coefficient, and what we're going to do is, we're going to fill this with a solid suspension and turn it into a gel.
So what's the function of the gel?
Don't say that Professor Sadoway said that he stopped diffusion.
Diffusion goes on, but it goes on at a slow rate.
So over the time duration of the experiment, for all intents and purposes, the pH changes imperceptibly in comparison to everything else that's going on.
It's all about time scales.
You realize, we are all aging.
We've been aging ever since I started this lecture.
But for all intents and purposes, we can view ourselves as being the age we were when we walked in the room.
And even over the semester, chances are.
All right?
Now if we meet back here in ten years' time, expect to see some changes.
Right?
But over the time scale of the experiment, we can say we take diffusion out.
So that means the high pH is here, the low pH is here.
Now what I want to do, what is it I'm trying to accomplish?
I want to separate an amino acid mix.
And just for grins and chuckles, we're going to have A1, A2, and A3.
Three different amino acids.
Which means just three of these things, but three different R values.
I'm going to pour them the top.
I've got this aqueous solution, got all three amino acids.
So what happens when they fall in here?
When they fall in here, they start diffusing, too.
Right?
Well, first of all, they hit a very, very high pH regime.
So what what's the immediate response to those amino acids in a high pH regime?
They're all amino acids.
They're all those rescuers.
And they say, Hi, pH.
It's proton deficient?
I can help!
And they all start shedding protons.
And so within moments, all of those amino acids have been totally-- as they say in California-- totally deprotonated.
And here it is.
So they're all sitting there, in net negative.
They're all sitting there in net negative, right up here.
But I just told you, I shut off diffusion, slowed it down.
But I want to get out of here before dinner.
So I don't want the protons moving, but I do want the amino acids to get moving.
So what do I do?
I put on my elecrochemist's hat, because as you know, electrochemistry is the highest form of chemistry, the most noble form of chemistry.
It happens to be the area that work in.
And so what we do is we electrify the ends.
And we'll polarize the top negative and the bottom positive.
Well, I just poured in a bunch of amino acids that became net negative, and now they're next to the negative electrode.
So what happens?
Now they're in an electric field, and they start moving.
They start moving in the electric field, and they're all net negative.
But what happens, is they move down here.
So this is medium pH.
Can you see?
I've established the pH gradient.
And as they move from high pH to medium pH, the proton deficiency isn't as intense.
So they're responding to their environments.
They're saying, you know, back there I had to get rid of all my protons.
But now, it doesn't feel so proton deficient, so I can start taking protons on.
They respond to their environments.
So here, they're all A minuses.
And if they ever get down here, they'll all be HAA pluses.
And so what are they in here?
They're sort of on their way to becoming HAs.
Because the pH is continuously variable.
So as they become less negative, what happens to the driving force by the negative electrode?
It's weaker.
So the protons were, first of all, they're anions.
They're net anions.
They go, whoa!
Negative!
And they start moving, and then they get less and less negative, and the field gets weaker and weaker.
And eventually, they take on enough protons that they become neutral zwitterions.
And what is the response of a neutral ion in an electric field?
Nothing.
It stops.
And so let's say, this happens to be the isoelectric point of amino acid number 1.
So it gets to the isoelectric point, and it doesn't move anymore.
And number 2 and number 3 keep going, and this might be the PI of have number 2, and this might be the PI of number 3.
And now I want only amino acid number 1 from this mix, I can now go in-- this is a gel.
It's put down on the glass plate, or you can put it in a column.
Now you where they are.
You come in-- because you've got a pH meter, you know where this is.
You know the PI.
Bingo.
That's where they are.
And if I want number 2, they're right here.
And number 3, they're right here.
So this process is called gel electrophoresis.
This is the term.
It tells you nothing about the process.
It tells you how to conduct it.
You make a gel, and you polarize the ends.
Electrophoresis is simply particles moving in an electric field.
You can paint a car this way.
Obviously the car, when the car is made of metal.
If you have a fiberglass body, you can't use electrophoresis.
The car is made of metal, you polarize the car, and the paint, the paint has got dipoles.
And so the paint sticks to the car.
And then you lift it out, let it drip, and then just get the right amount of paint proportional to the charge you put on the car by electrophoresis.
What's that got to do with this?
It doesn't talk about the focusing here.
So the mechanism here is called isoelectric focusing.
So isoelectric focusing is the principle by which we can separate the amino acids from one another.
The gel electrophoresis column is a device in which we do so.
And let's be modern, so we'll give it a three-letter initialization.
IEF, isoelectric focusing.
Isn't that cool?
Yeah.
I thought so.
Anyway.
OK. Everybody's still zoned out from turkey or something.
No response.
This is brilliant.
It's beautiful, and it integrates all of this material, and deadpan.
All of this withholding love.
You've been with your families, I can tell.
All of the old family dynamic, see?
It's here.
You brought it with you.
Leave it!
Leave it at the airport.
OK. Now.
So we've been talking about the amino acids.
We said we're going to use these as building blocks for proteins.
Let's build some proteins now.
So protein synthesis.
So we're going to make macromolecule.
What are we making?
We're making a macromolecule.
Protein is a macromolecule.
You notice I didn't say polymer, because a polymer going to have-- every element along the backbone is going to be the same.
But I'm telling you that we can change the choice of R at every station.
So it's a macromolecule, but it's not necessarily a polymer.
But we do know, if we're going back a macromolecule, we've got two choices.
Addition or condensation polymerization.
So I look at the basic structure-- let's put it back the way we found it, so that people don't leave thinking that we have two carboxylic acids.
So the most the most general form doesn't have a titratable R group, but it does have an R up here.
So I want to take one of these and link it to another one.
So how do I do it?
Well, right off the bat, addition polymerization is no good.
Because even if I had enough energy in my body to form a radical, what could I do with this?
There are no double bonds here!
There are no double bonds here.
I can't break a double bond and then allow this to bond to its neighbor.
So axiomatically, I have to go with condensation polymerization.
OK?
Macromolecule formation by condensation reaction.
I don't even want to say the word polymerization.
By condensation reaction.
OK.
So let's look at an example here.
An example will be this.
I'm going to write, here is the COO R. So I'm going to show this, and then over here-- well, maybe I'd better show some fine structure.
I'll give you a little more fine structure.
So we'll put the double bond up here, O minus.
And then over here, I'm going to put H3N plus.
So here's the neutral zwitterion, agreed?
And I've got some kind of an R group up here.
And I'm going to put next to it a second one of these.
So here's the alpha carbon, and we'll go COO.
And then over here, I'm going to unpack this nitrogen.
It's got one, two, three hydrogens, and it's net positive.
So there's the neutral zwitterion.
So now by condensation reaction, what we'll do is we'll take this oxygen off of the left amino acid, and we'll take two of the hydrogens off of this amino acid, and we will eject a water molecule.
We'll eject the water molecule.
And that's safe.
It's compatible with the body.
And now what's left?
This carbon is naked.
One, two, three.
There's a bond sitting here that used to go to the oxygen.
And this nitrogen, it's sitting naked too.
And so now we can do, is we can make a link between this carbon and this nitrogen.
And this bond is called the peptide bond.
This is the peptide bond.
OK?
And we see that the atomic mass of the dipeptide is less than the sum of the masses of the individual components, because we've lost the water.
So I think I've got some slides here.
What's here?
Oh.
This is just the various, pK1 and pK2.
And then this side chain here, this is the value of the pK should you have a titratable R group.
That's why there's the third column here.
So when you see this-- you see, they're all roughly around two for the highly acidic regime, and all around high nines, low tens for the alkaline regime.
Roughly the same.
All right.
Now connotation polymerization.
This is the formation of nylons.
You see how you bond this mer to the adjacent mer?
Look at this.
The nitrogen, that hydrogen is sticking up.
There's the nitrogen bonding to the carbon.
There's a long pair, and nitrogen-carbon.
Amide bond, peptide bond, same thing.
The same bond in the backbone of nylon is the same bond in the backbone of protein.
Only, the biologists call this the peptide bond, the polymer scientists call this the amide bond.
Think about that when, you know, the holidays are coming, you think you're really special, and you know, the proteins in your body are really bound the same way that nylon is, you know?
All right.
So now let's consider-- and here's some examples.
Here's a peptide.
So it's a long chain here.
So it starts, here's the amino group, H3N.
It's got glycene, and then leucine, and valine, glutamine, et cetera.
So there it goes, all the way around, around, around, around, around.
And here's another one.
There's the amino group, and then, boom, boom, boom.
What's happening here?
All of these click stops along here are this.
H2NCHCOOH.
And the only thing that's changing is the R group.
If this were nylon, it would be the same thing, all the way down.
That's how this differentiates itself.
All polymers are macromolecules.
All macromolecules are not polymers.
These are not polymers, but they are macromolecules.
And what else do you see here?
Look at this.
Cysteine happens to have a double bond.
A residual double bond in its R group.
Well, where you see a double bond, you go, I can break the double bond, and then I can make a another bond.
And that's what happens.
So we break the double bond with the sulfur, and now we make a disulfide linkage.
What do you think the mechanical properties of this protein are?
Mechanically, what do you think this one's like?
It's rubbery.
This is an elastomer, because you can move the top chain vis-a-vis the bottom chain until these disulfide linkages are flattened.
And then if you let it go, it springs back.
It's the mechanical properties of, in this case, this is insulin.
Now I want to give you a sense of the variety here.
How much information-- because ultimately, it's all about reproduction, right?
That's the number one function of any living organism, is to reproduce.
That's what makes it life.
Right?
So it needs an instruction set.
And the instruction set doesn't come in the box.
It comes inside the organism.
And this is the language.
This is all we have to work with.
So here's a simple example.
This is, how many combinations do you have of dipeptides?
So when you start with two amino acids.
So this is phenylalanine and aspartic acid.
So if I told you, you have A and B, and how many AB mixtures can you come up with?
You'd say, well, you've got AA, you've got BB, and you've got AB.
You know, it's sort of like calculus, right?
If you square A plus B, you'd get A squared, plus 2AB, plus B squared.
But because of chirality, AB is distinguishable from BA.
This is cool.
This is, again, math serving us, instead of we serving math.
All right?
So A plus B squared is A squared plus 2AB plus B squared.
But no, no, not in biology.
Because AB is chiral, and therefore distinguishable from BA.
Now you understand chirality.
So here's-- you get four.
Because see, in this case, phenylalanine peptide to aspartic acid is distinguishable from the one on the right.
They're distinguishable.
They're different.
OK, so that let's roll with that.
Let's say, so if I start with two amino acids, and I want to make dipeptides, well, that's trivial.
That's 2 squared, is 4.
So I got four different dipeptides, starting with two amino acids.
Now, suppose I said, how many different dipeptides can we make, period?
Well, I know I've got a library of 20 amino acids that are found in proteins.
So that means the number of dipeptides would be then 20 squared, which is 400.
But that's hardly a macromolecule.
So let's get into the macromolecule business.
So suppose I took a library of twenty amino acids, and I said, how many kilopeptides could I make?
What's a kilopeptide?
A kilopeptide is something that's got 1000 peptide bonds in it.
So that's going to be 20 to the power of 1000, which is 10 to the power of 1300.
And this is small.
n equals 1000.
I'm not stretching the point by pulling something exceptional out.
A polymer, if it were polymer, 1000 units long, you'd say, that's sort of fair to middling length.
I'm not pulling out the longest polymer ever known.
I didn't go to the Guinness Book of Records for this.
This is just plain vanilla mer length.
So look.
How much information can be contained?
That's in a single element.
OK.
So later on we're going to return to this, and appreciate how we can get to such information density.
And this I think I alluded to on Wednesday.
These are the three-letter initializations, and I'll post this at the website.
So these are the airports that correspond to the three-letter initialization.
Then I showed you the one-letter abbreviation for each of the peptides.
You know, it's kind of strange, because the amino acids are what they are.
And most airports, they're not like the main ones.
It's not like Boston or New York.
It's weird places like Australia and Texas and stuff.
You know?
So-- hey, you've got to have a sense of humor!
Come on.
All right.
And there's glutamic acid.
There's no airport that has GLU as its three-letter-- OK.
So here we are.
So now we want to do is want to talk about protein structure.
And so we're going to look at this as material scientists.
We're not going to look at it the way chemists, capital C-chemists, do.
And I think by looking at it from a structure perspective, you get some insights into what we can do with these things.
So protein structure.
And I'm going to use the terminology that the biologists use.
So first of all, they have something they call the primary structure of the protein.
And it speaks to composition, the composition.
Well, what's the composition?
We know that every protein is made of amino acids, and every amino acid is virtually identical, except for the R group.
So when you say the composition, you're really saying, what is the R sequence?
It's the amino acid sequence down the backbone.
Because you've polymerized this thing, or macromolecularized it.
Amino acid sequence, which along the backbone.
Which is really the R groups.
The sequence of R groups, which is going to be the sequence, OK?
All right.
So here you can see.
There's an amino acid sequence.
Number two is the-- these are really-- you know, like the polymer people had conformality.
You know, they had nice, powerful words.
The first order is the primary, and then the second one is the secondary.
I'm trying to make fun of the biologists, and you're sitting there, you know, with that same Thanksgiving dinner table stoicism.
OK.
So what's the secondary?
The secondary structure of proteins speaks to packing.
And what's the gambit here?
Why do they pack in different ways?
Well, just as we saw, you can have a long chain, straight chain.
Sometimes things full back on themselves.
It's still a straight chain, but there's a curve, a bend in it, and so on.
What proteins are going to do in order to pack, is to try to maximize hydrogen bonding.
And this is not even between proteins.
This is a protein forming on itself.
And you could argue, why do you maximize hydrogen bonding?
Because when you make more bonds per unit volume, you decrease the energy of a system.
You've been taught that over and over again.
So these are long chain molecules, but they still can loop back on one another.
And if you've got a choice of forming a dipole-dipole interaction, a induced dipole-induced dipole interaction, or a hydrogen bond, which is going to give you the most decrease per unit bond?
It's hydrogen.
So if you can form hydrogen bonds, that gives you the most impact per bond.
If you don't have hydrogen bonding capability, then you go for dipole-dipole.
And if you don't have dipole-dipole, then it's just the weak van der Waals, or the London dispersion.
So that's why this thing goes for hydrogen bonding.
Because it wants to.
It wants to decrease the energy.
So I have to show you a few cartoons here.
All right.
So here's a sketch taken from one of the readings.
So there's alpha carbon.
This is the carboxylic acid, right here.
And this is the peptide bond between one carbon and a nitrogen in an adjacent unit.
All right?
So there's the peptide bond.
The interesting thing is that even though this is sp3 hybridized, data show that all six atoms lie in a plane.
So what happened last time when we came up with this conundrum?
sp3 hybridized, but the data showed it all lies in a plane.
That was benzene, remember?
And how do we skate with the puck on benzene?
Resonance.
We said it resonates between two structures, and sometimes the nitrogen lies in the plane.
Sometimes the alpha carbon lies in the plane.
But thanks to resonance, all six of these can lie in the plane.
OK.
So now how do we get to hydrogen bonding?
So there's the resonance.
All right.
Good.
So you see, this is free to rotate, right?
Once we accept that these six atoms line in a plane, and these six atoms line in a plane, and 109 degree angle is fixed, but there's that degree of rotation, which is what gave us the original C17H36 folding back on itself, only we're going to fold back on ourselves in groups of six.
So we've got a deck of cards here.
I've got six here, six here, six here.
So what am I going to do?
I'm going to try to make this blue plane lie relative to the yellow plane in such a way as to maximize hydrogen bonding.
Speaking of energy deficit, we need battery.
All right, here we go.
So now we're going to maximize.
Look.
You see, if I rotate this blue plane in just the right way, I can set up to have a hydrogen bond form between this hydrogen and this oxygen, and on the orange plane, an oxygen and a hydrogen.
So now it becomes a matter of figuring out, what are these angles?
You see, there's a phi and a psi here.
What are the angles phi and psi to give the maximum hydrogen bonding between these two six-packs of atoms?
And from that, you get secondary structure.
It's all about maximizing hydrogen bonding.
After that, if you understand what I just said, the rest is trivia.
So here's what happens.
Here's one structure.
And this was first discovered, if you like, revealed, by Linus Pauling, 1951.
Linus Pauling and Corey, 1951, came with the realization that if the protein spirals in a helix, it will line up the maximum number of hydrogen-oxygen hydrogen bonding opportunities, as opposed to going straight.
So it doesn't go perfectly straight, it coils.
And this is two different cartoons trying to show.
So you can see peptide bonds, and then every so often you get a hydrogen-oxyge bond.
Because it spins around, and the idea is, as the helix forms, you can think of it as-- have you ever gone up a spiral staircase?
And you think of each subsequent turn of the spiral as a new gallery.
And what you've got is hydrogen bonds between the galleries.
So you're going up, and there's a hydrogen bond here, and you keep turning, and there's a hydrogen bond here, and there's a hydrogen bond here.
That's the way to get maximum hydrogen bonding.
There it is.
There's another example.
Maybe this is a nicer one.
And there's 3.6 residues per turn.
That means, per turn is 3.6 R groups, which, it's called residues because historically, when people did the chemical analysis, the residue contained the substituent.
That's how they determined what the identity of the protein is.
They can't just look at it.
And you can't use spectroscopy.
Maybe now you can, but in the old days-- I mean, these are all, look.
Low mass.
Hydrogen, oxygen, nitrogen.
They're not distinguishable.
They're all low-Z elements.
So they had to do wet chemistry, and they would actually peel off the R groups, which were called the residuals.
And so 3.6 residuals per turn give you this alpha helix structure.
there's a second structure that works.
The second structure is between two chains.
So in this case, instead of going coiling by yourself, and forming the galleries, what you do is you take two chains side by side, and you pace them so that you get maximum amount of hydrogen bonding.
I showed you this with nylon, how two strands of nylon bond together.
They can form hydrogen bonds.
Same thing here.
And now, since these are jerking in a motion of six in a plane, six in a plane, six in a plane, six in a plane, what you end up with is this pleating.
It's hard to see here.
Maybe if your eye can follow along the top here.
Here's sort of a cream-colored one, orangey-cream, and it's moving sort of southwest to northeast.
And here's one that's sort of a grayish, and it's moving from northwest to southeast.
And then up, and then down.
And zig and then zag.
And this is called the pleated sheet.
This is the beta form.
The alpha form is the helix, and the beta form is the pleated sheet.
Same thing up here in a different kind of model.
Which I look at that and I go, I don't get any information from this thing at all.
But they put it in the books.
Who cares.
This is where you learn something, right here.
Once you appreciate that those six atoms have to be in a plane, and how to maximize hydrogen bonding.
So let's put that down.
This is all Linus Pauling.
So secondary.
So you know that six atoms in plane, which leads to resonance.
And now, from there, you just needs the genius of Linus Pauling, and then you conclude that to maximize hydrogen bonding, you form the alpha helix with hydrogen bonding between galleries.
All right?
And then the second one is the pleated sheet, or people just call it the sheet.
But to me, the sheet means nothing!
It's the fact that it's pleated.
So I'm going to ignore what the books say.
Pleated, pleated sheet.
OK.
So this is hydrogen bonds between galleries of the same protein.
Whereas this is hydrogen bonding between macromolecules.
So if they're both pleated, it's going to work.
Or of course, that could be folding back on itself.
Remember how I showed you the first day the chain, how the chain wanted to zig-zag, to crystallize?
Do you think protein is liable to fold?
You've heard of protein folding?
Why does it fold?
Because it wants to maximize hydrogen bonding!
And when it folds, it's not going to maximize hydrogen bonding unless it is in the beta pleated sheet arrangement.
Otherwise, what's the point of folding?
Because you can't make the link.
Because I'm the oxygen, the hydrogen's way over there.
If that hydrogen were right over here, we could form the bond.
And then the third one is random.
So let's take a look.
I think I've got some images here.
Oh, here's another one, showing the-- yeah.
I'm trying to, I'm really working hard.
Here's six, here's six, here's six, and there's the common carbon.
OK?
And now you can see the hydrogen bonds forming between.
So this could be one protein, this could be another protein, or this could be the same protein that's folding back on itself in a pseudo-crystallization ploy.
Yeah.
All right.
So if you look in the biology literature, you'll see drawings like this.
You've seen these, huh?
You look at them and go, what are they talking about?
Now you know!
This is the secondary structure.
Let's start over here at the amino end.
So you're moving along, and it's just sort of random, and now, all of a sudden, look.
Round and round and round.
What are you looking at?
For this run of length, it's in the alpha helix form!
And then it kind of goes random for a little bit, and then it goes green.
And this green area is pleated sheet.
So if the pleated sheet goes this way, and then it goes random, and then the pleated sheet goes this way, what's going to happen?
there's going to be hydrogen bonding here, hydrogen bonding here, hydrogen bonding here.
And you know, how is it that this thing decides, oh, somewhere around here, I think.
I think I'll just go into an alpha helix arrangement.
Why is it alpha helix here, and not alpha helix here?
Because of the instant choice of R groups!
The instant choice of R groups will dictate whether it can turn around.
Because if you've got glycene with its dinky little hydrogen on one side, you're compact and you can move around.
But some of these other pendant groups are big, and you can't just shove them one into the other.
They'll repel.
They won't fit.
So this dictates-- so in essence, the secondary structure is set, the table is set by the primary structure, isn't it?
This is so cool.
Then there's the tertiary structure.
Because random isn't, as they say in California, it's not totally random.
There's some order to it.
So let's look at the order that comes.
See this?
Look over here.
You see up here?
This is supposed to be random coil.
But look at it.
It's kind of dentate there, isn't it.
See, it's sort of tooth-like.
Do you think that the artist just did that?
Why is it like that?
Because of the molecular forces!
So let's think on it.
Oh!
This is a little piece from Guys and Dolls.
1951.
The same your Pauling enunciated this.
I bet you think that when people talk about the chemistry between folks, that it's a '60s phenomenon.
You know, the hippies, drug culture.
No.
It goes way back.
Goes to the '50s.
Listen to this.
This Sky Masterson, and he's going to tell Sarah-- she's the one that's the, she plays the-- what do you call it-- she works at the Salvation Army mission.
And he's a gambler, which in those days was really, you know, like low-life.
And he's flirting with her, and she, of course, has nothing to do with him, and eventually, they fall in love, of course.
But anyway, so she's told him how she's going to meet the man of her life.
And she's very linear.
He's got all of these characteristics, and so on and so forth.
Real linear.
And this guy's a gambler, so he's going to tell her how he's going to find the love of his life.
[BEGIN FILM PLAYBACK] Do you want to hear how a gambler feels about the big heart drop?
No!
Well, I'll tell you.
Mine will come as a surprise to me.
Mine, I leave to chance and chemistry?
Chemistry?
Yeah, chemistry.
Suddenly I'll know when my love comes along.
I'll know-- [END FILM PLAYBACK] So there you have it.
From 1951.
Chemistry, again, chemistry!
Chemistry is the central thing.
OK. Now let's go forward.
All right.
So now let's figure out why-- and now I want to answer the question, why is this forming a tooth-like structure?
Why?
There's a reason for everything.
All right.
So here we are, moving along a length of random coils.
So we're in zone three here, random.
But it's not completely random.
So this is the tertiary structure, and it has to do with interactions between the R groups.
So let's look at number one, zone one here.
We've got two cysteines, and cysteine both have sulfur in them.
It turns out that the sulfurs can form a covalent bond here.
This is between two R groups along a certain length of chain.
And once that covalent bond forms, this chain now can't just go flopping any which way, because it's being held in place.
It's like you put a prop there.
Over here, what do we have in zone two?
We have one R group that's got a hydroxyl.
Another one's got an oxygen.
There's a hydrogen bond forming between side groups.
And in order to form that hydrogen bond, can you see that this coil is actually bent?
You see, there's a kink here.
This coil should be much longer, if it were straight, distended.
But it's actually compressed a little bit in order to facilitate the formation of this hydrogen bond.
Which means now, this coil will stay in this configuration.
And then over here, what do we have?
Here we have a substituent group that's net positive, here we have a substituent group that's net negative, and we have an electrostatic force.
This is coulombic.
This is plus attracts minus.
And they form.
And over here, this is very interesting.
What do we have here?
We have a whole bunch of R groups that are all non-polar.
And this thing's sitting in an aqueous solution.
And what do you know about non-polar groups in water?
They're hydrophobic!
So I've got this one hydrophobic group sticking out over here.
It's just going to have to take it.
Because it's bound to the backbone.
But I've got another one over here, and the backbone is flexible.
And I've got another one over here.
Can you see that if the backbone folds around, I can collect all the hydrophobic entities, and present to the water the hydrophilic opposite side of the backbone.
Because if the R group here is hydrophobic, look what's on the other side.
These guys.
So if I can take all of these and cluster them, then I sort of minimize the energy.
Because that there is a repulsive energy term, right?
The water doesn't like the non-polar R group.
So let's minimize their contact, you know?
If you've got two siblings that are fighting, you put them at the other end of the table.
You don't sit them next to one another.
So here's what you do.
You cluster all of this stuff, and that causes big hairpin turn.
So this is called hydrophobic interactions.
And that's the tertiary structure.
So you see all of those.
That's great.
OK.
I see we're getting close to the witching hour.
So maybe I'll teach you how to do your laundry and send you on your way.
So this interaction I'm just showing you right here is the same thing that's happening when you do your laundry.
Now, what's a detergent?
A detergent is this long molecule that's got-- you see this zig-zag here?
That's the aliphatic.
That's the hydrocarbons chain.
And it is hydrophobic.
The soil on your clothing is usually some dirt, but it's covered by some grease and oil.
So if you just soak it in water, the water can't get at the grease and oil.
So what you do, is you put these molecules in that are very long, and then they've got this COO ending, right?
And now what happens?
This can form hydrogen bonds to the water, whereas these long hydrophobic tails can bond to the grease and oil.
And then you put this in the washing machine, and what's the washing machine do?
Mechanically agitate.
So you've got these tails, hydrophobic tails, stabbing the grease and oil, bonded to the water, shake the living daylights out of this, and release the grease and oil, which then will release the soil, rinse the whole thing, and you've got clean clothing.
So it's the bonding here.
What you've got is an amphipathic molecule with a hydrophobic tail and a hydrophilic head that allows you to do your laundry.
All right.
Get out of here.
We'll see you on Wednesday.
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All right.
It's 11:05.
OK, let's get started.
Last day we talked about protein structure.
We talked about composition, which is the primary structure of proteins, and that's the instant R group sequence, as you go down the backbone.
The secondary structure proteins is all about packing, and we saw that there were three different structures, the alpha helix, the beta pleated sheet, and the random coil.
And the gambit here is to maximize hydrogen bonding.
And why you choose alpha, beta, or random coil depends upon the instant R group sequence, which is why in some instances, you can't form the coil.
Because the R groups won't allow it.
And then lastly, we talked about tertiary structure, and that's conformation, and that talks about our group interactions, and ultimately explains protein folding.
Today I want to talk about denaturing of proteins, which is disruption of secondary and tertiary structures.
But before we do so, I wanted to show you.
I found another movie reference to chemistry.
This comes from 1957, and the movie is called Silk Stockings.
It's based on an old Cole Porter play.
And this stars Cyd Charisse and Fred Astaire, who were both fantastic dancers.
And I want you to see in this film clip how she carries herself.
And she actually sings this.
She didn't have somebody dub it over.
So the singing is OK.
It's not great, but you've got to see her posture.
She's so graceful.
And Fred Astaire is a terrific dancer, but he doesn't dance in this sequence.
And last thing you need to know here is, this is set during the Cold War.
So she's playing Ninotchka Yoschenka, a good Ukrainian name, and she's a Soviet representative.
And she's fighting with Fred Astaire, who is an American agent.
And they're talking about the differences between the Soviet and the American systems.
So that's the background.
It's 1957.
I guess we better get some volume up here so we can hear it.
And she will explain to you-- and remember, just parenthetically, that my particular interest in chemistry is, of course, electrochemistry, which is the highest form of chemistry.
Now she'll explain it to you.
[BEGIN FILM PLAYBACK] -If you studied, Kamachev, you would know what I am talking about.
-Who's Kamachev?
-He was one of our greatest scientists.
He has proved beyond any question that physical attraction is purely electrochemical.
-You don't say.
-Kamachev has proven it!
-For 30 years, he worked.
-I happen to have worked on the same subject for about the same amount of time, and I have very good reasons for believing otherwise.
-Facts are facts [SINGING] -When the electromagnetic of the he-male meets the electromagnetic of the female, if right away, she should say, this is the male, it's a chemical reaction, that's all.
And though you Fascists may answer with kisses, the same applies when your Mr. And Mrs. Hey-diddle-diddle with middle-class kisses, it's a chemical reaction, that's all.
Say in love, with you, I fall.
And in love, with me, you also fall.
Though the uninstructed faction calls it mutual attraction, it's a chemical reaction, that's all.
[END SINGING] -You don't believe in Kamachev?
-No, ma'am.
[END FILM PLAYBACK]
See the posture?
It's just unbelievable.
Fantastic.
OK.
So let's talk about denaturing of proteins.
So we want to talk about how we can alter the structure.
And we do so by invasive means.
So let's get to the next slide.
So here's the tertiary structure, and I can point to that.
The first thing I wanted to draw attention to, is we can disrupt that structure by change in temperature.
So temperature is the first agent that we can use in disrupting, or denaturing, protein.
And the whole idea here is to break bonds.
But we're only going to break secondary or tertiary bonds.
We're not going to do anything to the backbone.
And a good example is what happens when we fry an egg.
The egg has protein in it.
One of them is ovalbumin.
It's about 90% water, about 10% protein.
And its native state, its natural state, conformation is a tight ball.
And the diameter of that ball is small enough that uncooked egg is transparent to visible light.
Egg white is transparent to visible light in its natural form, but when we heat beyond the denaturing temperature, that changes the bonds in here, and it unpacks.
And when it unpacks, two things happen.
First of all, the length scale changes to be great enough that it scatters light.
So now the egg white appears white.
And secondly, those various chains entangle, and that gives it its rubbery character.
So all of that is happening as a result of the increase in temperature.
So second way that we can operate, is to change pH.
And what that does, is it changes hydrogen bonding and electrostatic interactions.
Now I draw your attention to zone two here.
In zone two, imagine if we take that proverbial drink of cola, and now the pH in here goes way, way down.
You can see this N has got a hydrogen bond over to here.
Now, if there's a proton excess, the proton can cap, because this is a proton attachment site.
If the proton caps this carboxyl, then this hydrogen bond is broken, and now this thing can unfurl.
Over here, you see, this is an electrostatic attraction.
Again, the carboxyl could be kept by a proton, and now the plus minus electrostatic attraction is lost.
So by change of pH, we could denature what we see here.
So there's a good example.
And in fact, I think we do this.
We pickle our foods.
And then this one I have, because of my interest in high temperature electrochemical processing of metals, I've been to Norway many times.
And this is a food that they will never let me eat, because they're afraid if I ever taste this, I'll never return to Norway.
But this is lutefisk, which literally means lye fish.
And lye has high pH.
And this is an example of denaturing protein by going to very, very high pH.
And then you soak it and draw back out and so on.
So there's a good example of food.
Now the third one I want to talk about is oxidizing reducing agents to denature proteins.
And what they do, they can either create or destroy the sulfide linkages, the sulfur-sulfur linkages.
So for example, you see at position number one.
Position number one, we can use oxidizing and reducing agents.
And a good example, that comes from hair.
So first of all, I want to talk to you.
Hair is protein.
And this first example is just, how protein can, one strand can bond to the other by hydrogen bonding.
So up here you have one strand and a second strand, and there are hydrogen bonds in your hair.
Now, what you do when you want to change your hair by blow drying is first you wet the hair.
And the water goes in and interrupts these hydrogen bonds between strands.
So now each of these cross-links that want to form instead terminate with a water molecule.
And now with the water molecules, now you can move the hair one strand versus the other, hold it in place, blow dry, get rid of the water, and form new hydrogen bonds.
So what, on the basis of what I've just told you, how can you account for a bad hair day?
What is a bad hair day?
A bad hair day is a day in which it's rather humid.
And so water can get in from the atmosphere and interrupt these bonds, and then the hair will relax, and go back to a native state that is unfavorable vis-a-vis what you tried to achieve with the blow dryer in the mirror.
So you want to combat this.
So what do you do?
Well, you bring out covalent bonds.
So here's two strains of hair, and there are disulfide linkages between the strands of hair.
So now with a reducing agent, reducing agent is going to be a proton donor.
And can you see, these disulfide linkages are capped with hydrogen, and now they're terminated.
And so now the two strands of hair are free to move relative to one another.
And now when you're at the beauty parlor or the hairdresser, now you can take this hair, for example, and wrap it around-- what do you call this thing?
It's a mandrel in the metallurgical word.
What do you call this thing?
Curling iron or something?
Yeah.
OK.
So here you are.
You wrap this around-- I know what it does.
I don't know the terms, all right.
So now you see, we've got the hair now wrapped around in this fashion, and now we use an oxidizing agent.
The oxidizing agent is peroxide here.
You know the derogatory term, peroxide blonde.
There it is, right here.
So the peroxide goes in, and now it removes these hydrogens.
The peroxide takes the hydrogens away and reforms the disulfide linkages, but now the hair is in this curled state.
You know, you might have the opposite.
Maybe in your native state, your hair, because of the way the R groups are formed along the length of your hair, you know, it's always, the ones with the curly hair want the straight hair, the ones with the straight hair want the curly hair.
And so, you know, maybe you could be running this sequence in reverse.
But in this case, somebody obviously wants curly hair.
Now they form the disulfide linkages.
So this is the object, here, of denaturing.
So now when we go over to here, we have a disulfide linkage, which is forcing this run of the random coil to stay straight.
But if we get, in this case, the oxidizing, the reducing, and what will happen is that we will break this.
So the reducing agents will destroy disulfide linkages, and the oxidizing agents will create disulfide linkages, and thereby we can distort the natural form of the protein.
And here's a fourth example.
There are many others, but I'll just give you four here.
In the fourth one, we can use detergents.
And last day, I showed you the example of how to do your laundry.
And so you've got this long aliphatic tail and the hydrophobic head, and you can imagine if you take such species and you've got this hydrophobic pocket, the species will operate in such a way as to pull these non-polar entities out, and then destabilize this hydrophobic pocket.
So depending on what the nature is of the protein, these various chemical and thermal actions can lead to denaturing.
So this destabilizes hydrophobic pockets.
All right.
So that's pretty good.
All right.
Well, I think this is where I want to leave it with the treatment of proteins.
And I want to move on to a second biomolecule, and that's the lipid.
We'll say a few words about lipids.
Lipids are not classified on the basis of their composition, but rather by their properties.
So they are defined their properties, which is kind of unusual.
Normally we define things in terms of their chemical compositions.
And in particular, they're soluble in solvents of low polarity.
Some of the books say nonpolar solvents, but even something that's polar but only mildly so will work, in solvents of low polarity.
So that's the example.
They're insoluble in water, they're oily to the touch.
And this includes things like fats, oils, these are all lipids.
Cholesterol, hormones.
These are all members of the lipid class.
And so let's look at some examples.
We've got some slides.
So if you start over here with the glycerol, this is a trialcohol.
And you can see the OH.
We're going to drop the H's, and we're going to put these long aliphatic chains on.
There's 14 carbons here, plus the two, so it's 16 carbons long.
And this also is seven plus seven, and then the two carbons.
And the difference here is that in this case, it's all straight chain.
In this case, we go for seven, then we put in a double bond.
And by doing so, when we get to the double bond, that forces everything to go into planar 120 degree arrangement.
So you go for seven however you want, and then there's this rigid 120 degree placement, and then you go for seven however you want.
Well, clearly, the one in the middle is going to pack better than the one on the left.
And so even though they have seemingly the same chemical composition, the one in the middle is a solid.
It's a fat.
The one on the right is a liquid, because it doesn't pack well enough.
The bonds aren't strong enough.
So these are both examples.
And this is called a palmitic acid.
And you can see here oxygen acting as a bridge, as an ester.
So oxygen, over and over again, acting in this way to give us the ability to make these other longer structures.
Now we can replace that oxygen bridge with a dibridge.
So instead of having just an oxygen here, we're going to put a phosphate.
So with the phosphate, we've got a phosphodiester linkage.
There's one ester, there's a second ester.
And why we doing this?
Because nature wants to get the spacing right.
And remember what I just said, because by the end, you're going to see that spacing is everything.
So we can use this, and we can even lose that oxygen, and put something else here.
So here's a-- I'm not expecting you to know these by heart.
I would give you the structure and tell you what it is, tell you its name, and then we can go on from here.
So what do you see here?
Well, this is called a phosphatide.
And this one is a phosphatidylethanolamine, because there's the amine here.
And look at this thing.
Well, you've got a phosphate in the center that's the bridge.
This is the glycerol, the three carbons.
In its most primitive form, we could just have hydroxyls here.
This is the fatty acid off the one end.
Long chains here, not to scale.
So these things should be way, way over to here.
And over here, we have an ethanol amine.
But look more closely.
Well, this is a hydrophilic head, this is twin hydrophobic tails.
Look, this is just hydrocarbon.
It's not soluble in water.
So now you've got this amphipathic molecule.
It looks like a detergent, doesn't it?
This long, hydrophobic tail could stab the grease, and this hydrophilic head can bond to the water.
And now if we shake things up, we'll release the grease that's binding the soil to your T-shirt.
Got the phosphate bridge here, and look even more closely.
There's a minus, and there's a plus!
This is zitterionic, on top of everything else.
So this can function as a zwitterion, too.
And everything we learned about zwitterionic chemistry and buffering, this thing can do.
That's cool.
All right.
So now I want to show what happens when I take a whole bunch of those.
All right?
So let's go back.
I want to keep this one up, and I want to want to do is take a whole bunch of those, and show how lipids can actually operate in order to build complex structures for us.
So what I'm going to do, is I want to represent that molecule.
So here's the hydrophilic head.
That's the amine, right?
And then we've got twin hydrophobic tails.
And that's your carbon sp3 chains.
And I don't want to write that so much.
So I'm just going to simplify that as simply hydrophilic head and hydrophobic twin tails.
So this now represents the twin tails.
Now if I put a whole bunch of these in water, what's going to happen?
First of all, this is hydrophilic.
And what if I were to just pour this into a beaker?
What do you think would happen if I introduced this into a beaker of water?
Well, this is hydrophobic, this is hydrophilic.
Can you imagine that they would all sort of line up like this?
We're trying to stick the hydrophobic tails up into the air.
Because they don't want to be down in here.
This is hydrophobic.
Doesn't want to be here.
So now suppose I put a whole bunch of these in water.
More than four, right?
With only four, that's the best can do.
What if I put some molar?
So then what it does, is it forms a pocket.
It's going to do this.
All the heads are going to find each other, and all of the tails are going to find each other.
Because now they're going to form a hydrophobic pocket.
So they'll do it this way, because they can-- can you see how the heads can sort of make a wall against the water?
And you go on and on and on, not to scale, and then finally you have some end effects here.
So what have I formed here?
I've got something, first of all, I've got a lipid layer here, and a lipid layer here.
So together, I have a lipid bilayer.
In here, there's a pocket.
And now, what I can do with this whole thing, is I can build a cell wall with this.
I've got a cell wall.
I've got an outside and an inside.
So what happens is, when we put all these things together, because of the clustering of the hydrophobic and hydrophilic, we say this thing is endowed with the property of self-assembly.
Very hot topic in material science.
And that leads to cellular structure, and I think we've got a nice cartoon here.
So this is taken from one of the readings.
So you can see this with the plurality of these.
They're actually showing the twin tails.
Now, this was taken from another text.
I like this one, because it shows the hydrophilic top, hydrophilic bottom, and the hydrophobic interior.
And then this thing is called an integral protein.
Why?
Because it's integrated into the cell wall.
And let's think about this protein for a second.
What must be the nature of the R groups in this vicinity of the protein, that it sits where it does?
The R groups around here must be dominantly hydrophilic, otherwise it would get dumped out of that zone.
And in here, it must be dominantly hydrophobic, so that it feels at home with all of these hydrophobic tails.
And over here, it must be hydrophilic.
Now imagine what happens.
I'm going to go back to that drink of Coca Cola.
Forgive me, no brand names.
Cola.
And now the this zone here gets flooded with protons, because the pH is dropping.
So that could cause conformational changes, because pH changes the conformation of the protein, and it could cause this thing to change shape.
It could unfold, or it might unfold in such a way as to open a channel down the center here.
And so in response to a change in pH here, we open up a channel, which means this is acting as a chemical gate.
This is how things are animated!
It responds.
And then once the proton concentration has been depleted and you're back to a more neutral pH, then this thing changes back to its old confirmation, and the gate closes.
It's that simple!
The secondary bonding explains animation.
That's what's happening.
And there's one other cool thing.
Look at this.
If they all have the same head, They're going to close pack!
It's fantastic.
Everything.
Everything you need to know, 3.091 here.
All right.
So what is the key to animation?
The key to animation is self-assembly.
And why do we have self-assembly?
Because we have these molecules with the hydrophilic head and the hydrophobic tail.
These molecules are called amphipathic.
OK.
So this has hydrophobic and hydrophilic components.
So if you take amphipathy, you end up with self-assembly.
That's the key.
All right, good.
All right.
I think that's all I want to say about lipids.
That's all you need to know about lipids, is lipids will give you cellular structure.
Now we've got a little bit of time left, and we're going to talk about nucleic acids.
So what do I want to say about nucleic acids?
Nucleic acids carry information.
They carry information that directs metabolic activity, including replication.
And something can be animated, but it doesn't count as a life form unless it replicates.
That's the characteristic of life forms.
So nucleic acids carry information that directs metabolic activity, including replication.
So these are macromolecules.
Not polymers, but macromolecules.
We'll take a look in a moment at their structure.
They're macromolecules, and the basic structural unit is called a nucleotide.
And it's got three building blocks.
Every nucleotide has three building blocks.
See, what we're doing right now, is we're really tying together all of that chemistry you've been learning the entire semester.
And it's fun to see it actually go to use.
So what are the three building blocks?
Every nucleotide has a sugar.
And we didn't study carbohydrates, but I'll just show you the structure the sugar.
It's got an amine.
You know what that is.
That's the NH something group.
And it's got a phosphate.
So the nucleotide has those three components, So let's take a look at the sugar.
There's really two types of sugars found in amino acids.
There's the ribose-- and the only reason I'm showing you this, is so that you'll understand the terminology.
There's the deoxyribose.
So if you look here, the sugar on the left has this five-fold symmetry, the five-fold ring.
And there's hydroxyls at the number one position, number two position, number three position.
The deoxyribose is missing the hydroxyl at the number two position.
That's the difference.
The reason I'm showing you this, is this is called deoxyribose, and ultimately this is the D in DNA.
So you'll be able to at least hold your own with your course seven major friends, if you have any friends in course seven.
And you can say, I know what the D is.
It's deoxyribose.
OK.
So then the amines.
The amines we've got, there's five of these.
And they split for RNA and DNA.
And they're shown here.
So it's interesting.
There's two of them that are called purines, because they've got this ring structure.
And then the six fold, and then the five fold.
So these are the purines, and then the pyrimidines have just the six fold structure.
There are three of those.
So A, G, C, U, and T. Those are the five different amino acids.
And the difference is that in DNA, you only have A, G, C and T, whereas in RNA, you have A, G, C, and U.
This is the chemistry.
If you take 7012, you'll figure out how all of this other stuff goes.
And this thing here is phosphate.
And why is phosphate present?
Because it's acting as the bridge and a spacer.
So let's take a look at the structure.
OK.
There's more of the amines in nucleic acid.
So this is what it looks like.
So you have a backbone that consists of sugar phosphate, sugar phosphate, sugar phosphate, shown here.
So in this case, you've got the deoxyribose.
So there's the sugar, there's the five-fold symmetry.
Then the spacer, the phosphate, then the sugar, and then the spacer, the phosphate.
And you can see that the sugars are acting as hangers for these amine groups.
And they're called bases, because in fact they act as Bronsted bases.
They're proton acceptors, and in the early days of molecular biology, people determine the chemical composition by wet assay, and these were determined to be Bronsted bases.
And so to this day, people refer to them as base pairs, et cetera, et cetera, even though now we know they're amines, et cetera.
So this is the structure going up.
And you see, at one end, you've got the three, and going up to the five.
The three is, you have the carboxylic acid end, and at the other end you have the amino end.
OK.
So there the structure.
And you have a choice here.
So again, the sugar hanger, the phosphate spacer, the sugar hanger, the phosphate spacer.
And you have a choice of one of these four.
Any one of these four is what's found in a DNA strand, as you go up the strand.
Now, how is information encoded?
Well, the information is encoded-- first of all, we have to recognize what the structure is.
Know that the structure, turns out that it forms a double helix, which is the first secondary structure, right?
It doesn't have.
The primary structure is this instant A, C, G, T sequence all along.
And then the secondary structure is, in order to maximize hydrogen bonds, this forms a double helix with a second chain.
And furthermore, the pairing is such that A always pairs with T, because they have two hydrogen bonding sites between them, whereas C pairs with G, because they have three hydrogen bonding sites between them.
And furthermore, look at the spacing here.
This always just stuns me.
If you look down the center of the strand-- so I've got two strands of nucleic acid.
The center to center spacing is 1.085 nanometers between A and T, and it's 1.085 nanometers to four significant figures between C and G. And you might say, well, is it possible that I could take two of these hydrogen bond sites and line them up to two of the three hydrogen bond sites?
And the answer is no, because the spacing is wrong.
You can't, these two are far enough apart that they won't line up here.
So it's guaranteed that it's always A to T, C to G. All right.
So now you see the double strand.
You have one strand moving up.
You see the pentagon is pointing up, and here you see the pentagon is pointing down, and we have the base pairs in between.
C pairs to G, and T pairs to A.
And down here is the basic end, the acidic end, the basic end, the acidic end.
And so the secondary structure is the double helix.
Why?
To maximize hydrogen bonds in between, and then to keep maximizing hydrogen bonds.
Hydrogen bonds between the galleries.
Maximize, maximize, maximize.
So there's what they call the base pairs, and et cetera, et cetera, et cetera.
This distance is all prescribed by the fact that we have the phosphate present.
OK. Now where's the information here?
I said this encodes metabolic information.
Well, suppose I want to direct protein synthesis.
I want to put amino acids in a sequence.
So I have to be able to call.
You know, I'm sitting here as nature, and I'm saying, I want to make this protein.
So I need this amino acid, and then I need this amino acid, and then maybe that amino acid, and put them in sequence.
So I need 20 different words to call 20 different amino acids.
Well clearly, clearly if I have just A, C, G, and T-- so if I have one-letter words, it won't work.
Because I need to call out 20.
I need 20 amino acids.
So how am I going to get to 20?
I don't have 20 of these.
I'm only using, there's only four of these.
So I say, oh.
I know what to do.
I can use this idea.
The number of words will equal the number of letters that I have in my alphabet, raised to the power, number of letters per word.
OK?
So if I use this idea, then I can say-- what if I have two-letter words?
So in other words, to call out an amino acid, I have to take two base pairs in sequence.
That means I'll have-- there's only four letters.
But if I have two letters per word, that's 16, which is still less than 20, and that's no good, because I can only call 16.
So what does nature do?
What if I had three-letter words?
Three-letter words, then, is a 64, which is greater than 20, and that works.
So now I've got, in point of fact, 61 of these three-letter words.
So I go down the sequence.
First base pair, second base pair, third base pair.
That triad represents one amino acid.
I've got 61 to call out 20, And I've got three left over.
For what?
Well, let's think about this.
See this?
How do you know to read this, defined by their properties?
You know to read from left to right.
You know that there's a space here between words.
So if I just give you this strand of DNA, where do I begin?
Do I take these three, or do I start counting from here or here?
So built into this, it tells you to read either from left to right, or right to left.
It tells you where to start, and it tells you where to stop.
These three-letter words, by the way, are called codons.
So we've got sixty-one to call out 20 amino acids, which means, some amino acids have more than one name.
There's one of them that has six different names.
There's six different codons that can call it a given amino acid.
For some of them it's only one, for some it's two, three, et cetera.
OK.
So that's the information.
Look!
Here it is!
There's the information!
It's all in there.
Just hydrogen bonds like this, hydrogen bonds like this.
It's all bonding.
So here's the codon.
So somewhere here, see, we take this, this, and this.
This triad represents one piece of information.
All right.
So let's go back and look at some of the history.
Who got us to this point?
This is fresh!
A lot of this happened in my lifetime.
All right.
So Oswald Avery-- this is before my lifetime.
He was working at the Rochester Institute.
And he was the first person-- he was born in Halifax, actually.
And he worked professionally here in the United States.
And he was the first person to recognize that nucleic acids store and transmit genetic information.
Up until the 1940s, people thought nucleic acids-- are you ready for this?-- are too complicated.
Their structure was too complex to contain information.
It's precisely the complexity that gives us the abundance of information!
The prevailing belief was that proteins contained the genetic information.
But people couldn't figure out how to make it work.
So he was the first to make the argument that the nucleic acids have the information.
And then the second giant is Erwin Chargaff.
Erwin Chargaff worked at Columbia in New York, and hospitals in the vicinity.
And he was a painstaking analytical chemist.
He did all of this work by analytical chemistry, and he gave us Chargaff's Rules, 1949.
And he said that in any biological system, the concentration of a-- remember, they didn't know the structure of DNA yet.
I'm going to lead up to how we get to the structure of DNA.
So we're going back, this is like a flashback in a movie.
Now we're going back to see how we got to the double helix, all right?
People knew that the concentration of A in any nucleic acid was equal to the concentration of T. They knew that there were sugar present, they knew there were amines present, and they knew there were phosphates present.
That's all they knew, based on chemical analysis.
And now he tells us that, of the amines, concentration of A always equals concentration of T, concentration of G always equals concentration of C. And then the sum of concentration A plus G is the sum of C plus T. That's Chargaff's rule.
And so here we are, Homo sapiens.
31, 31, 19, 19, a little bit of statistical error, et cetera.
But look!
Corn!
Yes!
Corn reproduces using this same code.
All living organisms on this planet use the same code.
Which makes sense!
If I eat corn, and I'm going to get nutrition from it, and it's going to help me build cells, and regenerate my body, and get energy from it, it has to be made of this same stuff, otherwise my body can't recognize it!
Because I'm a biological machine.
This is not a fireplace in here, right?
If it's a fireplace, you can drink something combustible and boom!
You've got energy.
But we can't derive energy from something combustible.
We can only derive energy from something like this.
So, you know, when you think you're really hot stuff, your DNA is not that different from that of an ear of corn.
So a little bit of humility might be in order.
All right.
And then the last person we want to talk about is Rosalind Franklin.
Rosalind Franklin worked at the King's College in London, and her specialty was x-ray diffraction.
And she made painstaking experiments where she could take a strand of DNA with a stainless steel needle and pull it out of an aqueous solution, and hold it in a chamber where the chamber was humidified so that the DNA didn't dry out.
And while it was humidified in this chamber, take an x-ray diffraction exposure for hours, because all of the recording was done by film.
By halide photography, and you had to have long exposure time.
This is one of the most famous images of the 20 century.
This is the image of the DNA from calf thymus, the beta structure.
And this is a Laue pattern.
So the symmetry of this pattern is reflective of the symmetry in the DNA structure, which people don't know yet.
And what you see is five positions.
One, two, three, four, five.
Four is missing.
There's no reflection there.
But it's where you would expect a position to be.
So it's sort of Bragg-like, isn't it?
Not all of the lines are reflecting.
On the basis of this image, this is what we learned.
We learned, first of all, that the Laue pattern is indicative of a double helix.
Based on X pattern.
Remember, they had no computers in those days!
All they had was a pencil and paper.
So to do the Fourier transform, and go back out of K-space, and discern what the pattern is-- you know, today it's trivial!
They did this all longhand.
Number two.
On the basis of that pattern, you learn that it's 3.4 angstroms between nucleotides along the backbone.
We don't know what the backbone is!
We just know that there's sugar, there's amine, there's phosphate.
There's different ways of putting this together to make a chain.
And the third thing that you learn, because the strong, clear lines indicate that the heaviest elements must lie outboard.
What does that mean?
If you've got a double helix, based on the first finding, what's the heaviest element?
Phosphorus.
Because otherwise you've got carbon, oxygen, nitrogen.
Nitrogen is atomic mass 14.
Phosphorus is atomic mass 30.
So it must be outboard.
Otherwise, if it were inboard, there would be other elements farther out, and you wouldn't get the clear lines.
Because the heavy elements are the ones that give the reflection, and if the heavy elements are obscured by lighter elements that are shielding them, you get a lousy reflections.
Blunted reflections, blurry reflections.
So all of that comes out of her findings.
So what happens next?
So there's the 3.4, and 10, and et cetera, et cetera.
All of that happens based on the pattern that I just showed you.
So I'm going to read to you what happens in those days, back in the 1950s.
There was intense competition.
You know, Linus Pauling had proposed a triple helix, and people in the UK were working hard on it, and so on.
A lot of competition.
So what happens is that Watson goes down to King's College in London, and he meets Rosalind Franklin in the hallway, and they get into a big argument.
And she really chews him him out, and he goes running away.
She was very tough.
And so he goes to Maurice Wilkins, who was her boss.
And I'm reading, now, from Freeman Judson's book.
"Wilkins told Watson as they went down the hall that Franklin had found that DNA fibers, when kept wet, yielded a different x-ray pattern, suggesting a second structure.
Fourteen months after the King's colloquium, despite repeated correspondence and conversations, visits, meals together between Wilkins, Watson, and Crick, the possibility of a second structure was news to Watson, he wrote."
Now, this is quoting from Watson's own book, The Double Helix.
'When I asked what the pattern was like, Maurice went into the adjacent room to pick up a print of the new form they called the beta structure.
The instant I saw the picture, my mouth fell open, and my pulse began to raise.
The pattern was unbelievably simpler than those obtained previously, the a-form.
Moreover, the black cross of reflections which dominated the picture could arise only from a helical structure.
With the a-form, the argument for a helix was never straightforward.
With the b-form, however, mere inspection of its x-ray picture gave"-- listen to this-- "several of the vital helical parameters."
I'm going to come back to that.
"The picture that Wilkins showed Watson was Rosalind Franklin's, without her approval."
So then Watson goes on the train back to Cambridge, and sits down with Crick, and in no time, they've put the structure together.
And this is a famous picture.
This is Crick.
This is Watson.
And I don't know if you can see really well.
These are lab clamps that they're using.
This is a lab stand.
They're using lab clamps to put the structure together.
So here's the paper.
The paper as it comes out, Molecular Structure of Nucleic Acids.
And this was the first image of DNA.
And I submit to you that if the earth were to end in a giant explosion, and we were to only send two images in a spaceship out to let some other race discover what we're all about, it would be this image, and the image of the atom.
That's all you need.
Because you have the simple atom, and you've got this, and everything else is just detail in between.
So there's the first image.
All right.
"We wish to suggest a structure for the salt of deoxyribonucleic acid.
The structure has novel features which have considerable biological interest.
The structure for nucleic acid has already been proposed by Pauling and Corey.
They kindly made their manuscript available to us in advance."
These guys never made their manuscripts available to Corey and Pauling.
"Their model consists of three intertwined chains with the phosphates near the fiber axis and the bases on the outside."
See, they got the positions wrong.
"Without acidic--" da, da, da, da, da, and so on.
Whatever.
Now here's-- remember, I've quoted from Watson's book.
"The previously published x-ray data on deoxyribonucleic acid are insufficient for a rigorous test of our structure."
Really?
"So far as we can tell, it is roughly compatible with the experimental data, but it must be regarded as unproved until it's been checked against some more exact results."
Remember, he's writing this after having seen Rosalind Franklin's x-ray pattern.
"Some of these are given in the following communications."
Yeah, the next paper is Rosalind Franklin's paper.
"We were not aware of the details of the results presented there when we devised our structure,--" That's a lie. "
--which rests mainly, though not entirely, on published experimental data and stereochemical arguments."
You know what?
once I've given you the data, and you know what the structure is, you can give me a very, very strong stereochemical argument.
But that's hindsight.
If you don't know what the structure is, you have no idea what the stereochemical arguments are, because you can justify anything.
So this is just absolute lies.
Who are the authors?
Watson and Crick, even though it's based on data that was taken from other people.
So here they are.
This is the acknowledgments.
We're indebted to-- blah, blah, blah.
The experimental results in ideas of Wilkins, Rosalind Franklin, and their coworkers at King's College.
And one of us, Jim Watson, he was the American, and I think there's an irony here, was aided by a fellowship from the National Foundation for Infantile Paralysis.
And I think there's a joke there, if you know anything about Jim Watson.
And now contrast that with this.
This is the publication of the human genome.
And look at all the names here.
Everybody who was involved in the enterprise was named.
This is not the paper.
This is just the names of the authors.
Now, to be literate, there's this one phrase.
Typical British understatement.
Remember, this was about the structure of DNA.
But there's this passage.
It has not escaped our notice.
This is the litotes.
Understatement, right?
"It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material."
That is the understatement all time.
We've got the structure of DNA, and parenthetically, I think this is how you direct replication in all living creatures, by the way.
So when this paper was published.
and they rolled out the human genome as a sort of a literary-- what's the word-- allusion, thank you.
The literary allusion was-- there's a passage in this paper that at one point begins "It has not escaped our noticed," da, da, da, da.
So if you're ever with some Course 7 people, and you want to really get under their skin, you know, you could be at a party or something, and finally, you say you know, it's not escaped my notice.
And they know what you're saying.
It'll make you less popular then you are now.
So let's fast forward.
This is Stockholm, December 10, 1962, the Nobel Prize ceremony.
Here is the King of Sweden.
This is Francis Crick.
This is James Watson.
This is Maurice Wilkins.
No Rosalind Franklin.
This is Kendrew, and I'm drawing a blank.
These two guys were getting the Nobel Prize in Chemistry for hemoglobin structure.
This, if you look carefully, is John Steinbeck, getting the Nobel Prize in Literature.
Over in Oslo is Linus Pauling, getting his second Nobel Prize, but for peace.
So ironically, he's getting his Nobel Prize.
They're getting the Nobel Prize not in Chemistry, not in Physics, but in Medicine.
You know, what's medicine?
I mean it's not really a science.
But anyway.
Here it is.
It's worthy of a Nobel Prize.
So what really happened here?
What really happened?
Why was there no Rosalind Franklin?
She was marginalized as a woman.
She was mistreated by these men.
It was 1955.
You realize in the 1950s-- and I'm not talking about the England of Charles Dickens.
I'm talking about the England after World War II.
Women were not allowed in the common room.
And the common room is where everybody congregates at 4:00 p.m. to drink coffee and exchange ideas.
So she couldn't go into the common room and so on.
And it was just a terrible, terrible story of misuse, of information, lack of attribution, and so on.
If this story were unfolded today, I guarantee you these guys would not be getting a prize, and in fact, the publication would probably be rescinded on the grounds of its fraudulent claims, that they were unaware of the previous information.
So after the paper was published, she was so brokenhearted that she resigned her position at King's College, and took a job at another college in London.
Her terms of severance included her signing a document in which she agreed to abandon all future work in biochemistry.
I mean, imagine that you leave your job, and they tell you what you cannot do!
I mean, obviously, if you were privy to intellectual property and so on, you're bound by confidentiality.
But no one can tell you, you can't go to work for somebody!
And so she changed fields, and made seminal discoveries in two other fields outside of the DNA work.
So why isn't she here?
She died of cancer in 1958, very young.
And you can't get the Nobel Prize posthumously.
So some people are quite outraged by this story.
But there's a symbolism here to have Wilkins.
It's basically saying that you two guys are not the authors of this.
And so some people say that Wilkins is standing in for Rosalind Franklin, and the Nobel committee at least acknowledged that.
So what's the message here?
The message here is, I hope nobody in 3.091 is ever embroiled in a controversy like this.
We have to acknowledge our collaborators.
All of us are here because somebody helped us.
Don't go away.
You haven't been dismissed yet.
This is important.
What I'm telling you is far more important than that structure.
I'm telling you how to stay out of jail.
See, here, you know, 50 years ago, you cheat and steal, you get a Nobel Prize.
Today you go to jail.
It is a Nobel Prize in Medicine, of course, but anyway.
So this is a book, if you're interested in Rosalind Franklin's story, it's fantastic reading.
Real page-turner.
This is James Watson's book.
If you read this, you know, have your you-know-what sensors on all the time.
And here's a picture of Rosalind Franklin on vacation in France.
And we'll leave with The Twist by Hank Ballard, who was another person who was marginalized, because he was black!
And so you couldn't listen to this music, and so there was a cover made.
The version of The Twist that you have heard is probably by Chubby Checker.
This is the original one by Hank Ballard.
It's gritty.
It's very gritty.
All right.
Treat each other well.
Get out of here.
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All right.
So today we're going to start the last section.
We're going to do three lectures on phase diagrams.
And I've given this the label here of Stability, Sustaining the Solid State.
We've talked a lot about the solid state in 3.091 as the vehicle for teaching the rudiments of chemistry.
One of the things we have not talked about, is what are the conditions that sustain the solid state?
How do I know whether something's going to be a solid, or a liquid, or a gas?
This is very important.
For example, if you're in a foundry, you're running places, making auto parts, you have to know what the solidification temperature is of the alloys, so you can get the parts out quickly and keep productivity high.
It's even important in failure analysis.
You know, when they were looking at things like the rubble from the World Trade Center by understanding phase stability, the thermal history was imprinted in the metal.
It's possible to determine what temperatures were achieved, and therefore, what the chain of events was that led to the collapse of those buildings.
Now you might say, well, isn't this straightforward?
I mean, for example, you just look up the melting point or boiling point.
Water, 0 degrees C. Melting point, boiling point, 100 degrees C. Well, suppose you decide to realize your life's ambition.
You're going to go and scale Mount Everest.
So you pay the $10,000, you get a license from the Nepalese Government, and you're at base camp at 20,000 feet.
You're sitting there, you've got your campfire going, you've got a hankering for a hard-boiled egg.
And you start, you put the eggs into boiling water and you cook them 10 minutes, you take them out, and they're still runny.
You say, I must've lost track of time.
So you cook them 20 minutes, and they're still runny.
And you cook them 30 minutes, and they're still runny.
And you eventually come to the realization that at 20,000 feet, the boiling point of water is below the denaturing temperature of egg yolk.
So now the boiling point is a function of pressure.
So you know, we could have one of these Iron Chef cookoffs, only the dish is souffle, only we'll have the contest at Flagstaff, Arizona, where the altitude is so high that the classical recipes won't work, because the boiling point, the atmospheric pressure, and everything conspire so that you're not going to be able to support the souffle.
So you're going to have to understand phase diagrams.
In fact, if you understand phase diagrams, that's your ticket to being a four-star chef in the kitchen.
Well, here's another place we can look at, where pressure has an important role.
Let's look under the hood of a car.
You're running an internal combustion engine.
So this is the engine block, and there's combustion going on inside.
We've got to keep this thing from overeating.
It could damage the metal.
In extreme, you could melt the metal.
So we've got cooling channels in here with water flow.
But then the water's going to heat up, and the water's going to boil.
So we've got, over here, the radiator.
So it's really a double cooling system.
The water cools the block, and then the radiator cools the water.
So how's that going to work?
Well, here, the temperature is on the order of about 90 degrees C, and here the temperature inside the engine block is much greater than 90 degrees C. It's several hundred degrees C inside that engine block, and we've got water going like this.
So we've got cooling water flow in this manner.
And now cold water goes in and hot water comes out, and then we get into the radiator, and we've got channels here, and we've either got a fan or some kind of air flow.
And so now, what's the gambit here?
I want to see what's going on inside this radiator.
So if this is the wall of the radiator, and in here I've got water, and over here I've got air, the idea is to have a high heat flux to get the energy out of the water and cool it down.
Now, if things get really, really hot in here-- let's say you're zooming along down the highway at about 90, I mean, about 65 miles an hour, and all of a sudden you come upon a collision, and you have to go to a dead stop, all that energy in the engine is now dumped, and the water could overheat.
And in the extreme, if it overheats, it can actually go into a boil.
And this is bad.
Because heat transfer is really good between a liquid and a solid, and it's really poor between a gas and a solid.
This is a low heat flux, and this is bad.
High heat flux, that's good.
This is the bubbles.
This is boiling.
And why?
Because what transfers energy?
It's the atoms.
And what's the atom density in a liquid versus the atom density in a gas?
You just have really crummy heat transfer.
That's why insulation involves dead airspace.
Because you have very poor atom density.
So what can we do in order to try to repress the bubble formation?
Well, we said if we went up Mount Everest, we went to high altitude, the boiling point went down.
And why did it go down?
Because the atmospheric pressure is lower.
So why don't we go the inverse here, and we'll put a pressure cap on the radiator, and make the pressure go up, and when the pressure goes up, it exerts a back pressure and represses bubble formation?
So by increasing pressure, we can increase the temperature, and get the temperature up.
So let's say p equals-- if you look on the top of the pressure cap, it will say 15 psi.
And 15 psi is almost 14.7 psi, which is exactly 1 atmosphere, which in SI units is, and I know this to six significant figures, 101325 Pascals.
Or, you know, from Torricelli, 760 millimeters of mercury, that's the column that's supported.
So we're at 2 atmospheres pressure in there, and with two atmospheres of pressure, we can raise the boiling point to 106 degrees C. And then what we can do, is now change the composition.
So we change the pressure, and I change the boiling point.
I change the composition.
add 50% ethylene glycol.
So now I've got a one-to-one mix, ethylene glycol and water.
And that takes the boiling point up to 130 degrees C, or 265 degrees Fahrenheit.
So another example of how we can manage systems.
So when I'm showing you by these several examples, is that the boiling point is a function of pressure and composition.
So we can tune the boiling point.
See we were tuning materials properties before.
Now we're tuning physical chemical properties, because that's what chemistry is all about.
It's about control.
It's about management.
Management to advantage.
So how do we know what to do here?
Can we predict this from first principles?
No.
Systems are still too complicated.
Our models aren't robust enough.
So instead, we turn to an archive.
So we have phase diagrams.
And the phase diagrams guide us in our search for better management of material systems.
And these are stability maps.
They tell us which phases are stable under which conditions of pressure, temperature, composition.
You can consider them as archives.
So that's what I want to talk about, because it's really important to know this stuff.
OK.
So what are we going to do?
So first of all, let's-- I don't want to write on that one.
This is interesting.
This could be a work of art.
I'd hate to desecrate it.
You like?
This is really good!
So I'm just going to leave that.
That's very good.
That's excellent.
OK.
So I've been talking about phase.
What is a phase?
Let's get the right definition here.
What is a phase?
It's a region uniform chemical composition.
Just to get it down.
it's a region in a material, region of uniform chemical composition, and it has these other properties.
Uniform chemical composition.
It's physically distinct.
And I'll show you by examples, but let's just get the definition down.
It's physically distinct, and in the extreme case-- so that means it's bounded.
It's physically distinct.
That's a fancy way of saying that you can put a boundary around it.
It's bounded.
And then the other property that it has, it's mechanically separable in the extreme.
And so that's in the case where you have a multi-phase system.
You can point to the different p. So what I'm going to do, is I'm going to give you some examples.
And so I'm going to let circle p equal number of phases.
I'll give you some examples.
Why am I using circle p?
Because just plain old p is already pressure.
So pressure, so this is kind of a Western motif, you know?
Like a Circle Bar Ranch or something.
So it's a p with a circle.
That's my font.
That's how I say number of phases.
So let's take a look at some examples.
So here's some examples of p equals 1.
So simple one is pure water.
Pure H2O liquid.
It has all of this.
Wherever I have the water, all right, here's some water, it's all of the same chemical composition.
It is physically distinct.
I can put a boundary around it.
And in this case, there's nothing to separate, because it's just a one-phase system.
White gold.
White gold is one phase.
Because we have gold.
Obviously we have gold.
If you market something as white gold, it has no gold, you'll go to prison.
And how do we make white gold go to its white color from the normal yellow?
We add silver and we add nickel.
And these are all FCC metals, and they substitute for one another on the lattice, and if you sample anywhere in the white gold, you get the same chemical composition, and so on.
Air is an example of a one-phase system.
It's a gas, and in decreasing order, we have nitrogen, oxygen.
So it's a solution.
We're breathing a solution.
Nitrogen, oxygen, argon are the main constituents.
There's some CO2, some sulfur dioxide if you live near a power plant, NOx if you're near a tailpipe, and so on and so forth.
And then the last one I'll give as an example is calcia-zirconia.
Or cubic zirconia.
Let's just call it cubic zirconia.
The holidays are coming, so we better get cubic zirconia up here.
That's the poor man's diamond.
Cubic zirconia, which consists of a solid solution of calcium oxide in zirconium oxide.
This is a solid solution, so it's continuous and chemical composition identical.
Now, contrast that with two-phase system.
So p equals 2 looks like this.
So let's look at, instead of simple water liquid, if I have ice cubes in water.
So now I have two different phases, and they're separated by state of matter.
State of matter is the issue here.
Because I have a solid-- this is solid-- and I'm going to put the boundary around them.
The composition of the solid is different from the composition of liquid.
They're both the same water, but you can tell it's a different crystal structure and so on.
And I can put boundaries around, and it's mechanically separable.
You can pull the ice cubes out of the mix.
Milk is two-phase.
Milk has a fatty phase and it has an aqueous phase.
The aqueous phase contains the minerals.
That's why you can drink skim milk and still get your calcium, because calcium, even the general public knows calcium is a mineral.
It's not elemental calcium, it's a calcium compound.
But it's ionic, it's soluble in water, and the fat contains the other parts.
You know from last day, this is one of the lipids.
And so you have fat globules that are mechanically separable and distinct from aqueous.
So they actually have a different chemical composition, different type of bonding.
Right?
This is largely nonpolar, and nonpolar doesn't like to dissolve in something that is polar and hydrogen-bonded.
So they phase separate.
I've been saying phase separate all along.
Now you know what we mean by phase separate.
Here's a third one.
This is really cool, because you won't get this in any other chemistry class, because only 3.091 talks about such things.
David, can we cut to the document camera?
I want to show you a piece of snowflake obsidian.
Obsidian is a volcanic glass.
So here's the piece of obsidian.
And what you're looking at is, the black area is the glassy part, and the white is actually some of the obsidian that has decided to devitrify.
And it is turning crystalline.
And so those are islands of obsidian crystal sitting in a matrix of obsidian glass.
If we go back to the-- I picked that up in Yellowstone Park.
It's fantastic.
There's a wall, just a wall of obsidian that comes down.
And I think I've got an image here.
There it is.
This one I pulled off the internet.
So this is crystalline phase, and this is the amorphous phase.
And the overall composition, it's a glass, it's a silicate.
And why isn't it transparent to visible light?
Because it's got iron in it.
The iron gives it the absorption and capabilities.
The band gap is invisible, so it's black.
But then when this crystallizes, it's got a different index, and it's mainly SiO2.
It's really the cristobalite phase of the obsidian.
So that's cool.
So this is actually differentiated on the basis of amorphous and crystalline of something that's virtually the same chemical composition.
So this could give you an idea of what happens in the area of phase separation.
OK. Now the second thing that I have to bring to your attention, is that these phase diagrams treat systems at equilibrium.
They're stability maps, and they treat systems at equilibrium.
So what is equilibrium?
Let's just get one time up here a definition of equilibrium.
Equilibrium is a condition which there is no net reaction taking place.
The system could be reacting.
If you're looking at the ice cubes in water, the ice could be dissolving, and some of the water could be freezing, but overall, with time, there's no net reaction.
It's a dynamic system with no net reaction.
And it's characterized as the lowest energy state.
Systems strive towards equilibrium.
If it's the lowest energy state, no net reaction then.
If a system is truly at equilibrium, the properties should be invariant over time, because there's no net reaction.
You can't prove equilibrium.
You can only demonstrate the absence of equilibrium.
Because at equilibrium, there's nothing, there's no net reaction.
And it could be attainable by multiple paths.
So in other words, if I have ice as the equilibrium phase, I can make it by freezing water, and get to the same ice at the given temperature and pressure.
Or I could sublime.
But if I specify the pressure and the composition, I should always end up with the same system.
So attainable by multiple paths.
Because it's a stable state, so it doesn't matter how you got there.
It's a state function.
And you know from your physics, a state function has the same value, regardless of how you arrived there.
And then the last thing is, I want to differentiate.
See, these are all single phase, but some of them are just one element, some of them are just one compound, and some of them have a whole mix.
So we have a way of quantifying the degree of chemical complexity by using the term called component.
It's a measure of chemical complexity.
It's the number of chemically distinguishable constituents.
And again, these are definitions.
I'll give you some examples so then the definition comes to life.
I like to think of it as the number of bottles you've got to pull off the shelf to make the final product.
The building blocks.
So for example, water is just water.
You could say, but it's hydrogen and oxygen.
But if you take 560 or some other thermal class, you'll learn that they're bound by mole ratio.
You don't have freedom of hydrogen and oxygen.
So it's just one bottle, whereas to make white gold, you need three bottles.
You need a bottle of gold, a bottle of silver, and a bottle of nickel.
So that's a three component system.
And I'm going to designate the components by c with a circle around it.
Because c without a circle is already composition, concentration.
OK.
So now we can put all of this together, and we'll show the difference between multicomponent and multiphase.
So we'll make a little table here, and here I'm going to talk about one-phase systems, and distinguish on the basis of components.
So if it's single phase single component, that's just water.
One phase, one component.
Simple liquid.
If it's two components one phase, that could be calcia-zirconia.
I need a bottle of calcia and a bottle of zirconia in order to make this two component single-phase system.
And then to have three components, that's the white gold.
It's still single phase, but I need the gold, silver, and the nickel, whereas if I come over here, I've got two phases.
That could be slush.
Slush, that's ice water.
It's just one component.
I just start with water, and drop it down to 0 Centigrade, and I'm going to end up with ice crystals in the water.
So that's one component.
I just needed one bottle.
Literally, one bottle.
All right.
Now what about two components, two phases?
Well, we saw earlier when we started thinking about solubility, we did the bilayer with carbon tetrachloride and water.
Remember, we dropped potassium permanganate and iodine, and the potassium permanganate dissolved in the water, and the iodine dissolved in the carbon tetrachloride?
Two different components.
I need a bottle of carbon tet and a bottle of water.
And they'll phase separate.
They don't dissolve in one another, because this is a nonpolar liquid, and this is polar hydrogen bonded.
That's the origin of phase separation.
It's all about bonding, and bonding is all about electronic structure, and that sounds like a good topic for a class in chemistry.
And then the last one here is obsidian.
The snowflake obsidian, in which the components are SiO2-- so I'm going to have three components and end up with two phases.
SiO2, there's some magnesia, and then the black comes from the magnetite.
Fe304.
So I've got three bottles off the shelf, and I end up with something that is crystalline and amorphous.
So that's where I get the two phases.
Three components, two phases.
It's great.
So now I want to look at some stability maps.
And I'm going to start with the simplest of all, which is the one we know best from human experience, and that's water.
So we're going to look at the one-- so it's a one component.
So for water, we start with c equals 1, and the example is going to be water.
One component phase diagram.
So we don't have a composition axis, because the only thing we can vary is pressure and temperature.
If we change composition, we've lost water.
So it has to be axiomatically, no variation composition.
All we have is that the coordinates of temperature and pressure.
So there it is out of the book, and I'm going to draw it a little bit differently, just to emphasize three of my favorite words.
Not to scale.
Otherwise you can't see some of the subtle features.
So overall, it's a y-shaped diagram.
And this is pressure.
Let's make it pressure in atmospheres.
Then our human experiences is at 1 atmosphere.
So at the highest temperatures, we expect to have vapor, and at the lowest temperatures, we expect solid.
And in between, we expect liquid.
And then if we look at the 1 atmosphere isobar, what's happening on this line?
On the one side of line it's liquid, and the other side of line, it's vapor.
At this temperature, liquid and vapor coexist.
I call that the boiling point.
So I'm going to put 100 here.
And we already learned from our little example of Mount Everest that this locust here, this line, is the liquid equals vapor two phase equilibrium.
And you see how it changes?
If I go to the top of Mount Everest, the pressure is less than 1 atmosphere.
which means the boiling point of water is below the denaturing point.
So this actually gives you the map.
Or if I put the pressure cap on, I go up here.
And you can use this in the kitchen again.
If you love wild rice, and you can't wait 45 minutes when you get home, use a pressure cooker.
What the pressure cooker does, is it allows you to have liquid up to 130 degrees Celsius.
See, you can't steam it.
You've got to soak it in water.
And you know the Arrhenius Law.
If you can get the temperature up by 30 degrees, you'll cut the cooking time in half.
So how do you get water, liquid, at 130 degrees?
Then you can cook quicker!
Pressure.
Seal it.
And then you've got liquid water way, way up here.
Now over here, this is solid goes to liquid.
So this is the freezing point line.
At 1 atmosphere, that's 0.
0 Celsius.
And you know, if you go skating-- so let's say the ice is minus 5 Celsius.
And you're here, but you put your weight on the blades, and the blades have small area.
So pressure is force per area.
So that puts you up here, and now you cross, and now you glide on water.
Film of water.
You ever watch a hockey game?
I grew up in Canada.
We used to say, well, the ice is fast.
Why is the ice fast?
Because you get down to temperatures where you just get the right slickness of water.
If the ice is slow, the temperature is so close that, you know, you get these big, bruising hockey players, and they're moving up into here, and it's like they're dragging.
And if you ever come from Minnesota or North Dakota, if you've ever skated when the temperature gets down about minus 20, minus 30?
It's a different experience, because you never get across this line.
Your blades are just, they can't get a good grip.
See?
Everything you need to know is here.
This is four-star chef, this is prize-winning hockey player.
We haven't got off the 1 atmosphere line yet.
OK.
So now I'm going to say, this is single phase, right?
It's just vapor.
This is single phase, and this is single phase.
It's all solid.
And along this line, it's two phase, isn't it?
Solid-liquid.
This is ice cubes, isn't it.
Anywhere along this line is ice cubes in water.
Look at this one.
This is liquid vapor.
This is solid, liquid-- all three of them coexist here.
This is p equals 3, which means, ice cubes floating in boiling water.
And this happens at only one point.
It's called the triple point, and its coordinates are-- this is not to scale.
So this is 0.01 degrees C. It's a 1/100 of a degree higher.
And the pressure here is 4.58 millimeters of mercury, which then you can convert using the 760 as 1 atmosphere.
Oh, and this is solid-vapor.
You can go directly from solid to vapor down here.
If you hang up your clothes on a line and it's minus 20 degrees outside, first thing that happens, the clothes freeze.
You come back five hours later, and the clothes are dry.
Well, what happened?
Did somebody take them in?
Run them around the clothes dryer, then hang them up for you?
No.
You're going directly from solid to vapor, down in here.
So here's the whole story of the phase diagram.
One last thing I want to point out.
Here, you're at a given temperature and it's vapor.
And then you squeeze, squeeze, squeeze, and now things get closer together, and you go from paper to liquid at constant temperature.
That makes sense, right?
I start from vapor, constant temperature.
I squeeze it, I get liquid, I keep squeezing, I should make solid.
Here's vapor.
I squeeze, I get solid, and I squeeze harder, and I get liquid.
That makes no sense at all, but that's what happens.
What does this tell me?
It tells me that the number of nearest neighbors in the solid is greater than the vapor.
Yeah, I get that.
But the number of nearest neighbors in the liquid must be greater than the number of nearest neighbors in the solid.
This is an exception.
When I see this negative slope, it means that ice cubes are going to float on water.
That's what I learn from this negative slope.
So let's put that down.
dp by dt when dp by dt is less than zero.
This is for solid equals liquid.
That means that the density of the solid is less than the density of the liquid.
All from here.
OK. Good.
So let's look at a few other phase diagrams.
I think I've got a few things up here.
Here's silicon.
Pressure versus temperature.
Its normal melting point is about-- see, now I said normal melting point.
After today's lecture, if somebody says, what's the boiling point of water?
You don't go, 100 degrees.
You say, at what pressure?
And they go, oh, I hate you.
I hated you before.
I'm going to use an adverb here.
Now I really hate you.
OK.
So this is the normal melting point of silicon.
It's about 1430 degrees Centigrade, or a 1700 something Kelvin.
And it also has this negative slope.
And what does that tell you?
That liquid silicon is denser than solid silicon, and it's a good thing it is, because we wouldn't have the microelectronics age if it weren't for that.
We couldn't have Teukolsky crystal growth and everything else that makes silicon dirt cheap.
You know why it's dirt cheap?
Because it's made from dirt.
That's why it's dirt cheap.
It's true.
Here's aluminum.
This is normal.
This is the normal solid-liquid equilibrium.
This is FCC metal, which is the closest packed.
Axiomatically, the liquid must be less well-packed.
All right?
So this is dog bites man, this is man bites dog.
This is the unusual one.
OK. What else do we have?
Nitrogen!
Here's nitrogen.
I drew this.
So there's the 1 atmosphere line.
Nitrogen boils at 77 Kelvin, and it freezes at about 63 Kelvin.
And it's got the positive slope that you'd expect, and so solid nitrogen is denser then liquid nitrogen.
And you can remember 77 Kelvin.
You want to remember a few of these things.
I wanted to point out to you that at atmospheric temperature, ambient temperature, you can remember 20-20.
20-20 vision.
So at 20 degrees C, it's roughly 20 millimeters of mercury, is the vapor pressure of water.
It's not exactly true, but it's close enough.
I think it's really 24.
But 20-20 is nice to know.
And what's the street address for MIT?
77 Mass Ave. That's the boiling point of liquid nitrogen, 77 Kelvin.
So I thought we'd do a few things here.
And I'm going to show you a little bit of fun and games with liquid nitrogen.
But first, I've been instructed I have to practice safe laboratory-- So I'm going to put on my lab coat.
Yes.
It's a nice lab coat.
Mm-hm!
Yeah, look.
You want to know-- let's go back to this.
They have to know where the-- this is not just any old lab coat.
This is a nice lab coat.
This is from France.
It's from MAISON DUTECH.
See?
Notice the collar.
It's not that typical, you know.
MAISON DUTECH.
OK. Let's go back to the phase diagram.
So Dave Broderick is going to help us here.
So we're going to have some fun.
I'm going to put on a face shield so I don't blind myself.
All right.
Here we go.
All right.
So let's get some liquid nitrogen.
Sounds different, huh?
Sounds terrible for me.
All right.
So we've got some liquid nitrogen here, and I'll pour it into a Dewar.
So what happens when you wet a sheet of paper with ink with liquid nitrogen?
What kind of bonding is there in liquid nitrogen?
Look.
No running, no running.
If it's liquid nitrogen.
I tell you, everything you learn here is so valuable compared all the other stuff you learn here-- All right.
So there's a couple of things.
So let's look at glass transition temperature.
So this is a latex glove.
You can see it's above its glass transition temperature, and the cross-links are snapping it back.
What I'm going to do, is take it down below its glass transition temperature, and turn it glassy and brittle.
So that it's below the TG.
Here's some paraffin.
Paraffin, which is also a polymer here.
Take the paper off.
OK.
So here's paraffin.
Look.
Look at the van der Waals bond.
All right?
This is van der Waals bond.
Now we're going to go below the TG.
It's liquid nitrogen!
It doesn't matter.
OK, you can hear it already.
So it's now below the glass transition temperature.
So what else have we got?
Oh, you've heard of the glass flowers at Harvard?
Well, we've got glass flowers here, so.
Here's a rose.
See, in the proteins, you know, it's above the-- you know, it's soft.
But now, we're going to put the-- mm-hm.
Here we go.
You ready?
Oh, come on.
It's not a kitten.
What's the matter?
It's still burbling here.
We have to wait until it's dead.
This is for science.
OK. Now it's below its glass transition temperature.
Oh, I'm sorry, David.
I should put this over on the other side.
But maybe it's the red, or maybe it's because it's a rose.
So here's the chrysanthemum.
Today is your day.
Let's go.
Get this thing moving!
Yeah.
So there you have it.
There you have it.
I think that's-- oh, there's one, may we go back to the slide?
Better put the gloves on for this.
I'll have to write apology letters.
OK.
So we're going to do next, is we're going to look at phase equilibriums.
There's the melting points and boiling point of oxygen, argon, and nitrogen.
And what you notice is that nitrogen boils at a lower temperature then oxygen and argon.
So what I'm going to do, is I'm going to pour some liquid nitrogen into this beaker.
And then inside the graduated cylinder, I'm going to let it sit here, and we're going to condense air.
And the result will be that we're going to start condensing liquid oxygen, and they'll be little bits of argon ice floating in liquid oxygen.
And liquid oxygen is faintly blue, and it's paramagnetic, as you know, because it's got the unpaired electrons, which means that it will levitate in a magnetic field.
I couldn't bring a big magnet in here, but we'll see if we can make some blue stuff.
And it's foaming, because this would be like pouring water into something that's-- I don't know.
300 degrees Centigrade.
But eventually, the heat transfer gets low enough that you have stability here, and-- yeah.
OK. Oh, that's terrific.
OK. What else have we got?
Let's go back to the slide.
God, I hate this thing.
OK.
So you've been looking at that nitrogen.
Let's look at what's next.
Ah.
Now if you solidify nitrogen here, you get a whole bunch of different polymorphs.
Different crystal structures.
So there are about five or six different crystallographic forms of solid nitrogen.
For that, we'd have to bring liquid helium.
4.2 Kelvin.
Very cool stuff.
Here's carbon dioxide.
Now, it also has the positive slope, so solid will sink in liquid.
But look, the interesting point here.
The triple point is 5 atmospheres.
At atmospheric pressure, you go directly from solid to gas.
It sublimes.
And that's advantageous.
When we use carbon dioxide as a coolant, when it gives up its enthalpy to cool, it doesn't then turn into liquid.
If you chill something with ice, with water ice, when it gives up its enthalpy, the object is now swimming in water.
And that's no good.
So hence we have the term dry ice, because we go directly from solid to vapor.
And we want to prove this.
So what we've got here is a big block of dry ice.
So this is at minus 78 degrees Celsius, and it's still stable here.
Well, it's subliming before your eyes.
It's smaller than it was when we started the lecture.
So what I'm going to do now-- oh, I'm not wearing that stuff.
I guess I'd better.
They're going to write me up on some log, teaching you bad practice.
But anyways.
All right.
So what we're going to do, is we're going to test the proposal that this actually goes-- so I've got some water.
Here, David, let's go out to the video.
So let's take some water.
I think we got water from France.
Goes with the lab coat.
This is Evian, OK?
So I pour the Evian into the Florence flask.
This is Florence.
Florence is round, and the Erlenmeyer is the flat one.
So let's move this guy.
You're going in the backstage for a while.
You just keep condensing All right.
So now, this is here.
Now what we're going to do, is we're going to put some dry ice in there, and see if we go directly from solid to a vapor.
Take a chip off the old block, here.
OK.
So now you can see that it's going directly from solid to vapor.
So that's carbon dioxide bubbles.
And we assume that the phase diagrams is correct.
I mean, another thing we could do, is-- I mean, I know what club soda tastes like.
I could just test it, right?
Yeah.
It's carbon dioxide.
Now!
Here's the other thing.
You know, Schweppes likes to pride itself on tiny bubbles.
They have a tagline, Tiny Bubbles.
Right?
So here's Schweppes.
And you can see Schweppes, with their wimpy little tiny bubbles.
Can you get that, David?
But we have, we have big bubbles.
I mean, this is pure.
You can't get any better than that.
I got Evian water, and I got a block of carbon dioxide.
I like making my own carbonated water.
I've taken it to a new level.
OK. What else do we have?
Well, let's see if we made any liquid oxygen here.
Try this.
I don't know if we got anything here.
Oh yeah!
Oh, wow.
Look.
OK.
So that liquid in there-- oh, I see.
You've got a separate thing.
I don't know if you can see.
It's condensing.
That liquid in there, that's liquid oxygen.
I won't drink this.
But I can inhale it.
Pure oxygen.
I could have a cigarette, and just put the cigarette in there.
What happens if I put the lit cigarette into the oxygen?
All that happens is that it burns quickly, that's all.
There's no explosion.
It burns quickly.
If you walk into a room full of oxygen, all that happens is you'll singe your nose, that's all.
There's no explosion.
It's just oxygen.
There's no fuel.
The fuel is just in tobacco, and there's not much.
It's not a problem.
A room full of oxygen?
Pfft!
OK. Let's go back to the slides.
What have we got here?
We've got some examples of other phase diagrams.
All right.
Yeah.
[
FROM 'PHANTOM OF THE OPERA']
Actually, it's quite tasty.
All right.
So.
Now, here's zirconia.
Now, here's the thing to know about zirconia.
Look at all the different phases.
And cubic zirconia is the one that has a high index of refraction, but it's stable only at temperatures exceeding 2000 degrees C. So that's no good.
So what I'm going to show you next day is that by changing the composition, and adding calcium oxide, we open up this zone here that's labeled cubic, and we get it stable all the way down to room temperature.
It's an example of changing composition to get the stability of a phase that we want.
This is carbon.
So down here, the normal form, as we've learned earlier, the normal form is graphite.
Diamond is formed at elevated temperatures and pressures.
And this is liquid carbon.
All right?
Now, to make artificial diamond, you can try putting graphite in an anvil, and squeezing, and squeezing, and squeezing, and coming up into here.
And then it becomes metastable, and then maybe, because the activation energy to jump from that sp3 hybridized state to sp2 is so high, the diamond is stable.
So you don't have to go running home to see if your diamond studs are now turning into graphite.
They won't.
Once you get into this zone, they'll stay.
The activation energy is too high.
You need more than 1/40 eV to do it, is necessary.
But in the 1950s, General Electric, its Schenectady research labs, reasoned that because carbon is so soluble in iron, they could dissolve carbon in liquid iron, and then raise the pressure and temperature, and then change to exsolve and cause carbon to precipitate out of molten iron.
That's the birth of artificial diamond.
That's how they did it.
Using this concept, but then dissolving it in molten iron.
[
'DIAMONDS ARE A GIRL'S BEST FRIEND']
I don't know how that got in there.
All right.
Here's one for mercury.
I looked it up, using the tools that you've been taught here in 3.091.
So this is a paper phase diagram for mercury.
And physicists drive me nuts, because they plot temperature versus pressure.
I want pressure versus temperature.
So I just turned it around.
And so here's the liquid line.
So you know that liquid mercury is going to float on solid mercury, and there's a plurality of solid phases.
And these are all different crystal structures.
Now I've got a treat for you.
I was in Barcelona several years ago, and I saw the mercury fountain that was made by Joan Miro in collaboration with Calder.
Calder has the big sail outside of East Campus.
So Calder collaborated with Miro, back in the '30s and '40s.
And this is in honor of the miners of Almaden, where there are cinnabar mines.
And the miners fought bravely against the Fascist forces of Franco, and Miro wanted to honor them.
So here's their museum.
It's a beautiful place up on top of an escarpment over Barcelona.
And what I'm going to show you is a mercury fountain.
[MUSIC PLAYBACK] This is liquid mercury.
The speed has not been altered.
This is a liquid at room temperature, density 13.5, metallic bonds.
Look at it.
Very high surface tension.
Look at the way it shimmers.
Look at this shot.
This has not been altered.
That's the way it pours into gravity field.
It's really fantastic.
And of course, mercury vapor is toxic, so the whole thing is in a-- [END MUSIC PLAYBACK]
Oh, that's the-- I cut that snippet out of the documentary.
OK, we don't want to see that.
So they have the whole thing inside of a polymethylmethacrylate cage, to keep the mercury vapor from affecting the people that work in the museum.
Here are some phase diagrams from hell.
This is bismuth.
And look at bismuth.
Bismuth is like water, and like silicon.
So the solid is less dense than the liquid.
But look at all of these different phases.
Here's sulfur.
Sulfur is crazy.
Look, phases in the solid phase, there's even phases in the liquid!
Different phases.
They're allotropes.
Here's sulfur.
Elemental sulfur, the disulfide, hexasulfide.
It makes rings, it makes long chains.
And these are called allotropes.
They're different forms of the same element.
So oxygen is O.
There's the diatomic molecule.
And the triatomic molecule, which is ozone, these are all allotropes of oxygen.
What else do we have?
OK.
This is polymorphs.
So this is alpha goes to beta.
See, different crystal structures.
You could see with your naked eye what's going on here.
Different crystal structures.
And then this goes to lambda, and then finally, they're pouring here.
And if you get a very high cooling rate, and you've got those long chains, what happens?
You don't form the crystal.
You form the amorphous form.
So they'll call this forming plastic sulfur.
We know better.
It's amorphous sulfur.
It's not plastic sulfur.
Here's water, a complete diagram.
Now, this is in kilobars.
So there's all the different phases of water.
There's the negative line.
That's the thing here.
But at very, very high.
Look at all the different phases of water.
And they have numbers on them.
If you read Kurt Vonnegut's Cat's Cradle, there's ice nine.
You get that going, it freezes the whole world, right?
But here's, I draw your attention on this one.
You see this point here?
This is ice seven.
At 100 degrees C, it's solid.
You need 25 kilobars.
So we could go over to the lab, take some water, put 25 kilobars on it, make solid ice at 100 degrees C. And it'll be stable.
Look!
This is minus 78.
It's been here for a long time.
So now I come into a cocktail party with this, I drop this ice cube into a glass of aqueous beverage, it sinks to the bottom, and the beverage heats up to the boiling point.
That's what this is telling you!
That'll get you popularity instantly.
How did you do that?
How did you do that?
And up here is the supercritical fluid, the last thing we haven't said anything about.
Look.
If you take a gas, and you keep squeezing it and squeezing it, eventually the gas molecules are close enough together that they're indistinguishable from the liquid phase.
Or if you start here, with the liquid, and you keep raising the temperature, the density goes lower and lower until the molecules are far enough apart that you can't tell this zone, whether it's a rarefied liquid or a highly compressed gas.
The properties criss-cross.
This is supercritical.
It has liquid-like transport properties, and vapor-like chemical properties.
So you can end up doing things that are very, very different.
Now, I'm going to give you an example of decaffeinating coffee by taking the coffee, going up into the supercritical regime, which allows you to extract the caffeine, and not brew the coffee.
So you want to just get the-- well, if you just heat it up, you'll make the coffee.
So here's an example.
It's called solvent extraction.
Historically, this solvent was methylene chloride, which is also used as paint stripper.
We don't do that anymore to coffee.
They did in the '60s!
They did.
They also used trichloroethylene.
It works really well to leech out the caffeine.
How much is left in there for you to taste in your coffee?
I don't know.
Maybe that's why people were so excited in the '60s to drink their decaffeinated Sanka.
I don't know.
But anyway, so you take the green beans in carbon dioxide as supercritical, and drop the caffeine level, and then you play with temperature and pressure to go over the KSP.
The caffeine salts out, and then you recycle the CO2.
OK.
I hope I've given you some introduction to phase diagrams.
And we'll see you on Monday.
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OK.
So.
Last day we started looking at phase diagrams, and we looked at unary phase diagrams, one component systems, and one component system P versus T, pressure versus temperature.
And this is the situation for water.
Water is an exception, because it's got this negative solid equals liquid coexistence curve, but otherwise, you have large, single-phase regions here, which I've designated circle p equals 1, and then along these lines, we have equilibria.
So this is liquid goes to vapor, this is solid goes to liquid, this is solid goes to vapor.
And we know that for water, if we take the 1 atmosphere, if this is pressure in atmospheres, the 1 atmosphere isobar, then that would put this point at 100 degrees Celsius, and this point at 0 Celsius.
And then we have the triple point, which is solid equals liquid equals vapor at 0.01 degrees C, and p equals, I think it was 4.58 millimeters of mercury.
So that's the triple point.
And we looked at a variety of other one-component phase diagrams, and I think we came to a pretty good understanding.
And then up here, we have the supercritical fluids.
So I've got a break in the line here.
And for water, you reach supercriticality at 374 degrees Celsius, and up here at 218 atmospheres.
So these are not phenomenal, these are not geological pressures at all.
And so that's if we want to the decaffeinating of coffee and so on.
And so what happens here, is you have a highly compressed vapor or a highly rarefied liquid, and you end up with solubilizing power of a liquid, but transport properties of a gas.
So this stuff has really good diffusivity.
So it can penetrate into interstices and do its work very quickly, to say nothing of the fact that you're up at 374 degrees Celsius or higher.
So today what I want to do, is I want to look at two-component systems.
So component number is circle c, c equals 2.
So that means now I have to deal with pressure, temperature, and lowercase c is composition.
I want to know how things vary with composition.
So let me give you an example.
Suppose I have two substances, A and B.
So here's substance A, and I've got its one-component phase diagram.
And in this case, I'm going to put solid-liquid vapor in the conventional setting, where the solid-liquid line is-- the coexistence curve has a positive slope.
So this is A, and let's say I've got B over here, and it's got its solid-liquid coexistence curve, and so we've got its melting point, and whatever.
So now the question I want to ask is, what happens if I mix A and B?
And let's just say, here's the melting point of A, so this is at p equals 1 atmosphere.
And here's the melting point of B.
And I want to ask the question, how does the melting point of the mixture vary?
So let's say I'm going to connect these, and I want to ask, what happens if I've got an AB mixture?
So I'm going to put lowercase c here, which is concentration.
So at the one end, I've got 100% A, and over here, I've got 100% B, and I know those melting points.
Question is, what happens when I mix them?
Does the melting point-- is it just a straight line?
Is it a linear variation?
Do you go through a local maximum?
Do you go through a local minimum?
Or do you go wild?
I mean, what happens?
So a question you might ask is, suppose I made a 50-50 alloy of 50% A and 50% B.
If I knew the end-member melting points, could I predict this?
Or if not, do I have an archive?
Remember the phase diagram compendium is an archive.
What is the value of the melting point at 50-50 or 75-25?
So that's the question I want to ask.
Now this is getting really messy, because now I have to plot pressure, and I have to plot temperature, and I have to plot composition.
So this is really crazy, right?
I've got pressure, and I'm going to have composition here, and I'm going to have to have now a third axis, aren't I?
I'm going to have to have a third axis.
I'm going to need a temperature axis.
And I've got to make a three-dimensional drawing.
And this is messy.
But we're in luck, because 3091 is solid state chemistry.
So as solid state chemists, we're more interested in what happens in the solid here.
Now, I've drawn these lines with a slope, but they're really not to scale.
It turns out that, you know, you can move 10,000 feet up or down in the atmosphere and have a huge variation in the boiling point because this line is shallow.
But this line is virtually straight up and down.
You have to go to geological pressures to change the melting point very much.
And so, as a result, this is almost insensitive to pressure, to first order, right?
So we can say solid equals liquid is almost insensitive to pressure, whereas liquid equals vapor is very sensitive very sensitive to pressure.
But we don't care about this so much.
So what I'm going to do is throw it away.
I'm going to throw it away, and just say, let's look at just T versus C. Temperature versus composition on the strength of the fact that the melting point and all of these things are mildly dependent on pressure.
All right.
So now I'm going to give you three different types.
There's many, many different types of binary phase diagram.
So now we're going to be looking at C equals 2.
So instead of a unary, this is going to be called binary.
Binary phase diagrams.
And they come in all shapes and sizes, but they can pretty much be classified in several bins.
And I've made the classification.
So classify binary phase diagrams according to their bonding.
It all comes back electronic structure and bonding.
Vary by bonding.
Because bonding, then, ultimately indicates solubility, and that's what we're looking at here.
We're going to look at how well A dissolves into B, and how well B dissolves into A.
That dictates solubility.
So it all comes full circle.
So I've made this up.
You're now going to find this in the book.
So I just decided, for ease of teaching 3091, is I'm going to just give the different types.
And so I call them type one, type two, and type three.
You're not going to find that in the books.
I made that up, because it's simple.
So type one.
What's type one binary phase diagram, according to me?
What is the type one?
It means complete solubility.
So A and B are completely soluble in one another as solid and liquids.
That's the first thing that you'll see on a type one phase diagram.
And the second one is, a change of state is present.
And you know the only change of state we care about here is solid goes to liquid.
So we're going to show solid goes to liquid, and the thing is totally, as they say in California, totally soluble as liquids and solids.
Now, here's some bonding rules that you can think about.
So what would be the characteristics of two substances A and B that would give complete solid solubility?
Well, one thing is they'd have to have identical crystal structures.
That would heighten the chances of this.
Identical crystal structures.
So if they're both FCC metals, you've got a better chance of making an infinite variation in solution composition than if one is an FCC metal and the other is a BCC metal because FCC will just sit on the lattice site.
But there's more fine structures.
Second one is similar atomic volumes.
So if we're going to have a substitutional solid solution, let's make sure we're replacing oranges with oranges, and not trying to put a grapefruit on a site that normally is occupied by a lemon, because there's going to be a size restriction there.
And the third thing is, even if they have identical crystal structures and similar atomic volumes, you can still run into trouble with respect to complete solid solubility if you have a large variation in electronegativity.
And I'll show you an example of that.
So if you have a small difference in electronegativity, it means there's a low propensity for polarity, and ultimately no chance of electron transfer.
So I think it's pretty cool.
So I wrote those down, but was disappointed to learn that for metals, they were enunciated about 75 years ago by the British metallurgist Hume-Rothery.
So I can't take credit for this.
This is one name.
It's one of these hyphenated British names.
He's Sir da-da-da-da Hume-Rothery.
So he has the Hume-Rothery Rule.
And he called such systems that mix as binaries forming isomorphous, meaning they have the same structure.
So let's take a look at the prototypical isomorphous phase diagram, and it looks like this.
We're plotting temperature versus composition.
We threw away the pressure coordinate.
So I got pure A on the left, pure B on the right.
Little c is concentration, so it varies from 100% A to 100% B.
And the vertical axis is temperature.
So one extreme, I've got the melting point-- this is a melting point of pure A.
And at the other extreme, I've got the melting point of pure B.
So that, we know.
And now the question is, how does melting point vary as a function of composition?
And for an isomorphous phase diagram, it looks like this: lens-shaped.
Up here is all liquid.
You see A and B mix in all proportions.
And down here is all solid.
And furthermore, it's an all-liquid solution.
It's a mixture.
And this is an all-solid solution.
And then comes the piece that if you take enough thermodynamics, you'll be able to rationalize.
I'm simply going to tell you without proof that when you have a multicomponent system, when c is greater than 1-- so we're in that situation.
Now, c is 2.
When c is greater than 1, it is impossible to move from one field, up here, this is-- I'd better put the label on it-- this is a single-phase field.
So up here, p equals 1, because it's homogeneous liquid solution, p equals 1.
When you're in pure materials, you can go from solid to liquid.
But when you're in a two-component system, you cannot move from one single-phase field to a second single-phase field without moving through a two-phase field.
And we'll get to that, what it means to move from one field of p equals 1 to another field of p equals 1 requires traverse or transit across a field of p equals 2.
And so what's in here has to be the end member.
So this must be liquid plus solid.
So I'm going to call it slush.
They have a slush in here.
The other terminology that people give this is, it's lens-shaped, right?
Looks like a lens.
So let's call this lens shape.
So we're going to use the Latin word for lens, which is lens.
But we want to make it adjectival.
So we're going to use the adjective-- what's the genetive form of lens?
Lentis?
So we will call this lenticular.
This is lenticular.
The phase diagram has a lenticular shape.
Looks like a lens.
And you know, just as over here, I showed you, every line represents a coexistence curve, right?
The lines are all coexistence curves.
These lines are coexistence curves.
The line up here is liquid equals vapor, so this one must be the coexistence of the two things on either side.
So what do I have here?
I have liquid.
Here I have slush.
So I'm going to write that.
So I'm going to write the equilibrium, liquid goes to liquid plus solid.
And this line, this coexistence curve, it's called the liquidus line.
This is the liquidus equilibrium, the liquidus coexistence curve, and so on.
And so what's the liquidus?
The liquidus is the lowest temperature.
So you pick a composition.
The liquidus is the lowest temperature at which you can have a single-phase liquid solution, OK?
So liquidus equilibrium.
And the liquidus is the lowest temperature at which all liquid is stable.
You go below that temperature at that composition, you start making solid, because slush requires that there be some solid present.
And then the lower line, it also is a coexistence curve.
So on the one side, I've got solid.
On the other side, I've got slush.
So I'm going to write that one down.
That solid goes to liquid plus solid.
And that's called the solidus equilibrium, or that's the solidus line, the solidus coexistence curve, and the solidus is the complement.
The solidus is the highest temperature at which you can have all solid present.
So you pick a composition, and I'll tell you what the highest temperature is at which you'll have a single-phase solid solution.
So solidus is the highest temperature at which all solid is stable.
All right.
Good.
And I'm going to look at a few of these things.
It's always fun to see if I'm on track with the Hume-Rothery rule.
So here's copper nickel.
They're both FCC metals.
So we've got copper melting at about 1085, nickel melting at about 1455.
And there's the phase diagram, lenticular phase diagram.
So up here is all liquid, and then-- this is all metallurgy terminology.
A solid solution of A and B, they call alpha.
Metallurgist looks at that, goes oh, it must be a solid solution, OK?
So there's p equals 2 in between, p equals 1, p equals-- Here's a ceramic system.
This is nickel oxide, magnesium oxide.
It's not metals, but they have identical crystal structures, very similar atomic volumes.
They're ions, so we have nickel and magnesium substituting for one another on the cationic sublattice.
So they must have very nearly equal sizes.
If I look at this, and I see a lenticular diagram, it means they must have similar atomic volumes, which means the dominant defect in here must be Schottky, not Frankel because Frankel needs a big difference in atomic volume.
If you have a big difference in atomic volume, we wouldn't have a lenticular phase diagram.
So here's-- mag oxide melts at 2800 degrees centigrade.
It's a great refractory.
You can hold molten iron in it.
And here's nickel oxide down here.
All right.
Here's an interesting one.
This is gold nickel.
Both FCC metals.
Shade your eyes from the lower part.
Just look at the upper part.
It looks lenticular.
It's what you'd expect, gold and nickel.
They're both, you know, card-bearing metals.
But you've seen already, with the cesium-gold, gold has a fairly high electronegativity.
And what happens is that you get over here, at about 33 atomic percent of nickel in gold, you have an atomic ratio that allows to have something that starting to approximate electron transfer.
So it's almost as though you have a lenticular phase diagram between pure nickel and this nickel-gold compound.
But up in here, it's nice lenticular stuff.
All right.
What's the next one?
Oh, yeah.
So now I want to go in and I want to start talking about what goes on inside here.
So what I'm going to do, is I'm going to blow this up.
And I'm going to start at 40%, 40 weight percent nickel, and I'm going to say, what happens if we take a crucible that's 40 weight percent nickel and copper, and we cool it from all liquid at 1300, down to all solid at 1200, pausing in that slush zone at 1250?
So we're going to take snapshots and say, what's the contents of the crucible look like at 1300, at 1250, and at 1200, and come out of it, have an appreciation for what all of this stuff means.
So let's start our experiment.
So we're going to have three crucibles here.
And we're going to start with 40% nickel in copper, and that's going to be the experiment we'll perform.
So I've got three crucibles.
And so if I look at the phase diagram, here I am.
This was 1300 degrees C, this was at 1250 degrees C, and then this was at 1200 degrees C. So at 1300, that's trivial.
1300 for this thing, it should be just all liquid.
And maybe it's copper.
I know at this temperature, it's going to be blinding white heat.
So it doesn't matter.
I could use this.
But I felt compelled, it's one of those rare opportunities to use colored chalk.
So I want this to sort of be copper-colored.
The meniscus is going to look like this, because the liquid metals have a very high surface tension.
They want to ball up.
So you're going to have a meniscus looking like this, and this is all single phase, all liquid.
And over here, I'm going to do the easy one.
If we get down to 1200, it's all solid state, right?
At 1200, I'm going to end up with something that's solid, and it's going to be polycrystalline, and these are all going to be grains of alpha.
All solid.
And I'm going to label all of these as alpha solid solution.
They all have the same composition, which is 40% nickel in copper.
And these are all 40% nickel.
Now, according to this phase diagram, when we get down into the center there, something else happens.
We end up in a two-phase regime.
And that two-phase regime is a regime in which certain compositions are forbidden because that's an equilibrium.
It doesn't allow us to have what's in between.
I think the next slide actually shows this.
So what I've designated as c2 is this 40%.
So c2 is now going to park at that point in the middle.
Whoops!
Want to get into that.
So here's where we are.
We're in the center of that two-phase regime.
So this is the solidus.
This is the liquidus.
And we're at 1250. t equals-- And we're at this value.
We started at c2, which is 40%.
But in this regime, 40% is forbidden.
If you stop at 40% at 1250, this says that the stable phases are a liquid and a solid.
But the liquid has less nickel in it, and the solid has more nickel in it.
It's like if this is a solubility limit.
In other words, if you started at pure copper, and you started adding nickel, you can keep adding nickel until you get to this concentration.
You try to add any more nickel, it's like adding too much sugar to water.
What do you have?
The stuff just falls to the bottom of the cup, doesn't dissolve.
That's this.
It's the solid.
Only instead of being pure nickel, it's still a copper nickel solution, but kind of rich.
So I'm going to call this c star.
It's the solubility limit on the liquid side.
And this one here I'm going to call c star on the solid side.
In other words, I could start from this side, and keep adding copper to nickel, and I make a homogeneous alloy until I get to this composition.
If I try to add any more copper, I jump across.
So we're here in the middle.
But now, isn't there something strange here?
It's saying I'm going to have a liquid that's nickel-poor and a solid that's nickel-rich.
And I'm just putting up here what the diagram says.
And we know that FCC metals, which is denser, the liquid or the solid?
The solid.
So down here I'm going to have alpha, OK, a bunch of grains of alpha solid solution, and up here, I'm going to have a liquid solution.
And the composition here is c star liquid, and the composition here is c star solid.
It's different composition from this.
Here the composition is equal to c2, or the 40% nickel.
So I've got disproportionation.
But the total mass, I can't have sources or sinks.
So I have to conserve mass.
So I'm going to ask you to just do the algebra.
You've got two equations and two unknowns.
I know what the n-member concentrations are.
They're given by the n's of this thing called the tie line.
The tie line ties the two ends of the two-phase region together, and then we just do a mass balance.
Well, fortunately, the metallurgists have thought about this for a while, so we don't have to go through and figure out, well, if two apples cost 15 cents-- we don't have to do that kind of thing.
We can just invoke the lever rule, and it will tell us how much of the liquid that's nickel-depleted and solid that's nickel-rich add up to this.
And so let's look at the lever rule.
Basically what's going to happen here is that the stuff that started off as 40% nickel, 60% copper, that's my initial mix, it's going to break into two layers.
It's going to break into a liquid layer, which, if you go to the phase diagram, looks like it's about 38% nickel and 68% copper, and that's what we're going to call c star liquid.
And then over here it's 45% percent nickel and 55% copper.
And that's what we call c star of the solid.
And so it's just a lever.
So there's c star.
This is the c2 that we want.
And we want to ask, what's the relative amount of the liquid phase and the solid phase?
And why they call it a lever rule is, look, if you cooled something that was at this composition, it would be 100% liquid.
And if you cooled something at this composition, it will be 100% solid.
And so it's going to be a constant variation across here.
So you use, you take this, and you know, and I want this amount, I'm taking this versus this, kind of thing.
It's a lever construction.
So we'll just put it down.
So the percent of the liquid in the crucible at 1250-- and then this is a general thing.
Whenever p equals 2, you'll use a lever rule.
Percent of the liquid is given by this one.
The composition of the solid at the end of the tie line minus the composition that you started at.
This is your bulk initial concentration, divided by the length of that tie line.
c star solid minus c star liquid.
And then multiplied by 100%.
So you get a percent, and the number here is going to be 45 minus 40 over 45 minus 32, and that works out-- going to multiply by 100, or else there will be complaints.
And then this turns out to be 38%.
So what that means is that if you take this amount of liquid here-- pardon me-- based on conservation of mass, if you drop and hold at 1250, 38% of that volume is now sitting here in the liquid, which means 62% of it by volume has turned into solid.
And the composition here is different from the composition here, but if you add up all the nickel in the crucible and sum it versus all the copper in the crucible, you still end up with 40% net nickel.
So this very powerful.
Because what you can do here is you can separate metal.
Because I started here with 40% nickel, 60% copper, and now I've got something-- the liquid face, I wrote here, the liquid phase is now 68% copper.
So can you see that if I'm clever about this, I could use this as a technique for enriching.
And if I'm thinking about recycling metals, maybe I need to know something about phase diagrams because it's a very simple way of concentrating impurities or concentrating the desirable stuff.
Let's look at one other thing here that's really very interesting.
Suppose I had something, instead of c2 at 40%, suppose I choose a c1.
So this is 40% nickel.
Now what if I take something that's 35% nickel, and I cool it down to 1250?
What happens?
If I park here at 1250 degrees, according to the phase diagram, that substance has to phase separate, and I'm going to end up with the same thing.
I'm going to end up with liquid and solid.
Now, isn't there a contradiction there?
How is it that whether I started with 40% or 35%, I end up with the same end members?
What's the missing piece?
The relative amounts will be different!
The relative amounts.
Here, look.
As I'm getting closer to the liquid, can you see that axiomatically, I'm going to make more liquid and less solid if I ended up with something over here?
But at 1250, those are the end members.
That's the only stuff that's going to be present.
Ah!
That's so good.
I think I've got some other stuff here.
Oh yeah.
Oh, this was-- yeah.
So p equals 2.
Lever rule.
Whenever you see p equals 2, you have two things you think of.
p equals 2 means phase separation.
And that phase separation means you're going to get different amounts, and the lever rule.
And why I have this Manchurian Candidate-- it's this movie.
I know there's a remake.
The remake is horrible.
The original, if you see the original, is about some fellows that were brainwashed.
They were American prisoners of war brainwashed in North Korea and they're brainwashed to become assassins on cue.
And the cue is the Queen of Diamonds.
When someone shows the Queen of Diamonds, they just go into automatic pilot, and they're supposed to assassinate the political figure.
So for you, your Queen of Diamonds is p equals 2.
When you see p equals 2, you go, phase separation.
Lever rule.
All right?
No guns, just this.
I just want the formula.
You go, phase separation, lever rule.
Whenever p equals two.
So what happens in here?
Phase separation.
Boom, boom.
Right there.
Oh, I'm not supposed to say boom, boom.
Phase separation.
Ta-da!
You know, something.
PC, whatever.
OK.
So let's keep going.
Now I want to look at a second type of phase diagram.
So the first type of phase diagram was complete solubility.
Now the second type of phase diagram has partial solubility.
So let's look at that one.
So i call this type two.
And you don't predict this stuff.
We would give you the phase diagram and simply ask you, you know, you're the specialist.
Tell me happens if I cool this to such and such a temperature?
You can tell me you get phase separation, and the composition of the two end members, and so on.
So the characteristics, the bonding characteristics here, are partial or limited solubility of A and B.
We're doing all of this for a binary system A and B, and no change of state.
So that means either always solid, or always liquid.
That's not to say that the A and B never become solid.
I'm just saying that a type two diagram is limited to a single state.
So let's take a look at the diagram.
The diagram looks like this.
We're going to plot temperature versus composition.
So pure B on the right, pure A on the left.
Concentration is the abscissa, and then the ordinate, of course, is temperature.
And so the shape of the diagram is this.
There's the coexistence curve.
It's called a synclinal coexistence curve.
So above, we have all solid.
And in here, we have two solids.
So using the metallurgical terms, this is alpha plus beta.
Two phase regime, OK?
So let's get those labels up right off the bat, so we know who's where.
So outside the coexistence curve, p equals 1.
Inside the coexistence curve, p equals 2.
And wherever I'm saying all solid, alpha plus beta, I could write all liquid, and then it goes to l1 plus l2.
So maybe we should do that, to show that we're multilingual.
So it's either all solid, or it could be all liquid.
And then this would be liquid 1 plus liquid 2.
And so why do we call it synclinal?
What's this?
This is an incline.
If I put two inclines and I synchronize them, I get a syncline.
And if I don't synchronize them-- these are two ladders.
I could prop them up.
That's called a syncline.
And if I put two ladders like this, if I'm really stupid and I fail physics, if I do this, they fall down.
So this is called an anticline, OK?
That's an anticline.
This is a syncline.
It's not a U-shape, or a hump, or something like that.
This is 3091.
This is a syncline.
Synclinal coexistence curve.
All right.
So what's on here?
What's this equilibrium, then?
Well, the equilibrium must be this equals this.
So that would be, in the case of the solid, it's the solid solution goes to alpha plus beta, or it could be the liquid goes to liquid 1 plus liquid 2.
That's the equilibrium.
You can think about it almost as a solubility.
So let's start over here on the left.
Let's pick a temperature.
Let's call this T1.
So I can put B into A, and I continue to get an all-solid solution, up to this concentration here.
What happens at this concentration?
I've hit a solubility limit.
And if I try to put any more B into A, I get a tie line-- lever rule.
In here is lever rule time, isn't it?
Lever rule.
p equals 2, OK?
So it's solubility limits.
Solubility limits.
All right.
So let's look at some examples.
Oh, here's this one, actually.
That's why I had a-- you know, if you go to lower temperature, gold, nickel-- look!
They actually phase separate.
That's shocking.
I always find this one shocking, because you think gold, nickel, nice FCC metals, they should substitute for one another.
Look what happens.
If you start putting nickel into pure gold at 700 degrees, you get the 10 weight percent, you put any more in, boom.
Right across here.
So this is going to be two phase.
So if you look at the solid, what's that going to look like?
If I'm at, say, 30%, now I look underneath and I've got a polygrain system, and I'm going to have an alpha and a beta, and an alpha and a beta.
I'm going to have two different grains.
And now, what's the relative amount of alpha and the relative amount of beta?
It's given by the lever rule.
OK. Let's look at a few other examples.
Here's hexane nitrobenzene.
And so they've even put l-alpha, l-beta.
They put the-- actually, that's probably a-- it's hard to read.
But I think it's really an F, but it didn't come through very well, so it's the beta fraction, the alpha fraction.
So you can see how you use the lever rule.
So up here it's all p. That's the not pressure, that's my circle p. Single phase, two phase.
Drop down to 290 degrees, it separates into two liquids.
With the hexane-- with two liquids, you'll actually have them floating on top of one another, right?
Well, maybe.
Depending if the density difference is tiny, you might get a dispersion.
I'll show you that in a second.
All right.
So p equals 2, lever rule.
OK?
Hexane nitrobenzene.
Look at this one.
This is surprising to me.
Potassium chloride, sodium chloride.
It's almost lenticular with a little bit of depression, but down here, it actually separates.
This is polymer.
It's a polystyrene, polybutadiene, depending on what the-- at low index, polymerization index, they mix.
At high polymerization index, the two are mixing.
You get two phase, single phase.
Changes the mechanical properties, too, doesn't it?
Now this one-- some systems actually have a lower critical point.
Oh, I meant to tell you.
You see, there's this temperature here.
Above this temperature, they mix in all proportions.
You can ever have phase separation.
This is the maximum temperature at which you have no phase separation.
So this is called the consolute temperature.
And it's usually an upper consolute.
The higher you go in temperature-- it's like saying, do you dissolve more sugar in cold water or warm water?
You dissolve more sugar in warm water.
And eventually, you get to a temperature high enough that you can mix in all proportions, right?
That's the consolute.
Some systems, for entropic reasons-- and again, if you take 560 or some of the other thermal classes, you'll understand this, but for entropic reasons, you actually have homogeneous solutions at low temperature, and at elevated temperature, you get phase separation.
That's funny, isn't it.
So this is water triethylamine, and they mix in all proportions at this zone here.
And then above a certain-- it's called lower consolute temperature.
It actually phase separates.
Doesn't matter.
All you care about is p equals 2, phase separation, lever rule.
The rest is details.
This is a real hoot.
This is water nicotine.
It has a lower consolute temperature and an upper consolute temperature.
So it's got a solubility bubble.
You see?
Out here, things are soluble.
In here, they're insoluble.
So they are emissible.
So we call this zone, this region here under the dome, so to speak, this it's called the miscibility gap, because things in there are not soluble in one another.
But with nicotine, you have a lower consolute and an upper consolute, so you have a miscibility gap that's a complete ring.
That's cool!
I like this.
All right.
So now I'm going to do a little experiment.
We have a few minutes, so I'm going to show you ouzo water.
I'm going to mix it, and I'm going to actually mix ouzo and water, and go into the two phase regime, and then out.
So this is what it is, all right?
It's oily stuff, you know.
If you're of Greek ancestry, you know what this stuff is.
All right?
But it's got licorice, it's got a fair bit of oil in it.
So what I'm going to do, is I'm going to come in here, and I'm going to start adding ouzo to water.
And they're both clear, colorless liquids.
OK, Dave.
Let's go to the-- All right.
So what we're going to do-- this is distilled water.
So I'm going to put some distilled water in here.
A little bit of distilled water.
And this is ouzo.
It's clear and colorless.
OK. Can we show you this?
Let's do this.
OK. Ouzo, it comes from Greece.
All right.
So it's also clear and colorless.
That was the point.
Not to show you the label.
See?
It's clear and colorless.
So this is clear and colorless, and the water, as you know, is clear and colorless.
So now what I'm going to do-- see, I don't have to wear a smock, or goggles, or anything.
This is great.
So what I'm going to do, is I'm going to add some of this stuff.
Clear and colorless.
OK.
It's turned milky.
Why has it turned milky?
Because we've now crossed into here, and the second phase is coming out.
But the density difference and surface tension differences are so slight, that instead of having the second phase float on the first phase, we have a dispersion, G. If you go to the dairy case, and you look at something called milk, you'll have the same thing.
Why is milk milky?
Because the fatty phase is a fine dispersion, and the particle sizes is dadadada, versus the wavelength of visible light, et cetera.
Now if my phase diagram is correct, and I'm in here, if I keep adding ouzo, I should eventually emerge on this side, and instead of having a milky dispersion, I eventually should come here, and now I'll have a homogeneous, single-phase solution.
But now it's going to be ouzo-rich, instead of water-rich.
You never thought phase diagrams were interesting.
You don't know.
There we go.
Je vous presente.
There it is.
So what we've done is, we've come across the thing.
Yeah, that's good.
Just put that there.
Now, David, please, back to the slides.
May I have the next slide, please?
That's a joke.
When you have these terrible speakers at conferences, they get up there, really nervous, and they stand up there, and the first thing they say is, may I have the first slide, please?
That's sort of a gag among scientists.
What's your opening statement?
May I have the first slide, please?
May I have the next slide, please?
OK.
So we've done this.
All right.
Now-- yeah, OK, this is just more.
All right.
So now I'm going to show you absinthe.
Absinthe is the same thing.
Absinthe also comes from the same family.
And there's a little culture here.
It contains wormwood.
The wormwood was up at around 200, 250 parts per million.
And what wormwood does, it's got a hormone they call thujone.
And thujone antagonizes the gamma aminobutyric acid, which moderates firing of the neural synapses.
Basically, we've got this GABA that regulates how-- our brains could work much faster than they do, but the GABA regulates.
If GABA doesn't work, you can start moving so fast that you get muscle thing, and you get epileptic, is one example.
But anyways, what it does, is if you drink this stuff, instead of being a stupid drunk, you become a very high-functioning, very alert drunk.
And I'm going to show you what the consequence is in art.
Anyways, the cultural piece here was that there was a destruction of the French wine industry in the 1800s due to a blight called Phylloxera.
It's like a beetle, and it ate the vines.
And in fact, the French wine industry was saved by the American wine industry.
Vines from New York state were brought over, grafts were made, and virtually all French wine today is on American stock with the exception, occasionally you'll see something says vieille vin.
Old vines.
There was some areas that were spared.
So what's that have to do with anything?
It has to do with the fact that when the Phylloxera hit, the price of wine went very high, and the people at the bottom of the socioeconomic ladder couldn't afford wine.
So they turn to absinthe.
Absinthe was easily produced.
Thirty years later, when the vines come back, the French wine industry wants to recapture the market.
So there's a Faustian bargain between the French wine industry and the Women's Temperance Union to try to disparage absinthe.
And there were a few major show trials that involved vicious murders in which it was alleged that the killer was deranged on absinthe.
And with this, they slandered the name of absinthe so badly that it was eventually banned and literally taken off the market.
I mean-- you know, we had the Rosalind Franklin commentary a few lectures ago.
I hope I never read about you doing something like this in order to help your start-up company take over market share.
That's not the way you're supposed to do it.
Anyway, so what-- now, how did they drink it?
They drank it by mixing-- so it's now back, but it's got very low levels of thujone, so it's-- David, may we go to this?
So we're going to make a mix called the louche.
It's a-- OK, here it is.
And this stuff here is beautiful.
Because it's a green color, a soft green color.
And the French even have a name for it.
They call it the green fairy, la fee verte.
So this is what absinthe looks like.
Beautiful.
Very nice.
And it's got the same kind of licorice smell to it.
And so you make the louche by one part absinthe and five parts water.
So this, at no upcharge, you get this glass.
It's a beautiful piece of, you know, late Belle Epoque glassware.
And the interesting thing is, these people were absolutely crazy, scientifically.
And so what they did is, the markings on this glass are such that if you put absinthe up to this ring here, and then you put water the rest of the way, you get exactly the five to one ratio for louche.
So you don't even have to measure.
You just put like so.
OK. Let's see.
I'm going to get this right!
It's science.
Here's our distilled water.
You see how it's milky?
So there's the louche.
And if you wanted to be a real hipster, you had this special Belle Epoque spoon.
And what you'd put on top of this would be a sugar cube.
You know, this is how it was done.
You have to know some culture.
You don't know.
You've led sheltered lives.
You pour it through here, like this.
So now I'm going to show you the art.
So David, may we cut to the slides again?
So there's the louche.
All right.
So here's the poster.
Here's the man.
He's pouring the water through the slotted spoon with the thing, and there's a very well-to-do lady, you can tell, she's got a beautiful hat, and she's well-dressed.
And he's inviting her to join him for a glass of absinthe.
L'absinthe oxygenee.
That's the name of the company.
The oxygenated absinthe.
C'est ma sante.
This is my health.
Van Gogh painted this.
This is Picasso.
The absinthe drinker.
There's the absinthe again.
This is Picasso after many absinthes.
You can see the slotted spoon and the sugar cube.
What else is there?
That's up to you.
All right.
When you saw Moulin Rouge, you may not have known all of this cultural history.
So now let's take a look at what's going on here.
There's the absinthe, and I think we've got a little-- [VIDEO PLAYBACK] -I don't even know if I am a true Bohemian revolutionary.
-Do you believe in beauty?
-Yes.
-Freedom?
-Yes, of course.
-Truth?
-Yes.
-Love?
-Love?
Love.
Above all things, I believe in love.
Love is like oxygen.
Love is a many-splendored thing-- [END VIDEO PLAYBACK] Love is like oxygen.
Chemistry is everywhere!
All right, so this is what happens eventually.
This is the poster.
This is from Switzerland.
October 7, 1910.
Gentleman, This Is the Hour!
And there's the clergyman with the Bible, and there's the green fairy.
And she has a wand.
She's been stabbed, but she's lying here with a wand.
The wand is an opalescent wand, because this stuff is opalescent.
Last thing I'll show you is some really, really cool chemistry that you must remember.
When Toulouse-Lautrec drank-- let's go back, David, to the thing.
So when Toulouse-Lautrec drank absinthe, and he drank lots of it, he didn't like the milky color.
So he wanted to make the milky color disappear.
So I'm in the two-phase regime, and I've got an oily phase.
And so how do I get rid of the oily phase?
What he did is he added cognac And why did he add cognac?
Not because he was a hard-drinking alcoholic.
It's because if you've got a fat phase here, and you've got an aqueous phase here, and if you add alcohol, you've got CH3CH2OH.
This can bond to the water by a hydrogen bond, and this aliphatic tail can stab the fat and bring them into solution.
That's why you have these recipes.
Do you ever wonder why the recipe says, add brandy, or add this, and then two steps later, it says flame?
You say, geez, I just put the brandy in, now it's vaporizing away.
Isn't that kind of stupid?
No, it's not!
This is what you're doing.
You're cosolvating.
If you're ever making a cream sauce or something, and all of a sudden, everything just curdles?
First you scream.
You go, ahhh!
Phase separation.
The second thing you do, is you get some of this, and you cosolvate it.
All right.
So let's see what Lautrec did.
Lautrec-- I might have to dilute the volume here.
We've got quite a lot in here.
So just to make the point.
So here's the louche, ad now we're going to add cognac.
He called this drink, Le Tremblement de Terre.
The earthquake.
He had a walking stick a meter long that was hollow, and it always had this in it.
Look!
Look at that.
Isn't that something?
So now it's this beautiful blonde, golden color.
It's clear.
So this is all phase separation and so on.
At the end, it's all about chemistry.
All right.
We'll see you on Wednesday for our wrap-up.
Good.
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OK, OK. Settle down, settle down.
All right.
Today is the last lecture.
You think you're happy?
I'm happy.
So what I'm going to do, I'm going to teach for a little bit, wrap it up, and then I've got some comments about the final exam, and then some personal observations, reflections on the class.
So of course, the main announcement is the upcoming celebration of celebrations, a week from yesterday.
The final exam will be Tuesday, 15 December, 9:00 to noon.
I know 9:00 a.m. is kind of early for some of you.
Use the buddy system.
Get to that room on time.
9:00 to noon in the Johnson Athletic Center.
I'll have office hours on Monday, 3:00 to 5:00, and Hilary will find a room.
We'll post it, and you can swing by, if you want.
I guess the other thing I should say is just a reminder about the course evals.
Please take a few minutes and fill out the course evals, if not for me and for your successors, but certainly for our TAs, so that I can write nice things about them when the time comes.
So last day we looked at binary phase diagrams.
And we looked at two different types of binary phase diagrams.
Type one, which is the lenticular, and type two, which is synclinal, and in both instances, we came upon two-phase regimes, and when p equals 2, that causes you to start thinking along the lines of phase separation.
Phase separation then invokes a tie line.
The tie line tells you the composition of the two phases present, and the lever rule tells you the relative quantities, the relative amounts, of the two phases present.
What I want to do today is talk about the last type of phase diagram that we'll speak about in 3091, and that's called type three.
I came up with these very original, easy to remember labels.
So type three.
What characterizes type three phase diagrams?
Well, we have partial solubility of A and B.
We're doing all of this.
A is one of the components, and B is the other component.
So partial solubility of A and B, and change of state.
And the result of this set of characteristics is freezing point depression of both components.
And that's different from the situation in type one.
In type one, you see, if we have B on the right, then adding at A to B, if this is pure B over here, and this is pure A over here, adding A to B causes the freezing point of B to fall.
But adding B to A causes the freezing point of A to rise.
In this type three, we're going to have freezing point depression of both B and A.
That's the difference between the type one and type three.
And actually, I'm going to approach this by a hybridization of sorts.
So I'm going to consider type three a hybrid.
It's a hybrid diagram of the lens and the syncline.
I'm going to blend the lens and the syncline, and show you how to do that.
So let me get one of those up here, so that we can look at it.
So what I'm going to do here, this is a great diagram to look at.
It's gold nickel.
And at the top, we have a type one, and at the bottom, we have a type two.
And what I'm going to do, is I'm going to mix them.
I'm going to hybridize them.
I'm going to bring this bottom one up and have it touch the lens.
And so we get is something that looks like this.
We'll plot temperature versus composition.
Left-- so here's temperature.
Composition.
On the left I have pure A and on the right I have pure B.
And we'll start here with the melting point of pure B.
And when I add A to B, I'm going to get the depression that we see in the type one diagram.
We want to start off like this, and use everything we've learned up until now.
So up here, we have all liquid.
This is all liquid.
And over here, this is a solid solution.
This is pure B, and this is B that has a little bit of A in it.
So we're going to use the old metallurgical definition and call this beta.
Beta meaning, beta is a solution of A and B, but it's very rich in B.
It's almost pure B.
So I'm going to call this a B-rich solid solution of A and B.
And we know from last day, you can't go from p equals 1, let's get those labels up here, you can't go from p equals 1 to another p equals 1 without going through a p equals 2.
So this is single phase and this is single phase, and that means this one here must be two phrase.
And what are the two phases?
The two phases are the phases that are on either side.
So this must be liquid plus beta in here.
And that means-- phase rule-- p equals 2, tie line, lever rule.
So that's what's going on in there.
And then the difference is, on this side, if this is t melting point of A we're going to start off with the same concept here.
That we're going to get freezing point depression of A, as well.
So we'll have the liquid.
Over here we'll have an A-rich solution.
So then I'm going to call that alpha.
So alpha is an A-rich solid solution of A and B.
And then that must mean that in between here, we have liquid plus alpha.
And these keep going.
And in addition, down at the low end, I'm going to start with a syncline.
I'm going to start with a syncline, coming up like so, and coming up like so.
See what I'm doing?
I'm taking the lens.
I'm going to meet the syncline.
So what's here?
This is alpha, this is beta.
What's inside the syncline?
This is miscibility gap, so this is alpha plus beta.
So let's keep the phase labeling here.
This is p equals 2.
And look at how things have worked out so nicely.
p equals 1, p equals 2, p equals 1, p equals 2, p equals 1, p equals 2.
So nice.
And what happens when all this intersects?
It intersects like so.
Try to do this, stretch it a little bit.
So these two lines connect.
They cross at this point.
They touch at this point, the don't cross.
They just touch.
And then this arc comes up like so, this comes down like so, and now we have a line across here, and this comes down like so.
OK?
So the syncline stops at this value, which is the minimum temperature dictated by the two liquidus.
We have a liquidus here and a liquidus here, and this gets you down to the lowest temperature.
And let's look at the phase mix going in this direction.
p equals 1, p equals 2, p equals 1, p equals 2.
What's happening at this point here?
This point, we have three phases in equilibrium.
We have liquid, alpha, and beta.
And this is the special point here.
This is called the eutectic.
This is the eutectic, which is the temperature at which we get liquid, all liquid, at the lowest value.
So I'm going to call this here, I'm going to call this point e, and e is the eutectic.
Lowest melting point on the diagram.
Comes from the Greek, meaning easy melting.
And at the eutectic, eutectic is special.
Here we have p equals e. It's a triple point.
The eutectic is a triple point.
OK.
So now let's take a look at some of the examples of such diagrams.
All right?
This is a good one now that winter is upon us.
This is the antifreeze diagram.
So on the left, I have pure water, and on the right, I have pure ethylene glycol.
And as you can see, if you add glycol to water, you get freezing point depression.
If you add water to glycol, you get freezing point depression with a eutectic here at roughly two-thirds.
This is in volume percent, because that's practical unit.
When you mix glycol with water, you do so by volume, not by mass, or certainly not by moles.
So if you mix two glycol with one part water, you'll end up with a eutectic of minus 69 degrees C, or minus 92 Fahrenheit.
So that ought to take care of your driving needs.
All right?
Now practically speaking, we put an equal volume mix.
So it's about a 50-50 glycol-water water mix.
And that gives you liquidus depression down to about not quite minus 40.
It's about minus 35 Fahrenheit, which ought to take care of most peoples' driving needs.
Remember, what we're trying to do is to prevent the liquid in the engine block from freezing.
because of if it does freeze, you know that ice has a larger volume than water, and that energy is so great that you'll crack the engine block.
So you have to prevent it from freezing, the water from freezing, and so you add the glycol.
If, by chance, the temperature dropped to minus 50, you'd still be OK, because you'd be in the slush zone, two-phase regime.
You'd have some water ice forming, but you'd still have slush.
And by the way, why does this not look like this?
It turns out that the amount of solid solution behavior of a glycol in water is vanishingly small.
So this point is slammed up over here.
So that's why you don't see the right side of the lens.
And likewise, on the other side, the amount of water that's soluble in glycol is vanishingly small.
So according to this scale, you don't see the beta or the alpha phases here.
It looks like it's pure, but it can't be, because of the phase rule here.
Now, you might say, well, why don't we run pure glycol?
Because somebody might say, I'll just run pure glycol.
Well, here's the problem with running pure glycol.
First of all, if you get beyond two-thirds, the melting point rises.
But then you could say, well, but pure glycol, the solid glycol, has a smaller volume than liquid glycol, so it wouldn't freeze and crack the engine block.
But there's another problem here, and that is that the viscosity of water is 1 in these units, centipoise, and glycol is 21.
So you'd probably burn out your water pump by running pure glycol, because it's so viscous, the water pump is going to be fatigued by running.
And here's the point.
This is the 50-50, minus 37 degrees Celsius.
OK. And so what I've done, is I've superimposed the solid-liquid diagrams with the liquid vapor diagram.
And this is one of the times when nature is kind.
So when you add glycol to water, water is on the left here, as you add glycol to water, you get freezing point depression, and you get boiling point elevation.
So adding glycol to water opens up the liquid range.
You want to stave off boiling, because boiling will because lousy cooling, and the engine will overheat, and you want to stave off freezing, which will cause the ice to form and crack the engine block, and you get both.
Then you put the pressure cap, and then jump this up to about 265 degrees Fahrenheit.
So this is a very good example of management of the properties of the coolant through chemical change.
And the binary phase diagram eutectic type is a critical piece here.
Again, apropos of the weather, if you get ice on the sidewalks, if this rain freezes, it's terrible here.
Those of you who are in the Northeast for the first time, this is bad.
Not with what you see now, but when this freezes, it's a skating rink, and you can't see it.
It's black ice.
You're walking along, next thing you're on your-- you're in the horizontal position, all right?
So we can add sodium chloride to water, and when we do, we cause this, a eutectic type diagram, which can give you freezing point depression and prevent ice formation, and if you get enough sodium chloride in, you can get the temperature down to minus 21 degrees Celsius.
And I've color coded all of these.
This is the liquidus, this is the solidus, and then this line here, I can't remember if we gave it a name last day, this is called solvus.
This is the solvus.
This is, we know, liquidus.
This is liquidus, and this one here is solidus.
Yeah.
OK.
So this is all color coded.
And notice that minus 21 Celsius is not too different from 0 degrees Fahrenheit.
Turns out, that's how the Fahrenheit scale was born.
Daniel Fahrenheit was a glassblower in Leyden who also made exquisitely accurate mercury thermometers.
And he used ammonium chloride, not sodium chloride.
And the lowest temperature he could get for all liquid with ammonium chloride, he pegged as 0 on his scale.
And it's roughly this.
It's a little bit different, but you can see the similarity.
Now, you need two points to define a scale.
So he took this mercury in bulb thermometer, put it in the eutectic mixture of ammonium chloride and water, said, that's zero, calls in his wife, tucks the mercury in bulb thermometer under her armpit, and says, that's 96.
And that's the birth of the Fahrenheit scale.
And that's why we have water freezes at 32, and it boils at 212, and all the other nonsense.
You know, it's 98.6, but here it's a little bit colder.
Now, why he chose 96 and 100 instead of 100 or something, you know, you could read in the history.
Maybe he wanted something that was a power of two.
I don't know what the reason for it was, but that's why we have the Fahrenheit scale.
OK. Now, maybe you're a little bit hypertensive, so you don't want to throw sodium chloride on your walk, so you can put potassium chloride.
But you won't get much of a freezing point depression, And point of fact, we usually salt our roads with calcium chloride, which gives you two advantages.
Number one, very, very low eutectic.
Number two, it's a lot cheaper than anything else.
So it's the cheapest one to use, and that's why we use it.
Christmas is coming, and I told you that phase diagrams would save the day.
We saw that the cubic zirconia was stable in pure zirconia only at temperatures above 2000 centigrade, which is useless for a ring, or other such pieces of jewelry.
But by the addition of calcium oxide, we stabilize the cubic form of zirconia all the way down to room temperature.
And now we've got the high index of refraction, clear, colorless gemstone that passes for diamond.
Poor man's diamond.
Cubic zirconia.
And you can see in this regime here.
And notice that cubic zirconia-- here's a eutectic.
Why I'm showing you this is that the zirconia- calcia system ultimately becomes eutectic.
And the eutectic is at about 2300 degrees C. But over here is the solid solution.
Notice, if it's a solid solution, that means it's p equals 1.
Single phase.
As it must be.
If it were two phase, the gemstone would be frosty, right?
You'd have all those internal surfaces, and they'd be reflecting.
So this is clear and colorless.
And what's it bounded by, this region?
Cubic zirconia plus calcium zirconate.
So p equals 1, single-phase field, p equals 2, dual-phase field.
And what's on the other side?
Over here it's tetragonal solid solution, single phase.
Here's tetragonal solid solution with cubic zirconia, two phase.
Here's cubic zirconia, one phase.
Here's cubic zirconia with calcium zirconate.
Two phase.
Same concept here.
p equals 1, p equals 2. p equals 1, p equals 2.
OK.
So now you know.
When you go in jewelry store, ask the guy, is it single phase?
He'll say, get out of here.
Actually, you know how they test?
The index of refraction is so high with zirconia.
It's a little bit less than diamond.
Diamond's up around three.
Cubic zirconia is over two.
One of the ways they test is, diamond has a very high band gap, about 5.5 electron volts.
But it's got the remarkable property, even though it's an electronic insulator, it's a very good thermal conductor.
Most materials that are poor conductors of electricity are poor conductors of heat.
But diamond, for reasons I can't explain with just the utility of a 3091 education, has a good phonon transmission properties.
And the way they tell cubic zirconia from diamond, is they actually put a couple of tips on either side of the stone, and they heat one tip, and measure the time lag for the thermal wave to get to the other tip.
And if it's cubic zirconia, it takes a long time, because cubic zirconia is an insulator.
If it's diamond, it's almost like metallic conduction.
And that's how they can tell a fake diamond from a real diamond.
Here's, you know, I've been drinking from these cans all semester.
And this is the alloy system, aluminum magnesium.
So we put about 1% magnesium into the aluminum.
So here's pure aluminum over here.
It goes down to eutectic.
This is pure magnesium over here.
They both melt at roughly the same temperature.
And by putting about 1% magnesium in here, it changes the metallurgical properties favorably.
Notice we don't want to put 4% or 5% magnesium in here.
Otherwise, as we get down to room temperature, we'll enter a two-phase regime, and that means we won't be able to process.
Because this can is made from a single sheet of aluminum.
The body of the can.
There's two pieces, the top and the body.
There's no third bottom here.
Start with a flat sheet of aluminum and it's punched deep, drawn down, up, down, and it gives you the whole shape.
And you have to have, you know, this is FCC.
You have to have the glide.
Dislocations are gliding, the slip systems are operative, and the magnesium gives you just a little bit of strength.
But if you get into the two-phase regime, the punch, instead of making deep drawing metal, will end up punching holes.
So what's the difference between drawing and punching holes?
It's all in the metallurgy.
So you can see, this is where we are.
OK. Ah!
Microelectronic devices.
How do we hold all those wires together?
With solder.
And this is the phase diagram for solder.
It's made of lead and tin.
And we use eutectic.
The eutectic here is about 62 weight percent tin at 183 degrees Celsius.
I took a class in high school called Industrial Arts.
It was shop.
We had it drilled into us, the eutectic point of soft solder is 374 degrees Fahrenheit.
I will know that on my deathbed.
Plus the wavelength of copper K-alpha radiation.
1.5418 angstroms.
Yeah.
So what's the magic here?
The magic here is, the solder must be low-melting enough that when you're-- remember, you've got your microelectronic device.
You've already got your chip, and you're putting wires.
You want to put enough heat in to solder the wires, but you don't want to have to raise the temperature so much that you can damage the rest of the device.
So you say, well, why don't you get something that's lower melting?
Well, if it's lower melting, you know from running your laptops and charging your cell phones, they get warm.
So it's a Goldilocks problem.
If the melting point of the solder is too low, you might encounter that temperature in service, and pop all of the joints.
If the melting point of the solder is too high, then you have to use so much energy to make the joints that you damage some of the microelectronic componentry.
So this has to be in the right temperature range.
The problem with this excellent, excellent solder is that it contains lead.
No one's going to get lead poisoning from using a device.
But the disposal of the device ends up introducing the lead into the landfill, and then if it soaks in an aqueous solution, you could get leeching, et cetera.
So there's a lot of research right now on trying to find other solders that operate.
And there are other solders, but they're very expensive.
So that, how do you find something that's functional the way this is, nontoxic, and cheap?
They can give you any two of those, but to get all three of them to line up?
Requires research.
You see the alpha phase, beta phase, everything there.
All right.
And here's the lead tin system.
And let's just take this one that happens to be 40%, and let's see what happens.
We'll go through in time.
So we start here, all liquid regime.
And so this bubble shows what you might see on a microscope slide.
So it's all white here, meaning that it's homogeneous single phase.
And then you drop into the two-phase regime, just below the liquidus, and you start seeing these blobs.
Those little blobs are the primary alpha.
That's the solid solution, right?
Because we're in the two-phase regime, so we've got a tie line here.
We get down to a lower temperature, there's more of this stuff.
There's more of this stuff for me, because it's forming in time.
It takes you time to traverse this.
This isn't parking and waiting.
This is just cooling something.
Imagine you're soldering, so you took the temperature up to here, now you're going to let it cool.
Once you cross the eutectic temperature, now it's alpha plus beta.
So now you get this lamellar structure.
Stripes of alpha, stripes of beta.
But the previous blobs that formed the primary alpha, they don't disappear.
The only way they can disappear is by solid state diffusion.
That's going to take a long time.
So I can look at the microstructure here, and I can predict-- not predict-- I can go back and calculate what the cooling rate must have been.
If I cool very slowly, I will have nearly the equilibrium structure.
And so that's why, when I told you about, say, analyzing the records from the World Trade Center, I can look at the metal, and look at the phase distribution, and tell you what the thermal history was.
And so for example, there was some cock-eyed theory saying that the steel was burning.
Somebody said that the steel melted, and so on.
You can look and you say, that did not happen.
Temperatures went fairly high, crossed into some different solid phases, went from BCC to FCC and then back.
But didn't melt.
Because if it melted, we would have seen this kind of structure.
So this is-- every time a plane crashes, this is the metallurgical investigation that is taking place.
The history is written in the microstructure.
So this is a micrograph that's actually taken from some solder.
This is 50 weight percent.
The slide was 40, but this happened to be 50.
It's the one I could find easily.
So 50 tin, 50 lead, and these dark zones are the primary lead-rich alpha.
And it's all metal.
So how do I get the dark and the light?
First we polish, because it's an optical microscope, which has a really crummy depth of field.
So you've got to make it flat, and you polish it, and now you've got a mirror, and you look, and all you see is your eyeball.
So what you do then, is you etch it with some acid.
And the lead will etch at a different rate from the tin.
And the contrast, then, is what you see here.
So this is all the blobs of primary alpha.
Then you go below the eutectic, and, you know, this cute little schematic of stripes-- this is the textbook version of it, but this is the real version of it.
And you see, it's a distorted lamellar structure.
Alternating stripes.
Because you're trying to make alpha and beta.
And so you're making alpha and beta, and the system's trying to make both of it.
So it makes it in striped form called lamellar microstructure.
So I can look at that and I can say, wow.
That must have been cooled at a fairly rapid rate, because you've got a large amounts of primary alpha.
If it were cooled at a very slow rate, the thing should be all lamellar structure, right?
If you're down here, there should be no alpha.
This is not the equilibrium phase.
So this alpha is trapped there.
It's a victim of its history.
So there it is.
That's cute.
That's a nice, it's a beautiful graphic.
See?
There it is.
All right.
So I'm going to show you some phase diagrams from hell.
This is, you're going to be on airplanes.
The airplane alloy is aluminum copper.
It's about 4% copper and aluminum.
Now remember, I just told you about 1% magnesium and aluminum to give you single phase.
In the case of the airplane, we want to make the wings strong.
So what we do, is we come down into here, which is a two-phase regime.
And if we're clever about it, we can control the particle size of that second phase.
And by controlling the particle size, we can cause grain boundaries, which can arrest the dislocations and make sure the wings don't fall off.
This is aluminum manganese.
This is the one that Danny Schechtman worked with to get the quasicrystals.
Look at all the phases.
He's running out of Greek letters by the time you get across this diagram.
Look at it.
But one thing holds.
The two-phase regimes are always bounded by single-phase regimes.
Look.
Here's liquid, here's a two phases, there's a single phase, two phase, single phase, two phase.
Beautiful.
All makes sense.
Now, what happens if you try to take a metal and mix it with a nonmetal?
They don't want to mix.
Look at the big miscibility gap here.
Molten iron doesn't want to mix with molten sulfur.
This is a liquid-liquid miscibility gap.
Over here, you can add some sulfur to iron, you get a eutectic, get a line compound.
And then over here, no dice.
Does not want to mix.
And look at all the solid phases.
Why do we have all these solid phases?
Different crystal structures.
OK.
So now what we're going to do, is I'm going to talk about application of phase diagrams to the making of champagne.
Because champagne relies on phase diagrams.
And then I'm going to make some general comments.
But I'm going to actually show you some champagne.
So before I get into my commentary, I'm going to get some champagne chilling.
So first I have to show you how to chill champagne.
First I will show you how not to chill champagne.
This is how not to chill champagne.
And it's not because, OK, we can put it in there.
Why?
Because, as I showed you last day, with respect to the cooling of the engine coolant by the radiator, if you have-- if this a bottle here, here's the contents, the precious contents.
And I have this with blocks of ice.
The contact between the ice and the bottle is sporadic, and I've got air in between, so the quality of the cooling is very poor.
So instead, what I do is I rely on the phase diagram.
So what do I do?
What do I know about the phase diagram?
if I pour water in here, I will thermostat the entire system, liquid plus solid, at 0 degrees.
Because the ice is probably, I don't know, minus 3, minus 4 Celsius?
But the air is 20, and it gets in there, and it's very few atoms.
It's crummy heat transfer.
Instead, if I flood this with water, the whole thing is thermostatted at 0 degrees C. So now I've got a good delta t, because I started here at about 22 degrees C. I've got a delta t, and it's a fixed law, right?
It's going to be what?
It's going to be the flux.
The heat flux is going to be minus thermal conductivity, dt by dx.
It's Fickian type motion of the thing.
So let's do it.
This is how you properly chill-- I'm told it also works for sodas and fruit juices.
This technique, I mean.
But I do know that it works with champagne.
Now that's working beautifully.
I can just feel the heat being extracted.
OK.
So now let's talk about champagne.
All right.
So we're going to talk about champagne and a great inventor, and the inventor is the Widow Clicquot.
This is the Veuve Clicquot champagne.
And veuve is the French for widow.
So Veuve Clicquot, her real name was Nicole-Barbe Ponsardin.
And in 1798, she married Francois Clicquot.
This is a big French winemaking family.
And he died and left her a widow at the age of 27.
She was very unusual for a French woman in the 18th century, because they were not involved in society.
They were certainly not involved in business.
This woman was different.
She decided to get under the hood, and she took control of the winery.
And I've written here, bold, imaginative management.
There's a fantastic book about her.
It's actually a very good book portraying late 18th century, early 19th century France.
And these are among her accomplishments.
Marketed champagne to all the great courts of Europe.
Champagne was a regional beverage, drunk only in the Champagne region.
And she created the myth, propagated the myth of champagne.
You use it for festive occasions, bought land in the best vineyards, quality control.
It's about the chemistry.
Fought fiercely against counterfeiting.
As she started getting champagne more recognition, people started making sparkling wine and labeling it as champagne.
She used the court system to take them down.
Established strict quality control procedures.
That's all about defending the brand, and what's the quality control?
It's the chemistry.
Produced the first rose champagne.
The Pinot Noir grape has a black skin.
Depress, and then take the juice, and away you go.
What she did, is she let the skin sit with the juice for long enough to give a just a rose tinge.
And then oversaw the invention of new technology.
Remouillage.
And that's where the phase diagrams come into play.
Phase diagrams give us champagne we enjoy today.
So what's the problem?
Champagne is cloudy.
Why is it cloudy?
Well, here's the chemistry.
Here's how you make, this is how you make all wine.
It's one-stop shopping.
You take the grape.
The grape has sugar in the grape juice.
And yeast that attacks the sugar is in the skin.
So the yeast attacks the sugar to make alcohol and CO2.
So ethyl alcohol and CO2, plus some insolubles.
There are some insoluble products of that reaction.
And there's two types of insolubles.
Those that will settle, they're called sedimentary, and those that won't settle.
They're called suspended.
And how do you get rid of the sedimentary stuff?
By mechanical separation.
You let the juice sit, this reaction is taking place, and the sedimentary stuff is turning into gunk, and it sits at the bottom.
And then periodically you siphon.
You siphon the juice off, and you leave all the crud at the bottom of the barrel, and then you go for a few more months, and then you siphon again.
Well, that works for the sedimentary stuff, but it won't work for champagne, because as you're siphoning, you're letting out all the gas.
You see, you're trapping this gas.
The champagne has CO2 that's naturally occurring.
It's not carbonated.
It's carbonated naturally.
Not carbonated, man-made carbonation.
It's natural carbonation.
So how do you separate the juice from the crud without losing the gas?
That's the problem.
And by the way, the other stuff that's suspended, they use tricks.
You know, here's the free body diagram.
Remember, I showed you this for milk.
So what they do, is they add things like eggshell, and the eggshell acts as a coagulant, so the tiny particles that will float agglomerate to become big enough that then they'll settle.
Tiny particles of a given density will float, large particles of the same density will sink.
So you need to coagulate, and do that without spoiling the taste.
Fantastic chemistry.
So she comes along with the answer.
By the way, the early champagne glasses were all cut crystal.
Why were they cut crystal?
Because the champagne was cloudy.
It tasted fantastic, but who wanted-- it looked like swill!
So how to make it clear?
So what they did, is they made champagne bottles with the deep ruts, and they still do to this day.
They have a deep rut here, because the idea was, if you opened it carefully, you wouldn't disturb so much of the crud.
But it gets churned up by the gas.
So what she reasoned was, let's turn him upside down, and let's have the crud settle on the underside of the cork.
And furthermore, 45 degrees is the optimal angle.
You know this from your Newtonian mechanics.
And then you turn them, so you spiral the bottle.
Turn them, and have all the crud collect on the underside of the cork.
I know what you're thinking.
How are you going to get the cork out?
Now you take the cork out, you're going to lose not just the gas.
You're going to lose the liquid, too.
Ah, that's where the phase diagrams come in.
So here's a phase diagram of ethanol water.
And it's eutectic.
Look at, way down here.
In fact, ethanol makes a really good antifreeze.
Look!
A little bit of-- go way down here.
Now, most wines are somewhere between 10% and 15% alcohol, that gives you very modest freezing point depression.
Maybe about minus 5 Celsius.
Which is why if you forget, you put a wine bottle in the freezer and you forget about it, and you've got your freezer really cranked to, say, minus 5 Fahrenheit, you'll freeze the wine, and it'll expand, because it's 90% water, and push the cork out, make a mess.
But remember, you only get down to about minus 5.
Remember, I showed you this one?
It gets down to minus 21.
So what she reasoned was, what if you took the bottle upside down?
You've got all the crud on the underside of the cork.
What if you took the bottle and put it into not water, but what if I put sodium chloride in here?
I still have my ice cubes, only the temperature is now minus 21.
And that's below the freezing point of the alcohol, and now I've frozen an ice plug in the neck.
So now there's an ice plug between the crud and the cork.
So now I can turn the bottle like this, just melt ever so slightly, and the pressure in the bottle will blow the ice plug, take with it the crud out, and then I quickly put the cork on top.
And that's how every bottle of champagne is made.
And it all comes from these phase diagrams.
See?
So you put it in brine.
Minus 21.
But this thing is frozen already.
That red is frozen.
And now I've turned it right side up, out it goes, takes the crud with it, put the cork back on.
So it's a combination of this phase diagram, this phase diagram.
Both eutectic phase diagrams would give us clear champagne.
Positively brilliant.
So there it is.
The invention and-- oh, by the way.
They call this-- the English word remouillage is riddling.
So they have people that actually go, and they, these wooden racks with the slots in them at 45 degree angles, and all these bottles are pointed downward, 45 degree angle.
Guys come in every day, quarter turn, quarter turn, quarter turn.
French are very traditional.
Few champagne wineries have motorized, so they have this thing called giropalette.
Gyrating pallet.
The California champagne wineries call it VLM.
Very Large Machine.
I kid you not.
I didn't make that up.
In California, the giropalette is called VLM, which is Very Large Machine, dude.
I guess.
That's what they say.
California.
All right.
So let's have a few words now about the final exam.
So you know, it's going to be Tuesday, 9:00 to noon, Johnson Athletic.
Three hours.
You have three full hours, not three times 50 minutes.
Three full hours.
But I'm not going to give you three times the work.
It's sort of a hybrid between intensive coverage since T3, so sort of like a T4 and a redemptory part.
It'll go back, extensive coverage of everything.
One aid sheet.
One aid sheet.
I'm saying it three times.
One aid sheet.
Do not come in there with a bunch of aid sheets.
Well, I didn't know!
I thought we could have-- no!
One aid sheet.
That's all you need.
And the TAs are smart.
You know, every year somebody tries to bring the origami and all that stuff.
And they get caught, and then-- you know.
Bring your periodic table, table of contents, calculator, a pen.
No headphones, no audio.
Please.
Because people are writing, they're trying to concentrate.
I don't want to get into a dispute over whether your music is too loud for your neighbor.
So let's just, out of courtesy, no audio.
You can go three hours without your music.
All right?
What else.
This is going to be comparable difficulty to the monthly tests.
Of course, read the entire exam.
You've got plenty of time.
Most people are leaving 11:15, 11:30.
It's not a race against the clock.
And make sure you show your work, justify your conclusions, solve algebraically.
Stay confident.
Do your own work, academic honesty, all that applies.
Now, your overall grade is going to be based on many factors, including the trend.
So we're going to look at how you performed throughout the semester.
You started off really strong, and you crash and burn on the final, or you've done the math and you say, I only need a 37 on the final, and you go in there, shooting for a 37, and you miss and you get a 35, and your total average is 49.5?
You go down.
Because you've got a 49.5 and you wrote a 35 on the final.
I'm just telling you.
Claim, you can have your exam papers back.
But don't come and see us before Christmas.
You can have them back starting IAP.
So starting January 4, you can have your exams back, if you want them.
And there's no time limit for appeal, so you don't have to be bursting into your TA's office on January 4 if you see that we've misgraded something.
You can come and see us.
There will be some security measures in place.
It has not escaped my notice that some people have attempted to erase answers that were written in pencil, and then bring in this newly brilliantly solved problem for regrading.
And that's, of course, qualifies as academic dishonesty, and I'll take you to the committee on discipline if you try it.
But just to inspire honesty on a random basis, we'll photocopy about 5% of the exams.
So if you want to try this, the question you've got to ask yourself is do you feel lucky?
Do you?
All right.
So now it's time for some personal observations.
So I wrote them down.
OK.
So first thing is, you've come a long way.
Syllabus is full.
You think about where we were on day 1, you know, this is equations, stoichiometry, all that baby stuff.
And you know, look.
So be proud of what you've learned.
Number two.
I wish you much success on the final.
Remember, no grade that I assign to you, or any of my colleagues assigns to you has any relevance to your value as a human being.
It's just a grade in a subject.
That's all it is.
What's the worst thing that can happen?
Suppose you get all Fs, and you're required to withdraw from the Institute.
So what?
So what?
You know?
You're all smart people.
I know you're smart.
I used to chair the Admissions Committee.
I know how smart you are.
If you're getting all F's, this is not the right place, it's not the right time for you to be here.
Look, I have no degrees from this place.
I feel great.
I've been teaching here at MIT for 32 years, and I've never enjoyed teaching as much as I have this semester.
This has been really a blast to teach, really.
I can't believe I said that with a straight face.
So of course it's not true.
I want to think my staff.
My TAs, sitting here.
The tutors who helped out.
Professor Demkowicz, who subbed for me that time, when I had the visit of President Obama.
Administrators, the audiovisual techs.
Dave Broderick, who's done a great job of making sure that we have sound and light.
And the academic media production services people, who have recorded the lectures so that on the rare occasion that you didn't make it the class, you're able to catch up.
And the last thing is that, I've got here, there's one more chemistry lesson.
You know, we've learned about the four different types of primary bonding.
Covalent, ionic, metallic, and in some instances, van der Waals.
But there's a fifth type of bond.
The fifth type of bond that I haven't talked about yet.
And it's the type of bond that's been forming in this room over the last 14 weeks.
Yeah, I know.
It's kind of sappy.
I didn't come up with this.
It came from a student.
So the fifth type of bond is-- it's not this one.
[MUSIC PLAYBACK] No, it's not this one.
It's a bond that's formed between people, and this bond has a very special property.
It can never be broken.
I know!
It came from a student.
It's very sappy.
I'll tell you the story.
I'll tell you the story how it happened.
I used to teach 321, which is a core graduate class in kinetics.
And I'd been teaching it for, I don't know, five, seven years.
And I was at a conference, and there's these two people standing there talking.
And I know both them.
One was from industry, and one of them was an alum.
So I go up, and the fellow in industry, says, hi, how are you doing, do you know so and so?
And I said, yeah.
He was at MIT.
I was his kinetics professor And the student, the alum, he says, excuse me, Professor Sadoway.
It's true that I was at MIT, but please do not say that you were my kinetics professor.
You will always be my kinetics professor.
I went, oh, man.
This is really something.
So now the tables are turned.
And, and, you know, I know you're going to go go on and do something remarkable in your life, and I hope it's remarkably wonderful and not remarkably stupid.
But you know, at some point, somebody may say to you, where did you learn your chemistry?
And you'll have to say, Sadoway at MIT.
So whether you like it or not, I will always be your chemistry professor.
So now, let's see.
I think it's time to open up the champagne.
So first thing, we have to teach you how to open champagne.
So there's the foil here.
So we'll take the foil off.
Take the foil off, very carefully.
And then there's the wire basket that keeps the cork on.
How many turns on the wire basket?
Everyone in 3091 knows.
Six turns.
One, two, three, four, five, six.
Six turns.
And so now, I don't want to point that at anybody, because the only thing between me and six atmospheres of pressure is that cork.
And Dave, if we can cut to the document camera.
The cover, the metal cover on this, has the image of the Widow Clicquot.
This is Veuve Clicquot, and this is the image of the Veuve, and she's on every bottle.
She was the one that gave us clarified champagne, so you thank her very much.
All right.
So now what we're going to do, is we're going to open this thing.
To open it, we point it 45 degrees and turn it ever so slightly.
Never point it at people.
No, really, point it away.
You'll put an eye out with this thing.
You turn it very slightly.
Actually, my hands are wet now, so I'm going to use a towel here.
Oh yeah.
Go like this.
You like the suspense, huh?
See that?
No pfff!
That's for football locker rooms.
It's very declasse.
OK. Now we have to have a glass.
So since we have clarified champagne, we're not going to use cut crystal.
We're going to use Baccarat, French crystal that's not cut.
Because we've got nothing to hide, right?
So now we're going to pour this.
And the rut now has some value.
The rut has value, because now what you can do is to pour it like so.
I've been waiting fourteen weeks for this.
All right.
So the surface tension, you see the carbon dioxide outgassing, because it's supersaturated.
Six atmospheres, not one atmosphere.
The bubbles are nucleating on the defects in the glass.
It's all about chemistry.
So there we go.
OK.
So I'm going to propose a toast to the class, to the 3091 class of Fall 2009, to wish you much success in your academic pursues, and much happiness in your personal lives.
And with that, I'm going to find out what's going on inside.
Look at those bubbles!
You know, the legend is, the blind monk, Dom Perignon, when he first drank champagne, said, I feel as though I'm drinking stars.
He's right.
Drinking stars.
OK.
This is to you.
The morphological unit of life is the cell, whether you're a flea or you're an elephant.
Right?
And we all have water as the major solvent.
We all have the same biological molecules that play a major role inside our bodies, like sugar or fat, or amino acids, or nucleotides.
We all have the same ways of making reactions work fast enough so they can succeed inside the cell, and controlling the specificity, and they're conserved from bacteria to humans.
We all use the same vitamins on your vitamin bottle.
And those are also conserved in general from bacteria to humans.
And we also all have the same major regulatory mechanisms, although as you go from prokaryotes to eukaryotes, things get more complex.
And so what we cover in the course is, in fact, most of the things we talk about are bacterial systems because they're better studied but can be extrapolated between the two systems.
So are these things ordered inside the cell?
And how do you study them inside the cell?
That's the major focus of a biochemist.
You want to understand things at the molecular level, but then you need to understand what you see at the molecular level, how does that relate to what's going on inside the cell?
And so you need to go back and forth between inside the cell and understanding the chemical and physical principles by which all the reactions work.
If you look at the vitamin bottle, you have all these vitamins you eat, like, what do they call them?
Biotin, riboflavin, pyridoxal.
So all of those small, organic molecules, and they all have chemical reactivity and they interact with the protein and allow the protein to do chemistry that can't normally be carried out by just the amino acid side chains of the protein in the active site of the enzyme.
So the vitamins you eat are not actually what's interacting with the protein, they need to be modified.
So they are the precursors to what are called co-factors that, again, expand the repertoire of what's found in enzymatic systems.
So you have a whole bunch of organic co-factors that actually can be made spontaneously.
If you throw in some simple molecules like cyanide and stuff, these things all self-assemble, so they're all from the primordial soup.
And in fact, many of these vitamins can do chemistry without any enzyme at all.
But they can't do it specifically.
And they can't do it rapidly.
So we have all these organic molecules that can self-assemble that expand the repertoire, but if you look at a vitamin bottle, you will see minerals.
And that's one thing that I think most biochemistry courses don't talk about, is metals.
And it's now proposed that between 35% and 50% of all the proteins in our body have metals that are essential for function.
And so these minerals that you eat, iron or zinc or calcium, all play a central role, again, in expanding.
They all help in many cases facilitate chemical transformations, and in the way we can understand by understanding basic chemical principles.
So the vitamin bottle, we come back to over and over again through the course of the semester, because all the enzymes in metabolic pathways have different kinds of co-factors that are required to make the enzymes function.
The way we teach the course is we show them that all of these chemical transformations can be described by 10 pretty simple steps.
And if you understand the basic chemistry of these simple steps, you can really understand almost all of the kinds of interconversions you see and basic metabolism.
And what you'll see is while this looks overwhelming, really, with a few central pathways, which is what we focus on in this course-- glycolysis, fatty acid oxidation, biosynthesis, sugar biosynthesis as well as degradation, the Krebs cycle, which feeds into the respiratory chain-- knowing those central reactions-- almost everything in metabolism feeds in and out of these pathways.
And so then, it really is a question of what is the environment like and how do you enhance breakdown of sugar under the appropriate environment versus synthesize sugar under a different environment.
So then it's a question of regulation.
So what you're going to learn in this course is really focused on central metabolism.
And it doesn't matter whether you study a bacteria or a human, the central metabolism is pretty much the same.
The thing that's different is the detailed regulation and the complexity of the regulation.
And we don't really talk that much about regulation in 5.07.
What we do is introduce you to five or six basic regulatory mechanisms that are used over and over again.
But then, regulation is really distinct, even between organisms.
And so then that becomes much more complicated.
And when you go off and become a biochemist, you've got to study your own system and figure out what the environment is.
We spend a lot of time looking at-- all proteins are made from amino acid building blocks.
And so one question I get, and students hate this, is, why do I need to know?
I make them memorize the side chains of the amino acids.
Why do I do that?
Because the side chains of the amino acids are the key to the way all the catalysts in your body function.
A catalyst simply enhances the rate of conversion of some small molecule into some other small molecule, and it can enhance the rate of the interconversion by 10 to the 15th fold.
So if you didn't have that catalyst, you couldn't do anything.
So amino acid side chains play a key role in catalysis.
So thinking about the chemical properties of the amino acid side chains is key to understanding all the transformations in your body.
So, what is the pKa of imidazole?
You remember that?
Huh?
I don't.
I don't
How can you not remember that?
[LAUGHTER] Anyhow, the pKa of imidazole is close to seven, OK.
So those physiological-- everything in the body is controlled.
The pH has to be controlled.
And so that means that you can protonate or deprotonate it.
So knowing that is key to thinking about how the chemical reaction is going to work.
The amino acid side chains now-- everybody thought there were 22 amino acids.
But now we know almost all these amino acids can be modified once they get into the protein, so that's called post-transational modification.
So now we have, probably, another 250 modifications.
And one of the modifications that is essential, not for the catalysis part of proteins, but for the structural part of proteins, is hydroxylation of the amino acid proline.
So if you don't hydroxylate proline, then you can't make this molecule called collagen.
And collagen is a structural protein.
It's 25% of all humans' protein, OK. And it's found extracellularly.
Gram per gram it has the strength of steel.
It has very complicated biosynthetic pathway.
It has amazing tensile strength.
It's found in cartilage and teeth and bone.
And a key component of collagen is this hydroxylated proline.
Because without it, you can't form the actual collagen structure.
So collagen is this long-- most proteins, if you look at the structures, they look like little balls.
They're globular.
But collagen is a fibrillar protein, so it's very long.
it's probably the longest protein, too.
It's 3,000 angstroms long.
And it has three chains initially, and they they're left-handed sort of helixes but not real helixes, and they have to wind around each other to form a right-handed helix.
This all happens inside the cell, then somehow has to get to the outside of the cell.
People are studying that now.
And then it forms additional fibrils, and they become insoluble.
And that provides the strength-- the extracellular structures provide the strength that maintain the cell's shape and viability of the cell.
And a key component of all that is hydroxylation of proline.
And how did they find that?
This goes back to, again, misregulation.
So, in the 1600s, or whenever they used to sail the ocean blue with no food, they didn't have enough vitamin C. So they didn't have any citrus fruit.
So vitamin C has the vitamin ascorbate, and ascorbate plays a key role in the chemistry of allowing the proline to become hydroxylated.
So it turns out, to get the proline to be hydroyxlated, you use, again, metal catalyzed reactions.
So here's iron-2.
And if that iron-2 reacts with oxygen, if it gets oxidized with iron-3, it can't react.
So the function of this vitamin is to keep the iron-2 in the reduced state.
So there's an example.
People study that.
Took them many, many, many years, like 200 years later, when they really understand the details of how this post-translational modification actually occurs.
And we understand a lot about that chemistry now.
And this was the first one of these modifications discovered, and now they're finding this modification everywhere.
So lots of amino acids turn out to be hydroxylated, not just proline.
So these modified amino acids now, because the technology we have is so mind boggling, we can find a needle in a haystack.
Whereas in the beginning we were just trying to figure out what the amino acids were, now we have very, very sensitive analytical methods which allow us to see all of these modifications.
Then the key question is, what is the function of the modification?
And that's not so easy in terms of thinking about regulation, anyhow.
And that's the focus of a lot of people's efforts right now.
When I went to graduate school, which was in the late 1960s, I didn't even know what an enzyme was, because back in those days I'd never had a biology course.
They didn't do biochemistry back in those days.
And I think I heard a lecture by Van Tamelen at Stanford.
He was interested in steroid structures, steroids that have four rings and it's the basis for sex hormones and for cholesterol biosynthesis.
They're all sort of made through a common biosynthetic pathway that we don't really talk about.
These are these five carbon units to form carbon-carbon bonds that I mentioned earlier.
Anyhow, what he showed was you could take 30 carbon strung together in a linear form.
And then somehow these 30 carbons folded up with one enzyme to form these four rings, putting in eight asymmetric centers in a single step at pH7 in 100% yield.
Once I saw that, I said man I don't want to be a chemist.
If you could see how nature designed this, and what is the basis for being able to make these molecules, it was sort of mind boggling to me.
And I think enzymes are like that.
What they do is they've evolved for billions of years to accelerate rates of reactions by factors as much as 10 to the 20th.
So that's really fast, like a bat out of hell.
And so nature has figured out how to control all of this.
But again, she has a limited repertoire of reactions that she catalyzes, but she's extremely good at it.
And so understanding people have studied enzymes forever because I'd like to understand the basic principles of catalysis.
And then if you understood those, can the chemist then take this understanding and translate it into the bigger repertoire of the periodic table you have to be able to do these transformations?
So the mechanisms of rate acceleration, which we talk about in some detail in a lecture, what causes these catalysts to work, and the way you describe how these catalysts work.
It doesn't matter whether you're using a small inorganic molecule or small organic molecules or a protein, the basic principles and thinking about catalysis is exactly the same, except nature has figured out how to do this better than anything man can do.
But again, she's limited in terms.
She's had a billions of years to evolve these catalysts, but then she's limited in the repertoire of reactions that she needs to catalyze.
So how can you not think enzymes are cool?
My lab works on the only cool enzyme in the world-- ribonucleotide reductase.
It's the only way in all organisms that you make the building blocks de novo that are required for DNA biosynthesis and repair.
So if you inhibit this enzyme, you have no building blocks.
You can't survive.
So from a practical point of view, it's the target of drugs they use therapeutically in the treatment of cancer.
And I think in probably not so distant future in the antibacterials because I think there are sufficient differences between humans and bacteria reductases that you could make specific inhibitors.
Why am I interested in it?
Because the chemistry is sort of unbelievable.
So I mean it was the first example where you learn, or you hear about-- you heard from John-- radicals.
They're reactive oxygen species and nitrogen species that you can't control.
They want to pick up an extra electron and form a stable octet.
And if you leave them to their own demise, they react with anything and destroy it.
Well nature has figured out how to harness the reactivity of radicals to do really tough chemistry with exquisite specificity.
And ribonucleotide reductases have been the paradigm for thinking about that.
And from bioinformatics now there are 50,000 reactions in metabolic systems that are going to be radical mediated transformations, yet we never talk about radicals in introductory courses.
So I think that's all going to change.
So why is it unusual?
Well, for the human ribonucleotide reductase, the key to making this work catalytically is the amino acid side chain tyrosine needs to be oxidized to a tyrosyl radical.
So automatically nobody believes that.
A tyrosyl radical in solution has a half-life of a microsecond.
In the active site of these enzymes, the half-life of the enzyme can be on the order four days.
And this radical, which is again one electron oxidized amino acid-- if you reduce it with an electron and a proton, the enzyme is completely dead.
So this was the first example of-- it would be another example of a post-translational modification that we talked about earlier-- modifying your amino acids.
And so nature has figured out a way.
How do you do this oxidation?
She has a little metal cluster right adjacent to where this tyrosine is.
And the function of this little metal cluster is to put this into the oxidized state, which is essential for the way the enzyme works.
So the other thing that's amazing about the enzyme is the chemistry.
There are two subunits.
The chemistry all happens in this subunit, but the tyrosyl radical is there.
And this oxidation-- normally when you do an oxidation the two atoms are sitting within a few angstroms of each other-- the oxidation happens over 35 angstroms.
So that's unprecedented.
It involves hopping radicals which no one has ever seen before.
And so that was another thing that was completely fascinating from a chemical perspective about how the system works.
The other reason that people in biology are interested in this, besides the fact that makes a building block for DNA, is that if you believe in an RNA world where we have a ribosome where a catalysis of peptide bond formation is all with the RNA, not with the protein.
How do you get from an RNA world to a DNA world?
The only enzyme that does that transformation making these building blocks are ribonucleotide reductases.
And there are many classes of ribonucleotide reductases-- one uses this tyrosyl radical-- but they all have the same active site and do the same chemistry, but they have different metal cofactors depending on where they evolved.
And the function of the metal cofactors in all cases, even though one's cobalt, one's iron sulfur cluster, one's manganese, one's iron-- the function in all cases is to generate a radical in the active site and then the chemistry is the same in all these things.
I'm always interested in listening to JoAnne's lectures, and she talks about how she got interested in biochemistry from a background in what I would call more physical organic chemistry by attending a lecture and being completely inspired.
And I am a toxicologist by training.
And how did that actually happen.
I have a similarly inspiring story, so I thought I would tell that.
I went to a lecture about 45 years ago, by a fellow by the name of [Richard] Evans Schultes.
He was the director of the Herbarium at Harvard-- and I'd been to it; it was very interesting-- and a professor there.
And he is arguably the father of the field called ethnobotany.
I went to this talk-- there were only about 10 people at the talk.
He-- because it was a small talk, he gave it sitting down, he asked permission to do that, he's a very polite man.
And it was very informal, only a few slides.
And the slides were mainly of him living with indigenous peoples, Native Americans in the Southwest, Amazonian native peoples, very interesting.
And he talked about, for example, he had just graduated, his undergraduate degree from Harvard.
His parents were eager to have him go to medical school.
He was very interested in botany and in native languages.
So he went to live with the-- I think it was the Kiowa people in the Midwest-- I think it was probably Oklahoma.
And he learned about peyote and a lot of other chemicals that were hallucinogenic.
And he realized, of course, that many of these kinds of compounds that were used in a lot of rituals, were also medicines at other concentrations, anesthetics and so on.
He then went down to South America, living with various tribes.
He'd be gone for up to a year at a time.
There are wonderful stories, well actually terrible stories, I guess, about him paddling his canoe for 40 days with malaria-- it was terrible-- to get to a hospital, things like that.
But anyway these are stories-- I was a young scientist at the time-- these were very influential to me.
And he talked about, I was down in South America and they-- and I found the plants from which the native people got their dart and arrow poisons.
And he said, out of that we isolated curare, and initiated the path toward the clinic of curare as a muscle relaxant.
He said that it didn't work so well.
I remember this lecture like it was yesterday.
He said that it was the beginning of World War II, we were cut off from the rubber plantations in the Philippines and other places.
And since South America had jungles, his form of, quote unquote "military service," involved trying to find sources in the jungle of latex that could be used to make rubber so that we could have an effective war machine.
So anyway, I went up to this fellow after he gave his talk.
And I said, look, I just got to do this kind of stuff.
This sounds really interesting.
And he said, what's your background?
I said, well, you know, I worked as a chemist.
I was a biology major but I worked as a chemist at an industrial consulting company during my undergraduate years.
And he said, oh, you got to go talk with Gerry Wogan at MIT.
He said that he isolates toxins from fungi that you find out in the jungles of Southeast Asia.
And, so anyway, that's how my career began.
As you know, I got to eventually meet Gerry Wogan and work with him.
He had already identified this toxin, aflatoxin.
But he needed somebody to figure out how it worked, and that's how I got my start.
So my interests have been in chemicals from the environment.
It could be a pollutant.
Or it could be a chemical that could be a precursor to a therapeutically useful molecule, and how do they interact with biological systems.
And that's what toxicologists and pharmacologists do.
I work in the field of genetic change.
In a perfect world, you would say, guanine would always pair with cytosine and adenine would always pair with thymine.
It turns out, however, that sometimes chemicals from the environment can react with our normal nucleotides and change their coding characteristics so that mistakes are made when polymerases try to read them.
These are mutations, and mutations cause genetic diseases.
My role in the field of toxicology is as a person, who is both a biologist and a chemist, who studies how chemical damage to our informational molecules is converted into changes in coding that results in genetic change.
I'll emphasize, of course, this is the basis for all genetic disease, but it's also the basis for evolution.
In other words, mutations happen naturally.
That means that we're not all-- we don't all look alike.
And that means there's diversity in a population, and that's because of mutations.
And evolution is a really good thing, because if we were all alike when the environment changed, then the chance of extinction might be very high.
If there's diversity in a population, that's actually a hedge that life uses in order to be able to make sure something's going to survive, because some members in the population, while they may be considered quote unquote "weaker" in the initial environment, when the environment changes-- they're the ones, for example, with hair on them, and survive the global winter that happens after the meteor strikes.
So, we are interested in chemical modification of DNA and RNA as it relates to the causation of disease, but we're also interested in the rates at which genetic change happens in a population, and how that's a good thing, and that it provides for the continuance of life.
One example of our work in evolution comes from recent work that we've been doing on HIV.
When a virus infects one of our cells, our cells respond by trying to limit the growth of the virus.
They try to kill it.
And one of the strategies that used by what's called our innate immune system is to induce a number of enzymes that start to rip apart the DNA bases.
They really take the amino groups off of cytosines and adenines in order to try to convert them into non-coding nucleotides or miscoding nucleotides.
What happens is, if you take a cytosine and you take away it's amino group, it makes it into a uracil, and then, rather than pair with a guanine, it'll pair with an adenine.
Because these are enzymes that kind of move along the viral genome, they create a huge number of mutations.
The process is called lethal mutagenesis, because what happens is, eventually the number of mutations is so large that you can no longer produce a functional protein or nucleic acid.
That's a natural strategy that we use.
And thinking about this, some years ago, given my lab's expertise in knowing about the structural modification of normal bases that makes them mutagenic, we wondered if we could contaminate the nucleotide pool of a cell with mutagenic nucleotides that would force a virus to mutate even quicker.
HIV, it turns out, almost goes extinct, but not quite.
In other words, our innate immune system, in one day, creates every single point mutation in the virus, but it also creates every single drug-resistant variant.
And it just doesn't push hard enough to be able to push the virus to extinction.
So, we wondered whether-- if we could push a little harder by using creatively designed molecules-- derivatives of cytosine, that would pretend sometimes they're a cytosine and sometimes they pretended that they were a thymine-- that we could push the virus over the top.
And we found that it did work, OK.
In other words, we were able to push the virus to a technical state of extinction using our understanding of the chemistry of the molecules that the cell has the capability to use in replication.
It may seem unwise to intentionally put mutagenic chemicals into people, and, obviously, we worked out a strategy to prevent mutations in people in the drug design process.
Specifically, it turns out there are pathways called DNA repair pathways that repair damage in our nuclear genomes, and they happen in the nucleus.
That's where the enzymes are located.
It turns out that the early stage in HIV replication involved replication in the cytoplasm.
There are no repair enzymes in the cytoplasm.
So, the virus has no defense against the mutagenic affects of these compounds.
We picked compounds that, if they were to get into our nuclei-- it turns out that our polymerases don't like them very much, anyway, but if they did get in, they're very rapidly repaired.
So, it creates what we call a therapeutic index which is very favorable in favor of killing the virus but not putting mutagenic chemicals into us.
At the end of the course, when there's so many pathways out there that things start to get really, really complicated, we try to increase the number of physiological scenarios, again, to try to get the students sufficiently interested in a pathway and thinking about how the whole pathway works, not just each individual step-- to get the student to like the example so that he or she will go right in there and study it.
And one of the areas that I think really important these days in basic metabolic biochemistry is in oncology-- cancer biology.
JoAnne mentioned that not that many years ago, academic programs wondered whether teaching metabolic pathways was really the best use of time, because so many other things were happening in the molecular biology revolution.
And certainly in 5.07 we decided to maintain emphasis in this, because we-- just because these are the first pathways that were studied doesn't mean they aren't important.
They're medically very important.
They were originally studied because of medical or economic reasons that-- fermentation, for example-- that these pathways were right in front and center.
They're always very topical.
But what happened in I'd say the last 10 years is that there's been a reawakening of the understanding of how metabolic pathways redirect themselves in order to accomplish disease goals-- for example, with a tumor.
And one of the things we teach is that a tumor is a lot like a running muscle.
And Matthew Vander Heiden, who is here in our biology department and teaches the comparable course is a real expert, a pioneer in this area.
But what we've learned is that cancer cells tend to be addicted to glucose.
They would much prefer glucose over any other fuel.
In fact, they-- when you think about what a cancer cell really is, it's a cell that's not terribly different from all our other cells.
It's acquired some small number-- we'll call driver mutations, maybe seven or so driver mutations.
There are many more passenger mutations, but maybe seven or so genetic differences that separate it from a normal cell.
But among the things that has happened to it is reprogramming of its metabolic needs.
And it has acquired the commitment to unremitting cell division.
If you think about what's needed for a cell to divide, probably the principal structural resource-- which, again, this would come from JoAnne's part of the course-- is membranes.
You've got to recreate the outside of the cell and all those organelles.
And what that means is fatty acid biosynthesis is absolutely going to be paramount.
One way that we now know that tumors work, if you understand all the pathways, is a tumor will consume glucose, it will convert that glucose to acetyl-CoA.
The acetyl-CoA takes a ride on the molecule citrate until it gets put out into the cytoplasm, and then in the cytoplasm it becomes a factory for producing-- out of acetyl-CoA, it produces the lipids that are necessary for membranogenesis.
That's absolutely critical for cancer cell development.
Now, the reason that's important is that understanding it, it tells us that, well, there are certain enzymes in the pathway.
One of them is called ATP citrate lyase.
Another one is called malic enzyme.
So, when you look at the charts, you'll see these things.
These are enzymes that are necessary for cell division, because they're necessary for fatty acid biosynthesis, and they represent good targets-- next-generation targets-- for cancer chemotherapy.
A normal cell, you can tell it not to divide, and it won't divide.
But if you can tell a cell that's committed to divide and you tell it not to divide, oftentimes that cell will kill itself.
So by blocking these critical points-- test points, we'll call them, in this metabolic chart-- you're able to selectively kill cancer cells-- we hope, anyway.
But this study of metabolic biochemistry has identified these new targets.
Another thing that's, I think, an important anecdote about cancer cells is that they need other cells in order to survive.
If you think about the problems that a tumor faces-- remember, it started from a normal cell that acquired new mutations, converted into a cancer cell, and then started to grow out into a tumor.
As it grows, it becomes more remotely placed relative to the blood supply.
OK, now, certainly there will be a response, and the blood supply will try to grow toward the tumor cells.
But the core of a tumor is hypoxic.
It doesn't have enough oxygen to do real respiration.
So, what happens is, the cancer cell tries to hard-wire the glycolysis pathway to be on.
And rough estimates are that through this-- initially, a hypoxia-inducing factor, which is a transcription factor-- you get an upregulation by 10 to 20-fold of almost all of the enzymes of the glycolytic pathway.
So, what happens is, the cancer cell becomes a specialist at glycolysis.
It does high-throughput glycolysis-- glucose to pyruvate.
But, again, we don't have the metabolic equipment to do a lot of-- we don't have the metabolic equipment to be able to do respiration if you're removed from oxygen.
So, the pyruvate gets converted into lactate.
Lactate gets pumped out into the blood, and, as I mentioned before, it acidifies the blood.
That's what happens with a working muscle.
But the tumor sends the lactate into the blood.
It's picked up by the liver.
The liver is a specialist in gluconeogenesis.
The liver then rebuilds the lactate into glucose, sends this back out into the blood.
So, the tumor is then able to re-eat the lactate that it sent out.
So a partnership emerges between the liver-- the gluconeogenic organ-- and the tumor.
So, the tumor finds that it's fed by-- unwittingly-- by the liver.
And the liver is contributing to providing the nutrition that's eventually going to kill the organism if something doesn't go in to stop it.
But, again, understanding the pathways can give you ideas with regard to how to intervene.
So, metabolic biochemistry is pretty critical to understanding this generation's cancer scientific agenda.
If you're a teacher and you're inventing a course for the first time, or revising it a lot, you sit down with your teaching partners and you put on the table all the ideas.
Teaching, you've got an ever-expanding universe of knowledge out there, and you have to cherry-pick the things that are going to be important.
It has to hang together.
One of the strategies JoAnne and I thought, when we went into the current iteration of teaching 5.07 biological chemistry, was to abandon completely, higher eukaryotes, namely us, because genome sequencing projects had sequenced so many bacteria that one could create an entire course in biochemistry that would be very meaningful, just focusing on microorganisms.
We spent a couple of days reading and thinking about it.
That would be the course JoAnne and I would teach, if it weren't for the fact that we actually have-- feel as though we have a commitment to students that are going to go on to medical school and therefore, if we avoided mammalian biochemistry, students wouldn't know anything about the mitochondria and organelles, and things like that that are associated with eukaryotes.
We wouldn't be able to make these connections to disease, the physiological scenarios.
Nevertheless, why were we so interested in bacteria?
What would be an interesting story, that I might be able to tell you, if we had taken that path.
So we have a fellow in biological engineering, named Eric Alm.
And he is an informaticist and an engineer, and an extremely good chemist.
He really knows his pathways.
When he looks at a cell, he thinks about what it is, but also where it came from, in terms of how it evolved from precursors, its family tree, so to speak.
One of the most interesting organisms that he's published on, not too long ago, is an organism called, Desulforudis.
And this would be a wonderful biochemistry course in itself.
He wondered, if he went out and dug up a cubic meter of dirt-- and outside MIT-- and if he did 16S RNA sequencing, how many living things would be there, many thousands, maybe 10,000.
Then he asked the question, what if you went down 100 meters, you know, maybe you see 1,000.
But what if you go down until there's really only one thing there.
And that's what he did, going down two miles into the ground.
And there was a single, species ecosystem called, Desulforudis.
And I remember seeing this paper, and I brought it over to JoAnne.
I was so excited because the last picture showed its metabolic network, its metabolic pathways.
It had everything.
And it makes sense.
It can't rely on other things.
For example, we can't make all of our amino acids.
We got to get them from food that we eat, or in our co-factors, some of our vitamins are made by the bacteria in our gut.
So, if all the bacteria disappear, we would too.
So we rely on other things, but Desulforudis doesn't rely on anything.
So when you look at it's biochemical networks, what you see is that it can fix nitrogen.
It can take N2 and convert it NH3, and then put that into amino acids, and it can make all of its amino acids.
It has a really good pentose phosphate pathway.
It actually uses radiation in a strange way to generate some of the energy that it needs.
It uses it to generate carbon monoxide.
Ultimately, that CO is going to form an acetyl group that will be able to generate all of the organic material inside the Desulforudis.
It's got all kinds of electron transport pathways.
So it's developed enormous versatility by being a single-species ecosystem.
So, again, this was a course where we had to make a compromise because of our clientele.
Teachers have to think about that.
We have to teach to what the people need in order to go on to the next step.
But as sort of a closing thought, I think that it would be wonderful for next-generation biochemists to really turn their attention to the microbial world, to teach this vast biochemistry and understand how bacteria effortlessly, swap biochemical pathways, pick-up entire biochemical pathways without even breaking a sweat.
Whenever they find themselves stressed, they just pick up a new pathway and they survive.
One of the physiological scenarios that we teach, and is probably very common in biochemistry courses, has to do with diabetes.
I'm diabetic, and my great grandfather was diabetic.
His daughter, my grandmother was diabetic.
Five of her six sons were diabetic, one of whom was my father.
So you see a very genetic disease here.
OK?
That's, in my case, type 2 diabetic, so adult onset.
So my cells are insulin-resistant.
So if I eat a meal that has carbohydrate or fat in it, my pancreas is probably responding.
My beta cells in my pancreas are probably OK, but the insulin just isn't able-- but producing insulin-- but the insulin just isn't able to signal my cells to be able to take up the glucose.
As a consequence, I'm somewhat in a technical state of starvation.
My glycogen reserves are therefore smaller than yours, and I can tell this in times.
For example, when I first became diabetic, I loved mountain climbing and hiking and so on, and I noticed that I would go out.
And I'd be totally exhausted for the first 10 minutes or so, and that's because I have very small glycogen reserves.
What I had to do is this metabolic switch over to booting up oxidative metabolism of fats.
That all had to be turned on, and it's hard.
It's a hard transition, but once it got going, as long as I stay aerobic, in other words, I don't go so fast that I go anaerobic, I can climb forever.
Because if I stay aerobic, I've got plenty of fat reserves.
In fact, it's probably a good thing that I'm burning them.
So I'm just a little bit different.
I like to teach this stuff in 5.07, in part because well there's probably not that many kids in the class, of 100, 130, or so, students would be diabetic, but it's about 6% to 8% of the population in the United States right now.
So everybody knows someone who is diabetic, or maybe your parents are, and what are the risk factors?
So we teach that.
There are many different treatments for diabetes.
I'll just use myself as a case study.
One of the things I did to myself is I measure my blood sugar at different times during the day, and my blood sugar this morning was 116.
And my blood sugar at about five o'clock this afternoon will probably be in the high 80s or 90.
My blood sugar after I have a meal might go up to 160, 170, and it shouldn't go that high.
And some mornings, like a couple of days ago, my blood sugar was 146, which is too high.
And that reflects two things, what I ate the night before, but the second thing is the one that's my problem, which is gluconeogenesis.
Gluconeogenesis technically means new synthesis of glucose from non-carbohydrate precursors.
For example, I could take the amino acid aspartate, and I could find a path by which I could convert that into glucose.
I could take lactate, and I could convert it into glucose.
OK, so during the day, I use my glucose, and I stay active, and I walk around.
I ride my bike and so on.
So my glucose level, it's pretty reasonable by the time I go to bed.
But then at night, when I stop moving around a lot, this gluconeogenesis process continues in me, and that's what causes my blood sugar to go up.
The medicine I take is called Metformin.
It has a number of targets, but one of them is one of the enzymes, called PEPCK, Pyruvate Carboxykinase, that's in the gluconeogenic pathway.
Let me say a word about gluconeogenesis, another word actually.
So we all have dinner, like say six to nine o'clock at night.
We go to bed, and there are certain organs in the body, the brain, our red blood cells, that require glucose.
They can't work with anything else.
So gluconeogenesis, by principally the liver, provides a constant stream of glucose to these organs that absolutely require it, like our brain.
Now, when we go to sleep at night, as time gets longer and longer and longer, after the meal, our glucose level, our natural glucose level, is going to start to fall off.
And the liver then compensates by increasing the amount of gluconeogenesis in order to keep our blood glucose at about 100 while we're not eating.
What I do, during the night, is to take this drug that will prevent the switch to produce more and more sugar by gluconeogenesis.
As I mentioned, I'm diabetic.
And when things that I noticed in my dad, before he knew he was diabetic-- he'd come home from the train at night, and my sister and I would rub his legs, because they hurt.
This is one of the problems with diabetics.
His feet hurt.
And I remember my sister Nancy saying once, gee.
Daddy smells like Mommy's nail polish remover.
Which was acetone.
And acetone is one of the three ketone bodies, the other two being Beta-Hydroxybutyric acid and acetoacetate.
So what happens in diabetes is, the diabetic doesn't have ample carbohydrate reserves within their cells as glycogen.
So they're doing a lot of fatty acid oxidation, rather than carbohydrate metabolism.
If you take a fatty acid, and you break it down to Acetyl-CoA, the last step is called Beta-ketothiolase-- backs up, then your two carbons of Acetyl-CoA will be four carbons of acetoacetate.
And then acetoacetate-- a third Acetyl-CoA can add to it to form what's called hydroxy-methylglutaryl-coenzyme A-- HMG CoA.
And then rearrangements can happen that give rise, ultimately, to the three ketone bodies-- acetoacetate, Beta-Hydroxybutyrate, and acetone.
So in situations where-- of starvation-- and diabetes is basically, my cells were in a technical state of starvation.
In a case of starvation, what happens is you get this backup of the lipid breakdown, or catabolism pathway.
And then these ketone bodies flush out into your blood.
In starvation situations, the ketone bodies will act as fuels for your brain and other organs.
It's a difficult transition.
But ketone bodies are-- in the case of diabetics, they're produced to excess, to such extent that they're actually quite harmful to you.
The reason a person smells of acetone is the intermediate ketone body-- the acetoacetate basically decarboxylates.
It's a Beta-keto acid, and so it's a spontaneous decomposition to form acetone.
Acetone is not, to my knowledge, biochemically useful as a kind of an energy source.
But it's a sentinel, or it's something that is smelled, so that it can be diagnostic of the disease.
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Thermodynamics, all right, let's start.
Thermodynamics is the science of the flow of heat.
So, thermo is heat, and dynamics is the motion of heat.
Thermodynamics was developed largely beginning in the 1800's, at the time of the Industrial Revolution.
So, taming of steel.
The beginning of generating power by burning fossil fuels.
The beginning of the problems with CO2 and [NOISE OBSCURES] global warming.
In fact, it's interesting to note that the first calculation on the impact of CO2 on climate was done in the late 1800's by Arrhenius.
Beginning of a generation of power moving heat from fossil fuels to generating energy, locomotives, etcetera.
So, he calculated what would happen to this burning of fossil fuels, and he decided in his calculation, he basically got the calculation right, by the way, but he came out that in 2,000 years from the time that he did the calculations, humans would be in trouble.
Well, since his calculation, we've had an exponential growth in the amount of CO2, and if you go through the calculations of -- people have done these calculations throughout times since Arrhenius, the time that we're in trouble, 2,000 years and the calculation, has gone like this, and so now we're really in trouble.
That's for a different lecture.
So, anyway, thermodynamics dates from the same period as getting fossil fuels out of the ground.
It's universal.
It turns out everything around us moves energy around in one way or the other.
If you're a biological system, you're burning calories, burning ATP.
You're creating heat.
If you're a warm-blooded animal.
You need energy to move your arms around and move around -- mechanical systems, obviously, cars, boats, etcetera.
And even in astrophysics, when you talk about stars, black holes, etcetera, you're moving energy around.
You're moving heat around when you're changing matter through thermodynamics.
And the cause of some thermodynamics have even been applied to economics, systems out of equilibrium, like big companies like Enron, you know, completely out of equilibrium, crash and burn.
You can apply non-equilibrium thermodynamics to economics.
It was developed before people knew about atoms and molecules.
So it's a science that's based on macroscopic properties of matter.
Since then, since we know about atoms and molecules now, we can rationalize the concepts of thermodynmamics using microscopic properties, and if you are going to take 5.62, that's what you'd learn about.
You'd learn about statistical mechanics, and how the atomistic concepts rationalize thermodynamics.
It doesn't prove it, but it helps to getting more intuition about the consequences of thermodynamics.
So it applies to macroscopic systems that are in equilibrium, and how to go from one equilibrium state to another equilibrium state, and it's entirely empirical in its foundation.
People have done experiments through the ages, and they've accumulated the knowledge from these experiments, and they've synthesized these experiments into a few basic empirical rules, empirical laws, which are the laws of thermodynamics.
And then they've taken these laws and added a structure of math upon it, to build this edifice, which is a very solid edifice of thermodynamics as a science of equilibrium systems.
So these empirical observations then are summarized into four laws.
So, these laws are, they're really depillars.
They're not proven, but they're not wrong.
They're very unlikely to be wrong.
Let's just go through these laws, OK, very quickly.
There's a zeroth law The zeroth law every one of these laws basically defines the quantity in thermodynamics and then defines the concept.
The zeroth law defines temperature.
That's a fairly common-sense idea, but it's important to define it, and I call that the common-sense law.
So this is the common-sense law.
The first law ends up defining energy, which we're going to call u, and the concept of energy conservation, energy can't be lost or gained.
And I'm going to call this the you can break even law; you can break even law.
You don't lose energy, you can't gain energy.
You break even.
The second law is going to define entropy, and is going to tell us about the direction of time, something that conceptually we, clearly, understand, but is going to put a mathematical foundation on which way does time go.
Clearly, if I take a chalk like this one here, and I throw it on the ground, and it breaks in little pieces, if I run the movie backwards, that doesn't make sense, right?
We have a concept of time going forward in a particular way.
How does entropy play into that concept of time?
And I'm going to call this the you can break even at zero degrees Kelvin law.
You can only do it at zero degrees Kelvin.
The third law is going to give a numerical value to the entropy, and the third law is going to be the depressing one, and it's going to say, you can't get to zero degrees.
These laws are universally valid.
They cannot be circumvented.
Certainly people have tried to do that, and every year there's a newspaper story, Wall Street Journal, or New York Times about somebody that has invented the device that somehow goes around the second law and makes more energy than it creates, and this is going to be -- well, first of all, for the investors this is going to make them very, very rich, and for the rest of us, it's going to be wonderful.
And they go through these arguments, and they find venture money to fund the company, and they get very famous people to endorse them, etcetera.
But you guys know, because you have MIT degrees, and you've, later, and you've taken 5.60, that can't be the case, and you're not going to get fooled into investing money into these companies.
But it's amazing, that every year you find somebody coming up with a way of going around the second law and somehow convincing people who are very smart that this will work.
So, thermo is also a big tease, as you can see from my descriptions of these laws here.
It makes you believe, initially, in the feasibility of perfect efficiency.
The first law is very upbeat.
It talks about the conservation of energy.
Energy is conserved in all of its forms.
You can take heat energy and convert it to work energy and vice versa, and it doesn't say anything about that you have to waste heat if you're going to transform heat into work.
It just says it's energy.
It's all the same thing, right?
So, you could break even if you were very clever about it, and that's pretty neat.
So, in a sense, it says, you know, if you wanted to build a boat that took energy out of the warmth of the air, to sail around the world, you can do that.
And then the second law comes in and says well, that's not quite right.
The second law says, yes, energy is pretty much the same in all this form, but if you want to convert one form of energy into another, if you want to convert work, heat into work, with 100% efficiency, you've got to go down to zero degrees Kelvin, to absolute zero if you want to do that.
Otherwise you're going to waste some of that heat somewhere along the way, some of that energy.
All right, so you can't get perfect efficiency, but at least if you were able to go to zero degrees Kelvin, then you'd be all set.
You just got to find a good refrigerator on your boat, and then you can still go around the world.
And then the third law comes in, and that's the depressing part here.
It says, well, it's true.
If you could get to zero degrees Kelvin, you'd get perfect efficiency, but you can't get to zero degrees Kelvin, you can't.
Even if you have an infinite amount of resources, you can't get there.
Any questions so far?
So thermodynamics, based on these four laws now, requires an edifice, and it's a very mature science, and it requires that we define things carefully.
So we're going to spend a little bit of time making sure we define our concepts and our words, and what you'll find that when you do problem sets, especially at the beginning, understanding the words and the conditions of the problem sets is most of the way into solving the problem.
So we're going to talk about things like systems.
The system, it's that part of the universe that we're studying.
These are going to be fairly common-sense definitions, but they're important, and when you get to a problem set, really nailing down what the system is, not more, nor less, in terms of the amount of stuff, that's part of the system, it's going to be often very crucial.
So you've got the system.
For instance, it could be a person.
I am the system.
I could be a system.
It could be a hot coffee in a thermos.
So the coffee and the milk and whatever else you like in your coffee would be the system.
It could be a glass of water with ice in it.
That's a fine system.
Volume of air in a part of a room.
Take four liters on this corner of the room.
That's my system.
Then, after you define what your system is, whatever is left over of the universe is the surroundings.
So, if I'm the system, then everything else is the surroundings.
You are my surroundings.
Saturn is my surroundings.
As far as you can go in the universe, that's part of the surroundings.
And then between the system and the surroundings is the boundary.
And the boundary is a surface that's real, like the outsides of my skin, or the inner wall of the thermos that has the coffee in it, or it could be an imaginary boundary.
For instance, I can imagine that there is a boundary that surrounds the four liters of air that's sitting in the corner there.
It doesn't have to be a real container to contain it.
It's just an imaginary boundary there.
And where you place that boundary becomes important.
So, for instance, for the thermos with the coffee in it, if you place the boundary in the inside wall of the glass or the outside wall of the glass and the inside of the thermos, that makes a difference; different heat capacity, etcetera.
So this becomes where defining the system and the boundaries, and everything becomes important.
You've got to place the boundary at exactly the right place, otherwise you've got a bit too much in your system or a bit too little.
More definitions.
The system can be an open system, or it can be a closed system, or it can be isolated.
The definitions are also important here.
An open system, as the name describes, allows mass and energy to freely flow through the boundary.
Mass and energy flow through boundary.
Mass and energy -- I'm an open system, right?
Water vapor goes through my skin.
I'm hot, compared to the air of the room, or cold if I'm somewhere that's warm.
So energy can go back and forth.
The thermos, with the lid on top, is not an open system.
Hopefully, your coffee is going to stay warm or hot in the thermos.
It's not going to get out.
So the thermos is not an open system.
In fact, the thermos is an isolated system.
The isolated system is the opposite of the open system, no mass and no energy can flow through the boundary.
The closed system allows energy to transfer through the boundary but not mass.
So a closed system would be, for instance, a glass of ice water with an ice cube in it, with the lid on top.
The glass is not very insulating.
Energy can flow across the glass, but I put a lid on top, and so the water can't get out.
And that's the closed system.
Energy goes through the boundaries but nothing else.
Important definitions, even though they may sound really kind of dumb, but they are really important, because when you get the problem, figuring out whether you have an open, closed, or isolated system, what are the surroundings?
What's the boundary?
What is the system?
That's the first thing to make sure that is clear.
If it's not clear, the problem is going to be impossible to solve.
And that's also how people find ways to break the second law, because somehow they've messed up on what their system is.
And they've included too much or too little in the system, and it looks to them that the second law is broken and they've created more energy than is being brought in.
That's usually the case.
Questions?
Let's keep going.
So, now that we've got a system, we've got to describe it.
So, let's describe the system now.
It turns out that when you're talking about macroscopic properties of matter, you don't need very many variables to describe the system completely thermodynamically.
You just need a few macroscopic variables that are very familiar to you, like the pressure, the temperature, the volume, the number of moles of each component, the mass of the system.
You've got a magnetic field, maybe even magnetic susceptibility, the electric field.
We're not going to worry about these magnetic fields or electric fields in this class.
So, pretty much we're going to focus on this set of variables here.
You're going to have to know when you describe the system, if your system is homogeneous, like your coffee with milk in it, or heterogeneous, like water with an ice cube in it.
So heterogeneous means that you've got different phases in your system.
I'm the heterogeneous system, soft stuff, hard stuff, liquid stuff.
Coffee is homogeneous, even though it's made up of many components.
Many different kinds of molecules make up your coffee.
There are the water molecules, the flavor molecules, the milk proteins, etcetera.
But it's all mixed up together in a homogeneous, macroscopic fashion.
If you drill down at the level of molecules you see that it's not homogeneous.
But thermodynamics takes a bird's eye view.
It looks pretty, beautiful.
So, that's a homogeneous system, one phase.
You have to know if your system is an equilibrium system or not.
If it's an equilibrium system, then thermodynamics can describe it.
If it's not, then you're going to have trouble describing it using thermodynamic properties.
Thermodynamics talks about equilibrium systems and how to go from one state of equilibrium to another state of equilibrium.
What does equilibrium mean?
It means that the properties of the system, the properties that describe the system, don't change in time or in space.
If I've got a gas in a container, the pressure of the gas has to be the same everywhere in the container, otherwise it's not equilibrium.
If I place my container of gas on the table here, and I come back an hour later, the pressure needs to be the same when I come back.
Otherwise it's not equilibrium.
So it only talks about equilibrium systems.
What else do you need to know?
So, you need to know the variables.
You need to know it's heterogeneous or homogeneous.
You need to know if it's an equilibrium, and you also need to know how many components you have in your system.
So, a glass of ice water with an ice cube in it, which is a heterogeneous system, has only one component, which is water, H2O.
Two phases, but one component.
Latte, which is a homogeneous system, has a very, very large number of components to it.
All the components that make up the milk.
All the components that make up the coffee, and all the impurities, etcetera.
cadmium, heavy metals, arsenic, whatever is in your coffee.
OK, any questions?
All right, so we've described the system with these properties.
Now these properties come in two flavors.
You have extensive properties and intensive properties.
The extensive properties are the ones that scale with the size of the system.
If you double the system, they double in there numerical number.
For instance, the volume.
If you double the volume, the v doubles.
I mean that's obvious.
The mass, if you double the amount of stuff the mass will double.
Intensive properties don't care about the scale of your system.
If you double everything in the system, the temperature is not going to change, it's not going to double.
The temperature stays the same.
So the temperature is intensive, and you can make intensive properties out of the extensive properties by dividing by the number of moles in the system.
So I can make a quantity that I'll call V bar, which is the molar volume, the volume of one mole of a component in my system, and that becomes an intensive quantity.
A volume which is an intensive volume.
The volumes per mole of that stuff.
So, as I mentioned, thermodynamics is the science of equilibrium systems, and it also describes the evolution of one equilibrium to another equilibrium.
How do you go from one to the other?
And so the set of properties that describes the system -- the equilibrium doesn't change.
So, these on-changing properties that describe the state of the equilibrium state of the system are called state variables.
So the state variables describe the equilibrium's state, and they don't care about how this state got to where it is.
They don't care about the history of the state.
They just know that's if you have water at zero degrees Celsius with it ice in, that you can define it as a heterogeneous system with a certain density for the water or certain density for the ice, etcetera, etcetera.
It doesn't care how you got there.
We're going to find other properties that do care about the history of the system, like work, that you put in the system, or heat that you put in the system, or some other variables.
But you can't use those to define the equilibrium state.
You can only use the state variables, independent of history.
And it turns out that for a one component system, one component meaning one kind of molecule in the system, all that you need to know to describe the system is the number of moles for a one component system, and to describe one phase in that system, one component, homogeneous system, you need n and two variables.
For instance, the pressure and the temperature, or the volume and the pressure.
If you have the number of moles and two intensive variables, then you know everything there is to know about the system.
About the equilibrium state of that system.
There are hundreds of quantities that you can calculate and measure that are interesting and important properties, and all you need is just a few variables to get everything out, and that's really the power of thermodynamics, is that it takes so little information to get so much information out.
So little data to get a lot of predictive information out.
As we're going on with our definitions, we can summarize a lot of these definitions into a notation, a chemical notation that that will be very important.
So, for instance, if I'm talking about three moles of hydrogen, at one bar 100 degrees Celsius.
I'm not going to write, given three moles of hydrogen at one bar and three degrees, blah, blah, blah.
I'm going to write it in a compact notation.
I'm going to write it like this: three moles of hydrogen which is a gas, one bar 100 degrees Celsius.
This notation gives you everything you need to know about the system.
It tells you the number of moles.
It tells you the phase.
It tells you what kind of molecule it is, and gives you two variables that are state variables.
You could have the volume and the temperature.
You could have the volume and the pressure.
But this tells you everything.
I don't need to write it down in words.
And then if I want to tell you about a change of state, or let's first start with a mixture.
Suppose that I give to a mixture like, this is a homogeneous system with two components, like five moles of H2O, which is a liquid, at one bar 25 degrees Celsius, plus five moles of CH3, CH2, OH, which is a liquid, and one bar at 25 degrees Celsius.
This describes roughly something that is fairly commonplace, it's 100-proof vodka 1/2 water, 1/2 ethanol -- that describes that macroscopic system.
You're missing all the impurities, all the little the flavor molecules that go into it, but basically, that's the homogeneous system we were describing, two component homogeneous systems.
Then you can do all sorts of predictive stuff with that system.
All right, that's the equilibrium system.
Now we want to show a notation, how do we go from one equilibrium state like this describes to another equilibrium state?
So, we take our two equilibrium states, and you just put an equal sign between them, and the equal sign means go from one to the other.
So, if we took our three moles of hydrogen, which is a gas at five bar and 100 degrees Celsius, and, which is a nice equilibrium state here, and we say now we're going to change the equilibrium state to something new, we're going to do an expansion, let's say.
We're going to drop the pressure, the volume is going to go up.
I don't need to tell you the volume here, because you've got enough information to calculate the volume.
The number of moles stays the same, a closed systems, gas doesn't come out.
Stays a gas, but now the pressure is less, the temperature is less.
I've done some sort of expansion on this.
I've gone from 1 equilibrium state to another equilibrium state, and the equal sign means you go from this state to that state.
It's not a chemical reaction.
That's why we don't have an arrow here, because we could go back, this way too.
We can go back and forth between these two equilibrium states.
They're connected.
This means they're connected.
And when I put this, I have to tell you how they are connected.
I have to tell you the path, if you're going to solve a problem.
For instance, you want to know how much energy you're going to get out from doing this expansion.
How much energy are you going to get out, and how far are you going to be able to drive a car with this expansion, let's say, so that's the problem.
So, I need to tell you how you're doing the expansion, because that's going to tell you how much energy you're wasting during that expansion.
It goes back to the second law.
Nothing is efficient.
You're always wasting energy into heat somewhere when you do a change that involves a mechanical change.
All right, so I need to tell you the path, when I go from one state to the other.
And the path is going to be the sequence, intermediate states going from the initial state the final state.
So, for instance, if I draw a graph of pressure on one axis and temperature on the other axis, my initial state is at a temperature of 100 degrees Celsius and five bar.
My final stage is 50 degrees Celsius and one bar.
So, I could have two steps in my path.
I could decide first of all to keep the pressure constant and lower the pressure.
When I get to 50 degrees Celsius, I could choose to keep the temperature constant and lower the pressure.
I'm sorry, my first step would be to keep the pressure constant and lower the temperature, then I lower the pressure, keeping the temperature constant.
So there's my intermediate state there.
This is one of many paths.
There's an infinite number of paths you could take.
You could take a continuous path, where you have an infinite number of equilibrium points in between the two, a smooth path, where you drop the pressure and the temperature simultaneously in little increments.
All right, so when you do a problem, the path is going to turn out to be extremely important.
How do you get from the initial state to the final state?
Define the initial state.
Define the final state.
Define the path.
Get all of these really clear, and you've basically solved the problem.
You've got to spend the time to make sure that everything is well defined before you start trying to work out these problem.
More about the path.
There are a couple ways you could go through that path.
If I look at this smooth path here.
I could have that path be very slow and steady, so that at every point along the way, my gas is an equilibrium.
So I've got, this piston here is compressed, and I slowly, slowly increase the volume, drop the temperature.
Then I can go back, the gas is included at every point of the way.
That's a reversible path.
That can reverse the process.
I expand it, and reverse it, no problem.
So, I could have a reversible path, or I take my gas, and instead of slowly, slowly raising it, dropping the pressure, I go from five bar to one bar extremely fast.
What happens to my gas inside?
Well, my gas inside is going to be very unhappy.
It's not going stay in equilibrium.
Parts of the system are going to be at five bar.
Parts of it at one bar.
Parts of it may be even at zero bar, if I go really fast.
I'm going to create a vacuum.
So the system will not be described by a single state variable during the path.
If I look at different points in my container during that path, I'm going to have to use a different value of pressure or different value of temperature at different points of the container.
That's not an equilibrium state, and that process turns out then to be in irreversible process.
Do it very quickly.
Now to reverse it and get back to the initial point is going to require some input from outside, like heat or extra work or extra heat or something, because you've done an irreversible process.
You've wasted a lot of energy in doing that process.
I have to tell you whether the path is reversible or irreversible, and the irreversible path also defines the direction of time.
You can only have an irreversible path go one way in time, not the other way.
Chalk breaks irreversibly and you can't put it back together so easily.
You've got to pretty much take that chalk, and make a slurry out of it, put water, and dry it back up, put in a mold, and then you can have the chalk again, but you can't just glue it back together.
That would not be the same state as what you started out with.
And then there are a bunch of words that describe these paths.
Words like adiabatic, which we'll be very familiar with.
Adiabatic means that there's no heat transferred between the system and the surrounding.
The boundary is impervious to transfer of heat, like a thermos.
Anything that happens inside of the thermos is an adiabatic change because the thermos has no connection in terms of energy to the outside world.
There's no heat that can go through the walls of the thermos.
Whereas, like isobaric means constant pressure.
So, this path right here from this top red path is an isobaric process.
Constant temperature means isothermal, so this part means an isothermal process.
So then, going from the initial to final states with a red path, you start with an isobaric process and then you end with an isothermal process.
And these are words that are very meaningful when you read the text of a problem or of a process.
Any questions before we got to the zeroth law?
We're pretty much done with our definitions here.
Yes
Was adiabatic reversible?
Adiabatic can be either reversible or not, and we're going to do that probably next time or two times.
Any other questions?
Yes
Is there a boundary between reversible and irreversible?
A boundary between reversible and irreversible?
Like something is almost reversible and almost irreversible.
No, pretty much things are either reversible or irreversible.
Now, in practice, it depends on how good your measurement is.
And probably also in practice, nothing is truly reversible.
So, it depends on your error bar in a sense.
It depends on what what you define, exactly what you define in your system.
It becomes a gray area, but it should be pretty clear if you can treat something is reversible are irreversible.
Other questions, It's a good question.
So the zeroth law we're going to go through the laws now.
The zeroth law talks about defining temperature and it's the common-sense law.
You all know how.
When something hot, it's got a higher temperature than when something is cold.
But it's important to define that, and define something that's a thermometer.
So what do you know?
What's the empirical information that everybody knows?
Everybody knows that if you take something which is hot and something which is cold, and you bring them together, make them touch, that heat is going to flow from the hot to the cold, and make them touch, and heat flows from hot to cold.
That's common sense.
This is part of your DNA, And then their final product is an object, a b which ends up at a temperature or a warmness which is in between the hot and the cold.
So, this turns out to be warm.
You get your new equilibrium state, which is in between what this was, and what a and b were.
Then how do you know that it's changed temperature, or that heat has flowed from a to b?
Practically speaking, you need some sort of property that's changing as heat is flowing.
For instance, if a were metallic, you could measure the connectivity of a or resistivity, and as heat flows out of a into b, the resistivity of a would change.
Or you could have something that's color metric that changes color when it's colder, so you could see the heat flowing as a changes color or b changes color as heat flows into b.
So, you need some sort of property, something you can see, something you can measure, that tells you that heat has flowed.
Now, if you have three objects, if you have a, b, and c, and you bring them together, and a is the hottest, b is the medium one, and c is the coldest, so from hottest to coldest a, b, c, -- if you bring them together and make them touch, you know, intuitively, that heat will not flow like this.
You know that's not going to happen.
You know that what will happen is that heat will flow from a to b from b to c and from a to c. That's common-sense.
You know that.
And the other way in the circle will never happen.
That would that would give rise to a perpetual motion machine, breaking of the second law.
It can't happen.
But that's an empirical observation, that heat flows in this direction.
And that's the zeroth law thermodynamic.
It's pretty simple.
The zeroth law says that if a and b -- it doesn't exactly say that, but it implies this.
It says that if a and b are in thermal equilibrium, if these two are in thermal equilibrium, meaning that there's no heat flows between them, so that's the definition of thermal equilibrium, that no heat flows between them, and these two are in thermal equilibrium, and these two are in thermal equilibrium, then a and c will be also be in thermal equilibrium.
But if there's no heat flowing between these two, and no heat flowing between these two, then you can't have heat flowing between these two.
So if I get rid of these arrows, there's no heat flowing because they're in thermal equilibrium, then I can't have an arrow here.
That's what the zeroth law says.
They're all the same temperature.
That's what it says.
If two object are in the same temperature, and two other object are in the same temperature, then all three must have the same temperature.
It sounds pretty silly, but it's really important because it allows you to define a thermometer and temperature.
Because now you can say, all right, well, now b can be my thermometer.
I have two objects, I have an object which is in Madagascar and an object which is in Boston, and I want to know, are they the same temperature?
So I come out with a third object, b, I go to Madagascar, and put b in contact with a.
Then I insulate everything, you know, take it away and see if there's any heat flow.
Let's say there's no heat flow.
Then I insulate it, get back on the plane to Boston, and go back and touch b with c. If there's no heat flow between the b and c, then I can say all right, a and c were the same temperature.
B is my thermometer that tells me that a and c are in the same temperature.
And there's a certain property associated with heat flow with b, and it didn't change.
And that property could be color.
It could be resistivity.
It could be a lot of different things.
It could be volume.
And the temperature then is associated with that property.
And if it had changed, then the temperature between those two would have changed in a very particular way.
So, zeroth law, then, allows you to define the concept of temperature and the measurement of temperature through a thermometer.
Let's very briefly go through stuff that you've learned before.
So, now you have this object which is going to tell you whether other things are in thermal equilibrium now.
What do you need for that object?
You need that object to be a substance, to be something.
So, the active part of the thermometer could be water.
It could be alcohol, mercury, it could be a piece of metal.
You need a substance, and then that substance has to have a property that changes depending on the heat flow, i.e., depending on whether it's sensing that it's the same temperature or different temperature than something else.
And that property could be the volume, like if you have a mercury thermometer, the volume of the mercury.
It could be temperature.
It could be resistivity, if you have a thermocouple.
It could be the pressure.
All right, so now you have an object.
You've got a property that changes, depending on the heat flow.
It's going to tell you about the temperature.
Now you need to define the temperature scales.
So, you need some reference points to be able to tell you, OK, this temperature is 550 degrees Smith, whatever.
So, you assign values to very specific states of matter and call those the reference points for your temperature.
For instance, freezing of water or boiling of water, the standard ones.
And then an interpolation scheme.
You need a functional form that connects the value at one state of matter, the freezing point of water, to another phase change, the boiling point of water.
You can choose a linear interpolation or quadratic, but you've got to choose it.
And it turns out not to be so easy.
And if you go back into the 1800's when thermodynamics was starting, there were a zillion different temperatures scales.
Everybody had their own favorite temperature scales.
The one that we're most familiar with is the centigrade or Celsius scale where mercury was the substance, and the volume of mercury is the property.
The reference points are water, freezing or boiling, and the interpolation is linear, and then that morphed into the Kelvin scale, as we're going to see later.
The Fahrenheit scale is an interesting scale.
It turns out the U.S. and Jamaica are the only two places on Earth now that use the Fahrenheit scale.
Mr. Fahrenheit, Daniel Gabriel Fahrenheit was a German instrument maker.
The way he came up with his scale was actually he borrowed the Romer scale, which came beforehand.
The Romer scale was, Romer was a Dane, and he defined freezing of water at 7.5 degrees Roemer, and 22.5 degrees Romer as blood-warm.
That was his definition.
Two substances, blood and water.
Two reference points, freezing and blood-warm, you know, the human body.
A linear interpolation between the two, and then some numbers associated with them, 7-1/2 and 22-1/2.
Why does he choose 7-1/2 as the freezing point of water?
Because he thought that would be big enough that in Denmark, the temperature wouldn't go below zero.
That's how he picked 7-1/2.
Why not?
He didn't want to use negative numbers to measure temperature in Denmark outside.
Well, Fahrenheit came along and thought, well, you know, 7-1/2, that's kind of silly; 22-1/2 that's, kind of silly.
So let's multiply everything by four.
I think it becomes 30 degrees for the freezing of water and 22.5 x 4, which I don't know what it is, 100 or something -- no, it's 90 I think.
And then for some reason, that nobody understands, he decided to multiply again by 16/15, and that's how we get 32 for freezing of water and 96 in his words for the temperature in the mouth or underneath the armpit of a living man in good health.
What a great temperature scale.
It turns out that 96 wasn't quite right.
Then he interpolated and found out water boils at 212.
But, you know, his experiment wasn't so great, and, you know, maybe had a fever when he did the reference point with 96, whatever.
It turns out that it's not 96 to be in good health, it's 98.6 -- whatever.
That's how we got to the Fahrenheit scale.
All right, next time we're going to talk about a much better scale, which is the ideal gas thermometer and how we get to the Kelvin scale.
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So last time we talked about the zeroth law, which is the common-sense law, which says that if you take a hot object next to a cold object, heat will flow from the hot to the cold in a way that is well defined, and it allows you to define temperature.
It allows you to define the concept of a thermometer.
You have three objects, one of them could be a thermometer.
You have two of them separated at a distance.
You take the third one, and you go from one to the other, and you see whether heat flows, when you touch one object, the middle object, between those two objects.
Let me talk to you about temperature scales.
We talked about the Celsius scale then the Fahrenheit scale.
The late 1800's were a booming time for temperature scales.
People didn't really realize how important it was to properly define the reference points: Fahrenheit's warm-blooded or 96 degrees, and Romer's 7.5 degrees.
Romer because he didn't want to go below zero degrees measuring temperature outside in Denmark Those are kind of silly.
But they're the legacy that we have today, and that's what we use.
In science, we use somewhat better temperature scales.
And the temperature scale that turns out to be well-defined and ends up giving us the concept of an absolute zero is the ideal gas thermometer.
So, let's talk about that briefly today first.
The ideal gas thermometer.
It's based on Boyle's law.
Boyle's law was an empirical law that Mr. Boyle discovered by doing lots of experiments, and Boyle's law says that the limit of the quantity pressure times the molar volume, so this quantity here, pressure times the molar volume, as you let pressure go to zero.
So, you do this measurement, you measure with the gas, you measure the pressure and the molar volume.
Then you change the pressure again, and you measure the pressure in the volume, and you multiply these two together, and you keep doing this experiment, getting the pressure smaller and smaller, you find that this limit turns out to be a constant, independent of the gas.
It doesn't care where the gas is.
You always get to the same constant.
And that constant turns out to be a function of the temperature.
The only function it is -- it doesn't care where the gas is.
It only cares where the temperature is.
All right, so now we have the makings of a good thermometer and a good temperature scale.
We have a substance.
The substance could be any gas.
That's pretty straightforward.
So now we have a substance, which is a gas, with a property.
So now the volume of mercury, or the color of something which changes with temperature, or the resistivity.
In this case here, our property is the value of the pressure times the volume, times the molar volume.
That's the property.
The property is the limit as p goes to zero of pressure times molar volume.
It's a number.
Measure it.
It's a number.
It's going to come out.
That's the property that's going to give us the change in temperature.
Then we need some reference points.
And Celsius first used the boiling point of water, and called that 100 degrees Celsius, and the freezing point of water and called that zero degrees Celsius.
And then we need an interpolation scale.
How to go from one reference point to the other with this property.
This property, which we're going to call f(t).
There are many ways you can connect those two dots.
If I draw a graph, and on one axis I have this temperature.
The idea of temperature with two reference points, zero for the freezing point of water, 100 degrees for the boiling point of water.
And on the y-axis I've got the property f(t).
It has some value corresponding to t equals zero.
So let's get some value right here.
There's another value connected to this property here, when t is equal to 100, a reference point here.
Now there many ways I can connect these two points together.
The simplest way is to draw a straight line.
It's called the linear interpolation.
My line is not so straight, right here.
You could do a different kind of line.
You could do a quadratic, let's say.
Something like this.
That would be perfectly fine interpolation.
All right, we choose to have a linear interpolation.
That's a choice, and that choice turns out to be very interesting and really important, because if you connect these two points together, you get a straight line that has to intercept the x-axis at some point.
Now what does it mean to intercept the x-axis here?
It means that the value of f(t) for this temperature is zero.
That means that at this point right here, f(t)=0.
That means the pressure times the volume equals zero, for that gas.
And if you're below this temperature here, this quantity, p times v it would be negative.
Is that possible?
Can we have p v negative?
Yes?
No, it can't be.
Negative pressure doesn't make any sense, right?
Negative volume doesn't make any sense.
That means that this part here, can't happen.
That means that this temperature right here is the absolute lowest temperature you can go to that physically makes any sense.
That's the absolute zero.
So the concept of an absolute zero, a temperature below which you just can't go, that's directly out of the scheme here, this linear interpolation scheme with these two reference points.
If I had taken as my interpolation scheme, my white curve here, I could go to infinity and have the equivalent of absolute zero being at infinity, minus infinity.
So, this temperature, this absolute zero here, which is absolute zero on the Kelvin scale.
The lowest possible temperature in the Celsius scale is minus 273.15 degrees Celsius.
So that begs the notion of re-referencing our reference point, of changing our reference points.
To change a reference point from this point here being zero, instead of this point here being zero.
And so redefining then the temperature scale to the Kelvin scale, where t in degrees Kelvin is equal to t in degree Celsius, plus 273.15.
And then you would get the Kelvin scale.
All right, it turned out that this thermometer here wasn't quite perfect either.
Just like Fahrenheit measuring 96 degrees being a warm-blooded, healthy man, right, that's not very accurate.
Our temperature probably fluctuates during the day a little bit anyways, it's not very accurate.
And similarly, the boiling point, defining that at a 100 degrees Celsius, well that depends on the pressure.
It depends whether you're in Denver or you're in Boston.
Water boils at different temperatures, depending on what the atmospheric pressure is; same thing for the freezing point.
So that means, then, you've got to define the pressure pretty well.
You've got to know where the pressure is.
It would be much better if you had a reference point that didn't care where the pressure was.
Just like our substance doesn't care where the gas is.
It's kind of universal.
And so now, instead of using these reference points for the Kelvin scale, we use the absolute zero, which isn't going to care what the pressure is.
It's the lowest number you can go to.
And our other reference point is the triple point of water -- reference points become zero Kelvin, absolute zero, and the triple point.
The triple point of water is going to be defined as 273.16 degrees Kelvin.
And the triple point of water is that temperature and pressure -- there's a unique temperature and pressure where water exists in equilibrium between the liquid phase, the vapor phase, and the solid phase.
So the triple point is liquid, solid, gas, all in equilibrium.
Now you may think, well I've seen that before.
You take a glass of ice water and set it down.
There's the water phase, there's the ice cube is the solid phase, and there's some water, gas, vapor, and that's one bar.
Where am I going wrong here?
The partial pressure of the water, of gaseous water, above that equilibrium of ice and water is not one bar, it's much less.
So the partial pressure or the pressure by which you have this triple point, happens to be 6.1 times 10 to the minus 3 bar.
There's hardly any vapor pressure above your ice water glass.
So this unique temperature and unique pressure defines a triple point everywhere, and that's a great reference point.
Any questions?
Great.
So now we have this ideal gas thermometer, and out of this ideal gas thermometer, also comes out the ideal gas law.
Because we can take our interpolation here, our linear interpolation, the slope of this line.
Let's draw it in degrees Kelvin, instead of in degrees Celsius.
So we have now temperature in degrees Kelvin.
We have the quantity f(t) here.
We have an interpolation scheme between zero and 273.16 with two values for this quantity, and we have a linear interpolation that defines our temperature scale, our Kelvin temperature scale.
And so the slope of this thing is f(t) at the triple point, which is this point here, this is the temperature of the triple point of water, divided by 273.16.
That's the slope of that line.
The quantity here, which is f (t of the triple point), divided by the value of the x-axis here.
So that's the slope, and the intercept is zero, so the function f(t), you just multiply by t here.
This is the slope.
f(t) is just the limit.
As p goes to zero of p times v bar.
And so now we have this quantity, p times v bar, and the limit of p goes to zero is equal to a constant times the temperature.
That's a universal statement.
It's true of every gas.
I didn't say this is only true of hydrogen or nitrogen, This is any gas because I'm taking this limit p equals to zero.
Now this constant is just a constant.
I'm going to call it r. I'm going to call it r. It's going to be the gas constant, and now I have r times t is equal to the limit, p goes to zero of p r. It's true for any gas, and if I remove this limit here, r t is equal to p v bar, I'm going to call that an ideal gas.
See, this is the property of an ideal gas.
What does it mean, ideal gas?
It means that the molecules or the atoms and the gas don't know about each other.
They effectively have no volume.
They have no interactions with each other.
They occupy the same volume in space.
They don't care that there are other atoms and molecules around.
So that's basically what you do when you take p goes to zero.
You make the volume infinitely large, the density of the gas infinitely small.
The atoms or molecules in the gas don't know that there are other atoms and molecules in the gas, and then you end up with this universal property.
All right, so gases that have this universal property, even when the pressure is not zero, those are the ideal gases.
And for the sake of this class, we're going to consider most gases to be ideal gases.
Questions?
So now, this equation here relates three state functions together: the pressure, the volume, and the temperature.
Now, if you remember, we said that if you had a substance, if you knew the number of moles and two properties, you knew everything about the gas.
Which means that you can re-write this in the form, volume, for instance, is equal to the function of n, p, t..
In this case, V = (nRT)/P.
Have two quantities and the number of moles gives you another property.
You don't need to know the volume.
All you need to know is the pressure and temperature and the number of moles to get the volume.
This is called an equation of state.
It relate state properties to each other.
In this case it relates the volume to the pressure and the temperature.
Now, if you're an engineer, and you use the ideal gas law to design a chemical plant or a boiler or an electrical plant, you know, a steam plant, you're going to be in big trouble.
Your plant is going to blow up, because the ideal gas law works only in very small range of pressures and temperatures for most gases.
So, we have other equations of states for real gases.
This is an equation of state for an ideal gases.
For real gases, there's a whole bunch of equation the states that you can find in textbooks, and I'm just going to go through a few of them.
The first one uses something called a compressibility factor, z. Compressibility factor, z.
And instead of writing PV = RT, which would be the ideal gas law, we put a fudge factor in there.
And the fudge factor is called z.
Now we can put real instead of ideal for our volume.
z is the compressibility factor, and z is the ratio of the volume of the real gas divided by what it would be were it an ideal gas.
So, if z is less than 1, then the real gas is more compact then the ideal gas.
It's a smaller volume.
If z is greater than 1, then the real gas means that the atoms and molecules in the real gas are repelling each other and wants to have a bigger volume.
And you can find these compressibility factors in tables.
If you want to know the compressibility factors for water, for steam, at a certain pressure and temperature, you go to a table and you find it.
So that's one example of a real equation of state.
Not a very useful one for our purposes in this class here.
Another one is the virial expansion.
It's a little bit more useful.
What you do is you take that fudge factor, and you expand it out into a Taylor series.
So, we have the p v real over r t is equal to z.
Now, we're going to take z and say all right, under most conditions, it's pretty close to 1, when it's an ideal gas.
And then we have to add corrections to that, and the corrections are going to be more important, the larger the volume is.
Remember, it's the limit of p times v goes to zero, so if you have a large volume with a large pressure, then you're out of the ideal gas regime.
So let's take Taylor series in one over the volume, it's going to be one over the volume squared, etcetera.
And these factors on top, which are going to depend on the temperature, are the virial coefficients, and those depend on the substance.
So you have this p B(t) here.
This is called a second virial coefficient.
And then, so you can get, you can actually find a graph of this B(t).
It's going to look something like this.
It's the function of temperature, as B(t).
There's going to be some temperature where B(t) is equal to zero.
In that case, your gas is going to look awfully like an ideal gas.
Above some temperature is going to be positive, below some temperature is going to be negative.
Generally, we ignore the high order terms here.
So again, if you do a calculation where you're close enough to the ideal gas, and you need to design your, if you have an engineer designing something that's got a bunch of gases around, this is a useful thing to use.
Now, the most interesting one for our class, the equation of state that's the most interesting, is the Van der Waals equation of state, developed by Mr. Van der Waals in 1873.
And the beauty of that equation of state is that it only relies on two parameters.
So let's build it up.
Let's see where it comes from.
Let me just first write it down, the Van der Waals equation of state.
p plus a over v bar squared times v bar minus b equals r t. All right, if you take a equal to zero, these are the two parameters, a and b.
If you take those two equal to zero, you have p v is equal to r t. That's the ideal gas law.
Let's build this up.
Let's see where this comes from, where these parameters a and b comes from.
So, the first thing we're going to do is we're going to take our gas in our box, let's build a box full of gases here.
We've got a bunch of gas molecules or atoms.
OK, there's the volume of a box here.
While these gas molecules or atoms through first approximation, are like hard spheres.
They occupy a certain volume.
Each atom or molecule occupies a particular volume.
And so, we can call b is the volume per mole of the hard spheres, volume per mole that is the little sphere that the molecules are.
So that the volume that is available to any one of those spheres is actually smaller than v. Because you've got all these other little spheres around, so the actual volume seen by any one of those spheres is smaller than v. So when we take our ideal gas law, p v bar is equal to r t we have to replace v bar by the actual volume available to this hard sphere.
So instead of v bar, we write p v bar minus b, equal r t. OK, that's the hard sphere volume of the spheres.
Now, those molecules or atoms that are in here, also feel each other.
There are a whole bunch of forces that you learn in 5.112, 5.111 like with Van der Waals' attractions and things like this.
So there are attractive forces, or repulsive forces that these molecules feel, and that's going to change the pressure that the molecules feel.
For instance, if I have, what is pressure?
Pressure is when you have one of these hard spheres colliding against the wall.
There's the hard sphere.
It wants to collide against the wall to create a force on the wall, and I have a couple of the hard spheres that are nearby, right, and in the absence of any interactions, I get a certain pressure.
This thing would but careen into the wall, kaboom!
You'd have this little force, but in the presence of these interactions, you've got these other molecules here that are watching this, you know, their partner sort of wants to do damage to themselves, like hitting that wall, and they say, no!
Come back, come back, right?
There is an attractive force.
There are no other molecules on that side of the wall.
So there's an attractive force that makes the velocity within not quite as fast.
The force is not quite as strong as it was without this attractive force.
So the real pressure is not quite the same because of this attractive force as it was, as it would be without the attractive forces.
The pressure is a little bit less in this case here.
So instead of this p here.
Now if I re-write this equation here as p is equal to r t divided by v bar minus b, just re-writing this equation as it is.
So the pressure is going to depend on how strong this attractive force is.
So the pressure is going to be less if there's a strong attractive force.
And the 1 over v squared is a statistical, is basically a probability of having another molecule, a second molecule in the volume of space.
So, if the molar volume is small, then one over v bar is large, there's a large probability of having two spheres together in the same volume.
If the molar volume is large, that means that there's a lot of room for the molecules, and they're now going to be close to each other, and so this isn't going to be as important.
So, a is the strength of the interaction, v bar is how likely they are to be close to each other.
And that's going to affect the actual pressure seen by the gas.
And a is greater than zero when you have the attraction.
And that gives use the Van der Waals' equation of state, with two parameters, the hard sphere volume and the attraction.
You don't have to go look up in tables or books.
You don't have to have all the values of the second virial coefficient, or the fudge factor, just two variables that make physical sense, and you get an equation of state which is a reasonable equation of state, and that's the power of the Van der Waals' equation of state, and that's the one we're going to be using later on this class to describe real gases.
Question?
OK, so we've done the zeroth law.
We've done temperature, equations of state.
We're ready for the first law.
We're just going to go to through these laws pretty quickly here.
Remember, the first law is the upbeat law.
It's the one that says, hey, you know, life is all rosy here.
We can take energy from fossil fuels and burn it up and make it heat, and change that energy into work.
And it's the same energy, and we probably can do that with 100% efficiency.
We can take heat from the air surrounding us and run our car on it with 100% efficiency.
Is this possible?
That's what the first law says, it's possible; work is heat, and heat is work, and they're the same thing.
You can break even, maybe.
So let's go back and see what work is.
Let's go back to our freshman physics.
Work, work is if you take a force, and you push something a certain distance, you do work on it.
So if I take my chalk here and I push on it, I'm doing work to push that chalk.
Force times distance is work.
The applied force times the distance.
There are many kinds of work.
There's electrical work, take the motor, you plug it into the wall, electricity makes the fan go around, that's electrical work.
There's magnetic work.
There is work due to gravity.
In this class here, we're going to stick to one kind of work which is expansion work.
So expansion work, for instance, or compression work, is if you have a piston with a gas in it.
All right, you put a pressure on this piston here, and you compress the gas down.
This is compression work.
Now the volume gets smaller.
p external here.
Pressure, the piston goes down by some volume l. The piston has a cross-sectional area, a, and the force -- pressure is force per volume area.
So the force that you're pushing down on here is the external pressure times the area.
Pressure is force per volume area.
That's the force you're using to push down.
Now the work that's it is calculated when you push down with the pressure on this piston here, that work is force times distance, f times I. f is p external times a, times the distance l. So that's p external times the change in the volume.
The area times this distance is a volume, and that is the change in volume from going to the initial state to the final state.
Now we need to have a convention.
We've got force.
Work is force times distance, it's p external times delta v, and I'm going to be stressing a lot that this is the external pressure.
This is the pressure that you're applying against the piston, not the pressure of the gas.
It's the pressure the external world is applying on this poor system here.
OK, but we need a convention here.
The convention, and then we need to stick to it.
And this convention, unfortunately, has changed over the ages.
But we're going to pick one, and we're going to stick to it, which is that if the environment does work on the system, if we push down on this thing and do work on it, to compress it, then we call that work negative work.
No, we call that work positive work.
All right, so that means we need to put a negative sign right here, by convention.
So if delta v is negative, in this case delta v is negative, OK, delta v is negative, pressure is a positive number, negative times negative is positive, work is greater than zero.
We're doing work on the system, to the system.
In this case here, work is positive.
If you have expansion on the other side, if the system is expanding in the other direction, if you're going this way, right, you're going to do work to the environment.
There might be a mass here.
This could be a car.
Pistons in the car, right, so the piston goes up.
That's going to drive the wheels.
The car is going to go forward.
You're doing work on the environment.
Delta v is going to be negative.
w is going to be negative.
Sorry, I got it backwards again.
Delta v is positive in this direction here, the work is negative.
So work on the system is positive.
Work done by the system is negative.
Convention, OK, this negative sign is just a pure convention.
You just got to use it all the time.
If you use an old textbook, written when I was taking thermodynamics, they have the opposite convention, and it's very confusing.
But now we've all agreed on this convention, and work is going to be with the negative sign here.
OK, any questions?
This is an example where the external pressure here is kept fixed as the volume changes, but it doesn't have to be kept fixed.
I could change my external pressure through the whole process, and that's the path.
We talked about the path last time being very important.
Defining the path.
So if I have a path where my pressure is changing, then I can't go directly from this large volume to this small volume.
I have to go in little steps, infinitely small steps.
So, instead of writing work is the negative of p external times delta v, I'm going to write a differential.
dw is minus p external dv, where this depends on the path, it depends on path and is changing as v and p change.
Now I'm going to add a little thing here.
I'm going to put a little bar right here.
And the little bar here means that this dw that I'm putting here is not an exact differential.
What do I mean by that?
I mean that if I take the integral of this to find out how much work I've done on the system, I need to know the path.
That's what this means here.
It's not enough to know the initial state and the final state to find what w is.
You also need to know how you got there.
This is very different from the functions of state, like pressure and temperature.
There's a volume, there's a temperature, than the pressure here.
There's other volume, temperature and pressure here, corresponding to this system here.
And this volume, temperature and pressure doesn't care how you got there.
It is what it is.
It defines the state of the system.
The amount of work you've put in to get here depends on the path.
It's not a function of state.
It's not an exact differential.
So the delta v here is an exact differential, but this dw is not.
That's going to be really important.
So if you want to find out how much work you've done, you take the integral from the initial state to the final state of dw minus from one to two p external dv, and you've got to know what the path is.
So let's look at this path dependence briefly here.
We're going to do two different paths, and see how they're different in terms of the work that comes out.
So we're going to take an ideal gas, we can assume that it's ideal.
Let's take argon, for instance, a nice, non-interacting gas.
We're going to do a compression.
We're going to take argon, with a certain gas, certain pressure p1, volume V1, and we're going to a final state argon, gas, p2, V2.
Where V1 is greater than V2, and p1 is less than p2.
So if I draw this on a p v diagram, so there is volume on this axis.
There's pressure on this axis.
There is V1 here.
There's V2 here.
There's p1 here, and p2 here.
So I'm starting at p1, V1.
I'm starting right here.
And I'm going to end right here.
Initial find -- there are many ways I can get from one state to the other.
Draw any sort of line to go here, right?
There are a couple obvious ones, which we're going to -- we can calculate, which we're going to do.
So, the first obvious one is to take V1 to V2 first with p constant.
So take this path here.
I take V1 to V2 first, keeping the pressure constant at p1, then I take p1 to p2 keeping the volume constant at V2.
Let's call this path 1.
Then you take p1 to p2 with V constant.
An isobaric process followed by a constant volume process.
You could also do a different path.
You could do, let me draw p v, there's my initial state.
My final state here, I could take, first, I could change the pressure, and then change the volume.
So the second process, if you take p1 to p2, V constant, and then you take V1 to V2 with p constant.
This is path number two.
Both are perfectly fine paths, and I'm going to assume that these paths are also reversible.
Let's assume that both are reversible, meaning that I'm doing this pretty slowly, so as I change, let's say I'm changing my volumes here, V1 to V2, it's happening, I'm compressing it slowly, slowly, slowly so that at any point I could reverse the process without losing energy, right?
It's always an equilibrium.
All right, let's calculate the work that's involved with these two processes.
Remember it's the external pressure that's important.
In this case, because it's a reversible process, the external pressure turns out to be always the same as the internal pressure.
It's reversible, that means that p external, equals p. I'm doing it very slowly so that I'm always in equilibrium between the external pressure and the internal pressure so I can go back and forth.
So, let's calculate w1.
The work for path one.
First thing is I change the volume from V1 to V2 The external pressure is kept constant, p1, so it's minus the integral from 1, V1 to V2, p1, dv.
And then the next step here is I'm going from -- the pressure is changing.
I'm going from V2 to V2 dv -- what do you think this integral is?
Right, so this is easy part, zero here.
This one is also pretty easy.
That's minus p1 times V2 minus V1.
p1 times V2 minus V1.
What that turns out to be, this area right here.
It's V1 minus V2 times p1.
This is w1 here.
OK, I can re-write this as p1 time V1 minus V2 and get rid of this negative sign here.
Now V1 is bigger than V2, so this is positive.
So I am compressing, I'm doing work to the system, positive work everything follows our convention.
Number two here, OK, the first thing I do is I change the pressure under constant volume, V1, V1 minus p dv, and then I change the volume from V1 to V2 and then this is p2, dv.
This first integral is zero V1 to V1, then I get minus p2 times V2 minus V1 or p2 times V1 minus V2.
Again, a positive number.
I'm doing work to the system to go from the initial state to the final state.
But it's not the same as w1.
In this case, I have p1 times delta V. In this case here, I have p2 times delta V. And p2 is bigger than p1.
w2 is bigger than w1.
The amount of work that you're doing on the system depends on the path that you take.
All right, how do I, practically speaking, how do I do this?
Anybody have an idea?
How do I keep p1 constant while I'm lowering the volume?
Change the temperature?
Change the temperature, right.
So what I'm doing here is I'm cooling, and then when I'm sitting at a fixed volume and I'm increasing the pressure, what am I doing?
I'm heating, right?
So I'm doing cooling and heating cycles.
So in this case here, I cool and then I heat.
In this case here, I heat and then I cool.
All right, so I'm burning some energy, I'm burning some fuel to do this somehow, to get that work to happen.
All right, now suppose that I took these two paths, and coupled them together.
So in this case, it's the amount of work is the area under that curve.
And in this case here, the amount of work is bigger, w2 is bigger, and it's the area under this curve.
Now, suppose I took this two paths, and I took -- couple them together with one the reverse of the other.
So I have my initial state, my final state, my initial state, my final state here.
And I start by taking my first path here.
I cool, I heat.
So there's w1.
So the w total that I'm going to get, is w1, and then instead of the path from V1 to, from 1 to 2 going like this as we had before, I'm going to take it backwards.
If I go backwards, to work -- everything is symmetric, the work becomes the negative from what I had calculated before, so this becomes minus what I calculated before for w2.
The total work, in this case here, is p1 times V1 minus p2 times V1 minus V2, it's p1 minus p2 times V1 minus V2.
This is a positive number, p1 is smaller than p2.
This is a negative number.
The total work is less than zero.
That's the work that the system is doing to the environment.
I'm doing work to the environment.
The work is negative, which means that work is being done to the environment.
And that work is the area inside the rectangle.
What you've built is an engine.
You cool, you heat, you heat, you cool, you get back to the same place, but you've just done work to the environment.
You've just built a heat engine.
You take fuel, rather you take something that's warm, and you put it in contact with the atmosphere, it cools down.
You take your fuel, you heat it up again.
It expands.
You change your constraints on your system, you heat it up some more, then you take the heat source away, and you put it back in contact with the atmosphere.
And you cool it a little bit, change the constraints, cool it a little bit more, and heat, and you've got a closed cycle engine.
We're going to work with some more complicated engines before.
But the important part here is that the work is not zero.
You're starting at one point.
You're going around a cycle and you're going back to the same point.
The pressure, temperature, and volume are exactly the same here as when you started out.
But the w is not zero.
The w, for the closed path, and when I put a circle there on my integral that means a closed path, when you start and end at the same point, right, this is not zero.
If you had an exact differential, the exact differential around a closed path, you would get zero.
It wouldn't care where the path is.
Here this cares where the path is.
So, work is not a function of state.
Any questions on work before we move on to heat, briefly?
So heat is a quantity that flows into a substance, something that flows into a substance that changes it's temperature, very broadly defined.
And, again, we have a sign convention for heat.
So heat, we're going to call that q.
And our sign convention is that if we change our temperature from T1 to T2, where T2 it's greater than T1 then heat is going to be positive.
Heat needs to go into the system to change the temperature and make it go up.
If the temperature of the system goes down, heat flows down heat flows out of the system, and we call that negative q.
Same convention is for w, basically.
Now, you can have a change of temperature without any heat being involved.
I can take an insulated box, and I can have a chemical reaction in that insulated box.
I can take a heat pack, like the kind you buy at a pharmacy.
Break it up.
It gets hot.
There's no heat flowing from the environment to the system.
I have to define my terms.
My system is whatever's inside the box.
It's insulated.
It's a closed system.
In fact, it's an isolated system.
There's no energy or matter that can go through that boundary.
Yet, the temperature goes up.
So, I can have a temperature change which is an adiabatic temperature change.
Adiabatic means without heat.
Or I could have a non-adiabatic, I could take the same temperature change, by taking a flame, or a heat source and heating up my substance.
So, clearly q is going to depend on the path.
I'm going from T1 to T2, and I have two ways to go here.
One is non-adiabatic.
One is adiabatic.
All right, now what we're going to learn next time, and Bob Field is going to teach the lecture next time, is how heat and work are related, and how they're really the same thing, and how they're related through the first law, through energy conservation.
OK, I'll see you on Wednesday then.
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As you can probably tell from the volume of my voice, I'm not Moungi.
I'm Bob Field, and I will be lecturing today as a special booby prize, because Bawendi is so wonderful.
Last time you heard about work and heat.
So there's p v work and it's given by the integral minus p external dv or just the integral from one to two of dw.
And this little slash here means an in inexact differential.
The reason for inexact doesn't mean it's a crummy measurement, it means that it's path dependent, and so the value of this integral depends on how you get from one to two.
w greater than zero means that work is done on the system, and it's really important to keep track of the signs of these things.
For heat, heat is defined, or the energy unit for heat, calorie, is defined as the amount of heat needed to raise one gram of water from 14.5 degrees c to 15.5 degrees c. That great deal of specificity implies that heat is also path-dependent and again we have the convention that if heat is added to the system, the quantity is greater than zero.
OK, so that's a quick statement of what happened last time.
Today we're going to talk about heat capacity.
The first law, a process that takes a gas from p1, V1, T to p2, V2, T examined for various paths.
And what you'll find is that the maximum work out is obtain for a reversible path.
We'll then look at the quantity, internal energy, which we define through the first law, and we think of it as a function of two variables T and V. And whenever we say something like that, we'll be able to write a differential, the partial of u, with respect to T at constant V, dT, plus the partial of u with respect to V at constant T dV.
So, if we write something like this, it implies this kind of equation.
And the main points of the latter part of the lecture is what are these quantities?
How do we measure them?
How do we understand them?
This one turns out to be the heat capacity, and this one turns out to be something that we measure in the Joule-free expansion.
So, that's the menu for today, at least that's what Moungi assigned me to cover, and I'll do the best I can.
So, let's talk about heat capacity.
Heat capacity relates the amount of heat that you add to the system to the change in temperature, and this is the relationship.
The heat-added, temperature, and this is a proportionality constant.
Heat capacity depends on path.
So, here are two kinds of experiments.
Here we have a fixed volume, and we have a little candle, and we're adding heat, and when we add heat, the pressure does what?
It increases and the temperature.
OK, we're you know, this is MIT.
You guys know something.
You should yell it out, not just whisper it.
You have the confidence -- I mean maybe up the street we whisper, but here we know it.
And, so here is a different kind of system where we have a constant external pressure.
We add heat to the system.
And so here the volume can change.
The temperature can change.
And so as we add heat here, the temperature goes up and the volume -- that was even quieter than last time.
OK, but you know it.
You understand it.
Now, the coefficient that relates the amount of heat in to the temperature change is obviously going to be different for these two cases.
in this case, where we have a fixed volume, we say dq is equal that Cv, heat capacity dT, where the v specifies what's constant for the path.
And so this is one path, constant volume.
This has a particular value.
Over here, we have dq=Cp dT, the heat, the proportionality between heat and temperature rise is given by this, the constant pressure heat capacity.
They're different quantities.
If we want to know the total heat added to the system, we can measure it, which is the straightforward thing, but sometimes you want to calculate in advance, or sometimes you want to calculate it on an exam.
So, we do an integral over a path, for the heat capacity along that path, dT.
So, for Cp and Cv, these are often quantities that are measured as a function of temperature, and one could, in fact, calculate this integral.
For most problems on exams, those quantities are constant, independent of temperature.
But there's no guarantee that they will be.
Now I get to tell a fun story.
The relationship between heat and work was initially proposed in the 1940's by Joule.
Now, there are two stories about Joule and how he came to this insight.
One is that he observed when people were machining cannon barrels, a lot of heat was generated, and there was a lot of work done.
And so maybe the work generated the heat.
Or there was a relationship between work and heat.
Well, this is ridiculous because people were making cannons long before 1840.
The other story, which is probably just as untrue, is that Joule, on his honeymoon, took his wife to a mountain resort, and they were sitting at the bottom of a waterfall, and he had this wonderful idea.
And that was the water at the bottom of the waterfall is probably going to be hotter than the water at the top of the waterfall.
So he grabbed his thermometer, and went and made a couple of measurements and discovered the first law of thermodynamics.
In fact, the water at the bottom of the waterfall is hotter than the water at the top of the waterfall.
And the idea was that gravity did work on the water and falling, and that work led to the generation of heat.
If you think about this, you probably can come up with several other ideas for how the water at the bottom might be warmer than at the top.
I mean it's flowing fast at the top, and it's sort of pooled at the bottom, and there is sunlight and all sorts of things that could make this coincidental as opposed to an insightful observation.
But it's a good story, Joule decided that there must be a direct relationship between work and heat.
They are the same quantity.
They are both forms of energy.
And so, we have this cartoon.
Again, we have an open beaker and a candle, and we're putting only heat into this beaker, and the temperature goes from T1 to T2.
And delta T is given by the heat, which has to do with how much of the candle burnt, divided by the constant pressure heat capacity.
Now you could do a similar sort of thing, but instead of having a candle, you have a paddle wheel, and the paddle wheel is spun by a weight that's dropping from here to here.
So, this weight is being spun-- now I'm really fantastic at drawing this mechanical device, but you can imagine that dropping a weight can cause a paddle wheel to turn.
And we know about gravity, and we know about work.
So, if you're doing physics, work is the integral of force, dx, x1 to x2.
Right, that's work.
Now, gravity, well force, is equal to mass times acceleration, and the acceleration due to gravity is g. The force, as this weight drops is constant, and so the work is just going to be m g h, where this is h. So, it's a trivial matter, by looking at what is the weight, and how far does it drop, to say OK, how much work is done by the paddle wheel.
And this is in an isolated, in an adiabatic container.
All the energy that is inserted into this, which might be turbulence initially, becomes heat, or becomes -- it raises the temperature.
And so, again, we see a temperature increase, and we know the work, and the temperature increase, it's a constant pressure thing.
And so we now have two observations.
The same temperature increase, work and heat, and we have a relationship between heat and work.
The first law of thermodynamics is -- I'm fine.
I don't -- I'm all set, I only use this chalk.
I brought it with me.
The first law of thermodynamics, written concisely is dw plus dq, two inexact differentials, integrated over any closed path, is zero.
This is an abstract and powerful mathematical statement of the first law of thermodynamics.
There are much better or more appealing expressions.
One is, du, u is called the internal energy or just the energy, is equal to dq plus dw.
OK, notice we have two inexact differentials and exact differentials.
This is a condition.
If you have a quantity which is constant over any closed path, that quantity is a thermodynamics state function.
So, this observation is equivalent to saying that there must be something that is path independent.
Therefore, we don't have a cross through the d. And the first law says, well heat and work are different forms of energy, and we can add them, and the path dependence of these two things is somehow cancelled in the fact that we have this internal energy.
So, we can also write delta u as integral from 1 to 2 of du.
That's u2 minus u1, and it's q plus w. So, these two quantities, again, are path dependent.
This is not.
That's the first law of thermodynamics.
It seems trivial, but it's really important.
It's almost as important as the second law of thermodynamics, which you'll see in a week or so.
There is a corollary to this.
If we say we have a system, and it's in its surroundings.
We can talk about du for the system well, that's q plus w. And we can say du for the surroundings.
Well the system got its work from the surroundings.
It either had work, got its heat from the surroundings, or it got worked on by the surroundings.
And so, we can immediately write this and then we can write du for the universe, which is system plus surroundings is equal to zero.
This is the corollary.
It's due to Clausius and it says the energy in the universe is conserved.
Fairly powerful statement, and that's another form of the first law of thermodynamics.
OK, enough for really great discoveries.
Now let's talk about some simple observations on isothermal gas expansions.
The purpose here is to look at a series of processes in which temperature is held constant, and we're going to calculate how much work we get from allowing a gas to expand under various conditions.
And what we'll discover is that when we allow the gas to expand reversibly, we get the most work.
This is neat.
This is important.
OK, so we have constant temperature, because it's isothermal.
And let's do a series of experiments.
First let's set the external pressure equal to zero.
So we have an experiment that looks like this.
We have a piston.
We have a pair of stops.
We have another pair of stops.
We start at p1, V1.
and p external is equal to zero.
In other words, this is vacuum.
And so what happens when we remove these stops is that the piston slams up against the next pair of stops.
It goes very fast.
You could calculate the time.
You could do all sorts of stuff, but what is important is that this piston in moving from here to here did no work on the surroundings because work is the integral of pressure dv, and the pressure external is zero.
It doesn't matter what the pressure internal is.
This is a point that is often confusing, because you can think, well maybe I could calculate what the internal pressure is even for this very rapid process.
It doesn't matter.
Thermodynamics is asking you, what work does this thing do on the surroundings or the surroundings do on the system?
And there is no work done on the surroundings because pressure is zero.
So, the work for this process is the integral, or minus the integral, V1, V2, p dV.
And it's p external and p external is zero, so there's no work.
OK, the next example is, well, when this piston slams up against these stops, the gas has expanded -- we're doing this isothermally, so this is in contact with the heat bath.
The gas has expanded.
It has a particular pressure and a particular volume.
Every time you do the experiment in equilibrium with the heat bath at T, you'll get the same p2 and V2.
One can know them.
And so we can then say, OK, let's do this second expansion.
and let's set p external equal to p2.
And we do the same sort of thing.
We have stops here.
We have p1, V1.
And we don't really need stops up here.
We have a weight on the piston, and that weight is chosen so that it acts as though p external is equal to p2.
We know the pressure is equal to force per area.
And so we know the force for a particular mass, and we know the area of the piston.
so we know how to do this, and we know this thing when it hit these stops, the pressure with p2, and the volume was V2.
Now we could put stops here at V2, but this piston would just kiss the stops.
It would stop there because we've chosen p2.
The work for that process is going to be minus V1, V2, p2, because that's what we chose p external to be, dV, and that's going to be minus p2, V2 minus V1.
That's the work.
V2 is larger than V1, and so this quantity here is positive, and we have a negative sign because the system did work.
OK, so now we can take the result from this and put it onto a p v diagram.
p1, p2, V1, V2 -- so here is the initial situation, and here is the final situation.
And the equation of state, pressure versus volume at constant temperature, is going to have some form, let's just draw it in there like that.
So that's an equation of state.
It's like the ideal gas law, and one could know that in principle.
OK, so we did work -- oh, I should mention p times V has units of energy or units of work.
Remember that, because you're going to find lots of thermodynamics quantities, and you're often going to be writing on exams, deriving things on exams, and you're going to almost always want the combinations of quantities to have units of energy.
So, dimensional analysis is extremely valuable in thermodynamics, and here is an example of it.
Anyway, the work associated with process number two is described by this box.
Because we did work at constant pressure, and so it's just volume difference times pressure.
OK, now we can do the process in two steps.
So we start out with our piston.
We have two weights on it.
So we have p1, V1.
We have stops.
We've chosen two weights, so that this is p3.
External pressure is p3.
It's higher than p2.
So we remove the stops, and the piston moves up, and now we remove one of these weights.
and when we remove one of the weights, this gives an external pressure p2.
So, the piston moves up again, one weight.
So here is p2, p external.
Here is p3.
And so we have the work coming from two steps.
And it's trivial to calculate what that will be, and when you do that, if we put p3 in on this diagram, what you end up finding is that you get an initial little bit of work corresponding to V3.
So we break up our work into three pieces, and we get more work out.
So the final process involves do it reversibly, or almost reversibly.
We want p equal to p external for the entire expansion, and p external is decreased steadily from p1 to p2.
How do we do this?
Well, we start out with our usual system, and we have a bunch of little pebbles on it.
Each weighing, say, one 100th of the weight needed to establish p external is equal to p1.
We remove a pebble, system expands.
We remove another one, system expands.
That's equivalent to doing the integral, and so, what we end up getting is that the reversible work is equal to minus integral V1, V2, p dV.
Because what we've done is we forced p, pressure here, to be equal to the external pressure.
We've changed the external pressure slowly, and again this is isothermal.
There is a heat bath here that keeps the temperature constant.
So, the system does work on the surroundings, hence the minus sign.
Now, if this is an ideal gas, we know that pressure is equal to nRT over volume.
So we can put that in here and do this integral.
We have minus V1, V2, nRT over V dV.
These are all constant.
It's isothermal.
We can take them out.
It's a closed system, so the number moles doesn't change.
The ideal gas constant doesn't change, temperature doesn't change, and so we just have minus nRT integral V1, V2, dV over V. Now, we are very gentle in this course with respect to knowing integrals, but this is one you have to know.
The integral of one over a quantity is the natural log.
And so we can write this, minus nRT log V2 over V1.
That should not take a breath.
You know that much about integrals.
OK, now we actually would like to simplify this or to write this in terms of not the volume change, but the pressure change.
So, we have V2 over V1.
Well, what we can write that using the ideal gas law twice, V2 is equal -- pV = nRT.
So V2 = (nRT)/p2, and V1 = (nRT)/p1.
So, the nRT's cancel, and we have p1 over p2.
And so, we can rewrite this as the work is equal to minus nRT log p1 over p2, or nRT log p2 over p1.
Now, p2 is less than p1, so this is a negative quantity.
The system has done work.
So the reversible process, we had this curve, and for the irreversible processes, we got this, and then this, whoops, and now we get the whole thing.
So for the reversible process, the work done is the integral under the pressure volume state function, the function of state.
OK, what time is it?
I'm going to actually get caught up, make up for but Bawendi's slow lecturing.
That doesn't mean it's better, it just means that I'm making up for a problem that he created.
OK, so what do we know so far?
Maximum work out is by reversible path.
Delta u is q plus w. So the maximum work out required the maximum heat in.
So a reversible process leads to requiring certain quantities to be maximized.
Now, we have u, q and w. We have a relationship between them.
Often, for a particular state change, it is easy to calculate two of these, but not the third.
And because there is an explicit relationship between u, delta u, q and w, you can always find the easy way to derive the change in internal energy or the heat or the work.
Even if the thing that you really want is an integral that would be difficult to evaluate.
OK, now, we're going to look at the internal energy, and we're going to pretend that it is explicitly a function of temperature and volume.
We could choose any two quantities, and, in fact, it turns out that these are going to prove, after we have the second law, not to be the best choice.
But it's allowed to say the internal energy is a function of temperature and volume.
When you say that, it implies that the differential is given by this pair of partial derivatives.
V dT, partial of u with respect to the volume holding temperature constant, dV.
So if you say this, it requires you to say this.
Now, the purpose of this exercise is to give you a little bit of practice in figuring out what these quantities are.
And do a little practice in manipulating these differentials.
So, we have, we're interested in the change in internal energy for various experimental constraints.
And so, one constraint is the process be done reversibly.
Well then, du is equal to dq reversible plus dw reversible, which is minus p dV, because p is equal to p external for a reversible process, and we can write that.
We could have isolated.
Suppose we're looking at the system isolated from the outside world.
Well, dq is equal to zero and dw is equal to zero because it's isolated.
So that implies du is equal to zero.
We could do an adiabatic process.
Adiabatic means there's no heat transferred in or out of the system.
We don't say anything about whether the system does work or has worked on -- implicitly here, we're talking about a closed system, so there's no mass leaving the system.
But if it's adiabatic, then dq is equal zero, and for an adiabatic process, then du is equal to dw.
OK, now this is a point we want to be careful.
Be careful.
Suppose we are doing an adiabatic process.
We can do it reversibly, or we can do it irreversibly.
So suppose we do it irreversibly.
Suppose we just remove stops, and the system slams up against the other stops.
Did no work.
Does that mean du is zero?
You bet it does not.
So, it's important to write this little thing on here, du is equal to the reversible work, not just the work.
You can have a process where you can measure an irreversible process, where you could calculate the work done.
It could be zero.
It could be something else.
But if it's not reversible, it's not du.
And you will be invited to make that mistake on an exam, I'm sure.
OK, so, the thing about a state function is that the function has a value for initial conditions and at final conditions.
And the difference between those is what you could measure.
If you measure something else, you won't get du.
Be careful, and this is going to be especially complicated and confusing when we get to quantities that have a more obscure meaning like entropy.
I mean we can, we can sort of understand why OK, the total energy, if we measure it, we measure a process which is not reversible.
Well it might not give the energy change for that process, but when we have a quantity which is more obscure, which the definition of that quantity requires a very specific prescription for calculating, you're going to get into trouble.
So exam this and be sure you understand that.
OK, constant volume.
Well for constant volume, dw is equal to zero.
Why?
How does it do, what is work?
Well it's p v work if the volume doesn't change.
There is no work.
So in this case du is equal to dq, and we put a little v on it to imply the work, the change in internal energy is equal to the heat added at constant volume.
Nothing reversible about it.
It's now, all we have to do is say we're going to have heat at constant volume.
OK, and now we return to this differential.
We want to ask the question, what are these two quantities?
How do we know what they are?
This should be particularly bothersome to you because, as you've already experienced in 5.60, there are a lot of partial derivatives.
there are a lot of variables.
There are a lot of things held constant.
It's easy to get lost in this sea of quantities, none of which have obvious meaning.
So now what we're going to do is start to extract what these things mean.
OK, so for a constant volume process, we can write du, partial derivative of u with respect to T at constant V, dT, plus partial derivative of u at constant V, dV.
OK, for constant volume, this is zero.
So this term is gone, and we rewrite this du, V is equal to du/dT, at constant V, dT v. So we, have a change in temperature done at constant volume, and we have a change in internal energy done at constant volume, and we rearrange this.
And we discover that du/dT at constant V is equal to du/dT at constant V. What, have I done something silly?
Oh well, yes I have.
dq v, so du v is equal to dq v and so what I should have written here, this is true, we can get a partial derivative by taking two total derivatives at the same pressure, at the same quantity, but what I really wanted to do is to write dq v is equal to the partial derivative of T, constant v dT v. OK, and now, we know the relationship between heat and a temperature change is given by a quantity, a heat capacity for a particular path, and here it is.
So what we've discovered from this relationship that du at constant volume is equal to dq v. We have discovered that this partial derivative that appears in the definition, the abstract definition of the differential for internal energy, is just equal to the constant volume heat capacity.
So, in this definition now, we have one term which we know.
It's something that we can measure.
We can measure the heat capacity at constant volume, and now we have another term, and if we can figure out how to measure it, we'll have a complete form for this differential which will enable us to calculate du for any process.
So let me write where I am now, or we are now.
du is equal to Cv dT, plus partial of u with respect to volume at constant temperature dV.
So, we've simplified the expression by replacing one of the partial derivatives by a quantity that we can measure.
And we like to know what about this.
So we need an experiment that will enable us to measure this quantity.
And that's where we get Joule, and I like to say it Joule's free expansion -- it's usually referred to as the Joule free expansion, which sort of implies that no energy flows, which actually is true, but it's an experiment proposed by Joule, and the experiment involves an adiabatic box.
So we have system insulated from the outside world, and we have two bulbs.
There's a valve between them.
And so we have gas and we have vacuum.
So the Joule free expansion involves opening this valve and asking what happens when this gas moves into the other bulb or distributes between the two.
well since the gas is expanding into vacuum no work is done.
Since it's isolated, no heat is added.
So du is equal to zero because dq and dw are both zero.
So we can now take this expression and rewrite it under the condition of du is equal to zero.
So we have du equal to zero, just a zero here.
Cv dT constant u plus the derivative du/dV at constant T, dV at constant u.
This is just a number.
We don't have to specify that it's measured at constant u.
It's just a number.
So now we rearrange this expression, and so we get du/dV at constant T is equal to minus Cv times dT u over dV u or minus Cv partial derivative of temperature with respect for volume a constant, free energy, a constant internal energy.
So, this is the quantity we want.
It's related to the heat capacity, the constant volume of heat capacity and something you could measure.
What happens as you expand into volume?
Does the temperature go up?
Does it not change?
Joule actually did this experiment, and he observed that for the gas expansions that he could do, that the temperature did not increase measurably.
So he made an incorrect conclusion.
Because something was small and unmeasurable, he said, well the best of my knowledge dT/dV at constant u is equal to zero.
And that implies that since the quantity we want is given by this quantity, which is zero times a constant, the quantity we want is also zero.
So it would imply that du was equal to only the first term Cv dT.
Now, this is an important lesson in what you do in science.
You make an observation.
You make an observation doing an experiment that is as accurate as you can do.
And so an experiment said the gas didn't increase its temperature when it expanded the vacuum.
And so the next thing you do is you call up a journal and say I've discovered a fundamental law of nature.
And, so you propose that there is no, that this derivative is zero, and that the internal energy is given simply by this quantity.
It turns out that this quantity here, which is called eta of J the Joule free expansion parameter, is not quite zero.
When you expand a real gas into vacuum, the temperature goes down.
So this is a very small number and for ideal gases, eta J is equal to zero.
This quantity is exactly zero for an ideal gas and we'll discover why eventually it has to do with what we mean by an ideal gas it turns out.
And so for many, many problems, especially on exams, especially on this first exam, you will be able to say that this is the relationship between internal energy and temperature.
That u is a function of temperature only.
It doesn't matter what other things change.
The value of the internal energy is only determined by temperature.
But this is only true for an ideal gas and it's approximately true for other things.
So, delta u is equal to zero for all isothermal ideal gas processes.
And it's approximately equal to zero for all real gas processes.
And so that means that delta u is always calculable from Cv(T) dT for any ideal gas change.
So even if work is done, it doesn't matter.
All you care about is what was the temperature change?
And this is always the easy way to calculate du, and it's often the easy way to calculate either q or w, using the first law.
So, since du is equal to q plus w, and for an isothermal process this is equal to zero, q = -w. For an ideal gas, for an isothermal process.
It simplifies your life enormously.
You don't have to calculate much.
OK, so it's time to stop, I did, in fact, get caught up.
I hope you didn't mind the enhanced velocity.
Now Bawendi can do the beautiful stuff on lecture number four.
Thank you.
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Last time you talked about the first law of thermodynamics.
And you talked about isothermal expansion, the Joule expansion.
You saw a very important result.
which is that for an ideal gas, the energy content is only dependent on the temperature, nothing else.
Not the volume, not the pressure, it just cares about the temperature.
So, if you have an isothermal process for an ideal gas, the energy doesn't change.
q plus w is equal to zero for any isothermal process.
And you also saw that du then could be written as Cv dT for an ideal gas always.
This is not generally true.
If you have a real gas and you write du is Cv dT, and your path is not a constant volume path, then you are making a mistake.
But for an ideal gas, you can always write this.
And this turns out to be very useful to remember.
OK, now most processes that we deal with are not constant volume processes.
So energy, which has this wonderful property here, du is Cv dt for constant volume process, which happens to be equal to d q, constant volume, because there's no change.
there's no work if you've got a constant volume process.
So du here is a very interesting quantity, because it's related to the heat that's going in or out of the system under constant volume process.
But as I said, we're not operating usually in a constant volume environment.
When I flail my arms around I generate work and heat.
This is not a constant volume process.
If I'm the system, what's constant when I do this?
Anybody have an idea?
What's the one function of state?
I'm the system, the rest are the surrounding.
What's the one function of state that's constant when I'm doing all my chemical reactions to move my arms around?
Temperature?
Pressure?
Pressure, right.
Pressure is constant.
What is the pressure at?
One atmosphere, one bar.
So the most interesting processes are the processes where pressure is constant.
When I had have a vial on bench top, and I do a chemical reaction in the vial, and it's open to the atmosphere, the pressure is constant at one atmosphere.
When you've got your cells growing in your petri dish, the pressure is constant at one atmosphere, even if they're evolving gas, pressure is constant.
So we'd really like to be able to find some sort of equation of state, or some sort of rather function of state that's going to relate the heat going in or out of the system with that function of state, because this isn't going to do it.
du only relates to the heat under constant volume.
And the heat is a really important thing to know.
How much heat do you need to put into a system, or how much heat is going to come out of a system when something is happening in the system?
All right, this is a really important quantity to know.
Your boiling water or whatever, you want to know how much heat do you need to boil that amount of water under constant pressure?
And this is where enthalpy comes in.
You've all heard of enthalpy.
H we're going to write it as the function of temperature and pressure.
And the reason enthalpy was invented was exactly for that reason, because we need some way to figure out how to relate the heat coming in or out of a system under a constant pressure process.
Because it's so important.
And I should add and also under reversible work, where the external pressure is equal to the internal pressure.
OK, so we're going to define enthalpy as u + pV, these are all functions of state here, So H is a function of state, and we're going to see now how this is, indeed, related to the heat flow in and out of the system.
If you have a constant pressure, reversible work process.
Let's take a system.
Under constant pressure T1, V1, going to a second -- this is the system, so let me write the system here.
And it's more dramatic if the system is a gas, p, T2, V2, And let's look at what happens to these functions of state, to H to u under this transformation.
OK, so let's look at delta u. Delta u is q plus w. That's the first law.
And this is a constant pressure path, so now I can write, this q is actually q under constant pressure.
Little p means the path is a constant pressure path.
And I'm doing reversible work.
So that w is minus p, dV where p is the pressure inside the system, minus p delta V. Rearrange that, delta u is plus p delta V is equal to q p. All right, so this is the heat flowing in or out of the system, and these are all functions of state.
This depends on the path.
It tells you right here, the path is constant pressure.
These don't depend on the path, right.
V doesn't care how you get there.
u doesn't care how you get there.
In this case, p is a constant because the path is constant.
So we can bring the p inside, delta u plus delta p V it's q p. Take this delta outside again, delta of u plus p V is equal to q p. And there you have it.
There is the H right there.
The u plus p V. Delta H is equal to q V. And this is the reason why enthalpy was invented, and why it's so important.
Because we want to know this.
So this for a finite change.
If you want to have an infinitestimally small change, you end up writing dh is dq sub p. It's not always equal to the heat.
It's only equal to the heat if your process is constant pressure reversible work.
OK, so this is the kind of, this is the kind of concept that needs to be branded into your brain, so that if I come into your bedroom in the middle of the night and I whisper to you delta H, you know, you should wake up and say q p, right?
Heat under constant pressure reversible work.
This should become second nature.
This is where the intuition comes from.
This is why people right tables and tables of delta H's.
Why you have delta H's from all these reactions, because this is basically the heat, and the heat is something we can measure, we can control.
We can figure out how much heat is going in and out of something.
This is what we're interested in.
OK, so last time you looked at -- any questions on this first?
Yes
[INAUDIBLE] from the T delta V to the delta p here?
What was the reasoning behind that?
p is constant here.
It's constant pressure.
OK, so now, last time you looked at the Joule expansion to teach you how to relate derivatives like du/dV.
du/dV under constant temperature.
du/dT under constant volume.
You use the Joule expansion to find these quantities.
Now these quantities were useful because you could relate them.
The slope of changes, with respect to volume or temperature of the energy with respect to quantities that you understood, that you could measure.
We're going to do the same thing here.
So if we take as our natural variables for enthalpy to be temperature and pressure, and we have some sort of change in enthalpy, dH, and it's going to be related to changes in temperature and pressure through the derivatives dH through the slope of the enthalpy in the T direction, keeping pressure constant, dT plus the slope of enthalpy in the pressure direction, keeping the temperature constant, dp, and these are knobs that we can turn.
We can change of temperature.
We can change the pressure.
These are physical knobs that are available to us as experimentalists.
And so when we turn these knobs on our system, we want to know how the enthalpy is changing for that system.
Because eventually they will tell us maybe things about how heat is changing further on.
OK, but in order to relate turning these physical knob to this quantity here, which we don't have a very good feel for, we've got to have a feel for the slopes.
If I keep the pressure constant.
I change the temperature, what does that mean?
What is dh/dT?
If I keep the temperature constant, and just change the pressure, dH is going to change, but how is it going to change?
What does this mean in terms of something I can physically understand?
That's the program now for the next few minutes.
What are these quantities?
What is dH/dT as a function, keeping pressure constant, what is dH/dp, keeping temperature constant?
All right, let's start with the first one, dH/dT, keeping the pressure constant.
And we're going to look at a reversible process to help us out, but the result is going to be independent of whether or not we have a reversible process or irreversible process.
Constant pressure, that means dp is equal to zero.
So for reversible process, constant pressure, what do we know?
This is already branded in your brain, right?
Reversible process, constant pressure dH=dq.
So we can write that down, dH=dq, constant pressure.
That's by definition of enthalpy.
That's why we created enthalpy.
What else do we know?
Well we can go look up here, looking at the differential, there are no approximations here.
This is just an equality.
I have a constant pressure process.
This term here is equal to zero.
That means that dH is also equal to dH/dT, constant pressure dT.
All right, so now I've got more dH/dT under constant pressure.
dH is equal to this, and it's also equal to this.
All right, these two are equal to each other.
Now, I know how to relate the heat flow to temperature change, through the heat capacity.
dq constant pressure is that heat capacity, and I have to tell you the path for the heat capacity.
So it's C sub p, the heat capacity under a constant pressure path, dT, all right?
So these two are equal to each other.
So, these two are equal to each other as well, which tells me that this derivative, dH/dT constant pressure is Cp.
So now I have my first of my two slopes, in terms of something that's related to my system, the heat capacity of the system.
Something I can measure and I can tabulate, and when I turn my dT knob here I know what's going to happen to the enthalpy.
So this is the first, this is the one.
So it's very similar to what we saw with the volume and the energy, where du/dV under constant temperature was equal to Cv in this case here, right.
and you're going to find that there's a lot of these analogies between energy and enthalpy.
You just change volume to pressure and basically you're looking at enthalpy under a constant -- anything that's done at a constant volume path with energy, there's the same thing happening under constant pressure path for enthalpy.
So you can guess the answer usually that way.
OK, so now we have the other one, dH/dp constant temperature.
How do we relate this to something physical?
Well, it's going to be an experiment, very much like the Joule experiment.
The Joule experiment was a constant energy experiment, right.
Here we're going to have to find a constant enthalpy experiment, and that is going to be the Joule-Thomson experiment.
That's going to extract out a physical meaning to this derivative here.
OK, the Joule-Thomson experiment.
This is going to get us dH/dp constant temperature.
What is this experiment?
You take a throttle valve, which consists of some sort of porous plug between two cylinders that is insulated.
There is insulation here.
Insulation on the bottom.
It's like a, think about a tube, a large tube, insulated tube, with a bunch of -- a frit inside here, which prevents flow of gas, which slows down the flow of gas from one side to the other.
It's a blockage in this tube here.
Then you put two pistons, one on that side here, and one on this side here, and the external pressure here, we're going to call that p1.
The external pressure here, we're going to call that p2, and we're going to do it slowly enough that the pressure on this side of the cylinder is in equilibrium with the external pressure, and the pressure on this side of the cylinder is in equilibrium with this pressure.
But not so slowly that these two are in equilibrium with each other.
So this is restricting the flow, so there's some sweet spot here.
When I'm pushing slowly enough here that the pressure here is equal to that one, but not so slowly that the air flow from here to here is fast, compared to how fast I'm pushing.
You got the picture here?
Any questions on that?
All right, then as I push through, I'm going to start with all of my gas on this side, and at the end, I'm going to have all the gas on the other side.
Let me first ask you this, is this a reversible or in irreversible process?
Right, let me add one more piece of data here which I said in words, but which is actually important to write down before doing the problem.
Is this a reverse -- any guesses?
How many people vote for that this is a reversible process?
I've got one vote back there, two votes, three votes, four votes.
Anybody else?
How many people think this is irreversible?
It's about a tie, and everybody else doesn't now.
All right, I'm going to give you ten seconds, fifteen seconds to make up your mind.
You're not allowed to be on the fence here.
You've got to decide, all right?
This is, you can talk to your neighbors, you know, do a little bit of thinking.
And I'm going to give you ten seconds to figure this out, what your vote is for that.
All right, let's try again.
How many people vote that this is reversible?
That looks like a majority to me.
Irreversible?
Let's look at the show of hands?
All right, so this is the majority here.
Good thing physics doesn't work on the rule of the majority, otherwise we'd be in big trouble.
Wow, let's walk through that.
I'm sorry to say that this is the wrong answer.
OK, why is that the wrong answer?
Well, just think, you know, think about it.
You're pushing through here.
p1 is greater than p2.
What does it mean for a process to be in equilibrium or reversible?
It means at any point you can reverse the direction of time, and it will look fine, right.
So now I'm pushing on this plug here, with p1 greater than p2.
I'm pushing, I'm pushing, I'm pushing.
p1 is greater than p2.
Then I want to reverse direction of time.
I want the arrow of time to go so that the gas goes from p2 to p1.
p2 is less than p1.
Is that going to work out?
If p2, the pressure in p2, is less than the pressure in p1, is the gas going to want to go from p2 to p1 and the whole thing reverse back?
You've got to put more pressure on one side than the other if you want to push that gas through the throttle, right?
So this is where the time scale issue comes into play.
I said let's do this slowly enough that this p1 is in equilibrium with this p1, but not so slowly that this pressure is equivalent to that pressure.
If I do it really, really slowly, so that everything is reversible, well I won't be able to do it, because p1 and p2 are different.
But suppose that I fix my pistons here, with p1 greater than p2, and I don't touch it, eventually p1 will be equal to p2.
I'll come to some sort of equilibrium.
So this is a system which is out of equilibrium.
If I stop, if I move slowly, if I move more slowly, then these two will want equilibrium.
So this is an irreversible process.
The Joule-Thomson experiment is irreversible.
OK, important -- if you are part of this group here, think about it and make sure you understand that.
All right, so this is the experiment.
Now what are we doing with that?
The initial state, let's look at what we are doing.
So initially, we're going to have our piston, there's the plug sitting here.
Our piston on the right side here fully out, and the piston on my right side, your left side, fully inside.
There's no gas on that side here.
So there's p2 sitting here.
There is p1 sitting here, and all of the gas is sitting on that side of the plug.
Then after we're done with the experiment, we'll have transferred all of the gas from one side to the other.
p1 here.
p2 sitting here.
And there's going to be some volume V2 and some volume V1, but are not necessarily the same.
Especially since the pressures are different.
we don't know yet about temperature so I don't know what to say about these volumes because I don't know what the temperatures' are going to do.
OK, so let's go through this and see what we would do which is to calculate the heat and the work.
This is well insulated.
So, what is the -- let's do the first one here.
What's q for this process here?
Anybody?
STUDENT Zero
Zero, right.
This is an adiabatic process.
It's well insulated.
Heat is not going in or out, adiabatic.
q is equal to zero.
So all we need to find out is the work now.
Let's divide it up into the two sides, the work going on on the left hand side, my left hand side or your left hand side, and the work going on on the right hand side.
So let's first look at the left hand side.
OK, so w, first of all, work is being done to the system on the left hand side here.
I'm pressing on the gas.
So I expect that to be a positive number.
The pressure is constant, p. The V goes from V1 to zero.
So we write down p1, V1.
On the right hand side, the work, let's call this left hand side, let's call this right hand side.
Here there's an expansion going on, so the system is doing work to the external world.
This piston is being brought out, so we expect the work to be negative, negative.
And we start out with zero volume.
We end up with a volume of V2, and the external pressure is constant to p2.
Minus p2, V2.
Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, which is p1 V1 minus p2 V2.
Which I can rewrite as minus delta pV.
Delta pV is p2 V2 minus p1 V1.
It's the pressure volume multiplied together at the final state, minus pressure volume from the initial state, with the minus sign here because it's the negative of that.
All right, what is delta u?
delta u is q plus w. q is zero.
delta u is just w. So this is also just delta u. delta u is minus delta pV, for the process.
OK, delta H is delta of u plus pV.
By definition, that's how we define enthalpy up here.
H is u plus pV.
Delta H is delta of u plus pV, which is equal to delta u, plus delta pV.
Now delta u is minus delta pV.
So I have minus delta pV plus delta pV.
This is equal to zero.
So this irreversible process, this Joule-Thomson process, is a constant enthalpy process.
Delta h for this process is equal to zero.
Adiabatic q equal to zero.
It's also delta H which is zero.
The two didn't necessarily follow, because remember, delta H is dq so p is only true for a reversible constant pressure process.
This is an irreversible process.
So a priori, this was not necessarily true.
It turns out after you do all the math, it turns out to be delta H equals zero.
All right, so this is the experiment.
How do we go from that experiment to the terms that we're trying to get, these slopes.
Remember, we're trying to get delta H, we're trying to get dH/dT constant pressure and dH/dp constant temperature.
OK, these are the two things were trying to get here.
OK, so let's write down, what we know here.
I'm missing something.
Oh, we already know one thing.
We already know this guy here.
We already did that.
OK, dH/dT constant pressure is Cp.
That was easy one.
So we already know that.
So now we can write or differential dH as Cp dT plus dH/dp, constant temperature, dp.
Now we want to find out what this guy is here.
Now for this experiment, this is a constant enthalpy experiment for the Joule-Thomson experiment, this is equal to zero.
So I can rearrange this to get this dH/dp in terms of things that I can either measure, like the heat capacity, or that I have control of, like dT and dp.
So in this case, dH/dp constant temperature is minus Cp dT/dp, and this is under constant temperature, no, not constant temperature.
Whatever the experi -- for that the experiment is.
For that experiment, the constraint, so we need a constraint here, right, we need a constraint here.
Right?
We need a constraint here.
The constraint isn't constant temperature because the temperature is going to be changing.
It's not constant pressure, because we have a delta p going on.
It's not constant volume either.
The constraint is the constraint of the experiment, and the constraint of the experiment is that the enthalpy is constant.
So the constraints we have here, is the constant enthalpy.
It's the constant enthalpy process that we're looking at.
This we can do experiments on.
It's tabulated in books, and this we can measure in the experiment.
Delta p here is the change in pressure from the left side to the right side, and we can put a thermometer, measure the temperature before the experiment, and measure the temperature after the experiment.
So this is something we can measure.
So now we have this derivative, in terms of physical quantities, things that we can measure.
Things that we can relate to the properties of the substance that we're doing the experiment on.
So this is basically delta T and delta p and the Joule-Thomson experiment.
And so Joule and Thomson did these experiments, and they measured lots of gases, and they found that, in fact, this was something that they could measure.
Sometimes it was positive, sometimes it was negative, and it was an interesting number.
And so they defined them, after many experiments, the limit of this, delta T delta p and the limit of delta p goes to zero as the Joule-Thomson coefficient.
So, basically dT/dp, constant enthalpy is equal to mu, by definition, Joule-Thomson, where mu Joule-Thomson is the Joule-Thomson coefficient, just like you saw last time eta sub j was the Joule coefficient for dT/dV under constant energy.
So there's, again, total analogy here between what we're doing with enthalpy to what you did last time with energy, replace p with T and H with u. Flip those two and you get the same thing.
OK, so now we have dH/dT is equal to Cp, and we can also write, then, dH/dp under constant temperature is equal to minus Cp mu Joule-Thomson.
We have our two derivatives in terms of physical quantities, which is going to allow us, then, whenever we have a change to go back and, when we have a change where we adjust the temperature and the pressure, we'll be able to know what the enthalpy change is.
OK, now let's take two cases.
Let's first start talking about ideal gases.
The last time you saw that for an ideal gas, the energy only cared about the temperature.
It didn't care what the volume was doing.
du/dV under constant temperature was equal to zero for an ideal gas.
And by analogy, we expect the same thing to be true here, because enthalpy and energy have all this analogy going on here.
So let's look at an ideal gas.
So for an ideal gas, we saw that u was only a function of temperature.
We also have the equation of state for an ideal gas, pV = nRT.
We can write our definition of enthalpy, h is u plus pV.
This only depends on the temperature.
pV= nRT.
So we have u only depends on temperature plus nRT.
The only valuable now on this side is temperature.
Pressure and volume have dropped out.
So enthalpy, for an ideal gas, only cares about temperature.
Pressure has dropped out of the picture completely here.
So there is no p dependence here.
H for an ideal gas is only a function of temperature.
This is not true for a real gas, fortunately, but it's true for an ideal gas.
So for an ideal gas then, dH/dp under constant temperature, that has to be equal to zero.
Because temperature is constant H only cares about temperature.
and that's equal to zero.
And if that's equal to zero, that means that the Joule-Thomson coefficient for an ideal gas is also equal to zero.
We're going to actually prove this later in the course.
Right now, you're taking it for granted.
Right now we told you Joule did all these experiments and he found out that for an ideal gas, that the limit in and ideal gas case was that the eta J was equal to zero.
Therefore, from experiments, u is only a function of temperature for an ideal gas, and therefore from these experiments, we come out with delta H dH/dp is equal to zero.
The Joule-Thomson coefficient is equal to zero.
Later we are going to prove it exactly.
but right now you're going to have to take it for granted.
So, if the Joule-Thomson coefficient is equal to zero, just like we wrote, du = Cv dT for an ideal gas, we're going to have dH = Cp dT for an ideal gas as well.
dH is Cp dT.
This term goes away.
dH = Cp dT.
That's the only thing that's left behind.
So if you know the heat capacity, you know the change in temperature, you know what enthalpy is doing for an ideal gas.
This needs to be stressed that this is the ideal gas case.
Now regular gases, real gases, fortunately as I said, don't obey this.
This is important because we use this all the time for when, when we do technology.
One example of Joule-Thomson coefficient being not equal to zero for a real gas that you've like experienced is if you take a bicycle pump, take a bicycle pump, and you're pumping up your tire, you're working pretty hard, so you're getting hot.
But if you touch the valve going into your tire, which basically measures the temperature of the air going into your tire, that is getting hot, right.
So if you've got to pump that tire really a lot, then you're going to you're going to really feel a lot of heat there.
The compression of the, basically it's an adiabatic compression.
You're taking the air inside of the pump and you're compressing it.
You're doing it so fast that there's not enough time for heat to come out of the gas that's inside the pump towards the walls of the pump.
So your time scale it just fast enough that this is basically an adiabatic compassion.
Your compressing it really fast, all right, so you're changing the pressure.
You're changing the pressure, and the temperature is going up.
dT/dp is positive.
dT/dp is positive.
dT/dp is positive, well that's mu JT.
dT/dp is mu JT.
So for a real gas like air, this is a positive number.
It's not zero.
Air is not an ideal gas.
It's one simple example of -- The fact that mu JT is not zero for real gases is how we are able to liquify things like hydrogen and helium, by compressing them and pushing them through a nozzle, and the expansion through the nozzle cools the gas.
Right, in this case here it wouldn't happen if it was an ideal gas.
Or in many kinds of gas refrigerators where you push a gas through a nozzle close to room temperature, what you find is that the gas coming out on the other side under lower pressure is cooler than the gas that went through on the other side.
Real refrigerators actually work with liquids that go into gases so use the latent heat of the liquid, so it doesn't really work like the Joule-Thomson expansion.
So this is real.
This is real, unlike the Joule coefficient which is very small so that most gases have tiny Joule coefficients.
So if you do a Joule experiment, you hardly measure a temperature change.
With real gases, here you do actually measure it.
You can feel it with your finger on your bicycle tire.
OK, so we're going to see this using a Van der Waal's gas.
Let's look at a Van der Waal's gas and see what happens in the Van der Waal's gas.
Any questions, first?
What we've been talking about, the Joule-Thomson experiment, constant enthalpy process?
OK, so let's take our Van der Waal's gas.
Remember the equation of state for Van der Waal's gas is not pV is equal to nRT, but p plus the attraction term.
And then V minus the excluded volume term is equal to RT.
Two parameters, this is the attraction between two atoms or molecules in the gas phase.
This is the repulsion, not the repulsion.
This is the fact that we occupy a finite volume in space, because they're little hard spheres in this molecule.
OK, in a few weeks, you're going to find out that we can calculate dH/dp from this equation of state, and you're going to find out that dH/dp from that equation of state is proportional to b minus a over RT.
This is going to be probably a homework at some point to do this.
For now, let's take it for granted.
Let's take it for granted that we know how to calculate this derivative from an equation of state like this.
But now we're going to use that.
OK.
So since dH/dp under constant temperature is proportional to minus mu JT, then we have that dT/dp under constant enthalpy than it is related to the negative of this, a over RT minus b All right, this is how the temperature changes when you change, when you have pressure changing.
So when you do an expansion or compression, my adiabatic compression of my bicycle pump is dT/dp.
All right.
Or when I do an expansion of hydrogen or helium at low temperature, through a Joule-Thomson experiment, when I want to liquify hydrogen or helium.
I want to cool a gas with a Joule-Thomson experiment, what temperature do I have to be at?
So this tells you that you have to be careful what temperature you're at, because depending on how high, how big this temperature here is, you could either be, have a negative dT/dp, if this first term is small enough, meaning if temperature is very high, then you end up with a negative term.
If the temperature is very small, then one over the temperature is large, and the first term wins and you have a positive number.
So there's some special temperature which is going to depend on the gas where the first term is going to be equal to the second term, where Joule-Thomson is zero, or it's going to behave like an ideal gas.
So when that is the case, we're going to call that temperature the inversion temperature or T inv.
We call that inversion because on one side you end up cooling if you compress.
And on the other side of that temperature you end up heating if you compress.
OK, so there's some temperature, t inversion minus b where the gas behaves like an ideal gas.
The Joule-Thomson coefficient is zero and that inversion temperature, you can solve for it.
Conversion temperature is equal to a over R times b. OK, so when you are at that temperature, everything looks like an ideal gas, as far as the enthalpy changes are concerned.
Now if you're at the temperature which is higher than the inversion temperature, in that case here, a over RT is small compared to b, and this is going to turn out to be negative.
So if you had a high temperature, this a small compared to b.
If you're negative, which means that dT/dp at constant H is less than zero.
So that means that if you compress something it's going to cool.
The temperature rises when the pressure drops, right?
Or in this case here, if I do my Joule experiment delta p is negative, p2 is less than p1, that means that delta T is positive, right?
So in this experiment here, this side is going to heat up.
So for materials where T inversion is low, lower than room temperature, then you would end up heating up in this expansion.
You're basically expanding the gas from one side to the other and the expansion causes the temperature to rise.
If T is less than T inversion, you have the opposite case, and dT/dp is greater than zero.
So in this experiment here, delta p is less than zero.
You need to have this whole thing greater than zero.
So delta T is less than zero as well.
So if you're below the inversion temperature and you do the Joule-Thomson experiment, you're going to end up with something that's colder on this side than that side.
Ideal gas would be the same temperature.
But now, so this is where the refrigeration comes in.
So if you take a gas, and you're below the inversion temperature and you make it go through this irreversible process, the gas comes out colder from that side than that side.
So the work that you're doing to expand, to go through this experiment, ends up cooling the gas.
OK, for most gases, T inversion is much greater than 300 degrees Kelvin.
Much greater than room temperature.
For more most gases, if you do this experiment at room temperature, you end up cooling the gas, and you can cool it measurably, which is why also, your bicycle pump, you know, you push down, you compress, this is an expansion here.
You're expanding from this side to that side.
Bicycle pump you are compressing the gas, which is the opposite, you end up heating up the air in the bicycle pump in your compression.
There are two exceptions to this rule that most gases have a T inversion which is greater than 300 degrees Kelvin, and that's hydrogen.
The T inversion for hydrogen turns out to be 193 degrees Kelvin, and the T inversion for helium turns out to be 53 degrees Kelvin.
And so when, there's a lot of liquid helium that's being used on campus.
We use a liquid helium.
And so in order to make a liquid helium, you can't take helium at room temperature and do this, because if you did, you would just heat it up, because the room temperature is above the inversion temperature, so Joule-Thomson would heat up the helium.
So you need first to take the liquid helium and cool it below 53 degrees Kelvin before you can do the Joule-Thomson to cool it even further to make liquid helium.
So you have to do it in stages.
You take your room temperature liquid helium, and you cool it with liquid nitrogen to 77 degrees Kelvin, the new, you're not quite there yet, unfortunately right?
Then you take hydrogen you cool it would liquid nitrogen to 77, then you can use your hydrogen gas.
Do a Joule-Thomson hydrogen gas which you first cool with liquid nitrogen.
Liquid nitrogen, 77, that's below 193, so you can do Joule-Thomson on hydrogen, cool the hydrogen to below 53, then use your cold hydrogen to cool the helium, and then you can do the Joule-Thomson on the helium to cool it further until it liquifies.
So that's what you do when you make liquid helium.
All right, any questions on this lecture or any of the concepts that we talked about?
The last thing that we're going to do today then is to look at a relationship which is going to turn out to be very useful.
It's a relationship for ideal gases which relates the heat capacities at constant pressure and constant volume.
Cp = Cv + R. This is very useful because often you just have lots of tables of Cp's but sometimes you want to know what the energy change is going to be for the ideal gas, and you know that du is Cv dT not Cp dT.
So, you need to know what Cv is, and if this is true always, then there's a very easy way to go from one to the other.
We're going to do it two ways.
Today we'll do the first way, and then next time we'll do the second way.
The first way is just to turn the crank on the math, and the second way is to do a little bit more imaginative about the process.
OK, let's just turn the crank on the maths.
What do we know about the Cp and Cv?
Well Cp, we already know how to relate it to dH/dT through the slope of the enthalpy, and we also related Cv to the slope of the energy with respect to temperature under constant volume.
And we also know that H is u plus pV.
Right, H is u plus pV.
So we're going to take the derivatives of both sides of this equation here, by, with respect to temperature, keeping pressure constant.
So we have dH/dT keeping pressure constant, is du/dT keeping pressure constant.
Got to keep track of what's being constant, kept constant, plus dPV/dT, keeping pressure constant.
dH/dT keeping pressure constant, that's just Cp, and then we have du/dT keeping pressure constant.
The p is constant here, comes out of the equation, so we have p and then we have dV/dT well it's an ideal gas, an ideal gas for dV/dT for an ideal gas is equal to R over p because pV is equal to RT, right.
So dV/dT is Rp.
So dV/dT is times R over p, the p's are going to cancel out here.
OK, so now it's very tempting at this state to say, oh there's the answer right here.
du/dT is Cv.
There I have it Cp is equal to Cv plus R, right?
But even though it looks like that's the right way to do it, it's actually not right because it turns out to be right by sort of by accident.
But here you've got pressure constant.
du, this is du, not H here.
du/dT is only equal to Cv when the volume is constant, not when the pressure is constant.
So if you're going to turn the crank on the math correctly, you're going to have to change this p into a V somehow.
Because this isn't Cv mathematically speaking.
We don't know what it is yet.
In order to change this from a p to a V, you have to use the chain rule.
So let's use the chain rule.
And then we'll be done.
OK, so u is actually a function of temperature and volume, which in this case here could be a function of pressure and temperature.
So if we want du/dT under constant pressure, you have to use the chain rule.
There's the pressure sitting right here.
It's going to be du/dT under constant volume, plus du/dV dV/dT under constant pressure.
All right, chain rule.
du/dT constant pressure is the direct derivative with respect to temperature here, which is sitting by itself under constant volume, keeping this constant but there is temperature sitting right here too.
That's where that term comes from, du/dV dV/dT.
Now, for an ideal gas, du/dV under constant temperature is equal to zero.
It doesn't care what the volume is doing.
It only cares what temperature is.
If temperature is constant, there's no change in energy.
For an ideal gas, this is zero.
It's not zero for a real gas.
Right, so this whole term disappears and for an ideal gas, it turns out that du/dT constant pressure is equal to du/dT constant volume, but this is equal to Cv for an ideal gas.
It wouldn't be true for a real gas, and this is a common mistake that people make for real gases to equate this.
This is only true for an ideal gas.
Since it's true for an ideal gas, then we can go ahead and replace this with Cv, and then we have Cp with Cv plus R, which is what we were after.
OK, next time we'll do the other way of getting to the same answer.
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OK, so we ended up last time, we talked about Joule-Thomson expansion, which is an irreversible expansion through a nozzle, through a porous plug, constant enthalpy expansion.
And I want to make sure everybody figured out that it really was an irreversible expansion.
Anybody have any questions on that, because when we voted, the majority of the class thought it was reversible Yes
Just a quick question on some of the constraints, like isothermal, isobaric, isovolumetric expansion.
For isothermal expansion, that means that delta u does not change, but delta q is equal to delta w?
So the question was, for an isothermal expansion, delta u does not change, therefore, dq is equal to dw.
The answer is that's true only for an ideal gas.
For a real gas, that won't be true.
Ideal gas only depends on the temperature, the energy only depends on the temperature.
Therefore, if it's isothermal, the energy of the ideal gas doesn't change.
For a real gas it depends on more than the temperature
Are there any other constraints similar to that [INAUDIBLE]
So, for an ideal gas, the isothermal is the easy one because the energy doesn't change.
If you have an isobaric you're going to have to calculate where the energy changes, and that's a calculation that's likely on the homework and very like on an exam as well, too.
In fact, we're going to do some of that today, OK, calculate delta u.
Any other questions?
Good question.
OK, so we ended up last time showing that for an ideal gas, that there was was a relationship between the heat capacity under constant pressure, and this is the molar heat capacity, that's why we have the bar on top, and the molar heat capacity for constant volume path.
They're related through the gas constant.
OK, this is only true for an ideal gas, and we went through that mathematically where the, with a chain rule.
And I wanted to do it a different way which is a little bit convoluted, but it introduces the idea of a thermodynamic cycle, and it's something that we're going to use a lot in the class.
Because anytime you're working with the function of state, and you're trying to connect two points together in a p-V or a T-V diagram, then it doesn't matter which way you're connecting those two points, as long as you're dealing with state functions it doesn't matter.
It only matters if you're looking at the work or the heat right.
And so sometimes, when you want to calculate what the energy is at the endpoint, the path of the experiment that you're doing may not be an easy path to calculate.
You may have to find a different path which is an easier one to calculate, and it doesn't matter which path you take, because in the end all you care is what the energy is at the end.
so we're going to use this concept of the path to go from the initial point to the end point.
As long as you're dealing with state functions it doesn't matter to get at the same results here.
So we're going to start with a mole of gas, at some pressure, some volume, temperature and some mole so V, doing it per mole, and we're going to do two paths here.
We're going to take a constant volume path.
So V is constant and we're going to change the pressure to some, a little change in pressure, V is constant here, and then temperatures is also going to change.
So this is a path, which is either a temperature change, with the pressure changes at the same time, going up or down, constant volume.
And since the temperature is changing, clearly Cv is going to have something to do with this.
So constant volume path, where the temperature is changing.
So somehow this is going to allow us to get Cv.
Cv is going to be related to this path.
And if I draw a diagram on a T-V diagram of what I'm doing here, so there's temperature axis.
There's the volume axis here.
I'm starting with some V1 here.
And this is going to be here V plus dV.
So I'm starting right here.
This is my initial point.
And the path that I'm describing then, let's assume that we're raising the temperature up is this path right here.
OK, this is my constant volume path going from T to T plus dT.
OK, we'll call this 1 here.
All right, now I want to make a path where out of that path, C sub p is going to fall out.
So that's going to have to be a constant pressure path.
Temperature is going to change but pressure is going to be constant.
So what I can do is I can take my initial point here and create a new, different path, where, which is going to be a constant pressure path.
So at the end of this path the pressure is constant.
The volume is going to change though, to some new volume dV and the temperature is going to change also and I'm going to make it the same endpoint temperature as the endpoint temperature here.
And out of this path, since the pressure is constant but the temperature is changing, I'm hoping that C sub p will fall out of that path of that calculation.
All right?
This should give us something about C sub p. So this will give us something about C sub v. This will give us something about C sub p. Now these two endpoints here are different.
They have a different pressure, a different volume.
They do have the same temperature though.
So the connection between this endpoint here, and that one here.
Same temperature.
It's going to be an isothermal path that's going to connect them.
So I'm going to connect them in an isothermal way, and as, since I'm dealing with an ideal gas, and as we saw from the first question here, isothermal always means delta u equals zero.
So I'm going to write that immediately here, delta u equal zero, because I know the answer to that path here.
All right, so let's go again what our paths are.
Path number 1 I'm going straight up in the V-T diagram.
Path number 2 on my diagram it's a reversible, this path number 2, it's a reversible constant pressure path.
So I'm going to be keeping the pressure constant, but my volume is going to go up.
My temperature is going to go up.
Basically I'm putting heat in this system.
That's going to be number 2 right here.
It's going to be the same temperature as before but the volume is V plus dV now.
That's 2, and then, so I'm heating up the system in this path here, and then to connect the 2 endpoints here, I'm going to connect them through a constant temperature path.
Basically I'm going to cool down the system so that the -- no, I'm not going to cool down, I'm going to compresses the system back to the smaller volume V1 and the constant temperature conditions.
So that's path number 3 here.
And I'm only going to worry about energy in this case here.
Because I know energy doesn't care about the path.
Energy only cares about where I am on the diagram.
And I know energy is related to Cv through Cv dT etcetera.
So if I worry about energy I have a pretty good chance of extracting out these heat capacities, right, and I don't have to worry about exactly which path and I can really mix things up.
So let's start the process.
So the first path then, the first path, constant volume constant V, so I'm going to, again, let's just worry about energy.
So du for this first path here, du for path 1, and write dq plus dw what we usually write.
Now it's constant volume when the volume is constant.
There's no work being done so I can immediately ignore this dw here.
Constant volume dq = Cv dT.
Now it's an ideal gas.
So in some sense, I did too much work by writing this, because I knew already du = Cv dT.
I could have written that down immediately, but I just went through the process of writing the first law, what's dw, what dq, Cv dT, etcetera.
I'm just writing what we already know basically.
Let's look at path number 2.
Path number 2.
OK, let's look at the dw, du again.
du for path number 2. dq like the first law down, dq plus dw, then -- what's dq?
Well it's a constant pressure path.
dq is related to the temperature, the change in temperature.
dT through the heat capacity, happens to be constant pressure, so this is exactly what we want.
Cp, I forgot to put my little bar on top here because it's per mole Cp dT that's my dq here.
The path is constant pressure.
And then the dw.
Well there's a change now in volume and pressure.
So I need, well the pressure is constant, but there's a change in volume.
Minus p dV is the change, is the work, right?
So I can write that down minus p dV for that path there.
OK, now but it's an ideal gas, so I know something about the relationship between dV and dT, because I've got dT here the dV.
I don't want to have so many variables around.
I've got three variables, T, p and V. I know I only need 2, so I can relate dV to dp through the ideal gas law.
P dV is equal to R dT.
pV = RT for 1 mole, so I just take dV here.
dT here because the pressure is constant, so dV is equal to R over p dT.
Then pressure is constant.
So I can instead of the dV here, I can insert this R over p dT.
The p's cancel out, and now I have Cp dT minus R dT.
This is equal to Cp minus R dT.
All right, let's complete the cycle now, the path number 3 here.
Path number 3 is a constant temperature path, and I already wrote the answer.
Constant temperature isothermal delta u is zero.
There's no thinking involved.
It's one of these things like I mentioned last time, if I tell you delta H in the middle of the night what you say, q Cp for reversible process.
Same thing here.
isothermal process, what do you say?
Delta u is zero, right?
It should be completely second nature.
So delta u is zero here.
Delta u sub 3 is, for an ideal gas.
I gotta put that constraint in there for an ideal gas.
Delta u is equal to zero.
So now I'm back at the same place.
I'm back here right.
So path number 1 went from i, let's call this path up here.
went to f, and this is how much energy change there was.
The sum of path number 2 and path number 3 get me to the same place, so the energy change by going through this time path, this intermediate point here back all the way to final state should be the same the red path.
So what I'm saying is that du through path 1 should be equal to du through path 2 plus du going through path 3.
All right, because energy doesn't care about the path.
It only cares about the end points.
OK, so now let's plug what is du through path 1?
It's Cp dT and the du through path 2 is Cp minus R dT and then du through path 3 is zero.
It's the isothermal process.
The dT is cancelled out and voila I have my answer.
Cv is equal, oh Cv plus R is equal to Cp whichever way you want to do it, it's a relationship that we had up here that we wanted to prove.
Now, I wanted to go to through this just to go through one cycle quickly because we're going to be doing these all the time, and the importance of the fact that the path doesn't matter, and you can always connect things together in a way, whatever you want.
Whatever is the easiest.
If you want to calculate a change in delta u.
Now if you look at my derivation here, there's one spot where I could have just stopped and proven my point without going through the whole thing.
So if you look at path number 2 here, the constant pressure path du is Cp minus R dT, du is Cp minus r dT for that path.
What is du for an ideal gas?
It's -- what else do we know that du is always equal to for an ideal gas?
Cv dT right.
So I could have written right here immediately equals Cv dT, and that was the end of my derivation.
I could have done that.
It would've been easier, but I wanted to go through the whole path right?
OK, any questions?
Cool.
All right, so we're going to be doing more thermodynamics processes.
We're building up to entropy and to engines, Carnot cycles, etcetera.
And so we need to build our repertoire of knowledge in manipulating these cycles like the one I drew on the board that's hidden now You've already seen the simple process, which is the isothermal process, and you've seen how to functionalize, or what the function is that describes the isothermal expansion.
Let's say we start from some V1 and p1 here, so high pressure, small volume and we end up with a high volume low pressure, under constant temperature condition.
So there's some path, isothermal, so T is constant.
And we know the functional form for this path, it's the ideal gas law, pV = RT, pV = nRT.
T is constant, so the functional form here is pV is equal to a constant.
All right, or p is equal to a constant divided by volume.
If you want to write a function that describes this line here, it's pressure as a function of volume related to each other with this constant.
So we know how to do this.
So this allows us to calculate all sorts of things to get these functional forms.
That's the isotherm here.
We also know how to calculate the work.
If it's a reversible process, we know the work is the area under the curve.
w reversible.
in this case here let's see this is it's a pressure is going down.
It's an expansion, so we're doing work against the environment, or to the environment.
So work is leaving the system.
So minus w reversible is going to be a positive number.
V1 to V2 and RT -- you've done this already last time or a couple of times ago and you end up with minus nRT log p2 over p1, OK?
So minus w reversible, we're doing work to the environment.
This should be a positive number.
p2 is less than p1, so p2 over p1 is less than one, log of something less than 1 is negative times negative.
So in fact this is indeed a positive number so we didn't make a mistake in our signs.
You always want to check your signs at the and because it's so easy to get the sign wrong, and I was hoping that it was right here because I wasn't sure, but it turned out to be right.
Always check your sign at the end.
All right so we know how to do this.
Now we are going to do the same thing with a different constraint.
Our constraint is going to be an adiabatic expansion or compression.
Adiabatic meaning there's no heat involved, and we're going to see how that differs from the isothermal expansion and compression.
OK, so this is the experiment then.
We're going to take a piston here which is going to be insulated.
It's going to have some volume, temperature to begin with, and then we're going to do something to it.
We're going to change the pressure above, right now there's a p external, which is equal to p on the inside.
OK, we're going to do this reversibly, which means we're going to slowly change the external pressure very, very slightly at a time, so that at every point we're basically in equilibrium, until the pressure reaches a new smaller pressure p2.
So that's going to results in an expansion where the new volume new temperature new pressure and an external pressure, which is p2 which is a smaller pressure.
It's really important that this, that you remember that this is the reversible case.
We're going to also look at the irreversible case briefly.
So on the p-V diagram, then, there's a V1 here a V2 here, a p1 here a p2 here.
This V2 is going to be different than this one here.
A priori we don't know what it is.
We don't know if it's bigger or smaller.
We kind of feel like it's going to be different because it's a different constraint.
This is isothermal.
This is adiabatic, there's no heat.
So there's going to be a line that's going to connect the initial point to the final point, and that line mathematically is not going to be the same as this one here.
And our job is to find out what is the mathematical description of this path, this line in p-V's case that connects these two point.
And again, I want to stress this is a reversible path.
If it was non-reversible, I would be allowed to put an initial point and a final point, but I wouldn't be allowed to put a path between them like this, connecting them together.
Because if it's irreversible, it's very likely that I don't know what the pressure inside the system is doing while this is happening.
It could be very chaotic.
The pressure could not be, might not even be constant throughout the system.
It could have eddies and all sorts of things.
So for an irreversible process, I wouldn't really be allowed to put a path there.
I can do that because it's reversible, and I can get a functional form out.
What we're going to find in this process is that the functional form from this is that p is related to the volume through a constant, but instead of being V to the first power as it was for the isothermal, it's now a V to some constant gamma.
What we're going to find is that gamma is greater than 1.
And if gamma is greater than 1, and you're changing the pressure to the same final point, what we're going to find then is the volume that you go to is going to be a smaller volume than for the isothermal.
We're going to go through these points, so don't worry about getting at all straight right now.
All right, so let's do this.
Let's try to show this.
Let's try to show that in fact this is the right functional form, that pV to the gamma is equal to constant is the right form that describes the path in an adiabatic expansion or compression.
So let's do the experiment.
Let's write everything that we know down.
This is where, the way that we should start every problem set or ever problem.
It's writing everything we know and what it means.
OK, so adiabatic.
All the words mean things mathematically.
Adiabatic means dq equals zero.
Let's write that down.
dq equals zero.
Reversible.
Reversible means that if I looked at dw it's minus p reversible.
It's minus p external dV, that's what it is, but in the reversible process p external is equal to p. That's the only time you can set p external equal to p is if it's reversible.
So this is the external pressure.
This is the internal pressure of the system.
You can only do this because it's reversible.
What else do I know?
It's an ideal gas.
So I can write du is Cv dT and pV is equal to RT.
OK.
So what else can I write?
Well from the first law, du is equal to dq plus dw, and I wrote down everything I knew at the beginning here.
I wrote dq equals zero I wrote what dw was.
So du is just minus p dV.
Now from the ideal gas, I have another expression for du, Cv dT.
So now I have two expressions for du.
That's a good thing, because I can them equal to each other.
So if you get these two guys together you get Cv dT is minus p dV.
So we're connecting temperature and changes in temperature and changes in volume together.
OK, now we don't really want this p here.
This is going to be a function, I only need two variables, and here I've got three variables down.
So I know p is going to be a function of T and V, and I know how to relate p to T and V because I know the ideal gas law.
So instead of p, here I'm going to put nRT over V. or RT over V bar.
So now, this expression becomes Cv dT over T is equal to minus R dV over V. And I also did a little bit of rearrangement.
Yes?
[INAUDIBLE]
Well, let's see.
du is, for an ideal gas, it's always equal to Cv dT.
Constant volume process tells you that if you have a constant volume, then dq is Cv dT.
This is what the constant volume path tells you, that you can equate Cv dT with dq.
The ideal gas tells you that this is always true here.
It's an ideal gas in adiabatic expansion dq is equal to zero.
Adiabatic expansion means you can't use this.
dq is zero.
The that is adiabatic.
It's not constant volume.
You're allowed Cv comes out here for this adiabatic expansion, which is not a constant volume only because this is always true for an ideal gas.
We didn't use a constraint that V is constant.
Just use this as an ideal gas here.
OK, so we've got dT here dV here.
We want to get rid of derivatives.
We want to integrate.
So let's take the integral of both sides, going from the initial point to the final point.
We've got Cv integral from T1 to T2, dT over T is equal to minus R from V1 to V2 dV over V. OK, we do take that integral, and that leads us to Cv log of T2 over T1 is equal to minus R log of V2 over V1, and now we know how to rearrange this, because the Cv times a log of T2 over T1 is the same thing as the log of T2 over T1 to the Cv power, right.
So by rearranging these expression of the logs here, this is the same thing is writing log T2 over T1 to the Cv power is equal to log of V1 over V2 to the R power.
And I've reversed the fraction here because of the negative sign here.
OK, get rid of the logs, and you get that T2 over T1 is equal to V1 over V2 to the R over Cv by taking this Cv and bringing it on the other side here.
OK, so now we've got the initial and final points, a relationship between the temperature and volume for the initial and final points.
Eventually that's not what we want.
Eventually we want the same relationship in the pressure and volume.
We want a relationship in p-V space, not in T-V space.
So we're going to have to do something about that.
But first, it turns out that now we have this R over Cv.
and remember, we have this relationship between R Cv and Cp here, which is going to be interesting.
So let's get rid of R in this expression here.
So R over Cv, R is Cp minus Cv.
We proved that first thing this morning, divided by Cv so this is Cp over Cv minus 1 and because I know the answer of this whole thing here, I'm going to call this thing here, Cp over Cv.
I'm going to call it gamma.
Because I know the answer, right.
I know the answer.
I'm going to call this gamma.
By definition I'm going to define gamma by Cp over Cv by definition.
OK, so that means that this is really instead of R over Cv.
it's really gamma minus one.
So now I've got this gamma placed in there, gamma minus 1, so I've ben feeling a little bit better.
I've gotten V to the gamma power almost to the gamma power, it's minus 1 here.
But it's almost right and I'm looking for something like this here, V to the gamma power pV to the gamma power is a constant to describe that line.
So I just got to get rid of the temperature now.
Somehow I've gotta get rid of the temperature and make it pressure instead.
OK, so let's -- let me get rid of this here, let's use the ideal gas law to get rid of the temperature.
So we have T is pV over R. So now if I look at my V1 over V2 to the gamma minus 1, that's T2 over T1.
I'll replace the ideal gas law for those two T's.
That p2 V2 over R, and then I have p1 V1 over R, the R's cancel out.
Well let's see, there's a V1 over V2 to the gamma minus 1.
There's a V1 over V2 here or V2 over V1.
That's going to get rid of this minus 1 here.
Then I'm going to get, be left with a p2 over p1.
So this guy here gets rid of this, and I have my answer as V1 over V2 to the gamma power is equal to p2 over p1.
Or I can re-write this as p1, V1 to the gamma is equal to p2 V2 to the gamma.
That means that p1 and -- where I am, the point number 1 and point number 2 were completely arbitrary.
They happen to be on the path.
So any point I pick on that path will be equal to p1, V1 to the gamma.
So if I take p times V to the gamma, anywhere on the path, it's going to be equal to the same relation from my first point.
This is going to be true for any point on the path.
As long as I'm on the path, pV to the gamma will be whatever it was for the first point, which is going to be some sort of constant.
All right, so I've proven my point.
I've proven what I was trying to do, which was to show that on this adiabat, everywhere on the adiabat, if you take a functional form that relates p and V together, it's going to have this relationship with gamma as related to the heat capacities.
Gamma is Cp over Cv.
OK, now there's more we can say because we know that's Cp is related to Cv through this R here, and R is a positive number.
So Cp is always bigger than Cv.
In fact, if it's an ideal gas Cv and Cp are well defined numbers, it turns out.
If you have a monotomic ideal gas, OK, so an atomic version, not a molecular ideal gas, then it turns out that Cv is equal to 3/2 R, and therefore, Cp which is just R plus this is five halves R, which means that gamma for a mono atomic ideal gas is 5/3.
So if you have a gas of argon atoms, you know what gamma is.
This is something that you're going to prove in statistical mechanics, and so we're not going to worry about where this comes from.
We're just going to take it for granted.
All right, so gamma is for ideal gas, is bigger than one.
In fact we have a number for it if it's an atomic ideal gas.
So what it means then is if I look at if this relationship here, gamma is bigger than 1, so gamma something bigger than 1 minus 1.
This is, so this is positive here.
It's positive.
So if I have an expansion where V2 is greater than V1, so V2 is greater than V1, so this is a number which is less than one, then I expect then T2 is going to be less and T1, T2 is less than T1.
OK, I have this number here to a power, which is positive, a positive power.
And its number is less than one.
This is going to give me something which is less than one.
So T2 is less than T1, and I have a cooling of a gas.
So whenever you have an adiabatic expansion, the temperature cools, it gets colder.
It gets colder, and if I look at my graph here, then if the volume of the expansion, if the temperature of the adiabatic expansion was colder, that means that -- this is the wrong graph here.
If I want to compare it with the isothermal expansion, which is sitting here.
All right, so gamma, the gas is cooling so V2 is going to be less than it what would be if the temperature kept constant.
So finally, on the same graph, I put down what an isothermal expansion would be, if my final pressure is the same pressure, my volume for an isothermal expansion will be bigger than for an adiabatic expansion.
So when I expand this gas adiabatically and it cools down, why do you think it might cool down?
It cools down because I'm doing work through the environment, but the energy has to go somewhere, right?
There's an interplay between the energy inside the gas which is the heat energy which is allowing me to do all that work to be outside, and so I'm using up some of the energy that's inside the gas to do the work on the outside.
And here, in an isothermal expansion, The bath is putting back the energy that the gas is expanding or using to expand, and so the energy is flowing back into the gas through the environment in the isothermal expansion.
In the opposite case, if you have a compression, then it's the opposite of expansion.
Compression you're expected to heat up, right?
So in the compression, you're going to have V2 is less than V1.
So that T2 is going to be greater than T1.
That's going to be your heating up of the gas.
And that really is the bicycle pump.
That example I gave last time with the bicycle pump was not quite the right example.
But this time this really is the bicycle pump.
That you're pushing really hard on it.
So you start with the bicycle pump at one bar, let's say, right?
And you compress the air in the bicycle pump.
You do it so quickly that the heat flow between the inside of the bicycle pump and the outside is too slow compared to the speed at which you compress.
But you don't compress it so quickly that you're not in there reversible process so you compress that if you were to feel the temperature of the air and you can feel it through the nozzle gives you an idea of the temperature of the air inside your bicycle pump.
You'll feel that it's warmer.
You've just done an adiabatic compression of the ideal gas, you can pretend there is an ideal gas.
And that gives rise to the heating that you can actually measure.
OK, any questions on this part here.
All right, so we've just gone through the reversible adiabatic, and I'm not going to do this in detail, but I want to go through the irreversible adiabatic fairly quickly, and see where the difference is here.
OK, so now we're going to do the same kind of experiment, but irreversibly.
An irreversible adiabatic.
OK, so the experiment is going to be to take my cylinder now, and the external pressure and the internal pressure are not going to be in equilibrium with each other during the process.
I'm going to start with my cylinder here, I'm going to put little pegs here so it doesn't fly up.
There's going to be some external pressure.
I'm going to set that equal to p2.
There is going to be an internal pressure where p1 is less than p2 and there's V1 and T1 here.
The whole thing is going to be insulated.
Then I'm going to release these little pegs here and my piston is going to shoot up now because p2 is less than p1.
So p2 is less than p1, the external pressures is less than the internal pressure.
So it's going to shoot up until the internal pressure and the external pressure are in equilibrium.
So they're both p2.
external is p2 and I have p2, V2, T2, on the other side.
OK, so in my diagram now, I have p1, V1 for my gas.
p2, V2 for my gas, so I know that I'm starting here and I know that I'm ending here, but I can't connect this path here.
I don't know how to do that.
It's an irreversible expansion.
I don't know what the pressure is doing in there, doing that expansion.
It could be quite chaotic.
It could be non-spacially constant.
OK, so but you still know a number of things.
We're now going to be able to write you know dw is minus p dV, but we're still going to be able to write a bunch of things.
We're still going to be able to write that it's an adiabat so that dq equals zero.
We're still going to be able to write dw instead of pV where p is the internal pressure.
dw now is actually much easier.
It's minus p external.
Well p external is actually p2, so that makes it actually easier.
dV we're still going to write that it's an ideal gas.
So du is still going to be equal to Cv dT, and we're still going to be able to use the first law, all these things don't matter where the path is.
Still know that du is dq plus dw.
dq is zero.
dw is minus p2 dV.
So this is what we know, just like what we wrote there.
We write what we know.
I'm not going to turn the crank here.
I'm going to give you the answer.
OK, you use the ideal gas law, etc., then you get a relationship that connects the pressure and the temperature, like here we got a relationship that connected the temperatures and the volumes together.
You can get a relationship that connects the temperature so this is T2 times Cv plus R is equal to T1 on Cv plus p2 over p1 times R. This is going to be the connect, what connects the pressures and the temperature.
And p2 is less than p1, so this number right here is less than 1.
So and Cv plus R, Cv plus something less than R, so T1 better be bigger than T2.
OK, so T2 is less than T1.
Same qualitative result as before.
You have an expansion.
It's and adiabatic expansion.
You get a cooling of the gas.
OK?
Now here's a question for you guys, which we're going to vote on, so you better start thinking about it.
Is the temperature T2 in this process smaller or larger than if I were to do the process reversibly with the same endpoint pressure.
So now, instead of having p external equals to p2 here, I put p1 and I let this whole thing go reversibly.
And I compare T2 irreversible to T2 reversible.
They're both less than T1.
They're supposed, there's a cooling happening in both cases.
One is going to be colder than the other, maybe.
Maybe they're going to be the same.
I don't know.
I'm going to let you try to figure that out qualitatively without doing any math.
See if you can figure it out.
OK, so which is colder?
I'm going to let you think about it for about thirty seconds or so.
You can talk to each other if you don't, can't figure it out.
All right, let's take a vote.
How many people say that T2 irreversible is colder than T2 reversible?
One, two, don't be shy, three four, anybody else, five.
OK, I've got five votes here.
Professor Nelson?
Oh, I'm abstaining
You're abstaining, OK, good choice.
OK, what about T1 cooler than T2?
OK, I've got a lot of votes here.
So the majority of the people, so this is much bigger than five.
OK.
So remember last time the majority was wrong, right?
The majority was wrong last time.
That doesn't mean that the majority is wrong here.
It could be right.
Let me give you another piece of information here that you already know.
Minus w irreversible, this is the work which is done to the environment by the system, minus w irreversible is always smaller than minus w reversible.
You learned that a little while ago.
OK, so this is a little hint.
Let's vote again.
Let's think about it for ten seconds why this could be interesting and relevant, and I want to ask if anybody changed their mind.
Anybody change their mind, change their vote?
Yes You want to change your vote?
To which one?
From T2 reversible is greater than T2 irreversible, saying that T2 reversible is [UNINTELLIGIBLE]
So the question then, which one is colder?
T2 reversible should be colder
T2 irreversible should be colder?
Yes
All right.
OK.
I'm going to keep you there then, the majority.
Yes?
I'm switching to [INAUDIBLE]
You're switching this way here?
So now we have six here.
Anybody else want to switch?
All right, let's take the example of, the extreme example, let's go to the extreme example where p external is really small.
And I have this adiabatic expansion where p external is really small.
it's kind of like the Joule expansion, an ideal gas.
What happened to the temperature in a Joule expansion in ideal gas?
Anybody remember?
Anybody remember?
Joule expansion, this is where we proved that delta u was only dependent on temperature.
Bob Field taught that lecture.
Nobody remembers?
Well, all right, delta u is equal to zero for that.
What happened to the temperature in that expansion?
It's an adiabatic expansion.
Well, we'll revisit that.
Let me ask you another question here.
So w, the work is less for the irreversible process than the reversible process.
There's less work done to the outside.
So there's less energy expanded by the system.
The energy expanded by the system is smaller for the irreversible process.
It's not doing as much work.
It's not pressing against as much pressure.
Right?
p external equals p2 is less than p external equals p internal.
So it doesn't have to do as much work.
It doesn't have to expend as much energy, therefore, the majority in this case is right.
The temperature is not going to cool as much for the irreversible process.
The reversible process is going to be colder because it has to expend more energy.
It's going to have a higher pressure.
The pressure is going to decrease along the way, but it's going to have to do more work because of this right here.
So the majority in this case is right.
Now the Joule expansion, the temperature doesn't change, T equals zero.
eta sub J, the Joule coefficient for an ideal gas, is equal to zero and eta sub J was dT/dV under constant energy.
All right, any questions?
Yes?
[INAUDIBLE]
The only -- a priori, that's true.
But it's hard to imagine an irreversible compression the way we can just imagine an irreversible expansion.
You have to do it awfully fast to get your irreversible compression.
I don't know how to do it really.
I think it's, but it's more complicated, but you're basically right, basically right.
That's why we always talk about irreversible expansion, because it's easy to write down an irreversible expansion.
A compression is a little bit more complicated.
Any other questions?
OK, great.
Monday Professor Nelson who is sitting here will be taking over for a few weeks and then you'll see me again after that.
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Well, today we're going to continue with the lecture that was started last time that had the scintillating and descriptive title "some thermodynamic processes".
And we're going to continue along those lines, and really there a couple of things that I want to do to follow up what you saw last time.
One is that you've got introduced and led through a couple of elementary thermodynamic cycles.
And I want to do a little bit more of that today.
And one of the -- there are really two reasons.
One is that as you saw last time doing a thermodynamic cycle can be extremely useful, because if you have one path along which some change occurs, but along which it's not necessarily easy to calculate all of the thermodynamic changes along such a path, it may be irreversible, lots of variables may change.
It can be easier instead of looking at the complicated path to devise a cycle that involves several easier steps, each one of which is facile for the calculation of what happens to the thermodynamic properties.
So in that case, it's much easier to go through several elementary step, and then the one complicated path you know because going around the entire cycle state functions, you know that they won't have any change.
So in that sense, going through a thermodynamic cycle can be useful because it helps you calculate thermodynamic qualities.
But the other reason to go through the thermodynamic cycles and really to develop great facility with them is because there are just an awful lot of things in nature and things that we build that run in cycles, where we want to calculate the thermodynamics, right.
If we build an engine for a car or anything else, it almost always is going to have some key element that's operating in a cycle, otherwise it won't keep going, right.
But it's not just engines that we might build.
It's biological engines, right.
Things that either produce or consume biological fuel to do different sorts of processes.
Also, of course of key importance to understand what the, in that case, biological thermodynamics are.
And again, there'll be key elements that run in cycles and understanding those can be extremely important in understanding how cellular function works.
OK, so what I'd like to do is go through one cycle, and then I'll suggest another for you to work through on your own.
So consistent with the title of the lecture, we've got a section from your last lecture notes entitled "some thermodynamic cycles".
And what we're going to do is go through a cycle, please note these office hours and locations, which I think wasn't specified before.
So we're going to go through a thermodynamic cycle, and here's what I want to calculate when we do this.
Delta u, delta H, familiar state functions, changes in their values, q, w, heat and work.
And I also want to introduce one new function.
I won't give it a name yet, but it's going to have the unusual form dq over T. OK, we'll call it special function.
It's so special that we'll call this quantity dS.
OK, we'll just go through this on a whim, and this is going to be foreshadowing something that'll take on tremendous importance in subsequent lectures, but I'm going to use the opportunity now to just introduce the behavior of it in going around a cycle.
If that's for me, tell them I'm busy please.
OK, so let's look at the cycle we'll go through.
We're going to look at pressure volume changes, similar to what you saw before.
This time, we're going to have initial and final volumes, V1 and V2 and different pressures where we might end up.
And we're going to look at two ways to go to a final volume V2.
One of them is going to end up at pressure p2 and the other is going to end up at pressure p3.
One of them is going to be a reversible isothermal path.
This is going to be our path we'll label A, reversible isothermal which is to say constant T, right?
And we're going to start at temperature T1, so since it's constant temperature, this is going to end up at temperature T1.
And then we're also going to consider a reversible adiabatic path.
This'll be our path B, reversible adiabatic.
And then we'll close this circle, this cycle.
This is going to end up at a different temperature by the way.
You saw this last time in a slightly different way.
Last time what you saw is we compared isothermal and adiabatic paths that ended up at the same final pressure, and what you saw is that therefore, they ended up in different final volumes.
So this is just a little bit different from that.
So this is going to end up at a different temperature, we'll call it T2.
And then we're going to close the cycle.
And so this is a constant volume path then.
This is path C, constant volume.
OK?
So now let's go around the cycle and just compare notes on what happens to the thermodynamic quantities as we do that.
So here's path A, it's isothermal.
And I didn't specify, but let's make sure to do so, we've got a, it's going to be an ideal gas.
OK, so A is constant temperature.
What does that mean for delta u in path A?
Zero right?
Ideal gas, no temperature change, right has to be zero because da is Cv dT for an ideal gas.
dT is zero, there's no temperature change.
OK, how about delta H, zero Cp dT for an ideal gas.
dT is zero.
All right, now, it's reversible.
So that means we can immediately write down an expression for the work.
Since it's reversible, it's negative p dV right?
And since it's an ideal gas then we can replace p, right, with RT over V, right?
We want to do that because we have too many variables here.
We've already got dV we'll get rid of p as an additional variable and replace it with V which is already in here.
So it's minus R T1 dV over V, right?
I can put in T1 because we're a constant temperature.
so now we can just integrate straight away and find that the work in path A it's just the integral of that quantity minus integral of R T1 going from V1 to V2 dV over V. so it's minus R T1 log V2 over V1.
All right, so let's just do a reality check here.
The way I've got this drawn, the volume is going up in the process.
It's an expansion.
So is this system doing work on the surroundings, or are the surroundings doing work on the system?
Somebody say it real loud.
Yes, the system is working on the surroundings, right?
It's pushing out against them.
Work is defined as the work that the surroundings do on the system.
So this is the negative number, right?
V2 is bigger than V1.
This is positive.
This whole thing is negative.
All right, q has to be just the opposite of this because we've already figured out that there's no change in u.
This is just q plus w. There's w, q has to be R T1 log of V2 over V1.
OK, so that means heat is being imparted to the system, right, from the surroundings in a manner that compensates exactly the amount of work done.
OK, All right, so these are the thermodynamic quantities that you're familiar with already.
Let's just quickly look at our special function.
It's not going to be hard to calculate because it's a constant temperature path, so that T in the bottom is just T1, right.
You've already looked at q, right.
So dq A over T1 it's just R log V2 over V1 right.
So there's our special function for this particular path.
OK, so that's path A.
Now let's compare to path B.
So if path B is an adibat, what's zero?
q, it's adiabatic, that means there's no heat exchanged between the system and the surroundings.
So what happens, q B is zero.
There is a change in temperature, right?
We've got, if we say it's a mole of gas, it's going from p1, V1, and T1 to one mole of our gas at p3, V2, and T2.
So unlike before now temperature is going to change.
So that means that delta u isn't zero this time.
du, it's an ideal gas.
So this is Cv dT and of course we can just integrate this straight away.
So delta u B is Cv times T2 minus T1, right.
and similarly dH is Cp dT right.
So delta H in pathway B is just Cp T2 minus T1 okay?
All right, what we've already got that q is zero.
Delta u is q plus w. Delta u isn't zero.
So this must be equal to work, right?
And now we can look at our special function, but since this is zero right dq B is zero, along path B, so our special function, dS, so we're going to go, dq B over T is equal to zero.
OK?
OK, finally let's look at our third path, this constant volume path, that's going to connect T1 and T2, right?
Constant volume, what's zero?
[INAUDIBLE]
A little more noise here.
What's zero?
Work
Work, great.
So, path C constant volume means our work in past C is zero, right?
Now, our temperature is going to change from T2 to T1, and we just saw what happens to u and H when that happens going from T1 to T2, and it's no different.
It's an ideal gas going along these path.
So we can immediately write delta u C is Cv times T1 minus T2.
Delta Hc C is Cp times T1 minus T2, right?
Work is zero.
Delta u is w plus q, work plus heat.
This is zero, this isn't.
This must be heat q B, right.
So again, there are our familiar thermodynamic quantities, but now let's also look at our special function.
It's not going to be zero this time because we have non zero heat exchange between the system and the environment, right.
So are integral of dq C over T, right, which is our heat, but in this case we can do this calculation easily because since work is zero, we can equate dq with du, right.
So we can write this as our integral from T1 to T2, du over T. du is just Cv dT, right.
So this is just Cv log T2 over T1.
OK?
So now we've completed the cycle.
So let's just compare.
Let's compare what happened in path A to what happened in paths B and C. Yes?
[INAUDIBLE]
Ah sorry, yes of course.
Thank you.
Yes, oh yes, so
[INAUDIBLE]
And it's right because, thank you again, it's because we're going from T1 to T2.
So I'm getting confused by the oh sorry, we're going from T2 to T1.
So it's in reverse of the order of the subscripts and I let this confuse me.
There it is.
There it is.
There.
Thank you, any other details which should be pointed out?
All right, then let's proceed.
So what we're going to do is our comparison of what happened on pathway A to what happened in combined pathways B and C, right?
So here are results for pathway A, right, for delta u zero delta H zero.
I didn't actually explicitly write it or did I?
Let's just write it again.
Here's our work in path A, and here's our heat exchange in path A.
Now let's look at the sum of B and C. So B plus C, here is delta uB Cv times T2 minus T1.
Delta uC Cv times T1 minus T2, the sum is zero right.
That's the some of delta u in these two paths, and if we look at delta u in just path A, it's zero.
And it's going to be the same without delta H. The only difference is it'll be Cp instead of Cv, but there it is for pathway B.
There it is for a pathway C. So the state functions that we're familiar with are doing what we expect they ought to be doing, right?
If you go around in a cycle, starting and ending at the same place, the state functions have to stay the same.
They only depend on the state the system is in.
They don't depend on the path, and that's what we've shown, right.
Similarly if we go from this point to this point through path A or if we do it through the combination of these two paths, and the change in u and H in any state function, those have to be the same.
It turns out they're zero in both cases, and that's what we've seen, right?
Now let's compare what happens to work and heat.
So here are expressions for work and heat.
They're opposites for the pathway A.
Here's pathway B and C. In C the work is zero.
In B it isn't, it's Cv times T2 minus T1, right.
The work isn't equal to the work in pathway A, right, because you know work is not a state function it depends on the path right, and there are different amounts of work done on the system, or done by the system on the surroundings in these two different processes.
They're both expansions.
They'll both have net work done on the surroundings, but not the same amount, right.
And of course it's going to be the same with heat.
We've already seen that delta u is zero.
So we know that in each case the heat is going to be the opposite of the work, but the work isn't the same in these two different ways of getting from here to here, right.
So let's just see it explicitly.
Here's our qA.
Here's heat exchanged in pathway A and in pathway B heat is zero, and in pathway C, here is qC it's Cv T1 minus T2.
So again, for both heat and work we don't get the same result.
Now let's look at our special function, right.
So here's path A.
We found that it's R log V2 over V1.
Pathway B is zero.
Pathway C it's Cv log T1 over T2.
All right, let's just look at that a little more carefully.
R log V2 over V1.
B plus C, Cv log of T1 over T2, right.
Just some weird combination of functions, right?
But, of course, it's not quite like that.
Why?
Because this path is an adiabatic path.
And you already saw last time there was this relationship between the temperature and volume changes along an adiabatic path.
So let me just write that down.
Adiabatic reversible path.
What you saw is that T1 over T2 was V2 over V1 to the power R over Cv.
Isn't that something?
So what does that mean?
We take this Cv and put in the exponent here, right.
And we put this R up in the exponent here.
What are we going to discover?
Those two things are equal right, T1 over T2 to the power Cv is V2 over V1 to the power R, so dS over path A equals dS over paths B plus C, okay?
So what this suggests is that whatever this funny special function is, at least for this one cycle that we've, tried it's behaving like a state function, right.
It seems like its change is independent of path.
Start here, end up here, do it through two different pathways, end up in the same place, right.
OK, now that's all the foreshadowing I'm going to give it for right now.
We will certainly come back to this very special function shortly.
Before I move on, I'm just going to put on the board another cycle, and I'm going to urge you to work through that on your own.
It's worked through already, let's see, I believe it's worked through in the notes, yes.
So if you need the help of the notes it's in there, but I would urge you to work through it on your own, and it's the following: let's start with the same path A, all right?
So we've got our reversible adiabatic path right.
And now, try working through what happens if we close the cycle in a different way, right, like this.
So here's a path that's constant pressure, and here is a path that's constant volume, similar to the constant volume path that we did before, but not between the same pressures, right.
So we can label this one D and this one E, and I urge you to just try going through that and verifying for yourself that again the state functions will behave the way state functions should, and you can see whether this special function once again behaves like a state function, right.
And of course you should expect to see that the work and the heat again won't behave like state functions which they are, then you'll see you have different results for those, depending on the path, OK?
Any questions about going through these cycles and using expressions like this and so forth?
Expressions like this you know turn up in various places, like in the equations sheet that appears at the end of exams, right?
Which means that you may not need to be committing it to memory.
On the other hand it means that you should, it's the sort of thing that you should be familiar with so that the need to use it is something that you'll conjure when you're working on a problem and there's a reversible adiabat, and you have temperatures and volumes that change that you might like to relate to each other.
Cycles that are suggested that you go through and aren't gone through in class also sometimes turn up on exams too, by the way.
So that said, let's move on to the next topic, which is thermochemistry.
So, so far what you've seen in most of the examples in the class are essentially mechanical kinds of changes, pressure volume sorts of changes, where the system is doing essentially a kind of mechanical work, pressure volume work on the surroundings or vice versa and heat is being exchanged, and you're calculating the basic thermodynamics just like we went through, of processes of that sort.
Now I'd like to start introducing something that's really central to chemical thermodynamics, mainly chemistry, right.
Let's start looking at chemical reactions, right, and understanding what the changes in thermodynamic properties are that occur when chemical reactions occur, and the immediate application is to calculate the change in enthalpy, which, as you've seen it, at constant pressure which is the condition under which an awful lot of chemistry is done, that's just the heat.
Which means we're going to calculate heats of reaction, right.
You're running some reaction.
It's in the atmosphere.
It's at constant pressure.
You know, you mix acid and base together and the thing heats up like crazy, right.
Or other things might react spontaneously.
You can feel something cooling off right.
I mean simple examples of these happen when you, you know, if you buy cold or hot packs.
You break the seal between them and feel the thing get cold, for example, right.
What's happening there?
Well, the selection of reactants has been done judiciously to provide either heat or to provide something that's cool.
And all of that is coming out of the heat of reaction, whether it's positive or negative.
Of course the biggest practical application is burning of fuel.
You might use the heat to, and convert it into pressure volume work, right?
So if it's an internal combustion engine, you'll do a reaction.
There will be the heat of reaction.
There'll be a volume change, depending on the conditions under which is occurs, and you can extract work through the reaction and so forth.
So let's just to see how you run through some of those calculations.
Now let me just ask who here took 5.111?
And who here took 5.112?
OK, now you've seen some thermal chemistry in 5.111 and 5.112.
So I I'm going to do some review of that, but I'm also going to call on the thermochemistry that I'm pretty sure you're familiar with.
So I won't go painstakingly over every element of the notes here.
Subsequently we'll go on into additional topics that you haven't seen, and I'll treat them in full detail.
OK, so let's do some thermochemistry.
What we want to do is we'd like to be able to predict for any reaction what's the heat of reaction.
And so what we're going to calculator is delta H. So for any kind of reaction we should be able to do that, right?
And we're going to treat constant pressure situations certainly at first and most of the time, right?
Constant pressure, right, reversible delta H is going to give us our heat of reaction, OK. And then if we can also determine delta u, then we know this, we know delta u is q plus w, then we can determine work as well, right?
And typically, we'll be treating at least some cases where we're dealing with ideal gases in which case we can easily get delta u.
And then we'll be able to in a very straightforward way get w. So it's a really powerful and simple formalism, once we set up what is needed the go through it, right?
So let's just consider any reaction.
The one that I've got written down for you is it's the reverse of the rusting of iron, right?
So iron left to own devices in the atmosphere in the presence of a little bit of water, and the atmosphere will start combining with the water form iron oxide.
There's an equilibrium between the two.
So we'll have an iron oxide species.
It's a solid.
It's combining with hydrogen which is a gas to give iron, also a solid and water, right?
So the way I've written this we're imagining the iron, the solid iron immersed in the water as a liquid, could be calculated otherwise as well, but in this case were imaging the water as liquid and going in this direction then it's forming iron oxide and evolving hydrogen gas, okay?
And our heat of reaction or enthalpy of reaction is defined as the enthalpy at constant pressure.
Isothermal conditions, temperature won't change and reversible work.
For Isothermal reaction constant pressure reversible, right?
Of course, they're all sorts of conditions under which a reaction could be wrong in the lab or outdoors or however, right.
But this is the way we're going to define delta H of reaction.
We want to have that definition clear because in fact we're going to, we might want tabulate heats of reaction, right, and of course want to know what the conditions are for the tabulated values apply.
And we're going to want to calculate them from other quantities, and again, we're going to need to know each case what are the relevant conditions?
OK, so what happens?
Well, we should be able to get this.
We should be able to calculate delta H. It's a state function.
If we know the enthalpy of the products minus the enthalpy of the reactants, right.
It's a state function.
And we can we can do this in principle except for one important detail, which is that enthalpy, just like energy u isn't measured on an absolute scale but on a relative scale, right?
You know, if you want to measure the potential energy of something in a gravitational field, you have to define the zero somewhere, right, because it's arbitrary.
You can set it anywhere you want.
It's the same with enthalpy.
Enthalpy is just u plus p V. So there is an arbitrary set point that needs to be defined, right?
Because what you actually measure in the lab are changes in enthalpy, just like what you measure when you look at energy change of some sort, you measure the change in energy, right.
The absolute number that you assign to it is something that's arbitrary.
You have to set what the zero is.
And so there's a well-understood convention for what the zero is.
What we define as zero is the enthalpy of every element in its natural state at room temperature and ambient pressure.
In other words, we choose a convention for the zero of entropy, so that then we can write entropies of products and reactants always referring to the same standard state.
And then we calculate changes, the convention is understood with respect to what is the zero, right.
And so our tabulated values, they'll all work.
They'll all refer to the same standard state.
And we'll be able to use the formalism that we set up.
So our reference point for H, it's 298.15 Kelvin, one bar and in that standard state our molar enthalpy is defined as zero for every element in its stable form.
OK, very important.
OK, now, given that reference point, all we need to do to get the value of enthalpy that we're going to use for each reactant and each product is calculate how much does the enthalpy change to form it from the elements in their stable form at room temperature and pressure, right.
So we're going to replace the H or we're going to put in for the H what we'll call our heat of formation or delta H of formation starting from the elements in their stable states at room temperature and pressure, for each of these things.
And we can do that because now we're going to have instead of just sort of H, it's going to be delta H of formation for each of these things, is going to appear in our calculation of H, but that's OK. Delta H of formation means the enthalpy of this compound minus the enthalpy of its constituent elements in their most stable state at room temperature and pressure.
But we've defined the enthalpy of those elements in their stable state at room temperature and pressure as zero, right?
So we're just subtracting, in effect, zero, right, from the enthalpy of the product, but of course it's important have that established because the heat of formation is something you could measure, right?
You could run the reaction, take solid iron, gaseous oxygen, form iron oxide, measure the heat of formation of it, tabulate it.
We know it.
We know it forever, right.
And for any of the other compounds.
So in other words, by defining that reference state, we can then figure out or measure heats of formation of a vast number of compounds.
We can tabulate them.
We can know them, and then when we have reactions that inter-convert different compounds, we can calculate the heat of reaction is just the difference between the heat of formation of the reactants, and the heat of formation of the products, right.
So let's just write that out.
So first of all, for example, right, molar enthalpy of hydrogen gas at 298.15 K is zero.
Hydrogen gas it's in its most stable state, right at room temperature and pressure.
That little zero, that little superscript means one bar always.
OK. A molar enthalpy of or whatever, iron, as a solid at 298.15 Kelvin is zero.
Iron as an element is a solid.
That's it's most stable state at room temperature and pressure, right, and so on.
And then we can figure out heats of formation.
Now in this particular reaction, I've got hydrogen gas, iron solid.
Those already are elements in their most stable forms at room temperature and pressure.
But this is a compound, right, it has some non-zero heat of formation from the elements.
So is water, right?
So I need to find out the heats of formation of the iron oxide and the water.
And if I do that then I can find out the change in enthalpy of this reaction.
It's just going to be the heat of formation of these three moles of water, minus the heat of formation of the iron oxide.
OK.
So how am I going to do that?
Well, I need to write the reaction that forms that compound from it's elements, right?
And I want to tabulate that for an enormous number of materials, right?
So for example, if I want to look at HBr, there's a simple case, right, hydrogen bromine.
What's my heat of formation?
Delta Hf.
It's going to have our little zero, right?
How do I calculate it?
Well, I need to write the reaction that forms this from its constituent elements.
What is it?
Well, 1/2 H2 as a gas, temperature, at one bar, plus 1/2 bromine, no actually bromine is a liquid at room temperature and one atmosphere.
They form HBr which is a gas at room temperature and one bar, right.
I can measure that.
And the heat of reaction for this, delta H of reaction is equal to delta H of formation, of HBr as a gas at our temperature and one bar, okay?
So that's how we determine our heats of formation.
We measure them for all these compounds.
And then we go back to reactions like this, and we can just very simply determine the heat of reaction because all we're doing is the following cycle.
We go from reactants to products and there's some heat of reaction, and here's the cycle that we have.
We go to the elements in their standard states, in both cases.
So we're really just doing our delta H is the negative sum of delta H of formation for the reactants, all right?
Because here what we're doing is we're going to take apart our reactants and form the elements from them, right?
So written this way, for example, I'm going to pull these things apart, and I'm going to have solid iron, and I'm going to have gaseous oxygen, and I know the enthalpy of that reaction.
That's just given by the heat of formation, the enthalpy of formation of iron oxide.
Here, now I'm going to take those same elements, and I'm going to make the products, right.
I know it's going to work because I already had this reaction, so all the, you know, I'm going to conserve atoms in the right way.
So here, I'm going to have delta H, is just the sum for all the products of delta heat of formation, right?
Here I'm going to put together all the products.
So this is a positive heat of formation.
This is the negative heat of formation, right?
In other words, I've got reactants, and I've got products.
What's delta H of reaction?
It's delta H of formation of the products minus delta H of formation of the reactants.
That's the important message, delta H of formation of the products, minus delta H of formation of the reactants.
Any questions?
All right, tomorrow I'll go through the remaining details.
The truth is after this, it's all arithmetic, right?
We're just going to go through it and execute it a little bit.
And then we'll move on and talk about how we'd actually make some of these measurements of delta H and compare them to what we calculate.
See you tomorrow.
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Well, so last time we did a little bit more work on thermodynamic cycles, and basically went around a cycle and looked at some state functions, delta u and delta H. Saw that around a closed cycle those zero because they're state functions.
So if you start and end at the same place, they've got end at the same place that they started.
There's no change in them, and then we also looked at some at non-state functions, work and heat, and saw that those aren't zero going around a cycle.
Of course you can do work in a cyclic process, and heat can be exchanged with the environment at the same time.
So we calculated those values too.
And we also looked at this other funny function that's our special function.
We looked at this integral dq over T. It's so special that we called this derivative dS, and saw that at least for the cycle we looked at, it also behaved as a state function behaves.
That is, going around the cycle, it had no net change, and we'll see this come back later on.
Then we went on to look at thermochemistry, and that's what I want to continue today.
So really, the main result that we saw last time is that we can describe the enthalpy of reaction, so if we have reactants going to products, and we'd like to know our heat of reaction or enthalpy of reaction, remember it's a constant pressure as we're considering it, so it's equivalent to heat of reaction, than the way we can do this is construct a cycle where we're going to express these in terms of their constituent elements.
We're going to all the way back to the atoms.
Right, so if we have the elements in their standard states, that is their most stable forms at room temperature and pressure, then what we're really doing is we're saying let's take the reactants and let's pull them all apart into their separate elements, they're separate atoms.
And then put them back together this way, and we'll calculate the enthalpy of this process, the enthalpy of this process.
It's a cycle, so that'll give us the enthalpy of reaction, all right?
So this delta H is our sum for delta H of formation for the reactants, right.
And this is our sum of the quantity for the products, right.
And in this way, then we can express our delta H of reaction is just going to be given a sum of the heats of formation or the difference between the heats of formation of the products and the reactants, right?
Now, of course, we have to weight this by the number of moles of each reactant to each product involved in the reaction.
So to write this a little more carefully, it's, you want to write the sum over i of nu i delta H naught of formation of i summed over the products.
This is per mole minus sum over i, nu i here over the reactants in the same quantity.
Delta H per mole f i, right?
So in other words, were taking each of the products and we're going to weight them by their stoichiomeric coefficients, and we're going to have in this delta H of reaction their heats of formation, and then we're going to have the negative heats of formation of the reactants, because that's being subtracted, okay?
So, just to go through a basic example, let's just look at the burning of methane.
So we've got methane gas, combining with oxygen gas, so you have carbon dioxide gas and liquid water, and we'll consider the whole thing at room temperature and pressure, right.
So let's just undertake this set of processes.
Let's decompose these to the elements.
So in other words, our methane gas, we're going to write the chemical equilibrium between this and the solid carbon.
That is to say let's specify graphite, it's not diamond, right, graphite is the stable form of carbon at room temperature and pressure.
And then, hydrogen gas two moles of hydrogen gas and we're going to have a delta H of formation.
This is for CH4 in the gas phase, right?
And then over here, we have our oxygen but 2 O2 gas going to, well 2 O2 gas -- the point is oxygen already is in it's stable form at room temperature and pressure, right?
That is the elemental form of oxygen, of course, is as oxygen molecules in the gas so there's no delta H formation for oxygen.
The gas is zero, right.
And then second phase of this, now we're going to take the elements that we've produced, and we're going to put them back to make products, right?
So there we've got carbon as graphite, and a solid plus an oxygen gas goes to make CO2 gas, right.
And we've got our positive delta H of formation associated with that.
And finally we've got hydrogen gas plus oxygen gas giving us two water molecules and this is in the liquid phase, and there's some delta heat of formation associated with that, okay.
So there is our set of individual processes that's going to constitute our cycle, right.
If we take the combination of all those, which is to say we just take both steps in this cycle, then we'll have our net reaction, which is to say we'll have this, right.
So what that says is our heat of reaction is our heat of formation per mole of CO2 gas, right.
Plus 2 times the heat of reaction or heat of formation per mole of liquid water minus heat of formation per mole of methane, right.
And things we can easily look up.
So in a simple way, we can determine, we can calculate what the heat of reaction is for something like this, right.
And you'll see in the back of your book, you'll see moderately extensive tables of heats of formation, and if you go online you'll see, you can find extremely extensive tabulated values of heats of formation, that have been measured for an enormous number of compounds.
And from that, then you can look at enthalpies of reaction for countless numbers of reactions, right.
So by using the tabulated data, we can really determined heats of formation for most reactions that you might contemplate, OK?
Of course, the cyclic steps that we've taken to do this apply not just for breaking down reactants and product into the elements in their standards states, but of course we could also look at whole sets of reactions and write cycles as well, right.
So for example, if we wanted to look at the reaction of something like carbon graphite plus oxygen gas to make carbon monoxide gas, well normally this isn't the way the reaction would work.
Normally if you just tried to make this happen what you would wind up with the CO2, not CO, right.
But one way to approach this is you could look at the individual reactions of going, you could go to CO2 and then you can go with CO2 and oxygen and combine, or combine CO and oxygen to get CO2, right?
So normally, CO2 is formed, but you could calculate the heat of reaction.
Of course you could go back to the elements, but you can also say well, let's just take a known heat of reaction for, or heat of formation for CO2, and then also combine that with the heat of reaction for CO plus oxygen forming CO2 and determine this that way, all right, OK, so once you have done this, once we've got the heat of reaction then there are a few simple results that are useful.
One is we know immediately whether when we run a reaction heat is going to be released in the process out to the environment, or whether heat is going to be taken in, right.
So, and of course if you ever used a hot pack or a chemical hot pack or cold pack, you've seen examples of the two of those, where the constituents are selected to give you either heat released or heat taken up, right.
So in particular delta H of reaction less than zero, that means a negative amount of heat, which means that heat is released, right.
Yes, that is it's exothermic.
Positive enthalpy of reaction, which is to say a heat of reaction is positive.
Remember the way we define heat is positive it means that there is heat that is coming from the environment into the system, right.
In other words heat is absorbed or taken up by the reacting system.
That's endothermic, okay.
All right, one more important aspect of this that we need to be able to deal with, and that is the heats of formation that you'll find tabulated typically are going to be given for room temperature and pressure.
And it's very common that you might want to know the thermodynamic condition for reactions, especially at other temperatures, very common, of course.
And it's not hard to see how the heat of reaction at room temperature can be related to they heat of reaction at other temperatures.
So let's just look at that.
So remember your fundamental relation, dH, if you write that as a function of temperature and pressure, partial of H with respect to T, constant pressure, dT, plus partial of H with respect to p at constant temperature, dp.
But if we're working at constant pressure then dp is zero.
So we just got the derivative with respect to temperature.
That's our heat capacity, right, Cp.
And we may have tabulated values of Cp for an enormous number of materials.
So now, if we look at the temperature dependence of delta H for reaction, really what we're going to need to do is look at how that heat capacity changes going from reactants to products, right?
So in other words, d delta H of reaction, with respect to temperature.
That's what we want to find out, right?
So how does delta H change if we change the temperature?
Well, it's just given by Cp dT for the reactants and the products, right?
For the products minus the reaction, that is it's delta Cp, we need the sum over the products.
Nu i, Cp i minus same sum over reactants, i, nu i Cp i, right.
Similar to what we saw before.
Now we need to integrate, right.
So if we integrate between some pair of temperatures, T1 and T2, we do this and we have this quantity, d delta H of reaction with respect to temperature, dT, all this is at constant pressure, right.
Let's just be careful to specify that.
Well, so of course, this is then just delta H of reaction at T2 minus delta H of reaction at T1.
This is looking good.
This is what we want to get, is our heat of reaction at some new temperature T2.
If we already know it at some initial temperature, usually room temperature, T1.
And it's just given by the integral from T1 to T2 of delta Cp dT.
Now in general, this is as far as we can go, but at least if the temperature change isn't too big, for most materials that heat capacities doesn't depend very strongly on temperature, right.
If you take some material, and you measure its heat capacity, right, how much heat do I have to put into it in order to change its temperature by a degree, right.
And I make that measurement at room temperature.
And then maybe I raise the temperature to whatever, room temperature, maybe 20 degrees hotter than room temperature.
And I again say OK, now how much heat do I need to raise this thing's temperature by 1 degree?
The amount of heat I need to put in to do that is not very different from what it was at room temperature right.
In other words the heat capacity didn't change very much.
So, the result of that is that at least for modest temperature excursions, it may be the case that we can simplify this and simply say it's just the difference in heat capacities times the temperature change.
Again, we can't always be assured of that, but it's often the case.
Any questions on any of this so far?
OK, now what I want to do is just describe a little bit of how do you measure all this stuff, right?
So what we've done so far is just lay out in principle how we should describe heats of reaction, and of course assuming that we've got everything in tables we can always look them up, but sometimes you want to make new compounds, right.
Sometimes you'd like to know what the energetics are involved there, right, and so it's useful to be able to actually measure something.
So how do we do It?
This is what's called calorimetry.
So we're going to make measurements to determine heat of reaction values.
Actually, calorimetry is used for all sorts of things.
It can be used to determine, for example, the energetics of phase transitions.
Maybe you're not looking at a reaction, but you've got some new compound, and you're looking at it go from liquid to solid or to gas.
And you can see what the heat involved in a process like that is as well.
So what happens?
Well, the objective in the case of a reaction, we have reactants at some temperature going to products at that temperature, and that is our heat of reaction at that temperature.
That's how it's defined, right, at constant temperature.
OK?
What do we actually measure?
Well, what we can really do is we can put the reactants in some container, put the whole thing in an insulated place, you know, hold things in a big insulated box.
Run the reaction.
Maybe it produces heat.
If it does, then the whole thing will heat up, right?
The stuff that's inside the reacting volume, and whatever's right around it there inside the big box, it's all going to heat up.
So I'll start at some initial temperature, T1.
I won't end up at the same temperature.
It'll be hotter.
How much hotter?
Well, that depends on how much heat was produced.
So in principle, if I measure how much hotter, I can determine how much heat was produced, and from that, I should be able to calculate delta H at T1.
So let's think about how that's really working.
I've really got reactants at T1, plus a calorimeter at T1, and I'm going to end up with products -- well let's let me just write that's over.
In fact, let's start by just putting a little sketch of the whole thing up.
So here's my reacting stuff, maybe it's a liquid.
It's going to take place in there.
It's going to be a constant pressure, it might be open to the air, or even if it isn't, there might be plenty of room, and it's a liquid anyway, so the pressure isn't going to change significantly.
I want to measure the temperature.
And the whole thing is insulated right.
I don't want to have to deal with heat escaping to the outside environment in a way that might be difficult or complicated to measure or calculate.
OK, so this, what I've sketched here would be a constant pressure calorimeter.
There's a reaction.
Normally this is used for a reaction in the condensed phases and liquid usually.
So this is a constant pressure calorimeter.
It's set up to be well-insulated so it's adiabatic.
That's the set up.
We're going to run the reactants, the reaction.
The reactants are going to turn into products.
Let's look at what happens.
So what happens is my reactants at T1 plus calorimeter at T1 turn into products at T2 plus my calorimeter at T2, right.
That's my process.
That's what really happens.
Now that, the enthalpy of that process isn't what I want, right, because the temperature has changed, also the calorimeter is heated up.
So I need to relate this to what I do want.
Let's label this one.
It is adiabatic.
It is taking place inside this thing, and it's a constant pressure, and we'll do it reversibly, right.
So that's what we've got.
Now, we can always take these things the products and the calorimeter at temperature T2 and cool them or warm them to get to products plus calorimeter at T1, right.
Let's label that two.
It's still at constant pressure.
It's reversible.
It's a temperature change.
Now to make that happen, it's not adiabatic, right.
If I wanted to do that, I'd need a heating element or something to cool, so I could make that temperature change happen, right.
Well, so now I can complete the cycle.
I've got reactants and calorimeter at T1.
Fewer products and calorimeter at T1, right.
Calorimeter doesn't change in this process.
So here I've got some delta H associated with changing the temperature.
This delta H though, this is what I want.
This is delta H of reaction at T1, right?
It's isothermal, constant pressure, reversible.
Just what I want.
That's how I've defined delta H of reaction.
This is what I'm after.
All right, so now let's see how we execute it, and do the calculations that allow us to calculate this.
So, one, what's delta H in step one?
Yes, exactly, it's adiabatic, right constant pressure.
it's zero.
Delta H1 is zero, right.
What we're really going to do in practice is we're going to measure, we're going to use our thermometer and say great, how much did the temperature change, right.
So now, then we're going to use what we're going to learn from step two in order to calculate this part, what we could call step three.
OK, so two, that's the crucial part.
It's constant pressure.
It's just given by the corresponding heat, and it's just a temperature change, right.
So we know how to do this, integral from well it's T1 to T2, depending on which way we go of Cp for the whole system dT.
OK?
It's just how much heat is involved when we change the temperature.
Now, the products have some heat capacity associated with them right, it takes a certain amount of heat if we make their temperature change, to either put it in or take it away, depending on which direction the temperature is changing.
Same with the calorimeter, OK.
In actual practice though, the calorimeter, is going to be this big hunk of metal, and really what's going to, there's going to be some stuff in here, right, they'll be a fluid, it will usually be some sort of oil.
There's a pretty big thermal mass.
Now we run the reaction and it produces heat.
Most of the heat is going to warm up or cool off all that oil and stuff.
There's a moderate amount of material in the actual reaction.
You want to design the calorimeter to fulfill that condition, Of course, you can't make it so enormous that even for, that for any ordinary reaction there's not even a measurable temperature change, right, because if you have your enough oil to fill up this room, it would take a huge amount of heat to change it by even a tiny, it's temperature but even a tiny amount.
So the calorimeter is designed sort of to scale to match the maybe the ordinary volume of reactants that you'll put in there, such that, pretty much all the heat's just going to heat up the calorimeter, and there's only a small amount that's going toward heating up the products, right.
So in general, or at least usually it'll be the case, that this is approximately equal to integral from T1 to T2 Cp of just the calorimeter dT.
And that generally is just given by the heat capacity the calorimeter times delta T, right.. Because the heat capacity of the calorimeter just like this thing is not strongly temperature dependent, OK.
So the point is we can calculate delta H associated with this process pretty readily, OK. That leaves three, right.
But delta H for step three must just be the opposite of this delta H because this was zero and we know that were going around a cycle, right.
And that's our heat of reaction.
So we're going to be able to do this, right?
OK, now this is one way to do calorimetry, and it's practical as long as all the materials are in condensed phases, solids and liquids.
If gases are involved it can work, but it can be tricky to keep the pressure constant, right, because that means that now the moles of gas might be changing, and that means in some way the volume has to be adjustable.
You'd need some sort of balloon or membrane that will allow that to happen.
Could be done, but easier is to just do the whole thing at constant volume, right, and just run the reaction that way and redo the calculation to be a constant volume rather than constant pressure calorimeter, right.
And it's not hard to do that.
So let's just look at what happens in that case.
It's almost the same.
So let's see, just to finish the job here though, all this says is delta H of reaction is just negative Cp for the calorimeter times delta T, all right.
so that's what we think we know in constant pressure calorimetry.
Yes, and if we have gases involved, it's pretty similar, but now what will have is something like this.
We'll have a reaction vessel that's sealed, it's constant volume.
That'll be inside our calorimeter.
It's insulated, and there's still a thermometer, so we can measure the temperature.
So this is still adiabatic.
It's insulated, but now it's constant volume, OK. Now, you know with constant volume, now it's not going to be delta H that's straightforward to measure, it's going to be dealt u, all right.
But it's going to be almost the same, right?
So let's just think about what happens here.
Now it's the cycle that we've got will still basically look the same.
That is its reactants at T1 plus calorimeter at T1, going to products at T2 plus calorimeter at T2, right?
And it's still adiabatic, but now it's constant volume.
And it's also reversal right.
So this is our process one.
Now, process two, we're going to end up here with our products, again at T1 plus our calorimeter at T1, right.
So now we have a constant volume reversible temperature change.
And so this part now isn't exactly what we want to determine delta H, because it's a isothermal constant volume, reversible products or reversible process that takes a reactant to T1 to a product of T1.
What these are going to give us are delta u values.
This'll give us delta u of reaction, right, T1, right.
It's a constant volume.
All right, so in the end, we're going to determine delta u here, and then in the end, we're going to have to relate that to delta H, but that's straight forward enough to do.
So let's look at how we'll analyze what happens.
So one adiabatic, constant volume process, right.
What's zero in that case?
In this case delta H was zero in the constant pressure example.
Now we've got a constant volume process.
What's zero?
It's u, because u is to q plus w right, heat and work, but it's adiabatic.
So there's no heat, exchange with the environment, and it's constant volume, so there's no p dV work, right.
So q is zero adiabatic.
Work is zero.
Delta u1 is zero.
And again just in the constant pressure case, what happens that we're going to measure in step one, which is to say in actually running the reaction is the temperature's going to change.
Right?
The whole thing's going to come to some new equilibrium temperature between the products and the oil or whatever's around it, and we're going to measure that.
OK, two, now it's a temperature change, right?
We know how to calculate delta u for a temperature change.
It's very similar here, but what's going to be different from the case with constant pressure?
There's a real important detail that's different if you want to calculate what happens at constant pressure when you change the temperature?
What happens at constant volume?
Looking at this expression, what's got to be different?
Cv, right.
We're not going to have the constant pressure heat capacity, we're going to have the constant volume heat capacity, right.
So delta u for step two, that's our q under constant volume conditions, integral from T1 to T2 Cv for our whole system dT.
Integral from T2 to T2, and now I can separate again the calorimeter from the product, and at least approximately, again, it'll be Cv for the calorimeter dT, which is to say then it's Cv of the calorimeter times delta T. Great.
Three, well delta u1 was zero.
Delta u2 was there Cv delta T, so now all we need since the whole thing is going around in a cycle, Since we know delta u3 must be negative delta u2, and that is our delta u of reaction, right?
So delta u of reaction is approximately equal to negative Cv for the calorimeter times delta T. So, this is what we're going to measure, and this is what we're going to determine.
And now we're almost done, except what we really want is delta H and not delta V, right.
So now we're going to use the fact that H is u plus pV.
Delta H is delta u plus delta pV, and now this is all at constant temperature in the end, right?
We've determined delta u for some temperature T1, right.
So what happens then we're going to use the ideal gas law.
So it's approximately delta u plus delta nRT.
That's a constant.
That's a constant.
So it's delta u plus RT, we can say T1 is the temperature we've used here, delta n of the gas.
In other words, what matters here in changing the pressure volume product?
What matters is we turned some reactants into some products.
How many moles of gas are there in each case, in reactants and products?
If that changes, of course you know that the pressure in there is going to change at constant volume if the amount of gas in there is changing.
And nothing else is going to make a significant contribution to it, OK?
So finally, then delta H of reaction for our temperature T1 is approximately minus Cv of the calorimeter times delta T plus R T1 delta n of gas.
Change of the number of moles of gas, right.
We're going to measure that, and now we're going to determine this.
In practice, we'll already know the heat capacity of our calorimeter, when we buy it, right?
So we don't really need to put in a certain amount of heat and change the temperature of the products and the calorimeter and so on.
What we need to do is just measure how much the temperature changed, OK.
It's worth getting just sort of roughly calibrated how big is this compared to the rest of this stuff.
That is, how different is delta u from delta H, all right.
And of course it's straightforward to do this, and I've written this out in the notes, so I won't re-write the numbers here.
But I gave an example looking at combining HCl and oxygen right.
So 4 HCl plus oxygen gas.
This two in the gas going to water and the liquid, plus chlorine gas at room temperature.
Well, what you find out is delta u of reaction is minus 195 kilojoules for the reaction as written.
And it turns out, as written, if you say OK, how much, what changed for the pV product?
Well here we've got four moles of gas, five moles of gas.
Here just two, so we changed the number of moles of gas by three.
All right, how much did it matter, right?
Well it matters.
It's measurable.
So now delta H of reaction turns out to be minus 202 kilojoules.
So, you know, seven out of a couple of a hundred, right, a few percent, kind of typical.
In other words, if you look at energetics of ordinary reactions where you're, you know, you're making and breaking covalent bonds, there's a fair amount of energy stored in those, right?
The additional change due to changing pressure volume is certainly measurable.
You don't want to just ignore it, but in cases like that, it's usually a small fraction of the total.
A few percent is typical, okay.
All right, let me just go through one numerical example of a calorimetry calculation, OK.
I won't put all the numbers up on the board because our time is running short, but I just want to outline it.
All I really want to do is calibrate you a little bit for what happens with ordinary calorimeter heat capacities that makes the calculation turn out to be relatively easy, right.
So let's just write it out.
Let's take iron sulfide as a solid, plus 11 halves oxygen gas to make iron oxide, also a solid, plus sulphur dioxide gas.
All right.
We'll start at T1 is 298 Kelvin.
T2, maybe we don't know it yet, right.
OK, so we know how to calculate what's going to happen, delta H and delta u, because we can look up the heats of formation and so forth of all the compounds, right.
So if we do that, what we discover is that delta H of formation whoops, something's wrong here.
This is sulphur -- what am I doing, boy, S and it's a two, sorry, and the heat of formation is minus 180 kilojoules per mole, that's oxygen, it's zero.
Boy, I'm, I don't know what is about this reaction that's vexing me but it's not that complicated.
Here it's minus 824 kilojoules per mole minus 297 kilojoules per mole, right.
So that's our input thermodynamic data.
So first of all, let's just do a heat of reaction calculation, right.
Taking the product minus the reaction, right.
So it's minus 824 plus 4 times minus 297 minus 2 times minus 180.
Right?
In other words I've got the stoichiometric coefficients in there and the values, and I'm subtracting the reactants from products wind up with minus 1652 kilojoules per mole.
Well, it depends on what we write, what we consider mole, right, maybe I should just write kilojoules as written.
OK, I'm going to skip the delta and get the change in moles of gas calculation.
It's straightforward to do.
What I really want to do is just give an example of what happens when you throw the thing, the material into a calorimeter and see how much the temperature changes.
So let's imagine we start with 0.1 moles of our iron sulfide, and then we have a stoichiometric amount of oxygen, and the whole thing is done in a constant volume calorimeter, and we see what happens.
Now the crucial element is what is the heat capacity of the calorimeter, right?
And it's, again it's a macroscopic pretty big thing, so typical might be 10 kilojoules per Kelvin, and that's pretty big, right?
Noticed that's not per mole, right.
I mean the calorimeter is a big thing filled the little oil or whatever is inside it, right?
And it's for that whole unit that you've got some heat capacity.
How much heat does it take the warm the entire thing up or the insides of the thing up by a degree?
It's that number right.
Ordinary heat capacities are in Joule's per Kelvin mole, not kilojoules, right.
And what that's telling you is probably your reactants and products, so the amount of heat that's involved in changing the air temperature is going to be negligible compared to what happens to the whole calorimeter.
OK?
So now if we say, OK, we we've done this calculation starting with 2 moles of this, but now we're going to be at 0.1 mole, so we're going to need to divide by 20, right.
So instead of minus 1652, it's going to turn out delta u is minus 1648 kilojoules, so, and then divide by twenty, we end up with minus 82.4 kilojoules, that is, that's the delta u of reaction for what happens if you'd put in not two moles of this, but 0.1 mole of this, right.
Practical amounts is the reason I'm using numbers like this, right.
So now, what's delta T?
Well here's Cv, right, so delta T is just our minus 10 kilojoules per degree oh sorry, it's our minus 82.4 kilojoules.
That's the heat released, divided by minus 10 or 10 kilojoules per Kelvin, right.
It's 8.2 Kelvin.
In other words, how much does the temperature of the whole thing change when you put an ordinary amount of material in there and run a reaction, right.
Well, what do you do?
You calculate how much heat is released in the reaction.
And then what's going to matter is what's the heat capacity of the whole, of the calorimeter?
I didn't even need to know that heat capacity of the product, right.
Because it's effect the thermal mass of the product is negligible compared to the thermal mass of the calorimeter.
Now in real practice, I'll do the calculation in reverse.
I'll measure how much the temperature changed in the calorimeter, right.
I'll know the heat capacity, and what I'll really be calculating is OK, how much heat must have been released in the reaction to make that temperature change happen?
And that in the case of constant volume, in this case that's my delta u, and then I'll add my little delta n term to get delta H. Any questions on calorimetry?
OK. See you Friday.
We'll finish on calorimetry and thermochemistry and then we'll start in on one of the really most difficult topics that we'll deal with all semester, which is a second law and our special function that we've seen just a little bit so far.
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So last time we just about finished up the discussion of thermochemisty, and we went through a description of calorimetry, how we actually make measurements of thermodynamic a quantities.
I should mention, which I didn't last time, that you know a calorimeter, apart from measuring heats of formation also is used to measure standard thermodynamic quantities like heat capacities.
You know how much heat does it take to raise the temperature of something by a degree?
Well, you put the material inside the calorimeter, and you have a heater and it's connected up to current source, and you can accurately measure how much heat you're putting in, because you can measure how much current is going into it, and you separately are measuring the temperature of the material, so it's a straightforward way of measuring the heat capacities, thermodynamic data of this sort.
So work on phase transitions, just ordinary properties of material.
Chemical reactions, all the thermochemistry involved is done using constant pressure or constant volume calorimetry in a pretty routine way.
So we talked about how to use the results of measurements like that to calculate reaction energetics, enthalpies of reaction.
And I just want to end the unit on thermochemistry with a real brief discussion of an approximate way of formulating heats of reaction, and that is in terms of bond energies.
Now you know you're probably never going to sit down and do a series of thermodynamic calculations using bond energies.
You could always look up heats of formation of compounds and from those determine very accurately heats of reaction, by looking at the heats of formation of products and the heats of formation of the reactants and subtracting them.
And since that data is available for a really wide range of materials, that's sort of the simplest approach and the most accurate one to determine reaction energetics.
On the other hand, if you just want to think about molecules, you know, you think about bonds.
You know if you say I've got methane or ethane, I'm going to use a fuel and burn it and capture the energy.
How much energy do we have in there?
Well the way you want to be thinking about that is all right, I'm going to take the starting material, the fuel, I'm going to break bonds, and I'm going to form some final product with new bonds.
Roughly what kinds of energetics are we talking about?
In biological systems, it's also super important.
You know you think about ordinary biochemical energetics, what's used is fuel in biology and what are the products that are formed?
And again, it's just super important to be able to think sort of semi-quantitatively in terms of bond energies, because you don't want to always be going and looking up everything in tables.
If you need to do an extensive series of calculations, it's very useful to have the tables.
But if you just want to think about how something is working, you know biological or system or an engine or another application, it's very useful to just be able to think in terms of bond energies.
So let's just go over bond energies.
And you've seen this in 5.111 and 2, so I'm not going to belabor it but they're just a little bit of examples that I want to go through, just so it's fresh on your mind.
So the idea, of course, is to be able to have an idea of how much energy is involved in forming or breaking a single bond.
So, if we just do this by example, we can take methane so it's a real simple case because we know it's just got four identical carbon hydrogen bonds, and we could say OK, let's just break them.
So we'll start with methane gas and go to one atom of carbon in the gas phase, plus four atoms of hydrogen in the gas phase.
The energetics in this case aren't given by the usual heats of formation.
You'd just say, if I want to know about methane, I can look up its heat of formation from the elements in their standards states.
but the difference here is these elements aren't in their standard states.
Carbon in its standard state at room temperature and pressure is graphite, a solid, it's not carbon atoms in the gas phase, and hydrogen in the standard state is hydrogen molecules H2, not individual atoms.
All right, so we'd like to have this bond energy for these things.
So let's just to make a cycle that will allow us to determine this.
So this'll be our first step in that cycle, or one way of getting there.
And this delta H of reaction is going to be by definition four times our bond energy for a carbon hydrogen bond.
Now let's go to be elements that are in their standards states at room temperature and pressure.
So let's go to carbon as graphite.
And let's go to two molecules of hydrogen gas, so this will be step two and this will be step three.
And this will allow us to work because now we're starting in standard states, so we can just use regular heats of formation.
And that's the key.
So if we look at step one delta H1 there as we've defined it, four times our bond energy.
Two, this is just negative heat of formation of methane, which we can look up.
And three, is similar.
Essentially it's the heat of formation of carbon in the gas phase and hydrogen atoms in the gas phase, commonly called heats of atomization, but these are also tabulated just like heats of formation of ordinary compounds.
In fact, in the appendix to your book that has thermodynamic data, you can find carbon gas phase atoms including heat of formation, and same with individual hydrogen atom as opposed to hydrogen molecules in the gas phase.
So those data are available.
So we can write delta Hc from atomization, and two times delta H of H2.
This is the way this is tabulated and this is molar.
And of course now we've written a cycle.
So we know that to get from here to here is the same as to go through these two steps.
In other words delta H1 is the sum of delta H2 and delta H3.
And so this says that our four times the carbon hydrogen bond energy is just given by minus the heat of formation of methane plus the heat of atomization of carbon, plus the two times the heat of atomization for hydrogen molecules.
And if we look up the numbers that are tabulated, what we discover is that our carbon hydrogen bond energy is 416.2 kilojoules per mole.
Very useful to have numbers like that stored in your head.
To have some basic idea of just ordinary chemical energetics in terms of bond energies.
Hydrogen bonds, you know roughly 20 kilojoules per mole, and so forth.
Just having those kinds of energetic enables you to think in a qualitative way about what's likely to happen in an ordinary chemical or biochemical situation.
OK, so this is in a sense the simplest case, because we've talked about the dissociation of a molecule that consists of all identical bonds.
But it's straightforward to go from there to additional bond energies.
So from this we've got a value for a sort of typical carbon hydrogen bond, right.
We can do the same thing now for ethane, and that's got a single carbon carbon bond, as well as a bunch of carbon hydrogen bonds, but now the we know the carbon hydrogen bond energy, we can use that number plus the relevant heats of formation to determine the carbon carbon bond energy.
And we can keep going and extend this to essentially tabulate bond energies for a very large number of different sorts of bonds.
So just as an example, ethane, so now we're going to have -- this I'll write it all out just some we can keep proper track of all the bonds that we're going to be breaking.
So C2H6 goes to two carbon gas phase atoms, and six hydrogen atoms.
So in this case, our, I can write this again as one delta H1 is six times the carbon hydrogen bond energy plus the carbon carbon bond energy that we'd like to determine.
And again we can go through now a similar cycle.
So again this now can be equilibrated with carbon graphite plus three molecules of hydrogen in the gas phase, just like before.
And so from that we can see that we'll end up with an equality between these, and negative delta H naught formation for C2H6 or ethane, plus two delta H of carbon atomization, plus three delta H for atomization hydrogen molecule.
So we've already seen we could look up these numbers and this, and we've already determined a carbon hydrogen bond energy.
So again, if we find all the relevant values, we can determine that our carbon carbon bond energy is 342 kilojoules.
And so on.
So we can keep going and again build up a bigger and bigger inventory of bond energies just to provide us with the kind of intuition that we'd like to have to be able to think about ordinary molecular energetics.
Any questions about bond energies?
All right.
Of course, if we know bond energies, we also could make estimates of heats of formation.
So if we have some new compound, we don't know its heat of formation, we know its structure.
We know what bonds are involved.
We could get an estimate for what we might expect for its heat of formation if we know those, and I think I'll just write this out on the board by example, but not work through the numbers explicitly.
You know, let's say we were going to make, you know, n pentane C5H12.
And of course again keeping proper track of all the carbon carbon and carbon hydrogen bonds.
But now in principle we know the carbon carbon bond energies and the carbon hydrogen bond energies.
So we should be able to figure out the heat of formation of a compound like this, even if we don't know it from a table.
And it's again through a similar sort of cycle so if we start with the atoms in the gas phase, and go to n pentane.
So this is now the bond energies which we basically know.
And again make our cycle of involving the elements in their stable form at standard temperature and pressure.
Then again, here's our one, two, three.
Here's our minus delta H of atomization, and here's our heat of formation.
And that's the point, is that now we can use what we've already elaborated for the bond energies and the known enthalpies of atomization, and we could determine the heat of formation.
If we do that, we wind up with a value of minus 152.6 kilojoules.
If we measure it.
If we look up in a table what the actual value is, we discover it's minus 146.4 kilojoules.
So it's a few percent off.
Why is it off?
Why isn't it right?
[UNINTELLIGIBLE]
Yes, exactly.
You know you can't use the C-H bonds from methane for every C-H bond, or the C-C bond in ethane for every carbon carbon single bond.
There'll be variations, usually of modest magnitude, in the values.
So this isn't going to be an exact calculation, but certainly this is close enough for allowing us to think qualitatively and semi-quantitatively about ordinary energetics.
Another illustration of it is if we look at neopentane -- so same chemical formula.
Same number of carbon carbon and carbon hydrogen bonds.
So according to the way we've done this calculation, we should get exactly the same number, because all we're doing is adding up the total number of bonds, and it's the same.
But what we discover is that for neopentane, delta H of formation, neopentane is minus 166.1 kilojoules.
So again it's different in this case in the other direction.
But certainly close enough that allows us to think reasonably about what might happen in ordinary situations.
Any questions about bond energies?
OK, now we're going to go to a new topic, and let me say, the topic we're going to start now, the second law, is really in a sense, this is the start of the whole rest of the term.
So far, what we've covered essentially is conservation of energy, how you think about energetics and what contributes to them.
That is heat and work.
How to account for them all properly.
What happens under constant volume or constant pressure conditions that led us to the definition of enthalpy right and so forth.
And once you get your arms around these ideas, and it takes some work to do that, but once you do, the actual execution of things can be pretty straightforward, and some of this can seem like just you know a lot of accounting.
It's not.
It takes some effort to become accustomed to it and familiar with how to properly think about heat and work and how they contribute to energy and so forth.
But given that amount of effort, a lot of this can be reduced to fairly straightforward calculational stuff.
What we're about to start is a little different from that in some sense, and really in some ways more difficult because what we're going to try to understand starting with this next set of topics is what determines whether something happens spontaneously?
We haven't really talked about that in any of this.
We've talked about whether reactions are endothermic or exothermic.
But as you're going to see that doesn't determine what happens spontaneously.
It certainly is one of the factors, but by no means the only one.
What determines what happens just by itself?
Or another way of saying that is, if I have some sort of system, it could be an engine.
It could be a chemical reaction.
It could be a biochemical system or some part of it.
What determines where the equilibrium is and the direction that things would have to go in order to reach or approach the equilibrium?
What is the equilibrium state of something?
Actually, what we've discussed so far really doesn't allow us to determine that, and there are a few easy ways to sort of bring that out.
One is you can imagine constructing some kind of cyclic process that you know it could remove heat like a refrigerator does from a cool region and export the heat to a hotter region, like the room, or outside a window if it's an air conditioner or something like that.
Just move the heat.
That's it.
Why not?
I mean you could do that and not necessarily violate the first law.
Or what about, here's a simpler case.
What if I just have a chamber with a divider, and I pump out half of it, and I put gas in this half, so there's gas and there's vacuum, and then I remove the divider.
So I certainly expect that what happens is the gas fills the volume.
Why?
The first law didn't tell me that that needs to happen.
Why doesn't it go back the other way, anyway, even without the barrier?
Why doesn't the gas just by itself all wind up over here?
Let's say it's a pretty dilute gas, so even if it does all end up in this half, you know the energy of any sort of unfavorable repulsions between molecules is negligible.
Maybe they even have weak attractions?
Why doesn't that happened?
It's a good thing for us that it doesn't happen.
Why in this room doesn't the air all just sort of collect over there somewhere?
Or just the oxygen molecules in the air and suffocate all of us.
The first law doesn't have anything to say about that actually.
Whatever the reason it doesn't happen, we're not going to find the answer to that in the first law.
So that's what the second law of thermodynamics and the development of entropy is all about.
It's to help us understand what is the direction of spontaneous change?
What governs that, and what's the equilibrium state that is reached when the spontaneous change occurs?
OK?
So the second law is going to give us basically a principle that'll tell us the direction of spontaneous change.
One way of thinking about that is it's going to allow it to sort of determine the direction of time.
You know if we see initial or if we see one state and we see another state, we'll be able or know, time must have been going this way, right because the spontaneous change would happen in this direction.
So, I'm going to put up a couple of just statements of the second law of thermodynamics.
I'll start with kind of an easy one.
So maybe I'll start with that here.
Here's a statement by Clausius.
The first law says that the energy of the universe is constant.
We're always conserving in one way or another.
And the second law is that the entropy of the universe is increasing.
Now we're going to define entropy presently.
But I'm writing this up just to give you some foreshadowing of how we're going to try to guide our thinking about spontaneous processes.
This tells us about conservation of energy.
This tells us about which direction things move in spontaneously, namely the direction that leads to an increase in this soon to be defined quantity entropy.
It's going to turn out to be a state function, and it'll help us understand the direction of spontaneous change.
You know, like the first law the second law too was developed before there was a very well defined understanding of the molecular nature of matter.
We're going to talk about the second law and entropy in macroscopic terms.
Later on in the semester we'll also discuss them in microscopic terms.
We'll do a little bit of what's called statistical mechanics, of microscopic, basically discussing the microscopic origins of entropy in terms of disorder of molecules.
But really, the thermodynamics of entropy and the second law evolved because people were trying to build things.
It was the Industrial Revolution.
people were trying to build engines, steam engines, and other kinds of devices, refrigerators other things.
And they couldn't understand -- first of all, why couldn't you make it perfectly efficient?
You know, you'd have a steam engine.
There'd be a source of heat.
You could extract work.
Never seemed like you could get it all and turn it all into work.
Why not?
And if not, what were the limitations on it?
What was the best efficiency you could achieve?
So in a very practical way people were motivated to think very hard about these kinds of questions.
So let's just look at some of the things that brought out the need for this kind of consideration.
So let me start by drawing one of the impossibilities.
We could try to build a heat engine.
Where we've got some temperature T1 which is hot, and some amount of heat, q1 goes into our system, and our system runs as a cycle.
It goes around and around, because in any practical situation we're going to need to keep going and repeating the process.
Some amount of work comes out.
We'll call it minus w, because work is always defined as the work that's done by the environment on the system.
So if I'm drawing the arrow this way, then I want to write it as minus w. Now this would be just swell.
This is it.
That's the whole heat engine.
This is running in a cycle.
So we know that, delta u is zero.
So that would mean that, you know, the amount of work out would just equal the heat in every time around.
Never happens.
Can't make it happen.
The only way it works is if instead there's an additional part.
Somewhere heat is also lost to a cold reservoir.
In other words, you know I burned some fuel.
I consumed something to produce my hot source, and I'm going to then try to turn it into work, which I can, but only partly.
There's going to be heat loss somewhere to a colder region, and I can't avoid it.
And it was discovered that this was always the case.
So this can be built where, just to be clear on these.
q1 is a positive number, because again that's heat going from the environment to the system.
This is our system running in a cycle.
So q1 is positive, minus w is positive, that is work is being done on the environment by the engine.
And minus q2 is positive.
And T1 is greater than T2.
That I can build.
Here's another thing we could try to do.
It's sort of the opposite.
We could try to produce a refrigerator but not have to do any work.
So let's go the other way, in other words, let's go from T2 which is cold we're going to go in this direction now, which means q2 will be a positive number and here's our system running in a cycle.
And now here's minus q1 now is going to be our positive number, and we're going to remove heat from q2 and move it, or from T2 a cold region and move it to some place that's hotter.
Also a pretty nice idea.
And also obviously impossible.
It only works if I put work in there to make it happen, then yes I can take heat out of a cold body and move it up into a hotter reservoir of some sort.
So here's a refrigerator.
Again T1 is greater than T2, and in this case q2 is greater than zero.
I'm removing heat, minus q1 is greater than zero, that is heat is going up this way.
Work is greater than zero.
Has to be.
Of course the thing is running in a cycle.
delta u is zero.
So that means that work must be minus q1 plus q2, right?
And work is greater than zero, so that's saying minus q1 right that's a positive number, must be bigger than q2.
In other words, I am taking some heat away from the cold region.
The heat I'm dumping into the hotter region is bigger than the amount that I taking away here, because it's also got this amount.
That's why, you know, if it's a humid or warm summer afternoon and you think oh what I a great idea, I've got this little fridge in my room.
I'm going to open it up and cool the room off.
Turns out not to work too well.
Because of course what the fridge is doing, it's cold inside.
So it's a machine that's removing heat from the inside and dumping it outside.
There are coils in the back of the fridge, and if you touch them they're a little bit warm and so on.
The heat is being lost to the room.
How much heat?
Well, the work that goes in is also added.
So the amount of heat that goes out into the room is bigger than the amount of heat that you're taking away from the space.
That's bad.
So if you try opening the refrigerator door to cool off your room at the same time as you're dumping heat into your room, all you discover is the net effect is to heat up your room.
Because you know the amount of heat that the thing is removing from somewhere on the inside there and struggling to do that because you've left the door to the fridge open, right, is not as big as the amount out heat that's coming out in those coils on the back.
It's not as big because that heat has not only the heat that you're removing from the inside, but also this amount of work.
It's all conserved because delta u is zero.
The thing is running in a cycle.
Try it sometime.
You can verify this experimentally in a very simple way.
It's not a very effective way to work, and it wastes a little bit of energy, but could be worth it in the interest of knowledge and discovery.
Let me write an alternate statement of the second law of Clausius.
All spontaneous processes are irreversible.
When I open that panel in the container, I've got gas on one side and vacuum in the other and it expands in there, it happens spontaneously, and it's irreversible.
It's not just going to go back.
And now let me write a mathematical statement of the second law.
And we'll celebrate it's importance by using this color.
Going around in a cycle, if I integrate the quantity dq reversible over T, I get zero.
Remember this, this is our special function.
Let me just keep going a little more though.
Let me also write another version.
If I write dq for an irreversible path over T, right, less than zero.
It's not the same.
This is a state function.
This isn't.
Remember of course that the heat that's exchange between the system and the environment depends on the path.
If there is a reversible path, and you see how much heat was exchange with the environment, and you look at that divided by temperature integrate over the whole thing, you discover that as long as the path is reversible, it only depends on the states at the beginning and the end of the path.
In other words if you have various ways you can go from state one to two that are reversible, it's always the same.
We saw that for just a couple of examples, a couple lectures back.
Remember, we looked at these thermodynamic cycles and just said, well let's just calculate those.
And those were, those all involved reversible processes.
So, in effect, we looked at different paths to go from one state to another.
We had a cycle, and one path went to here, and the other path went back.
We discovered that, in fact, this one zero around the cycle or the way we expressed it at the time is the value of this was the same regardless of the way we got from one state to the other, whether we took a straightforward route or a more circuitous route, as long as they were reversible.
We're going to define dS as dq reversible over T, and since we're dealing with a state function delta S going from state one to two of dq reversible over T is S2 minus S1 state function.
And S as state function is the entropy, and of course the us, it will always be special.
All right, now, the second law, you know it's difficult to understand and to get your arms around, because you know, do I really calculate this?
I actually have to find a reversible path in order to do the calculation.
Normally, you know, when you calculate delta u and delta h and so forth, you have various ways to calculate them that can be pretty straightforward.
But here, you actually have to determine the heat exchanged between the system and the environment, and you know that does, in general, depend on the path.
But it turns out that this quantity, you know, the differential of that heat over the temperature, also depends on path unless it's a reversible pass.
Any reversible path will lead to the same results.
It doesn't depend on path among reversible paths.
What that means in practice is if we say OK, start here and end here and calculate the entropy, you actually have to sort of figure out a reversible way of getting from here to here so that you can do that calculation.
So it's really very different from ordinary functions that you've been dealing with.
A lot of people, you know, because the second law is difficult to understand, you know you can easily go online and see thousands of violations of the second law.
It's still easy to look, to go find claims of perpetual motion machines.
Some of them cost a lot of money.
You can buy them, see how they work.
But and also, all sorts of schemes for producing energy.
Go online and see, right.
You're just, they're just crazy ideas about getting energy from nothing.
Getting energy from the heat that's just already present in the air.
I mean there's heat in the air, all this motion, it must have energy.
We should be able to extract it and use it.
Turns out we can't do that.
It has to look something like this.
OK.
I want to look at some more statements of the second law.
To do that I just want to indicate one definition which is just to be clear what these things mean.
These are what I've called heat reservoirs or hot or cold bodies.
Here's what I mean to be more specific.
A heat reservoir is basically a very large thermal mass at the same temperature.
And the real idea is that you can bring heat out of it as much as you want and the temperature won't change.
It's an idealization.
right If you have something that's big enough, and it's at a constant temperature, of course you can take heat away, but it's just so enormous that you won't measurably change it's temperature.
In practice you might be able or approach that limit.
OK, now let's look at some more definitions or statements of the second law.
This is from Kelvin of Kelvin temperature scale fame.
It's impossible to convert heat into work with a system that operates in a cycle.
I'm going to write this all out.
I know you've got it in your notes too, but it's super important.
To convert heat into work with a system that operates in a cycle, without at the same time transferring some heat to a colder reservoir.
OK, a couple of things to notice.
One is, you know, in some sense the second law is kind of negative, right?
Must not have been done in the new age.
It tells us what we can't do.
And that really is in a sense what it's all about.
It's telling us what are our limitations of what can be accomplished.
Very important part of this statement -- in a cycle.
Of course any practical engine has to keep going, so there's going to be a cycle in some way or another, but it's important because, you know, there are some -- if I'm not operating in a cycle, I can convert heat into work just fine.
You know, I could have a heat source, got a candle, and I've got a piston, maybe I've got a weight on it.
I've got something stopping it.
And then I remove the stoppers and let it go, and because of the heat, there's expansion.
It pushes up, right, and it'll be up higher somewhere.
Worked!
Heat turned into work.
I don't have a cold body somewhere.
But I only did this once.
There was just one stroke of the piston.
If I want to continue it and run in a cycle, somehow I've got to have a place where the heat goes.
So this sort of diagram describes effectively the way the Kelvin statement works.
This works fine and like we illustrated before, it doesn't work fine if I don't have this part of it.
I won't be able to just take all the heat and convert it to work, like I, in principle, could do, if I don't need to run it in a cycle.
Here is yet another statement by Clausius And we'll end with that one.
Maybe I'll just read this one, so we can not go over the time.
So, it's impossible for any system to operate in a cycle that takes heat from a colder reservoir, transfers it to a hotter reservoir, without at the same time converting some work into heat.
In other words, it's a statement kind of similar to the Kelvin statement, but applying to the case of moving heat from a colder to a hotter region.
Can't do it, unless some work is also converted to the heat that gets released into the hot region.
I think we'll stop there.
Next time what we'll do is lay out a very well-defined kind of engine called Carnot cycle that allows us to very explicitly go through and say OK, let's just work out the steps of isothermal and adiabatic expansions and contractions.
In other words, how does the engine really work?
And just calculate it going around in a cycle and see exactly how all this is born out.
And we'll see that next time.
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Well, last time we started in on a discussion of entropy, a new topic, and I started out by writing out some somewhat verbose written descriptions that had been formulated that indicate certain limitations about things like the efficiency of a heat engine, and what can be done reversibly and so forth.
And these verbal descriptions lead to some pictures that I put up and I'll put up again about how you might try to accomplish something like run an engine or move heat from a colder to a warmer body.
And what some of the limitations are on things like that.
I'll show those again, but what I want to do mostly today is try to put a mathematical statement of the second law in place that corresponds to the verbal statements that we saw last time.
So, just to review what we saw before, we looked at a heat engine where there is some hot reservoir at some temperature, T1, and it's connected to an engine running in a cycle.
We could, write the work that we generate that comes out as negative w. And then there's a cold reservoir at some lower temperature T2.
So the way I've got things written here, q1 is positive.
Heat's flowing from the hot reservoir to the engine that runs in a cycle.
Work is coming out.
A positive amount of work is being generated and is coming out, so minus w is positive, and this minus q2 is positive, that is heat is also flowing into a cold reservoir.
And what we saw last time is that this was necessary to make things work, this part of it in particular.
You couldn't just run something successfully in a cycle and get work out of it, using the heat from the hot reservoir, without also converting some of the heat that came in to heat that would flow into a cold reservoir.
All the heat couldn't get successfully converted into work.
That would be desirable, but it's not possible.
And, similarly, if we try to run things backward and build a refrigerator, so now we have our cold reservoir, and we're going to remove heat from it, and pump it up into a hot reservoir.
And to do this, we saw that you have to put work in, right, you can't just remove heat from a cold reservoir, move it up to a hot reservoir, without doing some work to accomplish that.
So here, q2 is greater than zero.
The work is greater than zero, and minus q1 is greater than zero, that is heat is flowing this way.
So these are just the pictures that we saw last time.
And then we had some statements of the second law of thermodynamics that I won't re-write up here, but, for example, Clausius gave us the statement that it's impossible for a system to operate in a cycle the takes heat from a cold reservoir, transfers it to a hot reservoir, that is acts like a refrigerator, without at some, at the same time, converting some work into heat.
Work has to come in to make that happen.
And we had the similar statement by Kelvin about the heat engine that required that some heat gets dumped into a cold reservoir in the process of converting the heat from the hot reservoir into work.
Fine.
Now what I want to do is put up a specific example of the cycle that can be undertaken inside here in an engine, and we can just calculate from what you've already seen of thermodynamics.
What happens to the thermodynamics parameters, and see the results in terms of the parameters including entropy.
So let's try that.
So, the engine that I'm going to illustrate is called a Carnot engine.
And the cycle it's going to undertake is called a Carnot cycle, and it works the following way: we're going to do pressure volume work.
So this is something that, by now, you're pretty familiar with.
So we're going to start at one, go to two and this is going to be in isotherm at temperature T1, and all the paths here are going to be reversible.
So that's our first step.
Then we're going to have an adiabatic expansion.
So this is an isothermal expansion.
Here comes an adiabatic expansion, to point three.
So at this point the temperature will change.
Then we're going to have another isothermal step, a compression to some point four.
So this is an isotherm at some different temperature T2, a cooler temperature, because this was an expansion.
We know in an adiabatic expansion the system's going to cool.
And now we're going to have another adiabatic step, an adiabatic compression.
And that's it.
That's going to take us back to our starting point.
So that's our cycle.
Of course there are lots of ways we can execute the cycle, but this is a simple one, and these are steps that we're all familiar with at this point.
So this is the picture, and this is a cycle that undertakes it.
So let's just look step by step at what happens.
So going from one to two, it's an isothermal expansion a T1, so delta u is q1, we'll call it, plus w1, for the first step.
Going from two to three, that's an adiabatic expansion, so q is equal to zero in that step.
And delta u is equal to what we'll call w1 prime.
So we'll use w1 and w1 prime to describe the work involved.
The work input during the expansion steps.
Step three to four isothermal compression, delta u is going to be q2 plus w2.
And finally, we're going to go back to one.
We're going to close the cycle with another adiabatic step.
q is zero.
Delta u is w2 prime.
Now, the total amount of the work that we can get out is just given by the area inside this curve.
We'll call it capital W. It's just minus the sum of all these things, because of course these are just defined as the work done by the environment to the system.
So it's minus w1 plus w1 prime plus w2 plus w2 prime.
The heat input is just q1, and we'll define that as capital Q.
And I just want to write those because what I really want to get at is what's the efficiency of the whole thing?
And in practical terms, we can define the efficiency as the ratio of the heat in to the work out.
We would like that to be as high as possible.
So it's just capital W over capital Q, which is to say it's minus all that stuff over q1.
Now the first law is going to hold in all of these steps, and we're going around in a cycle.
So what does that tell us about a state function like u?
What's delta u going around the whole thing?
I want a chorus in the answer here.
What's delta u?
Zero
Excellent.
MIT students, yes!
OK, so the first law, we know that the, going around the cycle the integral of delta u is zero.
And that has to equal q plus w, summed up for all the steps.
Which is to say q1 plus q2 is equal to minus w1 plus w1 prime plus w2 plus w2 prime.
So that means we can rewrite this efficiency.
We can replace this sum by just the some of q1 plus q2.
So sufficiency is q1 plus q2 over q1.
Or we can write it as one plus q2 over q1.
Now that already tells us what we know, which is that the efficiency is going to be something less than zero, right?
Because we've got one and then we're adding q2 over q1 to it, but we've got negative q2 is a positive number.
Heat is flowing this way.
So q2, the heat in this way is negative.
q2 over q1 is negative.
q2 of course is positive.
That is heat flowing from the hot reservoir to the engine.
So that efficiency is something less than one, and we'd like to figure out what that is.
So let's just state that.
Well, now let's go back to each of our individual steps and look, based on what we know about how to evaluate the thermodynamic changes that take place here, let's look at each one of the steps and see what happens.
So from one to two it's isothermal.
And now we're going to specify, we're going to do a Carnot cycle for an ideal gas.
It's an ideal gas.
It's an isothermal change.
What's delta u?
You know, it's like lots of courses and all sorts of things that you're seeing, they're always the same.
What delta u?
Zero.
PROFESSOR NELSON?
Right.
Delta u is zero, and it's also equal to this.
So that says the q1 is the opposite of w1.
It's an isothermal expansion, so dw is just negative p dV.
So it's, this is just the integral from one to two of p dV.
And it's an ideal gas, isothermal, right.
RT over V. Were going to make it for a mole of gas, so it's R times T1, and then we'll have dV over V. So that's just the log of V2 over V1.
Let's have one mole, n equals one.
Okay two going to three that's this, it's adiabatic, right.
So delta u is just equal to the work but we also know what happens because the temperature is changing from T1 to T2.
So delta you is just Cv times T2 minus T1.
du is Cv dT.
It's an ideal gas, and that's equal to w1 prime.
Now, this is a reversible adiabatic path, so there's a relationship that I'm sure you'll remember.
Namely, T2 over T1 is equal to V2 over V3.
Don't be confused by the subscripts.
We're talking about quantities both starting at two and going to three.
So it's V2 and V3.
They're different volumes.
The temperature though at two is the same as it was at one, so it's still a temp at T1, and this temperatures is at T2.
Anyway T2 over T1 is equal to V2 over V3 to the power gamma minus one.
So we're going to kind of store that for the moment.
Now going from three to four right, so we have another isothermal process for an ideal gas, so I won't try to make you sing again so soon.
Delta u is zero.
And so just like here, now q2 is minus w2, that's integral going from three to four p dV.
So it's R T2, right, now we're at a lower temperature times the log of V4 over V3.
So that's our work in this path and heat.
And finally going from four back to one.
So we've already seen that q is zero.
So we know that delta u is just Cv times T1 minus T2.
Now we're going from T2 up to T1.
And this is equal to w2 prime.
And now, just like we had before, again we've got a reversible adiabatic path, so T1 over T2 has to equal V4 over V1, to the gamma minus one, okay.
All right, now, if this is the case, of course, this is just the inverse of this.
So this just must be the inverse of this.
We can combine these two to see that V4 over V1 must equal V3 over V2.
So now we have a relationship between the ratios of these volumes that are reached during these adiabatic paths.
Now, let's just look at our efficiency.
So we saw that efficiency is one plus q2 over q1.
Let's just use our expressions that we've found for q2 and q1.
We're going to have q2 over q1.
R is going to cancel.
We're going to have the ratio of temperatures and the ratio of these logs.
So it's T2 over T1 times the log of V4 over V3, over the log of V2 over V1, but we just arrived at this relationship between these volumes.
And this of course tells us that V4 over V3 is V1 over V2.
Right.
I'm just inverting these.
So here it is.
Here's the V4 over V3, oops, sorry.
V2 over V1, this is equal to the inverse of this.
So the ratio of the logs is just minus one.
So this is just, sorry, I forgot about the one here.
One plus this, but this is the same as one minus T2 over T1.
That's our expression for the efficiency.
All right, isn't that terrific?
Now, we have an expression.
We can figure out the efficiency.
All we need to know is the temperatures.
What's the temperature of the hot reservoir?
What's the temperature of the cold reservoir?
We're done.
That's the efficiency.
As we expected, it's less than one, and of course if we build an engine, we want it to be as high as possible, as close to one as possible.
What that means is we want to run the hot reservoir as hot as possible and the cold reservoir as cold as possible.
In principle, this value, this efficiency, can approach 1 as the low temperature approaches absolute zero.
If this were to be an absolute zero Kelvin, then we could we can have something, wait a minute.
Sorry, it's T2.
As T2 goes to zero, the cold reservoir, then this goes to zero and our efficiency approaches one.
So that would be the best we can do.
And that basically make sense, right?
The whole thing is being powered by sending heat from the hot to the cold reservoir.
The colder that cold reservoir is, the hotter that hot reservoir is, the better off we are.
The more work we can get out of it.
The closer the efficiency will get to one.
So in some sense, the first law would suggest you can sort of break even.
That is, it gives us the relationship between energy and work and heat.
It might suggest that you could convert all the work to heat.
The second law says that really, you can't do that.
The only way you could possibly contemplate it is to be working at absolute zero Kelvin.
Guess what the third law is going to tell us?
You can't get to zero Kelvin.
We'll see that shortly.
But this is those closest you could come at least by trying to do that to having your efficiency approach one.
Now, we've seen that q2 over q1 is equal to negative T2 over T1.
So let me just rewrite that as q1 over T1 is minus q2 over T2.
Or in other words, q1 over T1 plus q2 over T2 is zero.
But q1 and q2, each one of those is just the integrated amount of heat that was transferred going along a reversible constant temperature path.
Which means that q1 over T1, that's this delta S thing that we saw before.
And what we can say about this is it's saying that if we go around in a cycle and look at dq reversible over T it's zero, because that's what these quantities are.
Now, of course, we can run the engine backward and build a refrigerator, and if you've got your lecture notes from last period, your 8-9, well, they're labeled 8-9 lecture notes, I made an attempt to define something which was a little bit misguided.
And so instead of defining efficiency the way you've got it written there, I'm going to define what's called something different for a refrigerator which is called the coefficient of performance.
So it's the following: so we can define the coefficient of performance written as eta as q2 over minus w. In other words, you know, it's how much heat can we pull out of the cold reservoir, for whatever amount of work that we're going to put in in order to do that?
Well of course, we'd like that to be as big as possible.
But of course, this is just q2 over minus q1 plus q2.
Just to be clear, so it's heat extracted over the work in.
So let me just rewrite that as, I just want to divide by q1 everywhere.
So I'm going to write this as q2 over q1 over minus one plus q2 over q1.
I want to do that because we know that q2 over q1 is negative T2 over T1.
So this is negative T2 over T1 over negative one minus T2 over T1.
I can cancel those, and then I'm going to multiply through by T1.
So this is just going to be T2 over T1 minus T2, that's our coefficient of performance.
So it's a measure of how well we can do to run the refrigerator.
So you know, what the second law is doing, in words, it's putting these restrictions on how well or how effectively we can convert heat into work in the case of the engine, or work into heat extracted in the case of a refrigerator.
And it follows qualitatively from just your ordinary observations about the direction in which things go spontaneously, right.
You know the heat isn't going to flow from a cold body to a hot body without putting some work in to make that happen.
And so, we'll be able of follow that further and see really how to determine the direction of spontaneity for a whole set of processes, really in principle for any processes, went analyzed properly.
So the first step to doing that is I want to just generalize our results so far for a Carnot cycle.
You might think well okay, but this is a pretty specialized case.
We've formulated one particular kind of engine, and seen how we can analyze what it does, come up with relations that seem of value for efficiency and other quantities.
How general is it?
Well, let's take a look.
So, what I want to do is make a new engine which really just consists of two engines, side by side.
One is our Carnot engine as we've seen it, and the other is just any other reversible engine.
So generalize our Carnot engine results.
So let's take our hot reservoir and draw it bigger.
We know it has to big anyway, since we can extract heat from it without changing the temperature.
And on this side, we're going to write out an engine, and we're going to say this is a Carnot engine.
so on the side of it we have q1 prime.
We're going to have w prime.
And we're going to have q2 prime, and we're going to draw the arrows in the positive direction in these cases.
Or in the defined directions I should say.
Here is our colder reservoir.
OK, so here is just an engine like what we've already seen, and I'm going to specify that this is a Carnot engine which is to say all the results that we just derived hold for this case.
And side by side, we're going to run another engine.
So it's got q2, and it's got a cycle w. And it's got q1.
So this is some other engine that runs using reversible processes.
So I can define the efficiency of each one of them.
The efficiency for the one on the left is minus w over q1, The efficiency prime for the one on the right is minus w prime over q1 prime.
Now I want to just assume that in some way they're different.
So let's assume that epsilon prime is greater than epsilon.
So in other words, this engine is running less efficiently than my Carnot engine.
It's also reversible.
And now since it's reversible, we can run it forward or backward.
So we can run this one backward and we can use the work that comes out as the input, well sorry, use this work that comes out of the one of the right to run this one backward, which is to say we'll move heat from cold to hot.
So use work out of right-hand side to run left-hand backward.
Why not?
We need work to come in, we might as well get it from here.
So, the total work that we can get out must be zero, out of the whole sum of them.
After all, we're taking the output work that we get from the right, and using it all to drive the left.
So that means that minus w prime must equal w. And w is greater than zero.
that is in this one we have the environment doing work on the system.
OK, now we've assumed that epsilon prime is greater than epsilon.
So we can just write out what those are, minus w prime over q1 prime is greater than minus w over q1.
But we know that minus w prime is the same thing as w. So w over q1 prime is greater than minus w over q1.
Which is the same thing just to be a little bit maybe pedantic because w over minus q1, and the only reason I'm writing that is to illustrate that this says that q1 must be less than q1 prime.
So q1 is less than zero.
Remember this one's running backward, we're pumping heat up.
q1 prime is greater than zero, it's running as an engine.
It's taking heat from the hot reservoir.
Well, that's interesting.
That says that minus q1 prime plus q1 is greater than zero.
Well, it's a pretty interesting result.
There's no work being done on, out of the whole thing.
This is providing work that's being used in here, but if you take the whole outside of the surroundings and this whole thing is the system, no net work, these things cancel each other, and yet heat's going up.
How did that happen?
Well it happened because we clearly must have a faulty assumption underlying what we've just done.
This can't possibly be true.
This is impossible.
So what this says is the efficiency of any reversible engine has to be one minus T2 over T1.
It's not a result that's specific to the one cycle that we put up.
And you know, you could have a reversible engine with lots and lots of steps, but you could always break them down into some sequence of adiabatic and isothermal steps.
So you know your cycle, you know, you could have a whole complicated sequence on a p v diagram of steps going back.
As long as it's reversible, you know what the efficiency has to be, and in principle, you could break it down into a bunch of steps that you could formulate as isothermal and adiabatic.
They might have to be formulated as very small steps, in order to do that, but you could.
What this says too, is this result that we found, right, now of course that's our integral dS is zero.
It's general.
Entropy S is a state function, generally.
Now remember, we went through before how it's a state function but to calculate it, you'd need to find a reversible path, along which you can figure this out.
But the fact is it's a state function, in a general way.
Now, if we go back to our Carnot cycle which is a set of reversible paths, it's useful to compare this to what happens in an irreversible case.
In a real engine, of course, you can approach the reversible limit.
Every step won't be perfectly reversible.
And of course it's not hard to see what happens.
Let's just take our reversible engine and modify it a little bit.
Let's imagine that instead of all the steps being reversible, let's just put in one irreversible step.
Let's do this.
Instead of this reversible isothermal step, Let's make it an irreversible isothermal step.
We can have a different isothermal step.
Of course in the reversible case, you're always pushing against an external pressure, which is essentially equal to the internal pressure.
Let's not do that.
Let's imagine maybe instead we just immediately dropped the pressure and let the system expand against the lower pressure.
Now we know we're going to get less work out of it in that case.
You've seen that before.
And the work in this case is the area inside here.
The work in this step is just the area under this curve.
So in some way we're going to have a difference here between the irreversible case and the reversible one.
So if we do that, then what I want to do is just see what happens to dq over T, so in our irreversible engine, one to two, what we know for sure is that minus w in the irreversible step, that's the work out, extracted in that step, is going to be smaller than minus w in the reversible case.
So w irreversible is bigger than w reversible.
And of course, in either case, delta u is q plus w, so it's q irreversible plus w irreversible, but of course, u being a state function it's the same in either case.
So it's the same as q reversible plus w reversible.
And we've just seen that this is bigger than this, but the sums are equal.
So this has to be less than this.
q irreversible is less than q reversible.
In other words, and irreversible isothermal expansion takes less heat from the hot reservoir than a reversible one does.
It makes sense, right, because you know we got less work out and delta u is the same right, so it must be that less heat got transferred.
So the expansion against lower pressure draws less heat from the hot reservoir right.
So now let's look at the efficiency of our irreversible engine.
So it's one plus q2.
Now q2 was in this step, and we're going to leave that reversible, right, but q1 is irreversible.
And this has got to be less than one plus q2 reversible over q1 reversible, which is to say it's less than our efficiency in the reversible case.
Why?
This is smaller than this, but it's a negative number.
So this negative number has a bigger magnitude than this negative number.
We're subtracting them from one and they're less than one, so this is bigger than this.
And all we know about this is that it's really for some irreversible reversible step.
All we really needed to know about it is that this is smaller right.
So it's going to be the case for any irreversible engine.
And that's the point.
So it's that the irreversible efficiency is lower than the reversible one.
But, of course, since we saw that this is smaller than it was in the reversible case, we can also write that dq irreversible over T is less than dq reversible over T. Which is to say if we go around a cycle for dq irreversible over T, that's less than zero.
It's only in the reversible case that dq over T around a cycle is equal to zero.
So to write that in a general way, it's actually formulated by Clausius.
Going around in a cycle the integral of dq over T is less than or equal to zero.
Never greater.
OK, what we'll see shortly is that this will allow us to see that for an isolated system the entropy never decreases.
It only can go up.
And in fact any spontaneous process will make it go up.
Only in the case of reversible processes.
Of course, then you can see that this will be zero.
Anything else, which is to say any spontaneous process, it'll be less than zero.
We'll see that next time, and then we'll generalize in a broader sense to look at the direction in which spontaneous processes go.
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Over the last couple lectures through our consideration of a number of somewhat difficult topics, including these Carnot cycles and specific Carnot engine to represent them, we've tried to define and understand a little bit about this very special function, this entropy.
Where we've got our dS.
It's dq reversible over T. And our treatment of reversible and irreversible cyclic processes, what we saw is that if we go around a cycle and look at dq reversible over T, then this is zero.
Now go all the way around a cycle, of course, this is just delta S for the cycle.
And that's consistent with the idea that entropy is a state functions.
So if we go around cycle the cycle, we have the same ending point, the starting point, same state, then the change in the state function has to be zero.
But we saw that for an irreversible path around a cycle, then we have that, this gives something less than zero.
And so that was expressed in Clausius inequality that includes both these cases.
And now I just want to go through a number of calculations of entropy.
How entropy changes for a number processes, both to just learn more about it in general terms and also just to see how it's calculated and what sort of changes in undergoes for different sorts of changes in state.
So the first very general example that I want to show is the following.
Let's say we've got an isolated system, and it undergoes some irreversible change.
So we'll start at one, and we'll go irreversibly to some new state, two.
OK, now we can always return the system back to the initial state reversibly.
If we do that, then we can't keep the system isolated.
So it won't be isolated anymore.
In other words, in order to arrange things so that you could have a return along a reversible path, it can't be isolated.
In some sense you know that, because when it was isolated it spontaneously irreversibly went from one to two.
And example would be ice melting at some temperature that might be higher than the freezing point.
So you've got solid ice, but the temperature is above zero degrees Celsius, so irreversibly it'll melt.
Systems isolated, it'll melt.
Then you could put the system in contact with a cold reservoir reversibly re-freeze the water to form ice again.
But of course to do that you couldn't keep it isolated.
You need to have it in contact with a low temperature bath.
But again, this is just completely general at this point.
It's some irreversible process returned then along a reversible path.
So let's just look at what happens.
So here's our path B.
Along path A, well it's an isolated system.
So the heat exchanged is zero.
And then, for the cycle based on the Clausius inequality expressed here, the integral around the cycle of dq over T in general, it's less than or equal to zero.
If there was an irreversible process involved, then in fact, it'll be less than zero.
I'll write it here in the more general case, but we know what it's really going to be with the irreversible step.
Well, so we can write this as the integral of dq irreversible over T, going from state one to state two.
Oh, let me rewrite this down here.
So around the cycle, dq over T, it's integral from state one to two of dq irreversible over T, plus the integral from two back to one of dq reversible over T. And of course, we know that that's less than or equal to zero.
But this, we know is zero, right, because this irreversible step is taken in isolation.
The system is isolated.
We've already seen that there's no heat crossing the boundary of the system.
So we just have this.
So of course this dq reversible over T from two to one, that's, what now we've a reversible path, so that's just the entropy at the delta S going from two to one.
It's S1 minus S2 it's delta S, we can write it as delta S backwards, right, we're going back along that reversible path.
Which is of course negative delta S forward.
Because around the whole cycle delta s has to be zero.
So what that says is delta S forward is greater than or equal to zero.
That's interesting.
What that's saying is -- remember, we didn't specify the process.
It's just for any completely general irreversible process.
For any such process, for any spontaneous process, this is going to be the case.
So in other words, for an isolated system, delta S tells us the direction of spontaneous change.
In particular, for the isolated system, delta S is greater than zero for something that's irreversible.
Delta S equals zero for something that's reversible, and delta S is never less than zero.
In a sense, this is a direct consequence of the Clausius inequality.
That's a pretty useful result.
So remember when we started this whole discussion, I tried to emphasize that the first law of thermodynamics told us about conservation of energy.
But it didn't tell us which way something would go spontaneously if you just left the system to its own devices under certain conditions.
This is telling us that.
We can tell which way the system will evolve in which direction it'll go.
This is for an isolated system and we'll generalize this in subsequent lectures.
Now, let's look at what happens if the system isn't isolated.
So now let's go from one to two not isolated.
So of course, delta S I'm going to specify for the system for reasons that'll soon be obvious is S2 for the system minus S1 for the system.
Of course it doesn't depend on the path, even though to calculate it, we need to find a reversible path.
Just to add emphasis to what we just did here, of course, it was the same thing.
In order to calculate delta S along that irreversible path for that irreversible process we needed to construct some reversible path.
And what we did is imagined the reversible path going backwards and then said, OK, fine then delta S for the irreversible process must be the opposite of delta S that we calculated along the reversible bath going backwards.
The point is that to calculate delta S in either direction, we needed to construct some reversible path.
All right, and of course it'll be the same here.
Now, in this case since the system isn't isolated, that means heat's going to flow across the boundary.
So that means heat is going between, it's being exchanged between the system and the surroundings.
So what's happening to the surroundings?
Generally, we've worry a lot about the system, but not much in recent discussions about the surroundings.
But delta S surroundings must be something.
We could calculate it.
And that's depends on the path.
Not because it isn't the state function in the surroundings, of course it is, but because depending on the path, the final state of the surroundings will be different.
After all, depending on how much heat got exchanged, the surroundings got more or less of that heat.
Work might have been done on the surroundings.
So the surroundings are going to respond, are going to change also, when the system changes in this way and it's not isolated.
And how the surroundings change will depend on what path was taken.
And you've seen that before in calculating things like pressure, volume, work along different paths.
Where you calculated delta u for the system and saw that it was the same for different paths but q and w were not the same for the system, and therefore, of course, not for the surroundings either.
So let's just consider an irreversible path.
And in order to think about what happens to the surroundings, let's take the entire universe which consists of the system and the surroundings -- that is a big isolated system, because it's everything.
So it's not somehow in contact with another additional body or universe.
We can treat the entire universe as an isolated system.
So, we're going to think about the entire universe.
We can think about it as an isolated system.
Great.
And we know that delta S for an isolated system, for a spontaneous process, is greater than zero.
That's right there.
So delta S for the universe which is delta S for the system plus delta S for the surroundings, and I'll specify since we're considering an irreversible process, specify that, it's greater than zero.
The whole thing happened spontaneously, and the whole universe is an isolated system.
So if we wanted to, we can write that delta S irreversible for the surroundings must be bigger than minus delta S for the system in order that there sum be positive.
All right?
Now, if we consider a reversible process, then of course we already know what happens for an isolated system in a reversible process, delta S is zero.
So delta S in this case of the whole universe, which is delta S for the system plus delta S for the surroundings in the reversible case must be equal to zero.
In other words, delta S of the surroundings in the reversible case is exactly the opposite of delta S of the system.
Delta S of the system is the same in either case because reversible or irreversible, we're specifying the system, goes between the same states, but what happens to the surroundings is different in the two cases.
So one way of recasting this statement or the Clausius inequality is the entropy of the whole universe never decreases.
Of course, in some sense, that's obvious from this statement, since we're considering the universe an isolated system.
So the entropy in any process can, it can, the change in entropy can either be zero or positive and never negative.
So entropy off the whole universe never decreases.
That's actually a really quite profound conclusion.
All sorts of philosophical and other consequences of it that are sometimes considered in things like evolutionary biology and other things.
And how things can change altogether an entire system of things.
Now, I'd like to just go through a few calculations of delta S for common processes.
We've seen a few general features of entropy.
Let's calculate it for a few things.
So calculations of delta S. And we know the general procedure.
Whatever the process is, we're going to need to find reversible paths, and along a reversible path then, we'll be able to calculate it.
One, let's just take two pieces of material at different temperatures.
The entire system is going to be isolated.
And then let's connect them.
And then let's connect them.
Put some material between them.
So now heat can flow from one to the other.
And you know what's going to happen which is if they start out at unequal temperatures, and you wait a little while, eventually they're going to wind up at the same temperature.
In other words, you know the direction of spontaneous change.
So let's just see what happens to delta S. So isolated system, so there's no work being done.
There's no heat that's being exchanged.
So delta u is zero.
What's dS?
Well it's just the sum of the changes in entropy for the two sub-systems the two separate pieces.
So it's dq1 over T1 plus dq2 over T2, I think there's a type in your notes.
it's plus not minus of course.
But in this case, dq1 is just the opposite of dq2.
Because whatever heat is flowing into T1, into this block, is coming out of this block, or vice versa, depending on which one is hotter.
So this is just equal to dq1 times one over T1 minus one over T2, which is T2 minus T1 over T1 times T2.
So that's dS, and then we could integrate it to get the delta S, but this is sufficient for what I'd like to illustrate, which is just the sign of things.
We know dS must be greater than zero for the spontaneous change that's going to occur.
So let's just make sure that common sense prevails here.
That says if T2 is greater than T1, right T2 is hotter.
Then dq must be positive, so this is positive, this is positive, dS we know has to be positive.
So, it says T2 is hotter.
This being positive means heat is flowing in from this one to this one, from the hotter to the colder body.
So it looks right.
If T2 is lower than T1, if this one is colder, then in order for dS to be positive, which it has to be, this better be negative.
In other words, again heat is going to flow from the hotter toward the colder, into the colder body.
So, as promised, the condition that dS must be greater than zero, guides us and tells us the direction that spontaneous change is going to occur.
We can tell just from this condition which way things have to evolve.
All right, let's try another pretty simple process.
Let's just take a gas in some volume V and over here is going to be vacuum of equal volume, and we're going to remove the barrier.
You know what's going to happen spontaneously, right?
The gas is going to fill the available volume once it becomes available.
So this, we'll make this a Joule expansion, and we'll make it an ideal gas.
So the process is one mole of gas at our initial volume and some temperature, and we'll make this adiabatic, and this goes to one mole of gas at a new volume, 2V and at the same temperature.
All right, so there's no work done.
It's adiabatic.
There's no heat exchanged.
Delta u is zero.
Everything's zero.
Almost everything is zero.
What isn't zero?
Entropy
Entropy.
It's going to tell us which way this whole thing goes.
So this is an irreversible process.
In order to calculate delta S we need to construct a reversible path along which this can go.
So here's a way to do it.
We could compress it back, isothermally and reversibly.
Just like the example I gave, the general example, where I mentioned the ice melting.
In that case, it won't be adiabatic.
We'd have to put it in contact with a heat source of some sort.
So reversible process.
We could compress it back to volume V, isothermally, and reversibly.
Let me say, there's no -- we wouldn't have to consider this process in reverse.
Of course, we could also consider the forward process in the presence of some heat source with isn't isolated and do this, but let's just consider the compression.
So, OK, now we'll have the -- well, once again, delta u is zero.
It's an ideal gas.
There's no temperature change.
But this time q and w aren't going to be zero.
They're going to be some finite numbers that are opposites of each other.
Because we're no longer isolated.
So let's just write out delta S to go backward is dq reversible over T, and that's minus dw over T, not zero anymore.
And where we're going is from 2V to V, we're compressing, and it's an ideal gas.
So this is the same as p dV, but it's p dV over T which is the same thing as going from 2V to V. Now we're going to replace this with RT over V, and the T's will cancel.
So it's R dV over V going from 2V to one V. So it's just R times the log of 1/2.
So now we've just calculated the value of delta S by constructing this reversible path.
Now this is delta S backward.
To do the compression, so of course delta S forward is just the opposite of that.
So, delta S forward is R log 2.
Reassuringly, that's a positive number.
So this process that we considered originally, this irreversibly process in this isolated system, remember, for the irreversible expansion we considered the system isolated, Happened spontaneously, and of course you know that's the case.
Now, I'll just mention, and later on we'll go into this in considerably more detail, but I'll just mention at this stage through a little bit of sort of foreshadowing that everything we've discussed so far about entropy is with a macroscopic picture.
We started with heat engines and Carnot cycles and macroscopic processes, reversible and irreversible processes and so on.
But there's a, and that's certainly the way it was all formulated originally.
Part of the power of thermodynamics is that it doesn't depend on a specific microscopic model of matter.
That said though, we have a pretty good microscopic model of matter.
We know about atoms and molecules, and in fact there is an entirely microscopic formulation of entropy that has to do with disorder and the number of available states, microscopic states available to a system.
So I'll just state what we'll see in much more detail later on.
Which is in microscopic terms, it's going to turn out that the entropy of a system can be given by R over Avogadro's number times the log of the number of distinct microscopic states that are available to a system.
Right so in this isolated system, when we double that volume, each molecule suddenly has twice as many states available to it.
You could formulate that in a lot of ways.
You can sort of divide up the volume into tiny little molecule-sized cubes and realize well, now, there are twice as many of them.
Now if I have one molecule, the number of states doubles.
If I have n molecules, the number of states goes up by 2 to the n. Because for each one of the molecules, I can select any one of those states, and then the next one I can select any one.
So you have this enormous increase in the number of states.
So delta S in this case looks like R over NA log of 2 to the power NA, it's Avogadro's number, for a mole of molecules.
This can come out and cancel this.
There's our R log 2 result.
But the point I'm making here is that could actually be derived from entirely microscopic consideration.
We haven't gone through that yet.
We've done entirely macroscopic thermodynamics so far, but we'll do a little bit of treatment of statistical mechanics, is what is called the whole microscopic formulation of entropy in thermodynamics.
And that's a really important fundamental result.
And we'll get to it in much more detail in a while.
But for now, I just want to continue calculating entropy changes for a few more kinds of processes.
So, how about if we have two different substances and we mix them, entropy of mixing.
So, we start with some number of moles of substance A in some volume VA, and some other number of moles of substance B in volume VB, separated.
And then we remove the barrier, of course we know they're going to mix.
So we'll have some total number of moles, nA plus nB, and they'll be filling some total volume, which is the sum of the two original volumes.
Certainly this is what we expect to happen spontaneously.
So are process is nA moles of A, gas, initial volume VA and T, plus nB of substance B, I think this is miswritten as A there in your notes.
Gas VB and T goes to nA moles of A plus nB moles of B.
Gas, total volume and T. And we'll have constant temperature and pressure.
Well, there was no p v worked done.
The temperature didn't change and it's ideal gases, so delta u is zero, but what about delta S?
Well again, to calculate it, we need to find a reversible path.
This is irreversible if we just remove the barrier and let it all go.
But we could at least imagine a reversible path, and we could actually construct this for the right substances.
We could imagine that we can find some sort of piston here.
And we're going to prepare to push it inward, with a membrane that's permeable only to b.
And over here we could set up the exact same thing, but this one's permeable only to a.
And then we could just push them to recover this state.
And we can do it reversibly.
So we could construct a reversible process that does the reverse of this.
And in this case, it'll be isothermal still, constant pressure, reversible compression of the two gases.
So, reversible compression and demixing.
All right, so what happens?
Well, of course, our delta S of demixing is going to be minus delta S of mixing.
Delta u of demixing is still zero.
Temperature didn't change, ideal gases, and there's some reversible work and heat.
So dw in this reversible case, is just minus pA dVA minus pB dVB, the infinitesimal amount of work done as these things are gradually moved toward each other.
So delta S for demixing, of course it's dq reversible over T, but just like we did right there, of course, in this case that's just negative dw.
So it's integral from V to VA, right, were compressing.
So for substance a, we're going to go back to there.
It was occupying the whole volume, and then it's going to end up in only part of the volume, VA. pA dVA over T. Same thing for B, going to volume B, pB dVB over T. And now we can substitute for pressure, right?
pV is nRT, so it's nA times R log of VA over V plus nB of R log of VB over V. Now, so this is a suitable answer, but I'm going to make if a little easier by putting in terms of mole fractions.
So, mole fractions, of course, XA is nA over n, and XB is nB over n and XA is also equal to VA over V and XB is equal to VB over V for ideal gases, right?
The molar volumes are the same.
We're at the same temperature and pressure.
So that immediately gives us the delta X of demixing is nR times XA log XA plus XB log XB.
And of course delta S of mixing is the opposite of this, minus nR XA log XA plus XB log XB.
Now XA and XB are mole fractions.
They are between zero and one, So their log rhythms are both negative.
There's a negative sign.
So delta S of mixing is positive.
That's reassuring.
That tells us as we expect that the mixing should happen spontaneously and irreversibly when we remove the barrier.
Any questions so far about how we're going about these calculations of delta S?
OK, let's just do a couple more examples.
Here's a really straightforward one.
What if we just heat stuff up or cool it down.
It happens all the time, but it's pretty important, right?
So let's just heat or cool at constant volume.
So, we're going to go from substance A at T1 and some volume going to substance A at T2 at some volume.
We can do this.
Delta S is integral of dq reversible over T. Going from T1 to T2, but we know how to do this, right?
It's the heat that we need.
It's just given by the constant volume heat capacity times the change in temperature.
So this is just T1 to T2, Cv dT over T. And that is just Cv times the log of T2 over T1 if Cv is temperature independent.
That's not always the case.
Usually if it's not too big an excursion of temperature then it's a reasonable approximation.
So that's straightforward.
By the way, notice that delta S is greater than zero if T2 is greater than T1.
1 In other words, if we heated it up, delta S is positive.
Notice also delta S is less than zero if T2 is less than T1.
Delta S is negative if we cooled it.
Now, under certain conditions we've seen that delta S can't be negative.
Why can it be negative here?
[UNINTELLIGIBLE]
Yes, I heard it.
It's because the system isn't isolated, right.
It was for the isolated system that delta S is always greater than or equal to zero.
And you know, for the whole universe entropy never decreases and so on.
In this case, though, you know we're taking a system and we're putting it in contact with a cold bath and cooling it down.
It's certainly not isolated.
Heat is being exchanged.
It's going from the system to the surroundings, to the bath.
So it's fine.
Delta S can be negative.
What would happen if we calculated delta S of a new entire system consisting of the original system plus the heat bath?
What would delta S be then?
We already kind of did that with those two blocks, T1 and T2 in an isolated systems.
What do you think is going to happen?
In other words, if we call our system the stuff that is put in contact with the heat bath, well OK, then we've seen delta S can be plus, can be positive or negative.
Now let's call our system the original system plus the heat bath, and we'll put them all in some isolating box, that's thermally isolating and so on.
And now we'll put our original system, which is now sort of a sub, part of the system, and it's now going to be in contact with the cold bath.
The bath that it's in contact with is colder than it is.
That's why its delta S is less than zero.
But what's delta S to be for the entire new system including the cold bath?
Wouldn't it greater than zero?
Yes, it's going to be positive, because it's an isolated system and something happened spontaneously in it, namely the original system got cooler and presumably the heat bath got at least a little bit warmer.
And it happened spontaneously, which immediately means delta S had to be positive for that entire assembly, of you know the original system plus the bath.
Here's another important sort of process.
Reversible phase change, we'll melt something or freeze it.
How about water -- liquid 100 degrees Celsius one bar goes to water.
Gas 100 degrees Celsius one bar.
Well in that case, you know what the heat is.
It's just the delta H of vaporization, right, at constant pressure.
So there's nothing really much to calculate here.
Delta S of vaporization for H2O at 100 degrees Celsius is just q of vaporization over the boiling temperature, which is delta H of vaporization over the boiling temperature.
So that was easy.
One last thing.
What if we want to calculate the entropy of melting or some phase transition not at equilibrium?
In this case it's at equilibrium because it's water and water vapor at 100 degrees Celsius and one atmosphere or one bar.
So it's at its boiling point.
But what if it isn't at its boiling point?
So what if instead, let's take H2O, liquid, at minus ten degrees Celsius and one bar so it's cold.
It's going to freeze.
So it's going to turn into solid water at minus 10 degrees Celsius and one bar, we're going to have it be isothermal.
Now you know that's going to happen spontaneously, and it's irreversible.
Irreversible, which means we can't just straight away calculate delta S along this path because it's irreversible.
But what we can do, just like we've done before for making cycles, we can make some other set of events that are all reversible.
We can construct a reversible pass that will get there.
How do we do it?
Well you know for the phase change, for the freezing to be reversible, it has to happen at zero degrees Celsius right.
You know that's where you have reversible, freezing and melting.
Reversible equilibrium between liquid and solid water.
So surely that's got to get involved here.
H2O liquid zero degrees Celsius, one bar, in equilibrium with H2O solid at zero degrees Celsius, one bar.
Whatever reversible path we construct that's going to go from liquid to solid water, somewhere along the set of steps, that better be one of them.
So now what we need to do is go from liquid at minus 10 degrees Celsius and one bar to liquid at zero degrees Celsius and one bar.
That's called heating.
We have to heat it.
And here we have to cool it.
And we've already seen all these three processes, right.
So for the phase change, it's reversible now.
This is just delta heat of fusion.
Great.
Because this is now reversible.
This is going to be reversible.
This is going to be reversible.
Great.
So dq reversible, it's going to be the heat capacity at constant pressure, the example we did before it was constant volume for the liquid dT.
Here, dq reversible is going to be Cp solid, dT.
Those won't be the same right?
It's one thing to say the heat capacity of something doesn't change over some small excursion in temperature, but it's another thing when the thing actually changes phase.
The heat capacity of the solid and the liquid won't be the same.
So this we can just finish up in a jiffy, because we're just going to add the three things.
So delta S is delta S of heating, minus, I think there's a plus that should be a minus there, minus delta S of fusion because it's going in the direction of freezing the way we've written it.
Plus delta S of cooling.
So integral from T1 to the melting point of Cp of the liquid dT over T, minus delta H of fusion over the melting temperature.
Plus the integral going from the heat of the temperature of melting to T1 of Cp of the solid dT over T. That's it, all right?
So delta S is minus delta H of fusion over T. That's what it would be for just the reversible phase change happening at zero degrees Celsius.
And then there's this additional part which is, we can write it from T1 to melting point of Cp of the liquid, minus Cp of the solid, dT over T. Now usually we can assume that these will be temperature independent in their own phases.
So we can usually write this as minus delta H of fusion over the melting temperature plus Cp of the liquid minus Cp of the solid log of T of fusion over T1.
Any questions?
What did we do?
We constructed a reversible path going from here to here and calculated delta S that way, and of course that has to be the same as delta S in the one irreversible step.
All right, more on entropy and its consequences for how to figure out what happened spontaneously next time.
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All right, well last time we finished our sort of introduction to entropy which is a difficult topic, but I hope we got to the point where we have some working understanding of physically what it's representing for us, and also an ability to just calculate it.
So we went through at the end of the last lecture, a few examples where we just calculated changes in entropy for simple processes like heating and cooling something or going through a phase transition where that process, that sort of process is relatively simple because there's no temperature change.
While the ice is melting, for example, you're putting heat into it, but the temperature is staying at zero degree Celsius.
And we saw, how to do these calculations, we need define reversible paths.
So it was extremely straightforward to calculate the entropy of say ice melting at zero degrees Celsius, because there the process is reversible, because that's the melting temperature.
But if we wanted to calculate the change in entropy of ice melting, you know at, once it had already been cooled to ten degrees above the melting point, to ten degrees Celsius, then in order to find a reversible path, we had to say, OK, let's first cool it down to zero degrees Celsius.
Then let's have it melt, and then let's warm the liquid back up to ten degrees Celsius so we could construct the sequence of reversible steps that would get from the same starting point to the same end point, and we could calculate the change in entropy through that sort of sequence.
So now that we've got at least some experience doing calculations of delta S and we're just thinking a little bit about entropy, what I'd like to do is to try to relate the state variables together in a useful way.
And and the immediate problem that I'd like to address is the fact that right now we have kind of a cumbersome expression for energy.
So you know we have u, we look at du, right it's dq plus dw, and you know, I don't like those.
They're path specific, and it would be nice to be able to do a calculation of changes in energy that didn't depend on path.
You know it's a state function.
So in principle it seems like it sure ought to be possible, and yet so far when we've actually gone through calculations of du, we've had to go and consider the path and get the heat and get the work and so forth.
And then we found special cases, you know an ideal gas where the temperature, where the change is only a function of temperature and so forth, where we could write this as a function of state variables, but nothing general that really allows us to do the calculation under all circumstances.
So let's think about how to make this better.
So, and what I mean by that is you've seen examples like, you know, some special examples you saw awhile back.
The case where du was Cv dT minus Cv and this Joule coefficient d v. But you know you still need to find those coefficients for each system.
This isn't a general equation that tells us how energy changes in terms of only functions of state.
Because of things like this and this -- what I'd really like is to be able to you know write du equals something.
And that something, you know, it can have T and p and whatever else I need.
It can have S, H, and of course differentials of any of those quantities.
dT or dp or dS, dH, you name it, but all state variables.
That's what I'd much rather have.
And then, you know, if I want to, if I've got something like that and I want to find out how the energy changes as a function of volume, so I'll calculate du/dV With respect to some selected variable hold constant, I can do it.
Right now, in a general sense, that's cumbersome.
I've got to figure out how to do that for each particular case that I want to treat.
So, let's see how we could construct such a thing.
Let's consider just a reversible process, at constant pressure.
So, OK, I've got one and I'm going to in some path wind up at state two and I'll write du is dq, it's reversible in this case, minus p dV.
And from the second law, we know that we can write dq reversible as T dS.
dq over T is dS or entropy.
So, we can write du is T dS minus p dV.
That's so important, we'll circle it with colored chalk.
That's how important it is.
You know it's a dramatic moment.
So let's look at what we have here.
Here's du and over on this side we have T, we have S, we have p and we have V. Suddenly and simply, it's only functions of state.
Well that was pretty easy.
So, what that's telling us is that we can write u this way, and you know, this is generally true.
We got to this by considering a reversible constant pressure process.
But we know u is a state variable right.
So this result is going to be generally applicable.
And it tells us a couple of things too.
It tells us that in some sense, the natural variables for u are these, right, it's a function of S and V. Those are natural variables in the sense that then it written as functions of those variables, we only have state quantities on the right-hand side.
Very, very valuable expression.
And of course coming out of that then, we can take derivatives and at least for those particular variables, we can see that du/dS at constant V is minus p. And du/dV at constant S is T. All right, those fall right out.
Now we can have a similar set of steps for H, for the enthalpy, so let's just look at that.
So H, of course, it's u plus pV, so dH is just du plus d(pV), and now there's our expression for du.
We're going to use it this way.
So it's T dS minus p dV plus p dV plus V dp.
And of course these are going to cancel.
So we can write dH is T dS plus V dp.
Also important enough that we'll highlight it a little bit.
So again let's look at what we've got on the right-hand side here, T, S, V, p, only quantities that are functions of state.
And of course, we can take the corresponding derivative, so let's also be explicit here, that means that were writing H as a function of the variables S and p. And dH/dS at constant p is T, and dH/dp at constant S is V. So now we've got a couple of really surprisingly simple expressions that we can use to describe u and H in terms of only state variables.
All right?
We also can go from these expressions, using the chain rule, to expressions for particularly useful expressions for the entropy as a function of temperature.
So, you know, from du is T dS minus p dV.
We can rewrite this as dS is one over T du plus p over T dV.
And now we can go back, you know, if we can go back to our writing of u in terms of, as a function of T and V, right.
So we can write here du as a function of T and V, Cv dT plus du/dV at constant T dV.
The reason we're doing that is now we can rewrite dS is one over T times Cv dT, and that's the only temperature dependence we're going to have.
The other part is going to be a function of volume.
So it's, we've got p over T plus du/dV at constant T dV, and what this says then is that dS/dT at constant V is just Cv over T. Very useful, not surprising because of the relation between heat at constant volume and Cv, right.
And of course dS is just dq reversible over T, but this is telling us, in general, how the entropy changes with temperature at constant volume.
We can go through the exact same procedure with the H to look at how entropy varies with temperature at constant pressure, and we'll get exactly an analogous set of steps that will be Cp over T, right.
So also dS/dT at constant pressure is Cp over T. OK?
Now I want to carry our discussion a little bit further and look at entropy a little more carefully and in particular, how it varies with temperature.
And here's what I really want to look at.
You know, we've talked about when we look at changes in u and changes in H, and we've done this under lots of circumstances at this point.
And at various times I've emphasized, and I'm sure Professor Bawendi did too, that when we look at these quantities we can only define changes in them.
There's not an absolute scales for energy or for enthalpy.
We can set the zero in a particular problem, arbitrarily.
And so, for example, when we talked about thermochemistry, we defined heats of formation, and then we said, well, the heat of formation of an element in its natural state at room temperature and pressure we'll call zero.
We called it zero.
If we wanted to put some number on it, and put energy in it, we could have done that.
We defined the zero.
So far, that's, well not just so far, that's always the way it will be for quantities like energy and enthalpy.
Entropy is different.
So let's just see how that works.
So, let's consider the entropy, we'll consider as a function of temperature and pressure.
First let's just see how it varies with pressure.
We're going to see -- what we'll do is consider its variation to both pressure and temperature, and the objective is to say all right, if I've got some substance at any arbitrary temperature and pressure, can I define and calculate an absolute number for the entropy?
Not just a change in entropy, unlike the cases with delta u and delta H, but an absolute number that says in absolute terms the entropy of this substance at room temperature and pressure or whatever temperature and pressure is this amount.
Something that I can do by choice of a zero for energy or for u or H, but here, I want to look for an absolute answer.
All right, so let's start by looking at the pressure dependence.
So we're going to start with du is T dS minus p dV, so dS is du plus p dV over T. Now let's look, T being constant.
OK, and now let's specify a little bit.
I want to make it something as tractable as possible.
Let's go to an ideal gas.
So then at constant temperature, that says du is equal to zero.
So dS at constant temperature is just p over T dV.
And in the case of an ideal gas, that's nR dV over V. And at constant temperature, that means that d(nRT), which is the same as d(pV) is equal to zero, but this is p dV plus V dp.
So this says that dV over V, that I've got there, is the same thing as negative dp over p. Right, so I can write that dS as constant temperature is minus nR dp over p. So that's great.
That says now if I know the entropy at some particular pressure, I can calculate how it changes as a function of pressure, right.
If I know S at some standard pressure that we can define, then S at some arbitrary pressure, is just S of p naught and T minus the integral from p naught to p of nR dp over p. All right, which is to say it's S of p naught T minus nR log of p over p naught.
Right, now normally we'll define p naught as equal to one bar.
And often you'll see this simply written as nR log p. I don't particularly like to do that because of course, then, formally speaking were looking at something that's written that has units inside as the argument of a log.
Of course it's understood when you see that, and you're likely to see it in various places, it's understood when you see that the quantity p is always supposed to be divided by one bar, and the units then are taken care of.
For one mole, we can write the molar quantities S of p and T, is S, S naught of T minus R log p over p naught.
All right, so that's our pressure dependence.
What about that?
We still don't really have a formulation for calculating this, or you know, defining it or whatever we're going to do to allow us to know it.
Well, let's just consider the entropy as a function of temperature, starting all the way down at zero degrees Kelvin, and going up to whatever temperature we want to consider.
Now, we certainly do know how to calculate delta S for all that because we've seen how to calculate delta S if you just heat something up, and we've seen how to calculate delta S when something under goes a phase transition, right.
Presumably, if we're starting at zero Kelvin, we're starting in a solid state.
As we heat it up, depending on the material, it may melt at some temperature.
If we keep keep heating it up, it'll boil at some temperature, but we know how to treat all of that, right.
So let's just consider something that undergoes that set of changes.
So, we've got some substance A, solid, zero degrees Kelvin, one bar.
Here's process one.
It goes to A, it's a solid at the melting temperature and one bar.
Process two is it turns into a liquid at the melting temperature and one bar.
Process three is we heat it up some more, up to the boiling temperature at one bar.
Process four is it evaporates, so now it's a gas at the boiling temperature and one bar.
Finally, we heat it up some more, so now it's a gas at temperature T and one bar.
And if we wanted to, we can go further.
We can make it a gas at temperature and whatever pressure we want.
That part we already know how to take care of, right.
Well, let's look at what happens to S, all right.
S, a molar enthalpy at T and p, where we're going to finally end up, is, well it's s zero at zero Kelvin and one bar or one bar is implied by the superscripts here.
And then we have delta S for step one, and delta S for step two and so on.
So all right, let's, we can label this six so we to all the way to delta S for step six.
Well, so we can do that.
It's S of the material at T and p is S naught at zero Kelvin, plus, here's for process one.
We heat it up from zero Kelvin up to the melting point.
Cp of the solid more heat capacity, divided by T dT.
We can calculate delta S for heating something up, right.
Plus, now we've got the heat of fusion to melt the stuff, so it's just delta H naught of fusion, divided by Tm right.
We saw that last time.
In other words, remember, we're just looking at q reversible over T to get delta S, and it's just given by the heat of fusion.
All right, then let's go from the melting point to the boiling point.
So it's Cp now it's the heat capacity, the molar heat capacity of the liquid, divided by T dT.
We're heating up the liquid.
And then there's vaporization.
Delta H of vaporization over T at the boiling point.
Then we can go from the boiling point to our final temperature T. Now it's the molar heat capacity of the gas over T dT minus R log p over p naught.
OK, so that's everything, and these are all things that we know how to do.
Just about.
OK, this one we're going to have to think about, but all the changes we know how to calculate, right.
So if we plot this, S, and let's just do this as a function of temperature.
I don't have pressure in here explicitly.
Well, it's going to change as I warm up the solid, soon we're really starting at zero Kelvin.
This stuff is all positive, right, so the change in entropy is going to be positive.
Entropy is going to increase as this happens, and then there's a change right at some fixed temperature as the material melts.
So here is step one.
Here is step two, right, this must be the melting temperature.
And then there's another heating step.
Well, strictly speaking, I'm going to run out of space here if I'm not careful, so I'm going to be a little more careful here.
One, two, three.
I'm heating it up a little more.
Entropy is still increasing, right.
So I've done this.
I've done this.
Now I've heated up the liquid.
Now, I'm going to boil the liquid, so it's going to have some change in entropy.
This must be my boiling point, and now there's some further change in the gas, and that gets me to whatever my final temperature is, right, that I'm going to reach.
Four and five, great.
So there is monotonic increase in the entropy.
OK, so we're there, except for this value.
That one stinking little number -- S naught at zero Kelvin.
That's the only thing we don't know so far.
So, for this we need some additional input.
We got some input of the sort that we need in 1905 from Nernst.
Nernst deduced that as you go down from zero Kelvin for any process, the change in entropy gets smaller and smaller.
It approaches zero.
Now, that actually was certainly an important advance, but it was superseded by such an important advance that I'm not even going to reward it by placing it on the blackboard.
Forget highlight, color, forget it.
Because Planck, six years later, in 1911, deduced a stronger statement which is extremely useful, and it's the following.
What he showed is that as temperature approaches zero Kelvin, for a pure substance in it's crystalline state, S is zero.
A much stronger statement, right.
A stronger statement than the idea that changes in S get very small as you approach zero Kelvin.
No, he's saying we can make a statement about that absolute number S goes to zero as temperature goes to zero.
For a pure substance in it's crystalline state.
So that is monumentally important.
So as T goes to zero Kelvin, S goes to zero, for every pure substance in its, and I'll sort of interject here, perfect crystalline state.
That's really an amazing result.
So what it's saying is I'm down at zero Kelvin.
Minimally, I've somehow cooled it as much as I possibly could, and I've got my perfect crystal lattice.
It could be an atomic crystal like this, or, you know, it could be molecules.
But they're all exactly where they belong in their locations in the crystal, and the absolute entropy is something I can define and it's zero.
So S equals zero.
Perfect, pure crystal, all right.
OK, well this came out of a microscopic description of entropy that I briefly alluded to last lecture, and again we'll go into in more detail in a few lectures hence.
But the result that I mentioned, the general result, was that S was R over Na Avogadro's number, times the log of this omega number of microscopic states available to the system that I'm considering.
Now normally for a macroscopic system, I've got just an astronomical number of microscopic states.
You know, that could mean in a liquid, different little configurations of the molecules around each other.
They're all different states.
Huge amounts of possible states, and the gas even more.
But if I go to zero Kelvin, and I've got a perfect crystal, every atom, every molecule is exactly in its place.
How many possible different states is that?
It's one.
There aren't any more possible states.
I've localized every identical lateral molecule in its particular place, and it's done.
And you know, if I start worrying about the various things that would matter under ordinary conditions, right, you know, maybe at higher temperature, I'd have some molecules in excited vibration levels or maybe electronic levels.
Maybe if it's hydrogen, maybe everything isn't in the ground state.
It's not all in the 1s orbital but in higher levels.
Then there's be lots of states available, right, even of only one atom in the whole crystal is excited, well there's one state that would have it be this atom.
A different one would have it be this atom, and so forth.
Already, there'd be an enormous number of states, but at zero Kelvin, there's no thermal lexitation of any of that stuff.
Things are in the lowest states, and they're in their proper positions.
There's only one state for the whole system, so that's why the entropy is zero in a perfect crystal.
At zero degrees Kelvin.
Now, there are things that may appear to violate that.
Now of course you can make measurements of entropy, right, so it can be verified that this is the case.
There are some things that would appear to violate that result.
You know, you can make measurements of entropies, and for example let's take carbon monoxide crystal, CO.
So let's say this is a crystal lattice, it's diatomic molecules.
It's carbon oxygen carbon oxygen carbon oxygen.
All there in perfect place in the crystal lattice.
Well it turns out when you form the crystal every now and then -- you know, let's put it in color.
Let's put it in the color that we'll use to signify something in some way evil.
No insult to people who like that color.
I kind of like it in fact.
OK, so you know, you're making the crystal, cooling it, started out maybe in the gas phase, start cooling it, starts crystallizing.
Gets colder and colder, but you know, carbon monoxide is pretty easy for those things to flip sides.
And even in the crystalline state, even though it's a crystal, so the molecules center of mass are all where they belong, still at ordinary temperatures they will be able to rotate a bit.
So even in the crystalline state, when it's originally formed, not at zero Kelvin, there's thermal energy around.
These things can to get knocked around, and the orientations can change.
Now you start cooling it, and you know, by and large they'll all go into the proper orientation.
Right, that's the lowest energy state, but you know, there are all sorts of kinetic things involved.
There it takes time for the flipping, depending on how long, how slowly it was cooled, and so forth.
May never happen, and then it's cooled, and then anything that's left in the other orientation is frozen in there.
There's no thermal energy anymore.
It can't find a way to reorient.
That's it.
All right, let's say we're down to zero Kelvin, and out of the whole crystal, we've got a mole of molecules.
One of them is in the wrong orientation.
Now how many states do we have like that that would be possible?
We'd have a mole of states, right.
It could be this one.
It could be that one, right.
Or in general, of course really there's a whole distribution of them, and they could be anywhere, and pairs of them could be, and it doesn't take long to get to really large numbers.
So the entropy won't be zero.
Entropy of a perfect crystal would mean it's perfectly ordered.
So that's zero.
But things like that not withstanding, and of course it's the same if you have a mixed crystal.
In a sense I've described a mixed crystal where the mixture is a mixture of carbon monoxide pointing this way, and carbon monoxide pointing this way.
But a real mixed crystal with two different constituents, well of course, then you have all the possible configurations where, you know, they could be here and here and here, and then you could move one of them around and so forth, there are zillions and zillions of states.
But for a pure crystal, in perfect form, you really have only one configuration, and your entropy is therefore zero.
OK, so now, we can go back and we can do this.
And at least in principle, even for things that don't form perfect crystals, we could calculate the change in entropy going from the perfect ordered crystal to something else with some degree of disorder and keep going and change the temperature and do all these things.
So the real point is that this is extremely powerful because given this, we really can calculate absolute numbers for the entropies of substances, at ordinary, not just at zero Kelvin, but using this which, you know, this is a really very straightforward procedure.
And in fact these things are really quite easy to measure.
You know, you do calorimetry, you can measure those delta H of fusions, right, delta H of vaporization.
You can measure the heat capacities, the things in the calorimeter.
You'll see how much heat is needed to raise the temperature a degree.
That give you your heat capacity for the gas or the liquid or the solid.
So in fact, it's extremely practical to make all those measurements, and you can easily find those values of the heat capacities and the delta H's tabulated for a huge number of substances.
So this is, in fact, the practical procedure then of protocol for calculating absolute entropies of all sorts of materials at ordinary temperatures and pressures.
Very important.
Now, one of those corollaries to this law is that in fact it's impossible to reduce the temperature of any substance, any system, all the way to absolute, exact, no approximations, zero Kelvin.
Because you can't quite get down to zero Kelvin.
And there are various ways that you can see that this must be the case.
But here's one way to think about it.
So, let's just write that first.
All right, can't get quite down to zero Kelvin.
All right, let's consider a mole of an ideal gas.
So p is RT over V. And let's start at T1 and V1, and now let's bring it down to some lower temperature, T2 in some spontaneous process.
We'll make it adiabatic so it's like an irreversible expansion.
I just want to calculate what delta S would be there in terms of T and V. So, well, du is T dS minus p dV.
dS is du over T plus p over T dV.
But we can also write du is Cv dT in this case, right?
So that says that p over T, that's R over V, and we can write, dS is Cv dT over T plus R dV over V. It's Cv, I'll write it as d(log T) plus R d(log V), and so delta S, Cv log of T2 minus log of T1, if T2 is my final temperature, plus R log of V2 minus log of V1, okay.
Or Cv log of T2 over T1 plus R log of V2 over V1.
Well, what does it mean when T2 is zero?
Well, I don't know what it means.
This turns into negative infinity.
So we're going to write it again as our either evil or at least unattainable color.
What's going to happen?
I mean you could say, well, we can counteract it by having this go to plus infinity, right.
Make the volume infinite.
In other words, have the expansion be in through an infinite volume.
Of course that's impossible.
That would be the only way to counteract the divergence of this term.
In practice, you really cannot get to absolute zero, but it is possible to get extremely close.
That's doable experimentally.
It's possible to get down to some micro or nano Kelvin temperatures, right.
In fact, our MIT physicist, Wolfgang Ketterle, by bringing atoms and molecules down to extremely low temperatures, was able to see them all reach the very lowest possible quantum state available to them.
All sorts of unusual properties emerge in that kind of state, where the atoms and molecules behave coherently.
You can make matter waves, right, and see interferences among them because of the fact that they're all in this lowest quantum state.
So it is possible to get extremely low temperatures, but never absolute zero.
Here's another way to think about it.
You could consider what happens to the absolute entropy, starting at T equals zero, right.
So we already saw that, you know, that we can go, the first step will go from zero to some temperature of Cp over T for the solid dT if we start at zero Kelvin and warm up.
Now, already we've got a problem.
If this initial temperature really is zero, what happens?
This diverge, right.
Well, in fact, what that suggests is as you approach zero Kelvin, the heat capacity also approaches zero.
So, does this go to infinity as T approaches zero Kelvin?
Well, not if Cp of the solid approaches zero as T approaches zero Kelvin.
And in fact, we can measure Cp, heat capacities at very low temperatures, and what we find is, they do.
They do go to zero as you approach zero Kelvin.
So in fact, Cp of T approaches zero as T approaches zero Kelvin.
Good!
But one thing that this says is, remember, that dT is dq at the heat in divided by Cp.
And this is getting really, really small.
It's going to zero as temperature goes to zero, right.
Which means that this is enormous.
What it says is even the tiniest amount the heat input leads to a significant change increase in the temperature.
So, this is another way of understanding why it just becomes impossible to lower the temperature of a system to absolute zero, because any kind of contact, and I mean any kind, like let's say you've got the system in some cryostat.
Of course the walls of the cryostat aren't at zero Kelvin, but somewhere in here it's at zero Kelvin.
You can say, okay I won't won't make it in contact with the walls.
You can try, but actually light, photons that emit from the walls go into the sample.
You wouldn't think that heats something up very much, and it doesn't heat it up very much, but we're not talking about very much.
It does heat it up by enough that you can't get the absolute zero.
In other words, somehow it will be in contact with stuff around that is not at absolute zero Kelvin.
And even that kind of radiative contact is enough, in fact, to make it not reach absolute zero.
So the point is, one way or another, you can easily see that it becomes impossible to keep pulling heat out of something and keep it down at the temperature that's right there at zero, but again it's possible to get very close.
All right, what we're going to be able to do next time is take what we've seen so far, and develop the conditions for reaching equilibrium.
So in a general sense, we'll be able to tell which way the processes go left unto themselves to move toward equilibrium.
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So, over the last few lectures we've worked and struggled so formulate the second and third laws of thermodynamics in addition to the first.
Last time we reach the third law which is telling us that we can't quite get to zero degrees Kelvin, but that as the temperature approaches zero degrees Kelvin, the absolute entropy of a pure substance in perfect crystalline form is zero.
And what that corresponds to, if you recall, is the idea that in a perfect crystal at zero degrees Kelvin then you have no disorder at all.
You have a perfectly ordered system.
And in that case, the entropy is absolutely zero.
And the bigger lesson from that is that entropy, unlike energy u or enthalpy H, we could define an absolutely number for it.
The zero of it wasn't arbitrary.
Unlike the case for energy like you've seen in lots and lots of disciplines, where you can arbitrarily set the zero in a way that makes it convenient for you.
Entropy really is not like that.
There is an absolute zero of entropy, and that's really what we learn.
So now, now that we've got all three of the major laws of thermodynamics, what I want to start in on is a discussion of what happens spontaneously.
We've seen that in just one specialized case so far, but we should in a more general way be able to tell when is a system at equilibrium, where there's no net change taking place, and when is it still undergoing spontaneous change towards some other state, presumably toward an equilibrium state.
How do we tell?
So that's the topic that I'd like to address today.
So, that's the big question, right?
So let's say that we've got our stuff in some state.
A, whatever it is.
And there's some other possible state, B, whatever it is.
And maybe if it some well-defined temperature and pressure.
What do we do to tell whether that change will happen spontaneously?
Do we calculate, you know, delta S, delta u, delta H?
What tells us whether or not the change happens?
Certainly in principle we know how to calculate this and other stuff for a change in state of this sort, for lots of changes of state.
But calculating it alone doesn't necessarily tell us whether or not it will just happened of its own accord.
And that's the issue that we'd like to be able to address.
Now, we have addressed this in some cases.
So, for example, we know that if we take, you know, gas A and gas B with a barrier between them, and we remove the barrier, they're going to mix.
And we saw that and went through it somewhat carefully and saw that if the system is isolated, for that case we do have a criterion that tells us whether change happens spontaneously.
Namely, it's delta S is greater than zero.
That tells us whether the change is spontaneous.
And we saw that in fact in this case delta S of mixing, we calculated it, saw that it is positive.
So, clearly, if we remove this barrier mixing takes place, and obviously you know that that happens from lots of experience.
In general, the second law gave us the Clausius inequality for spontaneous change.
Namely dS is greater than dq over T. I suppose we could specify the surroundings temperature.
We saw that in general dS is greater than or equal to dq over T. if it's a reversible process then the equality holds, but if it's irreversible, which means it happens spontaneously, then dS is greater than this.
Just to illustrate the kind of issues that we're up against here, let me just consider a few different chemical reactions, all of which happen spontaneously.
So, I just want to write a few examples down with a few values for delta u or delta H or delta S, and see whether we can get any clues from what we see.
So here are some spontaneous chemical changes.
Here's one.
If we take hydrogen peroxide in the liquid state, it can break down to form water and oxygen.
If we look at the thermodynamic quantities, the enthalpy and the entropy of the reaction, what we find is delta H is minus 209 kiloJoules and delta S is plus 132 joules per Kelvin.
So it seems like there's a favorable change in entropy going this way.
That is, you've got lower energy on the right and also higher entropy.
Higher entropy basically because you're forming molecules of gas where there weren't any before, and there's more disorder in the gas phase than in the liquid.
That is, the gas phase molecules have more freedom to roam.
Okay, that's one example.
Here's another.
Let's just take hydrogen and nitrogen in the gas phase and form ammonia.
Well, here we get, we find that delta H is negative 92 kiloJoules.
Delta S is negative 198 joules per Kelvin.
This one also happens spontaneously.
One thing that makes it pretty clear is that certainly delta S or the sign of it alone is not dictating the outcome here.
All right, here's a third example.
Let's take salt, solid, and dissolve it in a bunch of liquid water.
Hopefully you've got experience saying that this happens, of course it does.
If we measure the thermodynamics, we discover that delta H is 4 kiloJoules, plus 4 kiloJoules.
Delta S is 45 joules per Kelvin.
So now we have a different sign for delta H and it still happens spontaneously.
So clearly, we've got signs and magnitudes of delta H and delta S, and if we wanted to put delta u there, similar things would happen.
They're all over the map.
And yet, these things are all spontaneous processes.
And I didn't specify the conditions, but if we were to do this under ordinary chemical conditions of some, you'd say room temperature and pressure, right, they all happen spontaneously.
OK, clearly we'd be much better off if we had some systematic quantitative way to tell whether something would happen spontaneously.
In other words, we need criteria for equilibrium under more general conditions than the ones that we've dealt with so far, than the one set of conditions that we've dealt with so far, which is isolated system.
Most chemical changes, most physical changes don't happen in isolated systems.
So let's start by writing down our definition of equilibrium.
It's very simple.
The equilibrium state is the one, and it's just one, in which there are no spontaneous changes that can take place to any other state.
Now that's under whatever constraints there are.
There's a box around it.
The temperature or the pressure are fixed, what have you.
But the point is, no spontaneous changes can occur to any other state.
So for example, when we remove the barrier and the gases mix, you know it's over.
Once the gases are mixed, there's not going to be any further net change in the system.
It's at equilibrium, under the new condition, that is with the barrier removed.
OK, so now let's try to formulate how to describe the equilibrium state and what dictates spontaneity.
So what we're going to do is consider the first and second laws.
So our first law, du is dq plus dw.
And our second law, dS is greater than dq over the temperature of surroundings for a change that happens spontaneously.
And now, I want to combine these two, which I of course can do.
I can substitute dq from this expression in here.
I also want to assume for our present purposes that there's only pressure volume work going on, which is to say I want to put p dV in here minus p dV for dw.
So for combining for p V work, what we see then is du has to be less than T surroundings dS minus p external dV.
Or we can rewrite this as du plus external pressure dV minus T surroundings dS is less than zero.
Now this is fundamentally important, and as you know that means that it warrants the exalted distinction of being put up in colored chalk.
And, in fact, since we're going to reuse this again and again during today's lecture, I'm going to put it over here and leave it sacrosanct for our further use.
du plus p external dV minus T surroundings dS is less than zero.
This Is our condition for spontaneous change.
Now this is a really quite useful expression.
For one thing what we have here are all functions of state and parameters that we can control like temperature and pressure.
So that's a big help.
And equilibrium happens when there isn't any possible change of state that would satisfy this.
In other words, you've got your system in some state.
You know it's in some state A, there are some other states around.
Here is what we calculate to tell whether it happens spontaneously.
So we now have a real usable criterion to help guide our understanding of whether things happen by themselves of their own accord or not.
Now, this is still a little bit cumbersome, in part because of the variables involved, including S. That is, most processes that we're concerned with, they'll happen with something held constant like pressure or temperature or maybe volume.
So this isn't the most useful form that we can have, but what we'll see shortly is that from this, we can then derive further criteria for essentially any set of variables or any set of external constraints, like constant temperature or pressure or volume and so forth that we might set.
And so that's what I now want to do.
So I just want to use that again and again, starting from that, for various different sorts of conditions and derive the criterion for equilibrium in each set of conditions.
So first, let's start with the one that we already know, and make sure that it still works, starting from here, mainly our isolated system.
So remember what that means?
It means no heat, no work.
Delta V is zero.
Delta u is zero.
So looking at this, du is zero.
dV is zero.
So all that's left is negative T dS is less than zero.
In other words, T surrounding dS has to be greater than zero, and of course temperature is always positive.
So dS for u and V fixed is greater than zero.
All right, so that's sounds right.
That's what we saw before.
When we have an isolated system, the criterion that determines whether something happens spontaneously is the entropy has to increase.
Now, what this means too is if we imagine a bunch of different states, and this is the entropy of them, so this could be any sort of variable but the point is there are a bunch of possible states around, whichever one has the maximum entropy, that's the equilibrium state.
In other words, you know we've got the two gases on either side of that partition.
We remove the partition, and they mix.
Well, the equilibrium state is the one with the gases completely mixed.
Of course there are lots of states that would have maybe local pockets of one substance in excess and another substance in excess somewhere else.
In other words, there would be lots of states nearby to the equilibrium state.
That is, the chain, they could they could be approached with very little change from the equilibrium state.
They aren't the equilibrium state.
The entropy in all of those states will be lower than the entropy of the fully mixed state.
So the point is, once you're at equilibrium none of the other states, they're accessible, the system could rearrange itself to form them, but there is no accessible state that has higher entropy than the equilibrium state.
OK, that's our familiar isolated system.
Now let's try moving to unfamiliar territory and extending what we know.
So, let's try constant entropy and volume.
And the motivation for choosing a pair like that is easy to see, if we look at our condition for spontaneous change or general condition.
Well, entropy and volume constant means dV and dS are equal to zero.
What does that say?
It means du is less than zero.
That's our condition.
So we immediately get du at constant S and V is less than zero.
That's our condition for spontaneous change.
In other words, if we don't have to worry about entropy or volume equilibrium is achieved when energy is at a minimum.
Now this is what you learn in elementary physics and in mechanics, right.
You're not worried about entropy.
So, you know, if you've got a hill or valley and there's a cart on wheels, it's going to go down to the bottom.
The spontaneous change lowers the potential energy in that case.
So this is a simple condition that's very familiar.
Now, the reason this condition always holds in ordinary mechanics is because you're never, in that case, concerned with a huge statistical population of particles where the disorder among them is an issue.
We're not worrying then about the fact that, well like in the case of gas molecules mixing, the macroscopic state of the whole thing, all those molecules, how many different microscopic configurations are there?
Remember, I mentioned then we'll go further later on into this, that entropy can be related to the extent of disorder.
That is, how many different possible configurations of all those molecules there would be for a particular state.
The reason the entropy of the mixed gases is the highest is because that has the most possible configurations.
If you start segregating the gases, there are fewer possible configurations that the whole system can be in because you're forcing a particular set of circumstances.
When you don't have to worry about criteria like that, ordinary, mechanical energy rules supreme, and that's dictating where equilibrium lies.
But of course, most chemical and biological systems aren't that simple precisely because you have to worry about many particles and their statistics and the way they might order or disorder.
So, and of course, you know, keeping entropy as a fixed variable for a system like that is extremely cumbersome.
As soon as you allow anything to mix, like you might if you want to do any chemistry, entropy changes.
If you change the temperature entropy changes and so on.
So let's go on.
Let's consider a few other examples.
Let's hang on for a little while longer to a set of conditions where we will maintain constant entropy, namely constant entropy and pressure.
So, the dS term is zero, but the other two are not.
So, du plus p dV is less than zero.
I can write this as d(u + pV) less than zero.
Normally I couldn't do that because this term would have p dV plus V dp, but we've specified the pressure is constant, so the dp part is zero.
And this is a quantity that you know, right?
What's u plus pV?
d
dH less than zero, criterion for spontaneous change.
In the case where S and p are held constant.
Now, once again, like I illustrated for entropy, and I could have done the same for energy here, you know, if we again look at a bunch of different states, and look at their enthalpy, well, like before, they'll be invariably lots of possible states.
And now, what is this saying, the equilibrium state is the one with the lowest possible enthalpy.
In the case here, that I just illustrated with the little cart going down the valley, would be exactly the same with regular energy, the equilibrium state is one of lowest energy, right.
And of course there are lots of nearby states.
The cart could be a little ways up the hill, and in this case, it's enthalpy, but again, there would be lots of accessible states.
But if the system is in equilibrium, none of those states has lower enthalpy.
It's already in the lowest enthalpy state.
That is the equilibrium state.
OK, well, now, let's get to the big ones.
That is, in real life, the variables that you'd normally control aren't some combination of entropy and these variables, but really their temperature, volume and pressure, any couple of those, might be what you'd really have under experimental control.
So now let's go to them.
Let's control T and V. So, all right, so now we're getting serious.
All right, well, there's our equilibrium criterion.
We're still going back to it.
It still holds.
So we've now got the dV part equal to zero.
So what this says is that du minus T dS is less than zero, and we can combine those to say d(u - TS) is less than zero.
And again, just like before, we can do that although this normally would say this has T dS and also minus S dT.
T is fixed.
So that part is zero.
So this is really the equivalent of this.
So when we did this here, we conveniently found that that quantity u plus pV is something we know and love, and we're familiar with it.
It's our enthalpy H. So we could write that criterion as dH less than zero.
So here, let's combine these to define a new quantity.
It obviously has importance because what it's going to say is that that's the quantity that defines equilibrium, that tells us about equilibrium, under the very important practical constraints of having fixed temperature and volume.
Realistically, the more likely constraints than either of those.
So let's -- we'll go to a brand new color, define A as u minus TS.
it's called the Helmholtz free energy.
OK, and then our criterion for equilibrium under these conditions is dA, V and T equal to the temperature of the surroundings, is less than zero.
OK, and once again, you know if we wanted to look at a bunch of states that could be accessed, well, we would find lots of states near by, in character to the equilibrium state.
The one that is at equilibrium, there is only one macroscopic state at equilibrium.
It has the lowest A.
In some sense, that's one reason to associate this as a kind of energy, just like mechanical energy u or enthalpy H, it's the minimum free energy state that is the equilibrium state under the relevant conditions.
Now, let's take the step to the biggest set of conditions of all.
What is it when you run a chemical reaction under ordinary circumstances, what's constant?
[UNINTELLIGIBLE]
A little louder
Pressure and temperature
Pressure and temperature, right.
You're running, you're shaking a beaker up here at room temperature.
So let's look at that set of conditions.
All right, there it is.
This is the condition for really the lion's share of chemistry, biology, and other kinds of changes we'll be concerned with.
So, there's our condition for equilibrium.
We don't get to set any of them to zero, right?
So, okay and we can handle that.
du plus p dV minus T dS is less than zero, but we do get to simplify in writing this as d(u + pV - TS) is less than zero, and just like we've seen before, yes, this has p dV and V dp, but the dp is zero because we're at constant pressure.
This has minus T dS minus S dT, but the dT part is zero because we're at constant temperature.
So the result is we can combine all of these as a single differential, and just like we've seen before, what that suggests is that we define another new quantity given by this expression.
And that is the last one we're going to describe.
And that is G, u plus pV minus TS.
The Gibbs free energy.
Notice, we could also write, let's rewrite that.
G is u plus the pV minus TS, but u plus pV is H. So we also can write this as H minus TS and u minus TS is what we just defined a minute ago as A.
So we can also write this as A plus pV.
And the main thing of crucial importance is what, by defining this in the way we have, what that's saying is that dG at constant p and T is less than zero.
There's our condition for equilibrium at constant temperature and pressure.
Boy, is that going to be important for the whole rest of the course.
So, and of course, I hardly need to emphasize further, but we could do the exact same consideration that we have for H and A, there's G. There's our equilibrium state.
It's the state that has the lowest Gibbs free energy.
All these things though are incredibly practical, useful criteria.
This is only defined in terms of state functions.
And just like we saw before for the case of entropy in an isolated system, now we have something we can calculate.
It's a state function, so we're at constant temperature and pressure, and now we want to consider some chemical change or a phase transition or you name it.
Does it happen of its own accord?
Well now we know what needs to be calculated in order to determine that.
So this one is so uniquely pervasive, let's just really explicitly write it all out.
For constant pressure and temperature delta G is less than zero, means A going to B is, all right, let's consider some process state A and state B.
If delta G is less than zero, it happens spontaneously.
If delta G equals zero, then we're already in equilibrium.
And if delta G is greater than zero, then it goes spontaneously in the other direction.
Any questions about any of this?
Let me just give a couple of examples.
If we go back to any of those chemical reactions that I wrote on the board before, right, well certainly we can calculate what delta G would be for each one of them.
Because we know how to calculate all the parts of it.
It's state functions, it's composed of state functions that we know how to calculate.
So we could tell.
When delta G is zero, you know, it doesn't mean that you've got all of one side, all reactants and zero products or all products and zero reactants.
There is some mixture of them.
What this will tell us is what mixture.
You know the stuff is in there in equilibrium, you know the hydrogen and nitrogen that will form ammonia.
And in the end, when it's at equilibrium, and you look and you'd make a measurement, right, you could do spectroscopy.
You could easily see how much of each thing is there.
It doesn't go all the way to absolutely 100 percent ammonia, zero hydrogen zero nitrogen if they were mixed together with the right ratios.
Doesn't happen.
There would be some of the reactants and some of the products.
In the biochemical reactions that are taking place in your body, there is equilibrium between a whole myriad of reactants and products, and thank heavens that gets maintained.
So that's what this will guide us through, and of course that's incredibly, incredibly important.
Here's another thing that's worth thinking about.
There's a balance here between ordinary energy or enthalpy and entropy.
Energy means, you know, chemical reactions happen, and you end up with something that might be exothermic, that is, the products are more stable then the reactants.
You burn methane, and it combines with oxygen to form water, to form CO2.
And if you work out the energetics as we've gone with thermochemistry, you discover there's a huge negative delta H. In other words, the bonds are much stronger.
CO2 is really a stable molecule.
Methane, there are certainly some solid bonds there, but breaking those to form CO2 and water, well it's worth it, right, energetically.
Still, the actual equilibrium depends on entropy also, not only on the energy.
And that's why, when I put up those three different reactions, and we saw the signs could vary.
It's because there's a balance between the two.
Energy may be favoring reaction in one direction, toward let's say products that have lower energy.
But at the same time, entropy is going to be favoring whichever side has higher entropy, has more disorder, and there's a balance that's achieved.
And that's why all those reactions, first of all, in some sense what I put up was kind of a trivial statement in actual fact saying they all happen spontaneously, because I didn't specify what we were starting with exactly, what concentrations we were starting with.
Even something quite unfavorable might happen at least a little bit spontaneously.
You'll have equilibrium.
In those cases, though, you'd have quite a reasonable equilibrium, spontaneously, that is there would be a lot of reaction that went if you simply started under practical conditions and let it go.
Even though the signs of the enthalpy changed, and the signs of the entropy changed because it's a combination of the two that matters.
Here's a really simple example.
Mixing of oil and water.
You know from experience if you've ever mixed them to make salad dressing, they don't mix too well.
And you may know that if you heat them up, they mix much better.
Why?
You know, we've done a bunch of thermochemistry, and we've kind of seen that the energy of mixing, your energetics don't change too much as a function of temperature.
What's changing?
Why does it mix better when you warm it up?
But, you know, looking at our definition of Gibbs free energy, here it is, right, or here.
Let me just say, actually if you calculated delta S for the mixing as a function of temperature, it doesn't change all that much.
You know, the amount of disorder upon mixing is not really sensitive to temperature.
What does change though?
T, and the entropy is weighted by the temperature, so the entropy matters more and more the hotter it gets.
And that's consistent with other things that we've seen, right?
Remember the whole thing about the perfect crystal at zero degrees Kelvin has zero entropy.
It's completely ordered.
Entropy doesn't matter anymore.
It'll go to the lowest energy state.
Raise the temperature, and now entropy plays a bigger role.
So the point is, this balance between energy that you could think of as say bond energies in chemical reactions, and entropy that you can think of in terms of disorder, how many different possible combinations or configurations of something wrong, will dictate where the equilibrium lies.
And knowing now how to calculate these free energies especially the Helmholtz and the Gibbs free energies, that's what's going to guide us in really calculating quantitatively, OK, where will equilibrium lie.
And before long, we'll start in on discussing chemical equilibrium, does deriving where they lie phase equilibrium?
Does stuff change phase to go from liquid to solid and so forth, right?
And where does that happen, at what temperature and pressure and so forth.
And it's always going to come down to calculating the appropriate free energy, and how it changes in the process.
So this is going to be a guide for us for essentially all that we're going to do in the rest of the term.
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Alright.
Well, we've been looking in the last couple lectures at a really important topic in thermodynamics.
Which is, how do you tell what's going to happen.
Which way does a process want to go?
Which way will it go spontaneously?
And if it goes in one direction or another, where does is lead?
In other words, what is the equilibrium state?
And this is just an incredibly important area that thermodynamics allows us to speak to.
So we started to see this.
Sort of direction of spontaneous change.
And where the equilibrium lies.
So what we did is, remember we started with the second law.
Right?
That dS is greater than dq over T. And for the spontaneous change which happens irreversibly That means that'll be dq irreversible.
It would be equal for the reversible case.
And we combine this with first law, which for the case of pressure volume changes we write as this.
And so what this gave us was a very, very useful general criterion for determining whether something happened spontaneously.
Namely, du plus p external dV minus T for the surroundings dS, is greater than zero.
Sorry.
It's less than zero.
And this is for any spontaneous change.
If it equals zero, then we're at equilibrium.
And if it's greater than zero, then the process goes the other way.
We would write the process in the reverse to have it be less than 0 and it would go spontaneously.
And based on this one result, we then looked under various constraints and said OK, what about looking at our variables, volume, pressure, temperature, other things, entropy, if we constrain those, what's the condition for equilibrium?
And that's what led us to a number of results to determine what quantities we even need to be looking at.
To figure out equilibrium.
And what the conditions were.
And so what we discovered were the following.
This one, which we already had seen, which is dS, is greater than zero.
Change in entropy is greater than zero, for an isolated system.
We also saw that dS for constant H and p was greater than zero.
du, regular energy, at constant entropy and volume is less then zero.
And u is minimized at equilibrium.
And this is the familiar result from ordinary mechanics, where you're not worrying about something like entropy for a whole collection of particles.
That is, you minimize potential energy and you see things falling under the force of gravity and so forth, going to potential energy minima in conformance with this result.
dH, S and p is less than zero.
So our H is u plus pV, as you know.
And H is minimized at equilibrium.
And this is, of course, with constant S V. This is constant S and p. But of course, the need to have entropy constrained is never going to be the most convenient one experimental.
There may be circumstances under which it's the case, but it's often difficult to control.
On the other hand, temperature, volume and pressure are variables that are much easier in the lab to keep constant.
To keep control over.
And so that led us to the definitions of other energy quantities, the Helmholtz and Gibbs free energy.
We discovered that the quantity dA, under conditions of constant volume and temperature, is less than zero.
And A is u minus TS.
And A is minimized at equilibrium, under conditions of constant T and V. And finally, and in many cases the most important of the results, because of the conditions it applies to, we saw that this Gibbs free energy is less than zero, that's our condition for spontaneous change.
Where the Gibbs free energy, u plus pV minus TS is H minus TS.
Also A plus pV and G is minimized at equilibrium with constant temperature and pressure.
And that's why the Gibbs free energy is just so enormously important.
Because so much of what we do in chemistry does take place with constant temperature and pressure.
So we have this condition that's established in a quantity that we know how to calculate.
That tells us the direction of spontaneous change for ordinary processes, chemical processes, mixing and you name it, under conditions that are easy to achieve in the lab.
OK, now what we'd like to do is be able to calculate any of these quantities in terms of temperature, pressure, volume properties.
That is, in terms of equations of state.
For any material.
Then we would really be able to essentially calculate anything.
Anything thermodynamic.
About a material.
Of course, that's assuming we know the equation of state.
We may or may not.
But because in many cases we can reasonably either model or measure equations of state, collect data for a material for its temperature, pressure, volume relations, then in fact if we can relate all these quantities to those, then in fact we really can calculate essentially all of the thermodynamics.
For the material.
So let's relate the thermodynamic quantities to equation of state p, V, T data.
And we can do that by going through and deriving what we'll call the fundamental equations of thermodynamics.
that'll provide these relations.
And at this point we know enough to do this in a straightforward way.
So if we start with a relation for energy, T dS minus p dV.
Where u is written as a function of entropy and volume.
And we've seen that that's generally the case.
It comes from the fact that dq reversible is T dS, and dw reversible is minus p dV.
And of course du is the some of those.
So, this is generally true.
Since these are all state functions.
That is, this is derived in the case for reversible paths.
But since these are all simply state functions and quantities, this is generally true.
Now we can use it to derive differential relations for all of the thermodynamics quantities.
So let's just go through and do that.
So H is u plus pV.
So dH is just du plus p dV plus V dp.
And now we're just going to substitute du in here.
And the p dV terms are going to cancel.
So we have the result that dH is T dS plus V dp.
Right?
And that shows us that H is written naturally as a function of entropy and pressure.
And now let's keep going.
A is u minus TS.
dA is du minus T dS minus S dT.
We're going to do the same thing.
Substitute this for du.
This time, the T dS terms are going to cancel.
So we have dA is minus S dT minus T dS.
That can't be right.
And it isn't.
Minus S dT, that's the p dV term that's left, minus p dV.
And it shows us that A is written naturally as a function of T and V. G, we can write in any of a number of ways.
Let's write it as H minus TS.
So dG is dH minus T dS minus S dT.
Here's dH.
We'll substitute that in, and the T dS terms are going to cancel.
So dG is minus S dT plus V dp.
And this shows that G is written naturally as a function of T and p. So these, which we will exalt and celebrate by our sparingly-used colored chalk, are our fundamental equations of thermodynamics.
So what they do is, they're describing how these thermodynamic properties change, in terms of only state functions and state variables.
Very, very useful.
And that's what it means, when we say well, it's natural then, to express say, G as a function of T and p, that's what we're saying.
Is that we can express its changes in terms of these variables.
Related only through quantities that are functions of state.
I don't need to know about a specific path here.
If I know about the states involved, I just need to know what the volume was in each of them.
Now, before, of course, in the first part of the class we started out looking at u and then looking at H not as functions of S and V or S and p, but as functions of temperature, mostly.
In general, temperature and volume or pressure.
And it doesn't mean that something was somehow wrong with that.
It certainly is, it still is going to be useful to do thermochemistry.
To ask questions like how much heat is released in a chemical reaction that takes place at constant temperature.
Not one of these variables.
And we can calculate that.
So it's not that we're somehow throwing away our ability to do that.
However, the thing to remember is, when you look at heats of reaction under those conditions it's all well and good.
But it doesn't tell you, this is the direction that the reaction is going to go.
It doesn't tell you, this is the equilibrium concentration that you'll end up with.
That doesn't come out of what we calculated before in thermochemistry.
What does come out, which is very useful is, if you do run the reaction, here's how much heat comes out.
And if you want to run a furnace and provide energy, that's an extremely important thing to be able to calculate.
Because you're going to run it and you'll probably find conditions under which you can run it more or less to completion.
But it doesn't tell you, by itself, which direction things run in.
Whereas under these conditions, these quantities, if you look at free energy change, for example, at constant temperature and pressure, you can still calculate H. You can still calculate the heat that's released.
This is what will tell you under some particular conditions what will actually happen.
Where will you end up.
Very, very important, of course, to be able to understand that.
Now, it's also very useful to look at some of the relations that come out of these fundamental equations.
And they're straightforward to derive.
So, all I want to do now is look at the derivatives of the free energies with respect to temperature and volume and pressure.
So for example, if I look at A, which we now have written as the function of T and V, of course, in general I can always write dA as partial of A, with respect to T at constant volume dT, plus partial of A with respect to V, at constant temperature dV.
And I know what those turn out to be.
It's minus S dT minus p dV.
So what does that tell me?
It tells me that the partial of A with respect to T at constant V is minus S. Right?
In other words, now I know how to tell how the Helmholtz free energy changes as a function of temperature.
Or as a function of volume.
dA/dV, at constant T, must be negative p. Things that I can measure.
So I can in a very straightforward way say, OK, well, here is my Helmholtz free energy.
If I'm working under conditions of constant temperature and volume, that's very useful.
Now, if I want to change those quantities; change the temperature, change the volume, how will it change?
Well, I can, for any given case, measure the pressure, determine the entropy and I'll know what the slope of change will be.
Similarly for G as a function of temperature and pressure, I can go through the same procedure.
That is, it's easy to write down straight away that dG, with respect to temperature at constant pressure is minus S. That is, this is, dG/dT at constant pressure.
And this is dG/dp at constant temperature.
So again with the Gibbs free energy, now I see how to determine, if I change the pressure, if I change the temperature by some modest amount, how much is the Gibbs free energy going to change?
Well, it's easy to see.
These two relations involving entropy are also useful because they'll let us see how entropy depends on volume and pressure.
And let me show you how that goes.
Now, you've already seen how entropy depends on temperature.
We've already seen that, going to write dS as dq reversible over T. And it's Cv dT over T at constant volume.
It's Cp dT over T at constant pressure.
So we know that dS/dT at constant volume is Cv over T, and dS/dT at constant pressure is Cp, over T. And we've seen that on a number of occasions.
So that tells us what to do to know the entropy as the temperature changes.
But now, what happens if, instead we look at what happens when we go to some state one to some other state two and it's the pressure.
Or the volume, that changes.
And by the way, just to be explicit about this, let's take this example, it means that delta S, if we undergo a change from, say, T1 to T2.
There's Cp over T dT.
So it's Cp log of T2 over T1, and we saw this before.
So now, instead, let's look at some process.
State one goes to state two.
Let's have constant T. And look at what happens if pressure goes from p1 to p2.
Or volume goes from V1 to V2.
And see what happens there.
We looked at pressure change before, actually, in discussing the third law, the fact that the entropy goes to zero as the absolute temperature goes to zero for a pure, perfect crystal.
But, actually, we didn't do that in a general way.
We just treated the one case of an ideal gas as the temperature is reduced.
But we can do this, generally, by using what are called Maxwell relations.
And all this is, is saying that when you take a mixed second derivative, it doesn't matter in which order you take the two derivatives.
So, let's, we're going to use this relationship.
And we're going to use these two.
So, using those, now, what happens if we take the second derivative of A, the mixed derivative, partial with respect to T and the partial with respect to V. So let's leave these off for a moment, and now let's try that.
And the point is that the second derivative of A, with respect to V and T in this order is the same as the second derivative of a with respect to T and V in this order.
It doesn't matter which order.
But that turns out to be useful.
So let's do this explicitly.
Which means we're going to take the derivative with respect to volume of dA/dT.
Now, the dA/dT isn't constant volume.
The derivative we're taking with respect to volume, when we take that it's at constant temperature.
But what is it?
Well, we already know what dA/dT at constant V is.
It's negative S. So this is negative dS/dV.
At constant temperature.
Now let's take it in the other order.
So d/dT of dA/dV, just like this.
The dA/dV is calculated at constant temperature.
We know it.
Then we can take the derivative of that quantity, when we vary the temperature, holding the volume constant.
But again, dA/dV dT, there it is.
It's negative p. So this is just negative dp/dT at constant volume.
These things have to be equal to each other.
Because these mixed second derivatives are the same thing.
But that's very useful.
Because this is what comes directly out of an equation of state, right?
You know how pressure changes with temperature at constant volume if you know the equation of state.
It relates the pressure, volume, and temperature together.
So from measured equation of state data, or from a model like the ideal gas or the van der Waal's gas or another equation of state, you know this.
Can determine how entropy is going to behave as the volume changes.
If we try that for an ideal gas, pV is nRT.
So dp/dT at constant volume, it's just nR over V. And that, now, we know must equal dS/dV, with a positive sign.
At constant temperature.
So now let's try looking at something where are V1 goes to V2.
The volume is going to change, and we can see how the entropy changes.
So, if one goes to two and V1 goes to V2, and it's constant temperature, just what we've specified there.
Delta S is S(T, V2) minus S(T, V1), T's staying the same.
So it's just the integral from V1 to V2 of dS/dV At constant temperature dV.
And now we know what that is.
So it's nR integral from V1 to V2 dV over V. So it's nR log V2 over V1.
There's our delta S. So we know how to calculate it.
Make sense?
Now, we can do the same procedure for the pressure change.
And all we do is, I'll just outline this, I think.
I won't write it all on the board.
But, of course, it's going to come from the fact that these second derivatives are also equal.
So d squared G dT dp is equal to d squared G dp dT.
In other words, the order of taking the derivatives with respect to pressure and temperature doesn't matter.
And what this will show is that dS/dp at constant temperature, here we saw how entropy varies with volume, this is going to show us how it varies with pressure.
Is equal to minus dV/dT at constant pressure.
And again, this is something that comes from an equation of state.
We know how the volume and temperature vary with respect to each other at constant pressure.
That's what the equation of state tells us.
So, again, I can just use that result.
So, if we do a process where one goes to two at constant temperature, and now the pressure, p1, goes to p2, well then delta S is just the integral from p1 to p2 of dS/dp times dS, so it's just this.
And so of course it's still pV equals nRT.
So now we just have nR over p dp.
Right?
So we're going to see the same story.
It's nR log of p2 over p1 for the process where there's a pressure change.
Any questions about this part?
So what we've done is take one step further.
We've used the fundamental equations that are hiding down here, out of sight but never out of mind.
And what we've done is look at the derivatives of the new free energies that we've just recently introduced, A and G. And then, the only thing we've done beyond that is say, OK, well now let's just take the mixed second derivatives, they have to be equal to each other.
And what's fallen out when we do that, because in each case, one of the first derivatives gives us the entropy.
Then the second derivative gives the change in entropy with respect to the variable that we're differentiating, with respect to which is either pressure or volume.
And the useful outcome of all that is that we get to see how entropy changes with one of those variables in terms of only V, T, and p, which come out of some equation of state.
And all we did, further, is take that second derivative.
That mixed second derivative.
And, of course, see that either way we do that we'll have an equality.
Now, let's go back to our older friends u and H. Which we've expressed now in terms of S and V, S and p. So, so far we don't have a way to just write off, relate them to equation of state data.
Which also would be very useful.
Here, A and G, we've already got as functions of these easily controlled, conveniently controlled state variables.
Let's look at those quantities.
u and H. And look at, for example, the V dependence of u.
The volume dependence.
And in particular let's look at, for example, du/dV at constant temperature.
Now, we can immediately see what du/dV at constant entropy is.
Experimentally, though, that's not such an easy situation to arrange.
Of course, this is a much more practical one.
But it doesn't just fall out immediately from the one fundamental equation for du.
But we can start there.
So, du is T dS minus p dV.
And I can take this derivative.
du/dV at constant T. And so, what is it?
Well, it's not just p because there's some dS/dV at constant T. This isn't zero.
There's some variation, dS/dV, at constant temperature.
That's going to matter.
This part, of course, is just minus p. But we just figured out what dS/dV at constant T is.
This is dp/dT at constant V. So that's neat.
So in other words, we can write this as T, dp/dT at constant V, minus p. Let's just check T, p, T, V, p. Right?
In other words, we just have pressure, temperature and volume.
Again, if we know the equation of state, we know all this stuff.
So again, we can measure equation of state data.
Or, if we know the equation of state from a model, ideal gas, van der Waal's gas, whatever, now we can determine u.
From equation of state data.
Terrific, right?
So let's take our one model that we keep going back to.
Equation of state, and just see how it works.
That is, ideal gas.
And see how it works with that.
Now, we saw before, or really I should say we accepted before, that for an ideal gas, u was a function of temperature only.
Well, now let's try it.
So, dp/dT, for our ideal gas, at constant volume, remember pV is nRT.
So this nR over V. And then, using the relation again, we can just write this as p over T. In other words, we're taking advantage of the fact that we now know that quantity.
In the case of the ideal gas, we just have a simple model for it.
More generally, we could measure it.
We could just collect a bunch of data.
For a material.
What's the volume it occupies at some pressure and temperature?
Now let's change the pressure and temperature and sweep through a whole range of pressures and temperatures and measure the volume in every one of them.
Well, then, we could just use that for our equation of state.
One way or another, we can determine this quantity.
For the ideal gas it's this.
So now our du/dV, at constant T is just T times dp/dT, which is just p over T minus p, it's zero.
Remember the Joule expansion.
And we saw that, you saw that the Joule coefficient for an ideal gas was zero.
So that you could see that for the ideal gas, u would not be a function of volume, but only of temperature.
But actually, when you saw that before, you weren't given any proof of that.
It was just that when the good Mr. Joule made the measurements, to the precision that he could measure, he discovered that for some gases it was extremely small.
At least, smaller than anything he could detect.
So it sure seemed like it was going to zero, under ideal gas conditions.
And that was the result that we came to accept.
Here, though, you can just derive straight away.
That for an ideal gas it has to be the case that there's no volume dependence of the energy.
Only a temperature dependence.
It's the same for H. Just like u, we'd like to be able to express it in a way that allows us to calculate what happens only from equation of state data.
But, again, our fundamental equations show us how it changes as a function of entropy and pressure.
So, dH is T dS plus V dp.
So let's look at dH/dp.
We know how to get it immediately if we keep entropy constant.
But we'd like to relate it to what happens if we keep the temperature constant.
So then, just like we saw, analogous to what saw just before, it's T dS/dp at constant T. Plus V. But now we've seen from the Maxwell relations that dS/dp is minus dV/dT, for constant p. Again, this is this quantity, one of these quantities that again we can determine from equation of state data.
Only V, p and T appear.
So it's minus T dV/dT at constant p, plus V. And so, again, this can come from equation of state data.
And if you do this again for an ideal gas, let me see.
So we have pV is nRT.
So dV/dT at constant pressure is just nR over p. But we can plug that in again just like we did before.
It's just equal to V over T. And so dH/dp under our condition of constant temperature is just minus T times V over T plus V, everything cancels, and that's zero.
That's our Joule - Thompson expansion.
That was a constant enthalpy change.
And again there, too, you saw an experimental result you were presented with that says, well at least to the extent that it could be measured, it was obviously getting very small.
For gases that approach ideal gas conditions.
Well, there you can see it.
Sure better have gotten small because in fact it has to be zero.
Now let's take just one somewhat more complicated case.
Let's look at a van der Waal's gas.
Let's try it with a different equation of state, that isn't quite as simple as the ideal gas case.
So, then p plus a over molar volume squared times V minus b molar volume V minus b is equal to RT, remember?
This was back from the first or second lecture in the course.
So, we can separate out p. It's RT over molar volume minus b minus a over molar volume V squared.
And then we can take the derivative with respect to temperature, it's just R over molar volume minus b.
So it's dp/dT at constant V is just R over V bar minus b.
Well, let's now look, given this, let's now look in that case, at what happens to u as a function of V. For the ideal gas, we know that u is volume independent.
It only depends on the temperature.
But for the van der Waal's gas, now it's going to be different.
And that's because this is different from what it is in the ideal gas case.
Namely, now du/dV at constant T, for the van der Waal's gas.
So it's this.
So it's RT over molar volume minus b.
Minus p, right?
But in fact, if you go back to the van der Waal's equation of state, here's RT over v minus b.
If we put it as minus b, that's just equal to a over V squared.
Equals a over molar volume squared.
But the point is, the main point is, it's not zero.
It's some number.
a over the molar volume squared.
a is a positive number in the van der Waal's equation of state.
So this is greater than zero.
In other words, u is a function of T and V. If we don't have an ideal gas.
By the way, just to think about it a little bit, it's a positive number.
What that means is, I've got my ideal gas in some container.
There's some energy, some internal energy.
Now I make the volume bigger.
I allow it to expand.
And the energy changes, it goes up.
In some sense it's less favorable energetically.
What's happening there, that a term in the van der Waal's equation of state, is describing interactions, favorable attractions, between gas molecules.
The b is describing repulsions.
Effectively, the volume changes.
The molar volume is being changed a little bit by the fact that if you're really trying to make things collide with each other, they can't occupy the same volume.
And that is being expressed here.
The a, though is expressing attraction between molecules at somewhat longer range.
So now, if you make the volume bigger, those attractions die out.
Because the molecules are farther apart from each other.
So the energy goes up.
Now, you might ask, well why does it do that, right?
I mean, if the energy is lower to occupy a smaller volume, then if I have this room and a bunch of molecules of oxygen, and nitrogen and what have you in the air, and there are weak attractions between them, why don't they all just sort of glum together and find whatever volume they like.
So that the attractive forces can exert themselves a little bit.
Not too close, right?
Not so close that the repulsions dominate.
Why don't they do that?
What else matters besides any of those considerations?
What else matters that I haven't considered in this little discussion?
Yeah.
What about entropy, right?
If I only worry about minimizing the energy, it's true.
They'll stick together a little bit.
But entropy also matters.
And there's disorder achieved by occupying the full available volume.
Many more states possible.
And that will end up winning out at basically any realistic temperature where the stuff really is a gas.
OK. Next time, what you're going to see is the following.
It turns out we can express all these functions in terms of G, we wouldn't need to choose G, but it's a very useful function to choose.
Because of its natural expression in terms of T and p. So we can write any of these functions in terms of G. Which means that we can really calculate all the thermodynamics in terms of only g. It's not necessary to do that, but it can be quite convenient.
And then we can say, OK, if we have many constituents, what if we have a mixture of stuff?
We can take the derivative of G with respect to how much material there is.
With respect to n, the number of moles.
And if there are one, and two, and three constituents with respect to n1, and n2, and n3.
Each individual amount of stuff.
What that's going to allow us to do is, if we say, OK I have a mixture of stuff, how does the free energy change?
If I change the composition of the mixture?
If I take something away, or put something else in?
And we'll be able to determine equilibrium under those conditions.
That's very useful for things like chemical reactions, where this constituent changes to this one.
And I can calculate what happens to G under those conditions.
And that's what you'll see starting next time.
And professor Blendi will be taking over for that set of lectures.
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Last time, you talked about the Gibbs free energy.
And the fundamental equations.
And how powerful the fundamental equations were in being able to calculate anything from pressure, volume, temperature data.
And you saw that the Gibbs free energy was especially important for everyday sort of processes.
Because of the constant pressure constraint.
And the fact that the intrinsic variables are pressure and temperature.
Well, it turns out that the Gibbs free energy is even more important than that.
And this is something that it took me a while to learn.
I had to take thermodynamics many times to really appreciate how important that was.
Even when I was doing research, at the beginning, I was a theorist, and I was trying to calculate different quantities of liquids and polymers and in these papers the first thing they did was to calculate the Gibbs free energy and I didn't quite appreciate why they were doing that.
And the reason is, is if you've got the Gibbs free energy, you got really everything you need to know.
Because you can get everything from the Gibbs free energy.
And it really becomes the fundamental quantity that you want to have.
So let me give you an example of how important that is, if you have an equation that describes the Gibbs free energy as a function of pressure and temperature, number of moles of different things.
Different things you can calculate.
So let's just start from the fundamental equation for the Gibbs free energy, dG is minus S dT plus V dp.
And let's say that you've gotten some expression, G, as a function of temperature and pressure for your system.
It could be, as we're going to see today that we're going to increase the number of variables here, by making the system more complicated.
So what I'm saying, now what I'm going to say now, is going to be more general than just these two variables here.
So you've gotten this.
So you have this.
You've got the fundamental equation.
You've got all the other fundamental equations, and from there you can calculate all these quantities.
For instance, you can calculate an expression for S. Because you know that S, from the fundamental equation, is just the derivative of G, with respect to T, keeping the pressure fixed.
So you've got your equation for G. That translates into an equation for S. You can get volume, volume is not one of the variables.
Temperature and pressure are the two knobs.
But you can get volume out, because volume is the derivative of G, with respect to pressure.
Keeping the temperature constant.
And you've got S now, you got volume.
Do you know where you, how did you define G in the first place?
We define G as H minus dS.
One of the many definitions of g. Reverse that, you've got now H as a function of G and temperature and entropy.
Well, you've got an expression for G, we just calculated, we just showed we could get an expression for S, which is sitting right here.
Temperature is a variable here, so now we have an expression for H. And you can go on like that with every variable that you've learned in this class already.
For instance, u is H minus pV.
Well, there's the H here.
We have that equation for H. We have an equation for V, coming from here.
So there's nothing unknown here.
If we have an equation for G here in terms of temperature and pressure.
Same thing for the Helmholtz free energy.
You can even get the heat capacities out.
Every single one of these interesting, useful quantities that one would want to calculate falls out from the Gibbs free energy here.
Any questions on that important step?
And, really, I can't believe how clueless I was when I started doing research.
Because I would go through the process of calculating G, and getting G, and et cetera, et cetera.
I wrote papers, you know, G equals blah blah blah.
And I didn't realize that that's why people wanted to know G. Anyway, I know better now.
So, there are a few things we can say about G that are fairly easy to calculate.
For instance, if I look at liquids or solids.
And I want to know how G changes with pressure.
So, I know that the volume here dG/dp, that dG/dp is the volume here.
So if I look under constant temperature, I pick my fundamental equation under constant temperature, and I want to know how G is changing, I integrate.
So I have dG is equal to V dp.
So if I change my, and I look at per mole, and if I change my pressure from p1 to p2, I integrate from p1 to p2, p1 to p2 here, final state minus the initial state is equal to the integral from p1 to p2, V dp per mole.
And what can I say, for a liquid or a solid, the volume per mole, over a liquid or a solid, is small and it doesn't change very much.
So V is small.
And usually these solids and liquids, you can assume to be incompressible.
Meaning, as you change the pressure, the volume doesn't change.
It's a good approximation.
So when you do your integral here, you get that G at the new pressure is G at the old pressure, then if this isn't changing very much with pressure, or not at all, then you can take it out.
It's just a constant.
Plus V times p2 minus p1.
And so, this is the incompressible part, you take it out.
The fact that it's small means that you can assume that this is zero, this whole thing is zero, that it's small enough.
And then you see that G, approximately doesn't change with pressure.
Tells you that G, for a liquid or solid, most of the time you can assume that it's just a function of temperature.
Just like we saw for an ideal gas, that the energy and the enthalpy were just functions of temperature.
And that's a useful approximation.
It's useful, but it's not completely true.
And if it were true, then there would not be any pressure dependents to phase transitions.
And we know that's not the case.
We know that if you press on water when it's close to the water liquid-solid transition, that you can lower the melting point of ice.
You press on ice, and you press hard enough, and ice will melt, the temperature is closer to melting point.
And we'll go through that.
So, that means that there has to be some sort of pressure dependence, eventually.
And we'll see that.
This is just an approximation.
What else can we do?
We can calculate, also, for an ideal gas.
Liquid and solid, we can do an ideal gas.
So for an ideal gas, again, starting from the fundamental equation, we have dG equals V dp.
We can do it per mole.
So integrate both sides, G(T, p2) is equal to G at the initial pressure, plus the integral from p1 to p2, the volume.
So instead of putting the volume, this is an ideal gas now, we can put the ideal gas law.
So V is really RT over p. RT over p dp.
We can integrate this.
Get a log term out.
G(T, p1) plus RT log p2 over p1.
And then it's very useful to reference everything to the center state.
p1 is equal to one bar, let's say.
So if you take p1 equals one bar as our reference point, and get rid of the little subscript two here, we can write G of T at some pressure p, then is G and the little naught on top here means standard state one bar plus RT log p divided by one bar.
And I put in a little dotted line here because very often you just write it without the one bar and bar.
And you know that there has to be a one bar, because inside a log you can't have something with units.
It has to be unitless.
So you know if you have something with bar here, you've got to divide with something with bar, and there happens to be one bar here.
So pressure p is G at its standard state plus RT log p. And this becomes a very useful, very useful, quantity to know.
OK, so G is so important.
And G per mole is so fundamental, that we're going to give it a special name.
Not to make your life more complicated but just because it's just so important.
We're going to call it the chemical potential.
So G per mole, we're going to call mu.
And that's going to be a chemical potential.
We're going to do a lot more with the chemical potential today.
And the reason why we call potential is because we already saw that if you've got something under constant pressure, temperature, that you want to use G as the variable to tell you whether something is going to be spontaneous or not.
So you want G to go downhill.
And so, we're going to be talking about chemical species.
And instead of having a car up and down mountains, trying to go down to the valleys, we're going to have chemical species trying to find the valleys.
The potential valleys.
To get to equilibrium.
And so we're going to be looking at the Gibbs free energy, or the Gibbs free energy per mole at that particular species, and it's going to want to be as small as possible.
We're going to want to minimize the chemical potentials.
And that's why it's called potential.
It's like an energy.
So, that's the end of the one component, thermodynamic background, before we get to multi-components.
So it's a good time to stop again and see if there's any questions.
Any issues.
OK.
So, so far we've done everything with one species.
One ideal gas, one liquid, one solid.
We haven't done anything with mixtures, except for maybe looking at the entropy of mixing.
We saw the entropy of mixing was really important, because it drove processes where energy was constant.
But most of what we care about in chemistry, at least in chemical reactions, species change.
They get destroyed.
New species get created.
There are mixtures.
It's pretty complicated.
For instance, if I take a reaction of hydrogen gas plus chlorine gas to form two moles of HCl gas, I'm destroying hydrogen, I'm destroying chlorine, I'm making HCl in the gas phase.
I get a big mixture at the end.
I get three different kinds of species at the end.
So the fundamental equations that I've been talking about, that we've been talking about, they're too simple for such a system.
Because they all care about one species.
Even more complicated, for instance, if I take hydrogen gas and oxygen gas and I mix them together to make water, liquid, for instance, not only do I have species that are changing, that are getting destroyed or created, in this case here the total number of moles is changing.
And the phase is changing.
Got all sorts of changes going on here.
And so if I want to understand equilibrium, if I want to understand the direction of time for these more complicated processes, I have to be able to take into account, in an easy way, these mixing processes, these phase changes, these changes in the number of moles.
And that's what we're going to talk about today.
We're going to try to change our fundamental equations to make them a little bit more complicated so that we can deal with these sorts of problems.
Because those are the real problems we need to keep track of.
And the ultimate goal, then, of changing our fundamental questions is to derive equilibrium from first principles.
To really understand chemical equilibrium, which you've all seen before.
You've all used the chemical equilibrium constant K, you've done problems.
But you've been given, basically, the equilibrium constant, and not really derived it, understood where it came from.
OK, another simple example here, which is actually the one that we're going to be looking at in the first case.
Where there's a change going on, is just to look at a phase change.
H2O liquid going to H2O solid.
There's a phase change, you can think of it as one species, the H2O liquid sort of changing into an H2O a solid.
It's the same chemical in this case here, there's no change in the molecules.
But it's still a change that we have to account for.
Another example that's also simple like this, that you all, I'm sure, have seen before, suppose I take a cell.
My cell here, full of water.
And then I put my cell, let's say I take a human cell.
My skin or something.
And I take it and I put it in distilled water.
What's going to happen to the cell?
Is it going to be happy?
What's going to happen to it?
It's going to burst, right?
Why is it going to burst?
Anybody have an idea why it's going to burst?
Yes
[INAUDIBLE]
That's right.
So the water wants to go from, you're completely right.
But let me rephrase it in a thermodynamic language here.
The water is going to go from a place of high chemical potential to low chemical potential.
And the cell can't take all that water in there.
The membrane's going to try to swell.
And eventually burst, right?
Same thing if you if you take a, go fishing, go to the Atlantic Ocean and then get a nice cod or something.
Bring it back and on your sailboat, you dump it in a tub of fresh water.
Is that cod going to be happy?
It's not going to be happy at all, right, because its biology is geared towards living in salt water.
And turns out that the chemical potential of water, in salt water, is lower than the chemical potential of pure water.
And so when you put the cod in there, the chemical potential of the water and the cod, is lower than the chemical potential of the fresh water you have on the outside.
And the fresh water wants to be at a lower chemical potential.
It rushes into the cod, and well, the cod does what the cell does, when you put it in distilled water.
It sort of bloats.
It isn't very happy.
OK, so but all these things are basically the same idea here.
Where you have a complicated change, where species are mixing, and things like this.
And it turns out the chemical potential is going to tell us all about how to think about that.
That's why the chemical potential is so important.
So we're going to go back to these two examples here many times.
So let's take the simplest example here.
Let's go back and derive some equations.
Let's take our simplest example that's not a one species system, but has two species.
Species 1 and 2.
And n1 and n2 are the number of moles of species 1 and 2.
And then we're going to see if I make a perturbation in my system.
I change the number of moles of 1, or the number of moles of 2.
How does this affect the Gibbs free energy?
That's the question we're going to post.
And our goal is to find a new fundamental equation for G that includes the number of moles of the different species as they change.
Because in chemistry they're going to be changing.
They're not going to be fixed.
So what we want is just purely mathematically formally, take the differential of the Gibbs free energy, which we know is dG/dT, keeping pressure, the number of moles of 1, the number of 2 constant, dT.
That is, dG/dp constant temperature, n1 and n2 dp.
And then we have our two more variables now, dG/dn1, remember this is just a formal statement keeping temperature and pressure and n2 constant.
dn1 plus dG/dn2, dn2 keeping temperature and pressure and n1 constant here.
I'm not writing anything new here, I'm just telling you what the definition of the differential is here, for G. We already know what some of these quantities are.
We know that this is the entropy, minus the entropy.
This here is the volume.
And I know the answer already.
But I'm going to define it anyways.
And we're going to prove it.
I'm going to define this as the chemical potential for species 1.
I'm going to define this, I'm going to give it a symbol, chemical potential mu, for species 2.
So that I can write my new fundamental equation as dG as minus S dT.
Plus V dp plus, and if I have more than two species present, the sum of all species in my mixture times the chemical potential of that species, dni.
The change in the number of moles of that species.
So, this quantity mu, that I've just defined, dG/dni, keeping the temperature, the pressure and all the n's, except for the i'th one constant, that is an intensive quantity.
Because G scales with size, scales, with size of system.
G is intensive.
n, obviously, scales with the size of the system.
Also intensive, and you take the ratio of two extensive variables, you get an intensive variable which doesn't care about the size of the system.
Which is a good thing.
For what we've been talking about.
Intensive.
If I'm talking about putting a freshwater fish and dumping it in the, putting it in the Atlantic Ocean, the chemical potential of the water in the Atlantic ocean better not care whether the Atlantic Ocean is huge or even huger.
It just cares about the local environment.
Just cares that it that wants to be in that freshwater fish.
So the chemical potential is intensive.
Just as we've written it down here for a single species system.
It's the Gibbs energy, free energy per mole.
Which we haven't proven yet.
We haven't proven yet here.
We've just defined it this way and we're going to prove that in fact the mu's are the Gibbs free energies per mole for each of the species in our system.
OK, so this is now our first new fundamental equation.
All we did was to add the sum here.
Now, we started the lecture by saying that if you have the Gibbs free energy, you've got everything.
And we wrote equations that are covered, were we can get S, we can get V, we can get H, we can get u, we can get A.
So now that we have a fundamental question for G, we've got our new fundamental equations for everything else.
Without really thinking too much.
We go back to our definitions.
Enthalpy is G minus TS.
So dH is dG minus d of TS.
dG, we've got our new fundamental equation for G. We plug it in here.
Expand things out with a T dS here, we can write immediately a fundamental equation for H. dH is T dS.
The beginning is going to look just like what you've seen before.
Plus V dp.
Plus an extra term, which is exactly the same extra term that we had in the fundamental equation for G. Exactly the same.
And every one of the other fundamental questions can be derived in the similar way from G. And they're going to be what you had before minus S dT plus minus p dV plus the sum of the mu i's, dni, and du is T dS minus p dV plus i mu i dni.
So immediately we can see that this mu, this quantity mu that we've defined as the derivative of G with respect to the n's, we can write many other equations for mu.
There are many other ways to derive it.
Because this is the differential for H. This is the first derivative of H with respect to entropy.
This is the derivative of H with respect to pressure.
And this is the derivative of H with respect to n. Just formally, that's what it is.
When you write a differential.
So formally, this is also mu i, is dH/dni, keeping, now, we have to be very careful, keeping the entropy and the pressure constant.
Because those are the variables.
Keeping the entropy, and the pressure, and all the other n's, constant.
We can also write it as dA/dni, keeping the temperature and the volume constant.
And all the other n's, or we can write it as du/dni, keeping that this entropy and the volume, and all the other n's, constant.
So we have many ways to write the chemical potential.
They give you all the same result.
So this is the formal.
Sort of the formal part of the chemical potential.
Now, what we really want to show is that the chemical potential really is connected to the Gibbs free energy per mole.
That's going to be the useful part.
Let me get rid of this here.
So I said earlier, at the beginning of the lecture, that the Gibbs free energy per mole was so important, we were going to call it the chemical potential.
And I said that here.
And then I said, well, we're going to define this here, the chemical.
But I haven't equated the two yet.
I haven't proven to you that in fact this quantity here, which we've formally defined as the derivative of G with respect to n, is the Gibbs free energy per mole, for this species.
In fact, what we want to show is that if I take the sum of all the chemical potentials, times the number of moles per species, that that is the total Gibbs free energy.
In other words, that the chemical potential for one species in the mixture is the Gibbs free energy per mole for that species.
Once we have that idea, then we'll be able to talk about the concept of chemical potential as this thing that we can use to look at equilibrium.
To look at going downhill for the species.
To see why the cell bursts and all these things.
Because now we understand that Gibbs free energy is so important for equilibrium.
We don't understand that quite yet, with the chemical potential.
So we got to make that relation here.
We need to go from the formal definition to a relation that we can understand better, because it includes the Gibbs free energy.
OK, so that's our goal now.
So let's see.
Let's formally do this now.
Let's define, let's derive this.
OK.
So remember, our goal in this derivation is to show that this is true.
Or that this is true, here.
And again, we're going to start with the simplest system possible.
We're going to start with a two component system.
And we can easily generalize to multi-component.
And in our derivation, what we're going to be after is, we're going to start with the Gibbs free energy, because that's where we always start with.
And we're going to remember that by definition, mu i is dG/dni, So if somehow in our derivation dG/dni falls out, that would be great.
Because we'll be able to replace this derivative with the chemical potential.
So the goal was to find something where this falls out, so we can replace it with the chemical potential.
We're going to start with the fact that G is an extensive variable.
So if I take G at a temperature and pressure times some scaling factor for the size of my system, lambda, number of moles of n1, lambda times the number of moles of n2, if I double the size of the system, lambda is equal to 2.
If I half it, lambda is equal to 1/2.
Because it's extensive, this is the same thing as lambda times G of temperature and pressure, n1, n2.
Just rewriting the fact that Gibbs free energy is an extensive property.
And lambda is an arbitrary number here.
Arbitrary variable.
Now I'm going to take the derivative of both sides with respect to lambda.
So I'm going to take d d lambda of this side here.
And d d lambda of that side here.
Now, lambda here is inside the variable here.
So I'm going to have to use the chain rule.
To do this properly.
So this is going to be dG/d lambda n1, there's lambda sitting in the variable lambda n1 here, times d lambda n1 d lambda.
Plus dG/d lambda n2 times d lambda n2 d lambda.
Using the chain rule.
And on this side here, lambda's sitting straight out here.
So this is very easy.
This is G(T, p, n1, n2).
Now, this is good.
Because this is what I'm looking for.
I'm looking for the derivative of g with respect to the number of moles.
Because that's the chemical potential.
That was my goal up here, to make sure in the derivation, somehow, this came out.
And so it's coming out right there.
Right here and right here.
Since the number of moles is lambda n1, that first derivative here is just the chemical potential of species 1 there.
So mu 1, then we have d lambda n1 d lambda.
Well, d lambda n1 d lambda, that's just n1.
It's lambda times the number n1 that doesn't have anything to do with lambda.
So this is n1.
This is the chemical potential of species 2.
Again, the derivative of lambda n2 with respect to lambda is just n2.
And there is G here.
It's a fairly simple derivation, but it gets us exactly what we want.
An association between this formal definition of mu, up here, directly from taking the differential.
How much more formal can you be, mathematically, here?
To associating this formal definition to the Gibbs free energy.
Number of moles times mu 1, number of moles times mu 2, this is the Gibbs free energy per mole of species 1.
Gibbs free energy per mole of species 2.
The sum of all the species of the Gibbs free energy per mole of species i times the number of moles of species i is G. Or, mu i. Voila, we've done it.
This is what we wanted.
The chemical potential is the Gibbs free energy per mole.
And in the mixture, it's the Gibbs free energy per mole of the individual species in that mixture.
And if you want to know what the total Gibbs free energy is, because if you have an equilibrium, what you care about is the total Gibbs free energy.
It's not the Gibbs free energy for one particular species.
What's going to tell you whether you have a minimum or not in your system, whether you're at equilibrium, where you're at the lowest state possible, is the total Gibbs free energy.
Now we'll be able to manipulate chemical potentials, of the individual species, to get at this number here.
Any questions?
This is really, we're going to see this over and over again now.
This chemical potential.
This idea.
And it's not an easy concept.
OK, let me give you an example, then, of the water melting.
And how the chemical potential comes in, now, instead of using the chemical potential.
Instead of the Gibbs free energy.
This is the phase transition.
But it's not very different than the cell bursting when you put it in distilled water.
So, we take a glass of water with an ice cube in it.
H2O liquid.
This is H2O solid.
And I'm looking at the melting process.
I'm looking at a process where I take a small number of molecules of water from the solid phase.
And I bring them to the liquid phase.
And I want to know, is this process spontaneous or in equilibrium, or not possible?
Is this process going to go on?
Is the direction of time that this is melting.
And I want to do this formally thermodynamically.
In terms of the chemical potentials.
That's going to be what we're going to be talking about.
So, formally then, what's going on is, I'm taking nl moles of liquid water, H2O liquid, which is in here.
Plus ns moles of solid water.
And I'm, this is my initial state.
My final state is nl, plus a small number of moles, dn, of H2O liquids, H2O liquid, plus ns minus dn, there's a conservation of the number of molecules here.
Whatever I add to the liquid has to come from the solid here.
Of H2O solid, of ice.
And to know if this is spontaneous or not, if this is done under constant temperature and pressure, what variable should we look at?
G, right.
We want to look at the Gibbs free energy.
So what is G doing during this process here?
What is delta G here?
Well, we have a way of doing it now, in terms of the chemical potentials.
Because we've just shown that this is the case here.
So G is the sum of the chemical potentials times the number of moles in the species.
Therefore, delta G is going to be equal to mu for the number of moles of l. Liquid, dn, number of moles of liquid, plus mu s dns.
So the change in G is going to be equal to the chemical potential times the change in the species, which is in a liquid form, plus the chemical potential of the solid.
Times the change in the species in the solid form.
Now, dns is equal to minus dnl.
This is what we did right here.
You take a certain number of moles from the solid form.
You put it to the liquid form.
That the dn up here.
And you've got to have the negative of it, up here.
So dns is minus dnl.
Which is minus dn.
Delta G is dn times mu l minus mu solid.
So now we can rephrase this, it's all rephrasing.
It's all basically the same thing.
But, we can rephrase this process by asking the question, is the chemical potential of the liquid greater than, equal to, or less than the chemical potential of the solid?
Of the water in the solid.
So the chemical potential of the water in the liquid phase is greater than the chemical potential of the water in the solid phase, mu l is greater than mu s, then delta G becomes positive.
In that case, delta G is greater than zero.
And that's not going to happen.
On the other hand, if the chemical potential of the water molecules in the liquid phase is smaller than the chemical potential of the water in the solid phase, mu s is bigger than mu l, this becomes a negative number.
Delta G is less than zero.
And this will happen spontaneously.
So that illustrates this idea, that the chemical potential of a species will want to go, so the species will want to go, where it can minimize its chemical potential.
So in this case here, when we have the spontaneous process of the water, of the ice cube, melting, you can think of it as these water molecules that are in the ice phase looking around.
They know what their chemical potential here is in the ice phase.
They're looking around, they're looking to see the water phase.
And they see that in the water phase, those water molecules have a smaller chemical potential.
They're happier.
And so these solid water molecules are jealous.
And they want to go in the water phase.
And the ice cube's going to melt.
And it all has to do with this difference in chemical potentials for the water.
And the same thing happens for the water molecules that are inside or outside of that cell that you put in the distilled water.
The water molecules in the distilled water have a chemical potential which is higher than the water molecules inside the cell.
And they don't want to be like that.
They want to change, the system wants to change, until the water molecules couldn't care less whether they're in the water phase, or outside or inside the cell.
The system is going to change until the water molecules have the same chemical potential everywhere.
Where they don't have to choose one place or the other.
And so that gives us, immediately, what we're going to need when we talk about equilibrium.
Equilibrium, chemical equilibrium, is going to be where the chemical potential of a species is the same everywhere in the system.
So at 0 degrees Celsius, one bar, which is the melting point of water, the chemical potential of a molecule of water in the ice phase and in the liquid phase is the same.
That's the definition of the melting point.
It doesn't care.
It could go either way.
It's an equilibrium.
You take an ice cube, water, liquid water, 0 degrees Celsius, one bar.
You come back three days later.
It's the same.
Come back a week later, it's the same.
It's an equilibrium.
Chemical potential of the water species is the same everywhere.
It's an equilibrium.
And I'm just repeating that because this is so important.
OK, any questions?
The last thing we're going to do is to illustrate also the importance of mixing to the chemical potential.
So I'm going to set up sort of an arbitrary system here.
This is kind of like the cell, or the fish, also, idea.
I'm going to put a system where on one side I have a pure gas, A.
On the other side, I have a mixture of A and B.
And here is going to be a membrane that only allows A to go through.
Only A can go through that membrane.
They're going to be partial pressures in here, p prime B, and p prime A.
For the gas pressures on that side, and pressure on that side is p sub A.
And my goal in this example here is to show that if I compare the chemical potential of a species in a mixture, where the temperature and the pressure total are T and p, and I compare that to the chemical potential of the same species when it's pure, when it's not mixed with anything else, under the same temperature and pressure conditions, that, in fact that that equals sign is not there.
That's not what I'm trying to show.
I'm trying to show that there's a less-than sign right here.
To show that if you take, again, this is the cell idea.
If you take the water and the distilled water, under constant temperature and pressure conditions, it's pure.
And it's looking at the cell.
And inside the cell is there, boy, is there a mixture of things.
There's salts, there are proteins, there are all sorts of things, right?
It's a mixed system.
The water in that mixed system, under the same temperature and pressure conditions, the chemical potential of that water molecule is less.
And that's just an entropy thing.
Entropy wants to increase.
It just wants to be in a place with high energy.
It's the Gibbs free energy.
Gibbs free energy has enthalpy and entropy incorporated into it.
The enthalpy's not doing anything.
It's all driven by entropy.
So this is what we're going to try to show.
And I'm not going to get to it today.
We'll start with it on Wednesday.
And I'll let your ruminate on this for the next few days.
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So let's start with some of the things we learned last time.
So there are two things that were important.
We learned that the chemical potential for a species is the Gibbs free energy for that species divided by the number of moles, this is Gibbs free energy per mole.
We learned that the pressure dependence of the Gibbs free energy gives you the pressure dependence for the chemical potential.
That it's equal to the chemical potential at one bar for an ideal gas plus RT log p. We also learned that a species will want to go to minimize its chemical potential, and we saw that for the cell bursting in salt, in distilled water.
Or an ice cube melting at a temperature greater than 0 degrees.
And what we want to prove right now is that if I take a species A, in a mixture, some temperature T, some pressure p, and I compare its chemical potential to the same species, A, when it's pure, the same temperature, the same pressure, what I want to argue is that the chemical potential in the mixture is always less than the chemical potential when it's pure.
The same conditions of pressure and temperature.
And, so we can do a little thought experiment.
Let's do a little thought experiment.
Let's make a box, and in our box we're going to have a partition.
And a flexible membrane here.
And on one side of the partition we're going to have a gas, B.
A gas, A, on this side here.
And gas A on this side here.
So let me just red chalk for A, And I don't think I have any other colors listed.
Yellow chalk for B here.
And everything's one bar.
Everything's one bar.
So one bar B here.
One bar A here, one bar A here .
And this membrane here only lets A through.
And this membrane is deformable, but it's elastic.
You can't deform it forever.
It has some strength to it, right?
So if you push on, it'll push back.
There'll be pressure associated with that.
So the next thing I do, then, in my experiment, shouldn't have done it here.
Next thing I do in my experiment is to break this partition here.
I'm going to break the partition.
And this will cause A and B to mix.
So now in my box I have my partition, my membrane here.
I've got A at one bar here, total pressure of one bar.
And on the other side I have A plus B, with a total pressure of one bar.
That's my initial point now.
What's going to happen?
What's going to happen is that molecules of A here are going to want to go through the membrane to go in this area here.
And there's two ways to look at it.
You can look at it the thermodynamic way, which is the way that we're going to want to be looking at it.
Which is that, from what we're going to prove, is that the chemical potential in the mixture is always less than for the pure substance.
Here we have a mixture.
One bar.
Here we have the pure substance at one bar.
So these molecules are going to look around and say hey, you know, I'm much happier here.
And they're all going to want to go in this area here.
As a result, the volume of this area, if you want to keep the same pressure on both sides, you're going to deform the membrane.
The membrane's going to get deformed.
It's going to bloat on that side here.
It's going to cause an increase in pressure on that side.
And the increase in pressure from the membrane sort of deforming and pushing back, is going to increase until the partial pressure of A, here, equals the pressure of A here.
At which point the flow of A from either side is going to be the same and you're going to be in equilibrium.
And at that point the chemical potentials of both sides are going to be the same.
So I sort of gave you the other way of looking at it.
Which is just purely in terms of pressures.
At equilibrium, the partial pressure of A here has to be the same as the pressure on this side here.
So that the flow of A on either side of that membrane, going from right to left or left to right is the same.
And that's going to cause this mixing to happen.
If you look at it from partial pressure perspective.
But really, it's a chemical potential idea.
Now, the chemical potential, as we saw, was the Gibbs free energy.
And the Gibbs free energy, you can write it as H minus TS.
So basically, in this process that I described, the enthalpy's not doing anything.
These are ideal gases.
They're not interacting with each other.
The only thing that's changing, that's driving the chemical potential, which is basically Gibbs free energy per mole, the only thing that's driving the chemical potential to be lower on this side here, is that entropy term.
It's the entropy of mixing.
So entropy of mixing is really super important.
When we're talking about systems where you have multiple components.
That's going to drive a lot of things.
And in fact that's going to drive equilibrium, as we're going to see a little bit later today.
Alright, let's quickly go through the math and prove this here.
So our goal, then, is to have a mixture, chemical potential of the mixture on one side, and the chemical potential of the pure material on the other side.
And so we're going to start by sort of a similar thing here.
We're going to have a box, let me redo my box here, we're going to have a box with, let me get rid of these one bars here.
We're going to put our membrane in the box.
And we're going to have a pressure, pA, on this side here.
And we're going to have a pressure pA prime for the partial pressure of A.
And pB prime for the partial pressure of B.
And the p total is going to be pA prime plus pB prime.
And I'm going to see at equilibrium, I'm going to write everything I know about equilibrium.
At equilibrium, I know that the partial pressure of A on that side here has to be equal to the pressure of A here.
The partial pressure, the pressure is basically the force of these molecules hitting that membrane per unit time, times the number of molecules hitting.
So we have the same flux of molecules going this way, is equal to the flux of molecules going the other way.
So at equilibrium, pA prime equals pA.
So.
What else can I say?
At equilibrium, in terms of the chemical potentials, I know that the chemical potential of the mixture, mu A in the mixture, temperature under pressure p total.
Is equal to the chemical potential of the pure system, same temperature under pressure is p sub a on that side.
These are the two things that I know at equilibrium.
So let's start to turn the crank.
And see if we can come up with, well, we basically already have this.
If we can massage the right side of our equation so that the pressure term, pA, is p total.
And then we'll have an equation that will compare the chemical potential of the mixture under the same conditions as the chemical potential of A in the pure state.
So, what can we use?
We can use Dalton's law here, which tells us that pA prime is equal to xA p total.
That's from Dalton.
And now, pA prime is equal to pA.
So this is also just pA. And we can plug that in here, and suddenly we've got p total included in here.
Let's just pass these out.
If you want to, thank you.
Thank you.
So, this is equal to mu A pure temperature xA p total.
So what else do I know?
I know that I've written something here, that mu is equal to mu naught, to temp, so this is at one bar.
RT log p. So I'm going to rewrite this as mu A pure temperature T at one bar.
Plus RT log xA p total.
Now, the log of xA times p total is the log of xA plus the log of p total.
It's equal to plus RT log p total plus RT log xA.
So I can lump these two things together.
I have mu A pure T plus RT log p. Well, that's just the chemical potential of A, in the pure state, at temperature T and pressure pT.
That's the equation here to relate pressure, at some variable pressure p, to what it would be at one bar.
Which is that.
Then we have the plus RT log xA sitting here.
Plus RT log xA.
So we're done.
We're done because on that side here I have mu A, the chemical potential of A in the mixture, temperature T, pressure p total.
I've got mu A pure temperature T pressure p total, plus a term.
Plus RT log xA.
Can xA be bigger than one?
xA's the mole fraction.
xA's always less than one.
Log xA is always less than zero.
This term here is always less than or equal to zero.
Therefore, this term is always less than that term.
Therefore the chemical potential in the mixture is always less than the chemical potential inside the pure material.
This is going to be important for the next part of the presentation.
Any questions?
This is what drives the death of saltwater fish in fresh water, right?
Because osmotic pressure is basically given by this basic idea here.
And it's all driven by entropy of mixing.
Sure you don't have any questions?
Speak up.
So now we have all the tools we need to look at chemical equilibrium.
When we have a mixture, and we're going to start with ideal gases.
But everything I'm going to say about equilibrium and ideal gases is valid for solutions.
An ideal gas, and we're going to be talking about ideal solutions.
And molecules in an ideal solution, an ideal solvent, are not very different than molecules in an ideal gas.
They don't interact with each other.
You use concentration instead of partial pressures, but pretty much all the ideas are the same.
All the concepts are the same.
The equations are basically the same.
You just do a little bit of replacement of variables.
But it pretty much is the same thing.
It's easier to think about it, to learn first in terms of the ideal gas, but it applies equally well to what you're more likely to use.
Which is solutions.
So let's look at the prototypical gas phase reaction that everybody writes down when they first do this sort of problems.
The Haber process.
Gas, temperature T, you take nitrogen.
You react it with hydrogen, gas, temperature T, p. And you make ammonia.
NH3 gas T, p. And we're going to ask the question, so I take nitrogen, I take some, hydrogen, I mix them together in a container.
And I make ammonia.
And I'm going to ask, after I reach equilibrium, what is the partial pressure of nitrogen hydrogen and ammonia?
Standard equilibrium problem.
You all, I'm sure you've all seen equations about equilibrium.
Equilibrium constant K, log delta G of the reaction, et cetera, et cetera.
But you probably don't have a really good intuition as to why it is what it is.
So the point of the class here is not to relearn log K is equal to minus delta G, reaction divided by RT.
The point is to learn how we get there.
How we get to that equation.
And what parts are important.
Specifically, how entropy of mixing really becomes key to equilibrium.
But before we get there, let me give you a little bit of history as to why this is such an important reaction.
This reaction, which you see in every chemistry class, was developed by Mr. Haber and Mr. Bosch.
Mr. Bosch made it large-scale around 1910.
And it's the reaction that you use to make fertilizer.
Ammonia is the feed stock to make nitrogen-based fertilizers.
So today, there are a hundred million tons of fertilizer, of nitrogen fertilizer made, using, essentially, the Haber process.
It's a huge, huge, commodity.
1% of the world's energy is taken up to make this reaction.
Almost 1% is about 3/4 of the world's energy is used on this.
In World War 1, Germany was making explosives out of nitrogen feed stock.
And it was getting its feed stock from Chile.
From saltpeter mines in Chile.
Chile was under British hands at the time.
Well, the British didn't let Chile sell saltpeter to Germany.
Germany had to find another way to make ammunitions.
And the Haber process, which had just been invented by Bosch-Haber, became the way that Germany made explosives.
Without this, Germany would have stopped the war in 1916.
Or even before then.
Way before then.
And this process was basically, and Haber and Bosch got the Nobel Prize, essentially, for this process.
For showing how to take chemistry, and doing chemistry and large scale processes under high pressure and high temperature conditions.
Haber got the Nobel Prize in 1918 and Bosch got it in 1931.
This process, arguably, you could say, was the birth of the dominance of Germany as a chemical industry.
Based on that, and the fact that they had to make liquid fuels out of coal.
The syn-gas, or the syn-fuel process.
Also invented during World War 1, where they had to use, they couldn't find any oil and they had to use their coal mines.
And obviously this process, the syn-fuel process, is coming back in vogue with the energy crisis around now.
So they developed a lot of knowledge about how to do large-scale chemical reactions.
And that was the birth of, or the explosion of, companies like Merck and Bayer and Bosch, and BASF, and all these German companies that dominate, basically, the chemical industry.
And when Germany lost the war, the US government confiscated the American divisions of these German companies.
And I'm sure you've heard of Merck as the company.
And you think of Merck as an American company that makes drugs.
Well, that's true.
It's a very large American company that makes drugs.
But there's another Merck, which is the German Merck.
Which also makes drugs.
But it makes liquid crystals.
And liquid crystals is where it makes all of its money right now.
And it's very confusing because they're both Merck.
And they both came from the Merck family from the 1600s that were pharmacists.
But after World War 1, Merck Germany was allowed to use the name Merck everywhere in the world except for the US.
In the US it's called EMD.
Merck USA is allowed to use the name Merck in the US and outside of the US it's called, let me remind myself, it's called MSD.
It's very confusing.
Bayer is the same way.
So Bayer was split up.
And there was Buyer in Germany and there was Bayer in the US.
You've got Bayer Aspirin from the US term Bayer.
And you've got all the other pharmaceuticals from Bayer Germany.
All the chemicals.
And a few years ago, Buyer bought Bayer, and now we have one Buyer-Bayer, depending on where you live.
And so it's very interesting to see the history of all these big chemical companies.
And that's why Switzerland and Germany are still the home of, it's all due to that reaction.
That's why you always see that reaction.
Fertilizers, at the birth of the chemical company, and it's a great example for equilibrium.
So that's the example.
Let's generalize it.
We're actually not going to work on this until the very end.
And we'll see it again.
But let's generalize.
Let's just take a mixture of gases with stoichiometries nu A, a gas, pressure, temperature plus nu B, B, gas, pressure, temperature, goes to nu sub C moles of C, it's a gas, with a temperature and pressure, plus nu sub D moles of D, which is a gas, temperature, pressure.
All ideal gases.
For now.
So this the setup here.
Now, when we write that reaction up here, what we're really writing is, in terms of the process, is take a container that contains A, plus a container that contains B, and go to a container that contains C, plus a container that contains D. And when we write delta G for that reaction here, the process that we're talking about is taking the reactions separately from each other.
That's the initial state.
And the final state is the product.
Separated in their own containers from each other.
And when we say delta G is less than zero, for this process, which means it's spontaneous, we mean the process to go from the separated reactions to the separated products.
But in reality, that's not what you do when you do an experiment.
In reality, you take these two containers and you mix them together.
A plus B.
And you let that react.
And at the end of the day, you have a big container with A plus B plus C plus D inside, all mixed up together in equilibrium.
And this is not the same as that.
It's not the same as that.
Because you've got entropy of mixing happening in all of this.
Forget about A and B interacting with each other.
Entropy of mixing is going to dominate equilibrium.
You've got this problem to deal with.
So that means that we're going to have to worry about, if we're going to want to know at which state the process is in equilibrium, you're going to have to worry about this issue right here.
It's not enough to know what delta G of the reaction is.
So for instance, if I plot, as function of the reaction, I've got the reactants on that side here.
And the products on this side here.
And I want to plot delta G as a function of the reaction.
Well, my initial delta G that I would write, to calculate delta G of the reaction, which is the delta G of the products, minus the delta G of the reactants, so initially I have delta G of the reactants.
That's, in those two boxes, that aren't mixed.
So I'm up here somewhere.
Delta G of the reactants.
And at the end of the process, in these two boxes, when I calculate this delta G naught reaction, I'm sitting somewhere here.
Let's say it's a downhill process.
Delta G naught of the products.
Well, the first thing that happens, when I take these two boxes and mix them together, is delta G naught of the reaction's going to go up or going to go down, or stay the same.
Anybody want to guess?
When I go from here to here, before there's any reaction that happens.
Shall we vote?
How many say that it's going to go down?
How many say that it's going to go up?
How many say that it's going to stay the same?
We have about three people that said it's going to go up.
And a lot of people that say it's going to go down.
And a few people that are abstaining.
OK. We're mixing things.
Anybody want to change their votes?
Something is happening here.
And this entropy term here, when A and B come together and start mixing up.
You're changing your vote.
Alright.
It's going to go down.
Now we have near-unanimity.
So the first thing that happens is, you're going to go down here.
You're going to have delta G naught of the reactants in the mixture.
And if it were to go all the way, if A and B were to disappear, it still would be mixed.
And so the end product here would actually be down here.
It wouldn't be up in the products.
It would be the mixture.
So that's the first thing we know.
That this delta G naught reaction is not the full story.
And along the way, here I have two species to begin with.
I've got two species to end with.
But I've got three different species here in the middle.
So as soon as I form a little bit of products, in this case here, or if I start from products and go the other way in the reaction and form the reactants, the first thing that's going to happen is I'm going to decrease the delta G. Just from the entropy of mixing.
And so if I plot my delta G as a function of reaction conditions, I'm going to get a bowing curve like that.
If the entropy wasn't there, then it would just be a straight line from one to the other.
The entropy of mixing of reactants and products wasn't there.
Actually, let me put it up here.
If entropy of mixing wasn't there, I would start from up here.
And as my stoichiometry changed, I would have a linear curve from here to there as a function of the process of reaction.
But the entropy of mixing is causing my initial state and my final state to go down.
And it's causing a bow in this here.
Because if I have equal amounts of A, B, C, and D, that's a lot of entropy of mixing there.
So equilibrium is actually somewhere down here.
It's where delta G of the mixture is at its lowest.
Delta G of the mixture is at its lowest.
Any questions?
Now we going to do the math.
We're going to see how that comes about.
Let me do it here.
So the question we're going to ask is, suppose that I've got my mixture here.
And I'm sitting somewhere on this curve here.
So I've got pA for partial pressure of A, partial pressure of B, partial pressure of C, and partial pressure of D in my mixture.
And I want to know, I've got this mixture of reactants and products.
Which way is the reaction going to go?
Is it going to go towards the products?
Is it going to go towards the reactants?
Or is it at equilibrium?
And to answer that question, I'm going to let the reaction react a little bit more to create a little bit more products.
Remove a little bit of reactants.
And see what the sign of delta G is for that process.
So I'm going to go from a moles of A, b moles of B, c moles of C and d moles of D in my mixture, which is that point on my graph here.
Partial pressure A, partial pressure of B, partial pressure of C, and partial pressure D, and I'm going to react it a little bit more.
a plus, a minus epsilon times nu A, where epsilon is a very small number.
Of A, b minus epsilon times nu B times B.
And I'm going to have, now, products being formed.
c moles plus epsilon times nu C, I've going to have the stoichiometry in there.
For every nu A moles of A that I lose, I create mu C moles of C. Times epsilon as the scaling factor.
And d plus epsilon times nu D moles of D. That's going to be my new mixture.
And I'm going to ask, as I go from this initial state to that final state, I'm sitting on that curve, which sign is delta G. How do I know what the sign of delta G is?
What is delta G for this process?
Not the reaction with the isolated reactants and products, but the reaction where everything is mixed together.
Where I'm going to have to worry about the chemical potentials of mixtures.
And I'm going to go from one mixture to another mixture, and that's going to be the key to telling me if I'm in equilibrium or not.
But we know how to do this.
We know how to do this.
So I want to calculate delta G for the process is delta G after minus delta G before.
That's delta G after minus delta G before.
And I know that delta G, rather, let's call it G after minus G before, the Gibbs free energy after minus the Gibbs free energy before.
And I know the Gibbs free energy is just the sum of the chemical potentials, right?
You take all the species, and you take all the chemical potentials.
You add it up together times the stoichiometry.
And that gives you the Gibbs free energy.
That's what we learned last time.
So if I want to find what the Gibbs free energy at the end here is, I look at the chemical potentials of A, B, and C and D times their number of moles.
So I look at a minus epsilon times nu A times the chemical potential of A, plus b minus epsilon times nu B times the chemical potential of B plus c, plus epsilon times nu C, the chemical potential at C, plus d, plus epsilon times nu D, the chemical potential of D. And then I subtract what G was before.
This infinitesimally small process.
Was a times mu A plus b times mu B plus c times mu C, plus d times mu D. And we're assuming that my small change, the small amount of A and B that get destroyed to form C and D is small enough that the chemical potential basically stays the same during this infinitesimally small change.
That's why I can use the same chemical potentials before and after.
OK, a lot of things drop out.
This term drops out from that term.
This term drops out from this term.
This term drops out from this term.
This is a minus, no, this is all plus.
That's fine.
There's the minus sign right here.
So then this term drops out from that term.
And I am left with epsilon times nu C mu C, the chemical potentials of the products minus the sum of the chemical potentials of the reactants, nu A, mu A plus nu B mu B.
That's the delta G for this small change.
Now, you remember way back, maybe from even just last semester if you've taken 5.112 last semester, or from last year, or from high school, that the partial pressure is going to the equilibrium constant.
Somehow we're going to have to get partial pressures in there.
But we know now how to go from chemical potentials to partial pressures.
It's written right here.
The chemical potentials of the, and we also know how to go from the chemical potential in the mixed species, in the, mixture to the chemical potential in a pure.
We saw that mu A in the mixture, temperature, pressure was equal to mu A pure temperature, pressure plus RT log xA.
So those are things that we're going to be using, to go from something that has chemical potential to something where we'll be able to get back pressures, partial pressures, and delta G of the reaction, because we're going to need the chemical potential of the pure stuff.
So let me go forward just a little bit.
Remind you, if I look at the delta G of the reaction, delta G of the reaction, in terms of chemical potentials.
Delta G naught of the reaction.
This is the delta G of the products minus the delta G of the reactants when they're pure, not when they're mixed.
So delta G naught of the reaction is nu C mu C at one bar pure plus nu D mu D pure.
Minus nu A mu A pure minus nu B mu B in the pure state.
That's what delta G naught of the reaction is in terms of the chemical potentials of all these species.
Everything's at one bar, and everything is pure.
Somehow this is going to have to come out of this.
So let's keep going.
And see how it falls out.
So for each one of these chemical potentials, I'm going to write it in terms of the pressure.
What I just covered up here.
Mu(T, p) is mu naught of T times RT log p. So now, delta G is going to be equal to epsilon, and I'm going to do a little massaging quickly.
And I'll let you take this, go home and see how I went from one state to the other.
The secret is to put in here the pressure dependence.
Nu C mu C naught, plus nu D mu D naught, minus nu A mu A naught, plus nu B mu B naught, plus RT log pC to the nu C, pD to the nu D, divided by pA to the nu A, pB to the nu B.
These log partial pressures all come from expanding out the chemical potential as mu naught plus RT log p. And then we recognize that this part right here, this part right here is delta G naught reaction.
So I have delta G now, for this process of taking reactants to products just a little bit, let me take it as a function of epsilon here.
Delta G naught of the reaction plus RT log of this ratio of partial pressures.
I'm going to call that Q. I'm going to call this thing here Q, which you've seen before.
The reaction quotient.
And this tells me that for this very small process, epsilon, very small, if delta G naught, delta G, of this process, is less than zero, then the reaction will keep going forward.
It means that I'll be on the side of the curve here.
I'm going to go down a little bit.
delta G naught is going to be great, it's going to be spontaneous.
I'm not in equilibrium.
I'm going to go towards the products.
If delta G is equal to zero, then I'm at the bottom of this curve here.
I'm here.
If I go forward a little bit, the slope is zero.
Delta G is zero, I'm in equilibrium.
And if delta G is greater than zero, then I'm going to back to reactants, and I'm sitting, then, on that part of the curve here.
And if I try to make more products I'm just going uphill a little bit.
So now, I've done this all for epsilon is equal to very small.
What you usually find written in books is where the epsilon equals to one.
Basically taking one mole of this process.
So you'll see it per-mole of this reaction.
But it's the same idea.
It's the same thing.
And in fact it doesn't really matter, because the quantity that matters is this delta G naught reactions plus RT log Q.
It's the sign of this quantity here.
This epsilon is just an arbitrary number.
I can pick it, really whatever I want.
So it's the sign of this thing that's important.
Let me go back.
So what you'll see, then, is delta G is equal to delta G naught reaction plus RT log Q.
Basically, taking epsilon is going to zero.
As determining where the equilibrium is going to be.
OK, any questions?
All driven by entropy of mixing.
Without entropy of mixing, we would be sitting on this curve here.
Delta G naught of the reaction would tell us that everything should go to completion.
Things don't go to completion, and that's a good thing.
Otherwise there would not be life on Earth.
We're basically a set of equilibria in a big membrane.
Which is our skin, right?
All these biochemical cycles are in equilibrium with each other.
And it's a complicated process.
And it's all driven by entropy.
Ultimately.
And other things, but entropy is very important.
Alright, so equilibrium now.
Equilibrium is when we have this delta G equal to zero.
That's when delta G naught of the reaction equals RT log Q.
And at that point, we replace Q with to equilibrium, and we call that the equilibrium constant.
And we're going to put a little p here, because it's in terms of the partial pressures.
And it's equal to the partial pressures of the products raised to their stoichiometry.
And divided by the partial pressures of the reactants, raised to the power of their stoichiometry.
And this, you've seen before, I'm sure.
At equilibrium.
And there's a minus sign somewhere here.
Because it's this plus that that's equal to zero, and there's the minus sign that I forgot to write down.
So we define, then, equilibrium constant this way.
And one of the things to note is that Kp as written here is actually unitless.
It doesn't look like it.
The way that I've written it.
Because I did a shorthand way of writing the pressures, when I wrote RT log p here.
There's the assumption, when you have the log of the pressure, that there's always one bar sitting behind there.
Underneath.
It's always referenced to a reference pressure.
There's always a reference pressure.
p naught dividing it by, because you don't want to have any units inside the log.
And this reference pressure, we took as one bar.
And it's pretty common to just ignore the fact that you've got one bar sitting in the denominator.
And so, actually, all these pressures here are divided by p reference divided by p reference, divided by p reference, divided by p reference, which happens to be 1 bar.
But it's there.
And that means that the bars on top and the bars on the bottom cancel out.
Which means that K sub p doesn't have any units.
It is unitless.
It's a number.
Straight number.
Very common mistake to make, to write it and forget that there's one bar sitting on the bottom here.
And you take all the bars to the new powers on top, and the bars to the new powers on the bottom.
And make units there.
And get Kp is equal to some number, to the, times bar to some power.
That would be wrong.
I know I've made that mistake before.
But you are not going to make that mistake, right?
Because you've been warned.
That this is a common mistake.
This is unitless.
Now, you can invert this to get Kp as a function of delta G. e to the minus delta G naught of the reaction divided by RT.
And those are the things that you use to go back and forth between the thermodynamic quantities, like G that you calculate, to equilibrium quantities like the K, to finding equilibria between something like the Haber process.
Now, there's other equilibrium constant that is used a lot.
Which is the form of the equilibrium constant not in terms of the partial pressures, but in terms of the mole fraction.
That's also an important one when you're looking at solution cases.
Because we can rewrite the partial pressures using Dalton's law.
pC is equal to the mole fraction of species C times the total pressure, I'll call it p. So if I replace every one of these partial pressures using Dalton's law, I get K sub p is equal to xC pC times p to the nu C, times xD pD to the nu D, divided by xA pA - no, not pA, total pressure.
Dalton's law.
To the nu A, xB p to the nu B. OK, so I've got p, p, p, p. They all come out.
And that's p to the minus delta nu.
Where delta nu is the difference in the number of moles of reactants minus the number of moles of products.
So delta nu is nu C plus nu D minus nu A minus nu B.
It's the moles of products minus moles of reactants.
And then I have xA to the nu A, xC to the nu C, xD to the nu D, xA to the nu A, xB to the nu B.
And that ratio, which is in terms of mole fractions, we call K sub x.
The mole fractions.
Because p, to the minus delta nu, times Kx.
Now, this was unitless.
This pressure to the minus delta nu sitting here, this has bars to the minus delta nu.
Kx has units.
It may not look like it because it's a bunch of mole fractions, and it certainly doesn't look like it has any units.
Mole fractions or ratio, but it's got units.
So let's rewrite Kx in terms of Kp.
Rewrite Kx as equal to p to the minus delta nu, K sub p. And the units are in terms of bars, let's say, bar to the minus delta nu.
Always want to check your units at the end of the day.
If your units don't work out after you've done a calculation, you're in trouble.
Any questions?
Because we're going to end today, at here.
Yes
[INAUDIBLE]
In the notes it says Kp and Kx are both unitless.
That is a mistake.
I'm going to fix that, and put it on the Web.
So it may be that there's a special case where the number of moles before and after are the same.
But generally that is not true.
Where did I say that?
Oh, yeah.
Both unitless.
Well, that is actually true.
Let me think through this
[INAUDIBLE]
About one bar, one bar, one bar.
That is actually true.
They're both, that is true.
They are both unitless.
Well, this is a big boob on my part here.
Because you are absolutely right.
Because there's one bar sitting here.
one bar, one bar, one bar, one bar, and the bars cancel out.
Good catch.
OK, the notes are right.
My notes are right.
Alright.
Next time we'll talk about the temperature dependence and the pressure dependence of equilibrium constants.
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I want to remind and clarify as needed the equilibrium constant Kp for gas phase reaction was the ratios of the partial pressures referenced to some reference pressure, which we usually take as one, one bar.
To the stoichiometry , pD divided by p naught, to the mu D where species C and D are products.
And the reactants are on the bottom.
And usually we don't write p naught, but it's important to remember that it's there.
And then we can also write this in terms of the Gibbs free energy for the reaction.
The standard Gibbs free energy, minus delta G naught of the reaction, divided by RT.
And what this tells us is that this is a number.
This is a number, there's no pre-factor here that has units.
That's a unitless number.
And it doesn't depend on the total pressure.
And the other thing to remember is that delta G naught for the reaction is the process of taking the reactants separated in separate boxes, separate containers, and the final product, the final step, is the products separated in individual containers.
That's what we write when we write delta G naught for the reaction.
We also looked at K in terms of mole fractions.
So if you replace all the partial pressures with the mole fraction times the total pressure, you get an expression for K sub x, which we define as the mole fraction of the products to the stoichiometric powers.
Which is also unitless.
This is true, it's unitless.
But, if you write it in terms of K sub p, the total pressure comes in here.
p total, divided by the reference pressure to the minus delta nu, where this is the change in the number of moles, in going from reactants to products.
And there's K sub p sitting here.
So unlike K sub p, which doesn't depend on the total pressure, K sub x does depend on the total pressure through this term right here.
So when we look at problems where we change the pressure, the total pressure of the system, this is going to stay the same.
Because it only cares about delta G naught for the reaction.
But this K sub x will depend on the total pressure.
And that's often a source of confusion in doing problems.
OK, any questions?
We're going to do an example where we change the pressure first.
So there are examples in the notes, and I'm going to skip the first one.
I'm going to go to the second one, which is the effect of the total pressure on the reaction.
And Le Chatelier's principle, for pressure.
And the example we're going to take is a fairly standard example, also.
Which is the reaction of N2O4, which is a gas, to 2 NO2, which is a gas, the kind of reaction that happens when you have smog and, fairly common in big cities.
The question we're going to ask is, what happens when we change the total pressure in this reaction here.
Which way does the equilibrium go?
Does it go to the right, does it go to the left?
Does it go to the products or the reactants.
And so, in order to answer that question, we're going to ask a slightly different question.
We're going to ask what is the molar ratio, what is the fraction, what is fraction of the reactant, the N2O4 that's reacted.
That has reacted.
And we're going to call that alpha.
It's the number of moles that have reacted divided by number of moles initially.
So we're going to need to find at equilibrium what is the number of moles of N2O4 that has reacted.
So we have to set up the problem.
And so the way that, the standard way of setting up the problem is to write the equilibrium, 2 NO2 gas.
And then on this line here, we write the initial conditions before we set up the equilibrium.
And let's say that we have n moles of N2O4 initially in the box.
And zero moles of the NO2.
So we have n moles here.
And zero moles here.
At equilibrium, let's write the number of moles.
A certain number of moles of N2O4 will have reacted, let's call that x.
So n minus x moles left.
For every x moles of N2O4 that's reacted, we create two moles of NO2.
So we have 2x here.
And then we're going to need the total number of moles, because we're going to be doing mole ratios, mole fractions.
So the total number of moles at any time is the sum of these two, n minus x plus 2x.
It's n plus x.
So if we're going to be writing our equilibrium constant in terms of mole fractions, we're going to need mole fractions.
So the mole fraction at any time is n minus x divided by the total number of moles, which we just calculated as n plus x.
And this is 2x divided by the total number of moles, n plus x.
And what we want is this ratio here.
We want the ratio of the number of moles reacted, which is x, that's the number of moles that's gone.
That have reacted.
Divided by the number of moles initially, which is n. That's what we want.
We want to see how that is going to change with pressure.
So we're going to deal first with Kp, because Kp doesn't depend on total pressure.
We're going to write that down.
Then we're going to go to Kx, somehow.
And that's going to depend on pressure.
So let's see what Kp is here.
So K sub p, you've got the products on top.
So it's the partial pressure of NO2 to the second power, divided by the partial pressure of N2O4 to the one power.
And everything is referenced to one bar, everything's in bar.
And in terms of the molar fractions, it's the total pressure squared, times the mole fraction of NO2 squared, divided by the total pressure to the first power.
So the square root on top gets divided by one factor of pressure.
So we have total pressure in front.
Divided by x to the N2O4, and that's p times Kx.
So let's plug in what these mole fractions are from our table here.
The mole fraction of NO2 is 2x divided by n plus x.
2x divided by n plus x to the square power.
Mole fraction of N2O4, n minus x over n plus x. n minus x over n plus x.
Multiply, square the top, 4x squared.
Divided by n plus x squared.
Things sort of cancel out here.
Rearrange.
4x squared divided by n squared minus x squared.
What we're really interested in is, we're not interested in x.
We're interested in x divided by n. So let's divide both the top and the bottom by n squared.
And we're going to get alpha come up.
So this is then p times 4 alpha squared divided by one minus alpha squared.
This is not, there's no total pressure here.
The only way, the only place, where the total pressure comes in, is right here.
This is just a number that doesn't care what the total pressure is.
Which is why we're using it.
And not K sub x, which cares what the total pressure is.
So now we can solve for alpha.
We can solve for alpha by rearranging this equation.
This is just a number.
And this is where the total pressure comes in.
So you rearrange that, and you get alpha is equal to 1 plus 4p divided by Kp, to the minus 1/2 power.
And if I rewrite that slightly to make it a little bit easier to see what's going to happen, when I change the pressure, 1 plus 4p divided by Kp, to the 1/2 power.
And that's what I'm after.
This tells me what happens at equilibrium to the amount of NO2 as I change the total pressure.
This is the only place where it comes in.
So now I can see that if I raise the pressure in my container, raise the pressure, this is in the denominator, so this fraction gets smaller.
Alpha gets smaller.
I raise the pressure, the fraction of material that reacts gets smaller.
Therefore, the reaction goes towards the reactants.
If I decrease the pressure, this is a smaller number here.
The fraction gets bigger.
Alpha goes up.
If I decrease the pressure and I compare what happens to the number of moles of reactants that react, more of it reacts.
Equilibrium shifts towards the product.
And this is Le Chatelier that you already know.
Le Chatelier's principle, for pressure.
The way it works is that Le Chatelier's principle states that, this chemical system wants to stay as close to what it was before.
It doesn't like change.
It doesn't want to have any change happen.
So if you increase the pressure, the chemical system says, hey, you know I'm not so happy that you're increasing the pressure on me.
I'd like to go back to a smaller pressure.
It doesn't like change.
Very conservative.
And the way to decrease the pressure is to decrease the number of moles in the container.
How does it decrease the number of moles?
Goes back to where there are fewer moles.
And that's on the reactant side.
How many of you know Lenz's law?
In magnetism.
The diamagnetic materials.
Right.
At least one person knows it.
It's the same idea.
You take a diamagnetic material in the absence of a magnetic field, and you slowly move it into a place where there's high magnetic field, what does the diamagnetic material do?
It orients its magnetic moment to reverse the field.
So that there's no field inside.
It starts out with no field.
Doesn't like change.
So Lenz's law says it's going to do whatever it can so that it retains no field inside.
Le Chatelier's principle is basically the same thing.
Equilibrium systems are very unhappy if you try to change them.
And that's what happens for Le Chatelier here with pressure.
Any questions?
So before we go to Le Chatelier's with temperature, and the van 't Hoff equation.
Let's do a little detour here and talk about equilibrium in solution, which is really as important, or if not important for a lot of you, then gas phase equilibrium.
Although gas phase equilibrium was where everything started.
And still a huge deal.
OK, so in equilibrium now, when we talk about equilibrium in solution, we still have to, still going to be the chemical potential.
It's still going to be looking at how chemical potential likes to go downhill.
And we're going to have to write chemical potential for a species, A, let's say, which is in solution.
At some concentration c sub A in solution.
And the concentration could be given in moles per liter.
Or it could be in grams per liter.
Or it could be in grams per 1000 grams.
Whatever your favorite unit of concentration is.
Use it.
Stick to it.
And in order to do equilibrium, we're going to have to reference it to, so this would be the species at some arbitrary concentration.
We're going to have to reference it to some reference concentration.
Just like we referenced everything to one bar before, as our standard pressure.
And we're going to take, usually you take one mole per liter, or one gram per liter, or one whatever.
One as your reference concentration.
And the reference concentration is going to disappear from the equation.
It's just like the reference pressure.
Disappear from the equation.
So we're going to reference this to some standard state chemical potential.
Where the naught refers now to the standard concentration.
And instead of having RT log p, now we're going to have RT log cA.
It's kind of like considering the molecules in the solution to act like an ideal gas.
Knowing fully well that behind, that underneath the cA, is this reference concentration of one.
One whatever is your favorite units.
Now, it's a little bit more complicated than for the ideal gas.
Because your solution may contain other things than your reactants and your products.
Especially if you're doing biology.
It could be a buffer.
It could be a buffer, it could have salt.
It could have a pH that's not equal to seven, whatever.
And so this reference, chemical potential, now needs to be referenced to a particular pH or salt concentration.
Or whatever the properties of your solvent are.
Or your solution are.
And that's the big difference.
In an ideal gas, it's reference to vacuum, basically.
There's nothing there.
In here, in solution you have all these molecules of solvent, molecules of salt, molecules of acid, or whatever, that are going to be around to buffer the pH.
And that's going to change what the chemical potential of a species is.
And if I change the pH and I've got a molecule that I'm interested in, it may not have an acidic moiety on it, but it could still care what the pH is, slightly.
And that would change what the reference potential is, chemical potential is.
So this is really important to remember.
And there are textbooks that are written on how to do this the right way.
We're not going to do that here.
We're just going to remember this is, we're going to assume that this is done correctly.
Once you take that as a given, that you have a way to have a reference chemical potential at a properly referenced pH and salt concentration, then you can go through the same analysis that we went to for partial pressures.
This looks just like the ideal gas, where the concentration replaces the partial pressure.
Or the pressure of chemical A.
And you can go through, then the same argument, where you take your reaction.
And you initially have some delta G for the reactants.
And you have some delta G for the products.
And the difference is the delta G for the reaction, delta G naught for the reaction, and then on this side here, you have a solution of A, so the reaction would be nu A times A, which is in a solution.
Temperature and pressure.
Plus nu B of reactant B, in a solution, temperature and pressure.
Going to a nu C, C solution, temperature and pressure plus nu D, D in a solution, constant temperature and pressure.
So this is taking a solution of A, in one container, a solution of B in another the container.
That's the initial point.
Mix them together.
Let them react.
Then you take the product, you put them in separate containers.
And that gets you the stuff that you have the reaction.
So when you mix A and B, you're going to have the same entropy of mixing.
You're going to lower the delta G. Of the solution.
And you're going to have the same curve that goes down like this.
To the mixture of products.
And just like for the gases, where we wanted to know what is the bottom of this curve which gives us the equilibrium, we can do the same thing.
Exactly the same thing, for solutions.
And so we start out with a mixture of the A and B in solution and C and D, reactants and products together.
And we let the reaction proceed a little bit.
And we look at the change in delta G, going from, say this point here through that point here.
This will be delta G of epsilon.
We ask, is this positive, negative, or zero.
And if it's zero, that means that we're in equilibrium, that we're actually sitting down here.
And that gives us the equilibrium constant.
So, just for the sake of completeness, let me just write down what we would do.
We would react it for a small amount.
And then we'd end up with nu C. So you would have the chemical potentials of the products minus the chemical potentials of the reactants, nu C mu C, plus nu D mu D, minus nu A mu A, minus nu B mu B.
And instead of these chemical potentials, you would write them in terms of the pure chemical potentials times their concentrations.
And then you'd end up with epsilon times delta G naught of the reaction, plus RT log, and then the concentrations.
And then you write them in a different way.
So if it's moles per liter, you usually write that with these brackets here.
That means concentration of A in moles per liter, I'd say.
So C to the nu C power, D to the nu D power, A to the nu A power, and B to the nu B power, in these concentrations.
Where this ratio of logs comes from expanding out the chemical potential here.
And there's the log term here.
Just like an ideal gas.
Then at equilibrium, this is equal to zero, you're at the bottom of that curve.
And you set these two things equal to each other.
And you get the chemical, you get your equation that you know well.
For the equilibrium constant.
And this time it's not K sub p, it's just K. And that's what you know from doing solution equilibrium.
And it's just like V. So the thing to remember, which is the slightly more advanced part, which you'll have to worry about at some point if you stay in some sort of biochemistry oriented field, is that you've got to reference your initial solution properly.
To get to the right equilibrium constant.
OK, any questions?
Alright, then now we can do the temperature dependence of the equilibrium constant in a general way.
Whether it be a gas or a solution.
It doesn't matter.
It's going to be the same thing.
So the question that we ask now is, suppose that I change the temperature of my equilibrium, which way is the equilibrium going to shift?
And you all know the answer already, probably.
But let's derive it out.
So we're going to want to know basically, we want to know what is dKp/dT.
How does equilibrium constant, or dK/dT, if you're doing solution, how does the equilibrium change with temperature?
What's the slope?
Is it positive, negative?
If we have this, we can integrate it out.
We can do an integral over temperature.
And get an actual change.
So that's our goal.
To find how the equilibrium constant changes with temperature.
What do we know?
Well, we know how Kp depends on temperature, through the Gibbs free energy of the reaction.
The Gibbs free energy, delta G naught, has a temperature dependence.
And then there's an RT sitting on the bottom.
There's another temperature dependence here.
Well, dKp/dT is sort of like, we could also ask what's d log Kp dT.
That might be an easier question.
It's basically the same question.
Especially since we have something which is log K, is equal to something.
So let's ask this question instead.
Let's ask, what is d log Kp dT?
Alright, so let's differentiate both sides.
d/dT, d/dT here.
Got to use the chain rule now.
Because we've got temperature as part of delta G Write it out.
So let's take the derivative with respect to the temperature on the bottom first.
We have delta G naught, which is a function of temperature, divided by RT squared.
The minus sign here disappears when you take the derivative on the bottom.
Minus one over RT, d/dT of delta G naught.
So this is a derivative of delta G, where zero means here one bar.
Fixed at one bar.
So really, d/dT, with delta G naught fixed on one bar, is the same thing as the partial derivative of delta G with respect to temperature, keeping p is equal to constant at one bar.
It's the same thing, just different notation.
And we know what this is.
In terms of other things that we can find in books, like delta H, or delta S. Because we can go to the fundamental equations.
To find out how delta G depends on temperature.
And our goal is to get rid of delta G, which clearly has a nice temperature dependence through the entropy term.
And to replace delta G with delta H, if we can.
Because delta h is going to be much less sensitive to temperature and it's, delta H is going to be over small temperature ranges, is going to be independent of temperature.
And we know the temperature dependence of delta H, because it's through the heat capacities.
So our goal is to get rid of this delta G here.
And to try to replace it with delta H, if at all possible.
So we go to the fundamental equation for G, dG is equal to minus S dT plus V dp.
And sitting right here is dG/dT at constant p. Which is what we have here.
So we get rid of this derivative of G. And replace it with S. So now, now we have d log Kp dT, and I mentioned already that I want to get rid of G. Because it has a strong temperature dependence.
And I want to somehow get H in there, which is not going to have a strong temperature dependence.
And delta G naught, I can write in terms of H and S, and T. Delta H naught minus T delta S naught divided by RT squared.
And then my derivative here, I have minus one over RT.
Partial of G with respect to T. p is equal to one bar.
Well, that's just delta S. p is equal to one bar, well, that's just delta S naught.
Times delta S naught.
And that's great, because now there's minus T divided by RT squared.
That's one over RT.
And somewhere I've lost a sign.
And there's my sign that I lost, right there.
This minus sign here.
d/dT of delta G naught is minus S. It actually includes this minus sign right here.
Which is great, because now things work out, because this becomes a plus sign.
And this and this cancel out.
And this becomes delta H naught over RT squared.
Great, so we have what we want.
We have how the equilibrium constant depends on temperature in a way which is very clear.
Where the top part is only very weakly dependent on temperature, usually.
And this is called the van 't Hoff equation.
And this will tell us what happens to equilibrium when we change the temperature.
So if you want to do a finite temperature change, now what you need to do is, you need to integrate.
You integrate both sides here.
From some T1 to T2.
From T1 to T2, and that tells you, then, that the log of the equilibrium constant at the new temperature is equal to the log of the equilibrium constant of the old temperature, T1, plus the integral from T1 to T2 of delta H over RT delta H naught, over RT squared.
And this could be slightly temperature dependent.
dT.
And this is the integrated van 't Hoff equation.
And if you're going to be designing a chemical plant where you have high temperatures and high pressures around, you better use that.
Because there is some temperature dependence in delta H, through the heat capacities of the reactants and the products.
And that could make the difference between your plant running nice and smoothly or your plant exploding.
And you don't want to have exploding plants around.
So for heavy-duty uses of this equation, you've got to do the integral properly.
But for most normal applications, like if you're doing biology, where the temperature changes by a few degrees, like today I have a little bit of a cold.
I don't have a fever, but I could have a fever.
So my biochemistry would change if I had a fever.
The equilibrium constant of all my reactions would change a little bit.
It's a small change in temperature.
I'm not going to explode.
And so you can then take the approximation.
In that case, the delta H naught, is independent of T. And this is fine over small temperature ranges.
And that's the one that you're most used to.
Is this approximation here, this approximate van 't Hoff equation.
Which is really fine for most cases that you're going to be dealing with.
So then, if that's the case then you can take your delta H. Ignore the temperature dependence and take it outside of the integral.
And now you can do the integral fine.
And then you have an analytic expression for the change in the equilibrium constant with temperature.
Log Kp at a new temperature, T2, is log Kp temperature T1.
And I'm carrying this little p around everywhere.
But really, it doesn't have to be there.
This could be solution.
I shouldn't really have written this for the specific case of partial pressures.
But it's equally valid for solutions.
Then we have delta H naught over R. And then we have the integral from T1 to T2, over one of RT squared.
And if you do that the right way, you get T2 minus T1 over T1 T2.
And that gives you the approximate van 't Hoff equation.
Which is fine.
And you'll know that it's fine in problem sets or exam, because we'll say assume that delta H is temperature - yes
You said that the [INAUDIBLE]
For K, if K is solution
Right
Yeah
So then, can you also use Kx?
Can you use Kx?
Well, as long as you keep the pressure, the total pressure, constant, then you should be able to use Kx.
Let me think about this.
Yeah, here you would have p, you have a log c so, you can use Kx, fine.
Because then you would have log p to the minus delta nu times Kx, log p to the delta nu minus Kx.
And the log of the multiplication is the sum of the logs.
And the logs will just fall out.
So it could be any K that you want.
Doesn't matter.
As long as you keep the pressure constant.
If you change the pressure, then you're in trouble.
So now we can see what happens when you do change the temperature.
If I have some equilibrium, and it's all going to depend on the sign of delta H. Whether the reaction is exothermic or endothermic.
And it's the same thing as Le Chatelier's for pressure, or Lenz's law.
The system doesn't want to have change happening.
So if you have something that's, delta H is less than zero, it's exothermic.
Exothermic, that means that it's putting out heat.
it wants to heat up its environment.
And if I take temperature and I raise the temperature, the system's not going to like that very much.
It doesn't want to get hotter, and going from reactants to products makes things hotter.
And if you go from products to reactants, that's the opposite.
Go from reactants to products, that becomes endothermic.
It sucks in heat.
You raise the temperature.
The system said no, no, no, I'm happy where I am at my original temperature.
I'm going to start sucking in heat, to try to get the temperature down.
And it's going to try to make more product.
More reactants.
So, equilibrium K is going to go down.
and the reaction goes towards the reactants.
And the opposite if you have something that's endothermic to begin with.
OK, delta H is positive here.
Delta H naught is positive.
You raise the temperature, delta H naught is positive, T2's bigger.
This is a positive number.
K becomes larger at higher temperature.
K goes up.
The reaction goes to products.
So if you think of it in terms of the system, the system is at some temperature.
You raise the temperature.
System doesn't like it.
Says, I want to go back to my original temperature, what can I do.
I can try to suck in heat that you're trying to put in the environment.
That's great because if I make more products, that's endothermic.
And I'm just going to make more products until I try to lower my temperature.
So I move the equilibrium to the products.
OK, Le Chatelier for temperature.
Any questions?
On equilibrium.
OK, let's do a quick example.
Because this was the example that we started out with, talking about, the Haber process.
This important industrial reaction that started the chemical industry, essentially.
That uses up 1%, or close to 1%, of the world's energy.
If you think about it, that's an amazing number.
1% of all energy produced in the world goes to one reaction.
One industrial reaction.
Just shows how important it is.
OK, why does it take so much energy?
We're going to find out.
We're going to find out why it takes so much energy to run this here reaction.
Alright, let's look at this Haber process.
Take some nitrogen gas.
Plus some hydrogen gas.
And this is usually done over catalysts, like an iron oxide catalyst or something.
To try to speed it up.
It doesn't change the thermodynamics.
As you know, and you'll hear again in this class, catalysts just affect the kinetics.
They don't change the thermodynamics.
So this is usually done over some catalyst to try to speed things up.
To make ammonia.
And ammonia becomes the feedstock for fertilizers, for almost anything that contains an amine in it, or a nitrogen in it.
If you're going to make proteins or whatever.
You've got to have ammonia somewhere in the process.
OK, delta H naught of the reaction, we're given all these numbers.
At 298 degrees Kelvin.
And they're in your notes, so I'm not going to go through them in detail.
Delta G naught for the reaction, we're given that number.
At 298 degrees Kelvin, that's minus 16, roughly minus 16 kilojoules per mole.
And we want to know, what is the equilibrium constant.
Room temperature.
So you know how to calculate that.
Minus RT log Kp, log K is equal to minus RT.
Minus delta G naught over RT.
So you put that in there.
You get Kp is equal to 860.
A number, no units.
It's a big number.
It's a big number, you've got the products divided by the reactants.
It means that the products are favored.
This is great.
What a wonderful reaction.
Shouldn't take energy to for us to do that, right?
It's a room temperature reaction.
Thermodynamics is great.
But even over a catalyst, this is a really, really, really slow reaction.
We'd still be waiting here for Mr. Haber to produce his first mole of amine, if you were doing it, or ammonia if we were doing it at room temperature.
It's just so slow.
Thus, not at all practical.
We're not going to run the world on room temperature Haber process.
But it turns out, if you raise the temperature, kinetics is wonderful in terms of the temperature dependence.
It's exponential.
Arrhenius rate law.
Great thing, you raise the temperature by a little bit.
Rates speed up, things go faster.
So if you were to raise the temperature from 298 degrees Kelvin to 800 degrees Kelvin, the rate speeds up.
You're going to need some energy.
As input here, to feed that.
Hence the 1% energy use.
Rate speeds up, that's great.
Things happen faster.
It becomes practical.
But, what happens if you raise the temperature?
Let's see.
This is an endothermic, or exothermic, negative sign.
And negative sign's exothermic.
I raise the temperature, K goes down.
I know how to calculate it here.
And if I want to be super careful, because it's a fairly large temperature range, I can even use the exact form of the van 't Hoff equation.
And what I find, if I do that, and putting the heat capacities for all these gases, I find that Kp does go down.
In fact, it goes down quite a bit.
It becomes 0.0007.
Two zero's.
Still really small.
That's not practical.
Not practical at all.
No good.
We can't run an industrial plant with this kind of yield.
There's just no way it's going to work.
So, you're a chemical engineer.
Or a chemist, like Haber and Bosch were.
And you're trying, you know by Le Chatelier, you know that it went in the wrong direction for you here.
And then you look at your reaction and you say, how many moles of reactants do I have?
3/2 plus 1/2, that's two moles of reactants.
And I've got one mole of product.
Two moles reactants, one mole of product.
Two moles reactant... What happens if I change the pressure?
If I change the pressure, if I increase the pressure, the system is going to say, no, I don't want the pressure increased.
It's going to go to where there's less moles.
And the less moles in the product area.
It's going to go to my product.
That's great.
I've got to increase the pressure.
Wonderful.
Let's start increasing the pressure.
Again, we need some energy to do that.
We're going to go from one bar to some higher pressure.
It's going to make our lives more complicated.
The plant might explode now.
If the pressure's too high.
All sorts of problems going to come into play.
But, let's do it.
Let's increase the pressure.
So, you increase the pressure from one bar to 100 bar.
And you calculate Kx.
Which is really what you want.
So Kx, in this case here, is equal to p times Kp.
Kp doesn't change.
Kp doesn't care what the total pressure is.
It's Kx that cares.
At one bar, Kx is equal Kp.
Kx is p to the minus delta nu.
Number of moles of products minus the number of reactants.
If I go from one bar to 100 bars, Kx goes from 0.007 to 100 times 0.007, which is equal to 0.7.
That's a lot better.
Kx is the mole fraction or the, of products divided by reactants.
And if I go to p is equal to 300 bars, then Kx goes to 2.1. three times 0.7.
This is great.
Now I'm really starting to make good products.
But I've got to go to 300 bar.
I've got to go to 300 bar, and 800 degrees Kelvin.
That is incredibly energy-intensive.
But it works.
That's why Haber and Bosch made this work.
And why Germany stayed in the war longer than after 1916.
The first world war.
Ended in 1918.
Nobel Prizes.
Merck, Bayer, all these german companies.
Because they figured how to do this at high pressure and high temperature.
Without blowing everything up.
OK. Any questions?
Alright.
The last topic is, so far we've seen equilibria where you had things that were well mixed.
equilibria of ideal gases.
Or in solutions, where your solutes, your solute molecules are mixing around.
And the entropy of mixing was really super important.
Our curve, our going down for delta G, was all because of the entropy of mixing.
Now, suppose that I have a heterogeneous mixture.
I've got some solids or some pure liquids that are refusing to share their environment.
And staying as pure materials.
So, for instance, if I have a beaker with some solid reactant on the bottom here.
And the products are in solution.
How do I deal with that equilibrium?
Well, you know the answer, but let's just do it out again.
So we're going to have nu A moles of A, of solid.
Not mixed in the solution.
Let's say we have multiple phases here.
Nu B moles of B, which is a gas.
Instead of a solution, let's do a gas phase reaction.
Nu C moles of C, which is a pure liquid.
And nu D moles of D, which is a gas.
So these two gases can mix.
But the pure solid and the pure liquid can't mix.
So let's think again, where does the equilibrium constant come from?
It comes from looking at this delta G of the mixture and letting it react a little bit more.
And taking out these chemical potentials for the species and the mixture.
Expanding it out in terms of log p or log concentration.
So we need to have this delta G in.
Let's take epsilon equal to one, to make our life simpler here.
So now we have nu C mu C of the pure.
The pure solid.
Plus nu D mu C of the gas.
Which is in the mixture.
Minus nu A mu A of the pure liquid, minus nu B mu B of the gas.
Which is in the mixture.
That's what this delta G is, when we allow the reaction to proceed for a little bit more.
We add a little bit of chemical potentials from the products.
Subtract a little bit of chemical potential from the reactants.
And then we expand it out in terms of the standard chemical potentials for everything being pure.
Entropy of mixing comes in here.
Comes in here.
Delta G of mixing comes in here.
And we end up with something that looks like nu C mu C naught, plus nu D mu D naught, minus nu A mu A naught, minus nu B mu B naught.
Plus RT log, and the only place where we have these log p's, or log concentration coming in, is for those species that were not pure.
And those are only these two guys here.
The ones that are in the gas phase.
D and B.
So we end up with partial pressure of D to the nu D power.
Partial pressure of B, to the nu B power.
The other two species don't come in there.
Because they started out as pure.
There's no mixing going on.
And there's no expansion of the log for them.
And so now, when we look at the Q for the reaction, the reaction quotient, it doesn't contain any of the pure species.
It only contains those species that are allowed to mix.
Those that are in the gas phase or in solution.
And so K, then, for this reaction only takes in those products like D. Or reactants like B, which are in the gas phase.
The pure solids or pure liquids don't come in.
They come in for delta G naught.
There's delta G naught sitting right here.
Delta G naught for the reaction is sitting right there.
So when you write your log K is equal to minus delta G naught over RT, the delta G naught has everything in it.
The pure stuff, the solution stuff, the gas phase stuff.
But the K only has the gas phase and solution stuff.
Alright, any questions?
Good.
Next time we'll do an example.
And then we'll go on phase transitions.
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So today we're going to do an example of using some of the equilibrium concepts that we've learned to drug design.
And it's going to be an example that's out in the literature.
And potentially a very big deal if you can do drug design this way.
And it's an example of how, remember how I told you that if you have the Gibbs free energy, you have everything and how clueless I was as a graduate student, and when I look back I think how silly it was for me not to realize how important it was.
Well, this is an example where people calculate delta G's or Gibbs free energy changes for binding of proteins to receptors on cells, or ligands.
And from that, design drugs that work much better than the real thing.
And it's all about calculating delta G. We're going to see how that is, and how that comes in with equilibrium.
So the paper that this is based on was published in 2002.
And since then there have been larger-scale trials, animal trials.
And I believe there have been human trials of this concept, of this drug that they have designed.
You've got the reference in the notes.
And we're going to have to do a little bit of review of biology first, to figure out what's going on.
And this particular process, this particular drug, is called the, I have it right here, granulocyte stimulating factor.
Where was it in here?
To give you the right name for it.
There it is.
It's a protein drug called GCSF.
Granulocyte Colony Stimulating Factor.
And it's there's wild tie for natural version of this protein.
That is generated by normal tissue.
And this protein goes to the blood, to the bone marrow.
And it binds the receptors on cells that are in the blood and the bone marrow, and stimulates the growth of white blood cells, and stem cells, and also acts to tell the bone marrow to pump out these stem cells into the blood.
And the problem is that if you have a chemotherapy patient, their bone marrow cells are largely destroyed through the chemotherapy.
And so they have a problem with the white blood cell counts, and with stem cell counts and all that stuff.
So one of the ways that you can try to reverse this problem is by stimulating, artificially stimulating the growth or the proliferation of these white blood cells.
And one of the ways you could do that is by giving the patient a lot of this protein, granulocyte colony stimulating factor, to stimulate the output of white blood cells from the bone marrow.
Whatever is left of the bone marrow.
To rebuild it up.
And so that's done with wild type, with protein.
You can make the wild type protein.
But if you had something that was basically the same thing but somehow mutated, where you made a few changes to the amino acid sequence, so that it worked a little bit better, then you might have a lot of people, a lot of patients.
And that's what the basis of this paper is, that's referenced here.
Is how to go about mutating this protein here to make it a drug that would be more effective then the natural protein.
So that you can give it to patients.
So now let me go back and tell you a little bit about this equilibrium that we're talking about.
So we've been talking about equilibria of molecules coming together.
And then creating new molecules, products and reactants.
Well, when you're talking about ligands and receptors on cells, the same sort of ideas come in.
The same equilibrium concepts come in.
So if you have a cell membrane, the cell membrane could have a receptor in it.
Which is a protein that extends through the membrane that's got some sort of pocket outside here.
The membrane keeps on going.
And you've got the blood on the outside here, or some of the tissue of the cells, and you've got the inside of the cell here.
And this receptor is waiting for some ligand.
A ligand being another protein that's floating around.
That's specific for binding to this receptor here.
And there's the ligand here.
There's the receptor here.
And every now and then one of these ligands will find a receptor and bind to it.
So we write, then, in equilibrium, ligand plus receptor goes to complex.
This is happening all the time on cells.
There are millions of different kinds of receptors, millions, billions of kinds of ligands.
Wild types, fake ones, et cetera et cetera.
When you've got a binding event, the cell knows that's something has bound, usually.
And that usually triggers a cascade of other things.
It signals a cell to do a lot of things.
So, for instance, in the case of the GCSF, you've got the receptor on the cell.
And you've got this granulocyte colony signaling factor, that would be the ligand here.
It binds to the receptor on the cell.
That triggers the bone marrow cells to produce more proteins, which signals more things to happen, and eventually down the way, after a bunch of signaling processes, out come a burst of stem cells or white blood cells.
It's a complicated process, and every step of the way needs to be done correctly.
And for this process at least, this protein here, this ligand protein, triggers the cascade.
And so this binding and unbinding of this protein then triggers the whole sequence of reactions.
So this equilibrium becomes super-important then.
OK. What else should we review about, I should have used a different board than this, but.
This is going to get covered up.
What else can we review about biology that would be interesting?
So, other things that we will need to remember, need to know, is that these receptors on the surface of the cells aren't static.
The cells recycle their receptors.
Things can happen.
Things get degraded.
And so the cell is constantly taking these receptors, bringing them back inside the cell, merging them with lysosomes, where the pH is five, or 5.5 compared to the outside of the blood, which is 7.4.
It breaks down the proteins into their amino acids, the cell can use the amino acids again to make more proteins.
Make better, make new receptors.
And that's how the cell recycles the receptors.
So basically, you end up with the cell taking the receptor.
Making a small sort of cavity around the receptor.
So there's the cell sitting here.
There's the inside of the cell.
There's the receptor being engulfed by the cell, now.
And it could have the ligand in on it, also.
There's the ligand sitting right here.
It's called an endocytotic event, or it's endocytosis of the receptor.
Into the cell.
And then once you're inside the cell, let's put the nucleus in the middle here, you've got things called lysosomes that are sitting nearby.
There's a lysosome here.
Lysosome.
And there's your little vesicle that contains your receptor.
Merges towards the lysosome, these two get together.
Then you have your ligand, I'm running out of colors here.
So there's the receptor and the ligand together in here.
And then the pH here becomes 5.5, things break apart.
And you've got to find another receptor, another ligand, to continue the process.
Everybody knows this biology, or you know it well enough.
OK, good.
So let's go do this example.
So this is the process of binding.
We can have a delta G associated with this, delta G a, delta G0 is minus RT log K sub a.
This is the association equilibrium.
And you can have the reverse process, where the complex gets broken up into receptor plus a ligand.
And then you have a delta G for this process here.
So this is delta Ga, let's call it.
This would be delta GD, which is the negative minus RT log K sub D. This is the dissociation process.
And the dissociation process, K sub D, is equal to R, the concentration of the receptors times the concentration of the ligand, divided by the concentration of the complex.
And the lower K sub D is, the smaller this number is, small means that you're mostly on this side here, mostly in the complex, the tighter the binding.
So small KD means tight binding.
So if I want, in principle, then, if I want to design a drug that's going to signal this event here, I want K sub D for this ligand that I'm going to design to be very small.
To be small.
Small enough so that it binds strongly and does its job, so I don't need too much of it.
Now, you need to do experiments to figure out what's going on.
So how do you do these experiments?
You need to be able to measure, then these K sub D's, experimentally, to see whether or not what you've designed on the computer, when you calculate delta G's on the computer you've got to know whether or not it's working.
And this is still an experimental science.
How does it work?
So, usually you do the experiment with the ligand concentration very high.
Very large.
So that the concentration of the ligand at any time is basically the same thing as the concentration of the ligand you put in.
So you overwhelm the system with ligands.
So that L is much bigger than C or R, and so it doesn't matter which way the equilibrium is.
L stays roughly the same.
So you know, throughout the experiment, what concentration is.
Then you get a bunch of cells.
In a well or something, or a 96 well plate with a bunch of different kinds of ligands.
And you can take your ligand, you can label it radioactively.
So you can take your ligand as some long protein.
And you can take an iodine 125, let's say, radioactive label on your ligand, on your protein.
So you can keep track of it.
The nice thing about radioactive labels, and which is why people use them in biology or bio-medicine.
We use them to look at, for instance, bio-distribution of things in animals.
We want to know where things are, and whether they all left the animal, or.
Because you can use a Geiger counter and you can count the events.
The radioactive events.
And that gives you extremely quantitative analysis of where things are.
So you take those radioactive ligand.
And, L0, very large concentration.
You expose this concentration to the cells.
The cells have some receptors on the surface.
There's a bunch of receptors on the surface of the cell.
And some fraction of the receptors will have the ligand bound to them.
So you incubate.
You let it wait a while.
Then you wash.
So you get rid of all the excess ligand that's on top there.
And then you take your petri dish, or your 96 well plate.
And you read the radioactivity that's coming from the cells.
And that signal, that radioactive signals, tells you exactly how many ligand you have on these cells.
You knew what the cell concentration was initially.
So that tells you what the concentration of ligands, of complexes, was.
So this experiment then gets you, experimentally, gets you the concentration of C, the complex.
So now this is something you know.
You also know what L0 was, because that's what you put in.
And that's enough for you to find out what KD is.
And if you know what KD is, then you know what delta G0 is.
And so generally, then, let's go ahead and do that.
So we know what L is.
We know what L0, C is, let me just do it on this board here.
So we start with rewriting KD as equal to R times L divided by the complex concentration.
And now L is basically L0, so we can use that approximation.
So we have L0 sitting here.
We still have the complex on the bottom, and that's something that we've experimentally discovered.
And the concentration of receptors, this is the concentration of receptors that don't have anything bound to them.
So it's these guys right here.
That's equal to the concentration of receptors, the total number of receptors, minus those receptors that have a ligand bound to them.
Complexes.
Something we can experimentally define, or discover.
So we replace this R here with RT minus C. And then we're all at equilibrium.
So we're going to put a little equilibrium sign under these C's here.
equilibrium.
And then we can rearrange this equation so that, people like to have plots that are linear, in a way that is a linear plot, where the slope of the plot is the inverse of the equilibrium constant.
So you do some massaging of this equation here.
And you rewrite it in terms of C over L0.
There's the equilibrium concentration of the complex.
This is the total receptor concentration divided by the equilibrium constant.
And then you have the equilibrium concentration of the complex divided by KD.
And so you plot, then, this ratio.
Which is an experimental ratio.
You've just measured this concentration of complexes using this radioactivity, radioactive labeling experiment.
You know what this is, because that's what you've put in.
This is your x-axis on your plot.
This is what you've measured.
And this is going to be the slope.
It's called a Scatchard plot.
After Mr. Scatchard So you get a straight line.
So we're plotting here the equilibrium constant of the complex.
And on this here we're plotting the ratio of the complex divided by L0.
And we get this straight line like this, where the slope is one over KD.
Minus one over KD.
OK, now there's a couple of things that you can look at that are sometimes useful.
In this analysis here.
You can also rewrite this equation up here in terms of the ratio of C equilibrium divided by RT.
So that's basically the ratio of receptors that have a ligand attached to them, divided by the total concentration of receptors.
So if most of the receptors are empty, then this is a small number.
Then most of the receptors are taken up, this is a number close to one.
It can't be anywhere, it can't be ever bigger than one, because the biggest number of complexes you can get is the total number of receptors.
So if you take this equation here and you massage it a little bit, one over one plus KD over L0, and that, you can also plot that.
As a function of L0.
How much ligands you put in.
And you find this is something that saturates at one.
So this ratio here is going to saturate at one.
This is C equilibrium divided by total number of receptors.
And so if L0 is small enough, if L0 is small enough, meaning that it's smaller than KD, so if L0 is much smaller then KD, then you can rearrange this ratio here so that C equilibrium divided by RT is roughly L0 over KD.
So that's linear, with a slope of one over KD.
So this slope here is one over KD is the slope.
And as you get L0 to be quite large, bigger than KD, then this saturate to one.
And this one over something very large become zero, and basically you end up with something close to one.
So you saturate at one.
So that's another way of doing this.
But this is really what we want to concentrate here.
The fact that you can get KD out of experimentally.
If you have KD, you have delta G. So now let's go back and think about this whole process here.
If we're going to design this drug here, this protein, artificially.
So what you want, then, is you want something that's going to bind strongly enough to receptor, to stimulate the growth of granulocytes, or the colony of granulocytes.
But when the cell decides to recycle the receptor, and destroy it, chew it apart in the lysosome here, you want this drug not to be degraded.
Because you have to keep injecting in the patient.
So you want this drug to release from the receptor, before the lysosome has a chance to chew it up.
So that means that you want the equilibrium constant at pH 5.5, you want that KD, to be much larger at 5.5 than you want it at 7.4.
You want strong binding at pH 7.4, and you want weak binding at pH 5.5.
That way the drug gets recycled.
Can find, go back to the blood.
Bind to a receptor again, generate the signaling events, gets engulfed by the cell.
Releases before it has a chance to be chewed up, et cetera.
The wild type will get chewed up.
The regular kind will get chewed up, and so that's why you have to, with the patients you have to keep giving them this drug over and over again.
Because it gets chewed up by the cells.
So that's the design principle that these authors had in mind when they started their study.
And so they knew what the sequence, what the amino acid sequence was, for this wild type drug, and they started doing point mutations.
Changing one amino acid here and there.
And cranking out the calculation to calculate delta G for binding of this protein to this receptor.
At pH 7.4 and at pH 5.5, and getting differences of delta G's.
So let me then review again.
What is the motivation here.
The motivation is to try to get the ratio, KD, at 5.5 divided by KD at 7.4, pH 7.4.
And you want this to be bigger than one.
And really as large as possible.
As possible, of course within limits.
So we want this to bind at pH 7.4.
And the point of comparison is the wild types.
Which at 7.4 has a KD of 270, roughly.
And at the ratio of pH 5.5 to 7.4 is 1.7.
So it's a little bit weaker binding at 5.5.
Remember, large KD means weak binding.
Small KD means tight binding.
So the fact that KD at 5.5 is bigger than KD at 7.5 means that it's slightly weaker binding at 5.5 than 7.4.
So this is sort of the baseline that the protein designers had to deal with.
So everything gets compared to this guy here.
So they don't actually, in a calculation they don't actually measure this.
What they do is, they look at delta G's.
And so they look at differences of delta G's.
They look at delta G0 for the binding of the ligand to the receptor.
At 7.4 minus the delta G0 at pH 5.5, let's call this the delta delta G. Since you know delta G is minus RT log K, this is equal to minus the delta of log KD, which is log KD at 5.5 divided by KD at 7.4.
So then they calculate delta delta G0, and it's completely related to this ratio that you measured in the experiment.
And so for the wild type, this delta delta G is just basically the log of this number here.
Is 0.53.
So now they go ahead and do their calculation.
And they find a couple of mutants.
Where, and they focus on delta delta G0.
They want something that's bigger than this.
And then they'll worry about whether there are more problems associated with it.
So they found two mutants, let's call them D110H and D113H.
I think it's a histidine mutation, point mutation at 110 and 113, where this ratio here was 8.3 and 17.
So, huge differences here.
More than a factor of 20, or factor of 30 difference in the change in the binding efficiency, at least according to these delta delta G, between pH 5.5 and pH 7.4.
That means that this protein here with the point mutation, with this one point mutation, if it binds as strongly to the receptor on the surface, as soon as the lysosome comes in and begins to decrease pH inside the cell, this ligand is going to come off.
And it's going to be able to float away.
Hopefully, before it gets recycled.
Before it gets chewed up.
So they can be used again.
So that's good.
Alright, so experiments were done.
And on the D110H and the D113H.
And KD was measured in the experiments.
And remember, the wild type is 270.
The KD's were at 370.
And 320, with some error bar.
And the different, the ratios of KD's were measured.
There were 4.4 and 6.8.
With some error bars.
And they turned out to match reasonably well, at least as far as comparison between experiment and theory.
Still not great for these sorts of calculations, because pretty involved.
A lot of approximations go on in there.
And it gives you a rough guide.
So this 17 here is probably not quite right, because it doesn't quite translate to 6.8 here.
This difference between 4.4 and 6.8 doesn't quite match the difference that they saw here in the calculation.
But it's the right direction, right?
And that's why you still need to do experiments.
This number here is a little bit bigger than this number here.
Which means that these don't bind quite as strongly.
Because KD small means strong binding.
That means the initial binding of the ligand to the receptor is not quite as good as the wild type.
That's often the case.
It's often not so easy to find something that binds as strongly as the wild version.
But, it's good enough.
And this ratio is really what clamps the deal in this case here.
And so, these mutants, in fact these two mutants are the ones that have undergone the animal trials.
That are in the pipeline.
This is a big deal.
This is a big deal because this is a huge, huge market.
There's a very large number of people that are affected with chemotherapy.
And this is basic thermodynamics here.
It's basic thermodynamic calculation of a complicated molecule with some fairly simple equilibrium concepts.
OK, any questions on this?
OK, well, we're ended really early today.
I don't have the phase transition ready, but Keith Nelson is going to start a phase transition next time.
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Today I want to start us off on a somewhat new topic.
Which is phase equilibria.
So, recently you've been looking at chemical equilibria.
What happens when substances change their chemical structure and interchange, and you've figured out how to calculate where equilibrium lies in the case of chemical reactions.
Calculate equilibrium constants, and so forth.
Now we'll introduce phase equilibria.
So this is a little bit different.
You still have a lot of molecules in one state changing to another state, but they're not changing their own molecular structure.
They're changing what phase of matter they're in.
From liquid to solid to gas.
And so the treatment is a little bit different.
And there's some new results that are somewhat different from what you've seen in the case of chemical reactions.
So we'll start with phase equilibria in just one component system.
So in other words, we're going to start with a single substance and see how it goes from solid to liquid to gas.
And we'll from there see what happens in multiple component systems.
Where, for example, then we'll be able to talk about solubility and mixtures of liquids and so forth.
And into which phase different constituents in a mixture will go.
But let's start here.
So, let's say we just have two phases at equilibrium.
And this could just look like ice and water.
You could have a little bit of ice floating in water.
So in this case you have solid and liquid.
That could be in equilibrium.
And this is going to be at some pressure and temperature.
Or, you could have liquid and gas in equilibrium.
So let's put a lid on our container.
Still have some specified pressure and temperature.
And now there's a liquid at the bottom.
And a gas at the top.
So, you could have equilibrium between liquid and gas phases.
And in some cases you have equilibrium between solid and gas phases, depending on the pressure.
And and a little piece of solid could sit at the bottom and there's equilibrium with the gas above it.
And so, to start, what quantity are we going to need to look at to tell whether the different phases are in equilibrium, or if they're not in equilibrium, in which direction are things going to go?
Are things going to change so that the ice will melt and we'll have water or, so the water will freeze and we'll have ice.
What quantity is going to tell us about all that?
Who thinks they can guess?
Give you a hint.
We're specifying pressure and temperature.
Nobody really knows?
What quantity is it that you've seen from which you can determine everything.
Yes.
The Gibbs free energy is going to tell us everything.
And we're going to frame the discussion in terms of the chemical potential.
That is, the Gibbs free energy for a mole.
And since there's only one constituent, one component at this point, we just have n. We don't have different mole numbers for different constituents, we'll see that later.
OK, so what's going to happen is, the value of the chemical potential in the different phases is going to tell us everything.
Because you know that at equilibrium, the chemical potential needs to be the same in all of these.
If not, let's say the potential energy here is lower in the solid than in the gas, nothing's going to stop the molecules in the gas phase from going into the solid phase.
That's what they'll do.
If the chemical potential of the water is lower than the chemical potential of the ice, then the ice is going to melt.
Because the molecules are going to want to go toward lowest possible accessible chemical potential.
And there's the liquid right there.
They have access to it, so they can, the ice will melt.
And you'll have just water.
So the main point here is that at equilibrium, mu is identical everywhere.
By everywhere, I mean in all phases that are at equilibrium.
And so if multiple phases are present, mu has to be the same in all the phases.
And if it isn't, then the molecules will go into the phase with the lower chemical potential.
And this is what is going to guide everything in our consideration of phase equilibrium.
So, OK, what's going to happen?
Let's consider our ice water equilibrium.
So of course, if mu of the solid is equal to mu of the liquid, which for water it should happen at zero degrees Centigrade.
And pressure of one bar, for example, not the only place it could happen.
But certainly at that condition.
Then the ice and water will be in equilibrium, they'll co-exist.
Now, what happens if you just want some ice water and you take some ice out of the freezer and put it in the cup, and you put some water in?
And you see them both there.
They're sitting there.
Maybe for a while.
And yet it's not at zero degrees Centigrade, at one bar.
It may be at one bar, you might be out in the room, but it might be room temperature.
Does that tell us that this isn't true?
What's going to happen?
The ice is going to melt.
It may take a while.
This doesn't tell us about the kinetics of it.
And it could take a while, gradually, from the surface inward, for the ice to melt.
So the fact that you might see something sitting there with two phases present, that alone of course, doesn't tell you the two phases are in equilibrium.
And it may take time for equilibrium to occur.
And what's going to happen if at some pressure and temperature, you have the mu s greater than mu of the liquid.
Then which way are things going to go?
What's going to happen?
You've got to be able to figure this out.
Of course.
The ice is going to melt.
The chemical potential of the liquid is lower.
And so molecules are going to move toward that phase.
And, of course, if mu solid less than mu liquid, then the water freezes.
All this behavior is something that we can summarize in a phase diagram.
So the phase diagram can tell us for all pressures and temperatures, what phase, or what phases, will be present in equilibrium.
And so typically, phase diagram might look like this.
Where here's our solid.
Here's our liquid.
Here's our gas phase.
So let's just look at what's happening here.
If we go to really low temperature, in general, what happens?
What phase does stuff enter when you're at the lowest accessible temperatures?
Solid, right?
Everything eventually freezes.
In fact, really I should make this, leave some room on my axis for that effect.
So of course, low temperature you get solid.
How about high pressure, what usually happens?
You squeeze something real hard.
Tends to go into the solid phase.
Now let's warm stuff up.
And depending on the pressure, you might go to the liquid phase.
Certainly you'll go to the gas phase at high enough temperature.
And that's what'll happen.
And now, if you start changing the pressure, let's say you're at high temperature and you've got just the gas present, in some big container.
Now you start squeezing it.
You just apply more and more pressure to the gas at constant temperature.
What's going to happen?
Forget the phase diagram.
You know what's going to happen.
You keep squeezing, what will happen eventually?
This isn't going to be part of your grade.
You can speak up loud, and even if you're completely wrong, nothing bad is going to happen to you.
Yeah.
It's going to liquefy, of course.
You're going to start getting liquid to form.
It's essentially Le Chatelier's principle, of course.
The liquid occupies less volume.
The density of the liquid is much higher.
So as you increase the pressure, the stuff wants to go from the gas phase to the liquid phase.
Yeah
[INAUDIBLE]
Well, with water you know that at ordinary pressure, you know where the gas-liquid coexistence would occur.
So let's say you're there.
Let's say you're at the boiling point.
And now, you already in some sense know that that's a point that's somewhere on the gas-liquid curve.
So now what's going to happen?
Well, if you from there keep that temperature constant.
You're at the boiling point.
And you could easily do this.
You've got a pot that's covered, right?
And it's boiling.
So there's liquid down here and there's gas here.
So what'll happen?
As you start squeezing, literally what'll happen is you'll squeeze out all the room for the gas.
And eventually you'll have all liquid.
And in fact, you know what'll happen at that point, if you keep squeezing and you keep the heat on.
The liquid will get hotter than the boiling point.
Normally that doesn't happen, if you're just out at atmospheric pressure.
Because it'll boil.
And instead of getting hotter, molecules will simply leave the liquid and go into the gas phase.
So that's what'll happen.
If you increase the pressure, you'll liquefy all the gas.
And obviously the opposite will also occur.
If you just allow the volume to expand and keep expanding, maybe you don't have too much liquid there, you wouldn't have to expand it too much.
You'll run out, all the liquid will vaporize.
And you'll just have gas at 100 degrees C. You could heat it more if you wanted, but even there.
So in other words, changing the pressure if you think about actually executing that in the lab, or in this case even in your kitchen, you'll get the result that you should see on the phase diagram.
So let's just look at this.
So we've already talked about boiling.
That's what's happening along this line.
The line, and I'll explain the end of this in a while, all these lines are coexistence curves.
That is, these are where you have equilibrium between two phases.
They coexist in equilibrium.
and it could happen for water, for that case, for 100 degrees C and one bar.
But that's not the only place.
It could happen it a whole bunch of temperatures and pressure.
Because as you vary the pressure, you're almost surely familiar with this, the boiling point changes.
So the boiling point up in Denver at high elevation is not the same as the boiling point here in Massachusetts right at sea level.
Because the pressure's lower up in Denver.
Let's just look at the others.
So, of course, this is going from solid to liquid.
So we're looking at melting.
And then there is solid-gas equilibria in many cases.
And this is sublimation.
And this endpoint is called the critical point.
And there's one point where all three phases are in equilibrium, called the triple point.
So let's look at the various cases.
For example, let's look at the melting line.
It could be water, it could be water and ice.
Solid going to a liquid.
Let's just look at them one at a time.
What you have there is mu of the solid at some temperature and pressure.
Is equal to mu of the liquid at the same temperature and pressure.
Now, just if we back off and look at this essentially mathematically, there's an equation here.
And there are two free variables.
Temperature and pressure.
The equation puts a constraint on them.
So there are going to be a bunch of solutions along some line or some curve.
So in other words, it's one equation in two unknowns.
Temperature and pressure.
So the solution is a line or a curve.
In other words, we could solve for pressure as a function of temperature, or temperatures as a function of pressure, along the coexistence curve.
And we'll do that shortly.
Now, if we're in one of these regions, then there aren't any equalities.
There aren't any constraints.
Because we don't have multiple chemical potentials that are equal to each other.
Right here we have three phases in chemical equilibrium.
Solid, liquid and gas.
So let's just look at that triple point.
Mu solid equals mu liquid equals mu of the gas.
We have two unknowns, or two variables, temperature and pressure.
And we have two equations.
What that means is there's only one solution, and that's what the phase diagram shows.
There's a unique solution, the triple point temperature and the triple point pressure.
So this has only one solution.
If we summarize, how many degrees of freedom, or how many free variables, we have, anywhere here, we can do that in the following way.
We can write F is 3 minus p. This is the number of degrees of freedom.
Or independent variables.
And this is the number of phases in equilibrium.
So, what's it telling us?
If we have just one phase in equilibrium, or way over in the solid or just gas or just liquid part of the phase diagram.
This is one.
We have two independent variables.
Which is exactly what we know to be the case.
In other words, in those regions, temperature and pressure can vary freely without constraints.
If we go to any of the coexistence curves, now we have two phases in equilibrium.
And then only one variable can be changed freely.
If we want to stay on the coexistence curve, let's say we've got gas and liquid in equilibrium.
It's boiling.
We could change the pressure some, but if we want to keep them in equilibrium, if we've changed the pressure, we'd better change the temperature accordingly to stay on the coexistence curve.
Otherwise everything will go up into the liquid or into the gas phase.
Because you'll go off of the coexistence curve if you were to change one of these variables and not the other.
In other words, there's a constraint.
So only one can vary freely.
And if all three phases are in equilibrium at the triple point, then you have no degrees of freedom.
Nothing can vary and there's only one place where that happens.
Now let's see if we can understand also, why does the whole thing have the overall structure that it does.
Why are some of the slopes generally steeper than others.
Turns out, we can get a qualitative understanding of phase equilibria in a pretty simple way.
So in other words, can we understand how p and T change with respect to each other along any one of these coexistence curves?
Can we understand the qualitative features of the phase diagrams?
And what we'll see is, we can get an equation for dp/dT at coexistence.
That's, in a sense, our objective here in order to get the sort of qualitative picture.
So let's say we are somewhere here.
It could be the gas-solid part.
And we have some T. And now we go to T plus dT, and out here is pressure.
How much does the pressure change when we stay on the coexistence curve?
Well, let's go through this.
Let's just consider any two phases, alpha and beta.
They could be solid, liquid, gas.
And the main condition, if we're going to be on the coexistence curve, is that the chemical potentials have to be equal.
Well, that's fine.
Now let's change the conditions a little bit.
Staying on the coexistence curve.
So, let's start again.
mu s, or mu alpha of T and p is equal to mu beta at T and p. Now let's let temperature go to T plus dT, and pressure go to p plus dp.
But since we're staying on the coexistence curve still, at the new temperature and pressure, mu alpha and beta have to be equal to each other.
So mu alpha is going to go to mu alpha plus d mu alpha.
There'll be some change in it.
Mu beta is going to go to mu beta plus d mu beta.
But we already have that mu alpha and mu beta at the beginning were equal to each other.
And these have to be equal to each other.
Because we've specified that we're staying on the coexistence curve.
So what that says is d mu alpha equals d mu beta.
And now, remember in general d mu is dG per mole, so our fundamental equation for G, it's minus S per mole dT plus molar volume dp.
So this equality is telling us that minus S alpha dT plus V alpha dp is equal to minus S beta dT plus V beta dp.
So we're just applying the fundamental relation to the changes in each of the phases.
But since the changes in mu are equal, so are these changes.
And now I'm just going to rearrange these.
In other words, S beta minus S alpha dT is equal to V theta minus V alpha dp.
And that immediately is going to give us our derivative dp/dT.
So, dp/dT, staying the coexistence curve, is just S beta minus S alpha over V beta minus V alpha.
Which is to say it's just, we can say delta S over delta V. Going from alpha to beta.
So we have a result that can guide our thinking into what the slopes of those coexistence lines are going to be.
And it's pretty useful to look at that.
By the way, we can actually express this in another way if we want.
And in some cases it can be advantageous.
And that is, we know that of course G is H minus TS.
So our mu alpha equals mu beta condition, or a d mu alpha is d mu beta condition, can be applied that way.
That is, H alpha minus TS alpha equals H beta minus TS beta.
So we can do the same rearrangements.
H beta minus H alpha equals T S beta minus S alpha.
And what that's saying, in other words, is that we can substitute for delta S, delta H over T. So we can write our relation for dp/dT in two ways.
We can also write dp/dT at coexistence is equal to delta H over T delta V. Going from alpha to beta.
These are both forms of what's called the Clapeyron equation.
And these are always valid.
We haven't made any approximations.
Yet.
Later, in going to the gas phase, we'll make some approximations about having an ideal gas.
We'll be able to treat this delta V in a simple way using the ideal gas law.
But for now, these are exact.
OK, so now let's look at what it tells us.
Let's look at the different pieces of the phase diagram and see what we can conclude.
Essentially we should be able to construct qualitatively, the features of the phase diagram using this relation.
So let's try to do that.
Here's p and T. We're going to start with an empty slate.
And now let's look at the liquid gas case.
So, let's look at our delta S delta V, going from liquid to gas.
Well, one thing you know, when you go from liquid to gas, what happens to the molar volume?
It increases.
By a little, by a lot?
By a lot, right?
The molar volume in the gas is enormous compared to the molar volume in the liquid.
So delta V is positive and very big.
Now let's think about delta S. What's it going to do, going from liquid to gas?
It's going to increase?
The disorder increases.
Now, it increases significantly.
But the liquid already is a disordered phase.
And the molecules and the liquid already have considerable freedom.
Certainly not near as much as they have in the gas phase.
But still considerable.
So, to be sure, delta S positive.
And it's not small.
So there's your condition.
It's not nearly as big as delta V is going to turn out to be.
So what's going to happen here is, if we look at the slope going from liquid to gas, it's positive.
But it's not enormous.
It's not very steep.
So dp/dT, for liquid-gas coexistence right, greater than zero, but small.
Not steep slope.
So somewhere, we're going to put our liquid-gas curve.
And I, where exactly we can worry about it.
We haven't put any numbers on this scale.
But the point is, wherever it is, it's not going to be a very steep slope.
How about solid to gas?
Well, so what's delta V going from the solid to the gas?
What happens to the molar volume?
What happens to the molar volume, going from solid?
Of course, by a lot, by a little?
By a lot.
It's an enormous increase.
Even bigger than before.
How about delta S, the entropy change?
Yeah, it increases.
And also by a lot, because now you're going from an ordered crystalline solid, usually, so you have a very, very, low entropy there.
Into a disordered phase.
Not just the liquid, but the gas phase.
So this is also very big.
Well, ok, so now we've got delta S over delta V. There's going to be a steeper slope.
Because this is bigger then for liquid to gas.
So typically, the solid-liquid part will have a steeper slope then the liquid to gas part.
Now, wait.
Sorry, sorry, this isn't right.
And that's because I've really badly misplaced things here.
That was solid to gas that we just discussed.
It has the steepest slope relative to liquid-gas.
OK, now let's think about solid to liquid.
Delta V for going from solid to a liquid.
What is it?
What sign is it?
How big is it?
Let's forgot about the sign for a minute.
How much does the volume, molar volume, change going from solid to liquid?
Yeah, not a very lot, right?
You melt the solid.
There's still a little piece of stuff down there.
Didn't increase in volume enormously.
Usually, it increases.
So generally delta V is greater than, or sort of equal to, zero.
It's not a big amount.
How about entropy?
Going from solid to liquid?
It increases, by a good amount.
You're going from an ordered to a disordered phase.
So delta S is greater than zero.
And so the result here is, now, if you look at dS/dV, or delta S over delta V, delta S is positive and you a pretty good amount.
Delta V is small.
The molar volume hardly changed.
And so what that's going to tell us is the slope is going to be steep.
And generally positive.
So now, we're going to really be on the increase.
Typical kind of phase diagram and individual coexistence curve.
And all these things, again, you could understand in terms of Le Chatelier's principle.
In terms of going from, say, solid to liquid if you know think about two pressures.
Go from p1 to p2, some higher pressure.
Where are you?
Well, if you want to stay on the coexistence curve you need to raise the temperature quite a bit in order to do that.
Because, basically, to keep on there, you know what's going to happen if you have, say, solid and liquid.
It's ice and water.
You know the freezing point, it does depend on the pressure, of course.
But since the change in the volume isn't that big, what happens is, a modest change, if you change the temperature by a little bit, maybe easier to think about it that way, you've got to change the pressure by a lot to stay on the coexistence curve.
Otherwise it'll have a great tendency to simply go into one phase or the other.
Now we've, in a sense, reconstructed the coexistence curve by our consideration of the Clapeyron equation.
There's another way we can usefully understand the qualitative features of the phase diagram.
And that's just by looking at mu itself.
Looking at the free energy itself, at different temperature.
So let's just do that.
Let's just consider mu of T at some fixed pressure.
Just qualitatively, what will happen?
So let's go back to d mu is minus S dT plus V dp.
And just look at the derivatives.
So d mu / dT at constant pressure is minus s. Minus the molar volume.
This is the one we're really going to look at.
Of course, we also could look at d mu / dp at constant temperature, and just the molar volume.
Let's just look at this.
We know that entropy is always a positive number, right?
Entropy at the lowest, in a perfect crystal, at zero Kelvin, could be what value?
Entropy of a perfect pure crystal at zero degrees Kelvin.
Zero, right?
And it's only going up from there.
So entropy is always positive.
So what that tells us is that, then, if we just look at mu versus T, it's negative S, for whatever phase.
So it's always negative.
So in some sense, you say why?
Well, it's because we're keeping pressure constant and we're raising the temperature and essentially the effect of the entropy is more important.
Is going up.
You could see it just by saying well, G is H minus TS.
And as you raise temperature, the product TS is increasing.
Which means the G is decreasing, because it's H minus TS.
So it's always a negative slope.
Let's look at it in the different phases.
So we have entropy of the gas, that's greater than the entropy of the liquid.
Which is greater than the entropy of the solid.
So that immediately tells us the relative slopes for the three phases.
So d mu gas / dT at constant pressure is minus S gas.
d mu liquid / dT is minus S of the liquid.
d mu solid / dT is minus S of the solid.
And the magnitude is much bigger in the gas than the liquid than the solid.
And so the same goes for the slope.
So now let's sketch this.
There's mu.
And there's temperature.
So what we've seen is for the gas, first of all, the slope's always negative.
And it's steepest by far, or steepest substantially, for the gas.
Next steepest for the liquid.
And least steep of all for the solid.
Now, there can be variation.
Because, of course, these do change as a function of pressure.
But this, certainly the slope, the magnitudes of the slopes are always going to be like this.
What could change as a function of pressure is where these happen to lie.
But this is certainly typical.
So what does it mean?
So, it means, remember, the stuff is always going to be in the phase with the lowest value of mu.
So at low temperature, down here, that's where the solid phase has the lowest chemical potential.
Now, let's keep going up.
Eventually they cross.
This is where, for this particular pressure, this is where the liquid and the solid are in equilibrium.
In other words, we've started, we've got the solid and we're raising the pressure.
Now it's going to melt.
Here's our melting point at this particular pressure.
And now we've got, if we keep raising the temperature, we're no longer in equilibrium, and we'll be in the one phase part of the phase diagram.
It's just the liquid.
Because that's the lowest chemical potential.
Let's keep raising the temperature.
Down here, of course, the chemical potential of the gas is way up there, right?
So it's not going to come into play now, though they're going to be in equilibrium.
Here is the boiling point.
So now let's keep raising the temperature.
And we'll be in just the gas part of the curve.
And of course that'll go on forever.
So that can just help us understand, in addition to the slopes, the positions, the relative positions, of where these equilibria occur.
Now if we start changing the pressure in particular, we could move this.
So that you could have a sublimation point where the gas and the solid are in equilibrium because this has moved over to here.
Any questions about the overall structure of the phase diagrams and these phase equilibria?
OK, let me just end with a few comments about one part of the phase diagram that we haven't talked about yet.
This.
What's going on there?
If we were to change, of course, I've got a finite amount of space on the blackboard, so I haven't drawn T and p out to infinity.
But if I could, this line would keep going forever.
There would always be some pressure and temperature at which you would have equilibrium between two different, distinct solid and liquid phases.
Turns out, that's not true for the gas and the liquid.
And there are a bunch of ways to see this.
But one is you could think of the symmetry difference.
When you go from a crystal, a crystal and solid, to a liquid or a gas, there's no question that you've changed phase.
Because there's a symmetry in the solid that's no longer maintained in the isotropic liquid or gas.
So clearly there's never going to be a point where these two phases become the same thing.
Become somehow indistinguishable.
But the gas and the liquid, I don't know, what's the difference between them?
The density, right?
The gas is more dense.
It's condensed.
Sorry, the liquid is condensed.
The gas isn't.
But that starts to change when you adjust the pressure and temperature accordingly.
And you can sort of see it coming.
What'll happen is, you change the temperature, the molar volumes of the liquid and the gas start to get equal to each other.
That's what'll end up happening.
And at that point, what's the difference between them?
There is no difference.
There's no symmetry difference.
So you can actually, instead of going through a phase transition, where there's a meniscus, you can see the top of the liquid when you heat it up.
There's boiling and then there's the gas.
You can actually go from the gas, let's go the other way, from the liquid to the gas phase, without ever boiling.
If you change the pressure as well.
So that you can actually walk all the way around the critical point.
What happens, is up here you don't have two distinguishable phases any more.
And actually, material out at this region has a bunch of very unusual behaviors.
Very useful behaviors, actually.
Chemical behavior can be really very, very, different in what's called supercritical, the supercritical region beyond the critical point.
Water behaves really unusually there.
Or it turns out, in that region, it turns out to be a terrific solvent for organic stuff.
Organic molecules which normally don't dissolve well in water.
And it doesn't dissolve salts very well.
That's pretty unusual.
Carbon dioxide, CO2, turns out to be a pretty good solvent for organics in the supercritical region.
That's what dry cleaning is.
Very nice, because know CO2, apart from its slow effect on our atmosphere, it's not poisonous.
It's not toxic.
So you can do dry cleaning without organic solvents.
All sorts of unusual behavior happens above the critical point.
Beyond the critical point.
There's also a whole range of behavior at the critical point.
Right at the critical point that's extremely unusual.
That you could think of as a consequence of the fact that the molar volumes are getting to be almost equal to each other.
That's unusual in a lot of ways.
And what ends up happening is, you can see if you go online, actually, if you just Google critical point, you can see amazing movies of stuff that's at the critical point.
Because what happens is, so you've got the liquid.
And you've got a meniscus, and you've got the gas.
And you can tell.
You look at it.
There's the liquid.
There's the gas.
But if you go real close to the critical point, the stuff doesn't really know what phase it's in.
Because it's losing the distinction between the two phases.
You start seeing globules of liquid up there in the gas phase.
And globules of the gas down in the liquid.
Because their densities - I mean, the only reason the liquid's at the bottom normally is because the density is higher.
If that stops being the case, it has no more right to occupy the bottom part than the gas.
And so they sort of start interchanging.
And if you get even closer, of course, what eventually happens is, there's just one phase.
You don't see distinct meniscus's any more.
But right near there, you see enormous fluctuations in the local density and in the location.
The physical locations of different phases.
And we'll see that come in another context later.
Because when we talk about two phase equilibria solutions.
You get this with you oil and water, you get critical mixtures also, where right at the special point, the meniscus goes away and the oil and water start interchanging.
And in those situations, a tiny amount of force can give you a really big response.
Because the two things are just teetering in the balance.
So a little bit of force pushes things one way or the other.
That's actually useful in a whole bunch of different cases.
Because that critical behavior happens in all kinds of phase transitions.
Magnetic phase transitions.
Ferroelectric phase transitions.
And in those situations, a tiny little magnetic field, for example, gives you a really big change in the magnetization.
If the alignment of spins.
And that's very useful.
You can get a very big material response and change.
You can switch something, for example, with a really small external field.
And in these cases, you can change the density really a lot by only a small change in pressure.
OK, next time we'll extend the Clapeyron equation and also go to two component systems.
So you actually, Professor Blendi will take over again on Friday.
And we'll see you then.
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Let's do a one minute review, and then move onto the Clausius-Clapeyron equation and see how far we can go on that.
So, reminding you then, what you learned last time about phase transitions is, you drew a phase diagram in temperature pressure space here.
And then you had a triple point somewhere, which was a unique point.
The triple point temperature, triple point pressure.
Then there was a gas, solid, call it coexistence line.
Which kept on going ad infinitum.
And the a gas liquid coexistent line.
With a critical point.
So that allowed you to go around.
You wanted to go from liquid to gas, you could actually go around this critical point, and never actually see a phase transition.
Then there was a solid liquid coexistence line.
Which usually has a negative slope, except for the two most important substances on Earth, which are water and silicon.
So for water and silicon, we have positive slope.
H2O and silicon.
This is why there's life on Earth.
Did you hear the story of why there's life on Earth?
This is the secret of life.
Without this, this is why we're here.
Because of this property.
So, the reason why these slopes come from the, so I'll tell you the secret of life.
But first let me remind you what the coexistence curve is.
dp/dT.
Coexistence is delta S over delta V, or delta H over T delta V. And that's the Clapeyron equation.
Now, the slope of the curves points us by, look at this solid liquid line, delta S, for solid liquid.
I looked at it up here.
It's going to be S liquid minus S solid.
And you know that's greater than zero because the entropy in the liquid state is much bigger than it is in the solid state.
You have a crystal, entropy's very small.
Liquid state, much more disorder.
Got a positive sign here.
Yes.
Questions?
[INAUDIBLE]
So, this is almost everything.
It has this slope here.
Which is a negative slope.
And water has a positive slope.
Right?
Is that wrong?
It's wrong.
Let's go through the argument.
All right.
All right.
Let's go through the argument.
So, then you have delta V, and delta V is V solid minus V liquid, this is per mole.
Now, almost every substance has V solid less than V liquid.
So this is negative.
So almost every substance, you have delta S divided by delta V is negative.
A negative slope.
You're right.
So I have it backwards.
I do have it backwards.
OK. See, I was really trying to make it so that my drawing was right.
Which is why I inverted the solid and liquid here.
So I really wanted to be right, but I ended up being wrong.
Liquid is less than solid.
Well this is water.
Why do I have it backwards in my notes.
I've got to correct that.
So this is most substances, is positive.
OK so, and most substances, OK?
Except for water and silicon.
Because for water and silicon this is reversed.
Because the molar volume of solid for water is bigger than that for liquid.
So this is bigger than here.
Thank you.
OK.
Thanks for catching it.
OK, so what happens if the molar volume for the liquid is bigger than the molar volume of the solid.
The density of water which is the inverse of the mole volume, so the density, the molar density, is equal to one over the molar volume.
So you've got the molar density for liquid is then smaller than the molar density for the solid.
It's one over, right?
So that's why ice floats.
Because it's less dense than liquid water.
So what do you think would happen in the winter if ice didn't float, in the Charles River.
It would go to the bottom.
It would freeze, solid and you would have no fish.
You would have no life.
It would freeze solid and that would be the end of life.
And you know, five hundred million years ago there would be no fish.
There would be no dinosaurs.
There would be no anything, there would be no humans.
That's the secret of life.
And the fact that this is also the case is very useful for making silicon.
For processing silicon.
Another the secret of life.
Where would we be without silicon?
Our civilization would be, we'd still be using stones and things, right?
OK, so it's super super important.
Alright, so this is a Clapeyron story.
Which we ended up getting right.
Even though I insisted on getting it wrong.
So what happens now is Mr. Clausius came around and he realized that you can make some approximations that are very useful.
So the first approximation is, you can realize that the molar volume of the gas is always much bigger than the volume of the solid or the liquid.
And so as a result whenever you have these delta V's, for sublimation, or delta V for vaporization, which is the volume the gas minus the volume of the solid.
Or the volume of the gas minus the volume of the liquid, you might as well ignore the volume of the liquid or of the solid.
And this can just be equal to roughly the volume of the gas.
So this is the first approximation he made.
So now, if you go back to the to the Clapeyron equation up here, with this approximation then dp/dT, the coexistence line, is delta H divided by T V gas.
For sublimation.
Or vaporization.
You don't have the delta V down there any more.
Then the next thing that, next assumption that he realized you can make was, well, all these gases could, they're like ideal gases.
So we can, instead of having the volume of the gas here, we can use the ideal gas law.
This is all, let's do all this per mole.
So V is equal to RT over p. You can plug this back in here.
And then you end up with the Clausius-Clapeyron equation.
So this is approximation now.
We put in RT over p for here.
We have p delta H, where this is either sublimation or vaporization divided by RT squared.
dp/dT, sublimation or vaporization, and this is the Clausius-Clapeyron equation.
Two important approximations that go in here.
And it's not valid for solids or liquid.
You have to have a gas in there because of the ideal gas for the approximation that goes in here.
And once you have that, you can integrate both sides.
You've got to put the temperatures on one side.
Put the pressures on the other side.
I'll go ahead and cover this up here.
So you have one over p dp/dT is equal to delta H over RT squared.
One over p dp/dT that's just like d log p. And that's another form that you're going to see often for the Clausius-Clapeyron approximation is d log p / dT is equal to delta H over RT squared.
And I'm freely dropping the sublimation and the vaporization because we know that's what I mean here.
It's only valid for those two lines.
So that's another form that you'll see.
And that's not dT here, that's d log p, d log p dp.
Yeah, that's right, d log p / dT.
And so now you can take the dT to the other side and integrate.
From point one to point two d log p, is from point one to point two, delta H over RT squared dT.
And you make the usual approximation, which we've made before.
That delta H is roughly independent of temperature, or very slowly changing with temperature.
And that's fine as long as you're in a narrow temperature range.
Which is usually the case when you have these Clausius-Clapeyron equation problems.
So you can take the delta H out of the integral, the R out of the integral.
And you can integrate both sides.
To give you log p2 minus log p1, is delta H times R, divided by R, delta H divided by R. Delta H over R times T2 minus T1 over T1 T2.
It looks a lot like and the temperature dependence equilibrium constant.
And in fact, but it's not, right?
And some people use that equation here instead of equilibrium constant.
Temperature dependence.
That's, can give rise to problems.
Because obviously they're not the same equation.
But roughly has the same form.
So, another way that you'll also find this written is very often p1 and T1 will be some reference.
Pressures and temperature, maybe one bar, let's say.
T1, 298 degrees Kelvin.
That's your reference point and you want to find out the pressure temperature dependence in an equation.
So if you rearrange your equations so p1, T1 are now some reference point.
They become numbers.
So you have log of a reference point, T1 here is a number.
So you often see this equation rewritten then as log p is equal to delta H over RT.
And I feel like I've got a minus sign missing somewhere.
Here.
Oh yeah, when you integrate this, the one over T squared, there's a minus sign.
Oh, I took care of that by doing T2 minus T1 here.
I think there's still a minus sign problem somewhere.
I think there's a minus sign problem.
Let me check up the notes here.
Minus one over T1 minus T2 over T1 T2.
Where's my minus sign?
T2 minus T1.
OK, so I got minus T1, minus T2 over T1 T2.
T2 minus T1.
This is fine, right.
There's no minus sign problem.
Plus a constant.
OK, so you'll often see it written like that.
And that gives you a relationship between the pressure and the temperature then, for a substance where all the reference point information is contained in this constant, C. Which is where the T1's and the p's come from.
OK, any questions?
We'll do a little example here of how this could be used.
So let's switch to an example now.
Let's go to the example, let's first do the example which is at the very end of the notes.
So this is an example of RDX.
RDX is a famous explosive, as I'm sure you know.
And it's plastic explosive.
It's got a melting point of 204 degrees Celsius.
481 degrees Kelvin.
And it's a problem for, so it's a very powerful explosive and you can't see it in X-rays.
So it's a problem at airports.
You've got be able to find if somebody's carrying a little piece of RDX in their luggage.
And so, the question becomes what do you need to know to detect it?
You want to find the vapor pressure.
You've gotta have, and the machine which basically is sensitive enough to detect molecules of RDX that are in vapor at room temperature.
So the way that you do that is, you do an experiment.
Where you measure the pressure, the vapor pressure, as a function of temperature.
And room temperature, it turns out if it's not volatile at all, if the vapor pressure is very, really tiny.
And so in order to get accurate numbers, you actually do your measurements at significantly higher temperatures than room temperature.
They do it close to the melting point, so you're actually looking at the sublimation of RDX.
You do it, let's say, at 400 degrees Kelvin, where the melting point is at 481 degrees Kelvin.
So slightly below the melting point you're looking at sublimation.
And you plot.
Log p versus one over T. And according to Clausius-Clapeyron, that should give you a straight line.
And in fact, that's exactly what you see when you take RDX and do that experiment.
You end up with a bunch of data points.
That's fall nicely on this straight line.
And then you can extrapolate to room temperature.
Somewhere here, let's say, 300 degrees Kelvin.
And find out what log p is, at 300 degrees Kelvin.
And what you find is that, and that's the graph that's on the last page of the lecture notes, that the vapor pressure of RDX at room temperature is 10 to the minus 11 bar, That's ten parts per trillion.
So it tells you that if you want to detect RDX with a sniffer machine at the airport, that machine better be able to tell you, find one molecule of RDX out of a tenth of a trillion other molecules, basically.
Which is a really hard thing to do.
So this gives you a design rule for these pieces of equipment.
And why they're so expensive.
OK, before we do the next example, let's see if any questions on Clausius-Clapeyron.
Yeah
[INAUDIBLE]
So, p as some exponential here?
You know, I don't recall seeing it as an exponential form.
But that doesn't mean it's not used the exponential form.
This is an easy way to do it, because then you've got a linear relationship.
And generally we like to see straight lines.
So this gives you a nice straight line.
OK, so now let's do a slightly more complicated example.
Which is, so most of the time you don't have pure material.
You have a mixture of some sort.
For instance, if I have a glass of water on the table at room temperature.
Above the glass of water, I have air, there's my H2O here.
And then I've got some air.
Above the glass of water.
And the air's inert to the water.
It's not reacting with the water.
So it's like an inert gas sitting on top of the water.
Then I want to ask the question, what is the vapor pressure of the water?
In the presence of this inert gas, the air.
It's not really the same as the problem that we've been looking at up here.
Right?
Because this gas liquid coexistence line, this diagram, and this Clapeyron equation, is all done for a pure substance.
And at room temperature, at room temperature, we're sitting squarely in the liquid phase.
We're not on the coexistence line.
At room temperature and one bar pressure.
So instead of having this system here, I looked at the system where I had the water, H2O, with nothing on top.
With no air on top.
Except the cylinder.
Now, I'd better make the surface of my water a straight line, otherwise I'm going to get in trouble.
There's the water sitting right here.
And I put a cylinder with one bar pressure on top.
My cylinder's going to sit squarely on the stump of the surface of the water.
This is not going to be any water vapor at one bar pressure on that cylinder.
At one bar, we're way up in the liquid phase.
I would have to decrease the pressure on that cylinder down to, I don't know, whatever the water vapor pressure is at room temperature here on this coexistence curve.
0.1 bar or something.
0.05 bar.
Something pretty small.
So that's what we've been working on.
This pure system.
And now we're going to be working on this system here, where we have air.
With a cylinder on top, one bar.
And we want to know what is the vapor pressure of H2O in this system here.
Is it zero?
Anybody want to guess, if it's zero?
We know there's vapor pressure there, right?
And a good guess is that that vapor pressure here, it's pretty close to the pressure that you would guess by using this diagram with the pure water.
But the key word is, it's pretty close.
It's not exactly the same.
So, let me ask you this.
Without looking at the notes, now is p H2O in this problem here greater than, less than, or the same as p H2O in the pure Clapeyron case?
How many people think that it's the same?
How about smaller than?
Smaller than?
Few people think that it's smaller than.
How about greater than?
A bunch of people don't know what to answer.
OK.
I think that if you think about it, what the difference is between here and there, you could probably reason which way it should go.
Which way the partial pressure of the water should go under one bar pressure, room temperature.
Compared to what you might expect it to be at room temperature with a diagram.
So, I'm going to give you a little bit of time just to think about it.
Without looking at the notes, what you think the right sign is here.
Then we'll do the problem and then we'll figure out if you were right or wrong.
OK. How many people say it's greater than?
One, two.
Two brave people here.
Two.
How many people say it's smaller than?
OK.
Smaller than.
So I'm going to say a large number, OK?
Many.
How many people think it's equal?
One person thinks it's equal.
Alright, let's do it out now and see how it comes out.
And then we can reason why it came out the way it came out.
Logically, without actually doing math.
Because there's a way to reason it out that you should be able to do.
OK, so this is what we have here.
Then the system is, we have this container.
Let me redo it instead of writing air.
So we have the liquid state here.
Some substance, and it doesn't have to be water.
That's a liquid, at some pressure p. And I'm going to say P capital, that's going to be the total pressure.
And the total pressure is given by the piston here.
p total, p T. And on top here I'm going to have two substances.
I'm going to have A, in the gas phase, that's the water vapor.
Some partial pressure, p sub A, temperature T. And I'm going to have the inert gas, with some partial pressure p, inert.
Such that the sum of the partial pressure of the inert gas, plus the partial pressure of A, is the total pressure.
One bar, in our case.
With the water with the air on top.
And the question, oh, let me make a few more definitions.
We're going to define p0 as the vapor pressure of pure vapor pressure of pure A.
So that's what you would read off the diagram here.
And what we want, the question we ask is, what is the partial pressure of A as a function of the total pressure?
That's what we're asking.
As I'm increasing the total pressure, meaning I'm putting more and more inert gas in there, which way is this going?
Is it going up, down, or staying the same?
And we've got a bunch of votes up here.
Most people think that it's going to go down.
Because I put more gas in there.
OK, I know that if P, capital P, is equal to pA, then pA is equal to p0.
Right?
There's no inert gas.
So that's the one extreme case here.
There's no inert gas, then the partial pressure's going to be what you read off that graph here.
So another way of saying what is p as a function of capital P, is to ask the question what is the slope?
I've got one point here.
If I can integrate the slope then I have this equation.
So the other way to ask the question is, what is dpA/dp, at a constant temperature?
Where I'm keeping the temperature constant.
And that turns out to be easier to calculate.
These slopes are always easier to calculate than the absolute things.
So we're doing an equilibrium here.
We have an equilibrium between the liquid state and the gas phase.
And we want to know something about this equilibrium as we change a parameter.
Equilibrium, what do we always start with when we have an equilibrium.
What's the first thing we think about?
We think about chemical potential.
Chemical potentials of A, in the gas phase has to be the same as the chemical potential of A in the liquid phase.
Well, I can't think of anything else to start with.
So let me start with that.
So let's start by writing the chemical potentials being equal to each other.
Because we know that's true.
So mu A, in the gas phase, temperature and with a partial pressure p sub A.
Is equal to mu A in the liquid phase, where the pressure on the liquid is capital P. What else do we know?
Well, we, what else are we going to need?
Before we do that, since what we're after is the derivative, let's take the derivative of both sides.
We're looking at a derivative with respect to pressure here.
Let's take the derivative outside with pressure and see what happens.
So let's take the derivative of this with respect to total pressure.
Well, this isn't total pressure.
This is the partial pressure of A.
But it's a function of the total pressure.
It's a function of the total pressure.
Partial pressure of A is total pressure minus p inert.
So, we have to use the chain rule.
So on this side here, we have d mu A / dpA.
So I'm taking d/dp of both sides.
Then by chain rule, dpA/dP.
Constant temperature.
And then on the right-hand side, I have d mu A / dP, constant temperature.
Well, d mu / dP, that may or may not seem familiar to you.
I never remember these things, but I always remember that there's, that these relationships are interesting.
Especially when you're talking about something like a chemical potential, which is really nothing but the Gibbs free energy.
With respect to pressure, when we know that the variables that we use for Gibbs free energy are pressure and temperature.
Right?
So this is the variable that goes with Gibbs free energy.
So this means that they were taking the derivatives of the Gibbs free energy with respect to one of its variables is something that we should know.
So we go back to writing what the Gibbs free energy is.
So dG is equal to V dp minus S dT.
Which is the fundamental equation for Gibbs free energy.
And this is the Gibbs free energy per mole.
That's just the chemical potential.
The same thing.
And this is the derivative with respect to pressure.
And this is the derivative with respect to temperature.
So this thing here is d mu / dp.
At constant temperature.
It's just the volume.
So now we have this equation, where we started out with the chemical potentials.
And we've got these d mu A / dpA, we've got d mu A / dP here.
That's just the molar volume of the gas.
Right, because this is mu of the gas phase here.
So this is the molar volume of the gas.
And I have dpa/dP.
And, this is very nice.
Because this is actually what I'm trying to get.
Trying to see how the partial pressure of a changes with the total pressure.
That's what I'm trying to calculate here.
I know what this is.
This is just an experimental number.
And then on this side here I have d mu A / dP.
This is the liquid phase.
Derivative with respect to total pressure, that's the molar volume of the liquid.
The liquid.
Great, I'm done.
dpA/dP total, constant pressure, is the ratio of the molar volume of the liquid of A divided by the molar volume of the gas.
What's the sign?
Volumes are positive, right?
This has to be positive.
There's no choice.
It can't even be zero.
It has to be positive, right?
So that means the slope is a positive slope.
As I increase the total pressure, p is going to go up.
If I integrate now, this, starting at, so the total pressure is equal to pA, I'm going to get a curve that looks like this.
OK, if I write on this axis p inert, so when p inert is equal to zero, total pressure is pA.
I'm going to start at p0.
And I'm going to go up from there, positive slope.
So.
Who was right?
A couple people are right.
So let's think about it in a different way.
Let's think about it in a different way.
So we did the math, and we found that actually it's, and that turns out to be a very small amount.
You do a very good job by just taking this curve here.
And using the pure substance.
In fact, in the notes we have an example of how much it changes for mercury.
So for mercury, at 100 degrees C, for pure mercury, the vapor pressure is 0.27 bar.
If you add enough inert gas to make it 1 bar for the total pressure, you go from 0.27 to 0.2701.
Not much of a change.
They do a really good job of just picking this.
And if you go to 100 bar, then you increase it to 0.28.
So it takes a lot of pressure on top, a lot of extra inert gas to really make much of a difference.
But, let's think about why it would make a difference, just logically.
So in the absence of this inert gas here, I just have the pure substance.
I add some inert gas, what happens to the entropy on top?
It's increasing, right?
Do systems like to have more entropy?
Yeah.
So, I have a lot of inert gas here.
A small amount of A.
What if I had a little bit more A, in the presence of the inert gas?
Which way would the entropy go?
It would go up a little bit more, you'd have more disorder.
If I have nothing, if I have no A here, up here and just the inert gas, then there's no entropy of mixing.
I add a little bit of inert gas, there's a little bit more entropy of mixing.
Eventually, I add more and more, entropy of mixing goes up and up and up, and eventually it comes back down to where I have pure A on top.
So the entropy of mixing up here is driving the increase in the pressure of, in the partial pressure of A. Entropy is this amazingly important term.
That's what we saw.
Entropy of mixing was responsible for equilibrium.
Without entropy of mixing, then everything would go directly to the products.
You'd never have an equilibrium.
Another secret of life.
So entropy of mixing here is driving this positive slope.
That's another way of thinking about it.
And that's why you can guess ahead of time what the signs of these things ought to be.
Without doing the math.
Alright, any questions?
OK.
So there's another sample problem with the notes, which is sort of like a standard problem.
Where you're given a Clausius-Clapeyron equation for a substance in this form.
So if you look at it, log p is minus some energy divided by RT plus some number.
In this case, it's negative, the delta H is negative for the substance that you have there.
And you're given it for the liquid phase, for the vapor pressure above the liquid and the vapor pressure above the solid.
You can, because the only way it doesn't work is if you're looking at solid liquid.
So, and then you're asked to manipulate this equation to find different properties.
For instance the triple point.
It's a very common thing to get, the triple point from this.
The delta H, it's a way to get delta H. Just like we saw for the RDX, right?
The data points formed a straight line.
And the slope gave you delta H, for RDX.
Delta H of sublimation for RDX.
So this is the way that this equation is used for problems.
I'm not going to do this problem.
Instead I'm going to go ahead and forge ahead to the next topic.
And start it.
This is the topic we're going to do it after spring break.
And and it really is a topic that follows this pretty straightforwardly.
Well, it's more complicated, but it's similar to this problem here.
So we saw that in real life yes, question?
[INAUDIBLE]
The exam.
Oh there's an exam is, which is, oh I forgot about the exam.
So let's see.
There's a problem set that's due today.
The exam will be up to the problem set that's due today.
Which means that what I talked about today is not going to be on the exam.
Because it's not covered in the problem set.
So let's say it goes up to Wednesday's lecture.
Which means you might still have a phase transition problem, right?
In fact, you're very likely to get a phase transition problem.
It's very likely to be the case.
Because it's an easy problem to put together.
But you don't have to worry about Clausius-Clapeyron.
You do have to worry about Clapeyron, but not Clausius-Clapeyron, OK?
So, for the exam, we're going to put on the Web last year's exam, and you guys are going to do reviews.
The usual thing.
Yeah?
[INAUDIBLE]
So that you don't see the answers.
Yeah, absolutely this can easily be done.
We'll do that.
Any other questions, about the exam?
Now, I don't remember.
Is there a problem set that, so I'm not assigning a problem set today, right, then?
Right.
OK. Good.
Because I hadn't loaded it up yet.
And so now I won't load it up.
OK.
Anything else?
OK, so let me tell you where we're going to go then, after the break.
So after the break, we're going to be starting to talk about things that are called colligative properties.
Colligative properties are the properties like the vapor pressure, lowering when you have a mixture in a liquid, instead of having a pure substance you have a mixture of substances here.
And then the vapor pressure of both substances above gets lowered.
Anybody want to guess why they got lowered?
What's the magic word?
Entropy of mixing, right.
Right.
There's more entropy of mixing if you've got a mixture in the liquid than if you have a pure gas up there.
So usually, so let me back up.
Usually it's when you have a solute, like salt in water.
Where water is volatile and the salt is not.
Right, a salt in water solution will have a lower vapor pressure than a solution of pure water.
Because of the entropy of mixing in a salt and water solution.
The salt and water solution we have a lowering, lower melting point.
And that's significantly lower.
Which is why we use it on roads, right?
That's a colligative properties.
Osmotic pressure.
Which you've already sort of heard about.
Why saltwater fish are unhappy in fresh water.
That's a colligative property.
So we're going to go through these colligative properties.
And the mainstay of these colligative properties is that we're going to be talking about mixtures in the liquid phase.
Here we talked about a mixture in the gas phase, changing some property.
From now on, we're going to be talking about mixtures in the liquid phase, pretty much.
And so our goal is going to be able to, is going to be predict the way that properties change as you have these mixtures.
OK. so the standard mixture that we're going to have, is to have a binary liquid gas mixture.
So that means we're going to have two substances.
A and B.
And at first, we're going to make it as general as possible.
So we're going to have liquid phase of A, the liquid phase of B.
So, for instance, this could be vodka, right?
Water and ethanol.
And water and ethanol are both volatile.
So we're going to have water and ethanol in the gas phase.
There, and it's a mixture, they're perfectly miscible in the liquid phase, A and B.
And there's going to be some fraction, molar fraction, of these things.
We're going to have a molar fraction in the liquid phase for A and B.
And some molar fraction in the gas phase, for A and B.
We're going to call it yA and yB, that's the molar fraction in the gas phase.
And there's no reason why the molar fractions in the gas phrase should be the same as that in the liquid phase.
And we know that.
We know that ethanol is more volatile than water.
And so if you have a gas of vodka, then you can smell the alcohol because it's coming out.
But the water is, vapor pressure is, very slow, very small.
It's the usual, it's pretty close to the usual vapor pressure.
And our question is, what we're going to be asking is, how many, if I give you the total pressure and the temperature here, what else do I need to know to find out all these molar fractions?
So let's first see what all the variables are that we have.
Usually, we have two variables.
The temperature and pressure, and that's enough to tell us everything.
But here now we have a mixture.
So we're going to need more than that.
We're going to need four variables.
We're going to need the temperature.
The pressure, and we're going to need the molar fraction of A in the liquid phase.
And the molar fraction of A in the gas phase.
We're not going to need any more than that.
Actually we're going to need less than that.
But those are the four variables that we can easily identify here.
And we don't need yB and xB, because the sum of the molar fraction has to be equal to one.
yB is one minus yA, xB is one minus xA.
So we don't need y.
So these are a priori the independent variables that we have to work with.
Now, we have some constraints here.
We have a constraint.
Our constraint is that we are the coexistence point.
There's a coexistence between the gas phase and the liquid phase.
And what does it mean for A liquid to be coexistent with A in the gas phase?
Chemical potential.
The chemical potential of a molecule of A in the liquid phase here is the same as the chemical potential of A in the gas phase.
So we have two constraints.
We have mu A for the liquid phase is equal to mu A for the gas phase, and mu B for the liquid phase equal to mu B for the gas phase.
So we have four variables, two constraints.
That means that the total number of independent variables we have is just two.
It's the number of variables four, minus the constraints.
OK?
Or you could say six variables if you include yB and xB, and four constraints.
Whatever we accounted, you just have two degrees of freedom.
So we have two degrees of freedom.
Which means that if you know the temperature and pressure, then you still know everything.
At the coexistence point.
You still know what the molar fraction is of your two substances in both phases.
So that's really powerful.
Or if you know just the molar fraction of A, and the temperature, then you know the pressure above.
And you know everything else.
So in this very complicated mixture here, you still only need to know a couple of things to tell you everything about this mixture.
And this example here is a specific example of something called the Gibbs phase rule.
Which you've seen before.
In this diagram.
In this diagram that we wrote here.
Where, let me write down what the Gibbs phase rule is first.
The Gibbs phase rule tells you the number of independent variables given a number of constraints.
So, it tells you that the number of degrees of freedom, the number of independent variables in a mixture, in a complicated system like this, is equal to C, which is the number of components.
So in this case, it would be two components.
You have A and B.
The components is the number of independent species that you have, minus the number of phases.
So here we have two phases, the gas phase and the water phase, plus two.
So for this example here, then we would have two components, C equals two.
Two phases liquid and gas.
A and B are the components.
And then plus two.
So two minus two plus two, F equals to two.
Two degrees of freedom, two independent variables.
If I'm looking at this diagram here, in this diagram here I have C equals one.
One component, right?
If I'm sitting where there's only one phase, say, the liquid phase or the gas phase, then p equals one.
Then F is one minus one plus two, is equal to two.
Two degrees of freedom.
And so in the pure phase, I am in a plane.
The pressure and the temperature are my two degrees of freedom.
I can freely move around the plane.
At a coexistence line, then the number of phases equals to two.
Then I have one minus two is minus one, plus two is one.
The degrees of freedom is one.
That's a line.
I have one degree of freedom.
I can move in the line, in the line in a plane.
Then if I change to the triple point, where the number of phases is three, then the number of degrees of freedom is one minus three is minus two, two plus two is zero.
I don't have any degrees of freedom, I'm stuck at one point.
So this is an example of the Gibbs phase rule, specific where C happens to be equal to one.
And then, this is a more generalized example.
So next time, then, what we'll do is we'll start by deriving the Gibbs phase rule.
Which is not so hard.
And then start on our way to the colligative properties.
And then you'll have an exam.
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Alright, so last time we started talking about more complicated mixtures.
To get into colligative properties.
And the example we gave was a binary system that has two components.
And two phases.
So we have a liquid phase on the bottom.
A gas phase on top, and two components, A and B, where compositions xA and xB are in the liquid phase.
Those are the mole fractions.
And yA and yB in the gas phase, the mole fractions in the gas phase.
And what we did last time was to begin to say how many degrees of freedom do we have, how many variables do we need to completely describe this mixture.
And it turned out that for this mixture, we only needed two variables.
Two degrees of freedom.
We started out with four variables; the temperature, the pressure, and the components in the composition in the liquid phase and the composition in the gas phase.
We need one of these, either xA or xB, and one of these, either yA or yB because they add up to 1.
So you start with four, but because you are in a mixture, an equilibrium, you have constraints.
You have the chemical potentials that have to be equal to each other.
The chemical potential of A in the gas phase has to be equal to the chemical potential of A in the liquid phase.
And the chemical potential of B in the gas phase has to B equal to the chemical potential of B in the liquid phase.
So you start with four variables.
Temperature, pressure, xA, yA, then you have the constraints that mu A, in the liquid phase has to be equal to mu A in the gas phase.
Because you're at equilibrium.
And mu B in the liquid phase has to be equal to mu B in the gas phase.
So four minus two constraints means you have two degrees of freedom.
And so if you want to know everything about this mixture, all you need is the temperature and the total pressure.
And that's enough.
It's kind of amazing.
Give me the temperature and pressure and these two components, like water and alcohol, for instance, and I can tell you everything about the compositions.
It's very powerful.
And then we said that this turned out to be a special case, or sub-case of the more general Gibbs phase rule.
where the Gibbs phase rule tells you that if you have C components, in this case here we have two, and the general case of C components with p phases, the number of degrees of freedom is C minus p plus two.
This is components.
And phases.
And where we're going to start today is by proving this.
And then we'll do more with this problem here.
So the first thing is for us to prove Gibbs' phase rule.
So Gibbs phase rule.
We start out with all the variables that we have.
And then we add the constraints.
And for the Gibbs phase rule, then we start with the two knobs that we can turn externally.
Which are the temperature and the pressure.
So we start with temperature and pressure, as two variables that we have.
And then we have a bunch of phases.
We have C phases.
And in each phase we have to describe the composition in that phase.
So, for each phase alpha, we have to describe its mole fraction.
So here we needed to describe xA for phase A.
And yA for phase A.
For component A in the gas phase.
So now we have alpha components.
And for each component we have to describe its mole fraction in that particular phase.
So we have to describe x sub i.
But we have a constraint on that.
The constraint is that the sum of all the mole factions has to be equal to one.
That's what it is, to be a mole fraction.
So the constraint is that x sub i, i from one to alpha, is equal to one.
So the constraint on the number of the components, instead of having, actually, i goes from one to C, because that's the number components.
So instead of having C different compositions that we need to take care of, we only need C minus one.
Just like here we only needed one, just xA.
We didn't need both xA and xB, because we knew the sum was equal to one.
So that means that we really need C minus one variables to describe the composition in one of the phases.
But we have p phases, so let me fix this.
We had C components.
We have p phases.
And so for each phase, we have to describe the composition.
So that means that we need p times C minus one variables to describe all the components and all the phases, plus these extra two.
So the total number variables then that we start out with before putting the fact that we're in equilibrium, is going to be two plus p times C minus one.
So that's a basic description of the system.
Then we put in the same constraints that we had up here.
In terms of the chemical potentials.
If I take any of the components, let's take the alpha component, or the i'th component.
Because it's in equilibrium, the chemical potential of that particular component has to be the same in all the phases.
So let's take the i'th component.
So yi in phase one has to be equal to yi in phase two, has to be equal to yi in phase three, has to be equal to yi in phase p. All the chemical potentials have to be the same across all the phases for that component.
And if you count the number of equations, the number of equals signs here, that's p minus one.
p minus one equals signs, constraints due to the fact that I have an equilibrium here.
So this gives me p minus one constraints.
And this is true for every single component.
Every single component has to have its chemical potential, its chemical potential equal throughout the phases.
So p minus one's constraint for one component times C components.
So for a total of C times p minus one total constraints.
My variables, my constraints.
So the number of degrees of freedom then becomes p times C minus one minus C times p minus one and then plus two.
And if you multiply this out, p times C gets rid of the p times C here.
And you get the plus C minus p plus two.
Which is the Gibbs phase rule, right?
Which I'm going to bring back up here.
C minus p plus two.
So just purely in accounting, just accounting for the variables and the constraints.
OK, any questions on how we did this?
Great.
OK.
So let's move on, then, to these ideal solutions.
Which is this case right here.
And the first thing we're going to do is look at the case where we have A is a solvent, and it's volatile.
And B is a solute, which is non-volatile.
So let's take the first case.
We have A volatile, that's a solvent.
Could be water.
And B is going to be the solute.
It's non-volatile.
And let's interchange solute or non-volatile to be consistent here.
So this could be sugar.
A sweet water.
So, Mr. Raoult looked at such solutions, and found that they could be well described by the following diagram here.
So if I plot on the x-axis here, from zero to one, I'm going to plot the partial molar fraction of the solute.
Partial molar fractions of the solute, that's going to be xB.
So when xB equals to zero, I have a pure water solution.
When xB is equal to one, I have pure sugar.
Then on the y axis, I'm going to plot the total pressure.
So when I have a pure water solution, the total pressure is going to be the same as the vapor pressure of water at a particular temperature.
Let's take temperature fixed at some value.
Temperature is fixed at some value.
I'm going to have the total pressure is going to be the partial pressure of the water.
And clearly, when I don't have any water left, all I have is the sugar, there's no pressure.
We're going to assume that it's totally non-volatile.
Obviously there's going to be a little bit of vapor pressure, even at room temperature.
Ten to the minus, I don't know, 13 torr or something.
But anyway, it's basically zero here.
And what Raoult said is that while these two points are just connected by a straight line.
Simplest way to connect two points.
Basically, he said that the pressure of, the vapor pressure of the water in this case, is equal to xA times pA star.
I could also plot this with xA, going from one to zero, since xA plus xB is equal to one.
Or I could write this as one minus xB times pA star.
Where a star, the star, means pure water.
So pure A.
And pA is the vapor pressure of the mixture.
Which in this case here is the total pressure, because I don't have anything else that's volatile.
And this is Raoult's law.
And the first thing that you can use this for is to look at your first colligative property which is vapor pressure lowering.
And let me point out another thing first.
So if a solution obeys that property here, it's called an ideal solution.
So if the water-sugar solution really did behave like a straight line like this, as you increase the amount of sugar, then it would be an ideal solution.
Many things are very close to it.
It's not such a bad approximation.
Especially in that region up here.
Around zero, when the amount of sugar is pretty small.
You expect to obey an ideal solution behavior.
Alright, so let's look now at a glass of water that has sugar in it.
We've got my glass of water, H20.
With some sugar dissolved in it.
And it's got a certain vapor pressure on top.
pA, or pH2O.
And I'm going to ask the question, what's the difference between the vapor pressure what it would be without the sugar in it.
What is pA star minus pA?
Obeys Raoult's law.
pA star, pA is xA pA star, plug that in here.
That's one minus xA pA star, one minus xA is xB.
xB pA star.
Well, the first thing I know is that this is a positive number.
xB is a positive number.
pA star is a positive number.
This is greater than zero.
That means that the act of putting some solute in my water decreases the vapor pressure of the water in the gas phase.
Pretty simple.
It's called vapor pressure law.
And we can actually take this one step further.
We can draw a phase diagram.
So our usual phase diagram here with temperature on this axis here and pressure on this axis here, and we've got our triple point sitting here.
And our gas-liquid line, with a critical point here.
So this is gas here, there's a liquid here.
And we've got our, this is going to be water.
So it's got a negative slope.
I think I've got it right this time.
And there's the solid phase here.
OK, this is for the pure stuff.
So now I can draw a similar diagram for sugared water.
And I just looked at this, assuming it obeys Raoult's law.
What we looked at here was the gas-liquid coexistence.
And we found that the gas-liquid coexistence point, the vapor pressure of the water, was less than if it were pure.
So that means that if I looked at the gas-liquid coexistence, which is this line right here.
In the presence of sugar, that whole line is going to get lowered.
The pressure, the vapor pressure, of the water is going to be lower when the sugar is there.
So the whole thing is going to get lower.
Now, if I have a solid, if the water is a solid, it's going to crystallize the water, the sugar is going to be separated from the water.
The sugar and the water are not going to really know about each other.
They're just going to have big chunks of pure water, with every now and then a molecule of sugar that's in there.
So you don't really expect to see any difference between the pure solid water, in terms of the gas-solid interface, and the pure liquid water.
There's no sugar in the gas phase.
And as far as the gas phase is concerned, that solid phase is pure water.
It's just seeing pure ice crystals on the surface.
Every now and then there's a molecule of sugar.
So you expect this to be pretty close here.
So you expect the triple point to go down.
And if I now, I go up to the solid-liquid coexistence, and go up like this, I can make a prediction about solid-liquid coexistence.
I can predict that if I'm looking at, let's put one bar here somewhere.
One bar.
Room temperature, sitting here somewhere.
One bar, room temperature, water is all liquid.
This is what it would usually, solid, liquid, that would be 273 degrees Kelvin.
That's the melting point of water.
So now if I put sugar in the water, and I follow my diagram here, what I find is that the new melting point of water is some new temperature T, T1, where T1 is less than 273 degrees Kelvin.
You all know this.
When you put something, like an impurity, a salt or sugar in the water, the melting point gets depressed.
It goes down.
Straight from here.
You can build it up straight from here.
When we get to the colligative properties, we'll attack it from the point of chemical potentials.
We'll equate chemical potentials and we'll look at the change in the chemical potential in the water, and the presence of the sugar and we'll do it rigorously.
But this is sort of the first indication that these colligative properties are all connected to each other.
They're connected to this diagram here.
OK.
So when I was a kid, and I knew that this was the case.
Everybody knows this is the case, right?
You put salt in the water and you expect all sorts of things to happen.
You expect the vapor pressure to go down.
And certainly you know that you add salt to the roads and the water melts, et cetera.
And my mother always told me that the reason why you added salt to the water was so that the temperature of the water would get higher when you try to boil pasta.
And for, I don't know, fifteen years I believed her.
Until I taught thermodynamics.
And I actually calculated the change in the temperature by adding a few pinches of salt to the water.
And you know how small that temperature change is?
You should calculate it.
It's really tiny.
It won't make any difference for the water.
Probably the difference in the temperature of the water will be the same as if I cooked here versus halfway up Mount Washington in New Hampshire or something.
Probably even less.
There's no difference.
The reason why you add salt is so that it tastes better.
That's the only reason.
It has nothing to do with thermodynamics.
Alright, any questions on this?
So we going to make it a little bit more complicated now, if there's no questions.
We're going to now have, instead of having one non-volatile solute, use we're going to have two, we're going to have this mixture.
Where both components are volatile.
Can we erase this?
Actually, I may not erase this.
Because I want to keep Raoult's law here.
OK, we're going to have a mixture of two volatile components.
For instance, water and alcohol.
Your your typical martini type situation.
Or margarita.
Whatever's your favorite drink.
So now we have xA, we have xB, we have yA, we have yB.
And we're going to assume that both obey Raoult's law with respect to each other.
So now if I draw a diagram, and I just focus on, so there's the total pressure sitting here.
And I just focus on one component and ignore the other.
Suppose I focus on A.
And I know that A in the presence of B, the partial pressure of A is going to obey Raoult's law.
And let's take A to be the more volatile component in this case here.
So pA star is bigger than pB star.
The partial pressure, pure A, had a particular temperature, T fixed is bigger than pB star, it's the more volatile of the two.
So if I were to plot pA as a function of xB, going from zero to one, so it'll be Raoult's law, I'm going to start at pA star.
And I'm going to go down linearly to xB equals one.
Now I can do the same thing for the partial pressure of B.
And assume that in the presence of A, just looking at this component B, now, and I'm looking at the partial pressure of B, and it's got its impurity in it, A.
B obeys Raoult's law with A in there.
But now if I want to keep the same x axis here, I can look at xA here, going from zero to one.
So I just basically flip the thing over.
So at xA equals zero that's pure B. pB star is less than pA star.
I'm starting here, on that side here.
pB star, and I'm going down in a straight line to zero.
pB is equal to xB pB star, straight line all the way to zero when xB is equal to zero.
So that's pB.
And that's pA. Now, the total pressure is the sum of the two.
pA plus pB.
So the total pressure in this system here, that I measure here, is just the sum of these two straight lines.
Let me do it in blue.
The total pressure, if you add up these two straight lines together, you got another straight line.
Two straight lines make a straight line.
p is pA plus pB is equal to xA pA star plus xB pB star.
Now, xA and xB are related.
xB is one minus xA.
And so really, this is only a function of one variable here.
Linearly, with respect to one variable.
OK, what does this diagram tell me?
So now, let's ignore how we built it.
And let's take a look at it.
So I have, going to take xB on the bottom here.
From zero to one.
We could choose either one, and I've got the straight line for the total pressure going from pA star to pB star.
T is fixed.
I need to tell you that T is fixed because I have two degrees of freedom, right?
I have two degrees of freedom.
My degrees of freedom are the temperature, which I'm fixing.
And the composition in the liquid phase, which I'm fixing.
So my two variables are xB and T. Those are my two degrees of freedom.
And I'm telling you what the total pressure is.
I'm giving you total pressure as a function of xB and T. By fixing T, I'm really telling you what the pressure is of the function of xB.
So the Gibbs phase rule then tells me that the pressure should be only a function of xB if T is fixed.
Should be a line.
Not necessarily straight, but it should be a line in this diagram.
To have coexistence.
So what this line is, then, this line is the line of points that tells me when I have coexistence between the gas phase and the liquid phase.
The coexistence line.
This is the gas-liquids coexistence line for the mixture.
It's not so different than this coexistence line up here.
So you can think of this as a phase diagram, kind of like you thought, you know this as a phase diagram.
It's got a coexistence line.
And then if I'm above this line, if I'm at a certain pressure here, I'm at this point here, this pressure well, let's go to a slightly higher different pressure here.
And this composition here.
I'm not on the line.
That means that I don't have coexistence between liquid and gas.
I'm above the line.
I'm at a pressure higher then where I would get coexistence.
The pressure is higher, means I probably don't have a gas any more.
Probably means that the total pressure pushing down on my mixture is too high to have a gas phase around.
It means that the top part here is a liquid phase.
So I'm in the liquid phase if I'm above this line and I have a coexistence between the gas and the liquid if I'm on the line.
Now, if I'm below this line here, well, I get in trouble.
I get in trouble because my first instinct would be, well this is just gas.
I go from liquid, and then I have coexistence.
Then I go into the gas phase.
But my x-axis here is the composition in the liquid phase.
There's no liquid around.
So I really can't say anything here.
This diagram is a little bit meaningless down here.
Because my x-axis is describing a phase which doesn't exist on the diagram.
On this diagram, here.
So when you see it, this diagram here with Raoult's law on it, you've got to remember that the part that's really interesting is above the line and at the line itself.
So you could do an experiment, then, on this line here.
T fixed.
This is the liquid phase.
So suppose I start with some high pressure up here somewhere.
At this point, let's call this point one.
We've got my container, my piston here.
p is equal to p1.
Again, I put the point in an awkward place.
Let's put it right there.
p1, and composition xB.
p1 and xB here.
And I'm going to slowly decrease the pressure.
I start in this composition.
The composition doesn't change.
I'm just decreasing the pressure.
Decrease the pressure, decrease the pressure, decrease the pressure.
And at some point, I get to that line.
Get to that line, what happens when I get to that line?
I start to see bubbles coming out of the liquid.
Vapor starts to form, because it wants to be in the coexistence line.
So I get to that line.
And I get a little bit of bubbles forming.
And a little bit of vapor.
That's the liquid here.
Little bit of gas.
And that's why this line here is called the bubble line.
So I've gone down, decreased the pressure.
So p has gone down now.
Decreased the pressure.
And if I keep on decreasing pressure, well, I'm not going to jump into this area of the diagram.
Because this area of the diagram means there's no liquid, it's a meaningless area.
If I keep decreasing the pressure, I'm still going to be in coexistence.
I'm just going to make more gas.
More vapor.
I'm going to transfer more of the liquid into the vapor phase.
I'm going to ride down that line.
I'm going to ride down the line.
I'm decreasing the pressure.
I'm not going to skip into here.
I'm going to ride down the line here.
We're going to keep decreasing the pressure even more.
p goes down even more.
Now, I make even more vapor.
There's my piston right there.
There's the gas phase.
There's the liquid phase.
Now I've transferred material from the liquid phase to the gas phase.
But I also changed the composition of my liquid phase.
And if I read my new composition, this is what I started out with.
xB.
My new composition in the liquid phase is going to be, let's call it xB prime.
There's going to be more solute in there.
Or more B in there, than what I started out with.
Now which one was the more volatile one?
A or B?
A was more volatile, right?
So as I decreased the pressure and I started bubbling off the material, which one's going to bubble off preferentially?
A is going to.
So it makes sense that I would concentrate B in there.
Decrease the pressure.
Bubbles that are rich in A come out.
Both A and B come out.
But more A comes out than B.
And what's left over is rich and then more than the less volatile material which is B.
So we're beginning to see how distillation comes about.
This is the first step in it.
In a distillation process.
So let's make this a little bit more complicated now.
Any questions, first?
Alright, now, suppose I wanted to know.
So I found out where the composition in the liquid phase is here.
And I know that should tell me immediately whether the composition in the gas phase is here.
So I have all, I have everything I need to know to calculate what the composition in the gas phase is.
So I have everything I need, then, to calculate and draw a diagram that looks just like this.
Except where my x-axis is the y's instead of the x's.
Going to make the same diagram, except now I'm going to use the gas phase as my reference point.
The composition in the gas phase.
So I need to get, so here I got p as a function of xB.
I want p, total pressure, as a function of yB.
So I can draw a diagram that looks just like this except now with the x-axis being the gas phase composition.
Let's turn the crank.
What do I know?
I know Dalton's law.
pA is yA times total pressure.
I'm trying mix up the partial pressures and the total pressure in all ways that I know how to write it.
And then hope that something's going to come out that's going to be helpful.
So that's Dalton's law.
I know Raoult's law.
pA is xA star times pA star.
And that's Raoult.
And I can write Raoult's law in a different way.
I can write as pB is equal to xB pB star, that's just Raoult's law for B.
One minus xA pB star.
And I'm looking for this.
I'm looking for the composition in the gas phase.
Or the total pressure as a function of the composition in the gas phase.
Let's start by finding the composition in the gas phase as a function of the composition in the liquid phase.
And the total pressure.
Rewrite yA is equal to pA over P. Total pressure is pA plus pB.
Sum of the two partial pressures.
pA over pA plus pB.
And I'm trying to get yA in terms of the xA's.
Or xB's, in terms of composition in the liquid phase.
And I know how to relate that to a constant.
Which is the vapor pressure of the pure material times the, and the composition of the liquid phase, Raoult's law.
This is xA pA star divided by xA pA star plus xB pB star.
And I know that xB is one minus xA.
So this is xA pA star times pB star plus pA star minus pB star times xA.
Where all I did was replace xB with one minus xA.
And rearrange my equation on the bottom.
So I've gotten the composition in the gas phase in terms of the composition in the liquid phase.
They're not independent.
I knew they weren't independent.
And this just shows me that math works.
OK.
So I've got that.
And I can invert this to get the composition in the liquid phase in terms of the composition in the gas phase.
It's not so straightforward, but you can get xA as a function of yA, as well.
You've got xA, if you reverse this you get xA is equal to yA pB star.
It looks kind of the same.
pB star plus pA star plus pB star minus pA star times yA.
xA as a function of yA.
Alright.
So now you can actually put everything together, starting from Dalton's law.
Because really what we want is the total pressure as a function of the composition in the gas phase.
So there's our total pressure here.
There's the composition in the gas phase.
There's pA. We've found composition in the gas phase in terms of composition in the liquid phase.
P is equal to yA over pA.
Which is equal to yA over xA pA star.
So if we could get rid of this here, in terms of yA, we'd be all set.
We'd get the total pressure as a function of the composition in the gas phase.
And there's what we need.
So we plug that in here.
And now we have the total pressure in terms of constants like these pure vapor pressures and the composition of A in the gas phase.
So let me find a way to write that somewhere.
Let me write it right here in this little box here.
Plug it in, you get something that's not a straight line.
Unfortunately.
Doesn't look like Raoult's law.
pA star pB star.
pA star plus pB star minus pA star times yA.
It's not a straight line.
It's a more complicated function.
What does it look like?
Looks like this.
I'm going to use the same sort of plot.
I going to plot the total pressure on this axis here, on the y axis.
I'm going to keep the T fixed.
Temperature is fixed.
So I'm going to find the total pressure as a function of the composition in the gas phase.
I'm going to put yB here.
yB going from zero to one.
So it's kind of looking like what I had before.
Total pressure is going to look the same.
If you want to do it in terms of B, you just have everywhere you see B replace it with A.
And everywhere you see A replace it with B.
You just need to change the two.
You know that if yB is equal to zero, that you have pure A.
So you know that this is going to be pA star here.
You know if you have yB equals to one in the gas phase, you have pure B in the gas phase.
You know that what you have is the vapor pressure of the pure B, in this case the pure water.
pB star, OK?
Where A is more volatile then B.
A would be the ethanol and B would be the water.
And it's not a straight line.
If it were Raoult, he would say just connect these two with a straight line.
But it's not.
It's not, we just figured out with an equation here.
And it turns out to be a line that looks like this.
It's a curved line that looks a bit like this.
OK, what does this line mean?
This line is also a coexistence line.
That's what we started out with.
We started out with Raoult.
That was part of our derivation.
Raoult's law tells you the composition at coexistence.
Tells you the pressure of, the total pressure, or partial pressure at coexistence and the composition.
So this is a coexistence line between the liquid and the gas.
Coexistence between liquid and gas.
But now, it's not as a function of the composition at the liquid phase.
It's a function of the composition in the gas phase.
Which was missing before.
So now if I do my experiment, and I'm, so what does this mean?
So if I have a point, if I have a pressure which is lower than the line here, then I have pure gas.
Pressure, low pressure, I have pure gas.
Don't have any liquids.
And that's fine, because I can, I know what that point is.
If I'm sitting below the line here somewhere, so I'm sitting here.
Some composition in the gas phase, a certain pressure.
That's fine.
It's a meaningful point.
So my container here has a piston on it.
And I'm starting out with this pressure, p1 here.
p1 pressing down.
And I got yB in this here.
And I increase the pressure, pressure goes up.
Composition doesn't change.
I'm just increasing the pressure.
Increasing the pressure, increasing the pressure.
Now, at some point the pressure becomes big enough that I start to see liquid forming.
Little drops of liquid forming in my container.
So now I have little drops of liquid that are forming in my container, that are going to start to rain down and form a little bit of liquid pooling at the bottom of the container.
And that's called the dew line.
The other one was called the bubble line, this is called the dew line.
So when I reach the dew line, I start forming liquid.
And again, just like in the previous case, if I keep increasing the pressure, I don't go into this area here.
First of all, this area's meaningless for this diagram here.
Because this area's pure liquid.
But the x-axis here doesn't tell me about the composition of pure liquid.
It's all yB.
It's all about the composition in the gas -- yes
[INAUDIBLE]
Between the dew line and the bubble line?
Yeah
This never, never world.
Where you're not allowed to inhabit.
We're going to go into that next time.
On how to use those two curves together.
OK, but the same thing happens here.
If I increase the pressure, I'm going to ride the coexistence line up.
I ride it up, and keep riding it up.
And I get liquid forming in here.
And I can find out the composition in the gas phase now.
The composition in the gas phase, as I increase the pressure, I get yB prime, there's yB that I started out with with.
With yB prime is less than yB.
And remember, B was the less volatile of the two.
As I increase the pressure, I am forming liquid.
The composition in the gas phase changes.
You decrease the mole fraction of the less volatile material in the gas phase.
The more volatile A becomes more prevalent in the gas phase.
And what's coming down is mostly, then, it's enriched in B.
You've done an enrichment of the less volatile material down in here.
And now you can put the two together, as the question was asked.
Now we can have, we mix the two together.
We mix our Raoult diagram with our dew line diagram.
We mix the bubble line and the dew line together.
In one diagram, which is going to be a powerful diagram to use.
Total pressure.
Temperature fixed.
You've got two materials, pA star is bigger than pB star.
So I'm going to have pA star here, pB star down here.
And if I choose, as my x-axis, to have xB going from zero to one, I can draw a coexistence line on this axis.
On this diagram.
That's the coexistence line in terms of xB.
And any point up here is well-described by this xB value, and the p value.
And as long as I have xB here, I'm not allowed to touch the bottom part of this diagram.
But now if I add another x-axis, I'm going to add yB.
From zero to one.
I draw a coexistence line in terms of yB.
Like this.
So this line is p as a function of yB coexistence line.
This line is total pressure of function of xB.
And as long as I'm on this line or below this line, then this axis makes sense.
That's how we build this diagram here.
It's got a bubble line on top, and the dew line on the bottom.
So I'm up here going down, I'm going to hit the bubble line.
If I'm down here and go up in pressure, I'm going to hit the dew line.
And next time -- well, next time we have an exam.
But on Friday, we'll figure out how to use this diagram.
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So.
In the meantime, you've started looking at two phase equilibrium.
So now we're starting to look at mixtures.
And so now we have more than one constituent.
And we have more than one phase present.
Right?
So you've started to look at things that look like this, where you've got, let's say, two components.
Both in the gas phase.
And now to try to figure out what the phase equilibria look like.
Of course it's now a little bit more complicated than what you went through before, where you can get pressure temperature phase diagrams with just a single component.
Now we want to worry about what's the composition.
Of each of the components.
In each of the phases.
And what's the temperature and the pressure.
Total and partial pressures and all of that.
So you can really figure out everything about both phases.
And there are all sorts of important reasons to do that, obviously lots of chemistry happens in liquid mixtures.
Some in gas mixtures.
Some where they're in equilibrium.
All sorts of chemical processes.
Distillation, for example, takes advantage of the properties of liquid and gas mixtures.
Where one of them might be richer, will be richer, and the more volatile of the components.
That can be used as a basis for purification.
You mix ethanol and water together so you've got a liquid with a certain composition of each.
The gas is going to be richer and the more volatile of the two, the ethanol.
So in a distillation, where you put things up in the gas, more of the ethanol comes up.
You could then collect that gas, right?
And re-condense it, and make a new liquid.
Which is much richer in ethanol than the original liquid was.
Then you could make, then you could put some of them up into the gas phase.
Where it will be still richer in ethanol.
And then you could collect that and repeat the process.
So the point is that properties of liquid gas, two-component or multi-component mixtures like this can be exploited.
Basically, the different volatilities of the different components can be exploited for things like purification.
Also if you want to calculate chemical equilibria in the liquid and gas phase, of course, now you've seen chemical equilibrium, so the amount of reaction depends on the composition.
So of course if you want reactions to go, then this also can be exploited by looking at which phase might be richer in one reactant or another.
And thereby pushing the equilibrium toward one direction or the other.
OK.
So.
we've got some total temperature and pressure.
And we have compositions.
So in the gas phase, we've got mole fractions yA and yB.
In the liquid phase we've got mole fractions xA and xB.
So that's our system.
One of the things that you established last time is that, so there are the total number of variables including the temperature and the pressure.
And let's say the mole fraction of A in each of the liquid and gas phases, right?
But then there are constraints.
Because the chemical potentials have to be equal, right?
Chemical potential of A has to be equal in the liquid and gas.
Same with B.
Those two constraints reduce the number of independent variables.
So there'll be two in this case rather than four independent variables.
If you control those, then everything else will follow.
What that means is if you've got a, if you control, if you fix the temperature and the total pressure, everything else should be determinable.
No more free variables.
And then, what you saw is that in simple or ideal liquid mixtures, a result called Raoult's law would hold.
Which just says that the partial pressure of A is equal to the mole fraction of A in the liquid times the pressure of pure A over the liquid.
And so what this gives you is a diagram that looks like this.
If we plot this versus xB, this is mole fraction of B in the liquid going from zero to one.
Then we could construct a diagram of this sort.
So this is the total pressure of A and B.
The partial pressures are given by these lines.
So this is our pA star and pB star.
The pressures over the pure liquid A and B at the limits of mole fraction of B being zero and one.
So in this situation, for example, A is the more volatile of the components.
So it's partial pressure over its pure liquid.
At this temperature.
Is higher than the partial pressure of B over its pure liquid.
A would be the ethanol, for example and B the water in that mixture.
OK. Then you started looking at both the gas and the liquid phase in the same diagram.
So this is the mole fraction of the liquid.
If you look and see, well, OK now we should be able to determine the mole fraction in the gas as well.
Again, if we note total temperature and pressure, everything else must follow.
And so, you saw this worked out.
Relation between p and yA, for example.
The result was p is pA star times pB star over pA star plus pB star minus pA star times yA.
And the point here is that unlike this case, where you have a linear relationship, the relationship between the pressure and the liquid mole fraction isn't linear.
We can still plot it, of course.
So if we do that, then we end up with a diagram that looks like the following.
Now I'm going to keep both mole fractions, xB and yB, I've got some total pressure.
I still have my linear relationship.
And then I have a non-linear relationship between the pressure and the mole fraction in the gas phase.
So let's just fill this in.
Here is pA star still.
Here's pB star.
Of course, at the limits they're still, both mole fractions they're zero and one.
OK.
I believe this is this is where you ended up at the end of the last lecture.
But it's probably not so clear exactly how you read something like this.
And use it.
It's extremely useful.
You just have to kind of learn how to follow what happens in a diagram like this.
And that's what I want to spend some of today doing.
Is just, walking through what's happening physically, with a container with a mixture of the two.
And how does that correspond to what gets read off the diagram under different conditions.
So.
Let's just start somewhere on a phase diagram like this.
Let's start up here at some point one, so we're in the pure - well, not pure, you're in the all liquid phase.
It's still a mixture.
It's not a pure substance.
pA star, pB star.
There's the gas phase.
So, if we start at one, and now there's some total pressure.
And now we're going to reduce it.
What happens?
We start with a pure - with an all-liquid mixture.
No gas.
And now we're going to bring down the pressure.
Allowing some of the liquid to go up into the gas phase.
So, we can do that.
And once we reach point two, then we find a coexistence curve.
Now the liquid and gas are going to coexist.
So this is the liquid phase.
And that means that this must be xB.
And it's xB at one, but it's also xB at two, and I want to emphasize that.
So let's put our pressure for two.
And if we go over here, this is telling us about the mole fraction in the gas phase.
That's what these curves are, remember.
So this is the one that's showing us the mole fraction in the liquid phase.
This nonlinear one in the gas phase.
So that means just reading off it, this is xB, that's the liquid mole fraction.
Here's yB.
The gas mole fraction.
They're not the same, right, because of course the components have different volatility.
A's more volatile.
So that means that the mole fraction of B in the liquid phase is higher than the mole fraction of B in the gas phase.
Because A is the more volatile component.
So more, relatively more, of A, the mole fraction of A is going to be higher up in the gas phase.
Which means the mole fraction of B is lower in the gas phase.
So, yB less than xB if A is more volatile.
OK, so now what's happening physically?
Well, we started at a point where we only had the liquid present.
So at our initial pressure, we just have all liquid.
There's some xB at one.
That's all there is, there isn't any gas yet.
Now, what happened here?
Well, now we lowered the pressure.
So you could imagine, well, we made the box bigger.
Now, if the liquid was under pressure, being squeezed by the box, right then you could make the box a little bit bigger.
And there's still no gas.
That's moving down like this.
But then you get to a point where there's just barely any pressure on top of the liquid.
And then you keep expanding the box.
Now some gas is going to form.
So now we're going to go to our case two.
We've got a bigger box.
And now, right around where this was, this is going to be liquid.
And there's gas up here.
So up here is yB at pressure two.
Here's xB at pressure two.
Liquid and gas.
So that's where we are at point two here.
Now, what happens if we keep going?
Let's lower the pressure some more.
Well, we can lower it and do this.
But really if we want to see what's happening in each of the phases, we have to stay on the coexistence curves.
Those are what tell us what the pressures are.
What the partial pressure are going to be in each of the phases.
In each of the two, in the liquid and the gas phases.
So let's say we lower the pressure a little more.
What's going to happen is, then we'll end up somewhere over here.
In the liquid, and that'll correspond to something over here in the gas.
So here's three.
So now we're going to have, that's going to be xB at pressure three.
And over here is going to be yB at pressure three.
And all we've done, of course, is we've just expanded this further.
So now we've got a still taller box.
And the liquid is going to be a little lower because some of it has evaporated, formed the gas phase.
So here's xB at three.
Here's yB at three, here's our gas phase.
Now we could decrease even further.
And this is the sort of thing that you maybe can't do in real life.
But I can do on a blackboard.
I'm going to give myself more room on this curve, to finish this illustration.
There.
Beautiful.
So now we can lower a little bit further, and what I want to illustrate is, if we keep going down, eventually we get to a pressure where now if we look over in the gas phase, we're at the same pressure, mole fraction that we had originally in the liquid phase.
So let's make four even lower pressure.
What does that mean?
What it means is, we're running out of liquid.
So what's supposed to happen is A is the more volatile component.
So as we start opening up some room for gas to form, you get more of A in the gas phase.
But of course, and the liquid is richer in B.
But of course, eventually you run out of liquid.
You make the box pretty big, and you run out, or you have the very last drop of liquid.
So what's the mole fraction of B in the gas phase?
It has to be the same as what it started in in the liquid phase.
Because after all the total number of moles of A and B hasn't changed any.
So if you take them all from the liquid and put them all up into the gas phase, it must be the same.
So yB of four.
Once you just have the last drop.
So then yB of four is basically equal to xB of one.
Because everything's now up in the gas phase.
So in principle, there's still a tiny, tiny bit of xB at pressure four.
Well, we could keep lowering the pressure.
We could make the box a little bigger.
Then the very last of the liquid is going to be gone.
And what'll happen then is, we're all here.
There's no more liquid.
We're not going down on the coexistence curve any more.
We don't have a liquid gas coexistence any more.
We just have a gas phase.
Of course, we can continue to lower the pressure.
And then what we're doing is just going down here.
So there's five.
And five is the same as this only bigger.
And so forth.
OK, any questions about how this works?
It's really important to just gain facility in reading these things and seeing, OK, what is it that this is telling you.
And you can see it's not complicated to do it, but it takes a little bit of practice.
OK. Now, of course, we could do exactly the same thing starting from the gas phase.
And raising the pressure.
And although you may anticipate that it's kind of pedantic, I really do want to illustrate something by it.
So let me just imagine that we're going to do that.
Let's start all in the gas phase.
Up here's the liquid.
pA star, pB star.
And now let's start somewhere here.
So we're down somewhere in the gas phase with some composition.
So it's the same story, except now we're starting here.
It's all gas.
And we're going to start squeezing.
We're increasing the pressure.
And eventually here's one, will reach two, so of course here's our yB.
We started with all gas, no liquid.
So this is yB of one.
It's the same as yB of two, I'm just raising the pressure enough to just reach the coexistence curve.
And of course, out here tells us xB of two, right?
So what is it saying?
We've squeezed and started to form some liquid.
And the liquid is richer in component B.
Maybe it's ethanol water again.
And we squeeze, and now we've got more water in the liquid phase than in the gas phase.
Because water's the less volatile component.
It's what's going to condense first.
So the liquid is rich in the less volatile of the components.
Now, obviously, we can continue in doing exactly the reverse of what I showed you.
But all I want to really illustrate is, this is a strategy for purification of the less volatile component.
Once you've done this, well now you've got some liquid.
Now you could collect that liquid in a separate vessel.
So let's collect the liquid mixture with xB of two.
So it's got some mole fraction of B.
So we've purified that.
But now we're going to start, we've got pure liquid.
Now let's make the vessel big.
So it all goes into the gas phase.
Then lower p. All gas.
So we start with yB of three, which equals xB of two.
In other words, it's the same mole fraction.
So let's reconstruct that.
So here's p of two.
And now we're going to go to some new pressure.
And the point is, now we're going to start, since the mole fraction in the gas phase that we're starting from is the same number as this was.
So it's around here somewhere.
That's yB of three equals xB of two.
And we're down here.
In other words, all we've done is make the container big enough so the pressure's low and it's all in the gas phase.
That's all we have, is the gas.
But the composition is whatever the composition is that we extracted here from the liquid.
So this xB, which is the liquid mole fraction, is now yB, the gas mole fraction.
Of course, the pressure is different.
Lower than it was before.
Great.
Now let's increase.
So here's three.
And now let's increase the pressure to four.
And of course what happens, now we've got coexistence.
So here's liquid.
Here's gas.
So, now we're over here again.
There's xB at pressure four.
Pure still in component B.
We can repeat the same procedure.
Collect it.
All liquid, put it in a new vessel.
Expand it, lower the pressure, all goes back into the gas phase.
Do it all again.
And the point is, what you're doing is walking along here.
Here to here.
Then you start down here, and go from here to here.
From here to here.
And you can purify.
Now, of course, the optimal procedure, you have to think a little bit.
Because if you really do precisely what I said, you're going to have a mighty little bit of material each time you do that.
So yes it'll be the little bit you've gotten at the end is going to be really pure, but there's not a whole lot of it.
Because, remember, what we said is let's raise the pressure until we just start being on the coexistence curve.
So we've still got mostly gas.
Little bit of liquid.
Now, I could raise the pressure a bit higher.
So that in the interest of having more of the liquid, when I do that, though, the liquid that I have at this higher pressure won't be as enriched as it was down here.
Now, I could still do this procedure.
I could just do more of them.
So it takes a little bit of judiciousness to figure out how to optimize that.
In the end, though, you can continue to walk your way down through these coexistence curves and purify repeatedly the component B, the less volatile of them, and end up with some amount of it.
And there'll be some balance between the amount that you feel like you need to end up with and how pure you need it to be.
Any questions about how this works?
So purification of less volatile components.
Now, how much of each of these quantities in each of these phases?
So, pertinent to this discussion, of course we need to know that.
If you want to try to optimize a procedure like that, of course it's going to be crucial to be able to understand and calculate for any pressure that you decide to raise to, just how many moles do you have in each of the phases?
So at the end of the day, you can figure out, OK, now when I reach a certain degree of purification, here's how much of the stuff I end up with.
Well, that turns out to be reasonably straightforward to do.
And so what I'll go through is a simple mathematical derivation.
And it turns out that it allows you to just read right off the diagram how much of each material you're going to end up with.
So, here's what happens.
This is something called the lever rule.
How much of each component is there in each phase?
So let's consider a case like this.
Let me draw yet once again, just to get the numbering consistent.
With how we'll treat this.
So we're going to start here.
And I want to draw it right in the middle, so I've got plenty of room.
And we're going to go up to some pressure.
And somewhere out there, now I can go to my coexistence curves.
Liquid.
And gas.
And I can read off my values.
So this is the liquid xB.
So I'm going to go up to some point two, here's xB of two.
Here's yB of two.
Great.
Now let's get these written in.
So let's just define terms a little bit.
nA, nB.
Or just our total number of moles.
ng and n liquid, of course, total number of moles.
In the gas and liquid phases.
So let's just do the calculation for each of these two cases.
We'll start with one.
That's the easier case.
Because then we have only the gas.
So at one, all gas.
It says pure gas in the notes, but of course that isn't the pure gas.
It's the mixture of the two components.
So.
How many moles of A?
Well it's the mole fraction of A in the gas.
Times the total number of moles in the gas.
Let me put one in here.
Just to be clear.
And since we have all gas, the number of moles in the gas is just the total number of moles.
So this is just yA at one times n total.
Let's just write that in.
And of course n total is equal to nA plus nB.
So now let's look at condition two.
Now we have to look a little more carefully.
Because we have a liquid gas mixture.
So nA is equal to yA at pressure two.
Times the number of moles of gas at pressure two.
Plus xA, at pressure two, times the number of moles of liquid at pressure two.
Now, of course, these things have to be equal.
The total number of moles of A didn't change, right?
So those are equal.
Then yA of two times ng of two.
Plus xA of two times n liquid of two, that's equal to yA of one times n total.
Which is of course equal to yA of one times n gas at two plus n liquid at two.
I suppose I could be, add that equality.
Of course, it's an obvious one.
But let me do it anyway.
The total number of moles is equal to nA plus nB.
But it's also equal to n liquid plus n gas.
And that's all I'm taking advantage of here.
And now I'm just going to rearrange the terms.
So I'm going to write yA at one minus yA at two, times ng at two, is equal to, and I'm going to take the other terms, the xA term.
xA of two minus yA of one times n liquid at two.
So I've just rearranged the terms.
And I've done that because now, I think I omitted something here.
yA of one times ng.
No, I forgot a bracket, is what I did.
yA of one there.
And I did this because now I want to do is look at the ratio of liquid to gas at pressure two.
So, ratio of I'll put it gas to liquid, that's ng of two over n liquid at two.
And that's just equal to xA of two minus yA at one minus yA at one minus yA at two.
So what does it mean?
It's the ratio of these lever arms.
That's what it's telling me.
I can look, so I raise the pressure up to two.
And so here's xB at two, here's yB at two.
And I'm here somewhere.
And this little amount and this little amount, that's that difference.
And it's just telling me that ratio of those arms is the ratio of the total number of moles of gas to liquid.
And that's great.
Because now when I go back to the problem that we were just looking at, where I say, well I'm going to purify the less volatile component by raising the pressure until I'm at coexistence starting in the gas phase.
Raise the pressure, I've got some liquid.
But I also want some finite amount of liquid.
But I don't want to just, when I get the very, very first drop of liquid now collected, of course it's enriched in the less volatile component.
But there may be a minuscule amount, right?
So I'll raise the pressure a bit more.
I'll go up in pressure.
And now, of course, when I do that the amount of enrichment of the liquid isn't as big as it was if I just raised it up enough to barely have any liquid.
Then I'd be out here.
But I've got more material in the liquid phase to collect.
And that's what this allows me to calculate.
Is how much do I get in the end.
So it's very handy.
You can also see, if I go all the way to the limit where the mole fraction in the liquid at the end is equal to what it was in the gas when I started, what that says is that there's no more gas left any more.
In other words, these two things are equal.
If I go all the way to the point where I've got all the, this is the amount I started with, in the pure gas phase, now I keep raising it all the way.
Until I've got the same mole fraction in the liquid.
Of course, we know what that really means.
That means that I've gone all the way from pure gas to pure liquid.
And the mole fraction in that case has to be the same.
And what this is just telling us mathematically is, when that happens this is zero.
That means I don't have any gas left.
Yeah
[INAUDIBLE]
No.
Because, so it's the mole fraction in the gas phase.
But you've started with some amount that it's only going to go down from there
[INAUDIBLE]
Yeah.
Yeah.
Any other questions?
OK. Well, now what I want to do is just put up a slightly different kind of diagram, but different in an important way.
Namely, instead of showing the mole fractions as a function of the pressure.
And I haven't written it in, but all of these are at constant temperature, right?
I've assumed the temperature is constant in all these things.
Now let's consider the other possibility, the other simple possibility, which is, let's hold the pressure constant and vary the temperature.
Of course, you know in the lab, that's usually what's easiest to do.
Now, unfortunately, the arithmetic gets more complicated.
It's not monumentally complicated, but here in this case, where you have one linear relationship, which is very convenient.
From Raoult's law.
And then you have one non-linear relationship there for the mole fraction of the gas.
In the case of temperature, they're both, neither one is linear.
Nevertheless, we can just sketch what the diagram looks like.
And of course it's very useful to do that, and see how to read off it.
And I should say the derivation of the curves isn't particularly complicated.
It's not particularly more complicated than what I think you saw last time to derive this.
There's no complicated math involved.
But the point is, the derivation doesn't yield a linear relationship for either the gas or the liquid part of the coexistence curve.
OK, so we're going to look at temperature and mole fraction phase diagrams.
Again, a little more complicated mathematically but more practical in real use.
And this is T. And here is the, sort of, form that these things take.
So again, neither one is linear.
Up here, now, of course if you raise the temperatures, that's where you end up with gas.
If you lower the temperature, you condense and get the liquid.
So, this is TA star.
TB star.
So now I want to stick with A as the more volatile component.
At constant temperature, that meant that pA star is bigger than pB star.
In other words, the vapor pressure over pure liquid A is higher than the vapor pressure over pure liquid B.
Similarly, now I've got constant pressure and really what I'm looking at, let's say I'm at the limit where I've got the pure liquid.
Or the pure A.
And now I'm going to, let's say, raise the temperature until I'm at the liquid-gas equilibrium.
That's just the boiling point.
So if A is the more volatile component, it has the lower boiling point.
And that's what this reflects.
So higher pB star A corresponds to lower TA star A.
Which is just the boiling point of pure A.
So, this is called the bubble line.
That's called the dew line.
All that means is, let's say I'm at high temperature.
I've got all gas.
Right no coexistence, no liquid yet.
And I start to cool things off.
Just to where I just barely start to get liquid.
What you see that as is, dew starts forming.
A little bit of condensation.
If you're outside, it means on the grass a little bit of dew is forming.
Similarly, if I start at low temperature, all liquid now I start raising the temperature until I just start to boil.
I just start to see the first bubbles forming.
And so that's why these things have those names.
So now let's just follow along what happens when I do the same sort of thing that I illustrated there.
I want to start at one point in this phase diagram.
And then start changing the conditions.
So let's start here.
So I'm going to start all in the liquid phase.
That is, the temperature is low.
Here's xB.
And my original temperature.
Now I'm going to raise it.
So if I raise it a little bit, I reach a point at which I first start to boil.
Start to find some gas above the liquid.
And if I look right here, that'll be my composition.
Let me raise it a little farther, now that we've already seen the lever rule and so forth.
I'll raise it up to here.
And that means that out here, I suppose I should do here.
So, here is the liquid mole fraction at temperature two.
xB at temperature two.
This is yB at temperature two.
The gas mole fraction.
So as you should expect, what's going to happen here is that the gas, this is going to be lower in B.
A, that means that the mole fraction of A must be higher in the gas phase.
That's one minus yB.
So xA is one minus -- yA, which is one minus yB higher in gas phase.
Than xA, which is one minus xB.
In other words, the less volatile component is enriched up in the gas phase.
Now, what does that mean?
That means I could follow the same sort of procedure that I indicated before when we looked at the pressure mole fraction phase diagram.
Namely, I could do this and now I could take the gas phase.
Which has less of B.
It has more of A.
And I can collect it.
And then I can reduce the temperature.
So it liquefies.
So I can condense it, in other words.
So now I'm going to start with, let's say I lower the temperature enough so I've got basically pure liquid.
But its composition is the same as the gas here.
Because of course that's what that liquid is formed from.
I collected the gas and separated it.
So now I could start all over again.
Except instead of being here, I'll be down here.
And then I can raise the temperature again.
To some place where I choose.
I could choose here, and go all the way to hear.
A great amount of enrichment.
But I know from the lever rule that if I do that, I'm going to have precious little material over here.
So I might prefer to raise the temperature a little more.
Still get a substantial amount of enrichment.
And now I've got, in the gas phase, I'll further enriched in component A.
And again I can collect the gas.
Condense it.
Now I'm out here somewhere, I've got all liquid and I'll raise the temperature again.
And I can again keep walking my way over.
And that's what happens during an ordinary distillation.
Each step of the distillation walks along in the phase diagram at some selected point.
And of course what you're doing is, you're always condensing the gas.
And starting with fresh liquid that now is enriched in more volatile of the components.
So of course if you're really purifying, say, ethanol from an ethanol water mixture, that's how you do it.
Ethanol is the more volatile component.
So a still is set up.
It will boil the stuff and collect the gas and and condense it.
And boil it again, and so forth.
And the whole thing can be set up in a very efficient way.
So you have essentially continuous distillation.
Where you have a whole sequence of collection and condensation and reheating and so forth events.
So then, in a practical way, it's possible to walk quite far along the distillation, the coexistence curve, and distill to really a high degree of purification.
Any questions about how that works?
OK.
I'll leave till next time the discussion of the chemical potentials.
But what we'll do, just to foreshadow a little bit, what I'll do at the beginning of the next lecture is what's at the end of your notes here.
Which is just to say OK, now if we look at Raoult's law, it's straightforward to say what is the chemical potential for each of the substances in the liquid and the gas phase.
Of course, it has to be equal.
Given that, that's for an ideal solution.
We can gain some insight from that.
And then look at real solutions, non-ideal solutions, and understand a lot of their behavior as well.
Just from starting from our understanding of what the chemical potential does even in a simple ideal mixture.
So we'll look at the chemical potentials.
And then we'll look at non-ideal solution mixtures next time.
See you then.
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And today I'll continue with both ideal and non-ideal, or real, liquid mixtures.
And let me just say the reason we're in some sense lavishing so much attention on this topic is because, after all, there's just an enormous amount of chemistry that happens in liquid mixtures.
Awful lot of biology.
Awful lot of just processing of stuff.
Distillation and everything else.
So there's so much chemistry that takes place in liquid mixtures that it is really important to have a sense of what the free energy and chemical potential of each of the species is doing in there.
Because of course this is what directly guides the chemistry that happens.
So I just want to start by finishing up something, a couple of things, from the topics from the last lecture.
One of them is, I derived an expression for the lever rule.
And I just wanted to make a little more explicit the result.
It may not have been completely clear.
So, I just want to go back to that.
We had looked at what happens if you start at some particular point, labeled one, and work your way up.
So we're raising the pressure.
So we're in the all gas region to start.
All gas region of the phase diagram.
This is xB and yB.
The liquid and gas mole fractions of B.
And the idea is that here is our initial value.
Now we're only in the gas phase.
There is no liquid.
So this is yB of one.
There is no xB at this point.
We raise the pressure.
And at some point, depending on where we end, we then look to both coexistence curves to determine the compositions of both the liquid and the gas phases.
So when we look out here, at the liquid phase.
That gives us xB.
At where I labeled point two, so there's two after we raise the pressure.
We started at one.
And then over here, we look at the gas phase coexistence curve.
so there's our yB value at two, right?
And I derived an expression for the ratio of the number of moles in the gas and liquid phases.
Because the idea here is, let's say you're working your way up on the curve.
If you just barely get to the coexistence curve, of course then you could see how much material you've got in the two phases.
But there's very little material still.
You've still got almost everything in the gas phase when you first reach here.
And as you work your way up in pressure, you'd have more and more liquid.
Of course, if you were to just keep going, you'd get into the pure liquid phase.
Typically, though, you might stop somewhere in the middle and have some reasonable amount of material in both phases, and you want to find out the composition in each phase.
And also, want to know how much material there is in each phase.
So the result that I derived was n gas at two over n liquid at two.
The ratio of total moles in the gas to the liquid.
What I derived was xA of two minus yA of one.
Over yA of one.
Minus yA of two, and I just want to, in some sense, finish up by making the obvious substitution, which is, of course these are mole fractions.
So I can just write xA is one minus xB.
And yA is one minus yB.
And I just want to put this in the terms that are written here in the phase diagram.
When I got to this result I pointed out that therefore you can use the lever rule.
And I just want to make that a little bit more explicit.
So using that relation, then we can see that ng of two over nl of two, total number of moles in the gas to the liquid is, now I'm going to substitute in to get yB of one minus xB of two over yB of two minus yB of one.
So all I've done here is substitute in these expressions.
And now I'm going to just switch both of these.
And so I'm going to multiply both sides by negative one.
So I can write xB of two minus yB of one over yB of one minus yB of two.
So that's the result that I want.
Because if I look at the two segments of this line, on either side of the pressure value I've reached, then what I see, of course, is that over here I've got xB of two minus yB of one.
That's this.
And this part is yB of one minus yB of two.
That is, it's this part.
And that's here.
And so that's the point, is that the ratio of moles in the gas and liquid phases is just given by the two segments of this on either side.
And so what this allows us to see very simply is alright, I'm going to raise the pressure at some point.
And depending on how much material I feel like I need to collect in the new phase, I can always determine that simply by reading off the phase diagram what the ratio of these two segment lengths is.
Now what I want to do.
What what we've done mostly is just go through these phase diagrams, see how to read them.
How to follow events on them as you change pressure in a diagram like this.
Or change temperature in diagrams like I also showed last time.
Now what I'd like to do is go a little bit further and just look at expressions for the chemical potential and the free energies.
So that we can, a little more quantitatively, see what's going on.
So let's just see what's happening.
So let's look at the chemical potentials.
In ideal liquid mixtures.
So everything is derived from the fact that when we have any of the constituents in both phases, the chemical potential must be equal in both phases.
Right?
So, we can write for A, mu A of the liquid at some temperature and pressure must equal chemical potential of A for the gas.
That's the partial pressure.
So if we have an ideal gas, and certainly if we're going to assume an ideal liquid mixture, we can safely assume that it's an ideal gas above it.
Then we can write mu A in the gas.
Is just mu A naught.
That's a function of the temperature plus RT log of pA over p0.
Nothing new here, this is just our expression for the chemical potential in the gas, with reference to a standard potential.
Usually one bar.
At whatever the temperature is.
And, of course, this is how it varies as the partial pressure of A in the gas phase varies.
So then we can just write our expression for the liquid.
At our temperature and pressure, it's given by mu naught in the gas phase.
Plus RT log pA over p0.
So it's the same thing.
Because I'm just taking advantage of this equality.
But of course, this is an obvious step, having written this.
But is a super important step.
That's what allows us to do this treatment in such a straightforward way.
We know a lot about the chemical potential of something in the gas phase.
Since the gas and liquid are in equilibrium, therefore we know the chemical potential in the liquid phase too.
So we can rewrite this by recognizing that if we just go to the limit where we only have pure liquid A, so pure liquid A.
Well, in that case, mu A naught or sorry, mu A in the liquid phase, mu A star is mu A naught in the gas phase.
Plus RT log pA over p0.
And so for the mixture, now all I'm going to do is just add and subtract terms.
So we can write mu A in the liquid at T and p, right?
So this is in the case of the limit of the pure liquid.
Now we're going to the mixture.
So it's just mu A naught plus RT log pA star over p0 minus RT log of pA star over p0.
Wait a minute, lost a term.
Plus RT log pA over p0.
And all I want to do now is combine terms.
To write mu A star liquid temperature and pressure.
Plus RT log of pA over pA star.
So this is a very convenient form for it.
And then, of course, we have an expression for pA, right?
From Raoult's law.
pA is just the mole fraction xA times pA star.
That's just true for the ideal mixture.
So now, finally, we can write that mu A in the liquid at T and p is just given by mu A star.
So that's for the pure liquid at that temperature and pressure.
Plus RT log of the mole fraction of A.
So that's a very simple expression for the chemical potential of A.
And of course the analogous expression will hold for any of the constituents.
In an ideal liquid mixture.
By the way, it's convenient because it looks just like the chemical potential in a mixture of ideal gases.
Except that we have liquid instead of gases, right?
Now, of course, since this is a mole fraction, it's always between zero and one.
That means this is always a negative number.
So what that means is that the chemical potential in the solution is always lower than the chemical potential of the pure liquid.
Very important result.
So it has all sorts of implications that we'll see.
One of them is osmotic pressure.
It means that if I have a biological cell, or some container with a membrane through which one of the constituents might pass.
So in other words, the let's say, component A, maybe it's just water.
It can pass freely, let's say, from the outside to the inside of the cell, or from one side to another of a membrane.
And on one side I've just got pure water.
And on the other I've got saline solution or whatever is in the cells, right?
What's the water going to do in that situation?
What's going to happen?
Anybody know?
So I've got, I just take fresh cells and plunk them into pure water solution.
What happens?
Yeah.
They burst.
Water rushes in.
Because the chemical potential is lower inside.
It's always lower in solution than outside.
So water rushes in, and the membrane expands.
Now, if the membrane is strong enough, at some point it may not burst.
And the pressure might start to go up.
Let's say the membrane is strong enough to resist and not burst.
Eventually, the water won't keep going in indefinitely.
That's because it won't be at the same temperature and pressure any more.
In particular, the pressure will have changed.
So you can build up some high pressure, what's called osmotic pressure.
Because of the fact that at the same pressure the chemical potential of the water's lower inside the cell or inside the enclosure with the membrane.
So water will keep filling.
And at some point the chemical potentials will equalize because of the change in pressure.
And at that point there'll be an equilibrium established.
We'll see that quantitatively a little bit later.
OK.
The other thing to notice is, this is familiar from the expression for a gas mixture.
What drives the gas mixture?
Remember back when we discussed mixing of gases and the fact that they would mix at all.
What makes that happen?
What's driving it?
Yeah, it's entropy, right?
And that's what's happening here too.
Of course, in the case of the ideal liquid mixture, there's no energetic interaction.
The molecules are non-interacting in this case.
But of course, entropy is going to want them to mix.
And that's what's resulting in the decrease in chemical potential.
Let's see that a little more explicitly by just calculating out the free energy change of mixing.
So, delta G of mixing.
So we're going to start with two separated liquids.
And then we'll remove the barrier.
And we'll have the two mixed together.
So of course, the free energy in either case is just the sum of the number of moles of each times the chemical potential of each.
We have expressions for that.
So starting G1 in this case.
It's n of A, number of moles of A.
Times the chemical potential.
So it's xA mu A star in the liquid.
Plus nB xB B mu B star of the liquid.
Here, G of two.
After we let them mix.
Then it's n -- wait a minute.
It's just n. I think I've got that wrong in the notes also.
Right, it's just a number of moles times the chemical potential in each case.
So it's the number of moles times xA.
Times mu A in the mixture.
Plus n xB mu B in the mixture.
But we've just figured it out our expressions for mu A and mu B, our chemical potentials of each constituent in the ideal liquid mixture.
So this is just n xA.
And here's our expression.
And of course these are going to cancel.
Let's write it out.
So it's mu A star plus RT log xA.
And then, mu B star plus RT log xB.
So our delta G of mixing is just the difference between these two.
So it's n RT xA log xA plus xB log xB.
Where have you seen that before?
Yeah.
The same thing that you had for the delta G of mixing for an ideal gas mixture.
Why?
Because we're not accounting for any interactions between the molecules in either of the phases.
And if the molecules aren't interacting, it's all entropy.
And the entropy term has the same form in either case.
In microscopic terms, it's just measuring the fact that there's more disorder in the mixture than in the pure liquids.
Just the same it was in the gas phase.
OK. We can see this even more explicitly if we just recall that G is V dp minus S dT.
So we can just explicitly calculate the entropy of mixing.
Delta S of mixing is just the partial of delta G of mixing.
Negative partial.
With respect to temperature.
At constant pressure.
So it's just minus n R xA log xA plus xB log xB.
So that's our entropy of mixing.
We can calculate our enthalpy of mixing.
Just delta G of mixing.
Plus T delta S of mixing.
But it's immediately apparent that these are just going to cancel when we multiply this by T. So there's no enthalpy of mixing.
Just as we expect when there are no energetic terms involved.
It's all entropy that's driving the mixture.
One more detail, it's straightforward to see that there's no volume change.
That is, if we take the derivative of delta G of mixing the partial derivative with respect to pressure, at constant temperature, of course, there's no explicit pressure dependence.
This is zero.
So again, in the ideal liquid mixture case, their molecules aren't interacting.
So there's no reason, when I open that barrier, that the amount of volume they occupy altogether is going to change.
Any questions, so far?
OK. Then, let's move on to non-ideal solutions.
By the way, just to return briefly to this topic of osmotic pressure, I just want to emphasize that result didn't need any kind of energy of mixing, either, right?
Just from the entropy term you would burst the cell or do whatever.
You end up with a larger pressure.
So in other words, if you have a situation where, here is A, pure liquid.
And here is A plus B liquid, so mu A star is greater than mu A.
We know that in all cases at the same pressure, the chemical potential of the mixture is lower than the chemical potential of the pure liquid.
What's going to happen?
Well, A is going to, oh, sorry, A over here is going to rush in.
It's going to get through this, if I've got a semi-permeable membrane, it's permeable to A but not to B.
Common situation, of course.
That's certainly the case for the cell membrane.
Water may flow easily in through the membrane, but all of the stuff, all the salt and everything else that's inside the cell generally won't.
So of course A then flows through this, what's called a semi-permeable membrane.
And you'll wind up with either a burst membrane, or something that looks like this.
Where now there is additional pressure here.
And you'll have some different mole fraction than you started with before.
Now let's go over to non-ideal solutions.
So let's just think microscopically for a moment about how this is going to work.
You know, normally, always, there are interactions between the molecules and the liquid.
The liquid is a condensed phase.
The molecules are in immediate proximity to each other.
So it's very different from the gas phase, where it can be a pretty realistic approximation to say, well, the molecules are essentially non-interacting.
The ideal gas law may turn out to be a very good approximation.
In the liquid, really, it's never the case that you don't have interactions.
So if we just sketch that.
If we start with a bunch of molecules in A, they're interacting.
So there's some interaction energy.
I'll call it AA, it's the interaction between two molecules of A.
In most cases it'll be less than zero.
That is, there'd be a weak attraction.
Same thing between molecules of B.
So that's my situation in the separated liquids.
I've got molecules in each container.
They're interacting.
There's some energy associated with that.
So here's the pure separated liquids.
And now, I'll open up the barrier and let the liquids mix.
And now suddenly, of course, A and B are going to interact with each other as well as other molecules of their own kind.
So now, suddenly, there's going to be some interaction energy.
uAB.
And very simplistically, we can envision that for some of the molecules, essentially, there would be exchange of this sort.
So, one neighbor of this molecule and one neighbor of this molecule will wind up exchanged for some pairs of molecules with the unlike species.
And essentially, likes interactions will be replaced by unlike interactions.
So there's some now change in energy involved.
There's a delta u.
And as outlined in this simple picture, it's just 2 uAB minus uAA plus uBB.
Real liquids are very complicated.
For liquids with relatively simple non-directional interactions, things like organic liquids that might interact van der Waal's interactions.
This might be a reasonable starting point.
And in general, if we're going to deal with relatively small deviations from the ideal liquid mixture, the ideal solution, then we can start with a model like this.
And this difference is what's going to determine how far the liquid mixture will deviate from the ideal solution case.
Of course, we could always just write a term like this.
It may or may not be easily expressed in this sort of way.
So let's just think about what the possibilities are.
Of course, simply, there are two.
The sign could be positive or negative.
And what that means is, you could have situations where it's positive because basically there's an energy of mixing now in these cases.
If that's not zero, unlike the ideal case, now unlike before, where we just had entropy driving the mixture, now there's an energy of mixing.
Could be positive or negative.
If it's positive, that means energetically speaking, the mixture is unfavorable.
In other words, the intermolecular interactions between like molecules are more favorable then the interactions between the unlike molecules, in that case.
So, delta u is greater than zero.
And our delta H of mixing is going to be close to delta u of mixing.
That is, we're going to figure that delta pV is not going to be something large.
And in that case, we can say delta G of mixing is 1/4 n. And that's only because there are 4 species involved here.
Times delta u.
In other words, I'm multiplying this molecular delta u.
This is the change in energy for this collection of four molecules.
Or of four molecular pairs.
Plus the other stuff that we've seen for the ideal case.
That is, the entropy term, of course, is still there.
xA log xA plus xB log xB.
And because we're treating the case where this is positive, that means this is bigger than in the ideal solution case.
So of course there are lots of examples of each of these.
In fact, this is the more common deviation.
So, and lots of examples.
An easy one is acetone.
And carbon disulfide.
So, if we look at that, and just see what the phase diagram is going to look like, it's the following.
So let's start with the ideal case.
So I'm going to make B the more volatile component.
That is, B is going to be CS2.
So here's p star CS2, which I'll call p star B.
And somewhere down here will be p star for A, that is, p star for acetone.
And this is going to be the mole fraction of CS2.
Or xB.
So let's start with the ideal case.
Not steep enough.
Pretty good.
OK. And now let's look at what's going to happen in the real mixture.
Well, because we've got a positive deviation, what's going to happen is the mixing is unfavorable.
The molecules aren't that happy any more about being in solution.
So what that means is, relative to the ideal solution, they'd rather go up into the gas phase.
They don't like being in the mixture.
Compared to what they would feel in the ideal solution, where there are no interactions.
Because the interactions between unlike molecules are unfavorable.
So what's going to happen, then, is the pressure is higher.
You have deviations from the ideal case, in the total pressure.
And in each of the individual partial pressures.
That's the total pressure.
pA plus pB.
In the ideal case it's this straight line.
In the non-ideal case, though, it's bigger than the straight line.
Because more stuff wants to get up into the gas phase.
And it's the same for each individual component.
So here's pA. And there's pB.
Any questions?
OK, so it's obvious just from looking at this that each of the partial pressures is bigger than it is in the ideal case.
And so, so is the total pressure.
So in other words, if I say what's the partial pressure of CS2, well, since it's not the simple ideal case, actually I don't know what it is in general.
It's not easy to calculate a priori what it's going to do all the way across the phase diagram for any mole fraction.
What I do know, though, is it's bigger than it would be for the ideal solution case.
In other words, it's bigger than x CS2 times p star CS2, the pressure over the pure liquid CS2.
Same for acetone.
It's bigger than x acetone p star acetone.
And since both partial pressures are bigger, the total pressure has to be bigger.
And of course, that's something you can read right off the diagram as well.
In other words, the total pressure is bigger than in the ideal solution case.
Any questions?
OK.
So the negative case, of course, is exactly the opposite.
So I'll just go through it quickly.
So, remember positive deviations came, let's put this back up there.
That came from the case where the unlike intermolecular interactions are not as favorable.
There's less attraction, more repulsion than in this case.
Of course, there are plenty of cases of the opposite sort.
Where the unlike intermolecular interactions are attractive.
More than the interactions between like molecules.
And in that case, you have the opposite result.
So you can have negative deviations.
In other words, delta u is less than zero.
Lots of examples.
I've got one in the notes, it's just acetone and chloroform.
So if we get rid of the carbon disulfide and put CHCl3 in there, that's more favorable.
Because there's weak hydrogen bonding between the unlike species.
And not between the like species.
So what happens is, you have acetone, it's this.
And you have chloroform.
And there's weak hydrogen bonding between them.
There's an attraction there.
That doesn't exist between either case of the like molecules.
So now there's stronger attraction between them than there is between molecules in either of the pure liquid cases.
So, of course, you know that the opposite result is going to happen.
In this case, now you mix them together, relative to the ideal solution where there are no interactions, now they want to be together.
They're clinging to each other through these hydrogen bonds.
What's that mean for the pressure up in the gas phase, in the solution?
They don't want to go there any more.
Stuff in the gas phase now wants to condense down into the liquid.
Because there they have favorable interactions.
So that's, of course, what'll happen.
And so then you just see the opposite sort of deviation from what I just illustrated before.
So then if we look at p star CHCl3, where are these lines going?
Well, may have been more artistic.
But I'm afraid I'm going to have to go with reality.
p star for acetone.
There's chloroform.
And now our deviation is going to be in the opposite direction.
So all the, both of the partial pressures and the total pressure are going to go lower than indicated here.
So p total is now less than, for the real case, is less than p total for the ideal case.
In the situation where you have attractive interactions between the unlike molecules.
Any questions?
OK. Now, in general, this is complicated.
To describe quantitatively throughout the entire phase diagram.
The interactions are complicated.
Lots of times, molecules interact with more than one neighbor.
So in general, we don't have a simple analytical expression for what the pressure is going to do as a function of mole fraction all the way from zero to one.
Of course, we have it in very simple form for the ideal solution.
Because everything's linear.
So it's very simple.
These are replaced by equalities and we know everything.
Now we don't.
But we can at least make some progress by looking at the limiting cases.
Where you have just very dilute solutions.
So in that case, you can anticipate what's going to happen.
So let's say we mix A and B.
And one of them is in very dilute concentration.
For the other one, for B, most molecules of B are just going to see themselves.
They hardly know that you've mixed any molecules of A in there.
For them, it's pretty much an ideal solution.
They think they're in an ideal solution.
Which is to say there's an entropy change in mixing.
But hardly any of them experience interactions with the unlike molecules.
For the dilute molecules, there is a change.
You can't use Raoult's law any more.
But at least we can look of the slope of the curve and say, OK, at least in the dilute solution case for a little while it's going to be linear.
We can understand it.
We can tabulate it.
So let's look at how that works.
Alright.
That limit is something that has a name, that's called the ideal dilute solution.
Not to be confused with the ideal solution.
Where both constituents follow Raoult's law, as we've seen before.
So ideal dilute solution.
We're going to just look at these limits.
So, if we start with the ideal case, so let's imagine we're going to look at carbon disulfide again.
Here's p star CS2, which is our pB star.
And this is x CS2, or xB.
OK, well, what happens if we've got almost pure carbon disulfide.
Well, what's going to happen is in this limit, like I was mentioning, up here the CS2 pretty much sees other CS2.
It thinks it's in an ideal solution.
It's going to look like that.
Now, let's go to the other limit.
So this is the limit x CS2 equals xB approaches one.
Raoult's law.
Two, the other limit.
x CS2 approaches zero.
We're down here.
So it's definitely not going to be the same.
Because now, every molecule, every CS2 molecule, is completely surrounded by acetone.
So they all feel the energy of mixing.
Because there are hardly any of them there.
So every one of them is, it's very rare for one to find a neighbor of its own kind.
And this is a case where there's a positive deviation from ideality.
That is, the interactions are less favorable between unlike species.
So, it's up here.
Doesn't like that.
It says, get me out here, I'd much rather be in the gas phase, compared to the ideal case, then surrounded by all these acetone molecules, right?
So that's what'll happen.
And now it's up to the artist.
But in some way there's going to be a set of values in between.
And the ideal dilute solution by itself isn't going to tell us about that.
It's telling us about the two limits.
Now, if we follow this, and I can see I'm going to be in trouble.
But I can do a little creative bending of the line here.
If I follow this slope up to here, it'll intercept somewhere.
That has a name.
It's called the Henry's law constant.
I can write it as K CS2, or KB.
So the point is, that at least for some range of concentrations, the variation is linear.
And what that means is, if I know this Henry's law constant, for CS2 mixed with acetone specifically, it would be different.
If it's mixed with something else, right?
Other kinds of interactions with different energies will have a different slope.
But if I know that number, it's still useful.
Because at least for that pair of constituents, I know the dependents for some range of concentrations.
So this is Henry's law.
p CS -- Oops, CS2 is x CS2 times K CS2.
Not quite as convenient as Raoult's law.
That number, I don't need to know what the other constituent is.
It's just a property of CS2.
It depends on the temperature, but not the other member of the mixture.
This one does, but still there's at least a substantial number of these things tabulated.
Especially for important mixtures.
So it becomes useful for dilute solutions.
And there's an awful lot of chemistry.
There's an awful lot of biology that happens in relatively dilute solutions.
OK.
I'll leave it at this point.
Next time we'll look at the total phase diagram, just like we did last time looking at both the liquid and gas coexistence curves, and seeing what the behavior is as a function of mole fractions.
And we'll talk about distillation and so forth in non-ideal liquids.
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So last time you learned about the ideal solution limits with Henry's law and Raoult's law.
And I just want to start with one reminder of what this is all about.
So let's say that you have a mixture of CS2 and acetone.
So this is a review of what you saw last time.
Acetone like this, CH3, CH3.
And if you take the CS2 as the solvent and the acetone as a solute, then when you plot as a function of the amount of CS2 here as a solvent, then you have the limit where the mole fraction of CS2 goes to one, meaning that CS is the solvent.
You expect Raoult to work if it's an ideal solution.
So that the vapor pressure of CS2 is related to the mole fraction of CS2 times the vapor pressure of CS2 if it was pure.
So you're sitting here somewhere.
This is the vapor pressure of CS2 pure and CS2 goes from zero to one.
And you expect Raoult to work.
So let's do Raoult.
Like this.
So in this area, if I were to plot the real curve, I would expect it to start out just like Raoult.
And then there's going to be some deviation.
Then if I take the other limit, where the mole fraction of CS2 goes to zero.
So now the CS2 is the solute.
And the acetone is the solvent.
what happens for this particular mixture is that Henry's law is no longer valid.
Raoult's law is not valid here.
Instead, you have to look at Henry's law.
And Henry's law tells you that for the solute now, not the solvent, so the CS2 here is the solute, very small amount of it.
Very small mole fraction.
That the vapor pressure of the solute is related to the mole fraction in the solution.
But instead of having the vapor pressure of the pure solute, you have another number.
Another constant.
Which we call K. Which is a Henry's law constant.
And this constant depends on what else in the solution.
It depends on the fact that you have acetone in there.
You would have a different constant if the acetone were replaced with chloroform, let's say.
So this constant here cares what the solvent is.
And it cares what the temperature is, also.
There's a lot of stuff that's built into this constant here.
It's a phinemelogical constant.
And in the case of the mixture of CS2 and acetone, the deviation from Raoult's law is positive.
Meaning that the Henry's law constant is greater than the vapor pressure of the pure CS2.
So that means that the molecules of CS2 would rather go in the gas phase.
If there's acetone around.
That means they don't like to mix so much.
They're not happy with each other.
So the CS2 molecules would rather escape what would normally be a Raoult's law type of solution to go in the gas phase away from being close to the acetone.
And as a result you get this positive deviation.
You get a higher pressure than you expect from Raoult's law.
So that the slope here This slope here is K CS2.
And that slope here is p star CS2.
This is a positive deviation.
And you saw an example where you if you mix chloroform and acetone, you get a negative deviation.
Chloroform and acetone are happy together because of the hydrogen bonding between the oxygen here and the hydrogen in chloroform.
And you get a negative deviation.
So let's start by doing an example with numbers.
Of how this might come about.
Any questions on this first?
I always got confused with Henry's law and Raoult's law.
And I always got confused as to when I was supposed to use Henry's law and when I was supposed to use Raoult's law.
Very confusing to me, initially.
The thing to remember is that if you're talking about something which in a very small amount, just the solute, like, well I'm going to give you an example which is a bad example, like sugar and water.
Where sugar is clearly the solute, but sugar is not volatile clearly this is a bad example.
But something which in a very small concentration, then Henry's law is going to apply.
Raoult's law is only going to apply for the solvent.
When you have a lot of stuff around.
On this side here is Raoult's law.
On this side here is Henry's law.
Henry's law, the deviation's from Raoult's law, are going to happen where the molecules in the solution mostly see molecules that are not like themselves.
If they mostly see molecules like themselves, since it's an ideal solution, then they don't know that there's other stuff around.
In this case here, the CS2 molecules don't really know.
Most of them don't know that there's any acetone around.
They couldn't care less.
So it obeys an ideal solution.
Here, they don't see any other CS2 molecules.
They mostly see acetone.
And so Raoult's law is going to fall apart.
Because Raoult's law assumed that you have a single component.
Or mostly one component.
OK. Let's do an example.
So in this case here, we're going to work the example out here.
So let's take two components, A and B.
Where B is, and let's look at something which is a dilute solution and non-ideal.
So there's going to be some deviation from Raoult's law.
And what we're told is at 50 degrees Celsius, that the vapor pressure of pure A is equal to 0.67 bar.
And the vapor pressure of pure B is equal to 1.2 bar.
So B is clearly the more volatile.
And we're also told, then, that at 50 degrees C we have a molar fraction of A in the liquid phase is 0.9.
And the molar fraction of B in the liquid phase is 0.1.
And the total pressure above the solution is 0.7 bar.
And eventually what we're going to try to find out is, we're going to change these concentrations.
Are we going to try to figure out how the total pressure above the solution changes.
That's maybe the goal of the problem.
When you change this to, eventually we'll change it to 0.95 and 0.05, and we'll want to know how does that change the total pressure.
So immediately we see that this is a solvent.
This is a solute.
This is volatile.
So, we're close to where this is zero.
So if we're going to apply any, sort of, Raoult or Henry's law problem to this, this is probably going to be Henry's law for B and Raoult's law for A.
We're going to be close to this area here for B.
We're going to be close to this area for A. OK.
So ah the first question is, what's the composition in the gas phase.
What is y sub B.
And it looks like this particular set of notes doesn't have the calculations.
I'm going to go to my other set of notes here that has them.
So, because I'm going to make a mistake if I go through it without my notes, I'm going to leave it up to you to do part a.
That's a good way out of making a mess.
I'm just going to tell you the answer.
Because the actual, the interesting part, is part b of the problem.
Part a is the easy part.
So this is something that you can do on your own.
So yB is equal to 0.139.
Is going to be answer to part b, given these two molar fractions and the total pressure.
So now, what we're going to do is, we're going to change the molar fractions and we're going to find out how that changes the total pressure.
So, second part is we're still at 50 degrees Celsius.
And we're going to change the molar fraction of A to 0.95.
And the molar fraction of B to 0.05.
And we're going to ask, what is p total now.
So we write what we know.
Which is that we expect p total to be the sum of the partial pressures of A and B.
A is the solvent, B is the solute.
It's a dilute non-ideal solution.
pA is going to be proportional to Raoult's law, because it's a solvent.
It's in large quantities.
So it's going to be xA pA star.
And xB is going, and pB is going to be proportional to the mole fraction again.
But because it's a solute, it's seen most the other molecules of A around, it's not necessarily going to be obeying Raoult's law.
It's going to obey Henry's law.
It's going to be close to the zero point.
There's going to be a deviation because of the interactions between A and B.
Which dominate now.
So instead of pA star, we have KB here, where this is the Henry's law constant.
Which depends on the fact that A is around.
We get a different constant if we have a different solvent around.
So we know this.
We know this because that was given to us at the beginning.
But we don't know this here.
We don't know what the Henry's law constant here is.
But we can get it.
We can get it from part a here.
Because we have all the information we need there.
So from part a, we already have a mixture where we're given the total pressure.
So from part a, KB is pB over xB.
And we calculated what yB is over there.
On part a.
So we know what the partial pressure of B is.
It's the molar fraction of B in the gas phase times the total pressure.
That's Dalton's law.
Then we divide by xB.
And then we plug in the number 0.139 times the total pressure up there is 0.7.
Divided by 0.1.
And we get that this is 0.973 bar.
So now we can plug this in to our equation here.
We know xA, we know xB, we know Raoult's law, the 0.67 pA star.
We know KB here.
We plug in all the numbers, and we get that the total pressure is now 0.685 bar.
This is a pretty typical example of a problem that combines Raoult's law, Henry's law and Dalton's law together.
And keeping track of these laws takes a little bit of thinking.
And it's very, very common on problem sets or exams to mix these up.
So it's good to go through examples like this and really keep track of where each law applies.
Dalton's law applies in the gas phase.
It's partial pressure in the gas phase.
Henry's law is the pressure.
If you know what's happening in the liquid phase, for a solute.
Raoult's law is the partial pressure in the gas phase, if you know what's happening in the liquid phase.
So Raoult and Henry tell you something about the pressure if you know the liquid.
Dalton only deals with what's happening in the gas phase.
OK, any questions on the problem?
Alright.
So we did this problem here, and now let's draw the phase diagram for our solute here, or for B, rather.
Zero to one.
And let's take a vote here.
So if Raoult's law worked for B, for the mixture of A and B, then we would have pB star here.
And everything would be great.
But Raoult's law doesn't work.
Close to molar fraction of zero.
Instead, Henry's law.
There's a deviation, due to the interactions between the molecules A and B here.
So we have two choices for the deviation.
We can have a positive deviation, or we could have a negative deviation.
Positive deviation, if the molecules of B would rather be in the gas phase, than be in the mixture in the liquid phase.
Meaning they don't like molecules of A.
They don't like to be mixed with molecules of A.
The energy of interaction is not favorable.
The enthalpy is not favorable, for mixing A and B.
In this case here, the enthalpy is favorable for mixing A and B.
So in this problem here, which do you think it is?
The blue one or the green one.
Let's raise hands here.
How many people think it's the blue one here.
Anybody else?
The right answer, we have four people.
How many people think it's the green one?
Four.
Five.
Six.
You didn't vote twice, did you?
Vote early, vote often, right?
Mayor Daley, the old Mayor Daley in Chicago.
And if you know somebody who's dead, vote for them.
OK, so this is the, I've got about five people here.
And a bunch of people didn't vote.
So it's a tie.
It's basically a tie here.
I think you should think about this.
Maybe talk to each other a little bit.
See if you can figure out, given all the numbers that are in this problem here, you know what, the pure pressure on the solvent is.
We change concentrations here.
Got Henry's law constant out.
And Henry's law constant is going to tell you whether or not that enthalpy of interaction is favorable or not favorable.
It's going to be a positive deviation or a negative deviation.
OK, shall we vote again.
How many people say it's the blue one?
How many people say it's the green one.
Thank you.
Alright.
Good.
Alright.
We have K sub B is less than p star, therefore it's a negative deviation.
The pressure is less than you'd expect.
The molecules like each other better than you expect.
There is a favorable enthalpy of interaction between A and B.
They like to interact.
It could be hydrogen bonding.
Could be a lot of different things.
OK, so now if I put the whole diagram together, which I shouldn't have erased here.
Right, so there's my diagram here again.
There's Raoult's law.
There's the deviation from Raoult's law here.
For one substance, let's say for B.
There's Raoult's law for A.
There's a deviation for A.
It could be like this.
And if I now add these two together, instead of having a diagram for, this is a pressure xB, yB diagram, pressure xB, yB diagram, instead of having Raoult's law sitting right here.
And then the yB part down here below, your usual sort of ideal solution phase diagram with the high pressure of the liquid on top and the gas on the bottom.
Now if you add the blue curve and the red curve together, you're not going to get a straight line like you did before.
When Raoult's law applied throughout.
Instead you're going to get a deviation from the straight line.
You're going to get a curve that's always below the straight line.
Because we have the negative deviation on both sides.
So you end up with something that's curved and deviates below.
And the same thing will happen for the bubble line.
It will also deviate below.
In some way.
Maybe not as much, maybe.
So this is what you usually see.
You see deviation of that whole phase diagram, like you're pushing down on it.
Or maybe it didn't pulled up.
If you have a positive deviation, then it's the opposite.
It gets, this is negative deviation.
And if you have a positive deviation, instead of having a usual phase diagram, you're going to get something that's looking like this.
Which is the Henry's law coming in.
Yes
[INAUDIBLE]
Yes
[INAUDIBLE]
Doesn't matter.
Good question, though.
OK, so this is what it usually looks like.
And everything you've learned so far can be applied to these diagrams here.
There's one exception, though, which is an important one.
Which is, sometimes, these non-ideal diagrams have a funny spot.
Which is that if I draw them, so there's my xB, yB, on the bottom.
Zero to one.
I have pressure here.
Sometimes, there's the dew line.
And the bubble line.
Sometimes, instead of having a nice separation between the bubble and the dew line, there's a special point which for some reason those two curves meet.
Non-ideality.
There's interaction between the molecules.
In the gas phase and the liquid phase.
This special point is an interesting one.
Because, well, if I'm away from this special point, let's say I'm sitting to the left of the special point, right here.
And I can distill this thing.
I can take my gas phase here.
And increase the pressure.
Say I'm sitting here in the mixture, gas phase mixture with x prime B, y prime B.
Sorry, x prime B.
And I go up and I increase the pressure.
Start making vapor.
Start making vapor, I can find out what the, I start making liquid.
I can find out where the composition in the liquid is by going over to the blue line here.
That tells me the composition in the liquid.
And reading off the composition of the liquid down here.
So this is actually the composition in the gas phase is yB prime.
yB prime is the gas phase.
I'm starting here.
I start to increase the pressure.
I begin the distillation process.
And I get a little bit of liquid out.
And I can read off the composition in the liquid.
xB prime.
And I know from this diagram that xB prime is less then yB prime.
So I'm enriching my liquid in A. I'm purifying my solution in A, right?
I can get pure A out eventually.
The fact of doing this, I can get pure A.
So I decrease the pressure, I get the pure gas phase of my new solution here.
And I increase the pressure.
And go over and make a little bit of liquid.
And capture that liquid.
And repeat the process over and over again.
And eventually I have pure A.
Now, suppose I'm sitting here at this special point.
And I'm starting with yB And let me call it a here.
yB of a. I'm sitting right here.
And I'm increasing the pressure.
I go all the way up here.
And I start making liquid.
.
What's the composition of liquid, compared to the composition of the gas phase?
What's y?
What's xB compared to yB of this point up here?
It's the same.
So I'm making liquid.
I got the same composition.
Doesn't help me.
If I make a gas again, I still have the same composition.
I can't distill.
I can't purify it.
So if I have this special mixture here, of A and B, at this point here, I form what's called an azeotrope.
And then you can't separate it out.
You can't separate out the mixtures into A and B.
You can't distill it at that point here.
Now, you can draw the same diagram in, instead of pressure, mole fraction, you can draw the same diagram in terms of temperature.
And the same thing happens.
You can have deviations from, this is the extra figures that are on the bottom page of the notes from last time here.
That we've replaced here.
You can have a positive deviation.
With a positive azeotrope.
And your T p diagram, T x diagram.
Or you can have, depending on the interactions between the two molecules, you can have something that has a negative deviation.
And a negative azeotrope here.
So in this case here, this is a particularly awful situation.
Because there's your liquid phase down here.
There's the gas phase here.
And now, suppose that you're doing some sort of distillation.
So you start here in the liquid phase, with some composition.
You make a gas.
Let's say you're doing a fractional distillation of some sort.
Your gas now has a new mixture.
So this is the composition in the gas phase.
You take that gas phase mixture, and you condense it in your condenser.
In your fractional distillation condenser.
So now I've condensed it.
And I heat it up again.
I've condensed it.
So I'm here somewhere.
And cool it down so that I have a liquid.
And I condense it.
And slowly I move over to the azeotrope point.
And once I get there, I'm stuck.
And the same thing happens if I'm on the other side.
I'm on this point here, I keep my solution, my mixture.
I make a gas, I read off the composition of the gas phase here.
I take my gas phase composition out and I re-liquefy it.
So I'm here, down here somewhere, again.
And I heat it up again.
And and I move to the azeotrope point.
That's what happens if you have a mixture of benzene and water, for instance.
You can't distill it.
You work your way down to this point here.
You can't use fractional distillation until you've separated it out.
So this is a real life problem.
And you have to do things like change the pressure.
Do your distillation under high pressure or low pressure.
To try to get this point to separate out.
So you can get past this point.
Or to try to get the deviation to become of the opposite sign.
Where you can start to do some distillation.
Any questions on azeotropes, before we get to colligative properties?
If you're doing organic chemistry, and you're trying to purify things, this is a real pain, right?
Pain in the neck.
So you've got to know, which things form azeotropes.
So we're going to start colligative properties.
There are four colligative properties, and you already know what they are.
And you've even seen the math behind them.
And in fact, you've seen the math for one of them already pretty clearly.
And these colligative properties are properties that you use all the time in your day to day life.
At least, three of them.
OK what are the four colligative properties?
There is vapor pressure lowering.
If you have a mixture and the vapor pressure of the mixture is less than the vapor pressure of the pure material, there's the boiling point elevation.
If you add a large amount of insoluble - not insoluble, non-volatile solute to a solution, you will depress, you will increase the boiling point of that solution like the example of the salt and water that I gave.
So if you need a lot of salt to increase the boiling point of water.
There's the freezing point depression.
That's what we use to melt ice in the winter.
By adding salt to it.
Then there's the osmotic pressure.
Which is the example of the freshwater fish in salt water, or the saltwater fish in fresh water.
Or the cells bursting when you put them in.
OK, so these colligative properties all come from the fact if I have a mixture, the chemical potential of a molecule in a mixture, in the liquid phase here, is less than the chemical potential of the pure guy.
Where A is the solvent, and B is some solute.
So this is always true if your number of moles of A is much larger than number of moles of B.
Mix a little bit of B in there, and the chemical potential of A in the liquid is going to come down.
Because of the mixing process.
Because of the entropy of mixing.
And that's going to give rise to these properties.
And, to do these, one more thing before we start.
We're going to use a special concentration for these guys.
We're going to use, call something molality.
Molality.
Molality is the number of moles of solute divided by kilograms of solvent.
So it's moles per weight of solvent.
As opposed to moles per liter, which is something that we're more used to.
Moles per kilograms of solvent.
It's pretty close to moles for liter for water.
Because a gram of water is a milliliter, roughly.
The density of water's close to one.
So let's start on these things.
So we've already seen this.
You've already seen vapor pressure lowering.
It comes from Raoult's law.
If I look at the change in the vapor pressure of A, pA minus pA star.
pA is equal to xA pA star.
xA pA star, that's Raoult's law.
Minus pA star, that's equal to one minus xA.
Minus pA star.
Which is minus xB pA star.
Which is less than zero.
So, that means that the vapor pressure above the solution is lowered.
So I should have said vapor pressure lowering.
And we also drew a phase diagram to show how this vapor pressure lowering was going to imply freezing point depression and boiling point depression.
So now let's go on and look at the freezing point depression.
Number two.
We're going to find that the change in the boiling point is going to be some constant K sub b, which is not to be confused with Henry's law's constant.
It's completely different.
This is a colligative property.
Times the molality of B.
That's going to be the answer we're going to get.
Amount of temperature change depends linearly on the molality of B, with a constant that depends on what the mixture is.
And what the pressure is.
Whenever we derive anything connected to phase transitions, we usually always start with the fact that the chemical potential changes.
And it has to be equal at equilibrium.
So that's what we're going to do.
We're going to start with the chemical potential equal of the liquid phase is equal to the chemical potential of the gas phase.
For the solvent at equilibrium.
And the liquid phase obeys Raoult's law.
So it's going to be the chemical potential of the pure material.
The pure liquid.
Plus RT log xA.
So this is where the mixing of the solute comes in.
And on the gas phase side, mu A, it's going to be the same thing as the chemical potential of the pure A.
Because we're going to take B to be non-volatile.
So the only thing in the gas phase is A.
Same chemical potential as if it were pure.
Because it is pure.
So now we solve for log xA is equal to one over RT mu A star in the gas phase minus mu A star in the liquid phase.
And this is just the Gibbs free energy per mole of A in the gas phase.
The Gibbs free energy per mole of A in the liquid phase.
So we can replace this with the change in Gibbs free energy from going from gas to liquid.
Per mole, which is just the Gibbs free energy of vaporization.
We're going from liquid to gas.
And I probably have a negative sign somewhere.
No this is right, this is the final state.
This is the initial state.
Divided by RT.
OK. That's not really what we want, though.
What we want is delta T. Here I've got some complicated equation that has T in it, but not delta T. But before we do that, before we transfer and try to get a delta T out of this equation, let's see if we can do any sort of simplifications here.
I don't like to carry logs around.
There are no logs around here.
So let's see if we can get rid of that log.
What is log xA?
Log xA, well, xA is close to one.
It's connected to xB.
And everything in here is connected to B.
There's a molality of B here.
The concentration of A doesn't come into play here.
I don't want it there.
So let me replace it with B. one minus xB.
xB is a small number.
It's a solute, it's in small amount.
We're close to being a pure solvent here.
So one minus something very small.
The log of that.
Well if you've worked with Taylor's expansions.
You probably know that that's close to minus xB.
The log of one minus x where x goes to zero is close to minus x.
So that's minus xB.
Now, minus xB, in terms of the moles, is the number of moles of B divided by the number of moles of B plus the number of moles of A, the total number of moles.
The number of moles of B is very small, compared to the number of moles of A.
So I can get rid of this.
So, and there's a minus sign here.
So this is approximately minus nB divided by nA.
Now, the only I can do is, I'm trying to get at molality here.
I said we're going to use molality.
Got moles.
I don't have kilograms of solvent here.
I have moles of solvent.
I'm not so happy with that.
So, in order to get something that has kilograms out, let's multiply up and down by capital M where capital M is that total mass of A.
Basically the number of kilograms of solvent.
Total mass of A.
So now, this fraction here, that's the molality.
That's what I'm looking for.
So this is minus the molality.
Then I have the total mass of A divided by the number of moles of A.
Well, that's the molecular weight of it.
Total mass divided by the number of moles is the molecular weight.
So this is the molality of B times the molecular weight of A. OK.
So now I can plug this instead of log xA.
And I get minus nB times nA is equal to delta G of vaporization divided by RT.
Let's rearrange a little bit.
And mB is equal to to delta G vaporization divided by nA times RT, with a negative sign.
Well, I've eliminated the log at least, but I still don't have a delta T in there.
I don't have a delta T. How am I going to get a delta T in there?
Got to get the delta T. Everything's very small.
So one way to get a small change in temperature is to take a derivative with respect to temperature.
That's going to get a delta T in there.
So I'm going to take the derivative of both sides with respect to T. And delta T is going to come out that.
I'm going to take d nB / dT, constant pressure.
Minus one over mA times R d delta G over T. This is delta G of vaporization.
dT, constant pressure.
And I wish I'd made my boards a little bit better.
Because I have to cover things up.
OK, now I have this derivative of the Gibbs free energy divided by the temperature.
This turns out to be something that you've seen before.
It's called the Gibbs-Helmholtz equation.
And it relates the temperature change in the Gibbs free energy with the enthalpy change.
The Gibbs-Helmholtz equation is that the d delta G over T dT is equal to minus delta H over T. And you can get that from just writing fundamental equations about G and H. And turning the crank.
And we've done that before.
And I'll leave that as an exercise for you to go back and figure out what, you can go back to your textbook and look at, you know under the index.
Gibbs-Helmholtz, and figure out why this is the case.
So now, we have this equation here.
This is going to be nice, because it's going to get rid of the d/dT.
On the right-hand side.
And so now our equation becomes d mB / dT is equal to delta H vaporization divided by mA R T squared.
Now, I can rewrite this in a slightly different way.
This is delta m over delta T, basically.
Delta the change in number of moles.
I mean, the molality of B divided by the change in temperature.
I'm getting very close here.
There's a change in temperature here.
There's a change in molality here.
I can rewrite this as delta T, then, is equal to molecular weight of A times RT squared divided by delta H vaporization times delta mB.
Now, I'm making a very small change in B here.
Starting at zero.
Basically I'm taking pure A and I'm dumping in a little bit of B.
That's what I'm doing when I experiment.
I'm dumping in a little bit of B in my pure A, and I'm trying to figure out how the temperature is changing.
So I'm doing delta m sub B is my small amount of B that I'm adding to the solution.
Minus what I started out with, which was zero.
So delta mB is basically m sub B.
Right?
That's basically what it is.
So let's get rid of this delta here.
What else can I say?
Well, I can say that the temperature change that I'm looking at is also pretty small.
Compared to the absolute temperature.
This is a small change.
I'm only adding a small amount of B in there.
So this temperature up here is not going to change very much.
Let's take it to be the temperature of the boiling point of the pure material.
T sub B star, of the pure material.
OK And there's my delta T here.
So this is of the form delta T is equal to molecular weight of A.
Times R. Times the boiling point of the pure material squared.
Divided by delta H of vaporization of the pure solvent.
Times the molality of B.
And this combination of properties of A, the molecular weight of A, the boiling point of pure A.
The heat of vaporization of pure A, this is K sub b up here.
Which I covered up.
It's of the form delta T is equal to K sub b times mB.
And this turns out to be always greater than zero.
Now.
If you want to look at freezing point depression, it's really easy.
In your equations, it's the same thermodynamic picture.
All you need to do, in all your equations, you replace delta G of vaporization with minus delta G of freezing.
When we did a vaporization, we're going from liquid to gas.
We had delta G of vaporization.
When we're doing freezing point depression, we're going from liquid to solid.
And so we have to use minus delta G of fusion.
Minus delta H of fusion.
s to l would be the positive delta G of fusion.
So we're going from liquid to solid.
We have to replace delta G of vaporization with delta G of fusion.
We replace delta H of vaporization, where minus delta H of fusion.
Replace the boiling point with the freezing point.
And replace K sub b with K sub f. And the math is exactly identical.
It all starts from the chemical potentials being equal.
And turning the crank at equilibrium, being equal between the solid phase and the liquid phase.
And turning the crank.
The solid phase is like the gas phase.
It's a pure A, pretty much.
Because the molecules of B are stuck.
And A doesn't know that B's around, in the solid phase.
They know that they're around in the liquid phase because the fusion fast enough.
.
And then we can write, if we turn the crank, we get an equation that looks like this.
The freezing point depression is equal to minus mA R Tf star squared.
Divided by delta H of fusion, for mole.
Times the molality of B.
And this number is always less than zero.
There's the negative signs in here.
OK.
Always depressed.
Any questions about those freezing point depression and boiling point elevation.
Yes
[INAUDIBLE]
Why did I drop it here?
Because the small change that I'm making in nB is going from pure A, where nB is zero, to a very small amount of nB.
That's why I'm allowed to say this.
And then just replace delta nB with nB.
I'm comparing every, my reference point is the pure material.
I'm just adding a little bit to start with.
It's a good question.
Any other questions?
OK.
I didn't get to osmotic pressure.
You'll get to that on Friday.
And also start statistical mechanics.
And after osmotic pressure, you'll be done with thermo.
It'll be statistical mechanics and then kinetics.
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Last time, under the tutelage of Professor Blendi, you started to see some of the consequences, the more immediate consequences, of some of the properties we've been discussing over the last few lectures of binary liquid mixtures.
And in particular he started showing you some of the colligative properties by which you see changes in the vapor pressure and the boiling point and so forth, that arise because you've got liquids in mixtures and this affects their chemical potentials.
And today I want to start by going through the fourth colligative property that's commonly seen, what called osmotic pressure.
And I mentioned this before.
Today we'll go through a quantitative calculation of its effects.
And then we'll start in on a different unit of the course.
We'll start in on statistical mechanics.
So let's continue with the colligative properties.
And osmotic pressure plays a super-important role, certainly in all of biology, because of course inside biological cells you have solutions.
Water is in there, along with a whole bunch of other components.
And so the chemical potential of the water is strongly affected by all the other constituents.
And this has profound effects on what's involved in maintaining an equilibrium between what goes on inside and outside of the cell.
And this is the case in all organisms.
And it's also exploited of course in all sorts of other contexts.
So what's happening with osmotic pressure is, you've got some situation where there's a semi-permeable membrane.
It allows one of the constituents but not others to pass through.
And typically that's water.
Although, of course, it could be anything.
So, in the lab the way this typically looks is something like this.
Here's your semi-permeable membrane.
Outside, we can imagine that there's pure liquid A.
Could be water, could be something else.
Inside, you've got A mixed with B, one or more other constituents.
And then there's some level of the liquid outside the container.
Of course, A can pass between, pass through, the membrane.
And so what ends up happening is, if you just look at the situation, what you immediately notice is that the liquid in here goes up higher than the liquid around it.
It's at a higher level.
It seems counterintuitive at first.
And in fact it can be enormously higher.
But that's what happens.
So there's extra height, h, of liquid inside the container relative to what's outside it.
And what that means is the pressure inside here must be higher, right?
Of course, the weight of this liquid is pushing down on what's down here.
So the pressure inside here is higher than the pressure outside.
But of course, from what you know already about liquid mixtures, this should in fact make sense.
Because what's happening is that if these things are at the same pressure, you know that the chemical potential of A is going to be lower in the mixture.
It's always lower in the mixture than it is in pure A.
So there's nothing to stop A from rushing in.
So A does rush in.
If the liquid level goes up, it raises the pressure down here.
But as the pressure rises, now the chemical potential changes and goes up.
So eventually, equilibrium is reestablished.
Because these aren't at the same pressure.
The additional pressure means the chemical potential of A goes up.
Of course, it goes down by virtue of being in the mixture.
And those two things will eventually reach an equilibrium where the chemical potential of A is the same everywhere, as it has to be.
So what we'll calculate now is, OK, how much does the pressure go up.
When that happens.
So let's take a look.
And so of course, there are all kinds of examples.
In a typical one, A would be water.
And B would be one or more sugars.
All sorts of other potential constituents also.
OK, so let's just consider the pressure at two places that are the same level.
The same height.
We'll label this one point alpha.
Let's put some color in here.
We'll have points alpha and beta.
And we'll make this distance l. So now let's just first calculate what the pressure is at both points.
It's going to be higher here, because of this additional liquid on top.
So, if we look at point alpha, well, let's start at point beta, actually.
That's the simpler one since it doesn't have the additional liquid.
And first of all, there's just atmospheric pressure.
So there's an external pressure.
Certainly, typically, it would just be atmosphere pressure.
It's pushing down here.
It's also there.
So that has to factor in.
So there's our external pressure.
And then there is the product of this height times the density of the solution.
Times the gravitational force constant.
And I'll work through that momentarily so you see that that indeed is the pressure.
That's exerted by the amount of liquid that's up above this level.
But before I do that, let's put the pressure at point alpha.
So p alpha.
And that's p external plus l rho g. But now, there's this additional height on top of that.
So there's also h rho g. There's an extra pressure.
We'll label this pi.
So we can write p alpha is equals to p beta plus h rho g. Which we'll say is equal to p beta plus pi.
Now, just to make sure we're all on the same page about these quantities being the pressure that's exerted by the liquid, let's just look at that.
So l times rho times g. It should be a pressure which is a force per unit area.
And in particular, this is going to be a weight per unit area.
Remember the weight is the force that a mass exerts.
And it exerts it because of the gravitational constant.
So what happens?
Well, this is just a length.
Rho is a density, right?
So it's mass per unit volume.
And of course, g is just our gravitational force constant.
So it'll be in units of meters per second squared.
So what's going to happen?
If we multiply l times rho together, this is units of length.
This is units of volume.
What we're going to get then is mass per unit area.
And that's what it is, of course.
It's distributing across the area of that surface where point alpha is.
So these two together, l times rho, give us mass per unit area.
So it could be in units of kilograms.
Per unit area, per meters squared.
And now we're going to multiply that by the gravitational force constant.
So now we're going to have all together kilograms per meter second squared.
In other words, it's force per unit area.
And so that's what we have.
And now, of course, we have the additional component.
Due to the height here.
Now, what do we know about the chemical potential in both parts?
The chemical potential has to be equal outside and inside the container.
So mu A at point alpha is just mu A of the liquid at pressure p plus pi and temperature T. And that has to be equal to mu A at point beta.
But point beta is just in the pure liquid.
So that's just mu A star of the liquid at pressure p and temperature T. So here's the chemical potential at point beta.
Here it is at point alpha.
Those two have to be equal to each other.
And now we're just going to use Raoult's law.
And what does that tell us?
We have RT log of xA plus mu A star of the liquid at pressure p plus pi and T So there is our chemical potential inside the container.
And that's equal to mu A star of the pure liquid at point beta.
And so we can just rewrite this as RT log xA.
Plus mu A star liquid at pressure p plus pi of T minus mu A star at pressure p is equal to zero.
OK., now we need the pressure dependence of the chemical potential.
But we certainly have an expression for that.
So we know that dG is S dT plus V dp.
The temperature is the same on both sides.
But we need to worry about what happens at different pressures.
So at constant T, dG is V dp.
So now if we look at the chemical potential, which is just the Gibbs free energy per mole, then d mu A star is VA star molar dp.
In other words, the difference in the chemical potential is, this changes as a function of pressure.
Is going to be given by the molar volume of a under these conditions in the pure liquid dp.
It's just the potential, the pressure dependence of G. It's not the kind of topic I would have thought would drive away a lot of people.
But, you never know.
OK. For the rest of you who are willing to bear with me, let's continue.
So now let's just integrate.
We need to know the change in mu A at a finite jump in pressure from inside to outside the container.
So we're just going to integrate.
So if I want mu A star at pressure p plus pi minus mu A star at pressure p, then I just have to integrate from p to p plus pi VA bar star dp.
And now I'm going to assume, and this is certainly a safe assumption, that this quantity this, molar volume, isn't going to change.
In other words, the molar volume of liquid A isn't going to change significantly, going from the pressure out here to the pressure in here.
On the grand scale of things it's a small pressure change.
We can assume that the liquid is incompressible over that small pressure change.
Which means that this is constant, since all we have then is simply VA bar star times pi.
So then, substituting back here, so now we've calculated this difference, it's a simple result.
So then we simply have that RT log xA plus VA bar star pi is equal to zero.
Now, you saw last time, and I'll work through this quickly again, that the log of xA can be written approximately as minus nB over nA if you remember, this came from writing that this is equal to the log of one minus xB.
But we're assuming that B is the minor constituent here.
So xB is a small number.
So that this is roughly equal to minus xB.
And that's just minus nB over nA plus nB.
And since nB is much smaller than nA, this is approximately equal to minus nB over nA.
So we're going to make this substitution.
Also, this quantity, this molar volume of A over the pure liquid, since the concentration of B is low, we can assume that this is just the molar volume of A in general.
So, in other words, we can write VA bar star.
We can just write it as VA bar.
That is, we're not going to worry about changes in the molar volume, either as a function of pressure or a function of concentration at the low concentrations that we're working.
And then, note that nA times VA per mole is just the total volume.
Oops, not bar.
Not the molar quantity.
This is the molar quantity multiplied by the number of moles.
So it just gives us VA, the total volume, occupied by A.
And using these two results, then we have that RT times negative nB over nA, that's this result.
Plus VA over nA times pi is equal to zero.
nA will cancel.
And finally, since almost all the volume is due to A, again because B is the minor constituent, we can approximate further that VA is approximately equal to V. So finally getting rid of the nA on the denominators, we're left with a simple expression.
pi times V is RT nB.
So we have a very simple expression which is called the van't Hoff expression, and look at how it resembles the ideal gas law.
This is the pressure times the volume equals the number of moles times RT.
Of course, really it's a change in pressure.
And this number of moles is the number of moles of a guest constituent in a solution.
But it has the same form that you're familiar with.
We also, sometimes it's convenient to substitute, since nB over V, that's just the concentration.
That's the number of moles over the volume.
And in that case, the expression is rewritten as pi is RT times c, where it's understood that this is the concentration of the solute.
OK, any questions?
Let's work through a numerical problem.
These things can seem a little bit, they can seem simple, but sometimes when confronted with a problem, sometimes it may seem less than obvious what to do.
And also the results can seem a little bit surprising.
So I just want to work through a simple numerical example to see how these play out.
So let's try, let's imagine we have 10 grams of unknown solid.
Dissolved in 1000 grams of water.
The vapor pressure is 25.195 torr at 27 degrees C. And the pure water vapor pressure is 25.000.
No, .200 torr at the same temperature.
So first, let's just calculate the molecular weight of the solid.
Now, this you I think, saw from last time.
So this is straightforward.
It's going to come from the vapor pressure lowering not from the osmotic pressure, which we'll do next.
So the delta p of H2O is minus xB times p star of H2O.
Alright, this is from last time.
The expression for the vapor pressure reduction.
Due to the concentration of the mole fraction of B.
And then xB is 0.005 torr over 25.2 torr.
Right?
And that's 1.98 times 10 to the minus 4.
So that's our mole fraction of B.
And then xB is nB over nA plus nB.
So, now we have this, it's 10 grams divided by the molecular weight, which is in grams per mole, over 1000 grams of water.
Divided by 18 grams per mole plus 10 grams divided by the molecular weight in grams per mole.
But since the solution is overwhelmingly water, we can get rid of this term and make the math simple.
Can you see that?
I guess barely.
So this turns out to be about 1.98 time 10 to the minus 4.
And if we solve for molecular weight, then we wind up with 907 grams per mole.
So that's using one of the colligative properties that you saw last time.
Now let's, though, finish up by calculating the additional pressure that we'll find.
So what's the osmotic pressure of the solution.
So now we're going to assume that we have that solution inside, in a situation like this where there's pure water outside that can go in.
And there's going to be excess pressure because of the water rushing into the solution.
So what happens?
And let's use a value of the density.
0.995 grams per centimeters cubed.
So, pi is RT times c. That's 0.08314 bar per Kelvin mole.
Times 300 K times c. c is nB over V. So it's 10 grams divided by 907 grams per mole, over 1010 grams over 0.995 grams per milliliter, I'm equating with centimeter cubed.
And that turns out to be 1.09 times 10 to the minus 5th mole per millileter.
Let's get that into liters.
So we'll multiply by 1000 milliliters per liter.
So it's 1.09 times 10 to the 5th moles per liter.
OK, yeah
[INAUDIBLE]
Let's see.
Oh, don't forget that, right
[INAUDIBLE]
Ah, yes, yes, yes.
And I think that's the case in the notes also.
Yep.
Thank you.
OK, so the concentration, just to calibrate us here, it's a pretty low concentration, right?
It's about 10 to the minus 5 moles per liter.
The excess pressure is 0.27 bar.
In other words, about a quarter of an atmosphere.
Uh-oh, what's happening?
[INAUDIBLE]
Where are we?
[INAUDIBLE]
Sorry, where exactly?
[INAUDIBLE]
Oh, yes.
Yes, sorry.
Yeah.
10 to the minus 2 moles per liter.
All right.
And so the pressure that results is a modest pressure, right?
It's about a quarter of an atmosphere of excess pressure.
2.7 time 10 to the 4 pascals.
So let's just calculate, given this small additional pressure, how high is the liquid going to go, above the level of the solution.
Anybody want to guess?
You know, a centimeter.
10 centimeters.
100 centimeters?
Give me an order of magnitude.
Who guesses one centimeter?
You got three choices.
One, 10 and 100, and you have to make one.
Who guesses one centimeter?
Who guesses 10 centimeters?
Who guesses 100 centimeters?
OK. Now, fortunately we don't do problem sets.
Or other scientific problems by popular vote.
Rather, we just solve them.
Whenever possible.
So, what's the height of a column of solution?
Well, that pi, remember, that's equal to rho times g times h, right?
And we know pi now.
So h is pi over rho times g. So it's 2.7 times 10 to the 4 pascal over rho, that's 0.995 grams per centimeters cubed.
Times 1 kilogram over 1000 grams times 100 centimeters per meter cubed.
I'm just getting this into units that will give me, that will be in meters, times 9.8 meter per a second squared.
And when I multiply all that out, it's 2.8 meters.
So that puny little pressure ends up being meters high.
Taller than any of us, by far.
Sorry, it's up to there.
And that's actually pretty typical.
And what that means is, it actually makes osmotic pressure a very, very sensitive measurement method.
If you want to determine a molecular weight, for example.
Because it's so high, that means, and of course you can measure that height pretty accurately.
Certainly within about a millimeter.
And so even relatively modest concentrations, it's still easy to see the effect and measure the impact.
And use that to calculate the properties of the solute.
OK, any questions?
Alright.
Now we're going to start something new.
So, so far what you've seen, and I hope come to appreciate, is the whole structure of thermodynamics is built on empirical macroscopic observation and deduction.
We observe things.
We formulate these broad thermodynamic laws.
All macroscopic.
It doesn't depend on any microscopic model at all.
And in fact much of it was formulated before there was a good microscopic model of matter.
Before the atomic theory of matter, and molecules, and so forth was well worked out.
And it's incredible how powerful that whole formalism is, right?
In some sense, although it may seem like neglecting what we now know to be an important part of nature, actually part of its power is its empiricism.
Right?
There are these very small number of fundamental laws from which everything else follows.
And at this point, of course, there's just enormous confidence in that small number of laws.
Because of their being verified in so many context.
But, at the same time, we do know about the atomic theory of matter.
We know there are atoms and molecules.
So it ought to be possible, at least in principle, to start from a purely microscopic approach to nature.
And just based on figuring out the microscopic properties.
And then saying, well, OK, my macroscopic stuff is just a collection of those microscopic entities.
I should also be able to figure out macroscopic thermodynamics.
I should be able to start from my microscopic picture and get to macroscopic thermodynamic results.
And in fact, that is possible.
And the theoretical formulation for it is what's called statistical mechanics.
Called that because, of course, you won't be surprised to learn that it's going to require a statistical treatment.
We're going to be dealing with moles of material.
But now we're going to be trying to think about their microscopic properties.
And so we'll be dealing not with n number of moles, but with 10 to the 24th number of molecules or atoms.
And you know we won't be able to keep track of every one of their individual states, all the time.
We may be able to do a terrific calculation of the quantum mechanics or the classical mechanics that describe the states that they could be in.
But there's no way in practice that we're either going to experimentally or theoretically keep track of all of that.
So we're going to, at some point, have to introduce statistics.
To take what we know about the microscopic properties and try to go from there to the macroscopic results.
The thermodynamics that we've seen so far.
And that's what statistical mechanics is all about.
So let's try to introduce a little bit of statistical mechanics.
Where we're going to go from what we know about microscopic properties all the way to macroscopic thermodynamics.
That's our objective.
So let's start by just trying to calculate energies of individual molecules, or individual particles.
And what's the probability that some molecule, one of the oxygen molecules somewhere in this room, is in a certain energy state.
Right?
And what our strategy is going to be, is to determine what are the probabilities that molecules are in certain states with certain energies.
And if we can determine that for all the possible states, then we can average over those.
So without keeping track of every individual molecule, we could then calculate, an average energy, which is what you would measure thermodynamically when you look at the whole collection and measure what we call u, right?
Of course, we're really averaging over disparate energies of lots of different atoms or molecules.
They don't all have the same molecular energy.
And we don't try to measure their individual energies.
So that's what we'd like to calculate.
And so we'd like to be able to know what are all these probabilities of different energy states.
And then from that, statistically averaging, what are the macroscopic average energies.
And other macroscopic quantities.
So, let's start there.
Probability that a molecule, must be specific as possible is in state i with energy Ei.
And right now we're not even going to specify the nature of the state.
We could be worrying about translational energy.
What state it is, how fast it's whizzing around the room.
We could worry about its vibrational or rotational energy, or electronic state.
For now, let's not even specify it.
Let's just say the only thing we're really specifying about the state is it has some energy that we presumably know.
Now, what we'd like is to know a functional form for the probability.
And we don't know one a priori.
But let's just think about two molecules.
And the probability that one of them has energy i, and another one has energy Ej.
It also, let me say, would be sufficient to think about even one molecule and say, we'll let's think about independent parts of its energy.
Let's say, translational energy in this direction.
And in an orthogonal direction.
What matters is, we're thinking about two independent energies.
Could be separate molecules that aren't interacting.
Or independent degrees of freedom on the same molecule.
But let's just consider it.
So a molecule is in state i, and that another molecule is in state j, with energy Ej.
So what we're going to calculate is the joint probability that the two molecules are in those states.
So we want a probability.
This one, we'll call Pi(Ei).
This would be Pj of Ej.
And this probability together will be Pij(Ei + Ej).
I could just write Ei comma Ej, but the energy's add.
And that's crucial.
Now, since the molecules are completely independent, this should just be the product of the separate probabilities.
Two completely independent events, or things whose probability we want to determine together.
So, that suggests a really simple functional form.
Because what we're saying here is we want a function of a plus b, which is equal to the same sort of function of a times the same sort of function of b.
In other words, when we multiply these functions, we get a function of the sum of those arguments.
What functional form does that?
Exponential.
Sure.
So in other words, e to the a plus b power equals e to the a times e to the b.
So it suggests that there's an exponential probability.
Suggests it.
Now, there's also some physical insight that we can apply to the problem.
First of all, we expect that states with higher energy in general are going to be less probable than states with lower energy.
If it's cold, not much thermal energy around, and you say, how much energy does a particle in an equilibrium collection of particles have, at thermal equilibrium, if there's not much thermal energy around, you don't expect it to be very likely that a particle has a huge amount of energy.
That one individual particle has it.
Could happen, right?
From some random collisions that just happened to bulk up the energy of that one particle.
But on a balance it's not going to be very probable.
Lower energies will be more probable.
And the other thing is, even in thinking about something like that, temperature comes in.
Immediately.
Right?
If you raise the temperature a lot, surely we have to expect many more molecules will start to become energized.
Will have more energy.
So, lower probability for higher energy.
But the probability of higher energy should go up if the temperature goes up.
In other words, the ratio of energy to temperature should be involved here.
So, P of Ei should go down as Ei goes up.
And should depend on Ei over T. Right?
So, we can start with this functional form.
And by the way, there's no reason we can't have, in general we would have some constants here, right?
We haven't changed anything to do that.
And now let's just use the little bit of physical intuition that we're thinking about here, to refine this just a little bit.
Let's write, we expect that Pi of Ei is some exponential to the minus C Ei over T, where C is some constant greater than zero.
So this will have the property, then, that as energy goes up, the probability goes down.
But it's scaled by temperature.
If I raise the temperature, that makes the higher energy states get more and more likely.
Well, this is our functional form for probability of a molecule being in a state with energy Ei.
And the only difference between this and what's written conventionally is the way the constant is labeled.
So really what we have is Pi of Ei is proportional to e to the minus Ei over kB T, or just k T, where kB is called the Boltzmann constant.
And it's just equal to R over Avogadro's number.
It's the gas constant per molecule, rather than per mole.
One way to try to rationalize this is, you've probably seen, you've all seen Arrhenius kinetics, right?
Arrhenius rate laws?
If you remember what that looks like, you get this rate constant.
Arrhenius rate constant is some constant A times e to the minus Ea over RT.
Remember that?
You've all seen that before?
And so what's happening here, Ea is what is an activation energy.
Remember, the idea is that you've got reactants and products, and there's some barrier you've got to get over.
The activation energy, before you can have reaction.
So the rate depends on surmounting that activation barrier.
This is really coming from the same idea as this, which is the probability of one of the molecules having this much energy.
Depends on e to the minus energy over RT.
This is per mole, so this is per mole.
The exact same relation.
So the idea in the context of kinetics is that the rate depends on how many molecules can get up here.
And what the probability is of their getting enough energy to go over the barrier.
And that energy, that probability, is given by this expression.
So in this form you've seen this kind of dependence before in a very explicit way.
Let me just take it one small but important step farther.
Which is that proportionality constant.
Couldn't we do better?
Couldn't we say, couldn't we figure out what exactly it is, not just what it's proportional to?
Well, let's try.
So as we've written it, we've got Pi of Ei is proportional to e to the minus Ei over kT.
We could write that as equals A e to the minus Ei over kT.
This is just some proportionality constant.
But we can determine what that constant is, because we know that if we sum over all the possible states, the molecule has to be in some state, right?
So that sum of all those probabilities has to equal one.
So sum over i of Pi of Ei, is equal to one.
Because the molecule must be in one of the states.
OK, now this is our expression for all those p i's.
So let's just write that way.
a is just a constant times the sum over all the i's of e to the minus Ei over kT.
And so there it is.
a is equal to one over the sum over i, e to the minus Ei over kT.
So now we can rewrite Pi of Ei is equal to e to the minus Ei over kT over, the sum over i, e to the minus Ei over kT.
Now, just so you're not confused, this i matters.
That's the i we're talking about.
What's the probability of it being in some particular state i with energy e i.
This i is just a dummy variable.
So just to be explicit, we could write this just as well as e to the minus Ei over kT times the sum over j, e to the minus Ej over kT.
It could be confusing, but this is usually the way that it's written.
Even though this dummy variable here has nothing to do with the particular choice that's made here.
So now we know.
And we can calculate what the probability is for a molecule to be in any particular state if we know the energy of that state.
Now, notice we need to know all the energies of all the other states, too.
Because this thing, in other words, this a, right, depends on summing over all of those.
But if we know those, then we can do it.
We can do the whole calculation.
And for a good number of cases, we realistically do know it all.
So from the proportionality relationship alone, of course, we can tell the ratio of chances that you're in one state or another.
Even without this a constant, we already could've said well, if what's the ratio of, the probability of being in state i with energy Ei to state j with energy Ej.
Well, it's just e to the minus Ei over kT over e to the minus Ej over kT.
It's e to the minus Ei minus Ej over kT.
But apart from just being able to get the ratio, we can get the absolute number.
What's the absolute probability that a molecule is in any particular state of a given energy.
Next time what you'll see is what can be calculated based on the results that you've seen, just the results you've seen so far.
From these quantities alone, it turns out you'll be able to calculate every single macroscopic thermodynamic quantity.
So you'll see some of that next time.
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So, last time we started in on a discussion of a new topic, with was statistical mechanics.
So what we're trying to do now is revisit the thermodynamics that we've spent most of the term deriving and trying to use in a microscopic approach.
So our hope is to be able to start with a microscopic model of matter, starting with atoms and molecules that we know are out there.
And formulate thermodynamics starting from that microscopic point of view.
In contrast to the way it was first formulated historically, and the way we've presented it also, which is as an entirely empirical subject based on macroscopic observation and deduction from that.
So, we looked at the probabilities that states with different energies would be occupied and we inferred that there would be a simple way to describe the distribution of atoms or molecules in states of different levels.
First of all, although I didn't state it explicitly, essentially assumed there is that if I've got a bunch of molecules, and I've got states that they could be in of equal energy, then the probability that they would be in one or another of those states is the same.
If the energies of the states are equal, the probabilities that those states will be occupied.
That they'll be populated by molecules, will be equal.
And then we deduced what's called the Boltzmann probability distribution.
Which says that the probability that a molecular state, i, will be occupied is proportional to e to the minus Ei over kT, where little Ei is the energy of that molecular state.
And then realizing that the probability that some state out there has to be occupied, we saw that of course if we sum over all these probabilities that sum has to equal one, right?
In other words, the sum over all of i of P of i is equal to one.
So that means that we could write not just that this is proportional to this, but we could write that Pi is equal to e to the minus Ei over kT divided by the sum of all such terms.
So now we know how to figure out how likely it is that a certain state is occupied.
And what it means is, let's say we have a whole bunch of states whose energy is pretty low compared to kT.
kT, the Boltzmann constant times temperature is energy units.
So if it's pretty warm, maybe it's room temperature, maybe it's warmer.
And there are a whole lot of states that are accessible because the energies are less than kT.
What does that mean?
That means that you could put a bunch of different i's here whose energies are all quite low compared to kT.
And they'll all have significant probabilities.
Now let's decrease the energy.
Go to cold temperature.
So kT becomes really small.
So that e to the minus energy over kT starts to, if this gets to be bigger than kT by a lot, then this whole thing is a very small number.
Suddenly, you get very few states accessible.
So if we just plot this distribution, of course it's easy to do, it's just a decaying exponential.
So Pi, which is a function of Ei.
It just looks like this.
And here are a bunch of states.
And if we have a classical mechanics picture of matter, then there would just be continuous states.
And if we have a quantum mechanical picture of matter, there might be individual states with gaps in between the energies.
Either way, what happens, what this is saying is, if we go out to higher and higher energies, then you have a smaller and smaller probability to be in a state like that.
And if you go to low energies, the probability gets bigger.
And it depends on temperature.
Because it's the ratio between the energy and kT that dictates the size of that term.
So let's say this is moderate temperature.
If we go to low temperature, it might look more like this.
Hardly any states can be occupied because even states of rather moderate energies, suddenly now those energies are much bigger than kT.
In other words, kT is measuring thermal energy.
It's saying there's not enough thermal energy to populate, to knock things into states whose energy is much higher than that.
So you have a very precipitous decay.
If you go to the other extreme, of very high temperature, then this will tend to flatten out.
Eventually it'll decay, but it may take a long time.
Because now kT is enormous.
So the energy has to get to be very big.
Before it's bigger than kT.
And as long as it isn't, then this exponential term is not small.
So there be a whole bunch of states that may be occupied.
In other words, a whole bunch of states that are thermally accessible at equilibrium.
Systems at thermal equilibrium, molecules are getting knocked around with whatever thermal energy is available.
Crashing into each other or into the walls of a vessel.
And there'll be a distribution of molecular energies.
And that distribution is skewed either low or high depending on the temperature.
So that's what that distribution, the Boltzmann distribution, is telling us.
This is often called a Boltzmann factor, because it's telling what the population of some particular state is.
OK. Now, this is just dealing with individual molecule states.
Then we said, OK, what about the whole system?
Well, the same kind of relation holds there, too.
In other words, if these are individual molecule energies, now if I look at the entire system, right well, still, there's no reason that same distribution doesn't hold.
And it does.
So in other words, Pi of Ei, that's the whole system energy, is e to the minus Ei over kT, over the sum over i, Ei over kT.
Now, this i doesn't refer to a single molecule state.
We're talking about a whole system.
It might be a mole of atoms or molecules in the gas phase, or what have you.
It refers to a system state where the energy, the state of every one of those atoms or molecules is specified.
So this is a system microstate.
Every molecular state is specified.
So for a mole of stuff, that means there might be 10 to the 24 or so molecular states that this single subscript is indicating.
But the point is, those states exist.
They have a total energy.
What's the probability that the whole system will be in such a state?
Still going to be proportional to the total system energy.
It turns out that these summations end up taking on an enormous importance in statistical mechanics.
And the reason is, as we'll see shortly, it turns out that every single macroscopic thermodynamic function can be derived by knowing just that.
Just these sums, what are called partition functions.
So of course they take on enormous importance.
So, we call them partition functions because what they're doing is, they're indicating how the molecules are partitioned among the different available levels.
The molecules or a whole system.
So the molecular partition function is labeled little q.
And the system partition function is labeled big Q.
It's called the canonical partition function.
And because they are going to take on such special importance, let's just look at some of their properties for different kinds of systems and in general.
OK, first of all, let's talk about units and values.
They are unitless.
This is an exponential function.
Here are units of energy and energy.
But this is a unitless was number.
It's just some number, right?
Could be 1.
Could be 10.
Could be 50.
Whatever, right?
Could be 10 to the 24.
Its magnitude tells you about more or less how many states are thermally accessible.
Because, look at it.
And then go back to the example that I showed you if lots of these terms are big, are significant because this is really a big number, right?
It's really hot.
So lots of states have energies lower than kT, which means this is not too small, for lots and lots of values of i, then this just keeps adding up.
And of course, same here.
So the number might be very large.
But if we're in the low temperature limit, maybe we're in such low temperature that only the lowest possible state is occupied.
And everything else, it's just too cold.
There's not enough thermal energy to occupy anything but the very lowest state available.
Well, in that case the lowest state, this would be one.
Essentially we could label this zero.
We could put the energy, the zero of energy there.
Everything else is really big compared to kT, which means this exponential gets to be a really small number for every state except one.
And this is just equal to one.
In other words, the magnitude of these numbers tells us about how many states are accessible, thermally accessible, to molecules or to a whole system.
So let's just go through a couple of specific examples to try to make that a little more concrete.
One is, let's start in the simplest case.
Which it'll turn out I just alluded to.
Let's consider a perfect atomic crystal at essentially zero degrees Kelvin.
Zero Kelvin.
So every atom.
Or even if it's a molecular crystal, every molecule, they're all in the ground state.
There's no excess thermal energy.
Every molecule is in its proper place.
Every atom, if it's an atomic lattice.
In other words, it's in the ground state.
That's it.
So again, we can place the zero where we like.
We'll place it there.
That one, for that one state, this will be equal to one.
And for everything else it'll be zero.
So Q is just the sum of e to the minus Ei over kT.
It's equal to e to the minus zero over kT plus e to the minus E1 over kT plus e to the minus E2 over kT.
But remember, T is really tiny.
It's almost zero degrees Kelvin.
So all these things are much bigger than that.
So this is vanishingly small.
This is vanishingly small.
The whole thing is approximately equal to one.
That's it.
Also, if we say OK, now what's the probability of the system being in a particular state.
Well, we have an expression for that.
So let's look at P0.
It's e to the minus E0 over kT over the sum.
Which we've just seen.
So it's e to the minus E0 over to kT divided by e to the minus E0 over kT plus e to the minus E1 over kT, and so on.
This is the only term that's significant.
So it's approximately equal to one.
Now, while I've got this written here, let's just make sure we've got something clear.
What if we hadn't arbitrarily set the zero of energy equal to zero?
I mean, it's arbitrary, right?
We can put the zero of energy anywhere.
And you might think, well, gee that's going to have a big effect on everything.
Well, it would have an effect on the actual number that we get for Q.
But what you'll see is that any measurable quantity that we calculate won't be affected.
It's only the zero of the energy scale that's going to be affected.
So for example, let's look at this probability.
It doesn't matter where we put the zero of energy.
This term is still going to be enormously bigger then the next one.
And the next one.
And the next one.
So in this sum only this term is going to matter.
It's going to cancel with this term.
So whether or not these individual terms are equal to one, which happens if we set this to zero, or whether they're equal to some other number.
Still, the probability that the system is in the lowest state is one, right?
That doesn't depend on where we arbitrarily put the zero of the energy scale.
The state is still going to be in the ground state, at essentially zero degrees Kelvin.
And it'll turn out to be that way with any property that we can measure.
Only the zero of this scale moves if we arbitrarily move it somewhere.
But the observable, measurable quantities that we calculate, they won't change.
Other than that scale.
So that's one simple example.
Now let's look at another example.
Let's consider a mole of atoms, roaming around in the gas phase at room temperature.
OK, so now what I want to do is just have a simple model for their translational motion.
And of course, we could do that either quantum mechanically or classically solve for that.
We're going to use an even simpler model.
And this model is going to be very, very useful for a lot of the things that we'll treat.
It's called a lattice model.
All it means is, we're going to divide up the available volume.
This room, for example, into zillions of tiny little elements.
Little volume elements.
Each one about the size of an atom.
The idea being, we're going to specify the state of the atom by saying where is it.
Is it in this lattice sight, in this one, in this one, in this one.
So it's a lattice model.
Might be an atom there.
Might be one there.
So we're going to divide the volume up.
So let's call our atomic volume little v. Our total volume big V. And an atomic volume, it's going to be on the order of 1 angstroms cubed.
Or 10 to the minus 30 meters cubed.
And our room, our volume macroscopic one, might be on the order of one meter cubed.
Ordinary sort of size.
So now let's figure out our molecular partition function.
Now, implicit in this, we're basically saying that the energy, the translational energy, is basically zero.
In other words, all these states have the same energy.
They're just located in different positions.
At any given instant of time.
And what that means is all these terms, all these Boltzmann factors, we just set them equal to one.
We'll set the zero of energy there, be done with it.
It's a simple model.
But it's going to give us the right order of magnitude that we're after.
So, how many states are there that are accessible?
Well, on the order of 10 to the 30th, right?
So q, little q, we'll call it little q translational, it's just discussing where things are.
Is on the order of 10 to the 30th.
And if we do a more careful treatment, if we treat the translational energy of atoms, either classically or quantum mechanically and solve it.
We'll still get, we still do get, about the same order of magnitude.
Now let's treat the whole system.
So, what happens?
What are our total possible states?
Because we have to add this up for every possible state.
Well, let's start with the first atom.
It has to be somewhere.
It has 10 to the 30th possibilities for where it can be.
Let's start with the second atom.
And put it somewhere.
Well, it has 10 to the 30th minus one, which is still pretty close to 10 to the 30th.
Let's let's go to the third atom.
And the fourth.
If we have about a mole of atoms, let's say 10 to the 24th atoms, that's still going to occupy a very small fraction of the site.
Only one in a million.
So we don't have to keep careful track.
Every one of them, we can just say, look, there are 10 to the 30th available sites.
Because we're not going to worry about a change in one in a millionth.
So what that means is, capital Q, trans for the system, is 10 to the 30th times 10 to the 30th.
In other words, let's now take the whole state.
Well, I could put atom one here.
I have 10 to the 30th possibilities.
Atom two, I can put anywhere else.
So the joint probability of that particular state, for just the two atoms, is 10th to the 30th times 10 to the 30th.
It's 10 to the 30th squared.
So this is going to keep going.
It's going to be 10 to the 30th to the Nth power.
Where N is the number of atoms, which is 10 of the 24th.
Huge number, right?
It is a huge number.
And that, too, if we treat the whole thing classically or quantum mechanically and work it all out.
We'll get that number.
Or something on that order.
Because you know there really is a simply astronomical number of states accessible to the whole bunch of atoms or molecules in this room.
It really is that big.
So in other words, capital Q is just an astronomical number.
And it is the case.
OK.
There's an important sort of nuance that we need to introduce.
And it's the following.
Turns out, it is an astronomical number.
But a tiny bit less astronomical than what I've treated so far would indicate.
So let's look a little more carefully.
What I've said is that Q translational is little q translational, that is, that 10 to the 30th number.
To the Nth power.
But there's one failing here.
Which is, when I got that, when I decided that if I have just the first two atoms I've got 10th to the 30th times 10 to the 30th possible states, the trouble with that is then if I keep counting all the possible atoms starting with each one, I'll double-count it, right?
In other words, what if I interchange those two atoms?
Well, those states are identical, right?
Indistinguishable.
And it's only distinguishable states that count.
When you specify these things, these are indicating distinct states.
In some way, at least in principle, measurably different.
If the atom's are identical, in other words, if it's a mole of the same stuff, I don't have a way of distinguishing between those two.
I have to correct for that.
And when I come to the third atom, I have to correct for all the possible interchanges.
Of course, it's three factorial.
And in general, it's N factorial.
So this result needs to be modified.
This is true for distinguishable particles.
In other words, if I had all different atoms, so I could label them all, then I don't have that correction.
Because then there's really a difference between that state and the state with the two atoms interchanged.
But if it's a mole of identical atoms, that's no longer the case.
So then, Q translational is little q translational to the N power divided by N factorial.
It's still going to be an absolutely enormous number.
But it's going to be a little less absolutely enormous than it was a minute ago.
Now finally, I just want to introduce a handy approximation to N factorial that's going to turn out to be very useful again and again.
And that's called Stirling's approximation.
For the log of a big number, ln N factorial is N log N minus N. If we take e to those, both sides, then we find that N factorial is equal to e to the minus N N to the nth power.
So this, then, is approximately equal to q translational to the Nth power.
Over N to the N times e to the minus N. Now let's put the numbers back in.
There's our 10 to the 30th to the 10 to the 24th power.
And now we're going to diminish that just a bit.
It's N to the Nth power.
So it's 10 to the 24th to the 10 to the 24th power.
Times e to the minus 10 to the 24th power.
So, we can cancel something here.
So we have 10 to the 6th to the 10 to the 24th power times e to the 10 to the 24th power.
e is about 10 to the 0.4th power.
So this whole thing is about 10 to the 6.4th power to the 10 to the 24th power It's still a pretty respectable number.
You could still say astronomical.
But maybe astronomical, but not quite to the edge of the universe.
Whereas the one before maybe hit the edge of the universe and then beyond.
So that's our second example.
And again, part of what I'm trying to do here is just introduce some ideas.
Things like this way of modeling positions and so forth.
But also, again, orders of magnitude.
How big are these numbers.
For different sorts of systems.
So let's do one more example.
Now, let's consider a polymer in a liquid.
And it has different configurations.
And they might be a little bit different in energy.
For example, some configurations might bring neighboring regions of the polymer into proximity where they could hydrogen bond.
And the point is that then the molecular energies involved will change a little bit.
Because of some sorts of interactions that are possible in some configurations.
And not in other configurations.
So what I'm trying to do is introduce a very simple framework through which we might be able to look at things like protein folding, or DNA hydrogen bonding.
Things like this.
And just in a simple way model how those work and what the forces are that drive them.
So we'll think about polymer configurations.
So let's look at a few configurations.
Here is going to be one.
I'm going to label this one here.
And then here are a few others.
So I've labeled them this way so that you can see how they are distinct from each other.
This one would have a possible interaction.
So we'll label its energy.
So this is molecular state i, here's going to be our energy, Ei.
And let's call this one negative e int, for an interaction energy that's favorable.
And these will be zero.
Let's just put that there.
Zero, zero, zero.
And let's also indicate the degeneracy.
How many states, different states are there with the same energy.
And that's called gi.
And here it's one.
And here are these three.
So that's the framework of our model.
Well, so what's our molecular partition function for this configurational degree of freedom?
Well, we can label it little q configurational.
So we're going to sum over these states.
e to the minus Ei for the different configurations over kT.
So it's e to the e int over kT.
Plus three times e to the zero over kT, we'll get all those other three terms.
So that's it.
It's e to the e int over kT plus three.
Remember, the interaction energy is negative e int.
It's a favorable interaction.
e int is a positive number.
So that's our result.
Now we've described the probabilities in terms of the states that can be occupied.
That is, we've added it up.
But of course, even the way I wrote it just out of convenience, I didn't actually write out each term in the sum.
In the notes I actually did that.
But of course it's not necessary to write e to the zero over kT plus e to the zero over kT and write that three times for each of these three.
Rather, it's convenient to group them together.
And the point I'm illustrating here is that in our expression, for the partition function, we've written that in terms of the individual states.
But we doing a sum over the individual molecular states.
But we could also sum over energy levels.
Including the degeneracy.
So we could say, let's not do the sum over every state.
After all, what if there are a hundred states that had the same energy.
Rather than just three.
Gets to be kind of painful, right?
Instead let's just sum over energy levels.
They're all going to be the same factor anyway.
And then we'll have, we'll just explicitly write the degeneracy in there.
So, we can write the partition function as a sum over energy level.
So, in other words q is the sum over i, e to the minus Ei over kT.
That's individual molecular states i.
But we also could write it as the sum over i, where this now is molecular energy levels i.
And then we need to incorporate the degeneracy gi e to the minus Ei over kT.
Same thing, of course.
But again, sometimes much more convenient to write things this way.
And of course, we can write the probabilities of the occupancies in the same way, too.
That is, we could talk about the Pi in terms of individual states.
e to the minus Ei over kT over q, the whole sum.
Or Pi summing over energy levels, gi e to the minus Ei over kT, divided by q.
Here, of course, this is bigger if it's degenerate, right?
It's saying that the probability of being in this energy level is three times the probability of it being in any individual one of the states.
But sometimes it's useful to keep track of things in this way.
What it shows you too is that, remember when we looked at, did I erase it?
I guess it's gone.
When we looked at this probability distribution for the occupancies of the levels, of course, what it says is at any temperature, the lowest level is the most probable.
For intermediate temperatures.
For low temperatures, for high temperatures.
At high temperature, it might be only a little more probable than the next one over, and the next one, and so forth.
But the very lowest state always has the highest probability.
But the lowest energy doesn't always have the highest probability because of degeneracies.
There might be many states with an energy up here.
And the probability of any one of them is only a little lower than the probability of the lowest energy.
That could be the case here.
Let's say, the interaction energy isn't enormously strong.
So there is some energetic favoring of this state.
Maybe there are 10% at room temperature.
Maybe it turns out there are 10% more molecules like this than in any one of these states.
But of course, these three altogether mean that there are many more molecules at this energy than at the lowest energy.
Now of course we could do the same thing for the canonical partition function.
Not just the molecular one.
So in other words, capital Q sum over i system microstates.
e to the minus capital Ei over kT.
But we could also write it as a sum over energies.
Sum over system energies.
Ei.
And then we have to include the degeneracy.
Capital Omega i e to the minus Ei over kT.
Degeneracy of the system energy Ei.
Little g here is the degeneracy of molecular energy Ei.
Now, same form.
Just an important difference, though.
This number, this little gi, is typically a small number.
It could be one, it could be a few.
This number is usually astronomical.
That's basically, that is, what we calculated here.
In other words, how many total system states are there with a particular energy.
Well, in many, many, many cases the answer is some astronomical number.
So this number might often be between one and ten.
This number might be 10 to the 24th.
It might be 10 to the 24th to a large power.
And, of course, that has a big effect on the way things end up working.
In statistical mechanics and in thermodynamics.
A lot of thermodynamics results are the way they are because you have so many possible states with a particular energy.
That that energy can be strongly favored just by virtue of the number of states that there are.
Just the way, in a very small way, this energy might be favored just because there are more states with it then there are states here.
But again, with the system, it's not a factor of three to one.
It might be a factor of 10 to the 24th to the power of something.
It might be just enormously larger than other possibilities that'll tend to put the energy at a certain place.
And just to continue, of course, the same thing goes for the system probabilities.
Pi, right, which is e to the minus Ei over kT divided by capital Q.
If this is a sum over states, or omega i e to the minus Ei over kT over Q, let's write this over, if now we're calculating the probability of an energy level.
And it's important to make the distinction.
Because in many cases, that's what we care about.
What's the energy of the system?
And we often don't care about exactly where's that molecule and that one, and that one, and that one, right?
The individual states that might be involved that would comprise that energy.
Now, what I want to do is start deriving thermodynamics.
Like I promised, we're going to be able to derive every thermodynamic quantity if we just know the partition function.
So now I just want to show that that might really be true.
So, the point is that from Q we're going to get all of our thermodynamics.
And let's start with the energy.
Remember u, right?
That's our system energy.
It's an average energy.
It's an average of the energy that we would get by looking at the states that the system might be occupying.
So we can write it as a sum over i.
Of Pi times Ei.
In other words, it's going to be determined by the energy of any system state times the probability that the system is in that state.
Add them all up.
Well, we know what that is.
It's the sum i of Ei, e to the minus Ei over kT divided by Q.
Now, I'm going to just make a simple substitution.
I'm going to use the term beta to mean one over kT.
I'm really just using that so I don't need to use the chain rule a billion times and doing derivatives.
So I'm going to write this is sum over i Ei e to the minus beta Ei over Q.
So Q, then, in this term is just a sum over i e to the minus beta Ei.
So.
Now I do want to take some derivatives.
If I take dQ / d beta, keeping V and N constant, then I get derivative with respect to beta.
Sum over i e to the minus beta Ei.
So that's going to bring out Ei, right?
So it's going to bring out minus Ei minus the sum over i, Ei e to the minus beta Ei.
That obviously looks like a handy thing.
Because that looks like that term.
Then, our average energy, remember, it's one over Q.
Sum of i, Ei e to the minus beta Ei.
So now, it's just minus one over q, dQ / d beta.
So that's just the same as minus d log Q / d beta.
And now I'm going to use the chain rule.
So it's minus d log Q / dT times dT / d beta.
All at constant V and N. Now I can easily get dT / d beta.
Because d beta / dT is just minus one over kT squared.
It's just the derivative of one over kT with respect to T. So finally, my average E, which is u, is just kT squared times d log Q / dT constant V, N. That's terrific.
In other words, if I have an expression for Q, I know the partition function, and I can calculate it at any temperature.
I just need to take log of it, take its derivative with respect to temperature.
Multiply it by k T squared and I've got my energy.
Not very complicated, right?
So in other words, macroscopic thermodynamic properties come straight out of our microscopic model of statistical mechanics.
Statistical thermodynamics.
Now I'm just going to state the next result, just because I want to get there and it'll be followed up more next time.
But it's the following.
You can see, of course, our Q is a function of V and N and T. It's a function of V because in principle the energies that are going into all this can be a function of volume.
What thermodynamic function is naturally a function of N, V, T?
Who remembers?
Gibbs free energy?
Helmholtz free energy?
Enthalpy?
Which one?
Nobody knows.
What's a function of N, V, and T. Or, V and T, is what we really formulate it as.
N was introduced later.
The Helmholtz free energy.
Thank you.
What that suggests is that actually the simplest and most natural connection between Q and macroscopic thermodynamics is to the Helmholtz free energy.
And the result that you'll see derived next time is A is just minus kT log of Q.
What a simple result.
And you'll see the derivations in your notes.
You can see it's a couple of lines, right?
But of course, if you know A, and you know E, you know everything, right?
Because S is in there.
And then other combinations can give S and G and mu.
And p. And anything else you want.
So that's the point, is that all of macroscopic thermodynamics follows.
And you'll see that elaborated more next time.
And in addition, a very simple natural expression for the entropy in terms of the states available follows, that we've alluded to before.
And now you'll see it played out.
Right.
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So you've been learning about statistical mechanics.
The microscopic underpinnings of thermodynamics.
And last time we ended up working with the canonical partition function and showing that once you have the canonical position function, you have basically every thermodynamic quantity that you've learned how to calculate this far in the course.
But before I start, on the lecture notes 25, there were a couple typos in there that actually didn't make any difference to the final result because they cancel each other out.
But it's been corrected on the web version.
And near the bottom of the page -- because we're going to be mostly using these notes today.
Near the bottom of the page, where it says A is equal to u minus TS is equal to u.
That should be a plus here.
dA/dT, volume.
And then further down the next line, blah blah blah blah blah.
Minus u over T squared.
And this should be a minus here.
One over T. dA/dT, V. And et cetera.
So luckily these two typos cancel each other out the result is correct.
But there it is.
OK.
So last time, then, you saw how from the canonical partition function, you could get something like the energy.
You wrote down an equation.
The energy is equal to k, T squared, d log Q / dT.
Under constant volume.
And number of particles.
And then you notice that the important variables are the volume, the number of particles, and the temperature.
And we know that every thermodynamic quantity has a set of natural variables.
For the Gibbs free energy it was the pressure and the temperature, the number of particles.
And for the volume of the number of particles and the temperature, we know that that's the Helmholtz energy.
So the natural variable that we would associate -- the natural thermodynamic variable we would associate with that set of constraints is the Helmholtz free energy.
So it becomes interesting, then, to figure out, how can we write the Helmholtz free energy in terms of the canonical partition function?
They seem to have the same set of natural variables.
And that's what you started doing and what we'll do today.
And that's where the typo comes in.
So let's write what we know about the Helmholtz free energy in terms of the energy u. I already wrote it up there.
u minus TS.
That's the definition of the Helmholtz free energy.
And from the fact that dA is equal to minus p dV, minus S dT, plus mu dN.
We can read out what S, here, is in terms of the Helmholtz free energy.
S is just dA/dT, constant V and N. So we can plug that in here.
u minus T dA, with a minus sign there, so that becomes a plus.
And hence the typo.
dT, constant V, and N. OK.
So now we have an equation that relates u and the partition function.
We want an equation that relates A and the partition function.
If we rearrange this slightly, we can get that u, then, is equal to A minus T, dA/dT, constant V and N. So the question that we could ask ourselves is -- is there a function of A that kind of looks like that?
And I know the answer, so I'm going to give it to you.
A function of A that looks like the sum of A minus something times the derivative, with respect to that something.
We should try to look at something like the derivative d(A/T)/dT, constant V and N. So if you take that derivative, you end up with one over T, dA/dT, minus A over T squared.
But if you look at this result, here, and you multiply by T squared, there's the A, and there's the T times dA/dT.
There's the A that -- we just have the sign wrong.
So we have -- multiply this by minus T squared.
Minus T squared, times minus T squared.
And you have the same thing as here.
So that tells us then, that u can be written as minus T squared, d(A/T)/dT, constant volume.
It's a nice way to relate those two energies.
And we have an expression for u in terms of the canonical function.
And we can then replace it in here.
And that gets us, then, that minus T squared, d(A/T)/dT, constant number and volume is equal to u.
And u, we saw, was equal to k T squared d log Q / dT.
V and N fixed.
And then we can start -- The t squareds disappear here.
And then we have d/dT on this side and d/dT.
Let's just take the integral.
You take the integral of both sides, and that gets us that then A over T is equal to k, log Q, plus the constant of integration.
And we can take that constant of integration to be whatever we want.
Energy is all relative to some reference point.
We can take it to be zero.
A is equal to k T log Q.
That's a pretty neat result.
There's the microscopic underpinning of things.
Where we know about atoms, and energies, and states, and even quantum mechanics, and all sorts of things goes into here.
All the microscopic information goes in here.
And there's a thermodynamic variable that only cares about the macroscopic state of matter.
It doesn't care that there are atoms there.
It just cares that you know the pressure, the volume, the temperature, or any couple variables.
And you can get a direct equality without any derivatives or anything.
Between the macroscopic and the microscopic.
So this is really pretty remarkable.
And once you have A, you have everything.
Just like before, we said once you have G, that Gibbs free energy, when we were talking about things that depend on pressure and temperature.
You have everything.
It's the same thing here.
We've got A, we've got u.
We have everything.
We can calculate every single thermodynamic variable from then on.
For instance, if we want to have the entropy, S is equal to minus A over T, minus u over T. Where did I get that?
I got that from way up here.
A is equal to u minus TS.
I solved for S in terms of A and u. I've got expressions in terms of the canonical function for A and u. Plug that in there.
Get k log Q, plus k T, d log Q / dT, constant number and volume.
et cetera.
You can get the pressure.
You can get the pressure from the fact that the pressure up here is a derivative of A with respect to volume.
So you take the derivative of A with respect to volume here.
You get the pressure.
If you want the chemical potential.
The chemical potential from the fundamental equation up here is the derivative of A with respect to the number of particles.
You take the derivative of A with respect to the number of particles, you get the chemical potential in terms of the canonical partition function.
So you've got everything.
You've got u.
You've got A.
You've got p. You've got S. You've got H. You've got G. Name your valuable, you've got it.
You want the heat capacity?
It can get you the heat capacity in terms of the partition function.
Any questions?
Alright so let's go on.
Let's look a little bit closer at the entropy, in terms of the microscopic theory here.
So let's start with S is A minus u over T. And let's see how far we can go.
Alright.
Let's write it in terms of -- let's write A and u in terms of things that we know.
And let me just go to the end to tell you where I'm going, and why I'm going to make certain changes in my math here.
So what we're going to get at the end is that S is this very nice quantity, which is minus k, pi log pi.
Where the p's are the microstate probabilities.
The probability that your state is in a particular -- yes?
[UNINTELLIGIBLE]
S is u minus A.
Yes.
u minus A.
Thank you.
I should read my notes.
OK.
So we're going to equate S here in terms of the probabilities of microstates.
And that's going to be -- remember how we talked about S is related to concepts of randomness, or order, or disorder.
So the number of possible microstates is related to the number of amount of disorder that you might have.
If you have a pure crystal, and every atom is in its place, then the number of microstates at zero degree Kelvin is one.
So the probability of being in that microstate is one.
And the probability of being in every other microstate is zero.
Alright, so there's a relationship that we're going to have here, which is going to be interesting.
Let's derive it.
That means that we're going to want to have this somehow pop out of the equation.
We're going to want to have this pop out of the equation.
If you remember, pi is the -- pi is e to the minus the Ei over kT, divided by the partition function Q.
So somehow we are going to have to get this to come out.
We're going to have to have these e to the minus Ei over kT's come out.
OK.
So let's try to get them to come out right away.
We know u.
A way to write u is the average energy.
Which means let's take one over kT out here.
And there's a -- let's divided by k here.
Let's get the k here.
kT here.
We're going to want to have a k come out here.
There's the k here.
So that's one way of getting it to come out.
And then the u is going to be the average energy.
e to the minus Ei, minus kT.
That's just writing the average energy.
So the energy times the probability of having that energy divided by the normalization.
Plus, well, A is just log Q.
A over kT is just log Q.
So we've managed to extract this guy out here.
OK. Now, this sum, here, this is sum over i. I'd really like to have this sum come all the way out.
So I've got to find a way to do that.
And it would be nice if I could find a sum here.
Maybe if I multiply by one.
If I write one in a funny way.
Get a log here.
You know, if I've got a couple logs here, maybe I can combine them to get a ratio.
So let's rewrite this Ei in a funny way, here.
Ei is -- I'm just rewriting it, but in a strange way.
Log e to the minus Ei over kT.
So if I take the log of e to the minus the Ei over kT.
I get minus Ei over kT.
The kT's disappear.
So I just get Ei is equal to Ei.
I'm just writing something that's pretty obvious here.
And then we're going to take that expression, and put it in here.
So now I'm going to be able to write S over k is minus -- So the kT here cancels out this kT here.
Sum over i. e to the minus Ei over kT.
Over Q.
That's that term right here.
And then I have the Ei, which is log e to the minus Ei over kT.
This whole thing is in this parenthesis here.
And then I have the plus log Q.
Plus log Q. OK.
So I have this nice thing here.
e to the minus Ei over kT divided by Q.
Well that's looking an awful lot like this pi here.
Which is what I'm trying to get out.
I'm trying to get these pi's coming out.
So that's a nice thing to have here.
Now if I could only have a pi coming out here, somehow, that would be great too.
And if I also got a sum here, that's a sum over i I could sort of combine everything together.
So I'm going to write one in a funny way.
One is equal to the sum of all probabilities.
That's obvious.
And I'm going to write this pi here in a form that looks like this.
Sum over all i, e to the minus Ei over kT, divided by Q.
Just writing one.
The sum of all probability is equal to one.
And I'm going to take this one, here, and I'm going to put it right in here.
Now log Q doesn't care on i, so it's just a number.
So that allows me to rewrite -- S over k is minus the sum over i. e to the minus Ei over kT, divided by the partition function.
Times the log of e to the minus Ei over kT.
Plus sum over i, e to the minus Ei over kT over Q.
Times log Q. OK good.
I'm going to take these summations.
Now everything is over the sum of Ei.
This is great.
And there's this factor here.
E to the minus Ei over kT divided by the partition function.
That appears in both.
I can factor that out.
and.
Then I have these two logs.
Log of this and log of that.
That I can also combine together.
This is the log of this.
And then there's a minus sign.
So if I take the -- it's going to be the log of this minus the log of that,.
It's going to end up with a ratio.
S over k is equal to minus Ei.
Taking the summation out.
e to the minus Ei over kT, over Q.
That's this term.
That term.
And then I have the logs.
Log of e to the minus Ei over kT.
This is a minus sign.
This is a plus sign.
That means I divide here by Q.
This is great.
Look.
This e to the minus Ei over kT over Q. e to the minus Ei over kT over Q.
What is that?
That's just pi.
This is pi.
That's the probability of microstate i.
And this is equal to minus sum over i, pi log pi.
There's the k, here.
S is equal to minus k, log sum over i, pi, log pi.
Another great result.
Now if you system is isolated.
If you have an isolated system, that means that the energy -- You've got your boundary.
The boundary doesn't let energy go in and out.
Doesn't let the number of particles go in and out.
Every single microstate is going to have the same energy.
If this system is isolated.
The only thing you're going to change is the positions of the particles.
Or their vibrational energy.
Or something.
But let's just stick with translation.
You're just going to change the positions.
You're not going to change the energy.
So, if the system is isolated, then the degeneracy of your energy is just a number of ways that you can flip the positions around.
Indistinguishable ways.
So the probability is just one over the number of possible ways of switching positions around for your particles.
So for an isolated system, all microstates have the same energy.
We can set that equal to zero as our reference point.
And the probability of being in any one microstate is just one over the number of possible ways of rearranging things.
So the probability of been in any one microstate is one over the number of my microstates.
They all have the same energy.
Where this is the degeneracy.
So now when you plug that in here, S is minus k. Sum over all microstates from one to blah, one over blah, log of blah.
This is a number.
It can come out.
The sum of one to omega.
Of one over omega.
Is omega times omega.
It's one.
The one over omega -- the log of one over omega is minus the log of omega.
The negative signs cancel out.
k log capital omega.
For an isolated system.
You've seen this before, probably.
This is called the Boltzmann equation.
And that is what is on his tombstone.
If you go to - I think it's in Germany somewhere.
Is it in Germany?
Do you guys know?
Austria
Austria, thank you.
I knew it was that part of the world.
If you go to Austria, to some famous cemetery, and go look for the tombstone that says S is equal to log omega.
That's where Boltzmann is buried.
OK.
So this picture of -- yes question?
Can you repeat the argument for making that one over omega go away?
Making the one over omega go away here?
Because you're taking the sum of all states from one, i equals one to omega.
So it's one over our omega plus one over omega plus one over omega plus -- omega times.
And this is just a number.
So it comes out.
It's not in the sum.
So the sum, here, is all by itself.
Any other questions?
OK. That's why we talk about entropy as being this fundamental property that tells you about the number of available states.
That's what it is.
You've got this connection now between this variable, which is sort of hard to really intuitively understand, when you're talking about thermodynamics.
And this is much easier to understand here, in terms of the available ways of distributing your energy, or your particles, in this case here.
In different bins.
OK any questions?
So the next topic is we're going to work a little bit with the partition functions.
And see how when you have systems that have multiple degrees of freedom, where each degree of freedom has a different kind of energy.
Let's say translation, rotation, vibration.
Then you can have a partition function for each of these degrees of freedom.
And whereas the energies of the degrees of freedom add up, the partition functions get multiplied.
So it's the separation of the partition functions into subsystem partition functions.
So so far, we've written for translation partition function.
That the system partition function is the molecular partition function to the Nth power.
If you have distinguishable particles.
And you have to divide by N factorial.
If you can swap particles without knowing the difference.
This is the number of ways of swapping N particles with each other, so it's indistinguishable particles.
OK.
So now let's say that -- Let's just make sure that -- Let's say that your energy of your system -- yes?
It says that when system's not isolated, [UNINTELLIGIBLE]
Where does it say that?
In the notes?
[UNINTELLIGIBLE]
Let me see the notes here.
What does it say?
So that sentence has to do with this guy here.
Basically it says that if you've got a huge number particles, the average energy is a given number.
And the fluctuations around that average are very small.
And so, the system behaves as if it's isolated.
So when you have a system which is not isolated, then energy can come in and out of the system.
So in principle, over time, you could have huge energy fluctuations.
As energy comes out, or energy comes in.
And if you have a countable number of molecules in your system, then if one molecule suddenly captures a lot of energy, then the whole system energy will go up a lot.
But if you have ten to the 24th molecules, if one molecule suddenly gains a lot of energy, the system energy doesn't care.
So small fluctuations -- or big fluctuations in the small number of molecules doesn't make any difference to the total energy.
And so you can still use this then.
It's good enough
So how long is that good enough [UNINTELLIGIBLE]
Well if it's accountable, a handful of things, and it's not valid.
If it's ten to the 24th and it's valid, and somewhere in between it breaks down.
Then I don't know what the answer is.
But usually if you have a thermodynamic system, then it's big enough.
That's what thermodynamics is about.
Where you don't really care that you have atoms there.
You don't even know you have atoms there.
So it's big enough.
Good question.
Alright, so now let's take our microstate energy here.
And our microstate energy is the sum of all the molecular energies Ei.
So it's the sum of all energies E sub, over all the atoms.
And each one of these energies, if it's a molecular energies, can be indexed by a quantum number of some sort.
So it would be the sum over all energies.
So quantum number for particle one, n1 is some sort of quantum number.
n2 is some sort of quantum number.
n3 is some sort of quantum number.
And then you have all the molecules.
En1 plus En2 plus En3, et cetera.
So this is the energy from molecule one, energy from molecule two, energy from molecule three, energy from molecule four.
And that little n tells you which energy state that molecule is.
And the sum of all these energies is your microstate energy.
As long as you can write this this way, then you're allowed to write this this way.
So that basically means that they're not interacting with each other.
They're independent from each other.
In this case here.
So now if I write Q in terms of the sum over all microstates Ei.
e to the minus Ei over kT.
I'm going to replace this Ei here with the sum over all these energies here.
And so the sum over all microstates, then, becomes the sum over all possible combinations of quantum numbers.
n1, n2, n3, n4, et cetera.
All the possible ways of getting molecule one in some state.
All the possible ways of getting molecule two in some quantum number state.
And then e to the minus -- and instead of capital Ei, I'm going to write the molecular energies.
En1, plus En2, plus epsilon n3 plus et cetera.
And then divide by kT.
Basically I'm going to prove that this is a fine statement to make, as long as you can write the energy as a sum of component energies.
OK so now this term here, e to the minus En1, only cares about this sum here.
En2, that's molecule number two, only cares about this sum here.
Molecule number three only cares about the sum over all possible quantum numbers connected to molecule number three.
So I can factor out all these sums into a factor of sums.
Is equal to the sum over quantum number n1.
e to the minus epsilon of n1, divided by kT.
Times n2 e to the minus epsilon n2 over kT, times n3 e to the minus epsilon n3 divided by kT, et cetera.
And now each one of these is basically the molecular partition function.
These are all the possible energies of that molecule.
And the sum over all possible energies times e to the minus E over kT is the partition function for the molecule.
So we have q for molecule one times q for molecule two.
And they're all the same.
There are N of them.
Plus q to the N. So just, in a way, clarified that the reason why we're able to write this system partition function, in terms of the molecular partition functions, with N of them, to the Nth power, is because we were able to separate out the energy here.
In terms of the independent molecular energies.
Where this is saying the molecules don't interact with each other.
And are independent from each other.
And then the one over N factorial comes in, so that you don't over count, for translation, the positions.
They're indistinguishable.
OK.
So now we can have -- Actually we're not -- This basic concept of the partition function multiplying each other, if the energies add, is not limited to going from the molecular partition functions to the system partition function.
You can also look at the molecular partition function itself.
And if the energy, the molecular energy, can be written in terms of a sum of energies of different degrees of freedom, for instance, the energy of a molecule could be the energy of the vibration, plus the energy of the translation, plus the energy of the rotation, plus the energy of the magnetic field, plus the energy of the electric field.
et cetera.
You have many energies that can add up with each other to create the molecular energy.
And what we're going to be able to write, then, is that this molecular partition function itself can be written in terms of a product of partition functions for the sub parts of the molecular energy.
So let me clarify that statement.
So if I can write my molecular energy, epsilon, is equal to a translational energy, plus a vibrational energy, plus a rotational energy, plus every other little energies that you can think of that are independent of each other.
Then using the same argument we used to show that Q is the multiplication of these molecular partition function, we can write that the molecular partition function, little q, is just the multiplication of the degree of freedom partition functions -- molecular partition functions.
The translational partition function times the vibrational partition function, times the rotational partition function, et cetera.
If the energies add, then the partition functions multiply each other.
And that's going to be powerful because when we look at something like a polymer or DNA or protein or something, in solution.
And we're going to be looking at the configurations possible for that polymer or that biopolymer, then we'll know that the energy of that polymer in solution is going to be -- we'll be able to approximate it as the energy of the configuration for that polymer.
The different ways that you can fold the protein, for instance.
Plus everything else.
The energy of everything else.
OK.
So if the configurational energy can be separated from the sum of all vibration energies of all the bonds in that polymer.
The way that polymer interacts with -- The way that the solution itself interacts with itself.
Then if we can do this and we can do this approximation most of the time, then we'll be able to take the partition function for the polymer, and write it as the configurational partition function times the partition function for everything else.
And we'll find that this part here will tend to factor out.
We won't have to worry about it.
And that this will carry all the important information that we'll need to know to see about changes in the system.
Changes in Gibbs free energy, changes in the chemical potential.
Everything will be related to this partition function.
This subsystem.
And because of the fact that you can factor them out, then this thing will end up dropping out.
And this will become the important factor.
OK. Let's do a quick example.
OK.
So this is the example of having a very very, short polymer.
Containing three monomers.
Which can be in two configurations.
And the energies are the same for these two configurations.
So the configurational partition function, which you would generally write as the sum of e to the minus Ei for that configuration, divided by kT, plus e to the -- So we would usually write it as e to the minus epsilon one over kT, plus e to the minus epsilon two over kT, plus et cetera.
They're all the same energy.
And there are two configurations.
The degeneracy is two.
So you can write this as the degeneracy of the configuration, times the energy of the configuration.
And you can set your energy reference to be zero.
You can choose whatever you want it to be.
And zero is a good number.
So that e to the zero is equal to one.
So that configuration partition function is just the degeneracy, which is equal to two, in this case here.
So now let's calculate the molecular and canonical partition functions for an ideal gas of these molecules here.
And it's usually interesting to use a lattice model as a guide.
And so in this lattice model here, you would divide space up into little cells.
Pieces in two dimensions, but in reality it would be three dimensions.
And then you place your molecules in lattice sites.
Something like this.
And then you end up counting the number of ways of arranging the molecules on the lattice.
And let's say that we have N molecules that are in the gas phase.
And the molecular volume, i.e.
the size of the lattice site is v. That's the molecular volume.
And N times v is the total volume.
That's the total volume occupied by the particles.
The total volume is the number of lattice sites.
V is the total volume.
And we're going to assume that all particles, all molecules have the same translation energy.
It's an adequate approximation.
We're going to set that equal to zero.
So all molecules at any position here has the same E translation.
And we're going to set that equal to zero.
So the translational partition function.
The molecular transitional partition function is -- Well there's only one energy.
It's zero.
So we only care about the degeneracy of that molecule.
That molecule could be in here, or it could be here, it could be here, it could be here, it could be anywhere, right?
The number of ways of putting that molecule on the lattice is the number of lattice sites available.
Which is basically the molecular volume.
Which is the total volume divided by the molecular volume.
The total volume is -- right.
So the total volume is the number of lattice sites times the volume of each lattice site.
So the total volume divided by the small volume is the total number of lattice sites.
And the number of choices of putting that one molecule is anywhere on the lattice.
That's your degeneracy.
So now if I look at the total molecular partition function, it's going to be the multiplication of the configurational partition function and the translational partition function.
At each site, the molecule could have two configurations.
So q, for the molecule, that's q translational.
I'm going to ignore all vibrations, rotation, et cetera.
I'm going to assume that there are two degrees of freedom.
Translational one, which is basically the positional one.
And then the configurational one, which is internal to the molecule.
So this one is V over v. That's the degeneracy.
Capital V over little v. The degeneracy of placing the molecule on the lattice.
And the configuration is the degeneracy of how the molecule folds.
Yes
[UNINTELLIGIBLE]
The notes are wrong.
So usually capital V is large and little v is small.
So in the notes, if we have it in reverse, we should fix that.
This is lecture 25, right?
So we have q.
No the notes seem to be right.
Total volume is capital V, molecular volume is little v. Where is it that wrong in the notes?
[UNINTELLIGIBLE]
Oh look at that.
My notes are different than yours.
My notes are right.
OK well it's obviously right on the web.
Because this is the latest version.
Alright so flip those big V's and little v's then.
Huh.
I thought they were from the same pile.
OK.
So this is your molecular partition function.
And then when you look at the system, the system partition function can also be separated into a translation and the configuration for the system.
We know what you need to do is take all the molecular partition functions, the transitional ones, and -- to the N factor.
The number of particles.
But now you have degeneracy.
You've over counted.
So you need to divide this N factorial.
You also have the system partition function for the configurations.
And that's q configuration to the N. Except here we don't need to divide by one over N factorial, because we're not over counting here.
The over counting only happens when you're placing identical particles in a lattice, and you can swap them without making a difference.
Here we talking about configurations.
When we're talking about configurations, we're not talking about placing the identical particles in different spots.
We're just looking at these two configurations here.
And then the next particle is two configurations.
The next particle is two configurations.
So this is really important.
That this N over factorial only comes into play when you're talking about the translational degree of freedom.
Not the other degrees of freedom.
And now the total system partition function is the multiplication of these two.
It ends up being capital V over v, to the N power, over N factorial, times two to the N. OK.
In general you could extend this analysis to include vibrations, rotations, energy in a magnetical field, electric field, et cetera.
Any questions?
Alright the next thing that you're going to do then is to use this concept, as a sort of example, as a way to begin to calculate things that you've already calculated before.
For instance, if you look at an expansion of an ideal gas, can we now calculate the entropy change.
Not based on thermodynamics, but based on the statistical mechanics.
On the microscopic description that we've just gone through.
And it turns out that that's what happens.
That you can do that.
You get the same answer.
Thank god you get the same answer.
Otherwise we'd be in big trouble.
So I'm just going to set up the problem, because I won't have time to do it.
And then you can do it next time, when Keith comes back.
So the problem is going to be the usual problem of having a volume V1 of a gas on one side, and a vaccuum expanding to volume V2, gas.
And asking what is the entropy change.
And you know that from thermo, that delta S in this case here is nR log V2 over V1, when the temperature is constant.
That's going to be our answer.
It has to be our answer.
But this time, instead of knowing the answer, we're going to calculate it from microscopics.
So what you do is you start out with your initial state.
You ask what is -- you write down the molecular volume V. The total volume here is V1.
You assume that all molecules have the same translational energy.
You set that equal to zero.
The system translational energy is equal to zero.
And so the entropy for this gas here is just the number of ways of placing the molecule in the lattice.
This model that we have of space being separated into little cells.
And so S is k log omega, where omega is the number of ways of placing the molecules in the lattice.
Which is basically k log V, over v. Where this is the number of lattice sites.
OK. And what you're going to do next time, then, is start from here.
Calculate what it is before.
Calculate what it is after, and turn the crank, and get to the right answer.
Then you're going to do the same thing for liquids.
And that'll be it for this simple statistical mechanics.
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And last time, you got to see how you can derive macroscopic thermodynamic results from the microscopic point of view of statistical mechanics.
So in contrast to the way we'd gone through the course from the beginning, starting from empirical macroscopic observation and ideas through macroscopic laws and then working out the consequences of them, in statistical mechanics you start from a microscopic model, build up the energy levels that are going to give you the terms in the partition function, determine the partition function, and what you saw last time is that given that, you can calculate everything.
You can calculate all the macroscopic properties that ordinarily come from the thermodynamic laws that were based on empirical macroscopic observation.
So today, I just want to continue doing some statistical thermodynamics.
Basically going through a few examples, just to see how it plays out when we calculate thermodynamic quantities, based on our microscopic picture.
So let's just look at a few cases.
And I'll start with a couple of examples that are entirely entropy driven.
And we've seen them before from the macroscopic point of view.
What we should hope, of course, is that we can derive those results from the partition functions that are appropriate here.
So, we'll look at entropically driven processes.
And the first one, the simplest one, maybe, to cover, is just free expansion of a gas.
So what happens if we've got particles in a gas that are enclosed in a certain volume.
And now we let the volume expand into vacuum on the other side.
Of course, you know what'll happen.
The volume will be filled and we've derived the thermodynamics for it before.
So let's see what happens in a statistical mechanical treatment.
So let's say, here's V1, there's our gas.
There's vacuum.
And now we'll open it up and just have the bigger volume, V2, and let the gas fill it.
So I think last time you got introduced to basically a lattice model for translational motion.
And I started in on this just a little bit the time before also.
So this is a simple way to describe the statistical mechanics of filling an open volume.
Filling a space with molecules.
And all it does is say that we're going to divide up our available volume into little bits that are basically the size of an atom or a molecule, or whatever the particle is in the gas phase.
And say, OK, there are maybe 10 of the 30th of those little volumes.
The little volume elements are going to be on the order of an angstrom cubed if it's an atom, a little bit bigger if it's a molecule.
And the available volume is on the order of meters cubed.
So that works out, an angstrom is 10 to the minus 10 meters.
And then the cube of that is 10 to the minus 30th.
So if you look at the total number of little volume elements of this sort, it's on the order of 10 to the 30th.
So, the total volume, capital V, let's make that distinction clear.
And in this case, all the states of the system have equal energies.
So in other words, it doesn't matter we're dividing our volume up into imagined little volume elements.
And for both the molecular and the system states, the energy is the same.
It doesn't matter whether the molecules are here, here, here and so forth.
They're not interacting in this picture at all.
So it doesn't matter how close or far they are from each other.
They can't be in the same volume.
So all the energies are the same.
And what that means is, in the partition function, which is a sum over all these terms with these Boltzmann factors.
That have e to the minus energy over kT.
But energy's all the same in every one of the terms.
So in order to determine the partition function, to determine Q, all we have to do is count up the total number of states that are available.
So, our energy, our molecular translational energy, we'll just set it to zero.
Same with our system translational energy.
We'll set that to zero.
All the microscopic available states, that is, if I take an individual particle and I say where it can be, all those states have the same energy.
If I take a microstate of the system, you know, the whole collection of the particles, all of those states also have the same energy.
So what that means is that little q is just equal to capital V over little v, right?
How many states are there?
Well, I've got my little volume, and then however many individual cells there are, that's the number of available states in this model.
So it's big V over little v, on the order of 10 to the 30th.
And then, big Q, the canonical partition function for the whole system, it's something that we've been through before.
It's this process of saying, well, you take the first atom or molecule, and it has any one of these possible states.
So it's on the order of 10 to the 30th possibilities.
Then where does the second one go?
Well, there are basically 10 to the 30th more possibilities.
And the third one has 10 to the 30th.
Since there are so many fewer atoms or molecules then there are volume elements, when we're dealing with the gas phase, we don't have to worry about the fact that, well the first million of the atoms filled some of the sites.
So the next ones have fewer sites available to them.
It's true, but it's such a small fraction that are ever filled that we don't need to worry about it.
So Q is just little q to the capital N power, the number of particles.
And then we've seen you have to divide by N factorial to avoid the overcounting of configurations that are in fact not distinguishable.
The whole idea that maybe there's an atom here and another one here.
That configuration is indistinguishable from the configuration with those two atoms reversed if we're dealing with a mole of identical atoms.
So, OK.
There's our capital Q.
And that's also our system degeneracy.
So the degeneracy, the little g for the molecular degeneracy, how many molecular states are there of a certain energy.
Well, it's the same thing as q.
And it's the same thing here for the system states.
The capital Omega is just equal to that.
Well, now let's look at what happens when we do this.
When we expand from volume V1 to volume V2.
So that capital V is going to change.
So capital V1 goes to capital V2.
And we know it's entropically driven.
Let's calculate the change in entropy.
So delta S is just k log capital Omega 2 minus k log capital Omega 1.
Now, in the process of just seeing how the development goes, how you can derive all these thermodynamic quantities from the partition function, one of the really most central results concerns the entropy.
How you can describe the entropy in terms of the probabilities that the different states are occupied.
And how, in the case like this, where the states all have the same probability of being occupied, then you have this very simplified form for the entropy.
Just, if S is k log capital Omega, where capital Omega is the degeneracy.
The number of system states of that energy.
An amazing result.
You told them about Boltzmann's tombstone and so forth.
So it's on there.
In fact, the ill acceptance of that result kind of led to the tombstone being erected when it did.
Boltzmann, depressed over the lack of acceptance of his theories, put himself to an early death.
And this was off a big part of the reason.
But we, generations later, have come to accept the results that concerned him so deeply.
So this is our change in entropy, when we just have this expansion of gas from V1 to V2.
So it's just k log omega 2, omega 1.
So now let's just put in our results for the volumes.
It's k log V2 over v, I'm exaggerating the size there.
To the N, over N factorial, over V1 over v to the N divided by N factorial.
So this is going to turn out to be a relatively simple case, because the factorials are going to cancel.
So, then we just have N k, log V2 over V1.
Terrific, right?
Now, remember, k, the Boltzmann constant, is just the ideal gas constant per molecule rather than per mole.
So this is the same thing as little N, the number of moles, times R, times log V2 over V1.
And that should be a familiar result.
That's the change in entropy in expansion, free expansion of a gas, from V1 to V2.
So now we've been able to derive that just based on this really simple molecular picture.
Based on that plus the result that Boltzmann fought and paid so dearly for, so that we would have it and understand it.
As before, of course note it's greater than zero.
If the volume, if we're expanding into a bigger volume than before.
The entropy goes up.
Also notice one of the other results that you've seen is that you could relate the partition function to A, right?
The Helmholtz free energy is minus k T log Q.
And S is negative dA/dT.
And so this would immediately give us the same result, because omega's going to go in here.
So we could get this result from another pathway, too, but a more simple and direct way is just to start directly from the Boltzmann result for entropy.
Any questions?
Alright, let's go one step more complex.
Let's now look at the entropy of mixing.
So now we've got two species.
Rather than one expanding into a free volume, and we're going to open a barrier between them and see them mix.
And again, it's still entropically driven.
And we should be able to calculate the entropy change that we saw before from a macroscopic perspective.
So, let's take a look.
So now, we have NA molecules of A in some volume VA. And NB molecules of B, in some volume VB.
And then we're going to open it up.
Then we've got capital N, NA plus NB.
And capital V. VA plus VB, right?
And the whole thing is happening at constant pressure.
Well, then let's look at the initial and final expressions for the entropy.
So, S1, at the beginning.
It's just k log omega A plus k log omega B.
And that's k log VA over little v to the NA power.
Divided by NA factorial times the same thing for B. VB over little v to the NB power over NB factorial.
So that's our initial entropy.
The sum of the two entropy contributions, from the two sides.
And S2, the final one, is just k log omega for the whole shebang.
And that's k log.
And now we have capital V over little v to the N power.
And now we have a combinatorics result that I think probably you're all familiar with from one context or another.
That is, now we have to divide by NA factorial times NB factorial.
In other words, the amount that we need to divide by to avoid overcounting is the product of those two factorials.
Just in case it's been a while since you've seen stuff like that, I've got a couple of pages further on the notes at the bottom of the page.
I've just broken that out for a really simple example where there are just two molecules of A, and two molecules of B represented by different little balls in lattice sites.
And I just worked it out for the total where there are only ten total in the mixture.
The point is, now what happens is, when you start filling a lattice, but it's not all the same, right?
So now you have A here.
And A here.
And B here.
And B here.
So, of course if you interchange A and B here, you don't have an identical state.
An indistinguishable state.
That's distinguishable and needs to be counted.
It's only interchanging B with B that you need to correct for, and interchanging A with A.
So of course you need to account for that in a way that's different from how it turns out if every one of the molecules is identical.
So it's the product of these factorials.
In the numerator.
So now delta S is then just this minus this.
So it's k log V over little v to the NA plus NB over VA divided by little v to the NA power VB divided by little v to the NB power.
And, let's see, maybe I'd better back up.
Let me know, that's all.
And what's happening is, again, luckily, these factorials are canceling.
So we just have k log V to the NA, V to the NB over VA to the NA VB to the NB.
And now we're going to go back and use the fact that the pressures are the same, all the way through.
And what that means is that the volumes must be in the same ratio as the number of molecules.
And what that means is that, for example, VA over V is the same as NA over N. And that's just equal to XA, the mole fraction of A.
And the same for B.
So we can substitute that.
And know our delta S just becomes k log, let's see, let me break this out from the beginning.
And just take minus k log on XA to the NA power minus k log XB to the NB power.
All I've done make these substitutions here.
For the ratios of the volume.
So it's just minus N k XA log XA plus XB log XB.
Look familiar?
Same thing we saw macroscopically before.
So, again, we can still use our microscopic model.
And continue to derive the macroscopic entropy changes.
And, of course, for many of these we can still get our delta G and so forth.
Of course, it's only entropy that's going to contribute.
OK, let's look at just one more entropy driven problem.
And that is, it's a little bit different from these.
Let's look at the entropy mixing in a liquid.
And the difference between the gas and the liquid is that in the case of the liquid, every single cell is filled.
It's not like with the gas.
And that makes a difference in how the states need to be counted.
Because here, remember how I said when we were filling this lattice.
Well, the first molecules goes somewhere.
And there are 10 to the 30th possibilities.
And the second molecule goes somewhere and there are 10 to the 30th possibilities.
And there are always 10 to the 30th possibilities because there's so few molecules that you never have to worry about the fact that it's getting filled up, so there would become fewer possibilities for that last molecule.
And the reason again is because there are so few molecules that essentially, all the cells are open and available even to the last molecule.
Because maybe there's a mole of molecules.
Maybe there are 10 to the 24 molecules.
And there are 10 to the 30th lattice sites.
So there might be one in a million lattice sites occupied, even at the very end of the procedure.
But of course for the liquid, that's definitely not the case.
All the available volume is filled.
So by the time you get to that last molecule, it has one and only one space it can go into.
So you have to count for the diminishment of the available sites as molecules are placed in them.
OK.
So it's a different kind of combinatorics problem that results.
So, OK. Let's look at liquid mixture.
So now, it's going to look like, and they're all filled up.
I sure used fewer sites maybe, but that'll be easier here.
We've got open circles.
So these two are going to mix.
And now we're going to have filled mixture.
So we're going to have a bigger volume, but it's still going to be filled and all the positions are random.
So we have to consider all the possible configurations that we have.
OK.
So we still should go through the same part of the procedure.
So let's call this A.
And this, B.
So at first, already the situation is different.
If we say what's the entropy of system A.
Well, in this simple model the entropy of the pure liquid is zero.
Now, of course, that's not realistic.
But it's going to turn out to be suitable.
Because the change in entropy that we go from here to the mixture.
The other contributions to entropy are going to more or less be the same from beginning to end.
What's going to be significantly different is simply the fact that in the mixture you have the random positions that can be occupied.
So that's going to be the term that matters, the contribution that matters.
So entropy is just k log Omega A.
There's only one system state.
All of the lattice sites are filled, and they're all filled with A.
So there's only one way to do it.
And of course, it doesn't matter if you interchange them.
They're indistinguishable.
So this is zero.
This is one.
There's only one state.
Same, of course, for B.
So that's our initial entropy in this simple model.
Well then, delta S of the mixture is just S of the mixture.
We don't need to worry about the original parts.
Now, this is going to be k log of N factorial over NA factorial NB factorial, like we've seen before.
That's not different from before.
Except that unlike before, we don't have the convenient and simple cancellation of the factorials.
That happened before, because they were there.
In other words, they were there in the entropy contribution in the gas phase as well, in the pure gas.
They're only there here in the liquid.
And that means we have to deal with them.
But it turns out it's not too bad to deal with them.
And we're going to use this Stirling's approximation that I introduced before.
So the log of N factorial is approximately equal to N log N minus N. So let's just break that out then, and use the Stirling's approximation for each of the factorial terms.
So then we have S for the mixture is N k log N minus N k minus.
And now we'll do the bottom, the numerator.
NA k log NA minus NA times k, that's this part.
And now we'll do this one.
So it's plus NB k log NB minus NB times k. But NA and NB, of course, are just N. So these will cancel.
So then we're left with NA plus NB times k log N minus NA -- And I just want to write this this way so that I can then separate the terms.
Minus NA k log NA minus NB k log NB.
And now I just want to combine the easily combined terms.
So it's NA k log N over NA plus NB k log N over NB.
Looks like we're home.
Now we just have N k XA log XA plus XB log XB right?
This NA over N is just XA.
NB over is XB.
And then I've divided the same thing to make XA's over here.
And take the total number of moles out.
Alright.
Look familiar?
Again, the same thing derived now in a way that's a bit different from what we did in the gas phase.
So now I've got the entropy of mixing even in a condensed phase the liquid.
And again, the reason this simple model works is because although a real liquid, certainly a pure liquid has a finite entropy, a substantial amount of disorder, that's present, that kind of disorder.
In other words, the fact that you don't really have the molecules sitting in sites on a lattice but there's disorder.
There's also rotational disorder if it's a molecule.
There are various contributions to entropy.
But those are all comparable in the starting and final states.
The thing that's changed is simply the interchanging of A and B molecules.
That's introduced a new kind of disorder that didn't used to be present.
A very big difference in the number of states at the end from the beginning.
And that's what we've calculated and captured here.
And that's why such a simple model still works OK. Any questions about these entropy driven cases?
OK. Now let's move on and talk about the cases which are more commonly encountered, where the states aren't all the same energy.
Of course, here in the liquid, just like in the gas, we haven't treated interactions between the molecules.
We've assumed that they're equal between A and B, as they are between A and A and B and B.
So in this case then all the energy to the states are the same and it's just a counting problem.
How many states are there available.
As soon as the energies are different, then of course we need to account for all those Boltzmann factors.
Those e to the minus energy over kT factors become part of the, they weight the counting.
And we have to do that.
So, what I want to do is go back to this simple polymer configurational model that we introduced before.
And this sort of model, in one form or another, we're going to see a few times in the rest of this treatment.
The reason, really, is because it's a simple and also realistic way of portraying a system that has a finite number of well-defined energies.
If you don't have that, of course you could do the treatment.
In a, sort of, classical mechanic sense, a continuum of states, all with different energies.
It's hard.
Because then those sum in the partition function become integrals over all the different configurations and possibilities.
It's much more complicated to do.
It's much more straightforward if you can identify discrete states and add over them.
Now, someday you'll take quantum mechanics.
And then you'll see that the states are discrete, even if we're talking about translation or rotation and so forth.
So you'll have discrete quantum mechanical energy levels and so forth.
You haven't had that yet.
And so that's not the starting point that we're going to use.
Instead, we're going to use a starting point that goes through the exact same formalism, which is just the discrete states available in molecules that have multiple configurations.
So we're going to have unequal energy states.
And what we're envisioning is molecular or polymer configurations.
So here we have our states.
And just like before, the idea is that if you have two sub-units that are in proximity, there's some kind of favorable interaction.
It could be hydrogen bonding.
Could be different.
But the point is, there's some sort of interaction there that reduces the energy.
And then there are other configurations that just don't have that.
Because the sub-units aren't in proximity to each other.
And it's not hard to work out that in this very simple model, these are the only configurations that are available.
The only distinct configurations.
So then, our molecular energy, E, we can define as zero here.
Before we defined it as minus E int, but I'm going to, it's a little more convenient to make this the zero of energy, and then this is plus the interaction for each of these states.
Of course, we can put the zero of energy wherever we prefer.
And then we have the degeneracy is one.
And in this case it's three.
So now we can just write out the configurational partition function for the molecules and also the canonical partition function for the system.
So q configurational.
And we're just going to sum over the states.
So it's e to the zero over kT, for the lowest state.
And then there are three states that are going to have this term, e to the minus E interaction over kT, where E int is a positive number.
In other words, the probability of any one of these states is a little bit lower than this state.
Because this state has lower energy.
Remember what I mentioned earlier, which is although the probability of any one of these states is lower than the probability of this state, the probability of this energy is likely to be higher than the probability of this energy.
Because there's only one of these states and the degeneracy here is three.
So there are three possibilities in which the molecule could have this energy, and only one this.
So if, basically, kT is bigger than E int, so in other words, if this term isn't very much smaller than one, then of course this will be bigger than this.
So of course this is one plus three e to the minus E int over kT.
And now we have our capital Q, our canonical configurational partition function.
And that's just little q configurational to the Nth power.
And like you've seen so far, in various cases where the molecules are separate non-interacting molecules, this is just the molecular partition function taken capital N times.
If all the molecules are behaving independently, then you sum over all of those states.
And you see that each one of these is just taken again and again.
As we've seen implicitly in all these treatments so far as well.
Now, in the translational case where you interchange particles.
Like particles, you have to divide by N factorial.
.
But the different configurations don't do that.
If I've got a system where a molecule over here is in this configuration and a molecule somewhere else is in this one, and now they change configurations, that's a distinguishable state.
So there's no N factorial involved here.
In the configurational partition function.
So, fine, then it's just one plus three e to the minus E int over kT to the Nth power.
That's Q.
And once we know Q, as you've seen, we know everything.
And so we can immediately start in deriving all of the thermodynamics, right?
And the place to start is A, Helmholtz free energy, or configurational free energy.
Because that's the simplest relation, just as minus kT log of capital Q configurational.
So it's minus N kT one plus three e to the minus E int over kT.
I'm going to rewrite A configurational in terms of beta rather than kT, just because there are going to be a lot of these factors.
So it's minus N kT one plus three e to the minus beta E interaction.
And that's our A.
Everything's going to follow from that.
Before I write some of the other results, one thing to notice is, it has N in it.
In other words, there's a free energy per molecule.
The total free energy is just something times N. And that's because all the molecules in this model are independently behaving.
So whatever their average free energy is, that's going to add up and give us the total.
They're all acting independently.
And in fact, we could have gotten this directly from minus kT log little q configurational.
And if we look at energy, regular energy, u, configurational, it's minus d log capital Q with respect to beta.
V and N. And I'll just write the result.
It's N times three E int e to the minus beta E int over one plus three e to the minus beta E int.
And once again, the feature I want to point out is that there's a factor of N here.
Again, there's an energy per molecule.
And so if we want, we can also just write the average E, little E, configurational.
It's just u configurational over N. It's the same as this without the factor of N. Not only that, again, we could get this directly from the molecular partition function up there.
From little q configurational.
We'd have the same result.
Exactly as it should be.
So if we wrote E as we've seen in the past, it's just the sum over that energy times the probability for each state.
And that's just the sum of Ei, e to the minus beta Ei over little q.
And if you put in the terms, you immediately get this.
Here's our little q, and this has the interaction energy brought out, multiplied here.
So the point is, in a case like this, where you have a bunch of independently behaving particles, the totals for these, for quantities like energy, are simply for the system energy, it's just the average particle energy times the number of particles.
The number of molecules.
And remember, before, I spoke a little bit about the fact that, well, if you look at one individual molecule and another and another, the energy will fluctuate.
It'll vary considerably.
Of course, this is a simple case with only two possible energies.
But in a case where there may be many more possible energies, the molecular energies may vary quite widely.
Still, there will be a well-defined average.
And then the system energy would simply be the total number of molecules times that.
Where they're all independent.
Where all the energies are independent of each other.
And, of course, the system energy fluctuates a great deal less than the individual molecule energies.
OK, so that's our energy term.
And I think I won't write out the results.
They're in your notes for entropy.
For chemical potential.
They're all there.
And again, the same point would hold.
They all scale with N. But I want to talk a little bit about the heat capacity.
The expression for it isn't so simple.
So let me just right Cv configurational.
It's du configurational / dT, at constant V and N. And again, the details are worked out in the notes.
So I won't write out the whole derivation of it.
And even writing out the results, I'm almost reluctant to do it.
But I'll put it up.
Three E int over k T squared, because there are some things I want to point out about it.
N, and then one plus three e to the minus beta E int minus E int e to the minus beta E int.
Minus e to the minus beta, I have to get rid of this.
Minus e to the minus beta E int times negative three E int e to the minus beta E int.
All over one plus three e to the minus beta E int, that whole thing squared.
And this is just, of course, the kind of messy results that comes from taking this derivative with respect to temperature and doing the chain rule to get it with respect to beta and so forth.
So it looks like a little bit of an intractable, or at least a little bit of a complicated, function.
And the detailed functional form is complicated.
But what I want to emphasize is that it has simple limits that are very easy to understand physically.
And that are important to understand for lots of systems.
So I just want to look at the limits of the heat capacity at low and high temperature.
And this is something that recurs in statistical mechanics, in an enormous number of systems where you have simplified limits.
And they're really important.
Because what's going to matter is this.
Maybe I shouldn't have covered up those configurations.
So there's a big difference in what happens when kT is much bigger than this energy.
Where you know that under those conditions - let's say it's hot, and this is a small energy difference.
Then the probabilities are essentially equal.
That the molecules are in this state or in any of these states.
Because the energy is so tiny compared to the thermal energy.
The molecules are continually being kicked around between all the states, among all those states.
So the high temperature limit, physically, is one that's simple to understand.
Where now you're really just going to revert to the cases that we've treated before, where all the energies are effectively the same.
Because compared to kT, they are.
And in the low temperature limit, now go to the opposite limit.
Let's say kT is much, much, smaller than the interaction energy.
So that now this term is really small.
Because this is much bigger than this.
So, what it means physically is all the molecules are in the ground state.
The probability of this is basically one.
The probability of being in any of these states is zero.
And that's also a simple result.
And there are lots and lots of cases where one of those limits really obtains.
If you look at molecules moving around in the gas phase or in a liquid.
And you say, well, OK, let's think about their rotational motion.
Rotational energies are small.
At room temperature, kT's much bigger than them.
You immediately go to the high temperature limit.
Now let's go to a small molecule and look at the vibrational energy.
In most cases, the vibrational frequency is pretty, molecules are pretty stiff.
And in many cases you can just say, look, forget it.
All the molecules are in the ground state.
And again, in the opposite, but also simple limit ends up holding.
Or molecular electronic states, right?
If you've got benzene or hydrogen atoms in, say, at room temperature, how many hydrogen atoms thermally are going to be up in the n equals two state, the 2p state or the 2s state?
Forget it.
There's not nearly enough thermal energy to do that.
So these simpler limiting cases play a huge role in simplifying statistical mechanics and the calculations from them generally.
OK, so let's just see what happens.
So, this we want, so this we can move down.
So our limiting cases, when T goes to zero, not going to work out what happens to all the betas, beta gets big in that case.
But the result is that Cv configurational goes to zero.
That's the limit where, like I've described, here is E interaction, or E int.
Here is zero.
And your kT is barely above zero.
So when all the molecules are here, none of them has enough thermal energy to be up here.
So why should the heat capacity be zero?
The heat capacity is du/dT.
It's zero because, here's kT.
Let's say I increment kT up a little bit.
I just heat the system a tiny, tiny bit.
An infinitesimal amount, to look at the derivative.
Once I do this, how many molecules are in this higher level now?
Still zero, right?
There's still not nearly enough thermal energy to have any molecules up here.
So what, was the derivative of the energy with respect to the temperature?
I changed the temperature.
The energy didn't change a bit.
And that means the heat capacity is zero.
Now let's look at the other limit.
High temperature limit.
Really, it doesn't need to be infinity.
It's really just T greater than, much bigger than, kT is much bigger than E interaction.
And in this case, kT is much less than the interaction energy.
So it doesn't need to be nearly that extreme.
Well, what you find out is the heat capacity is zero.
Now, it's zero.
Because we're in the following limit.
Now kT is way up here.
Compared to kT, E interaction is way down here.
So, what happens?
It means that this is so hot, that term, The e to the minus E int over kT, forget it, it's one.
And so the probability, in other words, that the molecules are in this state or this state are essentially equal.
So now let's say I raise the temperature a little bit higher.
What happens to those probabilities?
What changes?
Right.
Nothing changes.
So what happened to the energy when I raised the temperature?
Of course, nothing happened to it.
Which means the heat capacity is zero.
Now, the first limit that I described, this one is almost universal.
For any system where you have quantized level, you can always eventually get to a low enough temperature that you're in the first limit.
Where the kT is lower than, by far, than the first excited level.
Everything's in the ground state.
And so you reach this limit.
This one, it's only cases where you have a finite number of levels.
Then you have the same high temperature limit to it.
In other words, the reason this result happens is because there aren't a bunch of other levels up here that eventually get to be comparable to kT.
And not all kinds of systems or degrees of freedom are like that.
This one is, because there's a finite number, four in this case, of configurations.
So there's just nothing above there.
There's another really important kind of degree of freedom like that.
And that's spin.
Think of proton spins.
It's plus 1/2 or it's minus 1/2, and that's it.
You can't put any more spin energy into it.
Just like you can't put any more configurational energy into the system than to be in this state.
That's it.
So in other words, the maximum possible energy is finite.
Of course, lots of other degrees of freedom are different from that.
If you think a molecule's rotating, they could always spin faster.
Vibrating, they can always vibrate harder, translate faster.
Those degrees of freedom won't have this limit.
They also will have a simple high temperature limit, but not zero, because, of course, if there are always more levels, and I keep increasing kT, then I'll have thermal energy to go into those higher and higher levels.
It'll still go up.
But for systems with a finite number of possible levels, and a finite amount of total energy, for degrees of freedom like that, once I get to a temperature higher than any of that stuff, then forget it.
You can't change the energy anymore thermally.
So your heat capacity is zero.
You can change the temperature and nothing further happens.
OK, next time we'll see what happens when you do have continuing basically unbounded possible energies.
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Last few lectures, and we started to see some of the consequences and looked at simple changes, transformations of expansion and mixtures of liquids and gases.
And saw how in the framework of statistical mechanics, we could derive the thermodynamic results that you saw before, based on an empirical framework.
On the thermodynamic framework we've been working with all term.
But you saw them derived from a microscopic perspective.
Now what I want to do is move to examples that are more commonly encountered in chemistry.
One that we did last time actually was very common, or is at least a prototype for something common.
That is we looked at what would be a simple model for a polymer with different configurations available to it.
And just tried to understand what the basic thermodynamics is that it would exhibit.
What the heat capacities would be in various limiting cases.
High and low temperature.
I want to continue that today.
But for an even more common case.
So the model that I want to lay out is indicated on your notes for today.
And essentially, you could look at it as double stranded DNA.
Any kind of polymer with a whole variety of configurations.
Ultimately what's going to matter is what is the set of energy levels available to the system.
And so here's what it looks like.
The way I've tried to depict it.
I'm imagining something like, covalently bonded chains that have links between them.
So of course, it's reminiscent of DNA with hydrogen bonding between the strands.
And the idea in this simple model is that you can have various numbers of these hydrogen bonds.
So this is the case that would be the most energetically stable.
That is, all the available hydrogen bonds are formed between neighboring pairs.
So this would be our lowest energy.
Now let's starts breaking some hydrogen bonds.
And what I'm imagining is they break starting at one end and working their way down.
So the next highest energy state would be like this.
So now these are still hydrogen bonded.
This is not.
And there's a cost in energy for breaking that hydrogen bond.
We'll call it epsilon zero.
It's a positive number.
So that this is higher energy than this.
And now we'll go to the next one, which could have two broken bonds.
Broken hydrogen bonds.
And so on.
And if we imagine that this is very long, there may be many many, members of the chain.
Then of course this sequence of structures with corresponding energies might go on for a very long way.
So this is a simple construction that gives us a set of states with distinct energies.
And so we have a series of configurational energy levels.
They're evenly spaced in this model.
So there's zero, E zero, two E zero.
And so on.
So those are available energy states.
And now, just based on that simple model, we should be able to figure out all the thermodynamics.
We should should be able to figure out what the equilibrium energy is at a particular temperature.
What the average energy is.
And so forth.
And we can do it in a straightforward way.
So we'll start with our molecular partition function, as always.
So it's q configurational.
And it's just a sum over our available states with their associated Boltzmann factors.
So sum over n of e to the minus En over kT.
Now the way I've modeled this, there are no degeneracies.
Each energy level has just one microscopic state of the molecule corresponding to it.
So there are no molecular degeneracies.
Turns out we'll be able to put this into a simple form.
So the first thing is, let's approximate that we could take the sum -- not to some finite level, which would be the case if there's a finite number of elements in the chain.
But all the way through infinity.
And the idea here is the following.
We'll assume that the chain at least has a great many members.
Even if it's finite.
And then at some point, this becomes a really small number.
Even if epsilon naught, the energy of one of the bonds, is less than kT.
Once we multiply it by a very large number n, it becomes very much greater than kT.
So the underlying assumption here is, well there will not be a significant number of molecules in the highest energy state anyway.
In other words, the terms higher than a finite but large value of n between there and infinity are going to be so small that those terms will contribute negligibly to the sum anyway.
And on that basis, we could make this approximation.
And the incentive to make this approximation and carry the sum to infinity is that then we can put this in what turns out to be a very simple closed form.
So now if we just write out a few of the individual terms in the sum, it's one plus e to the minus epsilon zero over kT.
Plus e to the minus epsilon zero over kT.
I'll write this as squared.
e to the minus two epsilon zero over k t. And so on.
So this has the form of a sum that looks like one plus x, plus x squared, and so on.
And because the sum goes to infinity, we can simply write it as one over one minus x.
Sort of a very simple result.
And now putting this back into x, it's one over one minus e to the minus epsilon zero over kT.
That's our q zero.
Our molecular partition function.
So this model yields a particularly simple result for q zero.
And then everything else follows from there.
So then we can write the canonical partition function capital Qn.
It's just q configurational to the capital Nth power, where capital N is the total number of these molecules.
So it's just one over one minus e to the minus epsilon over kT to the Nth power.
And then we can start writing out the results for the various thermodynamic properties.
So A configurational is minus NkT, log q configurational.
That is, remember, A is minus kT log capital Q.
And this is just going to come straight out.
So it's minus NkT log of little q.
Minus NkT, log of one over one minus e to the minus epsilon over kT.
Or positive NkT, log of one minus e to the minus epsilon over kT.
So a pretty simple form for the Helmholtz free energy.
And I'm not going to write out all of the individuals thermodynamic terms, but I'll write a few of them.
So mu, the chemical potential for these configurations, is just dA/dN.
With T and V constant.
And so the only capital N dependence, the only dependence on the number of molecules is multiplicative.
Right here.
So this just gives us kT log of one minus e to the minus epsilon naught over kT.
And the important point to realize here that we mentioned last time too, in the examples we considered then, especially the last one, is that because the only place that N figures in is as a multiplicative factor.
What this is telling us is that we just have a chemical potential, of Helmholtz free energy per molecule.
The molecules are all independent.
The total energy doesn't depend on any interaction between this molecule and its neighbor somewhere else.
So terms like this are simply additive.
So if we work out the chemical potential, it's just one over N times A.
And I'll just write the result for u configurational.
Because I do want to look at the heat capacity.
So it's Nk T squared, d log of little q configurational / dT.
With N and V held constant.
And it turns out to just be N epsilon zero.
One over E zero over kT minus one.
So now we can look at the heat capacity.
Cv configurational.
So it's du configurational / dT.
With constant N and V. And taking this derivative with respect to temperature.
Gives us an expression of the following form.
It's Nk epsilon zero over k T squared.
e to the E zero over kT, over e to the E zero over kT minus one squared.
Just taking the derivative in the usual way.
Now, this looks kind of complicated for the heat capacity.
But let's take a look, just like we did last time for the simpler case that we treated then, let's take a look at the high and low temperature limits of what happens to the heat capacity.
So at high temperature, well we can start by just looking at the energy.
That has a form that's fairly simple.
So when temperature is large, this is small, and we can Taylor expand it.
So then we're just going to have one plus epsilon zero over kT minus one.
Which means the ones will cancel.
And we end up with a fairly simple result.
So u configurational, in that case, is just NkT.
The epsilon zeroes are going to cancel also.
And what that says is that the heat capacity -- and of course we could take the limit of this, but we can just take the derivative of this with respect to temperature more easily.
So Cv configurational.
It's just N times k. In other words, the heat capacity in the high temperature limit is a constant.
Last time we treated a simpler case, where there were only four configurations altogether available, to this sort of simple polymer model that we drew.
Unlike this present model where we're saying there are essentially infinite number of configurations and different energies available.
So many that the highest ones we're never even going to access, because they'll be much higher than kT.
Here it's different, so now we have -- You know before, when we had a limited number of total states, remember what happened in the heat capacity at high temperature.
What was the limiting case?
Finite number of states.
What's the heat capacity at high temperature?
This is going to be on the exam.
What's the heat capacity at high temperature, if there's a finite number of states available?
Zero
It's zero.
What was the low temperature limit of the heat capacity?
Zero
It was also zero.
Right.
And the idea was for a system with a finite number of states.
The idea was -- so let's say, the way we had it before, there was one state with energy zero.
And we had three states with some energy epsilon zero.
And that was it.
Those were all the molecular states available.
So we looked at the two cases.
In one case, when the high temperature limit where kT is much bigger than any of this stuff.
In that limit, the molecules are just equally likely to be in any of these states.
Because there's much, much more thermal energy than this energy difference.
And the molecules are constantly getting kicked around by the available thermal energy.
So if you raise the temperature a little bit more, it doesn't make any difference.
The molecules already are evenly distributed among the states.
There's no additional configurational energy.
So du/dT is zero.
It can't increase any more.
So in that case, the high temperature limiting heat capacity is zero.
And in the other case -- In the low temperature limit, now let's say kT is really low.
It's much less than epsilon zero.
So now we'll redraw this as zero.
And put the epsilon zero states up here.
And kT is here.
Well when it's like this, the temperature is so low that there's not nearly enough thermal energy to populate any of these states.
And if you change the temperature by a little bit, it's still not enough thermal energy to populate any of these states.
So once again, du/dT is zero.
You change the temperature and the configurational energy doesn't change.
So the heat capacity is zero again.
This is why it's so informative to measure heat capacities.
Because you can learn an awful lot about the intrinsic structure of the material.
What are the energy levels available to it?
What do they do?
You can learn a tremendous amount about that by making measurements of the heat capacity.
Well, now we're in a different case.
We have a whole set of evenly spaced energy levels.
Never ends.
So now let's look at the high temperature limit.
But we're never in a limit that's higher than the highest level, because we're assuming it goes on forever.
So it's up here somewhere.
There's kT.
So what happens?
Well if you raise the temperature, there still are higher lying levels that can be populated, and that will get populated.
So du/dT isn't going to be zero in the high temperature limit, in this case.
But it stops changing at some point.
Because nothing's very different about this than having kT be, let's say, up here or up here.
So in the high temperature limit, yes, the energy does change with temperature.
But it changes in the same way at any temperature.
In other words, the energy is just linearly varying with temperature.
And the heat capacity is a constant.
du/dT doesn't change anymore, once you're in the high temperature limit.
Now without me writing any expression on the board, what's the low temperature limit of the heat capacity going to be in this case?
Going to be on the exam
Zero?
Zero, that's still going to be the same, right?
Put kT way down here.
That's just like this, right?
Not near enough energy to populate even though lowest, you know, the first level above the ground state.
Change the temperature a little bit, still not enough thermal energy to get up there.
So the energy doesn't change with temperature in that low temperature limit.
So you can immediately see what's going to happen at low temperature.
Any questions?
Yeah?
[UNINTELLIGIBLE] will still be n k t, and then it's just -- [UNINTELLIGIBLE]
So, of course, that's a limiting case here, right?
That happened because in the limit of high temperature, then this exponent is really small, right?
So then you can Taylor expand it.
In the limit of low temperature, this is really big.
This is nearly zero.
So e epsilon zero is much bigger than kT in that case.
This is big.
This is negligible, right?
Oh.
Wait a minute.
Something's making me real unhappy.
I can't have this right.
Ah.
Yes.
It needs to be zero, is what it needs to be.
What am I thinking?
Of course.
It's one over a huge number.
It's zero.
OK.
So of course you can't expanded it.
It's just, this is much bigger than this.
This is an enormous number at that point.
So in other words, what it's saying is the configurational energy is zero, because everything is stuck in the ground state.
And it stays zero if you vary the temperature.
So you get the zero limiting value for the heat capacity, and the energy itself is also zero.
Other questions?
OK. Let me just make a few comments about the entropy.
So I didn't write it out before.
And I'm tempted not to do it again, but I suppose I'll do it.
So it turns out to be Nk, minus log of one minus e to the minus epsilon zero over kT.
Plus epsilon zero over kT.
Over e to the epsilon zero over kT, minus one.
And this comes from combining the terms for A and u.
It comes from minus A over T. Plus u over T. And what I want to do is look at its limiting cases also.
In particular, what happens here in the limit of high temperature.
And what happens turns out to be k log kT over epsilon zero to the Nth power.
Or put the N over here.
OK. And what this is telling us about is the number of available states.
Roughly, how many states are there that are accessible to the system at some temperature.
So in other words, think of it as -- let's put the N back there.
k log of kT over epsilon zero to the Nth power.
And think of it as k log capital omega.
Where that would be the degeneracy.
Now all the states are in equal energy, but remember for the whole system, remember we discussed this before.
How you have a very, very narrow distribution of system energy states at equilibrium.
So you can think of this as the degeneracy of the system states that are actually going to exist at a particular temperature.
So if we look at the limiting value of the partition function, it's just kT over omega.
Or the same thing for capital Q. kT over epsilon zero, sorry.
To the Nth power.
So you have a very simple expression.
And so again, what this is doing, is it's giving us a measure of about how many states are available.
And it's particularly informative to look at that for the molecular partition function.
What it's telling us is, if I look at kT over epsilon.
And these levels, remember, are evenly spaced.
So here's epsilon zero, two epsilon zero, three epsilon zero, and so on.
It says, you know, if kT is about ten times bigger than epsilon zero.
So this is ten.
It's telling us, roughly how many states does the system have thermal access to.
The molecular state.
It has about ten states.
So going over to our picture of the set of structures, you could have anywhere up to about ten bonds broken.
Now, the individual molecules are going to be in a whole range of states.
Some of them will have fewer than that, and some of them will have more than that number of bonds broken.
But on average, it's going to be about that number.
And then, if you look at the whole system, the number of states available, of course it's astronomical.
Because you have to take each molecule, and say, well it could be in any one of something on the order of ten states.
And then whole set of N other molecules can be in the states they might be in.
Change the first one, and do it again.
So you have this astronomical number of system states.
But remember, like we discussed once before, it'll turn out that although the individual molecule states vary considerably with energy, the system states, which are averaging over some astronomical number of molecules, where capital N is something like ten to the 24th.
Once you average over that many individual molecules, you find that there's very, very little fluctuation in the energy.
In the system energy.
The individual molecule energies vary considerably.
Realistically the variation of the molecular energies -- that variation is going to be comparable to the energy itself.
To the average energy.
So if the average energy is roughly you know, the ten epsilon zero, you say okay, how much might it vary?
Well, there are going to be some molecules that have only a couple of bonds broken, and some that might have 20 bonds broken.
The variation will be on the same order of magnitude as the average itself.
But then you average over capital N of them.
And then you immediately discover that there's very, very, very little variation.
And in particular, what happens then, is, you know, for molecular average energy, epsilon zero, and variance, or standard deviation about the same magnitude.
System average energy is u, and it's going to be capital N times.
Well let's say that -- average -- Oh sorry, let me not put the zero There.
It's just average energy.
And the system average energy is just N times the molecular energy.
But the system variance is going to be on the order of the square root of N times epsilon.
If you've done statistics, then you've seen that sort of result.
You do a bunch of samplings a capital N number of samplings, and the variation looks like the square root of that number.
So what ends up happening then, if you look at the relative variation, you might look and then say, well it's pretty big.
This is ten to the 12, right?
It's still a huge variance.
But let's compare it to the average.
So this is ten to the 24th.
This is ten to the 12th.
It's on the order of ten to the minus 12.
An incredibly tiny fractional variation in the system energy.
You could never -- There would be no practical way to measure it.
And of course that is consistent with what you would expect.
If you say let's measure the configurational energy of a bunch of molecules in a liquid solution, or molecules in a gas floating around.
And it's a mole of them, that total average energy is not going to fluctuate significantly, even though individual molecules that you pick out of that whole system might have quite widely varying energies.
And that's the point.
So it's an incredibly small variation.
Now you can derive this.
I didn't derive this, of course.
I asserted that this is the case.
And it's probably familiar to some of you if you've seen some statistics.
But the way you would derive it is, you know in addition to calculating the average energy, the average of E, you can also calculate the average of the energy squared.
And you can calculate the standard deviation that way.
You calculate the average of E squared.
And then you minus the average of E, the quantity squared.
Take the square root.
That's your root mean square.
And that's what leads to this result.
So it's pretty straight forward to calculate it also.
Alright.
So any questions about just the extent of variation of the individual molecule energies or the system energies?
Okay.
Then, now what I'd like to do is look at another kind of energy that's going to turn out to have exactly the same set of levels that we just derived from this simple model.
And some of you may have seen this before.
Many of you may not have.
And for right now, I'll just assert it.
But it will illustrate why this is so useful.
It turns out that if I've got molecular vibrations -- you know, there's nitrogen and oxygen in the air.
If I look at those nitrogen vibrational energy levels.
Or oxygen energy levels.
They look like this.
Evenly spaced, non degenerate energy levels.
So the model that we've constructed here, based on this simple two-chain polymer, actually gives us a set of energy levels that maps directly onto the vibrational energy levels of a molecule.
So all the results that we've just seen apply, not just for conformations of a polymer, but for vibrations of a molecule.
So everywhere where it says configurational, you can just write in vibrational.
And you'll still be right.
because of course, what matters is what are the states that are available to the system, and what are their energies?
After that, the formalism of statistical mechanics takes over, and calculates partition functions and thermodynamic functions.
The input into that is states.
And their energies.
Well, we have the exact same set of states and energies.
So we immediately arrive at a super important case and its results for molecular vibrations.
And not only molecular vibrations, but vibrations of a crystal lattice.
Acoustic vibrations of a glass.
Or even a liquid.
So there's an enormously wide ranging set of results that we've derived by starting at this simple picture.
And when you take quantum mechanics, you'll see that indeed you do get this set of evenly spaced energy levels.
You may well have seen the results before, and you'll see it derived there.
OK.
Given that, then I want to talk a little bit further about the heat capacity.
So we've seen the limiting cases for the heat capacity.
Namely, the low temperature limit is zero.
That's the case whenever you have quantized levels.
You can always get kT lower by far than the lowest excited state.
So that everything is stuck in the ground state.
How low that is depends on how far apart the states are spaced.
But there's got to be some temperature somewhere that's down there.
So we've seen the heat capacity limiting cases.
Let me just sort of draw them again.
So here's kT much less than epsilon zero.
And up here, kT is much bigger than epsilon zero.
Now I'm just going to sketch the full temperature dependence.
In other words, I'm going to connect those two limits that we've seen.
So here's what that looks like.
And I'll write it as vibrational.
So we know it's got to be zero at the low temperature limit.
And we know that it's this constant value in the high temperature limit.
It's Nk.
And in some way, it smoothly connects.
And if you look at the actual scale, here -- so here's kT over epsilon zero.
Here's the limit of low temperature.
Actually, you don't have to be very high above epsilon zero to be in in this high temperature limit.
So if you look at the plot that's on your notes, this is already -- it's not quite leveling off precisely.
But it's already pretty close.
When kT is just two times epsilon zero.
It's already almost at the limiting case.
Now this has tremendous significance.
So a long time before quantum mechanics was developed, people made measurements of the heat capacities of materials.
And they were familiar with the fact that when you got to ordinary temperature -- room temperature, the heat capacity was temperature independent.
And that was understandable for reasons we'll discuss shortly.
Basically, in that case, the heat capacity -- it's not just Nk for one vibrational mode.
Of course, that's what I describe for a single vibrational mode of, say, a molecule.
If you have a crystal lattice with N atoms in it.
Let's say it's an atomic crystal.
Each atom has three degrees of freedom.
It can move in each of three independent directions.
And there are N of those atoms.
There are 3N total degrees of freedom.
That's how many vibrations the lattice has.
So in fact, you have to multiply this by 3N.
So what would be seen is, you'd have 3R, the gas constant, per mole, in other words.
And if you looked at Cv over 3R, then this value would be one.
Very useful to see that.
To figure that out.
And people understood it.
What people didn't understand is this stuff.
Why did it do that?
That was a great mystery.
Why did the heat capacity go to zero at low temperature?
And the reason they didn't understand it is because the model for vibration was really simple.
The lattice is a bunch of masses and springs.
I know the vibrational energy.
And I can calculate it.
That is, I know the classical vibrational energy.
Remember, the reason it goes to zero is this.
It's all because of the fact that the energy levels are discrete, quantized levels, with gaps in between them.
If I've got classical mechanics, then that means it's vibrational energy.
The energy just gets bigger if the amplitude get bigger.
It's not discrete, it's continuous.
There are always energies in there.
In that case, there's never a situation like this.
Where kT is lower than the first available excited level, and everything's in the ground state.
That never happens in classical mechanics, because there are always levels there.
But it does happen in quantum mechanics.
And Einstein recognized that this was a way to explain this low temperature limiting heat capacity.
So actually if you look at early development of quantum mechanics, really it was all predicated on statistical mechanics.
And the idea that, well, that you could then do the statistical mechanics with quantized levels, just the way we've done it.
And what you immediately discover is, gee, it all makes sense.
You go to low temperature, everything's stunk down here.
And suddenly you've got zero heat capacity, because you changed the temperature a little bit, and everything is still stuck back here.
So that's the vibrational heat capacity of a solid.
That's what it looks like.
And at moderate temperature, this doesn't have to be very much bigger than epsilon zero.
You're already in, essentially, the limit.
The high temperature limit.
And now for molecules, that limit isn't usually reached at room temperature.
Here's a kind of calibration.
kT at room temperature is about equal to 200 wave numbers.
So molecular vibrations, you know, you've taken IR spectra they're typically on the order of 1,000 wave numbers or so.
kT isn't bigger than the vibrational energy.
But a crystal lattice, you know the vibrations are the acoustics vibrations.
And those are much lower in frequency.
If you have an atomic crystal, it just has the sound vibrations at all the different wavelengths that are available.
They're never very high.
So it's easy to get into the high temperature limit, in that case.
Where you basically see a temperature independent heat capacity.
OK. By the way, this was actually used commonly to determine the molecular weights of molecular crystals.
Because you've got a factor of the number of moles in there.
If you ask how big is the heat capacity?
Well it does depend on how many moles of material you have.
Because it depends on how many atoms.
That control how many modes there are in the crystal.
How many vibrational modes.
Because each atom has 3N degrees of freedom.
So if you just weigh the whole crystal, and then you measure the heat capacity, you know how many moles there are, and now you know the weight, you can figure out the molecular weight.
OK.
So that's the low temperature limit.
And now I want to talk a little bit further about the high temperature limit.
And in particular, I want to talk about both the heat capacity and the energy that we've seen.
Namely, this thing.
Of course it's really the same result for the energy and the heat capacity.
They're obviously intimately connected.
So these are the high temperature limits.
Now it turns out that that high temperature limit doesn't just obtain for vibrations.
But it's also the case for molecular rotations, translations.
All these low energy degrees of freedom.
So let's see why.
It has a name, that result.
It's called the equipartition of energy.
Sometimes it's called the classical equipartition of energy theorem.
And what it says is one half kT per degree of freedom equals energy in the high T limit.
And in particular, for translation, so E translational is 3/2.
Well, NkT for N atoms.
Or molecules.
E rotational is, now it depends how many rotational degrees of freedom there are.
If I've got a linear molecule, there are only two.
It can rotate this way, and it can rotate this way.
If I've got a non linear molecule, parts all over the place, it has three unique axes.
It can also spin this way.
And that's a rotational degree of freedom.
Of course the linear molecule wasn't like that.
Nothing is moving when you do that.
So it's either NkT, linear.
Or 3/2 NkT.
Non linear.
Great.
And then E vibrational is NkT per vibrational mode.
That's because vibrational energy is potential and kinetic energy, and it's 1/2 kT each.
Why does all that stuff happen?
Turns out, you can see why pretty easily.
All those degrees of freedom.
Those classical degrees of freedom.
If you say, let's write out an expression for the energy.
Classical.
You know it's 1/2 m v squared.
Or it's 1/2 k x squared.
Or for rotational energy it's 1/2 I omega squared.
You see a functional form emerging here?
That looks similar in all these cases?
You know, 1/2 times some constant, times some variable squared.
Well, now let's look at our expressions for the energy.
Average energy.
We're going to sum over all the energies of epsilon i, e to the minus Ei over kT.
Over sum over i, e to the minus Ei over kT.
That's just how we originally derived the average of energy.
In other words, it's the probability of each state times the energy of that state.
Summed up over all the states.
Well, now we have an expression for the energy.
It's one of these things.
It's something times a variable squared.
So, we'll use some general functional form, a y squared.
And now, let's assume we're in the high temperature limit.
And we've seen what that means.
It means kT is big, compared to the separation between the energies.
When that's the case, there are lots of these terms in the sum.
We can convert them to integrals.
We can forget about the fact that the energies are discrete.
We can say look, they're so close together, compared to kT.
That we can turn the sums into integrals.
So then we have integrals instead of sums.
And here's our energy.
It's a y squared.
Whatever that is.
It could be this, it could be this, it could be this.
And it's not going to matter.
e to the minus a y squared over kT over, integral over e to the minus a y squared over kT.
dy, dy.
OK. And here's what's going to happen.
If you do this integral by parts.
This one.
What ends up happening is it gives you this integral times a certain number.
This is going to come out of the integral.
And so it will turn out.
And it's straightforward to do it.
It's in the notes.
It goes through.
But I'll just write the result here.
The result is that you get, in this high temperature limit where you've gone to the integral form, you get 1/2 kT.
And it will always be the case.
All you need to know is the form of the energy.
As long as that's the case -- Remember y is just a variable of integration here.
That's not going to be preserved.
This comes out because of what happens here.
So what that's telling you is whenever you have an energy of this form, and you're in the high temperature limit, then you get this classical equipartition of energy result.
So, for translation, of course there are three separate degrees of freedom.
For velocity in the x, y, or z direction.
For rotation, if it's a linear molecule, let's say there are two separate degrees of freedom.
You have to keep track of how many degrees of freedom there are.
But that's all you have to do.
That's enormously powerful.
It means that without doing anything, I know the average translational energy of the molecules in this room.
Because of course I'm certainly in the high temperature limit.
With respect to translational energy levels, they're really closely spaced.
Same with rotations.
Now at room temperature vibrations, forget it.
I have essentially no vibrational energy.
I'm at the low temperature limit for the molecular vibrations of nitrogen or oxygen.
Those are high in frequency.
So much higher than 200 wavenumbers.
But for each molecule, I have 3/2 kT of translational energy.
Linear molecules, I have kT of rotational energy.
Without doing any work at all.
And since that applies to most molecules at room temperature, it's an incredibly useful, very, very general result.
OK. Next time we'll do a little bit of chemistry, and look at phase transformations.
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So, the first thing that we did is look at a simple polymer model not too different from polymer models that we've seen before.
That is, models for molecular configurations.
The difference in the case that we treated last time, though, was that instead of having just a handful of possible configurations with their just, designated energies, we had essentially an infinite sequence of levels.
All of evenly spaced energy.
And we went through that and saw what the thermodynamics worked out to be.
And also with the high and low temperature limits of the thermodynamics turned out to be.
And the interesting consequence of that is that that particular choice of configurations and energies associated with them maps exactly onto the vibrational levels of an ordinary molecule.
So, of course, when we do the statistical mechanics in the end, everything depends on the partition function.
Right?
So we're always writing some molecular partition function.
And we've got a sum over our energies for all the states.
And so the only input information we need to know is, what are the levels and what are their energies.
So, although we started yesterday with a model of polymer confirmations, once we decided what the energies were, which was this sequence of energies starting at zero and going with E, and 2E, and 3E, I think we labeled them E0, and so on, that's all that mattered as far as the statistical mechanics was concerned.
And so, since we started, We formulated this by imagining different confirmations for a polymer, but in fact these are the vibrational energies of any molecule too.
Of any vibrational mode.
So the results are very generally applicable.
And so we saw what the thermodynamics worked out to be.
And what the limiting cases were.
And a couple of them, that I'll just review, are, one that's very important in lots of settings, is that the low temperature limit when you're down it basically zero degrees Kelvin, that's where you've got kT way down here.
So there's not nearly enough thermal energy for anything to get out of even the lowest level.
And so in that case, everything is in the ground state.
And then the heat capacity, which is this often a very simple thing to measure, you change the temperature.
And you see that there's no change in the total vibrational energy.
In other words, the limiting low temperature heat capacity is zero.
So we saw that for Cv for vibrations.
And the limit that T goes to zero was zero.
And that's because we were in this limit, where everything is in the ground state.
Which is to say that u vibrational energy in that limit is zero.
Unlike the case we treated earlier, where we had a limited number of levels.
Because we treated a simple four-unit polymer with only four possible confirmations, in this case we have essentially an infinite number.
So even in the high temperature limit, we don't have them, there's not a maximum total energy.
If the temperature keeps going up, you'll get more and more energy.
Because you can keep populating higher and higher levels.
But the way that happens stops changing.
Because the levels are all evenly spaced.
So if the temperature, instead of being down here, is somewhere up here, now for sure if we change the temperature, the energy will continue to change.
But what we've found is that in the high temperature limit, we had the equipartition of energy result.
And that is, u vibrational in the high temperature limit, was just, for N molecules, it was NkT.
1/2 kT for each kinetic energy degree of freedom.
1/2 kT for each potential energy degree of freedom.
And since vibrations have each of those, potential and kinetic energy, it's kT for each molecule, for each degree of freedom.
That's an incredibly simple, useful thing.
That means, if I've got a molecule in the gas phase and it's in the high temperature limit for translation, I know each translational degree of freedom will contribute for each molecule 1/2 kT.
Or 3/2 kT, in all three dimensions.
So I know the translational energy of a mole of molecules in the gas phase at room temperature without doing anything all.
It's 3/2 NkT.
Rotations, if it's a diatomic molecule, will be two different degrees of freedom.
For rotation.
In two orthogonal planes.
And I'll have 1/2 kT for each.
1/2 NkT for N molecules.
For vibrations, if I'm in a high temperature limit, then it'll be kT for each vibrational mode.
For molecules, the vibrational frequencies are usually high, so that you're not in the high temperature limit.
But for materials where you have vibrations, acoustic vibrations, those are low frequency.
You can be in the high temperature limit.
And easily are.
So the result for energy for vibrational energy in the high temperature limit, was NkT.
And so that means that the heat capacity in that same limit at high temperature was just Nk, derivative with respect to temperature.
Very simple thing to measure, again.
So that's what we saw last time in the case of both the conformational model that we treated and the vibrational energies of molecules onto which that same model maps.
Now what I want to do, is look a little further.
And let me actually write the partition functions too.
Also an important result was that the partition function itself, q vibrational, in the high temperature limit, was just kT over E0.
Also very simple result.
E0 is h nu 0 for vibrational mode with frequency nu 0.
And of course, given the partition function you can calculate everything.
So in the high temperature limit we can easily calculate things.
In the low temperature limit, it's just one, right?
There's only one allowed possible state in the low temperature limit.
Everything is in the ground state when it's cold enough.
So the sum of our states just gives us exactly one term that's of any reasonable value.
That is, the term with this being the lowest energy.
Everything else, the energy is much bigger than kT.
This is a vanishingly small number then.
So just write q, vibrational, in the low temperature limit, is just one.
OK, that was the case for vibration.
And remember, these low temperature limiting cases, that's a common case for nearly any degree of freedom.
As long as it's quantized.
At some point, you'll get this situation.
Where the temperature is low compared to even the lowest excited level.
Of course, what'll happen in the high temperature limit might vary, depending on the structure of the energy levels.
In the case of vibrations, it's like this, it'll turn out rotations do the same thing.
But it's not always as simple as this.
OK, now what I want to do is just go through the next step of what we can treat, given the statistical mechanics that we know.
Which is chemical equilibrium.
Why can't we just calculate equilibrium constants based on what we've seen so far.
If we can calculate the partition functions for molecules, and they undergo reactions and they're in equilibrium, we should be able to calculate the equilibrium constants.
From first principles, just based on the statistical math we've seen so far.
So let's try to do that.
So it just means that, again, just as always, if we know all the energy levels and what all the possible states are, there's no reason we shouldn't be able to set our sights on a calculation of that sort.
So let's just sketch out what the levels are going to look like for simple chemical events.
So I just want to draw an energy diagram, here's energy.
And, like we often do with chemical equilibria, I'm going to set the zero of energy at the separated atoms, and I'm imagining I've got reactants and products.
So let's make this our products.
Then over here we'll have our reactants and make the energies different.
Alright.
So there's some amount of binding energy, right?
There's a bond dissociation energy going from the lowest available level in each molecule to the dissociation limit where we pulled the atoms apart.
So now I'm going to draw vibrational energy levels inside the molecule.
Let's imagine, it wouldn't need to be this, but let's imagine it's just diatomic molecules.
So there's one vibrational mode in each.
Just the stretching mode.
And we've already seen the levels are evenly spaced.
So there's going to be a bunch of evenly spaced levels.
Actually, it stopped being quite evenly spaced once this stops being a simple harmonic oscillator, but for all the low lying levels it's pretty close to that.
So those are the available levels.
And there's a dissociation energy.
So minus E, and I'll call this for the products, Ep.
And it's just minus D0 for the products.
Right, in language, terminology that we've seen before.
And it's the same thing for the reactants.
And really, if this were more than a diatomic molecule, maybe there would be a bunch of vibrational modes.
But it wouldn't matter.
This would just represent all the vibrational energies.
There's some lowest state available.
So now we've got the dissociation energy for the reactants.
So those are the energies it takes to separate the molecule.
But those aren't the only states available to molecules.
Of course, they could have extra vibrational energy.
They're not always in their ground states.
And we're going to calculate partition functions.
In general we would sum over all the available states.
And then just calculate the probability of being in the state.
One way to look at this, since if there's chemical equilibrium between the species, it means the molecules can interconvert.
What that really means is that any of the molecule has access to any of these states.
Of course, we like to group all these states over here.
Because they correspond to a particular chemical structure.
And we like to group these states over here, because they correspond to this different chemical structure.
From a statistical mechanics point of view, it's just states and levels.
And we could just calculate the probability of any molecule being in any one of the states.
But it's useful to keep them separated like this.
So let's do that.
And let's look at the difference here, the difference in dissociation energy.
Delta D0.
Let's not put a double arrow.
Let's define the sign of it.
This way, going from products to reactants.
OK, now let's just look at how equilibrium should work.
So now, let's just take a generic reaction.
Little a, our number, our stoichiometric number, A plus b, and B goes to c C, d D. And we know that delta G0 is minus RT log of Kp.
And that delta G0 is just the free energy of the products minus the free energy of the reactants.
So it's c times G C0 plus d G D0 minus a G A0 minus b G equals B0.
So we need to know the free energy of each one of the species.
And now we know how to calculate that from first principles, through statistical mechanics.
So we know that G is A plus pV.
A is minus kT log of capital Q.
And I'm going to assume that we're in the gas phase.
And it's an ideal gas, so I'm going to replace pV by NkT.
And what's Q?
Well, we know what Q is.
It's q, little q, to the N power translational over N factorial.
For atoms, this is all it would be.
But we have molecules.
So we have individual other degrees of freedom besides translation.
So let me just label those internal, q internal.
And that's also to the N power.
And q internal, if you say, what are those degrees of freedom, well it's the electronic energy.
it's the vibrational energy.
It's the rotational energy.
And we'll deal with those shortly, but let's just separate it this way for now.
Just we can keep track of where the N factorial belongs.
And now we're going to use Stirling's approximation for the N factorial.
So our log of Q, which we need up here, is just N log of q trans q internal minus log of N factorial.
And that's just equal to N log q trans q internal, minus N log N. Plus N. And of course, this I'm going to put down here momentarily.
And now our expression for G is up there.
So it's minus kT log of Q plus NkT.
So these factors of N are going to cancel.
So when I take minus kT log of Q, that's going to be minus NkT over here.
That's going to cancel the plus NkT here.
So all I'm going to have left is this term and this term.
Which I can combine.
So it's just minus NkT log of q translational q internal over N. Now these, really, are just the same as q, it's just minus NkT log of little q over N. I didn't need to separate that into this product.
I only wanted to do it to be clear where we were getting this factor of N factorial from.
So now we have an expression for G. And if we know G for all the species involved in a chemical reaction, we should be able to calculate the chemical equilibrium.
So let's do it.
So now to do it, let's look a little more closely at what these internal degrees of freedom are.
Because that's where these important details are going to come in.
Obviously the equilibrium is going to depend on the energetics.
How much different are the bonding energies or the dissociation energies and the molecules involved.
And also, how much different or the other molecular energy levels.
The vibrations, rotations, and so forth.
So, q internal is the product of rotational, vibrational, and electronic partition functions.
Remember how we saw that if you can write the energy as a sum of energies, then the partition functions are multiplied.
Because, of course, if this is the sum of a rotational plus vibrational plus electronic energy, then of course I can just separate out these things.
These are in the exponent.
I can write it as a product.
Same with translations.
I already have separated that.
So now let's look at how these things behave.
Well, for the electronic case, there's really only one electronic state of interest in general.
And that's the lowest state.
In cases you're extremely familiar with, if it were the hydrogen atom, it would be down in the 1s orbital.
And it would take a huge amount of energy to get up into the 2s or 2p orbital.
At ordinary temperatures you never have that.
All the atoms would be in the ground state.
Now, for most molecules, it don't take as much energy as for the hydrogen atom.
If you have benzene, for example, the ground electronic state, the lowest electronic state, is quite far below the first excited state.
Not as much as the hydrogen atom going from 1s to 2s to 2p, but still by much more than ordinary thermal energies at room temperature.
And what that means is, again, only the ground state term is going to matter.
Now, sometimes we would define that is this as the zero of energy.
But since we're doing chemical equilibria, the zero of energy is up here.
So that epsilon, that energy term, is this amount.
The dissociation energy.
This is minus dissociation energy.
So we're going to have a positive number there.
So q electronic is just e to the D0 over kT.
So that's easy enough.
We also know what the vibrational partition function is.
We saw it last time.
It's this thing we could write as a simple form.
One over one minus e to the minus E, I'll call it E vibe, over kT.
It's h nu 0, where nu is the vibrational frequency.
Again, in the case of many molecules, the vibrational energy is pretty high compared to kT.
So even this often simplifies to be just one.
In other words, all the molecules, in many cases, are basically all of them are in the ground vibrational state at room temperature.
Certainly if you look at molecules of high frequencies, if you look at the hydrogen molecule, H2, the vibrational frequency is about 4,000 wave numbers.
Remember I mentioned last time, kT at room temperature, T is 300 Kelvin.
Then kT corresponds to 200 wave numbers.
It's a factor of 2/3.
Much less than 4 wave numbers.
In other words, the vibrational energy is much lower than kT.
So, again, everything would be in the lowest level.
And this just simplifies to one.
In many cases, it's reasonably close to one.
For high vibrational frequencies, nu 0.
Now, we haven't talked about rotation.
But you probably have just an intuitive feeling that at ordinary temperatures, if I do this, if I wave my hand in the air, molecules that I happen to intersect are going to start spinning faster.
In other words, ordinary thermal energies do populate some number of rotational levels of molecules.
Molecules probably aren't going to start getting squished together and vibrate harder when I do something like this.
So molecules might generally still be in the ground vibrational levels, thermal energy isn't enough to raise them in many cases.
But rotation, for sure.
They're not all in the lowest level.
Turns out that, in fact, the energy separation between rotational levels is very small compared to kT at room temperature.
So just like we saw for the high temperature limit for vibrations, it turns out that for q rotation in the high temperature limit, we have the same situation where we have kT over, now it's a rotational energy.
And it's typically on the order of about one to ten wave numbers for small to medium sized molecules.
Remember, too, last time when we talked about in the vibrational levels and said, well, if you're at kT and it's this big, it's basically telling you roughly how many levels do you have thermal access to.
Because we saw the same results for vibrations.
It's not so different for rotation.
And it turns out that at ordinary temperatures, you might have access to a few tens to a few hundreds of rotational levels, depending on whether they're closer to one or ten wave numbers or a little higher.
Again, room temperature is 200 wave numbers.
So, that means that, remember q is a unitless number.
You're counting the states, weighted by the energy.
Weighted by the Boltzmann factor.
And for rotations, it's a number that might be on the order of 100 or so.
Order of magnitude estimate, that's all.
And of course we've seen the translational partition function is on the order of 10 to the 30th, right?
Enormous number.
In other words, in the simple lattice model that we've use to describe translation, just breaking up the available volume into little pieces.
Counting.
Well, OK, you have something on the order of 10 to the 30th possible locations.
And if you treat this properly, quantum mechanically, for the translations, there's actually a magnitude of the number is similar.
Now I just want to go through a very simple example.
Just treating a particular generic reaction.
And look at what the equilibrium constant is.
Working it through, given what we've seen so far.
So I just want to simplify it by having all the stoichiometric coefficients equal to one.
It's not a great complication, if we don't do that.
But it's a little bit simpler.
So let's take a molecule that's A-B plus C-D. And now let's break bonds and reform them.
So we get A-C plus B-D, right?
So we're going to do something simple.
And since I've got all the stoichiometric coefficients equal to one, we've got the same number of molecules as reactants and products, that'll make things a little bit easier.
Just in the sense that if we look at the contribution of the translational energies at room temperature, they're going to be the same.
For the reactants and the products.
Any of the molecules is going to have a translational partition function on the order of 10 to the 30th, at room temperature and an ordinary volume.
And nothing's going to change from reactants to products.
And for vibrations, let's assume, as is often the case, that the vibrational frequencies are fairly high.
So all the partition functions for the vibrations are equal to one.
Like we've got written up there.
For all four of the molecules.
So then nothing is going to change in the vibrations.
Going from reactants to products.
Finally, the rotations.
So, for the rotations our partition function is something on the order of, I think I meant to write 100, not ten here.
As our order of magnitude.
That's what it's going to be.
It's a number on that order.
And if we assume that the masses of the atoms involved are comparable, then we can cheat a little bit and say that's also, those numbers are also, going to be comparable for the reactants and the products.
We don't have to do that.
We could put it in, and rather easily calculate it.
But I'm going to make life simple in this way and just work through how to carry through the calculation.
And I think it'll be clear how to put in different partition functions for those quantities if they are different from each other.
OK, so let's try it.
Then, our delta G0, let's go back over here.
Here's delta G, let me just put the relationship up.
So we've got an expression for G. Which is, let's start from back there.
It's minus.
And now for each one of the substances, it's Ni kT log qi over Ni.
I'm just going to put this in molar terms.
So it's minus little ni RT.
Log of qi over capital Ni.
And I only want to do that because I want to express this in free energy.
This is for substance i.
Free energy per mole.
So it's just minus RT log of qi over Ni.
That's G per mole.
That's Gi naught per mole.
Plus RT log pi over p0.
Where p0 is an atmosphere, basically.
One bar.
So that means Gi0 bar is minus RT.
And I'm going to take this minus this and combine them.
So it's log of qi over Ni, that's this part.
And now I've got log of pi over p, and I'm just going to use the ideal gas law.
So I'm going to use pV is nRT.
So then I've got Ni kT over my factor of p0, times the volume.
So that's Gi0 for each one of the substances.
The Ni's cancel.
And I've got minus RT log of qi kT over p0 times V. Now, qi is just the total molecular partition function.
It's this product of q trans times q rotational times q vibrational times q electronic.
And now I need delta G. So delta G0, is just minus RT.
And now it's just, I'm going to take this for each one of the substances, for the products, minus the reactants.
And I'm going to combine the log terms.
This is the same, and I'm going to have this in every term.
So what's going to happen?
Well, all this stuff is going to cancel.
Now, that doesn't necessarily happen.
That happens because of the stoichiometric numbers being the same.
Otherwise I'd have to take them to that the power of the stoichiometric number.
But in this case they're just going to simply cancel.
So I'm going to have left is the ratio of my partition function for molecule AC.
Partition function for molecule BD.
Partition function for molecule AB, and for molecule CD.
Products, reactants.
Pretty simple, because I know how to calculate all this stuff.
And again I'd suggested some simplifying assumptions for what those partition functions are.
But it wouldn't be hard to calculate each one of them if I just had all the relevant energy levels.
To make this simple, we're going to assume that the rotational energies are the same for all the molecules.
It wouldn't need to be, it wouldn't be hard, to plug in different values.
We're going to assume we're in the low temperature limit for vibrations.
So q vibrational for each one of these things is equal to one.
The translational contributions, are equal for all of them.
So the only thing that's different is the electronic contribution.
And that's different.
Because, of course, you have different electronic energies.
The binding energies for the products and the reactants aren't in general going to be equal.
And in most cases, it is the energetics that dominate the equilibrium constants.
Of course, that is, it's the electronic energies that usually dominate.
Of course, the other things do matter.
But it's not unusual for the electronic energies to dominate.
So here's delta G0.
Of course, this is minus RT log of Kp.
So there's our expression for our equilibrium constants.
So it's just this product.
So let's just put in the partition functions for the electronic part.
It's e to the D0 over kT for molecule AC.
Times e to the D0 for molecule BD over kT.
Divided by e to the D0 for AB over kT, times e to the D0 for molecule CD over kT.
So the whole thing is just e to the D0 AC, plus D0 BD minus D0 AB minus D0 CD over kT.
Or in other words, it's e to the delta D0 over kT.
That's the whole story.
So it's a super-simple calculation to execute.
And for lots of molecules, we know the dissociation energies.
These are measured, determined, either thermochemically or spectroscopically for a great many molecules.
So in fact, we have the information we need to do that calculation.
And if we want to worry about the vibrational and rotational levels, typically we have that information, too, spectroscopically.
So we know what the relevant energy levels are to do the whole calculation.
So what it means is, we can calculate equilibrium constants for ordinary chemical reactions just from first principles.
Knowing the energy levels of the available states of the molecules.
Any questions?
In your notes I've included another example.
It's a little bit of an extra example.
You can go through it if you'd like.
It turns out to have a kind of attractive closed form solution.
Which is the case of a chemical reaction like an isomerization inside a crystalline solid.
So you can imagine you've got a crystal of some molecule, many crystals, thermally over time, they might decompose.
They might have reaction processes that can take place.
So I've imagined you've got uni-molecular reactions that can occur.
And in that case you start with a pure crystal of one species.
And you end up with a mixture of some number of the original species, and some number of new species.
And there's an equilibrium constant that'll say where that should be.
So again, it'll depend on the energies.
On the dissociation energies of the different species.
And there's also a kind of mixing term, because effectively now you've got the different species located at different places in the crystal.
Those are distinguishable states, in the solid.
And you need to count those.
But the one other example I want to work through is a little bit different from chemical equilibria.
It's phase equilibria.
Why shouldn't we be able to calculate phase diagrams?
So in the earlier part of the course, you went through and saw the macroscopic thermodynamic treatment.
The equilibrium constants and chemical equilibria.
So of course you saw the whole delta G0 as minus RT log Kp and so forth.
And you've seen now we can actually calculate all that from first principles.
What about phase equilibria?
What you saw before, we just drew phase diagrams.
And they had boiling points or lines of boiling.
Or in other words, liquid-solid and liquid-gas and solid-gas equilibria.
And what we did is, given where those lines fell, you can calculate things.
But where do the lines fall?
What's the boiling point or the sublimation temperature of some material at a particular pressure?
Well, we didn't offer any prescription for calculating that.
You had to take that from measurement and then given that, you could use the Clausius-Clapeyron equation and so forth and look at the way things behaved if you moved along a line in a phase diagram.
But there was no prescription offered to calculate where exactly would that line be on the phase diagram.
But again, if you know all the energies of the possible states, in the solid, in the liquid and the gas, statistical mechanics shows us that we can calculate the equilibrium between those.
Which is to say we know we can calculate where those lines belong.
So I want to just go through maybe one of a couple of relatively simple cases.
I want to go through the case of a solid-solid equilibrium.
Let's imagine we have two solid phases and an equilibrium between them.
And this problem was on the problem set.
How many of you did it?
Everybody did it.
Did you all get to the end of it?
Who got to the end of it?
Some of you got to the end of it.
If you did, congratulations, it's not so trivial to work on for the first time.
So I'll just briefly go through that problem.
And show how it works.
So, it's similar, of course, to the case that we're doing now.
To the case we just did, of chemical equilibria.
The difference, though, is that it's cooperative.
The solid, at least in the way I formulated that problem, you either have the crystal in phase one or phase alpha, or in phase beta.
And there's nothing in between.
There are lots of systems like that.
Not just crystals, you could go from different forms of DNA.
Act like that.
Where you have super coiled DNA and regular DNA.
And you actually can have some of the DNA in each.
But it's actually very cooperative.
So there's a huge tendency to either be all in one, or be all in the other.
Or at least very very, large pieces of it like that.
Lots of other systems act like that.
In other words, unlike the case where we're thinking about chemical equilibrium among molecules in the gas phase, these two molecules over here crash into each other and they react.
Nothing else cares.
The other whole mole of molecules does whatever they were doing.
And a little later some other pair of molecules happen to crash into each other and maybe react or maybe don't.
In other words, there's no cooperativity at all.
What happens to some pair of reactants and products, everything else is completely independent of it.
That's not the case when you have, in many cases, a phase transition.
You can see it by eye, often.
If you do something like super cool water.
This is a kind of fun experiment.
If you take dry ice, and you've got a little bit of water.
Right or a little pot of water.
And you put some dry ice into it.
So you can actually get the water to be below the freezing temperature and it won't freeze yet.
Then you drop a crystal in there and boom, it all freezes.
Very cooperatively.
Lots of phase transitions behave like that.
So let's just see how it works.
In this case, it's really similar to what we just did.
We can treat it, here's phase alpha.
Here's phase beta.
And the crystal, the interactions between molecules or the atoms in the crystal are different in the two phases.
So effectively, the binding energy is different.
In other words, if you say, now let's think of the energy it would take to evaporate all the atoms or molecules and let them loose in the gas phase.
That's the analog of dissociation for the molecule.
You're pulling everything apart from the crystal and separating them all.
So that's what we'll call these energies.
This'll be minus E alpha.
And this will be minus E beta.
And then there are vibrational energies.
So the lattice has a bunch of vibrational energies.
We can assume they're evenly spaced.
And they're going to be different.
They're not the same in the two different crystalline forms.
That's typically the case.
You have a phase transition from one crystal to another.
The lattice vibrational frequencies aren't the same any more.
Speed of sound is different.
Things are different.
And that's all.
That's the only thing that's different.
So these are binding energies per atom.
So now, let's just try to calculate Q.
And what we are supposed to get out of this is, what's the phase transition temperature?
That's the thing that we didn't have any prescription for calculating before.
We just said, here it is.
Here's the boiling point.
Here's the melting point, or whatever the phase transition was.
Or if we drew a phase diagram, here's the line.
But now we're going to try to calculate where that is.
Or what those temperatures are.
So, Q for either phase it's just e to the E over kT.
Just like we saw before for the dissociation energies of molecules.
To the Nth power.
No N factorial any more.
There's no translation.
Times the vibrational part.
This is the electronic partition function.
And then there's the vibrational part.
And that's one over one minus e to the minus h nu, what's called E over kT, for the lattice I'm assuming it's just one frequency.
So this energy is h nu E. And of course, it's different for the alpha and beta phases.
And I'm going to assume we're in the high temperature limit.
Which is often the case for lattice vibrations.
So it's e to the, little E over kT.
For our electronic part, to the Nth power.
And we've seen what the high temperature limit is.
It's kT over h nu E. And that's also, that to the 3N power.
I should have written this here.
In other words if you say, how many vibrations are there in the lattice, well if there are N atoms, each atom in the gas phase would have three degrees of freedom.
Translational degrees of freedom.
In the crystal, it can't freely translate.
But those degrees of freedom are still there.
The atoms can all move.
So now, those are the lattice vibrations.
When the atoms try to move, they vibrate against each other.
So how many different modes are there?
They may be degenerate.
They may all have the same energy.
But those modes are all still there.
Actually, they're the acoustic modes of the crystal.
And there are 3N of them.
One for each translational degree or freedom that each atom had.
OK, so that's it.
A, minus kT log of Q, so it's just minus NE, and the kT cancels.
Minus 3NkT log of kT over h nu E. Second part.
And then we can calculate the chemical potential.
It's just the dA/dN, d constant T and V. So this goes away.
It's just minus little E. This goes away, it's minus 3kT log of kT over h nu E. Well, basically we just finished.
At the phase transition temperature for the two phases, those things have to be equal.
And that's all there is to it.
So, at, we'll call it Tc, the phase transition temperature, I guess I called it T1 here, mu alpha is equal to mu beta.
So this stuff for alpha is equal to this for beta.
That's it.
So what does it say?
It says E beta minus E alpha, these terms, equals 3kT1 log of nu E beta over nu E alpha.
Solve for T1.
That's it.
It's E beta minus E alpha over 3k log nu E beta over nu E alpha.
There is our phase transition temperature.
That's the whole story.
So if we know the electronic energy that binds the crystal together, usually something that can be measured easily, so it's known for many materials.
And if we know the lattice vibrational frequency.
Also something that's pretty routinely measured, we're done.
So again, from a very simple first principles approach, we can calculate that phase transition temperature.
We could do it for a solid-gas equilibria too, if we said OK, let's think of sublimation of the solid.
And now we know how to calculate the chemical potential in the gas phase.
We can have those be equal.
That's actually worked through in your notes.
Again, it's an extra problem.
If you would like to take a look at it, it's a straightforward calculation.
The point is, we can calculate those phase equilibria.
Liquids, I will say, are harder.
Just because it's harder to define and know all the energies available.
Doable, approximately.
But the point is, we can actually now locate all those lines that we sort of arbitrarily drew on the phase diagrams.
And make sense of what the temperatures and what the pressures are, for that matter, where they occur.
Any questions?
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So, now we'll start on the last of the main topics in the course.
So we've finished most of our macroscopic thermodynamics, and our microscopic approach to it through statistical mechanics.
And now, our final topic is kinetics.
Kinetics is really a very different kind of topic.
Because unlike thermodynamics, thermodynamics tells you all about equilibrium properties.
A huge part of the work of this term has been to figure out what equilibrium is.
What is the equilibrium state, given some situation.
You've got some phases present.
Different then, you could go from solid to liquid to gas.
Or you've got different chemical constituents together that can react and go back and forth.
What are the equilibrium concentrations?
But we haven't worried at all about how long it might take to get there.
And that's what kinetics does.
Kinetics is concerned with rates of reactions, primarily.
How long it takes to reach equilibrium.
And of course it's super-important.
Because if you look at that window glass, it's not in equilibrium.
It's silicon dioxide, the equilibrium state would be a crystal.
It would be crystal and quartz.
Nevertheless, none of us is very worried that on the moment it's likely to suddenly, spontaneously turn into a crystal and be opaque and scatter light and do all that sort of stuff.
And in lots of other situations, certainly if you look at any living biological system, including yourselves or your friends or any other one, it's certainly far from equilibrium.
And you probably would hope for it to stay that way.
So there's an enormous amount of chemistry and processes we're concerned with.
Which depend intimately on kinetics in order to work the way they work.
They also do depend on thermodynamics and where equilibrium states are.
But that doesn't mean they necessarily reach equilibrium states.
So we're going to go through kinetics, starting with the simplest examples and working our way up to more complex cases.
And just see how we can describe elementary chemical reaction rates and processes.
So that's our concern now, is dynamics.
How long things take to get to equilibrium.
And actually just like macroscopic thermodynamics, kinetics does take an empirical approach to the topic.
In other words, it's based on experimental observation.
Of macroscopic rates.
How long it takes a collection of stuff, a mole of stuff to change chemically.
And undergo reaction.
And so forth.
We often infer molecular mechanisms based on kinetics.
And it's hugely important and valuable to do that.
But it's also important to recognize that what kinetics can do is show us how we can formulate microscopic mechanisms that might be consistent with our macroscopic kinetics models.
But the kinetics models by themselves don't prove the mechanism.
And there are all sorts of examples where mechanisms that were proposed and accepted because they were consistent with macroscopic kinetics results turned out to fail.
There are other ways to prove mechanisms.
You might be able to design direct spectroscopic observations of intermediates and so forth.
And in that case, it often becomes possible to distinguish between different mechanisms.
Which might all satisfy the macroscopic kinetics equation.
So we use kinetics to infer a mechanism but not necessarily to prove it.
Not generally to prove it.
We also use kinetics to describe an enormous range of time scales.
So, the fastest things that we're sometimes concerned with that might take place on femtosecond time scales, 10 to the minus 13 seconds or so, or 10 to the minus 15 seconds, even.
So, if we look at the range of time scales we might be concerned with, might go anywhere from about 10 to the minus 15 seconds at the fastest, and might go to enormous scales on the other end.
All the way out to maybe 10 to the 10 seconds, which is on the thousands of years.
Could be longer than that, sometimes even millions of years.
And the formalism that we'll set up applies equally to the full range of time scales.
So it can describe an enormous amount of chemical activity.
More commonly, for stuff that we're going to compare to, you might measure in the lab.
Common time scales range from roughly 10 to the minus 6 seconds out to about 10 to the 5th seconds.
That is, microseconds to about a day.
But there are plenty of examples of going either faster or slower, depending on the need and the experimental equipment available.
Let's define a few terms.
Let's talk about how we're going to just formulate chemical reaction rates.
So, let's just imagine any simple reaction of this sort.
So A plus B are going to go to C. There's some rate at which it happens.
Note, by the way, it's an arrow in one direction.
It's not an equals sign like we've seen before, or a double arrow.
So the point I'm emphasizing here is when we talk about reaction rates, unlike equilibria, we're talking about a particular direction.
Later on, we will talk about reversible reactions.
But there, too, the arrow going this way refers only to one of the chemical reactions that can take place.
Namely, in this case, the changing of what was the product now, would be the reactant, C, back into A plus B and it has nothing to do with the rate this way.
Measured independently and so on.
So the rate, we can look at the rate of disappearance of A.
So it's just negative d[A]/dt, where the brackets indicate concentration.
Usually in moles per liter.
If it's in a gas, then it would be pA, of pressure.
So negative d[A]/dt is our rate.
And note that that's generally a positive number.
We're looking at reactions going, if it's a reaction going in this direction.
A is gradually disappearing.
So we're going to define our rate this way.
To be a positive number.
The rate for C is going to be plus d[C]/dt.
It's also going to be defined in a way that makes it positive.
Because in this case, since the reaction's going this way, we're looking at the appearance of C. Now, by stoichiometry , in this case, of course whenever a molecule or a mole of A disappears, a molecule or a mole of C appears.
So because of that, in this case, the stoichiometry tells us that that d[C]/dt is equal to negative d[A]/dt.
And also equal to negative d[B]/dt.
And any of those can be used to define the rate of reaction.
Now, this is a particularly simple case because I've chosen the case where all of the stoichiometric coefficients are equal to one.
So now let's just look at any case that's different.
Let's look at 2A plus B goes to something else, 3C plus D. And just look at the reaction rates that we might see there.
So here now, the appearance of C is going to be three times as fast as the appearance of D, for example.
And also three times as fast as the disappearance of B.
So if we write negative d[B]/dt, we expect that's going to be negative 1/2 d[A]/dt.
In other words, A is going to disappear twice as fast as B.
Every time a molecule of B reacts, two molecules of A do.
And that's going to be plus 1/3 d[C]/dt, and every time that happens three molecules of C get formed.
And it's going to be plus d[D]/dt.
One molecule of D gets formed.
And so, the reaction rate could be defined in terms of any of these.
But the important thing is to keep track of stoichiometry so that the rate as it pertains to each constituent is accounted for correctly.
So to generalize, if I have little a of A, and little b of B, going to little c of C, and little d of D, then the rate of reaction can be written as minus one over a d[A]/dt, or minus one over b d[B]/dt, or one over c d[C]/dt, or one over d d[D]/dt.
So, experimentally, lots of measurements of reaction rates have been made.
And now to start on what's seen empirically, basically the following result is extremely common.
The rate is equal to some constant times the concentration of A to some power alpha, times the concentration of B to some power beta, and so on.
For all reactants.
Multiplied together, each concentration taken to some power.
Notice no products.
Again, we're only looking at a reaction going one way.
And if we look at the rate, this is typically what's found.
Alpha is called the order of reaction, with respect to A.
Beta order with respect to B.
And so on.
And little k is our rate constant which, just to make clear, since we've been using it extensively in the past few lectures, is completely not equal to the Boltzmann constant.
Absolutely no connection between them.
Now, alpha and beta, what they are, what they turn out to be, is typically what's determined as a result of kinetic measurement.
Typically they're small integers.
So, just sort of a typical example, if you look at the reaction of NO and O2, to make NO2, simple reaction, and you look at minus d[O2]/dt, use that as a measure of the reaction rate.
What's found is that it's a constant times NO concentration squared, times O2.
Simple, right?
One of the exponents is two, the other's one.
Doesn't seem so surprising mechanistically.
But it's not always the case.
Even when you have small integers, it's not always the case that the most obvious mechanism you would infer is the real mechanism.
And sometimes those exponents don't turn out even to be integers.
But here's another example.
If you just look it CH3CHO going to methane and carbon monoxide, seems like it would be a pretty straightforward thing, too.
So if you measure, for example, the rate of appearance of methane, what you discover is that it's equal to a constant times the concentration of the starting material to the 3/2 power.
Not obvious mechanistically why that should be the case.
Usually it's telling us something.
It's telling us that the reaction mechanism is complicated.
It's a multi-step process.
Sometimes there could be chain reactions and so forth.
So again, seeing things like this certainly helps us to infer molecular mechanisms.
Again, they don't prove molecular mechanisms.
But they certainly can be very helpful in suggesting them.
And then other means can be used to try to prove them.
Including, above all, direct observation of the intermediates that you would expect on the basis of one mechanism or another.
Now let's just go through some elementary examples of kinetics one at a time.
So let's start with the simplest possible case.
Actually, a very rare case, but one that'll help us just set up the formalism of, and the mechanism for us to proceed.
So let's talk about zero order reactions.
Actually very rare.
So, what this means is something like A goes over to products.
Make a measurement of d[A]/dt, and discover that it's k time A to the power of zero, that is, it's just k. There's no dependence on the concentration of A.
Or anything else.
And you have some rate of its disappearance.
So here's an example of it.
You could have oxalic acid, and it just turns into hydrogen carbon dioxide and carbon dioxide.
Things break apart, forms these products.
Make a measurement of the disappearance.
Seems to have nothing to do with the concentration of it.
At least under certain conditions.
Now, turns out that there's another element that helps understand this.
Turns out that light is needed, it's a photochemical reaction.
And then it's easy to see how this can happen.
Let's say we have an abundance of the starting material, and not very much light.
So every now and then photons bleed in, and every now and then when they're absorbed, it leads to dissociation.
In that situation, you'll be limited by the photons, but not by the concentration of the molecules.
And in fact, strictly speaking although this is zero order in terms of the chemical constituents, it's not zero order in the photons.
So in some sense, in this sort of situation, the photons should be considered one of the reactants.
You could write this as this plus h nu plus one photon, goes over to these products.
And then if you measure the rate and things are under circumstances like I described, you would discover that in fact yes you'd be photon limited.
The rate would depend on how many both photons are coming in.
Still, ordinarily, chemical rate equations aren't formulated in those terms.
So in the usual formulation, this would still have the appearance of a zero order reaction.
Now, how do we write and formulate a solution?
So it's simple.
We've we've written a differential equation here.
It's a pretty straightforward one.
Minus d[A]/dt is just a constant.
So we can solve it.
And typically we'll solve it by rewriting in integral form and then doing the integration as long as we can do the integration.
So from here we can write the integral from starting concentration [A]0 to some other concentration, [A], d[A] is equal to minus k integral from zero to t dt.
So all we've done is integrate on both sides.
And we've assumed a starting concentration.
We've assumed an initial condition.
And we've also assumed an initial time, which usually we can just call zero.
And this is, of course, something we can solve for straight away.
So we just have that [A] minus [A]0 is negative kt minus zero, which is minus kt.
So [A] is minus kt.
Plus [A]0.
In other words, the concentration of [A] at any time is given by the initial concentration, minus the rate constant times time.
It decays linearly in time.
So we can sketch that.
This is our initial concentration.
And then it's just going to decline with time after that.
And there's our solution.
In lots of cases, it's useful to define what's called a half-life.
It's just useful.
Because it provides some timeframe, a single number that's a timeframe on which the reaction occurs.
So, the half-life is just the time that it takes for half of the reactants to disappear.
t 1/2 time to react half the reactants.
OK, so in this case it's straightforward to see when that happens.
In other words, that's the time at which this concentration of A is just equal to [A]0 over two.
So we have [A]0 over two is minus kt 1/2 plus [A]0 or t 1/2 is equal to [A] over 2k.
[A]0 over 2k.
So there's our half-life.
So if we go over here, we can put that in.
[A]0 over two, and this time is our half-life.
So that's zero order reactions.
And, again, zero order reactions are rare.
But the procedure we're going to used to solve for kinetics is outlined in this way.
And we'll use that again and again.
Let's look at first-order kinetics.
Let's go over here.
Now, first order reactions are quite common.
Much, much more common than zero order.
So here, you have A goes to products.
That's the simplest case.
But this time if we measure d[A]/dt, we discover that it's equal to a constant times the concentration of A.
There's an important point to note here.
What are the units of k, in this case.
What do they have to be?
Yeah, reciprocal seconds right?
The equation has to work.
This is concentration per second.
This is concentration.
This better be per second.
Let's just, before we move on completely, look at this.
What are its units?
Here's the equation.
What are the units of k?
No, not unitless, because, look at this.
This is some concentration unit, typically moles per liter.
So this is moles per liter per second, right?
It's disappearance of some concentration for time.
That's got to be here.
All that is here.
Moles per liter per second.
So in every case, the units of k, the rate constant, have to be figured out on the basis of the specific rate equation.
Doesn't have the same units.
When the kinetics are of different order.
Now, let's solve this using the same approach as before.
Namely, this is still a pretty straightforward differential equation.
So let's just integrate both sides.
So that is going to tell us the integral from [A]0 to [A].
But now we have [A] on this side.
So here we just had a constant.
And effectively, I didn't write it out.
But we effectively wrote this, rewrote this, as d[A] equals minus k dt, and then went from here to the integral.
We're going to do the same thing here, except now there's this.
So really we're going to have d[A] over [A] is minus k dt.
That's what we're going to integrate on both side.
We need to have the variables distinct on each side.
So we have d[A] over [A].
Equal to minus k integral from zero to t dt.
And so, of course, you know how to do this integral.
It's going to look like log of [A].
And again, it's taken at [A] or, at [A]0.
So we have log of [A] over [A]0.
We're going to have log of [A] minus log of [A]0 coming out.
And that's equal to minus kt.
So now the kinetics are quite different.
We have [A] is equal to [A]0, e to the minus kt.
Very important, very common sort of result.
It's saying we start with a certain amount of material, and there's an exponential decay of it.
So very different from kinetics here, which are just linear.
And again the kinetics here, if you imagine that situation where you've got starting material and bleeding in gradually are photons, presumably at the same rate, then sure that material you're going to see the disappearance of it linearly with time.
Just depending on the rate at which the photons are coming in.
Here it's very different because, presumably, A is required in order to do this reaction.
It'll depend on how much.
Because of course if there's more of it, you'll just have more at any given time decaying over to products.
So you have an exponential decay.
So let's plot that.
Here's [A], let's make that [A]0.
There it is.
Now, let's look at what happens to the product.
So let's imagine that it's A going to B, so of course minus d[A]/dt is just equal to d[B]/dt, and the rate of appearance of B has to match the rate of disappearance of A.
Let's assume that we don't have any of B present at first.
So let's make [B]0 equal to zero.
Well then, [B] just has to equal [A]0 minus [A], right?
All the stuff that's left, all of the A that has disappeared, that's given by this difference.
Is just equal to B.
So that's just [A]0 minus concentration of A, but that's just given by that.
Which is [A]0 e to the minus kt.
Or in other words, [B] is just equal to [A]0 times one minus e to the minus kt.
So at t equals zero, this is zero and this is one.
In other words, this is going to be zero at first.
And then it's going to grow in with the same exponential form that this decayed.
So, [B] is going to do the exact opposite.
It's going to be like this.
So this is [B] of t and this is [A] of t equals [A]0 e to the minus kt.
Now, it's always useful to, whenever possible, to plot these things linearly.
Find a way to plot these as straight lines.
And of course in this case it's straightforward to do that as a log plot.
So if we take the log of both sides, we know of course that the log of [A] is just minus kt, plus the log of [A]0, so now let's look at that.
Make this the log of [A]0 and this is the log of [A] on the axis.
That'll be time.
So there's just some linear decay now.
And the slope is minus k. So experimentally, of course, this'll be done typically as a simple way of determining it.
Now, we also can usefully look at the half life, in this case, in the case of first order kinetics.
So we have an expression in general for [A].
So if we let [A] equal [A]0 over two, at t equals t 1/2, that tells us that log of [A]0 over two divided by [A]0 is minus kt to the 1/2.
But this is just the log of two, or log of two is over k is t to the 1/2.
And so t to the 1/2 is just 0.693 over k. So we can write that just generally, completely independent of whatever [A] is, also independent of what [A]0 is, and that makes sense.
So the point is, if you have something that just spontaneously decays into products, maybe it's just gradual chemical decomposition of something.
Spontaneously, without the participation of other constituents.
There's a general half-life that can be associated with that.
That'll just be related directly to the rate of the decay.
So it's knowable and measurable in a simple form.
And again, the half-life is always useful.
Because it just gives a simple, one-number measure of the rough timescale for things to change.
So if we go back to this plot, and then once again here's our half concentration.
And here's our t 1/2, just a useful way of summarizing in a simple way, what happens.
Yeah?
[INAUDIBLE]
Ooh.
Well, I didn't think about it.
But it isn't necessary that that'll happen.
Sure seems like it must be.
When one is half decayed, the other must be half formed as long as there wasn't any B present at first.
So yeah, thank you.
Well, given that, something needs to move.
But let's pretend like I got it right at this intersection.
Even though it doesn't look that good.
So really it should be there.
Thank you.
So the single most common example of first order kinetics of this form is radioactive decay.
You've got some radioactive isotopic that can spontaneously decay into some other nucleus.
And of course, this is measured and half-lives have been tabulated for lots of cases of this sort.
So a simple example.
Let's look at carbon-14.
It's got a nuclear charge of six.
It can decay into nitrogen-14 through the loss of an electron.
Happens.
First order kinetics.
Now, in the atmosphere, what happens is this will end up, the 14C ends up getting replenished.. Because from cosmic rays, what can happen is in the atmosphere, you can have your nitrogen plus a neutron will come and form 14C plus a hydrogen atom.
So in fact, the overall concentration in the atmosphere of 14C tends to be constant over time.
But stuff that's formed down here on Earth, with carbon, its content of 14C decays over time, and it doesn't get replenished.
So, let's think back some long time ago.
Here's a tree.
we can make it pretty.
So it's got some concentration of 14C that came from carbon dioxide in the atmosphere.
That's what it started with.
That's our starting point.
At some point later, through natural occurrence or human intervention, that tree became horizontal.
Then, and although we're looking back here, let's call this our t equals zero.
And our 14C concentration at the time is our concentration at zero.
Now let's say, shortly after that, either right after because of deliberate action, or shortly after because it was just discovered, somebody decides to build something using that tree.
So, early human craft.
And then let's say lots later, depending on your definition of lots, modern human, which can be denoted by this style of hat, makes exciting discovery.
Terrific.
And would like to know how old is it.
How long ago did all that stuff happen.
Let's assume this is also approximately t equals zero.
Well, so a log of [14C] over [14C]0.
So this carbon dating, all it's really doing is measuring that.
It gives a number for it.
And we know this because we're assuming it hasn't changed any in all those years.
In the atmosphere.
It's different down here, because the tree or the boat wasn't replenished.
Not nearly as many cosmic rays fell on it.
So that's minus log of t. And it turns out the t 1/2 is 5,760 years.
Amazing, that this can be known down to ten years.
But it is.
So that says, k his 0.693 divided by t to the 1/2, which is one over 8,312 years.
So there's our answer, then.
The time, how long ago that happened, is just minus 8 years times the log of [14C] in the artifact.
Divided by the log of, by [14C] in the air.
And the assumption again is that this is the same now as it was back then.
And there's a bigger number than this, so it's a positive number overall.
So in a fairly straightforward way, we'll use first order kinetics to determine the lifetime of something like this.
Because we know the rate constant.
And everything else follows.
Let's see.
Next there's going to be second order kinetics.
But let me just stop here and say just a word about the exam on Wednesday.
So I've handed out an information sheet about it.
But I don't have anything very different to say this time than I've had to say about previous exams.
Solve lots of problems.
Go over the homework.
Go over practice problems.
Try last year's exam as a sample exam when you're ready to do it.
Try it under test conditions.
There's nothing on the exam that you're going to look at and say oh my God, how was I supposed to know that we ought to study that.
There won't be any surprises.
Most of the exam questions so far, not all have been easy, but I think they've been more or less plain vanilla in the sense that they're right more or less down the middle of what we're trying to teach.
And not very many are taking off little peripheral elements of the class.
And it's not going to be any different here.
So if you can just do the problem solving and get familiar enough with it that you're just good at it, so that you can do it with a reasonable speed, you're going to be fine.
Also try to work on just understanding the underpinnings.
Especially of statistical mechanics.
That's where when you have these true-false or multiple choice questions, these sort of thought exercises.
Those are really hard.
Because they go a little bit beyond problem solving.
If you could do the problem solving, you're going to do fine.
But it's always useful if you can also formulate things.
And for that, it's never easy to just tell you here's what you have to do, to get good at this.
Because you have to think about it.
And it's always hard.
But to be sure, if you can just try when you review the statistical mechanics, especially, and you review the expressions that you use and how you solve problems with them, the next step in studying is to try to think, OK, do I really understand where that expression came from.
And why it makes sense physically.
And if you can do that, then so much the better.
Then you'll be in an even stronger position.
The info sheet gives a handful of equations that we do expect you to come in with those on your fingertips.
There are not very many.
Mostly we'll provide expressions that you'll need, and the important thing is that you know, understand, where they apply.
And how to use them.
So, good luck on Wednesday.
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OK.
So last time you started kinetics which is a completely different topic from thermodynamics.
They're related and we'll see a relationship at some point.
And you did first order kinetics.
And today we're going to move on and go ahead and do more complicated kinetics and hopefully get to some interesting stuff in a couple lectures.
Right now we just have to do the review of stuff that you've probably seen before.
And so we're going to go reasonably fast.
So you saw first order reactions.
Today we're going to do second order reactions.
At least begin with second order reactions.
Second order kinetics.
Of the form, and then there are two kinds.
There's first order, rather, second order, in one reactant of the form A goes to products.
And then you have first order in two reactants first order in two reactants.
So I'll have the form A plus B goes of products.
Where that's first order in A and first order in B, and this is second order in A.
So you can think of this as A plus A goes to products if you want.
And there's some rate constant k, associated with this reaction.
And we do the rate analysis.
We write the rate of this process.
Minus dA/dt.
And for the purpose of writing on the board, and you might want to do this also in your homework, when you're tired of writing.
I'm going to skip the brackets.
The little brackets that we usually put for concentration.
I'm going to skip those, because it's just too much work to write them.
And you can understand that A is a concentration of A.
So this is equal to k A squared.
Second order on A.
And the units for k, it's important to keep track of your units, at the end of the calculations, often you want to make sure your units work out.
So the units for k are going to be of this, A is in moles per liter.
The units for k, you're going to have to be able to match the units on this side here.
So the units for k are going to be liters squared per mole squared per second.
To make the units match.
Then you integrate this, on both sides, from zero to t, or from A0, the initial rate, to A.
Or from zero to t and this doesn't go like this.
So integrate with A0 to A.
You've got dA over A squared, you put all the A's on one side, all the t's on the other side, from zero to t dt with a minus k on this side here.
And then you get your rate equation, integrated rate equation for this.
Which gives you one over A is equal to kt plus one over A0.
So this gives you A as a function of time.
And this is a convenient way to write it, because it's linear in time.
So you always try to get things to be linear in time.
Because then you can plot them as a straight line.
Plot on this axis here you plot one over A. on this axis here you plot t as a function of time.
And then you get a straight line where the slope is k. Gives you the rate, and the intercept is one over A0.
Yes
[INAUDIBLE]
k is moles per liter squared per second.
You're right, liters per mole per second, yes.
That is correct.
because it has to work.
Otherwise it doesn't work.
So I need to turn on my brain.
OK, think.
Liters per mole per second.
Thank you.
And the other thing that you want to know is the half-life.
What is the half-life.
So you set 2 A0.
So A0 over two, you look for the time where you get to A0 over two, so you put A0 over two in here.
And that's going to be equal to k times t 1/2 plus one over A0.
So you solve for the half-life and you get a half-life of one over k A0.
So the half-life is inversely proportional to the amount of stuff you started out with.
Unlike the first order reaction, where the half-life was independent of the amount that you started out with.
So this was the easy one.
The next one is a little bit more complicated.
Which is when your first order in each of two reactants.
Then your rate, your differential rate equation, looks like this.
Because the k times A times B, and now you don't know what to do with B, a priori.
So you want to rewrite this equation a little bit differently in terms of the amount of A that's used up.
So you define A, x is equal to A0 minus A, this is the amount of A that's used up.
This is what you started out with, this is what you're left with.
And so the difference is what's being used up.
And dx/dt is minus dA/dt.
And by stoichiometry, what you've used up, of A, is also what you've used up of B.
Because for every A that reacts, you have to use up one mole of B.
For every mole of A that reacts, you use up one mole of B.
And so you also have x to B0 minus B.
So, you can plug this in here.
And get a differential equation which is purely in terms of one variable, which is x.
Right here, it looks like it's in terms of two variables, which makes it complicated.
By doing this change of variables, you see that A and B actually related.
Because of the reaction stoichiometry.
So you can rewrite that as dx/dt is equal to k times A0 minus x times B0 minus x.
And now you have a differential equation in one variable, which with some tricks you can solve.
So we want to have the integrated equation.
So we take an integral of both sides.
We put in all the x's on one side.
All the times on the other side.
We'll go from x equals zero to x, dx, A0 minus x times B0 minus x is equal to k from zero to t dt.
And now you have to dig back into the last time you took integration calculus, which for me was about two centuries ago.
And figure out how to do this integral here.
And the trick for doing this integral is to use partial fractions.
So you use partial fractions to do this integral here.
Which means that if you take this ratio, one over A0 minus x times B0 minus x and rewrite it as some number, n1 divided by A0 minus x, plus some number n2 divided by B0 minus x, you solve for n1 and n2, and you find that n1 here is equal to one over B0 minus A0 and n2 is minus one over B0 minus A0.
And so you plug, now, this in here.
And instead of having this complicated denominator, you have a sum of two integrals that you know how to do.
Because they're basically of the form one over x.
And in doing this, you also realize that you have to be careful because when A0 is equal to B0, when you have the same amount of A and B, then things blow up and you're in trouble.
So that's going to be a special case.
So always look out for special cases for these things.
So you assume, then, that B0 is not the same as A0.
And then you can go forward with solving it.
So you integrate, and at the end of the process, I'm not going to go through it, it's really complicated, you get something that looks like this.
A0 minus B0 log of A B0 over A0 B.
And you have your equation here.
Now, we don't really have a good way to plot it against one variable.
And the usual thing is look at specific cases.
And limiting cases.
And there's one limiting case we already brought up.
Which is when you started with the same amount of material.
What does it look like if you have the same amount of material to begin with?
So if you have A0 equal to B0, you start out with the same amount of stuff, but if you start out with the same amount of stuff, then this doesn't look so different from, at least mathematically, from this one right here.
If you start out with A0 is equal to B0, and for every mole of A that you use up you use a mole of B, then throughout the whole reaction, the concentration of A and the concentration of B are going to be the same.
So for the whole reaction, if you started with this, A is equal to B for all times.
That makes it easy.
Because now you can go back and instead of writing A times B, you can write A times A, which is A squared, which is what we had here.
And then you have the whole thing solved.
So in that case here you just have minus dA/dt is equal to k A squared, and one over A is equal to kt plus one over A0.
You don't even have to do any math, you just look at it.
So that's one case.
Another case is if one of the reactants is in much higher concentration than the other reactant.
And that's called flooding.
You basically flood the system with one reactant.
And that's something that we'll use again, hopefully by the end of class today.
Let's say that we take, so this is another limiting case.
Let's say we take A0 to be much bigger than B0.
So we flood the system with A0.
As a result, the concentration of A doesn't change very much in my pot.
It's hugely concentrated in A, there's a little bit of B around.
Again, with the process.
If I use all of the B up, the difference in A is going to be very small.
So at the end of the process, I'm basically still going to have A0 left in the pot.
So during the whole process, during the whole time period, I might as well assume that A is equal to A0.
And that makes my life much easier.
Because now if I write my differential equation in terms of B instead of A, so the rate of destruction of B, k A times B, instead of writing A here, it's pretty much constant for the whole time.
I'm just going to write A0.
So now, if k times A0 is a constant, and this looks awfully like a first order reaction.
So I can solve for it.
And I get that, so I can just write the answer because I've done this already.
I don't have to do it again.
The concentration of B then goes like B0 e to the minus k prime t, that's the first order reaction.
Where k prime here is this new rate constant, this new number, which is k, times A0.
So that's easy to solve also.
So always go to the limiting cases, because they tend to easy.
And if you were to go to the full solution and put in this limiting case, then you'd find that you get the right answer this way as well.
You can directly go to the easy way of doing it, or you can go through the whole process of solving it and putting the approximation in there.
And do the cancellations and get this.
But this is much easier.
Just writing the answer down is always much easier.
So it's a pseudo first order reaction.
We call this a pseudo first order reaction.
So we're done with the simple stuff now.
Any questions on first order and second order reactions?
So the next step is, you've got a reaction.
It could be a gas phase reaction, it could be a solution phase reaction.
There's some quantity, some property, of the solution that's going to change.
That's going to allow you to follow it as a function of time.
And that property could be many things.
It could be spectroscopic.
It could be that there's an absorption in the visible that changes as the concentration of one of your reactants changes.
Or one of the products could have an absorption band that you could follow in time.
Or you could using infrared spectroscopy to follow it in time.
Or if you have a reaction in the gas phase, and you have more products or less products than the reactants, and the pressure is going to change in time if you have a finite volume.
So there's usually some quantity that you can use to follow the reaction in time to extract out data.
And then from that data, that you want to know what are the kinetics of this reaction.
Because eventually you're going to try to find a mechanism.
You're going to try to find a mechanism that's consistent with the data.
So you get data.
Then you want extract out of the rate constants and orders.
So let's assume that you've found a way to get data.
And now you've got to analyze your data.
And suppose that you've found a way to measure the reactant concentration as a function of time.
And let's say that in the first case, the simplest case is that you have one reactant.
So you have one reactant, A.
So A goes to products.
And you've managed to extract A as a function of time.
Well, the obvious thing to do is to plot A versus time and see what it fits like.
So you take A versus time, and you plot it.
And you know that if you plot log A versus time and it's a straight line, well, that's going to be a first order process.
Plot log A versus time in a straight line, you know that's going to be first order.
If it doesn't go to a straight line you know it's not first order.
So then you go ahead and plot one over A versus time.
And if it's a straight line, you know it's second order.
And if it's not second order, it's not a straight line.
It's not a straight line, you look for some other order.
So that's one way to do it.
And you've got to have enough points on your graph, because if I were to plot a, let's say this is my data point, if I have a second order process, at the beginning it's going to look an awful lot like a first order process.
It's not until after a while that it's going to start to deviate.
So you've got to have enough points down in time to make sure that you can differentiate between a straight line and a line that's not straight.
And usually, that's often a mistake that experimentalists make.
They look at the beginning.
They say, oh, it's a straight line, work is done.
Go home.
But usually you need to have a good amount of the reactant consumed before you can tell the difference between first and second order.
So you'll have an opportunity to do this on the homework.
This kind of exercise of extracting the order of a simple reaction.
Another way to do it if you have a simple reaction is to look at half-lives.
That would be the half-life method.
If you can measure A as a function of time, then you know when you've gotten A over two.
So, you know that if I look at the half-life versus the concentration, the initial concentration, of my reactant, that tells me something about the order.
Because we saw that for our first order, t 1/2 was independent of the initial concentration.
And that for a second order, t 1/2 was proportional to one over the inverse of the initial concentration.
So if you plot t 1/2 versus A0, or have a few A0 versus t 1/2s, then you can tell the difference between first order and a second order reaction, and see which one fits.
Sometimes to get even more solid numbers, because from here you can also extract k. If you have a bunch of points, of t 1/2 versus A0, you can extract k, the rate constant.
Not just the order, but also the rate constant out of this data.
You can use multiple lifetimes if you have enough data.
Multiple lifetimes.
So you can define, you can define a t 3/4.
Which is the amount of time it takes for the concentration of A to be 1/4 of what you started out with.
So 3/4 is gone.
And then you can put that into your first order rate law.
So when you have log of A over A0 is equal to minus kt, you get that t 3/4, we can solve for t 3/4, and you get that that's equal to two log two, over k. And then you can do the same thing for a second order process.
So this is first order.
You plug in t 3/4 and A is equal to 1/4 A0, and you solve for t 3/4 for a second order process.
And you get that this is equal to three over A0 times k. So it's the same functional form as these two.
But there's a pre-factor that's different here.
So here there's a two that comes in there.
And here there's there's a three that comes in there.
And so there's an obvious way to tell, then, if you have the t 1/2 and the t 3/4 signs.
Basically, you follow, instead of having many reactions, instead of having many reactions to do, with different A0's here you can do one reaction.
If you do one reaction and you watch the reactant go away, and you time it.
When 1/2 of it is gone, that's one time, then you keep going, like 3/4 is gone, that's another time.
Then you can take the ratio of those two times.
Of t 3/4 versus t 1/2.
t 3/4 versus t 1/2.
And if it's a first order process, the ratio here is just two.
And if you take t 3/4 over t 1/2 and it's a second order process, the ratio is three.
So with one experiment, then, you can extract out the order.
You can't extract out, well, you can extract out the rate constant.
If you know the order then you know which equation fits, and you can extract out the the rate constant, with a big error bar.
You're always better off doing many multiple lifetimes of different A0's or many of these just to get more statistics in the result.
So this is the simple process.
And you always try, if you have something complicated, you always try to bring it back to a one component process by doing something like flooding.
So if you have five different reactants, if you make four of them in very large quantities and keep one of them in very small quantities, then all the four that are in large quantities are basically constant over the process.
And you basically are looking at only one reactant going away.
And then you can use these methods to figure out what the order is for that one reactant.
So, questions about simple one reactant sort of processes.
It's pretty straightforward.
So now, let's say we have more complicated reactions.
There are two ways that we can deal with that.
The first, I already mentioned, which is the flooding which we'll get back to.
And another way is called, so we have some complicated reactions.
Complex reactions, with multiple reactants, that's A plus B plus C goes to products.
And there could be some stoichiometry in front of there.
So one of the ways to deal with that is called the initial rate method.
We want to find out orders and rate constants.
Initial rate method.
So if I look at minus dA/dt, one of the reactants or the rate of the reaction near time t equals zero.
That's the initial rate.
Right as the reaction starts.
I mix everything together and I, just as after I mix it together, I watch the process of A disappearing or B disappearing or C disappearing.
And so in reality what I'm doing is minus delta A / delta t near t equals zero, where delta t is a small interval.
So there's not much change in delta A. Experimentally that's what I do.
And that's pretty much, essentially getting this number out.
And we're going to call that the initial rate.
And the initial rate is k, and at the beginning I haven't used up anything.
So all the initial concentrations are there A0 to the alpha, B0 to the beta, C0 to the gamma.
Et cetera if you have more reactants.
So you measure this.
You measure this R0, and then you repeat the same process with a new concentration of one of those three reactants.
So with A0 prime, let's say.
And then you get a new R0, R0 prime.
Then you take the ratios of these R0 and R0 primes, R0 divided by R0 prime.
So R0 is k A0 to the alpha, B0 to the beta, C0 to the gamma, then you have k A0 prime to the alpha.
B0 to the beta, C0 to the gamma, and the k's you disappear.
The B0's disappear.
The C0's disappear.
The only thing that we've changed is the concentration of A0 to begin with.
So the ratios of these initial rates is the ratios of the A0's to the alpha power, to the order in terms of A0.
So if you're clever about your choice of ratios, then you can get alpha pretty easily.
So if you choose, so if you now choose A0 prime to be equal to 1/2 A0, then you measure R0 over R0 prime.
And if you get one, then you know that that alpha's zero.
Alpha has to be zero here.
Then you know alpha is zero, that gives you the order.
It's a zero order reaction.
If you get that R0 over R0 prime is square root of two, or rather 1/2, then a square root of two, square root of 2, then you know that alpha is 1/2.
And that's a half order.
We haven't seen any half order reactions yet, but we will.
Those are indicative of a complicated mechanisms.
But that's what you would get out of this experiment.
If you get that it's equal to two, if you get that this ratio is equal to two, then you know that alpha is equal to one, et cetera.
So it's a pretty easy way to get the order.
Then you repeat the experiment.
Now instead of changing A0, you keep A0 constant and you change B0.
Or, you can use flooding or isolation, which is the next process to get the other orders.
And eventually you get the rate constant.
Once you have all the orders, and you have R0, then you have the rate constant.
OK, so that's one way of doing it.
And a second way of doing it is the way we already mentioned.
To solve the second order reaction in two components.
Which is to flood the reaction with everything except for one.
So this is called flooding or isolation.
Basically, you isolate one reactant and watch it.
You're trying to get back to a system, which is a simple system.
Which is a system of one reaction.
So let's say you take A0 to be much smaller than all the other species.
You flood with B and C, and you isolate A0.
And then your rate minus dA/dt, it's going to be A to the alpha.
And instead of B, well, during that process B's going to say pretty much constant.
Because it's hugely concentrated in B.
You can replace B with B0.
You can replace C with C0.
So now you have an effective constant, an effective rate constant, and then you have a process which is effectively, or pseudo, one reactant.
Then you can use these methods here, you can plot A versus time, you can find path lines, et cetera, to gather the order for it.
Then you can get alpha and k prime.
And if you can change B0 and C0, then you get k out of this.
So this is basically all fairly straightforward, just tedious experimentation to get all these numbers out.
OK, any questions on this?
You'll get experience on the homework.
There's likely to be a question on the final where you're given data and asked to extract out orders and rate constants.
So let's move on now.
So, so far we've looked at first and second order elementary processes, and we've looked at taking data and extracting out rate and rate constant.
And the next step is to build mechanisms.
So a mechanism is when you take a complicated reaction, like A plus B plus C goes to D plus E. And you break it up into elementary steps.
What's an elementary step?
An elementary step is a step which happens in a single reaction.
So I could hypothesize that this complicated reaction happens in three steps, where I need to have a molecule of A and a molecule of B collide with each other to first form an intermediate F. Then I want a molecule of F plus a molecule of B to collide to form intermediate G plus a product D. Then have the intermediate G plus reactant C collide together to form the product E. So this set of elementary steps, where at each step you have a collision of two or three molecules together, three is not so common but two is very common, those elementary steps are called the steps of the mechanisms.
And these elementary steps you can define something called molecularity, which is the number of species that you need to collide with each other in one of these elementary steps.
So the molecularity here would be two, you need two molecules to react.
Here it's two, here it's two.
If I have an elementary step which is a zero order in one reactant, then the molecularity would be one.
Or I could have A plus A, the same molecules have to collide with each other.
Molecularity would be two.
And the molecularity and the order of the reaction are connected.
So if you have something which is a molecularity of one, then it's going to be a first order reaction.
One reactant is just sitting by itself, falls apart, like in radioactive decay.
Molecularity of one, that's a first order of process.
If I need to have two molecules come together, then it's a second order process.
If I have to have three molecules collide at the same time together, molecularity of three, then it's going to depend on the concentration of all three at the same time.
That's called a ternary reaction, and those are really quite rare.
Termolecular reactions, you need to have your concentrations very, very high to statistically get an event happening where all three molecules collide together.
So three body reaction is hard and anything higher than three body is essentially impossible.
So that limits your choices, which is nice.
So that's the mechanism.
And so what we're going to do next is go through some mechanisms.
Some simple mechanisms and build up the complexity.
Any questions about mechanisms here?
What we're doing here is, we're formulating a framework.
Where we can go back and look at things that are more complicated, like chain reactions or explosions or enzymatic reactions, and know when to apply approximations and et cetera.
So we basically, here, are just laying down the ground rules.
So let's go to our first example of a mechanism, a more complicated reaction.
And what we're going to do is we're going to extract out integrated rate laws out of all these mechanisms.
And see what it looks like, as a function of time.
So the first one we're going to do is called parallel reactions.
Simple mechanism.
In this case here I have one reactant, and that reactant has a choice.
You can think of it as a radioactive decay, an atom decaying in two different channels.
So it can decay into B, or it can decay into C. There are two rate constants, k1 and k2.
So you can write it like this, or you can write it as A goes to B plus C. This is how you would write a reaction.
And this is how you would write your mechanism.
A goes to B and A goes to C. Each elementary step, these are the elementary steps and this is the complex reaction, each elementary step is unimolecular.
It's a first order process.
So in all of these examples, the first thing you do is you write your rate law.
The rate at which A gets created or destroyed.
And there are two paths.
It gets destroyed and into B, with a rate which is proportional to the concentration of A, and it gets destroyed into C, proportional to the concentration of A.
So you write down all the ways that A can get destroyed.
There are two ways here.
Two channels.
This one happens to be fairly easy to solve.
k1 plus k2 times A, and you've seen this before.
It's minus dA/dt as a constant times A.
That's a first order process.
So you can just write down the answer.
You don't need to do any math here.
You recognize that we just call this one k prime.
And that the rate is A as a function of time.
This is A0 e to the minus k1 plus k2 times t. And everything you've learned about plotting first order processes et cetera, is applicable here, where the rate constant is the sum of these two.
So that's for the reactant.
The products are also interesting to plot as a function of time, to see how they are related to each other in terms of their concentrations.
So let me go through this also.
When things get more complicated we'll quickly go and make approximations.
But for now, we can still do everything exactly.
So you write down your rate law for the product.
dB/dt is equal to k1 A. dC/dt is equal to k2 times A.
The formation of B depends linearly on A.
The formation of C depends linearly on A, because they're both first order processes.
To make B and C. And you integrate.
You integrate here from zero to B, dB.
Is equal from zero to t, A dt, k1.
A is a function of time.
And we've already solved for that.
It's this exponential up there.
So you plug in here A of time.
And you turn the crank and you integrate, and it's an exponential.
So it's not so hard to integrate.
And you get that B is a function of time is k1 times A0 over k1 plus k2 times one minus e to the minus k1 plus k2 times the time.
Things are already starting to get a little bit more messy in the math.
And then to get C, you actually don't need to do anything.
Because you notice that the only difference between B and C here is replacing k2 with k1.
So don't worry about doing any math.
Just write down the answer.
k2 A0, you interchange k1 and k2 at every step.
One minus e to the minus k1 plus k2 times the time.
The only difference is up here in the k2 term.
And those are your equations for k1 and k2.
And what you find, is the ratio of B to C is a constant.
If I divide B by C, everything cancels out except for the k1 and the k2 here.
Is equal to k1 over k2.
And that is called the branching ratio.
The branching ratio, because there are two branches out of the reactions.
And this gives you the ratio of which one is more likely to happen than the other one.
And so if k1 is much bigger than k2, the rate is per unit time.
The units of k1 are per second or per minute or per hour.
So if this is big, if k1 is big, then mostly you're going from A to B, and only a little bit of C is formed.
And the ratio of B and C is always constant.
And so you can plot, then, you can plot the result.
You can sketch out the result.
So you know that A is going to come down exponentially.
Time on this axis here, concentrations on this axis here.
So this is A as a function of time.
That's that equation up here, in exponential decay.
And the quantity of A.
And B and C are going to come up in time, also, with this exponential format here.
B is going to saturate at this ratio right here, k1 A0 divided by k1 plus k2.
C is going to saturate at this quantity here.
So they're going to start both at zero.
And B is eventually going to go to k1 A0 over k1, plus k2.
And C, eventually, will go to k2 A0 over k1 plus k2.
And this, and the ratio of these two lines at every point is k1 over k2.
So, a simple question, for instance, that you might be of the type that you might be asked to look at is, suppose that k1 is 1/10 of k2.
Which would you expect?
Would you expect to have, let's see.
So C is in green here.
C, B.
Or do you expect, so A comes down.
B comes up.
C comes up like this, or do you expect the last choice, I'm going to put the last choice here, so this is, let's call this choice number one.
Choice number two.
Choice number three.
So k1, the rate k1 is 1/10 of the rate k2.
That tells you something about the branching ratio.
So do you expect this one here to be the right one?
How many people think this is the right one?
What about this one here?
How many people think this is the right one?
One person.
What about this one here?
So the branching ratio is the ratio of the two.
It's 1/10.
So this is approximately, in my sketch, poor sketch, granted, but this is approximately 10 times bigger than this.
So that's the ratio that you'd expect.
And it's the right, here k1 is the rate into B.
It's slower than the rate into C. So, you got it right.
So for more complicated, we have a more complicated process, we're going to ask you the same sort of stuff.
And it won't be as straightforward.
Any questions on this beginning here?
Next time we're going to finish with the parallel first and second order processes.
And hopefully we'll get done with the complex reactions and mechanisms and move on to, I forgot what's next on the list.
But some explosions or chain reactions.
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So, any questions from last time?
We started doing reaction mechanisms.
We did first order parallel reactions.
So one of the things that we'll be stressing is that the mechanisms that we write down are ways to try to understand a complicated chemical process.
And you take data, you can find intermediates sometimes.
And you get rate laws from the data.
You infer rate laws.
And from these rate laws, you think up of a reaction mechanism and you make sure that what you measure is consistent with your mechanism that you make up to explain the complicated chemical process.
And one really important thing to remember is that when you come up with a mechanism and you've got a bunch of data that supports it, it's great.
But it doesn't mean that you've solve the problem.
It doesn't mean that you've proven that the way that the chemistry proceeds is really the way it proceeds.
Because there's always the possibility that there's an intermediate somewhere in there that you haven't been able to measure.
And a lot of people got tripped up over the course of the last century, of making up mechanisms, getting a lot of data that supports the mechanism and missing something really important.
So it's a common mistake, that people think they've proven a particular chemical mechanism just because they have data that supports it.
It's up to a certain point that you know what you have.
And we'll see examples of that today, probably.
And then certainly next time when we start making approximations in our kinetics.
And see where things can really go wrong if you don't have enough data.
If you don't try to fish out really, really, tiny intermediates somewhere along the way.
So last time we did then parallel first order reactions.
And today we're going to march through and do parallel first and second order.
So one is first order, the other one is second order.
Things are going to get more complicated.
And what you're going to get is a flavor of how to solve the problems.
Setting up the problems.
We're not going to do a lot of the algebra here on the board, because otherwise we would spend the next three weeks doing algebra.
And that would be no fun at all.
That doesn't mean that you're not going to be doing any algebra on the homework.
Because part of it is figuring how to do it.
So, we have a reaction where A goes to B plus C. And the first thing we did was, the first mechanism we wrote was a parallel mechanism where we had A goes B and A goes to C. That's the mechanism.
And one other way to write it is A goes to B or C, in a way that is more, sort of describes the branching process, that when we talk about a branching ratio, the ratio of amount of B to C, this looks like the branching out of two different paths.
And this time we're going to do where this is first order and this is second order.
With the rate k1 and a rate k2.
And the way that you do all these problems, you have to be very systematic about it.
The first thing you do is you have to write down all your rate laws.
And make sure that you include everything.
So you start by writing the rate law for the destruction of A.
And then the rate law for the appearance of B and for C. So for a, we have dA/dt.
And again, I'm dropping the brackets around the A, because I don't want to carry around all these extra symbols.
So we destroy A by making B.
So that's going to be first order.
And then we also destroy A by making C, and that's second order.
So we have two paths out of A.
To get rid of A.
Then we write down the appearance of B, dB/dt.
It's only through one path, first order in A, and then the appearance of C, only one path which is second order.
And these are all our differential equations.
And now the trick is, how do you solve these three differential equations in a way that you get meaningful information out.
Well, the first thing to do is to, so these two here depend on A.
So the appearance of B depends on A, the appearance of C depends on A.
But this one here only depends on A by itself.
So this is the first one you're going to start out with.
Because you can put all of the A's on one side and all the time on the other side and integrate.
So then you solve, and you can rewrite this as minus dA/dt is equal to k1 times A times one plus k2 over k1 times A.
And then you put all the A's on one side, all the t's on the other side.
And integrate from A0 to A minus dA over A times one plus k2 over k1 times A.
That is equal to k1 from zero to t dt.
And then you have to solve this integral here.
And there's a reason why I factored out the A here.
It's because this is of the form you can use to use the trick of partial fractions, which we mentioned this before.
So you use partial fractions to solve this.
And I'm not going to do it on the board.
You turn the crank, you basically write one over A times one plus k2 over k1 A is equal to n1 of A plus n2 over this part here, and you solve for n1 and n2 and you plug it back in, and you redo your integral, et cetera.
And you get your answer.
Which I'm going to write down because I'm going to use it later.
k1 A0 over e to the k1 times t, k1 plus k2 A0.
So one of the things we immediately see is even though this mechanism isn't very complicated, just two paths, one first order, one second order, the solutions start to get not so simple pretty quickly.
So it depends, there's an exponential on the bottom here.
Depends on the initial concentrations and both rate constants are in there.
The first thing you want to do is you want to look at limiting cases.
Just like before.
We did, and the first limiting case we're going to look at, so interesting to look at is this guy right here.
k1 and k2 A here.
If one is bigger than the other, then things will cancel out.
And we can get some intuition.
So the first limiting case we're going to look at is where k2 A0, this term right here, is much smaller than k1.
Now, we want to make sure that in these limiting cases that when I write something like this, that the units actually make sense.
That I'm not saying, three apples are less than four oranges.
So k1, it's first order, so the units are one over second.
k2, it's a second order rate constant, so the units are one over second molar, times moles per liter.
So the moles per liters cancel out and I have one over our second is less than one over second, so things are great.
It I'm making the right kind of approximation here.
What else is this approximation saying?
This is saying that the rate into B, so this one is the faster one.
k1, the rate into B, is the rate into, is k1, is faster than k2 times A, which is basically the rate into C. So this is saying that the rate into B, to form B, is faster than into C. So we can write down a sketch.
We can sketch what we expect this approximation to be, then.
And we know it's going to look something like this.
We know that A is going to come down, exponentially.
Without any structure to it.
It's going to go either into B or into C, in some branching fractions.
And we know that the rate into B is faster than into C, so B is going to come up like this.
And C is going to come up, but slower.
And there's going to be a ratio between these two guys.
So we know the slope here is going to be slower then the slope for B.
And we can check that out, within our approximation.
We can write, at t equals zero, find out what these initial slopes are.
For the creation of B and C. So dB/dt is the slope of the creation of B at the beginning.
At t equals zero.
And that's just the rate law.
dB/dt at t equals zero is k1 A0.
And dC/dt, the initial slope, t equals zero, is k1 A0 squared.
Or k2 A0 squared, rather.
So I'm going to write it as k2 A0 squared, like this.
We said k1 is much bigger than k2 times A0.
There's the same A0 here.
So we see, just by writing this down, that our intuition that because the rate into B is much faster than into A, that it should look like this.
Well it's borne out just like this very simple sort of looking at the initial rate where the slope here is much smaller than the slope here.
And then we can take the approximation in here.
And see what it implies.
So let me do that here, let me just use my green chalk here.
So k2 A0 is much less than k1.
So this is going to go away.
k2 A0 is much less than k1.
Well, this is building up from zero.
So this is always bigger than one here.
So I can get rid of this as well.
Because there's the k1 sitting here.
And k1 is much bigger than k2 A0.
Once I got rid of all these two guys, then the k1's cancel out.
And this is looking very nice.
Because now I can take that e to the k1 t and put it upstairs.
A is equal to approximately A0 e to the minus k1 times time in this approximation.
It looks like it's first order coming down, with a rate constant k1.
It looks like basically the branching into C is nonexistent as far as, at least in this approximation, as far A is concerned.
It's going to look like this thing here.
It's going to look like first order A goes to B with rate constant k1.
So if you didn't know, if you didn't know that there was another substance, C, that was being formed along the way, if you didn't measure it, if you didn't do the analytical chemistry and sort of fish it out, and you just looked at the two major components, A and B, in this case here, you'd measure a rate out of A, it would look like it's first order.
It's great.
Got my mechanism.
A goes to B, period.
And you would know there was a minor component that was being formed at the same time.
OK, let's do another approximation.
Let's do the other case, where k2 A0 is much bigger than k1.
And, in this case here, you have to be a little bit more careful.
You have to add that you're going to look at this at early times.
And we're going to see why early times is important here.
In just five minutes.
And early times, and how you define early times.
What does it mean for things to be close to time equals zero?
Well, we have to have a reference time scale.
The reactions, I'm going at a certain rate.
And one second may be very slow.
Or it may be very fast.
Depending on the rate of the reaction.
So k1 here defines the time scale of the problem.
It's one over second, unit of one over second.
And early times means that the times that I'm looking at compare to this rate, it's very small.
So k1 t is less than one.
That means that the time, during the time period that I'm looking at, hardly anything has happened to the branching of A to B.
That that's what I mean by early time.
So B is hardly being built up yet.
k1 is our reference time.
k1 is units of one over second.
This has units of seconds.
So it's all working out fine.
In other words, I would have no idea how to define early times.
I could say, it's in a year, it could be, if you're looking at plutonium decay, it's a 100,000 year half-life.
A year would be an early time.
So it would be really fast, compared to the process.
But if you're looking at a reaction that takes a few nanoseconds than a year would be very, very long.
So you really need your reference time somewhere in there.
So this approximation means that essentially, no B is being created.
While we're looking at the process.
And so we might then very well expect that if we look at A and C, that the answer is going to be that we're going to see something that's second order.
So we expect that the form of A, the way to write it is going to be one over A equals k1 t plus A0.
So let's go ahead and take our complete solution and write it in the way that we might expect the answer to be.
In terms of one over A rather than A here.
So let's just invert it.
e to the k1 times time, k1 plus k2 A0, minus k2 A0 over k1 A0.
This should be a minus sign here.
There should be a minus sign here.
It didn't matter for our approximation because we got rid of it anyway, but it would be a proper sign here is minus.
And so now we need to make our approximation and figure out what this here.
So if we take the approximation that k1 t is small, less than one, then what that should trigger in your mind if that if you have e to the something that small you should use a Taylor series.
And expand this into a Taylor series.
So we're going to see this multiple times.
So we take e to the k1 t and expand it out as one plus k1 t plus et cetera.
We're going to keep the first order terms in there.
So we're going to plug this approximation into here.
That's this approximation there.
And then we're going to expand it out.
So plug it in here, and, keeping it to first order in time, and multiplying it out, and I'm not going to do the intermediate steps, I'm just going to give you the stuff after doing the multiplications, we get k1 t squared.
So that's basically k1 times one plus k1 t, and then you have plus k1 k2 A0 times t. That's this term here.
k1 k2 A0 times t times that.
And the k2 A0 times one gets subtracted out from this minus k2 A0 here.
So that's all we have on top.
And then there are terms of higher order in time.
But we're not going to worry about this.
We're going to stick to first order in time.
Divided by k1 A0.
And now you do your cancellations.
We have our approximation at k1 times time is much less than one.
So here we have k1 times k1 times time.
So this term here means it's much less than this term here, because of our approximations.
So we can get rid of this term here.
Get rid of that.
There's no reason for us to get rid of this one here.
Because we've got k2 A0 there, which is much larger than k1.
And so when you expand this out, all the k1's cancel out here.
So so k1 cancels out that k1, we have one plus one.
So this is basically one over A0.
And then we have plus k2 A0 time over A0, so this ends up being equal to one over A0 plus k2 times time.
Exactly what we expected.
That if you put in the right approximation in the math, it behaves as your intuition would tell you.
Which is that the branching into B nonexistent.
And that you're looking at everything going to C. It looks second order in C. But in this case here, it's only early times.
So if you keep going, what happens?
If you keep going in time, beyond the time scale of this first order rate constant, and you wait long enough, then the amount of A keeps decreasing.
Keeps getting smaller and smaller.
So if you start the clock again, a little bit later, where A is much smaller, then k2 times the amount of A you have there, is no longer going to be much greater than k1.
At some point along the way, k2 times A is going to be much less than k1.
So at some point along the way, it's not going to be this approximation any more.
It's going to be the previous one that we did.
The one up there.
So at early times we start with something that may look second order for A.
Time goes on, time goes on, time goes on, A gets depleted.
A gets depleted, and the rate k2 times A, which is sitting right here, this part here becomes smaller and smaller.
And pretty soon this part wins.
And it becomes first order.
So what you'd expect to see, then, is if you were to plot as a function of A, you expect to see something that starts out as first order and then as A gets depleted, switches over to second order.
Something that's second order here.
And first order.
This is the concentration of A.
So if you were to plot it, here you're plotting the concentration of A.
If you were to plot the log, of the concentration of A, at long times you'd expect it to be first order.
So something linear.
But at early times you'd expect it to be second order.
So you'd expect to see something that's nonlinear, then switching to something linear.
And if you were to plot it as one over A, then you'd expect to start off with something that's linear, at second order, early times.
But instead of keeping, being second order, as you deplete the amount of A, the branching into B becomes important.
And you become second order.
So this is second order at the beginning.
And this becomes first order.
So, any questions here?
The importance of doing this problem here was to lay out basically a systematic way of looking at the problem.
You lay out your rate equations.
Like this.
You solve what you can, in this case you solve for A.
And it's complicated.
So you want to learn something about what the data might look like.
And you look at limiting cases.
Two obvious limiting cases.
One rate is faster than the other.
You pick the first one first, then go through it.
And then use your intuition.
Every point along the way, your intuition can tell you what you expect the result to be.
So in this case here, our intuition told us that if we took the rate into B to be much faster than that into C, than we expected to see something that was largely first order.
And when you we the approximation in our exact solution, in fact it does look like it's first order.
So always use your intuition.
Because the math is going to get pretty hairy.
The algebra might make it complicated.
And it's really easy to make a plus sign into a minus sign somewhere along the way.
Just like I did here.
So if I hadn't fixed my mistake here, and I had gone and put in my Taylor series here, I would have gotten in trouble.
Because this minus sign would have been a plus.
And these k2 A0's would not have canceled out.
And I wouldn't have gotten the right result, which my intuition had told me should be a second order process.
And so if you have a problem, let's say you're on an exam and you're doing algebra, you're trying to, and you end up with a result that doesn't match what your intuition tells you, and you don't have time to fix it.
Tell us.
You know, I know this is wrong because I know it's supposed to be second order.
But I don't know where I went wrong.
And that's really important.
Just tell us.
That means that you're thinking.
Let's get a little bit more complicated.
Any questions?
Consecutive series reactions.
These were parallel reactions and now.
Basically what we're doing here is we're building up a toolkit of simple mechanisms.
And then we'll be able to put these mechanisms together to make something more complicated.
So the next kind of mechanism, series mechanism, series reactions, so we have our reaction.
Which is A goes to C. And the mechanism that's been thought for this reaction is that there's an intermediate.
That first you have A goes to B, some intermediate, which gets used up to form the final product, C, with a rate, k2, here.
And we can either write the mechanism in two steps.
Or we can write it as A goes to B goes to C, like this.
And see both ways of writing it.
And we want to solve this.
And we want to look at special cases, and we just want to understand the implications of this mechanism.
So the first thing to do, just like we did before, is to write all the rate laws.
Before we got going solving anything.
So we start with A, minus dA/dt.
There's only one way that gets used up.
Through the formation of B in a first order process.
Easy.
This is easy to solve.
We know the answer is going to be exponential in it.
For B, dB/dt is equal to, well it can be formed through my first order in A, so that's k1 times A.
But it also could get destroyed by forming C. So we've got to get all the paths out of A, here on the right side of this equation.
So it can be destroyed in a process, which is first order in B.
So we're going to take both k1 and k2 to be first order initially.
Then we'll make one of them second order and things will become very complicated and we'll throw up our hands.
But for now, let's not throw up our hands quite yet.
OK and for dC/dt, there's only one way it can be formed is first order by destroying B.
And we have a couple of differential equations here.
This differential equation for C, depends on B here.
The differential for B depends on A here.
So that means that things are going to get a little bit more complicated.
The easy one to solve is always the first one here.
That's first order.
So we just write down the answer.
A is equal to A0, e to the minus k1 t. We know the answer to that one.
Then we have to work on B.
So we want to find out integrated rate laws for every one of these chemical species.
And always there are tricks involved.
Partial fractions, whatever.
So we write down B now.
dB/dt plus, let me rearrange it a little bit, plus k2 times B, putting the minus k2 B on the other side here.
Is equal to k1 times A.
But I'm not going to keep this A there, because now I know what A is in terms of time.
And a constant.
k1 times A0 e to the minus k1 times time.
OK, that's my differential equation for B.
Got to solve this thing.
Well, the last time I did differential equations in college was a century ago.
And I wasn't kidding.
It really was last century.
In fact, it was last millennium.
That was a long time ago.
So how do you do this?
Well, the trick here, there's a trick.
And the trick is to multiply the whole thing, both sides, by e to the k2 t. And you multiply this side by e to the k2 t and you multiply this side by e to the k2 t, to solve this equation.
So once you do that, then you have e to the k2 t times dB/dt plus k2 times B.
This side here.
And then you have is equal to k1 times A0 e to the minus, e to the k2 minus k1 times time.
OK, so the reason why this trick works is because this guy right here can be nicely written as d/dt of B times e to the k2 t. It's in nice simple form, so when you integrate, it'll be very easy to integrate.
And then the other side, you still have k1 A0 e to the k2 minus k1 times the time.
So now you integrate both sides with respect to time, The integral of something d/dt with respect to time, it's just going to be itself right here.
So it's very simple.
So you integrate both sides from zero to time, dt.
Zero to t, dt on both sides.
And on this side here, you end up with B, e to the k2 times time minus B, e to the k2 times zero.
Just taking the upper limit minus the lower limit.
And that's just equal to B0, at time to equal zero.
And so, and at time to equal zero, B0 is equal to zero because we haven't made any intermediates.
So this goes away.
And then we have an equals sign.
And on the other side, we have k1 A0 over k2 minus k1, it's just the integral of this thing, here with a k2 minus k1 goes in the denominator.
e to the k2 minus k1 times time minus one.
So now you have your integral form for the amount of B that gets created at any time.
And I want to keep that on the board.
Because I'm going to use it later.
k1 A0 over k1 minus k2 times e to the minus k1 times time minus e to the minus k2 times time.
OK, we're two thirds of the way through.
Now we've got to find C. And for the last component, you don't want to be doing this kind of stuff.
You don't want to waste your time solving differential equations, because stoichiometry can help you out.
At the end of the day, you know that everything that started out as A, in terms of quantities, is going to end up at C. You know that if you end up with a concentration A0 at the beginning, at the end of the process the concentration of C, at infinite time, is also going to be A0.
If you know that there's a correlation, there's a stoichiometry that relates A and C together.
And B.
You have conservation of mass here, basically.
Conservation of atoms.
That has to come into play So, for C, we're just going to do algebra instead of calculus.
So let's write down the stoichiometry here.
The amount of C at any time is what it's going to be at the end.
Which is A0, very end of the process.
It's going to be A0.
But before we get to the end, there's still stuff that's left in A and B.
And that's going to decrease the amount of C. So minus whatever's still in A and B.
That's the stoichiometry.
That's one way of writing it.
You could also write it in a different way.
Which is that the amount of C is equal to the amount of A that you have used up.
A0 is what you started out with.
A is what's left over.
So this is the amount of A used up.
It's not all quite in C yet.
Some of it is stuck in B.
So it's the amount of A used up, minus the amount that's stuck in B.
Those two are the same.
I hope.
A0 minus A, yeah, that's right.
Minus B, yeah, that's right.
So this is used up.
And then stuck in B.
That's the hard part.
Putting down the stoichiometry.
Once you've done the hard part, then if you solve the other two then it's easy.
You just plug in.
It's just algebra.
And you get something complicated.
C is equal to complicated.
It's in the notes, I'm not going to write it down.
So we've solved the problem exactly.
And now you're going to do an experiment.
And the experiment you're going to do is not going to follow this whole thing, exactly.
It's going to look at limited cases.
That's what you can do.
And so the first thing to look at is to look at the initial times.
The very beginning of the process, how are things appearing.
And then we'll look at the late times.
So initial times.
Near t equals zero.
Well, every one of these things has an exponential in it.
And whenever we see exponentials and we look at something at the early times, times that are faster than one over k1, or faster than one over k2, it tells us use a Taylor series.
So use the approximation e to the minus kt is approximately equal to one minus kt plus kt squared over two plus et cetera.
And keep as many terms as you need in the Taylor series to get something sensible.
So if everything comes out of zero at the end, you know you haven't kept enough terms in t. Things have canceled out too much.
And you want something that is time dependent.
Because you've got a change in time, right?
It's a very common mistake to say only keep one here and then everything cancels out you get zero, and you say, well, B is equal to zero at all times.
That's nonsensical.
And I don't know a priori how much time to keep.
So let's go to second order.
So if you plug this Taylor series into A, and get something that's time dependent, you only need to go to first order.
One minus k1 times time plus et cetera.
Plug it into b, plug your Taylor series into these two guys here, get k1 A0 over k2 minus k1, you only need it to go to first order.
The one's cancel out but the times don't cancel out.
Minus k1 plus k2, that is, unless k1 and k2 are the same.
So we're making the approximation that k1 is different than k2.
Times the time.
And when we look at C, this complicated thing that I didn't write down, if you were just to stop at first order in C, you get C equals zero.
You know that can't be right.
You know that C is going to increase in time.
It can't be equal to zero at all times.
So we have to go to second order for C0.
And it's approximately equal to A0 times k1 k2 over two times the time squared.
Let me rewrite this a little bit closer in, because I need some room here.
So c is approximately equal to A0 k1 is k2 over two times time squared.
OK, and then we can start plotting out what this is going to look like then.
So we have the concentrations time on the axis here.
Let's start with the concentrating of A.
A is dropping down linearly in time.
It's dropping down linearly in time here.
B is increasing linearly in time, at the beginning.
So B is going up linearly in time, like this.
And C is increasing quadratically in time.
So it's pretty flat, coming up first.
And then it starts to curve up.
It's our first approximation.
Then, we go to late times.
t equals infinity.
So t equals infinity where am I going to write this?
Have to move this up.
Let's go, t goes to infinity.
Well, here I know that I'm allowed to go to zero.
Or allowed to go to constant, because infinity is forever.
So at t infinity, I know that I'm going to use up all of A.
So I know that A is going to be equal to 0.
I know I'm going to use up all of B.
So I know B is going to go to zero.
Somehow.
And I know that C is going to go to A0.
So C is going to go to A0 at infinity.
So now on my graph here, I know that at infinity, A is going to go down to zero like this, somehow.
Probably exponentially, because it is exponential in time.
So it's going to go to zero exponentially.
B is going to go down to zero exponentially.
There are the exponentials right here.
So B is going to go down to zero exponentially.
Something like this.
And C is going to saturate to A0.
Exponentially rise up to A0.
So this was A0 right here.
It's going to go like this.
So then I connect the dots.
And I know that C is going to start quadratically.
And then it's going to rise up an exponent like this.
I know that B is going to start off linearly, and then it has to go to zero.
So it has to go through some sort of maximum.
And A is going to go down exponentially, exponentially like this.
So the interesting part of this diagram is that B goes to a maximum.
There is some point, there's some time, that we can call t sub B max, where B is a maxima.
And that's an interesting point.
That's something that we can try to figure out experimentally what it is.
And we could get some parameters, we could maybe extract some information.
There's a maximum B, concentration of B, that gets formed.
And how do you solve that?
You set dB/dt equal to zero.
So you have your equation for B right here.
You set it equal to zero.
and you solve.
You get your maximum time.
It's made up of rate constant.
You get your maximum concentration, which is also made up of rate constants.
And you can get rate constants out of this.
Question
[INAUDIBLE]
Yes
[INAUDIBLE]
That is a coincidence.
That is my poor sketching.
It should probably be somewhere like, in the middle here.
Actually, it completely depends on the rates.
And we're going to see limiting cases where this maximum can occur here, or it can occur somewhere down here.
So this is the complete case.
So, limiting cases.
the important limiting cases.
We've made an approximation along the way that k1 is different than k2.
So that has to be a case that needs to be worked out.
And it's likely to be easier than the full solution.
So one of the limiting cases is k1 is equal to k2 and I'll leave that for homework.
You can do that yourselves.
It's good exercise.
What about another limiting case?
Well, you have two other limiting cases.
One, you've got two rates.
So, one is bigger than the other.
No information on this board here.
You start off with one case.
So k1 greater than k2.
k1 greater than k2.
What does it mean?
k1 greater than k2.
It means that the rate determining step, or the limiting step, in this reaction, is the second step.
k2 is very slow compared to k1.
And sometimes I like to think of these things as pipes.
And connecting vessels.
So let's say we have the amount of liquid in the top vessel is the concentration of A.
The amount of liquid in the middle vessel, let me get my color scheme to be consistent here.
Then we have a middle vessel, which is the concentration of B, and a final vessel, which is the concentration of C. And we start out with some sort of amount of A in here.
Started with a lot of A, and then these vessels are connecting with pipes.
So there's a very thick, big pipe that connects A and B together.
The rate is fast, going from A to B.
And the rate going from B to C is slow, so there's a skinny pipe connecting B and C together.
Then I turn on the system, I set time to equal zero, poof, what happens?
There's the big pipe connecting A and B.
The whole amount of a here gets transferred to B, suddenly.
Then it gets stuck here.
And then it dribbles out from B to C. So what do we expect, if we were to plot our quantities as a function of time.
What we'd expect then, then we're going to make sure that the math works out, is that A is going to decrease really fast.
It's going to go, poof.
First order in time with a very fast rate.
And all of this is going to go straight into B.
So we expect B to go linearly.
Remember, at the beginning it's linear.
Linear goes up.
Almost all the way up to A0.
Because it can't hardly get out.
Reaches its maximum, then it goes from B to C through this skinny, skinny pipe.
First order rate, with a rate k2, so we're exponentially going down, slowly, slowly, slowly, with a rate k2.
And then C starts quadratically.
And then very quickly, once everything's into B, A is forgotten.
A doesn't matter any more.
Basically what you're looking at is a transfer of B into C through this little pipe.
And you know that's going to be a first order of rate.
First order of process, B to C's first order of process with rate k2.
So you expect, then, C to go up in the first order process to A0 with rate k2.
There's the rate k2 here.
And the rate k1 here.
So when you turn the crank on your approximation in the math, what you'd better see is that at the end, B as a function of time should look like a first order process with rate k2.
C as a function of time should look like a first order process with rate k2, going to A0.
And you can write the answer, you should be able to write the answer, for C, it's going to be C is approximately, it's going to go to A0, there's no other choice.
Everything that was in A gets transferred to B0, so at time is infinity it's going to be A0, and it's going to be a first order process with rate k2.
One minus e to the minus k2 times times.
I don't even have to do the math.
I know that's the answer.
If you don't believe me, then do the math.
You should do it.
But that's the answer.
It has to be the answer.
And if you look at B, the answer has to be that it's approximately looking like B is disappearing with a rate constant that's k2, e to the minus k2, and the maximum here is very close to A0.
So I'm going to be putting A0 here.
Close enough.
So as long as I ignore the very initial time where A suddenly dumps into B, then everything after that very early time here is just looking like B goes to C. This is not zero, this is t. And A just disappears quickly.
And we can write it as A is equal to A0 e to the minus k1 times the time.
That's an easy approximation.
And you've just got to make sure that this fits.
So the other limiting case, which I'm also going to leave as a homework, because it's so straightforward, is to have k2 greater than k1, and you should make sure that you can predict it.
and that the math works out.
OK, any questions?
Next time we'll do reversible reactions.
And some more approximations.
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So last time we were starting more complicated mechanisms and we looked at series reactions.
A goes B, which is an intermediate.
Goes to C with rates k1 and k2.
And we solved the problem exactly.
Which turned out to be a little bit complicated.
And then we were looking at special cases.
And I left the special case of k1 equals k2 to do as a homework problem.
And then on the board we did k1 much greater than k2.
Which looks like, if k1 is much greater than k2, the rate into k1 is much greater than k2, so using the analogy of buckets and pipes, it's a big pipe between k1 and between A and B.
And a little pipe between B and C and you started with a bunch of liquid on top.
And you basically dump it into here.
And then you slowly drip into C here.
So the first thing that happens is that A gets dumped, it becomes B very quickly.
With a rate constant k1.
In a first order fashion.
So A is A0 e to the minus k1 times time.
Then B comes up sharply.
And then drops down with rate constant k2 slowly.
So B is approximately A0 e to the minus k2 times time.
And then C has a very small quadratic rise, followed by the exponential rise to saturation that you'd expect.
So C is approximately a0 times one minus e to the minus k2 times time.
The rate constant k2 dominates the long times.
And everything it's looking like it's pseudo first order at long times.
So you can do this either by thinking about it, solve the problem by thinking about it, which is perfectly fine.
Or you can do it by solving it exactly, as we have done last time.
And then putting in the right approximations.
Which in this case was Taylor's approximation, if I remember right.
And know just by looking at the equation and putting the right approximation, you get it out.
And a third approximation, which is the opposite one, k2 much less than k1, so k1 much less than k2, the way that that's going to look, we can do it by inspection.
So basically if we think of buckets and pipes, that's a very thin pipe connecting the bucket for A and B, so we drip from A to B, then we have this very fat pipe connecting B and C. So as soon as a little B gets in there, it immediately becomes C. So, the rate determining step is the first one, A goes to B.
And so k1, we expect that to be the rate that dominates the long time.
And we never expect to get any significant amount of B to pile up in the middle bucket, because this pipe is much bigger than that one here.
Therefore, what we expect to see is a slow decay in A, with rate constant k1 dominating the long times, A is approximately A0 e to the minus k1 times the time.
We don't expect B to amount to very much at all.
It'll rise up a little bit at the very early times, and it'll pretty much stay very, very small.
Not quite constant.
And in fact if you look at the full result, and you put in your approximations, and I'll let you do that, you would find that B is essentially related to A through a constant term where k1 is very small compared to k2.
So this is a very small number here.
B is equal to a very small number times A.
So it basically follows A, but at a very very, low level.
And then you would find that C rises up in approximately e to the minus k1 times t, so with rate constant k1.
One minus e to the minus k1 times the time.
So the dominant rate is k1, and in this case here this is an approximation that we're going to see again.
We're going to revisit this approximation here when we get to more complicated mechanisms, it's going to become this steady state approximation.
Where the amount of the intermediate, or the concentration of intermediate, is low at all times.
And doesn't change very fast.
The slope here is very slow.
Because k1 is so slow coming down, and B is related to A, which means it's related to this slow decrease in A, the amount of B is small and the rate of change in B is also very small.
And hopefully we'll get to that very soon in this lecture.
OK, so that's a summary of what we did last time.
Any questions?
So the next thing now is to go to the next mechanism.
Which is, so far we've looked at reactions that proceed all the way.
We haven't looked at equilibrium yet.
But we know thermodynamics is all about equilibrium.
So now we're going to connect kinetics and thermodynamics.
We're going to look at a mechanism which is an equilibrium mechanism.
Equilibrium or reversible reactions.
Where you have A goes to B, with summary constant k1, but you can have the reverse process B goes back to A with the rate constant k minus one.
Or you can write is as A k1, k minus one to B, where k1 could be k forward and k minus one could be k backwards.
Also written as k sub f, and this is also written as k sub b.
So now we have more information than we had before.
Because now we know this is an equilibrium problem.
And we already know about equilibrium from thermodynamics.
And we know that at equilibrium, in addition to all the rate laws that we're going to write down to solve the problem, we also know that at equilibrium, the equilibrium constant is related to the equilibrium concentrations.
To the ratio of the equilibrium concentrations.
This is something additional that we have.
So let's write down everything we know.
And our goal is to find out the time, the dynamics of this problem.
Suppose you start at some time t equals zero, at a certain amount of A0 in your pot, a certain amount of B0, how do you get to equilibrium.
What's the rate, what does it look like?
So the first thing is to write down the rates.
So the forward rate minus dA/dt, which is the forward rate, R sub forward is equal to k1 times A.
And the backwards rate, minus dB/dt, which we can write as R sub backwards, that's k minus one times B.
At equilibrium, we have no dynamics.
Or no ensemble dynamics.
The pot looks constant.
It doesn't look like anything's going on in there.
The concentration of A doesn't change.
The concentration of B doesn't change.
They're constant at the equilibrium concentration.
So that implies that the rate of formation of B must be equal to the rate of destruction of B.
Or, in other words, that the rate forward at equilibrium has to be equal the rate backwards.
Otherwise, there'd be changes in concentrations.
So if that's true, then that means that at equilibrium, k1 A equilibrium has to be equal to k minus one B equilibrium.
Which means that we can get this ratio of B to A.
And see that this is equal to k1 over k minus one.
So we've just, this is pretty major here.
Looks pretty simple, but we've taken an equilibrium property here.
And related to kinetics.
To a rate property.
So here we have kinetics.
Here we have thermo.
And this is going to turn out to be, this kind of ratio between forward rates and backwards rates, related to equilibrium constants is going to be valid, not just for this very simple mechanism we have here with just two species.
But it's going to be valid for more complicated.
And we have two species reacting to form another species, or bimolecular stuff, et cetera.
OK, so now we've laid the groundwork, let's look at the dynamics.
Let's look at the time evolution.
So that means that we need to take, well this is not quite right.
I shouldn't write minus dA/dt.
This is minus dA/dt rate forward in the absence of reverse process.
It says just the destruction of A and not the creation of A.
So if you look at the full kinetics, the full mechanism, this also, the rate backwards is minus dB/dt, in the absence of forming B back through the reverse process.
So both of these are in the absence of their reverse process.
The full kinetics, for the destruction of A.
A is the rate backwards, or the rate forwards, which is k1 times A, and then the rate backwards, which is the creation of A.
That's the two rates together.
Minus k minus one times B.
There's the formation of A here, this is the destruction of A.
This is the full differential equation for the full mechanism.
Not just one part of it.
This one here is just for the first step, and that one here is for the second step.
This is what we need to solve.
But now we have a lot of help by what we've already done.
So, what we can do is well, the problem here is that's it looks hard because there's B included in here.
This is not just A, there's B in here.
But there's only two species.
And stoichiometry is going to help us out.
Remember last time, or a few times ago, we said the last species is the easy one.
Because it's always related to the concentration of all the other species by stoichiometry.
And the last one here is B.
So we know that B is related to A.
B is whatever you started out with.
Plus whatever you used up for A.
That's one way of writing it.
So you start with this amount, and you create this amount.
By having A react.
This is what you started with, A, and this is what's left.
The difference has to go into B.
So stroke stoichiometry give you a relationship, which you can use to plug into your differential equation here.
And now, after rearranging your differential equation, putting B in there, you can rewrite this as k1 plus k minus one times A minus k minus one times B0 plus A0.
Something which depends on A, something which is constant here.
OK, and what else do we know?
We know something about equilibrium.
We know something about when we reach a state where the rate forward is equal to the rate backward.
When the rate forward is equal to the rate backwards, the rate of change of A is zero.
A is not changing.
The concentration of A is not changing.
So at equilibrium, dA/dt is equal to zero.
And just like in this example here, this is going to turn out to be related to an approximation called the steady state approximation, which we're going to see later this morning.
Writing at equilibrium that the change in the concentration of one of the species is equal to zero, we're going to use that as an approximation also later, when we get to more complicated mechanisms.
This is going to be the equilibrium approximation.
And this is going to be the steady state approximation.
Let's keep going with our, we're solving this guy here.
So at equilibrium, dA/dt is equal to zero.
And then we can replace our concentrations here with, and so if you write equal to zero here, then we can write it as equilibrium right here.
And we'll be able to solve for A equilibrium in terms of k1, k minus one, B0, and A0.
And then if you do that, you get A equilibrium is equal to k minus one over k1 plus k minus one times B0 plus A0.
And we're going to need that.
Because now we're going to be able to plug this, so there's A0 plus B0 sitting here.
And we're going to replace A0 plus B0 in terms of A equilibrium.
There's a minus sign here.
And between this A and this A equilibrium here, so by rewriting, by plugging in A B0 A equilibrium instead of B0 plus A0, we can rewrite that equation as dA/dt is equal to k1 plus k minus one times A minus A equilibrium.
This is nice because, this quantity, A minus A equilibrium, describes the difference between where are you are and where you want to be.
It's a nice variable to have.
It's going to change in time and at infinite time, this is going to go to zero.
A is going to go to equilibrium.
A equilibrium is a constant.
Now we have a differential equation that relates A to this difference.
But A equilibrium is a constant.
So there's nothing that forbids us from just writing minus A equilibrium here.
d of a constant dt is zero.
So I'm just basically subtracting zero here.
So and I have something of the form dx/dt is equal to constant times x.
And I know how to write that.
That looks just like a first order process.
I know the solution is A minus A equilibrium is equal to my initial A, my initial state.
A0 times e to the minus k1 plus k minus one times time.
And I've solved the problem.
I've described how, as a function of time, how A, the concentration of A, gets equilibrium.
And if you want to know B, it's just the same thing.
You replace B here and B0 here.
You put k minus one here, k1 here, but it's the same thing.
So if you were to sketch it out as a function of time, there's, eventually you'll get to the concentration A equilibrium.
If you're above it, you're going to decay in a first order fashion, with a rate constant k1.
A minus A equilibrium is A0 minus A equilibrium e to the minus k prime times t, where k prime is k1 plus k minus one.
And this is when A is greater than A0, A equilibrium.
A0 is greater than equilibrium.
And if you start below, you're going to come up like this.
A0 is less than A equilibrium.
And both rates come into here, in a very simple way.
Just the sum of the two.
So experimental, if you want to find out these rates, you can measure their equilibrium concentrations.
Measure k equilibrium by obtaining the equilibrium concentrations of B and A.
And that gives you the ratio of k1 plus k minus one.
And then you can just start the process with some concentration of A.
Which is different in equilibrium.
Then watching it in time, extract out, observe and measure k1 plus k minus one.
In this kinetic equation.
At kinetic relation.
And then you have two results, two pieces of data.
You've measured k prime, which gives you the sum of the two.
And you've measured K equilibrium, which gives you the ratio of the two.
And you've got your two rate constants out.
OK, any questions for the equilibrium problem?
We're slowly -- yes
[INAUDIBLE]
There is not yet a relationship between k1 and k minus one.
It depends on the problem.
And we're going to study the, when we get to potential barriers, we look at, we'll look at this issue.
OK, it's a good question.
We're going to do that probably next time.
So let's get going, then, with reversible reactions.
So now we have, let's make it a little bit more complicated.
Let's, instead of two species, let's make it three species.
So we have A plus B goes to C, where the rate constant k1, and then C goes to A plus to B with a rate constant k minus one that you can rewrite as A plus B goes to C, k1, k minus one, and then you write down your, you want to solve this.
So the first thing you do is your write down your differential equations.
dA/dt is equal to, put all the rates in there.
k1 A B minus k minus one times C. And at equilibrium, we set that equal to zero.
So you can write equilibrium here, equilibrium here, equilibrium here.
And when you bring this term over to the other side, you find, then that at equilibrium the ratio of the forward rate to the backward rates is equal to C equilibrium divided by A equilibrium.
The equilibrium, which, you know is the equilibrium constant.
So again, for this slightly more complicated problem, the ratio of the forward and backward rates are related to the equilibrium constant.
So now you want to solve this.
Well, we're not even going to try.
Because it's going to be too complicated.
So as soon as you get away from first order kinetics and go to second order kinetics with multiple steps, it's a mess.
So instead you try to find approximations.
In this case here there's one we can use.
Or a limiting case, at least.
One we can use, which is an obvious one, which is flooding.
As a limiting case.
If we want to isolate a small part of the problem, we can use flooding.
Because we can overwhelm the system with either A or B.
Take B0 much greater than A0 and C0.
So over the course of the reaction of the process, the concentration of B hardly changes.
And so when we write our kinetic equation, our rate law, k1 A B minus k minus one C, we could put a little naught here, because we know the concentration of B is not changing.
And so now we're left with the problem we had before.
Which was a reversible first order process which we have just solved.
And you go through the experiment and you extract out k1 times B0 and k minus one.
You do the experiment again with a different concentration of B0 to start out with.
And you can extract out k1.
So this is a process of getting the rates by using a simple approximation here.
Simple limiting case.
Any questions?
So we've just finished putting all the building blocks together now.
Let me remind you what these building blocks are.
We have a bunch of approximations under our belt.
Flooding, we've looked at rates being much faster than others.
And we have three simple mechanisms, we've looked at parallel reactions.
We looked at series reactions.
And we've looked at reversible reactions.
So a complicated mechanism will basically put these three building blocks together in a series.
And they'll all be happening at the same time somehow.
And obviously, since it was, since we threw up our hands, just with a simple reversible process like here, it's clear that we're going to throw up our hands a lot.
When we write down these mechanisms.
So we're going to need approximations.
We're going to need something that we'll automatically go to, and say this is going to be too hard.
I'm not even going to try.
Let's do an approximation.
And two approximations that we're going to talk about are ones I already mentioned at the beginning.
Which is the steady state approximation, where one rate, the rate to go into the intermediate is very slow, or the rate to get out of the intermediate is very fast.
And the equilibrium approximation where the system sets up a very fast equilibrium and you're allowed to use thermodynamics to help you out in solving the problem.
These are the two approximations that we'll use when we put these things together.
OK, so let's look at the first, simple, more complicated mechanism.
To see where these approximations come in.
So the first one is going to be a series.
We're going to start putting these things together.
First one is a series reversible.
Those two together.
Series reversible.
So we have first, a reversible process.
Everything is first order here.
k minus one goes to B.
And then B goes to C with some rate constant k2, all first order.
If we were to turn the crank, we'd say, oh, I've got to write down all my rate laws here.
Minus dA/dt is all the ways that I destroy and create B, so there's a k1 times A minus k minus one times B. dB/dt, all the ways that create and destroy B is that you create it through destruction of A. I destroy it through the backwards rate to make A.
And I destroy it by making the product, C. There's the intermediate.
And then for C, that's pretty simple.
There's only one channel into C, and that's the destruction of B to form C, k2 B.
All the rate laws.
I write everything I know here.
Couple of differential equations.
We know it's going to be hard to solve.
So, let's remind ourselves of what I just erased.
Which was the case where the intermediate concentration was always very small and didn't change very fast.
And the process was dominated by k1 here.
That was the rate limiting step.
Remember that our first example this morning.
So this is where the concentration of B is roughly constant over time.
It's small.
And over any small time period, it's roughly constant.
Which means that dB/dt is roughly equal to zero.
If you look at a long term, obviously it's going to change as you go from this point to that point.
In fact, there's a relationship between A which is changing in time and B.
But A is also changing in time very slowly.
Because the rate k1 is very small.
So dB/dt, which is related to this rate k1, is going to be very slowly changing in time.
We can approximate it at zero.
What are the ways in terms of matching these rate constants that we can get to the approximation.
To a diagram that looks like that?
But we don't want B to pile up.
That means we have to have the rates out of the intermediate to be much faster, at least one of the rates out of the intermediate to be much faster than the rate that creates the intermediate.
So the different ways of creating that approximation are if, and the length of the arrows now is going to be proportional to the rate.
We want it to be very hard to make B.
And as soon as we make B, we want it to go away.
So one of the ways to do that is to have the reverse rate much faster than the initial rate.
And we could have a slow rate into C, that's fine.
We could have, again, a very slow rate into B.
We could have a slow reverse rate and a fast rate into C. Perfectly fine.
We could have both.
Fast rate out of B through A, or out of B through C. As long as this first rate into the B is small compared to one of those two rates, we're never going to pile up B.
It's going to go away as soon as we make it.
So in all these cases, k2 plus k minus one is much bigger than k1.
When we have that situation, then we can make the approximation up here.
That the rate of change in B is very small.
Almost zero, which means it's basically a constant.
And instead of writing B in this case here, we're going to write it as B steady state.
Which is basically a constant in terms of solving the differential equations.
And that's going to make our life much easier.
Because instead of having to solve these coupled differential equations, we're just going to have to solve coupled algebraic equations.
Which is really messy, but less hard.
If you're a bean-counter than this is heaven.
So now we're going to put steady state, this is going to be a constant.
And this is going to be equal to zero.
And then here this is going to be steady state here.
This is going to be steady state here.
And the first thing to do now is that we've eliminated this differential equation.
We now have an algebraic equation where we can solve for B steady state.
So if you recognize that you're in this situation here, forget about trying to solve.
Immediately go to the process of putting in a constant for B.
Setting dB/dt for the intermediate equal to zero.
And then starting to turn the crank on the algebra.
So let me go ahead and turn the crank here.
Go through the steps, which is basically typical of these problems.
So you turn the crank.
You solve for the steady state concentration.
You get, in terms of A, over k minus one, plus k2.
Then you plug this back in here.
And you get a new differential equation, minus dA/dt is equal to k1 times A minus k minus one times B steady state, so we plug this in here.
And you get, so there's A sitting here.
A sitting here.
It's going to be of the form, effective rate times A.
It's going to be basically a first order form.
Which is what we'd expect from sketching the diagram just like that.
So you get minus dA/dt is an effective rate.
k1 times k2 over k1 plus k2.
Times A.
This is k prime.
So if you were to do an experiment under these conditions, you'd find that A behaves, it's basically a pseudo first order problem.
With a funny rate that contains all the elementary rate constants as part of it.
And when you do the same thing for C, dC/dt, you find that depends on A.
It's k1, you plug these steady states in here.
k1 k2 divided by k1 plus k2 times A.
This is k prime again.
So the problem, and it's first order.
So the problem looks like effectively you're going from A to C. Forget about the intermediate.
With an effective rate k prime.
Where k prime contains these rates.
So this is steady state approximation.
It's the prototypical problem.
Questions on steady state, before we go to the next approximation.
Alright.
So now, as promised, the next approximation is that where are set up the fast equilibrium.
And you can use thermo to help you out.
Equilibrium approximation A goes to B.
This is a fast process.
And then k2 out of B is a slow process.
Little arrow here, and two big arrows here.
You put your A in your flask.
Immediately you set up the equilibrium.
And you slowly dribble out of that.
If you want to do it as a function of buckets and, so you have a big pipe connecting two buckets that are just offset from each other.
So there's, in this case here B is favored over A.
Because it's a little bit lower than this guy here.
And then there's a little bucket, there's a little tube, comes out of here.
So there's a little dribble into C over time.
First thing that happens is you set up your equilibrium.
And then you slowly extract stuff through a little tube into C. So what do you expect this to look like?
Well, you expect A to really slowly come out.
Because the rate limiting step is the rate from B to C. So k2 is going to be the way that A is going to come out.
You expect it to slowly go away.
You expect B to get created very fast, to some equilibrium amount.
And then also to follow the same rate k2, slow k2, to disappear.
And you expect C to come up.
You're basically in a first order process to saturation of A0.
So when you expect the dominant rate to be k2, and the fast dynamics to happen at very, very early times, and then everything to follow first order kinetics.
So now let's do the math and make sure that it agrees with what we've just assumed that it's going to look like.
So equilibrium is happening.
You can assume that at any time after the initial very fast process of getting equilibrium.
So after some initial time, the ratio of B over A is always going to be a constant.
The dribble out of B here is not going to be fast enough to change that ratio.
As soon as you get a little bit of B out here, immediately the ratios, the amounts rearrange to keep the ratio constant.
So now when I look at the rate of the reaction, the rate of formation of C, k2 times B, well, in terms of A, it's k2 times K equilibrium.
Times A.
Which is k2 times k1 over k minus one times A.
And you can immediately see that C is behaving, or the reaction is behaving, like a first order process.
With an effective rate which contains all three rates in this ratio here.
So it looks like A goes to C, with some k prime where this is k prime.
It looks like A goes to C with an effective rate constant k prime.
OK.
Questions?
We're going to do some examples.
And then we're going to do chain reactions next time.
We're one lecture behind.
Alright.
Let's do some reactions.
Some examples.
I'm going to skip the first example, which is the example with the chaperone, in the notes.
And I'm going to go directly to the gas decomposition.
The two examples are basically very similar in their use of the steady state approximation.
So I'll let you read over the first example.
So this example here is going to give us the Lindemann mechanism, for which Mr. Lindemann got a Nobel Prize many many, years ago.
Basically, it's looking at decomposition of a gas phase molecule.
Where you have a molecule, A, that breaks down into products.
And the observation before Mr. Lindemann got around, was that it looked like a first order process.
It looked like A is just falling apart on its own into these products.
Mr. Lindemann got around and said, well, why would A just want to fall apart.
There's something a little bit odd about that.
Stable molecule, why would it want to fall apart.
And so he hypothesized a mechanism.
And it went like this.
That A actually collides with another molecule, which could be a bystander.
Could be a chaperone molecule that just happens to be there.
Or it could be another molecule of A.
There's a collision.
There's a collision that creates a vibrationally excited version of A.
Of A sitting there.
It collides.
It suddenly starts to be really vibrationally excited.
Bonds vibrate all over the place.
Atoms wiggle.
There's the collision partner that goes away.
So kinetic energy is transferred from M and A to the vibrational loads of A.
In a process which is reversible.
k1, k minus one, because this excited A, this vibrationally excited A, could also collide with a molecule and cool down.
The energy in the vibrations could be transferred back to kinetic energy for a reverse process.
But if you wait long enough, or if this vibrationally excited A, with these atoms widely moving around, that has more of a chance of falling apart than this guy here.
So A star could fall apart into products.
With a rate k2.
So he said, this is actually an important step.
It's not that A just suddenly falls apart.
But this has to happen.
So all the observations that supported a first order process, which basically only supported one elementary reactions, might not be right.
So let's assume his mechanism, and let's look at what we need to do as an approximation.
And where that leads us in terms of trying to experimentally confirm that this mechanism is plausible.
Not prove it, but just to make sure that it's consistent with, the data's consistent with the hypothesis.
So let's compare our rates here.
So we have a collision.
A and M collide, and there's a certain probability on the collision that kinetic energy will be transferred to vibrational energy.
That turns out to be a moderate probability.
So k1 is reasonably fast.
The reverse process, you have something that's vibrationally excited colliding with a molecule.
There's a probability that that excitation is going to get turned back into kinetic energy.
And then these two molecules will fly apart with high velocity.
And that tends to be much more highly probable then the first process.
So this is faster.
Then the excited molecule's waiting around.
And there's a probability that there's it's going to just fall apart.
And that turns out to be really slow.
So we're going to assume that the last step is a slow step.
So what does it look like?
Well, we have the creation is fast.
So the intermediate is fast.
But the destruction through the reverse process is much faster.
We have at least one step, out of the intermediate, which is faster than the step into the intermediate.
That's all we care about.
It doesn't really matter that this is slow in terms of using the steady state approximation.
We just want the step getting out of the intermediate, a backwards step in this case here, to be faster than the state into the intermediate.
So this allows this, this fact that this is faster, that allows us to use a steady state approximation.
So now we can go ahead and solve the problem.
So we write down all the rates minus dA/dt.
It's k1 A.
Let's start with the rate of the action.
Let's take a product.
Let's take C. Let's say this goes to product C here.
So dC/dt is equal to k2 times A, that's the rate of the reaction.
And we're going to look at this rate of reaction.
And see how it changes.
What it looks like under limiting conditions.
The rate of reaction.
Let's look at the intermediate.
Because this is what we're going to solve first, because this is going to turn out to be an algebraic equation.
Because we're going to apply the steady state approximations.
Intermediate d A star / dt is, we can create it through the forward process.
We destroy it through the backwards process.
A star times M. And we destroy it through the final process.
A star.
We apply the steady state approximation, steady state, steady state.
We solve for A steady state algebraically.
k1 A M over M times k minus 1 plus k2.
And then we plug that back into the rate of the reaction.
The rate of appearance of C. Which is what we're measuring experimentally.
We're looking at the products.
Measuring the appearance of products.
So they're destruction.
This is also equal to the destruction of A.
And so we plug this into the, we write our rate of reaction.
Which we measure, experimentally.
We find this is k1 times k2.
Times, we plug in for, and here, what we do here, we solve A in terms of A steady state.
A is equal to something A steady state.
Well, let's see, A is equal to M k minus one plus k2 times, well this is actually, I have this wrong up here, this is d A star.
So we don't need to do this here.
The appearance of C is, A star here.
So we plug in A steady state here.
A star steady state, and this is A star steady state.
The intermediate is what we're solving a steady state for.
So this is k1 k2 A M over M k minus one plus k2.
We just have the extra k2 that appears when we multiply A steady state star with the k2 there.
And then we can look at the limiting cases.
There are two limiting cases, which we can tell by looking at the denominator.
If we have one term bigger than the other.
Only two choices here.
So the first case is where M k minus one is much bigger than k2.
Experimentally, what does that mean?
It means that this in the gas phase now.
It means that regardless of what these rates are here, we've managed to make the concentration of M high enough so that this becomes true.
There's always an M that's going to be high enough so that the multiplication of these two is going to be bigger than that.
What does that mean?
It means that the pressure is high.
Concentration of M is high.
This is a high pressure case.
And then the other one is M k minus one is much less than k2.
M is very small.
Low concentration, low pressure.
The partial pressure of M is very low.
This could be A. M could be A, or it could be some other molecule that's in the mix.
And by high pressure we mean something on the order of one bar, let's say.
That by low pressure we mean something around 10 to the minus 4 bar.
So now we can go and put these limiting cases in our equation here.
k minus one is much bigger than k2, so we can ignore k2 here.
And then we have something that looks like the M's disappear.
We have something that looks like k1 k2 over k minus one times A is the rate of reaction.
First order.
Looks like it's first order in A.
This is what people had observed.
Great.
The problem is that they didn't know to go to low pressure.
They were doing all their experiments at atmospheric pressure.
They were getting a first order rate.
They thought they had solved the problem.
It's just A falling apart by itself.
Until you go to low pressure, where now this term dominates the denominator.
Or rather, this one dominates the denominator.
And now you have something of the form k1 A times M. This dominates, this term goes away.
The k2 goes away and you have k1 times A times M left over as the rate of the reaction.
Second order.
There are two species here.
A times M. M could be A or it could be some chaperone.
A squared.
Second order process.
So if you're at low pressure, you see something at second order.
You've discovered something about the mechanisms you didn't expect.
And then you get a Nobel Prize.
Alright, next time we'll do chain reactions.
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So we're going to start directly today and look at chain reactions.
And explosions.
So chain reactions, how many people have heard of chain reactions?
Lot of people have heard of chain reactions.
Have seen chain reactions done, in kinetics.
No.
Alright, well, then I'll do chain reactions.
Chain reactions.
Chain reactions are reactions that feed on themselves.
And there are three parts to a chain reaction.
There's an initiation step, which is where your reactant makes an intermediate.
And that's usually a slow step.
There's a propagation step, where the intermediate reacts with your reactant, becomes a feedstock.
And then there's some intermediate plus maybe a first product.
Then there's that second intermediate, can further react to re-form the initial intermediate.
And that's the loop.
That's the loop of the chain reaction.
Plus maybe a second product.
And then there's a termination step.
Which is where when your intermediate, I1, becomes a stable product.
Or I2 becomes a stable product.
Basically, where the intermediates come out of the loop.
So if you look at this loop here, you have your reactant that comes into the loop.
And then go around the loop.
You create I1, and maybe a product comes out.
Maybe I1 gets terminated, but most likely it keeps going around the loop.
Creating I2, which reacts with, the reactant at some point gets included in the circle.
And I1 comes again, and I2, and I1 comes, and I2.
And every time you form I1 or I2, you may form some of the product.
So this is the initial feed.
Then you go in circles.
Eating up R, spitting out products.
Keep going in circles.
And at some point the termination's going to win, and the circle stops, and the chain has stopped.
And there are two kinds of chain reactions.
There's a stable chain and an unstable chain, where you get your explosion.
Stable chain, or also called stationary, chain reaction, is where the amount of these intermediates, I1 and I2, are constant in time.
So you don't build up any intermediate.
So I1 constant in time, which is why it's called stationary.
Then you have the unstable, or non-stationary chain.
Where in that case your intermediate increases in time.
So in the example I gave here, you destroy I1 in the first step, you create it in the second step.
Comes back, you destroy it, you create it, you destroy it you create it.
For every I1 that you start out with, you end up with an I1.
There's no change in the concentration of I1.
But if instead I put a two here, than my reactant creates an intermediate I1.
I destroy it.
Create a new I2, the I2 decomposes, creates two I1's.
So after the first cycle, let's look at the concentration of P2 here, after the first cycle here I have one I1, and make one P2.
But I make two I2's.
So now I have two I1's coming in here.
So I have twice this reaction.
I make two I2's.
These two I2's form two products.
And now four I1's, two times two.
These four I1's go through the process.
I get four P2's.
So third cycle, I get four P2's.
And I get two times four, eight I1's.
These eight I1's go through, and I get eight here.
For cycle, et cetera.
So I end up with something that goes like, so this is two to the zero power.
This is two to the first, two to the second, two to the third, et cetera.
Eventually you get two to the 25 or whatever after so many cycles.
And if these are gas phase products, so if there's an exothermic process where heat gets created at any of these steps here, then you end up with a problem.
The thing goes out of control.
Sometimes that's a good thing.
If that's what you want to do.
And sometimes it's a bad thing.
Depends on what you want to do.
OK, so we're going to show two examples of these chain reactions.
One is stable chain and the other one is unstable chain.
And go through the process of how you solve for them.
So the first example is taking, so let's do a stable chain.
Take acetaldehyde, CH3CHO, that decomposers to form methane and carbon monoxide.
That's the reaction.
And here are the observations.
There are two observations.
The first one is that methane and carbon dioxide are not the only products that are formed.
In fact, there's a trace amount of ethane and hydrogen, dihydrogen are also formed.
That's a clue to something funny is going on here.
That there's a mechanism that's more complicated than just this decomposing in a first order process.
And the other thing is that the rate, if you do an experiment and you measure, somehow you do a half time experiment, or an initial rate experiment, or one of the ways that we talked about early on, you measure the rate of the reaction and you find that it's proportional to something to the 3/2 power.
That's pretty weird.
It's not a simple mechanism.
It's pretty weird.
And so those two observations are clues, usually, that you've got a chain reaction.
The fact that you have a trace amount of extra things, that's usually due to a termination reaction.
It's not a major product, it's a minor product that's the termination of the chain.
And then this funny power here, in the kinetics.
Of the rate of reaction, that's also usually a chain reaction.
So now we have to figure out if there's a mechanism that supports those observations.
And I'm going to tell you possible mechanism that supports those observations.
And you can do experiments to show that you can fish out certain intermediates.
It's not proved that that's the ultimate mechanism, but it's at least consistent with these observations.
More hidden chemistry.
OK, so this is the mechanism that's proposed.
And then we'll figure out whether this proposed mechanism fits the data.
So, the proposed mechanism is that the initiation step consists of the acetaldehyde, CH3CHO, decomposing into a radical.
Chain reactions often have radical processes also.
Because the radicals love to make chains.
To the methyl radical, and then the CHO radical.
So this is the first step.
It's the decomposition step, it's very slow.
That molecule doesn't really want to fall apart.
Then there's a propagation step.
And that's the methyl radical plus the acetaldehyde, your reactant, forming with the system rate k2.
Forming CH3CO radical, plus methane.
So that is their product right here, coming out of the first step of the propagation.
And then this new intermediate here, so there's this one here would be I1.
The methyl radical.
And the CH3CO radical here, this is I2.
And you have to use that up in the second step.
CH3CO.
That decomposes on its own.
Rate k3, to form the first intermediate back, the methyl radical, plus the second product, CO. And then this intermediate then feeds the first step back.
And you've got your cycle.
And every point you spit out your two products.
And this could go on forever.
But it doesn't, because at some point those two methyl radicals, two of those methyl radicals can get along, can collide.
And in a very easy way.
So we've got a few termination steps.
You can form two times CH3 radical ethane, C2H6.
Rate k4, let's call this k4.
And then there are other side reactions that can happen.
For instance, this radical here, which we haven't used up.
That can also react.
It's just sitting around, slowly building up as this thing, very slowly building up, because the first step here is very slow.
You don't need very much of this methyl radical to start the chain.
Then you start chewing up the reactant.
But you're still building up a little bit of this guy here.
This is a side reaction.
It's not a termination reaction, because that radical's not involved in the chain propagation step.
But that radical can react with some other species, M, and it could be any of the other species.
Because it's just a chaperone.
It's just a, to help this thing to decompose.
CO plus H radical plus the chaperone back.
And then that hydrogen radical here can itself react with your reactant.
CH3 acetaldehyde, CH3CHO, with a rate k6.
To form hydrogen gas plus your intermediate, CH3CO.
And this intermediate can feed the step right here.
OK, so in terms of the observations, then, we got the right products.
Methane, and carbon monoxide.
We've got the right amount of trace stuff, the ethane coming out from the termination step.
And hydrogen gas coming out from the side reaction.
And I should probably have written CO also forming from the side reaction - well, CO is a real product, so it's happening both from the side reactions as well as from the main reaction.
So we've got all the species taken into account.
And now we've got to show that this mechanism here, in fact, gives rise to this 3/2 power.
And I know I won't be able to do it on one board.
I want to keep all up, so let me use up.
So the first thing, as we've been doing in kinetics this whole time is, you've got to write down all your rates.
Once you've got that down, and you haven't made a mistake, then you basically have the problem solved.
So you write down all your rate laws.
Eventually, what we're interested in showing is the rate of the reaction, is that power.
So that's write down what the rate of the reaction is.
The rate of reaction is the rate of formation of one of the products.
Let's say the methane.
So we write down that rate.
How was it formed?
It's formed through the k2 step.
It's formed through this step right here.
Which is a second order in CH3 radical times CH3CHO.
acetaldehyde.
We're going to keep that in the back of our mind.
Because we're going to need this again.
Because we want to show, at the end, that in fact this is, this rate is proportional to this funny power.
There's the intermediate sitting here.
So what's the rate for this intermediate here is CH3, dt.
There are many places that it's coming from.
It's being created in the first step, and so the hard part here is bean-counting.
Making sure that you've found all the places where the species is being created.
k1 times acetaldehyde, CH3CHO, and then it's destroyed through the second step, minus k2 CH3, better get your signs right also.
So there's the destruction, there's the formation.
Times acetaldehyde, CH3CHO.
Then it gets created again, through the second step of the propagation, rate k3.
Radical CH3CO.
And then it finally gets destroyed in the termination step.
We've got to put that in, even though it may be a very small amount compared to these other amounts there.
So the termination step is k4.
And it's a second order process, CH3 squared.
We've got another intermediate.
We want to write down all our intermediates here.
Because they're all related, it's the intermediates sitting here.
So there are going to be a couple of differential equations.
So we've got to write down the equation for that CH3CO radical.
And again, you look at your mechanism.
And you find out where it's coming from.
It's being created in this step.
Destroyed in this step.
It's also being created in this step.
But we're going to ignore that side reaction for now.
Just to make our lives a little bit easier.
k2 CH3.
You've got to put the termination step in, but the side reactions are not so important.
CH3.
Because if you don't put the termination step in, then your reaction's just going to go on forever.
That's just not physical.
Minus k3 CH3CO.
Creation and destruction.
OK, so now what do we do?
Well, we know it's a chain reaction.
A stable chain reaction.
Because we don't create more intermediates than we started out with.
The first step is always very slow.
Initiation at this step is slow.
So we're not making very much of this methyl radical.
We're not making very much of it.
And we're not making any more of it here.
So during this whole time that the chain reaction is happening, this methyl radical here is in small quantities.
Small quantities and it's not going to change very much at all.
It's pretty much stationary.
Magic word is stationary.
Stationary means it's not changing.
Not changing, so which approximation does that belong to?
That's a steady state approximation.
We've got a steady state here.
Chain, which is going blah, blah, blah, blah, stationary.
Steady state.
All sounds the same.
So that's the clue here.
This is going to be a steady state approximation problem.
So we're going to put a steady state here.
We're going to put a steady state here.
We're going to set this equal to zero.
We're going to put a steady state right here.
We're going to put a steady state here.
We're going to set this equal to zero.
And suddenly the problem becomes a tractable algebraic problem.
Now we have, to solve this, let's do it in one more place.
Let's put it right here.
So now we have two equations, two algebraic equations.
Two unknowns.
These two intermediates.
The concentration of these two intermediates are the two unknowns.
Two equations, two unknowns, we can solve it.
It can be messy.
But not necessarily tricky.
Just turn the crank.
And computers could do this.
So, let me just write the answer.
Because I don't want to go through the algebra.
After you turn the crank, what you want is to get this out.
You've got your two equations, two unknowns.
You turn the crank.
And you get that form for the concentration of the methyl radical two k4.
CH3CHO turns out to be to the 1/2 power.
When you do that.
There's the funny power coming up.
That's going to be here.
And then this is what you want to put into here.
So, and there's the acetaldehyde, sitting right here.
There's one acetaldehyde to the 1/2 power, times acetaldehyde to the one power, that's the whole thing to a 3/2 power.
So in fact, the rate is proportion to the 3/2, according to our mechanism.
And all these rate constants go into it as well.
And what we find is that the rate of reaction is equal to k2 times square root of k1 over two k4.
CH2CHO to the 3/2 power.
Basically, k prime to an effective rate times that.
It's a typical problem.
Any questions on this one right here?
Yes
: [INAUDIBLE]
Here?
Right here
: [INAUDIBLE]
You know, according to my notes I don't have it, yes, you're absolutely right.
There's a two here.
Thank you very much.
It's arbitrary.
Where the two comes from, I didn't want to get into this.
Sometimes it's a matter of convention.
When you have a second order reaction, A plus B going to C, you write your rate of the reaction, as let's write it here, k times A times B.
Sometimes we have 2A goes to C. You have a rate k2, and the rate of the reaction is, you can write it as dC/dt, which is minus k2 times A squared.
But if you write dA/dt, because of the stoichiometry.
dA/dt concentration for, so you have that C is equal to A0 minus A, twice.
So dC/dt is equal to minus two dA/dt.
So minus 1/2 dC/dt is equal to dA/dt.
So dA/dt is equal to minus 1/2 k2 A squared.
And instead of having the 1/2 here, what we do is we redefine our rate constant.
So that's k2 prime is equal to twice k2.
You put a two here, a prime here.
Get rid of this here.
And there's the two sitting right here.
Purely convention.
Not something to really worry about.
You've just got to have it right at the beginning, and keep that two in, if you're going to put it in there.
Does that make sense?
It's the stoichiometry, it's because of this two sitting right here.
Then you have a choice.
You can either say the rate of the reaction is the rate of formation of this thing.
Or you can say the rate of reaction is the rate of destruction of A.
And then if you say that this is the rate, and you don't want to have any factors in here, then the two shows up here.
If you want to say this is the rate, you don't want any factors in here, then you're going to have a different rate constant.
Purely arbitrary choice.
Good question, though.
Any other questions?
OK.
So, there's another thing about the chain reactions, a bit of nomenclature that's interesting to talk about.
It's the chain length.
And that nomenclature is due to the fact that many of these chain reactions are used for polymer reactions.
Where you use radical polymerization and then you have, basically, in a polymer reaction you'll have an initiator and then you'll add a monomer through a radical process.
And another monomer will add through a radical process.
Another monomer will add, another monomer will add, another monomer will add, and then you're going to terminate.
And then you end up with a chain.
A polymer chain.
So your product, instead of being molecular products, the product is adding a monomer to the chain of the growing polymer.
And then it's meaningful to talk about chain length.
So that's where the nomenclature comes from.
It's a number of cycles of the reaction before you get a termination step.
So we can define the chain length as the number of propagation steps per initiation step.
That gives you the chain.
Each one of these is a propagation step.
And there's the initiation step here.
And this could also be termination step.
In in a stable chain, or termination step.
There's a termination step here.
For every initiation step in a stable chain, you've got a termination step.
Because you're not building up intermediates.
So you create it.
At the end of the chain you terminate it.
You have as many initiation as termination steps.
So you have a choice here, in the algebra.
You can use either one.
So another way, instead of doing numbers here, you can also rate.
This is also the rate of propagation divided by the rate of initiation.
You just take the derivative of this with respect to time of the top and bottom.
Same thing.
So we can also define it as rate of propagation divided by rate of, let's say, termination.
Or initiation, whichever is your favorite one to use to make it simple.
So, for instance, in our example here, the rate of propagation is the rate, also the rate of product formation.
Since we have a product formation.
Rate of propagation is the rate of forming this methane here.
So in our example here, the rate of forming the methane is the rate reaction here.
This is this right here.
So you can plug this here on top.
I may not write it down because I don't want to spend time writing too much down.
So this guy goes right in there.
And then the rate of termination is going to be, the rate of termination is, well, let's not use termination because that has intermediates in it it.
Let's use initiation.
Since we have a choice.
Rate of initiation.
Either one works fine.
Initiation is k1 times the acetaldehyde.
k1 times CH3CHO.
So there's the 3/2 power, which, where is it?
There's 1/2 plus, so 1/2 plus one is the 3/2 power here.
So there's CH3CHO to the 3/2 power.
CH3CHO to the 3/2 power, then some effective rate k prime.
And you end up with chain length going to the 1/2 power.
So there's some k double prime, so effective rate times CH3CHO to the 1/2 power.
That's the length of chain.
It depends on the concentration of the reactant.
And the typical chain's on the order of a couple hundred or something like this, usually.
You always have a termination step at at some point.
Maybe a few thousand if you're doing polymer chemistry and sometimes millions, if you're very good at doing polymer chemistry.
OK, any questions on the stable chain?
OK, let's do the explosion now.
And so this is the typical example.
Of an unstable chain.
It's the formation of water from hydrogen and oxygen.
2 H2O.
That's the reaction.
Gas phase reaction.
And the observation is that that, well, you know the observation.
You put a spark in a balloon that contains hydrogen and oxygen and get an explosion.
So we're going to have to account for that.
Sometimes you don't get an explosion.
Sometimes it kind of peters out.
You have a drop of water coming out.
Not bad.
And there's a temperature dependence to it also.
So, whatever mechanism we put in there needs to account for all the pressure dependence, the stoichiometry dependence, and the temperature dependence.
So this is the mechanism that's proposed.
Initiation.
You have your hydrogen molecule.
Spark, or something slow that decomposes it to two hydrogen radicals.
This could be a spark plug, and every time you make a little bit of the radical.
Then you have a branching step.
Where your radical reacts with the oxygen molecule to form a hydroxyl radical.
Plus an oxygen.
Oxygen atom plus your hydrogen gas, your reactant, reacts to form more hydroxyl radical plus the initial intermediate.
And then the hydroxyl radical, I don't know why I have parentheses here, hydroxyl radical reacts with the reactant again, the hydrogen gas, k3, to form more of the initial intermediate plus water, your product.
So there are three steps to the branching here.
You get one intermediate, one of the initial intermediates here coming in.
Forms a second intermediate.
OH radical.
Two OH radicals.
And the OH radicals form the hydrogen.
There's also radical hydrogen that's formed right here.
So you have this formation.
So because you're forming two hydroxyl radicals here, in this step here you're actually forming the equivalent of two hydrogen radicals.
So, one hydrogen atom here.
After going through the branching, you end up with three hydrogen radicals.
So you go from one to three in one cycle.
And the second time around, you go from three to nine.
The next time, you go from nine to 27.
Pretty soon you've got a couple of billion hydrogen radicals.
And that's the problem right there.
Or the opportunity.
The opportunity to get something that goes out of control.
And there are termination steps.
That we need to take into account.
The hydrogen radical can hit the wall.
Before it finds something to react with.
And gets stuck.
So if the pressure is low enough, you know that's going to happen.
You're going to make a hydrogen radical.
Hardly any gas molecules around.
It's going to find a wall, it's going to get stuck.
It's very reactive.
So that's going to terminate the process.
Hydrogen radical could re-find an oxygen and another chaperone, that's a termolecular process.
That's an unlikely process.
But nevertheless, it could happen.
With another termination step, to form the HO2 radical plus M. And then that HO2 too radical can react in a side reaction to form some products, but not be part of the chain.
Yes
: [INAUDIBLE]
How do we get three hydrogens?
Because there are two hydroxyl radicals here.
And so in this step here, you actually have two hydroxyl radicals to form two, so I could multiply this by two.
But you usually don't do that.
So this already tells you what conditions you're going to get, with something where the termination is fast, compared to the propagation.
Because you have low pressure.
If you can't get that hydrogen radical to react with something before hitting the wall, you're going to be OK. You're not getting an explosion.
This tells you that a very high pressure, where the probability of having a three-body collision becomes high enough, then you're going to terminate the chain as well.
Those hydrogen radicals are going to find oxygen molecules and some other molecule and form this radical before you propagate the chain.
At least some reasonably high pressure.
So that's the clue here.
What's the strategy?
The strategy is to assume that everything's going to be fine.
That is, stable.
That's going to be the strategy.
And then we're going to do a proof by contradiction, basically.
We can assume that the chain is stable.
That we can use the steady state approximation.
We're going to apply the steady state approximation.
We're going to find where that breaks down.
And where it breaks down means that it wasn't right.
That the intermediate concentration was too high.
And that means that we're going to get an explosion.
That's the logic.
So strategy is a proof by contradiction, basically.
Proof by contradiction.
So we're going to assume that the propagation, or the intermediates, are in small quantities.
We're going to assume that we can use the steady state approximation, the d[O]/dt steady state is equal to zero.
That the hydroxyl radical, d[OH]/dt, steady state is equal to zero.
And that d[H]/dt steady state is equal to zero.
These are going to be our assumptions.
That all the intermediates are in small quantities.
And they don't change very much.
And now we're going to write down, you write down the rates that correspond to these guys.
And then you solve.
So I'm only going to do this first one here.
So you write dO/dt, you see where it gets formed.
It gets formed in the first step here.
So it's going to be k1 times H radical times O2, and it gets destroyed in the second step here, minus k2 times O times H2.
You set the steady state approximation, you get a steady state.
Steady state here, you set that equal to zero.
You solve for O, radical O, steady state.
And you find that it's k1 times H times O2 times k2 H2.
Then you do the same thing for the other radicals.
You solve for OH steady state.
And you get some expression which is in the notes.
I'm not going to write it down, it's going to get too messy.
But the expression that I will write down is the final expression, which we're going to be taking into account.
So, this expression here, for the oxygen atom.
It has the radical in there.
So it's not, we'd like something that's just in terms of the rates and either products or reactants.
And this hydroxyl radical, the concentration turns out, it will also be a function of the concentration of the hydrogen radical.
So what we really want, I don't want to cover this up, let me go on this board here.
So when you solve for this, the hydrogen atom is where you get something that's free of intermediates.
And has just rates in it.
There's a termination rate here going in the denominator.
Another termination rate in the denominator.
Times O2 times M, where M could be anything, really.
Minus, and there's the creation rate of the first propagation step.
O2 here.
So now we have to see whether or not our approximation was valid.
We made an approximation.
We said the steady state approximation was valid.
If it is valid, then this concentration, this pressure here, needs to be small.
Needs to be small at all times.
If it's not small, then steady state is not valid, we have a branching non-stable chain and we've got an explosion.
So let's look at the different possibilities here.
What are the possibilities?
Well, the first possibility is that as we saw intuitively, when we wrote down this equation here, the first possibility is that we're at low pressure.
Where at low pressure, then, k1 times oxygen is very small.
Because the concentration of oxygen is small, the pressure is small.
k5 times the pressure of oxygen times the pressure of M is small also.
So k5 T O M, O2 times M is small.
And they're both, then, much smaller than k4 times the time.
k4 termination.
The hydrogen radical hits the wall before finding anything else.
End of chain.
So in the equation here, these two guys are small compared to here.
RI, the initiation rate, is very small.
You control that by your spark.
Once a second, once a minute.
Could be really, slow.
Steady state is good.
We don't have an unstable chain.
No explosion here.
Medium pressure.
You've got a medium pressure where two k1 times oxygen concentration is approximately.
You're building up your concentration.
This is getting bigger here.
This is a constant.
This is getting bigger.
This is also getting bigger, but you've got this rate here and this rate here.
At some point, at some point, this rate here, two k1 times oxygen concentration, is going to be on the order of k4 T plus k5 T times oxygen times M. At some point it's going to be like that.
And then you're in big trouble.
You're in big trouble, because you've got a denominator here that's really tiny.
Could be even, if you hit it right on the nail, you get zero in the denominator and then you're in real trouble from the math perspective.
But from the physical science perspective, it's not being in trouble.
Is just that your approximation is wrong.
The approximation that you were in a stationary state was wrong.
And so that means that in this case here, if you want to have an explosion, you've got it.
So you adjust everything until you're in a situation.
And then finally, you keep increasing the pressure.
And then if you increase it high enough, then this guy here, which goes as the square of the pressures, is going to overwhelm this.
Which goes linearly with pressure.
And k5, the termination step, O2 times M, is going to be bigger than two k1 times the pressure of O2.
And then you've got something small again.
This is where that's hydrogen radical here is able to undergo a termolecular process before finding a reactant.
Then steady state is valid.
OK. At very high pressures, you get in trouble again, because this hydroxyl radical, this HO2 radical, so at very high pressure.
then what happens is that you have this side reaction, HO2 radical plus hydrogen goes to H2O plus hydroxyl radical.
So there's a probability that this radical here finds a hydrogen molecule.
And then feeds into the propagation through this hydroxyl radical that's the result.
And then that keeps the chain propagating.
So at very high pressure, you're in trouble again.
And you get an explosion again.
So you can plot this as a function of the pressure.
And we're going to put pressure, actually, on the y axis here.
Let's put pressure here.
Put temperature here.
And so for a given temperature, let's call it temperature T prime, at low pressure there's no explosion.
Then there's a range of pressures.
Where there's an explosion.
Medium pressures.
And then a high pressure termination wins again.
And then if it's really high you get explosion again.
You do this for, as a function of temperature.
And what you find that is a diagram, looks like a phase diagram, that looks something like this.
Where at high temperature, you get an explosion at low temperatures, and this steady state is valid.
OK, any questions on this?
Don't try this in your kitchen.
Go to the desert if you want to do this.
Or don't do it either.
OK.
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So, let me start here with the temperature dependence of k. And this turns out to be extremely important.
And it's due to Arrhenius in 1889.
And Mr. Arrhenius is famous for many reasons.
Not just for his rate law.
So I learned recently, turns out that he was also one of the first people to calculate the potential effect of carbon dioxide in the atmosphere.
And and its potential effect on global warming.
This was the Industrial Revolution and people were starting to use fossil fuels at increasing amounts.
And back then, already, some people started to get worried.
And so he did the calculation.
It was a very crude calculation.
And he extrapolated, assuming that the rate to fossil fuel use would keep going, he got his calculation pretty much right.
What he didn't get right was the amount of fossil fuel that people would be using.
And so when he did his calculation, which is again a very crude calculation, he got that we would get in trouble at about 2,000 years from the time of his calculation.
So he said, no problem, 2 years, we've got a lot of time.
And ever since then people have redone these calculations and more and more sophisticated.
But his crude calculation was good enough.
And as people redo the calculations, the time that they say we're in trouble to the time of the calculation gets closer and closer together.
And the reason for that isn't that their calculations are wrong, it's that the rate at which we put out carbon dioxide just keeps getting faster and faster and faster.
So it's interesting to go back to these calculations through the last century and a half and see that.
Anyway, so Mr. Arrhenius predicted global warming.
And he wrote his rate equation, k is equal to e to the minus Ea over RT.
Famous Arrhenius rate equation.
If you plot it as log of k is equal to log A minus Ea over RT, you find that it looks like a straight line.
With an intercept at log A, where A is the pre-factor here and Ea is going to be the activation energy.
And the slope is Ea, minus Ea over RT.
R is the gas constant.
T is the temperature.
And if you put in some typical numbers, activation energies typically are on the order of a few tens to hundreds of kilojoules per mole.
Let's say 50 to 300 kilojoules per mole.
And typical activation factors, typical pre-exponential factors, are depending if it's a first order or second order.
Let's say, first order 10 to the 12th, 10 to the 15th, per second, so the A carries the units for the rate here.
So if you have A, so this is for first order, if you have a second order, then the units of A will be different.
And typically they're going to be on the order of 10 to the 11th or so, 10 to the 10th, 10 to the 12th, one over molar per second, for second order.
And it's interesting to do sort of a back of the envelope example to figure out how important this rate is.
And one of the interesting things to look at is, suppose that I'm at some temperature and I want to know how do I change the temperature.
How much do I need to change the temperature to double the rate.
And let's say my temperature's at, say, T1 is on the order of 300 degrees Kelvin, room temperature.
Room temperature, and let's take a typical Ea on the order of 100 kilojoules per mole.
Room temperature, or body temperature are roughly the same, right?
These things are.
And so the question is, what does T2 need to be to double the rate connected with an activation energy which is fairly typical of 100 kilojoules per mole.
So we want to know, what is, if I have k2 over k1 is equal to two, what's the new T2.
Alright so A e to the minus Ea over R T2 divided by A e to the minus Ea over R T1, we want that equal to two, the A's cancel out.
So log k2 over k1, which is equal to log two, is equal to Ea over R, one over T1 minus one over T2.
We know what this is, we know what this is, we solve for T2 and we find that T2 is 305 degrees kelvin.
It's a pretty small change to double the rate.
So it's pretty important that your body temperature doesn't change very much.
If you have a fever and your temperature goes up, rates start going up.
In your body, you could cause a lot of problems if the rates of some reactions go up by a factor of two.
So, very sensitive to temperature.
Rates are very sensitive to temperature.
What are these things physically?
Ea and and the pre-exponent factor.
Let's take a look at that.
So physically what's going on is, you have your two, let's say it's a bimolecular reaction, A plus B goes to C. So mechanistically what we think is happening is that the molecules come together and collide, right?
You have A and B getting together and colliding.
And the hypothesis is that when they collide they form a complex.
They sort of bind together momentarily.
And form this complex that has all the kinetic energy, or part of the kinetic energy of this collision as part of it.
So A and B are bound together in some highly excited complex.
Which then falls apart to give the product by rearranging its atoms around.
And if you plot the energy of this process as a function of the reaction, we call this the reaction coordinate, then we're going to start at some delta G, or some energy.
For the reactants.
We're going to end up with some energy for the products.
And it could be higher or lower.
Let me make them higher here, just because I don't have enough room on the board.
Products.
So if this is an endothermic reaction, that's the energy for C, this is the energy for A plus B, and then along the way we have this complex that captures some of the energy of the collision.
Up here somewhere.
And this is the energy, then, at A B star, which we call the activated complex.
So for the reaction to happen, then, we have to go over the barrier and then back to the product.
And this distance from the energy of the reactants to the top of the barrier, that's Ea.
That's the energy of activation.
How much extra energy you have to put in there to go over the barrier.
So it's very clear that as you increase the temperature and you increase the amount of energy that the reactants thermally have, or in terms of their kinetic energy, as you raise the temperature you get more and more kinetic energy, you're going up higher and higher in the Boltzmann distribution.
And the number of reactants that can make it over the barrier clearly goes up exponentially.
As the temperature goes up.
So this is Ea for the forward reaction.
Clearly there's going to be an equivalent activation energy for the backward reaction.
The two are related.
The difference between the forward and backward reactions, so E backwards minus Ea forward gives you the delta G for the reaction itself.
So, Ea backwards minus Ea forwards gives you delta E for the reaction.
Where E could be enthalpy or it could be free energy.
So that's the physical origin of this Ea, this activation energy.
Now, A, the pre-exponential factor, that's the rate of attempt for the molecules to try to go over the barrier.
So it has units that make sense for that.
And this A is different than this molecule here.
Rate of attempt.
How many times per unit time, or how many times per second, do these two molecules try to collide?
They never collide, they never try, they'll never make it over.
So it's one over second, or one over mole second.
So this is the rate of attempt.
Then e to the minus Ea over RT is the probability of success, of making it over the barrier.
So the rate of attempt times the probability of success gives you the rate per unit time of making it over the barrier.
Any questions on the dissection of this Arrhenius rate law?
Alright, so here's some examples, a few examples that you know of.
Let's look at OH minus plus methyl bromide.
Displacement reaction going to this OH minus attacks the carbon right here.
To form an activated complex H, H, H, with the OH coming in like this.
Let me go like this.
OK, there's the intermediate right here.
The activated complex in your reaction.
Which then falls apart.
It can fall apart two ways.
The OH can be spit out again, forming back to the reactants, or the bromine can be spit out, forming the product.
Which in this case here is methanol.
H, H, H, OH, plus Br minus.
Typical example.
So, given this picture here, which you're probably already somewhat familiar with, we can then move on to talk about how to affect this barrier.
How to change the rate.
Oh, I should say a couple more things.
One more thing, too.
Now, the rate of, so clearly the Ea's are related.
Because the difference of the two Ea's is the delta E for the reaction.
But the pre-exponential factors, these A's, for the forward reaction and the backward reactions aren't necessarily related.
You can imagine that in one case you have, first of all they don't even have to have the same units.
You could have a bimolecular reaction one way.
And a unimolecular reaction the other way.
So you can't really say anything about the forward versus the backward rate this way.
All you know is about the energies.
OK, catalysis.
So now we have this barrier we have to overcome.
And, so suppose that you have two motivators.
Suppose you have an equilibrium case which is slow.
And a reaction which is slow in both directions, k1, k minus one.
And you don't want to wait for this to happen.
And this could be in a biological environment.
Or it could be an industrial process, like the Haber process.
You don't want to wait.
And so one of the ways that you can speed it up, you know is by changing the temperature.
You change the temperature, the rate goes up.
You change it by five degrees, the rate goes up by a factor of two.
You can make a huge change by changing the temperature.
But, so if you raise T, speeds up.
But, K equilibrium also changes.
And we saw that when we did the Haber process.
We raised the temperature.
The rate speeds up, but the equilibrium switches to the reactants.
That's no good.
So changing the temperature is not always the best thing to do if you want to change the rate.
Instead, what you can do is use a catalyst.
if you find one.
A molecule that reacts with your reactant to form a product, that's B, spitting back that molecule C again without using it up.
And now with the activation energy, we can understand what the catalyst does.
What the catalyst does is speeds up those rates, k2 and k minus one, by changing the activation energy.
By making the E smaller.
So now I have a catalyst, and I can make this energy smaller.
So this would be A plus A C activated, getting ready to spit out B.
So in my example here where I have A plus B goes to products, so I would have A, some combination of A, B, and C together, to give out the products.
In this case, the difference in energies doesn't change.
All that you're changing is that the hump, in both ways.
Equilibrium constant doesn't change.
Just the rate changes through the Arrhenius rate law.
And this is extremely powerful.
Especially in biology.
So let me give you some examples here.
This is sort of a typical example of increasing rates.
Let's say you have the reaction, hydrogen peroxide, goes to water plus oxygen.
If you take a little bit of hydrogen peroxide and you put it on your skin, it starts to bubble.
But if you let it sit on the bottle nothing happens to it.
Put it on your hair, your hair turns white.
Blond.
But again, if you just let it sit in the bottle very little happens.
Well, it happens but very, very slowly over time.
So if you look at the rate of this reaction here, if the rate, moles per second, with no catalyst at all, the rate is 10 to the minus 8 molar per second.
Which is very slow.
And the activation energy in kilojoules per mole, in this case here is 71 kilojoules per mole.
Now we can start adding catalysts.
We can start adding inorganic catalysts like hydrogen bromide.
Increase the rate at 10 to the minus 4.
So this creates a complex with the hydrogen peroxide, which lowers the barrier to 50 kilojoules per mole, a small amount of lowering.
But because the energy is in the exponent there, it makes a big change in the rate.
Yes
[INAUDIBLE]
The rate is independent of the catalyst concentration
[INAUDIBLE]
The rate would have to change.
You're right.
That's a good question and I'm not prepared to answer it.
I'm going to have to think about this.
Let's pretend now that we are at per unit concentration of the catalyst.
And then, and I'll look into it.
OK, so this is what happens with an inorganic catalyst.
And instead if you use a generic biological catalyst, an enzyme catalase, it's a sort ubiquitous enzyme which is why you have it on your skin, et cetera, then this rate become 10 to the 7th.
And the activation energy drops to eight kilojoules per mole.
So your question really has to do with the units of A.
Of the pre-factor right there.
OK, and so there are all these examples of reactions.
That are very important biologically.
Where with an inorganic catalyst, an organic catalyst, you can change the rate by maybe a few orders of magnitude.
But as soon as you put in an enzyme, the rate changes by ten orders of magnitude.
Or in this case eight plus seven, fifteen orders of magnitude.
Humongous change in the rate.
So you've probably done some enzyme catalysis before.
But it's probably a good idea to quickly do it again.
Because it's just so important.
And it ties together our approximations that we've learned about.
So enzyme catalysis, so enzymes can be either, could be heterogeneous catalysis, can be homogeneous catalysis.
Enzymes are ubiquitous in the biological environment.
They serve to regulate the cellular activity in very complicated ways.
The cell will up-regulate or down-regulate the concentration of enzymes as it needs to make more or less products.
And there's whole cascades of events that happen in this way.
And they're in very small concentrations.
But they play an extremely important role.
Because they're also extremely specific.
So you can have an enzyme that will only act on one part of the biological cycle.
And affect it by 10 orders of magnitude or 15 orders of magnitude.
But not affect any other protein that's around.
And that's amazing.
And it turns out that a lot of the diseases of old age like, what I'm about to face, or am facing already, you, not yet, but, are the result of some of these enzyme up-regulation, down-regulation, beginning to break down.
So we have all these reactions going on that need to be essentially perfect.
When you have DNA replication or protein folding, or things like this.
And errors are made.
Errors are made all the time in these processes.
And errors cause diseases.
And so, even as a baby, your biological processes make errors.
But you have these processes, these enzymes especially based on enzymes, that can go in there and sort of fix things.
Fix things and make sure that you don't end up getting Alzheimer's at age six months.
But as we get older, for some reason, these repair processes lose their bearings.
Just like we lose our bearings.
And and they can't repair things any more.
They don't do it very well.
And that causes all sorts of diseases.
Cancer is probably one of the diseases, due to the lack of being able to repair problems.
Alzheimer's.
All sorts of dementias.
MS. Just name a chronic disease and it's probably due to a problem with up-regulation or down-regulation of some proteins, some enzymes, that are due to a repair process.
So anyway, these enzymes are big proteins.
10 to the 4th, 10 to the 6th molecular weight proteins.
On that order or so.
They tend to be fairly large in size.
On the order of nanometers.
Let's say ten nanometers to 100 nanometers.
That could be, that's a little bit on the big side.
Probably closer to ten nanometers.
Ten nanometers is kind of big, for any sort of biological molecule.
And they always end with their name ase.
Like catalase, uriase, rnase.
Esterase, clips ester bonds.
Your liver is full of esterases.
Because it likes to break things down into smaller and smaller pieces and lots of ester bonds and things that are not very biologically interesting.
And that's one way of the liver breaking things down.
So the way it works is that you have your reactants, which in the biological language are called your substrate, come in, into an enzyme.
Gets bound up in a pocket of, there's the substrate, reactant.
There's the enzyme here.
Forms a complex.
And then product gets spit out.
And the product floats off.
And does its thing.
It probably gets bound to another enzyme, which makes another product.
Et cetera, and the cascade goes on.
The signaling cascade goes on.
And this enzyme is in very small concentration.
The product goes away, so that's going to stay as a very small concentration as well.
So let's observe experimentally, then.
Experimentally, what's seen is that the substrate makes products in the presence of the enzyme.
With a rate dP/dt, that's, at t equals zero this is the initial rate is proportional to the concentration of the enzyme.
This is initial rate.
And in the language of enzymatic kinetics, this is called the velocity.
dP/dt is also called the velocity.
Velocity, moles per unit time.
And this would be called, then, the initial velocity of v initial.
v initial's proportional to the enzyme concentration.
And what else is seen?
For fixed, if I fix my concentration of enzyme, I look at the velocity over time, let's say minus dS/dt, which is dP/dt, I find that that's proportional to S. For small S. Small concentration of substrate.
And then at large concentration, it's a constant.
So it's not a straight line.
In fact, I don't want to do it on this board here.
I'll do it on this board here.
Plot the concentration of substrate on this axis, and I plot the velocity, or the rate, of the reaction on that axis here.
What I find is that it's a constant.
So it's going to saturate, the rate is going to saturate to some number.
And it's going to start linear with substrate at the beginning.
So it's going to be a straight line at the beginning.
And eventually it will saturate.
That sort of slope.
And the saturation point, that's the maximum velocity, or maximum rate that it can have.
So we call this v max.
And this part here, the velocity is proportional to S. And somehow we have to find, explain this.
Using what we know from kinetics.
So we have to come up with a mechanism, and then solve the mechanism.
And make sure that it reproduces the data.
So we're not the first ones to do this, obviously.
Michaelis and Menton did this many years ago.
For this mechanism.
And the idea is, you have the enzyme plus a substrate react, k1, k minus one, to form a complex.
Enzyme substrate complex.
But unlike what we've drawn before in terms of this hump, where there's an activated complex which is not stable, in this case here this enzyme substrate combination lasts a long enough time that it's basically a stable complex.
So we're going to write this as a real intermediate that lasts long enough for you to be able to fish it out.
And characterize it.
And then eventually, that, k2, k minus two, goes to product plus enzyme.
So the enzyme is a catalyst that forms a long-lived complex.
And if you were to draw this, then, in our energy diagram, where we have the reaction coordinate here, you start out with your enzyme plus substrate.
Go up, and you form your intermediate up here.
ES.
And I put a little dimple in there, because it's stable enough that it's not on the top of the hump, but it lives long enough.
And it comes back down to form the product.
Plus the enzyme.
Without the enzyme in there, this hump would be way, way up there.
Would be maybe a factor of ten higher.
So this really lowers it a lot.
Now we can start to solve this mechanism.
And I forgot one arrow.
There which is the arrow going back.
Which you usually don't see, but we might as well keep it there.
For the sake of completeness.
So what do we know?
We know that this intermediate concentration is very small.
The enzyme concentration itself is very small.
Intermediate is very small.
And it doesn't change very much.
Not changing much.
So that means that we need to use a steady state approximation.
So let's write down the rate for this intermediate, d[ES]/dt.
It gets formed through the forward process.
I'm going to put my brackets back in because E and ES would look the same otherwise.
Gets destroyed.
Through the backward ways.
And I'm going to use a steady state approximation.
So I'm already going to start adding steady state here every time I see an intermediate.
Get it destroyed to make products.
It gets recreated through the backwards reaction from the products.
And I'm going to set that equal to zero for the steady state approximation.
Now, we don't really want to have this E floating around here.
Because this is something that is very hard to measure.
It's much easier to measure the initial concentration of the enzyme.
What you put in there, instead of the amount that's not being bound up.
So we're going to solve, instead, in terms of [E] is equal to [E]0 minus [ES], where this is the initial concentration and this is the amount that's binding substrate.
And this is the amount of free enzyme here then.
And when you do that, and you plug in here, and you plug in here, you get your result.
Which is that [E] steady state, [ES] steady state, is this ratio.
k1 times [S] plus k minus two times the product divided by k minus one plus k2 plus k1 times [S] plus k minus two times the product, times proportional to the initial concentration of enzyme.
And then you can take your steady state approximation for the intermediate and plug it back into your rate equation.
So the velocity, we defined as the rate which is minus d[S]/dt, which is k1 times [E] times [S], this is the destruction of the substrate minus k minus one, times [ES] steady state, the backwards process.
So we put in, instead of this [E] we put in [E] is equal to [E]0 minus [ES] steady state.
And then we put in for [ES] steady state, we put in what we found here.
We turn the crank on the algebra.
And we find that the velocity, then, is k1 k2 times the substrate concentration.
Minus k minus one times k minus two times the product concentration.
The whole thing times the initial enzyme concentration.
And then k minus one plus k2 on the bottom.
Plus k1 [S] plus k minus two times the product.
So, as I mentioned, this product here usually just floats away.
And so locally, where you're doing the, and then it gets used by the next step in the cycle.
So this is pretty much zero.
We can pretty much ignore this.
We can ignore this guy here too.
Because it's not, you don't have any steady state or an equilibrium.
You have a steady state but not an equilibrium situation.
And so in that case here, you can rewrite this then as k1 k2 times the substrate times [E]0 divided by k minus one plus k2 plus k1 times the substrate.
And now we can look at these experimental observations and see whether they match our mechanism.
If we can understand something.
So let's look at the initial rate.
Initial rate is supposed to be proportional to the substrate.
Initial rate is supposed to be proportional to the substrate.
So the initial rate, k1 k2 plus k1, if at early times, plus one.
So the initial rate is going to be, somehow I've got the product missing here.
No, I got it backwards here.
This is not the initial rate.
This is the rate where [S] is small.
The initial rate is when you don't have any products made.
Or when the product concentration is very low.
And because we're making the assumption that the product concentration is basically equal to zero here, this is basically the initial rate here.
So by taking the product equal to zero here, we're also saying that this is the same thing as the initial rate.
So this is what it looks like here..
The initial rate here.
And this is, this, you rewrite as v initial is equal to k2 times [S] times [E]0.
You divide by k1 up and down.
And you have this k minus one plus k2 over k1 and then plus [S] sitting down there.
And you define this ratio of rates, k minus one plus k2 over k1 as the KM, the Michaelis constant, by definition.
And this is an interesting ratio.
This is the rate, k minus one is the rate of destroying the enzyme substrate complex by going back to the reactants.
k2 is the destruction of the complex by going to the product.
And k1's the creation of the complex.
So this is the rate of destruction of the complex divided by the rate of creation of the complex.
So if the rate of destruction of the complex is much faster than the rate of creation, meaning that this is a large number, then you're not going to pile up any complex.
It's going to be destroyed as soon as you create it.
So if KM is large, then the concentration, [ES], is going to be very small.
Compared to [E]0.
But if the rate of destruction of the complex is small compared to the rate of creation, you create complexes, you create complexes but you don't keep up in terms of destroying them, in terms of making products, or going back to reactants.
And so you end up saturating your enzyme.
Every enzyme ends up having substrate bound to it.
So when KM is very small, then [ES] goes to saturation.
Basically, when KM is very small, then you're limited by the rate, the second rate in the process, of the enzyme falling apart.
To form the product.
You have to wait until that happens.
Because then the product just floats away.
And that becomes your rate limiting step.
Another way that you also will see this written is as this, then, is equal to, we'll define k cat is equal to k2.
k cat times the enzyme concentration times the substrate, [E]0, divided by KM plus but the substrate concentration.
So let's look at a few limiting cases then.
First limiting case is, suppose, that [S] is large.
Let's take [S] to be much larger than KM.
Because you look at the denominator, and you see it's this ratio of rates but there's this concentration that's important here.
So in one case KM is going to dominate, and in the other limiting case the substrate concentration is going to dominate.
So let's say that the substrate concentration dominates.
Meaning K sub M is small.
In which case we already saw.
If K sub M is small then you reach saturation.
And in the equation, then, the velocity, KM is small, [S] is large, the [S]'s cancel out and the velocity is equal to k cat times the initial substrate concentration.
Both of these are constants.
The velocity is constant.
And the rate is constant, it's that limit up here.
That's experimentally seen.
And the rate is depending on the initial substrate concentration.
Initial enzyme concentration.
All the enzymes have a substrate in there.
The more enzymes you have to begin with, the more intermediates you're going to have, the faster you going to make products.
And then it's going to depend on this on the rate, k cat, which is just k2, which is the rate of formation of products.
That becomes the rate limiting step here.
The other special case is if substrate concentration is very small compared to KM.
And in that case here, and the other thing that we're going to do is, we're going to, because we now understand this as this maximum rate up there, we're going to call this v max.
k cat times [E]0, we're going to call it v max.
And so when [S] is very small, v, then, is equal to v max, which is k cat times [E]0.
This is v max here now.
So we can rewrite this as v max times substrate divided by KM plus the substrate concentration.
Another way of writing it.
Capital K. v max times the substrate concentration.
And we have KM plus [S].
But [S] is very small.
So we drop it.
And this is then proportional to the substrate concentration.
And then that's small substrate concentration compared to KM, we're sitting right here.
Where it's linear, where the rate is linear.
And there's a third place on the graph which is interesting.
Which is when the substrate concentration is equal to KM.
In that case there, you plug [S] equal to KM, and you end up with the velocity then is equal to v max divided by two.
So when [S] is equal to KM, you are halfway up.
There's v max over two and there's KM sitting here.
When [S] is equal to KM, you're at v max over two.
And so enzymes, then, are labeled by their KM's.
Because then it becomes very important to know how strongly they bind the substrate.
Sometimes you want the enzyme to bind it very strongly.
Sometimes you don't, you want it to be fleeting.
Depends on the role that the enzyme plays.
Now there's a way to plot this that extracts out these important numbers.
k cat and KM.
And that's the Lineweaver-Burk plot.. Lineweaver-Burk plot.. And I just looked up this morning to see if Mr. Lineweaver was still alive.
And as far as I can tell he's still alive.
He was 97, in 2003.
So as of 2007 he was still alive.
He's getting up there.
One of the most cited papers that you have in your notes is the in Jack's, was the paper that showed how to go from this curved line to a straight line by plotting one over v versus [S], instead of v versus [S].
So if you take your equation and massage it, one over v KM over v max times [S], plus one over v max, we haven't done anything except rewrite the equation in terms of one of v versus one over [S].
So it becomes linear.
In one over [S], and there's one over v sitting here.
And you get a straight line.
With an intercept here that's one over v max.
And if you keep going, extrapolate out, you get this point here to be minus 1 over KM and the slope is KM over v max.
And v max was equal to k cat times [E]0, so you get k cat out of this.
So it turned out to be a very useful plot.
It's very easy to plot a straight line, especially before computers.
In the age of computers.
And the referees, there were six referees that got this paper and pretty much turned it down because they didn't think there was any new chemistry in it.
Which is true, there's no new chemistry.
It's just a way of rewriting the plot.
But it was very important nevertheless.
OK, any questions on catalysis?
Enzymes?
Arrhenius?
Alright, next time we'll oscillating reactions and recap the course.
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So last time, there was a question that was asked about the order of magnitude rates.
You asked the question about the concentration of the catalysts.
So the rates that I gave you, which were the rates of reactions are basically order of magnitude.
They're estimates and odd numbers that, they obviously depend, as you pointed out, depend on the concentration of the catalyst.
Depends on the concentration of the reactants.
Depends on where you are on those concentrations.
As we saw when we did enzyme catalysis, is that depending on KM, the Michaelis constant, and the substrate concentration, if your substrate's concentration is much higher than KM, you are in the maximum velocity limit, where the rate depends on k2 times the initial concentration of enzyme.
In that case the rate of the reaction depends just on the concentration of enzyme, and not on the concentration of substrate.
But if you are at the small concentration of substrate relative to KM, then your rate depends linearly with the concentration of substrate.
And the same thing happens with the non-enzymatic catalyst, that there will be some sort of mechanism that goes along with it.
And most likely it's going to be second order in the catalyst.
And the reactant if it's just the two body process.
But it could be more complicated.
So definitely those concentrations will go in there.
But a priori, it's not clear how they will go in there.
But the orders of magnitude are roughly right.
And the basic idea is that you change your rate of reactions by 15, 18 orders of magnitude with the right catalyst, especially if it's a biological catalyst.
One concept that I didn't go through last time, which is also quite useful, it's called the turnover number.
Turnover number.
And this is useful for any kind of catalyst.
But for enzymatic, or enzyme catalysts, the turnover number is k cat.
The number per second.
It's a rate per second.
How many times does a reactant get, the number of times that a reactant turns into a product per second, per catalyst.
So let me write it out.
So it's the number of product formed per second, per second, per enzyme, per molecule of enzyme.
OK, so if one enzyme is going to be there, there's going to be some substrate going in.
Coming in, the pocket coming out.
Coming in, the pocket coming out, so the number of cycles of this process per second.
is the turnover number.
And so if you are where the concentration of substrate is very much larger than the Michaelis constant, where are you are at the saturation limit, then this turnover number doesn't depend on the substrate concentration.
It's just k cat.
So that's why k cat is important.
As this turnover number in the maximum velocity limit.
Because in the maximum velocity limit, then the process that's dominant is the second process, ES goes to enzyme plus product.
The first process here is not the rate limiting step.
This is the rate limiting step.
Because the substrate concentration is very high.
So this first part goes very high, goes very fast.
And the number of product formed per second, per molecule of enzyme, the number of molecules formed for second, that's the rate of product formation.
Rate of product formation per second.
Divided by the concentration of the enzyme.
If you normalize by the number of enzyme.
So this becomes rate of product formation in moles per second, divided by moles of enzyme.
It's the same thing as what we're saying here in words.
And so the rate of product formation is dP/dt, and you divide that by the concentration of enzyme, which you can write as E0.
dp/dt, and the maximum rate limit is k max.
It's a constant divided by E0.
And k max is k cat times E0 divided by E0.
The E0's come out.
And this is k cat.
Which is basically k2.
So that's the turnover number.
Any questions on this turnover number?
You see that a lot in catalyst literature and especially the enzymes.
I have to get out these lecture notes.
Send these back.
OK, so the last topic I want to talk about is oscillating reactions.
It's sort of like an interesting topic.
And it's applicable not just to just chemistry, but it's basically playing with the differential equations.
And you'll see this sort of stuff later if you keep doing science that involves coupled differential equations.
It's just all over the place.
So normally we have an equilibrium process.
A goes to B, B goes back to A.
And if you start out with something which is out of equilibrium, then you will eventually reach equilibrium at a rate which is dependent on those two rates.
k1 plus k minus one.
So B will to come up to some B equilibrium.
And A will, if you start out with a lot of A, and a little bit of B, and concentration of A, will come down smoothly, monotonically, to the equilibrium state.
But sometimes, if you're out of equilibrium, and this doesn't have to be chemistry.
It could be anything that's out of equilibrium, your emotion's out of equilibrium.
Stock market is out of equilibrium.
Population dynamics are out of equilibrium.
The weather.
Whatever.
So, you're out of equilibrium, let's say this is equilibrium for A, this is equilibrium for B, let me try to keep them somewhat the same as here, I messed up.
You're out of equilibrium, you start down here for A.
And you, instead of going linearly down or monotonically down, exponentially down to the equilibrium, instead you overshoot.
And you keep going back and forth.
Like a spring.
And then B does the same thing.
Overshoots, comes down.
Overshoots, comes down.
Like a pendulum.
A lot of things work like that.
The heart is something that works like that, right?
It beats.
It's not in equilibrium.
Good thing it's not in equilibrium.
So the heart is one example of something that oscillates, back and forth, back and forth.
And so there are actually many examples of complicated processes that involve chemistry, that aren't at equilibrium, that instead go back and forth, in and out, of on one side or the other of equilibrium.
Now, there's an important feature of, if you want to build a chemical process that looks like this.
It's often very useful to have a particular step in the mechanism that's called an autocatalysis step.
And that provides feedback.
It's like, if my microphone here, the speaker which is here, I were to stand right in front of the speaker and my microphone would pick up the volume from the speaker, it would get amplified, and I get feedback.
And that would be a bad thing.
So this is a form of feedback here.
You need some sort of feedback that the reaction knows that you're building up, that this is going too fast.
And then there's a feedback process that brings it back up.
And a feedback process that goes back and forth.
And the way you get this feedback is by having a step in the mechanism that has both a molecule as part of the reactants and as part of the product.
It's not the same as a chain reaction, where we talked about an intermediate being recycled and building up.
This is an actual reactant and an actual product that's part of this step here.
So this is an autocatalytic step.
And let's just see what that step looks like.
Suppose we just have that step.
So we want to find out, as a function of time, if you start out with some A and B, what happens to the concentration of B as a function of time.
The time dependence of that.
Clearly it's going to build up.
But how is it going to build up?
So we need to solve for B as a function of time.
So let's go ahead and write down our rates.
The rate of the reaction is d[A]/dt, is, let's put a rate constant here.
k, k times [A] times [B].
Now, [B] and [A] are related to each other through stoichiometry.
The concentration of B is the concentration of B that he started out with, [B]0, then every time you destroy an A, you create 2B.
So it's plus two times the amount of A that you've destroyed, [A]0 minus [A], [A]0 is what you started out with.
This is what's left over.
So the difference is what you've destroyed.
But every time you destroy an A to form 2B, you also destroy a b.
So you have to subtract away the B that you've destroyed.
Which is [A]0 minus [A].
This is the 2B that you created by destroying A, and this is the B that you destroy by having to destroy A, for the reaction.
You've got to bean-count correctly.
Keep track of everything.
So [B], then, is just, I'm going to drop those brackets.
B0 plus A0 minus A.
So you can plug that in here.
And is equal to k times A, times B0 plus B0 minus A.
And now you have a differential equation that only contains A and time.
B0 is a constant.
And the way you solve that is by partial fractions.
Use partial fractions, solve for this.
And I'm not going to go through the math of solving for it.
It's not that interesting.
Let me give you the answer.
You end up with [B] as a function of time, is this function A0 plus B0 over one plus A0 over B0 times e to the minus k times A0 plus B0 plus time.
And that's where the time component comes in.
Whoops, I'm sorry.
Usually I don't have this on, let's turn this off.
That's where the time dependence comes in.
And clearly, at t is equal to infinity, t equals infinity just goes to zero.
So you have B as A0, plus B0, everything goes through to the product.
t equals to zero, this is equal to one.
And B is equal to just B0.
Alright, and so the curve looks something like this.
This is what it turns out to look like.
We're starting with B0 here.
There's an induction period where very little happens.
It's like the dormant stage.
So, locusts that are sitting in the ground for seventeen years waiting for something to happen.
Dormant stage, the induction period.
And suddenly something starts to happen.
There's an inflection point.
And you get to A0 plus B0.
The locusts wake up.
And go to maximum concentration.
So there's an s-like shape that has an inflection point and an induction period.
That's pretty typical of an auto-catalytic reaction.
OK so let's apply it to a process now.
And again, this is going to be coupled differential equations.
Broadly, broadly applicable, the example that I'm going to give you.
This mechanism has been used for all sorts of things.
So what do we have here?
We're going to have an island.
In the ocean.
Island here.
And it's going to be sunny, so let's get some yellow chalk here.
There's the sun.
And every now and then it's going to rain.
We need some white chalk for the rain.
There's a cloud that comes in.
That's going to be one of the reactants.
And then some seeds blow in from the continent, and grass starts to grow on the island.
Grass.
And then there's a shipwreck and a couple of rabbits get on the island.
And now we have rabbits.
Rabbits.
A little tail on the back.
I hate rabbits.
I have a little vegetable garden, a little tiny vegetable garden.
And there are rabbits.
And I have to fence in my vegetable garden and make it impermeable impenetrable to rabbits.
And rabbits are very clever.
And my vegetable garden looks like a fortress.
It's got a green fence.
Heavy duty things.
It's terrible.
Rabbits are awful.
They're very cute but they're awful.
Alright, so now what happens.
Well, given the rain and the grass, it turns out that this island can only support 25 rabbits.
If there are more rabbits than that, then they all eat too much.
If there are less rabbits then that's fine.
But 25 is about the maximum that can support on this island.
The rabbits, unfortunately, are not that smart.
They don't know that.
They don't know that only 25 rabbits can live there.
So, what happens is that the Year one, Year one there are ten rabbits.
And the food condition is great.
Lots of food, number of rabbits.
Lots of food.
So the bunnies, the rabbits make bunnies.
Lots of rabbits.
Year two, they kind of went overboard.
Now there are 50 rabbits.
The food is lousy.
Terrible.
Well, there's a feedback here.
Feedback and rabbits start to die off.
Don't reproduce as much.
Year three, we're back to ten.
Great food.
Rabbits don't notice.
They multiply.
Food is terrible.
And they just don't learn.
Humans are a little bit like that.
Except our cycle isn't one year, it's usually on the order of 30 years.
It's like the memory of a generation disappearing.
And we have to relearn the mistakes of the previous generation.
Science is like that too.
Science goes in cycles.
20, 30-year cycles.
Topics that were out of fashion 20 years ago, 25 years ago, come back.
And suddenly everybody's excited about them.
And all the ideas, of course we make progress.
We go a little bit further than we did 20 years ago.
But all the ideas that were out of fashion 20 years ago come back.
People get really excited, and they relearn all the mistakes that people made at the beginning.
Technology has improved, we know more, and therefore we go a little bit further this cycle than we did the previous cycle.
But it's really interesting to look at history of science and see this sort of cycle.
So let's write a mechanism.
Let's write a mechanism here.
So we have rain plus grass makes more grass.
At the rate k1.
That's autocatalytic.
It provides some feedback.
Then we have grass plus rabbits make more rabbits.
OK, let's do more because I don't know how much, more, more.
Also autocatalytic.
Then we have rabbits.
Eventually, unfortunately, rabbits become dead rabbits.
Rate k2 k3, and that's the termination of the process.
OK, so now let's put some chemistry on this.
Let's assume that instead of having objects, live objects here, we have chemical molecules.
So A plus B goes to 2B.
That's the first step.
Then we have B plus C, C is the rabbits.
And then C goes to some sort of products.
Soil.
This is a very famous mechanism.
It's called the Lotka-Volterra mechanism.
It's also called the predator-prey mechanism.
In my case here, I made it sort of warm and fuzzy.
I didn't have a predator in there.
But you could change this.
Instead of having rabbits and grass, we could have rabbits and foxes.
Where A is the grass, if you take A to be the grass.
B to be the rabbits.
And C you be the foxes.
Works the same.
You have grass plus rabbits makes more rabbits.
Rabbit plus foxes make more foxes.
Eventually the foxes die off.
Same idea.
So now, let's solve this.
Let's assume that A, the rain here, it's constant.
There's a steady supply of rain.
It doesn't go away.
That makes sense in my example here.
So the concentration of A is kept constant.
In order to get oscillations to keep going, that turns out to be important.
To have one of the reactants just keeping being reintroduced in the system.
Because it gets used up.
And when all the reactant gets used up, then you're done.
So in the case of the heart, we have to keep feeding ourselves to produce energy.
Otherwise the heart would stop.
OK, now we want to know what is the concentration of B and C as a function of time.
So we write down our kinetic equations.
dB/dt gets produced through the first step, k1 A times B, gets destroyed in the second step.
k2 B times C. dC/dt, C gets produced in the second step, k2 B times C. Gets destroyed in the third step.
k3 times C, got to do your bean-counting correctly.
And the strategy here that we're going to use to solve the problem is to assume that we're at steady state.
We're going to find the solution at steady state.
Then we're going to perturb away from steady state a little bit.
And see whether this creates an oscillation.
First we need to find out what the steady state solution is.
So first, let's find steady state.
Then perturb away from steady state.
We're looking at a harmonic response, if you've heard that term.
When it's going to look like a spring, close to its equilibrium state.
That means that we're going to set it these two things equal to zero, for this steady state.
Things are not changing.
Concentration of B and C are not changing.
A is not changing on purpose here.
And so all these guys here are then steady state concentrations.
If we set it equal to zero.
We solve for A in this.
For C in this process here.
The B's cancel out here.
We divide by B.
And you solve for C as a function of A.
So this one here gives you C steady state is equal to k1 over k2 times the concentration of A.
And this one here, now you put in the C and you solve for B, B steady state, is equal to k3 over k2.
So the next step is to perturb away from equilibrium.
Or from the steady state, rather.
This is not equilibrium, it's not equilibrium because we keep adding A.
We have to put some input into the system.
So here we are.
We've got some concentration of B here at the steady state.
We have some concentration of C at the steady state.
And now we're going to add some delta B to the system, or delta C. Or both.
Delta C, we're going to ask the question, given our system here, how is it going to respond?
It has a number of choices.
It could respond by going back to the steady state in a monotonic fashion.
Without overshooting.
This is kind of like an overdamped system.
It's like having a shock absorber on your car.
You go over a bump, the car doesn't oscillate up and down.
It's sort of damped.
It's a damped oscillator.
It goes back to the steady state.
Or, you could, there we go, there's B, C. Or you could perturb, and it could just stay where it is.
That could happen too.
That's an inelastic response.
Kind of like the supply and demand with oil.
The price of oil goes from $10 a barrel to $120 a barrel.
The use of oil doesn't seem to be affected very much by the increasing price.
You're still using just as much oil.
There's an inelastic response.
What we're looking for is something that has an elastic response, or a harmonic response.
We perturb it, it tries to get back to steady state.
But because of feedback, and not being over-damped, if it's an oscillator, it goes up and down.
Like a scale that is not well, well, like a car that has bad shock absorbers, all it has is the springs.
Up and down.
So what we want is, we want to solve for dB/dt as a function of time.
We want, well, I'm going to do everything in green.
Since I lost my white chalk.
We want delta B, delta C as a function of time.
So, that means that we start with B is B steady state plus delta B, C is C steady state plus delta C. And we put that in our equations.
For dB/dt and dC/dt.
So now dB/dt is the same thing as dB steady state plus delta B dt, which is d delta B / dt.
Because this is a constant here.
And that's equal to k1 times A, times B steady state plus delta B.
Minus k2 times B steady state plus delta B times C steady state plus delta C. And then you have the same thing for dC/dt, it's going to look exactly identical.
d delta C / dt.
Thank you much, this is magic.
k2 times B steady state plus delta B, times C steady state plus delta C, minus k3 times C steady state plus delta C. So these are two differential equations.
And there are coupled differential equations.
Because delta B is occurring here, here, and here.
So the time derivative of delta C depends on delta B.
And the time derivative of delta B depends on delta C. It's a coupled system.
And if you actually expand this out and take away all the terms that cancel out, you end up with something that looks much simpler but is just as hard.
d delta B / dt is equal to minus k3 delta C and d delta C / dt is equal to k1 A delta B.
And then you really see that it's coupled, because the time derivative of delta B depends on delta C. And the time derivative of delta C depends on delta B.
And there's the feedback sitting right here.
And that's what you do.
You learn how to solve in 18.03.
So I'm not going to solve it here, I'm just going to give you the solutions.
Because that's what we're interested in.
After all, you're not expected to learn how to solve this equation.
Although you should be able to put the solutions in and make sure that they do work out.
And so the solutions are harmonic solutions.
Delta B is a function of time is equal to delta B0.
The amount of the perturbation times the cosine of omega t, where omega is the frequency of the oscillation.
Minus some pre-factor here, k3 over k1 A to the 1/2 power.
Times delta C0 sine of omega t. And then delta C looks the same.
The same frequency.
Delta C0, cosine of omega t. Plus k1 over A times k3 to the 1/2 power.
Delta B0 sine of omega t, where omega, the frequency of the oscillation, is these rates.
k1, k3 and A.
So obviously, keeping the concentration of a constant is important because we don't want a time dependence here.
The rain is constant.
And then if you do that, then you have this oscillation.
Up and down and up and down with this frequency here.
You get exactly what we're trying to get.
Which is this process right here.
So let me give you an example of this.
And the best way to do is to actually go and see a movie of a reaction that looks like this.
This is the reaction that was optimized and created for demonstration purposes.
So it's a pretty complicated reaction.
There are ten steps to the mechanism.
And if we gave you that mechanism to solve for the final exam, you would still be here probably two weeks form now.
It's pretty complicated.
You'd need a computer for sure.
So, ten steps, well, that's what's known.
There are probably more than that.
It's probably more complicated than what we really know in terms of fishing out intermediates.
And the basic reaction is, you start out with an oxide of iodine, which is clear.
You create I2, which is going to look gold.
In solution, and then it creates Ii minus, which is going to look deep blue.
And then it'll feed back and start again.
So we're going to have this cycle of products.
And amongst those ten steps, or more than ten steps, there are a few autocatalytic steps, which provide the feedback.
So now I have to start it.
Let's go ahead and read.
Alright.
So here we go.
So, there are two solutions, A and solution C. Actually, we mix three solutions together.
The potassium iodide is what is going to give rise to the different colors.
And now we're in the gold phase of the reaction.
It's important to stir, otherwise it wouldn't, and then it turns deep blue.
And then there's an induction period, where you wait for a while.
And then eventually it should go clear.
So it goes clear.
And then it turns gold again.
And then deep blue, and there's the oscillation.
And then a long induction period, and then it goes through that cycle over and over again.
And the recipe can be found pretty easily.
And if you ever are in a lab, have access to doing this, it's a pretty easy reaction to put together.
The only problem is that unless you use fresh hydrogen peroxide, it doesn't work.
Because hydrogen peroxide tends to go bad over time.
If you're going to do this, buy your hydrogen peroxide immediately before you try to do the experiment.
And then it'll work.
So what happens is that as long as there is a steady supply of hydrogen peroxide in here, it'll keep oscillating.
But as you've used up the hydrogen peroxide, this is going to start to die.
And eventually it'll have this pale blue solution.
And it's like the heart stopping.
If you want to start it up again, add more hydrogen peroxide and it'll start up again.
OK, any questions?
There's like an infinite number of these oscillating reactions that people have discovered.
And optimized, that you can find in different places.
Questions?
Questions about the final.
I didn't get to do a review, but you'll get that done with the TAs.
Yes
[INAUDIBLE]
Because there's enough of the equivalent of A, of the rain here.
So it'll oscillate and there'll be some damping.
So what will happen in this reaction is that the induction period will get longer, and longer, and longer.
Because the w, this term right here, the frequency which depends on the concentration of A, that well get slower and slower.
And eventually it'll just die.
OK?
Well, active learning is the idea that students, to really learn something, to really understand something have to be actively involved and that just sitting passively and listening to a lecture really doesn't help students develop the higher order cognitive processes that they need to really, really understand something.
So you can listen to something, you can watch a movie, you can watch TV, and you can generally get the plot.
But if you're asked to recall specific details or to even explain a particular nuance associated with the TV show or movie, you can't really do it.
And that's what happens often in a lecture, is that students will sit in the lecture, they'll write down what's being said, but they're not really engaged with the material.
So active learning is this idea of, people say minds on, always hands on sometimes.
So students have to be actively with their mind thinking about the material, applying what's being said, and given opportunities within the lecture to apply what's being taught or the topic at hand.
And then active learning, strictly speaking means that just one particular individual is active.
Interactive, we tend to parse that a little bit and say interactive learning would mean the student has been active in his or her own mind in thinking about the material, but then is also interacting with others, peers or potentially the faculty member or TA in order to further develop understanding, construct meaning for the topic.
I always start maybe the second session of the class.
The second class meeting is a discussion of what we know about how people learn.
So a discussion of the literature and the research on human cognition and learning.
And if you take a constructivist point of view or a constructionist point of view, which really says that as I said before, to understand, people have to make meaning of a topic, they have to construct their own meaning.
And we show the research that really shows that this is true.
For higher level cognitive processes, people have to be actively engaged.
And there's research to show that.
We also show the classroom-based research, so [?
Freeman's ?]
2014 paper that was a meta-analysis of this 225 other studies that showed that in courses, in college-level courses where active learning was used, there was a 12% decrease in the failure rate.
And they normalized it to all of the important factors that they should be normalized to-- the experience of the instructor, the size of the class, the type of the institution, the position that the class is situated within the larger curriculum.
And across the board it was shown that there was a 12% decrease on average of the failure rate.
And they make a comment in the paper that if that had been a clinical trial of a drug and 12% of the people on the drug had [?
shown ?]
marked improvement, they would have had to stop the trial and give everyone the drug.
So this idea that there's a 12% decrease in the failure rate in courses that use active learning, to me is pretty compelling that we should all be using active learning.
So whenever possible, because our students are MIT students, we use data, we use the research, and we try to find really good research, solid research that shows the way people learn and then how to support that with specific classroom practices.
So many of the students haven't had the experience of being in a class where active learning was used, so they don't really understand it.
So when we start to talk about it as a way of teaching, they may not really get it.
So throughout the course, from the first class all the way through, I try to use several different types of active learning exercises each class.
So the students themselves are actively engaged with the material from the first day.
So I may have them break into pairs and discuss a particular topic or identify something they didn't understand from the pre-class readings.
And then after three minutes, they can either share their comments with someone else or maybe we just report out to the larger group.
If they just report, if they just write down and then report back, that's a pretty good example of active learning.
It's pretty simple, it's what you might call low-hanging fruit in terms of what it takes for a faculty member to do that.
So I'll do that, I'll have them do that.
And then I step back and say, OK, why did I ask you to think about it?
Why did I ask you to take three minutes before we had this discussion?
So I try to deconstruct the exercise for them, showing them the advantages for the learner.
The beach ball strategy is a way to get students who might not want to answer questions or be a little less outgoing in their answering of questions to answer questions.
So there's a few different ways to use the beach ball, actually.
And one is you just pose a question that has enough variation in the types of responses.
So you wouldn't say OK, what's 2 plus 2 as an example for the beach ball question.
You would have to have something that was a little bit richer.
And then you throw the ball at one student or up into the lecture hall, if it's a big lecture hall.
And whoever catches it has to respond.
And that response, you share it with the group, comment on it if you want to comment on it.
And then that person who gave the first response throws the ball to somebody else.
And then whoever catches the ball that time responds, et cetera.
You can use it coupled with a picture prompt or a graph.
So if you show a graph or an image and you say, tell me something you observe about this image, and then you throw the ball and the person catches it first says one thing they observe, and they throw it and the next person says something that they observed about the picture prompt.
So that gives a foundation and it opens up space for lots of responses.
The beach ball does a few things.
One is that as the instructor, you don't have to be the one that calls on the students all the time, so the students don't feel like you're picking on them, that you're the one that's cold calling on them.
And in fact you can throw on the beach ball completely blind the first time, close your eyes, throw the ball, and so nobody really feels that you've called on them.
And then subsequently, it's other students that are throwing the ball, so there's no issue that you're the one that's picking on the students.
The other thing is it's kind of fun, you're throwing the ball around, so it's hard to get upset, it's hard to get as a student nervous that you're going to have to answer the question.
And in fact, you could dodge the ball if you really didn't want to answer the question.
So it's a great way to get students talking.
And in general, it's pretty fun.
And at first I thought oh, I can't really use this with grad students, it's too goofy, it's too baby or silly.
But actually, every time I use it with grad students, postdocs, everybody seems perfectly fine with it.
I think it's more to do with the nature of the question.
If you just ask stupid questions and throw the ball around, it'll be stupid.
But if you ask good questions and it's a way for students to give you their answers, then it seems to work totally fine.
That's the general way.
Sometimes I'll actually write questions on the beach ball and then throw the ball into the class.
And then students are loosely supposed to answer the question that's in front of them when they catch the ball.
So that's another way to do it.
But it's a nice, inexpensive, fairly flexible technique.
The technique of debate is a really interesting and useful technique.
I make sure to arbitrarily assign students the side that they'll be taking in the debate because it's pretty easy to debate something you feel passionate about.
It's not so easy to debate something that you don't feel passionate about, or where you feel the other side is correct.
In terms of learning, I think it's a great exercise because it forces people to, at least superficially, understand the stated arguments for the side they're taking and then find a way to at least be semi convincing about those arguments.
And then I give them a period of time to prepare, to sort of prepare their arguments about how they're going to defend their positions.
And then I have them voice their arguments for the particular side that they've been assigned.
So in crafting the arguments, pro if you're a con, you really have to start to understand what goes behind those arguments a little bit better, and so that's where the learning takes place I think.
The other thing is that the students get a little bit competitive, so they want to construct a good argument.
They want to sound convincing.
And so that extra little bit means that they really do put in the effort to construct a good argument and to make it believable.
If I were making recommendations to someone who had never used a debate before I would say that you should, first of all, don't pick anything that's really emotional to start out with.
So people can have a pretty strong opinion about it, but it's not something that's really deep.
So I would stay away from something too touchy.
The other thing is that it should be something where it's really not a slam dunk one way or the other.
So you want to make sure that the topic that you choose really does have two reasonable positions.
You don't want to give one team just a terrible thing to have to argue.
The other thing you want to make sure is that you give students enough time to craft their arguments and to debate the points.
So that if you cut the debate short after three minutes, it's just not going to help.
So you have to set aside a pretty good chunk of time.
Another active learning technique that can be very, very effective in supporting student learning is the jigsaw.
And it gets its name from the activity involved when you do a jigsaw puzzle.
And usually, most people when they do a jigsaw will look for the-- they'll find all the edge pieces, or they'll find all the blue pieces, or they'll find all the pieces that have a certain image on them.
You do the homogeneous aspects of the exercise.
So you get all the edge pieces, the blue pieces, the particular image pizzas.
But eventually, you have to take those pieces and put them together.
You have to link them up with pieces that don't look quite like those pieces, so that's the heterogeneous component of it.
And so what you do in a jigsaw is you ask students to become experts, if you will, homogeneous experts, in a particular area, on a particular topic.
And you can do this by giving them a reading in class, or you can do this by asking them to reflect on something that you've addressed in a previous class or that they've addressed in a previous homework.
But they sit together, or they group together in this, what I call, homogeneous group, where everyone is talking about the same particular topic.
And they just discuss that topic and they become experts in that homogeneous topic.
And then you can have three or four different topics, whatever, depending on the size of the class.
And you have these local groups of experts, like the edge pieces, and the blue pieces, and the picture pieces.
And then after giving them a period of time, maybe five minutes, 10 minutes, depending on the class, you break them up into heterogeneous groups.
So you get one with one representative from Group A, goes with one representative from Group B, and one representative from Group C, and one representative from group D. And now you have a new heterogeneous group that has an A, a B, a C, and a D in it.
Or if we use the puzzle analogy, you have a group that has some person that's the edge expert, some person that's the picture expert, and some person that's the blue piece expert.
And they're all in a group.
And now they share what they know individually with each other and then they combine that expertise, they synthesize that expertise, to make something bigger to come up with a more global understanding of the problem.
And then each group does that in their own way.
So the heterogeneous groups then form a more holistic, or a more heterogeneous, solution to the problem, as if they're putting together the different kinds of puzzle pieces.
And then again, as with the other activities, we ask them to then report out.
Before that I usually will circulate in the room the same way I do with the pair share to hear what each of the groups are saying, to see if any groups are having particular issues, or if they have some particular breakthroughs that I want to make sure they share with the class as a whole.
There are some logistical issues.
One, you want to make sure that the students really can become these homogeneous experts.
So that if you have a group of four people, that they've all really come up to speed on the topic.
So ideally, you would give a pre-class assignment and you'd say, yes.
I need everybody to read this article and to be able to explain this point, this point, and this point, and you would assign that to that particular group of students.
And then you'd do the same with a few other topics.
You're assuming then that the students have read the paper, or have read the article, whatever it is you want them to read, and that they understand it to the level that you hope they understood it.
Sometimes you don't have time for that or it just doesn't work out, so you give them something to read within class.
And again, that can be tricky.
Students may not have the same level of understanding, so groups may be a little bit spotty.
The other thing that can happen with the jigsaw, which is a really silly problem but sometimes it just trips you up, is if you have, depending on the number of students, you have to kind of think on your feet about how many students are in the homogeneous group versus how many students are in the heterogeneous group.
If it's a perfect square, it works fine.
So if you've got nine students, or 16 students, or 25 students, you know you have five groups in one situation and then five groups in the other, that means you have to have the same number of groups as heterogeneous topics.
So if I went A, B, C, D, I have to then have four groups because I had to have four A's, four B's, four C's, and four D's.
So sometimes just the number of topics and the number of students that you have in your class makes it kind of hard to make groups that don't have extra people or aren't short a particular expert.
And that's just something you want to plan beforehand.
And if somebody doesn't come to class, if you know you have 25 students in your class and someone doesn't show up, then you have to kind of deal with it on the fly.
But that can be the trickiest thing with the jigsaw, is just getting the number of students right.
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All right, great.
I looked over some of the Mud cards we had from last time.
And there were some really good questions, some really good points, and one lingering one from the previous time that I forgot to address.
Someone said, could we please just move the desks into an arrangement where we could see each other.
It was a fabulous suggestion.
Today we can't do it.
But I'm going to try to do it in the future whenever the logistics allow, because I think you're absolutely right.
A little bit more of a U, or a circle, would be great for this class.
So thank you for that suggestion.
Another one was that we didn't get to motivation theory.
And I think I mentioned this last time, the idea that there's a huge amount of stuff that's been written and published on motivation and theories of motivation.
And I'd be more than happy to discuss this as the semester goes on.
And it might be a great topic for one of those lunches that somebody was going to organize, one of those informal lunches that someone was going to organize.
Or maybe we can carve out some space at the end of the semester and pop it in, take something out and pop it in, because it's really fascinating.
It's really interesting, but it doesn't quite fit anywhere in particular.
So thank you for bringing that up.
And keep bugging me about it.
The other thing is someone asked if all those learning theories that we discussed were equally valid.
And yes, they are equally valid.
I mean, there are situations where being behaviorist could help.
There are situations where being constructivist could help.
It depends on what you're trying to have students learn.
So you want to make sure that you tailor the way that you're teaching to support what it is you want students to learn.
And we'll get into that in a lot more detail today when we talk about developing learning outcomes, and then also when we talk about tailoring your instructional activities to support those learning outcomes.
So this class, this particular class today is a balance between developing course design, developing intended learning outcomes, which for me is at the heart of it.
If you can come up with some good intended learning outcomes, you basically have designed your course.
And I'll show you how.
I mean, you can't just ignore it, but that's the hard work, and then the act of writing constructing a syllabus.
So once you've got those learning outcomes, you have an incredible anchor for the rest of your course.
And there was just a very timely haiku.
I put it on the wiki.
But there's this Twitter feed, academic haikus, which if you want to waste any time and read them, it's a way to waste time.
But there was one here, "that question you have?
I know where the answer is-- on the syllabus.
So it was, like I said, it came out yesterday.
I thought it was pretty timely.
And it's true.
If you write a good syllabus, it should really clear up a lot of, not the content, not the understanding level issues students have, but the issues with respect to when's the test, do we have to know this, how will this be graded, what's important?
All those things should be in your syllabus.
So that's why that's there.
So here's a little slide, a graph that I saw a few years ago, and I thought it's a really good way to start this discussion.
So on the y-axis we have sort of an assessment, a proclamation of students understanding.
They got it, they didn't get it.
And I think the word in French is [FRENCH].
So for some reason I find it very cool to say [FRENCH].
Got it.
I don't know.
So whether they get it, like whether after the tests, you've given a test, and you go, man, that student really got it.
That student really gets this material, and whether they passed the exam or failed the exam.
So the size of the circle is more or less the number of people that fall into those quadrants.
So anybody surprised by this plot?
Yes, Rachel
[INAUDIBLE]
OK. All right.
Well, there's the segue.
So we want, we don't want the green, but we more or less expect the green, right?
If you get it, you pass the exam.
If you pass the exam, you get it.
That's that, and we like to have lots of people have that happen too.
And then we also acknowledge that somebody isn't going to get it, and they're not and they're not going to pass the exam.
And that kind of maybe makes us sad, but it's not surprising.
And then anybody can kind of have a bad day or kind of a fluke.
So sometimes, sure, somebody will fail the exam, and actually really understand the material.
Generally speaking, that doesn't happen, and then there's often a reason for it.
But it's that red circle that Rachel mentioned that is troublesome, at the least, troublesome.
So why might that happen?
Why so many red circles?
Yeah, Dave
Because the test was set up for basic memorization, or you could figure out with a lucky guess with multiple choice.
So you could easily pass just with good test-taking skills or a good memory
Excellent.
So you pass with sort of good test taking skills.
Now I want us to hang on to that.
There's an underlying assumption in Dave's comment.
So I'd like to hang on to that thought, but let's hear a few other conjectures and see.
Rachel, why do you think you might have passed it?
So often I feel like TAs are either really good or really terrible when it comes to the exam reviews, and they're mainly good.
They often told us too much of what was going to be on the exam.
And so then you kind of just knew what the test was going to be if you could well on it if you [INAUDIBLE]
OK. That's interesting.
Yes, Gordon
Yeah, it's possible that [INAUDIBLE] tests [INAUDIBLE].
When it comes to true life situations and something that [INAUDIBLE] knowledge that was gained from that material, then [INAUDIBLE]
OK.
So Gordon is sort of kind of flirting with this idea of you as the instructor, this real life knowledge, this is what it means.
They have to have this practical working knowledge for us as the instructors to say they got it, right?
But the exam isn't measuring it.
And that's exactly what, that's what Dave was saying as well.
This idea that the exam is measuring one thing, but it's not really measuring whether they got it.
When we go home and think what does it mean to be a good material scientist, or what does it mean to be a good chemist, we don't think, oh, it means that you can define what a mitochondria is, or you can balance a simple equation.
And yet, we put that on the exam.
It's also a bit of the idea that students are good at taking tests.
They're good at gaming systems.
They're good at listening to TAs or TAs may have not given out appropriate information.
So it comes from both sides.
It comes from the students and it comes from the professors.
But you have control over this.
You shouldn't be writing an exam that people can get a good grade on, but then when you go home and say, oh, they got a good grade on this test, but they didn't get it.
You need to define what it means to get it in your field, to get this subject, whatever it is that the test is about.
And then you need to write a test that measures that.
And too many times we don't do it.
We're kind of rushed, or we don't really know.
We haven't really thought about how to measure whether somebody gets it.
So the first step in all this is to define what it is that means that they got it.
And that's the first step, and that's the hardest step.
But once that's done, the rest is easier, and it will save you from this.
OK?
OK.
So that's the topic of today's class.
And if you needed a little more motivation, I'd like you to read this paragraph, and I'd like you to tell me what it's talking about.
But I'm just going to give you about 2 and 1/2 minutes to read it.
And then we'll discuss what it means.
What do you think this is talking about?
Yes, David
Yeah, I think it stresses about how we can arrange by say, maybe schedules or activities that put them into groups.
Then maybe if we form a group, how we plan [INAUDIBLE]
OK.
So arranging schedules or activities.
OK. OK. Something else?
When you're planning, how to go about teaching a module, [INAUDIBLE]
Planning teaching.
What else?
Anything?
It's like a mailroom operation or something
Mailroom.
Something from the back, maybe?
You know you're smiling
I couldn't think of anything
All right.
Well, then don't say anything
The first is the course content, the course outline, and what you are going to teach the students, [INAUDIBLE]
OK.
So students, grouping students-- Adam, right?
Adam.
OK.
So content, arranging content.
Anybody in the back?
Flyer?
Yes, Gordon
I think you talked about sorting materials
Sorting materials.
So we'll go with maybe sorting.
So if I tell you-- so these are all great guesses.
They're as good as what I'm going to tell you the answer is.
So this was written to describe the process of doing your laundry.
Like doing the wash. You take the clothes and you put, maybe you put the dark clothes in one bin and the white clothes in another bin.
And then you fold them and you put them away.
It's impossible.
This happens almost every time I show this little piece of writing.
People don't generally get it.
They get the fact that it has to do with sorting, and some planning, some arranging things, but they don't get that it's about the laundry.
If you skim it-- you don't have to totally reread it-- but if you skim it with the idea that it's about doing the laundry, you probably can understand it better, maybe.
So now I'm going to ask why did I show you this now, today, before the third, the beginning of the third class?
Yes
I think it's sort of about arrangement of course content [INAUDIBLE]
So it could be about how you arrange the content in the syllabus.
[?
Hina.
?]
I think it's easy to describe methodology as a person more in a different field.
But if there's no context for the audience about what this is, it doesn't matter that you've laid out everything.
No one knows [INAUDIBLE]
Right.
It's very detailed at a certain level, but it's still really hard to understand, right?
Any other ideas?
Yes
I think it's interesting.
There's a lot of value judgements in it.
Like, it's really easy.
And one can never tell with comments like that.
Because I'm trained to figure out what they're saying, it's actually just frustrating me more, because I don't know what they're talking about.
And it's really easy--
All right.
So that's a great point.
So we're talking about writing learning outcomes, about organizing the course around writing learning outcomes.
If you're using phrases, or you're using language that's just, there's just like throwaway words in there or value judgments, it's not doing your students any good, right?
And also, that if you have this beautiful course that's beautifully organized even, wonderfully organized, but if you don't tell them what it is you're trying to do with it all-- what it is they're going to be able to do with it all, whatever it is you expect them to do-- they're not going to be able to follow you.
So imagine a procedure as fairly straightforward as doing the laundry.
Now imagine, OK, you're going to be able to select the right, the appropriate method to integrate equations, so methods of integration, by separation, by card or substitution or whatever.
So if you don't tell them what it is you're expecting them to do, even if you're very good at explaining what it is they're going to do, they won't get it.
So this is my sell for why when you write a learning outcome, you have to write a learning outcome that's clear and that's understandable to the students, that doesn't have a lot of extra words and that tells them exactly what they're supposed to do.
So I have some learning outcomes for today and I hope that they're clear.
You'll be able to describe the components of the constructive alignment and backward design processes.
And those, constructive alignment anyway, was in the pre-class reading.
You'll be able to classify the content of a course that you might teach.
So what do I expect students to do with this content, to be able to do with this content?
You'll be able to create measurable, specific, and realistic learning outcomes for a class you will teach, and you'll be able to state the components of a syllabus.
So I feel that those are pretty understandable.
You may disagree with me, or maybe you want to wait.
So constructive alignment, it was in the pre-class reading, right?
Anybody want to just give a brief summary of it?
I mean it's kind of up here, but in your own--
I think constructive alignment is about aligning assessment, method of assessment, with course content.
[INAUDIBLE] assess [INAUDIBLE] type of assessment so that what the teacher has in mind, the course content or the content I want them to get, to test it with the assessment process.
Then we would also do the assessment, and then come up with the intended outcome
Right.
They'll achieve the learning outcomes.
And if you do this right, you get rid of that red circle on that graph.
You get rid of students who did well on the assessment but don't really know what's going on, because you have aligned the assessment with what it was you wanted them to know or be able to do.
So that's important.
I just think this is a very, very compelling image to keep in your head, that these three things have to be linked.
And as I said before, this, you will see that the top part, writing the learning outcomes is the tricky part, it's the hard part.
It's the part that requires a lot of thinking and work.
When you get a good learning outcome, the other two things almost take care of themselves.
OK?
Did anybody have any, does anybody have any questions about something from the reading associated with constructive alignment or some other idea from the reading?
Yes, Gordon
I just wanted to make a contribution.
I think that this constructive alignment is one of the things that the intended outcome should measure.
So I think maybe that's confusing [INAUDIBLE] and also how we can test and be sure and confident
Right.
Right.
I think the measurability part for scientists and engineers is kind of intuitive.
Like why would you say something if you couldn't tell whether or not it was true, or not from a scientific point of view.
What's the point of saying, making a proclamation about what somebody should be able to do if you can't measure it in the end?
So it's comforting to say, OK, I'm going to make learning outcomes that are measurable.
But it also can be a bit challenging.
But we'll do that.
We're going to do that today.
Some other comment?
Nina
I guess as far as [INAUDIBLE] Do people write learning outcomes that have no constructive alignment process, like choosing to write learning outcomes, are you going through the prcoess?
It seems decoupled a little bit in the reading, but it seems like you can't have one without the other
You owe it to yourself-- and I don't know what people do, and I guarantee you people write learning outcomes without thinking about whether they can assess them or measure them, I guarantee that happens in real life-- you owe it to yourself to try very hard not to do that.
OK?
And I think the important thing to do is, often we say, oh, my god, this is the best assignment.
This is the greatest question.
It's this project, and they're going to do this, and they're going to whatever.
They're going to build a race car.
And if you sit down and say, well, what do I want them to get out of that?
If you can't articulate it in a way that's consistent with your class, consistent with the course as it sits in this series of their courses, or as it sits in the institution, then you either need to rethink the assignment, rethink the project, or think a lot harder about what learning outcomes that activity advances.
That's a very good point.
It's really kind of iterative.
And I think it might be easier to see with this diagram.
This is another-- so John Biggs is from the UK.
And so there's sort of the UK camp of people and they like John Biggs.
And then there's Wiggins and McTighe.
Grant Wiggins and Jay McTighe are from the US.
They wrote primarily for the K through 12 audience, but it's completely applicable.
And they're from the US, so they have kind of a US following.
It's just how it works.
It's essentially the same thing.
It's identify the results, so the intended learning outcomes.
Figure out how you're going to get them there.
So the black is Wiggins and McTighe and the red is Biggs.
So plan the learning experiences that are going to support them, achieving the learning outcomes, and really help you measure whether they got them or not.
And then determine what evidence will be that acceptable evidence that they achieved the learning outcomes and what are the assessments.
So real often we do have this project that we want.
Oh, this is a great problem.
Often in thermo-- I taught a lot of thermo-- there's like, oh, they're going to do this double loop, double reactor cycle with the refrigerant and a heating cycle.
And it's going to be coupled and they're going to use the output from one for the input in the other and all this crazy stuff.
Like, this is a great assignment.
And you kind of, I know this is a great assignment.
If I can't back up from this assignment-- whether or not I put it here or here-- if I can't back up and say after a successful completion of this assignment, you will be able to describe the components of the Brayton cycle, you will be able to define the increase in entropy for a system with x boundaries, If I can't articulate those things, then I need to either change the assignment, or, well, then I need to change the assignment.
OK?
But you can start with the activity, and then back up and say, well, what learning outcomes does that support, and then, am I OK with that?
Is that relevant to this class?
So it's often a very iterative, the arrows kind of go this way, but it's often iterative.
So I had asked that you come to class with some topics from a class you're likely to teach.
That was on the assignment page.
If not, start thinking now.
And so you want to have, for most classes, most undergraduate classes, there's a bunch of topics.
You might have 10 topics, 20 topics.
And I don't care right now about the scale of the topics, whether they're really big.
You probably don't want to go with like first law, second law.
That's probably a little bit too big, but on the order of 5 to 15 topics.
Jot down the topics.
And then I have a chart here.
So the chart has two sides, and the columns are Bloom's Taxonomy, which you don't have to worry about that specifically, but you can look at the words that are associated with it.
And then the second column is what students should be able to do with each topic.
So let's say my topic is first law for closed systems.
So I might think about, OK, do I want them to be able to define, to just state the first law for closed systems?
Do I want them to be able to use the first law for closed systems in order to solve certain kinds of equations?
Do I want them to be able to derive the first law?
Do I want them to be able, whatever it is.
So you're going to look at the columns, the categories on the left side, and you're going to think about your topic.
And you're just going to decide which box in the middle column to put the topics into.
You don't have to worry about writing anything.
You're just going to distribute the topics to the level of, it's really cognitive processes that are associated with the things on the left.
Remember, understand, apply, analyze, evaluate, and create.
So we'll take about five minutes.
And you just do that by yourself.
And just fill in that middle column of the table with the topics from your course.
Do you want more explanation, Rachel?
It's starting to make you think about this part of the beginning circle.
But you don't want to write any sentences yet.
But it's starting to help you think about what are they going to do with these topics.
What do I want them to do?
Is it remember, or is it create something using these things?
Where is it?
Those are two ends of the spectrum.
And in your chart, the things that at the top are the more remember, right?
It's more about kind of a recall information.
And then the things in the back of the chart tend to be more higher order cognitive processes, like synthesize, apply, create.
Yes
[INAUDIBLE]
So don't worry about the third column.
Don't worry about the last column yet.
We're going to do that later.
So all you have to do is just say where each topic falls loosely.
OK.
Does everybody have a chart?
Do you want us to write as many topics as possible?
Write as many topics as possible, but distribute them through the chart.
Right?
Not everything is going to be remember, right?
So some things will be remember, some things will be apply, some things will be analyze, some things will be create.
OK?
If you have any questions, just raise your hand.
All right.
So what we're going to do is I'm going to ask you to just keep those with you.
We're going to do another activity later on using those as a starting point, but I want you to just keep that sorting with you.
Any comments or questions about the experience, about what you just did?
Was it easy?
No.
It can sometimes be a challenge.
Well, what the heck you were supposed to do with this anyway?
What do I really want them to do with it?
This is the beginning of that process whereby you define exactly what it is you want students to do, know or be able to do.
And then you can start to think about how you're going to assess whether or not they can do it or not.
But it's very different if I were to say I want everybody to be able to state the first law of thermodynamics versus I want you to be able to use the first law of thermodynamics to define chemical, mechanical, and thermal equilibrium in a simple composite system.
That's different.
That's very different.
So that's what we're starting to do.
So now we have kind of the seeds.
We thought about which topics kind of we want to have, what we want to do with each of the topics.
But now we really have to turn them into learning outcomes, into real statements.
By the end of this class, you will be able to, by the end of this course, you will be able to.
And just as an aside, remember that sometimes you'll notice that for every class, every session we have, I have learning outcomes.
But there's also overarching learning outcomes for the course.
So I will use them interchangeably, but they'll be at a slightly different scale.
The ones for the class are a little bit more fine grained.
As was mentioned in the reading, and as Gordon mentioned, they have to be specific, measurable, and realistic.
If they don't have those properties they're not going to be particularly useful, or as useful.
All right.
And we'll see some examples of that in just a sec.
So here's what they're not.
They're not topics.
So you came with topics.
You will leave today with learning outcomes.
But topics are not learning outcomes.
The first law of thermodynamics is not a learning outcome.
You need a verb with it.
What do you want me to do, what do you want your students to do with the first law of thermodynamics?
What do you want them to know about the first law of thermodynamics?
So they're not topics.
They're not things that you will do.
It's easy to say, I could have a list a mile long.
I will talk about the first law of thermodynamics.
I can check that off pretty easily.
I mean I just did.
I just talked about the first law of thermodynamics.
It says nothing about what you learned about it, what your students learned about it.
So it's not about you.
Get over it.
That was a joke.
It's not value statements.
As Adrian pointed out, some of these think this is easy, this is good, this is useful.
Any of those things don't belong in a learning outcome.
And it's not your hopes and dreams for your students.
I mean, ultimately, it is.
You hope that they will be able to use the first law of thermodynamics, da, da, da.
But it's really about what they're doing, about what they're going to do.
So, no.
Here's a little example.
This is just from like an earth and planetary course.
Students will understand plate tectonics.
How's that?
Is that OK for a learning outcome?
Why not?
It's too broad
It's not very specific.
What the heck about plate tectonics?
What else?
Measurable
Thank you.
OK, the measurable part.
This word understand, strike it from your brains.
It's a property that we love.
We love understand.
But you can't flippin' measure it.
You can't go in there and measure it.
You can measure lots of things that tell you whether students understand, but you can't measure understanding.
And I think we'll see that with some more examples.
I don't think anybody is earth and planetary here.
So how about this one?
Students will be able to interpret unfamiliar tectonic settings based on information on volcanic activity and seismicity.
Again, I don't think any of us are geologists.
But the idea that I've told, or the person has told us what we're going to use to interpret these situations, and then we're going to make predictions based on those interpretations, I mean, that's an assumption.
So it's specific and now it's measurable.
Even though we're not geologists, I suspect we can think of how we could write a test question that would ask students to do this.
And then we could see whether they did it or not.
So that's measurable.
And probably, by doing this, by having students do this, we get a pretty good measurement about whether they understand plate tectonics.
But it's much more specific and it's much more measurable.
From here, this is from the University of Minnesota, which has a nice website on learning outcomes.
They're not from your point of view, as we mentioned.
They're not what you're going to do.
I'm going to introduce students to the major turning points and processes in North American history.
You can say anything you want, but it doesn't mean the students learned anything.
So it's not about you.
It's what they're going to do.
They're going to list and describe the turning points and processes in North American history.
I'm going to create an understanding of the formal constructs of physical design.
That one is like, this one is, it's like, I don't even know what it means, right?
Create an understanding of the formal constructs of physical design.
I don't even know what it means.
And so it's like that laundry example.
Your students aren't going to know what it means, and it's not going to help them.
It's not going to help them access the material.
So the ones on the left are all teacher-centered and they're not very useful.
The ones on the right are student-centered and they are generally useful.
But there's some things wrong with the things on the right
Understand
Understand, yes.
So what would you write instead of understand the formal constructs of physical design, whatever the heck that means?
What might you use?
Identify
Identify.
OK. That might be like a list, right, or identify or list, pick it out of the lineup kind of thing.
This is a nice one.
Explain to an intelligent non-expert, an INE.
And so explain this to somebody who's not in the field.
If you could do that, it probably means that you get it.
Somebody I know used to say, explain this to my four-year-old brother.
Explain-- what was it-- angular momentum, to my four-year-old brother.
I would argue if you can explain it to a four-year-old kid, you probably understand it.
So you can use that.
Use formal constructs to design something.
I mean, that's kind of an inversion.
How about the one on the bottom.
Understand gender, race, ethnicity, socioeconomic class, understand how they have shaped Americans' lives.
Completely unmeasurable, right?
So what would you write instead?
Describe
Describe how, sure.
Anything else?
[INAUDIBLE] it seems like you're trying to get [INAUDIBLE]
Right.
So that's a great point, is that we don't know where this faculty member put this on their table.
We don't know whether it really just states some ways that, list some ways that these things have affected Americans, North Americans' lives, or Americans' lives-- which is kind of a remember thing, just make a list-- or whether it's do some analysis on some situation.
We don't know that, but you as the instructors get to decide that for your classes.
So that's an important thing.
You get to call all the shots.
You get to make the decision.
We can only guess about what this person meant, and therefore, his or her students can only guess about what he or she means.
So you have on your table this hierarchy of Bloom's Taxonomy.
It's often shown in this triangle, this pyramid.
There's a handout which has exactly the same strata on it.
And it just has some handy dandy verbs associated with it.
But Benjamin Bloom came up with this classification of cognitive processes way back when, in the 1950s.
And it is this idea that it is this pyramid, that at the base of the pyramid are these ideas like remember.
And those are basic cognitive processes, remember, list.
And then it moves up the pyramid to things that require a higher level of cognitive processing.
Rather than just pulling facts out of storage and stating them, you're going to do something with those facts.
So as you go higher and higher up the pyramid, note that he chose-- really unfortunately-- to call the second level understand.
However, there are words that can help you.
If it really is understand that you want to measure, there's words that describe that a little bit better.
They're in your table, and they're also in that second table that I handed out.
So arrange, list, label.
For understand-- describe, relate, recognize, explain, those things.
Gordon
Just a little [INAUDIBLE].
When it comes to understand, maybe if you put it like [INAUDIBLE]
Understand to what?
Understand to an [INAUDIBLE].
Maybe that's what we can measure.
Understand to this level or to this [INAUDIBLE]
But how, can you give me a specific example?
I wish I could go back to the last slide [INAUDIBLE] can describe [INAUDIBLE]
OK.
So I see where you're going.
Does someone have a suggestion?
So think about that laundry.
Think about that description of doing a laundry.
Can somebody take what Gordon said and get where he wants to go, but get there in perhaps a more direct way.
Katherine
[INAUDIBLE]
Exactly.
So if you're going to do that, and I see where you're coming from, understand so that you can describe, in order to describe, or to the level of describe.
Just forget the understand part, and just say describe, or define, or whatever it is you want.
But yes, I mean, that first step is saying, what do I mean by understand?
And when you say that, then it generally rewrites itself.
And so the more direct you can be, generally, the better.
And keep that laundry example in your head.
Yes
You're asking a student [INAUDIBLE]?
So I would say, well, let's just open that up.
What do you think about that, of asking students?
So do you understand this stuff about learning outcomes?
I think as teachers, when you, in the front of the class, you're teaching students and you say do you understand, they say yes.
But you see from psychology you know who doesn't understand
So David says as I ask you that, do you understand?
Most people are going to say, hey, I understand, sure.
Don't think I'm an idiot.
I understand.
Right?
So I completely agree.
Students are not likely to say they don't understand.
What else?
That's difficult for the teacher to [INAUDIBLE]
Well, so do you want to say something about that?
Yeah, I agree.
I think the part that we do understand as nonspecific is the learning understanding [INAUDIBLE].
As an instructor you have to know, understand what you're trying to get at, which I think is where the clicker question is beautiful, because you know already what you're trying to evaluate that they're catching in class in real time, if you can see who's getting at the nuance of what you're teaching
Yes.
I think that's an excellent, excellent point, that many of these activities that we do, these small group activities, or the clicker questions or other activities that students are doing in class are helping them learn.
But they're also formative assessments of whether or not they get it or not, whether or not they understand.
And you as the instructor can see it.
If I walk around, if I listen, if I look at who answered what question on the clickers, or on the raising your hand, multiple choice questions, whatever methods I used here, I get a better, a much better sense of who understands or doesn't understand.
If all I'm doing is lecturing I can't tell.
And I can ask do you understand, but they're not likely to tell me.
Which is kind of the other point is that sometimes-- and I think you were alluding to this, Gordon-- that sometimes students don't know whether they understand or not.
So we need to give them the opportunities to find out whether they understand or not.
And arguably, if you're just lecturing and you're a great lecturer-- you're eloquent, you don't stumble, the writing on the board is awesome, all this stuff-- students will think they understand maybe when they don't, because they're not having to confront any misunderstanding, and it all seems very nice.
So these activities where you actually have students do something, say something, where you ask them a specific question, that's when you can determine whether or not they understand or not.
And because they are not likely to say they don't understand, either because they just don't want to admit it, or because they really don't know whether they don't understand.
So you need to give them measurements to find out whether they get it or not.
Yes
Sometimes, for example lecture and the activity has [INAUDIBLE].
For example, we have lectures and the teacher just [INAUDIBLE] and then we have expectations as we do exercises.
So do you recommend that we also do some activities during the lecture or just [INAUDIBLE]?
No, do them during lecture.
I mean, the model that we're trying to do here is the model that you should do even if you have a hundred or 200 people in your class.
You can use clickers.
You can use pair share discussions.
You can do all sorts of things to break it up, and it's effective.
There's a paper coming up for next week, a meta-analysis, active learning strategies, how they actually, students learn more in a lecture if you actually ask them to talk to each other, break up, share, et cetera.
And I think when we talked about misconceptions, remember that example with the coin?
How many students got it, could draw the free body diagram before and after?
I mean, before and after in a straight lecture, there's hardly any learning gain.
But if you incorporate active learning, the learning gains go way up.
So that's Bloom's Taxonomy.
And an interesting take on Bloom's Taxonomy, there's a woman, Kathy Schrock, who-- it's a nice website and the slides we posted so you have the link-- but she decided, she has this theory that it's not really so pyramidal.
That everything really is dependent, these cognitive processes are dependent on each other at a certain level.
And so she's drawn it as a gear, where everything sort of supports the act of creating, which is up at the top.
But the other ones are sort of around the side, but they all contribute to the ultimate cognitive process of creating.
So that might be a nice way to think about it.
And then she's taken the specific words, like analyzing, and said, OK, what's an instantiation of analyzing, outlining, deconstructing, organizing, structuring, surveying, whatever?
So it's the same idea, it's just a different graphical representation.
So what you've done is you've taken your topics, put them in those cells in the table.
And I just want to point out kind of how as the instructor you have a lot of power about where you're putting those, where you've decided to put those.
And it may be different for a first year undergraduate class, or a second or third year graduate course.
It could be the same topic, but you're going to put them in a different category.
So that's why when I asked you to bring a topic, it's for a specific class.
It's for a specific group of students.
So let's say we take the idea of interstitial sites.
So I have a crystalline structure, it's got an order of periodicity of atoms.
And then I have spaces between the atoms, those are the interstitial sites.
And so I could say you'll be able to identify the interstitial sites.
So I could say where's the tetrahedral site in a face-centered cubic or something, whatever?
And students just really have to find it.
I could say calculate the maximum size of an interstitial atom that could occupy that space, an ion that could occupy that space in a particular crystal structure.
So those are two very different levels, but it's the same topic.
X-ray diffraction, I could say apply Bragg's Law to calculate d-spacing.
So that's very much a turn the crank kind of thing.
I know the equation for Bragg's Law, I stick things in, I solve for the unknown.
I could then also ask you to index unknown diffraction patterns, meaning identify an unknown crystal structure based on the diffraction pattern.
That's a completely different activity, completely different set of cognitive processes than just using Bragg's Law.
But you have to decide what it is.
And arguably, if all you've had students do during the class is sort of problems that they applied Bragg's Law and they solved for d-spacing-- so they solved for the incident angle or they solved for the wavelength or whatever it is-- you really can't throw this at them on the exam, right?
Or conversely, if you spent the whole time, they've indexed, they've been in the lab indexing unknown diffraction patterns, and then this is the question you ask them on the exam.
Well, then you have people that perhaps get it without really, perhaps pass the exam without really getting it.
So that's just something to keep in mind.
The other thing to note is let's say this is my intended learning outcome-- calculate the maximum size of the interstitial atoms in a variety of crystal structures.
How do I know whether students get that or not?
What's that?
It's easy
It's very easy.
I give them a crystal structure and I say calculate the maximum size of the ion that can fit in the space.
So I have figured out how, the measurement is easy, it's cake.
It's totally straightforward.
Now I haven't told them, OK, I'm going to give you this crystal structure and I'm going to ask you this spacing, this ion size.
I haven't given it away, but I have told them what's expected of them.
And my colleague wrote these for VSEPR theory.
It's the same topic, and it's different.
Identify the common geometric shapes found in simple molecules.
Explain the assumptions of this theory.
Apply the theory to predict 3D structures.
Compare and contrast the geometry of a certain molecule as predicted by two different theories.
Evaluate the accuracy of each theory for a particular set of compounds, and then create some recommendations.
So it's the same topic, but you decide what it is that students should be able to do.
OK?
So I have some, just a little exercise for us.
So they should be specific, measurable, realistic.
I'm going to put them up here and we're going to say, you're going to tell me what's wrong with them.
So number one, t-tests are like a statistical analysis.
So is anybody a t-test kind of person?
We can skip that one.
How about gaining appreciation for the use of linearization techniques?
What's wrong with it?
It's not specific
It's not specific
It's not measurable
It's totally not measurable, exactly.
I mean, how do I know that you appreciate it?
Don't do it.
All right.
Great.
Have an intuition for the most effective method of integration for a given problem
It's vague.
It's not measurable
So I've heard it's not specific, it's not measurable, and it's not vague, I mean, and it's vague.
So right, so have an intuition, I cannot measure that.
I have not figured out a way to measure intuition.
I can tell whether you can do it, right, but I can't tell you whether you have an intuition for it.
Well, how would you rewrite it?
Actually, how would you rewrite 2?
Use [INAUDIBLE]
So it could be use the techniques.
Another suggestion?
Make something using the evaluation techniques
Calculate something.
Make a first pass estimate using a particular linearization technique
Describe when you would use a linearization technique
Describe when you would use it, describe why you would use it, all of those things
List those areas where [INAUDIBLE]
List the areas where it's useful, yes.
Great.
All right.
For number 3, the gain an intuition, we decided it's vague, it's not measurable.
So how would you rewrite it?
Compare.
Compare the areas integration technique [INAUDIBLE] Solve a particular problem using different kind of [INAUDIBLE]
OK.
So solve a problem, and then we need a little more.
Did you have something?
Identify the most effective method of--
Let's hear what Rachel had to say.
Rachel
Yeah.
Learn how to identify the simplest [INAUDIBLE]
Right.
So you probably don't want to say learn how.
You would say be able to identify.
Right?
Gordon
[INAUDIBLE]
Right.
I mean, you could just say integrate--
[INAUDIBLE]
Integrate a problem using whatever.
Select a technique to integrate a particular equation.
And again, those things end up being more measurable.
So 4, provide problem solving tools and strategies.
What's wrong with that one?
It's too wide
It's too wide
It's instructor centered
It's instructor centered.
What else?
But it's measurable
Well, I can measure that I provided it.
Here, I provided these handouts, right?
OK.
Check.
It happened.
Whether or not you could do anything with them, I don't know at this point, right?
That's the problem
It doesn't tell us what kind of problem and what kind of tools we're going to use.
Just who gets to choose the strategies
Right.
Right.
It's very, very general, totally not specific.
So what could we do?
I mean, we don't know what this instructor had in mind, but what could you do if this was your class?
What would you say?
What about develop instead of provide?
Develop.
And again, it would have to be student centered, so the student will be able to develop problem solving tools and strategies.
That's for x, exactly.
We want to make it a little more specific and a little more focused, because that's pretty broad.
I mean, I have developed a problem solving strategy that lets me get out the door without tripping.
Does that mean I satisfied your learning outcomes?
Probably not.
So in order to do something-- Use thermodynamics to solve engineering problems
That's not specific as well
It's not specific at all
Engineering is [INAUDIBLE]
Yeah.
I mean, it's crazy broad on a number of fronts, right?
I mean, it's like engineering is huge.
Thermodynamics is pretty darn big, too.
So which part of thermodynamics, which part of engineering?
Build an SAE race car
[INAUDIBLE]
What?
It's not realistic
Right.
It's pretty specific.
Build a functioning SAE race car.
It's very specific, and I can totally think of the metric of how I would measure whether students could do it.
I would just ask them to build the race car.
But they're never going to do it in a class, in one class period, in one course.
It's unrealistic, exactly.
Learn to use Laplace transforms to solve differential equations
[INAUDIBLE]
It's funny.
I mean, you can imagine the instructor saying you will learn to use that.
But let's go, Katherine, work your magic with this one
[INAUDIBLE]
Right.
You're my go-to person for when you want to get rid of all the extra words.
I'm going to call on you.
Right.
Use Laplace transforms.
That you can measure.
You can give them an equation and see if they can use Laplace transforms to solve it.
And yes, you might want to be a little more specific about the kinds of differential equations they use, and know how to upper diagonalize a matrix.
Katherine
.
Upper diagonalize a matrix
Upper diagonalize a matrix.
That's all you have to say, right?
Forget the extra words.
And that is really, I have just written the assessment question for that.
I know it's totally measurable.
I give you a matrix and I ask you to upper diagonalize it.
I haven't given it away.
I haven't told you you will upper diagonalize this matrix, right?
Memorize these equations that you're going to need to write down in order to upper diagonalize it.
But I have told you that that's what's expected of you
I have a comment on it.
It's pretty useful when a student is trying to evaluate the use of the class.
For example, the career fair tomorrow [INAUDIBLE].
There's a million [INAUDIBLE] to get jobs.
OK, how are they going to evaluate me?
And if you could come in and say, this happened and this is what we did, as opposed to having, or have them evaluate me on some random knowledge.
And then I don't know that I could respond, but I do know stuff.
So this is really helpful in kind of a broader sense
Right.
Great.
No, I'm glad.
And I think it's so true.
And people often think, well, you're giving it away.
But you're not giving anything away.
You're just telling them the expectations.
You're not giving them the problems, you're not teaching to the test, but you're just telling them what they're going to be able to do at the kind of high level, or medium level.
You're telling what they're going to be able to do.
I will say-- so ABET, the Accreditation Board for Engineering Teaching-- every accredited engineering program has to list, they have A through K outcomes.
And if you look it up, ABET-- and I'm sure you see us as accredited-- if you look up, they have to, they used to do a lot more bean counting.
But what they would do is the outcomes would be critical thinking and problem solving skills, and they were a little bit broad.
But every department that got accredited had to show how various courses contributed to the outcomes.
So somewhere in your department should be a pretty detailed list of how 6 double 00 whatever, did this.
Or 6 yada, yada-- I know your grad students-- but did this.
The problem is they don't have to do it for grad student courses.
But it might be helpful if you looked at that, because I would imagine that sometimes the recruiters would be looking at that information, so that might be a commonality.
But every program in the US that's accredited has to have, show at least how they've addressed those outcomes.
So you might want to take a look at that.
But yeah, if it was a little more explicit it would certainly be better.
And I'm glad it's helpful.
So I think we can see, you can start to see how there are better and worse learning outcomes, that if they're not measurable they're not so useful.
If they are measurable, they're quite useful.
So now what I'd like to do is revisit the worksheet.
And what we can do is we're going to take about five minutes where you just sit quietly and write out specific intended learning outcomes.
So you'll be able to upper diagonalize a matrix, you will be able to use Laplace transforms to solve this type of differential equation, whatever you've written in this column.
And note that if you put it in this column, let's say you put it in the apply column, then when you write the learning outcome it's going to be you will be able to interpret data using blah, blah, blah.
You will be able to model the vibration, model a car suspension system using linearization techniques.
They're going to be whatever, wherever you put it in the box.
These are some of the verbs, these are the verbs you're going to use in your learning outcome.
Does that make sense?
And I have even more verbs for you if that's helpful.
Did I pass those out?
You got those, right?
Yeah
So there's more verbs.
It's the same thing.
I mean, it's consistent.
So take five minutes by yourself and do that.
And then you're going to get into pairs and you're going to discuss.
You're going to trade worksheets and give each other feedback.
All right?
So five minutes of quiet and then we'll pair up.
So if you can pair up, you don't necessarily have to pair up with people that are sitting next to you.
You might want to get a different perspective, so feel free to get up and move around.
You can talk to somebody that wasn't in your group.
So maybe if you go back and talk with Alex, maybe.
We need one group of three.
Oh, are you a group of three?
So now just trade sheets of paper with your partner, or if you're a threesome you can figure out how to do it.
And then review the learning outcomes critically.
Guys, just one sec.
So just review your partner's learning outcomes critically.
Make sure that they're measurable, realistic, and specific.
If your partner has used the word understand, help him or her work through that to get a better word.
So you're going to be a critical friend here.
And then you guys can discuss how to improve it.
And then I'm going to pass around some flip chart paper and each one of you will write, maybe just write your name at the top, and write two or three learning outcomes on the flip chart paper.
And then we can stick them up around the room, and then everybody can see everybody else's, all learning outcomes.
[INTERPOSING VOICES]
Was there a question?
Oh, no.
I just think--
I think [INAUDIBLE]
How are you guys doing?
Doing OK?
[INTERPOSING VOICES]
So you can use this zone if you need it.
As you're looking at other people's learning outcomes, think about how clear, how easy it would be, or not easy, to come up with an exam question, or a project, or an assessment or measurement of whether or not the student reached, attained the learning outcome.
You can put them up here, Ina and Shau.
OK. And as I said, make sure that you've read everybody else's.
I'm going to say a few words about them.
I think we may have a record.
This is like the first time ever that no one has used the word understand, which I think you're all to be commended for.
So David, make sure you don't write understand.
You build me up just to knock me down
[INAUDIBLE]
You can, or you don't have to.
It doesn't matter.
Does anyone have any comments, either about the exercise itself, or about somebody's learning outcomes, that you see around?
Again, the measurement, the thing you want to look at when you look at your own learning outcomes and those of others, is are they specific, measurable, and realistic?
And would you know if a student achieved those learning outcomes?
So I would say about 90% of these are very specific, which is great.
I mean, there's some incredibly specific ones here.
The more specific, the easier it is to measure.
So that's one thing.
That's a useful thing.
You want to make sure you don't get so specific that there's only like one question you could ask.
So that's the balance.
Sometimes you can't tell that unless you're in the discipline.
Do we have other comments about any of these learning outcomes, something you don't get, something that looks particularly interesting?
Yes, Gordon.
Perhaps you can sit down if you're done reading
I'm looking around and I'm seeing [INAUDIBLE].
I don't know how it's going to be.
How are we going to test that?
Well, I guess you have to come up with a reaction that you're confident they haven't seen before.
So if you can do that, then you can ask them to propose a mechanism for it, I suppose.
If the person that wrote it wants to-- I know who wrote it-- but if the person that wrote it wants to talk a little bit about it, that would be great.
If not--
OK, So a lot of people are going to just try to memorize each specific outcome.
But it's easier to just identify your activity group and find patterns within them.
So if you can try [INAUDIBLE] They can't just [INAUDIBLE]
Right.
And this is a great point.
I mentioned it to one of the groups.
But sometimes we'll say explain how blah, blah, blah, works.
And that's a great thought, that, oh, they're going to be able to explain it, right?
But many times they can fall back on a memorized explanation or sort of kind of a canned procedure.
So if you really want them to be at the level of explain, you will have to kind of make some things up to get them out of, to get them away from the ability to just pull something completely from memory.
So this is an example of how you might do that.
Other observations or questions?
I also [INAUDIBLE] something that all [INAUDIBLE]
So perhaps the person that wrote it wants to comment on it
OK.
So [INAUDIBLE].
I think what we have, just [INAUDIBLE].
So we write and compile.
So you write a code and then you compile it
Right.
And this is a great example of how having people that aren't in the field read them can help clarify things for your students, who are by default not experts.
So perhaps to an expert, it totally makes sense, of course, code.
But to somebody that's not an expert, maybe that's confusing.
Maybe the use of code twice was confusing.
To an expert, it's totally readable.
It's totally understandable.
Perhaps to a novice it isn't.
You're laughing
It's funny, because we're both [INAUDIBLE]
Yeah, that was kind of random.
Well, that's a perfect example, right?
It's a perfect example of how getting maybe other people that aren't in your field to look at them can help make them more accessible to your students.
OK.
This is great.
Oh, Hina
I had, it's on a different topic.
But really, I'm talking about buzz words in learning outcomes.
I think with engineering, you're really excited about telling people about real world problems.
I guess that's the reason why.
I guess it's more of an opinion question.
As you write your learning outcomes I think in some ways it's better to be specific about the real world problems that they're going to attack.
Because I've seen that on syllabuses all the time.
I'm very excited about it.
I think it's a way to teach people in your class.
It's going to help them know what tools they're getting
Right.
I mean, real world problems in particular really do kind of encapsulate a whole pile of hopes and dreams.
So it may help to be a little more specific.
You can use it in your learning outcomes, but then maybe describe it in a little more detail in your syllabus, the kinds of real world problems, or why we care about real world problems.
They're messy.
They're ill-defined.
What about them is so important?
What about them is so important?
Try to minimize the buzz words.
But at some level, they may not know about something until they're completely through with the course, and you may have to use that word.
So there's a bit of a balance.
What I'm going to do before, I'll probably bring all these back to my office, take pictures of them, and put them up on the wiki so you'll have a record of them.
I wanted to just as we go out here, my point is really that the hard part is these learning outcomes.
And that once you write the learning outcomes, as we said before, you kind of know how you're going to test whether students got them, and you know what you're going to do to help them get them.
The syllabus itself is just an articulation of that.
It's a description of the course, an articulation of the learning outcomes stated, like the laundry example, with the title.
This is what you're going to do.
By the end of the class this is what you'll be able to do.
It's kind of a promise that that's what you're going to do.
It motivates students to take the class, to stay in the class, to engage with the class.
It keeps you on track.
So at the beginning of the semester maybe I'll be very clear, but four weeks in, and you're a little bit fuzzy, because you've been dealing with all sorts of stuff.
And you kind of go, what the heck.
What's this in here for?
Well, if you've lined everything up at the beginning using your learning outcomes as the anchor, then you don't have to worry about that.
You just follow the map.
So it really, really helps you and the students get through the semester in a logical way.
And it tells the students what they can expect of you, what you can expect of them.
So it's motivational, structural, and it's evidence, hopefully of what they did, but certainly of what you hope to do with them.
So for the post-session assignment, you'll have to write up more learning outcomes from your course.
You can use these as the basis for sure.
Hopefully, that table will help you as you write more of them.
And I will go in and give you feedback.
I want to make sure that everybody knows that you should read each other's postings.
So a lot of you had great examples from the first class about how you would use these learning theories in your courses and how you teach.
They were fabulous examples.
Make sure you read those postings of others.
That's one reason why it's up on a wiki and not just you handing it in to me.
So you have the Mud cards.
Fill them out if you have any more information that you want.
And otherwise, we'll see next week.
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A general comment is by and large, I thought you guys did a great job.
Your learning outcomes were for the most part specific, measurable, and realistic.
The biggest problem I think I had was with-- the biggest problem that you had on average was the idea of the learning outcomes really have to be measurable.
So if you use a word like appreciate, it's not measurable.
So that's something to remember.
If you use words like discuss, it's potentially measurable.
But you need to make sure that you set up a venue for students to discuss.
You can't just say you'll be able to discuss that and assume that they'll be able to discuss it, That If that's in your learning outcome, you have to set up some venue for them to discuss and you to know what they're discussing.
And then the two biggest things that come up, or the two most problematic words that come up-- and I compliment you all, because you did a great job.
Nobody used understand.
So that was awesome.
But the other thing, there's words like explain and describe, which are perfectly good words to use in a learning outcome.
But you have to make sure that when you say, students will explain the ideal gas law.
They'll be able to explain the ideal gas law.
Well do you know that they're really explaining it in the true sense of that word that they actually internalized what it means, and they're really explaining it?
Or are they just spitting it back, spitting back, maybe, what they read in a book, a definition that they read in the book?
So you have to make sure that you've set up a situation wherein they really are explaining.
And they're not just remembering.
So it's a great word to use.
But you have to make sure that we're using in the right context.
Same thing for describe.
You can imagine that has the same issues.
Describe da, da da, da, da.
Well are you sure they're doing the describing?
Or are they just remembering somebody else's description and writing it down?
One good way around that is to tweak the parameters a little bit.
To say, in a world where there is no oxygen, what do you expect this organism to look like, given that it has to run really fast or something?
So come up with alternate realities that that actually force students to transfer and then re-explain.
That can be helpful.
That can be really, really helpful.
I remember once in a kinetics class, we had been deriving the critical nucleus size for growth.
And it's all based on a radius, a critical radius.
And then on the exam, the guy said OK, they don't grow as sphere's.
They grow as rectangular boxes.
So then all of a sudden, we had to look at the ratio of the length to the height of the box or something.
So that really wasn't just we were just remembering a solution or remembering a method.
We had to actually think about what it meant and then reapply it.
So just think about how you can get students to really describe, explain, derive even, because you can memorize a derivation.
But if you really want them to derive it, what does that mean?
How are you going to test that?
So those are generally the biggest problems people have with learning outcomes.
But by and large, you all did a great job.
If you came in late, I'll give you your learning outcomes at the end to class.
If you looked at the Wiki-- and this is a test to see who looked at the Wiki-- I had suggested you bring a web enabled device if you had one.
And that's because I wanted to set up something called Back Channel The topic today is on active learning.
And so this is just one method that might help your students, especially if you have a big class.
So we don't really need it in this class, but I just wanted to give you the opportunity to play with it a little bit.
It was designed for conferences actually.
But it works pretty well, I think, in classes.
So if you login to here, this is the URL.
If you go to our Wiki page bookmarked, there's a link on the Wiki page, if you just want to link to it that way.
And then at any point in the class, you can type a question in.
And if you're on the page, you can see other people's questions.
You can upvote and downvote their questions.
And I will take a break in the middle, when you're doing some other things, and read what people have posted and then get back to you.
So it's an opportunity for students to interact who may be a little more shy, who may not want to.
I don't think we need that in this class.
I don't think that there's any one of you that isn't willing to share your ideas, which is great.
But this may be something that's of interest to you.
Before we jump into the topic today, active learning, there were a couple of mud card questions that I just wanted to talk about.
One was sometimes students may nod off during class.
And if it's a big class, there's probably not a lot you can do about it.
If you do enough active learning, though, that will keep them-- that will make them wake up.
Especially pair activities or activities where they have to talk to one or two other people, or two or three other people, it's pretty hard to be asleep during those.
And you'll see.
We're going to do something called a lightning round later on.
And it's virtually impossible to be asleep during that.
So you want to make sure that it's worth the student's time to actually be in class and be awake for it.
And active learning can play a huge role in that.
Sometimes people are tired.
They stayed up too late for whatever reason.
And you can't control everybody in your class.
But you want to make sure that you're doing everything you can do to keep them engaged.
And if it's just one student every now and then, you probably don't want to take it personally.
It's probably not you.
If the whole class is falling asleep, then it's probably a good opportunity to rethink your strategies.
Somebody asked if there were good times of day to have a lecture.
And part of it is you.
Part of is your schedule.
Sometimes you're allowed.
You can pick that.
And sometimes you really don't have much control over when you choose to teach your class.
People do say that students, undergrads, people between the ages of 18 and 21 are not particularly interested in being up at 8:00 or 9:00 in the morning.
But again you may not have any control over that.
Generally speaking, that's a pretty low time for people.
So if you want to take that into consideration, you might.
But it depends as much on you as the students.
If you're going to be really tired at 4:00 but they're just waking up, then don't teach it at 4:00.
Maybe don't teach at 4:00.
And the last one was guiding readings.
If you just put a lot of readings up, students might not know where to focus their attention until after the class.
So there are reading questions, reading guides that you can post.
Read such and such and try and think about how you might explain this, that, or the other thing.
And then there's also pre-class questions you can ask that they might derive, that they may have to pull from the readings to answer.
So the concept of just in time teaching, the idea that students do a reading and then they actually have to answer questions on the reading before they come to class and submit their answers.
So there's a whole procedure around that.
It can be very, very effective.
You can do it for a large class.
There's strategies and techniques to streamline the amount of time you put into it.
But it does force them to do the reading.
And it does guide, at some level, what they read.
And that's a great great approach.
And I'm happy to talk more about that at another time or later or offline.
I did go to the Teaching with Technology conference, the Teaching Professor Technology Conference.
And they were giving out little ribbons to stack on your badge.
And this was a prominent one.
It's in the syllabus, which I thought was just crazy timely because a couple of weeks ago.
So if you want one of these to stick somewhere.
And then they all said when I tweet, which was sort of weird.
But anyway I brought those back.
Anybody want some.
Sometimes my computer doesn't wake up.
So let's pretend that instead of the class active learning today, I decided I was going to teach you how to juggle.
And I said, here's the equation for how to juggle.
And here's the variables, F, D, V N, and H, which are ball in the air, ball in an hand, time hand is vacant, number of balls, number of hands.
And this is this handy dandy equation, which if you balance it out, it describes the juggling process.
And for those of you that are visual learners, I have a graph here, a yellow ball, a blue ball, and a pink ball.
And then you can see that just shows the path of the balls.
If you remember from physics, there's the equations of motion in the x direction in the y direction.
And so I think I probably did a pretty good job of teaching you how to juggle.
No, did a terrible job of teaching you how to juggle.
I didn't even come close to teaching you how to juggle.
And if I were to give you a test, let's say, and I were to measure whether or not you could juggle, I guarantee that some of you would be able to juggle.
And I could say, I did a great job teaching how to juggle.
But I didn't.
If you knew how to juggle coming in, presumably you still know how to juggle.
I did no harm, I think.
However if you didn't know how to juggle coming in, you don't know how to juggle now.
So why is this here now?
Why did I bother to do this stupid thought experiment right now?
Because this is clearly how most classes are taught.
Obviously it would be easier to have a ball and show us how to juggle, active learning as opposed to just showing the theory behind it and ending there
And I think a lot of times we really do, in science and engineering classes, we really put up a lot of equations.
And then it's not like we expect somebody to do something physically with them.
But we expect them to be able to manipulate the equations, transfer the equations, apply the equations.
And we never really talked about that.
We talked about the equations.
So it's a real call in my mind, for us to think about what it is we really want students to be able to do.
And then make sure that they're doing stuff to teach them that helps them learn how to do that.
And I know you all read the readings and that they talk a lot about active learning.
There's a few things about techniques for active learning.
There's a few things about the why's of active learning.
There's some data for why active learning works.
But I'm guessing that there's a range of buy in at this point from you guys.
Eh, I don't think it's worth it.
Or I don't really know.
So for the first 20 minutes, 25 minutes of class, what I'd like you to do is I'd like you to pretend that you were part of a department, the STEM department.
Call it whatever field you want.
And there is a proposal that your department should commit funds to train professors and develop resources, the resources necessary for the adoption and ongoing use of active learning techniques in all courses.
And the target date is 2017.
You've got a year to get all courses up to speed with active learning stuff introduced into these classes and with a plan for how to do that.
So what I'm going to do is I'm going to just count off one, two, three, one, two, three.
And I'm going to ask you to form a group.
If you're a one, go with the one group.
Two go with the two group.
A three go with the three group.
And you're going to have to physically move.
So just remember your number.
One
Two
Three
One
Two
Three
One
Two
Three
One
Two
Three
One
It doesn't matter that they're not the same size.
Can we get all the ones, let's say all the ones up here?
All the twos here.
And all the threes there.
And I'll tell you what to do in just a sec.
So if you're group number one, you have to adopt the position that you are pro.
You are going to push for the adoption of this resolution.
And you have to come up with a set of arguments and a set of reasons why it's a good thing.
I don't care what you think personally.
I don't care if you think active learning is the worst idea ever.
For the purposes of this debate, you need to pretend you're totally into it.
And you need to come up with some good arguments why.
Two are against it.
Again, I don't care how you feel personally.
But come up with some good solid arguments why you should not do this
We are against it?
Against, you do you not want it.
No.
Group number three, you got a little bit of time.
But get to know each other.
And you're going to hear the arguments.
You're going to hear a case presented by the pro people, the con people.
And then you are going to deliberate and decide what's going to happen in your department.
You guys are the deciders.
It's an American political joke.
But we'll let that go.
And you guys get this lovely microphone.
So I'm going to give you 10 minutes for this.
So be ready to share your views.
You've got about, maybe, two or three minutes of a spiel of proof.
And then we'll do one exchange of debate.
And then the deciders will decide.
[SIDE CONVERSATIONS]
I'm going to ask someone from each of the groups to make a statement that puts forth the side that you're supposed to defend, if that is the side you're supposed to defend.
So we'll start with the pros.
So someone from your group ready to talk pros?
All right, Michelle
So we are pro active learning.
And part of the basis for thinking that active learning's important is that we believe that without active learning, the professor usually just resorts to lecturing.
And we think that lectures are not effective.
Students don't learn simply because the material is presented to them.
Lecturing wastes resources, because you could be lecturing to 1,000.
But only 15 students may be reached.
And if you use active learning instead of lecturing, it can make sure the students are paying attention.
You can check that they're understanding.
You can know what they are getting and what they're not getting.
We also feel that active learning is a life skill.
And that if students learn to learn by do things and they learn by talking to each other and by going through a process and solving problems, that they can take this skill to outside of the classroom later in society.
Then that the society that you have students who are used to being active learners and who know how to solve problems that way.
And we also think that the end, the big point that we have is that you remember things that you use.
If you learn something, if it goes in one ear and out the other, if you don't actually use it, you're not going to remember it.
If you're not forced to actually work through a problem and really think about it and use it in a classroom, that you're much less likely to actually hold on to that information.
So these are all reasons why we think that we should promote active learning in our department
OK, thank you very much.
The con, or I guess I should say the not whatever.
Anyway, group two
Do you want to do it?
OK. We start with a basic question.
The question is, why do we want to change what is working?
We are training our students.
They are getting job.
The professor are doing well.
They are collecting grants.
The original [INAUDIBLE] is OK.
So why do you want to change what is working?
So that is the first question.
Now the first thing is that, if you want to go about this, it's going to consume a lot of money.
It's going to take a lot of money.
And then it cuts into your funds.
The one funds for other such, want to buy equipment for the laboratory, want to go hire scientists, want to go hire post doc.
So why do we want to change?
Where do we get these funds?
We divert funds for the training of professors.
And they're going to be little bit of no funds for all this such work.
And secondly, the efforts to train professors is likely to be very huge.
Professors are used to this method of teaching.
And as you saw, they will not want to change.
There's going to be a lot of time changing them.
And we're not even sure, if we try to change them, whether they're going to be able to do it at the end of the day.
Thirdly, we are not sure that the stakeholder would be happy.
What about the alumni that are very much used to the method of teaching?
What about the parents?
are they now going to go against us?
Will they not see that the professor no longer, just come to the class, give them group work like this without teaching, without writing?
Then the professor is no longer teaching.
And they are just going to collect their salary.
And then, there's not to be constrained on time to cover syllabus.
When I saw I spend a lot of money being in group work, individual, or this and that.
So how do you complete your syllabus?
You have about 20 topics to cover this semester.
And by the time you start active learning, how are going you cover about 50% of them?
Then you push all of that up [INAUDIBLE].
So it's not going to work.
We should just drop the idea.
What have we value very last class?
You have your class of 500.
How do you want active learning to be effective?
How do you want to treat them?
It's good here, because we are just about 15 or less than 20.
By the time you have 500 students in a class, it's going to be very difficult for you to have active learning, difficult for the professor to control classes, and of course, difficult for you to really get something concrete out of the whole thing.
So my final guys and professors, I think it's not necessary.
the.
System is working.
Let it continue to work this way.
Thank you
So now what I'm going to do is I'm going to let each group ask one question or a set of clustered questions to the other group.
The other group can answer them.
And then the deciders can ask a couple questions.
And then we'll let them deliberate.
So group number one, like to ask some questions to group number two or make some comments?
Go ahead
Yeah, I listened to you speak.
And I want to ask you.
You are saying that the system is changing.
It is changing because the status quo isn't delivering.
So that is why we are adopting active learning.
And this is proven from a research perspective.
There is actually letting more students learn more materials faster and more effectively.
So what do you think about that?
I'll answer it by saying the different research shows us that the student that is good is good.
[INAUDIBLE] So why can't we just concentrate on getting those students?
And what your presentation didn't say about funds, was the whole idea is to fund the proposal.
And if we divert funds to this new program, and that's a very big issue.
You have to concentrate on the question that is the diversion of funds away from research.
So if we can get the new students and they can learn by [INAUDIBLE], and it's working great-- go ahead
So you want to throw away the baby with the bath water?
If you have 20 students in a class, and you have five active learners, and then you want to discover 15 that wouldn't.
This program is geared towards each one of the students in the class to participate.
And I think it's more cost effective.
If you could get all the 20 students to learn and to learn actively, I think that would pay itself off at the end of the--
No, the problem there is that you have to face the proposal.
This is 2017.
It's a lot of change in a very short time.
The proposal itself is already against--
Our department could be three people big.
We don't know how much the department is.
[INTERPOSING VOICES]
But then the issue there is that you're going to retrain professors.
Professors are really stuck in their ways.
And you spend all this money.
And it's very short time.
There's lots of people from the labs, and money from the labs in this very short time to train them on something we don't know if they're ready to buy into it
So I think your assumption is that-- you're like, we're deferring funds from labs and resort.
But you never once said you're diverting funds from anything related to the students.
You're thinking of it from the professor perspective.
But really professors are here to teach their students.
And you're like, oh, we're taking money away from the research, from the lab equipment, from all these different things that sound great to the professor but have literally nothing to do with the students learning experience or the student's experience on campus.
And from the student's perspective, that's really frustrating, when that's all the professors care about.
And that's what makes a difference between a good professor and a crap professor is one who cares.
And I think, if you want a professor to care, oftentimes they're willing to make the effort that is needed to change their teaching style or do whatever it is that's necessary to have their kid interested and engaged in the same material that this professor loves.
So you keep talking about diverting funds.
But you never once mentioned diverting these funds away from this one teaching initiative to a different teaching initiative.
So just keep that in mind
So the problem is that when we are looking at the proposal itself, the time, 2017, this is already the end of 2015.
So that means you are going to do this, by 2017, that means you're going to do it by 2016.
We're not against active learning in the [INAUDIBLE].
[LAUGHTER] [INTERPOSING VOICES]
--the proposal.
We are against-- we need these two group and against this proposal.
So for this proposal, we are against this proposal, because you know how much it's about the funds and the resources.
And that is why we are against the proposal.
And also you pick whatever [INAUDIBLE], we're not even sure that it is going to work
I'm going to take, because you guys actually did ask a question, so I'm going to count that to you Did you have another separate one?
AJ, you look like you might have some questions
Yeah actually I was just curious, so if this initiative is really great, it looks like a lot of this research has been done since '80s.
Why isn't it adopted?
Why are we having this conversation now in 2015?
I think we should look back at this issue of the lecturing system that we are using now.
It's static.
When there was no printing, lecturers then, they get the material.
They reach the students.
So lecturing is a Latin word that means reading.
Now since then, now we are printing.
We have information readily available that people can get on at their home and their own and develop to something.
Too So there is a need for us to change to a new system, because the old system that more facilitated lecturing has changed.
So then we have to change
I'm going to turn it over to the deciders who now have to discuss.
They've been resting for many minutes.
So now they have to.
So you have questions for each of the groups?
Yeah
Yeah, thank you.
Great arguments on both sides.
I found it interesting, when you talked about the larger classes, so my first question is for this group.
So how would active learning apply to a large class?
The proposal is that it's for all courses.
And so are there courses where this doesn't lend itself to?
Even in the classes with 700, 800 students here, they still do active learning for the freshmen, i.e.
Clicker questions.
Oftentimes professors have to talk to your partners.
So they'll pose a question.
Everyone will click on their answer.
And then they'll be like, now go back and talk to the person sitting next to you.
And then you'll vote again in two minutes.
So I think that's just one example of how you can engage a class of a lot of students
And to add to that, actually it's also been found out that when peers speak to themselves, they even learn the material better than when the expert, who may have lost touch to what it could be like to learn from a student's perspective.
So peers speak to themselves and teach themselves even faster than the expert who is in front of them
Yeah I have comment for the against group.
You mentioned that you don't require to create or active learning, because some students are very smart.
They get it right.
But don't you think that education is a right of every child?
And they deserve to have a good shot at it.
So why do you want to pull one child out of here, because he's not smart enough?
I just to answer, I think we are pushing for the best education.
We believe that is the traditional way.
We think that active learning can happen outside of the classroom.
This is the way it's worked for 300 plus years over at Harvard, one of the best universities, MIT for 100 plus years.
We think that essentially it's working.
And the reason we're having this conversation right now is because it has worked.
So we want to continue that legacy.
We think that every system is going to have problems.
And if you analyze any system, you're going to find issues.
And there are certainly some smaller inefficiencies within the lecture.
But by and large, it's trusted.
And it works
I want to add on what adding.
So I think we are not pushing those less smart kids away, because they can actually learn from each other outside of class.
We just want to say that doing the lecture is more efficient if the professor lecturing at them, giving all the information.
As a footnote to that-- sorry-- if professors come to the class and just give theory and go away, and the smart kids get it, then how do you measure the professor's performance?
How do you think about, say, something like students information on educational quality?
How do you measure that?
You don't really need the professor to be extremely active and do active teaching in the class.
The full student can actually get what they need.
You have a bunch of homeworks.
You have a bunch of test questions, exam questions, that student, whether they like it or not, they will surely be forced to read those things.
And by the time they come back, they come back learning.
Lot of graduate from MIT, they are l really doing well.
About 25% start up their own business.
Probably like 70% get jobs when they finish.
What are you talking about?
The system is working.
And That's what we're saying.
And their value is a great.
The evaluation is done on the type of textbook, the mode of lecture, the [INAUDIBLE] lecture.
So these are the things that I try to evaluate in the evaluation process.
And it has been great.
And we want it to continue
So we addressed the measuring impact.
And that goes for both.
But for the active learning, how do we quantify how much active learning needs to go into these different classrooms in the department?
So how will we ensure the appropriate amount has been implemented in different classrooms?
Basically, how much active learning needs to go into a classroom to create change or to strengthen our student learning?
How much has to go in or how you're going to make sure that the professors are doing it?
Both, I guess, is the question.
If there's a professor that's more on the lecture side that might throw one thing pair share in and call it a day.
So how do we ensure the appropriate active learning techniques and the quantity are there for the classrooms?
I think we'd have to go back to the research and see how much, whether it's 30% of the time needs to be spent on active learning activities, or to see what has been proven, because active learning has been proven to be effective.
But we would have to go back and make sure that we set a standard that, if it is 30% of the time or 70% of the time.
And I think oversight from the department would be important, in that the leadership in the department will have to sit in on classes.
And as part of the training of the professors, there has to be training.
But then there has to be someone to observe them in the class, give feedback.
The training has to be ongoing.
It's not just a three hour seminar on this is with active learning is.
And this is what you should do.
I think it's going to be a long term investment in the professor training that someone's going to have to check up and give them feedback.
And then they can adjust from there
I'm sure that we could [INAUDIBLE] all day.
But I'm going to give the deciders two and a half minutes, three minutes, to talk to each other and come up with a decision.
All right, the committee has met.
And now they're going to determine your fate in this department
Does this mic work?
Is this on?
Do I need to use the mic.
You can hear?
Well thank you again for your arguments.
Certainly the panel felt that the pro arguments in terms of the benefits of active learning are great.
And we definitely support that.
However, given the time frame, and this proposal is very aggressive, it would be extremely disruptive.
Also the amount of funds that we think that it would take to implement this would be very high without any proven way to measure success.
So what we are willing to do is commit funds to a pilot program, where we can then measure what the success will be and then make it a decision next year as to whether or not we invest for the rest of the departments.
[INTERPOSING VOICES]
So I hope you found this exercise useful.
I just want to debrief on the exercise itself, why we bothered to do it.
It took probably 45 minutes of time to do it.
But I made a decision that would be useful.
So what did we get out of this exercise?
What was the point?
Whether you think that's it's worth it or not, it's really important time.
Also, it seems this confrontation is happening a lot in the department, especially [INAUDIBLE] as they go through the program.
I assumed this test would be debated quite a bit, internally and externally
Other reasons we engaged in this?
I think it gives everybody the opportunity to think deeply about the subject matter.
And it also, even if you are against or for something, you have to have a deeper knowledge to be able to bring out the good points, certain issues that might actually make see [INAUDIBLE]
[INAUDIBLE], you want to add to that?
The topic is active learning.
And what we just demonstrated is active learning in practice
Yes, exactly.
And the idea to Gordon's point, the idea that it can be very, very useful for you to have students take an arbitrary position and argue it, because they will learn things, they will think about things, that they would not have thought about otherwise.
And it's usually not a good idea to say OK everybody that thinks this, go over here.
Everybody that thinks this, go over here, because all that does is reinforce what people think already.
And maybe it's not completely correct.
So this is often a very good technique, when there's two clear viewpoints or two clear sides.
And you want students to really engage with that.
And exactly, we could have had a discussion.
Or I could have had PowerPoint slides that said, here are the good things.
Here are the pros with active learning, bum, bum, bum, bum, bum.
Here are the cons of active learning, bum, bum, bum, bum, bum.
And that would've taken-- just to come out of my neutral position here-- that would have taken probably four minutes.
Instead we took 35 or so, 40 minutes.
But I think you probably engaged with some of the ideas and concepts a little bit deeper, because you had to come up with it yourself.
So yes, all those reasons.
And it does come up often.
People are discussing it.
And it's not so clear cut.
There's not a right answer.
And you're going to want to say, OK, well we'll do it by 2019.
Or we'll do it.
But we won't do that.
There's going to be deals and all those kinds of things.
And that's OK.
It's OK that there's not a right answer.
But that was the point of this exercise.
So that served as our discussion for active learning.
But you guys did it.
I didn't do anything
And that buttresses the point that active learning is the way to go
Thank you.
Yes, this is very meta.
This is very meta here.
Yeah and I do have to feel a little bit bad when you're like, what?
Then we're going to do all the work.
And the lecturer's just going to sit back and not do anything.
I took it personally.
But no, no, it's OK. One thing I wanted to bring up is if you haven't read the Scott Freeman article, it's in the reading list for today.
It shows-- let me just bump to it-- it shows essentially these results.
He did a meta analysis of 225 studies on active learning or on classes that that used active learning.
And they showed a 12% decrease in the failure rate in classes that used active learning.
And so I was telling the pro group that what say in the paper is that if it had been a clinical drug trial, and 12% of the people on the drug were having a better outcome than the people on the placebo, they would have had-- or the people on the control-- they would have had to stop the trial and give everyone the drug.
So in the sense of why we're still lecturing, that's to me one of the biggest compelling reasons for why we should be doing more active learning.
12% of the students are going to pass the class, on average, normalized.
And they actually considered both experienced lecturers, who were just lecturing, people that were considered to be good lecturers that were just lecturing, as well as graduate students who were first time teachers but were using active learning.
So they looked at all sorts of-- they mixed up all sorts of outcomes.
And active learning appears to, the data's pretty conclusive.
And this isn't the only study that shows it.
So there are definitely issues.
There are definitely implementation issues.
They're definitely sticking points.
But it's something that I think we can't ignore
I just want to add that it's not only that we have 12% decrease in the failure rate.
We also have big shifts in the letter grade students get.
For instance, more people shift from grade C to grade B or from B to A.
And B to see a test like that.
So that's really significant also
So I just want not have learning outcomes.
But I did want to have the discussion before I even introduced them.
So I think you'll be able to explain the impact of active learning exercises in the classroom.
You've already started to do that.
And then you'll be able to identify and develop active learning exercises that you'll want to use.
And you'll be able to think about how you might use them.
I think, pretty attainable learning outcomes.
I'm going to show the slide a lot throughout the rest of the time.
But remember, it's constructive alignment.
We've seen this diagram before.
But the beauty of some of these active learning techniques is that their formative assessments of students' understanding, of students' learning.
And so they are actually activities that happen in the class.
But they're also measures, informative, formative measures of whether students are learning or not.
So if somebody had come up with a really terrible argument, or it was clear that nobody had read any of the papers on the active learning research, then I would know that from this debate.
I know what you've read and what you haven't read, and what you're thinking about and what you're not thinking about.
And I get a good sense for whether or not it you know it and how much and how well you know it.
And so do you.
And we'll see that a little bit more in these other situations, where if you get a clicker question wrong or you get a multiple choice question wrong or that you can't participate in the lightning round, then you know that you've got to step up to the plate.
Or you need to try something different.
Or you're not really getting it.
So you get information.
The learner gets information.
And the teacher gets information
I just thought that was really powerful in the readings, as well.
They mentioned that, that students who, when they get in the group sections, if it's clear they really are getting a lot of things wrong, they know they need to step up.
And that's in a different way, I think, than getting a bad grade on a paper.
And you know the average distribution or whatever.
Or the professor says you're wrong.
But when you know that the people sitting right next to you are getting it and you're not, that really, I think, will push the student forward to do something
It's news you can use.
It's information.
And then you're not being penalized.
You're not in trouble.
It's not affecting your grade if you act on it, if you take the information and use it.
So that's why, I think that's the beauty of it.
There are many, many active learning techniques.
And if you looked at the website from University of South Florida has just, I think, it's close to 200 techniques for what you might do.
I'm doing a bit of a-- I call it the foie gras method.
You know how they make foie gras?
They force feed the goose.
So I'm going to give a lot.
And we're going to do a lot of different active learning things in this session.
It's a little bit comical.
I know it's kind of funny how many we're trying to cram into two hours.
But I'm going to try to do that.
So we'll do some of these.
And we can talk about any one of them in particular.
But a nice one to get students in the frame of your class is to they come in.
And they have five minutes before you do anything to write down what they thought was the most interesting part of the reading or answer a certain question related to the reading or solve a problem or whatever it is.
You give them a prompt.
You sit down.
You focus them.
And then they're off.
I do want to just focus on the word active for just a second.
It may help you to think about active and interactive learning.
There's all sorts of ways to chop it up.
And at some level, it's semantics.
But at some level, it isn't.
So students can be active if they're sitting by themselves thinking about something and writing something down or solving a problem.
That is active learning.
They don't have to be talking to each other.
You can do active learning in a class of 3,000 people.
Stop and think.
Answer this question.
That's active learning.
That's way more than happens in a lot of classes.
Interactive learning generally implies that students are talking to each other.
That's the most general categoriz