The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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? The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at OCW.MIT.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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? The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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? The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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? The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu 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. 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 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. 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 ocw.mit.edu 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. 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 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. 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 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