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Okay. Let's get started. Welcome back.
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This is the seventh lecture of the condensed matter, of course.
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In the last lecture, we introduced a rather important,
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simple model of vibrations in one dimension the anatomic harmonic chain, chain of masses and springs.
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Every spring is the same. Every mass is the same extremely simple model, but introduce a lot of very important concepts to us.
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We solve for all of the normal modes of the chain.
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When we quantised the system, we discovered the meaning of phonons with the quantisation of modes where we introduce the concept of crystal momentum.
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The idea that if you shift away vector by two pi over a, you get back the same exact physical wave that you started with.
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We introduced the concept of a bronze zone arranged in space,
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usually taken from minus pi over a two pi over a where all of the wave vectors are physically different waves.
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And if you go outside the broken zone, you just start repeating the waves that you have already described.
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And we also manage to calculate the heat capacity of this chain. Exactly.
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And we compared it to the Devi theory and the Einstein theory that we had looked at previously.
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Now, one of the simplifying assumptions we used going into this model was that every single mass and every single spring in this chain were the same.
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And in real materials you typically had different types of atoms.
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For example, in sodium chloride you'll have a sodium atom and a chlorine atom, and these will be different.
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So maybe a better model of a typical material would look something like this.
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You'd have a light coloured atom and a dark coloured atom and drawing everything in one dimension again.
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And then a light coloured atom and a dark coloured atom and a light coloured atom, dark coloured and so forth.
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Now let me introduce some nomenclature that's going to be useful.
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The idea of a unit cell is the repeated, repeated motif.
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So, for example, we can take our unit cell to be this box.
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And if you repeat this box over and over and you stack the boxes together with no space in between, you will reconstruct the entire chain.
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Another important word that we introduced last time is lattice constant, which we defined as well.
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We usually call it a and we define it as the distance between distance between we in equivalent atoms.
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So for example, Gladys Constantine. If you like this, we'll call it a dislike for Adam to dislike called Adam.
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You may also notice that that lattice constant also happens to be the size of the nasal.
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Right like this this distance here is also a or if I drove properly it was.
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Now one side comment is that the the unit cell is actually not unique.
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Let me draw another picture of this chain. I could have drawn the initial sell equally well like this.
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So that's something we should probably even write down because it's important in itself, not unique.
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And if I had chosen this as a unit cell, I could stack duplications of that together and build up the entire chain just as well.
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Now,
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what we're going to do today in a real calculation is we're going to try to calculate the normal modes of a train of a chain that looks like this.
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And the reason we're doing this is not just to add complexity to a problem we already solve last time.
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In fact, by having different types of atoms in our chain, we're going to see that some fundamentally new and different things occur.
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So that's why it's why we're doing it. So let me draw the chain.
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This is known as the alternating chain, frequently known as the altering chain, sometimes known as the diatomic chain.
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And it looks kind of like this. There's a dark atom, then a light atom.
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Dark atom and a light atom and dark atom and light atom.
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And they're connected together with springs. And so forth.
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And we have a lattice constant A which is the distance between identical.
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We see the difference between light and dark. And to make clear enough, you see that.
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Okay, so a is a distance between equivalent atoms.
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So between light atom and light atom. And now we have a choice as to what model we want to write down.
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And there's two common choices that people make in solving these two simple problems, but they are similar.
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One choice is is frequently made is to make the mass of the light atom, the mass of the dark atom different.
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We're not going to solve that. You can actually solve that for homework. That one's a typical choice.
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But we're going to solve instead is we're going to make the two different springs this spring Kappa one and this Spring Kappa two different.
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And we're going to let them alternate back and forth. Kappa one, capita Kappa one and two and so forth.
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And hence the name altering chain, because the springs alternate between Kappa One and Kappa to the physics of these two chains.
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Whether we're letting the masses alternate or the springs alternate is extremely, extremely similar.
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The reason we're doing this one is because it's algebraically just slightly easier to keep track of than the one you going to do for homework,
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which is slightly algebraically more complicated. So let's write down some coordinates.
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So for example, let's let the light coloured atoms have positions x, n and the dark coloured atoms have position y n.
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So for example, that's like this one, the x one.
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This one will be y one. It's going to be x2.
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This will be y two and so forth.
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And now the way we solve problems like this, these mass and spring problems, is we start by writing down Newton's equations of motion.
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So, well, okay,
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one more definition is the Delta X and delta Y are the deviations from equilibrium emissions from equilibrium position from from equilibrium.
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Okay. And we're going to write down Newton's equations in terms of these Delta X and delta Y.
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So maybe actually, maybe I'll put it on another board over here.
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So mass delta x double dot.
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Mass times acceleration equals. We have to be a little bit careful here.
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XA atom, a light atom on the right, it has a Kappa two capital.
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And then on the right of an x, Adam is a y atom with the same number.
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So Delta Y and minus delta x n.
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And then on his left is a Kappa one, plus Kappa one, and on its left is a delta y atom with the index one lower right is Delta X.
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And does everyone agree with that? That on the left of x two is y one.
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So I have two. The capital one spring has the lower index.
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Yes. Good, happy, good. And then delta Y and double that.
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Okay. So on the right of a Y is an X with the index one higher and that's a Kappa one spring.
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So Delta X and plus one minus Delta Y n and on the left, to the left of y one is x one.
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So it's an x with the same index, delta, x and minus Delta.
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Why? And I think I got that right. Okay.
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So those are our equations of motion. And the way we're going to solve this is the same way we did last time.
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We will write down a wave on that wave onwards.
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So there is some disagreement on what the word Anzacs actually meant in German.
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And I actually saw someone stop me after a lecture, a German speaker said, No, no, no, you were right about what that means.
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And then someone else said, No, no, no, you're wrong. And then I asked a couple of German speakers and they all disagreed.
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And then actually someone pointed out that that is actually an English word.
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It is in the OED. You can look it up. It means it means mathematical.
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Yes. So so I'm going to say it's in there.
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It really is. Okay. So anyway, we are going to write down our wave on that.
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So Delta X and is sum and is the I omega t minus i k and a and delta y n is oops I called x rather and a y in the eye omega t minus high k and a.
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And what we really mean as last time is we mean to actually take the real part of this whole expression.
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And because of that, we're allowed to choose omega a greater than or equal to zero.
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But K can be either sign, either sign to represent either left going or right going waves.
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Now, as with last time, we realise that if you take K and you shifted to K plans to pi over A, you get back exactly the same wave.
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And the reason for that is because if you look at the exponential factor, even minus i k plus two pi over eight times n is.
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Times a times, eight times and is equal to exactly the same thing as in the IK and a and a.
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So it matches that those two are exactly the same.
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So if you shift k by the two pi over a, you get back exactly the same way form rather important fact we found out last time.
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Just a couple other things we're going to use periodic boundary conditions,
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periodic boundaries because we use a system of length L equals N times A, so this means we have n unit cells in its cells.
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And if we have a system of size l that means the case must be on a periodic boundary condition, must be two pi over l times p where p is an integer.
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So the spacing between different l different possible values of k are two pi over l.
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So now if we want to count the number of different k's.
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We did exactly the same calculation last time, different case.
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That equals the range of possible case. That's two pi over a before we start repeating,
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since you can shift by two pi over a and get the same way back and the spacing between k's is two pi over l that gives us l over a or n.
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So the number of different k's is equal to the number of unit cells.
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In the system. A rather general rule that we will use.
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Many times. Okay. So given our wave on that up there, we can solve this set of Newton's equations.
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It's by plugging in and doing a little bit of algebra.
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And then I can see that the algebra I'm about to do is a little bit messy, but bear with me.
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So here goes. First of all, plug in for Delta x n.
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So two derivatives gives me minus omega squared m then the wave is a x in the I omega t minus i k and a and that equals
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well k equals kappa two times delta y n which is a y in the i omega t minus i k and a and then let's do the other capital.
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That's the other term over here, the other Y term.
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So that's plus kappa one e to the I omega t minus i k and then it's and minus one a is an index and minus one and then must be the two x terms.
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So we have minus kappa one plus Kappa 2ay here is in die why a x in the i omega t minus i k and a everyone happy with that.
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Did I do that correctly? No one reacts, seems to work.
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Okay. And then the other equation is minus omega squared and a y in the eye, omega t minus i, k and a equals.
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And let's do the two x terms first.
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So we have kappa 1ax in the I omega t minus i k and then the coefficient and in the second equation of the index of the x with kappa one is one.
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So it's end plus one a here and then we have plus kappa to a x the omega t minus I, k and A and then the two,
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then the two the two y terms minus kappa one has kept to a y in the eye, omega t minus K.
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And a notable thing about this equation is this the longest equation I'm going to write on the chalkboard all year long, and now it's done.
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So we never have to do that again.
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But it's not actually kind of complicated as it might look because there's a bunch of exponential factors which drop out the counts on both sides.
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So we have two equations and we can simplify it into one matrix equation and write it this way minus omega squared times the vector x,
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a y equals some big matrix times the vector x, a y, and the big matrix is what is it?
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It's a minus kappa one plus Kappa to and then minus Kappa one plus Kappa two on the diagonal, Kappa one plus two.
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And then the off diagonal has Tabata plus Kappa one.
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Either the I k it's that part of the exponential didn't cancel it comes from the plus one and then the Kappa two plus Kappa one, the minus a.k.a.
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Come coming from the minus one and the exponential. Does everyone good with that agree?
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Okay. All right. So what is this? This is an eigenvalue equation or the eigenvalue is this guy over here minus seven mega squared.
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And the way you solve an eigenvalue equation is by moving the omega squared to the diagonals over in this side,
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then you set the determinant equal to zero.
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You get the characteristic determinant or characteristic equation, or sometimes called the secular equation.
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We do that the one step and omega squared minus capital one plus capita all squared
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minus capital plus capital one in the i k a absolute value squared equals zero.
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And we can solve that rather rapidly.
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Omega squared equals capital one plus capital plus or minus absolute value capital plus capital one.
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Okay. And that's our solution.
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Now, one thing you may notice is that for each K, there are two possible solutions of Omega, the plus solution in the minus solution.
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Right. So does that mean we'll write that down over here?
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Each K has two normal modes.
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Modes? We can call them omega plus and minus.
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Okay. And how many cases do we have? The number of cases are equal to the number of unit cells, and each K has two times two normal modes.
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So number of normal modes. Number of normal modes equals two times number of human cells.
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Unit cells equals the number of masses.
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And this we should have expected because we had that number of masses as the number of degrees of freedom we have.
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So when we solve this problem, we should add exactly that many normal modes.
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And indeed we do. So that's that's good. Good.
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We're happy with that. All right. Good.
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Now, so it is useful to actually plot this thing, and I'm going to sketch it out first and then we'll sort of justify why it looks the way it does.
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So herring and fat. Here's K, here's omega.
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Okay. Pi over A is minus pi over a, minus pi over a.
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So that's the the brown zone outside of the brown zone. It's everything just repeats.
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So we should have a mode that looks like this and then a higher frequency mode maybe looks like this.
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Okay. And then you could actually reproduce this periodically outside.
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I mean, you could you can make it go onwards as you like, because as you shift everything by two pi over a, you just get back the same wave.
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But we're going to really focus on the bronze zone because within the bronze zone here, every wave you find is different from every other wave.
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Now, it's useful here to probably look a little bit more carefully at this plot and see why it looks the way I've drawn it.
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So let us look at some convenient points. Maybe at K equals zero.
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That's a good point to look at. Well, what happens with cake zero?
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So over here we have this absolute K to plus k one either the a.k.a that goes to just a K to plus k one.
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So M omega squared is either two times k one plus k to or zero.
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And so I've drawn that here, here zero at cake zero.
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And here is the higher end. You want me to draw the higher one also.
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So omega plus k equals zero is square root of two Kappa one plus Kappa two over m.
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I get that right? I think so. So that point here, we'll give it a square root of two.
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Kappa one plus Kappa two. Over. And so we have a long human with the high energy mode.
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Now the low energy mode will come back to the high energy a moment, but the low energy,
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low frequency mode is one that we should expect should have been there,
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because we should expect that there should be sound waves somewhere in the system.
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And when there are sound waves,
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we know the sound waves should have a spectrum which is has frequency linear in wave vector and comes down to zero at zero wave vector.
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Now to convince ourself that that's what's going on with this equation.
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We have to actually do a little bit of more math and expand the dispersion curve near K equals zero.
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So let's see how we do that. So let's take ak2 plus k one of the i.
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K. That quantity there is.
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Well, yeah. That thing there can be written as square root of capital squared plus capital one squared plus two capital one kappa two cosine of k.
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And just by multiplying that out and if k is small, we're going to replace this cosine by one minus k squared over two.
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So then what we have is I guess we can write it like this square root of this
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whole thing of two plus kappa one squared that takes care of the one term.
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When I put together this, this and this, I get capital one plus capital squared.
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And then what's left over is twos council, capital one, capital K squared.
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I get that. I want to leave that. Yeah.
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Okay, then this conveniently factors out to give her a capital plus capital one times the square root of one plus capital one.
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Capital squared over capital one's capital square.
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And if you remember square root of one plus x, I get a sign wrong.
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I got a sign wrong. That's minus. That's minus.
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No one gets chocolate for that is one minus X over two.
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So so then we have we can rewrite this as capital plus capital one minus capital one.
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Capita K squared over two.
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Capital one plus capital. Does that look right?
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Did I miss anything? Yeah.
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Capital kept to thank you. Cavity.
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Yeah. Yeah. Good. Thank you. Whoever said that, I owe them chocolate, so.
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Yeah. Okay. Okay, good. So I don't have one today. I was going to bring one, but I ate it, so.
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All right, so if I plug that into into here,
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we see that the Kappa one plus Kappa two with the with the minus solution cancels this this
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capital one plus capital also will cancel this and we'll end up just getting m omega squared.
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Omega squared equals capital one kappa two K squared over two,
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Kappa one plus Kappa two or equivalently omega square root Capital One, Kappa to a squared over, uh, to Kappa one plus Kappa two.
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I guess there's an m down there and then absolute k, all right.
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And I realise that's a lot of math. We're not going to do too much math that heavy this year, but I have to do it once in a while.
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Okay, so this we recognise as the sound velocity, so we just derived the sound velocity of this wave.
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The sound velocity is the slope of this curve here.
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Now usually actually I should use the proper nomenclature.
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Usually people refer to this as the acoustic mode and they tend to call it the acoustic mode,
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even all the way out to the zone boundary out here at very high K,
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even though it's really only sound when it's when it's long wavelength or small K, they call the whole branch here the acoustic mode.
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Now, you remember last time when we derived the sound velocity.
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We also were able to derive the same sound velocity by using a hydrogen panic argument.
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So let's see if we can do that again this time.
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Remember from hydrodynamics we have the v sound should equal square root of one over the mass density times the compress ability.
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The mass density is where there are two masses per lattice constant a and the compress ability while we derive last time,
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the compress ability should be one over the spring, constant times a and the spring constant.
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Well, here the the unit cell has two springs in it and the spring constant for two springs in series is Capital One,
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capital over capital one plus capital. Does that sound familiar?
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You remember that from first year. Okay. So if we plug this and this and this into here, in fact, we get exactly the same result.
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V sound equals square root capital one kappa to a squared or to kappa one for step two.
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So the hydrogen AMOC calculation gives you exactly the same result as we got by actually solving the system completely.
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Now let's talk about this mode up here.
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This mode is known in comparison to the acoustic mode.
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This mode is known as the optical mode, optical mode, high frequency mode, occasional zero.
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And we should probably I mean, I should probably tell you why the name optical mode.
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So the word optical mode comes from experiments on unreal.
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Real materials. When you shine light on the material and you induce vibrations by by shining light on it.
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Now, if you think for a second,
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what's the property of light that we know similar to sound light has frequency which is proportional to its wave vector.
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The only big difference is that C is huge compared to sound velocity.
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So if I drew the light dispersion curve on the same plot, it would be an extremely steep slope line like this.
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Probably so steep I couldn't even. Right now, if you imagine shining light on a material and creating vibrations in quantum mechanics,
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we should think about absorbing a photon and creating a phonon.
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So imagine a process where you absorb a photon and create a phone on.
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In order for that to happen, you have to conserve both energy and momentum.
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So you have to match up both the frequency and the wave vector of the light with the frequency in the wave vector of a phone on.
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The only place that can happen is right here. And the reason you're never going to match the acoustic mode because the velocity mismatch is so badly.
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But with the optical modes, since the object optical has finite frequency, even a cable zero,
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there is a point where the optical mode frequency matches life frequency and the optical
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mode momentum matches the the light momentum or even crystal momentum in this case.
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Now, so that is an area of where why people call this the optical mode.
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Whenever you have interactions with light and vibrations, it's inevitably the optical mode.
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But I should be a little bit honest that that process I described to you by which you absorb a photon and you emit a phonon actually does not occur.
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And the reason it doesn't occur is because there is a conserved quantum number,
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the spin of the photon, which I mean, photons have spin phonons do not.
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So if that process were to occur, you would violate angular momentum conservation.
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So that is, you know, definitely bad. You don't want to violate angular momentum conservation.
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So, in fact, no such process exists by which a single photon is absorbed and a single phonon is created.
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However, there are more complicated processes that can occur involving photons and phonons.
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For example, absorb two photons and emit some photons.
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That's okay, because the angular momentum of the two photons can cancel and you can still have
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conservation and is still the same principle that is very hard to conserve,
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both energy and momentum in any sort of interaction between phonons and photons,
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unless you're getting optical modes into the game because you need to get high frequencies into the game somehow with phonons.
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And the only way to get high frequencies with small wave vectors in Phonons is to use the optical mode.
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So that's where the name comes from, actually. More generally, let me give more general nomenclature here.
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Acoustic mode. Generally, acoustic mode is any mode where Omega is proportional to K at small K and optical mode.
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Optical mode is any mode where omega goes to a constant, not equal to zero at small K.
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Okay, so let's think about this, this small K regime a little bit more seriously.
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So we found the eigen value is the frequencies, but let's look at the eigenvectors.
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So the matrix were actually diagonals and is that big matrix there?
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And let's look at it at take zero. So at zero the matrix we're interested in is of the form, I guess it's minus Kappa one plus Kappa two times.
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What is it. One minus one, minus one one.
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Something like that. I get the minuses in the wrong place and I think it's right.
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Okay, so it looks something like this because when when the wave vector goes to zero, the exponential, the okay both go to one.
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And so you have a matrix of the form one, one minus one, minus one one.
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If I pull out minus one plus kappa two. So the eigenvectors we have two possible eigenvectors one is for the omega equals zero solution.
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We have an eigenvector x, a y equals one one.
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Whereas for the the optical mode solution, the other eigenvector the high energy high frequency eigenvector we have x a y equals one minus one.
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So what is this telling us?
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What this is telling us is that for the acoustic mode near k equals zero, the atoms are actually moving with their neighbours.
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They both are. The atoms in the unit cell are moving in the same direction at the same time they're moving in concert they are.
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Whereas for the optical mode, the higher frequency mode, the two atoms in the unit cell are moving opposite each other.
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And it is actually quite natural to explain then why it is that the optical mode is so much higher in frequency,
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because an optical mode you're compressing the springs maximally,
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whereas with the acoustic mode you are hardly compressing the springs at all because everyone is moving in the same direction.
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These sort of trends occur more generally.
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So General, if if there are m if there are m atoms in the unit cell.
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Atoms in units cell. We will have emojis that each k one is acoustic.
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That's the mode where everyone moves in the same direction at k equals zero, and then the remaining and minus one of them are optical,
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meaning they don't all move in the same direction and k zero in d dimensions.
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The dimensions there are d times m modes at each k at each k.
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These are optic are acoustic and the remaining D times and minus one are optical.
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Okay. Why is this? Well, we're in the total count of the modes is a total count of the number of degrees of freedom.
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If there are m, there is always the same number of ks per as as the number of unit cells in the system.
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Each each value of each unit cell has m masses in it and they can move in d dimension.
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So it is total of D.M. degrees of freedom per unit cell.
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So we expect the M modes for each K.
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Now of them there can be d of them there acoustic and that corresponds to all of the masses moving in any of the three possible directions.
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So we know what these things are. We have discussed them before. One of them is longitudinal and two of them are transverse.
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So for each K, one of them is moving in the direction of K and then two of them are moving perpendicular to K in three dimensions.
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Okay. There's one more point on this picture that we should probably look at more carefully,
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which is these interesting points here near which are at the so bronze zone boundary.
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Pi over a when we put it over here. So consider consider these boundary.
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These boundary. K equals pi over a.
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So what do we have then? Well, then this factor capital plus in the I k a kappa one is actually absolute k two minus k one.
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And so the frequency is omega squared is one over mk1 plus kappa one plus capital plus
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or minus absolute kappa one minus Kappa to a cap of two minus capital one does matter.
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So absolute value. And that means that we have two possible solutions.
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Omega equals either square root of two kappa one over M or square root of two Kappa two over M.
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So we'll mark them here. So this one we'll call Square Root of two Kappa one over M, and this one will call the square root of two Kappa two over M.
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And here I've assumed we have Kappa one is greater than Kappa two otherwise.
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We are around the higher end. You want to squeeze on top and these modes actually never cross each other.
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You can convince yourself by looking at the form of of the of the frequency these two modes can never actually cross.
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So what is going on at this at this brand's own boundary, at the brand's own boundary, the wave form delta x, delta y equals x,
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a, y, you know, the i omega t either minus i k and a this e to the minus i k and a becomes minus one to the n.
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And that tells us that what's going on is that alternate unit cells are moving out of phase with each other.
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They're moving in the opposite direction as each other. So now is see if we can get this to work.
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Uh oh. No, no, no.
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Stop, stop, stop, stop, stop. Stop.
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Know, uh. Oh, dear.
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Does this happen? Only on Mondays. Here we go. Screen up. Okay.
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Oh, but that's. Oh, wow. Amazing. Okay, so we can still do this while it's going on.
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Okay, so this program with is the thing written by by Mike Glaser and you can click alternating
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chain and longitudinal longitude means that everything oscillates in the single line.
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You can change the K here by changing the slider and you can see that the the dispersion is plotted down here is plotted from
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zero to pi instead of zero instead of minus pi over zero to pi instead of minus pi over a two pi over a like I've drawn here.
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So it is the same picture. So first let's start by taking K very, very small.
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O And also you have these little clickers up here which will change the masses on the springs that's making the masses the same,
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and C one and C two, or we have what we call Kappa one and Kappa to the two spring constants.
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So we have two different spring constants, Kappa one, a cap of two.
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And here's the dispersion, the acoustic mode down here, the optical mode up here, here we've clicked acoustic.
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So we're seeing actually the acoustic mode here. And you see that in acoustic mode is very low frequency.
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And basically all of the let's move this up a little bit so it happens a little faster.
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You can see that basically all the masses are moving in concert with each other, sort of slow sloshing back and forth.
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Hydrodynamic oscillations you can turn viewed as one big fluid.
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Everything moves together. Now if we go to K equals zero, it becomes zero frequency.
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But we can click on Optic here and we'll get the optical mode. You see, the optical mode is each mass moving opposite the other mass in the unit cell.
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And you can see that this is something that should be high frequency because it's compressing the springs maximally.
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Now if we then change the K vector to the brand zone boundary, what we have here is,
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is as predicted, that the alternate units cells are moving opposite each other.
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So this unit cell, these two masses are moving opposite of these two masses.
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And you can see actually why it is that only one of the two spring constants enters in the
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frequency because the spring between these two masses actually isn't getting compressed at all.
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It's only the spring between the unit cells that are getting compressed and not the spring within the unit cell.
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It's getting compressed now if you switch than from looking at the optical mode at the zone boundary to the acoustic mode of the zone boundary,
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it looks extremely similar, but it's a lower frequency here.
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But you'll notice this the other spring, this now getting compressed is still you know,
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you have one unit cell moving in one direction or another unit cell moving in the opposite direction.
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But now it's the opposite spring, the key to spring, which is being compressed rather than the Kappa one spring being compressed.
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Okay, so I highly recommend that people download this program and mess around with it.
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It's got a lot of other fun features. This is what we can do on a boring Saturday night or something.
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Okay, so let me redraw this picture here.
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Maybe maybe stop this for a second. Okay.
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Redraw this picture here. All right.
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I don't know why it's still still doing that, but. Okay. All right, good.
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Me redraw this picture, except I'm going to draw slightly differently.
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Here's K again, here's Omega and I'm going to put Pi over here and minus pi over here and then two pi over here and then minus two pi over here.
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I'm going to draw the acoustic mode just like I had it before. And then the optical mode, what we had before was something that looked like this.
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00:40:11,010 --> 00:40:19,680
But I am perfectly allowed to take this piece here and shift it by two pi over a and plot it instead here.
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00:40:24,270 --> 00:40:28,650
Okay. So I did research at this piece and I shifted it by two pi rate to move it over here.
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00:40:29,040 --> 00:40:33,120
And then I can take this piece here and shift at two pi over a this way.
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And put it here. Okay. So now I'm going to erase.
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I know people hate when you do a race of transforms, going to raise the peace in the middle.
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And I leave these pieces here. So this piece and I have this piece.
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Okay. So I just took this piece and moved it by two pi over this way, and this piece moved by two pi over three that way.
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And I got this picture here. Now, this way of drawing things is known as the extended zone scheme.
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Extended Zone Scheme.
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And it has a nice advantage that there is only one K, there's only one mode at each frequency at each wave vector.
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K So whereas over here we have two modes at each wave vector.
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K Here we only have one, but we have used twice as much range of K and that's sometimes convenient to do just some nomenclature which is useful.
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This range here is known as the first boron zone.
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Boron Zone. And this region here is known as the second grand zone, along with this region here, second brand zone.
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So we took the optical mode and we moved it out of the first boron zone and put it into the second run zone for the extended zone scheme.
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This way of drawing things over here in comparison to the extended zone scheme is known as
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the reduced zone scheme because everything has been reduced into a single bronze zone.
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Now, why is it that it's so convenient to to spread everything out so that there is only one mode at each wave vector?
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Well, a case where this becomes extremely useful to do is the following case.
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Suppose consider Kappa one very close to Kappa two.
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So in that case, what's going to happen is that these.
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This little gap here is going to get extremely small. So let's see what that happens, what that looks like.
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So here's K, here's pi over a year to pi over a year minus pi over A and here is minus two pi over eight.
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So it's going to look like it's kind of this.
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But like this. Like that is a very small gap at the bronze zone boundary here.
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Because Capital One and Capital are very close to each other. Now, what happens when Capital One actually equals capital?
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Well, then what we have is the gap closes. It becomes a single, single connected line.
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But more importantly, we recover the monetary McCain Right.
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If Capital One equals capital, then every mass and every every spring is the same as every other spring.
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Every mass is the same as every other spring. So we get back the modern atomic chain.
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But the lattice constant has changed.
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The new lattice constant is only half as big as the old lattice constant,
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because the white and black masses now actually are the same, the same object now.
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So we don't have to make a unit.
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So with two things, we now make a unit cell with only one thing in it. Right.
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So the bronze on the bronze on the back.
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Now ranges from K is goes from minus pi over a prime to pi over a prime.
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00:44:08,250 --> 00:44:15,210
Okay. And that is the same as two pi over a minus two pi over a two to pi over a.
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00:44:16,630 --> 00:44:25,670
Okay. So this entire range here, one, if the two masses become the same, this entire range here becomes the first branch zone.
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00:44:26,100 --> 00:44:32,660
First because of monotonic chain. My atomic chain.
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00:44:33,140 --> 00:44:42,110
And then if you make Capital One capital slightly unequal, you open a small gap at the new own boundaries at PI over a.
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But really what's underlying is a single mode for the atomic chain from here to here and here to here, just a single connected node.
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And you open up make when you make capital one in capital slightly different, you open up a small gap at the bronze zone,
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at a bronze on boundary of the new smaller branch zone associated with the new larger unit self.
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Happy. Okay. In the last few minutes, I want to discuss a completely different topic,
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something we didn't we didn't get to last time, which was or the time before, which is van der Waals bonding.
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Van der Waals. You may remember this guy from the van der Waals equation of state that you probably studied in stap mac van der Waals bonds,
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also known as molecular bonds. Molecular or fluctuating dipole.
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Fluctuating dipole bonds.
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Bonds. And these are Vancouver's bonds are actually quite different from either ionic or covalent or
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hydrogen or metallic bonds in the sense that these bonds occur for inactive chemical species,
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inactive species. And what I mean by inactive species is that no electron is transferred between atoms, no electron is shared between atoms.
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Things like noble gases are classic cases of Andreev bonding.
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The noble gases are filled shell. It doesn't share its electrons.
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It doesn't donate electrons. It does not covalently bond.
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It does not, ironically bond. But it can still van der Waals bond.
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Now, another example of an inactive species are molecules such as nitrogen to nitrogen by itself as an atom is very active.
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But when you put two nitrogen atoms together, they form a nitrogen molecule, which is extremely inactive.
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You can think of it as being filled shell of a filled molecular shell, and it also does not share.
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Once you have the molecule, the molecule is very stable.
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It doesn't share its electrons with anything else and it doesn't donate its electrons to anything else.
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So nonetheless, even though you're not donating electrons and moving electrons around, you can still running out of space.
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You can still form a bond.
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The way the bond occurs is as follows Imagine that you have, say, a nice, noble gas atom over here and maybe another nice noble gas atom over here.
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And I suppose at some moment this atom has some dipole vector p want a polarisation?
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Now if you remember from your in m a distance away, there will be an electric field due to that dipole moment.
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In this case pointing in that direction and e will be equal to minus p one over four pi epsilon, not our cubed.
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And if it was, if I had drawn the angles differently, I conveniently chose R to be perpendicular to the moment there.
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But if I didn't make them perpendicular, there'd be some cosines and signs and stuff like that, which I'm leaving out now.
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When this atom experiences this electric field, it develops a polarisation, it gets an induced polarisation.
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P two which is chi some constant times.
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E This is this constant here is known as the polarised ability, polarised ability or electric susceptibility, electric susceptible.
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And you can calculate this for simple atoms like hydrogen, you can calculate the susceptibility in perturbation theory.
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And maybe if you had done that as an exercise, as an exercise in the book, they asked you to do that using quantum mechanics anyway.
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Now what we have is we have two dipole moments and you remember that the energy of two dipole moments is minus P,
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one P, two over four pi epsilon, not cubed.
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So the total energy is then just plugging these things all together.
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We get minus p one squared chi over four pi epsilon not r cubed all squared.
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And in particular this is negative and goes out to the sixth.
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So the force goes as one over to the seventh.
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Now that is more or less how the calculation goes that you have a polarisation,
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one atom induces a polarisation in the other atom and the polarisation is a track.
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But you might say, well, wait a second, this whole argument was predicated on the statement that the polarisation is not equal to zero to begin with,
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but in a spirit, the symmetric atom like like helium, the expectation of the polarisation is zero.
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00:49:32,310 --> 00:49:37,320
Why is that? Well, you remember that the the polarisation of an atom, here's the nucleus,
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here's the electron electrons running around the atom in some sort of spherical orbit.
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00:49:44,550 --> 00:49:52,680
So the, the position are of the electron is related to the polarisation just by minus eight times R or something like that.
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00:49:53,820 --> 00:49:59,160
So it's really, we're just saying that the average position of the electron is actually at zero,
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00:49:59,370 --> 00:50:06,330
but you'll notice that what actually comes into this equation is not the polarisation or the dipole moment P, but it's P squared.
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So P squared is not equal to zero so that you can get a nonzero dipole or attractive force, even though the average dipole moment is a zero.
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If you like to think about fluctuations in quantum mechanics as being some sort of dynamical fluctuation or fluctuation in time,
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what we really mean is that it's a sort of a course picture,
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a crude picture of what quantum mechanics is, but you can sort of think of it as at some moment in time,
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the electron is on one side of the nucleus, so there's a dipole moment.
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The other atom responds to that and oriented dipole moment in the opposite direction than the attract.
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00:50:41,490 --> 00:50:52,760
And then at some later time, there's a question. Yeah.
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If there's an angle between. You're saying that if there's an angle between you two, there would be cosine theta and so forth and things like that.
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00:50:59,600 --> 00:51:03,110
Yeah, but the force will still be attractive. You can.
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And actually there's, there's an exercise in the book that works you through it.
446
00:51:06,230 --> 00:51:09,680
And you can and you can get all the angles right. And it will always come out attractive.
447
00:51:10,100 --> 00:51:13,940
Good question, but not worth the chocolate bar. So but it's still a good question.
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00:51:13,970 --> 00:51:18,380
Thank you. Anyway, the.
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00:51:19,280 --> 00:51:22,819
So it's basically you get a little bit of fluctuation on one atom.
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00:51:22,820 --> 00:51:25,280
It causes a fluctuation in the other atom and they attract.
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00:51:25,550 --> 00:51:30,200
And no matter which direction the fluctuation is, the responding fluctuations in the opposite direction and they always attract.
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00:51:30,530 --> 00:51:37,250
So you can still get a nonzero van of force. Now the van de Vos force is much weaker than covalent ionic or hydrogen bond,
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00:51:37,520 --> 00:51:46,429
but it's still strong enough to cause important effects such as, you know, it holds together things like argon at low temperature.
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And when argon becomes a solid a more interesting case of where Vandervoort forces show up is with this guy.
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If you've ever been to a tropical climate, this is a gecko, a little lizard.
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They also they sell car insurance in the United States. I don't I don't know why that's true.
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This is the mascot of the Gecko Geico Gecko Company.
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Anyway. Geckos are amazing little creatures because they can crawl up almost completely flat glass walls with no trouble at all.
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And the thing that actually sticks them to the wall turns out to be Van der Waals forces.
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They have you know,
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they have a very flat foot and they stick their foot onto the glass and there's enough van der Waals force between their foot and the glass.
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And they're fairly like little creatures that they can actually walk up the wall because of that.
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All right. I apologise. I went over. I will see you on Wednesday.