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This is a. The Six Lecture of the condensed matter course.

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In the last lecture, we discussed bonding, ionic bonding and covalent bonding,

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and there were three other types of bonding that were on a list that we were going to cover eventually.

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Van der Waals bonding, metallic bonding and hydrogen bonding. And for various reasons, I'm going to push those off until later.

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We'll try to pick them up where they fit in. Instead, what we're going to do is we're going to go back and reconsider the motion of atoms.

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We're going to simplify our life an awful lot today, and we're only going to consider the motion of atoms in one dimension.

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So for the purpose of today, we live in a one dimensional world. We're going to summarise everything we know about bonding.

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Very simply, we're going to imagine we have two atoms some distance apart.

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It's called a distance X and there will be some potential V of X, which represents the force between the two atoms and the potential V of x.

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As we discussed last time, it probably looks something like this v of x.

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Something like that. So it has a attractive bonding force.

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And then if the atoms get too close, the potential shuts off to infinity.

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Okay.

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So what we're usually going to do is we're going to expand around the bottom of this minimum in a quadratic way and approximate it as a parabola.

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So we'll write V of x is some V not v not is the bottom v not here.

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Plus it's called this distance here. From here to here, let's call that x equilibrium.

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That's the the bottom of the well. In other words, the distance at which the atoms would most like two sets the minimum of the energy.

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So we'll give a quadratic term here, x minus X equilibrium squared plus dot, dot, dot.

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Okay. So the the energy is minimum. If the two atoms are separated by this equilibrium distance and then it's some sort of parabola,

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if you if the atoms are either farther apart or closer together.

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Okay. Now, we should be a little bit cautious in doing this,

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because occasionally by approximating something as a parabola, you throw the baby out with the bathwater.

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And in particular, if you're worried about thermal expansion, it's very important to keep the dot,

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dot, dot terms, the fact that in fact the potential is not a parabola.

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So let's think about that for a second. How does thermal expansion happen?

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Well, if you're at low temperature, the atoms, you sort of think of a like a particle in the bottom of a potential.

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Well, but what we're really talking about is the distance between the two, the two atoms.

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But you can sort of think about that distance oscillating back and forth, just like it was a particle in the bottom of a well.

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So it oscillates back and forth. The atoms get farther apart and closer together.

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They oscillate back and forth and pretty much the average distance stays at equilibrium.

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But if we give the atoms some higher amount of energy here, a higher temperature,

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then the atoms can oscillate in to here, but out all the way to here.

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Okay, because the potential is steeper on the inside than it is on the outside.

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Generally the atoms will be able to make it far.

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This x max here and this X-Men.

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X-Men will differ a different amount from x equilibrium in particular, x max plus X-Men over two will be greater than x equilibrium at t equals zero.

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So the average distance that the atoms are from each other when they start oscillating will start to increase.

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And this comes from the fact that when the when the atoms oscillate, they can push in a little bit,

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but the potential is really steep, so they can't push in that much.

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But then when they oscillate out, the potential is softer so they can go much farther distance out.

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So this is what gives you thermal expansion,

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the particular form of the potential function that's softer as the atoms go away from each other and gets very,

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very steep when the atoms get close to each other.

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But as long as we're not considering things like thermal expansion, it's okay to just truncate our potential at quadratic order.

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And we have a pure Hooke's law type spring between our atoms.

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So far so good. Everyone happy with that? Okay.

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So what we're going to do with the rest of the lecture is we're going to take a very simple model of atomic vibration,

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actually extremely simple model of atomic vibration, which is known as the one atomic, one atomic harmonic chain.

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Harmonic chain, which is potentially the most important model we're going to study all year,

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not only because it happens to show up on the final exams very frequently.

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So one atomic means that there is only one type of atom.

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Harmonic means that it's going to be just simple springs between the atoms and chains mean we're going to have a lot of these atoms.

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And the reason it's a very simple model, but the reason it's so important is because it introduces a lot of ideas.

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It will come back over and over again throughout the term.

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So this is what it looks like.

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We have a bunch of atoms and again, we're living in one dimension, the lined up in a row, each atom has some mass m all of them identical.

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And then there's a spring constant, all of the spring constants identical, identical between the two atoms, Kappa, Kappa Kappa and the Kappa.

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The spring constant comes from the expansion of the of the bonding potential, the harmonic expansion that we used up above.

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Okay. The equilibrium distance between the two atoms.

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X equilibrium here. Let's call it a.

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For the purpose of of argument generally put it over here.

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This distance A is known as a lattice constant.

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And generally we're going to use the word lattice constant frequently later in the term.

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Lattice constant generally generally means distance between between identical atoms.

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And in this case, all of our atoms are identical. So it's just the distance between the atoms here.

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Let's define some positions here. So let's call this 1x1.

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Maybe this one will be x2x3 that.

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So generally say that x of an is position of Adam and.

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And let's let xe and the superscript zero be the equilibrium position.

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Position of Adam and. So that's you've imagined letting the chain come to rest and you measure the positions they're all spaced by a distance.

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A So X and zero is just we can let it be.

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If we let the zero s atoms start at position zero, then we can just set zero to be n times a each atom separated by a distance a from the next.

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And that quantity we're actually interested in is the deviation from the equilibrium position, which we'll call Delta X.

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So delta x sub n is x, sub n minus x and not.

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And what I'm going to try to do is we're going to try to figure out the vibrations of this chain using completely classical physics to begin with.

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And we'll worry about quantum physics later.

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So you've probably done a couple of spring and math problems probably in your first year, and it's pretty straightforward.

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What you're supposed to do is you're supposed to write down Newton's equations for all of the masses.

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So that's pretty easy. We just have an ethical summary. So F on the atom and is mass times the acceleration delta x double dot.

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And so what is the force on an X on on the mass.

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Well okay, it has a force from the atom to its right, so it's Kappa Hooke's Law,

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Delta X and plus one minus Delta X and and then it has a force from the atom on its left, Kappa Delta X and minus one.

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Minus Delta X and. I guess we can simplify that a little bit by writing.

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It is Kappa Delta X and plus one plus delta x and minus one.

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Minus two. Delta x and. And.

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What we'd like to solve for is we would like to solve for the normal modes.

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Want, want normal modes of this chain.

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Normal mode to remind you means all atoms oscillate.

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At. At a common frequency. At common frequency.

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And if you remember how you did this in in your first year courses,

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it ended up being an eigenvalue problem with sort of a matrix, the dimension of the number of of masses you have.

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Now, we might have an infinite number of masses here to deal with if we have a very long chain.

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So it looks like an infinitely large eigenvalue problem and that might be sound a little bit frightening,

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but it turns out that solving problems like this is really easy and we're going to do it again.

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We're going to use the same kind of trick over and over this year, and the trick is to guess the answer.

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And fortunately the guesses are all the same. So it's easy to guess. The guess is you use what is known as a wave on dots.

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On dots is a German word that means something like Guess so.

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We're going to guess that the solutions are wave forms. Someone who speaks German is probably saying, no, that's not what it means.

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Is that what you're saying? Yeah, I have no idea.

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What does it mean? Rule of thumb or something? It means work.

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Okay. I stand corrected. Thank you. Well, in physics, we frequently we use it to mean.

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To mean. Guess what? So I don't know where it came from.

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All right. So so the trick is we are going to guess that that the that the the form of the of the oscillations are simple waves.

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So we'll write down a for a wave form delta X is some constant a you know,

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the I omega t minus i k times x and not k here is the wave vector and omega is the frequency.

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Now, this might be a little bit confusing because what we've written down is we've written down something complex,

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whereas we know the positions we're interested in are actually real.

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But this is just like when you study circuits in your first year, you're trying to think about currents that are oscillating,

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and instead of writing signs and cosines, you write down some sort of complex expression.

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And what you really mean is that you're supposed to take the real part at the end of the day.

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And that's what we really mean here is at the end of the day, you should take the real part.

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And the reason we do this is because it's always easier to work with exponentials than it is to work with science and cosines.

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Now, because we're going to take the real part,

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we could put an overall minus sign up in the exponent and still get the same answer so we can fix that.

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Omega is always greater than or equal to zero without any loss of generality.

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Because if you just change the sign of everything in the exponent, once you take the real part, you end up getting the same result.

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But we must keep track of the fact that K can have either sine k either sine.

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Which corresponds to either a left going wave or a right going wave.

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The other thing, I guess we should probably substitute in here the value of X and zero, which is end times A.

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Okay. So then all we have to do is we have to take this wave and that's plug it in to Newton's equations up there and see what we get.

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All right. Well, if we plug it in on the left hand side,

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we get minus M omega squared times A in the I omega T minus i k and a right to derivatives brings down minus omega squared.

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And then on the right hand side, we get a kappa a kappa a there's an eye omega t which is common to all the terms.

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And then we get either the minus I k and plus one a plus either minus i k and minus one a minus to in the i.

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K and a. Good.

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Now we can cancel out a whole bunch of things from this equation to simplify our life.

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So we get minus m omega squared equals kappa in the i k a plus minus i k a minus two.

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Then we can use a trig identity.

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This thing is actually two times a cosine and I actually move the m to the other side as well and we get omega squared is kappa

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over m two minus two cosine k another trig identity that we can replace one minus cosine k a as sine squared of k over two.

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So that gives us a total of four with. Yeah. Two on the four K over time sine squared.

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Okay. Over two and then I'll just take the square root of this whole equation.

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We get the final result. Omega is two square root of Kappa over m absolute value of sine k over two.

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And there we have it. We've, we solve for the frequency of the normal modes of our chain.

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Given the way. Vector K. Now, it's probably worth plotting the answer.

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Just go. So what is it look like?

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So here we have omega of k vertically and we have K horizontally.

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Like this and we'll put some points on let's make this point pie over a maybe this point over here is minus pi over a

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and we have the sine absolute value of sine and it looks like this absolute value of sine kind of looks like this.

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That. Okay, well, this must be the same height.

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And the height of this, uh, curve here is to square root of k k over M, and it has its peak at pi over A and at minus pi.

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Okay, so this curve is known as a dispersion curve.

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Dispersion. Dispersion.

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Term just means omega as a function of K.

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Now, somewhere in this picture, we should expect that there should be sound waves.

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We're talking about oscillations of of some solid.

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I mean, we have this we have our picture of a solid, which is just atoms stuck together in springs and somewhere in there.

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There should be something that looks like sound waves. Where are the sound waves?

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Well, sound wave is a very long wavelength phenomenon. The definition of sound is it has to be a long wavelength oscillation.

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So we should really be looking down here at small K to look for sound waves, just a as a, you know,

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a sort of pneumonic that the sound that you hear has wavelength somewhere between like a centimetre and several metres.

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That's what you can you can hear depending on how good your ears are.

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Whereas the spacing between atoms that we're talking about this pi over a a year one is so this the wavelength

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associated with this pi over a wave vector is on the order of the distance between atoms with a appears here.

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So two pi over the way vector will give you a number on the order of the spacing between atoms.

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So most of this picture is very, very, very small wavelength compared to what we usually think of as sound.

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Sound is for very, very small wavelengths right down near the middle, near K equals zero.

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So let's actually try to figure out what the sound will be doing.

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The way we figure out what sound is doing is to go to K very close to zero.

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We expand this thing, we expand the sine and we end up getting what do we get?

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Well, the one half cancels the two and we get square of Kappa over m r times a times absolute K And you remember the definition of the sound velocity

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is that frequency should be sound velocity times absolute K And so we've derived sound velocity is square root of kappa over M times.

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A There we have our sound velocity now.

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If you took the fluids course last term, you'll remember that sound can also be defined in terms of things like compressed ability of a fluid,

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and we should be able to get the same velocity of the sound from the formula that you derived last term, which is square root of one over density.

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Mass density here. Mass density times the data, the compressed ability.

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Okay. So what's the mass density of our chain? That's easy. It's just one mass per distance, eh?

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And what's the compress ability? Well, first we have to figure out what the definition of compress ability is.

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It's minus one over V, the volume, the pressure.

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But we're in one dimension, so that gets replaced by minus one over a length d length de force.

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So if you think about one little piece of the chain, the length is one over a length is A and D length.

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The force from Hooke's law is minus one over Kappa,

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so the compress ability is just one over Kappa A and then if we plug in these two expressions into that, well, okay, let's do it here.

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This is then using this mass dancing, this compressed ability.

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It's one over a, mass over A and then of Kappa one over Kappa A, and that gives us,

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in fact, the same result that we had over here, square root of Kappa over M times.

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A So it agrees with the expression we would get from a hydrodynamic picture of what sound waves should be.

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Okay, so everything all sorts of seems to work pretty well, but that is way down here in this regime,

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whether where the dispersion is linear at higher k, it's not linear anymore.

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It starts to curve. Now think back to what it was that Debye was saying.

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Debye was saying he was going to just assume that the frequency of his sound modes was always linear in K,

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and then he cut off the spectrum at some maximum frequency.

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Well, it starts out linear. It doesn't stay linear curves, but indeed it does have a maximum frequency above some frequency.

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There's no sound modes left. We found all of the normal modes of the system and this is all of them.

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And they have only and they only have frequencies between zero and this frequency here.

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So Debye was right in that respect that there really is a maximum sound wave frequency.

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It's not well, we wouldn't call sound, but vibrational frequency of of this chain.

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It's worth looking a little bit more closely at this point here, which is the highest frequency excitation, the highest frequency normal mode.

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So that occurs at K equals PI over A.

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And how are we going to look at that more closely? Well, let's let's write down what the wave form would be.

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It's a heat of the eye, omega t minus ei pi over a times a and or it would be a in the eye omega t times minus one to the n,

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which means that every alternate, even versus odd mass, is moving in the opposite direction.

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And that's the highest frequency you can possibly get. Okay. At this point, it's worth seeing a movie.

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So let me. Oh, here we go.

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See if it works today. Oh, beautiful. Okay, so this is a program we can download from my website.

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It was written by Mike Glaser. It works on Windows, it works under wine and Linux.

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I think it probably works on on Mac, but I'm not sure about that.

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So the it the purpose of it is to show you what oscillations of one of these chains actually looks like in detail.

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So here we click anatomic chain and we're going to click longitudinal, which means, oops, my gosh, change the speed.

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There we go. So you can change the speed of oscillation as well to make it easier to see.

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So what we have here in this corner over here, you have this dispersion curve and you can change k back and forth from high K to low K.

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So let's start with a very fairly low K and make the speed fairly fast so you can see what's going on here.

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Okay. And you can see at fairly low k come on.

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Yeah, you can see that this is sort of a long wavelength oscillation where everyone sloshes left,

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then slashes to the right and back and forth and back and forth. Now as we increase K,

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it oscillates faster and faster and the wavelength gets get smaller and smaller and until eventually it's oscillating really, really fast.

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And I'm going to have to slow down the speed so you can actually see what's going on when you get to PI over a this point here,

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you can see that every other mass is going in the opposite direction.

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And if you think about it for a while, you'll realise that this is there's a good reason why this is the highest frequency that you can possibly get,

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that you can't do any better than having every other mass opposing its neighbour,

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that if you want to get really high compression or high frequency, this is the highest frequency you could possibly get.

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Okay, so I'll. Leave that for you guys to play with.

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I recommend messing around with it a little bit. Oh.

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Are we going to have problem turning back on? All right, anyway.

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Good. So the most important thing that we that we deduced in this calculation is that the dispersion is periodic in K.

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It is proportional to assign and you'll see that the dispersion just keeps going

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exactly like itself over and over with sort of a periodicity know from here to here,

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you just keep replacing over and over. It looks exactly the same. This is not a coincidence.

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There's a general law that if you have something that's periodic in real space,

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real space like a chain which has a periodic periodicity, delta x equals a.

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If you take our whole chain and you shifted over by A, you get back exactly the same chain again.

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If you have a something that's periodic in real space. I should also say that real space is also known as direct space.

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Direct space. That will imply that you have a dispersion, which is periodic in K and K space, also known as reciprocal space.

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With a periodicity. Delta K equals two pi over a and hence the name reciprocal.

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Because here the A's upstairs and here the A is is downstairs.

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Now the periodic here's some some more words that are going to be useful. The periodic unit here is known as the Bruen zone for a one zone.

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Maybe I'll write that definition a little bit because it's rather important.

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The periodic unit in case space. In case space.

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Equals the brand's own. And I apologise not only to German speakers but to French speakers.

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Now that this word Bruyne, Louis de Bruyne is a word that English speakers have a terribly hard time pronouncing correctly,

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and probably anyone who speaks French will cringe at how I butcher it.

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But I'm not the only one. Every English speaker butchers this name.

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I really apologise about that. I almost failed French in high school, so I've never gotten any better at it.

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But I think it's something like brie wine. That's my best guess about how it's supposed to be.

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Okay, now, why is it this is a very important principle that we have periodicity in case space?

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Why is it we have periodicity in case space? Well, let's find out.

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We have our waveform delta x n is a, b, i, omega T minus I, k and A,

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and then let's take K and shift it by two pi over a shifted by this entire length.

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From here to here, we'll take some particular K and move it by two pi over A and see why we get back the same thing.

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Well, let's do that shift in the make a T and then we get minus i k plus two pi over a times

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and a that equals a in the iomega t r minus i k and a times e to the minus I to pi.

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And, and this thing here is one. So in fact, we're getting back exactly the same wave form that we started with.

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The reason the dispersion curve looks the same at this point here and this point here is because they're describing exactly the same wave.

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The wave is unchanged by shifting K by two pi over a.

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This is an extremely important principle that will come back over and over and over.

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Now, this might be something that you'll find disturbing to begin with.

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Ah ha ha ha. Okay, now we have a problem here.

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Okay, now I'll see if I can show you this. Come back. Okay.

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Let me show you. Something else.

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See if it comes back up eventually. Yes. Okay, good.

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So you might ask, well, if K is the same as K plus two pi over a, what's the wavelength.

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Right. I mean, we usually think of the wavelength as being two pi over a two pi over cake with the wavelength two pi over k because lambda.

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So if k is the same which isn't physically the same wave here and here, which one's the wavelength?

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Well, the way to understand that is from this picture here.

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So. Delta x the x here is plotted on the vertical axis just to make it more clear.

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So I plotted two waves here. One of them, the solid line is at K are actually mean.

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The dashed line is K and the and the solid line is at K plus two pi over a OC.

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The two curves agree at the position of the masses.

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And we're only talking about the displacement of the masses. The waves doesn't mean anything in between the position of the masses.

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The wave form is meant to describe what the masses are doing.

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You can't ask what is the value of the wave in between the masses?

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It doesn't have a value. The wave is meant to describe what the waves are doing, what the what the masses are doing.

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So it doesn't actually mean anything whether you extend the the curve in the solid line way or the dash line way in between the two masses.

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What only all that matters is what the masses you know, how much the individual masses are displaced.

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This is a phenomenon known as aliasing. And if you get this in your head, you'll understand pretty well why it is at this point.

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And this point are actually describing the same wave, even though they have different case.

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Okay, good. All right. All right.

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So. So a couple a little bit more nomenclature, that's fairly useful.

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The real space lattice lattice.

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Let me define it as all points. Points equivalent to x equals zero.

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So in our picture there that would be x and equals m times a.

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Because you can shift each point by an entire spring and you get back to something that looks exactly the same.

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We also have the reciprocal space lattice, reciprocal space subspace lattice, which is all points in case space points in space.

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Equivalent to K equals zero.

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And for this particular model, it would then be let's call those points g sub m equals two pi m over a.

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All those points are equivalent to archaic or zero shifted by two pi units of two pi over a.

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Now it's actually something that is kind of interesting and will be important later on this year is the the exam X

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and is always equal to one for all of the points in the direct lattice and all the points in the reciprocal lattice,

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you can choose any combination of those points and we'll always give you one.

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In fact, we're going to use that when we go to higher dimensions. We're going to use that property as a way to define the reciprocal lattice.

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So sort of put it in the back of your your head.

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So one thing that's actually really useful for us to do at this point is to count the total number of waves that we enumerated.

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We solve for normal modes of the system. So how many of them are there?

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How many? How many normal modes? Normal modes?

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We have. Well, we know that if we're counting different normal modes, we only have to look within the broken zone.

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Then if we go outside the bronze zone, we can always we're describing something that's equivalent to another point inside the bronze zone.

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If we take this point here, for example, it's equivalent to this point here.

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So we only have to describe things within the bronze zone. We have to figure out how many modes there are within the bronze zone.

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So we'll do the usual thing. Use periodic boundary conditions, periodic boundaries.

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So we're putting the springs in a in a in a big circle. Let's let the circle have a length l which is and so and masses around in a circle length l.

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So k then has to be two pi over l times some integer p.

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P is an element of integers. Usual way that happens when we put everything in a periodic box.

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So the total number of modes, number of modes is then the total range of k's that we're looking at two pi over.

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A So that's the entire range of K's we're looking at divided by the spacing between modes to PI over L.

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So what we're doing is we're dividing up this broad zone into lots of little pieces.

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Whose spacing is two pi over out. Okay.

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And so that is L over A or N.

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So the total number of normal modes that we solve for is end.

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There are normal modes because there are end masses to begin with. This should not be surprising.

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You probably found this when you study mass and spring problems before.

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If you start with and masses you inevitably get and normal modes.

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And this is actually exactly what Devi had guessed before,

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that he wanted to have just the same number of modes as he had degrees of freedom in his system.

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Okay, so so far everything we've done is completely classical so far.

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So now we're going to move on to quantum mechanics. Quantum.

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So in quantum mechanics, there's a very simple rule, which is that if you have a normal mode, a classical normal mode of frequency.

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Frequency Omega. When you go to quantum mechanics, that becomes ion states with energy.

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H bar east of NW equals H Bar Omega and plus one half.

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So what is this? What does this end telling us? This NW is telling us you pick a particular normal mode that you're looking at.

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And this RN is basically telling you the amplitude of that mode.

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Then if you if you let it oscillate a lot, it has a lot of energy.

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If you let it us a little bit, it has only a little bit of energy. So it's a difference between big and small end.

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But the key thing is that the energy is quantised in integer units of this bar omega.

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Now this is an important definition that one quantum.

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A quantum. A quantum of vibration.

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Of vibration. Is a phone on is known as.

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A phone on. Is a phone on. A phone on. This is entirely analogous to a quantum of light being a photon and its energy.

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Energy is h bar omega of K depending on which which K we're actually considering.

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So I now I have to I have to rant for a second about my my normal pet peeve.

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So my rant is that if you open up a lot of books,

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they will tell you that a phonon is a quantum of vibrational energy or a photon is a quantum of light energy.

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And I think that's actually a bad definition.

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I think you shouldn't specify energy because it's true that a photon carries light, but it also carries angular momentum and carries momentum.

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So why did you specify energy and not just and not just specify there's a quantum of light.

325
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Similarly, with phone on here, a phone on carries carries energy, but it also carries momentum and other things.

326
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So I'm just going to say a quantum of vibration is the phone on.

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So it'd be a little bit more specific about what I mean here by by a phonon we pick a particular normal mode or a

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particular oscillation frequency of particular wave vector K and then we say if it's in its ground state and equals zero,

329
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we say there are no phonons filling that state.

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If you excite rn up to one, we say there's one phonon in that mode.

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If we excite and up to two, we say there's two phonons in that mode.

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It's exactly similar to what you do with photons and light.

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Now, of course,

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these phonons have to be bosons because you can put phonon is Bose on is a boson because you can put more than one phonon in a given mode,

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given k mode. And it's energy.

336
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Energy of the phonons in a particular K mode is h bar omega k.

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So this is the energy of the phonons in a particular K mode times the Bose factor.

338
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There's another indication that we're talking about bosons here.

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Bose factor. This factor plus one half.

340
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So the bonus factor here tells you the expected number of phonons in the particular mode at a given at a given temperature.

341
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Okay. Is everyone comfortable with this? It should look a lot like like what you did with photons earlier.

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Yes. Yeah. Okay, good. So now what we can do is we can use this information to actually calculate the quantum mechanical heat capacity of our chain.

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Now, our chain is a rather simplified chain. It's a simplified model of what's going on.

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It has you know, it has these a number of simplifications that, you know,

345
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the springs are only between near his neighbours and the perfectly harmonic springs.

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But given those simplifications, what we're going to end up with is an exact answer for the heat capacity.

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No approximations at all. So the exact answer for the heat capacity, the total energy in the chain is the sum over all of the modes.

348
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K equals minus pi over a two pi over a of h bar omega.

349
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That mode those factor beta h power and make a k plus one half where the keys are taken in in steps of steps of two pi over l.

350
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Now, the usual way that we have before, so many times we can replace that sum with an integral decay over two pi times the length of the system.

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And we only have to integrate from minus pi over eight to pi over a because once we're outside the Brian zone,

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we're just re describing some of the modes that we have already previously describe.

353
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And if we count the total number of modes in the system, total number of modes, total number of modes.

354
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Number of modes.

355
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It would be well, okay, we can write it as l over two pi L over two pi integral four minus pi over a two pi over a decay of the number one.

356
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So we're just going to integrate over all the number of modes called the total number modes that will

357
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give us what L over A or the number of atoms in the system is again exactly similar to what divided.

358
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He cut off his integration over modes such that he would have exactly and modes in the system no more, no less.

359
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So we use we use our omega of K that we derived wherever it is.

360
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Well, off the top of the board, I'll write it again. We use two square root of Kappa over m absolute value sign of K over two.

361
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And if we plug that in to that, some of which we can turn into an integral,

362
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we will get the exact amount of energy in that system at any given temperature and you can differentiate it to get the heat capacity.

363
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The by used omega k equals v sound times absolute value of k which agrees pretty well

364
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at small k but doesn't agree at large k but it was a pretty good approximation.

365
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Einstein You can describe Einstein the same way. Einstein used all his oscillators omega k r at the same frequency omega zero.

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That will give you the Einstein model basically saying there's an oscillators and they all have the same, same frequency at any rate.

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This enables us to write down at least at least you can algebraically write it down

368
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an exact expression for exactly the heat capacity of our chain quantum mechanically.

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There's one more rather important concept that we need to introduce today, which is, again,

370
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coming back to this idea that Kay and Kay Plus are a member of the reciprocal lattice two pi over a represent the same wave.

371
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I mean, the fact that we we, you know,

372
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we like to think of these phonons as being particles the same way we think of photons as being quantum mechanics

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is great because you can think of things as particles and waves and waves and particles can go back and forth,

374
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however, however you like.

375
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So sometimes we like to think about these phonons as being waves, and sometimes we like to think of out of them as being particles and particles.

376
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We like to think about them as, you know, some object that's moving and carries some amount of of momentum.

377
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And in fact, sometimes in some models, these phonons can be more complicated models than this one.

378
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But in your phone asking can scatter into things in their way.

379
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They can scatter into other phonon scatter off of other phones, that can scatter off electrons, they can scatter off of light, for example.

380
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So we'll discuss some of those things later in the term.

381
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And the thing you might you might think that there would be a conserved momentum in the scattering process,

382
00:41:36,810 --> 00:41:42,629
but how is it possible we can conserve momentum in a scattering process if K isn't

383
00:41:42,630 --> 00:41:48,420
even well-defined up to this change that K and K plus g represent the same wave.

384
00:41:48,810 --> 00:41:54,900
So what one does is one defines define what is known as crystal momentum.

385
00:41:55,620 --> 00:41:59,579
Crystal Crystal. Well, okay, if it's K, it's the usual thing.

386
00:41:59,580 --> 00:42:03,720
If it's K, it's way vector. If it's H marking its momentum, crystal momentum.

387
00:42:04,230 --> 00:42:19,350
So I'll put in the H bars crystal momentum, which is defined as h bar k modulo two pi over a.

388
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Now what's modulo mean modulo is a word which means up two additive terms of additive terms.

389
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Of. So, for example, 12 mod five equals seven mod five modulo five because up to an additive term of five, 12 and seven are the same.

390
00:42:45,370 --> 00:42:49,419
Two mod five is also the same. So in other words,

391
00:42:49,420 --> 00:42:59,460
where we're confessing that momentum is not conserved up to this factor of two pi over a that you can take and give factors you know,

392
00:42:59,470 --> 00:43:04,750
additive terms of two pi over a freely and this might disturb you because you might you know,

393
00:43:04,750 --> 00:43:10,209
you might really like this thing called momentum conservation but we really need to

394
00:43:10,210 --> 00:43:14,200
think back to why is it that we wanted momentum conservation in the first place?

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00:43:14,200 --> 00:43:22,359
If you took the symmetry in relativity course last term, you'll remember that momentum conservation comes from a symmetry of space,

396
00:43:22,360 --> 00:43:25,510
and the symmetry of space that it came from was translational symmetry of space.

397
00:43:25,810 --> 00:43:29,590
And so you can translate, you know, your attention in space to anywhere you want.

398
00:43:29,740 --> 00:43:33,430
And all the laws of physics remain the same in our model here.

399
00:43:33,430 --> 00:43:41,559
We don't have that anymore because we're only allowed to translate by a our translation group is smaller than in, you know,

400
00:43:41,560 --> 00:43:47,290
in out in the vacuum of space here we were only allowed to translate by a and because of that

401
00:43:47,290 --> 00:43:52,569
we get a less strong conservation law that instead of having momentum completely conserved,

402
00:43:52,570 --> 00:43:56,660
we only have momentum conserved up to two pi over a week.

403
00:43:56,890 --> 00:44:00,820
So that's what's gone gone on here. It's very similar to this idea of North as Theorem.

404
00:44:00,820 --> 00:44:04,510
It's not exactly north this theorem into Did people tell you about Amy North?

405
00:44:04,510 --> 00:44:08,770
There she was. She was amazing mathematician from the early part of the 1900s.

406
00:44:09,790 --> 00:44:12,939
Northeast Theorem, strictly speaking, only applies to continuous symmetries.

407
00:44:12,940 --> 00:44:16,030
And this is a discrete symmetry, but it's very, very similar.

408
00:44:16,210 --> 00:44:21,700
And the reason we're getting less strong momentum conservation than we had in, you know,

409
00:44:21,730 --> 00:44:28,480
in in when you think about regular particles in space is because we don't have the same translational symmetry of space.

410
00:44:28,480 --> 00:44:32,830
We only have a discrete translational symmetry. Okay, we have a few minutes left.

411
00:44:33,700 --> 00:44:43,960
So I'm going to I'm going to add in subject that I promised you I would address, and that subject is hydrogen bonding.

412
00:44:44,470 --> 00:44:48,190
So this is a totally different subject. I'm just picking this up because it's only could take us 5 minutes.

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00:44:48,700 --> 00:44:58,150
Hydrogen bonds. The general idea of a hydrogen bond is that hydrogen is a very special element because it's basically just a proton and an electron.

414
00:44:58,570 --> 00:45:04,360
So you might imagine having a big atom like so here's a big atom like fluorine,

415
00:45:05,380 --> 00:45:11,410
and then it bonds to a hydrogen atom, more or less ironically, but maybe slightly covalently.

416
00:45:11,620 --> 00:45:16,570
So the hydrogen atom is sitting out here, it's a proton and its electron is more or less sitting over here.

417
00:45:16,810 --> 00:45:22,870
Its electron has been sucked to the the fluorine atom, and what's been left behind is a bigger proton.

418
00:45:23,110 --> 00:45:27,220
So this is our hydrogen nucleus, and what's left is a better proton.

419
00:45:28,720 --> 00:45:37,050
A proton. And that's extremely special because it's the only time in chemistry where you ever get a better proton.

420
00:45:37,060 --> 00:45:39,520
It's not screened by some electrons around it.

421
00:45:39,910 --> 00:45:45,340
So whenever you have chemical bonding, usually some of the electrons are moved around, but there's always some shell left over,

422
00:45:45,490 --> 00:45:50,950
except in the case of hydrogen, where the one electron can be removed from it and you're left with a proton.

423
00:45:50,950 --> 00:45:53,410
And that's why hydrogen bonds in very special ways.

424
00:45:53,710 --> 00:45:58,090
So what can happen then is if you have another atom over here, say, oxygen or maybe another fluorine atom,

425
00:45:58,450 --> 00:46:05,080
this bearer proton will then make a very strong polarisation of this fluorine atom and attract it.

426
00:46:05,230 --> 00:46:15,400
And this is the hydrogen bond, h bond, which is the bonding of the hydrogen to another atom via this strong dipole force.

427
00:46:15,760 --> 00:46:19,810
So let me show you a classic picture of this, how this shows up.

428
00:46:21,380 --> 00:46:27,740
So ice is sort of the classic example of that where you have hydrogen sitting on the surface of oxygen to make your H2O.

429
00:46:28,010 --> 00:46:32,570
But the oxygen has essentially removed the electrons from the hydrogen to a very large extent,

430
00:46:32,780 --> 00:46:38,090
leaving behind a bare hydrogen, their nucleus of a proton,

431
00:46:38,420 --> 00:46:46,879
that their proton polarises this other electron over here and attract this other electron here on this oxygen and attracts the oxygen,

432
00:46:46,880 --> 00:46:53,060
making a very weak dotted bond to this big red oxygen sitting over here.

433
00:46:54,270 --> 00:46:59,370
And this you know, it's a weaker bond. It's a much weaker bond and a covalent bond or an ionic bond.

434
00:46:59,610 --> 00:47:04,290
But it's strong enough to freeze ice, you know, below below zero.

435
00:47:04,830 --> 00:47:08,010
So it's a you know, it's a strong enough bond to do some interesting things.

436
00:47:08,310 --> 00:47:15,090
Another classic case where hydrogen bonds enter is in biology and particularly in DNA.

437
00:47:15,360 --> 00:47:21,570
If you took the biology segment last term, you'll recognise this as being DNA if you didn't.

438
00:47:21,790 --> 00:47:26,540
Well, now you're seeing it. This is DNA and you'll see it.

439
00:47:26,550 --> 00:47:28,920
So DNA is made up this this double strand.

440
00:47:29,220 --> 00:47:36,090
And right down the middle of the double strand where the two strands zipped together, there's a hydrogen bonded to an oxygen.

441
00:47:36,210 --> 00:47:41,240
This bond is the hydrogen bond. The hydrogen is really bonded to the nitrogen over here.

442
00:47:41,250 --> 00:47:47,910
And it has a weak bond to the oxygen over here. And you'll see that there's a similar hydrogen bond from a hydrogen here to a nitrogen here.

443
00:47:48,120 --> 00:47:51,450
Similarly, hydrogen bonds right down here and hydrogen bonds down here, right here.

444
00:47:51,660 --> 00:47:56,549
So it's a hydrogen bond that holds together the DNA right down down the zipping.

445
00:47:56,550 --> 00:48:01,380
So. So when you pull the strands of DNA apart, you're breaking hydrogen bonds.

446
00:48:01,620 --> 00:48:05,759
So I think that's all pretty much you need to know about hydrogen bonds.

447
00:48:05,760 --> 00:48:09,900
And I guess we will finish. We'll pick up again tomorrow.