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Okay, in this video, i'm going to be talking about problem number 20 of chapter 6.
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There are four parts to this problem, so let's go ahead and start doing part a.
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Oops, so for part a, we are going to be expanding the expression 1 over 1 minus x.
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And the way we're going to do it is by using a tailor expansion, because that's the easiest way to do it.
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Now, let me remind you of what a taylor expansion is.
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A taylor expansion just says that a function of x plus a small parameter.
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In other words, if you have a function, f of x, but then you don't evaluate it at x, but you evaluate it at x plus a small number, epsilon.
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The value of the function at that point can be written as f of x plus epsilon times f derivative at x plus one half epsilon squared times f double derivative at x plus dot dot dot okay now the dot dot just means all the higher order terms that's what it means there's epsilon cubed epsilon to the fourth power abson to the fifth power and all that okay um now how are we going to apply it to this well, it's very simple.
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Let's define a function to be x inverse.
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In other words, 1 over x.
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Okay.
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So we can go ahead and apply the taylor expansion formula to this because we know that f of x plus a small number is going to be 1 over x plus a small number, which is which we can.
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Can use this expansion formula here to evaluate.
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So that's the procedure that we're going to be doing here.
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And it's a very straightforward procedure.
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Let's go ahead and calculate all the derivatives of the function f.
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So the first derivative is going to be minus x to the power of negative 2.
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And the second derivative is going to be 2 times x.
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To the negative 3.
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All right.
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So now we're going to use this expansion to evaluate this.
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And if you do that, we get, first of all, f evaluated at 1 is going to be 1, plus negative x times f prime at 1, plus 1 1 1 1 1 1 1 1 1 1 negative x squared times f second derivative at 1 plus dot dot dot.
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All right.
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Now this is going to be 1 minus x times, according to this formula, we have negative 1 plus 1 1 half times x squared times, according to this we have 2 plus dot dot dot.
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This is going to be equal to 1 plus x plus x squared plus dot, dot, dot.
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And that's what we're trying to prove here.
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And, uh, and that's all for part a.
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Uh, we have proved the expansion of one over one over one minus x.
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Let's go ahead and start doing a part b where we need to evaluate the partition function for a single oscillator.
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Now, the partition function is going to be a sum over boltzman factors.
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Excuse me.
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It's going to be a sum over boltzman factors.
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All right.
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It's going to be a sum over both of the main factors, which is an exponential of negative energy of some particular state over tau, where we're summing over all the states, all the energy states and stuff.
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Now, let me remind you what all the states are.
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For a simple harmonic oscillator, a quantized one, not a classical one, but a quantized one, the energy levels are discreetly spaced.
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And the spacing between each level, all right, the spacing between each one is h -bar -o -mega, or which is equal to h -f, if you're not familiar with h -begra -o -mega.
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So in other words, if you think of this as a discrete energy unit, then each oscillator can have any random amount of, like any random number of discrete energy units, each of which is h -bore -omega.
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In other words, what i'm trying to say is, if you have an oscillator, then the energy of that oscillator could be 1 -h -bar -omega, 2 -h -bar -begger, 3 -h -bar -bore -omega, 4 -h -bar -begger, all right? or 0 -h -bar -begger if your oscillator has no energy.
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So these are the energy states, and the reason i'm talking about these states is because these states are going to go into our...
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These states are going to go into our partition function.
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These are the states that we're summing over.
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So let's do that.
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So if i, like 0, 1, 2, 3, 4, these are just numbers.
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If i give these numbers a name, let me call it s.
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So we're just going to be summing over s.
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All right.
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S could be 0 all the way up to infinity.
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And the balsman factors are e to the negative s, h bar, omega, over time.
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Temperature.
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That's the balsman factor.
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Now i'm going to do a very, very cute trick.
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I'm going to define a new variable called x.
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X is going to be, let's see, x is going to be e to the negative h bar omega over tell.
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So this thing, this partition function can be written as s equals zero all the way to infinity of x to the power of s.
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All right.
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That's just algebra.
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That's just algebra.
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Now, the reason i did that, the reason i did that, is because if you look at this, this is equal to 1 plus x plus x squared plus x cubed plus dot, dot, dot.
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Now what is this? this, according to the formula that we proved in part a, this is equal to 1 over 1 minus x.
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All right, that's why i pulled this trick here, is because it's so that we can use the formula that we proved in part a.
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Now let's write it out.
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We have 1 over 1 minus e to the negative h bar omega over temperature.
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That's the partition function for part b.
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Okay.
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Now let's go ahead and start doing part c.
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So for part c, we have a formula that we can use to calculate the average energy.
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And that formula is equal to minus 1 over 0.
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X times the derivative of z with respect to beta.
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All right.
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Now, we have the z here.
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Let me write z as a function of beta, okay? beta is 1 over tau, by the way, where tau is kt.
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So z is equal to 1 over 1 minus e to the negative beta, uh, h bar omega, right? so this is z.
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You can differentiate this function.
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With the respective beta and plug it in here, and then divide that by z, and then multiply it by negative 1.
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So that's just going to be very straightforward calculation, so i'm not going to go through it here, because we're very short on time.
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Let me just give you the result.
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And you are more than welcome to verify this result yourself.
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The result is going to be that the average energy of the system, which is just one oscillator, is going to be h bar omega times e to the negative beta h bar omega over e to the negative, e to the negative, e to the one minus e to the negative beta h bar omega.
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That's the average energy.
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All right.
09:09
That's the average energy.
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Now, of course, this average energy, let me write it in another way, in a simpler way, perhaps.
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I can multiply the top and bottom by e to the beta.
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H bar omega, right? i can do that.
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That's algebra.
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If i do that, i have one times e to the beta h bar omega minus one.
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You see, that's simpler, right? because now we just have one exponential compared to two.
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So that's simpler.
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Anyway, that's part c.
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That's the average energy of one oscillator.
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Now, what is part d? well, well, part d is just asking about what is the total energy of n oscillators.
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The difference between part c and part d is that c asks about one oscillator, d asks about n oscillators.
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Now, assuming that the oscillators are very independent, namely, for example, if you have like two oscillators and they don't know each other, they don't know about the existence of each other, then the total energy is just going to be two times the un -ocillators.
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Energy of each oscillator, right? so that's the idea here.
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But instead of two oscillators, we have n oscillators.
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So the point is that the average energy of n oscillators, an independent oscillators, is going to be n times the energy of each, right? we know what the energy of each is.
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It's going to be h -bar -omega times 1 over e -to -the -beta, h -bar -omega minus 1, right? we got this result from the previous part, from part c.
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Okay, that's part d.
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Let's go ahead and do part e.
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Now, part e wants us to compute the heat capacity.
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Now, the heat capacity, let me remind you, is defined as the derivative of e -bar with respect to temperature.
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All right, that's what the heat capacity is defined as.
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Now, now, let me write out this.
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Derivative here, d by the t of e bar, which i'm going to copy it copy from here.
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We got it from the previous part.
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N, h bar omega times 1 over e to the h bar omega over kt minus 1...