00:01
Okay, in this video, we're going to be talking about problem number 22, 6 .22.
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All right, let's start with part a.
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For part a, we want to show this identity here.
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So let me take the right -hand side of the identity, and then write it in the following way, 1 over 1 minus x times 1 minus x to the m plus 1.
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This is equal to we're going to use the result that we have from a previous problem i think it's problem 6 .20 but i'm not sure you can check for yourself if you're interested but i can write this one over one minus x thing as 1 plus x plus x squared plus all the way up to an infinity all right that's an infinite sum times 1 minus x to the power of n plus 1 now this is very simple because what you can do is you can foil you can take the 1, multiply everything by the infinite summation, and take the x to the m plus 1, multiply everything by the...
01:08
I'm just foiling here, okay? there's nothing fancy going on here.
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So let's foil.
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We have 1 plus x plus x squared plus dot -da -da, all right? that's one times...
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That's just one times everything here.
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Min is x to the power of n plus 1 plus x to the power of m plus 2 plus x to the power of m plus 2, plus x to the power of n plus 3 plus dot, dot, dot.
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All right.
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Now, if you stare at this long and hard, you'll see that this is just equal to 1 plus x plus x squared plus all the way up to x power, x to the power of n.
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Because all the things that come after x to the power of n, they're gone.
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Why? because they got subtracted, okay? thanks to this term.
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Okay, if this term does not exist, we wouldn't have been able to subtract off all the terms that come, that come after x of the power of n.
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Okay.
02:04
So that was part a.
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For part b, part a was very straightforward.
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Let's go ahead and take a look at part b.
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For part b, we want to have an expression for the partition function.
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The partition function is just a sum over bolzman factors.
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Okay, what's a boltman factor? well, a boltzman factor is negative energy of some particular state over tau.
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We're summing over all the states.
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Now, what are the states that were summing over in this case? well, in this case, we're just talking about the rotational states, right? the z component of the magnetic dipole moment is quantized.
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It's quantized.
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Therefore, there is a number j, which could be equal to one -half or three -half or one or whatever, and this number determines the quantized.
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Quantization.
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There's the energy levels are quantized.
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And if you look at the prompt, you're going to see that the z component of the magnetic dipole moment, that is just negative j times the number delta mu, or a negative j plus one delta mu, or dot, dot, dot, dot, dot, or all the way up to j delta mu.
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In other words, if you measure the z component of the magnetic dipole moment of the particle, you'll see the that it could not be continuous.
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It has to take on discrete values, and these are the discrete values.
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Of course, with each of these values of the z component of the dipole moment, there is an energy associated with it because this dipole moment interacts with the external field, external b field, if there is one, of course, which there is in this case.
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So what we are going to do is we're going to write the energy of each state as negative mu .b, which is just negative mu z times b, because b is in the z direction.
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This result, if you don't know where it comes from, it comes straight out of enm.
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Okay, that's just like introductory enm.
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There's no, there's no, nothing fancy here.
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So let's do, let's write everything that we talked about into this partition function.
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So we have a sum over some number n.
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This number n could range from negative j all the way up to positive j.
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We have this summation here.
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If both one factors negative beta, beta is just 1 over tau.
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T is just kt, if you don't know, times energy of the state, which is negative mu zb.
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Now, this is equal to a sum over n of e to the beta mu zb.
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Now, what is muz? what is it? if you look at the prompt, you'll see that muz, you don't even need to look at the prompt.
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I already wrote it here.
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Muz could take on these discrete values, and these values are determined by two numbers, j and delta mu.
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So, so, um, so we have the summation of n from negative j to j of both from factors of e to the power of beta times b times n delta.
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Mu all right that's just mu z that's what it is so what is this equal to well we want to define a new variable b equal beta delta mu b all right we are defining it we don't have you don't have to define it but defining it this way make things a little bit less messy so this is equal to if you use the new definition of b you can write the summation as sum of rand of e to the power of uh let's see where our b is.
06:11
We have beta here, delta mu here, and b here.
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So that's little b.
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The three things come together, the form little b.
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So we have e to the power of n b.
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Now, let's define a new variable called x.
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X is e to the power of b.
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Now, the reason i'm doing it is because i can write this as a sum over end of x to the power of n.
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Now, what is this equal to? this is just x to the power of negative j plus x to the power of negative j plus 1 plus dot dot plus x to the power of j.
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All right.
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Now, what is this? this is just x to the power of negative j times 1 plus x plus all the way up to x of the power of 2j.
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All right.
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Now, i'm going to use a result that we obtained from part a.
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And then if you define this thing as m, we just have x to the power of negative j times 1 plus x plus dot -da -dot plus x to the power of m.
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What is it? if you use the result from part a, you'll see that this is equal to x the power of negative j times 1 minus x to the power of m plus 1 over 1 minus x.
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Okay, that just comes straight out of part a.
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That's the whole point of doing part a is because it's so that we can use it here.
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So now now we need to do a little bit of manipulation here okay we're going to manipulate this we're going to massage it this expression i can write it as this okay because that's what it is all i did was i took the x of the power negative j and wrote it on the denominator because i can now now you can write this as 1 minus e to the power of b 2j plus 1 over e to the power of bj 1 minus x okay because x is equal to and i remind you x is equal to x e to the power of b and m is 2 j okay so if you substitute everything in here you're going to get what i get here all right that's just algebra there's nothing complicated going on here so but that's not the goal the goal is to show that this is this is equal to the expression that we're given in part b, which is a fraction of cinch something over cinch something.
08:53
We want to show that this is equal to cinch b of j plus one half over cinch of b over two.
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So how do i show that these two things are equal? well, very simple.
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I'm going to use the definition of the hyperbolic sign and write this as e to the power of b j plus one half minus e to the power of negative b j plus one half there is a over two here and let's see for the denominator we're going to do the same thing for the denominator we have b over two minus e to the power of negative b over 2 over 2 okay the 1 over 2 cancel um now now what i'm going to do is i am going to multiply both the numerator and the denominator by this number e to the b over 2 so now i just get e to the power of b, j plus 1, minus e to the power of negative bj over e to the power of bj, excuse me, e to the power of b minus 1.
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So now i am going to do another thing, which is multiply top and bottom by e to the bj.
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Okay, if i do that, i'm just going to get e to the power of b2j plus 1.
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Minus 1 over e to the power of b j plus 1 excuse me that's not what i wanted to write i can write this denominator as e to the power of b j times e b minus 1 now i can write this as 1 minus e to the power of b 2j plus 1 i am multiplying top by negative 1 that means i'm going to have to multiply the bottom by negative 1 as well so uh if i if you do that you have e to the bj 1 minus eb now i if you look at what we got before if you look at this and you compare it to this what do you see you see the same thing right you see the same thing because uh there's an x here which is which is e to the power of b okay you see the same thing, that means they're equal.
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That means we're done for part b.
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So we have successfully derived an expression for part b for the partition function, and then we showed that this expression in terms of boltzman factors and stuff is in fact equivalent to the expression that we're given, which is a fraction of cinch something over cinch something.
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They're equal.
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They're equivalent because of the definition of the hyperbolic sine function.
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Anyway, that was a lot of algebra.
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Let's go ahead and move on to part c.
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We want to show that the total magnetization of a system of n particles is equal to this long expression, which i'm not going to bother writing down.
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But let me remind you of something.
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The total magnetization of a system of n non -interacting particles is equal to n times the magnetization of one particle, the thermal average, of course, of the magnetization of one particle.
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Okay? so if you want this, you need to first get this.
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But this is very equal to get.
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Excuse me, very easy to get.
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Why? because a thermal average is just a sum of the variable that we're looking for times the probability of that particular state.
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That's just standard expression for the average value of some observable.
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We all know this from quantum mechanics or whatever.
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So what is the variable that we're looking for? well, the variable that we're looking for is, in fact, the z component of the dipole moment, the magnetic dipole moment.
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That's what we call the magnetization.
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That's what magnetization means.
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It's the z component of the magnetic moment.
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It doesn't have to always be the z component.
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Sometimes if you're just talking about the magnetization in a general context, then we're just talking about the magnetic moment.
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But in this case, since the b field is in the z direction, we're only interested in the z component of the, magnetic moment that that's why we're talking about mu z here anyway the probability what is the probability well the probability of each of these states that were something over is just the bolzman factor over the partition function all right now what's the bultman factor well the baltman factor is e energy of the state over tau this is of course a function of state it is not a constant now uh what is the energy of the stage, right? what is it? well, it is, in fact, e to the power of negative.
15:06
The energy of each state is just negative.
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Let me remind you that we have obtained an expression for the energy of these states from part a or something, but we have it.
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It is negative, mu, z, b, okay? over tell.
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Mu zb just comes from mu dot b but b is in a z component so you just have mu z b now this is equal to mu z x over z times e to the beta mu z b all right um now now i'm going to write this as one over z times mu z e to the beta we're of course summing over all the states.
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Now, if you look at this, this expression that we got, what is it? there's a very cute way that you can write this expression because this is just equal to muz, excuse me.
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You can pull out a factor of muz from the exponent if you just differentiate this exponential.
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You can do it.
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Before i, before i, before i, i, i, i, the derivative trick, let me first rewrite the muz as n delta mu.
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That's what mu z is, where n is the variable that we're summing over, which can range from negative j to positive j.
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So that's mu z.
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I'm going to write the same mu z on the exponent here.
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So we have beta, b, n delta, mu.
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Now, i'm going to write the same muz.
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Now, because delta mu is a constant, i'm going to pull it out of the summation, so we have delta me over z times a sum of n.
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N, e, let me remind you of beta, b, and delta mu.
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This combination of three variables, they make up a new variable called b that we have defined before.
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So we have nb here.
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Now i am going to pull the derivative trick, okay? because how do you pull, how do you how do you, how do you pull a factor of n out of the exponential? you differentiate the exponential with respect to b.
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That's how you get a factor of n in front.
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All right, that's very cute.
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That's very, very, very cute.
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So we just have an sum over n of derivative with respect to b of the modified version of the baltimore.
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I guess you can call it that.
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And b.
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If you don't believe me, just go ahead and differentiate.
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If you differentiate this exponential with respect to b, you're going to get a factor of vent different.
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Now, a sum over derivatives is a derivative over sums.
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So, in other words, i just want to pull the derivative outside of the summation.
18:27
Because i can, because that's calculus 1.
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Now, the reason i did that is because if you look at this, what is it? just ask yourself, what is this thing? what is it? isn't it just a partition function? that's just the partition function.
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Okay? this is just a partition function.
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You substitute all the weird variables back in, like b.
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If you substitute b back in, b is beta, b delta mu, you're going to see that this is just a partition function.
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It really is.
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So now you see.
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See a very interesting pattern because we have 1 over z times the derivative of z.
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What does that remind you of? it should remind you of the natural log function.
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Okay? it should because this is just the derivative of the natural log of z.
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Okay? if you don't believe me, just just do the derivative yourself.
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Um, so now, now the reason this is, uh, good, the reason i'm bothering doing this is because because we have an expression for a z and it's a very convenient expression too and we drive this expression from part b so i'm going to pull it out of part b.
19:54
Part b says that z is cinch of b of j plus one -half over cinch of b over 2 okay it just comes straight out of part b so now this thing, this thing i'm going to write as log of cinch of b .j plus one half minus log of cinch of b .mpt of that's b over 2, all right? because the natural log of a fraction, you can write it as not, you can write it as a natural log of this, minus natural log of that.
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That's the property of natural log now now i can i now i'm gonna differentiate this thing first and then differentiate that thing and then subtract them okay because i can that's because the derivative of a plus b is derivative of a plus derivative of b that's calculus 1 so now we have one number cent of b of j of j plus 1 half kosh of b of j plus 1 half multiplied by j plus 1 half because chain rule minus 1 over cich of b over 2 kosh b over 2 times 1 over 2 because chain rule okay now this is just equal to these things together, they form a co -tangent, a hyperbolic cotangent of b of j plus one -half times, of course, j plus one -half minus hyperbolic cotangent of b over two times one over two.
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Now, if you look at the expression that we have, we have, we have, we look at the expression that we have obtained, it is exactly the same expression as the one given in part c, except for the extra factor of n in front.
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In other words, there's supposed to be a factor of n here.
22:32
Well, our expression doesn't have it because we're just calculating one particle here.
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This is the, the thermal average muz for one particle.
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Okay.
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If there's end of them, you just multiply this thing by and you'll get the total value.
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So, so m where there's n particles is equal to n times average m for one particle where average m for one particle is just this expression, which is the same thing as the one that we're supposed to be getting.
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So this is all.
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This is all for part c.
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Let's go ahead and move on to part d.
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Let me put a qed here just for completeness.
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You want to show that the magnetization has the expected behavior as t goes to zero well uh you know as t goes to zero as t goes to zero remember b goes to infinity because because b as we defined it is proportional to beta.
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What is beta? beta is just 1 over t times a factor of ultimate constant, but that's not important...