 So as I mentioned in the derivation of Boltzmann, we got to this point where we had to account for the difference between cylinders, columns, whatever you call them with different volumes. And that didn't really work in my simple formulation when I only took energy into account. Now, what if I assume that I have two states, A and B, but I allow them to have different volume? Well in one case, the likelihood of finding something in state A if state A is ten times larger and state B is ten times larger, there will be ten times more things that I can find in state A, right? So maybe I should start comparing what is the probability of being in state A multiplied by the volume of state A? Relative to the probability of being in state B multiplied by the volume of state B. That follows by the loss. If I what is the probability of getting a six when I throw a dice? Well, that's one in six. The probability of getting a five is also one in six. The probability of getting five or six, that is the upper third. I can just add them up. So here I can multiply by the number of states that I have in A. That works. Here's statistics. That would then equal V A multiplied by the exponent. I will get rid of the constants now just to save myself some writing. So minus E A divided by kT divided by V B E raised to minus E B divided by kT. That doesn't really look simple. But you'll have to follow me here that and believe me that I say that this is a good idea. I'm going to use something very simple in terms of logarithms law that I can write any number as the exponent of the logarithm of that number of functions. So V equals E multiplied by L and V. Sorry. E raised to L and V. If I do that here, what I get is that I have an exponential of minus E A divided by kT. And then we need quite a bit of space here and another exponential minus E B divided by kT. And now I have an exponential here multiplied by an exponential. But if I have two exponences, that means that I'm allowed to add the argument inside the exponences, right? So the minus sign there just applies for the first term and then I have plus the logarithm of V A there. And here I have plus the logarithm of V B. Still doesn't look simpler. Let's try one more thing. I'll go to the next line here. Maybe it's going to be better if I can put these on the same fraction. So I'll take this one, the logarithm of V A, and multiply it and divide it by kT. It's worth a shot. As long as I follow the laws of mathematics, I can do anything I want. E minus long expression here. I divide everything by kT and then I have E A and then I have a count for the minus plus sign here. So I need to have a minus minus. And then that should be kT multiplied by the logarithm of V A. And the same way here, I'm going to have E raised to minus E B minus kT ln V B and that part divided by kT. This is what we should be pairing our hair. It's just getting worse and worse and worse. Please stop adding more things to the equations. No, this is beautiful. It's completely beautiful. This, my friends, compare that to the other equation you had on the previous slide with the Boltzmann distribution. So what do I have? Well, if I could just have this entire thing, this must have units of energy and this one must also have units of energy, right? So in principle, I have the same thing as on the last concept. I have a quotient between two exponentials raised to minus something with unit of energy divided by kT. But I have this extra component here. What we do as physicists, I'm not going to be able to get further there. I'm going to need to have them. I can't just simplify them away. But what I can do as physicists, when it's something complicated and I don't want that to confuse me too much, I can just define that. So the logarithm of the volume here, that appears to be a property of the state and then we have k, that's a constant, but temperature on the other, temperature is going to change. So it's good. Let's not, let's keep the temperature separate. So what I'm going to introduce is that let's say that k multiplied by the logarithm of v, let's call that s. I can call it absolutely anything. You can call it b if you want to, but you can have a really difficult time justifying your definitions every time you meet another physicist. So I would strongly recommend to call it s. If we do that, again, remember that we're what I'm after. I'm after understanding the relative population of two states that I can have different volumes. Then this simplifies to e raised to minus eA minus t sA divided by kT. And another e raised to minus eB minus tSb divided by kT. I know that you don't think this looks simple yet. So I'll take an even simpler step. Let's simplify that entire expression. The reason why I didn't put a single letter for the entire expression, we still had that eA on the previous slide, right? And that eA or eB, that describes the difference in energy. This second component somehow describes the difference in the property of the population of the states that is not related to energy, but these volumes that are introduced. So it makes sense to be able to separate them, but under some conditions you might prefer to stick them together and we can do that. So I'll just call that whatever letter comes off to e in the alphabets. Let's say f. So I say that this is a quotient e raised to minus fA divided by kT divided by e raised to minus fD divided by kT. And this looks exactly like the Boltzmann distribution, but it's not really the Boltzmann distribution. It's kind of different. Instead of e, I have f. We're going to call this free energy. The reason why we call this free energy you don't know yet, but later you can prove actually, and there might be, I think I have a study question with a link. You can prove that this, well obviously it has units of energy and it corresponds to the energy that is available to do work. And that's why we call this free energy. When you're at the local minimum and you're no longer moving, you are at the local minimum in free energy. But what this derivation means, I can just use the simplification of it. I've said that f equals e minus Ts. So for any state, this concept f is the energy of the state minus the temperature multiplied by this letter s of the state. And the s of the state is then proportional to, that's the Boltzmann constant, the logarithm of the volume, or pretty much the number of holes or the number of microstates. Because this is a problem. We like to think of a state, say, if the water is in the ice or the liquid state. But here I was talking about microscopic state. The extreme here would be quantum mechanics, right? So we kind of having very small states inside a system and then small states, sorry, large states as if in a room. To keep those apart, I'm going to try to stick to the nomenclature microstates and macro states. But you will frequently see me skipping that because it gets too cumbersome to make that distinction all the time. This is seemingly the simplest equation in the entire class and I'm going to promise you that it will frequently be the most complicated one for you. You have probably already identified, at least many of you, that the s here is entropy. But that's where you go wrong. We don't know what entropy is yet. This comes out purely from statistical mechanics. Entropy is not something that is complicated to understand. That's where you go wrong if you introduce it that way. It's a pure letter that we assign to something that is proportional to the logarithm of the number of microstates in a system. Now we might then try to interpret that what that means, but start from the definition, don't start from the meaning. And I would argue that if you just do that it's going to turn out that entropy is actually something very easy. There are other things that you think that are easy on the logarithm that's going to turn out to be fairly hard. Let's look at them.