 So we've seen how lattice models can be useful in a number of different circumstances, but we don't have to use lattice models just to describe where molecules are and how they move from place to place. We can also use lattice models for other circumstances like how energy transfers from one location to, from one molecule to another in a system. So this is useful for examples like, let's say I have a block of material, maybe a block of metal doesn't really matter what it is, at some elevated temperature that's in thermal contact let's say with another material that's at a colder temperature. So we know what's going to happen in this system is that energy is going to be transferred in the form of heat from the hotter system to the colder system, the one with high temperature to the one with lower temperature. So lattice models can help us understand why that happens and I'll point out that for the moment I'm going to be fairly loose with the terms energy and heat and temperature. We don't have definitions for those yet, we'll treat those a little more precisely later on in the course, but right now I'll just use our intuitive understandings of things being hot, things with high temperature and, and heat in the form of energy flowing from hot to cold. Alright, so the lattice model that we can use to understand this type of behavior, let's make a lattice not for the positions of our molecules but for the energy that those molecules have. Let's say that in my system, in my very oversimplified system, molecules can have an energy of zero or they can have an energy of one, they can have a little energy or they can have more energy. And let's say I've got on the hot side to exaggerate things, I've got a few molecules that have a lot of energy and on the colder side, I've got a few molecules that have less energy. So I'm using n equals six molecules on each side of this system and again I've just simplified the hot and cold to having more energy or having less energy. One side of the system has more energy than the other, my diagram is meant to indicate here are all the individual molecules that have certain amounts of energy, six molecules have energy one on the left side of the system, six molecules have energy zero on the right side of the system. That's the micro state of the system, I've shown you each individual molecule and how much energy it has on this energy diagram. The macro state of the system would be on the left, the description is all the molecules, all six of the molecules have one unit of energy on the right, all six of the molecules have zero units of energy. So we can calculate the multiplicity, calculate the number of ways of writing down a micro state that matches the macro state I just described, six on the left with one energy, none with zero energy on the right, none with one energy, six with no energy. So the ways I have of choosing which six of my molecules have one unit of energy on the right, that's just six choose six, I'm sorry on the left. On the right side, how many ways are there of choosing which zero of my molecules are in the excited state and therefore which of them are in the ground state? That's six choose zero. So as with the lattice models for gas molecules or liquid molecules, we end up with these binomial or multinomial expressions to describe the multiplicity, in this case for this very distinct macro state where all the molecules are excited on one side, all the molecules are low energy on the other side, six choose six is just one, six choose zero is just one. There's only one micro state, the one I've drawn that matches that description. That is of course not the most probable state of the system. We can write other micro states that are more likely that have a higher multiplicity. Let me point out that there's one additional constraint we have to start paying attention to when we're talking about energy transfer from one place to another is that energy only transfers. I can't create energy, I can't give molecules more energy in the system without that energy coming from somewhere because energy is conserved. So if I want to draw a different state of the system where I want to maybe elevate one of the molecules on the right hand side from the ground state up to the excited state, that energy has to come from somewhere. One place it can come from, the only place in this system it can come from is by dropping one molecule down to the ground state on the other side of the system. So here's a micro state where I've got five excited state molecules on the left side, only one on the right side. The total energy of the system is conserved. Notice I had a total of six units of energy in this system. I've got five on the left, one on the right in this system. This system energy is still six, the energy is conserved. I haven't created or destroyed any energy. I've just moved it from this molecule to this molecule. The multiplicity in this case, my macro state would be among the six molecules on the left. I need to choose which five of them are in the excited state. And on the right side, I choose which of my six molecules, which one of them are in the excited state. So my multiplicity now, six choose five is six, six choose one is also six. I've got a multiplicity of 36. So not surprisingly based on the examples we've seen so far, this macro state is more probable than this macro state. But of course, this is not the most probable macro state. I can draw another one, I can draw several more. I can promote two molecules or three or four, any number of molecules I want, as long as I pay for them with the appropriate demotions over on the other side, it will not surprise you when I tell you that the most probable macro state, the one with the highest multiplicity, is going to be the one where half of my molecules are in the excited state on the left, half of them are in the excited state on the right. The multiplicity in this case, of my six molecules on the left, I have to choose which three are in the upper state. So six choose three on the left. Likewise, six choose three on the right. Let's work those vectorials out. So six choose three is six times five times four times three, two, one. Over, let's go ahead. So that's six factorial over three factorial. These, this three factorial cancels this three factorial. That factor of six cancels this factor of six. And what I've got is 20 for six choose three. So 20 times 20. So do this times another six choose three. And I end up with 20 times 20 is 400. So the multiplicity of this macro state is 400. That's greater than this multiplicity. If I wrote out all the possibilities, that would be the macro state with the highest multiplicity. So like I said, it's not surprising to learn that the lattice model helps us confirm our intuition that the most likely state, the most random state, the one with the highest multiplicity, the highest probability will be the one where the energy is equally distributed across both halves of the system. I've conserved energy. There's still a total of six units of energy in this system. But now half that energy is on the left, half that energy is on the right. However, the lattice model can help us understand a few things that are not quite as intuitive. So let's take a case that's one step more complicated. Let's take a system with three energy levels rather than two. So let's take a very similar example. When I start out, I've got six molecules all the way up here in the most excited state on the left. And on the right, none of the six molecules have any energy. So I, again, have n equals six molecules on each side. Now I have a total of 12 units of energy that I need to conserve. Six molecules times two is 12 units of energy. The multiplicities, very similar in this case. But now since I have three different states, the macro state, the way I describe the macro state of the system is on the left, I've got six molecules. I don't care which ones. But six in the energy state two, none here, none here. So my multiplicity is going to be six choose 600. It's going to be a multinomial coefficient rather than a binomial coefficient. That's multiplying the multinomial coefficient on the right side, which would be of my six molecules. I need zero in the upper state, zero in the middle state, and all six of them in the ground state. Those multinomial coefficients, six factorial over six factorial, zero factorial, zero factorial, that's just one. Again, the most separated, the most ordered state, again, has a multiplicity of one. If we conserve and, well, we'll skip the step by step demotion of molecules one at a time, you might say, all I need to do is get half of the energy from the left over to the right. If I take these six molecules and drop them each into the energy equals one state, losing six units of energy in order to pay for elevation of these six molecules up to that same state, that's the case where I've split the energy equally between the left and the right halves of the system. The macro state would be all six molecules in the equals one state on the left, all six molecules in the equals one state on the right. My multiplicity is going to be of my six molecules, zero in the upper state, six in the middle state, zero in the lower state on the left, same thing on the right. And again, those multiplicities, those multinomial coefficients, six factorial over six factorial gives me one. So that's perhaps surprising. I've equalized the energy on the left and the right, and somehow it's no more likely than this state. If you think about it in the right way, it's not surprising at all. This is just as ordered a state, just as few ways of drawing this micro state that matches my macro state, all of them in the middle, as there are of writing this macro state. A much more probable state, a state with much higher multiplicity, would be for my six molecules, let's put two of them up here, two of them here, two of them here. So I've spread the molecules, spread the energy out not only among the left and right halves of the system, but I've spread the molecules out between different energy levels as well. So same thing on the left, same thing on the right. I've made the system much more disordered by spreading the molecules out between more states. Notice that the energy is still conserved. Every molecule in this state has an energy of 2. So I've got 1, 2, 3, 4 molecules with energy of 2. So that's 8 energy units for the molecules in the upper state, only 4 energy units for the molecules in the middle state. So the total energy, 8 plus 4 is 12. So essentially, I can get from this state to that state by promoting two molecules and dropping two molecules. Same thing, promote two, drop two. The multiplicity of that third state that I've drawn, macro state would be two in the upper, two in the middle, two in the lower, on the left, and the same on the right. So my multiplicity is going to be 6 choose 222, multiplied by another 6 choose 222. Those factorials, if I take 6 factorial over 2 factorial, 2 factorial, 2 factorial, I won't work those out. But after the cancellation that happens there, each one of these factorials, 6 times 5 times 4 times 3 times 2 divided by 8, that works out to be 90. 90 squared works out to be 8100. So significantly more likely to find the system in this state than in either of these two states. And again, what we've discovered is that it's not necessary just to spread the energy out between the left and right halves of the system, but also spread the energy out between different energy levels as well. As one final example, that's going to teach us something else surprising about the way energy spreads throughout the system. Let's do one more example, a lot like this one. Again, I'll take an energy level diagram with three different energy levels, 0, 1, and 2. Now, though, I'm going to make energy a little less available, a little more scarce. So I'll continue with six molecules on each half of the system. But now I'm only going to have six units of energy to go around instead of 12. So initially, let's put the six molecules on the left in the equals 1 state. On the right, let's put them in the equals 0 state. So maybe ask yourself what you think is going to happen in this system. This system has a multiplicity of 6 choose 0, 6, 0. And on the right side, 6 choose 0, 0, 6, which is going to work out to be 1, as usual, in this ordered state. You can ask yourself, what you think the most likely distribution of these energies is going to be in the state that turns out to be most probable. Go ahead and pause the video and write down the state you think is going to be most likely. I'll wait a couple seconds. So if you chose this microstate, clearly, that's got a higher multiplicity, this microstate is no good for this system. That one doesn't conserve energy. If I were to write down two molecules here, two molecules here, two molecules here, same thing on the left as the right, that's going to be a state with a very high multiplicity, but it violates the conservation of energy. That state uses 12 units of energy, and I can't get there because I only have six units of energy. I'd have to promote more molecules than I can afford to pay for. So that's no good. I can't use that state. If we go back to our approach of just one at a time, lifting some molecules up and paying for those with some other molecules that fall down. So my macrostate is 3 in the equals one state, 3 in the equals zero state on both sides of the system. That's spread the energy out considerably more than where I started. It's got the same amount of energy on the left and the right. That's beginning to sound pretty good. The multiplicity, so what have I got? I've got 6 choose 3. So if I stick to the same order I've been using, it's 6 choose 0, 3, 3 times 6 choose 0, 3, 3. So that's the same calculation I did right here. 6 choose 3. That works out to be 20 times 20 is 400. You might have chosen that as the most likely distribution of energies among these different states. Certainly more probable than the one we started with. It's less. It's got a lower multiplicity than this one, but that one's illegal. We're not allowed to use that one. But it turns out there's one that's even better than this distribution. Suppose I consider the following one. Suppose instead of 3 in the ground state, 3 in the middle state, and none up top, let's take 4 in the ground state, 1 in the middle state, 1 in the upper state. So the way I've generated this state, for example, I can drop one molecule from the middle to the bottom and simultaneously promote one up to the top. Or I can just add up the energies and say 0 plus 1 plus 2. That's a total of three units of energy on the left. Here's another three units of energy on the right. So there's certainly a legal state, a valid state. When I calculate the multiplicity of that state, among the six particles on the left, I've got 1 in the upper, 1 in the middle, 4 in the ground state. Multiply that by another 6, choose 1, 1, 4. Those factorials, 6 factorial over 4 factorial is 30. 6 times 5 is 30 divided by another 2 factorial gives me. Did I do that right? 6 times 5 is 30 divided by 2 is 15. Yes. So no, no factorial. 6 factorial divided by 4 factorial is just 6 times 5 is 30. 30 times 30 is 900. That's the right math. Multiplicity in that case turns out to be larger than the multiplicity in this case. That may be a little surprising. This is, in fact, the macro state, 4 in the ground state, 1 in the middle, 1 in the upper state. That's the macro state with the highest multiplicity. If I try all the different possible combinations, this is the one that turns out to be highest multiplicity and therefore most likely. The surprising thing is I don't have a perfectly even distribution of energy throughout the system. I've got more in the ground state than in the upper states. And that turns out to be because energy is relatively scarce in the system. So it's going to be useful to understand in a more general way, how can we predict what is the most likely distribution of the energies between the systems? At the moment, let's just be happy that we've got a way of using lattice models to describe something like energy transfer as well as motion of molecules around the system, like an expansion of a gas or mixing of fluids. And notice along the way that the results we get are perhaps a little surprising, counterintuitive, not quite what we would have predicted if we used a relatively naive expectation for whether the energy is going to flow. So we'll look in more detail at how to predict the most likely distribution of the energies in a system like this. But first, we're going to have to take a little side trip in order to calculate these multiplicities for systems with numbers of molecules bigger than just a handful of molecules like the six we've considered here. We're going to start having to do factorials of numbers larger than it's convenient to do calculations with. So that's our next topic is to think about how to do factorials of very large numbers.