Alright, now that we've had a chance to explore the utility of partition functions for predicting gas properties, let's take a deeper dive into the partition function itself. And in particular, the ensemble partition function. First let me note, sort of philosophically, that the ensemble partition function capital q. Is to statistical mechanics what the wave function, usually abbreviated as capital Psy is to quantum mechanics. That is, it plays a central role and it allows one to determine all of the properties that you might be interested in, for a macroscopic system. Just as psi, the wave function contains all the information necessary to determine quantum mechanical properties for microscopic systems. So the Q, that we've actually been working with so far, and I've just called it a partition function, which is fine But we can be a bit more specific. It should in fact be called the conical partition function. And the ensemble that we've been working with, you remember the very large water cooler, is the canonical ensemble. And what that means, that adjective canonical, is it's telling you which variables were we holding fixed in our ensemble. So we were holding fixed the number of particles. The volume and the temperature. We will look at some additional ensembles later on, but the canonical ensemble is a very important and useful one. So, Q is needed to compute these macroscopic properties. We've played around with it a bit when we've been handed certain ensemble partition functions. But for an arbitrary system, what you need in order to have all the relevant energy levels is all of of the eigenvalues, that is all of the solutions to its N-body Schroedinger equation. Where N is the number of particles which is presumably a very, very large number. So that is rarely practical to compute, where rarely practical means essentially never. It's, it's just about impossible. Happily Q, capital Q, can be approximated based on results for individual molecular energy levels as opposed to needing to know about all the energies for an interacting system. Under favorable circumstances. So let's look more carefully at this issue of the ensemble partition function in terms of a molecular partition function, little q. So given a system of distinguishable, that is I can apply labels to them and tell them apart. Non-interacting. And so that's like an ideal gas. Doesn't int, interact; the individual molecules have no interaction with each other. Identical. So every molecule is the same molecule. Given that system of distinguishable, non-interacting, identical particles the ensemble partition function can be written as a product of the individual molecular partition functions. Let's look at why mathematically. So the ensemble function is sum over all accessible energy states e to the minus the energy of that state. However, what are the accessible energy states? Well, the molecules don't interact with one another. So the total energy is just the sum of all their individual energies. So I've got that here, I've got the energy for molecule one. And I'm able to label them, molecule one, molecule two, molecule three, out to molecule N, divided by k t. Okay, well this is the exponential of a sum, so that's equal to the product of exponentials, and if I consider all the possible energies that every one of these molecules can take on That'll give me the product of a sum now over, my subscript here actually contains the label for the molecule. So this is for molecule 1. I'll run over index j e to the minus all the possible energies for that molecule. And the same for two, and the same for three, and so on, out to N. So products out of this sum will make every one of these things appearing in this sum. And they're all the same molecules, so this term in brackets is exactly equal to this term in brackets, and so on. They're all the same, they're just differ by the label on the molecule. So it is the partition function, because that's what this is. Right? It's a sum over accessible energy states, e to the minus energy, divided by kT. That is a partition function. Just happens to be the partition function for the molecule. So it is, molecular partition function, which depends on V and T raised to the nth power. So here's where the n dependance comes in. The molecule doesn't depend on n. There's one molecule. The ensemble partition function depends on N because we're to the Nth power. Right, a key feature here is, the non interacting aspect means that this energy is just the sum of individual energies. And the identical character of the molecules let's us go from this product to this exponential. Finally, again, emphasizing little q only requires you to know about the allowed energies of an individual atom or molecule, not about interactions between them. Alright, well, let's pause for a moment and I'll give you a chance to work on a problem, and then we'll return. Let's talk about indistinguishability. So this result for the ensemble partition function, that it's simply the molecular partition function raised to the Nth power. It's very pretty and nice and reasonably easy to work with. But it's only sometimes correct and the problem is, that atoms and molecules are typically indistinguishable. So, gas molecules, liquid molecules, they're in a homogeneous mixture, they're moving around, there's no way to label them, Unless you want to change the number of neutrons in their nuclei, but then you'd need a different kind of partition function we're talkin about everything being the same, identical means indistinguishable. And so just to provide a concrete example to allow us to talk about things, imagine that you have two particles. Each has an energy, and there are only three energy levels available to each of these two particles, call them e1, e2. I've actually got epsilon here. I'll use a Greek letter for an individual atom or molecule, epsilon 1, epsilon 2, and epsilon 3. So how many different ways can they be arranged if they're distinguishable? Well if there distinguishable molecule one can take on any of three energy levels, and molecule two can take on any of three energy levels, so there's 3 times 3, there's 9 possibilities, right? And so the partition function. And this is the partition function expanded in sort of a graphical way, so the ensemble partition function is the sum over all possible energy states, but the possible energy states are just the sums of the two individual energy states. So, I can write that as a exponential of a sum, as a product of exponentials I'll just separate that into a product of 2 summations. Each of those is an individual partition function. This is just what we did before, except you know, step by step in, sort of gory detail. Q squared, Q to the N. Alright, that's, that's this upper result here. And so, what are the possible states? 1, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3, 3, 1, 3, 2, 3, 3. Not really any rocket science, yet. However, if we think about the energies of some related states, one and two have the same energy as two and one. All I've done is reordered the labels on the particles. And the same for one and three and three and one. And the same for two and three and three and two. So if the particles are indistinguishable, I'm not allowed to put those labels on, then I need to throw away the duplicates. So that takes my nine terms down to six therms. And so I'll keep the one where the first label's smaller than the second label. So I got 1, 1, 1, 2, 1, 3, 2, 2, 2, 3, 3, 3. Six things left. So let's bring those 6 things to the top of the next slide. Now what if no 2 particles can be in the same state. So that's known as fermion statistics or fermionic behavior. So fermions are particles that our universe is made up of. That follow a certain rule, and that rule is no two fermions can be in exactly the same state, and sometimes people would say characterized by the same quantum numbers. So, if that's true, and what are typical fermions, electrons, for example are fermions, but in any case, when that's true, more states drop out. I can't have 1-1. I can't have 2-2, and I can't have 3-3. So my original nine reduced to six are now down to three. So here are the three states I'm allowed to have left. One two, one three, two three. Unique for indistinguishable fermions. And, incidentley, even for bosons And so the boson is the other kind of particle in the universe besides the fermion, where it's okay for two bosons to be in exactly the same state. However, even if they are allowed to, it'll be very unlikely to find two particles in the same state if The number of available states vastly exceeds the number of particles. So we're going to see that that is typically true. But for now just sort of accept, right if I've got a zillion boxes and 5 particles to throw into the, the odds that any 2 land in the same box are going to be very, very small. So in that case, pretty much all of the over counting of the unallowed terms in this simpler expression for the partition function comes from failure to consider the permutational symmetry in the labeling of the particles. Right? And let me make that more clear with an equation. So, here I have the ensemble partition function exponential of a sum of a whole bunch of energies for individual particles. Now, if all the energy levels are different for the individual particles. That is there is every one of these epsilons is a unique number. Well, out of the n possible numbers, because there's n particles, this first one can take on n values, can be any one of those. Second one of course, you've now used up one of your n values, so it can take on n minus 1 values. And finally of course by the time you are down to the last particle. It only takes the energy level that's left. There's only one way to do that. So there are n factorial different ways then, to make the same contribution to the ensemble partition funcion. And if we choose to take q as the product of the molecular partition functions, that is capital q the ensemble as that product. Each being allowed to take on the various values will over count that term. Will make it again, and again, and again with different permutations. How many permutations? N factorial of them. So we should divide by n factorial in order to remove that that non-allowed distinguishability. And come up with an ensemble partition function capital Q. And we actually employed this. I told you I'd derive it, and now we have. in, and it is appropriate then, for Fermions and bosons, when they are indistinguishable particles. So let me return to this question of, is it possible for two particles to be in the same state, and the same energy level. And the rule that one can use is, if this expression here, number of particles divided by volume, times Planck's constant squared, divided by eight times the mass times Boltzmann constant, times the temperature, all raised to the three halves power. If that expression is considerably less than one then Boltzmann statistics are valid the division by n factorial to compute the ensemble partition function is valid the particles in the system will obey Boltzmann statistics. Let's just explore how often that might happen that this inequality is satisfied. And let's just take a simple system. Molecular hydrogen as a gas, at one bar pressure. And 300 Kelvin. So, under those conditions. We can replace N over V using the ideal gas equation of state as P over kT. And that allows us, then, to derive. The pressure in SI units would be 10 to the 5th Pascal. The Boltzmann's constant is expressed here, and the temperature. And that all adds up to 2.414 times 10 to the 25th per cubit meter. Meanwhile, if we evaluate the quantity in parentheses, we have h squared over 8 m k t. Here's Planck's constant squared, 8, the mass of molecular hydrogen is 3.35 times 10 to the minus 27th kilograms. Here is Boltzmann's constant, here is 300 kelvin again. And the net when raised to the three, has power 2.486 times 10 to the minus 31st cubic meters. Multiplying these two quantities together, we end up with the full expression is equal to 6 times 10 to the minus 6th. Right, so six one-millionths, which certainly is a great deal less than one. Then hence, we would expect Boltzmann's statistics to be satisfied for hydrogen gas at this pressure. And, indeed, we see, in order to get up to a value closer to one, given that pressure appears here linearly. And this is about hm, it's about a, let's round it and say it's one one hundredth thousands. That would be ten to the minus fifth. You would need to go to a hundred thousand times higher pressure. That'd be one hundred thousand bar. So that's a whole lot of pressure, and obviously everyday systems are not under that kind of pressure. Alright we've worked with the ensemble partition function. That's great and now have a better feel for that en-factorial in the denominator and its dependence on the molecular partition function. So next we're going to look at the molecular partition function and talk about its construction.