So we could also write, then, completely equivalently, that the partition function

depends not on n, v, and beta. But on n, v, and t, because Boltzmann's

constant is a constant, so beta varies with temperature so if we hold beta

fixed, we're holding the temperature fixed, that we sum over states e to the

minus e sub j, now not multiplied times beta, but divided by kT.

And I'll always indicate in formulae that it's Boltzmann's constant with a

subscript capital b, but I'll nearly always just say k.

I won't say kb, it takes too long. So let's think about this this partition

function. We see the mathematical form, but again

it's nice to have some intuitive feel. What, what does it mean?

What does the value of the partition function tell you.

Well, just for simplicity in order to think about the meaning of the partition

function, let's take the ground state of the system, that is the lowest possible

energy. Let's take that to be non-degenerate, so

there's not multiple ways to make that energy, there's just one state with

energy zero. And we will define that energy to be

zero. It's arbitrary where you set it we'll

just pick that because it's convenient. Well in that case when I sum over the

states the very first state will be the ground state, and I'll get e to the minus

0. E to the 0 is 1.

So I'm going to pull that out 1. And then I'll keep all the remaining

states, their excited states, because I said my ground state was non degenerate.

And so same expression on the inside except that I know that e will be above 0

because it's an excited state. So notice a few features then of the

partition function. As I let the temperature go to 0 I will

be dividing by 0 in this expression. So it'll become infinitely large.

I get e to the minus infinity, that's 0. And so as the temperature goes to 0 the

only term that survives in the partition function is 1.

Temperature goes to 0, q goes to 1. Now what about if T goes to infinity?

Well in that case, I will be dividing something by infinity, so it will be 0.

I'll get e to the minus 0, that's 1. So for every single state I'll add a 1.

So 1, 1, 1, 1, 1, 1 I just count all the states.

So as the temperature becomes infinite, q goes to the total number of accessible

states. So the partition function can be thought

of as an effective measure of the accessible number of energy states, from

only one at absolute 0 for a non-degenerate ground state to everything

that can be accessed at. Infinitely high temperature.

So, with that in mind I'm going to let you work on a conceptual question having

to do with the partition function, and then we'll return to consider this

further. Alright, let's take one further look at

the behavior of the partition function and what I want to look at is its

behavior with respect to the density of states.

I have that in quotes it's something you'll hear physical chemists talk about

with some frequency. What, what does the density of states

mean? It means, how closely spaced are the

relative energy levels. So you might remember in Week 1 we talked

about the relative spacing of energy levels, and we mentioned that electronic

energy levels are usually spaced quite far from one another.

Closer are vibrational levels, still closer are rotational levels and finally

very, very closely spaced indeed are translational energy levels.

But what happens as these levels get closer and closer and closer to one

another? Well, let's do the easy case first.

Let's make them get very far from one another, that is the density goes to zero

because they're so far separated. Well if the density of states is going to

0 that means this first excited state is very high in energy so I get e to the

minus a very large number, and I'll just mark it some fixed temperature now.

We looked at the dependence on temperature before.

Now we're looking at the dependence on energy.

So e to the minus a very large number is 0.

I get nothing. I'm back to q equal 1.

I've got one accessible state. On the other hand, as the density of

states becomes infinite, so that means the first excited state is at 10 to the

minus 94th joules, for instance, some incredibly small number, and they're all

really closely spaced. Well in that case, because this is such a

small number, I'll really be taking roughly e to the minus 0.

10 to the minus 90 is pretty close to 0. And so again, this sum will do a whole

lot of counting of states before the energy gets large enough that it actually

starts to exponentially die off. And so, as the density goes to infinite,

Q will go the total number of states, which itself will be infinite in that

case. It will be a continuum of states.