# Partition function vs config integral

1. May 22, 2013

### aaaa202

In classical statistical physics we have the partition function:

Z=Ʃexp(-βEi)

But my book says you can approximate this with an integral over phase space:

Z=1/(ΔxΔp)3 ∫d3rd3p exp(-βE(r,t))

I agree that x and p are continuous variables. But who says that we are allowed to make this kind of discretization and what values are we choose for 1/(ΔxΔp)3 except for them being small?

I think my book is definately hiding something from me, and using a rather lame argumentation in doing so.

2. May 22, 2013

### daveyrocket

For the most part, it doesn't matter. Everything that matters with the partition function comes from derivatives of log Z, which means that constant factors like 1/(ΔxΔp) will drop out. I've seen some books argue that (ΔxΔp) could be taken as approximately h or hbar, since that is sort of the "scale" of quantum effects.

3. May 23, 2013

### aaaa202

But ln(Z1) is not equal to ln(Z2) when Z1=Ʃexp(-βEi) and Z2=∫d3rd3p exp(-βE(r,t)) ?

4. May 23, 2013

### daveyrocket

Yeah, it's an approximation. So it won't be exactly equal.

5. May 26, 2013

### aaaa202

still dont understand. Lets say you want the derivative of Ln(Z). Then you are saying that:
∂(Ʃexp(..))∂β ≈ ∂ln(∫exp(...))/∂β

6. May 26, 2013

### daveyrocket

Yes, because choices of Δx and Δp drop out of such derivatives.

Large sums can often be approximated by integrals (are you familiar with the Riemann integral). For a typical partition function this approximation is not going to be very good at low temperatures, which you can see it in the math just from the fact at low temperature the terms for higher energy states have a very tiny contribution to the summation. But at high enough temperatures, where many terms contribute to the summation an integral can be a good enough approximation.

Last edited: May 26, 2013
7. May 26, 2013

### daveyrocket

Maybe this will be a little more clear. The Riemann definition of the integral is something like this:
$$\int_0^\infty f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_{n = 0}^{\infty} f(n \Delta x) \Delta x$$

Rewriting
$$\frac{1}{\Delta x}\int_0^\infty f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_{n = 0}^{\infty} f(n \Delta x)$$

You can make this into an appoximation by dropping the limit:
$$\frac{1}{\Delta x}\int_0^\infty f(x) dx \approx \sum_{n = 0}^{\infty} f(n \Delta x)$$

This is often done the other way; the computation of an integral numerically on a computer is approximation as a summation.

8. May 26, 2013

### aaaa202

okay makes sense. But since we are computing a partition function which sums over all states. How do we know what value to assign to Δx, or is this insignifanct as you say, because we are taking the log of Z?

9. May 26, 2013

### daveyrocket

Well, one argument is that ΔxΔp should be about h, since that is a quantity that gives a scale relevant to quantum effects. But yeah, it doesn't actually matter because we take the logarithms.