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Partition function vs config integral

by aaaa202
Tags: config, function, integral, partition
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May22-13, 03:38 PM
P: 1,005
In classical statistical physics we have the partition function:


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.
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May22-13, 05:39 PM
P: 185
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.
May23-13, 06:09 AM
P: 1,005
But ln(Z1) is not equal to ln(Z2) when Z1=Ʃexp(-βEi) and Z2=∫d3rd3p exp(-βE(r,t)) ?

May23-13, 09:15 AM
P: 185
Partition function vs config integral

Yeah, it's an approximation. So it won't be exactly equal.
May26-13, 07:39 AM
P: 1,005
still dont understand. Lets say you want the derivative of Ln(Z). Then you are saying that:
∂(Ʃexp(..))∂β ≈ ∂ln(∫exp(...))/∂β
May26-13, 02:54 PM
P: 185
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.
May26-13, 03:02 PM
P: 185
Maybe this will be a little more clear. The Riemann definition of the integral is something like this:
[tex]\int_0^\infty f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_{n = 0}^{\infty} f(n \Delta x) \Delta x[/tex]

[tex]\frac{1}{\Delta x}\int_0^\infty f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_{n = 0}^{\infty} f(n \Delta x) [/tex]

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

This is often done the other way; the computation of an integral numerically on a computer is approximation as a summation.
May26-13, 07:04 PM
P: 1,005
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?
May26-13, 09:25 PM
P: 185
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.

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