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In David Chandler's 'intro to statistical mechanics' he states that for an ideal gas at high-temperature

$$ \langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}} $$

Which I can believe from intuition, but he losses me on the derivation.

Starting with the general form for the occupancy of a boson or Fermion gas:

$$ \langle n_j \rangle=[e^{\beta (\epsilon_j-\mu)} \pm 1]^{-1} $$

Then at high-temperature, $\beta \rightarrow 0$, [ page 101 (b) ]

$$ e^{\beta(\epsilon_j-\mu)}>>1 $$

So the assumption can be made that

$$ \langle n_j \rangle=e^{-\beta (\epsilon_j-\mu)} $$

Which would make sense if $e^{\beta(\epsilon_j-\mu)}>>1$, but it sure seems like that $ e^{\beta(\epsilon_j-\mu)}\approx 1$ in that case. Is there another assumption that's made here?
He does say:

Note that if this eqn is true for all $\epsilon_j$, then $-\beta \mu >>1 $

so, $\mu \rightarrow -\infty$ at high-temperature? That doesn't strike me as something that's obvious.

The rest of the derivation:

$$ \langle N \rangle = \sum \langle n_j \rangle =\sum e^{-\beta(\epsilon_j-\mu)}=e^{\beta \mu} \sum e^{-\beta \epsilon_j} \\ e^{\beta \mu} = \frac{\langle N \rangle}{\sum e^{-\beta \epsilon_j}} $$

using $\langle n_j \rangle = e^{-\beta(\epsilon_j - \mu)}$

$$ \langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}} $$

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There are indeed other assumptions in the derivation you quote. Namely, Chandler considers the classical limit of an ideal quantum mechanical gas with average particle number

$ <N> = \sum_j <n_j> = \sum [ e^{\beta (e_j - \mu)} ± 1 ]^{-1} $

(plus for Fermi-Dirac, minus for Bose-Einstein statistics). In the classical limit (low density) there are many more single particle states than particles. Thus $ <n_j>$ is much smaller than 1, which implies

$ e^{\beta ( \epsilon_j - \mu)} >> 1 \ \ ($ i.e. the "$± 1$" in the eqn above becomes irrelevant)

In other words, the distinction between FD and BE statistics vanishes in the classical limit.

The second part of your question concerns the chemical potential $\mu$, which is shown to be just the standard Boltzmann factor in the classical limit. You don't need to consider any limits on $\mu$, just the fact that $<n_i> \ \rightarrow \ \exp(-\beta(\epsilon_i - \mu))$ in the classical limit.

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  • $\begingroup$ For sure, I'm okay with the $\pm$ being irrelevant if $e^{\beta(\epsilon_j - \mu)} >> 1$. but how is it much greater than 1 if $\beta \rightarrow 0$ $\endgroup$ – Ben Aug 21 '14 at 23:24
  • $\begingroup$ You're right. The chemical potential is just the standard boltzman factor, cause the gas would still behave classically at high temperatures. I think it does help to consider the limits on $\mu$ though. $\endgroup$ – Ben Aug 21 '14 at 23:49
  • $\begingroup$ @Ben H. at high temperature and low particle density, $<n_i>$ is much smaller than one. Its reciprocal is therefore much larger than one and approximately equal to $\exp(\beta(\epsilon_i-\mu))$ $\endgroup$ – Felix Aug 22 '14 at 19:06
  • $\begingroup$ I agree, intuitively that makes sense. I just wasn't seeing how it came out of the math $\endgroup$ – Ben Aug 23 '14 at 0:45
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Actually, it does make sense that $\mu \rightarrow - \infty$

Given the ideal chemical potential for in ideal gas:

$$ \mu = -k_B T\ln \left( \frac{V}{N} \left(\frac{mk_B T}{2 \pi \hbar}\right)^{3/2} \right ) $$

so

$$ \mu \beta \sim - \ln(T) \\ \:\\ \therefore \lim_{T \rightarrow \infty} -\mu \beta >> 1 $$

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