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I have a little question I can't seem to find the answer to. It is as follows:

When does Fermi-Dirac statistics reduce to Maxwell-Boltzmann statistics?

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up vote 1 down vote accepted

The Fermi-Dirac distribution

$$ f = \frac{1}{1+e^{(E-\mu)/kT}} $$

where, $E$ is the energy, $\mu$ is the chemical potential and $kT$ is Boltzmann's constant times temperature reduces to the Boltzmann distribution

if $E \gg \mu$ then $e^{(E-\mu)/kT} \gg 1$ and so you can write $$ f \approx e^{-(E-\mu)/kT}. $$

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The statement $E\gg\mu$ does not make much sense. Physically it is the value of $T$ that matters. – Peter Kravchuk May 23 '13 at 14:38
From Atkins, Physical Chemistry, p. 800: "For energies well above $\mu$, the 1 in the denominator can be neglected...". – TMOTTM May 23 '13 at 15:00
Well, by $E\gg \mu$, TMOTTM clearly meant $E-\mu\gg kT$. – Luboš Motl May 23 '13 at 15:36

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