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?


1 Answer 1


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 $$ f \approx e^{-(E-\mu)/kT}. $$

  • 2
    $\begingroup$ The statement $E\gg\mu$ does not make much sense. Physically it is the value of $T$ that matters. $\endgroup$ Commented May 23, 2013 at 14:38
  • $\begingroup$ From Atkins, Physical Chemistry, p. 800: "For energies well above $\mu$, the 1 in the denominator can be neglected...". $\endgroup$
    – TMOTTM
    Commented May 23, 2013 at 15:00
  • 6
    $\begingroup$ Well, by $E\gg \mu$, TMOTTM clearly meant $E-\mu\gg kT$. $\endgroup$ Commented May 23, 2013 at 15:36

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