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When does Bose-Einstein distribution function reduce to Maxwell-Boltzmann distribution function in statistical mechanics?

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    $\begingroup$ it can be asked as "under what conditions the Bose-Einstein or Fermi-Dirac distribution reduces to a Classical distribution like Maxwell -Boltzmann distribution" $\endgroup$
    – drvrm
    Commented Dec 13, 2018 at 21:26

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Recall the formula for the expected energy level occupation number in Fermi-Dirac & Bose-Einstein statistics:

$$\langle n \rangle = \frac{1}{e^{(\epsilon - \mu)/kT} \pm 1}$$

where $\epsilon$ represents the energy quanta per particle, $\mu$ is the chemical potential, $k$ is the Boltzmann constant and $T$ is the temperature. The $+$ in the denominator corresponds to fermions (Fermi-Dirac) while the $-$ corresponds to bosons (Bose-Einstein).

If the exponent term $(\epsilon - \mu)/kT$ is much greater than $1$, we can neglect the $1$ in the denominator and approximate:

$$\langle n \rangle \approx \frac{1}{e^{(\epsilon - \mu)/kT}} = e^{-(\epsilon - \mu)/kT} = \langle n_{MB} \rangle$$

which is precisely the expected occupation number for distinguishable (Maxwell-Boltzmann) particles.

Note that, if $\epsilon \approx 0$, then we find the approximation is accurate for states where $\mu \ll -kT$, which is equivalent to stating that the number of available single-particle states is much greater than the number of particles in the system—meaning that particles rarely attempt to occupy the same state, and the properties of state-sharing that distinguish bosons and fermions become irrelevant.

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    $\begingroup$ Equivalent formulation: "the thermal wavelength is much less than the interparticle spacing" David Tong: Lectures on Statistical Physics, p. 81 damtp.cam.ac.uk/user/tong/statphys.html $\endgroup$ Commented Dec 15, 2018 at 17:36
  • $\begingroup$ In your last paragraph, you mention that $\mu << -kT$. This should mean that $e^{-\frac{\mu}{kT}} << \pm 1$ in your very first equation. This then cannot give the Boltzmann distribution. May be I'm missing something? Also, you mentioned "which is equivalent to stating that the number of available single-particle states is much greater than the number of particles in the system". I think this is a very interesting statement. Could you please elaborate how you arrived at this conclusion. $\endgroup$
    – Matrix23
    Commented Dec 23, 2021 at 15:31

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