# When does the Bose-Einstein distribution function reduce to the Maxwell-Boltzmann distribution?

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

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

## 1 Answer

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.

• 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 Commented Dec 15, 2018 at 17:36
• 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. Commented Dec 23, 2021 at 15:31