91Ó°ÊÓ

The Boltzmann distribution is:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}}$

The Bose-Einstein statistics:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}-1}$

The Fermi-Dirac statistics:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}+1}.$

Let us first list down the occupation numbers $\mathcal{N}$associated with a specific type of particle. For classical distinguishable particles, the Boltzmann distribution writes:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}}$ ……. (1)

For bosons, the Bose-Einstein statistics gives:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}-1}$ ……. (2)

For fermions, the Fermi-Dirac statistics gives:

$\mathcal{N}\left(E\right)=\frac{1}{B{e}^{E/k_{B}T}+1}.$ ……. (3)

Here B is a normalization coefficient, E is the energy,$k_B$is Boltzmann's constant, and T is the temperature.

Now in order for the occupation numbers to be much less than one: $\mathcal{N}\left(E\right)<<1$, the term $B{e}^{E/k_{B}T}$must be very large. Ideally if$B{e}^{E/k_{B}T}→∞$ , then it must be true that$\mathcal{N}\left(E\right)→0$ . In such a scenario, the terms $-1$and$+1$ in eqns. (2) and (3) become irrelevant, and all coincide to the Boltzmann distribution:

$\mathcal{N}\left(E\right)≈\frac{1}{B{e}^{E/k_{B}T}},For\mathcal{N}\left(E\right)<<1$