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The Bose-Einstein statistics:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈭�1}$

The Fermi-Dirac statistics:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}+1}.$

The normalization condition writes:

$\mathrm{N}={鈭�}_0^{\mathrm{鈭�}}\mathcal{N}\left(\mathrm{E}\right)\mathrm{D}\left(\mathrm{E}\right)\mathrm{dE}$

Where, dE is the energy range, and $\mathrm{D}\left(\mathrm{E}\right)$is the density of states which in this problem, takes the form:

$\mathrm{D}\left(\mathrm{E}\right)=\frac{2\mathrm{s}+1}{\mathrm{鈩�}{\mathrm{蝇}}_0}$

Here,$\mathcal{N}\left(\mathrm{E}\right)$ is the occupation number, and depending on the type of particle, follows the form:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}卤1},$

Let $\mathrm{z}=\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$, then we have$\mathrm{dz}=\frac{\mathrm{dE}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈬�\mathrm{dE}={\mathrm{k}}_{\mathrm{B}}\mathrm{Tdz}.$

Performing a change of variables, the integrand becomes a function of$\mathrm{z}$:

$\begin{array}{c}\mathrm{N}=\frac{2\mathrm{s}+1}{\mathrm{鈩�}{\mathrm{蝇}}_0}{鈭�}_0^{\mathrm{鈭�}}{({\mathrm{Be}}^{\mathrm{z}}卤1)}^{鈭�1}\left({\mathrm{k}}_{\mathrm{B}}\mathrm{Tdz}\right)\\ ={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{鈩�}{\mathrm{蝇}}_0}{鈭�}_0^{\mathrm{鈭�}}{({\mathrm{Be}}^{\mathrm{z}}卤1)}^{鈭�1}\mathrm{dz}\end{array}$

Notice that ${鈭�}_0^{\mathrm{鈭�}}({\mathrm{Be}}^{\mathrm{z}}卤1)\mathrm{dz}$$\mathrm{is}\mathrm{a}\mathrm{familiar}\mathrm{integral}\mathrm{with}\mathrm{an}\mathrm{analytical}\mathrm{value}\mathrm{of}:$

${鈭�}_0^{\mathrm{鈭�}}({\mathrm{Be}}^{\mathrm{z}}卤1)\mathrm{dz}=卤\ln (1卤1/\mathrm{B})$

$\begin{array}{c}\mathrm{N}={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{鈩�}{\mathrm{蝇}}_0}(卤\ln (1卤1/\mathrm{B}\left)\right)\\ =卤{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{鈩�}{\mathrm{蝇}}_0}\ln (1卤1/\mathrm{B})\end{array}$

Solving for $B$, we thus obtain its expression depending on which type of particle it is. For fermions, ${\mathrm{B}}_{\text{FD}}$is thus:

$\begin{array}{c}\mathrm{N}={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{鈩�}{\mathrm{蝇}}_0}\ln (1+1/{\mathrm{B}}_{\text{FD}})\\ \ln (1+1/{\mathrm{B}}_{\text{FD}})=\frac{{\mathrm{N丑蝇}}_0}{2\mathrm{s}+1}\frac{1}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\end{array}$

Setting$蔚=\mathrm{N}\mathrm{鈩�}{\mathrm{蝇}}_0/(2\mathrm{s}+1)$we get:

$\begin{array}{c}\ln (1+1/{\mathrm{B}}_{\text{FD}})=\frac{\mathcal{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{e}}^{\ln (1+1/{\mathrm{B}}_{\text{FD}})}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ 1+1/{\mathrm{B}}_{\text{FD}}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \frac{1}{{\mathrm{B}}_{\text{FD}}}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈭�1\\ {\mathrm{B}}_{\text{FD}}=\frac{1}{{\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈭�1}鈰�\end{array}$

For bosons,${\mathbf{B}}_{\text{BE}}$ is thus:

$\begin{array}{c}\mathrm{N}=鈭�{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{鈩�}{\mathrm{蝇}}_0}\ln (1鈭�1/{\mathrm{B}}_{\text{BE}})\\ \ln (1鈭�1/{\mathrm{B}}_{\text{BE}})=鈭�\frac{\mathrm{N}\mathrm{鈩�}{\mathrm{蝇}}_0}{2\mathrm{s}+1}\frac{1}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\end{array}$

Setting $\mathcal{E}=\mathrm{N}\mathrm{鈩�}{\mathrm{蝇}}_0/(2\mathrm{s}+1)$we get:

$\begin{array}{c}\ln (1鈭�1/{\mathrm{B}}_{\text{FD}})=鈭�\frac{\mathrm{蔚}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{e}}^{\ln (1鈭�1/{\mathrm{B}}_{\text{BE}})}={\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ 1鈭�1/{\mathrm{B}}_{\text{BE}}={\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \frac{1}{{\mathrm{B}}_{\text{BE}}}=1鈭�{\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{B}}_{\text{BE}}=\frac{1}{1鈭�{\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\end{array}$

Plugging the expression for${\mathrm{B}}_{\text{FD}}$into eq. (3), the occupation number associated with a system of fermionic oscillators having intrinsic spin is thus:

$\mathcal{N}{\left(\mathrm{E}\right)}_{\text{FD}}=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}{{\mathrm{e}}^{\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈭�1}+1}$

Finally, plugging the expression for ${\mathrm{B}}_{\text{BE}}$into eq. (3), the occupation number associated with a system of bosonic oscillators having intrinsic spin is thus:

$\mathcal{N}{\left(\mathrm{E}\right)}_{\text{BE}}=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}{1鈭�{\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}$

$\begin{array}{c}{\mathrm{B}}_{\text{FD}}=\frac{1}{{\mathrm{e}}^{\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}鈭�1}\\ {\mathrm{B}}_{\text{BE}}=\frac{1}{1鈭�{\mathrm{e}}^{鈭�\mathrm{蔚}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\end{array}$