chemical_potential_equilibrium

\begin{matrix}
Z = \frac{1}{N!} \left[ 2 V \int_0^{\infty} \frac{4 \pi  p^2 dp}{\hbar^3} \exp \left(- \frac{p^2}{2 m k T} \right )\right]^N =
\\
= \frac{1}{N!} \left[ \frac{8 \pi V}{\hbar^3} \left(2 m k T \right )^{3/2} \int_0^{\infty} \xi^2 \exp(-\xi^2) d\xi \right ]^N =
\\
= \frac{1}{N!} \left[ \frac{2 V}{\hbar^3} \left(2 \pi m k T \right )^{3/2} \right ]^N \approx \left[ \frac{2 V}{N \hbar^3} \left(2 \pi m k T \right )^{3/2} \right ]^N
\\

\mu = - k T \frac{\partial}{\partial N} \ln Z = - k T \ln \left[ \frac{2 e V}{N \hbar^3} \left(2 \pi m k T \right )^{3/2} \right ]

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\textup{When the particles are in chemical equilibrium with photons}

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\mu + m c^2 = 0

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m c^2 = k T \ln \left[ \frac{2 e V}{N \hbar^3} \left( 2 \pi m k T \right )^{3/2} \right ] \Rightarrow
m c^2 = k T \ln \left[ \frac{2 e V}{N \hbar^3} \left( 2 \pi m k T \right )^{3/2} \right ]

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n = 2 e \left( 2 \pi \right )^{3/2} \lambda^{-3} \exp \left( -\frac{1}{\theta} \right ) \theta^3

\end{matrix}