8.13. The Fermi energy of the gas is given by the obvious relation

\[
\begin{equation*}
N=\int_{0}^{\varepsilon_{F}} a(\varepsilon) d \varepsilon \tag{1}
\end{equation*}
\]

At the same time, the quantities \(N\) and \(U\), as functions of \(\mu\) and \(T\), are given by the standard integrals

\[
N=\int_{0}^{\infty} \frac{a(\varepsilon) d \varepsilon}{e^{\beta(\varepsilon-\mu)}+1} \text { and } U=\int_{0}^{\infty} \frac{\varepsilon a(\varepsilon) d \varepsilon}{e^{\beta(\varepsilon-\mu)}+1}
\]

At low temperatures we employ formula (E.18), with \(x=\beta \varepsilon\) and \(\xi=\beta \mu\), to obtain

\[
\begin{align*}
N & =\int_{0}^{\mu} a(\varepsilon) d \varepsilon+\frac{\pi^{2}}{6}(k T)^{2}\left\{\frac{d a(\varepsilon)}{d \varepsilon}\right\}_{\varepsilon=\mu}+\ldots \\
& \simeq \int_{0}^{\varepsilon_{F}} a(\varepsilon) d \varepsilon+\left(\mu-\varepsilon_{F}\right) a\left(\varepsilon_{F}\right)+\frac{\pi^{2}}{6}(k T)^{2}\left\{\frac{d a(\varepsilon)}{d \varepsilon}\right\}_{\varepsilon=\varepsilon_{F}},  \tag{2}\\
U & =\int_{0}^{\mu} \varepsilon a(\varepsilon) d \varepsilon+\frac{\pi^{2}}{6}(k T)^{2}\left\{a(\varepsilon)+\varepsilon \frac{d a(\varepsilon)}{d \varepsilon}\right\}_{\varepsilon=\mu}+\ldots \\
& \simeq \int_{0}^{\varepsilon_{F}} \varepsilon a(\varepsilon) d \varepsilon+\left(\mu-\varepsilon_{F}\right) \varepsilon_{F} a\left(\varepsilon_{F}\right)+\frac{\pi^{2}}{6}(k T)^{2}\left\{a\left(\varepsilon_{F}\right)+\varepsilon_{F}\left[\frac{d a(\varepsilon)}{d \varepsilon}\right]_{\varepsilon=\varepsilon_{F}}\right\} . \tag{3}
\end{align*}
\]

Comparing (1) and (2), we obtain for the chemical potential of the gas

\[
\begin{equation*}
\mu \simeq \varepsilon_{F}-\frac{\pi^{2}}{6} \frac{(k T)^{2}}{a\left(\varepsilon_{F}\right)}\left\{\frac{d a(\varepsilon)}{d \varepsilon}\right\}_{\varepsilon=\varepsilon_{F}} \tag{4}
\end{equation*}
\]

which leads to the desired result for \(\mu\).

Next, substituting (4) into (3), we obtain the remarkably simple expression

\[
U \simeq U_{0}+\left(\pi^{2} / 6 \right) k^{2} T^{2} a\left(\varepsilon_{F}\right)
\]

whence

\[
\begin{equation*}
C_{\mathrm{V}} \simeq\left(\pi^{2} / 3 \right) k^{2} T a\left(\varepsilon_{F}\right) . \tag{5}
\end{equation*}
\]

It follows that

\[
\begin{equation*}
S=\int_{0}^{T} \frac{C_{\mathrm{V}} d T}{T} \simeq\left(\pi^{2} / 3 \right) k^{2} T a\left(\varepsilon_{F}\right) \tag{6}
\end{equation*}
\]

For a gas with energy spectrum \(\varepsilon \propto p^{s}\), confined to a space of \(n\) dimensions,

\[
a(\varepsilon) d \varepsilon \sim p^{n-1} d p \sim \varepsilon^{(n / s)-1} d \varepsilon
\]

By eqn. (1), the Fermi energy of the gas is given by

\[
N=\int_{0}^{\varepsilon_{F}} A \varepsilon^{(n / s)-1} d \varepsilon=\frac{s A}{n} \varepsilon_{F}^{n / s}=\frac{s \varepsilon_{F}}{n} a\left(\varepsilon_{F}\right)
\]

Substituting this result into (5), we get

\[
\begin{equation*}
\frac{C_{\mathrm{V}}}{N k} \simeq \frac{n}{s} \cdot \frac{\pi^{2}}{3}\left(\frac{k T}{\varepsilon_{F}}\right) \tag{7}
\end{equation*}
\]

cf. eqn. (8.1.39), which pertains to the case \(n=3, s=2\). See also Problem 8.11.