Specific heat capacity of the free electron gas (Fermions)

3.9 Specific heat capacity of the free electron gas (Fermions)

We again apply the Eq. (3.7) to (3.9) for the calculation of the particle number and the energy:

\begin{equation*} N(T,V,\mu) = \int_{-\infty}^{+\infty} D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon \end{equation*}

(3.46)

and

\begin{equation*} E(T,V,\mu) = \int_{-\infty}^{+\infty} \epsilon D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon \qquad . \end{equation*}

(3.47)

\(f_{\beta \mu}(\epsilon)\) is the Fermi statistics; obviously we find:

\begin{equation*} N(T,V,\mu) = \int_{-\infty}^{+\infty} D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon = \int_{-\infty}^{\epsilon_F} D(\epsilon) d\epsilon \label{n_electron} \end{equation*}

(3.48)

Multiplying both sides of Eq. (3.48) with \(\epsilon_F\) we get

\begin{equation*} \left(\int_{-\infty}^{\epsilon_F} + \int_{\epsilon_F}^{+\infty}\right) \epsilon_F D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon = \int_{-\infty}^{\epsilon_F} \epsilon_F D(\epsilon) d\epsilon \label{int_int} \end{equation*}

(3.49)

Since we calculate the derivation with respect to temperature, we can subtract a constant from the inner energy.
We get

\begin{equation*} \Delta U = \int_{-\infty}^{+\infty} \epsilon D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon - \int_{-\infty}^{\epsilon_F} \epsilon_F D(\epsilon) d \epsilon \label{du_int} \end{equation*}

(3.50)

Combining the equations (3.49) and (3.50) we find

\begin{equation*} \Delta U = \int_{\epsilon_F}^{+\infty} (\epsilon - \epsilon_F) D(\epsilon) f_{\beta \mu}(\epsilon) d\epsilon - \int_{-\infty}^{\epsilon_F} (\epsilon - \epsilon_F) \left(1-f_{\beta \mu}(\epsilon) \right) D(\epsilon) d\epsilon \qquad . \end{equation*}

(3.51)

The first term describes the excitation of an electron from the energy \(\epsilon_F\) to \(\epsilon\) and the second term the excitation from \(\epsilon\) to \(\epsilon_F\).
Finally we get

\begin{equation*} c = \frac{d\Delta U}{dT} = \int_{-\infty}^{+\infty} (\epsilon - \epsilon_F) D(\epsilon) \frac{\partial f_{\beta \mu}(\epsilon)}{\partial T} d\epsilon \qquad . \end{equation*}

(3.52)

Only around the Fermi energy \(\frac{\partial f_{\beta \mu}(\epsilon)}{\partial T}\) differs from zero; we therefor substitute \(D(\epsilon)\) by \(D(\epsilon_F)\) and take

\begin{equation*} x := \frac{\epsilon - \epsilon_F}{k T} \qquad , \end{equation*}

(3.53)

leading to

\begin{equation*} c \approx k^2 T D(\epsilon_F) \int_{-\infty}^{+\infty} \frac{x^2 e^x}{\left( e^x+1 \right)^2} dx = \frac{\pi^2}{3} D(\epsilon_F) k^2 T \qquad , \end{equation*}

(3.54)

Since for free electrons

\begin{equation*} N(\epsilon) = const. * \epsilon^{\frac{3}{2}} \qquad , \end{equation*}

(3.55)

we get

\begin{equation*} \frac{\partial N}{ \partial \epsilon} = const. * \frac{3}{2} \epsilon^{\frac{1}{2}} = \frac{3}{2}\frac{N}{\epsilon} \qquad . \end{equation*}

(3.56)

Therefore

\begin{equation*} D(\epsilon_F) = \frac{3 N}{2 \epsilon_F} = \frac{3 N}{2 k T_F} \qquad , \end{equation*}

(3.57)

and

\begin{equation*} c = \frac{\pi^2}{2} N k \frac{T}{T_F} \qquad . \end{equation*}

(3.58)

For room temperature and a typical Fermi temperature of several \(1000^o\)K follows

\begin{equation*} c \approx \frac{1}{100} N k \qquad . \end{equation*}

(3.59)

Thus at room temperature the heat capacity of electrons is not important.