D.62 Fermi-Dirac in­te­grals at low tem­per­a­ture

This note finds the ba­sic Fermi-Dirac in­te­grals for the free-elec­tron gas at low tem­per­a­ture. To sum­ma­rize the main text, the num­ber of par­ti­cles and to­tal en­ergy per unit vol­ume are to be found from

\begin{displaymath}
\frac{I}{{\cal V}} = \int_0^\infty \iota^{\rm {f}}{\cal D}{...
...^{\rm p}\iota^{\rm {f}}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}
\end{displaymath}

where the Fermi-Dirac dis­tri­b­u­tion and the den­sity of states are:

\begin{displaymath}
\iota^{\rm {f}} = \frac{1}{e^{({\vphantom' E}^{\rm p}-\mu)/...
...(\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{{\vphantom' E}^{\rm p}}
\end{displaymath}

and the num­ber of spin states $n_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2s+1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 for sys­tems of elec­trons. This may be rewrit­ten in terms of the scaled en­er­gies

\begin{displaymath}
u = \frac{{\vphantom' E}^{\rm p}}{k_{\rm B}T} \qquad
u_0 = \frac{\mu}{k_{\rm B}T}
\end{displaymath}

to give

\begin{displaymath}
\frac{I}{{\cal V}} = \frac{n_s}{4\pi^2} \left(\frac{2m}{\hb...
...{u=0}^\infty \frac{(u/u_0)^{1/2}}{e^{u-u_0}+1}{\,\rm d}(u/u_0)
\end{displaymath}


\begin{displaymath}
\frac{E}{{\cal V}} = \frac{n_s}{4\pi^2} \left(\frac{2m}{\hb...
...{u=0}^\infty \frac{(u/u_0)^{3/2}}{e^{u-u_0}+1}{\,\rm d}(u/u_0)
\end{displaymath}

To find the num­ber of par­ti­cles per unit vol­ume for small but nonzero tem­per­a­ture, in the fi­nal in­te­gral change in­te­gra­tion vari­able to $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(u/u_0)-1$, then take the in­te­gral apart as

\begin{displaymath}
\int_{-1}^0 \sqrt{1+v}{\,\rm d}v
- \int_{-1}^0 \frac{\sqrt...
...+1}
+ \int_{0}^\infty \frac{\sqrt{1+v}{\,\rm d}v}{e^{u_0v}+1}
\end{displaymath}

and clean it up, by di­vid­ing top and bot­tom of the cen­ter in­te­gral by the ex­po­nen­tial and then in­vert­ing the sign of $v$ in the in­te­gral, to give

\begin{displaymath}
\int_{-1}^0 \sqrt{1+v}{\,\rm d}v
+ \int_{0}^1 \frac{(\sqrt...
...+1}
+ \int_{1}^\infty \frac{\sqrt{1+v}{\,\rm d}v}{e^{u_0v}+1}
\end{displaymath}

In the sec­ond in­te­gral, the range that is not killed off by the ex­po­nen­tial in the bot­tom is very small for large $u_0$ and you can there­fore ap­prox­i­mate $\sqrt{1+v}-\sqrt{1-v}$ as $v$, or us­ing a Tay­lor se­ries if still higher pre­ci­sion is re­quired. (Note that the Tay­lor se­ries only in­cludes odd terms. That makes the fi­nal ex­pan­sions pro­ceed in pow­ers of 1/$u_0^2$.) The range of in­te­gra­tion can be ex­tended to in­fin­ity, since the ex­po­nen­tial in the bot­tom is ex­po­nen­tially large be­yond $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. For the same rea­son, the third in­te­gral can be ig­nored com­pletely. Note that $\int_0^{\infty}x{\,\rm d}{x}$$\raisebox{.5pt}{$/$}$$(e^x+1)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi^2$$\raisebox{.5pt}{$/$}$​12, see [40, 18.81-82, p. 132] for this and ad­di­tional in­te­grals.

Find­ing the num­ber of par­ti­cles per unit vol­ume $I$$\raisebox{.5pt}{$/$}$${\cal V}$ this way and then solv­ing the ex­pres­sion for the Fermi level $\mu$ gives

\begin{displaymath}
\mu = {\vphantom' E}^{\rm p}_{\rm {F}}
- \frac{\pi^2}{12} ...
...ight)^{2/3}
\frac{\hbar^2}{2m} \left(\frac{I}{V}\right)^{2/3}
\end{displaymath} (D.39)

This used the ap­prox­i­ma­tions that $\mu$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ ${\vphantom' E}^{\rm p}_{\rm {F}}$ and $u_0^{-2}$ is small, so

\begin{displaymath}
u_0^{-2} = \left(\frac{k_{\rm B}T}{\mu}\right)^2 \approx
\...
...ht)^{-2/3}
\approx 1 - \frac{2}{3} \frac{\pi^2}{8\,} u_0^{-2}
\end{displaymath}

The in­te­gral in the ex­pres­sion for the to­tal en­ergy per unit vol­ume goes ex­actly the same way. That gives the av­er­age en­ergy per par­ti­cle as

\begin{displaymath}
\frac{E}{I} = {\vphantom' E}^{\rm p}_{\rm ave} = \frac{3}{5...
...{\rm {F}}}\right)^2 {\vphantom' E}^{\rm p}_{\rm {F}}
+ \ldots
\end{displaymath} (D.40)

To get the spe­cific heat at con­stant vol­ume, di­vide by $m$ and dif­fer­en­ti­ate with re­spect to tem­per­a­ture:

\begin{displaymath}
C_v = \frac{\pi^2}{2\,} \frac{k_{\rm B}T}{{\vphantom' E}^{\rm p}_{\rm {F}}} \frac{k_{\rm B}}{m} + \ldots
\end{displaymath}