The Sommerfeld expansion is an expansion in powers of $\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$, which is assumed to be small. In this section I kept all terms through order ${\left(\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.\right)}^2$, omitting higher-order terms. Show at each relevant step that the term proportional to localid="1650117451748" $T^3$is zero, so that the next nonvanishing terms in the expansions forlocalid="1650117470867" $\mu$and localid="1650117476821" $U$are proportional to localid="1650117458596" $T^4$. (If you enjoy such things, you might try evaluating the localid="1650117464980" $T^4$terms, possibly with the aid of a computer algebra program.)

We have been given that the Sommerfeld expansion is an expansion in powers of $\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$, which is assumed to be small and we kept all terms through order ${\left(\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.\right)}^2$, omitting higher-order terms.

We need to show at each relevant step that the term proportional to role="math" localid="1650117659810" $T^3$is zero, so that the next nonvanishing terms in the expansions for$\mu$and$U$ are proportional to$T^4$.

The total energy given by the integral $7.54$is:

$U={\int }_0^{\infty }\epsilon g\left(\epsilon \right){\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$

$U=g_0{\int }_0^{\infty }{\epsilon }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$

Integrating by parts, we get:

$U=\frac{2}{5}{g}_0{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\overline{n}}_{FD}{|}_0^{\infty }+\frac{2}{5}{g}_0{\int }_0^{\infty }{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$ (Let this equation be ($1$))

If we substitute $\epsilon =0$, the integral becomes zero due to the dependence of the term on ${\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$and the first term vanishes.

On simplifying, the equation ($1$) becomes:

$U=\frac{2}{5}{g}_0{\int }_0^{\infty }{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$ (Let this equation be ($2$))

We know that ${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}$

Taking derivative with respect to $\epsilon$, we get:

$\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=\frac{\partial \left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}}\right]}{\partial \epsilon }$

On simplifying, we get:

$\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{\left(e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1\right)}^2}$

Let localid="1650119493586" $\raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT=x$}\right.$, then we can write as :

$dx=\frac{d\epsilon }{kT}$

Substitute $dx,x,\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }$in equation ($2$), we get:
$U=\frac{2}{5}{g}_0{\int }_0^{\infty }{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{kT}\frac{e^x}{{(e^x+1)}^2}kTdx$

$U=\frac{2}{5}{g}_0{\int }_0^{\infty }{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$

We need to change the boundaries of integration, so:

$$\begin{gathered} \epsilon \to \infty x\to \infty \\ \epsilon \to 0x\to -\frac{\mu }{kT} \end{gathered}$$

As $kT\ll \mu$, so we can put $-\infty$as the lower limit of integral, so the integral will become:

$U=\frac{2}{5}{g}_0{\int }_{-\infty }^{\infty }{\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be ($3$))

Now by expanding the term ${\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$about $\mu$using Taylor series, we get:

${\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{2}(\epsilon -\mu ){\mu }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{16}{(\epsilon -\mu )}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{128}{(\epsilon -\mu )}^3{\mu }^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{15}{8}{(\epsilon -\mu )}^4{\mu }^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}...$

Substitute $\epsilon -\mu =kTx$, we get:

${\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{2}\left(kTx\right){\mu }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{15}{8}{\left(kTx\right)}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{16}{\left(kTx\right)}^2{\mu }^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{128}{\left(kTx\right)}^2{\mu }^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}...$

Substitute the value of ${\epsilon }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$in equation ($3$), we get three integral say $I_1,I_2,I_3$:

$U=\frac{2}{5}{g}_0(I_1+I_2+I_3+I_4+I_5)$ (Let this equation be ($4$))

where, $I_1={\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{-\infty }^{\infty }\frac{e^x}{{(e^x+1)}^2}dx$

$I_2=\frac{5}{2}kT{\mu }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{-\infty }^{\infty }\frac{x{e}^x}{{(e^x+1)}^2}dx$

$I_3=\frac{15}{8}{\left(kT\right)}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{-\infty }^{\infty }\frac{x^2{e}^x}{{(e^x+1)}^2}dx$

$I_4=\frac{5}{16}{\left(kT\right)}^3{\mu }^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{-\infty }^{\infty }\frac{x^3{e}^x}{{(e^x+1)}^2}dx$

$I_5=\frac{5}{128}{\left(kT\right)}^4{\mu }^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{-\infty }^{\infty }\frac{x^4{e}^x}{{(e^x+1)}^2}dx$

On simplifying $I_1,I_2,I_3,I_4,I_5$, we get:

role="math" localid="1650601109788" $$\begin{gathered} I_1={\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {I}_2=0 \\ {I}_3=\frac{5{\pi }^2}{8}{\left(kT\right)}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {I}_4=0 \\ {I}_5=\frac{7{\pi }^4}{384}{\left(kT\right)}^4{\mu }^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Substituting $I_1,I_2,I_3,I_4,I_5$in equation ($4$), we get:

$U\approx \frac{2}{5}{g}_0\left[{\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5{\pi }^2}{8}{\left(kT\right)}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{7{\pi }^2}{384}{\left(kT\right)}^4{\mu }^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

We have proved that the proportional term to $T^3$is zero. We can also evaluate the further terms using computer algebra program, if wanted.