91Ó°ÊÓ

We have been given that the Sommerfeld expansion is an expansion in powers of $\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Î\mu }}_F$}\right.$, which is assumed to be small and we kept all terms through order ${\left(\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Î\mu }}_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$\mathrm{μ}$and$U$ are proportional to$T^4$.

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

$U={∫}_0^{∞}\mathrm{Î\mu }g\left(\mathrm{Î\mu }\right){\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$

$U=g_0{∫}_0^{∞}{\mathrm{Î\mu }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$

Integrating by parts, we get:

$U=\frac{2}{5}{g}_0{\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\stackrel{¯}{n}}_{FD}{|}_0^{∞}+\frac{2}{5}{g}_0{∫}_0^{∞}{\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}d\mathrm{Î\mu }$ (Let this equation be ($1$))

If we substitute $\mathrm{Î\mu }=0$, the integral becomes zero due to the dependence of the term on ${\mathrm{Î\mu }}^{\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{∫}_0^{∞}{\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}d\mathrm{Î\mu }$ (Let this equation be ($2$))

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

Taking derivative with respect to $\mathrm{Î\mu }$, we get:

$\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}=\frac{∂\left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}}\right]}{∂\mathrm{Î\mu }}$

On simplifying, we get:

$\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}=\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{\left(e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1\right)}^2}$

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

$dx=\frac{d\mathrm{Î\mu }}{kT}$

Substitute $dx,x,\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}$in equation ($2$), we get:
$U=\frac{2}{5}{g}_0{∫}_0^{∞}{\mathrm{Î\mu }}^{\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{∫}_0^{∞}{\mathrm{Î\mu }}^{\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:

$$\begin{gathered} \mathrm{Î\mu }→∞x→∞ \\ \mathrm{Î\mu }→0x→-\frac{\mathrm{μ}}{kT} \end{gathered}$$

As $kT≪\mathrm{μ}$, so we can put $-∞$as the lower limit of integral, so the integral will become:

$U=\frac{2}{5}{g}_0{∫}_{-∞}^{∞}{\mathrm{Î\mu }}^{\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 ${\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$about $\mathrm{μ}$using Taylor series, we get:

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

Substitute $\mathrm{Î\mu }-\mathrm{μ}=kTx$, we get:

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

Substitute the value of ${\mathrm{Î\mu }}^{\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={\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}_{-∞}^{∞}\frac{e^x}{{(e^x+1)}^2}dx$

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

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

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

$I_5=\frac{5}{128}{\left(kT\right)}^4{\mathrm{μ}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}_{-∞}^{∞}\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" $$\begin{gathered} I_1={\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {I}_2=0 \\ {I}_3=\frac{5{\mathrm{π}}^2}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {I}_4=0 \\ {I}_5=\frac{7{\mathrm{π}}^4}{384}{\left(kT\right)}^4{\mathrm{μ}}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

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

$U≈\frac{2}{5}{g}_0\left[{\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5{\mathrm{π}}^2}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{7{\mathrm{π}}^2}{384}{\left(kT\right)}^4{\mathrm{μ}}^{\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.