91Ó°ÊÓ

We have been given that the energy integral $7.54$is$U ={∫}_0^{∞}\mathrm{Î\mu }g\left(\mathrm{Î\mu }\right){\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$.

The Equation $7.67$, $U=\frac{3}{5}N\frac{{\mathrm{μ}}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{{\mathrm{Î\mu }}_F}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{3{\mathrm{Ï€}}^2}{8}N\frac{{\left(kT\right)}^2}{{\mathrm{Î\mu }}_F}+.....$and equation $7.68$, $U=\frac{3}{5}N{\mathrm{Î\mu }}_F+\frac{{\mathrm{Ï€}}^2}{4}N\frac{{\left(kT\right)}^2}{{\mathrm{Î\mu }}_F}+.....$are given.

We need to Carry out the Sommerfeld expansion for the energy integral $7.54$to obtain equation $7.67$. Then we have to plug in the expansion for $\mathrm{μ}$to obtain the final answer, equation $7.68$.

The given total energy integral $7.54$is

localid="1650054588139" $U={∫}_0^{∞}\mathrm{Î\mu }g\left(\mathrm{Î\mu }\right){\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$

localid="1650054613027" $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 will become zero$\left(0\right)$due to the dependence of the term on ${\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$and the first term vanishes. If we substitute with $∞$, the exponential term in the denominator will grow faster than ${\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

So the equation ($1$) reduces to :

$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$}\right.}}+1}$

By 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$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1)}^2}$ (Let this equation be ($3$))

Let $\raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.=x$, the equation ($3$) becomes :

$dx=\frac{d\mathrm{Î\mu }}{kT}$

Substitute $dx,x$and $\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}$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^{∞}{e}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be ($4$))

As we need to change the integration boundaries also, 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 the lower limit of integral in equation ($4$) as $-∞$,

The integral in equation ($4$) becomes :

$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 ($5$))

Now expand the term ${\mathrm{Î\mu }}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$using Taylor series about $\mathrm{μ}$,

${\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{15}{8}{(\mathrm{Î\mu }-\mathrm{μ})}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+...$

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

${\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.}+...$

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

$U=\frac{2}{5}{g}_0(I_1+I_2+I_3)$ (Let this equation be ($6$))

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$

Simplifying three integrals $I_1,I_2,I_3$, we get :

$I_1={\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

The second integral $I_2$is odd integral, the integration of integral is from $-∞$to $∞$, so the integral $I_2$is directly zero .

$I_2=0$

$I_3=\frac{15}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\frac{\mathrm{π}2}{3}\right]=\frac{5{\mathrm{π}}^2}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

By Substituting the values of the three integrals $I_1,I_2,I_3$in equation ($6$), we get

$$\begin{gathered} U≈\frac{2}{5}{g}_0\left[{\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5{\mathrm{π}}^2}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right] \\ U≈\frac{2}{5}{g}_0{\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+g_0\frac{{\mathrm{π}}^2}{4}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Substitute $g_0=\frac{3}{2}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Î\mu }}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.$, we get:

$U≈\frac{3}{5}\frac{N}{{\mathrm{Î\mu }}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{\mathrm{Î\mu }}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2{\mathrm{μ}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Set $\mathrm{μ}={\mathrm{Î\mu }}_F$in second term, we get

$U≈\frac{3}{5}\frac{N}{{\mathrm{Î\mu }}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{\mathrm{Î\mu }}_F}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2$ (Let this equation be ($7$))

The equation ($7.66$) is given by:

${\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\mathrm{Î\mu }}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[1-\frac{{\mathrm{Ï€}}^2}{12}{\left(\frac{kT}{{\mathrm{Î\mu }}_F}\right)}^2\right]}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

By expanding the terms in the brackets, we get:

${\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\mathrm{Î\mu }}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{\mathrm{Ï€}}^2}{24}{\left(\frac{kT}{{\mathrm{Î\mu }}_F}\right)}^2\right]$

Substitute the value of ${\mathrm{μ}}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$in equation ($7$), we get:

$U≈\frac{3}{5}\frac{N}{{\mathrm{Î\mu }}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\mathrm{Î\mu }}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{\mathrm{Ï€}}^2}{24}{\left(\frac{kT}{{\mathrm{Î\mu }}_F}\right)}^2\right]+\frac{N}{{\mathrm{Î\mu }}_F}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2$

$U≈\frac{3}{5}N{\mathrm{Î\mu }}_F\left[1-\frac{5{\mathrm{Ï€}}^2}{24}{\left(\frac{kT}{{\mathrm{Î\mu }}_F}\right)}^2\right]+\frac{N}{{\mathrm{Î\mu }}_F}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2$

$U≈\frac{3}{5}N{\mathrm{Î\mu }}_F-\frac{1}{8}N{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_F{\left(\frac{kT}{{\mathrm{Î\mu }}_F}\right)}^2+\frac{N}{{\mathrm{Î\mu }}_F}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2$

$U≈\frac{3}{5}N{\mathrm{Î\mu }}_F-\frac{N}{{\mathrm{Î\mu }}_F}\frac{{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2+\frac{N}{{\mathrm{Î\mu }}_F}\frac{3{\mathrm{Ï€}}^2}{8}{\left(kT\right)}^2$

$U≈\frac{3}{5}N{\mathrm{Î\mu }}_F+\frac{N}{{\mathrm{Î\mu }}_F}\frac{{\mathrm{Ï€}}^2}{{\mathrm{Î\mu }}_F}\frac{{\mathrm{Ï€}}^2}{4}{\left(kT\right)}^2$