Question

Carry out the Sommerfeld expansion for the energy integral $(7.54)$, to obtain equation $7.67$. Then plug in the expansion for $\mu$to obtain the final answer, equation $7.68$.

Short answer

The final answer, equation $7.68$is$U\approx \frac{3}{5}N{\epsilon }_F+\frac{{\pi }^2}{4}\frac{N}{{\epsilon }_F}\left(k{T}^2\right)$.

Step-by-step solution

Given information

We have been given that the energy integral $7.54$is$U\hspace{0.17em}={\int }_0^{\infty }\epsilon g\left(\epsilon \right){\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$.

The Equation $7.67$, $U=\frac{3}{5}N\frac{{\mu }^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{{\epsilon }_F}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{3{\pi }^2}{8}N\frac{{\left(kT\right)}^2}{{\epsilon }_F}+.....$and equation $7.68$, $U=\frac{3}{5}N{\epsilon }_F+\frac{{\pi }^2}{4}N\frac{{\left(kT\right)}^2}{{\epsilon }_F}+.....$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 $\mu$to obtain the final answer, equation $7.68$.

Simplify

The given total energy integral $7.54$is

localid="1650054588139" $U={\int }_0^{\infty }\epsilon g\left(\epsilon \right){\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$

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

So the equation ($1$) reduces to :

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

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

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

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

Substitute $dx,x$and $\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }$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 }{e}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be ($4$))

Finding the values of integrals.

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

The integral in equation ($4$) becomes :

$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 ($5$))

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

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

Substitute $\epsilon -\mu =kTx$,

${\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.}+...$

Substitute the value of ${\epsilon }^{\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={\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$

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

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

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

$I_2=0$

²3=5π28(kT)2μ12

Substituting the values of integrals 

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

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

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

$U\approx \frac{3}{5}\frac{N}{{\epsilon }_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{\epsilon }_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{3{\pi }^2}{8}{\left(kT\right)}^2{\mu }^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Set $\mu ={\epsilon }_F$in second term, we get

$U\approx \frac{3}{5}\frac{N}{{\epsilon }_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{\epsilon }_F}\frac{3{\pi }^2}{8}{\left(kT\right)}^2$ (Let this equation be ($7$))

The equation ($7.66$) is given by:

${\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\epsilon }_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[1-\frac{{\pi }^2}{12}{\left(\frac{kT}{{\epsilon }_F}\right)}^2\right]}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

By expanding the terms in the brackets, we get:

${\mu }^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\epsilon }_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{\pi }^2}{24}{\left(\frac{kT}{{\epsilon }_F}\right)}^2\right]$

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

$U\approx \frac{3}{5}\frac{N}{{\epsilon }_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\epsilon }_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{\pi }^2}{24}{\left(\frac{kT}{{\epsilon }_F}\right)}^2\right]+\frac{N}{{\epsilon }_F}\frac{3{\pi }^2}{8}{\left(kT\right)}^2$

$U\approx \frac{3}{5}N{\epsilon }_F\left[1-\frac{5{\pi }^2}{24}{\left(\frac{kT}{{\epsilon }_F}\right)}^2\right]+\frac{N}{{\epsilon }_F}\frac{3{\pi }^2}{8}{\left(kT\right)}^2$

$U\approx \frac{3}{5}N{\epsilon }_F-\frac{1}{8}N{\pi }^2{\epsilon }_F{\left(\frac{kT}{{\epsilon }_F}\right)}^2+\frac{N}{{\epsilon }_F}\frac{3{\pi }^2}{8}{\left(kT\right)}^2$

$U\approx \frac{3}{5}N{\epsilon }_F-\frac{N}{{\epsilon }_F}\frac{{\pi }^2}{8}{\left(kT\right)}^2+\frac{N}{{\epsilon }_F}\frac{3{\pi }^2}{8}{\left(kT\right)}^2$

$U\approx \frac{3}{5}N{\epsilon }_F+\frac{N}{{\epsilon }_F}\frac{{\pi }^2}{{\epsilon }_F}\frac{{\pi }^2}{4}{\left(kT\right)}^2$