Q. 7.29 - An Introduction to Thermal Physics

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 $7.68$.

Step-by-Step Solution

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 1: 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 $7.68$.

Step 2: Simplify

The given total energy 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 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 ($4$))

Step 3: Finding the values of integrals.

As we need to change the integration boundaries also, so :

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

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 .

$I_2 = 0$

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

Step 4: 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$

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

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

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

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