Although the integrals ($7.53$and $7.54$) for$N$and $U$cannot be

carried out analytically for all T, it's not difficult to evaluate them numerically

using a computer. This calculation has little relevance for electrons in metals (for

which the limit $kT<<EF$is always sufficient), but it is needed for liquid ${}^{3}He$and

for astrophysical systems like the electrons at the center of the sun.

(a) As a warm-up exercise, evaluate the$N$integral ($7.53$) for the case$kT={\epsilon }_F$

and $\mu =0$, and check that your answer is consistent with the graph shown

above. (Hint: As always when solving a problem on a computer, it's best to

first put everything in terms of dimensionless variables. So let $t=\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$role="math" localid="1649996205331" $,c=\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$

, and $x=\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$kT$}\right.$. Rewrite everything in terms of these variables,

and then put it on the computer.)

(b) The next step is to vary$\mu$ holding$T$ fixed, until the integral works out to

the desired value,$N$. Do this for values of $\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$ranging from $0.1$ up to $2$,

and plot the results to reproduce Figure$7.16$. (It's probably not a good idea

to try to use numerical methods when $\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{${\epsilon }_F$}\right.$ is much smaller than $0.1$, since

you can start getting overflow errors from exponentiating large numbers.

But this is the region where we've already solved the problem analytically.)

(c) Plug your calculated values ofµ into the energy integral ($7.54$), and evaluate

that integral numerically to obtain the energy as a function of temperature

for$kT$up to $2{\epsilon }_F$ Plot the results, and evaluate the slope to obtain the

heat capacity. Check that the heat capacity has the expected behavior at

both low and high temperatures.

We have to find that evaluation the$N$integral ($7.53$) for the case$\frac{kT}{{\epsilon }_0}$and $\mu =0$, and checking the answer is consistent with the graph shown above.

To find no. of particles,

$$\begin{gathered} N={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d\epsilon \\ g\left(\epsilon \right)=\frac{3N}{2{{\epsilon }_F}^{3/2}}\sqrt{\epsilon } \end{gathered}$$

putting these things, $$\begin{gathered} N={\int }_0^{\infty }\frac{3N}{{2{\epsilon }_F}^{3/2}}\sqrt{\epsilon }\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d\epsilon \\ 1={\int }_0^{\infty }\frac{3}{{2{\epsilon }_F}^{3/2}}\sqrt{\epsilon }\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d\epsilon \end{gathered}$$

We assume , $t=\frac{kT}{{\epsilon }_F},c=\frac{\mu }{{\epsilon }_F},x=\frac{\epsilon }{{\epsilon }_F}\to dx=\frac{d\epsilon }{{\epsilon }_F}$

We get , $$\begin{gathered} 1=\frac{3}{2}{\int }_0^{\infty }\frac{\sqrt{x}}{e^{\left(x-c\right)/t}+1}dx \\ 1=\frac{3}{2}{\int }_0^{\infty }\frac{\sqrt{x}}{e^{x/t}+1}dx \end{gathered}$$

using Matlab , RHS integral can be calculated , it comes out to be $1.017$, so $\mu$is not equals to 0 when ${\epsilon }_F=kT$.

We need to vary$\mu$holding $T$fixed, to get the desired value,$N$, for values of $kT/{\epsilon }_F$ranging from $0.1$up to $2$,and plotting the results to reproduce Figure $7.15$

Here, for specific value of $c$and , we well find integral , also temperature will be fixed, so we assume the value of$t=1$. We will find the value of $c$,so that we get correct result of integral which is $1$,to find this we will use Matlab. With respect to the two variables,$x$ and$c$, equation 1 will be defined, we will integrate with respect to $x$from $0$and $\infty$ , so here function will be in variable only, we will calculate the value of $c$ , the interval will be from $-1$to $0.1$ ,

So, $c=-0.0215$.

Then , We will find constant with different temperatures value, here t will be changed and same instructions will be done . The interval of solutions, will also be changed from $-1$to $2$. We will change$t$from $0.1$to $2.0$. The following graph will be obtained.

We are asked to check that the heat capacity has the expected behavior at both low and high temperatures.

By plotting data in Excel, we will add trend line to actually see the equation which fits into data ,

Here, the equation between $t$and $c$from graph can be seen ,

$$\begin{gathered} c=-\left(08469\right)t^2-\left(0.2019\right)x+1.0412 \\ U={\int }_0^{\infty }\epsilon g\left(\epsilon \right)\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d\epsilon \end{gathered}$$

We can find the value of energy density here as;

$$\begin{gathered} g\left(\epsilon \right)=\frac{3N}{{2{\epsilon }_F}^{3/2}}\sqrt{\epsilon } \\ t=\frac{kT}{{\epsilon }_F},c=\frac{\mu }{{\epsilon }_F},x=\frac{\epsilon }{{\epsilon }_F}\to dx=\frac{d\epsilon }{{\epsilon }_F} \end{gathered}$$

Finally , $\frac{U}{N_{{\epsilon }_F}}=\frac{3}{2}{\int }_0^{\infty }\frac{x^{3/2}}{e^{{\displaystyle \raisebox{1ex}{$\left(x-c\right)$}\!\left/ \!\raisebox{-1ex}{$t$}\right.}}+1}dx$

We can find heat capacity by doing partial derivative of total energy with respect to to temperature.