Problem 7.69. If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate $\mu \text{for}T>T_c$.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define $t=T/T_c$,$c=\mu /k{T}_c,\text{and}x=ϵ/k{T}_c$. Express the integral that defines , equation 7.22, in terms of these variables. You should obtain the equation

$2.315={\int }_0^{\infty }\frac{\sqrt{x}dx}{e^{(x-c)/t}-1}$

(b) According to Figure

the correct value of $c$when $T=2T_c$is approximately $-0.8$. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary $\mu$, holding $T$fixed, to find the precise value of $\mu$for $T=2T_c$. Repeat for values of $T/T_c$ranging from $1.2$up to $3.0$, in increments of $0.2$. Plot a graph of $\mu$as a function of temperature.

The equation for density of states of fermi energy level:
$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{\frac{(\epsilon -\mu )}{kT}}-1}d\epsilon$ (Equation-1)

For, critical temperature

$k{T}_c=n\epsilon$

Here,

$k$= Boltzmann constant,

$n=$number of nearest neighboring dipole

$\epsilon$=interaction energy.

Relation between fermi temperature and fermi energy

$t{\epsilon }_F=kT$

Relation between chemical potential and fermi energy

$\mu =c{\epsilon }_F$

Substituting the values of $t{\epsilon }_F=kT,\mu =c{\epsilon }_F$in Equation-1

$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{\frac{\left(\epsilon -{\epsilon }_r\right)}{u_r}}-1}d\epsilon$

$={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{\frac{\left(\epsilon -c_p\right)}{t_r}-1}}d\epsilon$

Substituting the values of $t{\epsilon }_F=kT,\mu =c{\epsilon }_F$in Equation-1

$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{\frac{\left(\epsilon -c{\epsilon }_F\right)}{t{\epsilon }_F}}-1}d\epsilon$

$={\int }_0^{\infty }g\left(\epsilon \right)\frac{1}{e^{\frac{\left(\epsilon -c{\epsilon }_F\right)}{t{\epsilon }_F}}-1}d\epsilon$

$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{{\epsilon }_F}{e^{\frac{\left(x{\epsilon }_F-c{\epsilon }_F\right)}{t{\epsilon }_F}}+1}dx$

$N={\int }_0^{\infty }g\left(\epsilon \right)\frac{{\epsilon }_F}{e^{\frac{(x-c)}{t}}+1}dx$ (Equation-2)

For bosons zero spin confined in a box of volume $V$, the function can be equated as,

$N=2.612{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V$

Density of state =

$g\left(\epsilon \right)=2.612{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V\sqrt{\epsilon }d\epsilon$

Substituting the value of ${\epsilon }_F=xk{T}_c\text{and}d\epsilon =k{T}_{c}dx$

$g\left(\epsilon \right)d\epsilon =\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V\sqrt{xk{T}_c}\left(k{T}_c\right)dx$

$=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi mk{T}_c}{h^2}\right)}^{\frac{3}{2}}V\sqrt{x}dx$

Substituting the value of $N=2.612{\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V$

$N={\int }_0^{\infty }\frac{2}{\sqrt{\pi }}\frac{N}{2.612}\sqrt{x}dx\frac{{\epsilon }_F}{e^{\frac{(x-c)}{t}}-1}$

$=\frac{2}{\sqrt{\pi }}\frac{N}{2.612}{\int }_0^{\infty }\frac{\sqrt{x}}{e^{\frac{(x-c)}{t}}-1}dx$

$2.315={\int }_0^{\infty }\frac{\sqrt{x}}{e^{\frac{(x-c)}{t}}-1}dx$

Therefore, the equation is proved.

The given mathematical function is used to evaluate the value of $c$when $T=2T_c$

$\text{Nintegrate}\left\{\frac{Sqrt\left[x\right]}{Exp\left[\frac{(x+0.8)}{2}\right]-1},\{x,o,\text{Infinity}\}\right\}$

The value of $c$is evaluated as $2.35$which differs 1.6% from the accurate value.

The integral equation we attained from part(a)

$2.315={\int }_0^{\infty }\frac{\sqrt{x}}{e^{\frac{(x-c)}{t}}-1}dx$

The following mathematical function is used to evaluate the value of $cwhenT=2T_c$$\text{Find root}\left\{2.315=\text{Nintegrate}\left\{\frac{Sqrt\left[x\right]}{Exp\left[\frac{(x-c)}{2}\right]-1},\{x,o,\text{Infinity}\},\{c,-0.8,-0.9\}\right\}\right.$

This function gives the value of $c=-0.820$.

The following mathematical function generate a table for all values of $c$.

$\text{mutable=Table}\left\{\left\{t,\text{Find root}2.315=\text{N integrate}\left\{\begin{array}{l}\frac{Sqrt\left[x\right]}{Exp\left[\frac{(x-c)}{2}\right]-1},\{x,o,\text{Infinity}\}\\ \{c,-0.1,-0.2\},[1,2]\{t,1.2,3,0.2\}\end{array}\right\}\right\}\right\}$

The graph shows the chemical potential ($\mu$) as the function of temperature.