Q. 7.69 - An Introduction to Thermal Physics

Question

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.

Step-by-Step Solution

Answer

(a) The equation was proved using the expression for density of states of fermi energy level Bose-Einstein condensation.

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

(c) The graph is plotted as shown below.

Step 1. Given Information

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

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

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

Step 2. (a) Solving the expression for density of state

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.

Step 3. (b)Checking the equation by putting values

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.

Step 4. (c) Plotting the graph

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 $c when T=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.