Q 7.22 - An Introduction to Thermal Physics

Question

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic $\left(ϵ\gg m{c}^2\right)$ so that their energies are $ϵ=pc$ (where p is the magnitude of the momentum vector).

(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by =

$\mu =hc(3N/8\pi V{)}^{1/3}$ 

(b) Find a formula for the total energy of this system in terms of N and $\mu$.

Step-by-Step Solution

Answer

The chemical potential is given by:

$\mu =\frac{hc}{2}{\left(\frac{3N}{\pi V}\right)}^{1/3}$

Total energy of the system is:

$U=\frac{3N}{4}{ϵ}_F$

Step 1: Given information

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic $\left(ϵ\gg m{c}^2\right)$ so that their energies are $ϵ=pc$ (where p is the magnitude of the momentum vector).

Step 2: Explanation

The allowable wavelengths and momenta for a relativistic particle in a one-dimensional box are the same as for a non-relativistic particle, and they are provided by:

${\lambda }_n=\frac{2L}{n} p_n=\frac{hn}{2L}$

Where,

n is positive integer

In the three dimensional box, the momenta are:

$p_x=\frac{h{n}_x}{2L} p_y=\frac{h{n}_y}{2L} p_z=\frac{h{n}_z}{2L}$

Energy for relativistic is:

$ϵ=pc$

But,

$p=\sqrt{p_x^2+p_y^2+p_z^2}$

Thus,

$ϵ=c\sqrt{p_x^2+p_y^2+p_z^2}$

Substitute with momenta 

$$\begin{gathered} ϵ=\frac{hc}{2L}\sqrt{n_x^2+n_y^2+n_z^2} \\ ϵ=\frac{hcn}{2L} \end{gathered}$$

Where,

$n=\sqrt{n_x^2+n_y^2+n_z^2}$ each n can be a positive integer, so we can visualise this as a lattice of a points in the first octant.

Because we have two spin stats, the total number of electrons is equal to the volume of an octant of a sphere with radius of n_max multiplied by factor 2.

$$\begin{gathered} N=2\times \frac{1}{8}\times \frac{4}{3}\pi n_{max}^3 \\ N=\frac{\pi }{3}{n}_{\max }^3 \\ {n}_{\max }={\left(\frac{3N}{\pi }\right)}^{1/3} \end{gathered}$$

the chemical potential is just the energy of the last level, which indicated by n_max, that is:

$\mu ={ϵ}_F=ϵ\left(n_{\max }\right)$

Substitute with $ϵ$ and n_max:

$$\begin{gathered} \mu =\frac{hc{n}_{\max }}{2L}=\frac{hc}{2L}{\left(\frac{3N}{\pi }\right)}^{1/3} \\ \mu =\frac{hc}{2}{\left(\frac{3N}{L^3\pi }\right)}^{1/3} \\ \mu =\frac{hc}{2}{\left(\frac{3N}{\pi V}\right)}^{1/3} \end{gathered}$$

Here,

$V=L^3$

Step 3: Explanation

(b) Due to the spin, the total energy equals the sum of the energies of occupied states multiplied by factor 2, i.e.

$U=2\sum _{n_x}\sum _{n_y}\sum _{n_z}ϵ\left(n\right)$

To convert this to spherical coordinates, multiply by the factor of the integration in spherical coordinates, which is $\left.n^2\sin \left(\theta \right)\right)$, as follows:

$U=2{\int }_0^{\pi /2}d\Phi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{n_{\max }}{n}^2ϵdn$

Substitute with $\epsilon$

$$\begin{gathered} U=\frac{hc}{L}{\int }_0^{\pi /2}d\Phi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{n_{\max }}{n}^{3}dn \\ U=\frac{hc}{L}\left[\frac{\pi }{2}\right]\left[1\right]\left[\frac{n_{\max }^4}{4}\right] \\ U=\frac{hc\pi n_{\max }^4}{8L} \end{gathered}$$

Substitute with n_max 

$$\begin{gathered} U=\frac{hc\pi }{8L}{\left(\frac{3N}{\pi }\right)}^{4/3} \\ U=\frac{hc\pi }{8L}\left(\frac{3N}{\pi }\right){\left(\frac{3N}{\pi }\right)}^{1/3} \\ U=\frac{3hcN}{8L}{\left(\frac{3N}{\pi }\right)}^{1/3} \\ U=\frac{3hcN}{8}{\left(\frac{3N}{\pi V}\right)}^{1/3} \\ U=\frac{3N}{4}\underset{{ϵ}_F}{\underbrace{\frac{hc}{2}{\left(\frac{3N}{\pi V}\right)}^{1/3}}} \\ U=\frac{3N}{4}{ϵ}_F \end{gathered}$$