91Ó°ÊÓ

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

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:

${\mathrm{λ}}_n=\frac{2L}{n}{p}_n=\frac{hn}{2L}$

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:

$\mathrm{Ï\mu }=pc$

But,

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

Thus,

$\mathrm{Ï\mu }=c\sqrt{p_x^2+p_y^2+p_z^2}$

Substitute with momenta

localid="1650014884234" $$\begin{gathered} \mathrm{Ï\mu }=\frac{hc}{2L}\sqrt{n_x^2+n_y^2+n_z^2} \\ \mathrm{Ï\mu }=\frac{hcn}{2L} \end{gathered}$$

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.

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

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

$\mathrm{μ}={\mathrm{Ï\mu }}_F=\mathrm{Ï\mu }\left(n_{\max }\right)$

Substitute with $\mathrm{Ï\mu }$ and n_max:

localid="1647741783682" $$\begin{gathered} \mathrm{μ}=\frac{hc{n}_{\max }}{2L}=\frac{hc}{2L}{\left(\frac{3N}{\mathrm{π}}\right)}^{1/3} \\ \mathrm{μ}=\frac{hc}{2}{\left(\frac{3N}{L^3\mathrm{π}}\right)}^{1/3} \\ \mathrm{μ}=\frac{hc}{2}{\left(\frac{3N}{\mathrm{π}V}\right)}^{1/3} \end{gathered}$$

Here,

$V=L^3$

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

$U=2\underset{n_x}{∑}\underset{n_y}{∑}\underset{n_z}{∑}\mathrm{Ï\mu }\left(n\right)$

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

$U=2{∫}_0^{\mathrm{Ï€}/2}d\mathrm{Φ}{∫}_0^{\mathrm{Ï€}/2}\sin \left(\mathrm{θ}\right)d\mathrm{θ}{∫}_0^{n_{\max }}{n}^2\mathrm{Ï\mu }dn$

Substitute with $\mathrm{Î\mu }$

localid="1650015025150" $$\begin{gathered} U=\frac{hc}{L}{∫}_0^{\mathrm{π}/2}d\mathrm{Φ}{∫}_0^{\mathrm{π}/2}\sin \left(\mathrm{θ}\right)d\mathrm{θ}{∫}_0^{n_{\max }}{n}^{3}dn \\ U=\frac{hc}{L}\left[\frac{\mathrm{π}}{2}\right]\left[1\right]\left[\frac{n_{\max }^4}{4}\right] \\ U=\frac{hc\mathrm{π}{n}_{\max }^4}{8L} \end{gathered}$$

Substitute with n_max

localid="1650015120881" $$\begin{gathered} U=\frac{hc\mathrm{Ï€}}{8L}{\left(\frac{3N}{\mathrm{Ï€}}\right)}^{4/3} \\ U=\frac{hc\mathrm{Ï€}}{8L}\left(\frac{3N}{\mathrm{Ï€}}\right){\left(\frac{3N}{\mathrm{Ï€}}\right)}^{1/3} \\ U=\frac{3hcN}{8L}{\left(\frac{3N}{\mathrm{Ï€}}\right)}^{1/3} \\ U=\frac{3hcN}{8}{\left(\frac{3N}{\mathrm{Ï€}V}\right)}^{1/3} \\ U=\frac{3N}{4}\underset{{\mathrm{Ï\mu }}_F}{\underset{ÁåŸ}{\frac{hc}{2}{\left(\frac{3N}{\mathrm{Ï€}V}\right)}^{1/3}}} \\ U=\frac{3N}{4}{\mathrm{Ï\mu }}_F \end{gathered}$$