Q 7.40 - An Introduction to Thermal Physics

Question

Starting from equation 7.83, derive a formula for the density of states of a photon gas (or any other gas of ultra relativistic particles having two polarisation states). Sketch this function.

Step-by-Step Solution

Answer

Hence, the formula for density of states of a photon gas is $g\left(ϵ\right)=\frac{8\pi V{ϵ}^2}{(hc{)}^3}$

Step 1: Given information

The equation 7.83 is 

$\frac{U}{V}=\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{{ϵ}^3}{e^{ϵ/kT}-1}dϵ$

Step 2: Explanation

The equation 7.83 is:

$\frac{U}{V}=\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{{ϵ}^3}{e^{ϵ/kT}-1}dϵ$

We can write the equation as:

style="width:30%" style="width:30%" $U={\int }_0^{\infty }ϵ\left(\frac{8\pi V{ϵ}^2}{(hc{)}^3}\right)\frac{1}{e^{ϵ/kT}-1}dϵ \left(1\right)$

Distribution function for Planck's constant is given as:

${\overline{n}}_{\mathrm{Pl}}=\frac{1}{e^{ϵ/kT}-1}$

Substituting this into (1)

$U={\int }_0^{\infty }ϵ\left(\frac{8\pi V{ϵ}^2}{(hc{)}^3}\right){\overline{n}}_{\mathrm{Pl}}dϵ$

Hence the energy density for Planck's constant is 

 $g\left(ϵ\right)=\frac{8\pi V{ϵ}^2}{(hc{)}^3}$

Using Python to solve this function, the code is:

The graph is: