Question

Number of photons in a photon gas.

(a) Show that the number of photons in equilibrium in a box of volume V at temperature T is

$N=8\pi V{\left(\frac{kT}{hc}\right)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

The integral cannot be done analytically; either look it up in a table or evaluate it numerically.

(b) How does this result compare to the formula derived in the text for the entropy of a photon gas? (What is the entropy per photon, in terms of k?)

(c) Calculate the number of photons per cubic meter at the following temperatures: 300 K; 1500 K (a typical kiln); 2.73 K (the cosmic background radiation).

Short answer

Hence, the number of photons per cubic metre is

$$\begin{gathered} {\left.\frac{N}{V}\right|}_{300}=5.458\times {\left.{10}^{14}{\mathrm{m}}^{-3}\frac{N}{V}\right|}_{1500}=6.82\times {10}^{16}{\mathrm{m}}^{-3} \\ {\left.\frac{N}{V}\right|}_{2.73}=4.11\times {10}^8{\mathrm{m}}^{-3} \end{gathered}$$

Step-by-step solution

Given information

Number of photons in a photon gas.

Number of photons in equilibrium in a box of volume V at temperature T is

$N=8\pi V{\left(\frac{kT}{hc}\right)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

Explanation

Because we have two polarisation modes, the number of photons equals the total of Planck distributions over all modes ${\mathit{n}}_x,{\mathit{n}}_y\text{and}{\mathit{n}}_z$, multiplied by a factor of two, that is:

$N=2\sum _{n_x}\sum _{n_y}\sum _{n_z}{\overline{n}}_{\mathrm{Pl}}\left(ϵ\right)=\sum _{n_x,n_y,n_z}{\overline{n}}_{\mathrm{Pl}}\left(ϵ\right)$

But, ${\overline{n}}_{\mathrm{Pl}}\left(ϵ\right)=\frac{1}{e^{ϵ/kT}-1}$

Consider a cubic box with a volume of V and a side width of L; the photon's permitted energy is:

$ϵ=\frac{hcn}{2L}$

Substitute this the above equation:

$N=2\sum _{n_x,n_y,n_z}\frac{1}{e^{hcn/2LkT}-1}$

Now we need to convert the total to an integral in spherical coordinates, which we can do by multiplying it by the spherical integration factor $n^2\sin \left(\theta \right)$which gives us:

$N=2{\int }_0^{\pi /2}d\Phi {\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{\infty }\frac{n^2}{e^{hcn/2LkT}-1}dn$

So,

$N=\pi {\int }_0^{\infty }\frac{n^2}{e^{hcn/2LkT}-1}dn$

Now, let

$x=\frac{hcn}{2LkT}\to dx=\frac{hc}{2LkT}dn$

Thus,

$N=\pi {\left(\frac{2LkT}{hc}\right)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

The volume of the box is $V=L^3$

$N=8\pi V{\left(\frac{kT}{hc}\right)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

Evaluate the integral:

${\int }_0^{\infty }\frac{x^2}{e^x-1}dx$

We get

$x=2.4041$

Explanation

The number of photons is:

$N=8\pi (2.404)V{\left(\frac{kT}{hc}\right)}^3\left(1\right)$

The entropy is:

$S\left(T\right)=\frac{32{\pi }^5}{45}V{\left(\frac{kT}{hc}\right)}^{3}k$

Hence, entropy per photon is:

$$\begin{gathered} \frac{S}{N}=\frac{\frac{32{\pi }^5}{45}V{\left(\frac{kT}{hc}\right)}^{3}k}{8\pi (2.404)V{\left(\frac{kT}{hc}\right)}^3}=3.60k \\ \frac{S}{N}=3.60k \end{gathered}$$

(c) To find the number of photons per cubic meter at T= 300 K, substitute the values

$$\begin{gathered} {\left.\frac{N}{V}\right|}_{300}=8\pi (2.404){\left(\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{300}=5.458\times {10}^{14}{\mathrm{m}}^{-3} \\ {\left.\frac{N}{V}\right|}_{1500}=8\pi (2.404){\left(\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{1500}=6.82\times {10}^{16}{\mathrm{m}}^{-3} \\ {\left.\frac{N}{V}\right|}_{2.73}=8\pi (2.404){\left(\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}\right)}^3 \\ {\left.\frac{N}{V}\right|}_{2.73}=4.11\times {10}^8{\mathrm{m}}^{-3} \end{gathered}$$