The Cosmic Background Radiation — Question 4

Question 4

A photon gas has a calculable mean energy and particle density

Use $n_\gamma=20.29T^3$ cm$^{-3}$ and $\langle E\rangle=2.701k_BT$. Let T=2.725 K and $k_B=8.617\times10^{-5}$ eV/K.

1. Find photon density and mean photon energy.

2. Calculate total photon energy in one cubic meter, using $1\ {\mathrm{eV}}=1.602\times10^{-19}$ J.

Question 4 - Solution

A photon gas has a calculable mean energy and particle density

(a) Method and calculation

$$\boxed{n_\gamma=20.29(2.725)^3\simeq410.6\ {\mathrm{cm^{-3}}}=4.106\times10^8\ {\mathrm{m^{-3}}}}.$$$$\boxed{\langle E\rangle=2.701(8.617\times10^{-5})(2.725)\simeq6.343\times10^{-4}\ {\mathrm{eV}}}.$$

(b) Interpretation and further calculation

$$U=n_\gamma V\langle E\rangle.$$$$\boxed{U(1\ {\mathrm{m^3}})\simeq4.17\times10^{-14}\ {\mathrm{J}}}.$$

Physical check

The mean photon energy is not the energy at either spectral peak. The result agrees with the independent blackbody energy-density law to rounding.