5.7 Black-body radiation

Take-home message: Black-body radiation is an example of a Bose gas

There are interesting properties of gases of bosonic atoms, but they are beyond the scope of this course. However black-body of cavity radiation can be considered as a gas of photons. The walls of the cavity act as a heat bath; however as photons don't exist in the walls (only energy does) the chemical potential is zero: $\mu=(\partial S_{\rm walls}/\partial N)_E=0$. The density of states for the photon gas is just twice the usual density of states (to allow for the two polarisation states) and the energy of a photon with wavenumber $k$ is $\varepsilon(k)=\hbar\omega=\hbar c k$. The total energy in the cavity is

$$\begin{eqnarray*} {\left\langle E \right\rangle }\!\!\!&=&\!\!\!\int_0^\infty\! ... ...riptstyle B}^4\over 15\hbar^3c^3}V T^4={4 \sigma \over c} V T^4. \end{eqnarray*}$$

This gives the Planck distribution of energy over the frequency range as well as the derivation of the Stefan-Boltzmann constant $\sigma$ in terms of fundamental constants.

(The factor of $4/c$ comes from the connection between the cavity energy and the emissivity of a perfect black body, which is given by the formula for effusion from a small hole ${\rm d}N/{\rm d}t = -(v/4) A n$.)

Below is a plot of the cosmic microwave background at a range of frequencies, obtained from a variety of experiments, compared with the Planck distribution for $T=2.7$ K (©Douglas Scott of the University of British Columbia). The agreement is excellent.

References

• Mandl 10.4
• Bowley and Sánchez 8.5