Planck photon distribution

For photons in thermal equilibrium at temperature $T$, the number density per frequency interval is

Blackbody radiation is a thermal equilibrium gas of photons. In units with $k_B=1$, its energy density isand its isotropic radiation pressure is $p=u/3$.

In a large periodic box of volume $V$, one polarization has $V/(2\pi {)}^3$ wavevector modes per unit $d^{3}k$. The two transverse photon polarizations therefore giveafter integration over directions.

For an equilibrium system with energy $E(T,V)$ and no conserved particle number, exactness of the entropy differential obtained from $dE=TdS-PdV$ givesIf $E=aV{T}^4$ and $P=o(T)$ as $T\to 0$, integration gives $P=a{T}^4/3=E/(3V)$.

Changing variables to $x=h\nu /(k_{B}T)$ in the Planck distribution giveswhere $\alpha$ and $\xi$ are temperature-independent constants.

In natural units, a thermal photon gas with two polarizations has

In an orthonormal tetrad, $J^{\hat{0}}=\int f{d}^{3}p$ is the photon number density and $J^{\hat{i}}=\int f{e}^i{d}^{3}p$ is its number flux. Integrating the Boltzmann equation gives ${\nabla }_{\mu }{J}^{\mu }=\int C[f]d^{3}p/E$ when momentum-boundary terms vanish. Collisionless propagation and photon-number-conserving scattering give zero divergence. Photon emission, photon absorption, electron-positron annihilation in cosmology, bremsstrahlung and double Compton scattering generally do not. This is a number current, not an electric charge current: photons are neutral.