Blackbody Thermodynamics and Radiation Pressure

The Planck spectrum fixes the energy in each frequency interval. Integrating over the spectrum collapses the radiation field to a handful of thermodynamic functions of $T$ and $V$ alone, because the photon gas has no independent particle number to carry. The energy density, pressure, entropy, and heat capacity all follow one power law in temperature, and their exponents differ only by the counting. These are the quantities a star's interior and the expanding universe are built on.

The Stefan-Boltzmann law

The total energy density integrates the Planck density over all frequencies. With $x=hf/k_{B}T$,

The dimensionless integral is a standard Bose integral,

so the energy density is proportional to $T^4$.¹

The factor $\frac{1}{4}c$ relating the interior energy density to the emitted flux comes from averaging the outward photon flux over the hemisphere of directions leaving a small hole in the cavity: a factor $c$ for the photon speed, a factor $\frac{1}{2}$ for outgoing directions, and a factor $\frac{1}{2}$ for the mean projection $⟨\cos \theta ⟩$ onto the surface normal.²

Radiation pressure

Photons carry momentum $p=\epsilon /c$. A photon striking a wall at angle $\theta$ to the normal and reflecting transfers momentum $2p\cos \theta$ along the normal. Averaging the momentum flux over an isotropic distribution of directions gives a pressure equal to one third of the energy density — the universal result for an ultrarelativistic gas, where $PV=\frac{1}{3}U$ because $\epsilon =cp$ makes $p\partial \epsilon /\partial p=\epsilon$.

Free energy, entropy, and heat capacity

Because $\mu =0$, the grand potential equals the Helmholtz free energy, and both equal $-PV$:

Every other function follows by differentiation. The entropy is

and the internal energy reconstructs correctly, $U = F + TS = -\tfrac{1}{3}aVT^4

• \tfrac{4}{3}aVT^4 = aVT^4$. The heat capacity at constant volume is

Both the entropy and the heat capacity vanish as $T^3$, so the photon gas obeys the third law: $S\to 0$ and $C_V\to 0$ as $T\to 0$. Unlike a material gas, there is no residual entropy and no additive constant, because the number of photons is itself a function of temperature.

The adiabatic law and cosmological cooling

A reversible adiabatic expansion holds the entropy fixed. Since $S\propto V{T}^3$, constant entropy requires

and combined with $P\propto T^4$ this gives the adiabatic relation

so the photon gas has adiabatic index $\gamma =\frac{4}{3}$, distinct from the $\frac{5}{3}$ of a monatomic ideal gas. The universe expands adiabatically to good approximation, its volume scaling as the cube of the scale factor, $V\propto R^3$, so $T\propto R^{-1}$: the blackbody radiation that filled the early universe cools in inverse proportion to expansion, and a Planck spectrum stays a Planck spectrum at a lower temperature.³

Photon number and the cosmic microwave background

The photon number density integrates the mode density times the occupation,

using ${\int }_0^{\infty }{x}^2/(e^x-1)\mathbf{d}x=2\zeta (3)=2.404$. Numerically $n\approx 2.03\times 10^7{T}^3{\mathrm{m}}^{-3}$ with $T$ in kelvin. Dividing the energy density by the number density gives the mean energy per photon,

somewhat below the peak-frequency quantum $2.82k_{B}T$ because the broad low-frequency tail pulls the mean down.

Summary

• The energy density integrates the Planck spectrum to $u=a{T}^4$ with $a=8{\pi }^5{k}_B^4/15c^3{h}^3$; the surface flux is $\sigma T^4$ with $\sigma =ac/4=5.670\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$.
• Isotropy and the linear dispersion give $P=u/3=\frac{1}{3}a{T}^4$, independent of volume, versus $\frac{2}{3}U/V$ for a nonrelativistic gas.
• From $F=-\frac{1}{3}aV{T}^4$ follow $S=\frac{4}{3}aV{T}^3$ and $C_V=4aV{T}^3=3S$, both vanishing as $T^3$ in accord with the third law.
• Adiabatic expansion holds $V{T}^3$ fixed, so $T\propto V^{-1/3}$ and $P{V}^{4/3}=\text{const}$; the same relation cools the cosmic background as $T\propto R^{-1}$.
• The number density is $n=16\pi \zeta (3)(k_{B}T/hc{)}^3\approx 2.03\times 10^7{T}^3{\mathrm{m}}^{-3}$, giving $411\mathrm{c}{\mathrm{m}}^{-3}$ for the $2.725\mathrm{K}$ microwave background, with mean photon energy $2.70k_{B}T$.