Black-body radiation ($u=a_{\mathrm{rad}}{T}^4,p=u/3$)

Black-body radiation is the electromagnetic radiation in thermal equilibrium with an ideal blackbody. It can be described as a photon gas: a cavity whose walls are all at the same temperature contains radiation with the universal Planck's law spectrum. A small opening in such a cavity approximates a blackbody emitter because incoming radiation undergoes repeated absorption inside. The radiation's equilibrium spectrum depends on temperature, rather than the wall material. ^[ref]

There are three useful quantities to keep distinct: directional spectral radiance $B_{\nu }$ or $B_{\lambda }$, total energy density $u$ in the cavity, and outward radiant exitance $j^{\ast }$ from its opening. Planck's law supplies the first; integrals over frequency and directions supply the others. For isotropy, every direction has the same $B_{\nu }$, so the full $4\pi$ solid angle givesHere $\hslash =h/(2\pi )$ is the reduced Planck constant, and the radiation constant is $a_{\mathrm{rad}}\approx 7.565733250\times 10^{-16}\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-4}$. The explicit Boltzmann constant makes this an SI formula; adopting natural units with $k_B=1$ recovers $u={\pi }^2{T}^4/(15{\hslash }^3{c}^3)$.

For a short integral outline, substitute $x=h\nu /(k_{B}T)$ and useThe expansion is a geometric series with nonnegative terms; repeated integration by parts evaluates each integral, and $\zeta (4)={\pi }^4/90$ is a value of the Riemann zeta function. This gives the coefficient above.

The photon gas also exerts radiation pressure, $p=u/3$. The factor $1/3$ comes from the angular average of the squared directional component of photon momentum, $⟨{\cos }^2\theta ⟩=1/3$. The thermodynamic derivation of blackbody radiation pressure gives the same result. In contrast, outgoing power involves one projected-direction factor over an outward hemisphere, yielding radiant exitanceThis is the Stefan–Boltzmann law with unit emissivity; $\sigma$ is the Stefan-Boltzmann constant. The factor $\pi$ follows from Lambert's cosine law, rather than the $4\pi$ full-sphere factor used for energy density.

For example, an ideal $6000\mathrm{K}$ cavity has $u\approx 0.980519\mathrm{J}{\mathrm{m}}^{-3}$, radiation pressure $p\approx 0.326840\mathrm{Pa}$, and outward radiant exitance $j^{\ast }\approx 7.34881\times 10^7\mathrm{W}{\mathrm{m}}^{-2}$. A stipulated $3000\mathrm{K}$ cavity has one-sixteenth each value because all three scale as $T^4$. These describe stored energy, mechanical pressure, and outgoing power per area, respectively. In a closed equilibrium cavity, equal opposing radiation balances the directional power flow even though $u$ and $p$ are nonzero.

The usefulness of black-body radiation is this common reference: Planck's law, Wien's displacement law, and the Stefan–Boltzmann law consistently connect spectral measurements, temperature, and total emission. The three illustrative curves in Planck's law are ideal equilibrium spectra; applying them to a material requires considering its emissivity and whether thermal equilibrium is a suitable approximation.

The assumption that energy exchange with a mode of frequency $\nu$ occurs in discrete steps $h\nu$. A thermal oscillator with $E_n=nh\nu$ has $⟨E⟩=h\nu /(e^{h\nu /(k_{B}T)}-1)$, recovering equipartition for small $h\nu /(k_{B}T)$ and exponential suppression for large values.

The bulk photon mode density per volume per angular frequency is ${\omega }^2/({\pi }^2{c}^3)$, including two transverse polarizations. Multiply it by the thermal oscillator energy $\hslash \omega /(e^{\hslash \omega /(k_{B}T)}-1)$ to obtain the displayed spectrum. Its integral is $u={\pi }^2{k}_B^4{T}^4/(15{\hslash }^3{c}^3)$. This is volume energy density rather than radiance or surface exitance; isotropic equilibrium radiation has outward black-surface flux $cu/4$. zero-point energy is excluded from this thermal spectrum.

Two wide perfect-conductor plates separated by $d$ support a gapless transverse electromagnetic mode whose electric field is normal to the plates. At $\omega \ll c/d$, multiply the single-polarization two-dimensional photon density of states by $\hslash \omega$ and the Bose-Einstein distribution to get the title's spectrum. This is a low-frequency statement; it need not describe the full thermal spectrum if thermally populated higher branches cross their cutoffs.

The spectral radiation energy density $u(\omega )$ is defined per unit angular frequency and physical volume. Thermal Planck law convention excludes the vacuum zero-point term. A density per ordinary frequency instead obeys $u_{\omega }=u_{\nu }/(2\pi )$, evaluated at $\nu =\omega /(2\pi )$; Einstein coefficients multiplying these densities must be converted correspondingly.

In a thin rectangular electromagnetic cavity, the window $\hslash c\pi /L_x,\hslash c\pi /L_y\ll k_{B}T\ll \hslash c\pi /L_z$ freezes the nonzero thickness indices while retaining many in-plane normal modes. The zero-thickness-index family has one polarization of an electromagnetic wave and density of states $dN=Akdk/(2\pi )=2\pi A\nu d\nu /c^2$. Multiplying by $h\nu /(e^{h\nu /k_{B}T}-1)$ and integrating gives $U_{2D}=4\pi \zeta (3)A(k_{B}T{)}^3/(h^2{c}^2)$. Thus thermal energy scales as area times $T^3$. The Riemann zeta function appears through ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$.

This is an intermediate low-temperature regime of an anisotropic finite cavity. When $k_{B}T$ also falls below the first in-plane excitation, thermal energy becomes exponentially small. If a different model excludes every zero thickness index, its first transverse mode has a finite gap and the proposed $T^3$ regime does not occur. In the three-dimensional continuum regime the corresponding density of states is proportional to volume times ${\nu }^2$ and thermal energy is proportional to volume times $T^4$.

A blackbody spectrum is the spectrum of ideal black-body radiation in thermal equilibrium, given by Planck's law. Thus Planck spectrum and blackbody spectrum name the same ideal spectrum.

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 $P=a{T}^4/3=E/(3V)$.