= Solution
A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by <temperature> alone. A cavity with a small aperture approximates this ideal. <Photon> number is not conserved in <thermal equilibrium>, so its <chemical potential> is zero. Combining the Bose occupation with the two-polarization electromagnetic <mode> <density> gives the spectral <energy> <density>
$$
u_\omega=\frac{\hbar\omega^3}{\pi^2c^3}\frac1{e^{\hbar\omega/(k_BT)}-1}.
$$
Changing variable to $x=\hbar\omega/(k_BT)$ shows directly why the total <energy> <density> is proportional to $T^4$:
$$
u(T)=\frac{(k_BT)^4}{\pi^2\hbar^3c^3}\int_0^\infty\frac{x^3\,dx}{e^x-1}
=\frac{\pi^2k_B^4}{15\hbar^3c^3}T^4.
$$
Three <powers> of T arise from the number of thermally populated <modes> and one from their characteristic <energy>. Isotropic radiation incident on a surface has outward hemispherical <flux> $cu/4$, giving
$$
\boxed{j^\star=\sigma_{\rm SB}T^4,\qquad
\sigma_{\rm SB}=\frac{\pi^2k_B^4}{60\hbar^3c^2}.}
$$
The factor 1/4 is the hemispherical angular average; multiplying <energy> <density> by c alone would overcount. The spectrum's <frequency> scale grows with T. Classical <equipartition> over all <modes> would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency <modes>. Real materials can have wavelength-dependent <emissivity>, and a body's net radiative exchange also includes absorbed environmental radiation.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1-tsp2013-blackbody.png]
{title=Original Planck energy spectra for three relative temperatures, showing the shift to higher frequencies and the growth of the integrated energy as temperature to the fourth power}
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