In a box of volume $V$, the <electromagnetic mode density in a box>, including the two transverse <polarization of an electromagnetic wave>[polarizations], is
$$
dN=2\frac{V}{(2\pi)^3}d^3k
=\frac{V}{\pi^2}k^2\,dk.
$$
Each <photon> has energy $\hbar\omega=\hbar ck$, and the given <Planck photon distribution>[mean occupation number] is $(e^{\hbar ck/T}-1)^{-1}$. Omitting the temperature-independent <zero-point energy>, the total energy is therefore
$$
E=\frac{V\hbar c}{\pi^2}
\int_0^\infty\frac{k^3}{e^{\hbar ck/T}-1}\,dk.
$$
With the <change of variables> $x=\hbar ck/T$,
$$
E=\frac{VT^4}{\pi^2\hbar^3c^3}
\int_0^\infty\frac{x^3}{e^x-1}\,dx.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 35A / i / Solution

{wiki=Polarization_(waves)}

The polarization of an electromagnetic wave describes the path traced by its transverse electric-field vector at a fixed point. A monochromatic wave may be linearly, circularly, or elliptically polarized; in vacuum it has two independent transverse polarization states.


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 Incoming links: Polarization of an electromagnetic wave