Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section B / B2

Black-body radiation is the equilibrium electromagnetic radiation associated with a perfect absorber, whose spectrum is fixed by its temperature and the available modes. Photons can be emitted and absorbed, so their equilibrium chemical potential is zero. The model here retains two transverse polarizations in an effectively one-dimensional cavity; the resulting pressure is a one-dimensional mechanical force, not a three-dimensional pressure in pascals.

Standing-wave wave numbers are $k_n=n\pi /L$, with one positive mode index and two polarizations. Since $ϵ=\hslash ck$, the continuum density of states isThe factor two counts polarizations; counting independent positive and negative traveling momenta in addition would double count standing modes. The thermal energy isThus the one-dimensional photon-gas thermodynamics givesReplacing the mode sum by an integral requires the stated temperature to be large compared with the level-spacing scale. This is thermal energy above the vacuum contribution; a separate boundary-dependent vacuum energy is not included.

Differentiating at fixed length gives the heat capacityThe linear temperature dependence reflects the constant one-dimensional mode density, in contrast to the cubic dependence of a three-dimensional photon gas.

Write $b=\pi k_B^2/(3\hslash c)$, so $U=bL{T}^2$. At fixed length, $dS=C_{L}dT/T$, giving $S=2bLT$ with the entropy constant fixed by the unique photon vacuum. The thermal Helmholtz free energy and generalized pressure follow:The entropy constant can also be checked from the exact discrete photon partition function; the continuum approximation should not be extrapolated blindly below its level-spacing regime. In one dimension $P=U/L$, whereas the isotropic three-dimensional relation is $p=U/(3V)$. Both follow from the number of spatial directions carrying momentum flux.