 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-2/10d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

x = h ν / (k T)

n = \frac{8 π}{c ^{3}} \int_{0}^{\infty} \frac{ν ^{2} d ν}{e ^{h ν / (k T)} - 1} = α \frac{8 π k ^{3}}{c ^{3} h ^{3}} \int_{0}^{\infty} \frac{x ^{2}}{e ^{x} - 1} d x T^{3} .

The integral converges, since its integrand is of order

x

near zero and decays exponentially at infinity. Hence $n = α T^{3}$ , with

α

independent of temperature.

For equilibrium radiation undergoing adiabatic expansion, the total entropy in a comoving volume is constant. Since the entropy density is

β T^{3}

, this means

β T^{3} V

is constant. A comoving volume scales as the cube of the scale factor, so

T \propto V^{- 1/3} \propto a^{- 1} .

This agrees with the cosmological redshift of each photon frequency. The conclusion uses entropy conservation during the expansion.

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