 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/33c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The photon number in thermal blackbody radiation is not fixed, so $μ = 0$ . With

x = E / (k_{B} T)

U = A V (k_{B} T)^{4} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = A V (k_{B} T)^{4} \frac{π ^{4}}{15} .

The integral is a temperature-independent finite constant (its explicit value follows from the gamma-zeta integral). Differentiating at fixed volume gives

C_{V} = \frac{4 π ^{4}}{15} A V k_{B}^{4} T^{3}, α = 3.

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