 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/9b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For a bosonic degree of freedom, the supplied integral gives

I_{B} = \int_{0}^{\infty} \frac{y ^{3}}{e ^{y} - 1} d y = \frac{π ^{4}}{15} .

For a fermionic degree of freedom, use

\frac{1}{e ^{y} + 1} = \frac{1}{e ^{y} - 1} - \frac{2}{e ^{2 y} - 1} .

Changing variable

u = 2 y

in the second integral gives

I_{F} = I_{B} - \frac{1}{8} I_{B} = \frac{7}{8} I_{B} .

Summing the result of part (a) over all particle species therefore gives

ϵ = \frac{4 π g _{*}}{( h c ) ^{3}} (k T)^{4} \frac{π ^{4}}{15} = \frac{4 π ^{5}}{15 ( h c ) ^{3}} g_{*} (k T)^{4},

g_{*} = bosons \sum g_{i} + \frac{7}{8} fermions \sum g_{i} .

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