 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/36a/b/iv/solution

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

For

T > T_{B}

, the ground-state occupation is negligible to leading order, so

N ≃ N_{ϵ}

. Comparing the number equation with its value at

T_{B}

gives

\frac{ϵ - μ}{T} = \frac{ϵ}{T _{B}}, μ = ϵ \frac{T _{B} - T}{T _{B}} .

For

T < T_{B}

μ

is close to zero and

N_{0} ≃ - \frac{k T}{μ}, N_{ϵ} ≃ \frac{M}{e ^{ϵ / (k T)} - 1} .

Using

M = N (e^{ϵ / (k T_{B})} - 1)

N_{0} = N - N_{ϵ}

and solving for

μ

gives

μ ≃ - \frac{k T}{N} \frac{e ^{ϵ / (k T)} - 1}{e ^{ϵ / (k T)} - e ^{ϵ / (k T_{B})}} .

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