 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/36a/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

a = \frac{1}{2} κ R^{2} ≫ k_{B} T,

then

e^{β a}

is large and the wall correction is exponentially small:

⟨ E ⟩ \sim 2 k_{B} T .

This is exactly the equipartition theorem result from the two kinetic and two harmonic-potential quadratic terms.

a ≪ k_{B} T

, use

\frac{a}{e ^{β a} - 1} = \frac{1}{β} - \frac{a}{2} + O (β a^{2}) .

Then

⟨ E ⟩ = k_{B} T + \frac{κ R ^{2}}{4} + O (\frac{κ ^{2} R ^{4}}{k _{B} T}) \sim k_{B} T .

The leading term is the kinetic energy. The particle is nearly uniform in the disk, so

⟨ r^{2} ⟩ \to R^{2} /2

and

⟨ V ⟩ \to κ R^{2} /4

, much smaller than

k_{B} T

. The hard wall therefore suppresses the positional equipartition contribution, as in the hard-wall correction to harmonic equipartition.

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