 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/35d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

d S = C (T) d T / T

S (0) = 0

when the ground state is nondegenerate, so below the transition

S (T_{c}^{-}) = \int_{0}^{T_{c}} \frac{α T}{T} d T = α T_{c} .

Above the transition,

S (T) = S (T_{c}^{+}) + \int_{T_{c}}^{T} \frac{β}{T ^{'}} d T^{'} = S (T_{c}^{+}) + β lo g \frac{T}{T _{c}} .

For a transition with no latent heat, entropy is continuous, and comparison with

S = β lo g T + γ

gives the directly testable relation

γ = α T_{c} - β lo g T_{c} .

Thus calorimetric measurements of

C (T)

and

T_{c}

verify

γ

. More generally, if a measurable latent heat

L

S (T_{c}^{+}) - S (T_{c}^{-}) = L / T_{c}

, and the formula becomes

γ = α T_{c} + \frac{L}{T _{c}} - β lo g T_{c} .

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