 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/35c/b/solution

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

Put

x = β μ B .

The one-spin partition function is

z_{1} = e^{- x} + 1 + e^{x} = 1 + 2 cosh x .

The spins are independent, so the spin-1 paramagnet has

Z = z_{1}^{N}

F = - \frac{N}{β} lo g (1 + 2 cosh (β μ B)) .

The canonical heat capacity can be obtained from the energy variance:

C = k_{B} β^{2} \frac{\partial ^{2}}{\partial β ^{2}} lo g Z .

Because

\frac{d ^{2}}{d x ^{2}} lo g (1 + 2 cosh x) = \frac{2 ( c o s h x + 2 )}{( 1 + 2 c o s h x ) ^{2}},

we obtain

C = N k_{B} (β μ B)^{2} \frac{2 ( cosh ( β μ B ) + 2 )}{( 1 + 2 cosh ( β μ B ) ) ^{2}} .

x \to 0

, and hence

- βF ⟶ N lo g 3, C \sim \frac{2}{3} N k_{B} (β μ B)^{2} ⟶ 0 .

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