 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-2/34e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

β = (k_{B} T)^{- 1}

and

Z = \sum_{j} e^{- β E_{j}}

A

, differentiating its normalized expectation gives

\partial_{β} ⟨ A ⟩ = - ⟨ A E ⟩ + ⟨ A ⟩ ⟨ E ⟩

. Taking

A = E

yields

\partial_{β} ⟨ E ⟩ = - Var E

. Since

\partial_{β} = - k_{B} T^{2} \partial_{T}

at fixed volume,

⟨(E - ⟨ E ⟩)^{2} ⟩ = k_{B} T^{2} C_{V} .

Differentiating

⟨ E^{2} ⟩ - ⟨ E ⟩^{2}

with the same rule gives minus

⟨ E^{3} ⟩ - 3 ⟨ E ⟩ ⟨ E^{2} ⟩ + 2 ⟨ E ⟩^{3}

, namely minus the third central moment. Thus

⟨(E - ⟨ E ⟩)^{3} ⟩ = k_{B}^{2} [T^{4} (\frac{\partial C _{V}}{\partial T})_{V} + 2 T^{3} C_{V}] .

These identities exhibit the second and third energy cumulants as successive derivatives of

lo g Z

For the monatomic ideal gas,

⟨ E ⟩ = 3 N k_{B} T /2

C_{V} = 3 N k_{B} /2

and

\partial_{T} C_{V} = 0

. The variance is

3 N (k_{B} T)^{2} /2

and the third central moment is

3 N (k_{B} T)^{3}

. Division by the corresponding mean-energy powers gives

\frac{⟨( E - ⟨ E ⟩ ) ^{2} ⟩}{⟨ E ⟩ ^{2}} = \frac{2}{3 N}, \frac{⟨( E - ⟨ E ⟩ ) ^{3} ⟩}{⟨ E ⟩ ^{3}} = \frac{8}{9 N ^{2}} .

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