 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/36a/e/solution

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

For relativistic one-particle energy

ϵ (p) = p^{2} c^{2} + m^{2} c^{4}

, replace

q_{1}

q_{1}^{rel} = \frac{V}{h ^{3}} \int_{R^{3}} e^{- β ϵ (p)} d^{3} p .

Classical particle counting still gives

Z = exp (e^{β μ} q_{1}^{rel}), ⟨ N ⟩ = e^{β μ} q_{1}^{rel}, (Δ N)^{2} = ⟨ N ⟩ .

Since

q_{1}^{rel}

is proportional to

V

Φ = - k_{B} T ⟨ N ⟩ = - p V

, and therefore

p V = ⟨ N ⟩ k_{B} T, \frac{Δ N}{⟨ N ⟩} = \frac{1}{⟨ N ⟩} .

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