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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/33c/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 33C / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

Let

x = E / (k_{B} T) > 0

and

r = e^{- x}

. The one-particle internal partition function is

z = (1 - r^{q}) / (1 - r)

, and independence gives

Z = z^{N}

. The mean level index is

⟨ j ⟩ = - \frac{d}{d x} lo g z = \frac{r}{1 - r} - \frac{q r ^{q}}{1 - r ^{q}} .

Thus the full internal entropy is

S = N k_{B} [lo g \frac{1 - e ^{- q x}}{1 - e ^{- x}} + x (\frac{1}{e ^{x} - 1} - \frac{q}{e ^{q x} - 1})] .

This describes the specified internal degrees of freedom, with no translational or mixing contribution added. For a positive level spacing, $S \to 0$ as $T \to 0$ , since the unique lowest internal level becomes certain, while $S \to N k_{B} lo g q$ as $T \to \infty$ , since all

q^{N}

internal configurations become equally likely. The formula includes

q = 1

, where the entropy is identically zero.

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