 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/35a/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

N_{\to} + N_{\leftarrow} = N, a (N_{\to} - N_{\leftarrow}) = L .

Therefore

N_{\to} = \frac{1}{2} (N + \frac{L}{a}), N_{\leftarrow} = \frac{1}{2} (N - \frac{L}{a}) .

Choosing which links point right gives the state degeneracy

Ω (L, N) = (N _{\to} N) = \frac{N !}{N _{\to} ! N _{\leftarrow} !} .

Put

x = L / (N a)

. Applying the Stirling formula to

S = k_{B} lo g Ω

and discarding terms subleading in

N

yields

S (L, N) = k_{B} N [lo g 2 - \frac{1 + x}{2} lo g (1 + x) - \frac{1 - x}{2} lo g (1 - x)], x = \frac{L}{N a} .

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