Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 35A / b / i

Table of contents
- Solution i

Solution

For the one-dimensional freely jointed chain, counting right- and left-pointing links gives

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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