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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 37A / d / Solution

Substituting the fixed-tension partition function of a one-dimensional chain into the given free-energy definition gives

G = - k_{B} T lo g Z = - n k_{B} T lo g (2 cosh (τ a)) .

Differentiation with respect to the parameter conjugate to extension gives

\frac{\partial G}{\partial τ} = - n k_{B} T a tanh (τ a) .

The same result follows directly from the expected value under

p_{i} = e^{τ l_{i}} / Z

:

L = \sum_{i} p_{i} l_{i} = \frac{1}{Z} \frac{\partial Z}{\partial τ} = \frac{\partial l o g Z}{\partial τ} .

Consequently

L = - \frac{1}{k _{B} T} \frac{\partial G}{\partial τ}, tanh (τ a) = \frac{L}{na} .

Solved by gpt-5.6-sol high.

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