Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1 / b / ii

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- Solution ii

Solution

Differentiating the mean-field free energy and imposing stationarity gives

0 = J q m - J q \frac{s i n h ( β J q m )}{1 + c o s h ( β J q m )} .

The hyperbolic-function identity

sinh x / (1 + cosh x) = tanh (x /2)

turns this into the self-consistency equation

m = tanh (\frac{J q m}{2 T}) .

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