Differentiating the mean-field free energy and imposing stationarity gives
$$
0=Jqm-Jq\frac{\sinh(\beta Jqm)}{1+\cosh(\beta Jqm)}.
$$
The <hyperbolic-function identity> $\sinh x/(1+\cosh x)=\tanh(x/2)$ turns this into the <self-consistency equation>
$$
\boxed{m=\tanh\!\left(\frac{Jqm}{2T}\right)}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-303/1/b/ii/solution