Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 35E (Statistical Physics)

Mean field replaces edge and triangle products by $m^2$ and $m^3$. There are $3N$ edges and $2N$ elementary triangles, so

Stirling and replacing the magnetization sum by an integral gives the stated form with $P(m)=-\beta (Bm+3J{m}^2+2K{m}^3)$, up to subexponential constants.

Stationarity gives $\mathrm{artanh}m=\beta (B+6Jm+6K{m}^2)$, or $m=\tanh Q(m)$ with $Q(m)=\beta (B+6Jm+6K{m}^2)$.

For $B=J=0$, $m=0$ remains stationary. At large $\beta K$, evaluating the free energy near a positive $m$ makes the negative cubic energy dominate the bounded entropy cost, so some $m>0$ has lower free energy than zero. At high temperature entropy uniquely favors zero. Continuity in temperature therefore forces a transition at an intermediate ${\beta }_{\ast }$.