Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 303 / 1 / b / Solution

Discuss the zero-field phase boundaries, taking $h=0$ and writing $r={\mathcal{A}}_2$, $q={\mathcal{A}}_4$, $s={\mathcal{A}}_6>0$. This qualification matters because a conjugate field generally rounds the zero-field continuous transition. Nonzero stationary points satisfy $r+q{m}^2+s{m}^4=0$.

For $q>0$, the continuous phase transition occurs atOn the $r>0$ side the only stable phase is $m=0$; on the $r<0$ side the equilibrium global minima are $\pm m_0$ with $m_0^2\sim -r/q$. The magnetization therefore grows continuously from zero as $r$ changes sign.

For $q<0$, the first-order phase transition instead occurs while $r$ is positive. At phase coexistence, $m=0$ and $m=\pm m_0$ are equally deep global minimum points. With $a=m_0^2>0$, stationarity and equality of free energies giveEliminating $r$ gives $-q{a}^2/4-s{a}^3/3=0$, henceAt this coexistence value, the sextic even Landau potential factors as $f(m)=s{m}^2(m^2+3q/(4s){)}^2/6\ge 0$. This verifies that the three equal stationary values are global minima. Above this coexistence value of $r$ the disordered phase wins; below it the ordered phase wins, with a finite jump in the phase-selected magnetization. The values $r=0$ and $r=q^2/(4s)$ are the disordered and ordered spinodal points, respectively, not the coexistence boundary: the latter follows from the repeated root of $r+qa+s{a}^2=0$.

The two transition loci meet at the tricritical point,Two independent control parameters such as $T$ and $g$ can tune these two conditions. In a generic local phase diagram their coefficient map has a Jacobian matrix of rank two at the intersection; unspecified functions ${\mathcal{A}}_2(T,g)$ and ${\mathcal{A}}_4(T,g)$ do not by themselves determine the geometric orientation of the boundaries. The sextic term stabilizes this intersection and the negative-quartic first-order side.