Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 303 / 1 / b

Assume the scalar order parameter can take either sign. The cubic term breaks $m\mapsto -m$ symmetry, and a positive quartic coefficient bounds the Landau free energy below. Stationary points satisfyso, besides the disordered point $m=0$, the possible ordered points areThe disordered point has curvature $A^{\prime \prime }(0)=A_2$ and is locally stable when $A_2>0$. Nonzero stationary points first appear whenThat condition alone does not make them the equilibrium phase: one must compare their free energies.

At a nonzero stationary point, substitute $A_2=-A_{3}m-A_4{m}^2$ to obtainFor an ordered state to coexist with $A(0)=0$, this requiresBoth curvatures at phase coexistence are positive and equal to $2A_3^2/(9A_4)$. The barrier is at $m=-A_3/(3A_4)$. In fact the potential at phase coexistence factorizes:The two distinct minima are therefore explicit, with a finite jump in the order parameter whenever $A_3\ne 0$.

To state the equilibrium inequalities for either sign of $A_3$, put $m=-\mathrm{sgn}(A_3)\rho$ with $\rho \ge 0$. This sign always has lower energy than the opposite sign of equal magnitude. ThenThe bracket's minimum is $A_2/2-A_3^2/(9A_4)$. It follows thatThe favored ordered minimum, when it exists, isThus a first-order phase transition is possible and occurs when the coefficients cross the coexistence relation. There is no continuous transition between $m=0$ and the equilibrium ordered state while $A_3$ remains nonzero and $A_4$ remains finite and positive. The ordered state becomes globally favorable while $A_2$ is still positive, before the disordered curvature can vanish. Although a stationary solution tends to zero at $A_2=0$, it is not the global transition branch. A continuous transition would require eliminating the cubic term, outside the stated $A_3\ne 0$ condition.

The spinodal points clarify the metastable region: for $2A_3^2/(9A_4)<A_2<A_3^2/(4A_4)$ the ordered minimum is metastable; for $0<A_2<2A_3^2/(9A_4)$ the disordered minimum is metastable. At $A_2=0$ the disordered state loses local stability. These local-stability limits should not be mistaken for the first-order transition in a cubic-quartic Landau potential.