Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 51 / 1 / f / Solution

A tricritical point occurs when the continuous and first-order transition lines meet. In the symmetric sextic Landau free energy this requires tuning two coefficients, $r=u=0$, while $v>0$ stabilizes the potential. For $u>0$, the zero-field transition lies on $r=0$ and is continuous. For $u<0$, the coexistence curve is $r=3u^2/(16v)$. The order-parameter jump satisfies $M^2=-3u/(4v)$ and shrinks to zero as the tricritical point is approached. The labelled phase diagram below uses $(u,r)$ as local coordinates; two independent physical controls, such as temperature and composition, can provide these coordinates.

Along the tricritical tuning $u=0$, the equation of state is $h=rM+v{M}^5$. At zero field the ordered solution is $M=(-r/v{)}^{1/4}$, and the critical isotherm has $M=\mathrm{sgn}(h)(|h|/v{)}^{1/5}$. The inverse magnetic susceptibility is $r$ above and $4|r|$ below. The minimized singular potential is $-|r{|}^{3/2}/(3\sqrt{v})$ below, so differentiating twice thermally gives a $|t{|}^{-1/2}$ singularity. These calculations giveThe correlation-length and correlation-function exponents follow from the same quadratic gradient term. A generic path with a nonzero positive quartic coefficient instead has ordinary critical behaviour; observing tricritical powers requires the extra tuning. The sextic coupling is marginal in three dimensions, giving tricritical upper critical dimension three and logarithmic corrections there.