Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 64 / 1 / d / Solution

A tricritical point is the meeting of a line of continuous phase transitions and a line of first-order phase transitions. In the symmetric scalar Landau free energy, it occurs at $r=u=0$ with $v>0$. Two independent controls are needed to tune both coefficients. The continuous transition lies at $r=0,u>0$, and the tricritical three-phase line found above has $r=3u^2/(16v),u<0$; both approach the tricritical point. At zero conjugate field the three coexisting states on this first-order line are the disordered state and the two opposite ordered states.

Setting $u=H=0$ makes the leading stabilizing interaction sextic. The nonzero minimum obeys $M^4=-r/v$, so the order-parameter critical exponent along this tuned direction is ${\beta }_{\mathrm{tri}}=1/4$. Substitution givesand two temperature derivatives give the heat-capacity critical exponent ${\alpha }_{\mathrm{tri}}=1/2$. At $r=u=0$, the equation of state is $H=v{M}^5$, giving the critical-isotherm exponent ${\delta }_{\mathrm{tri}}=5$.

For magnetic susceptibility, differentiating $H=rM+v{M}^5$ gives ${\chi }^{-1}=r+5v{M}^4$. This is $r$ above the transition and $-4r$ below, hence ${\gamma }_{\mathrm{tri}}=1$. The quadratic fluctuation kernel still has a mass proportional to $|r|$, giving ${\nu }_{\mathrm{tri}}=1/2$ and ${\eta }_{\mathrm{tri}}=0$. Thus the mean-field tricritical exponent calculation yieldsDetuning $u$ gives tricritical crossover scaling; one must specify the approach to the tricritical point when quoting these critical exponents. The sextic interaction has upper critical dimension three, so sufficiently strong long-distance fluctuations change these mean-field predictions below three dimensions.