= 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 $h=rM+vM^5$. At zero field the ordered solution is $M=(-r/v)^{1/4}$, and the critical isotherm has $M=\operatorname{sgn}(h)(|h|/v)^{1/5}$. 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 give
$$
\boxed{(\alpha,\beta,\gamma,\delta,\nu,\eta)_{\rm tri,MF}=(\tfrac12,\tfrac14,1,5,\tfrac12,0).}
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-51-tricritical-phase-diagram.png]
{title=Zero-field sextic Landau phase diagram with continuous and first-order boundaries meeting at the tricritical point, and dashed spinodal limits}
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