 Solution

At the tricritical point, both the quadratic and quartic even coefficients vanish, while the sextic coefficient remains positive. Two even control parameters must therefore be tuned, in addition to setting the symmetry-breaking conjugate field to zero. The continuous phase transition line and first-order phase transition line meet, the coexistence order parameter jump shrinks to zero, and the tricritical wings end there. The balance governing the equation of state changes from quartic to sextic stabilization.

Along the tricritical thermal path $u=0$, $r=at$, $h=0$, minimizing $V$ gives $M^4=-r/v$ on the ordered side. At $r=u=0$, the equation of state is $h=v{M}^5$. ConsequentlyThese mean-field critical exponents require the tuned tricritical path; a path with fixed $u>0$ instead has the ordinary exponents derived above. The corresponding equilibrium Landau free energy is $f_{\min }=-(-r{)}^{3/2}/(3\sqrt{v})$ below the transition. Its first thermal derivative is continuous but its second derivative diverges, giving the tricritical mean-field approximation a heat-capacity exponent $\alpha =1/2$. Fluctuations are negligible asymptotically only above the tricritical upper critical dimension three; at three dimensions logarithmic corrections can accompany these powers.