Mean-field tricritical exponent calculation ($\alpha =1/2,\beta =1/4,\gamma =1,\delta =5$)

For the sextic Landau free energy $V=r{M}^2/2+v{M}^6/6-hM$, $v>0$, with $r$ linear in reduced temperature, the equation of state is $h=rM+v{M}^5$. Its ordered zero-field minimum satisfies $M^4=-r/v$, the inverse magnetic susceptibility is $4|r|$ below the transition and $r$ above, and the critical isotherm satisfies $h=v{M}^5$. The minimized singular free-energy density is $-|r{|}^{3/2}/(3\sqrt{v})$ below and zero above. These facts give the displayed mean-field critical exponents, obeying both the Rushbrooke scaling relation and Widom scaling relation. Both the quadratic and quartic coefficients must be tuned; a generic path with a positive quartic coefficient instead has ordinary critical behavior. Fluctuations change the result below the tricritical upper critical dimension three, and marginal corrections can add logarithms at three.

For $F=F_{\mathrm{reg}}+a{m}^2/2+c{m}^6/6$, $c>0$, $a=a_0(T-T_c)$, the ordered global minimum gives $F-F_{\mathrm{reg}}=-(-a{)}^{3/2}/(3\sqrt{c})$ and $C_{\mathrm{sing},-}=T{a}_0^2/(4\sqrt{c}\sqrt{-a})$. On the disordered side the minimum is exactly $m=0$, so the polynomial's heat capacity is zero. The tricritical scaling label $\alpha =1/2$ can be used with zero upper-side amplitude, but no nonzero upper-side power law follows from this saddle alone. A smooth background such as $F_{\mathrm{reg}}=-C_{0}T\log (T/T_{\ast })$ gives a finite upper-side heat capacity $C_0$ without changing the phase diagram.

For $F=F_0+b(T-T_c)M^2+d^{\prime }{M}^6$ with $b,d^{\prime }>0$, the ordered minimum has $M^4=b(T_c-T)/(3d^{\prime })$. Its singular free energy is $-2b^{3/2}(T_c-T{)}^{3/2}/\sqrt{27d^{\prime }}$. Twice differentiating in temperature gives the displayed ordered-side heat capacity singularity, with mean-field critical exponent $\alpha =1/2$. This path keeps the quartic coefficient tuned to zero; a generic path with a positive quartic term instead crosses over to the ordinary quartic result.