Heat-capacity jump in quartic Landau theory ($\Delta C=T_c{b}^2/(2d)$)

For zero field, $F=F_0+b(T-T_c)M^2+d{M}^4$ with $b,d>0$ has ordered minima $M^2=b(T_c-T)/(2d)$ below $T_c$. Minimizing gives $F_{\min }=F_0-b^2(T-T_c{)}^2/(4d)$. Since $C=-T{\partial }_T^2{F}_{\min }$, the ordered-phase excess heat capacity is $T{b}^2/(2d)$ and its jump relative to the disordered side is $T_c{b}^2/(2d)$. The value is a total heat capacity or a heat-capacity density according to the normalization of $F$.

For density $F=\alpha (T-T_c)M^2/2+b{M}^4/4-{\mu }_{0}HM$, the zero-field ordered minimum is $-{\alpha }^2(T-T_c{)}^2/(4b)$. Its constant-field heat-capacity density jump is $\Delta c_H=T_c{\alpha }^2/(2b)$. In an Arrott plot, let physical slope be $S={\mu }_0/b$ and horizontal intercept be $X_0(T)=a(T)/{\mu }_0$. ThenThis makes numerical sensitivity explicit: uncertainty in the intercept slope is squared. A volumetric free energy determines a volumetric heat-capacity jump; total or molar values need further sample information.