= Solution
Write $t=T-T_c$, take $\alpha_2=at$ with $a>0$, and include the magnetic term $-Bm$. At $B=0$, minimizing the <Landau free energy> gives $m=0$ for $t>0$ and
$$
m^2=-\frac{\alpha_2}{\alpha_4}\propto(-t)
$$
for $t<0$, so the order-parameter <critical exponent> is $\beta=1/2$. Substitution gives a singular free-energy density proportional to $-t^2$ below $T_c$, whose second temperature derivative has a finite jump, so the heat-capacity exponent is $\alpha=0$. Above $T_c$, the equation $\alpha_2m+\alpha_4m^3=B$ gives the <magnetic susceptibility> $\chi=(\partial m/\partial B)_{B=0}=1/\alpha_2\sim t^{-1}$, hence $\gamma=1$. At $t=0$, $B=\alpha_4m^3$, so $m\sim B^{1/3}$ and $\delta=3$.
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