= Solution
Along any $\langle111\rangle$ direction the quartic term is $(b+c/3)M^4$. Minimizing the amplitude below $T_c$ gives
$$
2a(T-T_c)M+4(b+c/3)M^3=0.
$$
For the ordered solution,
$$
\boxed{M^2(T)=\frac{a(T_c-T)}{2(b+c/3)},\qquad
M(T)=\sqrt{\frac{a(T_c-T)}{2(b+c/3)}}\quad(T<T_c).}
$$
Its <free energy> is $F_{\min}=F_0-a^2(T_c-T)^2/[4(b+c/3)]$. The denominator is positive by the supplied stability assumptions. The square-root onset is the <Landau theory> exponent; sufficiently close to a three-dimensional critical point, fluctuations can change the actual critical exponent without changing this saddle-point calculation.
Back to article page
