= Solution
Set $B=0$, $h=\beta Jqm$, and $A=1+2\kappa$. The required <power series> is obtained from
$$
1+2\kappa\cosh h=A+\kappa h^2+\frac{\kappa}{12}h^4+\frac{\kappa}{360}h^6+O(h^8).
$$
The quadratic Landau coefficient is
$$
a_2=\frac{Jq}{2}-\frac{\kappa(Jq)^2}{T(1+2\kappa)}.
$$
Its vanishing gives the line of continuous transitions
$$
T=\frac{2\kappa Jq}{1+2\kappa}.
$$
The quartic coefficient is
$$
a_4=-T(\beta Jq)^4\left[\frac{\kappa}{12A}-\frac{\kappa^2}{2A^2}\right].
$$
At a <tricritical point>, both $a_2$ and $a_4$ vanish. Since $A=1+2\kappa$, the condition $a_4=0$ gives $A=6\kappa$ and hence $\kappa=1/4$. Substitution into the critical line yields
$$
\boxed{\kappa_{\rm tri}=\frac14,
\qquad T_{\rm tri}=\frac{Jq}{3}}.
$$
The sextic coefficient is positive there, so the sixth-order term stabilizes the free energy. Equivalently, $g_{\rm tri}=T_{\rm tri}\log4$.
Back to article page
