= Solution
At zero field, write the right-hand side of the <self-consistency equation> as $G_T(S)=g(4\beta JS)$, with $g(x)=2\sinh x/[1+2\cosh x]$. The origin is always a solution, and the slope there is
$$
G_T'(0)=4\beta J\,g'(0)=\frac{8\beta J}{3}.
$$
The critical crossing occurs when this equals one, giving
$$
\boxed{k_BT_c=\frac{8J}{3},\qquad T_c=\frac{8J}{3k_B}.}
$$
This is a <continuous phase transition> in the <mean-field approximation>. For $x>0$,
$$
g''(x)=-\frac{2(7+2\cosh x)\sinh x}{(1+2\cosh x)^3}<0,
$$
so the positive branch is a <strictly concave> function, while $g$ is odd and bounded by one. For $T\geq T_c$, the slope at zero is at most one and <strict concavity> puts $G_T(S)$ below $S$ for every $S>0$, leaving only the unmagnetized solution. For $T<T_c$, it initially lies above $S$ but has $G_T(1)<1$; hence there is one positive intersection and, by <spin inversion symmetry>, one negative intersection.
The curvature of the <Helmholtz free energy> in the <mean-field approximation> is $4J[1-G_T'(S)]$ per site. It is negative at $S=0$ below $T_c$ and positive at the nonzero intersections, identifying the ordered minima and the unstable zero solution.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2022-mean-field-self-consistency.png]
{title=The <self-consistency equation> develops two nonzero <order parameter> branches below the mean-field critical temperature}
The <critical temperature> and branches here belong to the <mean-field approximation>. In the exact finite system, <spin inversion symmetry> forces zero average <magnetization> at zero field; selecting a broken-symmetry phase requires the <thermodynamic limit> with a vanishing field of specified sign.
Back to article page
