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A critical exponent describes a power-law singularity near a continuous <phase transition>, such as $\xi\sim|t|^{-\nu}$ or $\chi\sim|t|^{-\gamma}$.

Critical exponent

The <Taylor series> at $x=0$ is
$$
\log(2+2\cosh x)=\log4+\frac{x^2}{4}-\frac{x^4}{96}+O(x^6).
$$
Substituting $x=Jqm/T$ gives
$$
f=-T\log4+
\left(\frac{Jq}{2}-\frac{(Jq)^2}{4T}\right)m^2
+\frac{(Jq)^4}{96T^3}m^4+O(m^6).
$$
The quartic coefficient is positive, while the quadratic coefficient changes sign at
$$
\boxed{T_c=\frac{Jq}{2}}.
$$
This is therefore a continuous mean-field <phase transition>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1 / b / iii / Solution

Using the field derivative while imposing the <self-consistency equation> gives
$$
\boxed{m=\frac{2\kappa\sinh[\beta(Jqm+B)]}
{1+2\kappa\cosh[\beta(Jqm+B)]}}.
$$
At $B=0$, $m=0$ is always a solution. Linearizing the right-hand side for small $m$ gives slope
$$
\frac{2\kappa\beta Jq}{1+2\kappa}.
$$
A continuous <phase transition> occurs when this slope crosses one, so, with $k_B=1$ and $\kappa_c=e^{-g/T_c}$,
$$
\boxed{1=\frac{2e^{-g/T_c}}{1+2e^{-g/T_c}}\frac{Jq}{T_c}}.
$$
On the side where the slope exceeds one, the nonzero solutions lower the mean-field free energy and the trivial solution is unstable. The equation has a physical continuous-transition solution only in the corresponding parameter range of this spin-one model.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 303 / 1 / a / iii / Solution

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A phase transition is a nonanalytic change in equilibrium behavior as a thermodynamic control parameter varies. A continuous phase transition has a diverging <correlation length> and no latent heat.

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 Incoming links: Phase transition