Write $t=T-T_c$. Analyticity gives $\alpha_2=at+O(t^2)$ with $a>0$ and $\alpha_4(T_c)=u>0$. For $t<0$, minimization gives $m^2=-\alpha_2/\alpha_4\sim-at/u$, so $\beta=1/2$. At the minima,
$$
F_{\min}=-\frac{\alpha_2^2}{4\alpha_4}\sim-\frac{a^2}{4u}t^2.
$$
Two temperature derivatives give a finite heat-capacity jump, hence $\alpha=0$. This is the ordinary <mean-field critical exponent>[mean-field] transition of <Landau theory>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 1 / a / i / Solution

For an ordinary quartic Landau transition, the heat-capacity and order-parameter exponents are $\alpha=0$ and $\beta=1/2$. At a tricritical sextic point they are $\alpha=1/2$ and $\beta=1/4$.


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