The ordered solution gives
$$
|m_{\rm eq}|\sim(T_c-T)^{1/(2n-2)},
$$
so the <order-parameter critical exponent> is
$$
\boxed{\beta=\frac1{2(n-1)}}.
$$
At $T=T_c$, adding a magnetic term $-hm$ gives $h=2nu m^{2n-1}$, hence
$$
\boxed{\delta=2n-1}.
$$
Above $T_c$, the zero-field susceptibility is $\chi=(2a)^{-1}$. Below $T_c$, the curvature at a minimum is $-4(n-1)a$, so both sides give
$$
\boxed{\gamma=1}.
$$
At the ordered minimum, $a=-nu m^{2n-2}$ and
$$
f_{\rm sing}=(1-n)u m^{2n}\sim-|T-T_c|^{n/(n-1)}.
$$
Since $f_{\rm sing}\sim|T-T_c|^{2-\alpha}$, the <heat-capacity critical exponent> is
$$
\boxed{\alpha=\frac{n-2}{n-1}}.
$$
These values obey the <Widom scaling relation> and <Rushbrooke scaling relation> at mean-field level.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 303 / 1 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-303/1/a/ii/solution