 Solution

The scaling hypothesis for critical phenomena says that the singular equilibrium free-energy density is a generalized homogeneous function of the thermal and field variables, after analytic backgrounds are removed. To derive the ordinary mean-field form, put $r=r_{t}t$, $u>0$ and neglect the higher powers near the transition. Rescale $m=\sqrt{r_t/u}|t{|}^{1/2}\psi$. Minimizing the quartic potential becomesThis is the mean-field scalar free-energy scaling form with $a=r_t^2/u$, $b=\sqrt{u}/r_t^{3/2}$. The two functions correspond to $t>0$ and $t<0$; the latter has two pure-phase branches and a cusp at zero field. The question's $>$ and $<$ subscripts distinguish these temperatures above and below $T_c$.

Differentiating with respect to $h$ gives $m=-{\partial }_h{A}_s\propto |t{|}^{1/2}$ on an ordered pure branch, and differentiating again gives $\chi =-{\partial }_h^2{A}_s\propto |t{|}^{-1}$. Thus the displayed scaling form independently yields $\beta =1/2$ and $\gamma =1$.

To allow anomalous powers, replace it byThen $\beta =2-\alpha -\Delta$ and $\gamma =2\Delta -(2-\alpha )$. Taking $t\to 0$ at fixed small $h$ gives $A_s\propto |h{|}^{(2-\alpha )/\Delta }$, so $1/\delta =(2-\alpha -\Delta )/\Delta =\beta /\Delta$. Eliminating $\Delta$ proves the Rushbrooke scaling relation and Widom scaling relation:These power-law relations assume the simple scaling form; marginal corrections or dangerously irrelevant couplings can require additional scaling variables.