= Solution
A further derivative gives the <magnetic susceptibility> within a selected <pure thermodynamic phase>:
$$
\chi(t,h)=\frac{\partial m}{\partial h}
=|t|^{2-\alpha-2\Delta}g_{m,\pm}'(h/|t|^\Delta).
$$
Assuming a finite nonzero zero-field amplitude on the branch considered, comparison with $\chi(t,0)\asymp|t|^{-\gamma}$ gives
$$
\boxed{\gamma=2\Delta+\alpha-2.}
$$
Together with part (a), this proves the <Rushbrooke scaling relation> $\alpha+2\beta+\gamma=2$ and the <Widom scaling relation> $\gamma=\beta(\delta-1)$. Below the transition, use the one-sided response of an ordered phase; differentiating through the cusp of the equilibrium <free energy> at $h=0$ would include phase switching rather than this critical response.
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