Choose $\zeta=t^{-\nu}$. The singular free-energy density scales as $t^{d\nu}$, giving the <hyperscaling relation> $\alpha=2-d\nu$. The order parameter gives
$$
\beta=\nu\Delta_\phi=\frac12(d-2+\eta)\nu.
$$
The susceptibility has exponent
$$
\gamma=\nu(d-2\Delta_\phi)=\nu(2-\eta).
$$
At $t=0$, choosing $\zeta=B^{-1/\Delta_B}$ gives
$$
\delta=\frac{\Delta_B}{\Delta_\phi}
=\frac{d+2-\eta}{d-2+\eta}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 3 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-303/3/a/iv/solution