Put
$$
K_d=\frac{\Omega_{d-1}}{(2\pi)^d}.
$$
For $\zeta=e^s$, differentiating a thin radial shell at $s=0$ gives
$$
\left.\frac d{ds}\int_{\Lambda e^{-s}}^\Lambda
\frac{d^dq}{(2\pi)^d}h(q)\right|_{s=0}
=K_d\Lambda^dh(\Lambda).
$$
Replacing bare parameters by running parameters after each infinitesimal step yields the <beta functions>
$$
\boxed{
\frac{d\mu^2}{ds}=2\mu^2
+4(N+2)gK_d\frac{\Lambda^d}{\Lambda^2+\mu^2},}
$$
$$
\boxed{
\frac{dg}{ds}=(4-d)g
-4(N+8)g^2K_d\frac{\Lambda^d}{(\Lambda^2+\mu^2)^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 303 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-303/3/c/solution