Let
$$
K_d=\frac{\Omega_{d-1}}{(2\pi)^d}.
$$
For a radial integrand $h(q)$, differentiating the thin shell at $\zeta=e^s$ 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 the bare parameters by running ones 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 / 2024 / iii / Paper 303 / 3 / a / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-303/3/a/v/solution