For $d=4-\epsilon$, define the dimensionless couplings $r=\mu^2/\Lambda^2$ and $u=\widetilde g=\Lambda^{-\epsilon}g$. Since $K_4=1/(8\pi^2)$, the leading <epsilon expansion> of the flow is
$$
\frac{dr}{ds}=2r+4(N+2)K_4\frac{u}{1+r},
\qquad
\frac{du}{ds}=\epsilon u-4(N+8)K_4\frac{u^2}{(1+r)^2}.
$$
There is a <Gaussian fixed point> $(r_*,u_*)=(0,0)$. Its thermal eigenvalue is $y_t=2$, so its <correlation-length critical exponent> is $\boxed{\nu_{\rm G}=1/2}$.

The interacting <Wilson-Fisher fixed point> is
$$
\boxed{u_*=\frac{2\pi^2}{N+8}\epsilon+O(\epsilon^2),
\qquad
r_*=-\frac{N+2}{2(N+8)}\epsilon+O(\epsilon^2)}.
$$
Linearizing the <renormalization-group flow> gives the thermal eigenvalue
$$
y_t=2-\frac{N+2}{N+8}\epsilon+O(\epsilon^2).
$$
Taking its reciprocal gives
$$
\boxed{\nu_{\rm WF}=\frac12+\frac{N+2}{4(N+8)}\epsilon+O(\epsilon^2)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 3 / a / vi / Solution

 Solution

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