For $d=4-\epsilon$, define dimensionless variables
$$
r=\frac{\mu^2}{\Lambda^2},
\qquad
u=\widetilde g=\Lambda^{-\epsilon}g.
$$
To leading order, with $K_4=1/(8\pi^2)$,
$$
\frac{dr}{ds}=2r+4(N+2)K_4\frac{u}{1+r},
$$
$$
\frac{du}{ds}=\epsilon u-4(N+8)K_4\frac{u^2}{(1+r)^2}.
$$
The <Gaussian fixed point> is $(r_*,u_*)=(0,0)$. Its thermal eigenvalue is $y_t=2$, so
$$
\boxed{\nu_{\rm G}=\frac12}.
$$
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 flow gives
$$
y_t=2-\frac{N+2}{N+8}\epsilon+O(\epsilon^2).
$$
Since the <correlation-length critical exponent> is $\nu=1/y_t$,
$$
\boxed{
\nu_{\rm WF}=\frac12+\frac{N+2}{4(N+8)}\epsilon+O(\epsilon^2).}
$$

Solution

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

 Solution

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