Since $\alpha_i=g_i^2/(4\pi)$, the stated <renormalization-group beta function> implies
$$
\frac{d\alpha_i}{d\log\mu}=-\frac{\beta_i}{2\pi}\alpha_i^2,
\qquad
\frac{d\alpha_i^{-1}}{d\log\mu}=\frac{\beta_i}{2\pi}.
$$
For $\beta_i>0$, the <running coupling> decreases at high energy: the theory is <asymptotically free> and becomes strong in the infrared, as in <Quantum chromodynamics>. For $\beta_i<0$, it is infrared free but grows toward an ultraviolet <Landau pole>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-305/4/a/solution