Integrating the <renormalization-group beta function> and defining $\Lambda$ by $\alpha_3^{-1}(\Lambda)=0$ gives
$$
\alpha_3^{-1}(\mu)=\frac{\beta_3}{2\pi}\log\frac\mu\Lambda,
$$
so
$$
\boxed{\alpha_3(\mu)=\frac{2\pi}{\beta_3\log(\mu/\Lambda)}}.
$$
The scale $\Lambda$ is the <QCD scale> in this one-loop approximation.

Solution

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

 Solution

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