Separating variables gives
$$
\frac{d}{d\log\mu}\frac1{\alpha_3}=\frac b{2\pi}.
$$
Therefore
$$
\frac1{\alpha_3(\mu)}
=\frac1{\alpha_3(\mu_0)}
+\frac b{2\pi}\log\frac{\mu}{\mu_0}.
$$
For $b>0$, define the <strong-coupling scale>
$$
\Lambda_{\mathrm{QCD}}
=\mu\exp\left[-\frac{2\pi}{b\alpha_3(\mu)}\right].
$$
Then
$$
\alpha_3(\mu)=\frac{2\pi}{b\log(\mu/\Lambda_{\mathrm{QCD}})}.
$$
It decreases logarithmically toward zero in the ultraviolet and grows toward the infrared, becoming nonperturbative when $\mu$ approaches $\Lambda_{\mathrm{QCD}}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 4 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-305/4/a/i/solution