= One-loop inverse gauge coupling evolution
{title2=$\alpha_i^{-1}(\mu)=\alpha_i^{-1}(\mu_0)-\frac{b_i}{2\pi}\log(\mu/\mu_0)$}
From $\mu\,dg_i/d\mu=b_i g_i^3/(16\pi^2)$ and $\alpha_i=g_i^2/(4\pi)$, the <chain rule> gives $d\alpha_i/d\log\mu=b_i\alpha_i^2/(2\pi)$. Taking its reciprocal makes the differential equation linear, proving the displayed relation. It assumes a fixed set of active fields and constant one-loop coefficients between the two scales; mass thresholds require piecewise matching. A negative coefficient gives <asymptotic freedom>.
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