# Callan-Symanzik equation for the QCD scattering cross section of the $e^{-}e^{+}\to q\bar{q}$ process

In Peskin and Schroeder (Section 17.2) it is stated without derivation that the scattering cross section for the $e^{-}e^{+}\to q\bar{q}$ process obeys the following Callan-Symanzik equation: $$\left[M\frac{\partial}{\partial M}+\beta(g)\frac{\partial}{\partial g}\right]\sigma(s,M,\alpha_{s})=0$$ where $M$ is the renormalization scale, $s$ is the center of mass energy and $\alpha_{s}=g^{2}/4\pi$. Why is it so? I do recognize the typical form of a Callan-Symanzik equation, but I cannot imagine how one would derive it by giving to the function $\beta$ the usual interpretation and hence being able to calculate the final expression of the cross section by plugging in the $\beta$ function of QCD.

I would argue as follows. Let's say that we want to define the cross section at an arbitrary renormalization scale. Then we put $$\sigma=\sigma(s,M,a_{s}(M))$$ as these are the only variables on which the cross section can explicitly depend (of course we are assuming that the QED renormalization is being held fixed). The cross section, however, depends on $M$ only formally; that is, physically the cross section cannot depend on the renormalization scale. So we must have
$$0=\frac{d\sigma}{dM}=\left[\frac{\partial}{\partial M}+\frac{\partial g}{\partial M}\frac{\partial}{\partial g}\right]\sigma$$ or equivalently $$\left[M\frac{\partial}{\partial M}+M\frac{\partial g}{\partial M}\frac{\partial}{\partial g}\right]\sigma=0$$ The second term in brackets is just the $\beta$ function of the theory, i.e. $$\beta(g)=\frac{dg}{d\log M}$$ so that $$\left[M\frac{\partial}{\partial M}+\beta(g)\frac{\partial}{\partial g}\right]\sigma=0$$ It wasn't that hard after all.