Consider the running of the strong coupling in QCD $$ \mu\frac{d}{d\mu}\alpha(\mu)=-\frac{1}{2\pi}\beta_0\alpha^2+\dots $$ there I have written the perturbative expansion of the beta function to leading order. The value cited in the literature is $$ \beta_0=11-\frac{2}{3}N_f $$ where it is indicated that $N_f$ is the number of fermions in the theory. Now, I wonder, are this $N_f$ fermions supposed to be massless, or all massive with different masses?
1 Answer
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The value of the coupling $\alpha_S (Q)$ depend on the energy scale $Q$ we consider. And $N_f$ here are the number of active fermions in the theory, which means fermions whose mass is smaller than $Q$. So, at some points of calculation, you can approximate some fermion's masses to be zero, but generally, $N_f$ is the number of massive fermions with $m<Q$ in the theory.