# Are the $N_f$ quarks in the coefficients of the beta function massless?

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?

• "cited in the literature" where? Can you please include at least one (reputable) reference? – AccidentalFourierTransform Sep 27 '18 at 18:41

## 1 Answer

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 in the theory.