# Asymptotic freedom in QCD

From renormalization group equation $$t \frac{d \bar{g}(t , g)}{dt} = \beta (\bar{g}(t , g)), \quad \bar{g}(1 , g) = g$$ (here $t$ is momentum scale factor, $g$ is initial coupling constant and $\bar{g}$ is an effective coupling constant) and $$\tag 1 \beta (g) = ag^{2}$$ we can get $$\bar{g}(t , g) = \frac{g}{1 - gln(t)a} .$$ For QCD $$\tag 2 a = c\left(-11 + \frac{2}{3}n_{f} \right),$$ where $n_{f} = 6$ is the number of quark types.

How to get that for QCD we asymptotically have $(1)$ and $(2)$?

## 1 Answer

First off, equation (1) should be $\beta(g)=ag^3$. This is not an asymptotic formula. It is the approximate expression for the $\beta$ function to one loop order. Expression (2) is an exact expression at this order. The simplest method to compute this would be perform the computation in the background field gauge. Take a look a Peskin chapter 16 section 6 where this method is developed. This is a very efficient method to compute the $\beta$ function as it avoids the evaluation of numerous Feynman diagrams.

• No, expression $\beta (g) = ag^{2}$ is right. – Andrew McAddams Sep 14 '14 at 8:39
• Yes you are right. I should have mentioned that both the expressions are correct modulo a numerical factor that can be absorbed in the constant 'a'. This can always be done by the redefinition of the coupling. Thanks for pointing out. – Orbifold Sep 14 '14 at 8:55