# How does the 't Hooft renormalization scheme work?

I recently learned there is a trick called t' Hooft transformation that allows us to define a new coupling constant $g_R$ from the usual one $g$ in such a way that the beta function for $g_R$ is two-loop exact. In other words, the beta function has no terms higher than $g_R^2$.

$$\beta(g) = \sum a_n g^n .$$ Then, we define $g_R \equiv g + \sum r_n g^n$ such that

$$\beta_R(g_R) = a_1 g_R + a_2 g_R^2.$$

How does this work, i.e. why aren't there terms $\propto g_R^3,g_R^4,\ldots$ in $\beta_R(g_R)$?

I found the following statements, but wasn't really able to understand what is going on:

"The nullification of higher order coefficients of the β-function is achieved by finite renormalizations of charge." (Source)

"t Hooft' has suggested that one can exploit this freedom in the choice of $g$ even further and choose a new coupling parameter $g_R$ such that the corresponding $\beta_R(g_R) = a_1 g_R + a_2 g_R^2$ (for the above case) and thus contains only two terms in its expansion in $g_R$."(Source)

"The 't Hooft transformation is based on the observation that the first two terms in the perturbation series for $\beta$ are scheme independent. These terms are therefore invariant under a certain class of transformations on the renormalized coupling parameter $g$, a class that includes the 't Hooft transformation in which P is exactly reduced to these two terms." (Source)

PS: In addition, to my confusion how this really works, I am confused about what it implies, e.g. for the radius of convergence of the perturbation series. See this question.

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. – Qmechanic Jun 12 '17 at 11:03