# How does the renormalization scale $\mu$ cancel in all finite observables?

In dimensional regularization, we must shift the dimensionless coupling $g$ by the renormalization scale $\mu$ (which has unit mass dimension): $$g \rightarrow \mu^{4-d} g \tag{1}$$ This is done to ensure that the action will have the correct dimensions.

Apparently, $\mu$ will not affect any observables. But I don't understand how this can be proven.

Edit: I am aware that we do not want to "true" theories to depends on the cut-off $\Lambda \rightarrow \infty$ or $\epsilon=4-d \rightarrow 0$, but this doesn't tell me what will happen with $\mu$.

Edit 2: Ok, maybe I understand it. We introduce the renormalization scale $\mu$ and this is an unphysical parameter. Therefore, we demand that that physical observable are not dependent on $\mu$. This procedure results in the renormalization group equations. Is this correct?

• your second edit is correct, it is made independant by construction. If it is not the case, then the theory is ill-defined. – bendaizer Mar 5 '14 at 10:27