# Running couplings

1. Aug 14, 2008

### ledamage

Hi there!

I'm struggling a bit with running couplings. Srednicki introduces dimensional regularization and the $\overline{\mathrm{MS}}$ scheme, then calculates a squared transition amplitude for some reaction in $\varphi^3$ theory. Eventually, he calculates the beta function for the coupling and solves the renormalization group equation. Clear so far.

In the transition amplitude, a factor $\ln(s/\mu^2)$ occurs, where $s$ is a Mandelstam variable and $\mu$ is the factor arising from dimensional regularization. Now he says, to avoid large logarithms, we should put $\mu^2 \sim s$ (which we can do since physics must be independent of $\mu$). Then, according to the beta function, the coupling runs with the involved momenta, the well-known behavior.

Now, $\mu^2 \sim s$ certainly is a convenient choice, but what if I choose $\mu$ to be constant (or something else), which I am free to do? Then, the coupling doesn't run at all. It seems to me that the running coupling is just a way of viewing things (here in particular by using the $\overline{\mathrm{MS}}$ scheme and $\mu^2 \sim s$)?

Another quick question: I read everywhere that renormalization is related to the behavior of the theory at small distances/large momenta. How are small distances and large momenta related? And what has renormalization to do with it?

2. Aug 14, 2008

### Avodyne

Correct. If you use a fixed $\mu$, then the coupling doesn't run. The expansion parameter governing perturbation theory then appears to be $\alpha(\mu)\ln(s/\mu^2)$, and if this is large, perturbation theory breaks down. The advantage of using the running coupling is that we see that the expansion parameter is really $\alpha(s^{1/2})$, which might be small even if $\alpha(\mu)\ln(s/\mu^2)$ is not. Indeed, this is the case at high energies in QCD.

Fourier transform.

With small distances = large momenta, nothing. As for the connection between behavior at small distances and renormalization, see section 29 of Srednicki.