# Asymptotic freedom

1. Jun 25, 2009

### Bobhawke

In every text book Ive seen, in order to show that QCD is asymptotically free the first coefficient of the beta function is calculated and shown to be negative. However, neglecting the terms that are higher order in the coupling is only allowed if we know the coupling is small, and we dont know when the coupling is small until we've calculated the beta function! This reasoning seems circular to me. Any explanation of this would be appreciated.

2. Jun 29, 2009

### Avodyne

The first term is sufficient if the coupling g is very small. Then we know it gets smaller and smaller as we go to higher and higher energies.

There is no evidence from higher-order corrections (or anything else) that the beta function ever turns around and becomes positive, but it is logically possible. Suppose this happened at g=g*. Then the physics depends on whether g>g* or g<g*. If g<g*, the coupling goes to zero at high energies (asymptotic freedom), but at low energies gets stuck at g=g*. It seems to me this would lead to unconfined quarks, but I'm not sure if this is known to be the case. If g>g*, then the coupling would become arbitrarily large at low energies, but stop at g=g* at high energies. This would seem to be inconsistent with the small measured values of g at high energies; I think these values are within the range where we are confident in the calculation of the beta function.

So, putting all the evidence together, there is a pretty strong case for a beta function that is always negative.