# Why are only logarithmic divergence relevant for the Callan-Symanzik equation? Intuitive understanding?

I may be wrong, but it seems that only logarithmic divergences need to be retained when using the Callan-Symanzik equation, finding running couplings, etc. Why is this the case? Is there some simple intuitive understanding for why the logarithmic divergences are most important for these applications?

Someone should certainly expand on my answer but heres the basic idea:

The Callan Symanzik equation basically tells us how the n-point functions evolve as a function of the energy scale we are looking at, so the divergences we look at should be sensitive to every length scale. Now lets look at the three kinds of divergences/convergences possible:

1) Convergent: these graphs are only sensitive to long wavelength, low energy physics.

2) Power Divergent: These graphs are only sensitive to very short wavelength physics. They're sensitive to the UV scale but nothing else.

3) Log Divergent: These are interesting because these kinds of graphs get contributions from all length scales (as is evident because we have a vanishing degree of divergence). So if we care about the evolution of the system at intermediate length scales (not in the far UV or far IR) only the logarithmic divergences will contribute. There is no scale associated with these divergences.

Of course, you can't just neglect power divergent terms. However, we have a regulator (dimensional regularization, DR) that automatically eliminates power divergent terms. To the extent that all consistent regularization schemes are equivalent, we would expect that nothing new can be learned by including power divergent terms. There are, however, cases where DR, while not wrong, obscures the physics or leads to apparently poorly converging expansions. See http://arxiv.org/abs/nucl-th/9802075 for an example in which the authors modify DR to retain power divergences (PDS''), and solve the associated RG equations.
• An interesting context for this sort of issue is the question: Does Newton's constant run? Does gravity modify the running of $\alpha$? See arxiv.org/abs/arXiv:1209.3511 for a nice discussion. Commented Oct 22, 2013 at 13:42