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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?

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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.

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  • $\begingroup$ Do you have a physical intuition though for why one might expect that you learn nothing new by including power divergent terms? The answer given by David Meltzer above is reasonable to me, but I'm wondering if there are other viewpoints on the matter. Thanks for the response. $\endgroup$ – user26866 Oct 21 '13 at 19:38
  • $\begingroup$ I don't think I understand David Meltzer's answer. I think there are two issues: 1) We tend to focus on renormalizable theories, so the divergences associated with physical parameter are logarithmic. 2) Ultimately, we don't care about the dependence of the coupling on some unphysical regularization scale. The important thing is that RG equations take that information and provide constraints on the momentum dependence of physical Green fcts. The interesting fact is that log divergences typically constrain higher order terms. $\endgroup$ – Thomas Oct 22 '13 at 13:27
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    $\begingroup$ 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. $\endgroup$ – Thomas Oct 22 '13 at 13:42
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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.

Edit (comment added to answer on suggestion of innisfree):
Power divergent terms simply give the relationship between bare masses (far UV) and renormalized masses, but they dont matter when we consider a flow between different scales. Convergent terms have no dependence on the cut off so if we change it, like we do in an RG flow, they won't effect the running of the coupling constants.

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  • $\begingroup$ So is it only an approximation that we ignore power divergences when using the Callan-Symanzik equation? $\endgroup$ – user26866 Oct 20 '13 at 21:19
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    $\begingroup$ No I do not believe its an approximation. Power divergent terms simply give the relationship between bare masses and renormalized masses, but they dont matter when we consider a flow between different points. Convergent terms have no dependence on the cut off so if we change it, like we do in an RG flow, they won't effect the running of the coupling constants. $\endgroup$ – David M Oct 20 '13 at 22:53
  • $\begingroup$ I think the above comment should be incorporated into your answer. $\endgroup$ – innisfree Oct 23 '13 at 9:00

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