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  What is the logic of regarding perturbative renormalizability is not a fundamental requirement?

+ 3 like - 0 dislike
2329 views

I read a statement in Becker and Becker's String Theory and M-Theory page 2. After pointing out the non-renormalizablity of GR by the dimension of gravitational constant, it is said:

Some physicists believe that perturbative renormalizability is not a fundamental requirement and try to quantize pure general relativity despite its nonrenormalizability. Loop quantum gravity is an example of this approach. Whatever one thinks of the logic, it is fair to say that despite a considerable amount of eort such attempts have not yet been very fruitful.

I am curious, what is the logic behind "perturbative renormalizability is not a fundamental requirement"?

This post imported from StackExchange Physics at 2014-06-03 17:04 (UCT), posted by SE-user user26143
asked Jun 2, 2014 in Theoretical Physics by user26143 (405 points) [ no revision ]
retagged Jun 3, 2014
Presumably that the non-renormalisability is an artefact of the perturbation technique being used, and that a non-perturbative treatment will produce finite results.

This post imported from StackExchange Physics at 2014-06-03 17:04 (UCT), posted by SE-user John Rennie
They're probably referring to the fact that one can view a given theory (such as Einstein gravity) as a low-energy effective theory in which case renormalizability isn't so relevant: en.wikipedia.org/wiki/Effective_field_theory

This post imported from StackExchange Physics at 2014-06-03 17:04 (UCT), posted by SE-user joshphysics
Then the theory goes to high energy scale, the possible other irrelevant contributions will increase. The leading term starts lose the prediction power....

This post imported from StackExchange Physics at 2014-06-03 17:04 (UCT), posted by SE-user user26143
A specific case of John's comment - they may be referring to asymptotic safety, also know as non-perturbative renormalizability: en.wikipedia.org/wiki/Asymptotic_safety_in_quantum_gravity

This post imported from StackExchange Physics at 2014-06-03 17:04 (UCT), posted by SE-user DJBunk

1 Answer

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To do perturbation theory, one needs a fixed point (unperturbed theory). These are commonly free theories, for which the fixed points are called Gaussian points (because free theories are quadratic in fields, so that the path integral integrand resembles a Gaussian function). But this does not need to be the case necessarily. GR is not perturbative renormalizable around its Gaussian point, but it might be a  fixed non-gaussian point around which one can carry out a renormalizable perturbation theory, as suggested by the Asymptotic Safety (AS) scenario. Whereas the existence of such a nontrivial fixed point is in principle possible, the counting of degrees of freedom connected with the BH entropy makes this scenario unlikely, as no local field theory in 4 dim such as AS  can reproduce the BH entropy. The Gross-Neveu model is a typical example of a QFT with a non-trivial fixed point.

In addition, if one believes that QFT is just a framework to describe low-energy physics, existing a more fundamental framework such as strings, then one shouldn't necessarily be worried because the underlying framework should take care of the breakdown of the perturbative expansion in the low-energy QFT.

answered Jun 3, 2014 by drake (885 points) [ revision history ]
edited Jun 3, 2014 by drake

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