Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have read in a few places that GR is renormalizable at one loop. (hep-th/9809169 for example, second sentence, although they don't seem to develop this point at all). Is this do to some hidden symmetry in the theory? Naively we need new counter terms at all orders, even one loop, in perturbation theory, right?

share|improve this question
    
Well, the analysis performed examines counterterms ... and the EH action requires an infinite number of counter-terms for two-loop corrections. See arXiv:0910.4110 for some details and references... –  Alex Nelson Jun 29 '12 at 5:43
    
Isn't GR divergent at the tree level if coupled to matter? –  Jerry Schirmer Aug 19 '12 at 18:59
    
DJBunk, could you un-accept Ron's answer so that he can delete it himself? –  dmckee Aug 19 '12 at 22:41
add comment

2 Answers

GR have nothing to do with renormalization. You mean: “one loop semiclasical quantum gravity”. The way this is worked out is well explained in the classical book by Birell & Davis. What is done is to consider only closed graviton loops. The metric tensor is decomposed in a classical part + small fluctuation part that is quantized only to first loop level. The Field equations of GR are solved with the Einstein tensor equated to the average value the quantum mass-energy tensor of some quantum fields that are present. This average mass-energy tensor has to be regularized some way because is formally infinite. The book B&D contains many worked examples of this one loop semiclassical quantum gravity. The paper is calling SQG Einstein's gravity, but GR is a well-tested theory while SQG is not. You are right in principle there must be counter terms to all orders. The problem is that GR can't be quantized to all orders because is fundamentally not renormalizable. This means that the theory actually truncates quantum GR to all orders after one loop, and renormalize it to only this one loop level.

share|improve this answer
2  
This is a collection of statements that have no relation to the question. One loop semiclassical quantum gravity is renormalizable with no matter, it is not renormalizable with arbitrary matter. The OP is asking why. –  Ron Maimon Jul 4 '12 at 7:01
2  
@Ron - my thoughts exactly. Thanks. –  DJBunk Jul 4 '12 at 13:09
    
I don't see the word "matter" or the word "why" in the op. –  Ernesto Ulloa Jul 4 '12 at 13:33
    
The proper question is then why is QGR renormalizable to one loop lv in the presence of arbitrary matter fields? –  Ernesto Ulloa Jul 4 '12 at 13:42
2  
Ron just expressed that is exactly what doesn't happen. –  DJBunk Jul 4 '12 at 13:50
add comment

The counterterms at one loop would be $R^2$ operators, because loops are counted by powers of $G_N = 1/M_P^2$. The tree-level Lagrangian is the Einstein Hilbert action $M_P^2 R$, so the one-loop counterterms for logarithmic divergences should be terms that carry no powers of $M_P$ in front. Simply from dimensional analysis, then, these are $R^2$ terms, of which there are three linearly independent choices: $R^2$, $R_{\mu\nu}R^{\mu\nu}$, and $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$.

Two combinations can be eliminated by field redefinitions (of the form $g_{\mu\nu} \to g_{\mu\nu} + c_1 R g_{\mu\nu} + c_2 R_{\mu\nu}$), and the third is a total derivative and so has no local physical effect. (Namely, it's the Euler density $R_{\mu\nu\rho\sigma}^2-4 R_{\mu\nu}^2+R^2$, also known as the Gauss-Bonnet term, whose integral is the Euler characteristic, a topological invariant). The upshot is that you have to go to $R^3$ terms before you get nontrivial counterterms, and these come from two-loop diagrams.

As far as I can find, the original source for the argument is this paper of 't Hooft and Veltman.

share|improve this answer
3  
Although this is the standard argument, I'm feeling a little unsatisfied with it at the moment, because it still allows a divergent renormalization of $M_P^2 R$. In dim reg such power divergences would be absent, but I'm not very happy with a regulator-dependent statement. I'm not sure what to think about this. –  Matt Reece Aug 18 '12 at 19:39
    
I'd like to know this argument--- where does it come from? Is it Veltman's? I was guessing. –  Ron Maimon Aug 19 '12 at 6:44
    
+1 This is the conventional argument, I already knew it. What is more (and I don't if this is well-known), one can also renormalize gravity at 2 loops trough non-local metric redefinitions. –  drake Aug 19 '12 at 18:06
1  
@RonMaimon: I learned the argument from hearing it repeatedly in talks about the possible finiteness of ${\cal N}=8$ supergravity, but it looks like the original source is the 't Hooft/Veltman paper I've now linked in the answer. (Maybe since 't Hooft seems to be hanging out here these days, he could clarify things...) –  Matt Reece Aug 19 '12 at 18:48
    
@drake: Please, could you explain how to do this? –  Ron Maimon Aug 20 '12 at 2:52
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.