Nonrenormalizability of GR

Main Question or Discussion Point

Can someone please explain to me why general relativity is not renormalizable? Because from how I understand it, the lagrangian density is just R/G, where R is some complex function that contains 2,3, and 4 point terms of the metric g (which is the quantized field). Since R has mass dimension of 2, and g has mass dimension of 0, this lagrangian is superficially renormalizable. Why is it not?

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Bill_K
The Lagrangian contains not just 2,3 and 4 point terms but terms to all orders.

No I realize what the definition of nonrenormalizability is. In renormalizing it, you need to add counter terms of all orders to the lagrangian. However, the BARE lagrangian only has 2nd, 3rd, and 4th orders and as far as I can tell IS renormalizable.

fzero
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No I realize what the definition of nonrenormalizability is. In renormalizing it, you need to add counter terms of all orders to the lagrangian. However, the BARE lagrangian only has 2nd, 3rd, and 4th orders and as far as I can tell IS renormalizable.
The simple argument for why Einstein gravity is nonrenormalizable is the following. The coupling constant $$\sim \sqrt{G}$$ has units of length. Divergences will therefore be proportional to terms with more derivatives than appear in the EH term. So the counterterms that would have to be added to the EH action are genuinely new terms. As we add counterterms, we continue to encounter new divergences proportional to terms which are not in the renormalized action. Therefore we must continue to add new counterterms ad infinitum. Therefore EH gravity is nonrenormalizable.

tom.stoer
Therefore EH gravity is nonrenormalizable.
It is not renormalizable perturbatively using an expansion in G~0. This would correspond to a Gaussian fixed point in the renromalization group flow and we all agree that this does not exist. But there are indications that the theory may be non-perturbatively renormalizable with a non-Gaussian fixed point G>0 which means that a certain truncation of the Einstein-Hilbert action using only a finite number of terms IS consistent in that sense. This is the so-called asymptotic safety program.

fzero
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It is not renormalizable perturbatively using an expansion in G~0. This would correspond to a Gaussian fixed point in the renromalization group flow and we all agree that this does not exist. But there are indications that the theory may be non-perturbatively renormalizable with a non-Gaussian fixed point G>0 which means that a certain truncation of the Einstein-Hilbert action using only a finite number of terms IS consistent in that sense. This is the so-called asymptotic safety program.
As I well know, but it's obvious that adding these details won't help the OP understand more fundamental issues and will probably serve to confuse things further.

tom.stoer
As I well know, but it's obvious that adding these details won't help the OP understand more fundamental issues and will probably serve to confuse things further.
The point is that strictly speaking that not renormalizable could be simply wrong. That's why it makes sense to point out that not perturbatively renormalizable but non-perturbatively renormalizable makes more sense. Many people confuse renormalization with perturbation theory, and one should try to correct such misinterpretations as soon as possible. That's the reason for my post, even though it seems to complicate the matter slightly.

By the way, here is a fairly simple argument that shows what's going on: think about the function f(x)=1/(1-x) and its Taylor expansion 1+x+x²+x³+... One could argue that the function f(x) does not exist for x>1 simply because the Taylor expansion is not defined for x>1. But of course that's wrong. f(x) does exist for x>1, and all what one has to do is to use a different Taylor expansion (or no Taylor expansion at all).

In the same sense quantum gravity as a "straighforward" quantization of the EH action may exist in terms of a different expansion for some G - but certainly not for a perturbation expansion using G=0.

Bill_K
No I realize what the definition of nonrenormalizability is. In renormalizing it, you need to add counter terms of all orders to the lagrangian. However, the BARE lagrangian only has 2nd, 3rd, and 4th orders and as far as I can tell IS renormalizable.
I'm talking about the bare Lagrangian. The Ricci scalar R contains gμν, which is the inverse matrix of gμν. When you do the perturbation expansion gμν = ημν + hμν, the expansion of gμν written out is an infinite series, and so contains all powers of hμν. You must include Feynman diagrams in which a vertex has any number of legs.

thanks for the answers, but what I'm confused about about is where that coupling constant actually appears. The free-space GR lagrangian is just R/G right? that means that the self-coupling constant (the 3 and 4 point g terms in R) is 1/G and has units of energy squared right?

I'm talking about the bare Lagrangian. The Ricci scalar R contains gμν, which is the inverse matrix of gμν. When you do the perturbation expansion gμν = ημν + hμν, the expansion of gμν written out is an infinite series, and so contains all powers of hμν. You must include Feynman diagrams in which a vertex has any number of legs.
*edit* sorry, I get what your saying, that makes sense. I did notice that the Ricci scala had a gμν and wasn't quite sure how to handle it. This is different than simply being nonrenormalizable though... That would mean GR is nonrenormalizable for two separate reasons??

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fzero
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thanks for the answers, but what I'm confused about about is where that coupling constant actually appears. The free-space GR lagrangian is just R/G right? that means that the self-coupling constant (the 3 and 4 point g terms in R) is 1/G and has units of energy squared right?
You actually expand

$$g_{\mu\nu} = \eta_{\mu\nu} + \sqrt{G} h_{\mu\nu}$$

so that the kinetic term for h is the standard one

$$\sim (\partial h)^2,$$

i.e., with only a numerical, dimensionless factor. This will make the tree-level interaction coefficients depend on $$\sqrt{G}$$.

oo right the kinetic term shouldn't have a dimensional factor. Thanks, that explains it. I am still a bit confused about Bill_K's response though. It seems like in addition to being nonrenormalizable (treating gμν as gμν), there are an infinite number of terms in the bare lagrangian?

fzero
I believe that's correct. Each of the higher dimensional operators comes with a higher power of $$\sqrt{G}$$ though, so at any order of perturbation theory, there are only a finite number of terms one needs to consider. However, the problem still remains that the counterterms that are needed are of a different structure than appears in the expansion of the EH action. It also happens that at any given level of perturbation theory, one needs to include counterterms arising from new types of curvature invariants, so there's no way to resum the renormalized Lagrangian as arising from a finite number of curvature expressions.