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I understand that quantum gravity is nonrenormalizable because there are an infinite number of counterterms. In QED the counterterms correspond to the loop corrections to the vertex function as well as the electron and photon self-energy. This corresponds to corrections between the bare vs measured mass and charge. What is the analog of this in quantum gravity? Why isn't there just a gravition self-energy correction, etc? What do each of the infinite counterterms correspond to? Are there diagrams corresponding to each one?

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Come on guys, I want to see a nice answer here too :-) – Dilaton Feb 14 '12 at 13:29

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The graviton counterterms correspond to the fundamental particle gravitational multipole moments, the precise form of the mass distribution in the viscinity of the particle. If you have an electron, the requirement of spherical symmetry doesn't restrict many of the multipole moments of the gravitational field you produce--- you can make an arbitrary mass (zero order moment), dipole (along the direction of spin), quadrupole (corresponding to an extended mass distribution), etc.

The reason is that the fundamental point solutions of gravity are black holes, which are not points at all, but extended. So that the consistent treatment of a gravitational theory requires a specification of all the multipole moments of the "point" sources, making them effectively extended. This is one way of understanding why string theory is necessary--- you need to find an extended system which describes all the gravitational excitations of the elementary particles, and it is required that this consistently connects with the excitation spectrum of a classical black hole for large masses.

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Awesome, this really answers the 'measurements' part of my question! I take it this is related to the fact indicated by Josh that the Feynman calculus for the theory involves n point graviton vertices (unlike gluons in QCD which have only 3 and 4 point vertices)? (now I'm just trying to pin down the 'feynman diagram' side to my question) – user1247 Feb 16 '12 at 9:09
Note: I had a hard time deciding who to award the bounty to here, but in the end I'm choosing your answer, because I assume your explanation is the physical reason why there are an infinite number of n point tree vertices in QG rather than, say, QCD. – user1247 Feb 19 '12 at 8:54

I assume you're talking about perturbative quantum gravity, i.e. trying to give a QFT treatment to the Einstein-Hilbert action by considering fluctuations around, say, a Minkowski background. Then, write out $g_{\mu\nu} = \eta_{\mu\nu} + \kappa\,h_{\mu\nu}$, where $\kappa \propto G_N^{1/2} \sim M_{\rm Pl}^{-1/2}$, for $G_N$ the Newton gravitational constant and $M_{\rm Pl}$ the Planck mass, and $h_{\mu\nu}$ is the metric fluctuation (the graviton). Then, your expansion of the Einstein-Hilbert action generates an infinity of interactions. There is a two graviton term, which is just the kinetic term, and then $n$-point interactions for $n = 3, 4, \ldots$, with a coupling proportional to $\kappa^{n-2} \sim M_{\rm Pl}^{-(n-2)/2}$. Because the coupling $\kappa$ in dimensionful, this generates power dependence (not just logarithmic) on energy in the amplitudes. For example, the two-to-two scattering amplitude scales like $\kappa^2$ and will thus have energy dependence $E/M_{\rm Pl}$ at tree level. Clearly this is irrelevant at particle energy scales, but the point is it grows rapidly with energy. Moreover, the one-loop correction to this term will scale like $(E/M_{\rm Pl})^2$.

So say you fix your scale and want to simply measure these renormalized coefficients using a presumably very sensitive device. Then, you would have to make a measurement for every $n$-point interaction in your expansion of the Einstein-Hilbert action. If you've never seen it written out, take a look at the quantum gravity chapter in Scadron's book "Advanced Quantum Mechanics" to see this process done. There are no diagrams in it, but you can also read the notes from 't Hooft's Erice lectures on perturbative quantum gravity for some of the issues implicit.

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Thanks for your answer. I read the 't Hooft lecture I think you are referring to. After eq2.11 he says: "Again, having just one extra adjustable parameter seems to be not so bad a price to pay for a theory with one-loop accuracy." He then makes the leap to "Thus, perturbative gravity generates new infinities at higher orders, requiring counter terms [...] at each order in G." He is referring to loop-order rather than N-point vertex (otherwise his statement about one-loop would be wrong)? Is this equivalent to what you wrote (a counter term for each n-point vertex)? Sorry if this is obvious. – user1247 Feb 15 '12 at 11:29
I'm glad you took a look. So, the comment you mention comes right after 't Hooft argues for the one-loop finiteness of on shell amplitudes in pure Einstein-Hilbert gravity. Then, he adds a massless scalar field minimally coupled to gravity and shows that at one loop level you develop the need for only a single counterterm. You would need to expand R^2 around Minkowski space ($g_{\mu\nu} = \eta_{\mu\nu} + \kappa\,h_{\mu\nu}$) as before to see the story I described above play out. Then you can your divergence of $n$ point graviton vertices due to the "phion" loop. – josh Feb 16 '12 at 5:51
"I assume you're talking about perturbative quantum gravity, i.e. trying to give a QFT treatment to the Einstein-Hilbert action by considering fluctuations around, say, a Minkowski background." BTW, is there a keyword I can use to refer to this (or when doing searches in the literature)? I want to make a follow-up question and use the most effective vocabulary. Maybe 'Canonical quantum gravity'? – user1247 Feb 16 '12 at 9:31
I think perturbative quantum gravity is the term. It need not be a canonical/Hamiltonian treatment. – josh Feb 16 '12 at 15:06
Actually my question is going to be about non-perturbative quantum gravity, but I want to make sure the reader understands I'm talking about QFT + E-H action rather than LQG or something. In other words the original/naive straightforward application of QFT to the E-H action in the 1960s or so that ended up discouraged due to nonrenormalizability. Do you know what I mean? I've also seen the term 'Quantum Einstein Gravity' used, but I'm not sure if that is supposed to be synonymous with 'Canonical Quantum Gravity'. – user1247 Feb 16 '12 at 15:33

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