Feynman rules are the basic tool to compute amplitudes in perturbation theory for a QFT. Here, I am trying to understand perturbation theory in GR around the flat space metric, in terms of Feynman rules. There are two basic questions one can ask here :

  1. What is the graviton propagator?

  2. What is the off-shell 3 point function for the GR vertex?

DeWitt has a collection of papers that contain this but the expression for the vertex is slightly obscure and very prone to errors when we expand the symmetrized terms by hand. Hence, can someone please write down the full off-shell 3 point vertex explicitly.

Also, he works in the de-Donder gauge, in which the 3 point vertex is sufficiently lengthy. Is there a gauge choice in which the Feynman rules for GR is simpler and less tedious. Why is the de Donder gauge choice so much more popular in the literature?


1 Answer 1


I'm working on gravity's three-point vertex recently. I might be wrong but here is my understanding.

  1. If you expand the Einstein-Hilbert action around flat metric to second order, you might get the action at the order of two gravitons. Adding the gauge-fixing term, you can get the whole two-point function. From this, you can derive the graviton's propagator. Just like the indices on the photon's propagator, the graviton's propagator has two tensor indices.

  2. Expanding the Einstein-Hilbert action to third order (see Berends in 1975 "On the high-energy behavior of Born cross sections in quantum gravity" Eq 7), replacing the graviton $h_{\mu\nu}$ with indices (see appendix in Goldberger, hep-th/0409156v2), you can get the 3-point vertex. You may want to use MMA permutation to help you go through all the permutations (6 in this case).

As to why de Donder gauge is so popular, I guess it might because this gauge is symmetric on $\mu\nu$, $\alpha\beta$ indices, namely, the in and out gravitons' indices. This gauge also guarantees the traceless fact of the graviton. I think there should be other gauge choices similar to this just like the $\xi$ gauge in QED. Hope this could help you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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