Having the interaction lagrangian for scattering between two scalars and the graviton $h$

\begin{equation} \mathcal{L}_{\phi \phi h}=\frac{\kappa}{2}\left(-\frac{1}{2} h_{\mu}^{\mu} \phi^{2} m^{2}+\frac{1}{2} h_{\mu}^{\mu} \partial_{\nu} \phi \partial^{\nu} \phi-h^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi\right). \end{equation}

I don't see how to get the following Feynman rule for the vertex, it doesn't seem straightforward to me \begin{equation} V_{\alpha \beta}^{\{\phi \phi h\}}\left(p_{1}, p_{2}, m\right)=i \frac{\kappa}{2}\left[\left(p_{1 \alpha} p_{2 \beta}+p_{2 \alpha} p_{1 \beta}\right)-\eta_{\alpha \beta}\left(p_{1} \cdot p_{2}-m^{2}\right)\right]. \end{equation}

Here the vertex:


1 Answer 1


In this case, there is a fairly quick way to see the answer, however in general you should not expect vertices in gravity to be something you can just immediately see. Usually there are a huge number of indices (6 for a 3 graviton vertex) with lots of integrations by parts to keep track of. Nevertheless, conceptually it should always be clear how to derive the Feynman rules, at least in principle, using standard methods. For example, you can start from the path integral, and take appropriate functional derivatives.

Anyway in your case, note that the interaction Lagrangian can be written as \begin{equation} \mathcal{L}_{h\phi\phi} = \frac{\kappa}{2} h^{\mu\nu} \left[- \left( \partial_\mu \phi \partial_\nu \phi + \partial_\nu \phi \partial_\mu \phi \right) + \eta_{\mu\nu} \left( \partial_\alpha \phi \partial ^\alpha \phi - m^2 \phi^2 \right) \right] \end{equation}

The motivation for symmetrizing the term $\partial_\mu \phi \partial_\nu \phi=\frac{1}{2}(\partial_\mu \phi \partial_\nu \phi + \partial_\nu \phi \partial_\mu \phi)$ is because this term is contracted with $h_{\mu\nu}$, which is symmetric; this is a useful trick for speeding up the calculation and making the symmetry more manifest.

In this form, hopefully it is at least plausible that if you work out the functional derivatives, the net result will be that the derivatives of the scalars become momenta, $\partial_\mu \phi \rightarrow i p_\mu$. The vertex has an overall $-i$ factor because of standard conventions for the normalization of the vertex.

One thing I don't understand is why there is a relative minus sign between the $p_{1\alpha} p_{2\beta}$ term and the $\eta_{\alpha \beta} p_1\cdot p_2$ term in the vertex... without working this out I don't know if this is because of a subtlety in the calculation or an issue with conventions. If your Lagrangian is written in a different metric convention than the source you got the vertex from, it would explain the minus sign, but I don't know if this is the right explanation.

It is definitely a good exercise to actually work out the functional derivatives and I strongly encourage you to do this, rather than take my handwavy justification at face value. But hopefully it's clear that the Lagrangian (in this case) can be written in a way where the index structure falls out fairly easily.

Lastly, I'll just mention in passing that in the past several years there has been enormous development in the general area of computing gravity scattering amplitudes more efficiently, without passing through Feynman diagrams. Here is a semi-random resource: https://www.cambridge.org/us/academic/subjects/physics/theoretical-physics-and-mathematical-physics/scattering-amplitudes-gauge-theory-and-gravity?format=HB


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