# Linearizing Gravity to ${\cal O}(h^3)$

I've seen the action of linearized gravity in many places. We basically have

$${\cal L} ~\sim~ \frac{1}{G_N}\left( - \frac{1}{2}h^{\alpha\beta} \Box h_{\alpha\beta} + \frac{1}{4} h \Box h + {\cal O}(h^3)\right)$$

in the gauge where the trace-reversed field is divergenceless. I'm doing some field theory, on linearized gravity backgrounds by treating $h_{\mu\nu}$ as a massless spin-2 field. I can't seem to find the ${\cal O}(h^3)$ terms in the Lagrangian anywhere. I know how to evaluate it, but it looks nasty.

Are there any known references that just lists the next to leading order terms in the above Lagrangian?

-

Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

I think the earliest papers where this was written down were by DeWitt. But for a reference easily available via the arXiv look at hep-th/9411092. Eq (2.17) has the expansion you want and eq. (2.18) even has it to fourth order in $h$.

-
Thank you so much!! This was exactly what I was looking for. – Prahar Sep 13 '12 at 15:13

Most of the difficulties of these calculations comes from the fact that people write the Lagrangian directly in terms of the metric perturbation $h_{\mu\nu}$.

It is much simpler to write it in terms of the difference of connections tensor $F_{\mu\nu}{}^\beta$, i.e., $$(\nabla_\mu-\bar{\nabla}_\mu)A_\nu = \mathcal{F}_{\mu\nu}{}^\beta{}A_\beta,$$ where $\nabla_\mu$ is the connection compatible with the full metric ($g_{\mu\nu}$) and $\bar{\nabla}_\mu$ compatible with the background metric ($\bar{g}_{\mu\nu}$ in your case Minkowski). The curvature scalar is naturally quadratic in $\mathcal{F}_{\mu\nu}{}^\beta$, hence, the higher order terms comes from the expansion of $\sqrt{-g}$ and $\delta{}g^{\mu\nu}$ in terms of the metric difference $\xi_{\mu\nu} = g_{\mu\nu} - \bar{g}_{\mu\nu}$. Using this method one don't need to choose a gauge a priori.

In our paper http://arxiv.org/abs/1206.4374 we develop the Lagrangian and Hamiltonian up to second order using the methodology described above. Additionally, we write the action in term of the kinetic quantities defined in geodesic space-time foliation. Using the results of this paper it is easy to generalize (in the Minkowski background) the Lagrangian to higher orders.

-
very interesting! +1 – lurscher Sep 18 '12 at 18:39
I have never seen this before! Very interesting! I'll look into it. Thanks! +1 – Prahar Sep 23 '12 at 16:59