# 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?

• Note to readers: when the background metric is a general vacuum solution of Einstein's equations there is an additional term of the form $R_{\alpha\beta\gamma\delta}h^{\alpha\gamma}h^{\beta\delta}$ where $R_{\alpha\beta\gamma\delta}$ is the Riemann curvature of the background metric. – Bruno Le Floch Feb 24 '17 at 19:36

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
• You write the full metric as $g_{mn}=\bar{g}_{mn}+\xi_{mn}$, and proceed to expand your geometric objects to quadratic order. But if you are expanding to quadratic order, shouldn't you write $g_{mn}=\bar{g}_{mn}+\xi_{mn}+\xi_{m}^{p}\xi_{pn}$ (with contractions performed by background metric)? – NormalsNotFar Nov 15 '18 at 6:40
• I ask because I am also trying to expand geometric objects to quadratic order, but in the vielbein formalism instead. But to allow for the vielbein to be invertible I must include terms of second order in its expansion i.e. I must write $e_m{}^{a}=F_{a}{}^{b}\bar{e}_m{}^{a}$ and $e_a{}^{m}=F^{-1}{}_{a}{}^{b}\bar{e}_m{}^{a}$ where $$F_{a}{}^{b}=\delta_a{}^{b}+\xi_a{}^{b}+\frac{1}{2}\xi_{a}{}^{c}\xi_{c}{}^{b}~,~~~~~~F^{-1}{}_{a}{}^{b}=\delta_a{}^{b}-\xi_a{}^{b}+\frac{1}{2}\xi_{a}{}^{c}\xi_{c}{}^{b}~.$$ – NormalsNotFar Nov 15 '18 at 6:44
• If I translate this back to the metric formalism then I get $$g_{mn}=\bar{g}_{mn}+\xi_{mn}+\xi_{m}{}^{p}\xi_{pn}$$ up to some coefficients. – NormalsNotFar Nov 15 '18 at 6:48