# Second-order Lagrangian of Einstein-Hilbert action

I'm having trouble deriving the equation (44) of https://arxiv.org/abs/1710.08863 .

The question is how to get the second-order lagrangian of the Einstein-Hilbert action, i.e.

$$$$\mathcal{L}_{GR}^{(2)} = \dfrac{1}{4} [ \partial_\alpha h \partial^\alpha h + 2\partial_\alpha h_{\beta \gamma} \partial^\beta h^{\alpha \gamma} -2\partial^\alpha h \partial_\beta h^\beta_\alpha - \partial_\gamma h_{\alpha \beta} \partial^\gamma h^{\alpha \beta} ]\tag{44}$$$$

where the usual decomposition $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$ has been used. This expression is the same as Carroll (Eq. 7.9) except for an overall minus sign (maybe different metric signature). It is very annoying that this computation is nowhere to be found!

My attempt: we just need to compute the second-order Ricci scalar. Since I can't find it anywhere in the literature, I figured I'd calculate it in this way:

$$$$R=g^{\mu\nu}R_{\mu\nu} \rightarrow R^{(2)} = g^{\mu\nu (1)} R^{(1)}_{\mu\nu} + g^{\mu\nu (0)} R^{(2)}_{\mu\nu}$$$$

and then we can use eqs.(7.6) and (7.153) of Carroll's book for the expressions of $$R^{\mu\nu (1)}$$ and $$R^{\mu\nu (2)}$$, respectively. However, after some computations, I find something different from the above lagrangian $$\mathcal{L}$$. Is this procedure wrong ?

Edit&Answer: Taking into account the Avantgarde's suggestion (i.e. including also $$\sqrt{-g}$$) and using eqs. (7.6) and (7.153) of Carroll's book, I managed to get Eq. (44). The point to pay attention to is the sign of $$g^{\mu\nu (1)}$$. So the correct expression is

$$$$\mathcal{L}_{GR}^{(2)} = \dfrac{1}{2}h R^{(1)} - h^{\mu\nu} R_{\mu\nu}^{(1)} + \eta^{\mu\nu} R_{\mu\nu}^{(2)}$$$$

The result will be equal to (44) above if the metric signature is (-,+,+,+), and will have an overall minus sign for opposite signature (as the case of Carroll), since $$R$$ changes sign.

• The quantity $g^{\mu \nu (1)} R^{(1)}_{\mu \nu}$ will contain lots of quantities of the form $h (\partial \partial h)$. But if you integrate them by parts, they will become of the form $(\partial h)(\partial h)$. Did the results still not agree after you did that integration by parts? Commented Feb 28, 2023 at 0:47
• @MichaelSeifert Is it legal to integrate by parts in order to get the lagrangian ? Commented Feb 28, 2023 at 8:50
• An integration by parts doesn't change the equations of motion, so at the classical level the two Lagrangians are equivalent. Effectively this is because integration by parts allows you to write (action #1) = (action #2) + (boundary term). Since we typically assume that the fields are fixed on the boundary when taking variational derivatives, the two actions lead to the same Euler-Lagrange equations. Commented Feb 28, 2023 at 13:12

$$(\sqrt{-g} R)^{(2)} = \sqrt{-g}^{(0)} R^{(2)} + \sqrt{-g}^{(2)} R^{(0)} + \sqrt{-g}^{(1)} R^{(1)}$$
• Sorry but I'am having trouble computing the last term in your expansion above, i.e. $\sqrt{-g}^{(1)} R^{(1)}$. Online I found that $\sqrt{-g} = 1 + (1/2) * h$ at first order. So, when I multiply this quantity by $R^{(1)}$, one also obtain a term which is not second order in total, namely $1*R^{(1)}$ . This is very strange, because each term in the expansion should be of second order in $h$. Should I just take the piece (1/2) * h ?@Avantgarde Commented Mar 22, 2023 at 17:27