# Einstein-Hilbert Lagrangian in linearized gravity

The Einstein-Hilbert Lagrangian is:

$$\mathcal{L}_{EH}=\sqrt{-g} R$$

where $$g={\rm Det}[g_{\mu\nu}]$$ and $$R$$ is the Ricci scalar. In linearized gravity $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$ and

$$\sqrt{-g}=1+\frac{h}{2}+\mathcal{O}(h^2)$$

and

$$R=\partial_\alpha(\partial_\mu h^{\mu\alpha}-\partial^\alpha h)+\mathcal{O}(h^2),$$

where $$h=h^\mu_{~\mu}$$.

Then

$$\mathcal{L}_{EH}=\partial_{\alpha}(\partial_\mu h^{\mu\alpha}-\partial^\alpha h)+\mathcal{O}(h^2)$$.

I am comparing this to this paper: https://arxiv.org/abs/hep-th/9411092 . Apparently they don't get an $$\mathcal{O}(h)$$ term (cf. their eqs. (2.15)-(2.18)). The first non-vanishing term they get is $$\mathcal{O}(h^2)$$. They say in the paper that they use the De Donder gauge which is $$\partial_\mu h^{\mu\alpha}-\frac{1}{2}\partial^\alpha h=0$$. This is very similar to the lowerst order term, EXCEPT the factor of $$\frac{1}{2}$$.

I am pretty sure I did the expansion of the Ricci scalar correct, since I find the same result in Carroll's book. I checked that the De Donder gauge condition usually has the factor $$\frac{1}{2}$$... So I really don't see why the first order term in the Einstein-Hilbert Lagrangian should vanish?

• You need to expand the Ricci scalar to second order to derive the quadratic Einstein-Hilbert action, since $\sqrt{-g}$ starts at order $0$ in $h$. As a warning this calculation is a mess. It is faster to derive the linearized Einstein equations. Then to derive the action, start with the most general Lorentz invariant action quadratic in $h$ and with $2$ derivatives (you should have 4 terms with 4 free coefficients after combining terms related by integration by parts), derive the equations of motion, and fix the coefficients in the action by matching against the linearized Einstein equations. Aug 22, 2022 at 17:31
• Thanks for the answer. My question is not "How can I derive the second order terms?" but more like "Why does the first order term vanish?"
– user255856
Aug 22, 2022 at 18:57
• The first order term only vanishes if the cosmological constant is zero. Generically the first order term when you expand the action around a background should vanish, if the background satisfies the Euler-Lagrange equations. Aug 22, 2022 at 19:16
• I think in my case it's okay to assume that the cosmological constant is zero. How can I see that the first order term vanishes? The Euler-Lagrange equations are $\frac{\partial\mathcal{L}}{\partial h_{\gamma \delta}}-\partial_\alpha \frac{\partial\mathcal{L}}{\partial\partial_\alpha h_{\gamma\delta}}=0$. But both terms that appear in the Euler-Lagrange equations are zero?
– user255856
Aug 22, 2022 at 19:45
• This has nothing to do with gravity. Suppose we have a generic field $\phi$ that we split into a background $\Phi$ and perturbation $\varphi$ via $\phi=\Phi+\varphi$. Then imagine expanding the action in powers of $\varphi$: $\delta S = \delta S^{(1)}[\Phi] \varphi + \delta S^{(2)}[\Phi] \varphi^2 + \cdots$. Now in the first term, $\delta S^{(1)}[\Phi]$ is the Euler-Lagrange equation, applied to $\Phi$. The notation might be unfamiliar, but just think about how we got to that point -- we wrote $\phi=\Phi+\varphi$ and expanded the action to first order in the variation $\varphi$. (...) Aug 22, 2022 at 21:47