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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?

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    $\begingroup$ 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. $\endgroup$
    – Andrew
    Commented Aug 22, 2022 at 17:31
  • $\begingroup$ 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?" $\endgroup$
    – user255856
    Commented Aug 22, 2022 at 18:57
  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Commented Aug 22, 2022 at 19:16
  • $\begingroup$ 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? $\endgroup$
    – user255856
    Commented Aug 22, 2022 at 19:45
  • $\begingroup$ 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 $\varphi$. (...) $\endgroup$
    – Andrew
    Commented Aug 22, 2022 at 21:47

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The linear order term is a total derivative. Thus it does not contribute to the Lagrangian.

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