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I am stuck in the derivation of $G$ (Einstein tensor) in the condition of weak field ($h$ small) where $g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}$ and $h^{\alpha\beta}=\bar{h}^{\alpha\beta}-\frac{1}{2}\eta^{\alpha\beta}\bar{h}.$ I calculated the Ricci tensor:

$$R_{\alpha\beta}=\frac{1}{2}\left(\bar{h}{^{\mu}}_{\beta,\alpha\mu}+\bar{h}{_{\alpha\mu,}}{^\mu}_{\beta}-\bar{h}{_{\alpha\beta,}}{^\mu}_{\mu}+\frac{1}{2}\eta_{\alpha\beta}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu\right)$$

and I now should calculate the Ricci scalar to get $G_{\alpha\beta} = R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} R$.

So

$$R = g^{\alpha \beta}R_{\alpha \beta} = \frac{1}{2}\left(\bar{h}{^{\mu\nu}}_{,\nu\mu}+\bar{h}{_{\nu\mu,}}{^{\mu\nu}}_{}-\bar{h}{_{\nu\,}}{^{\nu}},^{\mu}_{\mu}+\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu\right).$$

I know the result should be $R = \bar{h}_{\mu\nu,}{^{\mu\nu}}+\frac{1}{2}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu$ but I really can't figure out how to do the calculus in the right way. (I'm very new to tensor calculus.) I know I did some errors in the contraction of $\alpha$ and $\beta$ but I don't undertand why.

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    $\begingroup$ Did you assume that $\eta^{\alpha\beta}\eta_{\alpha\beta}$ is$1$? It isn’t. $\endgroup$
    – G. Smith
    Oct 21, 2020 at 17:53
  • $\begingroup$ You have an incorrect factor of 1/2. The Ricci scalar is just the contraction of the Ricci tensor, not 1/2 of it. $\endgroup$
    – G. Smith
    Oct 21, 2020 at 17:55
  • $\begingroup$ @G.Smith 1/2 is there , but I don't understand how to work out the last formula for $R$. Did I do any mistake in the contraction of $\alpha$ and $\beta$? $\endgroup$
    – G.To
    Oct 21, 2020 at 17:55
  • $\begingroup$ Can someone find the error? Check-my-work questions are off-topic on this site. $\endgroup$
    – G. Smith
    Oct 21, 2020 at 17:59
  • $\begingroup$ Did I do any mistake in the contraction of 𝛼 and 𝛽? Yes, at least one, and I pointed it out in my first comment. $\endgroup$
    – G. Smith
    Oct 21, 2020 at 18:00

1 Answer 1

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Note that $\partial_\mu\partial_\nu A^{\mu\nu}=\eta_{\alpha\mu}\eta_{\beta\nu}\partial^\alpha\partial^\beta A^{\mu\nu}=\partial^\mu\partial^\nu A_{\mu\nu}$, upon renaming indexes on the last step. By similar reasoning, $A^\mu_{\;\,\mu}=A_{\mu}^{\;\,\mu}$ Also, note that $\eta^{\mu\nu}\eta_{\mu\nu}=\delta^\mu_{\mu}=4$, so the last term should be $2\partial^\mu\partial_\mu h^\nu_{\;\,\nu}$. Correcting and using the identities, we have

$$ \begin{align} R &= \frac{1}{2}\left(\partial^\mu\partial^\nu\bar{h}_{\mu\nu}+\partial^\mu\partial^\nu\bar{h}_{\mu\nu}-\partial^\mu\partial_\mu\bar{h}^\nu_{\;\,\nu}+2\partial^\mu\partial_\mu \bar{h}^\nu_{\;\,\nu}\right) \\ &=\partial^\mu\partial^\nu\bar{h}_{\mu\nu}+\frac{1}{2}\partial^\mu\partial_\mu \bar{h}^\nu_{\;\,\nu} \end{align} $$ as expected.

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