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For the past few days I've been busy with Relativity. I've been reading through the book "Spacetime and geometry" by Sean Caroll and currently I'm at chapter 4. In this chapter they make the following calculation in equation (4.42)

\begin{equation} \begin{split} R_{00} &= R^i_{0i0} \\ &= \partial_i[\frac{1}{2}g^{i\lambda}(\partial_0g_{\lambda 0} + \partial_0g_{0\lambda} - \partial_{\lambda}g_{00})] \\ &= -\frac{1}{2}\delta^{ij}\partial_i \partial_j h_{00} \\ &= \text{...}, \end{split}\tag{4.42} \end{equation}

where we also have assumed $g_{00} = -1 + h_{00}$ since in the Newtonian limit we assume the gravitational field to be weak and therefore can treat it as a perturbation of Minkowski space. I'm strugling how we go from second to the third line and thus where the Kronecker delta comes from. I know that in the second line the partial derivatives $\partial_0$ ought to vanish since we assume the Newtonian limit for which the fields are static, however I don't get how the third term in this second line is computed. Any help with this would be greatly appreciated :))

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We typically take the Newtonian metric to be of the form $$ ds^2 = g_{\mu\nu}dx^\mu dx^\nu = -(1+2\phi)dt^2 + (1-2\phi)\delta_{ij}dx^idx^j, $$ where $\phi$ is the Newtonian potential. In your notation, this means $h_{00} = 2\phi$ and $h_{ij} = 2\phi\delta_{ij}$ and $h_{0i} = g_{0i} = 0$.

This means that on your second line, $\lambda = 0$ sets the entire right hand side to zero, while $\lambda = j$ gives

\begin{aligned} R_{00} &= -\partial_i\left[\frac12 g^{ij}\partial_j g_{00}\right]\\ &= -\partial_i\left[\frac12 (\delta^{ij} - h^{ij})\partial_j h_{00}\right]\\ &= -\frac12 \delta^{ij}\partial_i\partial_j h_{00} \end{aligned}

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  • $\begingroup$ oh ok i see, thank you very much for your time! $\endgroup$
    – luki luk
    Nov 9, 2023 at 20:29

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