A Derivation of Einstein's equation

1. Jun 6, 2016

JonnyG

Going through Carroll's book, he is deriving Einstein's equation by looking at what it should reduce to in the Newtonian limit. Part of this process is in calculating $R_{00}$ (the $00$ component of the Ricci tensor). So he lets $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ where $h$ is some small perturbation. So \begin{align*} R_{00} &= R^i_{0i0} \\ &= \partial_i[\frac{1}{2}g^{i \lambda}(\partial_0 g_{\lambda 0} + \partial_0 g_{0 \lambda} - \partial_\lambda g_{00})] \\ &= \partial_i(-\frac{1}{2} g^{i \lambda} \partial_\lambda g_{0 0}) \\ &= -\frac{1}{2} \partial_i g^{i \lambda} \partial_\lambda g_{00} \\ &= -\frac{1}{2}\delta^{ij} \partial_i \partial_j g_{00} \end{align*}

I don't understand how he gets from the second last step to the very last step?

EDIT: Is this correct: $g^{i \lambda} \partial_\lambda = \partial_i$ and hence $\frac{-1}{2} \partial_i g^{i \lambda} \partial_\lambda g_{00} = \frac{-1}{2} \partial_i \partial_i g_{00} = \frac{-1}{2} \delta^{ij} \partial_i \partial_j g_{00}$ where $\delta^{ij}$ is the Kronecker delta?

Last edited: Jun 6, 2016
2. Jun 6, 2016

Orodruin

Staff Emeritus
Keep only terms which are linear in h and use the fact that $\eta$ is constant.

3. Jun 6, 2016

JonnyG

Thank you. The calculation worked out.