# 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.