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A Derivation of Einstein's equation

  1. Jun 6, 2016 #1
    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. jcsd
  3. Jun 6, 2016 #2

    Orodruin

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    Keep only terms which are linear in h and use the fact that ##\eta## is constant.
     
  4. Jun 6, 2016 #3
    Thank you. The calculation worked out.
     
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