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EFE's question regarding Ricci scalar

  1. Jan 1, 2012 #1
    Quick question about the EFE's. When writing the einstein tensor [itex]G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}[/itex], and using the definition of the Ricci scalar [itex]R=g^{\mu\nu}R_{\mu\nu}[/itex], how does this not give you problems when you expand out [itex]R[/itex] so that the second term becomes [itex]-\frac{1}{2}g^{\mu\nu}R_{\mu\nu}g_{\mu\nu}=-2R_{\mu\nu}[/itex] when evaluating the trace, giving you the EFE's as [itex]R_{\mu\nu}=-4πGT_{\mu\nu}[/itex]?
    Any help would be appreciated.
  2. jcsd
  3. Jan 1, 2012 #2

    George Jones

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    In any one term, an index can only appear at most twice, so [itex]-\frac{1}{2}g^{\mu\nu}R_{\mu\nu}g_{\mu\nu}[/itex] is not legal. Maybe you want to write [itex]R=g^{\alpha\beta}R_{\alpha\beta}[/itex], or maybe you want do do something like
    \left( R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} \right) g^{\mu \nu} &= 8\pi T_{\mu\nu} g^{\mu \nu}\\
    -R = 8\pi T^\mu_\mu.
    Using [itex]-R = 8\pi T^\alpha_\alpha[/itex] (after relabeling to avoid the same symbol being used as both a free index and a summed index) in [itex]R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}[/itex] gives [itex]R_{\mu\nu} = 8\pi \left( T_{\mu \nu} - \frac{1}{2}T^\alpha_\alpha g_{\mu\nu} \right)[/itex], another useful form of the EFE.
  4. Jan 1, 2012 #3
    Thanks, I always thought there was some index trickery involved in resolving this, but I never knew about multiple repeated indices being disallowed.
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