EFE's question regarding Ricci scalar

1. Jan 1, 2012

unchained1978

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

2. Jan 1, 2012

George Jones

Staff Emeritus
In any one term, an index can only appear at most twice, so $-\frac{1}{2}g^{\mu\nu}R_{\mu\nu}g_{\mu\nu}$ is not legal. Maybe you want to write $R=g^{\alpha\beta}R_{\alpha\beta}$, or maybe you want do do something like
\begin{align} \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. \end{align}
Using $-R = 8\pi T^\alpha_\alpha$ (after relabeling to avoid the same symbol being used as both a free index and a summed index) in $R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$ gives $R_{\mu\nu} = 8\pi \left( T_{\mu \nu} - \frac{1}{2}T^\alpha_\alpha g_{\mu\nu} \right)$, another useful form of the EFE.

3. Jan 1, 2012

unchained1978

Thanks, I always thought there was some index trickery involved in resolving this, but I never knew about multiple repeated indices being disallowed.