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In my lecture notes, there is the following consequence of Einstein's equations, (it follows on to have multiple consequences, so I don't think there's a mistake in the notes):

$$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi GT_{\mu\nu},$$

$$\implies -R = 8\pi G g^{\mu\nu}T_{\mu\nu} = 8\pi GT.$$

However I can't see why this is the case, my working is as follows.

Take the trace of Einstein's equation through multiplication by the inverse metric:

$$R_{\mu\nu}g^{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}g^{\mu\nu} = 8\pi GT_{\mu\nu}g^{\mu\nu},$$

$$\implies R - \frac{1}{2}R = 8\pi GT,$$

$$\implies \frac{1}{2}R = 8\pi GT.$$

Help would be much appreciated in deciphering where I've gone wrong.

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    $\begingroup$ $g^{\mu \nu} g_{\mu\nu}\neq 1$ $\endgroup$ – nwolijin May 19 at 11:29
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You had a mistake in this part of going from the first to the second line.

$${g_{\mu \nu }}{g^{\mu \nu }} = 4, \text{not} ~ 1.$$

This follows from the definition of the inverse metric, i.e. ${g_{\mu \lambda }}{g^{\lambda \nu }} = \delta _\mu ^\nu $, and $\delta _\mu ^\mu =1+1+1+1=4$. For higher $D$-dimensions, you can also find:

$${g_{\mu \nu }}{g^{\mu \nu }} = D.$$

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