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I am reading some lecture notes for GR and it is currently showing how we are going to derive the field equations using a metric for a massive free particle with a metric

$$g_{00}=1+\frac{2\phi}{c^2}, \qquad \phi=-\frac{GM}{r} $$

We noted several things

$$ \nabla_{u} ({R^u}_{v} - \frac{1}{2}{\delta^u}_{v}R) =0, \\ \partial_{u}{T^u}_{v} =0 $$

From this we deduced, $$ {R^u}_{v} - \frac{1}{2}{\delta^u}_{v}R = \alpha {T^u}_{v},$$

and therefore,

$$ R_{uv} - \frac{1}{2}g_{uv}R = \alpha T_{uv},$$

However I have a problem with the last line, why has the delta function disappeared without affecting the metric, should the metric not be $g_{uu}$?

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  • $\begingroup$ unfortunate repeating the exact same indices; to me it seems it was multiplied by $g_{wu}$ and summed over $u.$ Then it would be pairs $wv.$ If that is not quite right, try variants. $\endgroup$
    – Will Jagy
    Dec 19, 2014 at 21:14
  • $\begingroup$ isn't that indexing $R^u_w$ causes confusion? instead use ${R^u}_w$ to keep perfect tracking of what happening with your rows and columns $\endgroup$
    – janmarqz
    Dec 19, 2014 at 21:18
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    $\begingroup$ Once the order all matches up properly, then you just replace the $w$ back into $u$ again. $\endgroup$
    – Will Jagy
    Dec 19, 2014 at 21:18
  • $\begingroup$ Thanks for the help, that makes a lot of sense! $\endgroup$
    – Rammus
    Dec 19, 2014 at 21:21
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    $\begingroup$ @janmarqz Yeah you're right sorry, I have never used used tensors in LaTeX, will always write them that way from now on. $\endgroup$
    – Rammus
    Dec 19, 2014 at 21:23

1 Answer 1

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Start from $$R_{\,\,v}^{u} - \frac 12 \delta_{\,\,v}^{u} R = \alpha T_{\,\,v}^{u}$$ multiply both sides by $g_{wu}$ to lower the inedx

$$g_{wu}(R_{\,\,v}^{u} - \frac 12 \delta_{\,\,v}^{u} R) = \alpha g_{wu} T_{\,\,v}^{u}$$ or $$R_{wv} - \frac 12 g_{wv} R = \alpha g_{wu} T_{wv}$$ It remains to ralabel $w\to u$ to get the last line.

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  • $\begingroup$ Okay, but in writing this solution you are assuming that $g_{wu}=g_{uw}$? If so you really ought to write that assumption, thanks. $\endgroup$
    – Electra
    Aug 18, 2022 at 6:02

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