# The metric and Kronecker's delta

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 $g_{uu}$?

• 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. Dec 19, 2014 at 21:14
• 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 Dec 19, 2014 at 21:18
• Once the order all matches up properly, then you just replace the $w$ back into $u$ again. Dec 19, 2014 at 21:18
• Thanks for the help, that makes a lot of sense! Dec 19, 2014 at 21:21
• @janmarqz Yeah you're right sorry, I have never used used tensors in LaTeX, will always write them that way from now on. Dec 19, 2014 at 21:23

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