- #1
MarkovMarakov
- 33
- 1
I can't see how to get the following result. Help would be appreciated!
This question has to do with the Riemann curvature tensor in inertial coordinates.
Such that, if I'm not wrong, (in inertial coordinates) [tex]R_{abcd}=\frac{1}{2} (g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})[/tex]
where [itex]",_i"[/itex] denotes [itex]\partial \over \partial x^i[/itex].
How does [tex]g_{ab,cd}=-\frac{1}{3}(R_{acbd}+R_{adbc})[/tex]?
______
So
[tex]-\frac{1}{3}(R_{acbd}+R_{adbc})=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}-g_{ba,cd}-g_{cd,ab}+g_{bd,ac}+g_{ac,bd}-g_{ab,cd}-g_{cd,ab})[/tex]
[tex]=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}+\frac{1}{3}g_{ab,cd}[/tex]
But how does [tex]\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}=\frac{2}{3}g_{ab,cd}[/tex]
____
...Have I made a mistake?
This question has to do with the Riemann curvature tensor in inertial coordinates.
Such that, if I'm not wrong, (in inertial coordinates) [tex]R_{abcd}=\frac{1}{2} (g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})[/tex]
where [itex]",_i"[/itex] denotes [itex]\partial \over \partial x^i[/itex].
How does [tex]g_{ab,cd}=-\frac{1}{3}(R_{acbd}+R_{adbc})[/tex]?
______
So
[tex]-\frac{1}{3}(R_{acbd}+R_{adbc})=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}-g_{ba,cd}-g_{cd,ab}+g_{bd,ac}+g_{ac,bd}-g_{ab,cd}-g_{cd,ab})[/tex]
[tex]=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}+\frac{1}{3}g_{ab,cd}[/tex]
But how does [tex]\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}=\frac{2}{3}g_{ab,cd}[/tex]
____
...Have I made a mistake?