# Index gymnastics-Maxwell's equations

1. Sep 3, 2013

### WannabeNewton

1. The problem statement, all variables and given/known data
Let $X^{ab}$ be an antisymmetric tensor such that $\nabla^{(a}X^{b)c} = 0$. Show that if $R_{ab} = 0$, then $F_{ab} = R_{abcd}X^{cd}$ satisfies Maxwell's equations $\nabla^a F_{ab} = 0, \nabla_{[a}F_{bc]} = 0$ where $R_{abcd}$ is the Riemann curvature tensor and $R_{ab}$ is the Ricci tensor.

3. The attempt at a solution
Well to start with, $\nabla^a X^{bc} = -\nabla^b X^{ac} = \nabla^b X^{ca} = -\nabla^c X^{ba}$ so $\nabla^a X^{bc}$ is antisymmetric in all its indices. Thus, $\nabla^a F_{ab} = R_{abcd}\nabla^{a}X^{cd} + X^{cd}\nabla^a R_{abcd} = -R_{b[acd]}\nabla^{a}X^{cd} -2X^{cd}\nabla_{[b}R_{c]d} = 0$ since $R_{b[acd]} = 0$ by the first Bianchi identity and $R_{ab} = 0$ by hypothesis.

Showing that $\nabla_{[a}F_{bc]} = 0$ should be just as easy but for some reason I can't seem to find a way to show it. We have $\nabla_{a}F_{bc} =\nabla_{a}X^{de} R_{bcde} + X^{de}\nabla_{a}R_{bcde}$ so $\nabla_{[a}F_{bc]} = \nabla_{[a}X^{de}R_{bc]de}$ since $\nabla_{[a}R_{bc]de} = 0$ by the second Bianchi identity. One can either try to show that $\nabla_{[a}X^{de}R_{bc]de} = 0$ directly or (equivalently) show that $\epsilon^{abcf}\nabla_{a}X^{de}R_{bcde} = 0$. I know I'm missing something very trivial but I can't seem to proceed from here either way. Thanks in advance for any help!

Last edited: Sep 3, 2013