Hello! I was rereading this nice thread and wanted to discuss a bit further (if you are willing to do so as well of course!
)
Mmm I am actually trying to show Maxwell's equations in vacuum i.e. ##g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0## and ##\nabla_{[\mu}F_{\nu \sigma]}=0## using the same Riemann tensor property i.e. ##\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}##, where ##F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}##.
Let's focus on ##g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0## first.
We get
\begin{align*}
\nabla_{\mu} F_{\nu \sigma} &= \nabla_{\mu} \nabla_{\nu} A_{\sigma} - \nabla_{\mu} \nabla_{\sigma} A_{\nu} \\
&= R_{\sigma \nu \mu \rho} A^{\rho} - R_{\nu \sigma \mu \rho} A^{\rho} \\
&= 2R_{\sigma \nu \mu \rho} A^{\rho}
\end{align*}
Multiplying both sides by ##g^{\nu \mu}## yields
\begin{equation*}
\nabla^{\nu} F_{\nu \sigma} = 2 R_{\sigma \rho} A^{\rho} = 0
\end{equation*}
Where I used the fact that, in vacuum, ##R_{\mu \nu} = 0##.
Does this proof look correct to you?
I am stuck on how to show that ##\nabla_{[\mu}F_{\nu \sigma]}=0## by means of ##\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}## though
I start expanding it out
\begin{align*}
\nabla_{[\mu } F_{\nu \rho]} &= \frac{1}{3!} \left( \nabla_{\mu} F_{\nu \rho} + \nabla_{\nu} F_{\rho \mu} + \nabla_{\rho} F_{\mu \nu} -\nabla_{\mu} F_{\rho \nu} -\nabla_{\nu} F_{\mu \rho} -\nabla_{\rho} F_{\nu \mu} \right) \\
&= \frac{1}{3} \left( R_{\rho \nu \mu \sigma} + R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma} - R_{\nu \rho \mu \sigma} - R_{\rho \mu \nu \sigma} - R_{\mu \nu \rho \sigma}\right) A^{\sigma} \\
&= \frac{2}{3} \left( R_{\rho \nu \mu \sigma}+ R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma}\right) A^{\sigma}
\end{align*}
But I do not see how to show that the last equation vanishes without having to explicitly write the Riemann tensor as a function of Christoffel symbols (which does not seem to be the best idea
).Regards.
JD.
