Proving an Identity "easily" as written by author, in GR electromagnetism The last day and, some days before, I found myself incapable of proving an Equation, while the author said it was "easily" deducted (Chapter 6, Page 127, 3+1 Ideal Magnetohydrodynamics- Éric Gourgoulhon)
In almost every other derivation, I had no problems but currently stuck with this.
Stress-energy tensor of the electromagnetic field is: ($F$ is anti-symmetric)
$$T_{\alpha \beta} = \dfrac{1}{\mu_0} \bigg(F_{\mu\alpha}F^{\mu}_{\,\,\,\beta}-\dfrac{1}{4}F_{\mu\nu}F^{\mu \nu}g_{\alpha\beta}\bigg)$$
While these equations hold:
\begin{equation}
\label{2}
\nabla_\mu *\!F^{\alpha \mu} = 0
\end{equation}
\begin{equation}
\label{1}
\tag{2}\nabla_\mu F^{\alpha \mu} = \mu _0 j^\alpha
\end{equation}
where $*F$ is the hodge dual operator, the first equation  is written in the simpler form  $$\tag{3}\nabla _\alpha F_{\beta\gamma}+\nabla _\beta F_{\gamma\alpha} +\nabla _\gamma F_{\alpha\beta} =0$$
So basically we use the last two equations, I need to prove that $$\nabla_\beta T_{\alpha}\,^\beta = F_{\mu \alpha}j^{\mu}$$
One can go right away to this:
$$\nabla_\beta T_{\alpha}\,^\beta = F_{\mu \alpha}j^{\mu}+ F^{\mu \beta}\nabla_\beta F_{\mu \alpha} -\dfrac{1}{4} \nabla_\alpha( F_{\mu\nu}F^{\mu \nu})$$
and even with using the equation $(3)$, we can't get rid of the 1/4. Basically, aside the 1st term, the right part should be vanishing, but I can't really establish it.
 A: First note that
$$\nabla_\alpha(F_{\mu\nu}F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F_{\mu\nu}(\nabla_\alpha F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})$$
$$=2F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})=-2F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu}).$$
Using this equation, the anti-symmetry of $F_{\mu\nu}$, and the fact that we can freely rename dummy indices, we get
$$F^{\mu \beta}\nabla_\beta F_{\mu \alpha} -\dfrac{1}{4} \nabla_\alpha( F_{\mu\nu}F^{\mu \nu})=F^{\mu \beta}\nabla_\beta F_{\mu \alpha} + \frac{1}{2}F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu})$$
$$=F^{\mu \nu}\nabla_\nu F_{\mu \alpha} + \frac{1}{2}F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu})$$
$$=-F^{\mu \nu}\nabla_\nu F_{\alpha\mu} + \frac{1}{2}F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu})=\frac{1}{2}F^{\mu\nu}(\nabla _\mu F_{\nu\alpha}-\nabla _\nu F_{\alpha\mu})$$
Using the anti-symmetry of $F_{\mu\nu}$, you can verify that the quantity inside the parenthesis is symmetric under the exchange of $\mu$ and $\nu$. Since it is contracted with the anti-symmetric $F^{\mu\nu}$, the contraction must vanish.
