Show that $\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$ In a theoretical physics homework problem, I have to show the following:
$$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$
Where $T$ is the Energy-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.
I know the following relationships:


*

*Current and magnetic potential with Lorenz gauge condition:
$$\mathop\Box A^\mu = \mu_0 j^\mu$$

*Energy-Momentum-Tensor:
$$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$

*Field-Tensor:
$$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$

*d'Alembert operator:
$$\mathop\Box = \partial_\mu \partial^\mu$$

*Bianchi identity:
$$\partial^{[\mu} F^{\nu\alpha]} = 0$$
So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in Penrose graphical notation, but I still cannot see how to tackle this problem.
Could somebody please give me a hint into the right direction?
 A: Let's look at different terms from differentiating $T^{\mu\nu} $.
The first from differentiating $ g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}$ is
$$\partial_\nu g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}=
 g^{\mu\alpha} F_{\alpha\beta} (\partial_\nu F^{\beta\nu})
 +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}=
 - \mu_0 F_{\alpha\beta} j^\beta
 +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$
The first term is exactly what you want, the second cancels against the stuff you get from differentiating $g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$:
$$\partial^\mu F_{\kappa\lambda} F^{\kappa\lambda}=2 F_{\kappa\lambda} (\partial^\mu F^{\kappa\lambda})=-2 F_{\kappa\lambda} (\partial^\kappa F^{\lambda\mu}+\partial^\lambda F^{\mu\kappa})
=-4 (\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$
where in the second equality sign we have used Bianchi identity and in the last equality we 
have used
$$
F_{\kappa\lambda} \partial^\kappa F^{\lambda\mu}
\underset{\text{relabel indecies}}=
 F_{\nu\beta}\partial^\nu F^{\beta \mu}
\underset{\text{antisym. of $F$}}=
F_{\beta\nu}\partial^\nu F^{\mu\beta}
$$
This exactly cancels the second term in the first equation.
