# Divergence of the energy momentum tensor

I need to prove that in a vacuum, the energy-momentum tensor is divergenceless, i.e.
$$\partial_{\mu} T^{\mu \nu} = 0$$
where
$$T^{\mu \nu} = \frac{1}{\mu_{0}}\Big[F^{\alpha \mu} F^{\nu}_{\alpha} - \frac{1}{4}\eta^{\mu \nu}F^{\alpha \beta}F_{\alpha \beta}\Big]$$ Here $F_{\alpha \beta}$ is the electromagnetic field tensor and $\eta^{\mu \nu}$ is the metric.
I've worked through all the steps. I need to know if my answer is correct.
Now we have
$$\partial_{\mu} \Big(F^{\alpha \mu } F^{\nu}_{\alpha}\Big) = F^{\alpha \mu} \Big(\partial_{\mu}F^{\nu}_{\alpha}\Big) + F^{\nu}_{\alpha}\Big( \partial_{\mu} F^{\alpha \mu}\Big)$$
The second term vanishes because of Maxwell's equation with $J^{\alpha} =0$.
$$F^{\alpha \mu}\Big( \partial_{\mu} F^{\nu}_{\alpha} \Big) =\frac{F^{\alpha \mu} (\partial_{\mu} F^{\nu}_{\alpha}) + F^{\mu \alpha} (\partial_{\alpha} F^{\nu}_{\mu})}{2}$$ In the second term, we let $\mu \rightarrow \alpha$ and $\alpha \rightarrow \mu$, since they're both dummy variables. But since $F^{\mu \alpha} = -F^{\alpha \mu}$, the expression becomes
$$\partial_{\mu}\Big(F^{\alpha \mu } F^{\nu}_{\alpha}\Big) = \frac{F^{\alpha \mu} \Big(\partial_{\mu}F^{\nu}_{\alpha} - \partial_{\alpha}F^{\nu}_{\mu}\Big)}{2}$$ Now, $$\Big(\partial_{\mu}F^{\alpha \beta}\Big)F_{\alpha \beta} = \eta^{\gamma \alpha}\eta^{\delta \beta} \eta_{\theta \alpha} \eta_{\phi \beta} F^{\theta \phi} \Big( \partial_{\mu} F_{\gamma \delta}\Big) = \delta^{\gamma}_{\theta} \delta^{\delta}_{\phi}F^{\theta \phi} \Big( \partial_{\mu} F_{\gamma \delta}\Big) = F^{\theta \phi}\Big( \partial_{\mu} F_{\theta \phi}\Big) = F^{\alpha \beta}\Big( \partial_{\mu} F_{\alpha \beta}\Big)$$
So differentiating the second term of the energy momentum tensor gives us,
$$\partial_{\mu} \Big(\eta^{\mu \nu} F^{\alpha \beta} F_{\alpha \beta}\Big) = -2 F^{\alpha \mu} \partial^{\nu} F_{\mu \alpha}$$
Where I replaced $\beta$ by $\mu$ and flipped the indices.
Combining the previous two expressions, we have
$$\partial_{\mu} T^{\mu \nu} = \frac{1}{2 \mu_{0}} F^{\alpha \mu} \Big[\partial_{\mu} F^{\nu}_{\alpha} - \partial_{\alpha} F^{\nu}_{\mu} + \partial^{\nu} F_{\mu \alpha} \Big]$$
Now consider the term in the square brackets:
$$\partial_{\mu} F^{\nu}_{\alpha} - \partial_{\alpha} F^{\nu}_{\mu} + \partial^{\nu} F_{\mu \alpha} = \eta^{\nu \theta}\partial_{\mu} F_{\alpha \theta} - \eta^{\nu \theta}\partial_{\alpha} F_{\mu \theta} + \eta^{\nu \theta} \partial_{\theta} F_{\mu \alpha} = \eta^{\nu \theta}\Big( \partial_{\mu} F_{\alpha \theta} + \partial_{\alpha} F_{\theta \mu} + \partial_{\theta} F_{\mu \alpha}\Big)$$
Where I flipped the indices in the second term.The term in brackets is zero because of Maxwell's other equation. Therefore,
$$\partial_{\mu} T^{\mu \nu} = 0$$

Are all the steps correct? Thank you for your help!