Electromagnetism in curved spacetime I am trying to follow a derivation outlined in Asenjo et al. 2017.
In equation 1, they define the covariant derivative of the field tensor,
$$ \nabla_{\alpha} F^{\alpha \beta} = 0 $$
From this they arrive at,
$$ \partial_{\alpha} [\sqrt{-g} g^{\alpha \mu} g^{\beta \nu} (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu})] = 0$$
Now, since $F^{\alpha \beta} =g^{\alpha \mu} g^{\beta \nu} F_{\mu \nu} $ and $F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}$, I can see the general methods and substitutions taken to arrive at this answer, but am confused on 2 points:
Why the switch from covariant to partial derivatives?
Where does the $\sqrt{-g}$ term come from? What is $g$?
 A: There are some aspects here:


*

*First, you are correct that $g^{\alpha\mu}g^{\beta\nu}$ simply raise the indices on $F_{\mu\nu}$.

*The field strength tensor is really defined as a second-order differential form, i.e. $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ with partial derivatives. That doesn't matter for computing the components, since the extra terms with Christoffel symbols cancel, but the formalism is much clearer. 

*Finally, about the $\sqrt{-g}$: This is a standard trick to rewrite (covariant) divergences. Observe that the covariant derivative (your first equation) can be expanded as
$$D_\alpha F^{\alpha\beta}=\partial_\alpha F^{\alpha\beta} + \Gamma_{\alpha\gamma}^{\alpha} F^{\gamma\beta}+ \Gamma_{\alpha\gamma}^{\beta} F^{\gamma\alpha}\,.$$
The last term drops out because $\Gamma$ is symmetric in the lower indices and $F$ is antisymmetric. The first Christoffel symbols is
$$\Gamma_{\alpha\gamma}^\alpha=\frac{1}{2}g^{\alpha\delta}\left(\partial_\gamma g_{\alpha\delta}+\partial_\alpha g_{\gamma\delta}-\partial_\delta g_{\alpha\gamma}\right)\,,$$
where the second and third term cance (can you see why?), so
$$\Gamma_{\alpha\gamma}^\alpha=\frac{1}{2}g^{\alpha\delta}\partial_\gamma g_{\alpha\delta}\,.$$
This is of the form $\text{tr}\left(M^{-1}\partial M\right)$ for the matrix $g$. Using the identity $$\ln \det M=\text{tr}\ln M$$  (see e.g. https://math.stackexchange.com/questions/1487773/the-identity-deta-exptrlna-for-a-general), we can rewrite this as
$$\frac{1}{2}g^{\alpha\delta}\partial_\gamma g_{\alpha\delta} = \frac{1}{\sqrt{-g}}\partial_\gamma \sqrt{-g}\,,$$ and your second formula follows from the Leibniz rule. (I may or may not have misse a minus sign somewhere.)

