Electromagnetism in curved spacetime

Question

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 $g$?

Answer 1

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.)