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 $g$?

  • 1
    $\begingroup$ If you mean what is the $g$ in $\sqrt{-g}$, it is the determinant of the metric tensor. $\endgroup$ – John Rennie Dec 12 '17 at 11:25
  • 1
    $\begingroup$ No offense, if you don't know what g is, then you should first read some standard textbook GR (easy read is d'Inverno's book or even Dirac's brochure), then this research article. $\endgroup$ – DanielC Dec 12 '17 at 11:45
  • 5
    $\begingroup$ Not all introductory literature uses the notation $g$ for the determinant of the metric. It may simply be that the OP is not familiar with this notation, not that he is unfamiliar with the notion of the metric and its role in general relativity. $\endgroup$ – Erik Jörgenfelt Dec 12 '17 at 12:08
  • $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient prior research. $\endgroup$ – AccidentalFourierTransform Dec 12 '17 at 14:41

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.)

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