Using partial derivatives for Maxwell's equations in curved spacetime

Replacing covariants with partials:

The source equation you cite involves a skew-symmetric tensor density.

You may know that if $\nabla$ is the Levi-Civita connection, you can calculate divergences of vector fields without using the connection coefficients/Christoffel symbols: $$ \nabla_\mu X^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(X^\mu\sqrt{-g}). $$

Now, let $\mathcal{J}^\mu=X^\mu\sqrt{-g}$ be a vector density. Then $$ (\nabla_\mu X^\mu)\sqrt{-g}=\nabla_\mu(X^\mu\sqrt{-g})=\nabla_\mu\mathcal{J}^\mu=\partial_\mu(X^\mu\sqrt{-g})=\partial_\mu\mathcal{J}^\mu, $$ so vector density fields can be differentiated partially.

However, as it turns out the situation is the same for arbitrary (k,0)-type antisymmetric tensor fields: Let $F^{\mu\nu}$ be such a field. Then $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\mu_{\ \nu\sigma}F^{\sigma\nu}+\Gamma^{\nu}_{\ \nu\sigma}F^{\mu\sigma}, $$ but the second term here vanishes because $\Gamma$ is symmetric in the lower indices, but $F$ is skew-symmetric in the upper indices, so we're left with $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\nu_{\ \nu\sigma}F^{\mu\sigma}=\partial_\nu F^{\mu\nu}+\partial_\sigma\ln\sqrt{-g}F^{\mu\sigma} \\ =\frac{1}{\sqrt{-g}}\partial_\nu(F^{\mu\nu}\sqrt{-g}). $$

Defining $\mathcal{F}^{\mu\nu}=F^{\mu\nu}\sqrt{-g}$ gives $$ \nabla_\nu \mathcal{F}^{\mu\nu}=\partial_\nu \mathcal{F}^{\mu\nu}. $$

Dependence of Maxwell's equations on the (pseudo-)Riemannian structure:

Remember that the fundamental field here is $A_\mu$, which is a covector field. The only Maxwell equation that doesn't depend on the Riemannian structure is $\partial_{[\mu} F_{\nu\sigma]}=0$, because you can replace the covariants here with partials do to skew-symmetry.

Also do remember that $F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu$ is also well-defined without covariants.

When we get to the source equation things change, however, because

• to define divergence we need upper indices, however $F$ is lower-indiced by nature, so we need the metric to raise indices;
• you can replace covariants with partials there only if you multiply with $\sqrt{-g}$ to create densities. Which, of course, depends on the metric.

So you can weasel out of using the connection (which is great for computations), but you cannot weasel out of using the metric, therefore Maxwell's equations are absolutely not topological.

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An aside on differential forms:

You should read about differential forms. I am trying to think of reference that should be quick-readable by physicist. Probably Flanders' book is a good one. Otherwise Anthony Zee's General Relativity and QFT books also contain differential forms but only in a heuristic manner. Sean Carroll's GR book also has an OK recount of them.

Basically, differential forms are totally antisymmetric covariant tensor fields. Instead of using index notation as in, say $\omega_{\mu_1,...,\mu_k}$ to denote them, usually 'abstract' notation is used as $\omega=\sum_{\mu_1<...<\mu_k}\omega_{\mu_1...\mu_k}\mathrm{d}x^{\mu_1}\wedge...\wedge\mathrm{d}x^{\mu_k}$, where the basis is written out explicitly. The wedge symbols are skew-symmetric tensor products.

Differential forms are good because they generalize vector calculus to higher dimensions, arbitrary manifolds and also to cases when you don't have a metric. Differential forms can be differentiated ($\omega\mapsto\mathrm{d}\omega$), where the "$\mathrm{d}$" operator, called the exterior derivative, turns a $k$-form into a $k+1$ form without the need for a metric or a connection, and generalizes grad, div and curl, all in one.

The integral theorems of Green, Gauss and Stokes are also generalized.

The point is, if you also have a metric, the theory of differential forms is enriched. You get an option to turn $k$-forms into $n-k$ forms ($n$ is the dimension of your manifold), and also to define a "dual" operation to the exterior derivative, called the codifferential. The codifferential essentially brings the concept of divergence to differential forms.

Written with differential forms, Maxwell's equations are given by $$ \mathrm{d}F=0 \\ \mathrm{d}^\dagger F=kJ, $$ where $\mathrm{d}^\dagger$ is the codifferential, and $k$ is some constant I care not about right now.

I am noting two things:

• The $F$ field strength 2-form is given by $F=\mathrm{d}A$, where $A$ is ofc the 4-potential. The exterior derivative satisfies $\mathrm{d}^2=0$ (think of $\text{div}\ \text{curl}=0$ and $\text{curl}\ \text{grad}=0$), so with potentials, the first equation is $\mathrm{d}F=\mathrm{d}^2A=0$, which is trivially true.
• The first equation contains only $\mathrm{d}$, which is well-defined without a metric. The second equation depends on the codifferential $\mathrm{d}^\dagger$, which does depend on the metric. There is your metric dependance!

Bence Racskó's answer is great! I want to add something. You're saying that:

"In the wikipedia article Maxwell's equations in curved spacetime, it states without calculation that despite the use of partial derivatives, the equations are invariant under arbitrary curvilinear coordinate transformations. Because of symmetry, it is not too hard to see that:"

$F_{ab} \, = \, \nabla_a A_b \, - \, \nabla_b A_a = (\partial_a A_b -{\Gamma^c}_{ab} A_c) - (\partial_b A_a - {\Gamma^c}_{ba} A_c) = \partial_a A_b - \partial_b A_a - {\Gamma^c}_{ab} A_c + {\Gamma^c}_{ba} A_c$

"And the last two terms cancel because the Christoffel symbols are symmetric in the lower indicies."

The Faraday tensor $\textbf{is defined:}$ $F_{μν} = \partial_{μ}Α_{ν} - \partial_{ν}Α_{μ}$. Let's say that we allow torsion. Then we cannot say: $\Gamma^{k}_{μν} = Γ^{k}_{νμ}$ since the lower indices don't commute now. $\textbf{This does not mean that the Faraday tensor will change.}$ My point is that the Faraday tensor takes the form $F_{μν} = \partial_{μ}Α_{ν} - \partial_{ν}Α_{μ}$ not because " the Christoffel symbols are symmetric in the lower indicies". It's because it is defined that way.