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.
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!
Best Answer
Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)
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You can use them indifferently in that case. However, NOT if you switch to non-cartesian coordinates (e.g. Spherical coordinates), because in that case the connection coefficients are not zero in general, even in the absence of curvature, and so the covariant derivatives may differ from the ordinary partial derivatives, even in flat space.
I would simply avoid mixing the symbols or, in the future, it will cost you some extra effort to undo the habit, when you learn about GR and curved spaces.
These are the definitions:
$\partial_\mu V^{\nu} = \frac{\partial}{\partial x^{\mu}}V^{\nu}$
$\nabla_\mu V^{\nu} = \frac{\partial}{\partial x^{\mu}}V^{\nu}+\Gamma^{\nu}_{\mu\alpha}V^{\alpha} $
The so-called connection coefficients are the $\Gamma^{\nu}_{\mu\alpha}$. Their definition consists on a certain combination of partial derivatives of the elements of the metric. In a flat space and cartesian coordinates you can ignore them: they are zero since the elements of the metric are all just constant numbers, $(1,-1,-1,-1)$. It doesn't mean however, that they are zero in Special Relativity in general: the diagonal elements of the metric tensor in spherical coordinates for instance, are functions of the coordinates, namely, $(1, -1, -r^2, -r^2 sin^2 \theta)$ although the space is flat.
If you are interested in getting used to covariant derivatives and tensor calculations in general, without investing too much effort, I suggest you the last chapter (specially the solved problems) of the classic, small book "Vector Calculus" (M.R. Spiegel) from the Schaum series. And, for getting a feeling of the geometrical meaning of the connection coefficients, google for "Parallel Transport". The Schutz book mentioned above has also a very nice explanation.