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In special relativity, Maxwell's equations may be written as \begin{align*} dF = 0, \\ \star\, d\star F = J. \end{align*} In four-vector notation, this translates to $\partial_{\mu}F^{\mu\nu} = J^{\nu}$ and $\partial_{[\lambda}F_{\mu\nu]} = 0$ where the brackets mean that there is a sum with indices cyclically permuted.

Generalizing this to curved spacetime requires us to use covariant derivatives instead of partial derivatives, so we instead write $\nabla_{\mu}F^{\mu\nu} = J^{\nu}$ and $\nabla_{[\lambda}F_{\mu\nu]} = 0$.

Is there a way to write this using differential forms? Is there any reference that talks about this?

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Differential forms are natural objects that depend solely on the smooth structure, therefore they are valid in any spacetime.

Note that the Hodge star $\star$ does, however, depend on a metric tensor. However, any metric tensor works with it. Therefore, Maxwell's equations in differential form notation are valid in any spacetime without modifications, assuming the Hodge star in the equations refer always to the metric tensor that defines the geometry of your spacetime.

Note that even in standard tensor calculus notation, Maxwell's equations can be cast in the form $$ \partial_{[\kappa} F_{\mu\nu]}=0 \\ \partial_\nu\mathfrak F^{\mu\nu}=\mu_0\mathfrak{j}^\mu, $$ where $\mathfrak F^{\mu\nu}=F^{\mu\nu}\sqrt{-g}$ and $\mathfrak j^\mu=j^\mu\sqrt{-g}$, which do not involve covariant derivatives.

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This is the OP—same person who asked the question. I wanted to explain my understanding. If anything is not clear, please tell me. I want this to be understandable to others if anyone ever reads this.

Is there a way to write this using differential forms (such that it reduces to equations involving covariant derivatives rather than partial derivatives)?

It turns out that the differential form equations written in the original post are already what you want. This seemed unintuitive to me, so I will show that the differential form equations reduce to the covariant equations.

In fact, I will give two proofs for this (although the second one is a bit redundant).



Proof 1. Given any point $p\in M$, choose "locally Minkowski coordinates" $(x^{i})$ centered at $p$. These yield $g_{\mu\nu}(p) = \eta_{\mu\nu}$ and $\Gamma^{\mu}_{\nu\lambda}(p) = 0$. In these coordinates, the differential form version of Maxwell's equations reduce to $$ \partial_{\mu}F^{\mu\nu}(p) = J^{\nu}(p) \quad\text{and}\quad \partial_{[\lambda}F_{\mu\nu]}(p) = 0. $$ We also have $\partial_{\mu} = \nabla_{\mu}$ at point $p$ in these coordinates, so this means the above is the same as \begin{align} \nabla_{\mu}F^{\mu\nu}(p) = J^{\nu}(p) \quad\text{and}\quad\nabla_{[\lambda}F_{\mu\nu]}(p) = 0 \end{align} in these coordinates. But by general covariance, a transformation to any other coordinates must preserve the form of the above two equations, so they hold for all coordinates where $p$ is an arbitrary point. $$\tag*{$\blacksquare$}$$



Proof 2. We will do explicit computations. In any given coordinates system, we write $F = \frac{1}{2}F_{\mu\nu}\,dx^{\mu}\wedge dx^{\nu}$. Then \begin{align} dF = \frac{1}{2}\partial_{\lambda}F_{\mu\nu}\, dx^{\lambda}\wedge dx^{\mu}\wedge dx^{\nu} = 0. \end{align} This implies that for a given set of distinct, fixed $\lambda, \mu, \nu$, \begin{align} \frac{1}{2}(\partial_{\lambda}F_{\mu\nu} - \partial_{\lambda}F_{\nu\mu} + \partial_{\mu}F_{\nu\lambda} &- \partial_{\mu}F_{\nu\lambda} + \partial_{\nu}F_{\lambda\mu} - \partial_{\nu}F_{\mu\lambda}) = 0 \\ \implies \partial_{\lambda}F_{\mu\nu} + &\partial_{\mu}F_{\nu\lambda} + \partial_{\nu}F_{\lambda\mu} = 0. \end{align} Then we will add & subtract the appropriate "Christoffel terms," giving us \begin{align} & \partial_{\lambda}F_{\mu\nu} - \Gamma^{\alpha}_{\lambda\mu}F_{\alpha\nu} - \Gamma^{\alpha}_{\lambda\nu}F_{\mu\alpha} + \Gamma^{\alpha}_{\lambda\mu}F_{\alpha\nu} + \Gamma^{\alpha}_{\lambda\nu}F_{\mu\alpha} \\ & \partial_{\mu}F_{\nu\lambda} - \Gamma^{\alpha}_{\mu\nu}F_{\alpha\lambda} - \Gamma^{\alpha}_{\mu\lambda}F_{\nu\alpha} + \Gamma^{\alpha}_{\mu\nu}F_{\alpha\lambda} + \Gamma^{\alpha}_{\mu\lambda}F_{\nu\alpha} \\ & \partial_{\nu}F_{\lambda\mu} - \Gamma^{\alpha}_{\nu\lambda}F_{\alpha\mu} - \Gamma^{\alpha}_{\nu\mu}F_{\lambda\alpha} + \Gamma^{\alpha}_{\nu\lambda}F_{\alpha\mu} + \Gamma^{\alpha}_{\nu\mu}F_{\lambda\alpha} = 0. \end{align} Using the symmetry of the Christoffel symbols $\Gamma^{\alpha}_{\mu\nu} = \Gamma^{\alpha}_{\nu\mu}$ and the antisymmetry of the Faraday tensor $F_{\mu\nu} = -F_{\nu\mu}$, the right two columns above cancel out. The remaining terms above reduce to $$ \nabla_{\lambda}F_{\mu\nu} + \nabla_{\mu}F_{\nu\lambda} + \nabla_{\nu}F_{\lambda\mu} = 0. $$


To get the other equations, we'll invoke two theorems. The first of these is that $$ g^{\lambda_{1}\kappa_{1}}g^{\lambda_{2}\kappa_{2}}g^{\lambda_{3}\kappa_{3}}g^{\lambda_{4}\kappa_{4}}\epsilon_{\kappa_{1}\kappa_{2}\kappa_{3}\kappa_{4}} = (\det g^{\mu\nu})\widetilde{\epsilon}^{\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}} $$ where $\epsilon_{\kappa_{1}\kappa_{2}\kappa_{3}\kappa_{4}}$ and $\widetilde{\epsilon}^{\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}}$ give you the sign of the appropriate permutations. The second theorem is that $$ \frac{1}{\sqrt{|g|}}\partial_{\lambda} \left(\sqrt{|g|}\, F^{\lambda \sigma} \right) = \nabla_{\lambda} F^{\lambda\sigma}. $$ Now we will apply the explicit formula for the Hodge star operator: \begin{align} \star\, F &= \frac{\sqrt{|g|}}{2!} g^{\mu_{1}\nu_{1}}g^{\mu_{2}\nu_{2}} \, \epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}} \left(\frac{1}{2} F_{\mu_{1}\mu_{2}} \right) \, dx^{\nu_{3}}\wedge dx^{\nu_{4}}, \\ d\star F &= \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\nu_{1}\nu_{2}}}{4} \right) \epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}}\, dx^{\lambda}\wedge dx^{\nu_{3}}\wedge dx^{\nu_{4}}, \\ \star\, d\star F &= \frac{\sqrt{|g|}}{1!} g^{\lambda\kappa_{1}}g^{v_{3}\kappa_{2}}g^{\nu_{4}\kappa_{3}}\, \epsilon_{\kappa_{1}\kappa_{2}\kappa_{3}\kappa_{4}} \cdot \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\nu_{1}\nu_{2}}}{4} \right) \epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}} \, dx^{\kappa_{4}} = J_{\kappa_{4}}\, dx^{\kappa_{4}}. \end{align} Now we equate the components and apply $g^{\sigma\kappa_{4}}$ to both sides: \begin{align*} \sqrt{|g|}\, g^{\lambda\kappa_{1}}g^{\nu_{3}\kappa_{2}}g^{\nu_{4}\kappa_{3}}g^{\sigma\kappa_{4}} \cdot \epsilon_{\kappa_{1}\kappa_{2}\kappa_{3}\kappa_{4}}\epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}} \cdot \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\nu_{1}\nu_{2}} }{4} \right) = J^{\sigma}. \end{align*} Applying our first theorem (for the determinant) gives \begin{align*} \sqrt{|\det g_{\mu\nu}|}\, (\det g^{\mu\nu})\, \widetilde{\epsilon}^{\lambda\sigma\nu_{3}\nu_{4}} \epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}} \cdot \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\nu_{1}\nu_{2}} }{4} \right) = J^{\sigma}. \end{align*} Note that $(g^{\mu\nu}) = (g_{\mu\nu})^{-1}$ so the prefactor is just $1/\sqrt{|g|} = 1/\sqrt{|\det g_{\mu\nu}|}$. To handle the Levi-Civita symbols, one can reason that $\widetilde{\epsilon}^{\lambda\sigma\nu_{3}\nu_{4}} \epsilon_{\nu_{1}\nu_{2}\nu_{3}\nu_{4}} = 2\delta^{\lambda}_{\nu_{1}}\delta^{\sigma}_{\nu_{2}} - 2\delta^{\lambda}_{\nu_{2}}\delta^{\sigma}_{\nu_{1}}$, and so \begin{align*} \frac{1}{\sqrt{|g|}}\cdot (\delta^{\lambda}_{\nu_{1}}\delta^{\sigma}_{\nu_{2}} - \delta^{\lambda}_{\nu_{2}}\delta^{\sigma}_{\nu_{1}}) \cdot \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\nu_{1}\nu_{2}} }{4} \right) = J^{\sigma} \\ \frac{1}{\sqrt{|g|}}\, \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\lambda\sigma}}{2} \right) - \frac{1}{\sqrt{|g|}}\, \partial_{\lambda}\left( \frac{\sqrt{|g|}\, F^{\sigma\lambda}}{2} \right) = J^{\sigma} \\ \frac{1}{\sqrt{|g|}}\, \partial_{\lambda}\left( \sqrt{|g|}\, F^{\lambda\sigma} \right) = J^{\sigma}. \end{align*} By our second theorem, this gives us the desired formula. $$\tag*{$\blacksquare$}$$

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