# Covariant form of Maxwell Equations in curved spacetime

The real world doesn't care about our choice of coordinate to describe nature. Maxwell equations in vectorial form are written with respect to an Inertial frame of reference as: \begin{align} \vec\nabla\cdot\vec{E} &= 4\pi\rho \label{Diff I}\\ \vec\nabla\times\vec{B} &= \dfrac{4\pi}{c} \vec{j}+\dfrac{1}{c}\dfrac{\partial\vec{E}}{\partial t} \label{Diff IV}\\ \vec\nabla\times\vec{E} &= -\dfrac{1}{c}\dfrac{\partial\vec{B}}{\partial t} \label{Diff III}\\ \vec\nabla\cdot\vec{B} &= 0 \label{Diff II} \end{align}

And the potentials:

\begin{align} \vec{E} &= -\frac1c \frac{\partial \vec{A}}{\partial t} - \vec\nabla\phi\\ \vec{B} &= \vec\nabla\times\vec A \end{align}

Those equations are valid in any inertial coordinate frame of reference. How about non-inertial frame? To answer this question and to cast Maxwell's Equations in ANY frame of reference, I think it's useful to use tensorial calculus. So:

In Special Relativity we write:

\begin{align} \partial_{\mu}F^{\mu\nu} &= \frac{4\pi}{c}j^{\nu} \tag{1}\\ \partial_{[\mu}F_{\alpha\beta]} &= 0\;. \tag{2} \end{align}

But here is my questions:

1. Those equations are written with respect to the Minkowski metric, so with Cartesian coordinates for the spatial coordinates. Those are covariant with respect to Lorentz transformations, but they are not valid in ANY inertial coordinate system. If I choose cilindrical or spherical coordinates, I can't use them. How does those equations transform in any other coordinate system (inertial or not)?

2. Before GR, so in flat spacetime, why don't we write Maxwell equations in a coordinate-free notation? For example why don't we use Covariant Derivative and a general metric to cast the equations in their most general form, like we do in general relativity?

Because in GR we need their general form to account for spacetime curvature, but here we would also need it to account for any inertial or non-inertial coordinate system in flat spacetime, and not only in Cartesian Coordinates.

$$\nabla_\mu F^{\mu\nu} = \frac{4\pi}{c} j^\nu.$$