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.


1 Answer 1


It seems like you already know the answer to your first question: to use the equations in a general coordinate system you have to replace the derivatives by covariant derivatives, getting

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

(The other equation is in fact the same whether you use covariant or regular derivatives.) As I've said before, all the formulas you know for the gradient, divergence and all that stuff in polar coordinates are just the covariant derivative.

As for the second question, in flat spacetime we can choose to use coordinates in which the Christoffel symbols are zero, so we usually do so and ignore the covariant derivative to make life simpler. But in curved spacetime you can't do that, so the covariant derivative becomes a necessity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.