In classical electrodynamics there's the electrostatic potential $\phi$ which can be differentiated wrt space to give the static electric field $\vec E$. Is this idea perhaps extended relativistically by instead differentiating the potential wrt a space-time interval to give the electromagnetic fields?

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    $\begingroup$ See e.g. Jackson: Classical Electrodynamics and come back for any more precise questions? $\endgroup$
    – Sanya
    Dec 11, 2016 at 20:52
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    $\begingroup$ Any book about classical field theory treats this subject. So does wikipedia and many other sites. $\endgroup$ Dec 11, 2016 at 21:03
  • $\begingroup$ I've deleted an number of comments that were developing into a rather personal argument. Keep it civil, please. $\endgroup$ Dec 11, 2016 at 21:17

1 Answer 1


As stated in the comments, you can find the answer to your question in any textbook about classical electrodynamics. I'll just quote the results here.

In relativistic classical electrodynamics, the electric and magnetic fields are deduced from the potentials $\phi$ and $\vec{A}$ just as you know, i.e. :

$$ \left\{ \begin{array}{c} \vec{E} =-\vec{\nabla}\phi - \dfrac{\partial\vec{A}}{\partial t} \\ \vec{B} = \vec{\nabla}\times\vec{A} \end{array} \right. $$

Then, $A^\mu \equiv (\phi/c, \vec{A})$ is a four-vector, i.e. it transforms from one inertial reference frame to another according to the Lorentz transformation. Defining $F^{\mu\nu}\equiv \partial^\mu A^\nu - \partial^\nu A^\mu$, the equation of motion for a test charged particle is : \begin{equation} m\dfrac{dv^\mu}{ds} = qv_\nu F^{\mu\nu} \end{equation}

  • $\begingroup$ quadrivector is a term I've never heard of until now. I've always used/heard/read four-vector $\endgroup$
    – Kyle Kanos
    Dec 11, 2016 at 21:11
  • $\begingroup$ you are totally right, i just the had the french word in mind (which is "quadrivecteur"). I edited. $\endgroup$ Dec 11, 2016 at 21:13
  • $\begingroup$ OK, so we use a four-potential, but shouldn't this be differentiated wrt a space-time interval, rather than space and time individually? $\endgroup$ Dec 11, 2016 at 21:24

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