I'm trying to derive the wave equations for the electric and magnetic fields in covariant (tensor) formulation. Starting with Gauss-Ampere law, $$ \partial_\alpha F^{\alpha\beta}=\frac{4\pi}{c}J^\beta $$ and Gauss–Faraday law $$ \partial_{\alpha}(\tfrac{1}{2}\epsilon^{\alpha\beta\gamma\delta}F_{\gamma\delta}) = 0 $$ I would like to derive the values of $\square \vec{E}$ and $\square\vec{B}$, or in general get to $\partial^{\nu} \partial_{\nu} F^{\alpha\beta}$. Is there an elegant way to do this? I did try using summation in each of the equations, but got quite confused with all the indices.

  • $\begingroup$ Do you want the wave equations in vacuum? So with $J^\beta=0$? Or do you want the wave equation for inside materials? $\endgroup$ Jan 26, 2021 at 8:23
  • $\begingroup$ Preferably inside material, which is more general (so I can plug in $J^\beta=0$ afterwards to get them in vacuum). $\endgroup$
    – occd2000
    Jan 26, 2021 at 8:26
  • $\begingroup$ Try simplifying $\partial^\alpha ( \partial_\alpha F_{\beta\gamma} + \partial_\gamma F_{\alpha\beta} + \partial_\beta F_{\gamma \alpha} ) = 0$. The thing in the bracket is the Bianchi identity (or Gauss-Faraday law). $\endgroup$
    – Prahar
    Jan 26, 2021 at 10:11

1 Answer 1


In the Lorentz gauge your first equation becomes the wave equation for the potential $$ \partial_\alpha \partial^\alpha A^\beta = \frac{4\pi}{c}J^\beta ~. $$ By deriving left and right hand side you obtain $$ \partial_\alpha \partial^\alpha F^{\gamma\beta} = \frac{4\pi}{c} \left(\partial^\gamma J^\beta - \partial^\beta J^\gamma \right) ~, $$ which is the required wave equation.


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