The Einstein equation can be derived from the idea that energy causes the curvature of spacetime. Hence we have on the right-hand side of our equation the energy-momentum tensor $T_{\mu\nu}$ and need on the left-hand side something which describes curvature. The unique object which we can write on the LHS is the Einstein tensor $G_{\mu \nu}$ since it has divergence zero and carries two indices: $$\Rightarrow \quad G_{\mu \nu} = 8 \pi G T_{\mu \nu}$$

Analogously, we can argue that we can derive the inhomogeneous Maxwell equations from the idea that electric charge causes electromagnetic fields. Hence, we have on the right-hand side the electric current $J_\mu$ and need on the left-hand side something which describes the electromagnetic field. A divergence-free object with one index, we can write on the LHS is $\partial^\nu F_{\mu \nu}$: $$ \Rightarrow \quad \partial^\nu F_{\mu \nu} = \mu_0 J_\mu $$ Is this choice unique in some sense, analogous to what we did for the Einstein equation or are additional terms possible?

  • 1
    $\begingroup$ Would you allow magnetic monopoles? Breaking of Lorentz symmetry? $\endgroup$
    – Qmechanic
    Oct 18, 2018 at 8:50
  • $\begingroup$ Related: physics.stackexchange.com/q/3618/2451 $\endgroup$
    – Qmechanic
    Oct 18, 2018 at 8:53
  • $\begingroup$ @Qmechanic I would so no, although it's, of course, interesting to see how Maxwell equations get modified in these cases. (Also, e.g. when electric charge wouldn't be conserved.) $\endgroup$
    – jak
    Oct 18, 2018 at 8:58
  • $\begingroup$ i would object to Einstein equations being unique. First of all, one may add additional scalar fields like in Brans-Dicke theory ( en.wikipedia.org/wiki/Brans%E2%80%93Dicke_theory ). Then one may consider more polynomials of the Einstein Tensor. Furthermore, one might consider theories with non-zero Torsion etc. $\endgroup$ Oct 18, 2018 at 10:47
  • $\begingroup$ You probably should drop the $q$ in your second equation. $\endgroup$
    – my2cts
    Oct 18, 2018 at 13:08

1 Answer 1


The origin of Maxwell's equations is completely phenomenological at a macroscopic level. In terms of fields only, as it was shown by Einstein in 1905, they encode special relativity as the underlying symmetry and, moreover, they are linear partial differential equations of first order in the fields. Moving to the Lorentz-covariant Minkowski spacetime approach, whose Lagrangian implementation necessarily brings into picture the 4-potentials as an equivalent description of the theory, the only admissible (i.e. second order in the derivatives of the potentials) kinetic term in the Lagrangian density is (up to a conveniently chosen numerical factor) $F_{\mu\nu}F^{\mu\nu}$. Encoding field-sources (stationary electric charges and moving electric charges/currents) in a tensorial object such as $j_\mu$ and assuming minimal coupling (here one can prove that a minimal coupling conserved current-potential is a must), one finds that:

$$ \partial^{\mu} F_{\mu\nu} = \kappa j_{\nu}$$

is the equivalent of Einstein's equations of GR, along with $\partial^{[\mu}F^{\nu\sigma]} = 0$ which is required by the antisymmetry of the Faraday tensor, in turn a consequence of having the 4-current conserved.

One further point. We can try to deform $ \partial^{\mu} F_{\mu\nu} = \kappa j_{\nu}$ to, let us say $ \partial^{\mu} F_{\mu\nu} + a \partial_\nu \partial^{\mu} A_{\mu} = \kappa j_{\nu}$, but at the price of losing both current conservation, and the link to the phenomenological equations in terms of $\vec{E}, \vec{B}$.

  • $\begingroup$ thanks! Do you have a reference or further information on how "one can prove that a minimal coupling conserved current-potential is a must"? $\endgroup$
    – jak
    Oct 18, 2018 at 10:04
  • $\begingroup$ In a far more general context (field theory only in flat D-dim spacetime), it is proven here: doi.org/10.1002/1521-3889(200111)10:11/… $\endgroup$
    – DanielC
    Oct 18, 2018 at 10:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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