Tensor notation of Maxwell's equation read

$$\partial_\mu F^{\mu\nu} = j^\nu.$$

So when we explicitly try to find the Maxwell's equation from the above tensor equation we only get gauss law and curl of B. The div.B=0 and curl of E are not present. What is happening here?? I have obtained the above tensor equation from the four maxwell equation but when i try explicitly write the equation component wise some how two of those equations dont appear? I know it has something to do with two of those equations not being equations of motion, but i m still very unclear about this.

  • $\begingroup$ Hello. As written in the answer of Mr Logan below, to derive Maxwell' s equations from the Faraday tensor you need both the relationships, the one you mention and the other in the answer, since in general $F_{ab} = 2u_{[a} E_{b]} + ε_{abc} B^c $. and $E_a= F_{ab} u^b $ $ B_a=ε_{abc} F^{bc} /2$. $\endgroup$ May 19, 2016 at 7:47
  • $\begingroup$ The equation you wrote show that there exist a fluid that produces the field, and the Bianchi shows the existence of a potential. $\endgroup$ May 19, 2016 at 7:49
  • $\begingroup$ Related Lagrangian question: physics.stackexchange.com/q/71611/2451 and links therein. $\endgroup$
    – Qmechanic
    May 19, 2016 at 9:07

1 Answer 1


As is written here the two remaining equations follow from the Bianchi identity which says that the anti-symmetrized derivative is zero, ie. $$ \partial_{[a} F_{bc]} = \partial_{a} F_{bc}+\partial_{b} F_{ca}+\partial_{c} F_{ab} = 0 $$ (remember the $F_{\mu\nu}$ is antisymmetric itself!)

  • $\begingroup$ Still very unclear.. Can u show some detailed math? $\endgroup$ May 19, 2016 at 7:16
  • $\begingroup$ @SiddharthaDam Alternatively consider that if $A_\mu$ is the vector potential, then $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. The Bianchi identity follows trivially if you expand this. Or, in differential forms notation $F=\mathrm{d}A$, and it is a well known and easily derivable identity that $\mathrm{d}^2=0$, and the Bianchi identity essentially states that $\mathrm{d}^2F=0$. $\endgroup$ May 19, 2016 at 11:04

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