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$. $\endgroup$ – Constantine Black May 19 '16 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. $\endgroup$ – Constantine Black May 19 '16 at 7:49
  • $\begingroup$ Related Lagrangian question: physics.stackexchange.com/q/71611/2451 and links therein. $\endgroup$ – Qmechanic May 19 '16 at 9:07

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? $\endgroup$ – Siddhartha Dam May 19 '16 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 $\mathrm{d}^2F=0$. $\endgroup$ – Bence Racskó May 19 '16 at 11:04

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.