0
$\begingroup$

I am trying to understand the motivation behind one of the Maxwell equations (say in vaccum rather than matter for simplicity).

As far as I understand in magnetostatics there is a theorem which says $$rot \vec B=\frac{4\pi}{c}\vec j\tag{1}$$ where $\vec B$ is a constant magnetic field, $\vec j$ is a current density.

This result cannot be true for time dependend magnetic fields as it contradicts to the continuity equation $\frac{\partial \rho}{\partial t} +div (\vec j)=0$. Hence Maxwell observed that if one modifies the equation as $$rot \vec B=\frac{4\pi}{c}\vec j +\frac{1}{c}\frac{\partial \vec E}{\partial t}\tag{2}$$ the contradiction is removed. Moreover for time independent fields (2) becomes (1).

This procedure is highly non-unique. In fact one can add to the right hand side of (1) any divergency free field which vanishes for time independent fields, e.g. $\frac{\partial \vec B}{\partial t}$. Are there any other a priori reasons leading to equation (2)?

I realize that a posteriori this equation was confirmed by all the experiments. Was Maxwell just lucky?

$\endgroup$
2
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/182652/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 19, 2020 at 17:30
  • $\begingroup$ @Qmechanic: The answer at your link does not answer my question. It is just explained that without the correction there is an inconsistency. But it is not explained why this specific correction is chosen among infinitely many other corections. $\endgroup$
    – MKO
    Apr 19, 2020 at 18:15

1 Answer 1

1
$\begingroup$

Adding a solenoidal field to (1) does not change the underlying problem namely that $\mathbf{J}$ is not solenoidal unless the charge distribution is static and hence the current is stationary. You still have the unresolved problem of a vector equation being solenoidal on the left side and non-solenoidal on the right side manifesting itself with the same problem of the flux between the plates of a capacitor. Since we (Coulomb. Ampere, Maxwell...) understand the physics of a dielectric-filled capacitor reasonably well it is reasonable to modify the equation so that some of the contradiction is removed by including the polarization current after which the real puzzlement that remains is what happens in a vacuum.

You may always add a lamellar field to $\mathbf{B}$ since the curl of the the sum does not change (curlgrad=0) but that lamellar field is not observable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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