Can the charge conservation law be derived from Maxwell equations?

[timeant]

From Maxwell's equations \partial_\nu F^{\mu\nu}=J^{\mu}, one can derive charge conservation. The derivation is
0\equiv \partial_\mu \partial_\nu F^{\mu\nu}= \partial_\mu J^{\mu} { \Rightarrow}\partial_\mu J^{\mu}=0.
However, a circular reasoning exists in it. For the sake of better understanding, we suppose F^{kl} is an antisymmetric n-dimenstional (n > 2) tensor. We consider the following equation
\partial_l F^{kl}= J^{k}, \qquad n=3,4,5,\cdots \qquad (\star)
Where J^{k} is known source. If the source is chosen as \partial_k J^{k}\neq 0 (e.g. J^{k} \propto x^k), then the above equation(*) has no solutions. Hence, \partial_k J^{k}= 0 is one of preconditions of existence about solutions of the above equation (*). If \partial_k J^{k}= 0 is considerd as a corollary of Eq.(*) (0\equiv \partial_k \partial_l F^{kl}= \partial_k J^{k} { \Rightarrow}\partial_k J^{k}=0), and at the same time it is one of preconditions of existence about Eq.(*)'s solutions. It must involve circular reasoning. Therefore, \partial_k J^{k}= 0 is NOT a corollary of Eq.(*) for any n. When n=4, Eq(*) is one of Maxwell equations.

Hence the charge conservation law can NOT be derived from Maxwell equations.

 

[mfb]

You just rephrased it, nothing circular.

Hence, $\partial_k J^{k}= 0$ is one of preconditions of existence about solutions of the above equation (*)

In other words: You can't write down the Maxwell equations without having $\partial_k J^{k}= 0$. That's the original point: The Maxwell equations require charge conservation. If you don't have charge conservation then there are no Maxwell equations, or they have to look differently.

You can also take the historic approach: The Maxwell equations are older than charge conservation.

 

[Dale]

I agree with @mfb. That is not circular reasoning, that is just two different ways to say the same thing.

 

[vanhees71]

This is a specific feature of theories with a gauge symmetry. There is a constraint on the sources, i.e., $J^{\mu}$ which must be necessarily fulfilled in order for the theory to be gauge invariant. That's why you can deduce the necessity of charge conservation from the equations of the em. field alone without referring to the dynamics of the charges. This implies constraints for the dynamics of these charges, i.e., the total Lagrangian of the coupled system of matter and em. field must be gauge invariant. So there is no circularity in the argument here. It's just the necessity for a gauge theory to be gauge invariant. Usually, if you break somehow gauge invariance explicitly the theory looses its physical consistency. This is particularly true for any relativistic field theory with massless fields of spin $\geq 1$, which necessarily is a gauge theory if you don't want to have continuous intrinsic "polarization states". That's also the reason, why the em. field has only two polarization states, determined, e.g., by the helicities $\pm 1$ (building the basis of left- and right-circular polarized modes), and not 3 as expected from a spin-1 field. Indeed, massive vector fields have 3 polarization states.

 

[timeant]

I insist on my opinion.
If \partial_k J^{k} = 0 is seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is one circular reasoning.
If \partial_k J^{k} = 0 is NOT seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is NO circular reasoning.

@vanhees71, charge conservation is determined by Dirac field's gauge invariant, not by Maxwell's ones. We can say: charge conservation is consistent with e.m. field gauge invariant.

 

[Dale]

I insist on my opinion.

Have you a professional scientific reference which shares your opinion? If not then your opinion is personal speculation.

If \partial_k J^{k} = 0 is seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is one circular reasoning.

The reason that you will not find this incorrect objection in the literature is that $\partial_l F^{kl}= J^{k}$ is already the inhomogenous Maxwell’s equations. So something that is a consequence of $\partial_l F^{kl}= J^{k}$ is a consequence of Maxwell’s equations.

What you have discovered is not circularity. It is merely the fact that you only need Gauss’ law and Ampere’s law to deduce charge conservation, not all four of Maxwell’s (non tensor) equations.

If you can arrive at charge conservation from a subset of Maxwell’s equations then you have still arrived at charge conservation from Maxwell’s equations. If the homogenous equations are unnecessary or redundant it does not imply circularity.

The proof you first wrote, using both the homogenous and the inhomogenous equations is convenient and easy, but it is not necessary. You can, as you showed, prove it from the inhomogenous equations alone. Doing so is still a proof from Maxwell’s equations. The fact that such a proof exists in no way invalidates the other proof. It is not necessary for there to be only one way to prove something.

 

[vanhees71]

I insist on my opinion.
If \partial_k J^{k} = 0 is seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is one circular reasoning.
If \partial_k J^{k} = 0 is NOT seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is NO circular reasoning.

@vanhees71, charge conservation is determined by Dirac field's gauge invariant, not by Maxwell's ones. We can say: charge conservation is consistent with e.m. field gauge invariant.

The Dirac field is just one possible kind of field allowing for a local representation of the Poincare group including space reflections and time reversal.

Whether you have a gauge theory or not is determined by how you couple the vector field to it, and of course, if you gauge the $\mathrm{U}(1)$ symmetry (multiplication of the Dirac field with a pase factor), you get using minimal coupling QED, and that's a gauge theory.

In other words: If you take the free Dirac equation you have global a $\mathrm{U}(1)$ symmetry, leading according to one of Noether's theorems to a conserved charge. Now you make this symmetry local by introducing a gauge connection and the corresponding gauge field. Of course then the charge is still conserved, but it already follows from the equations for the gauge field alone, i.e., in order to have a gauge theory it is necessary to have charge conservation.

From the point of view of gauge symmetry the homogeneous Maxwell equations are Bianchi identities following necessarily from the properties of the gauge connection/curvature.

This is also clear from the point of view of the Maxwell equations for the Faraday tensor, i.e., the electromagnetic field, because due to the homomgeneous Maxwell equations you can introduce the four-potential, and the four-potential is not unique for any given physical situation but all four-potentials connected by a gauge transformation are equivalent.

 

[PeroK]

I insist on my opinion.
If \partial_k J^{k} = 0 is seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is one circular reasoning.
If \partial_k J^{k} = 0 is NOT seen as a consequence of \partial_l F^{kl}= J^{k} for any n, then there is NO circular reasoning.

It might be useful for you to define in mathematical-logical terms what you mean by "circular reasoning".

Note that in mathematical terms in general if proposition A implies proposition B, then proposition B could be viewed as a precondition for A. In the sense that if B fails then you can't have A.

In short, it appears that your objection would apply to any corollary.