electromagnetism - Classical EM : clear link between gauge symmetry and charge conservation

In the case of classical field theory, Noether's theorem ensures that for a given action $$S=\int \mathrm{d}^dx\,\mathcal{L}(\phi_\mu,\partial_\nu\phi_\mu,x^i)$$ that stays invariant under the transformation $\{ x^i\rightarrow x^i+\delta x^i \,;\, \phi_\mu\rightarrow \phi_\mu+\delta \phi_\mu\}$, namely a local symmetry, then there exist conserved quantities named currents.

If it exist a set of variables $\{\epsilon_r,X_r^i,\Phi_r^\mu\}$ (where $\epsilon_r\rightarrow 0$) such that $\delta x^i=\epsilon_r\,X_r^i$ and $\delta \phi^\mu=\epsilon_r\,\Phi_r^\mu$, then these currents are the quantities : $$ J_r^{\,\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\Phi_r^\mu+\left[\mathcal{L}\delta_\sigma^{\,\nu}-\frac{\partial\mathcal{L}}{\partial(\partial_\nu\phi^\mu)}\partial_\sigma\phi^\mu\right]X_r^\sigma $$ These are conserved in the sens that $\partial_\nu J_r^{\,\nu}=0$, which is true on-shell, i.e. provided equations of motion are verified.

Now I'm trying to apply this result in the case of classical EM field theory. Considere the associated lagrangian : $$ \mathcal{L}_{EM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu j^{\,\mu} $$ where $A_\mu=(\phi,-\textbf{A})$ is the 4-potential and $j^{\,\mu}=(\rho,\textbf{j})$ is the 4-current, describing classical charge sources. $F_{\mu\nu}$ is the EM tensor.

In that case, equations of motion are : $$\partial_\nu F^{\nu\mu}=j^{\,\mu}$$

Considering the global gauge symmetry of the classical EM field theory $A_\mu\rightarrow A_\mu+\epsilon\,\partial_\mu\chi$ where $\chi$ is an arbitrary gauge function, I'd like to apply the Noether's theorem in the case where $X_r^i=0$. Refering to this P.S.E post, then the associated conserved current reads : $$ J^\mu=F^{\mu\nu}\partial_\nu\chi\quad\text{with}\quad\partial_\mu J^\mu=0 $$ Fine, now I have a few questions :

• Is it actually really legit to apply Noether's theorem to global gauge symmetry? I mean, is Noether's theorem still hold even if I have a set $\{\epsilon_r,X_r^i=0,\Phi_r^\mu\}$, i.e. no transformation on coordinates?

Related : physics.stackexchange.com/questions/13870/gauge-symmetry-is-not-a-symmetry.

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  In that case, what does the $J^\mu$ quantity physically means? One can compute, using equations of motion and the anti-symmetry of $F^{\mu\nu}$: $$ \partial_\mu J^\mu=\partial_\mu F^{\mu\nu}\,\partial_\nu\chi + F^{\mu\nu}\,\partial_\mu\partial_\nu\chi=j^{\,\nu}\,\partial_\nu\chi+0 $$ I would rather expect something like $$ \partial_\mu J^\mu=\partial_\mu j^{\,\mu}=0,\;\forall\chi $$ In that case I could understand that charge conservation stems directly from global gauge symmetry. There is something I'm clearly missing. Or should I trick by computing $\int\mathrm{d}x\,\partial_\mu J^\mu$ and perform an integration by part?

  
  


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  I can't see anywhere clearly appearing the fact that the physics of classical EM theory is invariant under Lorentz transformation. I would expect that such invariance would be a really strong symmetry of the system.

  
  

Answer

I) OP's model $$S[A]~=~\int\! d^4x~ \left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+J^{\mu}A_{\mu}\right),$$ is mentioned as point 4 in my Phys.SE answer here. Let us (to be self-contained) repeat: $J^{\mu}$ are treated as passive non-dynamical classical background matter sources. In other words, only the gauge fields $A_{\mu}$ are dynamical variables in this model. Before we even get started, we have to ensure local (off-shell) gauge symmetry of the action $S[A]$ up to boundary terms. This implies that the classical background sources $J^{\mu}$ must satisfy the continuity equation $d_{\mu}J^{\mu}=0$ off-shell. Thus a conservation law is forced upon us even before we apply Noether's Theorems. Note that global gauge symmetry is an empty statement in this model.

II) It seems that OP's problem with applying Noether's first theorem for global symmetry is related to the fact that he does not include the matter theory in his action. One should consider the full action $S[A,\Psi]$ of both gauge and matter fields.

Electric charge conservation then follows from global gauge symmetry of the full action $S[A,\Psi]$, cf. e.g. this and this Phys.SE posts.

III) Since global gauge symmetry is a subset of local gauge symmetry, then charge conservation in principle also follows from local gauge symmetry. But from a purist/minimalist point of view, it is overkill to use local gauge symmetry to prove charge conservation.

Local gauge symmetry is the realm of Noether's second theorem, cf. e.g. this Phys.SE post. The off-shell conservation law associated with the second Noether current is a triviality for electrodynamics.