Why does charge conservation due to gauge symmetry only hold on-shell?

Question

While deriving Noether's theorem or the generator(and hence conserved current) for a continuous symmetry, we work modulo the assumption that the field equations hold. Considering the case of gauge symmetry: to my understanding, it's a redundancy in the "formulation" of a theory itself. So, shouldn't it lead to quantities which are conserved irrespective of whether the field equations hold?

Answer 1

Whether your current $j^\mu$ is conserved off-shell depends on your definition of $j^\mu$. If you define it via the Dirac and other charged fields, it will only be conserved assuming the equations of motion.

However, if you define $j^\mu$ via $$ j^\mu = \partial^\nu F_{\mu\nu}, $$ i.e. as a function of the electromagnetic field and its derivatives, then $\partial_\mu j^\mu=0$ holds tautologically because it is $$\partial_\mu j^\mu= \partial_\mu\partial_\nu F^{\mu\nu} =0$$ which vanishes because the $\mu\nu$-symmetric second derivatives are applied to a $\mu\nu$-antisymmetric field strength tensor. The possibility to make the local conservation law tautological is indeed linked to the existence of a gauge symmetry. Why? Because it's the equation of motion one may derive from variations of the fields that are equivalent to gauge transformations: the vanishing of the variation of the action under such variations is guaranteed even without the equations of motion, by the gauge symmetry, so the corresponding combination of the currents, $\partial_\mu j^\mu$, has to vanish identically.

This logic also guarantees that the Dirac and other charged field coupled to electromagnetism will have equations of motion that guarantee the local charge conservation.

An analogous statement exists in the case of the diffeomorphism symmetry: $$\nabla_\mu G^{\mu\nu} = 0$$ also holds tautologically for the Einstein tensor $G$ defined in terms of the metric tensor and its derivatives.

Answer 2

The above answer is partly wrong, to my understanding. (the Noether current for local gauge transformation is not $J^\mu = \partial_\nu F^{\mu\nu}$)

The Lagrangian $$ \mathcal{L} = F^{\mu\nu} F_{\mu\nu} $$ is invariant off shell under local gauge symmetry $$ A_\mu(x) \mapsto A_\mu (x) - \partial_\mu \theta(x) $$ The Noether current is $$ J^\mu = \frac{\delta{L}}{\delta \partial^\mu A_\nu} \delta A_\nu = F^{\mu\nu} \partial_\nu \theta(x) $$ But $J^\mu$ is conserved only on-shell: $$ \partial_\mu J^\mu = \partial_\mu F^{\mu\nu} \partial_\nu\theta(x) + F^{\mu\nu}\partial_\mu \partial_\nu \theta(x) $$ The second term vanishes identically(off-shell), because of anti-symmetricity of $F^{\mu\nu}$. But the first term is proportional to the equation of motion for the photon: $$ \partial_\mu F^{\mu\nu} = 0 $$ which is not tautological!!!

The reason is, by checking the proof for Noether’s theorem, the local gauge symmetry is not the ‘local version’ of the global gauge symmetry. The later, which is used in the derivation of Noether’s theorem, reads: $$ A_\mu (x) \mapsto A_\mu(x) - \epsilon(x) \partial_\mu \theta(x) $$ which is NOT a local gauge transformation!!!