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?

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    $\begingroup$ Yes. In fact, that's precisely what happens if you try and go through applying Noether's theorem to a gauge symmetry. You obtain that the conservation is off-shell too. $\endgroup$
    – genneth
    Commented Oct 20, 2011 at 8:09
  • $\begingroup$ In Classical Electrodynamics a charge $q$ is defined to be constant whatever happens to the particle. Of course, the motion equations are compatible and should be compatible with this definition. There is no physical symmetry behind it. Of course, one can invent equations incompatible with the charge conservation, for example, a diffusion equation with a sink. If no sinks/sources are implemented, the charge will be conserved. $\endgroup$ Commented Oct 20, 2011 at 10:19
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    $\begingroup$ One can express $\frac{dq}{dt}$ from the equations. If you use the true solutions in such expression, the charge will not depend on time. If you put arbitrary functions of time in this expression, there is no guarantee that $\frac{dq}{dt}=0$. Off-shell "solutions" may have such a drawback. $\endgroup$ Commented Oct 20, 2011 at 10:31
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    $\begingroup$ Related: physics.stackexchange.com/a/13881/2451 $\endgroup$
    – Qmechanic
    Commented Jan 17, 2013 at 23:07
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    $\begingroup$ @genneth Your statement is incorrect; see arxiv.org/abs/hep-th/0009058 for a detailed discussion. Conservation of charge charge follows from either the gauge field's or the matter field's equation of motion, but you do need to assume one. There are certainly "completely off-shell" field configurations in which charge is not conserved. $\endgroup$
    – tparker
    Commented Jul 21, 2017 at 3:09

2 Answers 2


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.

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    $\begingroup$ There is at least a stretch in your "proof" (tautology=gauge invariance). The equation $j=\partial F$ is not a definition of a charge but that of fields. So one can easily write down a $j$ incompatible with charge conservation. ;-) $\endgroup$ Commented Oct 20, 2011 at 11:04
  • $\begingroup$ Ah, okay. So charge conservation just reproduces (Bianchi) identities and then we can use the field equations to get it's implications for the source fields. Thanks @Lubos! $\endgroup$
    – Siva
    Commented Oct 21, 2011 at 4:55
  • $\begingroup$ @VladimirKalitvianski , If $j$, in the equation $j=\partial F$, is incompatible with charge conservation, then the Gauge invariance is broken. The variation of the action with respect to a gauge transformation is given by $\partial (\partial F-j) - eom´s$, so if the equations of motion are satisfied the charge must be conserved in order to preserve the gauge invariance $\endgroup$
    – Nogueira
    Commented Jun 1, 2017 at 21:27

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!!!

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    $\begingroup$ What is the interpretation of such currents? Is there any literature where they are discussed and used? $\endgroup$
    – Nikita
    Commented Feb 4 at 18:50
  • $\begingroup$ @Nikita Sadly I have not found this expression of current in any textbooks. It’s just a computation by hand. Nor do I know the interpretation. It’s said by other answers in PSE that Noether currents of local gauge symmetries are not gauge invariant so that it is not physical. $\endgroup$ Commented Feb 5 at 5:32
  • $\begingroup$ Maybe you can found the discussion of this current in this PSE answer. $\endgroup$ Commented Feb 5 at 5:40
  • $\begingroup$ current in your answer is gauge invariant... $\endgroup$
    – Nikita
    Commented Feb 6 at 6:39
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    $\begingroup$ Oh....In the answer I gave link above, he proved the charge to be zero. I understand its proof but I wonder that since charge corresponds to quantum generator of the transformation, how can it be zero? $\endgroup$ Commented Feb 6 at 10:21

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