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This is a popular question on this site but I haven't found the answer I'm looking for in other questions. It is often stated that charge conservation in electromagnetism is a consequence of local gauge invariance, or perhaps it is due to some global phase symmetry. Without talking about scalar or spinor fields, the EM Lagrangian that we're familiar with is: $$ \mathcal{L} = -\frac{1}{4}F^2 - A \cdot J + \mathcal{L}_\mathrm{matter}(J) $$ The equation of motion for $A$ is simply $$ \partial_\mu F^{\mu \nu} = J^\nu $$ From which it follows that $J$ is a conserved current (by the antisymmetry of the field strength). But what symmetry gave rise to this? I'm not supposing that my matter has any global symmetry here, that I might be able to gauge. Then so far as I can tell, the Lagrangian given isn't gauge invariant. The first term is, indeed, but the second term only becomes gauge invariant on-shell (since I can do some integrating by parts to move a derivative onto $J$). If we demand that our Lagrangian is gauge-invariant even off-shell, then we can deduce that $\partial \cdot J = 0$ off-shell and hence generally. But we can't demand that this hold off-shell, since $J$ is not in general divergenceless!

For concreteness, suppose that $$ \mathcal{L}_\mathrm{matter}(J) = \frac{1}{2} (\partial_\mu \phi) (\partial^\mu \phi) \qquad J^\mu \equiv \partial^\mu \phi $$ Then we find that $\phi$ (a real scalar) satisfies some wave equation, sourced by $A$. The equations of motion here constrain the form of $J$, but off-shell $J$ is just some arbitrary function, since $\phi$ is just some arbitrary function. Then it is clear that the Lagrangian is not gauge-invariant off-shell.

And this is a problem, because when we derive conserved quantities through Noether's theorem, it's important that our symmetry is a symmetry of the Lagrangian for any field configuration. If it's only a symmetry for on-shell configurations, then the variation of the action vanishes trivially and we can't make any claims about conserved quantities.

So here's my question: what symmetry does the above Lagrangian have that implies the conservation of the quantity $J$, provided $A$ satisfies its equation of motion? Thank you.

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  • $\begingroup$ For the fact that global gauge symmetry implies electric charge conservation (in e.g. QED or scalar QED), see physics.stackexchange.com/q/48305/2451 and links therein. $\endgroup$ – Qmechanic Jan 20 '16 at 18:18
  • $\begingroup$ My Lagrangian has no global symmetry: $\phi$ is a real scalar. $\endgroup$ – gj255 Jan 20 '16 at 18:27
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    $\begingroup$ This is late, but such a Lagrangian can't exist. $\endgroup$ – knzhou Jul 19 '16 at 23:04
  • $\begingroup$ The symmetry is $\phi \to \phi + a$ $\endgroup$ – Prahar Jul 20 '16 at 21:04
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You should probably read up on the Stueckelberg action and the Affine Higgs mechanism it sends you to. Your boldface supposition "I'm not supposing that my matter has any global symmetry here, that I might be able to gauge" is unwarranted for the specific model you propose, $J_\mu=\partial_\mu\phi$.

There is a global symmetry, $\phi \to \phi+\alpha$, whose Noether current is, of course, the above.

You then almost gauged it, by giving it a charge 1, $A_\mu \to A_\mu+\partial_\mu\alpha (x)$, and $\phi \to \phi+m\alpha(x)$, where I'm adding the requisite dimensionful charge m so α is dimensionless, and the coupling term is $-mA_\mu\partial^\mu \phi$. This is almost gauge invariant, but not quite: there is a "seagull" term missing, as the variation of this coupling term has a net remainder $-m^2 A_\mu \partial^\mu \alpha$. (Stueckelberg, got rid of it by adding a mass term for the photon, $\frac{m^2}{2} ~~A_\mu A^\mu$ with the appropriate dimensionful double charge, $m^2$.)

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  • $\begingroup$ Thank you for the links, I had not come across the Stueckelberg action before. However, I cannot help but note that whilst my Lagrangian is almost gauge invariant, it isn't. And yet I still have current conservation. The bigger picture here is that I will get current conservation for arbitrary $J$, composed of arbitrary fields. It doesn't seem like one can make the general statement: there will always be some local transformation of fields that leaves the Lagrangian invariant. Indeed, in this very case, it doesn't seem there is one, as $\phi \to \phi + \alpha$ doesn't work.. $\endgroup$ – gj255 Jul 25 '16 at 14:12
  • $\begingroup$ Indeed if I modify my Lagrangian with a $\phi^2$ or $\phi^3$ term, then this transformation doesn't even yield a global symmetry any more.. $\endgroup$ – gj255 Jul 25 '16 at 14:16
  • $\begingroup$ I fear you've veered off into odd territory. The canonical rules are: If there is a global symmetry, there is a conserved current; and coupling gauge fields to that current systematically, one may upgrade the global symmetry to a local one; nothing more. I'm not sure there can be further conjectures to be confirmed or falsified here. $\endgroup$ – Cosmas Zachos Jul 25 '16 at 15:57
  • $\begingroup$ If you add your quadratic and cubic terms, the current you wrote is not conserved anymore, on shell. If you augment it, by fiat, to the one that is on-shell conserved, its charge need not transform the fields along a valid symmetry, and you only reorganized some eqns of motion into a form suggestive of an underlying symmetry, which you do not have off shell. Introducing fictitious auxiliary fields, as in supergravity, might possibly help you manufacture a symmetry which all fields obey, including the auxiliary ones. $\endgroup$ – Cosmas Zachos Jul 25 '16 at 16:13
  • $\begingroup$ I agree with your canonical rules, and if that's all there is to it, I'm happy. But I find it strange that people say 'charge conservation arises from so-and-so symmetry' when it appears straightforward to invent Lagrangians with charge conservation but no such symmetry. If I add my quadratic and cubic terms, the current I wrote is still conserved, through the equations of motion of the vector field: $\partial_\mu F^{\mu \nu} = J^\nu$. $\endgroup$ – gj255 Jul 26 '16 at 9:08
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In a relativistic field theory, a conserved quantity according to Noether's theorem is given by a conserved current (density) $J^\mu$, i.e. $\partial_\mu J^\mu=0$. Hence, the contradiction you suggest does not really exist. The symmetry corresponding to conservation of electric charge is indeed the global part of the $U(1)$ gauge symmetry.

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  • $\begingroup$ But my Lagrangian doesn't have a U(1) gauge symmetry. $\endgroup$ – gj255 Jan 20 '16 at 18:06
  • $\begingroup$ But that is not true. You state that it is gauge invariant if $\partial_\mu J^\mu=0$. According to Noether's theorem, this holds for a conserved quantity, even off-shell. $\endgroup$ – Frederic Brünner Jan 20 '16 at 18:08
  • $\begingroup$ My Lagrangian isn't gauge invariant off-shell though. $\endgroup$ – gj255 Jan 20 '16 at 18:10
  • $\begingroup$ You write that it is gauge invariant off-shell if we demand $\partial\cdot J=0$. We do not have to impose it, this is always true for a conserved quantity. $\endgroup$ – Frederic Brünner Jan 20 '16 at 18:14
  • $\begingroup$ Yes, so I only get gauge invariance if I demand the very thing I'm trying to prove. I want to find some symmetry that tells me $\partial \cdot J = 0$. It can't be gauge invariance, because the only way my Lagrangian is actually gauge invariant is if I suppose this relation holds. $\endgroup$ – gj255 Jan 20 '16 at 18:28
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You're right that that Lagrangian isn't in general gauge invariant. In addition to making the $A^\mu$ terms gauge invariant, the $\mathcal{L}(J)$ term must also be gauge-invariant. And not just the equations of motion for $J_\mu$, either - the specific algebraic expression for $J_\mu$ in terms of fundamental fields must be such that if you literally plug in the gauge transformation $\lambda(x)$ and simplify, all the terms involving $\lambda(x)$ must cancel out through pure algebra. For example, in QED $J_\mu$ isn't a single fundamental field, it's a composite field composed of the spinors $\Psi$: specifically, $J_\mu = e \bar{\Psi} \gamma_\mu \Psi$, and the $\Psi$'s also transform nontrivially under the gauge transformation, in such a way that $\partial_\mu J^\mu$ is a simple algebraic identity.

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