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 sense 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. Consider 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 $X_r^i=0$. Referring to this Phys.SE 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: Gauge symmetry is not a symmetry?.

  • 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?

  • 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.

  • 3
    $\begingroup$ Strictly speaking, Noether's theorem applies only to global symmetries, and what you wrote down there is not a global symmetry since $\chi$ depends on spacetime. I don't understand your last question - the Yang-Mills Lagrangian is manifestly Lorentz invariant, and the quantity associated with the Lorentz symmetry is, as always, the stress-energy tensor. $\endgroup$
    – ACuriousMind
    Jan 14, 2015 at 17:11
  • 1
    $\begingroup$ Thanks for your comment. There is something I don't understand : so if here $A_\mu\rightarrow A_\mu +\partial_\mu\chi$ is not a global symmetry, then why the transformation $x^i\rightarrow x^i+\delta x^i\,,\,\phi_\mu\rightarrow \phi_\mu +\delta\phi_\mu $ should be called global symmetry since it's coordinate dependant too? Has the difference between local and global symmetry something to do with the fact that conservation laws can be derived on-shell or off-shell? $\endgroup$
    – dolun
    Jan 14, 2015 at 19:04
  • $\begingroup$ For the last question, I was just feeling that Lorentz invariance would "mix" with gauge invariance to ensure charge conservation, but I was surely mistaking. $\endgroup$
    – dolun
    Jan 14, 2015 at 19:05

2 Answers 2


Noether's (first) theorem is not applicable to local symmetries, such as gauge symmetries. For local symmetries, one should instead use the less-well-known second theorem of Noether. In general, one can state both her first and second theorem as follows:

Let $\mathscr{L}(\phi):= \frac{\partial L}{\partial \phi} - \partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$ denote the Lagrange expression for the field $\phi$. Let $\phi_i$ be the fields of our theory. Then, for any variation of the fields ($\delta\phi_i = \delta_0 \phi_i + \partial_\mu\phi_i\delta x^\mu$) and coordinates $$ \sum_i \mathscr{L}(\phi_i)\delta_0 \phi + \sum_i \partial_\mu (L\delta x^\mu + \frac{\partial L}{\partial(\partial_\mu \phi_i)}\delta_0\phi_i) = 0$$ holds if $\delta S = 0$.

From this, Noether's first theorem with the usual conserved current follows by taking the variation of the fields to be a global symmetry transformation, i.e. $\delta \phi_i = \sum_a \frac{\partial \delta \phi_i}{\partial \epsilon^a}\epsilon^a$ for the continuous parameter $\epsilon \in \mathbb{R}^N$ of the symmetry group, and recognizing that, on-shell, $\mathscr{L}(\phi_i) = 0$.

Noether's second theorem is what you get when the $\epsilon^a$ are allowed to depend on spacetime. If we allow such local transformations, then we have $\delta \phi_i = c_{ai}\epsilon^a + d^\mu_{ai}\partial_\mu\epsilon^a$ for some constants $c,d$. (In general, the local transformation may be allowed to depend on higher derivatives, too, but in the context of gauge theories it won't (and even there, the only derivative-dependent transformation is that of the gauge field itself)) Then, the statement of Noether's second theorem is

$$ \sum_i \mathscr{L}(\phi_i)c_{ai} - \sum_i \partial_\mu (\mathscr{L}(\phi_i)d^\mu_{ai}) = 0$$

In the case of electromagnetism, you have (at least) two ways to use these variations on Noether's theorem to get charge conservation: You can use the global gauge symmetry together with the matter field Euler-Lagrange equations, or you can use the local gauge symmetry together with the gauge field E-L equations (you have to be a bit tricky to do it the latter way - see the reference at the end).

The naive conserved current from the question, $J^\mu = F^{\mu\nu}\partial_\nu \chi$, has no physical significance, since it is not gauge-invariant, and hence not observable.

A wonderfully concise tour of the different versions of Noether's theorem in the context of gauge theories is given in "Noether's Theorems and Gauge Symmetries" by K. Brading and H. R. Brown.


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.

  • 1
    $\begingroup$ Thanks for your answer. So technically speaking if I'm well understanding : if I want to prove charge conservation I have to include matter field $\Psi$ in the model, right? Actually, background sources $J^\mu$ tautologically satisfy continuity equation just because they are sources. Is that right? $\endgroup$
    – dolun
    Jan 29, 2015 at 18:04
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jan 29, 2015 at 18:56

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