Noether charge of local symmetries

Assume that the Lagrangian density

$$\tag{1} {\cal L} ~=~ {\cal L}(\phi(x), \partial \phi(x), x) $$

does not depend on higher-order derivatives $\partial^2\phi$, $\partial^3\phi$, $\partial^4\phi$, etc. Let

$$\tag{2} \pi^{\mu}_{\alpha} ~:=~ \frac{\partial {\cal L}}{ \partial (\partial_{\mu}\phi^{\alpha})} $$

denote the de Donder momenta, and let

$$\tag{3} E_{\alpha}~:=~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} - d_{\mu} \pi^{\mu}_{\alpha} $$

denote the Euler-Lagrange equations. Let us for simplicity assume that the infinitesimal local quasi-symmetry$^1$ transformation

$$\tag{4} \delta_{\varepsilon} \phi^{\alpha}~=~ Y^{\alpha}(\varepsilon) ~=~Y^{\alpha}\varepsilon + Y^{\alpha,\mu} d_{\mu}\varepsilon $$

is vertical$^2$ and that it does not depend on higher-order derivatives of the infinitesimal $x$-dependent parameter $\varepsilon$. [It is implicitly understood that the structure coefficients $Y^{\alpha}$ and $Y^{\alpha\mu}$ are independent of the parameter $\varepsilon$. If the theory has more that one symmetry parameter $\varepsilon^a$, $a=1, \ldots m$, we are just investigating one local symmetry (and its conservations law) at the time.] The bare Noether current $j^{\mu}(\varepsilon)$ is the momenta times the symmetry generators

$$\tag{5} j^{\mu}\varepsilon + j^{\mu,\nu}d_{\nu}\varepsilon ~=~j^{\mu}(\varepsilon) ~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}(\varepsilon) ,$$

$$\tag{6} j^{\mu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha}, \qquad j^{\mu,\nu}~:=~ \pi^{\mu}_{\alpha}Y^{\alpha,\nu}. $$

(Again, it is implicitly understood that the structure coefficients $j^{\mu}$ and $j^{\mu\nu}$ are independent of the parameter $\varepsilon$, and so forth.) That the infinitesimal transformation (4) is a local quasi-symmetry$^1$ implies that variation of the Lagrangian density ${\cal L}$ wrt. (4) is a total space-time divergence

$$ d_{\mu} f^{\mu}(\varepsilon) ~=~ \delta_{\varepsilon} {\cal L} ~\stackrel{\begin{matrix}\text{chain}\\ \text{rule}\end{matrix}}{=}~ \frac{\partial {\cal L}}{ \partial \phi^{\alpha}} Y^{\alpha}(\varepsilon) + \pi^{\mu}_{\alpha}d_{\mu}Y^{\alpha}(\varepsilon) $$ $$\tag{7} ~\stackrel{\begin{matrix}\text{Leibniz'}\\ \text{rule}\end{matrix}}{=}~ E_{\alpha}Y^{\alpha}(\varepsilon) + d_{\mu} j^{\mu}(\varepsilon). $$

Here$^3$

$$ \tag{8} f^{\mu}(\varepsilon) ~=~ f^{\mu}\varepsilon + f^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} f^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon $$

are some functions with

$$\tag{9}f^{\mu,\nu\lambda}~=~f^{\mu,\lambda\nu}. $$

The full $\varepsilon$-dependent Noether current $J^{\mu}(\varepsilon)$ is defined as$^3$

$$\tag{10} J^{\mu}\varepsilon + J^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} J^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~J^{\mu}(\varepsilon) ~:=~ j^{\mu}(\varepsilon) - f^{\mu}(\varepsilon), $$

where

$$\tag{11}J^{\mu,\nu\lambda}~=~J^{\mu,\lambda\nu}. $$

Eqs. (7) and (10) imply the $\varepsilon$-dependent off-shell Noether identity

$$ \tag{12} d_{\mu} J^{\mu}(\varepsilon) ~=~ -E_{\alpha}Y^{\alpha}(\varepsilon) . $$

The $\varepsilon$-dependent off-shell Noether identity (12) is the key identity. Decomposing it in its $\varepsilon$-independent components leads to the following set (13)-(16) of identities,

$$ \tag{13} d_{\mu}J^{\mu} ~=~-E_{\alpha} Y^{\alpha} , $$

$$ \tag{14} J^{\mu} + d_{\nu} J^{\nu,\mu}~=~-E_{\alpha} Y^{\alpha,\mu} ,$$

$$ \tag{15} J^{\nu,\lambda}+J^{\lambda,\nu}+d_{\mu}J^{\mu,\nu\lambda} ~=~0 , $$

$$ \tag{16} \sum_{{\rm cycl}.~\mu,\nu,\lambda}J^{\mu,\nu\lambda} ~=~0, $$

in accordance with Noether's second theorem. Eq. (13) is just the usual off-shell Noether identity, which can be derived from the global symmetry alone via Noether's first theorem (where $\varepsilon$ is $x$-independent). As is well-known, the eq. (13) implies an on-shell conservation law

$$ \tag{17} d_{\mu}J^{\mu}~\approx~ 0, $$

or more explicitly written as

$$ \tag{18} \frac{d Q}{dt}~\approx~ 0,\qquad Q~:=~\int_{V} \! d^3V ~J^0. $$

(Here the $\approx$ sign denotes equality modulo Euler-Lagrange equations $E_{\alpha}\approx 0$. We have assume that the currents $J^i$, $i\in\{1,2,3\}$, vanish at the boundary $\partial V$.)

The remaining eqs. (14)-(16) may be repackaged as follows. Define the second Noether current ${\cal J}^{\mu}(\varepsilon)$ as$^4$

$$ \tag{19} {\cal J}^{\mu}\varepsilon + {\cal J}^{\mu,\nu}d_{\nu}\varepsilon +\frac{1}{2} {\cal J}^{\mu,\nu\lambda}d_{\nu}d_{\lambda}\varepsilon ~=~ {\cal J}^{\mu}(\varepsilon)~:= ~ J^{\mu}(\varepsilon)+ E_{\alpha} Y^{\alpha,\mu}\varepsilon. $$

It satisfies an $\varepsilon$-dependent off-shell conservation law

$$ d_{\mu} {\cal J}^{\mu}(\varepsilon) ~\stackrel{(12)+(19)}{=}~ -E_{\alpha}Y^{\alpha}(\varepsilon)+d_{\mu}(E_{\alpha} Y^{\alpha,\mu}\varepsilon)$$ $$ \tag{20}~\stackrel{(13)+(14)}{=}~ - \varepsilon d_{\mu}d_{\nu} J^{\nu,\mu}~\stackrel{(15)}{=}~\frac{\varepsilon}{2}d_{\mu}d_{\nu}d_{\lambda} J^{\lambda,\mu\nu}~\stackrel{(16)}{=}~0 . $$

One may introduce a so-called superpotential ${\cal K}^{\mu\nu}(\varepsilon)$ as$^3$

$$ {\cal K}^{\mu\nu}\varepsilon+{\cal K}^{\mu\nu,\lambda}d_{\lambda}\varepsilon~=~{\cal K}^{\mu\nu}(\varepsilon)~=~-{\cal K}^{\nu\mu}(\varepsilon) $$ $$~:=~ \left(\frac{1}{2} J^{\mu,\nu}-\frac{1}{6}d_{\lambda}J^{\mu,\nu\lambda}\right)\varepsilon+ \frac{1}{3} J^{\mu,\nu\lambda}d_{\lambda}\varepsilon-(\mu\leftrightarrow \nu)$$ $$ \tag{21}~\stackrel{(14)+(16)}{=}~ \left( J^{\mu,\nu}+\frac{1}{3}d_{\lambda}(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda})\right)\varepsilon+ \frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\lambda}\varepsilon$$

A straightforward calculation

$$ d_{\nu}{\cal K}^{\mu\nu}(\varepsilon) ~\stackrel{(15)+(21)}{=}~J^{\mu,\nu}d_{\nu}\varepsilon -\varepsilon d_{\nu}\left(J^{\nu,\mu}+d_{\lambda}J^{\lambda,\mu\nu}\right)$$ $$ \tag{22}+\frac{\varepsilon}{3}d_{\nu}d_{\lambda}\left(J^{\lambda,\mu\nu}-J^{\mu,\nu\lambda}\right) +\frac{1}{3}\left( J^{\mu,\nu\lambda}-J^{\nu,\mu\lambda}\right)d_{\nu}d_{\lambda}\varepsilon ~\stackrel{(14)+(16)+(19)}{=}~{\cal J}^{\mu}(\varepsilon)$$

shows that ${\cal K}^{\mu\nu}(\varepsilon)$ is the superpotential for the second Noether current ${\cal J}^{\mu}(\varepsilon)$. The existence of the superpotential ${\cal K}^{\mu\nu}(\varepsilon)=-{\cal K}^{\nu\mu}(\varepsilon)$ makes the off-shell conservation law (20) manifest

$$ \tag{23}d_{\mu}{\cal J}^{\mu}(\varepsilon)~\stackrel{(22)}{=}~d_{\mu}d_{\nu}{\cal K}^{\mu\nu}(\varepsilon)~=~0. $$

Moreover, as a consequence the superpotential (22), the corresponding second Noether charge ${\cal Q}(\varepsilon)$ vanishes off-shell

$$ \tag{24}{\cal Q}(\varepsilon)~:=~\int_{V} \! d^3V ~{\cal J}^0(\varepsilon) ~=~\int_{V} \! d^3V ~d_i{\cal K}^{0i}(\varepsilon) ~=~\int_{\partial V} \! d^2\!A_i ~{\cal K}^{0i}(\varepsilon)~=~0, $$

if we assume that the currents ${\cal J}^{\mu}(\varepsilon)$, $\mu\in\{0,1,2,3\}$, vanish at the boundary $\partial V$.

We conclude that the remaining eqs. (14)-(16) are trivially satisfied, and that the local quasi-symmetry doesn't imply additional non-trivial conservation laws besides the ones (13,17,18) already derived from the corresponding global quasi-symmetry. Note in particular, that the local quasi-symmetry does not force the conserved charge (18) to vanish.

This is e.g. the situation for gauge symmetry in electrodynamics, where the off-shell conservation law (20) of the second Noether current ${\cal J}^{\mu}=- d_{\nu}F^{\nu\mu}$ is a triviality, cf. also this and this Phys.SE posts. Electric charge conservation follows from global gauge symmetry alone, cf. this Phys.SE post. Note in particular, that there could be a nonzero surplus of total electric charge (18).

--

$^1$ An off-shell transformation is a quasi-symmetry if the Lagrangian density ${\cal L}$ is preserved $\delta_{\varepsilon} {\cal L}= d_{\mu} f^{\mu}(\varepsilon)$ modulo a total space-time divergence, cf. this Phys.SE answer. If the total space-time divergence $d_{\mu} f^{\mu}(\varepsilon)$ is zero, we speak of a symmetry.

$^2$ Here we restrict for simplicity to only vertical transformations $\delta_{\varepsilon} \phi^{\alpha}$, i.e., any horizontal transformation $\delta_{\varepsilon} x^{\mu}=0$ are assumed to vanish.

$^3$ For field theory in more than one space-time dimensions $d>1$, the higher structure functions $f^{\mu,\nu\lambda}=-J^{\mu,\nu\lambda}$ may be non-zero. However, they vanish in one space-time dimension $d=1$, i.e. in point mechanics. If they vanish, then the superpotential (21) simplifies to ${\cal K}^{\mu\nu}(\varepsilon)=J^{\mu,\nu}\varepsilon$.

$^4$ The second Noether current is defined e.g. in M. Blagojevic and M. Vasilic, Class. Quant. Grav. 22 (2005) 3891, arXiv:hep-th/0410111, subsection IV.A and references therein. See also Philip Gibbs' answer for the case where the quasi-symmetry is a symmetry.

In its most comprehensible derivation, Noether's procedure derives the current by considering the global symmetry transformation whose parameters $\epsilon$ are made to depend on the spacetime coordinates. Because $\delta S$ has to vanish if $\epsilon$ is constant, the actual variation $\delta S$ in the generalized case has to be proportional to the integral of spacetime derivatives $\partial_\mu \epsilon$ multiplied by some coefficients $J^\mu$, the currents. By integrating by parts, one may then show that the current obeys the continuity equation if the equations of motion are satisfied.

Now, when the symmetry is actually local, the "generalization" of the global transformation isn't a real generalization: it's a symmetry by itself. So because the action is locally symmetric, $\delta S$ vanishes for any configuration $\epsilon(x^\alpha)$, including a non-constant one, which means that all the coefficients $J^\mu$ actually vanish themselves, as you said. These conditions (constraints plus equations of motion) can be equivalently obtained from the variation of fields like $A_\mu$.

Because the currents are classically vanishing – or, using a more general description in quantum mechanics with an extended Hilbert space, they have to annihilate the physical states in quantum mechanics – it really means that the local symmetry is gauge. You don't need to assume this fact; we have just derived it. So you can't avoid it, either.

Lets consider a local symmetry continuous. Look at an infinitesimal element. By that I mean the following transformation of fields: \begin{equation} \phi_a(x)\to\phi_a(x)+F_\alpha[\phi]g^\alpha(x)+F_\alpha^\mu[\phi]\partial_\mu g^\alpha(x)+\ldots\tag{1} \end{equation} Here $\phi_a$ is the whole set of fields, $F$ are some functionals specific to the symmetry, and $g^\alpha$ is the set of infinitesimal parameters that can depend on the spacetime point $x$. Note that I consider only local symmetries (as opposed to non-local), so $F$ can depend only on $\phi$ and a finite number of its derivatives at the point $x$. I also suppose, for simplicity, that this is a symmetry of the Lagrangian (i.e. no total derivatives pop out).

Now, note that the rhs of $(1)$ is just a specific variation of fields, that leave the action invariant. By using the same technique as in deriving the EOMs, we can write this invariance as $$ \delta S=\int d^dx R_\alpha(\phi,\phi',\phi'',\ldots) g^\alpha(x)=0, $$ which leads to identities $$ R^\alpha(\phi,\phi',\phi'',\ldots)=0. $$ If we do this using the explicit form of the Lagrangian, these identities should be tautologic. However, if we do not use the specific form of the Lagrangian, but rather porceed in a manner similar to the derivation of the Euler-Lagrange equations, we will get identities involving derivatives of the Lagranginan. As it is pointed out in the article Jia Yiyang refers to, one can combine these with the Euler-Lagrange equations to get general on-shell identities involving derivatives of the Lagrangian, which reduces to the usual current conservation when it comes to a specific Lagrangian. However, I have not read the article carefully, so I may be missing some additional thoughts.

But this does not fully address the question. Note that by fixing a specific $g$ one may forget about the local nature and treat \begin{equation} \phi_a(x)\to\phi_a(x)+\epsilon\left\{F_\alpha[\phi]g^\alpha(x)+F_\alpha^\mu[\phi]\partial_\mu g^\alpha(x)+\ldots\right\}\tag{2} \end{equation} as a global symmetry parametrised by $\epsilon$. This includes the traditional global symmetry as the specific case $g=const$. One may proceed with the traditional procedure described by Luboš Motl to derive the Noether's current. We know that there is at least one choice of $g$ corresponding to a good conserved current.

Proceed as follows. Allow $\epsilon$ to depend on $x$: $\epsilon=\epsilon(x)$. Note that this does not correspond to a local symmetry anymore [unless there is no derivatives in $g$ in $(2)$]. Then derive the first variation of action. If $\epsilon$ is constant, it is a symmetry, so the linear term should depend on $\partial_\mu\epsilon$ only: $$ \delta S=\int d^dx \partial_\mu\epsilon J^\mu[g]. $$ Here I have indicated that the current depends on the choice of $g$. If we assume that the equations of motion are satisfied, then the first variation of action is zero. This means that this integral vanishes, and by integration by parts and choosing $\epsilon$ in form of appropriate local bumps, we show that $J^\mu[g]$ is conserved.

Haha, just kidding! If we assume that the equations of motion are satisfied, then the first variation of action is zero. -- is a lie. EOMs imply vanishing of the first variation only if the variation of fields has compact support/decays sufficiently rapidly at infinity/etc. This is because the EOMs are connected with the first variation by a number of integrations by parts. This is no problem for the above, because it suffices to consider only $\epsilon$ with compact support to show that $$ \int d^dx \partial_\mu\epsilon J^\mu[g]=0\tag{3} $$ implies $$ \partial_\mu J^\mu[g]=0. $$ However, this little subtlety is extremely important for our problem. If $(3)$ was true for any $\epsilon$, the choice of $\epsilon=\delta\:\theta(x^0-t),\,\delta\ll 1$ would lead to $$ \int d^{d-1}x J^0[g]=0\tag{4}. $$ That is, it would imply that the associated Noether's charge is zero on equations of motion! But we know that it is not true for the electric charge! That is because $g=const$ together with this neat choice of $\epsilon$ leads to a variation of fields that does not decay rapidly enough.

But lets us call the local symmetries only those choices of $g$ that decay rapidly at infinity, and global symmetries are those $g$ that are not so well-behaved. Consider a local symmetry given by $g$. Now $g$ decays radily at the infinity, and take $\epsilon=\delta\:\theta(x^0-t),\,\delta\ll 1$ (or, if you are being pedantic, some smooth approximations), which is bounded. So, the variation $(2)$ is now well-behaved and we have $(3)$ satisfied! This means that for local symmetries the Noether charges are always zero. For global symmetries we do not have such nice field variations, so the charges are allowed to be non-zero.

Example: Take the lagrangian $$ L=i\bar{\psi}\gamma^\mu(\partial_\mu+iA_\mu)\psi-\frac{1}{4q^2}F_{\mu\nu}F^{\mu\nu}, $$ and for $(2)$ the transformation $$ \psi\to\psi+i\epsilon g\psi\\ A_\mu\to A_\mu-\epsilon\partial_\mu g. $$ The associated current is $$ J^\mu[g]=-\bar\psi\gamma^\mu\psi g+\frac{1}{q^2}F^{\mu\nu}\partial_\nu g. $$ Now $g=-q$ corresponds to the traditional electric current. You can check that it is conserved by using the traditional current conservation and the EOM $\partial_\mu F^{\mu\nu}=q^2\bar{\psi}\gamma^\nu\psi$. When you try to compute the charge, you get, schematically $$ Q=\int d^3x\left(\rho g+E\cdot\nabla g\right) $$ you use $E\cdot\nabla g=\mathrm{div}(gE)-g\mathrm{div} E=\mathrm{div}(gE)-g\rho$, and you now have the integral $$ Q=\int_{S^2} gE_n dS $$ over a very large sphere $S^2$. If $g$ decays rapidly at the infinity, you have zero. If $g$ has non-zero averages over large spheres, then the integral converges only if these averages approach a constant value, and in that case the $Q$ is given by the Guass-Ostrogradsky theorem as the total electric charge times this constant.

So, my conclusion is:

• The charges associated to local symmetries are either zero, proportional to the global charge(if the transformation does not 'decay' rapidly enough), or divergent (senseless). The reason that the global symmetry is allowed to have a charge is its 'bad' behavior at the infinity.