If our Lagrangian is invariant under a local symmetry, then, by simply restricting our local symmetry to the case in which the transformation is constant over space-time, we obtain a global symmetry, and hence a corresponding Noether charge.

Because, however, this Noether charge didn't come from just any old symmetry, but in fact, a local symmetry, we might be able to say something special about it. In particular, I believe that the Noether charge should vanish, but I don't know why. If this is indeed the case, how do we prove it?

(Note that I don't want to make the assumption that local=gauge (i.e. non-physical).)


5 Answers 5


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

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


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

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$^3$

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


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

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.

  • $\begingroup$ Comment to the answer (v6): The superpotential ${\cal K}^{\mu\nu}$ in eq. (21) can only be defined for space-time dimensions $d\geq 2$. $\endgroup$
    – Qmechanic
    Commented Aug 8, 2013 at 19:19
  • 3
    $\begingroup$ It seems that (18) is a special case of (24) for $\epsilon=\mathrm{const}$ on shell. How can (24) be vanishing if (18) is not, as indicated in your second to last paragraph? $\endgroup$
    – Friedrich
    Commented Jun 11, 2014 at 13:34

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.
  • $\begingroup$ @PhilipGibbs I hope that I have answered your question concerning the special role of the global symmetry. $\endgroup$ Commented May 28, 2013 at 16:28
  • 2
    $\begingroup$ Yes this is equivalent to my understanding. The current is not zero, it is the integrated charge that is zero when the symmetry is localised to a region within the boundary. I may yet post my own answer if I have time. $\endgroup$ Commented May 30, 2013 at 17:02
  • 1
    $\begingroup$ Took me 2 years to notice this answer, better than mine, +1. $\endgroup$
    – Jia Yiyang
    Commented Aug 8, 2015 at 9:45
  • $\begingroup$ Hello! Amazing answer. I have two questions: 1) The fact that the total charge Q is zero only means that you have a zero net amount of charge in the universe but you could still use this conservation law in the lab, it doesn't mean that a conserved charge doesn't exist. In other words, there's still a continuity equation that is perfectly valid. 2) It is usually said that local symmetries give Noether currents that are conserved off-shell. Can we see that from the current you derived? $\endgroup$ Commented Dec 3, 2019 at 6:17

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.

  • 3
    $\begingroup$ Are you saying that $J^{\mu}=0$ for any gauge theory? Doesn't that mean there are no electric currents or charges? $\endgroup$ Commented May 27, 2013 at 14:50
  • 6
    $\begingroup$ @Philip Gibbs: $J^\mu = 0$ if the current $J$ is associated to a local symmetry. The conserved charge that is interpreted as the electric charge actually comes from the residual global $U(1)$ symmetry of the action. Even if we gauge-fix the action, there should still be a residual global symmetry, and this does give a conserved current. The trick Luboš Motl is referring to ends up with something like $$\int d^4 x \partial_\mu \epsilon (x) J^\mu$$ which is obviously only zero in general if $J^\mu = 0$, as Luboš explained. But we can restrict this to $\epsilon (x) = \epsilon$, in which case $\endgroup$
    – user25032
    Commented May 27, 2013 at 15:25
  • 3
    $\begingroup$ (continuation of allicoder's comment) "the expression is zero, and we can integrate by parts to obtain the result that $\partial_\mu J^\mu = 0$. So only the current corresponding to global variations of the fields is conserved; the rest must be zero." $\endgroup$
    – David Z
    Commented May 27, 2013 at 17:19
  • 5
    $\begingroup$ @DavidZaslavsky, This integral is zero on the equations of motion, because its the first variation of the action. If you use $\epsilon=const$ to claim that it is zero, you cannot then deduce that $\partial_\mu J^\mu=0$ at every point, only the integral, which is tautologic. $\endgroup$ Commented May 27, 2013 at 23:33
  • 8
    $\begingroup$ Luboš, your analysis is not completely correct. As Philip Gibbs says, with your reasoning we shouldn't be able to recover the conservation of the electric current. Note that Noether's theorem starts with an specific 1-parameter group of field transformations. Only then it generalizes the parameter. When you derive $U(1)$ current conservation, you first assume that it is global (hence no change in $A$), then you write the transformation formulae $\psi\to e^{i\phi}\psi,\,A\to A$, and only then you let $\phi$ to depend on $x$. Compare to $\psi\to e^{i\phi}\psi,\,A\to A-\partial\phi$. $\endgroup$ Commented May 28, 2013 at 0:01

First I will say something about the paranthetical point at the end of the question. A local symmetry is defined as invariance under transformations that are parameterised by field variables, i.e. functions of co-ordinates in space-time. Any local symmetry implies a redundancy in the equations of motion (by Noether's second theorem). Whether this makes them "non-physical" is a matter of interpretation and debate.

The answer to your question, "Is the Noether charge zero" is "yes" for a spacetime that is spatially closed and orientated. In that case the total charge of the universe is zero if the charge is derived from a local symmetry. This is true for energy in GR as well as charges in Yang-Mills theory.

It is not true for global symmetries. A simple example of this is the non-relativistic Schrodinger equation which has a global U(1) symmetry that ensures the total probability $\int \mid\phi\mid^2 dx^3$ is conserved and positive so it can be normalised to 1. This works under cyclic boundary conditions which give a finite closed space but the total probability is still one, not zero.

You might wonder what happens when you add gauge fields to the Shrodinger equation with the usual gauge connection for a non-relativistic charged particle. This is possible and the conserved charge is still given by the same positive definite expression. How can the charge then be zero in a closed space? The answer is it cant and there are therefore no solutions to the gauged shrodinger equation with a non-zero wavefunction $\phi$ in a closed space. This is not the case in an open space, and it is not the case for the Dirac equation or for the Klein-Gordon equation where the charge density is not positive definite.

It is important to appreciate that the Noether charge in a closed universe is only sure to be zero when the field equations apply. It is not identically zero or trivially zero or zero off-shell as some people claim. There is however an alternative non-Noether expression for the charge that does give zero off-shell. I will come back to this point later.

If space is not closed and infinite then one of two things can happen. If there is charge spread over unbounded regions of space then total charge does not make sense. It may be infinite or indeterminate. If charge is limited to an isolated bounded region of space then the total charge is a well defined quantity and does not have to be zero.

The hard part of your question is "how do we prove it?", i.e. how do we prove that the charge is zero in a closed universe? None of the other answers have given the complete solution to this, perhaps because they interpreted the question differently, but also because you don't see the full proof in many places. It requires Noether's first and second theorems plus a third theorem that was added by Felix Klein that is called the "boundary theorem"

Euler Langrange equations

We start from the principle of least action

$S = \int {\cal L}(\phi_a, \phi_{a,\mu}) d^4x$

where $\phi_a$ are the field variables and for simpicity we assume that the Lagrangian only uses derivatives up to the first. The action is stationary under small changes ${\delta}{\phi}$ to the field variables that are fixed at the boundary. (There are implied sums over indices $a$ and $\mu$)

${\delta}S = \int (\frac{\partial {\cal L}}{ \partial \phi_a}\delta\phi_a + \frac{\partial {\cal L}}{ \partial \phi_{a,\mu}}\delta\phi_{a,\mu})d^4x = \int R^a(\phi_b,\phi_{b,\mu},\phi_{b,\mu\nu})\delta\phi_a d^4x = 0$

where using partial integration we find

$R^a(\phi_b,\phi_{b,\mu},\phi_{b,\mu\nu}) = \frac{\partial {\cal L}}{ \partial \phi_a} - \partial_\mu\frac{\partial {\cal L}}{ \partial \phi_{a,\mu}} = 0$

Noether's first theorem

Suppose the Langrangian is invariant under a transformation of the field variables ${\delta}\phi_a = \epsilon \Phi_a(\phi_a, x^\mu)$ for small $\epsilon$. This may be one of many possible symmetry transformations but we only need to think about one. For simplicity we are only considering internal symmetries which do not transform co-ordinates. The invariance implies an identity.

$\delta{\cal L} = \epsilon(\frac{\partial {\cal L}}{\partial\phi_a}\Phi_a + \frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi_{a,\mu}) = 0$

The Noether current is defined as

$J^{\mu} = -\frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi_a$

This expresses the current as a sum over contributions from each field. Its divergence is given by

$J^{\mu}_{,\mu} = -\partial_\mu\frac{\partial {\cal L}}{ \partial \phi_{a,\mu}}\Phi_a - \frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi_{a,\mu} = R^a \Phi_a = 0$

The last step is not an identity so the divergence of the Noether current is only zero with the compliance of the field equations. Since the 4-divergence of the current is zero we know that it is conserved.

Noether's second theorem

The second theorem is about the implications of local symmetry. The changes that generate the symmetry will be assumed to depend linearly on a field $\theta(x^\mu)$ and its derivatives

$\Phi_a = \Phi^0_a\theta + \Phi^\mu_a\theta_{,\mu}$

We can choose $\theta(x^\mu)$ to vary while keeping it zero at the boundary. Integration by parts can be used twice, first to remove the derivative on $\Phi_a$ and second to remove the derivative on $\theta$,

${\delta}S = \int R^a\Phi_a d^4x = \int (R^a\Phi^0_a - (R^a\Phi^\mu_a)_{,\mu})\theta d^4x = 0$

Since this is true for all compact $\theta$ it implies the identity

$R^a(\Phi^0_a - \Phi^\mu_{,\mu}) - R^a_{,\mu}\Phi^\mu_a = 0$

This means that the equations of motion are not indepedent which reflects the redundancy implied by the local symmetry. Another result from this is that the divergence of the term that was sent to the boundary using the two integrations by part must be identically zero and we can check this, define

$J_0^\mu = J^\mu - R^a\Phi^\mu_a\theta$


${J_0^\mu}_{,\mu} = J^\mu_{,\mu} - R^a_{,\mu}\Phi^\mu_a\theta - R^a\Phi^\mu_{a,\mu}\theta - R^a\Phi^\mu_a\theta_{,\mu} = R^a\Phi_a - R^a\Phi_a = 0$

Which shows that the Noether current can be written as the sum of two terms, the first of which has a divergence that is identically zero and the second of which is zero when the equations of motion are satisfied

$J^\mu = J_0^\mu + R^a\Phi^\mu_a\theta$

Sometimes people like to redefine the current to be just the first term and then they say that the current is trivially conserved (or conserved off-shell) because the divergence of this term is zero without the need to use the equations of motion. However, it is important to understand that only the original form for the Noether current describes it as a sum of charges arising from different fields and this expression for the current only gives a conserved charge when the field equations hold.

Klein's boundary theorem

To complete the proof that the charge over a closed space is zero we need one last result that was found by Felix Klein based on the results of Noether. Klein was able to go one step further and integrate the charge density over a volume space to reduce it to an integral over only the 2 dimensional boundary. First look more closely at the variation of the Lagrangian under small gauge transofrmations.

$\delta{\cal L} = \epsilon(\frac{\partial {\cal L}}{\partial\phi_a}(\Phi^0_a\theta+ \Phi^\nu_a\theta_{,\nu}) + \frac{\partial {\cal L}}{\partial\phi_{a,\mu}}(\Phi^0_{a,\mu}\theta + \Phi^0_a\theta_{,\mu} + \Phi^\nu_{a,\mu}\theta_{,\nu}+\Phi^\nu_a\theta_{\nu\mu})) = 0$

For the Lagrangian to be gauge invariant this expression must be identically zero for all functions $\theta(x^\mu)$. This can only be true if the separate coefficients of $\theta$ and its first and second derivatives are identically zero.

$\frac{\partial {\cal L}}{\partial\phi_a}\Phi^0_a + \frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi^0_{a,\mu}=0$

$\frac{\partial {\cal L}}{\partial\phi_a}\Phi^\mu_a+\frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi^0_a + \frac{\partial {\cal L}}{\partial\phi_{a,\nu}}\Phi^\mu_{a,\nu} = 0$

$F^{\mu\nu} = \frac{\partial {\cal L}}{\partial\phi_{a,\mu}}\Phi^\nu_a$ is antisymmetric

Using the earlier definitions we have

$J_0^\mu = J^\mu - R^a\Phi^\mu_a\theta = -\frac{\partial {\cal L}}{\partial\phi_{a,\mu}}(\Phi^0_a\theta + \Phi^\nu_a\theta_{,\nu}) - (\frac{\partial {\cal L}}{ \partial \phi_a} - \partial_\nu\frac{\partial {\cal L}}{ \partial \phi_{a,\nu}})\Phi^\mu_a\theta$

and then using the identities we simplify this to

$J_0^\mu = (F^{\nu\mu}\theta)_{,\nu}$

I used the notation $F^{\nu\mu}$ to show how this relates to the case of electromagnetic charge, but this analysis applies to any internal local symmetry. With the expression for the current in the form of the divergence of an antisymmetric tensor it follows immediately that its divergence is identically zero, but now we can also integrate the charge density which is the zero component of the current $\rho_0 = J_0^0 = (F^{\nu 0}\theta)_{,\nu}$ which is the divergence of a three component vector. The charge inside a volume is then expressed as an integral of $F^{\nu 0}\theta$ over the boundary surrounding the volume. In a closed space this volume can be taken to be the whole of space which has no boundary. Therefore the total charge is zero, which is what we wanted to prove. Note once again that for the charge given by Noether's first theorem this charge is only zero when the equations of motion are applied.


Firstly I'd like to explain that one doesn't need to assume that local symmetry implies gauge symmetry(i.e. unphysical degrees of freedom), we can see it through the equation of motions:

A local symmetry of the Lagrangian implies a local symmetry of the solution of the equation of motion, e.g. in the simplest case, if $x(t)$ is a solution, then some $y(t)=x(t)+\epsilon(t)$ is also a solution. Notice that so far the above argument can also be applied to a global symmetry, now here comes the difference: because of the arbitrariness and the locality of $\epsilon(t)$, we can also make $x(t)$ and $y(t)$ satisfy the same initial conditions, not just the same equation of motions, that is to say "initial conditions+equations of motions" does not uniquely fix the solutions, so quantities like $x(t)$ and $y(t)$ certainly cannot be physical in any common sense.

Secondly, about the conservation laws, a global symmetry gives a conserved quantity, this is guaranteed by Noether's first theorem. As for a local symmetry, there's so called Noether's second theorem, which does not directly give a conserved quantity but gives a set of identities. However in certain systems these identities indeed lead to conservation laws, and in fact "stronger" conservation laws than those derived from global symmetries. For example the QED lagrangian, the fermion current conservation law $\partial_\mu J^\mu=0$ can of course be derived by assuming the equations of motion for fermion fields to hold, but alternatively, it can be derived by using these identities from Noether's second theorem and assuming the equation of motion of the gauge fields to hold. In conclusion, the fermion current is conserved even when the fermion field equation of motion is not satisfied, so it is an "off-shell" conservation law, and it is in this sense a "stronger" conservation than those derived from global symmetries. You can refer to this article for more details.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.