41
$\begingroup$

I'm confused about Noether's theorem applied to gauge symmetry. Say we have

$$\mathcal L=-\frac14F_{ab}F^{ab}.$$

Then it's invariant under $A_a\rightarrow A_a+\partial_a\Lambda.$

But can I say that the conserved current here is

$$J^a=\frac{\partial\mathcal L}{\partial(\partial_aA_b)}\delta A_b=-\frac12 F^{ab}\partial_b\Lambda~?$$

Why do I never see such a current written? If Noether's theorem doesn't apply here, then is space-time translation symmetry the only candidate to produce Noether currents for this Lagrangian?

$\endgroup$
4
  • 7
    $\begingroup$ For local symmetries, one uses Noether's second theorem, while Noether's first theorem is for global symmetries, cf. physics.stackexchange.com/a/13881/2451 , physics.stackexchange.com/q/66092/2451 and links therein. For electric charge conservation from global gauge symmetry, see e.g. physics.stackexchange.com/q/48305/2451 and links therein. $\endgroup$
    – Qmechanic
    May 12, 2014 at 15:08
  • 1
    $\begingroup$ @Qmechanic Hi, do you still have this opinion? As per the below answer I dont't see why I would have to be redirected to Noether's second thereom when the above is what I get from noether's first theorem. Do you mean that the first theroem doesn't provide any significant information in the case of a local symmetry? Thanks. $\endgroup$
    – user28174
    Sep 13, 2016 at 11:39
  • 2
    $\begingroup$ As a guiding principle, Yes. Of course, one can rederive many of the consequences of Noether's second theorem for local symmetries via Noether's first theorem, but that is usually not the most systematic approach. These consequences are already present in Noether's second theorem (& Klein's boundary theorem). Concerning the proof of electric charge conservation, see also my Phys.SE answer here and links therein. $\endgroup$
    – Qmechanic
    Sep 14, 2016 at 21:22
  • $\begingroup$ The G-invariance of the YM lagrangian does not imply any conservation laws for the following reason. The Lagrangian is actually a scalar product on Lie(G) and, if G is compact, then this product is Ad-invariant, i.e. invariant under global gauge transformations. It is a trivial result of the representation theory of Lie groups, then, that this scalar product will be also skew-hemritian w.r.t. the action of ad, the induced representation of Ad. Therefore the only conserved currents are arbitrary 4-divergence vectors which are conserved independently of the YM equations or the for of L. $\endgroup$ Oct 24, 2016 at 23:18

4 Answers 4

28
$\begingroup$

Indeed, nothing is wrong with Noether theorem here, $J^\mu = F^{\mu \nu} \partial_\nu \Lambda$ is a conserved current for every choice of the smooth scalar function $\Lambda$. It can be proved by direct inspection, since $$\partial_\mu J^\mu = \partial_\mu (F^{\mu \nu} \partial_\nu \Lambda)= (\partial_\mu F^{\mu \nu}) \partial_\nu \Lambda+ F^{\mu \nu} \partial_\mu\partial_\nu \Lambda = 0 + 0 =0\:.$$ Above, $\partial_\mu F^{\mu \nu}=0 $ due to field equations and $F^{\mu \nu} \partial_\mu\partial_\nu \Lambda=0$ because $F^{\mu \nu}=-F^{\nu \mu}$ whereas $\partial_\mu\partial_\nu \Lambda =\partial_\nu\partial_\mu \Lambda$.

ADDENDUM. I show here that $J^\mu$ arises from the standard Noether theorem. The relevant symmetry transformation, for every fixed $\Lambda$, is $$A_\mu \to A'_\mu = A_\mu + \epsilon \partial_\mu \Lambda\:.$$ One immediately sees that $$\int_\Omega {\cal L}(A', \partial A') d^4x = \int_\Omega {\cal L}(A, \partial A) d^4x\tag{0}$$ since even ${\cal L}$ is invariant. Hence, $$\frac{d}{d\epsilon}|_{\epsilon=0} \int_\Omega {\cal L}(A, \partial A) d^4x=0\:.\tag{1}$$ Swapping the symbol of derivative and that of integral (assuming $\Omega$ bounded) and exploiting Euler-Lagrange equations, (1) can be re-written as: $$\int_\Omega \partial_\nu \left(\frac{\partial {\cal L}}{\partial \partial_\nu A_\mu} \partial_\mu \Lambda\right) \: d^4 x =0\:.\tag{2}$$ Since the integrand is continuous and $\Omega$ arbitrary, (2) is equivalent to $$\partial_\nu \left(\frac{\partial {\cal L}}{\partial \partial_\nu A_\mu} \partial_\mu \Lambda\right) =0\:,$$ which is the identity discussed by the OP (I omit a constant factor): $$\partial_\mu (F^{\mu \nu} \partial_\nu \Lambda)=0\:.$$

ADDENDUM2. The charge associated to any of these currents is related with the electrical flux at spatial infinity. Indeed one has: $$Q = \int_{t=t_0} J^0 d^3x = \int_{t=t_0} \sum_{i=1}^3 F^{0i}\partial_i \Lambda d^3x = \int_{t=t_0} \partial_i\sum_{i=1}^3 F^{0i} \Lambda d^3x - \int_{t=t_0} ( \sum_{i=1}^3 \partial_i F^{0i}) \Lambda d^3x \:.$$ As $\sum_{i=1}^3 \partial_i F^{0i} = -\partial_\mu F^{\mu 0}=0$, the last integral does not give any contribution and we have $$Q = \int_{t=t_0} \partial_i\left(\Lambda \sum_{i=1}^3 F^{0i} \right) d^3x = \lim_{R\to +\infty}\oint_{t=t_0, |\vec{x}| =R} \Lambda \vec{E} \cdot \vec{n} \: dS\:.$$ If $\Lambda$ becomes constant in space outside a bounded region $\Omega_0$ and if, for instance, that constant does not vanish, $Q$ is just the flux of $\vec{E}$ at infinity up to a constant factor. In this case $Q$ is the electric charge up to a constant factor (as stressed by ramanujan_dirac in a comment below). In that case, however, $Q=0$ since we are dealing with the free EM field.

$\endgroup$
8
  • 1
    $\begingroup$ Thank you very much. Do you mean then that Noether's (first) theorem is applicable here, although it's not a global symmetry? $\endgroup$
    – user46348
    May 12, 2014 at 18:30
  • 1
    $\begingroup$ I provided a proof, what should I do further? ;) $\endgroup$ May 12, 2014 at 18:32
  • 1
    $\begingroup$ I know... It's just that it conflicts with what others said here. Then can we say that as long as the symmetry is continuous, Noether's theorem applies? Also, why don't I see this current written anywhere? $\endgroup$
    – user46348
    May 12, 2014 at 18:36
  • $\begingroup$ @user46348 I believe, but please correct me if I'm wrong, that Noether's first theorem in general does not apply to gauge transformations, it only applies to global transformations. However, apparently Noether's first theorem happens to work for $\mathrm{U(1)}$ electromagnetic theory. $\endgroup$
    – Hunter
    May 12, 2014 at 19:02
  • 1
    $\begingroup$ Isn't this $J^a=\frac{\partial\mathcal L}{\partial(\partial_aA_b)}\delta A_b=-\frac12 F^{ab}\partial_b\Lambda$ not a gauge invariant operator...? $\endgroup$
    – wonderich
    Jul 2, 2019 at 19:24
5
$\begingroup$

The current

$$J^a=\frac{\partial\mathcal L}{\partial(\partial_aA_b)}\delta A_b=-\frac12 F^{ab}\partial_b\Lambda$$

is a conserved current. It is indeed a direct consequence of the Noether theorem. However, this current does not represent any physical observables since it is not gauge invariant.

This is explained in great details in this paper.

$\endgroup$
3
$\begingroup$

I asked myself the same question a while back, and this is what I came up with:

I assumed that the Lagrangian of the Maxwell action is of the form: $$\mathcal{L} = \mathcal{L}(A_\mu,\,\partial_\nu A_\mu)$$ Then I assumed that under variations of the type $A_\mu \rightarrow A_\mu - \frac{1}{e}\partial_\mu\alpha$ where $\alpha\equiv \alpha(x)$ a local gauge parameter, the Lagrangian remains invariant: \begin{align} \delta\mathcal{L} &= \frac{\partial \mathcal{L}}{\partial A_\mu}\delta A_\mu + \frac{\partial \mathcal{L}}{\partial(\partial_\nu A_\mu)}\delta(\partial_\nu A_\mu)\\ &= \frac{\partial \mathcal{L}}{\partial A_\mu}({\textstyle \frac{-1}{e}\partial_\mu \alpha})+ \frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}({\textstyle \frac{-1}{e}\partial_\mu \partial_\mu \alpha})=0\,. \end{align} The in the first term, I use the equation of motion to replace $\frac{\partial\mathcal{L}}{\partial A_\mu}=\partial_\nu (\frac{\partial\mathcal{L}}{\partial (\partial_\nu A_\mu)})$ to get \begin{equation} \phantom{wwwwww}\delta\mathcal{L}=\partial_\nu\big(\frac{\partial\mathcal{L}}{\partial (\partial_\nu A_\mu)}\big)({\textstyle \frac{-1}{e}\partial_\mu \alpha})+\frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}({\textstyle \frac{-1}{e}\partial_\mu \partial_\mu \alpha})=0\,. \end{equation} Now, since $\alpha(x)$ is an arbitrary smooth function, first derivative is independent of second derivative. So the two terms vanish separately.

  • Vanishing second term implies the tensor $\frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}$ is antisymmetric (as it contracts with symmetric $\partial_\mu\partial_\nu \alpha$). So just define it $\frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}:=F^{\nu\mu}$

  • Vanishing first term implies $\partial_\nu \big(\frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}\big) = \partial_\nu F^{\nu\mu} = 0.$

Therefore, I conclude that Noether's theorem for gauge transformations is not a conservation law, but an equation of motion. And, the requirement of gauge symmetry imposes that $F^{\mu\nu}$ is antisymmetric.

$\endgroup$
1
$\begingroup$

If you're using a variational principle this way - which is what Noether's Second Theorem covers - using it with an already-given Lagrangian, then you're seriously under-utilizing the theorem! The second theorem is a model-building technique for deriving action principles and reduces mostly to triviality, once the model is already set up.

The right way to use it is to start out by making no assumptions about what the Lagrangian density is - other than it have the desired symmetry property and then conclude that it has the form that you originally wrote in by hand and started out by assuming, or a generalization thereof.

When used this way with gauge fields, the results are applications of instances of the Utiyama Theorem.

Suppose we have a dynamics given by an action principle with the action integral $$S = \int 𝔏(q,v,A,V) d^4 x,$$ where the Lagrangian density $𝔏$ depends on fields $\left(q^A: 0 ≤ A < m\right)$ and $\left(A^a_μ: 0 ≤ a < n, 0 ≤ μ < 4\right)$ and their gradients, $$v^A_μ ≡ ∂_μ q^A,\quad V^a_{μν} ≡ ∂_ν A^a_μ.$$

For simplicity, we assume that $𝔏$ has no explicit dependence on the coordinates $x = \left(x^μ: 0 ≤ μ < 4\right)$. It won't change the analysis in any way, though a more general analysis that also for variations on $x$, too, should also account for variations on $dx$ and work directly with the Lagrangian four-form $L = 𝔏 d^4x$, instead of just with the Lagrangian density $𝔏$.

In here and the following I will use index notation for the components and the Einstein summation convention on indices.

Suppose we have a gauge symmetry $$δq^C = τ^C_{aB} λ^a q^B,\quad δA^c_μ = -∂_μ λ^c + f^c_{ab} λ^a A^b_μ,$$ that leaves the Lagrangian density invariant. We assume that the coefficients are constant and that their respective matrices $$τ_a = \left(τ^C_{aB}: 0 ≤ B,C < m\right),\quad f_a = \left(f^c_{ab}: 0 ≤ b,c < n\right),$$ satisfy the matrix relations $$τ_a τ_b - τ_b τ_a = f^c_{ab} τ_c,\quad f_a f_b - f_b f_a = f^c_{ab} f_c,$$ and that $$f^c_{ab} = -f^c_{ba}.$$

We also assume the symmetry is transparent with respect to differentiation so that we can write down the corresponding transforms for the gradients: $$δv^C_μ = ∂_μ\left(δq^C\right) = τ^C_{aB} ∂_μλ^a q^B + τ^C_{aB} λ^a v^B_μ,\\δV^c_{μν} = ∂_ν\left(δA^c_μ\right) = -∂_ν∂_μ λ^c + f^c_{ab} ∂_νλ^a A^b_μ + f^c_{ab} λ^a V^b_{μν}.$$

For convenience, define the following partial derivatives of the Lagrangian density: $$𝔣_A = \frac{∂𝔏}{∂q^A},\quad 𝔭^μ_A = \frac{∂𝔏}{∂v^A_μ}, \quad 𝔍^μ_a = \frac{∂𝔏}{∂A^a_μ},\quad 𝔊^{μν}_a = \frac{∂𝔏}{∂V^a_{μν}}.$$ Then the variational on the action can be written as: $$δS = \int δ𝔏 d^4x = \int \left(δq^A 𝔣_A + δv^A_μ 𝔭^μ_A + δA^a_μ 𝔍^μ_a + δV^a_{μν} 𝔊^{μν}_a\right) d^4x,$$ and we assume $δS = 0$, and indeed that $δ𝔏 = 0$.

The invariance is with respect to gauge transforms where the $λ^a$ may be any sufficiently-smooth function of the position. By "sufficiently smooth", we'll assume that they are continuous up to at least the second order of derivatives so that $∂_μ∂_νλ^a = ∂_ν∂_μλ^a$. Then, the integrand in the variational $δS$ must equate to zero - order-by-order in $λ^a$, starting with the highest order derivatives - the second order: $$0 = δS = \int \left(⋯ + \left(-∂_ν∂_μ λ^a\right) 𝔊^{μν}_a\right) d^4x\quad⇒\quad 0 = ∂_ν∂_μ λ^a 𝔊^{μν}_a = \frac{1}{2} ∂_ν∂_μ λ^a \left(𝔊^{μν}_a + 𝔊^{νμ}_a\right),$$ using the above symmetry condition on the second-order partial derivatives of $λ^a$. By the arbitrariness of $λ^a$, this implies: $$𝔊^{μν}_a + 𝔊^{νμ}_a = 0\quad⇒\quad 𝔊^{νμ}_a = -𝔊^{μν}_a.$$ Thus, the total differential of $𝔏$ can be written as: $$d𝔏 = ⋯ + dV^a_{μν} 𝔊^{μν}_a = ⋯ + \frac{1}{2}d\left(V^a_{μν} - V^a_{νμ}\right) 𝔊^{μν}_a,$$ using the just-derived anti-symmetry of $𝔊^{νμ}_a$.

Therefore, $𝔏$ may depend on the gradients $V^a_{μν}$ only through the combination $$F^c_{μν} ≡ V^c_{νμ} - V^c_{μν} = ∂_μA^c_ν - ∂_νA^c_μ.$$ Thus $$d𝔏 = ⋯ - \frac{1}{2}dF^c_{μν} 𝔊^{μν}_c.$$ The respective variational in $F^c_{μν}$ is: $$δF^c_{μν} = δV^c_{νμ} - δV^c_{μν} = f^c_{ab} \left(∂_μλ^a A^b_ν - ∂_νλ^a A^b_μ\right) + f^c_{ab} λ^a F^b_{μν}.$$

Next, for the first order derivatives of $λ^a$, after substituting the variationals $δF^c_{μν}$ for $δV^c_{μν}$, we have: $$0 = δS = \int \left(⋯ + \left(τ^C_{aB} ∂_μλ^a q^B\right) 𝔭^μ_C + \left(-∂_μ λ^a\right) 𝔍^μ_a - \frac{1}{2} f^c_{ab} \left(∂_μλ^a A^b_ν - ∂_νλ^a A^b_μ\right) 𝔊^{μν}_c\right) d^4x.$$ Using the anti-symmetry of $𝔊^{μν}_c$, we can write the last term as $$- \frac{1}{2} f^c_{ab} \left(∂_μλ^a A^b_ν - ∂_νλ^a A^b_μ\right) 𝔊^{μν}_c = -f^c_{ab} ∂_μλ^a A^b_ν 𝔊^{μν}_c.$$ Thus, $$0 = \int ⋯ + ∂_μλ^a \left(τ^C_{aB} q^B 𝔭^μ_C - 𝔍^μ_a - f^c_{ab} A^b_ν 𝔊^{μν}_c\right) d^4x.$$

By the arbitrariness of $λ^a$, the bracketed multiples of $∂_μλ^a$ must vanish, thereby giving us an expression for $𝔍^μ_a$: $$𝔍^μ_a = τ^C_{aB} q^B 𝔭^μ_C - f^c_{ab} A^b_ν 𝔊^{μν}_c,$$ and reducing the total differential of $𝔏$ to: $$\begin{align} d𝔏 &= dq^A 𝔣_A + dv^A_μ 𝔭^μ_A + dA^a_μ 𝔍^μ_a - \frac{1}{2} dF^c_{μν} 𝔊^{μν}_c\\ &= dq^A 𝔣_A + dv^A_μ 𝔭^μ_A + dA^a_μ τ^C_{aB} q^B 𝔭^μ_C - f^c_{ab} dA^a_μ A^b_ν 𝔊^{μν}_c - \frac{1}{2} dF^c_{μν} 𝔊^{μν}_c. \end{align}$$ Using the anti-symmetry of $f^c_{ab}$ and of $𝔊^{μν}_c$, we can write $$f^c_{ab} dA^a_μ A^b_ν 𝔊^{μν}_c = \frac{1}{2} f^c_{ab} \left(dA^a_μ A^b_ν + A^a_μ dA^b_ν\right) 𝔊^{μν}_c = \frac{1}{2} d\left(f^c_{ab} A^a_μ A^b_ν\right) 𝔊^{μν}_c.$$ Using integration by parts, and a little index-switching, we can write: $$dA^a_μ τ^C_{aB} q^B 𝔭^μ_C = d\left(τ^A_{aB} A^a_μ q^B \right) 𝔭^μ_A - dq^A τ^C_{aA} A^a_μ 𝔭^μ_C.$$ Thus, $$d𝔏 = dq^A \left(𝔣_A - A^a_μ τ^C_{aA} 𝔭^μ_C\right) + d\left(v^A_μ + τ^A_{aB} A^a_μ q^B\right) 𝔭^μ_A - \frac{1}{2} d\left(F^c_{μν} + f^c_{ab} A^a_μ A^b_ν\right) 𝔊^{μν}_c.$$

This shows that the dependence of the Lagrangian density on the field gradients reduces to a dependence on their gauge-covariant derivatives, and that there be no other dependencies on $A^a_μ$. Correspondingly, we rewrite $v$ and $F$ as: $$ v^A_μ = ∂_μv^A ⇒ v^A_μ = ∂_μq^A + τ^A_{aB} A^a_μ q^B,\\ F^c_{μν} = ∂_μA^c_ν - ∂_νA^c_μ ⇒ F^c_{μν} = ∂_μA^c_ν - ∂_νA^c_μ + f^c_{ab}A^a_μA^b_ν, $$ and adjust the differential coefficient of $q^A$ accordingly: $$𝔣_A ⇒ 𝔣_A - A^a_μ τ^C_{aA} 𝔭^μ_C,$$ so that the total differential of $𝔏$ now becomes: $$d𝔏 = dq^A 𝔣_A + dv^A_μ 𝔭^μ_A - \frac{1}{2} dF^c_{μν} 𝔊^{μν}_c.$$

With these adjustments, all dependency on the first-order derivatives are removed from the variationals and we have: $$δv^C_μ = τ^C_{aB} λ^a v^B_μ,\quad δF^c_{μν} = f^c_{ab} λ^a F^b_{μν}.$$

For gauge fields $𝔍^μ_a$ is the current density (corresponding to Maxwell's charge density $ρ$ and current density $𝐉$) and $𝔊^{μν}_c$ is the response field (corresponding to Maxwell's $𝐃$ and $𝐇$). The expression derived, above, for $𝔍^μ_a$ shows that there is a gauge-dependent self-interaction involving the response fields and potentials, as well as an extra contribution coming from the $q^A$ field and the response field $𝔭^μ_A$ for the corresponding field strength $v^A_μ$. This gets factored into an adjustment on the expressions for the field strength $v^A_μ$ for $q^A$ and $F^c_{μν}$ (which corresponds to Maxwell's $𝐄$ and $𝐁$) for the potentials $A^a_μ$ (which correspond to the electric/scalar potential $φ$ and magnetic/vector potential $𝐀$), so that the field strengths are now non-linear in terms of the potentials $A^a_μ$ and fields $q^A$.

Finally, at order-zero in $λ^a$, we now have: $$0 = δS = \int \left(τ^C_{aB} λ^a q^B\right) 𝔣_C + \left(τ^C_{aB} λ^a v^B_μ\right) 𝔭^μ_C - \frac{1}{2} \left(f^c_{ab} λ^a F^b_{μν}\right) 𝔊^{μν}_c d^4x, $$ which leads to the following identity: $$τ^C_{aB} q^B 𝔣_C + τ^C_{aB} v^B_μ 𝔭^μ_C = \frac{1}{2} f^c_{ab} F^b_{μν} 𝔊^{μν}_c,$$ which is actually just the continuity equation for the current density.

Edit: The continuity equation, derived here, is off-shell; i.e. independent of any application of the (Euler-Lagrange) field equations. That's a key point of the second Noether Theorem. There's an off-shell continuity equation. How is it to be interpreted? As a pre-condition for the sources in field equation. In particular, the current density associated with the sources (here: the $q^A$ field) must first satisfy the continuity equation before they can quality as sources for the field equations. Otherwise, the field equations are ill-posed.

The Euler-Lagrange equations with the original $𝔣_A$ were: $$∂_μ 𝔭^μ_A = 𝔣_A,\quad ∂_ν 𝔊^{μν}_a = 𝔍^μ_a.$$ With the modified $𝔣_A$, the first of these becomes: $$∂_μ 𝔭^μ_A = 𝔣_A + A^a_μ τ^C_{aA} 𝔭^μ_C.$$

The continuity equation for the current density follows by the anti-symmetry in $𝔊^{μν}_a$: $$∂_μ 𝔍^μ_a = ∂_μ ∂_ν 𝔊^{μν}_a = 0.$$ If you substitute for $𝔍^μ_a$ and use the equations for the modified $𝔣_A$ and for the adjusted $v^a_μ$ and $F^c_{μν}$, you will get the equation just derived from the order-zero variation in $λ^a$.

Further symmetry principles can be used to narrow down on the range of possibilities for Lagrangian density $𝔏$. In particular, requiring $𝔏$ to be locally Lorentz-invariant reduces it to a function of the fields $q^A$ and of the invariant combinations formed from the field gradients, e.g. $g^{μρ}g^{νσ}F^a_{μν}F^b_{ρσ}$, $g^{μν}v^A_μv^B_ν$ and $g^{μρ}g^{νσ}v^A_μF^b_{νρ}v^C_σ$. So, the application of the theorem - which for gauge fields is an instance of the Utiyama Theorem - to all the symmetries, tightly constrains the range of possible Lagrangian densities.

$\endgroup$

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.