Noether's theorem and gauge symmetry

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.