Noether's theorem and gauge symmetry 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?
 A: 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.
A: 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.
A: 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.
