Least action principle and gauge invariance

Question

In classical electrodynamics using variational "least" action principle we arrive at equation like

$$ -\int d^4x \left(j^\nu - \frac{\partial F^{\mu\nu}}{\partial x^\mu}\right) \delta A_\nu = 0 $$ (in some well-chosen units), where $j$ is 4-current, $F$ is EM field tensor and $A$ is EM potential. Then assuming that $\delta A_\nu$ are arbitrary we get Maxwell's equations.

But part of $\delta A_\nu$ can be attributed to a change of gauge: $$ \delta A_\nu = \frac{\partial \delta f}{\partial x^\nu} + \delta B_\nu $$ where $ \delta f = \delta f(x)$ is some function and 4-vector $\delta B_\nu$ has 3 independent components and cannot be further decomposed in the same way.

So part of our variation does not include actual change of the field, its energy, momentum, etc? Because of this if I didn't know better, I'd assume we must exclude $\delta f$-part by posing some condition on $\delta A_\nu$ (I don't which one btw?) before doing variations. This would result in additional terms with lagrange multipliers in action and eventually modify Maxwell's equations.

Since this is incorrect, and Maxwell's equations hold, I'd like to understand if this is just a part of the postulate or there is a flaw in my reasoning and we must compare all possible variations of $A$?

UPDATE

Leaving aside the current term $j^\nu =0 $, we can start from the lagrangian expressed in $\vec{E}$ and $\vec{B}$ since these quantities are gauge-invariant:

$$ L=-\frac12\left(\vec{B}^2 - \vec{E}^2\right) $$

Gauge-invariance is equivalent to two constraints:

$$ \nabla\times\vec{E} + \frac{\partial \vec{B}}{\partial t}=0,\qquad\nabla \cdot \vec{B} = 0 $$

which should be multiplied by lagrange coefficients and summed up with initial lagrangian. And the resulting thing should be varied, giving $$ \vec{B}\cdot\delta\vec{B}-\vec{E}\cdot\delta\vec{E} + \vec{\lambda}\cdot \left( \nabla\times\delta\vec{E} + \frac{\partial \delta\vec{B}}{\partial t} \right) + \mu \nabla \cdot \delta \vec{B} = 0 $$ Changing the order of differentiation and combining terms in $\delta\vec{E}$ and $\delta\vec{B}$ gives two equations:

$$ \vec{E} = \nabla \times \vec{\lambda},\qquad\vec{B} = \nabla \mu + \frac{\partial \vec{\lambda}}{\partial t} $$

that are equivalent to the second pair of ME (with zero current) $$ \nabla \cdot \vec{E} = 0, \qquad \nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t} $$

This is hardly anything new, and I'm not quite sure how to incorporate non-zero current here, but I wanted to clarify my question which is: why we vary potentials with no constraints, when actual fields may be unaffected by such variation. If we vary fields themselves, at least with zero current, we can obtain equations equivalent to the ones we know. It'd be interesting to try to get GR equations by varying $R\sqrt{-g}$ with respect to components of the Riemann tensor.

Answer 1

1. To ensure gauge quasi-symmetry of the Maxwell action $$ S~=~\int\! d^4x ~{\cal L}, \qquad {\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \pm j_{\mu}A^{\mu}, \tag{1} $$ we must assume that the background sources $j^{\nu}$ obey the continuity eq. $$d_{\nu}j^{\nu}~=~0\tag{2}. $$ Here the signature of the Minkowski metric is $(\mp,\pm,\pm,\pm)$.

2. Varying the Maxwell action (1) wrt. the $4$-gauge potential $A_{\mu}$ leads to the $4$ Maxwell eqs. $$d_{\mu}F^{\mu\nu}\pm j^{\nu}~=~0\tag {3}$$ with source terms.

3. Now we get to OP's question: Since there is a gauge symmetry $$\delta A_{\mu}=d_{\mu}\Lambda \tag{4},$$ there should only be 3 independent Maxwell eqs. with source terms. This is indeed the case since taking the divergence $$d_{\nu}\left(d_{\mu}F^{\mu\nu}\pm j^{\nu}\right)~=~0\tag {5}$$ on both sides of the Maxwell eqs. (3) leads to a triviality, due to eq. (2) and antisymmetry of $F^{\mu\nu}$. In fact, varying the Maxwell action (1) wrt. the gauge transformation (4) precisely leads to the vanishing divergence (5). See also this related Phys.SE post.