Does a 4-current J determine a unique maxwell-faraday F tensor up to isometry? Maxwell's equations on a pseudo-Riemannian manifold $(M,g_{ab})$ say,
$$d_a F_{bc} = \nabla_{[a}F_{bc]} = 0,$$
$$\nabla_a F^{ab} = J^b,$$
where $d_a$ is the exterior derivative, $\nabla_a$ is the covariant derivative compatible with $g_{ab}$ (i.e. $\nabla_a g_{bc}=0$), and $F_{ab}$ is antisymmetric ($F_{ab}=-F_{ba}$).
Let $J^a, F_{ab}$ and $\tilde{J}_a, \tilde{F}_{ab}$ be two solutions to Maxwell's equations in this form. Suppose that $J^a$ and $\tilde{J}^a$ are equivalent up to isometry, in that  $\varphi_*J^a = \tilde{J}^a$ for some isometry $\varphi:M\rightarrow M$.
Question: Does it follow that $F_{ab}$ and $\tilde{F}_{ab}$ are equivalent up to isometry, in that $\varphi_*F_{ab} = \tilde{F}_{ab}$?
Wald's General Relativity (1984, Chapter 10 Problem 2) shows a sense in which the answer is yes in the source-free case ($J^a=\mathbf{0}$). I wonder what is known about the non-source-free case.
 A: No, you also need an initial condition for the field, and a boundary condition for the field.  A plane wave solution to the Maxwell equations has the same 4-current as vacuum, after all.
A: Ok, here's my own (pedantic) argument that the answer is "No." Three background steps:


*

*Take $(M,g_{ab})$ and $(\tilde{M},\tilde{g}_{ab})$ to be two pseudo-Riemannian manifolds related by an isometry $\varphi:M\rightarrow\tilde{M}$. I will consistently use a tilde ($\tilde{}$) to refer to objects on the second manifold and no tilde to refer to objects on the first. Let $\varphi_*$ be the pushforward and $\varphi^*$ the pullback of $\varphi$.

*Let $F_{ab}$ and $J^a$ be any solution to Maxwell's equations on $(M,g_{ab})$ that is not source free, i.e., $\nabla_a F^{ab} = J^b \neq \mathbf{0}$.

*Let $\tilde{E}_{ab}$ and $\mathbf{0}$ be a non-trivial solution to Maxwell's equations on $(\tilde{M},\tilde{g}_{ab})$ that is source free, i.e. $\nabla_a E^{ab} = \mathbf{0}$ and $E_{ab}\neq\mathbf{0}$.
Now, we can state the counterexample: let $\tilde{F}_{ab}$ be the tensor field on $(\tilde{M},\tilde{g}_{ab})$ defined by,
$$\tilde{F}_{ab} := \varphi_*F_{ab} + \tilde{E}_{ab}.$$
It is antisymmetric and closed, since both $\varphi_*F_{ab}$ and $\tilde{E}_{ab}$ are, and so it provides a solution to Maxwell's equations. Let the 4-current associated with $\tilde{F}_{ab}$ be given by,
$$\tilde{J}^b := \tilde{\nabla}_a \tilde{F}^{ab}.$$
Then, $J^b$ and $\tilde{J}^b$ are related by an isometry:
$$\varphi_*J^b = \varphi_*\nabla_a F^{ab} = \nabla_a \varphi_*F^{ab} = \nabla_a(\tilde{F}^{ab} - \tilde{E}^{ab}) = \nabla_a\tilde{F}^{ab} = \tilde{J}^b,$$
where the second equality applies the fact that $\varphi$ is an isometry, the third the definition of $\tilde{F}_{ab}$, and the fourth the fact that $\tilde{E}_{ab}$ is source-free. And yet, $\varphi_*F_{ab}\neq \tilde{F}_{ab}$ by construction.
How curious.
