# 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.

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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.

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Just for clarification, what do you mean by "initial and boundary conditions" in this context? – soulphysics Aug 8 '14 at 15:16
@soulphysics: Let's say we solve this in terms of the vector potential $A_{a}(x,t)$. Then, to find a unique solution to Maxwell's equations, we need to know the value of $A_{a}(x,0)$ and ${\dot A}_{a}(x,0)$, as well as either a falloff condition for $A$ at infinity, or $A(0,t)$ on some surface. – Jerry Schirmer Aug 8 '14 at 15:32
It's not clear to me how fixing a vector potential helps the uniqueness problem. Since $F_{ab}$ is closed and antisymmetric, I can always (locally) write it as the exterior derivative of a vector potential, $F_{ab}=d_a A_b$. And of course there are many other equivalent vector potentials $\hat{A}_{a}=A_a+\nabla_a\chi$ for some scalar field $\chi$ (i.e. related by a gauge transformation). But all of them give rise to the same Maxwell-Faraday tensor $F_{ab}$. So, how does this provide a further constraint? – soulphysics Aug 8 '14 at 16:04
@soulphysics: because a plane wave has a different maxwell tensor from vacuum. The two solutions aren't related by a gauge difference. They are different physical states, but they both have $j^{\mu} =0$. – Jerry Schirmer Aug 8 '14 at 16:06
This is standard for the solution of a system of PDEs. You need an initial condition, and a set of boundary conditions. – Jerry Schirmer Aug 8 '14 at 16:07

Ok, here's my own (pedantic) argument that the answer is "No." Three background steps:

1. 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 $\varphi$.

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

3. 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.

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I'm probably missing something obvious, but where did you use the hypothesis $J^b \neq 0$? Or does this hold without that assumption? – Chris White Aug 8 '14 at 17:15
@ChrisWhite -- no, I think you're right. That assumption isn't necessary. – soulphysics Aug 8 '14 at 18:44
So then you're saying "no" to a situation where Wald said "yes"? I feel there's something subtle I'm missing still, perhaps in the particular way Wald actually formulates the question ($D_a E^a = D_a B^a = 0$ iirc), but I haven't fully thought this through. – Chris White Aug 8 '14 at 19:05
I believe Wald has more initial conditions than just a 4-current $J^a$, in specifying the value of $E^a$ and $B^a$ on a hypersurface (assuming then, also, that the spacetime is globally hyperbolic). – soulphysics Aug 8 '14 at 19:55