Are the EM fields of charges with different velocities subject to independent Wigner rotations under Lorentz boosts?

Question

I will attempt to answer the following sub-questions in addressing the main question which is the title of my post:

1. In the case of one electric charge of constant non-zero velocity in frame $\Sigma'$, is there a Wigner rotation of that charge's field when boosting to another frame $\Sigma''$?

2. In the case of one electric charge at rest in frame $\Sigma$, if one boosts to a frame $\Sigma'$ where the charge has a constant non-zero velocity, is there a Wigner rotation of that charge's field when boosting further to frame $\Sigma''$?

3. In the case of multiple electric charges with unequal constant non-zero velocities in frame $\Sigma'$, is the field of each electric charge subject to the same angle of Wigner rotation when Lorentz boosted to $\Sigma''$? In other words, can we be sure that the net electromagnetic field of these electric charges can be subject to a well-defined angle of Wigner rotation?

My attempts to answer:

1. Yes, the Wigner rotation of that electric charge's field depends on its non-zero velocity in frame $\Sigma'$.
2. Yes, there is a Wigner rotation if there are two non-colinear Lorentz boosts ($\Sigma \rightarrow \Sigma'$ then $\Sigma' \rightarrow \Sigma''$, where $\Sigma$ is the charge's rest frame).
3. No, the rest frames of each electric charge are different, thus the Wigner rotation of the electric field of each charge when boosting from $\Sigma'$ to $\Sigma''$ cannot be generally the same for the fields of all the charges.

Therefore my answer to, "Are the EM fields of charges with different velocities subject to independent Wigner rotations under Lorentz boosts?" would be yes.

However, if I understand correctly, the answers are actually supposed to be:

1. No, there is no Wigner rotation if there is only one Lorentz boost ($\Sigma' \rightarrow \Sigma''$)
2. Yes, there is a Wigner rotation if there are two non-colinear Lorentz boosts ($\Sigma \rightarrow \Sigma'$ then $\Sigma' \rightarrow \Sigma''$, where $\Sigma$ may be any arbitrary inertial frame).
3. Yes, the Wigner rotation results from two non-colinear Lorentz boosts and therefore has nothing to do with the instantaneous rest frames of electric charges.

This gives us what I think should be the standard conclusion.

However, suppose I told you that an electric charge was moving in some inertial frame of reference $\Sigma'$ and gave you the minimum information ($q$ and $\mathbf{r}(t)$) needed to specify its electric and magnetic fields in $\Sigma'$. Suppose, furthermore, that I do not specify to you whether or not this inertial frame of reference was arrived at by boosting from the charge' instantaneous rest frame, or for that matter, any other arbitrarily conceived inertial frame (symbolically: $? \rightarrow \Sigma'$). How then would you be able to calculate the angle of Wigner rotation of this charge's electromagnetic field without this information (symbolically: $? \rightarrow \Sigma'$ then $\Sigma' \rightarrow \Sigma''$)? Also, how would you do this for an ensemble of electrical charges having arbitrary motions?

TL;DR What is the correct answer to "Are the EM fields of charges with different velocities subject to independent Wigner rotations under Lorentz boosts?" and how is that answer justified?

Answer 1

Yes to (3). If you already know the fields $F'^{\mu\nu}_i$ for each particle $i=1...N$ in the frame $\Sigma'$ just add the fields up $F'^{\mu\nu}_{Total}=\sum_{i=1}^{N}(F'^{\mu\nu}_i)$ which you can do because Maxwell's equations are linear in $F'^{\mu\nu}$. Then transform $F'^{\mu\nu}_{Total}$ like a tensor to $F''^{\mu\nu}_{Total}$ using the boost from $\Sigma'$ to $\Sigma''$. This you can do because the EM field tensor $F'^{\mu\nu}$ (which contains $\vec{E'}$ and $\vec{B'}$) transforms like a tensor under Lorentz transformations.

Alternatively, you could boost each particle's 4-momenta $P^{\mu}_i$ by it's own $LorentzTrans_i$ to $P''^{\mu}_i$ in $\Sigma''$. Then calculate each particle's field $F''^{\mu\nu}_i$, and add them up to again get $F''^{\mu\nu}_{Total}$. Notice that $LorentzTrans_i$ is a combination boost/rotation when it is the product of non-colinear boosts.

Answer 2

I think you're making this unnecessarily complicated. Any two reference frames with the same origin are related by an element of the Lorentz group, and any element of (the connected part of) the Lorentz group can be written as the product of a Lorentz boost and a spatial rotation. The angle of the spatial rotation in that decomposition is the Wigner rotation angle. Charges and EM fields aren't relevant unless you use them to fix one or both of the reference frames.

The answer to "How then would you be able to calculate the angle of Wigner rotation [...] without this information" is that you can't, because you don't have enough information. It's the same as asking how you could calculate the angle between two lines in the plane if you only know one of them. The angle is a relationship between two lines, not a property of one line that could be calculated independently of the other.

Charge distributions and EM fields transform in a certain way under the action of the Lorentz group, and this is independent of any particular factoring of group elements. The decomposition into a boost and a spatial rotation is not really natural anyway since "boost" and "spatial rotation" aren't Lorentz invariant concepts.