If I have a moving charge observed in frame $S$, may a Lorentz boost from $S$ in a direction not parallel to the charge's velocity in $S$ result in an electric field that has a different magnitude of the component in the direction of that boost? In other words, is a single Lorentz Boost w.r.t. a frame where a charge is moving really not a pure Lorentz boost?

And what if in frame $S$ there was another charge that is not moving? Does it mean that whether or not a Lorentz boost is pure is not really an intrinsic property of that boost, but rather dependent also on the frames in which the electric field of one or more charges are not "transformed" (i.e. the source charge's/charges' rest frame(s))?

  • $\begingroup$ Pure boost - no rotation of coordinates $\endgroup$ Jul 14, 2019 at 16:21
  • $\begingroup$ Also, the electric field projected in the direction of a pure boost shouldn't change as a result of that boost. $\endgroup$ Jul 14, 2019 at 16:23
  • $\begingroup$ If it is not a pure boost, but a boost + rotation, I am wondering if that projected E-field will change. $\endgroup$ Jul 14, 2019 at 16:25
  • $\begingroup$ Please make it more clear, is your particle moving in x direction w.r.t. S, while S' is moving in y axis w.r.t S? Also fields are measured in particle's frame or S? Which you want to transform to S' frame? $\endgroup$
    – Paradoxy
    Jul 14, 2019 at 17:41
  • $\begingroup$ Let's have the charged particle move in the $x$ direction in $S$. The observer is either stationary in $S$ or moving at some arbitrary velocity in $S$ (stationary in $S'$). Both positions are arbitrary. All velocities are steady. Fields are measured by an observer at constant velocity. Perhaps think of two such observers making an observation at the moment that they cross paths at the same point in space time. Do those observers see the same E-field projected along their difference of velocity? They should, but does that follow from applying non-collinear boost w.r.t. the charge frame $S'''$? $\endgroup$ Jul 14, 2019 at 20:31

1 Answer 1


A pure Lorentz boost does not change the component of the electric field or the magnetic field along the direction of the boost, regardless of how the charges producing the field are moving.

See Wikipedia for the field transformation law.

  • $\begingroup$ I can attain a moving charge by boosting my frame from its rest frame. If I boost again, but in a different direction, how would it be determined whether or not that is a pure boost? How is that second boost fundamentally different than a single boost from a frame where a charge is moving? $\endgroup$ Jul 14, 2019 at 16:36
  • $\begingroup$ Two pure boosts in different directions do not produce a pure boost. Does that explain what is puzzling you? $\endgroup$
    – G. Smith
    Jul 14, 2019 at 16:40
  • $\begingroup$ If I boost from a frame where the charge is moving, is there a dependence on an implied prior boost from the rest frame of that charge? $\endgroup$ Jul 14, 2019 at 16:42
  • $\begingroup$ I can’t understand what you are talking about. Two Lorentz transformations (boost+rotation) compose to give another Lorentz transformation. Two pure boosts compose to a pure boost only if they are in the same direction. These are mathematical facts about coordinate transformations and have nothing to do with rest frames of charges. $\endgroup$
    – G. Smith
    Jul 14, 2019 at 16:47
  • $\begingroup$ If I boost from the rest frame of a charge then again in a different direction, the electric field projected in the second boost's direction should be able to change after the second boost, correct? Yet, after the first boost, there isn't any "memory" of the initial rest frame. So the second boost is just a boost from the frame where the charge is moving - except now we say it is a pure boost? $\endgroup$ Jul 14, 2019 at 16:57

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