Consider for a moment you have two bodies, a ferromagnetic mass A and a self-contained electromagnetic coil B and an observer. The masses are stationary relative to each other in a frictionless medium with no other forces present and seperated by distance d. The observer is next to the coil. B is initially off, but then turned on. The laws of electromagentism tell us that B will exert an attractive force on A. The laws of mechanics tell us that there will be an equal and opposite reaction force applied to the coil B. The observer times how long, in his reference frame, it takes between B being switched on and B moving in response to the mutual attraction. Relativity tells us that the magnetic field will propagate between the bodies at c, so what is the delay? Is it $$t = 0$$ or $$t = \frac d c$$ or is it $$t = \frac {2d} c$$ $t = 0$ does not make sense because that would imply the field propagates faster than light. $t = \frac d c$ seems wrong, because otherwise the "fact of interaction" between A and B would have to propagate back to B faster than light. $t = \frac {2d} c $ seems more correct from a relativistic view point, but raises questions as to whether momentum can be temporarily 'borrowed' due to relativistic delays - which feels more like quantum mechanics!

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    $\begingroup$ Regarding your feelings about this, two things to consider: (a) Newtonian mechnsics intuition doesn't necessarily work with electromagnetics, and (b) the EM field itself carries momentum, so you may picture the sequence of interactions as B->field->A->field->B, and in each case, momentum is conserved $\endgroup$
    – Toffomat
    Nov 22 at 12:55
  • $\begingroup$ Thanks @Toffomat. So, does this mean B actually accelerates twice? Once at t=0 as a result of transferring momentum from B to the field, then again later at t=2d/c when field arrives back to B? $\endgroup$ Nov 24 at 17:55
  • $\begingroup$ Well, I guess it is more complicated. If iI undertsand your setup correctly, A is not magnetised initailly, so B does feel a $\vec{B}$. Then B turns on, generates a magnetic field than spreads outwrad, but at this point, B wouldn't know where to move, so it stas put. (Of course, "B turning on" is a process that takes finite time.) When the magnetic field reaches A, it will magnetise A (again a prcess that takes time) and exert a force on A; the magnritastion of A generated a $\vec B$ that spreads back to B and accelerates it $\endgroup$
    – Toffomat
    Nov 25 at 7:53
  • $\begingroup$ Just realised where one error/oversight in my thinking was: the net wave momentum of the spreading magnetic field as a vector is zero, because it always spreads as a symmetrical shape, so B will not move even though it has given the field momentum whose magnitude could presumably be measured or calculated as a scalar. The field becomes assymetrical after interaction with A, thus allowing for movement in both A and B. $\endgroup$ Nov 26 at 11:40

1 Answer 1


The delay is obviously $\frac {2d} c$, which is the time it takes for a change in the electromagnetic field to propagate from B to A and then back to B again. Or, if you think in terms of particles rather than waves, it is the time taken (in B's reference frame) for photons to travel from B to A and then back to B again. This is exactly the same as if the observer at B shone a light towards a mirror at a distance $d$ and looked for the reflection of the light in the mirror.

Not sure what you mean by "borrowing" momentum. Remember that the electromagnetic field (or the photons, if you prefer) carries momentum, so I don't think there is ever any need to "borrow" momentum.

  • $\begingroup$ What I mean by 'borrowing' if t = 2d/c, is that from the view of a second observer at roughly equal distance, A will appear to move before B, so for a short interval the conservation of momentum will appear to be violated. $\endgroup$ Nov 22 at 13:41
  • $\begingroup$ However, if the field / photons can carry momentum, that implies t=0, i.e. the momentum will transfer instantly from B to the photons, which will carry it and transfer it to A, making it move d/c seconds later. That would allow conservation of momentum (just include the momentum of the photons) however violates speed of information needing to be less than c - because properties of A and d affect magnitude of the force. $\endgroup$ Nov 22 at 13:53
  • $\begingroup$ @StephenParry "...violates speed of information needing to be less than c" - information can travel at c, and in this case it does because photons travelling at the speed of light carry momentum from B to A and back again. $\endgroup$
    – gandalf61
    Nov 22 at 14:45

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