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Reaction-at-a-distance: Might charged plates immediately repel?

  1. Sep 23, 2013 #1
    Imagine that we have a pair of parallel plates, [itex]A[/itex] and [itex]B[/itex], separated by some distance as in Fig. [itex]1[/itex] below.


    At time [itex]t_1[/itex] we simultaneously charge both the plates. This could be done by previously sending a light signal to a charging apparatus at each plate from a source located at the mid-point between them.

    According to standard electromagnetic theory a retarded electric influence travels at the speed of light from [itex]A[/itex] to [itex]B[/itex] and vice-versa.

    At time [itex]t_2[/itex] the electric influence from [itex]A[/itex] produces a force at [itex]B[/itex] and vice-versa.

    There are two points that I would like to raise about this description:

    1. There are no reaction forces. It is as if a pair of boxers each punched the other on the nose simultaneously but neither felt a reaction back on their boxing glove.
    2. Once the electric influences have left the charged plates at time [itex]t_1[/itex], and before they have produced forces on the opposite plates at time [itex]t_2[/itex], they must exist "somewhere". That somewhere is the electromagnetic field.

    Now consider the picture described in Fig [itex]2[/itex] below which includes both retarded and advanced interactions.


    Again at time [itex]t_1[/itex] we simultaneously charge both the plates.

    Now as well as a retarded electric influence that travels at the speed of light from [itex]A[/itex] to [itex]B[/itex] we also have an advanced electric influence which travels backwards in time from [itex]B[/itex] to [itex]A[/itex]. Thus the force at plate [itex]B[/itex] at time [itex]t_2[/itex] is balanced by an equal and opposite force on plate [itex]A[/itex] at time [itex]t_1[/itex] (and vice-versa).

    Now as soon as we charge the plates up we measure electric forces on them.

    At first glance it seems that we have "action at a distance" but in fact we only have "reaction at a distance". In terms of spacetime, each plate at time [itex]t_1[/itex] is linked with the opposite plate at time [itex]t_2[/itex] in a manner that is consistent with the principle of locality provided we include advanced interactions.

    As there is no delay between charging the plates and measuring forces then there is no time interval during which the influences could be said to be in transit in the form of an electromagnetic field.

    Thus in this picture we have:
    1. Reaction forces
    2. No electromagnetic field

    Could one perform such an experiment to see if charged plates immediately repel each other?
    Last edited: Sep 23, 2013
  2. jcsd
  3. Sep 23, 2013 #2


    Staff: Mentor

    Advanced fields violate causality.

    Your experiment could be done in principle, but in practice measuring the timing of mechanical forces will be subject to delays from the speed of sound in the plates and force transducers. These will likely dominate the delay and make distinguishing advanced from retarded forces quite difficult.

    However, there is no need for that anyway. We can simply measure the EM fields directly to detect advanced or retarded waves directly. That measurement has been done, the waves are retarded only.
    Last edited: Sep 23, 2013
  4. Sep 23, 2013 #3
    Could you point me to a reference?

    As electromagnetic waves are always absorbed eventually somewhere it would be difficult to distinguish Wheeler-Feynman theory and the emission of standard retarded electromagnetic waves. In fact the authors showed that their theory would give the same results as the standard retarded field theory.

    I thought that it might be easier to distinguish the two theories in the case of forces between static electric charges. If indeed there is a reaction force at an "emitting" charge then it would be difficult to explain this in terms of a local interaction with the static electric field as such fields do not carry any momentum.
  5. Sep 24, 2013 #4


    Staff: Mentor

    All of section 3.4 here:

    In all of these tests the speed of light was measured to be +c, not -c.

    I doubt very strongly that this is a true premise. Certainly, we don't have evidence supporting it today, and to get evidence supporting it would require an infinite amount of time.

    In their theory only the total field is detectable, and that is retarded. You can certainly postulate the existence of the aether, advanced waves, the Easter Bunny, Santa Claus, magical pink unicorns, and so forth as long as you make them undetectable. I don't bother with such postulates.
  6. Sep 24, 2013 #5
    Thanks for the link above.

    I did not see any tests for the speed of propagation of an electrostatic field.

    Should it be possible in principle to measure the time difference between charging a conductor A and measuring the appearance of its static field at B some distance away?

    I guess one would need fast electronics as one could only measure static fields over a distance of meters.
  7. Sep 24, 2013 #6


    Staff: Mentor

    Of course not. Electrostatic fields don't propagate, by definition. That is what the "static" in "electrostatic" means. Static fields don't propagate or appear or disappear, by definition.

    I had assumed that you understood that. Only electromagnetic fields propagate, that is what was tested in the cited references.
  8. Sep 24, 2013 #7
    So when I charge up the conducting plate A is there is an immediate electric field at B in the direction normal to plate A?

    I thought standard theory says that the field will take time to propagate from A to B?
  9. Sep 24, 2013 #8


    Staff: Mentor

    No. When you charge up a plate the field is electromagnetic, not electrostatic. The electromagnetic field takes time to propagate, as shown in the references.

    Last edited: Sep 24, 2013
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