# Reaction-at-a-distance: Do charged plates immediately repel each other? Imagine that we have a pair of parallel plates, $A$ and $B$, separated by some distance as in Fig. $1$ above.

At time $t_1$ 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 $A$ to $B$ and vice-versa.

At time $t_2$ the electric influence from $A$ produces a force at $B$ 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 $t_1$, and before they have produced forces on the opposite plates at time $t_2$, they must exist "somewhere". That somewhere is the electromagnetic field.

Now consider the picture described in Fig $2$ below which includes both retarded and advanced interactions. Again at time $t_1$ we simultaneously charge both the plates.

Now as well as a retarded electric influence that travels at the speed of light from $A$ to $B$ we also have an advanced electric influence which travels backwards in time from $B$ to $A$. Thus the force at plate $B$ at time $t_2$ is balanced by an equal and opposite force on plate $A$ at time $t_1$ (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 $t_1$ is linked with the opposite plate at time $t_2$ 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?

• (Note: Haven't read through this in detail.) Is this in any way different from Wheeler-Feynman absorber theory? – Michael Brown Sep 23 '13 at 12:28
• I guess it's a generalisation of Wheeler-Feynman theory to explain static electric forces by direct interaction rather than assuming an electromagnetic field. – John Eastmond Sep 23 '13 at 16:57
• In a frame in which both plates are at rest we only have static electric fields which do not hold any momentum so it's hard to see how one could explain the forces on the plates as a reaction to a local momentum change in the fields around each plate. (If indeed there are any reaction forces at the plates). – John Eastmond Sep 23 '13 at 17:07
• By "reaction force" are you trying to find a Newton's 3rd Law pair of forces, or are you talking about something else? – Bill N Aug 25 '15 at 18:21
• NOTE: this is another old question. – John Duffield Aug 25 '15 at 18:21

Firstly, I think you've confused several points in your analysis of the situation. Working with the first picture, by standard electrostatic analysis, we see that there must be a standard Coulomb force between the plates for all times $t\ge t_2$, not just at $t=t_2$. This point is not too important, but I thought I'd mention it.
Continuing to work with the first picture, there will be an additional force over a short duration after time $t_2$ caused by the "retarded electric influence" (which is in fact basically radiation) that you mention. Now you're worries are 1) that there is no reaction force at A around time $t_1$ to account for this and 2) where the momentum is between $t_1$ and $t_2$. Now, in charging up the plates, the charges must undergo acceleration, which causes radiation, or as you call it, the "retarded electric influence", which is where the momentum resides between $t_1$ and $t_2$, resolving 2) from above.
Moreover, in since we are working with standard electromagnetic theory, we must account for the Abraham-Lorentz force - i.e., these charges interact with their own retarded fields and experience a reaction force. This force on $A$ is such that it exactly counterbalances the momentum transferred in the form of radiation, in accordance with Newton's second law. This momentum then propagates at the speed of light to the plate $B$, reaching it at $t=t_2$, and the force experienced by plate $B$ exactly balances the momentum absorbed from the radiation, once again in accordance with Newton's second law.
On point 2), I disagree with your analysis. Firstly, the statement "No electromagnetic fields" is patently incorrect - even if we ignore what happens around time $t_1$ and $t_2$, at a much later time there is still an electrostatic field due to the charges. What you perhaps meant to say is that there is no propagating, time-dependent disturbance in the electromagnetic fields as in the first picture. But even this is not true - the charging of the plate $A$ does indeed lead to an immediate force on $A$ - but that is a reaction force corresponding to "electric influence" that is about to travel across to $B$, and should obviously be have no delay. What necessitates some disturbance in the electromagnetic fields, or radiation, between the plates is that at time $t_2$ there will be a separate force on $A$ when the disturbance from $B$ reaches $A$, and so point 2) is incorrect.
Thus, we see that there is no phenomenological discrepancy between the two pictures. If this is still confusing, first try working through both pictures with just a single plate - in this case there will just be a reaction force at time $t_1$, and momentum flows out to infinity - no long duration electrostatic force like I described in the first paragraph, and no later force due to radiation coming from the opposite plate. Then, in the situation described in your question, you see that you must separately account for those as well, which gives you the full picture.