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In the case of a non-accelerating point charge "A" of stable velocity, its static field is treated as though it is instantaneously present at a distance, i.e. a second point charge "B" will react to the field as though it is centered on the ACTUAL location of charge A.

This is true of gravity as well, as the planets do not appear to be attracted to the retarded position of the Sun but instead the actual position, as the Sun is moving through space at a constant velocity.

Now, if my point charge "A" is accelerated, then it makes sense (and is experimentally verified) that the EM radiation given off by the accelerating charge will propagate at the speed of c. So far so good. My question is this:

If photons are the messenger particles of the EM force, and my point charge "B" feels the effect of charge "A"'s static field and responds as though that field is centered on the actual position of charge "A", presumably this effect is still being mediated by virtual photons, which cannot exceed 'c' - so how is this explained?

For clarification, when we look at the Sun we SEE it as it was roughly 8 minutes ago due to the time delay of light traveling at c to reach us, so we SEE its retarded position, however our orbit is centered on its ACTUAL position. Talking about electric fields is just easier for me, but the effect is the same - how is this possible if messenger particles must obey the c speed limit?

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As I don't yet know QFT, I'll try to answer classially, without virtual particles or any similar entities.

There's actually no part of the field which moves faster than light — even for static fields, even for those of uniformly moving sources. What happens instead is that's how the field itself evolves.

To better illustrate this phenomenon, let's consider an elastic wave on a membrane, resulting from a moving source. The equation describing such a system is isomorphic to equation for scalar potential of electromagnetic field, so this analogy seems quite fair. At time $t=0$ we'll create a source moving at $0.9c$ where $c$ is the speed of wave on the membrane, then, when the source is near center ($x\approx5.8$), we remove the source. (Don't pay attention at reflections from boundaries — these are a computational artifact of artificial boundary.)

enter image description here

What we get is that the static field (i.e. the distortion of the membrane) will move with the source, propagating away from its point of creation at $c$. The static part, which is inside the propagation circle, changes in time in such a way that it looks as if it follows the source. The part which moves at $c$ is the change of field.

When we remove the source, however, the distortion — the static part of field — will still continue moving as if the source were still there. And only after $\Delta t=d/c$ will the observer located at distance $d$ from the point where the source disappeared notice that the source is no longer there — only at that time the change of field will reach the observer.

So, the conclusion: it's not that the static field propagates at FTL speeds — it's instead the way the field of a uniformly moving source evolves in time itself — even if there's no longer any source present.

Due to the reasons mentioned in the wikipedia link you provided in the OP, the same is true for electromagnetic and gravitational waves. Even if the source vanishes, the field will still look as if the source moves it its expected (actual if it didn't vanish) position until the wave of change reaches the observer.

You might now have a question: if the field evolves in such a way as if the source which created it moves with some velocity, but doesn't require that source at all, how does the field know the speed of that source? It appears that this information is encoded in the shape of the static field. For a stationary source, the field is, as known from Coulomb's law for electrostatics and Newton's law for gravitation, spherically symmetric. But if the source moves, the field looks flattened in the direction of motion of the source. This is the consequence of length contraction and can be easily seen if you do Lorentz transformations on the observer, starting from Coulomb field.

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  • $\begingroup$ Excellent, thank you - even absent the QFT component this description helps me wrap my head around it a bit, if I manage to find a formal QFT description I'll send it on your way but this was genuinely helpful. $\endgroup$ – JPattarini Jan 8 '15 at 21:00

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