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When electron accelerates, there occurs a propagated ripple on it's electric field. But when it moves constantly, does the field "follow it", i.e. changes instantly? How does it deals with the fact that nothing can travel faster than speed of light?

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    $\begingroup$ No, nothing changes instantly. Liénard–Wiechert potential is what you're looking for. $\endgroup$ – Avantgarde Oct 22 '18 at 6:53
  • $\begingroup$ @Avantgarde, then why it doesn't considers as an em wave? $\endgroup$ – user210398 Oct 22 '18 at 6:54
  • $\begingroup$ @Artur A moving electric charge looks exactly like a magnetic field (or, more precisely, a magnetic field is the relativistic transformation of a moving electric field). An electric field that varies over time produces a magnetic field that varies over time and vice-versa, so an EM wave is only stable when the original variable electric field produces a variable magnetic field which itself varies in exactly the right way to produce exactly the same changing electric field you start with — this in turn requires the electric field to look like a sine-wave when you graph it over time. $\endgroup$ – BenRW Oct 22 '18 at 7:21
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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/436008/2451 $\endgroup$ – Qmechanic Oct 22 '18 at 12:20
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    $\begingroup$ @BenRW "A moving electric charge looks exactly like a magnetic field" - It certainly creates a magnetic field, but it also has an electric field, because it's an electric charge. Not only that, but, for an observer who is stationary in the frame relative to which the moving charge's velocity is measured, the electric field measured by that observer will change with time. As such, I'm not sure that your statement gets the right idea across. $\endgroup$ – probably_someone Oct 22 '18 at 13:04
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Yes, in a sense, the field "instantly" moves together with it's source (if this source moves uniformly).

There is no aberration of forces. For example, an illuminated charged sphere and you are both approaching each other in uniform inertial motion along paths that do not collide. Light coming from the sphere will appear approaching you at relativistic aberration angle $\sin\alpha = v/c$. However, the electromagnetic force of attraction to the sphere does not experience aberration. It points directly toward the actual position.

So, theoretically you can always know the actual position of the charge and follow it.

It is very simple to understand it by swapping frames. Just think about a charge "at rest" , which disseminates the field and an observer (a test particle), who moves in this field. It makes clear, that the direction of the electrical force exerted by the sphere on the test particle points directly toward the actual position of the sphere.

That does not mean that the force propagates infinitely fast. The force on a test particle at any given instant is due to the electromagnetic field in the immediate vicinity of the particle at that instant.

Aberration of Forces and Waves

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  • $\begingroup$ So, I am observing the constant moving charge, being in 999 999 999 meters distance relative to this charge. When charge changes it's position I instantly know about it, right? $\endgroup$ – user210398 Oct 22 '18 at 18:11
  • $\begingroup$ Yes, if this charge MOVES UNIFORMLY. The page I have linked provides comprehensive analysis. This question is very rarely addressed, although it is very interesting. $\endgroup$ – Albert Oct 22 '18 at 18:14
  • $\begingroup$ What can You say about Liénard–Wiechert potential? $\endgroup$ – user210398 Oct 22 '18 at 18:15
  • $\begingroup$ en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential, chapter Universal Speed Limit. " A charge moving with a constant velocity must appear to a distant observer in exactly the same way as a static charge appears to a moving observer, and in the latter case, the direction of the static field must change instantaneously, with no time-delay. Thus, static fields (the first term) point exactly at the true instantaneous (non-retarded) position of the charged object if its velocity has not changed over the retarded time delay. This is true over any distance separating objects." $\endgroup$ – Albert Oct 22 '18 at 18:23
  • $\begingroup$ Perfect. If you yet say about "near field" and how does it related, to the topic, it will be greate $\endgroup$ – user210398 Oct 22 '18 at 18:28
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Suppose that I take a huge piece of paper, many light-years in size, and inscribe on it a set of lines that all intersect at one point. I could make the angular spacing uniform, in which case this could be a picture of the field lines of a charge in its rest frame, but for the purposes of answering your question this doesn't actually matter. The angular spacing can be random.

If the paper is moving toward the right, then an observer far away, looking at the part of the paper near them, can look at the paper and see a line pointing at a certain angle, and then if they check back an hour later, they will see a line pointing at a different angle. There is nothing in this that violates relativity, and no information is being propagated from the center of the paper to the distant parts of it.

What would violate relativity would be if we could grip the center of the paper, change its state of motion, and have the effect be instantly observed far away. That would be analogous to suddenly changing the motion of the charge. If you do that, then the change propagates outward at $c$.

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  • $\begingroup$ Honestly, bad understood what did You write, because of bad engilsh, and, perhaps it is the reason why I disagree, so maby I just didn't understand. Bad You actually wrong. You say that it will be a relativity violation if we had get an information about system(center of paper) instantly, being in some distance, but we actually do. I hope I just didn't understand the second paragraph, because it seems out of sense at all. Why You wrote about wating hour. You write an observer, then write "near them(observer(s?))". $\endgroup$ – user210398 Oct 22 '18 at 18:07
  • $\begingroup$ @user210398: Sorry, but I can't understand your comment. $\endgroup$ – user4552 Oct 23 '18 at 13:45
  • $\begingroup$ The paper example illustrates the fact that a system moving along with all its parts in motion at the same speed is no great mystery. This is a helpful point in the present context. On the other hand, it takes some care to prove that electric field lines from an inertially moving charge behave similarly to lines of ink drawn on a paper; that is not self-evident at all (and it would not work for more complicated problems). $\endgroup$ – Andrew Steane Oct 23 '18 at 14:44
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No, the field doesn't change instantly.

While the charge accelerates, the field "ripples". When the acceleration is done, and the charge travels at constant velocity, the field is stable after that.

The ripple moves out at lightspeed, and behind the ripple the stable field moves out at lightspeed too.

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The source event which caused the field at some location at some time is indeed in the past of that field event, not at the present location of the charged particle, so this does merit some thought. Here is a quote from section 7.3 of Relativity Made Relatively Easy: "... the field seems to 'know' where the moving source is now. ... It is as if the source gives its 'marching orders' to the field in the form 'line yourself up on my future position, assuming that I will continue at constant velocity'." This is what happens, and it is brought about by ordinary light-speed-limited communication from the source event to any particular field event. One could argue from other considerations that it must come out like this, but it is very nice to see it all hanging together when one carries out the calculation of the solution based on Lineart-Wiechart potentials.

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