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As you see, this is the electric field generated by a point charge moving at constant speed v. I know that when $v$ -> 0, $E$ is just the Coloumb Law. But how do you interpret $E$ when $v$ -> $c$ ?

Can I just interpret it as the field of electromagnetic wave, because it moves at the speed of light?

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  • $\begingroup$ @BMS The question given to me is very vague. I don't think I need to do the Taylor expansion, right? $\endgroup$ – Lawerance May 20 '14 at 5:58
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    $\begingroup$ The homework question asks about a limit, whereas the title of the question refers to a charge moving at c. These are two different things. It's not possible for a charge to move at exactly c. All charged particles have mass, and massive particles can't move at c. $\endgroup$ – Ben Crowell Jul 24 '14 at 18:30
  • $\begingroup$ But it is easy to imagine e.g. a massless Dirac field with electric charge. $\endgroup$ – Robin Ekman Aug 28 '14 at 9:45
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Moving charges don't emit electromagnetic waves unless they are accelerating. A charge moving at a constant speed will just have a field that moves with it, but not a propagating field (i.e., electromagnetic waves).

Sources: Introduction to Electrodynamics by David J Griffiths 4th edition, section 10.3.2 and example 10.4 (pages 459-461), or the Wikipedia page for the Liénard-Wichart Potential.

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  • $\begingroup$ I know this, but how to make a connection about this field with the speed of light, as stated in the problem? $\endgroup$ – Lawerance May 20 '14 at 6:18
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    $\begingroup$ When $v^2/c^2$ is small, the $\sin^2 \theta$ doesn't matter much, but when $v^2/c^2$ gets close to 1, the $\sin^2 \theta$ will become very important. You'll find that the field will be small ahead and behind of the particle and much larger along the sides, where $\sin^2 \theta$ is closer to 1. I'll leave the math to you, but the effect is that the fields get flattened in the plane perpendicular to the charge's travel. $\endgroup$ – krs013 May 20 '14 at 6:27
  • $\begingroup$ I got your idea. Basically, because $\sin^2\theta$ is symmetric, we can just first consider 0-90 degrees. And the rest is just arguing the theta. But why this is somehow predicted as stated in the problem? Is it because of the contribution of retarded position of the charge? $\endgroup$ – Lawerance May 20 '14 at 7:33
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    $\begingroup$ I think they expect it to be intuitive because it is analogous to length contraction of space for fast-moving reference frames. $\endgroup$ – krs013 May 20 '14 at 7:37
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    $\begingroup$ Could you please mention the title and the author of the text book form which you cropped this piece of information? $\endgroup$ – R004 Jul 24 '14 at 4:41

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