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I have a question about whether I'm thinking about the field of a moving charge along the axis of its motion correctly, in cases of both acceleration and also constant velocity. I'm trying to develop an intuition that avoids equations unless they are absolutely necessary.

Purcell's text and online simulations of the field lines of accelerating charged particles show a circular kink in the field emanating from a moving charged particle when the charged particle accelerates. I have attached a screen capture from an online simulation where I had an initial stationary charge accelerate briefly to 0.2 c in the upwards direction, and then continue to move at that speed. It seems that the kink diminishes to 0 along the axis of motion. This seems to be consistent with the behavior of dipole antennas that don't emit radiation along the axis of the dipole.

However the diagram also seems to show that on the +y axis, there is a sudden narrowing / widening of the field lines. Is this sudden narrowing/widening just a visual artifact of the lack of field lines very near the axis?

Even if this is just a visual artifact, there is still the gradual narrowing of field lines corresponding to an increase in field strength that occurs as the charged particle moves upwards at constant speed. To a test charge on the y axis that is being approached by the moving charged particle below, the electric field must be increasing as it gets closer, and my intuition is that the charged particle's movement parameters (distance, velocity >0, acceleration = 0) can't be transmitted faster than the speed of light. Yet if there is no radiation along the y axis, how is the test particle detecting the current position and state of motion of the charge?

Is the answer connected to the fact that in a reference frame moving upwards along with the charged particle so that it appears stationary, then it is the test charge that is moving? In this frame, the charged particle is stationary and could have had an arbitrarily long time to suffuse space with its field before the test particle came moving along? enter image description here

lamplamp
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  • Possible Duplicate : Electric field associated with moving charge. – Frobenius May 03 '21 at 04:08
  • @Fobenius, thank you for link, may I ask if my rationale in the last two paragraphs is correct for how a test charge far away senses the current position of a charge moving at constant velocity? Purcell refers to the "flight plan" of the moving charge, and I'm not sure if my argument is a legitimate way of understanding this. – lamplamp May 03 '21 at 18:16
  • If I understand your comment correctly , your problem is not how we have this scheme of the field of a moving charge first accelerating from rest to a speed $\upsilon$ and then moving uniformly. Your problem is causality : if the electric field of a uniformly moving charge points to the present position then a test charge at a field point A knows instantly the current position of the charge. This sounds at first like instantaneous transmission of information! – Frobenius May 03 '21 at 22:41
  • But if you consider the Figure-01 and my equation (05) in the link with $\phi=0$ \begin{equation} \mathrm E =\dfrac{ q }{4\pi \epsilon_{0}}\dfrac{\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)}{r^{2}} \tag{05}\label{eq05}
    \end{equation} there are infinitely many pairs $(\upsilon,r)$ satisfying it. And this under the assumption that the observer with the test charge on field point A $''$knows$''$ (where from ???) that the charge is moving uniformly in order to use equation (05). So the velocity and position of the moving charge are undetermined.     – Frobenius May 03 '21 at 23:04

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