Relativistic field of moving charge. Why is it symmetric?

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SUMMARY

The discussion centers on the relativistic field of a moving charge and its symmetry. Participants reference Feynman's Lectures on Physics, specifically chapter 26, to analyze the field's behavior in inertial frames. The key conclusion is that the field generated by a moving charge does not conform to the classical Coulomb field due to the necessity of satisfying Maxwell's equations. The misunderstanding arises from assuming that the field can be treated as a simple retarded Coulomb field without accounting for relativistic effects.

PREREQUISITES
  • Understanding of Maxwell's equations
  • Familiarity with relativistic physics concepts
  • Knowledge of electric fields and Coulomb's law
  • Basic grasp of Lorentz transformations
NEXT STEPS
  • Study the implications of Maxwell's equations on moving charges
  • Learn about the Liénard-Wiechert potentials for moving charges
  • Explore the concept of retarded potentials in electromagnetism
  • Investigate the effects of relativistic speeds on electric field distributions
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Physicists, electrical engineers, and students of advanced electromagnetism seeking to deepen their understanding of relativistic effects on electric fields generated by moving charges.

Luis Babboni
Hi people!
First of all, sorry for my poor english.

I read in many places and I did the calculus and I agree that the field of a moving charge have this aspect:
campo%20carga%20en%20movimiento.jpg

(Taked from Feynman´s Lectures on Physics chapter 26th.)

But my problem is in that my intuition says me that it must be something like this:
campo%20carga%20en%20movimiento%202.jpg

I´m talking just about intensity in the same direcction of v, I still not work with perpendicular to v intensity.

My reasoning to say that is the following:

When in an innertial system where Q1 and Q2 are at rest both at distance L of to Q0 that is moving to right, cause Q2 "sees" the field produced by Q0 when Q0 was at, say, Q0´ and Q1 "sees" the field produced by Q0 when Q0 was at, say, Q0´´. And cause distance Q1-Q0´´ is smaller than distance Q1-Q0 and distance Q2-Q0´ is greater than distance Q2-Q0, the intensity of the field at Q2 must be smaller than intensity of the field if Q0 is at rest in the same system than Q1 and Q2 (that are the field you see in the next picture) and the intensity of the field at Q1 must be greater than the field if Q0 is at rest in the same system than Q1 and Q2. Ant then you could see something like I draw in the second picture.

Cause I always talk about an inertial system where Q1 and Q2 are at rest and L is a proper distance in it, I do not see the necesity of any Lorentz transformation...

Of course I´m wrong! :-D
But could someone explain me where is my mistake?

Thanks!

campo%20carga%20en%20movimiento%204.jpg
 

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Your charges are not at rest. In particular the charge that generates the field you want to look at. Your underlying assumption seems to be that the field will be the regular Coulomb field, just retarded based on the position that is a distance such that a light signal would just propagate from there to the point. This is wrong. Such a solution would not satisfy Maxwell’s equations.
 
Thanks, it seems i need to put an eye There then. I like the observation that the charge moving is the one that generates the field .
 

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