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We know that a static charge only produces an electric field and a charge in motion (be it constant velocity or accelerated) creates both magnetic and electric field. But recently I came across that "changing electric field changing magnetic field, and a charge moving with uniform velocity doesn't create a changing magnetic field, and hence electromagnetic waves aren't emitted". Now I am confused since I can't find a clear explanation of accelerated vs uniform velocity in case of charge anywhere.

Let's assume a source charge in space and the corresponding electric field associated with all points. Now if the charge starts moving (irrespective of uniform velocity or accelerated), since the position of the source charge is changing, the electric field associated with all the points in space are experiencing a change in both direction and magnitude and hence electric field is changing. So according to "changing electric field creates changing magnetic field theory" shouldn't electromagnetic waves be generated in this case? I really am finding it hard to understand the concepts. It seems that the electromagnetic waves are generated, only if the field lines are disturbed according to the following picture, but I am finding it hard to reconcile between the two concepts.

enter image description here

Kyle Kanos
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madness
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  • We need a source for the "changing electric field ... waves aren't emitted" quote. It seems either misquoted or wrong. Also, the image link is broken. – Dale Oct 25 '23 at 18:31
  • This gets at a key conceptual difficulty, and how people try to answer this will depend on your level of understanding of other things. I think we can agree that in the frame of reference of an observer moving along with the charge, the fields due to the charge are not changing since that observer only sees a stationary charge producing an E-field. According to that observer we can see that clearly no EM waves are produced. All other inertial observers agree with this (principle of relativity). Justifying this takes us into some difficult math. So, how technical an answer would you like? – gleedadswell Oct 25 '23 at 19:26

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