EM Wave creation in different frames of reference

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In the discussion on electromagnetic wave creation from different frames of reference, the key point is the distinction between stationary and moving observers regarding a charged particle. Observer S, stationary relative to the particle, perceives no electromagnetic wave, while Observer A, who moves back and forth, experiences the particle as accelerating and decelerating. This acceleration from Observer A's perspective would indeed create an electromagnetic wave propagating outward. However, the two observers are not equivalent; Observer S experiences changes in tension in a spring due to the particle's motion, while Observer A does not. The conversation highlights the necessity of special relativity for accurately calculating electromagnetic radiation speeds relative to different observers.
ealbers
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OK, I'm sure I'm just not thinking about this the right way, so please point out my simple mistake.

Imagine a charged particle and 2 observers.

Observer S is stationary relative to the charged particle, so sees no magnetic field from it.

Observer A moves back and forth in front of the particle, accelerating, then decelerating over and over, back and forth.

Would not Observer A see the charged particle as accelerating and decelerating, from their point of view/frame of reference?
Would not the acceleration of a charged particle from Observers A perspective create a EM wave which would propagate outwards?

Now from Observer S's perspective, there is just a charged particle and the other Observer moving around, no EM wave is created.

What am I missing??
Thanks for your time!
E
 
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Because S and the particle are moving back and forth (as opposed to inertially, in a straight line with constant speed) they know that they are being accelerated at each turnaround. S and A are not equivalent observers; if they're both holding a mass on a spring, S's spring is alternately tensing and relaxing while A's is not. Thus, they both know to do the calculations for an accelerated charge I stead of one at rest.

(Be aware, however, that you still will find a problem if you try calculating the exact speed of the electromagnetic radiation relative to the two observers using classical physics. That's why we need special relativity)
 
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