EM wave generation from different frames of reference Take 2 observers and a charged particle,
Observer S is stationary to the particle and sees the electric field and no magnetic field as he is stationary to the particle.
Observer A decides to accelerate and decelerate over and over back and forth in front of the particle.
From A's frame of reference, it can be said, by the equivalence principal, that the particle is accelerating and decelerating in A's frame of reference and therefor should emit a EM wave.
So A sees a Em wave come off the particle.
But wait, S does not see any such thing, other than A moving back and forth...
What am I missing??
Thanks!
Eric
 A: I'm guessing (and hoping) that you understand that both these observers cannot be in a freefall (aka inertial, or Lorentz) frame at the same time. That is, if observer $S$ is in a freefall frame, then $A$ will measure his/her own proper acceleration with an accelerometer. So you can see that you can't apply, say, the Lamor formula (or its relativistic generalization the Liénard–Wiechert potentials) to the charge's motion as seen by $A$ and deduce there is radiation.
However, there is a simpler version of a question very like yours which, to my knowledge, has not been fully resolved.
Simply drop an electron onto the surface of the Earth and let it sit there. Its proper acceleration is $g$, so, by the equivalence principle, it should radiate. I'm not expert with the current state of contemplation of this problem, so we probably need another answer to complete the answer to your question. But what I can say is that the power we'd expect to see radiated from charge sitting on the Earth's surface is fantastically small. From the Lamor formula:
$$P=\frac{q^2\,a^2}{6\,\pi\,\epsilon_0\,c^3}$$
I make the radiation from an electron in this situation to be $5.5\times10^{-52}{\rm watts}$. To put this into more meaningful units, this is $7\times 10^{-39}$ of the electron's rest mass per second. That is, it would take $3\times10^{20}$ ages of the universe to radiate the equivalent of its own rest mass. So we wouldn't expect to be able to measure this effect on Earth even though, as far as I can reason, it should be real. Owing to its miniscule size, it would have to be checked by astronomical measurements of radiation from much higher gravity phenomena.

It seems the wording above is causing some confusion, best illustrated by AnnaV's question:

An electron is an elementary particle and cannot lose part of its mass. Anyway why would it radiate its own mass, and not of the earth's gravitational field which is supply in the $g$? similar to Hawking radiation? 

Sorry for the bad wording: I'm not saying it's radiating its own mass, I'm comparing the radiation rate to its own mass, just to try to give some meaning to the figures. Otherwise, $5\times 10^{-52}{\rm watts}$ is a bit meaningless. As you say, you can interpret the radiation energy as being similar to Hawking radiation (Unruh radiation) and coming from the gravitational field.
