Q:

Charged particle as observed from an inertial and a non-inertial frame of reference

A charged particle fixed to a frame $S^\prime$ is accelerating w.r.t an inertial frame $S$. For an observer A in the $S$ frame, the charged particle is accelerating (being attached to frame $S^\prime$) and therefore, he observes it to radiate. However, for the non-inertial observer B standing on the non-inertial frame $S^\prime$, the charged particle is at rest, and therefore, does not radiate at all. 
A infers that the charged particle radiates but B infers it doesn't. Can both inferences (mutually contradictory) be correct simultaneously? If yes, how? 

A:

First, I'll note that unless the non-inertial frame has a changing acceleration, there is some doubt as to whether it radiates at all.
https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field
Assuming there is changing acceleration, yes the particle radiates, and this can be observed in both frames.  Seeing it radiate would be a way of determining that you are in fact in a non-inertial frame, assuming your inner-ear was not already telling you that.

A:

The notion of an ideal point charge is actually not completely well-defined in classical E&M due to issues with the infinite self-energy, back-reaction, etc., so there is no definite answer to this question.  See the extensive discussions at Does a constantly accelerating charged particle emit EM radiation or not? and Does a charged particle accelerating in a gravitational field radiate?.

A:

The one thing that all will agree about is the proper time $ d\tau $.
Energy is not conserved but action it is, because it's expressed in terms of proper time.
In order to see what is to be seen in the other reference frame one must do a coordinates transformation of a 4-vector:
$$ (V')^{\mu} = \frac{\partial X^{\mu}}{\partial X^{\nu}} V^\nu $$
or
$$ (V')_{\mu} = \frac{\partial X^\nu}{\partial X^{\mu}} V_{\nu} $$
if things depend on passing through multiple references 
$$ (V'')^{\mu} = \frac{\partial X^{\mu}}{\partial X{^{\nu}}} (V')^{\nu} $$
$$ (V'')^{\mu} = \frac{\partial X^{\mu}}{\partial X^{\nu}}\frac{\partial X^{\nu}}{\partial X^\sigma} V^\sigma $$
$$ ... $$
What one might see as red-shifting of some sort of degree due to gravitational field someone else would not see the red-shifting if he was on a direct course towards the gravitational field generator. 
But again a 3rd observer would see both the red-shifting of 1st frame and blue-shifting of the 2nd frame if he was in between the 2 above "frames" standing - "at rest".
So for the electromagnetic tensor $ F_{\mu\nu} (=\frac{\partial \phi_\mu}{\partial X^\nu} - \frac{\partial \phi_\nu}{\partial X^\mu}) $  one must perform a coordinate transformation:
$$ (F')_{\mu\nu} = \frac{\partial X^{\alpha}}{\partial X^{\mu}}\frac{\partial X^{\beta}}{\partial X^{\nu}} F_{\alpha\beta}$$
This transformations are proper time based and they express the real "value" seen in the "other" reference frame.
If one does the math for the partial derivatives for Minkowski metric one will get the Lorentz transformation of a general boost which can be applied to the electromagnetic tensor.
$$ (F')_{\mu\nu} = \Lambda_{\mu}^{\alpha}\Lambda_{\nu}^{\beta} F_{\alpha\beta}$$
A charge will be influenced differently ($ \frac{dp^\mu}{d\tau} = e F^\mu_\nu U^\nu $) by $ (F')_{\mu\nu} $ than by $ F_{\mu\nu} $ depending on the relative motion of the generator of $ F_{\mu\nu} $ .
The field once generated it is "independent" of the source and one also must use the "retarded" position/velocity of the source at the appropriate time in the "past".
Note: here $ F^\mu_\nu = \eta^{\mu\sigma} F_{\sigma \nu} $
One could stop for a while and read Feynman's Lecture (for a better description) :
http://www.feynmanlectures.caltech.edu/II_26.html

