Radiation from accelerated charged in a co-accelerated reference frame

Question

I was wondering if I sit on a accelerated charge particle I will not observe radiation since the particle is at rest according to me. But my friend observes it from outside and he will see radiation since it is accelerated according to him.

Please clarify my doubt.

Answer 1

It is quite easy, despite all the difficulties the comments are warning us about, to express this question in a mathematically tractable way: (and to actually answer it!)

1. Solve the Maxwell equations for an accelerated particle, while remaining in an inertial frame yourself. Of course you get radiated fields.
2. Go into the accelerated frame with the particle and look at those fields. We then only need to understand how to transform $E$ and $B$ fields.

Part 1) is a bit technically challenging, but with retarded Lienard-Wiechert potentials (or Jefimenko equations) it can actually be solved. For step 2), however, the usual EM textbooks only give coordinate transformations between inertial frames (e.g. here in [Jackson]). Luckily, we can circumvent this problem. At each point in the particle's trajectory we can take the inertial frame in step 1) comoving with the particle at that moment, and with the particle in the origin. So we always have the particle just at the turning point of its trajectory in that frame and with the given acceleration $a$. This means we always get exactly the same field as a solution! The transformation to the accelerated coordinate frame then has to take place for exactly the same situation each time. And whatever your transformation rules are, they should give the same result, if applied in exactly the same situation. Therefore, seen from the particle, the field is always the same.

But we are not yet satisfied. We want to actually see this by computing things, therefore we have to choose exactly how we want to transform the fields. We could use: a) an inertial frame moving instantaneously with the accelerated particle (but only at one point in time), or b) you could use [Rindler coordinates], or c) any wild coordinate transformation where different points in space-time have different accelerations and only your own position accelerates exactly with the particle. The problem is that special relativity only allows a), which we don't trust as our choice, while general relativity allows any wild transformation of class c). Fortunately, a) and b) are not very different, since Rindler coordinates are essentially just a continually updated inertial frame transformation.

Whatever you choose, you will find some transformed field, and then you still have to answer what you call "radiation". There exist quasi-static solutions with pure sinusoidal time-dependence and no radiated power, like fields enclosed in a cavity. But of course we do not expect that here and it turns out we can ignore that problem. In view of all this, the best way to interpret the question now is:

If we solve the field in an inertial frame with Lienard-Wiechert, and then transform to Rindler coordinates, will we see time-dependent fields? The answer in that case is NO!

To show it we use the [Lienard-Wiechert potential], to be precise [this equation] for ${\bf E}({\bf r},t)$ in the link, and the retarded time $t_\text{ret}$ for a uniformly accelerated particle along the $z$-axis: \begin{align} t_\text{ret} &={\small \frac14} \left( 2t - \frac{1+R_t+x^2}{1+t+z} + \frac{ (1+t+z)(1+t^2-2x^2-2z-z^2) }{1+R_t+x^2} \right), \\[8pt] & \text{with}\ \ R_t=\sqrt{ 4x^2 + (x^2-t^2 + 2z+z^2)^2 } \end{align} For some extra insight we let the acceleration begin at a finite time $t_\text{start}<0$. This also solves some infinity problems with a particle that starts accelerating at $t=-\infty$. We can then look at the field either in the lab frame or in Rindler coordinates.

Below is a computed result for the $E$-field strength of a particle at first traveling at constant speed and then at constant acceleration. The first movie is seen from the lab frame, the second one from the observer comoving with the particle. For details see [Lienard-Wiechert.f90] and [Lienard-Wiechert.zip]. It should be clear from the second movie that sufficiently long after the acceleration starts, the field seen from the particle's frame is stationary, although it is clearly not the Coulomb field; it is quite different at the bottom near the Rindler horizon and also elsewhere it is slightly asymmetric. So it is not the usual static EM solution!

It should also be clear that starting to accelerate after uniform motion (the "jerk" or "jolt" as it is called) does create time-variation in the particle's frame in this calculation, in a kind of wavefront moving out. But one might object that at that there we use Rindler coordinates in a non-uniformly accelerating frame, which might be wrong. In any case before or after this wavefront passes it should be correct.

Movie 1. $E$-field strength seen in the lab frame, using $c=1$ and parameters: $a=1, t=-6\dots 6, x=-0.06\dots0.06, z=-0.04\dots 0.04$. The acceleration starts at $t=-3$.
$$ \quad $$

Movie 2. $E$-field strength, now seen in Rindler coordinates, for observer moving with the particle.

Answer 2

if I sit on a accelerated charge particle I will not observe radiation since particle is at rest acc to me.

If the particle is deflected in the magnetic field, for example, you either continue to ‘sail’ straight ahead without control - and must still consider the particle to be accelerated - or you also accelerate with a drive, but then you are no longer in a rest system.

In either case, you will therefore see that the particle emits EM radiation.