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

Question

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?

Answer 1

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.

Answer 2

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?.

Answer 3

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

Answer 4

B doesn't infer that the charge doesn't radiate. The charge itself feels a back reaction due to the emitted radiation, which is due to the interaction it has with its own electromagnetic field. A rigorous treatment of electromagnetic self-force was given recently, see here.

Older approaches led to inconsistencies, but it's bad practice to hide behind these inconsistencies to argue for bad physics. A reasonable physics point of view in the absence of a rigorous mathematical treatment should always have been that the charge radiates in all frames because you can always invoke some physical effect of the emitted radiation that is manifestly clear in all frames. One can invoke conservation of energy, by putting the charge on a rocket and then considering a finite rocket burn and then arguing that the energy of the emitted photons have to be accounted for.

Another physics arguments that doesn't depend on mathematical subtleties is to invoke observer C who travels close to the speed of light, so close that the emitted radiation is blue shifted to hard gamma rays and C is fried by these gamma rays. Then how does B explain what happened to C?

Answer 5

This is not complicated.

• The question of whether the particle radiates must be resolved in its frame of reference.

• The question of whether the radiation is observed must be resolved in the observer's frame of reference.

Changing the observer or adding more or different observers does not impact whether or not there is radiation.

Acceleration is not determined relatively, you can tell if you are in an accelerating frame without ref to an external frame because you feel force.

Answer 6

A charged particle fixed to a frame S′ 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′) and therefore, he observes it to radiate.

Yes, he observes it to radiate, because it does radiate. Don't forget the wave nature of matter. Or pair production. You made that electron (and a positron) out of an electromagnetic wave. And don't forget Compton scattering either:

^{Image courtesy of Rod Nave's hyperphysics}

You can effectively "shave a slice" off the incident photon's kinetic energy and give it to the electron, whereupon the electron moves. The inverse Compton is the same in reverse. Bremsstrahlung isn't all that different. The electron is real, the photon is real, the interaction is real, the motion is real.

However, for the non-inertial observer B standing on the non-inertial frame S′ the charged particle is at rest, and therefore, does not radiate at all.

Not so. The observer knows he's not at rest because he can feel the acceleration. He's keeping pace with the electron, so he knows that's not at rest either. Plus he can detect that photon. The electron is real, the photon is real, the interaction is real, the motion is real. What isn't, is the frame. That's an abstract thing.

A infers that the charged particle radiates but B infers it doesn't. Can both inferences (mutually contradictory) be correct simultaneously?

No. The accelerated particle radiates*. Do not let abstract things that are not there cloud your understanding of the things that are.

*_{A falling electron doesn't radiate. But that's another kettle of fish altogether}.