Does a charged particle orbiting Earth radiate?

Question

I found this question, for which the answers I did not understand:

Does a charged particle accelerating in a gravitational field radiate?

So I want to dumb it down.

1. From my backyard, can I see photons emitted by charged particles bent by Earth's gravitational field?

2. What about the folks in ISS? Can they see it?

3. What about another particle, traveling in a close parallel with the charged particle. Does it see any radiation?

I guess this should be measurable, not as flashy as an Aurora, but at least detectable by the sheer number of particles passing by on a bent path.

Some plain explanation would be nice, but the ppl on the other post did their best. Now I just want to know if from different frames of reference, if I put a sensor there, can I detect something or not?

Answer 1

From my backyard, can I see photons emitted by charged particles bent by Earth's gravitational field?

In principle yes, in practice no. Supposing an electron in a circular orbit with a period of 90 minutes, the radiated power calculated naively from the Larmor formula is less than $10^{-51}\text{ W}$. The spectrum of the radiation is multiples of the orbital frequency; using $E=hf$, that translates to one photon per $10^7$ years or less.

What about the folks in ISS? Can they see it?

In principle, yes.

What about another particle, traveling in a close parallel with the charged particle. Does it see any radiation?

In principle yes, because spacetime in the vicinity of the orbits is curved, so the argument from the equivalence principle that they would see nothing doesn't apply. But the equivalence principle does apply approximately, so the amount of radiation you'd expect to see when co-orbiting is even smaller. I don't know how to calculate it.

Another answer says

[I]t can not be radiation in the meaning of a flow of energy coming from the charge according to the value of the Poynting vector in the lab. Otherwise an elliptical stable orbiting charge would be a source of endless EM energy to the Earth.

As Sten said in a comment, that argument is incorrect. The radiated energy comes from the decay of the orbit. The answer that Sten linked in the comment is worth reading.

Answer 2

From my backyard, can I see photons emitted by charged particles bent by Earth's gravitational field?

My answer tries to deal with E and B fields instead of photons. Of course these fields would change in time due to the change of the distance between the charge and the site. But it can not be radiation in the meaning of a flow of energy coming from the charge according to the value of the Poynting vector in the lab. Otherwise an elliptical stable orbiting charge would be a source of endless EM energy to the Earth.

An observer moving in the same geodesic of the charge would measure only a static electric field, so it is clear that there is no radiation.

For an observer in another geodesic (as the ISS), the E and B field would also change in time. But again it could not detect radiation in the meaning of a continuous flow of energy from the charge.

Maybe it is interesting to compare the situation of a charge moving at a constant velocity with respect to an inertial frame. When Lorentz transforming the fields (from the frame of the charge, where there is only a static E field), it can be proved that any component of the changing E and B fields, at any point P out of the charge, is a solution for the wave equation, ex: $$\frac{\partial^2 E_x}{\partial t^2} = c^2 \nabla^2 E_x$$ However we can not say that there is any wave passing by P.

The general solution of the 3D wave equation is not restricted to functions like $f(\mathbf{k\cdot r} - \omega t)$. It is only one of other possible solutions.

Answer 3

If a charge particle is accelerating (not by gravity) then it would radiate as a result of Maxwell's equations. If one charge particle and a neutral particle are accelerating due to free fall (gravitational field) then they will accelerate equally and the charge particle would not radiate. This seems paradoxical because these two scenarios can differentiate between acceleration due to gravity and acceleration by other means, violating the equivalence principle.

I can only state one thing about acceleration due to gravity and acceleration due to some other means: That acceleration due to gravity is just gravitational field which is just the curvature of space-time itself, and any particle irrespective of its mass and electric charge would follow the same trajectories. If we think this way then there's no radiation from a charge particle in a gravitational field. While acceleration by other means would mean a charge particle would radiate.