Will a stationary charge in a uniform gravitational field radiate?

Question

Imagine a charged particle Q standing still in a uniform gravitational field g (i.e. on a table on the surface of a planet). Assume observers K1 and K2 when K1 is in free-fall and K2 is co-stationary to the charge Q (sitting at a chair next to the table).

K1 is, (at least locally) an inertial reference frame (due to the Equivalence Principle). So since it sees the charge Q accelerating towards itself with acceleration g; K1 will conclude that this charge should radiate.

Even though K2 is a non-inertial reference frame, the charge Q radiating non-stop will clearly violate his the law of conversation of energy.

We have two impossibilities here, so the obvious question is does the charge radiate, or does it not ?

(PS:I have to say I'm quite surprised to see this question was never explicitly formulated and asked in this platform, so I am sorry if it turns out to be a duplicate after all.)

Answer 1

K1 will conclude that this charge should radiate.

It depends on what we mean by radiation. In the frame of K1, the motion of test charged particles will be as if there was, in addition to gravity, a time-variable EM field, so in this sense there will be EM radiation.

But this by itself does not mean this EM field will have character of expanding spherical waves coming out of the charge Q. Nor does it mean that there will be a systematic positive rate of loss of EM energy from the neighborhood of the charge to infinity. This is a property of so-called retarded solution of Maxwell's equations for given charge and current distribution, which has proven itself accurate in Earth frame, but in now way is necessarily correct in the frame of freely falling observer.

In this sense, there may be no radiation, with complete consistency with presence of radiation of the first kind. There may be some radiation of the second kind, but so far in the description there is nothing that would suggest it.

the charge Q radiating non-stop will clearly violate his the law of conversation of energy.

If there is a time-dependent EM field, but it does not have character of spherical waves expanding to infinity, there is no apparent reason why it should violate conservation of energy.

Actually, conservation of energy in the frame of the falling observer is a little confusing subject already for another reason. Consider kinetic energy of the Earth in the freely falling observer's frame. The Earth is gaining speed and thus its kinetic energy is increasing. So is the kinetic energy of the accelerating charge on its surface. Where is this kinetic energy coming from? How do we save the energy conservation?

We may introduce fictitious inertial forces acting on the Earth and the charge that make them move as they do, from the point of view of the falling observer. There are no gravity or inertial forces near the observer (local inertial frame), we got rid of those, but we have to consider the changing kinetic and potential energies of bodies far from our little inertial frame. This can be done by introducing mentioned inertial forces, which can be viewed as gravity field in the observer's frame (this kind of force imparts acceleration to a body without regard to its mass). These inertial forces can be understood as a field pervading the whole space, ready to supply the right amount of energy just to save the conservation of total energy.

So, if we are able to save conservation of kinetic and gravitational potential energy in freely falling frame by modifying the gravity field (or adding a new field of inertial forces), it is not hard to see the same could be done to account for the changing EM energy in the vicinity of the charge: the gravity field in the observer's frame moves the bodies/charges and supplies/extracts energy to them in exactly such a way as to preserve local conservation of energy. Part of this energy can change into kinetic energy of bodies around, part of it changes their gravitational potential energy and part of it can change into oscillating EM energy.

Answer 2

You are asking about a somewhat famous paradox of a charge in a uniform gravitational field. There are few resolutions for this paradox that you can read up on the Wikipedia.

Answer 3

You cannot have knowledge of the charged particle, be K1, and K2 at the same space time. That is what space time is. You would be in three space times simultaneously, then, travelling at infinite velocity and infinite acceleration among all three. That is an impossibility. . it makes no difference if you accelerate or if the charged particle accelerates. Neither of you can tell which. The charged particle sees you accelerate and, Of course, does not radiate. You see the charged particle accelerate and radiate. There is no experiment that you can do to conclude anything else. You must follow Lorentz transformation rules to move in space time. Such rules must be rigorously followed (and that ain't easy). Square roots of 1 minus velocity squared divided by the speed of light squared pop up all over the place. Algebraic substitutions of cos squared and sin squared (remember-- velocities are vectors V does not equal vx+vy+vz. When you correctly apply Lorentz Transformations to each space time, everything works out. That is how space time was arrived at. It is simple algebra and geometry as a consequence of equivalency and the speed of light! It all works out!