# What reaches the ground first? A charged particle or a neutral particle in a constant gravitational field?

This question is probably a duplicate of questions like these: Does a charged particle accelerating in a gravitational field radiate? or Does an classical charge that is stationary in a gravitational field radiate?

But unable to understand the mathematical solution on the radiation, I ask a corollary question:

Let us consider two identical (macroscopic) balls of mass $$m$$, one has an electric charge $$q$$ and the other one is electrically neutral. If both are held at a distance $$h$$ from the ground and later released to fall in free fall in a constant gravitational field, which ball reaches the ground first for an observer standing on the ground?

The classical point of view would describe that the radiating ball would reach the ground later because some of the potential energy is converted into radiation, while for the neutral particle all potential energy is converted into kinetic energy.

Of course, other answers have dealt with the question of whether or not radiation is emitted, but I am interested on whether the balls take the same time to reach the ground or not. I know that in a free falling frame, both balls will not radiate and both will "reach" $$h=0$$ at the same time. But what about from the perspective of somebody on the ground? I know that simultaneity might be lost between frames when dealing with relativity. Is this the case here?

Radiation from acceleration in a gravitational field of the Earth would be a tiny effect under most circumstances. Much greater effect would be the (nondissipative) force acting on the charge from the surface charges on the conducting ground induced by the very same free falling charge.

This situation is essentially equivalent to the classical problem for the method of images: a charge near conducting plane (with the ground being that plane).

Since the motion of a charge under reasonable assumptions would be nonrelativistic, to the first approximation fields would be almost electrostatic and our charge $$q$$ would feel an additional force: $$F_\text{add} = \frac{1}{4 \pi \varepsilon_0} \,\frac{q^2}{4 h^2},$$ where $$h$$ is the distance toward the surface of the earth. This force would be directed toward the Earth and under most reasonable assumptions would be much greater than any radiation reaction (which would be a very small relativistic effect). As a result, the charge moving under this force would reach the ground faster than a free falling uncharged mass.

The question also contains some misconceptions, so let us try to clarify some issues:

The classical point of view would describe that the radiating ball would reach the ground later because some of the potential energy is converted into radiation, while for the neutral particle all potential energy is converted into kinetic energy.

Charged mass in addition to gravitational potential energy also has electrostatic potential energy, and in our situation the additional force comes from the variation of this electrostatic energy with height. Even if we do consider radiation in this setup, radiated energy would also come mainly from electrostatic energy.

I know that in a free falling frame, both balls will not radiate and both will "reach" $$h=0$$ at the same time.

In most situations involving curved spacetime, conducting objects, dielectrics or multiple charges, radiation cannot be treated as a local quantity. So neither Larmor formula nor Einstein equivalence principle are applicable to the question of what the charge would be radiating. Instead, one has to solve both equations for the motion of the charge and Maxwell's equations in curved spacetime and/or material medium. Radiation would then emerge as an energy flux in the asymptotic region (and thus it is not something that could be measured by a single observer either moving with the charge or located on the ground).

• While this is surely relevant it fails to answer my question. Suppose there is no ground. There is a virtual floor (with no charges) that moves with constant acceleration towards the two balls, would a ball arrive before the other for an observer that moves with the virtual floor? – Mauricio Feb 17 at 10:16
• @Mauricio: If the charged and uncharged masses were initially at rest, they would remain at rest in their own reference frame and they pass through this virtual floor simultaneously. But that is not what has been asked in original question. Gravitational field of the Earth globally is not the same as accelerating floor, and it is global description that counts. – A.V.S. Feb 17 at 18:38
• … If we imagine the Earth's gravitational field is created by some 'dark matter' that does not interact with the charge directly, this gravitational field would be equivalent to a large diffuse 'blob' of weakly polarizable dielectric, then there would be tiny additional attraction for the charge toward the Earth's center. – A.V.S. Feb 17 at 18:40
• Suppose that there is no ground. You have a constantly accelerated test observer that is moving towards a charged and a neutral ball, do both ball cross the observer reference plane at the same time? – Mauricio Feb 19 at 13:58
• @Mauricio: This depends on how you are “preparing” your balls. If you accelerate your balls in an elevator with constant acceleration for a long time and then simultaneously release them, then the charged ball would be falling toward the elevator floor faster. However, if both balls were at rest in inertial frame, they would reach accelerating floor simultaneously. In other words: how the charge moves is determined not only by its initial position and velocity but also by EM field around it. – A.V.S. Feb 19 at 18:32