Question

If you drop a proton and a neutron in a gravitational field, they both fall, but the proton has a charge and accelerating charges radiate energy, so that leaves less kinetic energy for the proton and by this reasoning, it should fall more slowly than a charge-free object.

The issue is discussed but not in the terms above in Peierls's "Surprises in Theoretical Physics" in the chapter "radiation in hyperbolic motion", but I didn't understand the chapter well enough (or at all) to apply it to my version of the question. Peirls also refers to Pauli's Relativity book (section 32 gamma) but while Pauli claims there is no radiation from uniform hyperbolic motion, he does say there is radiation when two uniform rectilinear motions are connected by a portion of hyperbolic motion. So I take it that would mean a proton at rest which falls for a second and then is somehow forced to maintain its newly acquired downward velocity from the fall without speeding up any further would have radiated.

Answer 1

This question is somewhat academic - and it has been controversial - but the more correct answer is No, it should not.

It is academic because the electrostatic force between two protons is about times stronger than the gravity between them. In reality, a proton will polarize any conductor on the ground and it will be attracted by a much bigger electrostatic force.

But if you guarantee that such effects don't exist, then protons and neutrons - as well as everything else - fall exactly by the same acceleration. This follows from the equivalence principle. The principle is much more general than most people think. In a freely falling elevator, there is no way to find out whether the elevator is freely falling in a gravitational field, or in empty outer space without gravitational field. Different accelerations would surely allow us to distinguish the two cases and it would clash with the equivalence principle.

Usually we don't have to ask "in what frame Maxwell's equations are valid". Are they more valid in a frame attached to the ground, or in a freely falling frame? The difference is tiny because the electromagnetism-induced accelerations are typically greater than the gravitational ones. Of course, the gravitational force always has to be added, too. But for example, for the protons in the LHC, gravity is negligible for the protons.

But you're asking about a "mixed effect" that requires both electromagnetism and gravity. And a cleaner way to describe it is to use the freely falling frame. In that frame, the metric tensor is as flat as you can get, and the proton's and neutron's accelerations coincide. In fact, even in a very tall but thin cylinder observed over a very long time, one may always set the metric pretty much to constant, up to terms that go to zero when the cylinder is really thin.

So classically, the field around the falling proton may be obtained in the freely falling frame, with a huge accuracy, and it will be just a field of a static proton - in this frame. Of course, if a field of a static proton is observed from another, e.g. relatively accelerating frame, it may look different. But there won't be any real loss of energy that would allow the two accelerations to diverge.

If you tried to calculate "how much energy" the proton is supposed to radiate, the usual methods that are available in the flat space would fail. In the flat space, you usually "attach" the charged particle to the region at infinity, and solve Maxwell's equations with a source. However, all these equations are affected by the existence of the gravitational field - that is getting weaker as you go further from the Earth. Because of this weakening of gravity, and the terms by which gravity influences electromagnetism, you will find out that the radiation is zero in the freely falling frame.

Quantum mechanically, one has to deal with the Unruh radiation etc. Frames that are relatively accelerating with respect to each other have different ideas what the ground state (vacuum) is. So an accelerating particle could interact with the Unruh quanta. This is another extra subtlety. I am confident that all actual measurements of the time needed to fall etc. has to coincide for protons and neutrons, even including ? quantum corrections.