Should a neutron fall faster than a proton?

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

Yes, it should.

There are a lot of subtleties involved in some questions like this (see some articles by Stephen Parrott available on arxiv.org if you want to start digging into this), but fortunately in this particular case the answer is clear.

The easiest way to see it is just to apply energy conservation, as you say. There's no doubt that, long after the proton has fallen, radiation is escaping to infinity, carrying energy with it. That energy can only have come from the mechanical energy of the proton.

It's not necessarily the case that the charged particle's acceleration is less than $g$ during its entire fall. The difference in acceleration can come only during the "jerks" near the beginning and end of the fall. In fact, that's what's predicted by the Lorentz-Dirac equation (the closest thing there is to a standard equation for the radiation reaction on an accelerated point charge). The proton takes a bit longer to get going, accelerates at $g$ for a while, and comes to a stop when it hits the floor. The difference in what happens to the charged and uncharged particles during the initial and final phases can account for the difference in energy.

The Lorentz-Dirac equation has all kinds of problems, but in this particular case its predictions are pretty much bound to be qualitatively right. The problems all come from treating the falling charge as a mathematical point, but if we treat it as a small sphere (which the proton is anyway!) of radius much less than any other length scale in the problem, the LD equation is a good approximation to the correct equation of motion.

Answer 2

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 $10^{36}$ 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 $\hbar$ quantum corrections.

Answer 3

The answer depends on the initial state of the electromagnetic field around the proton. It cannot be zero everywhere because the proton has charge.

The electromagnetic field would of course affect the trajectory of the proton and determine whether it fell faster or slower than the neutron.

Since you have not specified what that field is the question is incomplete and cannot be answered.

Answer 4

In addition to my original answer, I've scattered bits of an argument through various comments in this thread, but I don't think I've tied them all together very clearly. More importantly, I think I made some mistakes. Let me try to say what I think is true, and justify it more carefully than I have.

Consider the situation in which you initially hold a proton and neutron at rest (relative to the Earth), drop them, and catch them after they've fallen a certain distance. I'm essentially certain that the the proton falls slower than the neutron, in the following specific sense: the speed of the proton just before you catch it will be less than the speed of the neutron (and also the proton will arrive later than the neutron).

Here's why. Even if the proton doesn't radiate during most of its fall, it does radiate for a brief period right when you drop it.

Before trying to convince you that this last statement is true, let me point out that there's certainly no equivalence-principle based argument against it. At most, the equivalence principle says that during the time the particle is in free fall, it shouldn't radiate. It doesn't say anything about what happens during the transition from non-free-fall to free fall.

Given this, it seems clear to me that the burden of proof is on anyone who says there's no radiation during the transition period. After all, we have a charge undergoing jerky motion. In the absence of an equivalence-principle argument, the default assumption would surely be that it radiates.

That's not a proof, of course. One thing that would count as a proof would be to calculate the fields and determine the radiated flux. This would be hard to do in the full Schwarzschild geometry, but a calculation in flat spacetime, replacing observers at rest with respect to the Earth with accelerated (Rindler) observers, wouldn't be hard. f I wanted to do it, I'd probably start with the basic formalism set up in a recent paper I just discovered by Maluf and Ulhoa.

[One might question whether such a Schwarzschild-Rindler substitution would be justified. I'll just say that the difference between the two is simply tidal forces, and I see no reason that they're relevant in this situation. If you like, change the mass and radius of the Earth to make them much larger, while keeping $g$ constant. That weakens tidal forces still further, but it's very implausible, at least to me, that it bears on the radiation question.]

But even without doing such a calculation, I'm confident that there is radiation during the jerk period. The reason is that the radiation reaction force can be shown to be nonzero during this period. The radiation reaction force is problematic for point charges, but if we model the proton as a small sphere of charge (which, after all, it really is!), with a radius much less than any other length scale in the problem, then calculation of the radiation reaction force is straightforward. You can look up how to do it in Jackson, and there's a bunch of more recent literature. Specifically, Rohrlich has many articles on the subject, but the monograph by Yaghjian is the most complete reference. Anyway, the conclusion is completely unambiguous: the radiation reaction force is nonzero during the jerk.

Let me conclude with a mea culpa: in some of my earlier comments, I think I said that there was radiation even during a constant-acceleration phase (although the radiation reaction force vanished then). I'm pretty sure I was wrong about that. (My belief wasn't quite as stupid as it might seem: that combination of radiation with no radiation reaction is precisely what happens when a charge accelerates uniformly in Minkowski space. It seems like a paradox, but it's not.)

Answer 5

This is a very good question. We think that from the equivalence principle that the proton and neutron should fall at the same rate. This is a good approximation to the situation. However, the equivalence principle holds for a local inertial frame in some small region of space. Within that region the electric field of the falling charge is a nice radially symmetric field. However, the field lines do not end there. They extend throughout the space and thread through the curved space or spacetime. As the charge approaches a radially symmetric gravity field the field lines will adjust to the curvature. This means the charged particle will be associated with a changing electric and magnetic field.

The electric and magnetic field elements $F_\mu~=~F_{\mu\nu}U^\nu$. The covariant constancy of the EM field $\nabla^\mu F_{\mu\nu}~=~0$ is used to derive $d^2F_\mu/ds^2$ with respect to some separation of the charge from another geodesic point --- which can be at $\infty$. The result is then a calculation of $d^2U^\nu/ds^2$ with the geodesic deviation equation. This is then equated to second derivatives of the fields $F_\mu$, which means radiation is emitted. This is an attenuating process.

This argument is “approximate,” for the attenuation acts as an acceleration, which adjusts this from a pure geodesic deviation. I worked this out in considerable detail a few years ago in response to a discussion on this matter. The analysis becomes rather involved. Yet the result is that the EM field response to the infall of the proton will act as a sort of "viscosity" that slows it down.

Addendum:

All one needs to do is to use the Larmor law $P~=~\mu e^2a^2/6\pi c$ and input the acceleration a = -GM/r^2 for Newtonian gravitation. A Newtonian bit of thought illustrates how a gravity field can pull a charge and induce a radiation. The radiation power $P~=~\int F\cdot dv$ $=~\int Pdt$ gives a radiation force (Abraham-Lorentz) law $F~=~\mu e^2{\dot a}/6\pi c$. A relativistic result should recover a Newtonian result. This radiation force is the resistance a charge experiences due to the emission of radiation. The time derivative of the acceleration is clearly dependent on the changing radial distance of the charge from the gravitating body.

Answer 6

Well, I'll put in my grain of salt, though I am certainly not a specialist of General Relativity.

Consider two protons in the gravitational field, not of Earth but of a non-rotating planet.

One is kept "stationary" by some support fixed on the planet. In terms of General Relativity, it is "accelerated" by the force exerted on the ground by the support. But nothing moves, no work is done.

The other one is in free fall along a circular trajectory, with just the right speed to stay at a fixed distance from the planet. Being in free fall it is not "accelerated" but it does go round and round.

My bet is :

A) the first one does not radiate. Where would it find energy to radiate ?

B) The second does radiate and his circular trajectory will, over time, get closer and closer to the planet, possibly on a trajectory not quite circular, losing potential energy, half of it being transformed into kinetic energy and the other half being radiated away. Exactly the same thing would happen if, away from any large mass, it was having a circular trajectory around an opposite charge (let us assume that this charge would be carried by an object of negligible gravity, but still much have heavier mass than the proton, so we do not have to go into "reduced mass" rigmarole).

This is my bet.

With this logic, a free falling proton just falling "down" would also radiate, and thus fall more slowly than a neutron.