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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.

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5 Answers 5

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.

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This question is about a conflict between the equivalence principle and the notion that charged particles must radiate. From what I understand, you are stating that the proton will radiate and that the equivalence principle is not valid here. I'm not saying you're wrong, but you need to provide a reason of why the principle is not valid. –  Bruce Connor Jan 18 '11 at 20:28
    
The equivalence principle is a slippery concept. Before saying whether I think the equivalence principle is being violated, I'd want to know precisely what statement of it we're using. In particular, the equivalence principle applies only to infinitesimal neighborhoods, and the question "is it radiating"? is best answered only by looking in the radiation zone (i.e., not in an infinitesimal neighborhood). Rather than quibbling about definitions, though, let me make some physical statements that I'm quite confident of (next comment). –  Ted Bunn Jan 18 '11 at 20:49
    
For definiteness, consider a weak, static gravitational field (say the Earth, modeled as the Schwarzschild metric). Suppose that a charged particle starts from rest (in the Earth's frame), is dropped, falls for a finite distance, and is brought to rest again. That charge does radiate, in the sense that the Poynting vector, integrated over a large sphere at some later time, is nonzero. Physically, that energy could be captured and used. If you believe that the charge followed the same path as an uncharged particle would in the same setup, then you can make a perpetual motion machine. –  Ted Bunn Jan 18 '11 at 20:54
    
Another way of looking at it. Let the charge be a small sphere (rather than a point charge). Let the 3 length scales in the problem have the obviously natural hierarchy: radius of particle << radius of Earth << $c^2/g$. Work in a non-inertial frame fixed to the Earth. With the given scale hierarchy, the Maxwell equations are, to an excellent approximation, the same as in flat spacetime, and the particle moves through the frame with constant acceleration. Standard methods can be used to calculate the radiation reaction force on the sphere. It's nonzero during the initial and final jerks. –  Ted Bunn Jan 18 '11 at 21:04
    
@Ted, Even though it is easier to deal with the radiation zone, it is not necessary. The energy flux going away from the particle is the same for any closed surface around it, even if I take an infinitesimal sphere around the proton. The statement here is the following: In the proton's frame, there is no radiation, and the ground reaches him at the same time it reaches the neutron (assuming they are close enough). In earth's frame, the two must reach the ground at the same time. The argument is on the next comment: –  Bruce Connor Jan 19 '11 at 1:44
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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.

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I agree with you that the accelerations of a charged and uncharged particle must be the same during the "steady-state" portion of the fall -- i.e., excluding a short window of time around the beginning and end of the experiment. But do you agree that, in an experiment involving dropping the particle, letting it fall for a finite time, and bringing it to a stop, there must be a difference? If not, then you could create a perpetual motion machine. Running out of room in this comment; will supply details in a subsequent comment. –  Ted Bunn Jan 18 '11 at 18:11
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If charged and uncharged particles fall precisely the same way during an entire finite-drop experiment, you could make a perpetual motion machine as follows: Let the charged particle fall a distance h from rest. Absorb all of the kinetic energy (as measured in Earth's rest frame). Use that energy to raise the particle back to its original height. Repeat. Meanwhile, have faraway radiation absorbers catch the emitted radiation. (This situation involves only weak-field, asymptotically flat gravity, so notions of energy conservation make straightforward sense.) –  Ted Bunn Jan 18 '11 at 18:14
    
Thanks to you and to everyone else who has answered my question so far. –  Donald Jan 18 '11 at 18:29
    
Oops. I sent that accidentally before I finished. Yes, I understand the question is academic and there would be an image charge whose effect would be far greater--my question is just whether the proton's acceleration in the gravitational field would create EM radiation that would slow it down. So far the responses have been interesting, but contradictory and I'm in the position of finding each response very convincing while I'm reading it. –  Donald Jan 18 '11 at 18:33
    
Donald-- Don't despair: This is a famous old puzzle, and it's confused lots of people over the years. The article by Parrott I mentioned in my answer, <arxiv.org/abs/gr-qc/9303025>; , is well worth looking at. In my strongly-held opinion, he's got the answer wrong: he assumes that the radiation reaction force during the constant-acceleration phase must be nonzero, based on an energy-conservation argument that does not appear valid to me. But he puts the problem in context very nicely and gives references to a bunch of useful literature. –  Ted Bunn Jan 18 '11 at 19:17
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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.

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good point but you might consider leaving a comment directly under the question as this is not an answer per se. –  Marek Jan 19 '11 at 20:41
    
@Marek, I thought about that and concluded that this is actually better seen as the answer to the question rather than just a point of clarification. –  Philip Gibbs Jan 20 '11 at 6:51
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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.)

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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.

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while I generally like your the approach you take in your answers, your language can seem a bit disjointed. This comment is meant not to discourage you from answering questions, but to encourage you to spend more effort in framing your replies. In any case +1 for your answer, in particular your very pertinent observation that the equivalence principle holds for a local inertial frame in some small region of space. This fact is often ignored in discussions involving the EP. –  user346 Jan 20 '11 at 19:29
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I sort of look here and make answers rather quickly. This is a marvelous way of avoiding actual work :-)). So sometimes I have to confess I rattle these off pretty quickly. –  Lawrence B. Crowell Jan 20 '11 at 19:53
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