# Runaway solutions and the equivalence principle

1. Aug 11, 2015

### bcrowell

Staff Emeritus
The references below describe two interrelated relativistic issues that come up when you try to describe radiation from point charges:
1. We normally expect that an accelerating charge would radiate, but this would seem to violate the equivalence principle.
2. Classical E&M can't describe point particles, because the energy in the field of a point particle diverges, and by relativistic mass-energy equivalence, this would give it infinite inertia.

Issue #2 is normally blamed for the pathological behaviors of the solutions to the equation of motion for a point charge according to the Lorentz-Dirac equation for the radiation reaction force, $F^c=(2/3)kq^2(\delta^c_b+v^cv_b)\dot{a}^b$, where $a=dv/d\tau$, and the dot means $d/d\tau$. These behaviors, described in Poisson, are runaway solutions and preacceleration (violations of causality in response to an external force).

Brown discusses the relationship between issues #1 and #2, but I'm getting confused by the details, and I'm posting to ask if anyone can help me to clear up my confusion. The basic idea is that if a charge is sitting on a tabletop, conservation of energy dictates that it shouldn't radiate. But the tabletop is actually accelerating compared to an inertial (free-falling) frame, so this implies that a charge undergoing constant acceleration shouldn't radiate. This is all consistent with the Lorentz-Dirac equation, which predicts zero reaction force when the proper acceleration is constant. After all, if there is no radiation, you shouldn't get a reaction force. So far, so good.

The first thing I'm confused about is the reasoning behind the runaway solutions. Poisson uses the low-velocity version of the equations of motion, under which the Lorentz-Dirac force becomes a Newtonian force three-vector proportional to $\dot{a}$. The equations of motion are then of the form $a=(\ldots)\dot{a}$, and this gives an exponential as a solution. But now suppose we try to do this without the low-velocity approximation. If you do this fully relativistically, then clearly the runaway motion has to look the same at any arbitrarily chosen moment in time, in the instantaneously comoving frame; therefore the proper acceleration has to be constant. But in that case the Lorentz-Dirac force is zero, so any runaway solution would have to violate the equations of motion. What's going on here?

My other confusion has to do with the Medina paper. In the abstract of the paper, he says that if the particle is not pointlike but has a certain minimum size, you don't get runaway solutions. (He states it as a condition on the mass, but it amounts to the same thing.) The runaway solutions exist because the particle is acted on by its own radiation reaction force, even in the absence of any external force. If there are no runaway solutions for big enough particles, then apparently the radiation reaction shows qualitatively different behavior for big particles than for pointlike ones. This seems strange, because I don't see how the resolution of the issues with the equivalence principle can be qualitatively different in these two cases. My understanding was that the resolution of this issue was that observers who are accelerated with respect to one another can disagree on whether a particular field pattern constitutes radiation or not. But when we judge whether a field constitutes radiation, we're talking about its large-distance behavior, and this doesn't seem to be compatible with the idea that there are two qualitatively different small-range behaviors.

Does anyone have any insights?

Brown, http://www.mathpages.com/home/kmath528/kmath528.htm

Medina, Radiation reaction of a classical quasi-rigid extended particle, J.Phys. A39 (2006) 3801-3816, http://arxiv.org/abs/physics/0508031

Morette-DeWitt, "Falling Charges," Physics, 1,3-20 (1964), http://www.scribd.com/doc/100745033/Dewitt-1964

Poisson, http://arxiv.org/abs/gr-qc/9912045

Rohrlich: The dynamics of a charged sphere and the electron Am J Phys 65 (11) p. 1051 (1997), http://www.lepp.cornell.edu/~pt267/files/teaching/P121W2006/ChargedSphereElectron.pdf

2. Aug 11, 2015

### atyy

The EP does not apply to radiation, because radiation is nonlocal (second derivatives etc).

To apply the EP exactly to electromagnetism, one has to work with the action.

http://arxiv.org/abs/0806.0464

3. Aug 11, 2015

### bcrowell

Staff Emeritus
Hi, @atyy -- Thanks for the link to the Gron paper, which I'd actually seen before. My two questions were more specific than just the general issue of whether falling charges radiate.

4. Aug 13, 2015

### vanhees71

I'm still not convinced of any "solution" concerning the self-consistency problem with point charges, but I'm pretty sure that finite extended bodies (in the most simple case a quasi-rigid charged body) can be formulated self-consistently and correctly. The key point is the proper definition of the energy-momentum tensor and the correct definition of the total energy and momentum to derive an equation of motion. To that end one has to take into account the inner stresses of the body that prevent it from flying appart due to the electric repulsion of the charge. This is known for more than a century now and is due to Poincare, von Laue, and many others. The point is to use the full relativistic expressions for all these quantities. Another source of confusion is to use "non-relativistic approximations" which then do not take properly into account the contribution of the inner stresses to the energy-momentum balance and the mass of the charged body. I guess, I'll write a section in my SRT FAQ article I'm writing, but that takes some time, particularly since I think I have to collect this from a lot of sources and many of those are not very satisfactory. There are some articles in AJP, and I also think the one by Medina you quoted (and which I didn't know until now), which give correct solutions of the problem for an extended charged object in terms of differential-difference or differential-integro equations. Particularly I don't think that the famous Wheeler-Feynman idea ("absorber theory") with advanced-retarded fields is a fully satisfactorily answer. I still don't buy (and perhaps understand) completely the argument concerning the non-violance causality.

Another question is this debate whether a falling body radiates. The point here seems indeed to be that the equivalence principle refers to local observations, while the bremsstrahlung is a phenomenon that refers to the electromagnetic field of a charged body as a whole and thus is not a local phenomenon. Here, I've never seen a satisfactory treatment in the literature and thus consider this question unsolved at the moment.

5. Aug 13, 2015

### stevendaryl

Staff Emeritus
I have never been clear how the equations of radiation for an accelerated particle relate to the intuitive explanation of radiation that is shown in most introductory E&M textbooks (such as the following, which I think is from Purcell):

In the picture, the arrows represent the electric field. For a charged particle at rest, or in uniform motion, the electric field points toward the current location of the particle. If a particle suddenly changes its state of motion (accelerates briefly and then continues at uniform velocity) at time $t$, then at a later time, $t+\delta t$, there will be a spherical shell of radius $c \delta t$ such that inside that shell, the electric field points towards the particle, and outside the shell, the electric field continues to point at where the particle would have been. (Because the information that its state of motion has changed hasn't reached this far.) In the transition between these two regions, the electric field is changing rapidly, and this rapidly changing field (with an associate magnetic field, which is not shown) is a shell of radiation.

With this heuristic way of understanding how acceleration produces radiation, it seems pretty obvious that a charged particle sitting on the surface of a planet will not radiate. Even though the proper acceleration is nonzero, the electric field of a charge on a planet is a static situation, and so there is no need for the electric field to adjust to the acceleration (and in this heuristic picture, radiation is nothing but this "adjustment" process).

As I said, it's hard to see how the heuristic picture of radiation connects with the equations for radiation...

6. Aug 13, 2015

### bcrowell

Staff Emeritus
This seems like a non sequitur to me. The argument implicitly assumes a frame of reference. We have the noninertial rest frame K of the particle sitting on the earth's surface, and alternatively the inertial frame K' of a free-falling observer. In which of these frames do you think the particle's field is a Coulomb field, and why? If anything, I would think it would assume its simplest form in the inertial frame, which is K'.

Why should staticity be defined with respect to K, and not K'?

7. Aug 14, 2015

### stevendaryl

Staff Emeritus
The frame K is a static situation (there is a time coordinate such that the metric and the configuration of charges are all time-independent), so I would not expect there to be any radiation, which to me is associated with nonstatic situations. The frame K' is not static, both the metric and configuration of charges are time-dependent, so the argument doesn't imply that there is no radiation in this frame.

It probably is not a coulomb field in either frame, but it seems to me that the electric field would be static in frame K.

But frame K' is definitely not simple at all, since both the metric and charge configurations are changing with time.

Things are clearly not static in frame K'. Heuristically, I would think that if there is a coordinate system in which the metric and charge configurations are time-independent, then there would be no acceleration-related radiation in that coordinate system. Assuming that the presence or absence of radiation is a coordinate-independent fact (I don't actually know whether that's the case, or not), I would think that if there exists any coordinate system in which the metric and charges are static, then there would be no radiation.

This heuristic (and I'm not actually claiming it to be correct; I was basically asking for an argument showing that it is or is not from someone more knowledgeable than me) would lead me to think that in static situations such as a particle sitting on the Earth, or a particle at "rest" in a Rindler coordinate system would not radiate.

8. Aug 14, 2015

### bcrowell

Staff Emeritus
No, the metric isn't time-dependent in K'. K' is an inertial frame, and the metric expressed in its coordinates is diag(1,-1,-1,-1). The charge distribution is time-independent in K, but that's merely because the charge is being accelerated by an external force from the ground. Similarly, I could apply a centripetal force to a charge, make it go in a circle, and pick a rotating frame in which the charge was at rest; that wouldn't make the problem into an exercise in electrostatics.

We could have a separate thread on this topic if you like, but this doesn't really have much to do with either of the two specific questions I asked in #1. The general idea of the application of the e.p. to this situation has been discussed multiple times on PF; you can find these discussions by searching on "falling charges" in PF's search interface.

Last edited: Aug 14, 2015
9. Aug 16, 2015

### stevendaryl

Staff Emeritus
That's not true, if you take into account higher-order effects. The metric tensor can only have that form if the curvature tensor is zero. If the curvature tensor is nonzero, then the metric can have that value at a single point, but not over an extended region.

What I said is true: a freefalling observer "sees" a time-dependent metric.

10. Aug 16, 2015

### stevendaryl

Staff Emeritus
That's a very good point.