From douglips:
It just depends on how you wish to define radiation. In your example – a dipole “compass” pointing at a moving charge – energy definitely moves from the charge to the dipole compass (since the compass begins moving to follow the charge as it goes by), but you wouldn’t normally call that radiation. It’s just the same as taking two point charges and holding them fixed and then releasing them – they each begin moving but not because of any radiative sorta thing. Just static forces. (Of course, the charge in your example will be changing speed as it pulls on (and is pulled by) the dipole ==> it’s accelerating ==> it radiates. Of course, this radiation is only a by-product and not the cause of the dipole’s motion.)
The two sources of difficulty in this whole issue are 1) radiation is not always clearly defined, and 2) once you define radiation, its presence will be observer dependent.
ZenBeam, in your tube-through-mass thought experiment, I don’t think it is so immediately clear that the system must radiate:
We first need an observer. I guess we’re just using someone far away. If the charges were in free space and forced along the prescribed path, the system would radiate w.r.t. our observer because the EM fields develop tangential “kinks” to maintain continuity. That is, the fields some distance away have components that the non-accelerating case doesn’t have because the field lines have to sort of catch up with the accelerating particle. It’s all due to the fact that the fields, as soon as they “leave” the charges, no longer feel the accelerating force.
In your case with the gravitational orbit, the field lines do not quit feeling the “force” when they leave the charges. They still feel the gravity. Alternately, the fact that the geodesics are curved fixes the continuity problem, and when the fields reach the observer, there are no unexpected components present.
This is not coming out nearly as clearly as I hoped. Hmmm. I’ll move on anyway.
Here’s an assortment of thoughts:
i) The original problem dealt with an accelerated observer and an intertial charge. One difficulty that crops up is that an accelerated observer is unable to put a good coordinate system over all of spacetime. (The region he can get to is often called the “Rindler wedge”, and so I’ll use this term.) For many classes of observers, this source cannot in general be placed entirely within the Rindler wedge, thus invalidating most simple radiation derivations (as they assume implicity either 1) spacetime can be covered entirely by a good set of coordinates, or 2) the source can be contained entirely within the Rindler wedge.) For carefully constructed sources and appropriately chosen observers, though, one can decide if radiation is detected. And the answer is that there is no radiation detected.
ii) One may hope to get around all the problems with defining radiation by just sticking with photons. However, the particle content of a quantum field theory is observer dependent. In particular, accelerated observers do not have to agree with non-accelerated observers when they count particles. (This hasn’t been an issue really so far since the discussion has stayed classical.)
iii) The equivalence principle tells us that a charge on the surface of a gravitating mass will radiate as seen by a freely falling observer. This begs the question, “Where’s the energy come from?”
First, the free-space version. A charge is strapped to an accelerating rocket. We (inertially) observe radiation and we conclude that the rocket must be burning a bit extra fuel to provide energy for the radiation. Now, the gravity version. We place the rocket on the launch pad and get it just right so that it supports the charge at a constant height from the earth’s surface. A high-diving observer sees radiation and decides that the rocket must be burning a bit extra fuel to provide energy for the radiation. If we used the ground instead of the rocket, we’d have to search for that extra bit, but it would be there somewhere. Maybe the table underneath the charge has slightly less thermal energy now.
I must admit that I’m not fully satisfied with this. But the conclusions one must now draw don’t seem too outrageous, I suppose. For instance, I have basically argued that if you take a massive object and coat it with charged particles, it will cool down. I don’t know why that would be so bad…
Incidentally, I have seen a completely different point of view on the energy conservation issue. All symmetries of a system have associated with them conserved quantities or “Noether currents”. Energy is the conserved Noether current associated with time translation symmetry. In a curved spacetime (or an accelerated frame), time translation invariance is not as straightforward as it is in Minkowski (flat) spacetime. The argument I’ve seen posits that that the symmetry in curved spacetime analogous to time translation symmetry in Minkowski space corresponds to a Noether current that is not our usual energy, and that the thing we normally call energy is not a Noether current at all in curved spacetime (and therefore needs not be conserved.) I’d need a relativist, though, (Chronos?) to determine where this guy’s argument lies on the crackpot scale.
It’s unsatisfying, however, because I would expect any radiation to depend only on the amount of charge and the gravitational acceleration at the surface. If the thermal energy in the mass were supplying the energy, I would expect the radiation to decrease with a decrease in the temperature of the mass. What if the charge and mass are increased, and the temperature of the mass is decreased to very near absolute zero, so there isn’t enough thermal energy to supply the radiation. We’ll shield the whole thing from the cosmic background radiation while we’re at it.