Classical charged particle's reaction to its own retarded field

[Drakkith]

If the particle is "emitting" the field at c, then from it's own rest frame it would not even see any squashness, right? Like shining a laser in front of you while traveling at 90% c, you wouldn't notice anything different.

 

[jjustinn]

If the particle is "emitting" the field at c, then from it's own rest frame it would not even see any squashness, right? Like shining a laser in front of you while traveling at 90% c, you wouldn't notice anything different.

OK...so I just spent half an hour typing out my response, and a gnat bumped my touchscreen and took the focus off of the textbox, so my backspace took me to the previous page...and I lost it ALL! aaaaarg.

Anyway, my eventual conclusion was this -- I think you're right; I think it's a special case of the moving magnet / conductor problem (http://en.wikipedia.org/wiki/Moving_magnet_and_conductor_problem).

The most basic example of this is current in a wire -- which is just a bunch of charges moving at a low, finite velocity, and therefore has both an electric field and a magnetic field. However, it's basic electromagnetism that if you go to the rest frame of those particles, they have only an electric field, *BUT* the effect on external objects is the same either way...in much the same way as the self-field seemingly magically cancels when the particle is moving -- and we know that therefore in its rest frame the forces must cancel, and the easiest way for that to happen would be a simple electrostatic field (assuming in both cases an extended spherical particle -- though I'm again bothered by what happens when you add Lorentz contraction).

However, I'm still not 100% convinced...I think the only thing that could convince me would be to actually sit down and do the calculations (or have the Physics Fairy come down and explain it all to me, which is what I was kind of hoping for ;), but after sitting here and trying to think how to even frame the problem in a calculable way, I kept running up against your point -- that in the particle's rest frame, there IS no motion, so there is no retarded position for it to be moving relative to.

 

[Drakkith]

That's what I'm thinking jjustinn, but I'm wondering if some QM rules trump this.

 

[jjustinn]

So I found a definitive reference -- it turns out the Feynman Lectures on Physics had a lot more detail on this than I remembered.

From Book 2, Chapter 28, section 4:

The picture is something like this. We can think of the electron as a charged
sphere. When it is at rest, each piece of charge repels electrically each other piece,
but the forces all balance in pairs, so that there is no net force.
However, when the electron is being accelerated, the forces will no longer be in
balance because of the fact that the electromagnetic influences take time to go
from one piece to another...
Both the magnitude and direction of the force depend on the motion
of the charge...

[With an accelerating charge w]hen all these forces are added up,
they don't cancel out. They would cancel for a uniform velocity, even though
it looks at first glance as though the retardation would give an unbalanced force
even for a uniform velocity. But it turns out that there is no net force unless the
electron is being accelerated.

I wish he had brought up the frame-of-reference argument here, because that would really drive the point home and would have been a less hand-wavey to demonstrate the same fact without having to show the calculations.

But in any case, there we are. A well-recognized source that explains it in plain English, and even explicitly calls out both the at-rest and constant-(non-zero)-velocity cases.

 

[Drakkith]

Ah ok, that makes sense. Thanks jjustinn!

 

[hanii]

what is the conclusion?? why the electron is not accelerated to its own field?

 

[jjustinn]

what is the conclusion?? why the electron is not accelerated to its own field?

For radiating charged particle moving at a constant velocity, in the frame where it's moving, the forces from its own field exactly cancel (with some assumptions on its geometry -- e.g. that it's a rigid shell, for example).

However, as Drakkith pointed out, another easier way to get the same answer without all of the calculations is to just do a Lorentz boost to look at the particle / its field in a frame where it is at rest -- in that case, it's a lot easier to see how the forces would cancel on a spherical particle...however, if you're like me, that's just a little too magical...but luckily, you get the same answer in either case.

As mentioned above, the Feynman Lectures do mention these results, but don't give the calculations -- but I did find another book that does them explicitly: http://books.google.com/books?id=m8... "constant velocity" charged particle&f=false (Introduction to Classical Electromagnetic Radiation by Glenn Stanley Smith, pg 437).

Of course, if the particle itself is accelerating, then this doesn't apply, and you get into the well-known Abraham-Lorentz forces, etc.