Classical charged particle's reaction to its own retarded field

[jjustinn]

I'm sorry, but I'm a bit lost here. Why is the charged particle ever in anything but a static field from its own point of view?

Because it takes the field a finite amount of time to travel / change.

Every time you bring that up it makes me more confused though. It seems to make good sense -- especially since it matches up with observations -- but I think the problem is that it's only valid with a hidden assumption -- that you can use the electrostatic / steady-state approximation, where the field is always as it would be if the particle had been "turned on" and at that position forever.

 

[Drakkith]

I thought it was that simple. But I definitely don't know. Perhaps the frame of reference idea doesn't quite apply here?

 

[jjustinn]

I thought it was that simple. But I definitely don't know. Perhaps the frame of reference idea doesn't quite apply here?

I'm definitely not 100% either. The particle obviously *does* have a rest frame; but I think the weirdness arises because in any frame, the field is "moving" relative to to the particle, since it's always changing at c, while the particle is moving at v.

If I'm thinking about this correctly, the field would be (very slightly, assuming v << c) "squashed", even in the rest frame, so that would be one very clear demonstration that it's different from a stationary charged particle.

That almost sounds like it's a violation of relativity, because it's a way to tell if a particle's moving or not -- but only with reference to an outside system (again, the retarded field). If you were really looking at the "instantaneous" field (and no retarded field) of a moving particle, there would be none (as opposed to the if the particle were stationary for a finite amount of time "t", you would have a radius of size c*t that could be associated with the particle at that position).

Any of those arguments pushing you closer?

 

[Drakkith]

I'm not following you jjustinn. Perhaps because I don't know enough about the subject. I don't see how that would be a violation of relativity. From another frame the charged particle could be moving, which would be a change in its field over time. Or so I thought.

 

[jjustinn]

...the field would be "squashed", even in the rest frame, so that would be one very clear demonstration that it's different from a stationary charged particle.
That almost sounds like it's a violation of relativity, because it's a way to tell if a particle's moving or not

I don't see how that would be a violation of relativity. From another frame the charged particle could be moving, which would be a change in its field over time. Or so I thought.

According to the equivalence principle, there should be no way to determine whether a system is at rest or moving at a constant velocity without going outside the system...but here, if I'm right you could tell how fast the particle was moving by the squished-ness of the field, even from its rest frame.

In retrospect, it seems obvious that it's not a violation, so maybe it was more confusing that I brought it up...but your suggestion is that in the rest frame, the entire field will look like the electrostatic field of a particle at rest, right? So that would seem to indicate that the entire field (including those contributions from the particle at previous times / positions -- which is actually, at any given instant, actually constitute the entire field).

Or am I just missing something?

 

[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.