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Classical charged particle's reaction to its own retarded field

  1. Jul 9, 2011 #1
    Classical charged particle's reaction to a retarded field

    This is something I've been curious about for a while -- every once in a while, I'll see some random reference to it in an article, but I never feel like it's the whole story.

    The situation is this -- you have a moving classical charged object; for simplicity, say it's a point charge, but it works just as well for extended objects. We'll also say that its charge was 'turned on' at t = 0, and that it's moving along the positive X axis at a constant 1 m/s.

    So, starting at t = 0, the charge is emitting electric / magnetic fields, according to the Maxwell equations. At t = 1s, the charge will be at x = 1m, but at that point, there is now a nonzero field from the charge t = 1 / c seconds prior, which is pointing in the positive X direction, so it would seem that the charge should be accelerating itself...so I feel like I'm missing something.

    It doesn't seem like this should count as a self-field, since how would the charge know that it's "its field"? Another way to get around this excuse would be to say that the charge has a second point charge stuck an small distance in front of / behind it, that can also be 'turned on' at an arbitrary time, and we don't turn it on until t = 1s...so you get the same effect, but it's definitely not a "self-field" in any sense of the word.

    So...what am I doing wrong here? My gut says that it's already worked into the Lorentz force / Maxwell equations, but I can't find that written anywhere.

    Thanks -
    Justin
     
    Last edited: Jul 9, 2011
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  3. Jul 9, 2011 #2

    Drakkith

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    I don't think a charged particle's field can influence the particle that produces it.
     
  4. Jul 9, 2011 #3
    That's what I was trying to address with the gedanken experiment in the third paragraph (e.g. a second particle attached to the first that doesn't turn on until t=1s) -- it seems like there's no way for a particle to "know" that any random field that was created at some point in the past was created by itself -- or by anything else for that matter...

    As a slight improve to the experiment that makes the point clearer, say we have two particles stuck together that can have their charge turned on and off. They've been off since t=-infinity, and get particle 1 gets turned on at t=0. Then, at t=1s, particle 1 is turned off and particle 2 is turned on. It's now a different particle -- but the dynamics should still be the same as if it was particle 1.
     
  5. Jul 10, 2011 #4

    Drakkith

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    A particle doesn't produce multiple fields or expel fields into space. A particle is surrounded at all times by its own field which never changes from its own frame of reference. How could it? Every frame of reference is seen as non-moving by its own observer. In the case of creating a charged particle, the moment it comes into existence its field would be moving off at c, even in its own frame. It could never be at a point where it's field is different.
     
  6. Jul 10, 2011 #5
    OK -- looking at it from the rest frame of the charge does make it seem trivial, and it explains why a second charge tacked to it wouldn't be any different (since it has to be moving at the same rate).

    However, that brings up some really weird behavior (from a nonrelativistic standpoint anyway, and we are working with nonrelativistic speeds) -- where at a given spacetime point, there WAS a field just before the charge got the, but it disappears when the charge gets there -- but only from the charge's moving frame.

    So, to take the experiment even further:

    The first charge is the same -- it's turned "on" at t = 0, and is moving along the X axis at 1m/s. At x = 1m, there's a second particle -- at rest -- which is not turned on until t = 1s.
    At t = 1s, both charges are at x = 1m, but now one is moving and one is not -- and now the second charge does feel the field, while the first charge does not (just as it did not at any of the previous instants on its way to t=1s / x=1m). Although now there's at least a plausible mechanism -- the second charge is at rest (or rather, moving with respect to the first), rather than the first charge somehow metaphysically "knowing" which piece of the surrounding electromagnetic field was "its own" and magically ignoring that component of the field.

    This still doesn't seem like the full story, but I think to get any further we'd need to abandon point charges...and at that point, I'm guessing we probably end up with something like Lorentz's electron theory.

    Unless I'm missing something?
     
  7. Jul 10, 2011 #6
    You don't seem to understand that a charge's field moves WITH the charge. If a charge is moving at 2 m/s, then the field ALSO moves at 2 m/s (but the change in the location of the field propagates at c). In order for a charge to interact with its own field it would have to move faster than the field can change - that is, it would have to reach an unchanged part of the field before it got a chance to change. The field can change at c, and the velocity of the charge must always be less than c - so where's the problem?
     
  8. Jul 10, 2011 #7
    You've got it backwards. Think about it -- the field moves at c, but the charge moves at less than c (MUCH less than c in this case -- 1m/s). For the charge to get OUTSIDE the influence of its field, it would need to move faster than c, not the other way around. If you don't believe me, take out a piece of paper and try it out. It's like if you talk while you talk, you can hear your own voice, because you're moving slower than the speed of sound.

    t = 0s: charge @ x = 0m. No field. Charge turned on for first time.

    t = 1s: charge @ x = 1m. Field goes out to 1c. 1c > 1m, so charge should be within its field from t = 0s.

    That's with delta-t = 1s, but you could obviously get the same result by taking the limit of delta-t -> 0. No matter what, the always be within its own retarded field.
     
  9. Jul 10, 2011 #8
    You're not understanding what I'm saying. A charged particle's electric field always appears stationary to the particle. The field always moves exactly with it in any reference frame, and because of this the particle is never "inside" of its own field.

    Pretend we're talking about fire-hoses: the hose emits a stream of water. What you're asking is akin to: "why doesn't the water emitted from the hose get the hose itself wet?" Because the hose itself isn't INSIDE the stream of water. In order for the hose to get wet, it would have to move at a greater velocity than that of the water it emits - which it can't do (but in this case not for the same reasons as a particle having to move faster than c).

    That might be a bad analogy, but it's the best I could think of.
     
  10. Jul 10, 2011 #9
    OK, so that's what Drakkith was saying in his second response, which again sounds reasonable -- though it does seem notable as a low-velocity relativistic effect, since a stationary particle at the same position / time WOULD experience a field there (and a rather strong one at that). But that's not the same as saying 'it would need to move faster than light to catch up with its field', which is definitely false.

    I think it's a decent analogy, and one that demonstrates that at low speeds, you'd expect a Galilean-looking transform -- where while moving and a stationary observers could disagree as to relative speeds, they won't disagree on whether there's water at (x, y, z, t) or not.

    To subvert the firehose analogy, it would be as if the hose didn't get wet, but if there were a second hose at the same instantaneous position but at rest, it WOULD get wet.

    I'm not saying that it's impossible -- it again seems like a much more satisfactory explanation than "a particle isn't affected by its own field" for the reasons laid out in prior posts -- but it's just not something I'd heard before and doesn't seem like the whole story...
     
  11. Jul 10, 2011 #10
    This conversation got me thinking of some better search terms, and I found an article that (I *think*) basically answers my question:

    http://www.phys.ufl.edu/~det/Talks/2005SelfForcePrimerCapra8.pdf

    It goes on to show how this is handled with perturbation theory.

    So -- I'm interpreting that basically means that the Lorentz force is ONLY applicable for a 'test charge', i.e. one that doesn't radiate. Or am I reading too much into it? It seems like if that were the case, it would at least get a footnote...
     
  12. Jul 10, 2011 #11

    Drakkith

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    I'm sorry, I don't know enough to comment on the info from the link you provided. Also, I don't understand why you are trying to insert multiple particles in your examples. It simply doesn't apply as far as I know.

    He means that the changes in the field move a c. USUALLY a field already exists, so you wouldn't need to say that the field itself moves at c. If a particle is moving along at .1 c, then over time the overall change in the field strength is moving with the particle at .1 c, but the instantaneous changes themselves move AT c. If it was moving at .9 c then the average change in the field over time and distance is moving with the particle at .9 c, and again each instant forward in time the change moves away from the charge at c. At such a high speed the field would be "warped" in front of the particle, but only from the frame measuring the particle at c. Does that make sense?
     
  13. Jul 10, 2011 #12
    I brought in the second particle to show that simply saying "a particle is not affected by its own field" is untenable; the easiest way I can think to demonstrate that is to show that you can get the same "self-field" effects with a totally different particle, which no one can argue "created" the field (and therefore has some metaphysical connection to it).

    Yeah, what you're saying totally makes sense and not in dispute at all (and in fact that's what creates the retarded self-field issue). But that's totally different than saying that the particle would have to go faster than c to be affected by its own field -- when it's just the opposite, that the particle would have to go faster than c to "escape" its own field...to get back to the firehose, if you fire it in front of you while you're moving, you're gonna get wet (the 'peeing in the wind' effect), unless you're moving faster than it's traveling ('c' for firehose water) -- and in which case (assuming an idealized, point-like firehose) it would be streaming out behind you in some weird manner.
     
  14. Jul 10, 2011 #13

    Drakkith

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    I am not familiar with "self field effects". Could you provide a link to a good description?



    I don't follow you. The particle never experiences a change in its own field. Again, from its own point of view it is always static inside its own field. From that example, no matter how fast you move, your fire hose is always spewing water out just as fast and just as far, with none coming back at you.
     
  15. Jul 10, 2011 #14
    Well, unfortunately, we're already the #1 google result for the topic ;)

    But here's some other stuff I've found; this one is a derivation of the actual equation of motion; the math is pretty heavy, but the intro / abstract kind of lays it out:

    http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.2391v2.pdf

    Another, less verbose article by one of the same authors:

    http://www.math.utk.edu/~fernando/barrett/bwald1.pdf

    Then there's the Wikipedia article on self-energy, which unfortunately goes all quantum field theory on the topic, which seems totally unnecessary:

    http://en.wikipedia.org/wiki/Self-energy

    Also, probably the best treatment I've read was in the Feynman Lectures on Physics...I remember it being mentioned several times throughout the lectures, but a quick flip through them hasn't turned it up...

    I agree that it obviously can't experience its own CURRENT field (since it would be infinite, for one), but it CAN/MUST experience its field from itself in the PAST (a.k.a. its retarded field). Just like the water that hits you from the firehose isn't the water that's coming out now, but rather the water from a few seconds ago that you've finally caught up with (assuming of course that you're driving fast enough that it doesn't hit the ground before then).

    If you're not convinced, try just working out by hand what the field would be from a moving charge at (x,t) = (0,0) and (v*dt, dt); unless you do something to manually change the field, there is going to be some retarded field from (0,0) at (v*dt, dt).

    Or, here's a totally different way of looking at it --this uniformly-moving charge is actually CHANGING the electromagnetic field surrounding it -- we're in agreement there, right? So therefore, it MUST be doing some work, and you can't do work by changing the frame of reference...

    But then again, maybe I'm just totally missing something. I've gone from absolutely convinced that you were right, to absolutely convinced that you were wrong, to just absolutely confused already today.

    An obvious problem with my interpretation, is that it seems like the field should *accelerate* the particle, since it seems like it should be pointing along the vector from the particle's retarded position (x=0) to its current position (x=vdt) -- but I guess I can handwave that away by saying that since it's a *moving* charge, there's a B field along the axis of motion, and perhaps that is what provides the "braking".
     
  16. Jul 10, 2011 #15
    I can't believe I didn't think of this earlier, but you can also pick up your copy of Jackson's Classical Electrodynamics and flip to chapter 16.

    Also, a relevant passage with some good background is here -- http://en.wikipedia.org/wiki/Abraham–Lorentz_force#Background

    But -- here's my problem...even with all of these great resources, they all (at least as far as I can tell) focus on *accelerating* charges particles; the charge in uniform motion is ignored...which makes me think maybe the reason is that a charge just can't be in un-forced uniform motion (and if it's forced, you know the trajectory -- so the question I guess would be what forces would it take to constrain a particle to uniform motion in its own field).
     
  17. Jul 10, 2011 #16
    Got it!

    http://books.google.com/books?id=m8... "constant velocity" charged particle&f=false

    The trick turns out to be assuming that it's a spherical shell (with the attendant problems -- Poincare stresses etc); in that situation, the self-field of a non-accelerating charged particle exactly cancels due to the geometry (though obviously with Lorentz contraction this breaks down), but it was always expected that classical intuition would break down at relativistic speeds.

    So, the good news is that there's no magic marker on fields keeping track of who created them so particles know which ones to ignore.

    The bad news? 20th century physics.
     
  18. Jul 10, 2011 #17

    Drakkith

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    Wait...are we talking about this?

     
  19. Jul 10, 2011 #18
    Close -- that's for an *accelerating* particle; I was referring specifically to a non-accelerating particle.

    In that last book I posted, they show how the self-field of a uniformly-moving *extended*/spherically-symmetric charge cancels.

    The Abraham-Lorentz force just arises because in non-uniform motion, the self-field does NOT cancel...so I guess there would a be a lot less confusion all around if they would mention in the Abraham-Lorentz article that there's a perfectly reasonable explanation for why it doesn't occur with uniform motion, but I *still* haven't seen that done anywhere.
     
  20. Jul 10, 2011 #19

    Drakkith

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    I'll take your word for it lol. I don't understand what they are talking about in that paper since it isn't complete at that link.
     
  21. Jul 10, 2011 #20
    The abstract? You can download the full paper in the PDF link on the right, though I found the powerpoint version a lot more enlightening (and neither explicitly mentioned constant-velocity particles)...but if you're familiar with the Abraham-Lorentz force, then that's basically the same thing; it's the charge's own field that causes that, too.
     
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