Classical charged particle's reaction to its own retarded field

In summary: 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
  • #1
jjustinn
164
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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
 
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  • #2
I don't think a charged particle's field can influence the particle that produces it.
 
  • #3
Drakkith said:
I don't think a charged particle's field can influence the particle that produces it.

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.
 
  • #4
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.
 
  • #5
Drakkith said:
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.

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?
 
  • #6
jjustinn said:
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?

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?
 
  • #7
Superstring said:
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?

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.
 
  • #8
jjustinn said:
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.

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.
 
  • #9
Superstring said:
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.

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.

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.

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...
 
  • #10
This conversation got me thinking of some better search terms, and I found an article that (I *think*) basically answers my question:

The Self-Force on a “charge” arises from the interaction of the “charge” with
its own “retarded field.” This includes the dissipative effects of radiation reaction
and may also have some conservative effects as well.

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...
 
  • #11
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.

You've got it backwards.
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?
 
  • #12
Drakkith said:
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.

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).

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?

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 going to 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.
 
  • #13
jjustinn said:
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).

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



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 going to 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.

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.
 
  • #14
Drakkith said:
I am not familiar with "self field effects". Could you provide a link to a good description?

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 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.

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".
 
  • #15
Drakkith said:
I am not familiar with "self field effects". Could you provide a link to a good description?

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).
 
  • #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.
 
  • #17
Wait...are we talking about this?

Abraham–Lorentz force is the recoil force on an accelerating charged particle caused by the particle emitting electromagnetic radiation.
 
  • #18
Drakkith said:
Wait...are we talking about this?

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.
 
  • #19
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.
 
  • #20
Drakkith said:
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.

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.
 
  • #21
jjustinn said:
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.

Correction: I am not familiar enough with the subject to really grasp it. :biggrin:
 
  • #22


i think that fields don't exist...these are just lines that tells us the force direction at various locations surrounding the charge...those are nothing but "effect" from the cause "turning on" of the charge..there would be no effect on the cause that produced the effect
 
  • #23


hanii said:
i think that fields don't exist...these are just lines that tells us the force direction at various locations surrounding the charge...those are nothing but "effect" from the cause "turning on" of the charge..there would be no effect on the cause that produced the effect

That's action-at-a-distance...while you could say that the electrons in the radio transmitter's antenna "caused" the electrons in the antenna of your car radio to move, it would be very hard to argue that the field in between "didn't exist" -- even ignoring the evidence from quantum mechanics (photons).

Or, what about light -- does that not exist? Is it the atoms in the sun stimulating your retina?

By that argument, if you reflected the light back at the source, it wouldn't be affected by it.
 
  • #24


jjustinn said:
That's action-at-a-distance...while you could say that the electrons in the radio transmitter's antenna "caused" the electrons in the antenna of your car radio to move, it would be very hard to argue that the field in between "didn't exist" -- even ignoring the evidence from quantum mechanics (photons).

Or, what about light -- does that not exist? Is it the atoms in the sun stimulating your retina?

By that argument, if you reflected the light back at the source, it wouldn't be affected by it.

the presence of the charged particle can be replaced by its field...refer feynman lectures on physics chapter 2 vol 1...in case of light...it is electromagnetic radiation...it is that radiation we feel...not electromagnetic field..we cannot feel electric and magnetic fields..can we??
 
  • #25


hanii said:
the presence of the charged particle can be replaced by its field...refer feynman lectures on physics chapter 2 vol 1...in case of light...it is electromagnetic radiation...it is that radiation we feel...not electromagnetic field..we cannot feel electric and magnetic fields..can we??

We feel the EFFECT that one particle has on another. HOW it gets from one to another particle is described using a "field". No matter how you describe it, the effects are still the same. In the case of light, we describe it as an oscillating electric and magnetic field. You most definitely are feeling the electromagnetic field from light.
 
  • #26


i think in case of light ..,we feel electromagnetic radiation...not electromagnetic field...we cannot feel the electromagnetic field of an electromagnet...but we can see the effect of its presence on other substances..
 
  • #27


hanii said:
i think in case of light ..,we feel electromagnetic radiation...not electromagnetic field...we cannot feel the electromagnetic field of an electromagnet...but we can see the effect of its presence on other substances..

The radiation IS a field. The reason you and I cannot feel a magnetic field from a magnet with our hand is because our hand is not magnetic. The magnetic field from the magnet has an opposite effect AS A WHOLE between the different charged particles that make up our atoms in our body. Since we cannot feel individual particles, only something that has a large net effect can be felt. If we jack the strength of the field up to extreme amounts we could definitely feel it. They have made a frog levitate inside a multi-tesla magnet, a similar effect could be produced for us.
 
  • #28
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 going to 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.

It wouldn't escape its own field unless it was already traveling faster than c by the time the electric field was 'turned on'. If it accelerated past c it would be catching up and interacting with virtual photons that were emitted by its own electric field prior to the time when it was moving less than c.

i think in case of light ..,we feel electromagnetic radiation...not electromagnetic field...we cannot feel the electromagnetic field of an electromagnet...but we can see the effect of its presence on other substances..

EMR is nothing more than an oscillating electromagnetic field. In other words light is an electric field. To say you think electromagnetic radiation exists but not electromagnetic fields is like saying you don't believe in dogs but you believe in chihuahua's.
 
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  • #29
ecneicS said:
It wouldn't escape its own field unless it was already traveling faster than c by the time the electric field was 'turned on'. If it accelerated past c it would be catching up and interacting with virtual photons that were emitted by its own electric field prior to the time when it was moving less than c.

Correct, and well put -- that was the paradox. Because that means a moving particle is *always* in its own (retarded) field, whether it's accelerating (and therefore subject to the well-known Abraham-Lorentz force) or moving at a constant velocity.

Lorentz solved this by making the charge an extended particle rather than a point (a rigid shell) -- and in that case, the self-field's force exactly cancels in the case of constant velocity. However, as we all know (and even Lorentz knew), a moving rigid sphere isn't compatible with special relativity / the Maxwell equations.

So, it still seems like there's no full classical (as in non-quantum) explanation for charges moving at relativistic speeds.

But more than that, it's the fact that no one even bothers to explain the *classical* explanation that bugs me -- since it leaves us all with the vague feeling that maybe a charge isn't affected by its own field or something.
 
  • #30
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?
 
  • #31
Drakkith said:
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.
 
  • #32
I thought it was that simple. But I definitely don't know. Perhaps the frame of reference idea doesn't quite apply here?
 
  • #33
Drakkith said:
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?
 
  • #34
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.
 
  • #35
jjustin said:
...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

Drakkith said:
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?
 

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