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Equivalence principle & accelerated charge

  1. Oct 17, 2008 #1

    atyy

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    This thread discusses an interesting point Jonathan Scott brought up at https://www.physicsforums.com/showthread.php?t=264782&page=2.

    A naive application of the equivalence principle (EP) would suggest that a free falling charge does not radiate, and that a charge on the surface of the earth does radiate. Both statements are false. What's going wrong? Some possibilities I've seen in textbooks and other threads are:

    1) The EP is only heuristic, since it assumes a uniform gravitational field which does not exist in General Relativity (Shooting Star).

    2a) The EP only applies locally (first order in Taylor series). Radiation from an accelerated charge is nonlocal and cannot be analysed using the EP.
    2b) The EP only applies in a free-falling frame shielded from outside influences (Rindler, Essential Relativity, OUP 2006).
    2c) The EP does not apply to radiation from an accelerated charge because it involves tidal forces which are second order (Martin, General Relaivity, Ellis Horwood 1988).

    3) "However, when you take into account the motion of the observer as well, the transferred energy depends on the relative acceleration, so that for example an observer in an accelerating rocket would not detect any radiation being emitted from a charge at rest within that rocket." (Jonathan Scott). See for example:
    Shariati and Khorrami http://arxiv.org/abs/gr-qc/0006037;
    Almeida and Saa http://arxiv.org/abs/physics/0506049.

    Question: Does (3) constitute a defence of the EP, or does it actually illustrate a limitation of the EP similar to those of (1,2a,2b,2c), since it appears to depend on Rindler coordinates, which are only locally inertial?
     
    Last edited: Oct 17, 2008
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  3. Oct 17, 2008 #2

    DrGreg

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    This is a subject I have no prior knowledge of.

    If the references you give in (3) are to be believed -- and I have no reason to disbelieve -- then whether or not an object is radiating depends not only on the object but also on the observer. So a free-falling (inertial) observer says a properly-accelerating charged particle radiates, but a co-accelerating local observer says the particle does not radiate.

    I suspect the uniformity or otherwise of the acceleration is irrelevant, provided all measurements are made locally.

    You have a misconception of Rindler coordinates: they are not inertial (locally or otherwise). They are the coordinates of a uniformly accelerating observer in flat spacetime (i.e. with no gravity), or, by the EP, the coordinates of a "stationary" observer in a uniform gravitational field (if there were such a thing in nature*). As such, I think(?) I'm right to say they are a second-order local approximation to a "stationary" observer in any "stationary" gravitational field (the first-order approximation being a momentarily stationary free-falling observer).

    If my understanding is correct, both statements are false according to a local "stationary" observer, but they are both true according to a local free-falling observer.

    ___
    *Uniform gravitational fields do not occur in nature, but I think I'm right to say they would occur theoretically if you had an infinite flat plane mass.
     
  4. Oct 17, 2008 #3

    Shooting Star

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    From https://www.physicsforums.com/showthread.php?t=264782&page=2
    Subject: On Movement

    This thread was locked, perhaps due to good reason. However, nobody said a word about the following post of atty, which made me very sad and surprised.

    From https://www.physicsforums.com/newreply.php?do=newreply&p=1841131
    Subject: Does a uniformly accelerating charge radiate?
    Moderation is not one of your virtues, atty -- that's evident. :smile: But what about the moderators?

    But I'm glad that you have started this thread. I only wish other "experts" would join and come to some sort of agreement here. This topic is raised in PF at least once a month, if not more.

    About the rocket mimicking g, have you thought about which portion of the rocket would have to accelerate at g, since a rocket is an extended object? Answer this and immediately you get an insight about whether charges in uniform acceleration should radiate or not, and what is the connection with the equivalence principle, and why people hate point charges.

    Moreover, when mentioning the points made by me in the thread “Does a uniformly accelerating charge radiate?”, you have misunderstood me at least to a certain degree. I had said that a hybrid of Newtonian mechanics and Classical EM, when applied in this way, is bound to give false answers. Just go through it once more.

    I'm sorry, I don't have time today, but definitely I would post some more on this topic tomorrow.

    Hi DrGreg, would you care to elaborate on both the statements? Infinite flat mass planes do not occur in nature too. Even if they did, using just SR, would we actually get uniform g-fields?
     
  5. Oct 17, 2008 #4

    atyy

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    Sorry, about that. I quoted you only because I wanted to give credit to where I learnt what I thought was a good point. Obviously, now that you are here, you can speak for yourself.

    Would I understand you correctly if I said, the equivalence principle is not wrong, but simply does not apply, because the situation is inherently nonlocal?
     
  6. Oct 17, 2008 #5

    atyy

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    To get the deflection of light using the EP, the free falling observer is treated as stationary in a (Galilean or Lorentz?) inertial frame. For the observer to detect no radiation, he is treated as an accelerated observer in a Lorentz inertial frame. Is this consistent?
     
    Last edited: Oct 17, 2008
  7. Oct 17, 2008 #6

    Shooting Star

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    You have to clarify by what you mean by local and non-local here. Also, that is another rather peculiar thing to say, isn't it? If it simply doesn't apply, then where's the question of its correctness coming into play? I am a bit confused as to what exactly you want to mean here.

    It is imperative that we restrict our discussion to basic relativity and classical EM (which we are doing, of course) and try to pinpoint the source of the whole problem from the time of its origin. (It is quite possible, for reasons outlined below, that something vital is missing if we discuss only these two scenarios. This is just a pessimistic but realistic outlook.)

    Who has derived and when, if at all, that a point charge will radiate when accelerating uniformly wrt an IFR? Who has derived that an extended charge will radiate when accelerating uniformly wrt an IFR? Is it necessary for the charge to be extended in order to radiate? Is it possible to reproduce this kind of a situation by placing a static point charge and an extended charge on the surface of a neutral gravitating body and would we expect the point charge and the extended charge both to radiate?

    Try to answer these questions. Most of all this has been discussed in the literature in great detail for the last 60 years. Unfortunately, there is no consensus after so much time. As I said, atty, we will definitely discuss more later.
     
  8. Oct 17, 2008 #7

    atyy

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  9. Oct 17, 2008 #8

    Jonathan Scott

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    I'm sorry I can't shed much light on the reason why I'm sure about the accelerated charge question.

    I remember that I raised the question about accelerated charges and gravity many years ago (while I was still a student) and one of my tutors convinced me that if you try to detect energy propagating from the charge using an aerial which is accelerating at the same constant rate (but which may have a constant velocity relative to the source), you cannot do so, although someone nearby in free fall would detect a flow of energy. I think I had previously read Feynman's approach and been a little hesitant about it, so this was presumably a different argument, probably based on conservation of energy and time dilation.

    There do seem to be various papers about it now, but since there's still confusion about point charges in Special Relativity (or indeed any sort of charge, although J W Butler cleared that up as far as I'm concerned) it's not surprising there's even more confusion in General Relativity. If I can find the time, I think I'll have another look at the subject. I don't have any notes from that long ago, but I might be able to remember something about the argument.
     
  10. Oct 17, 2008 #9

    atyy

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    The most extensive recent treatment in the literature seems to be by a series of papers by Eriksen and Gron, Electrodynamics of hyperbolically accelerated charges I-V, in Annals of Physics (2000-2004). Unfortunately not available on arXiv. Here are some quotes form their work:

    I: hyperbolic motion - motion with constant rest acceleration

    I: With a reservation concerning the region of observation, the main result is that the existence of electromagnetic radiation from a hyperbolically moving charge or a freely falling charge, as observed in a stationary reference frame, but vanishing in the rest frame of the charge, represents no violation of the principle of equivalence.

    V: In his book Surprices in Theoretical Physics Peierls [6] has devoted a chapter to the topic "Radiation in hyperbolic motion." With reference to the cited result of Boulware he writes: "The radiation which Fulton and Rohrlich showed to be emitted all goes into the part II of Minkowski spacetime which has no equivalent in the Rindler frame. Thus, we have no violation of the equivalence principle locally. Because that part of spacetime which can be described both in the inertial frame and the Rindler frame carries no radiation in either description. There is radiation in that part of Minkowski spacetime which has no equivalent in the Rindler frame, and for it the question of equivalence therefore does not arise. .........."

    V: It may seem somewhat surprising that there exist no coordinate frame which covers region II, in which the accelerated charge is at rest and the metric static. The reason is that with a static metric there is a Born rigid connection between the reference particles, i.e., they represent a rigid set of particles with uniformly accelerated motion in the X-direction. In order that two particles which are moving with increasing velocity, shall have a constant distance in their instantaneous rest frames, their Lorentz contracted distance in the laboratory frame has to decrease. This is possible only if the rear point has a larger acceleration than the front point.
     
    Last edited: Oct 17, 2008
  11. Oct 18, 2008 #10

    atyy

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    Rindler (Relativity, OUP 2006) provides a technical prescription on how to decide from within GR itself whether the equivalence principle fails: "A LIF at some event P corresponds to an orthonormal geodesic system of coordinates. ... Suppose the SR law involves second partial derivatives of a tensor, for example Fu,pa. When we generalize this to Fu;pa, that will not reduce to Fu,pa at the LIF pole. For at any point where there is curvature, we can make at most the Christoffel symbols but not their derivatives vanish. In such cases, the equivalence principle just cannot be fully obeyed: the inevitable tidal forces, described by the derivatives of the Christoffel symbols, make themselves felt even in the freely falling cabin."

    I think that might settle it if someone does the calculation for the scenario being discussed. :smile:
     
  12. Oct 19, 2008 #11

    atyy

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    The solution seems to be essentially as stated by standard texts such as those of Wolfgang Rindler and JL Martin.

    Parrott, Radiation from a Uniformly Accelerated Charge and the Equivalence Principle
    http://arxiv.org/abs/gr-qc/9303025

    Harpaz and Stoker, Radiation from a Uniformly Accelerated Charge
    http://arxiv.org/abs/gr-qc/9805097

    Harpaz and Stoker, Radiation from a Charge in a Gravitational Field
    http://arxiv.org/abs/physics/9910019

    Burko, Liu and Soen, Self force on charges in the spacetime of spherical shells
    http://arxiv.org/abs/gr-qc/0008065
    “Because the self force couples to the charge of the particle in a way which depends on the type of charge, the worldline of a particle which carries one type of charge deviates from the worldline of a particle which carries a different type of charge. The reason for this failure of the Equivalence Principle is obvious: the Equivalence Principle relates to the local neighborhood of the particle.”

    Pfenning and Poisson, Scalar, electromagnetic, and gravitational self-forces in weakly curved spacetimes
    http://arxiv.org/abs/gr-qc/0012057

    Shankar and Whiting, Self force of a static electric charge near a Schwarzschild Star
    http://arxiv.org/abs/0707.0042

    Chiao, The Interface between Quantum Mechanics and General Relativity
    http://arxiv.org/abs/quant-ph/0601193
    “By contrast, a finitely charged object experiences a nonvanishing electromagnetic force due to radiation damping, which is an effectively viscous kind of force. This implies that a finitely charged object is undergoing approximately, but not truly exactly, free fall. Hence there is no reason to believe that a finitely charged object would follow a neutral object along the same geodesic, and the equivalence principle is therefore not violated.”

    Grøn and Næss, An electromagnetic perpetuum mobile?
    http://arxiv.org/abs/0806.0464
    “Could there be a nonvanishing field reaction force on a freely falling charge? An argument against this possibility is the following. According to the principle of equivalence no local measurement should be able to distinguish between being at rest in flat space and being freely falling in curved space. As observed by the comoving neutral object there is no force acting upon the charged particle. Hence it should remain at rest relative to the neutral object, and this is a reference independent result. This argument has, however, a weak point. The principle of equivalence has a local character. The mentioned equivalence is only valid as far as the measurements does not reveal a possible curvature of space.”
     
  13. Oct 20, 2008 #12

    DrGreg

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    Agreed, an infinite flat mass plane does not occur in nature, but we can calculate what would happen if there were such a thing and I believe that if you calculate the metric in this situation (in a coordinate system stationary relative to the mass) you get exactly the same metric as for a uniformly accelerating observer in empty space. I believe I read this somewhere (so it's possible I've got this wrong.) This is according to GR. I don't know what {SR + Newtonian Gravity} would say. Note also that in relativity "uniform acceleration" does not mean "the same acceleration everywhere"; it means "Born rigid acceleration", i.e. acceleration that maintains constant "rest" distance between the accelerating particles, which turns out to be proper acceleration inversely proportional to distance (and independent of time). So when I say "uniform gravitational field" I mean gravity which gives rise to "uniform acceleration due to gravity" in the above sense.

    As for my first statement, my understanding is, after having read the references quoted in post #1 and assuming they are correct (I'm taking this on trust, not having read the papers in detail), the explanation being offered implies that:

    - a free-falling charge does radiate as measured by a local observer on the earth's surface
    - a free-falling charge does not radiate as measured by a local free-falling observer
    - a charge on the surface of the earth does not radiate as measured by a local observer on the earth's surface
    - a charge on the surface of the earth does radiate as measured by a local free-falling observer

    These are claims made within, or following from, the references quoted. I can't claim to have read and understood the reasoning in those papers.
     
  14. Oct 21, 2008 #13
    why radiation has to depend on observers? if this is true, then for one observer,energy of that charge is changing since it is radiating and for other observer the energy of charge is same since it is not radiating..then this would imply that conservation of energy is also different for different observers..isn't it?
     
  15. Oct 21, 2008 #14

    Jonathan Scott

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    If the charge is accelerating relative to the observer, then the energy of the charge and its field is changing anyway relative to the observer because of the changing kinetic energy.

    From what I remember from years ago I suspect (but I'm not sure) that the outward or inward flow of energy from the apparent "radiation" may be equivalent to the change in the energy distributed through the field as the charge is accelerated or decelerated relative to the observer. If I had the time and patience, I should be able to calculate and check that, but unfortunately I don't at the moment!

    Even if this isn't actually the correct explanation, it demonstrates how energy can appear to flow in one frame but not in another.
     
  16. Oct 22, 2008 #15

    DrGreg

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    Energy is a relative concept and depends on the observer.

    If a charge accelerates relative to an observer, its kinetic energy relative to the observer is changing. This is true even if the charge moves inertially and the observer is undergoing proper acceleration, in which case we also have to include potential energy (analogous to Newtonian gravitational potential). If there's an external force that is causing the charge to accelerate relative to the observer, the energy it supplies also has to be accounted for. When you consider all sources of energy, kinetic, radiative, potential, then it should all add up.

    I don't claim to fully understand the details of what's going on here, I am regurgitating what is said in the various references that have been thrown around in this thread.
     
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