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EM fields from an accelerated charge: something doesn't add up?

  1. Jun 13, 2013 #1
    I have seen on several books that the expression for the E field generated by an accelerated charge, at enough distance and in the non-relativistic aproximation, is something like that (taken from Jackson):

    [Broken]


    where "β with the dot above" is the acceleration divided by c, n is a vector pointing from the charge to the calculation point, and R is the distance to the charge. Books use this expression (and another one for the B field) to calculate de power irradiated in different directions with the Poynting vector, assuming that we have a radiation field, but... wait a moment!

    According to that expression, if the acceleration of the charge oscillates, so does the field, but if the acceleration is constant or varies monotonically, the E field does the same (consider for example an electron accelerating in a rectilinear motion). Does it have any sense?. Can a radiation field be described by a field that is constant (or at least not oscillating)?. There should be something I have misunderstood.

    By the way, there is a tendency to interpret always the Poynting vector as an energy flux density in that direction, but this cannot be done in general. Consider for example a static charge near a magnet: the Poynting vector is not zero, but there is no energy flux.
     
    Last edited by a moderator: May 6, 2017
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  3. Jun 13, 2013 #2

    phyzguy

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    Can a charge accelerate uniformly forever?

    How do you know there is no energy flux? Consider a point charge at the center of a current ring generating a magnetic dipole. The Poynting vector would tell you there is a circulating energy flux, with equal amounts of energy entering and leaving a given volume so that the energy density is contant in time. Why can't this be the case?
     
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  4. Jun 13, 2013 #3

    WannabeNewton

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    I'm not sure if this is meant to act as motivation for the OP and/or other viewers or if you are asking this yourself...but what the hey :)! It can't accelerate uniformly forever if it is to be radiating because the radiation reaction self-force will change its acceleration. This is assuming there is no external agent putting in work to keep it forced along a constrained trajectory of uniform acceleration. Would you agree with that?
     
  5. Jun 13, 2013 #4

    phyzguy

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    The OP was saying that the formula can't be right, because a constant acceleration leads to a constant E-field. But a constant acceleration means an infinite input of energy, as you pointed out. At some point the acceleration must end or reverse diection, so there is no contradiction. Am I missing the point?
     
  6. Jun 13, 2013 #5

    WannabeNewton

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    Yeah I was agreeing with you mister :)!
     
  7. Jun 13, 2013 #6

    phyzguy

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    OK, got it! Thanks.
     
  8. Jun 13, 2013 #7
    Of course one charge cannot be accelerated forever, but I don't see it as a convincing explanation. The acceleration can be constant during some time, and according to that expression the field E at a large distance would be nearly constant during some time too, or at least not oscillanting. Is it radiation?

    Or consider a line of electrons, one behind the other in linear motion, and when each electron arrives at a certain region experiences a strong decceleration in a short distance until it stops. We know electrons emit radiation when they are stopped. Let's calculate the field E. As the acceleration is always in the same orientation and sense, the induced field at a large distance according to that expression is always in a same orientation and sense too. Since there is always some electron in the stopping region, there is always some electron producing a field in the same direction, as long as the electron flux is constant. The electron flux could be cuasi-continuous, so the E field could be almost constant too. According to that, the stopping of an electron current would produce at large distances an E field that have always the same sense and direction.

    It seems very different from the typical electromagnetic wave.
     
    Last edited: Jun 13, 2013
  9. Jun 14, 2013 #8
    Well, when you say wave if you are thinking of a pure sine or cosine function then that is not going to be the case here. It doesn't need to oscillate for it to be considered a wave. Any propagating disturbance is a wave. For eg., anthing of the form ##f(\vec{x}.\hat{n} - ct)## is a plane wave. A function like f(r - ct) is a spherical wave. You can have cylinderical waves as well. You can think of other shapes for the wavefront too. These expressions would typically be attenuated by a factor of ##\frac{1}{r}## for waves produced by localized distributions due to conservation of energy.
     
  10. Jun 15, 2013 #9

    Jano L.

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    The component of the electric field given by your formula depends also on distance ##R##. This changes, and the electric field changes too - it is not static. The field won't be described by a simple plane wave well near the particle, but the farther we get, the better is the approximation.

    We call this electric field radiation field, because it describes well the EM field at long distances from the particle. Of course, the field won't be sinusoidal.
     
  11. Jun 15, 2013 #10

    clem

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    The formula you give is a reduction of the full formula and neglects the velocity.
    You also ask " Can a radiation field be described by a field that is constant (or at least not oscillating)?"
    Even if a is constant, v is not. This leads to Larmor radiation and bremsstrahlung even for a constant acceleration.
     
  12. Jun 15, 2013 #11
    Of course, but if the point is far from the charge, the variation of R could be negligible and E would be "almost static" during some time. Besides, if we don't have a particle but a stream of particles that are stopped in a very small region, it seems that the important distance to calculate the total field is the distance to the stopping region, not the distance to each one of the particles.

    OK, but according to that expressión, once the wavefront pass by the point, in all the points behind the wavefront that are far from the charge the field would be constant while the acceleration is constant, or in the case of the stream of electrons described above, while there are some electrons in the decceleration region. The stream could be maintained in time, and during all this time, all the points where the "radiation" have reached experience a constant E?. If we put a conductor there, could we see direct current or constant polarization or any "electrostatic behavior"? Is that the effect of radiation?

    Are you saying that this formula cannot be used to explain (even qualitatively) the Larmor radiation and the bremsstrahlung, and it is necessary to use the exact relativistic solution?. Many books derive the Larmor formula of radiated power using the non-relativistic aproximation.


    And perhaps this is a bit off-topic but...
    I think in that case the "energy flux" has no physical reality, since it has no observable effects at all. And I am not the only one. See for example: http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/jjeans.html [Broken]
     
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  13. Jun 15, 2013 #12
    In this case there's still wave propagation. The radius at which the E-field goes from zero to constant is always moving away from the charge, right? So there's a propagating disturbance, which sounds like a wave to me... Also, the Poynting vector still shows power flow even in the constant area. Even if you don't like the idea of the Poynting vector as localized energy flux, it's integral over a closed surface does represent power flow into or out of the volume. If you integrate it over a closed surface enclosing the accelerating charge you'll find that there is indeed power flowing away from the charge, no matter how far away the surface is from the charge because of the 1/R fields. As I understand it, that's the definition of radiation fields: they create non-zero power flow out of a volume which includes the source no matter how big you make that volume. So yes, the fields are static in time, which means some of the effects might eventually look similar to electrostatics problems, but this is still a radiation problem. (and the moment when the E-Field jumped from zero to constant would certainly exhibit some non-static effects). If you really insist on seeing oscillating fields, you could take the Fourier transform and represent the field as a sum of a whole bunch of different oscillations that just happen to add up in such a way that they produce a constant field over a certain interval, but I like the power flow explanation better personally.
     
  14. Jun 15, 2013 #13

    clem

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    Quote by clem:
    "The formula you give is a reduction of the full formula and neglects the velocity.
    Even if a is constant, v is not. This leads to Larmor radiation and bremsstrahlung even for a constant acceleration."
    Quite bt Gruxg:
    "Are you saying that this formula cannot be used to explain (even qualitatively) the Larmor radiation and the bremsstrahlung, and it is necessary to use the exact relativistic solution?."

    I did NOT say "that this formula cannot be used to explain (even qualitatively) the Larmor radiation and the bremsstrahlung,". I made two separate statements in two separate sentences.
     
  15. Jun 15, 2013 #14

    phyzguy

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    If you believe Feynman (and I do), this is not a physically meaningless question. Feynman reviews this case in some detail in "The Feynman Lectures on Physics" Vol 2, Chapter 27 (also Vol 2 Chapter 17). The circulating EM energy does have observable consequences, since it carries angular momentum. So whether or not there is a circulating EM energy is an experimentally answerable question. Feynman implies in his analysis that this experiment has been done and it has been found that there really is a circulating EM energy. I don't know for certain if this experiment has really been done, but I think one would find that the circulating energy is really there. Does anyone know if this test has been performed?
     
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  16. Jun 16, 2013 #15
    Ok, that picture is completely coherent. The problem is that it doesn't look like what we actually observe in real bremsstrahlung. If we take the Fourier transform of a nearly constant signal as you propose, it would be strongly peaked at zero frequency, but in actual experiments we get a broad continuous spectrum reaching frequencies as high as X-rays, and I am not aware of any effects that looks similar to electrostatic problems. I think the reason is twofold: first, my hypothesis of a continuous stream of electrons stopped with constant acceleration is over-idealized and does not describe the fine details of how electrons are deccelerated in real experiments, and in this case the fine details are important; besides, as stated by Jackson (chap 15, comment in "references and suggested reading") "Bremsstrahlung can be described accurately only by a proper quantum-mechanical treatment".

    I have taken a glimpse to Feynmann book and in that particular point I don't find his discussion much convincing, but I will think on it. What I have found quite interesting is that when Feynmann talks about the field energy of a point charge (in chap 28), he says that "classical theory of electromagnetism is an unsatisfactory theory all by itself. There are difficulties associated with the ideas of Maxwell's theory which are not solved by and not directly associated to quantum mechanics". He was referring to some issues regarding the "self-force" or "radiation reaction" in a charge emitting radiation.

    By the way, I have been looking at the derivation of the Poynting theorem. It considers a volume with charges and currents inside, and what leads us to identify the Poynting vector as somethig related to power is the fact that, in the resulting expression, the volume integral of J·E is the power by the field forces, which equals the variation in kinetic energy of the system of charges. However, if we apply it to study the emission of radiation by one charge, it seems to me that the assumptions break down since the variation of energy is not due to the E field, it would be due to the "self-force". So, if we accept the Poynting theorem as an empirical law, no problem, but from a strict theoretical point of view it seems doubtful if we can apply it to calculate the power emitted by a radiating charge.
     
    Last edited: Jun 16, 2013
  17. Jun 16, 2013 #16
    Hi,
    It turns out that the Poynting vector is proportional to the momentum density as well, and Feynman in his book appears to be referring to its interpretation as a momentum density rather than its interpretation as a energy flux density in the particular experiment that you are referring to.
    The only thing which can be experimentally verified at the classical level is the integral of these quantities, for eg. if you choose a different expression for energy density as long as its integral gives the same value it would not conflict with experiment. Now, as per GR energy density would also produce spacetime curvature so if we could measure this then it would provide us will conclusive evidence about whether our expressions for energy density/flux have physical significance but I doubt if such an experiment has been performed so far (At least I havent known about it.).
    Feynman in his book actually does talk about the ambiguity of the field energy. See section 27-4 page 27-6. To quote:
    "How do we know that by juggling the terms around we couldn't find another formula for 'u' and another formula for 'S'?"
    "There are in fact an infinite number of different possibilities for 'u' and 'S', and so far no one has thought of an experimental way to tell which one is right!"
    I feel that its a matter of personal preference, you could choose to think of the fields and associated concepts of energy and momentum as a mathematical convenience which helps in getting the right results, or you could choose to think of them as something real and corresponding to an actual flow of energy, in which case there is always the possibilty that a future experiment may prove the viewpoint to be wrong or right.
     
  18. Jun 18, 2013 #17
    First of all, I'll admit that I don't know much about bremsstrahlung, but if my vague recollection serves me correctly, it describes the radiation released from a sudden deceleration (or "braking") of a charged particle, right? That doesn't sound like an acceleration that's constant for all time to me, that sounds more like an acceleration that's zero for most of the time, and then suddenly peaks at some point in time, which is (more or less) like a delta function, whose fourier transform yields a flat spectrum. So to me and my limited knowledge of bremsstrahlung, the existence of a (more or less) flat spectrum isn't surprising.

    Edit: Just to maybe clarify, I'm not saying that the constant acceleration model is the problem, but rather the problem is that you've assumed that it's constant for an infinite amount of time. The key thing (to me) is that for braking radiation, the acceleration is going to happen over a short period of time (regardless of what exact form it takes), and that means we should expect a broad spectrum. Fourier transform theory tells us that localization in time => broad frequency spectrum, and vice versa (you've probably seen that in the context of the uncertainty principle).
     
    Last edited: Jun 18, 2013
  19. Jun 20, 2013 #18
    Thegreenlaser, I think you are right, but if you don't have one electron but a continuous stream of electrons in a stationary situation, since the aceleration cannot be really a dirac delta and since there is always the same number of electrons deccelerating in the same place, at first it may seem that this situation should be more or less equivalent to a continuous acceleration, regardless if the electron being deccelerated is always the same electron or not. This assumption is obviously not true, I suppose the reason is that any stream composed by discrete particles cannot be really continuous, or perhaps it has something to do with the fact that the emission is not coherent.
     
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