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I The stability of matter: is Maxwell really wrong?

  1. Jul 14, 2018 #1
    First of all: excuse me for my not good english written language.
    I come to the question with an example.
    Let us suppose two electrons are in space with initial velocity equal to zero.
    Given the Coulomb force such electrons will be submitted to an equal and opposite accelerations (for the action reaction Newton law).
    I expect that the complessive radiation emission at great distances will be null.
    So the system will not loose energy.
    But it is said that according to Maxwell theory this is not possible.
    Where is the truth ?
    I could give many other similar examples where, even if there are electric charges in accelerated motion, the system will not transfere electromagnetic energy to space.
    Is really Maxwell's theory wrong at an atomic scale ?
    Thanks for your considerations.
     
  2. jcsd
  3. Jul 14, 2018 #2

    Dale

    Staff: Mentor

    I think you should do some actual calculations. I am not convinced that either of your assertions here are correct
     
  4. Jul 14, 2018 #3

    Nugatory

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    Staff: Mentor

    Why do you expect that there will be no electromagnetic radiation emitted in this case?
    It is not, but you do have to be careful about analyzing atomic-scale interactions using classical mechanics - and if you're thinking about electrons as small charged objects moving under the influence of the Lorentz force and Newton's laws you're using classical mechanics. There's no problem with applying Maxwell's theory in that situation; the problem is that at the atomic scale electrons do not behave like small charged objects moving under the influence of the Lorentz force. Instead, you have to use quantum mechanics, which gives an entirely different picture of the electron and its interaction with the electromagnetic field and in which the classical notion of "accelerated charged particle" is absent.
     
  5. Jul 14, 2018 #4
    Thanks for your considerations.
    Let us suppose I do not know quantum mechanics and I only know Maxwell theory. So from a phylosophical point of view I do not see problems in considering electrons as particles.
    I ask myself: is it possibile to conceive motion configurations such that the electromagnetic emissions at great distances is zero ?
    This is my phylosophical position.
    I made calculations and I can assert that the cancellation of the total emission is nearly perfect.The reason is quite easy to understand: it is well known that the emitted electromagnetic field depends on the acceleration vectors; in our system the two acceleration vectors are opposite and equal in modulus.
    So when I add the two electric vectors in every point at great distance from the system the result is with great and good approximation zero.
    In an another way: the two emissions interfere each other in a destructive way.
    In another way: the electric dipolus moment is Zero.
    Now it is possibile that in a secondary approximation I have neglected something but what is certa in is that the electromagnetic emission will be enormously attenuated.
    From an esperimental point of view (by analogy): take Two speakers and drive them in phase opposition.
    The speakers are near each other.
    The re will be no sound.
     
  6. Jul 14, 2018 #5

    Dale

    Staff: Mentor

    That is just hand waving, not a calculation.

    If you were to actually calculate it then by Poynting’s theorem the result would be that energy is conserved. The energy in the field would decrease as work is done on the charges. Your hand waving “phylosophy” doesn’t negate Poynting’s theorem.
     
  7. Jul 27, 2018 #6

    Demystifier

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    Science Advisor

    While there is some kind of "attenuation" you qualitatively talk about, it turns out that the small radiation that remains is precisely what is measured. This small radiation is proportional to the square of the distance between the charged particles. The distance is indeed small, but it is not zero. If it was exactly zero (which it is not), then your qualitative argument that there should be no radiation would be correct.
     
  8. Jul 27, 2018 #7
    Dear Demistyfier,
    thanks for your considerations.
    Now, answer me please the following problem (which is only a little different respect to the previous).
    Take two electric charges on a straight line.
    The charges have opposite sign.
    The distance between the charges is fixed and the charges move with arbitrary acceleration along the straight line.
    Are you sure this system radiates at great distances ?
    If it is so I can deduce that an arbitrary body with no electric charge, if accelerates, radiates.
    Quite surprising and...impossible!
    So if you consider little distances from the system surely an electromagnetic field exists but at great distances there will be no appreciable radiation.
     
  9. Jul 27, 2018 #8
    I mean equal and opposite charges.
     
  10. Jul 27, 2018 #9

    Dale

    Staff: Mentor

    You are still guaranteed, by Poynting's theorem, that energy is conserved. Any energy that goes into the field increases the amount of work required for accelerating the charges. Again, I recommend actual computations over hand-waving assertions.
     
    Last edited: Jul 27, 2018
  11. Jul 27, 2018 #10

    Demystifier

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    It doesn't radiate, because in this case the charge is zero and all electric multipole moments are time-independent.
     
  12. Jul 27, 2018 #11
    Thanks Dale.
    I agree with you: energy theorem is always true.
    And I do not see problems as regards this aspect: if I make work on the system what I expect is an increase of the near field energy.
    As regards the calculations: I only made just estimations.
    I must be clear: what I claim is that the emission is much smaller if compared to the emission of one single charge submitted to the same acceleration.
    So what emerges is precisely this: Maxwell's theory allows to think possible an attenuation (although as you say not a perfect annullation) of the emitted radiation.
    This is for me the crucial point.
     
  13. Jul 27, 2018 #12
    Dear Demystifier,
    your reply is just as music for me.
    The reason is that between the two examples I have shown there is no substantial difference: if I change the sign of the charge and the sign of the acceleration vector the situation is the same.
    Please see Landau, Theory of fields (vol.2), Lenard-Wiechert E, B fields.
    Thanks again.
     
  14. Jul 27, 2018 #13

    Dale

    Staff: Mentor

    Of course, yes. Many different configurations achieve this, e.g. a Faraday cage.

    Your original post was very different than this new claim, I have no objection to this new one.
     
    Last edited: Jul 27, 2018
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