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Narrow EM wave beam - how it propagates

  1. Jul 29, 2010 #1

    htg

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    Consider a narrow beam of EM waves that propagates in the Z direction and is concentrated near the YZ plane, so its intensity fades rapidly as we move away from the YZ plane in the X direction. Let the E field be in the X direction.
    Consider a cube with edges parallel to the X, Y and Z axes, respectively. Let the cube be placed at some distance from the YZ plane, and let its side length be much smaller than the wavelength.
    It seems that the flux of E through a wall closer to the YZ plane will be bigger than the flux of E through a parallel wall furter from the YZ plane, so it seems to violate the Gauss' law.
    Why doesn't it?
     
  2. jcsd
  3. Jul 29, 2010 #2

    Dale

    Staff: Mentor

    This is not correct, the net flux is 0 at all times.
     
  4. Jul 31, 2010 #3
    I think that part of the problem with this notion is the hypothetical E field. Not even a laser beam has a Poynting vector which points strictly in the z direction, without some spreading. This being the case, I don't believe that the E field, pointing strictly in the x direction, can be achieved in a real world EM wave. Indeed, it is inconsistent with Maxwell's equations, and if we believe that all instances of the electromagnetic field must be consistent with those equations, then it seems to be a non-issue. In general, all EM waves must be consistent with the wave equation, which in turn can be derived from Maxwell's equations. As I mentioned in a previous thread, I was sympathetic with the notion that Gauss' law might fail in certain cases. But I found that not to be the case with a relativistically oscillating point charge inside of various shaped surfaces.
     
  5. Jul 31, 2010 #4

    jtbell

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    It does violate Gauss's Law, because you constructed it so that it would. This means that it is impossible to produce this configuration of E field in practice, as far as we know.

    A real beam of light in which the amplitude of the E field decreases towards zero at points away from the axis must also diverge or converge (become wider or narrower). See for example the http://en.wikipedia.org/wiki/Gaussian_beam" [Broken].
     
    Last edited by a moderator: May 4, 2017
  6. Jul 31, 2010 #5

    htg

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    Even though the beam diverges, it seems possible to place my cube so that two of its facets will be perpendicular to E. It still seems to violate Gauss' law.
     
  7. Aug 1, 2010 #6

    Dale

    Staff: Mentor

    What would make you think this? If the beam diverges then by definition the field lines are not parallel to each other. If they are not parallel then it is geometrically impossible for two faces to each be perpendicular.
     
  8. Aug 1, 2010 #7

    Delta²

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    It is futile to try to find EM wave beams that violate Gauss Law because if they do so they violate Maxwell Equations too and thus the EM wave cannot be a wave that can be produced in real world( because we assume that all real world EM waves satisfy Maxwell's equations).
     
  9. Aug 1, 2010 #8

    Dale

    Staff: Mentor

    Yes, exactly. And well said. Since Gauss' Law is one of Maxwell's equations any field which violates Gauss' Law by definition does not obey Maxwell's equations and is therefore not an electromagnetic field.
     
  10. Aug 1, 2010 #9

    htg

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    It is good to develop proper intuition. I am doing it by trying to understand why my beam will not violate the Gauss' law.
     
  11. Aug 1, 2010 #10

    htg

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    You are right, but I can slightly distort my cube to make some of its edges parallel to E. If the beam intensity decreases fast enough, I would have a violation of the Gauss' law.
     
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