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Do lightwaves exist? and what exactly is light?

  1. Aug 1, 2010 #1
    1) We draw the classic light wavetrain as a sine(x) function, but is it possible (even classically) for a 'single' wavetrain of light to exist. By 'single' I mean a varying electric/magnetic field occurring only along a one dimensional line in space. Or is there a fundamental reason why light must always exist as a wavefront over an extended region of space?

    2) If it could exist, what would the electric field be on points just to either side of the line? zero? I normally think of electric fields being 'caused' by a charged particle. In this case the electric field reduces as you move away from the particle, in other words the particle is the source of a field that extends some distance away. For a light wave though, it would seem that the fields exist only 'along' the one dimensional line itself? Is this true?

    3) In fact, for a single wavetrain (or photon of light), what is 'causing' the electric magnetic fields to vary exactly? Is it simply an unexplained physical phenomena on which we base other physics or is the existence of the varying electric/magnetic fields explainable in more fundamental terms. I imagine a magnetic field being produced by an accelerating charge. When we think of a 'single' light wave, do we imagine a more fundamental charged particle wiggling up and down somehow to produce the magnetic field?Does this have any relationship to quantum spin or not?
     
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  3. Aug 1, 2010 #2

    Dale

    Staff: Mentor

    If you read the caption for those drawings usually they are referring to an infinite plane wave. The drawing is shown that way because of the limitations of drawing a 4D tensor field on a 2D piece of paper. Any arbitrary EM wave can be decomposed into a superposition of plane waves. So, you can think of a plane wave as being a "single wavetrain" although there is no need to do so and no particular significance to it.

    I do not believe that it is possible for a field to remain along a single one-dimensional line and still satisfy Maxwell's equations, but I could be wrong.


    My understanding is that the existence of the EM fields is explainable in terms of an underlying symmetry. Per Noether's theorem each symmetry corresponds to some conserved quantity so a symmetry and a conservation law are two sides of the same coin. In the case of EM, the symmetry is called "gauge invariance" meaning roughly that you can change the potential without changing the field, and the conservation law is the conservation of charge.
     
  4. Aug 2, 2010 #3
    The electric and magnetic field vectors cannot vary discontinuously in charge-free space. Thus your "single ray" wave is in violation of Maxwell's equations. From a slightly different perspective, the electric field of a charge with arbitrary motion can be computed, and there is no source charge motion that can create your "surgical" one-dimensional wave.
     
  5. Aug 17, 2010 #4

    htg

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    Really? Can you show me how to decompose a narrow (naturally, somewhat divergent) beam into plane waves? Just give me the idea.
     
  6. Aug 17, 2010 #5

    Dale

    Staff: Mentor

    Sure. Just write the expression for the scalar and vector potentials and then do a Fourier transform in all four dimensions. Each point of the potentials in the Fourier domain represents a plane wave.
     
  7. Aug 17, 2010 #6

    htg

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    No. If you do the Fourier transform of the H and E fields as functions of the spatial coordinates (x,y,z) and time (t), then you get a decomposition into functions of the form exp(iwt+i(k1 x + k2 y + k3 z), which is very different from decomposing a beam into plane waves, presumably of the same frequency as the frequency of the EM wave considered.
     
  8. Aug 17, 2010 #7

    Dale

    Staff: Mentor

    What is the expression for a plane wave? Hint:
    http://en.wikipedia.org/wiki/Plane_wave
     
  9. Aug 17, 2010 #8

    htg

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    The problem is that this way you get a decomposition into plane waves of all possible wavelengths, not just the wavelength of the wave considered.
     
  10. Aug 17, 2010 #9

    Dale

    Staff: Mentor

    Yes, this is correct. A localized wave, in time or space, will have an infinite range of frequencies or wavelengths respectively.
     
  11. Aug 17, 2010 #10

    htg

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    You should not call it a decomposition into plane waves. It suggests that it is a decomposition into principally physically realizable plane EM waves, which is not the case.
     
  12. Aug 17, 2010 #11

    Dale

    Staff: Mentor

    Since when is being "physically realizable" a requirement for a set of basis functions or even implied?

    The fact is that each point in the Fourier domain represents a plane wave and any EM field can be Fourier transformed. Therefore, any arbitrary EM wave can be decomposed into a superposition of plane waves.
     
  13. Aug 17, 2010 #12

    Born2bwire

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    One should also note the fact that any plane wave is physically unrealizable as it requires an infinite source.
     
  14. Aug 18, 2010 #13

    htg

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    As I said, decomposition of an EM wave into plane waves suggests that it is decomposition into EM waves, which is not the case.
    You can calculate the Fourier transform of almost anything (using distribution theory), but this is not a decomposition into WAVES.
     
  15. Aug 18, 2010 #14
    I don't understand what you mean.
    We have a theoretical model of a physical process that we call an EM wave.
    That theoretical model allows us to de-construct it (the model) into fundamental parts and also to reassemble it.
    The fundamental parts are sometimes called plane waves.

    Are you saying that because the physical corollary is never observed in nature, we cannot call the fundamental parts of the model by the same name as the full construct?
     
  16. Aug 18, 2010 #15

    htg

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    Waves are actual physical phenomena - there are E and H fields, and EM waves propagate with the velocity c in vacuum. There is a clear relation between frequency and wavelength.
    This relation is violated by elements of the Fourier decomposition, so they are not EM waves.
     
  17. Aug 18, 2010 #16
    No it's not violated.
     
  18. Aug 18, 2010 #17

    htg

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    Yes it is, the Fourier transform of a wave of given frequency, which is localized in space (e.g. a beam) contains elements corresponding to all wavelengths, not just the wavelength of the physical EM wave.
     
  19. Aug 18, 2010 #18
    That's just semantics. The distinction is too subtle to be worth losing sleep over. If you shave the hair off a cat, it's still a cat. Why should we go to the trouble of inventing new name for 'whatever-it-is'?
    I'm perfectly happy with calling them waves. I've never met anyone else who has a problem with it.
     
  20. Aug 18, 2010 #19

    Born2bwire

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    That's not true. Look at the Weyl Identity which decomposes a point source into a superposition of plane waves. The Fourier transform of the spherical potential here is an integral over all directions of plane waves, not frequencies, as there are inherent restrictions on the wavenumbers over the coordinates.

    Even for the situations where we do have decompositions over a range of frequencies, each dk_a element of the integrand represents and independent plane wave. The integral is taking a superposition of these plane waves and the physics are perfectly valid. This is a vital part of electrodynamic theory because we rely on the change of basis to solve a myriad of problems.
     
  21. Aug 18, 2010 #20

    htg

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    Look at the Fourier transform of a propagating beam of EM wave of given wavelength.
    The Fourier transform contains all wavelengths, so the Fourier components are not EM waves.
     
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