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Phase invariance of e.m. waves

  1. Jan 20, 2008 #1
    Many textbooks derive the formulas which account for the Doppler shift and for aberration of light from the invariance of the phase of an electromagnetic wave. Do you know an explanation for the invariance?
     
  2. jcsd
  3. Jan 21, 2008 #2
    I would explain it in this way, don't know how much correct it is: if I send n pulses of light or anything else (balls, objects, ecc.), I have to count exactly n pulses of it in every reference frame (I don't know how to call it; "invariance of the number of objects"?); in the same way I will have to count n maximums (for example) of an EM wave in every ref frame; this means that the phase of the EM wave must be invariant. I don't know how to formalize it better.

    It's a good question. I have ever thought phase is a very important and deep concept in relativity and QM and so, maybe, it's one of the concepts that could link relativity to QM.
     
    Last edited: Jan 21, 2008
  4. Jan 21, 2008 #3
    Thanks for your help. Your solution is of help. I put the following question (not statement)
    Are the dimensionless combinations of physical quantities which appear at the exponent of e or in the argument of a trigonometric function relativistic invariants?
     
  5. Jan 21, 2008 #4

    jtbell

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    Staff: Mentor

    A wave's propagation vector [itex]\vec k[/itex] (whose magnitude [itex]k[/itex] is the wavenumber [itex]2 \pi / \lambda[/itex]) and frequency together form a four-vector:

    [tex]k = (\omega / c, k_x, k_y, k_z)[/tex]

    Position and time of course also form a four-vector:

    [tex]r = (ct, x, y, z)[/tex]

    Therefore their four-vector "dot product" is a Lorentz invariant:

    [tex]k \cdot r = \omega t - k_x x - k_y y - k_z z = \omega t - \vec k \cdot \vec r[/tex]
     
  6. Jan 21, 2008 #5
    I would say no. For example

    [tex]e^{i\omega t}[/tex] is not invariant, while

    [tex]e^{i(\vec k \cdot \vec r - \omega t)}[/tex] is invariant.
     
  7. Jan 21, 2008 #6
    I think a good question would be, why is the wave vector a four-vector? I think it inherents from the wave equation (the fourier transformed wave equation).
     
  8. Jan 21, 2008 #7
    phase invariance

    I rephrase my question
    Are the combinations of physical quantities that appear in formulas that account for a real effect as arguments of e or of trigonometric functions relativistic invariants?
    Example
    radiactive decay exp(-t/T)
    Plancks distribution law exp(-hf/kT)
    and the phase of the e.m. wave in discussion.
    Does your counter example account for something that hapens in nature?
    Regards
     
  9. Jan 22, 2008 #8

    pam

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    The first paragraph on page 2 of arXiv:0801.3149v1 reads:

    " If the plane-wave disturbance in Eq. (1) is observed
    from a different reference frame, the phase of the wave
    should remain invariant quantity. This claim is clarified
    by the fact that the elapsed phase of the wave is proportional
    to the number of wavecrests that have passed the
    observer, and thus it must be frame-independent, and
    hence, a Lorentz scalar. Alternatively, the same
    conclusion follows by considering optical interference experiments
    from different inertial frames, where the phase
    is the quantity that determines the interference pattern."
     
  10. Jan 22, 2008 #9

    pam

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    A relatistic invariant must be a four-scalar.
    t and f are neither in any expression.
    The arguments of trig frunctions and exponentials must be dimensionless,
    which is not the same as invariant.
     
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