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Classical light theory

  1. Apr 10, 2003 #1
    According to classical light theory the energy in an electromagnetic wave is proportional to the amplitude of the wave squared (A²). Doesn't doppler effect disprove this idea ?

    Picture an oscillating charge going up and down, emitting two waves in opposite directions perpendicular to the oscillation. The two waves will have the same energy, equal in amplitude and wavelength.
    Now, if we were to apply a doppler effect to this oscillation, what would happen ? The emitted energy on each side would stay the same, as would the amplitude. The wavelength on the other hand, would increase in one direction and decrease in the other.

    Well, all is fine then - The energy is proportional to the amplitude since it does not change and the wavelength does.... ... right ?

    .. Wrong. Spatial distribution is not accounted for in this conclusion. As a result of the doppler effect, the energy in the shorter wavelength wave is now distributed in less space, which gives a higher amount of energy per unit "volume space", which I believe is the property actually measured. Two observers comparing measurements of the two waves would argue that the energy is proportional to the frequency, since the energies and frequencies mismatch equally much, and the amplitudes are the same.

    My conclusion is that, in classical light theory, the energy in an electromagnetic wave should be proportional to the amplitude squared and the frequency.

    E = kA²f ? :smile:
  2. jcsd
  3. Apr 10, 2003 #2


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    I presume you would accomplish this by changing frames of reference (either yours or its) so that you perceive the oscillator as moving as well as oscillating along one dimension?

    Don't forget that the magnitudes of the EM field (i.e. the amplitude of the wave) is affected by changing reference frames, and that conservation of energy does not apply between reference frames.

    I haven't done the calculations on what I think would actually happen, but these are effects you need to consider in your argument.

  4. Apr 10, 2003 #3
    Yes, of course.

    How is the amplitude affected, specifically ? About the energy, I'm not talking about the difference between reference frames, but the difference between the emitted waves. The total energy emitted forwards should equal the total energy emitted backwards in any referenceframe, correct ? [?] (not sure)
  5. Apr 10, 2003 #4
    I am not to sure how you envisage the event you describe. It was once pointed out to me that it is the distance along the curve of the wave thar remains constant therefore the doppler shift between two waves, should increase one amplitude and decrease the other, keeping the distance along the curve the same in both waves.
    However I should point out that I have no formal training in physics so you would need to check this mathematically.
    But it does seem logical that if the energy of the wave is unchanged then the amplitude must increase if the wave-length is shortened and vice-versa. If the energy of the wave is altered the situation is different but then of course, you have done something more than a doppler shift.
  6. Apr 10, 2003 #5


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    Ok I know the trick. It's a pitfall of relativity of simultaneity!

    In the (average) reference frame of the oscillator, we measure the "front" and "back" waves simultaneously to see that they're equal.

    When we switch frames, the two events that were simultaneous in the oscillator's frame are no longer simultaneous, so we should not expect that measuring the total energy of the front and back waves yield equal results.

    As for the change in amplitude, I don't have my link to the translation of Einstein's paper, so I don't have a reference as to how EM fields transform when you apply a Lorentz transformation.

    The reason the energy density of an EM wave is proportional only to the amplitude squared is because electromagnetic energy density U is:

    U = (E2 + B2)/2

    in units where c = &mu0 = &epsilon0 = 1

    The average value of E2 is proportional to Emax2, it doesn't matter how fast E is varying, so the average energy density doesn't depend on &omega.

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