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EM waves

  1. Aug 10, 2003 #1
    I have an idea that I've tried to express on this forum before but without success. There's a lot of reasoning behind the idea so it won't do no good just blurting it out. I want to lead you and see if you reach the same conclusion... So here goes...

    First question... In classical physics, how would the behaviour of the electromagnetic wave emitted by a charge differ depending on these three types of oscillation patterns ?

    As you can see, all oscillation patterns have the same frequency, which would give all three emitted EM waves the same wavelength, correct ?

    Now, if all waves would have the same wavelength, would they differ in some other way perhaps ? Because they're emitted by charges oscillating in different ways, one (I) would expect them to.
  2. jcsd
  3. Aug 13, 2003 #2
    Somethin' wrong with my question ? :smile: Anyone understand it ?

    Never mind _how_ they would differ, just answer the question _if_ they would differ..
  4. Aug 13, 2003 #3


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    I'm guessing that the different waveforms would interfere with each other in different ways.
  5. Aug 13, 2003 #4
    ok, thnx :smile:. I agree...

    Next question...

    Let's say an oscillator is traveling back and forth, a relatively long distance, in your direction. This would cause a basic doppler effect on the EM waves you see. Now, if instead of traveling b&f, the oscillator was not only oscillating vertically, but also horisontally (in your direction), would this still be considered dopplereffect ?

    If you were to hold a prism in front of you, would the horisontal oscillation result in you "seeing" more than one beam ? Even if one half of the period was blueshifted and the other one redshifted ?
  6. Aug 13, 2003 #5
    Well, (I not sure I know what I'm talking about here ... ;) )

    But, you didn't specify what would happen to the magnetic component waveform. Stay the same, or be a mirror image of the electric component?

    I would expect some kind of "disruption" between the electic and magnetic component, especially in the second waveform as the electric component would be expressed more.
  7. Aug 13, 2003 #6
    Don't let the patterns fool you :smile:. If the field potential isn't symmetric, the oscillation won't be either. But in either case the EM waves' E vector is always proportional to its B vector.
  8. Aug 13, 2003 #7


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    The waveforms which are not perfect sin waves (the last 2) can be represented as a Fourier sum of sine waves. The energy distribution will show contributions at higher frequencies thus the transmitted waves will be a superposition of these waves.

    What is your question by the way?
  9. Aug 13, 2003 #8
    What I'm asking is if an EM wave, emitted by a charge oscillating vertically and horisontally, could produce multiple discrete lines when passed through a prism.
  10. Aug 16, 2003 #9
    A sinusoidal oscillation will produce a "pure" wave of one frequency, so that only one spectral line will be produced. Any other type of oscillation will produce harmonics so that a discrete spectrum will be produced.

    That is an exact answer, by the way.
  11. Aug 21, 2003 #10
    Ok, thnx Tyger.

    If this was the understanding of 19th century physicists, why were they so puzzled by the discreteness in atomic spectra ?

    Ok, enough questions... I'll just spit it out...

    There are two things I need to discuss for you to understand my idea - waveforms produced by an oscillator, and the principle of reinforcement.

    When you perform spectroscopy on an element there are not one, but billions of oscillators, all vibrating in random directions. So when analysing the spectrum emitted one needs to consider all waveforms 360° around the oscillator. With the basic sinusodial oscillator, you'll get a continuous spectrum of waveforms stretching from it's non-dopplereffected waveform, emitted perpendicular to the direction of oscillation, to it's most dopplereffected waveform, emitted in the direction of oscillation. How wide this spectrum will be is dependent on the amount of dopplereffect, which in turn is dependent on the speed of the oscillator (i.e. the temperature).

    I've tried to illustrate this with >>> pic Nr 1 <<< and >>> pic Nr 2 <<< ...

    In pic 1 I want to show that a sinusodially oscillating charge will produce a non-sinusodial waveform at every angle except perpendicular to the oscillation. When viewed from coordinate X2, the oscillation will be split into a vertical movement and a horisontal one (which produces the dopplereffect), unlike the X1 coordinate, where there's only a vertical oscillation with no doppler.

    In pic 2 I've used a snapshot from a flashmovie to illustrate the different waveforms around the charge ...

    [ !OBS! Due to the alignment of the waves there is an alternate view with "depth" in pic 2. This is not intended. The image is meant to be a 2D cross-section. ]

    Here is an image illustrating the connection between temperature and spectrum. Let me know if I haven't explained this well enough ...

    The next thing we have to discuss is the principle of reinforcement. I don't think this is the official name for it though ...

    Getting a wave to higher amplitudes require that you add energy to it in sync with it, i.e. you "reinforce it" in phase. This is pretty basic wave theory so I'll just leave it at that. How this relates to my spectrum of waveforms is very simple. As I said earlier, every waveform except the perpendicular one will be dopplereffected. So instead of a single pure sinuswave, each wave will have one half of it's period blueshifted and the other half redshifted. Although this dopplershift affects the outcome in a prism, it does not affect the overall frequency, no matter which direction the charge is oscillating. This means that, if a blue- or redshifted part of the wave has a wavelength that doesn't match the overall frequency, the recieving oscillator will not be reinforced ! And this my friends, is the key to understanding the hydrogen spectrum classically....

    Since all redshifted parts have wavelengths longer than that of the overall frequency, none of them will be reinforced efficiently.

    Mathematically speaking, only blueshifted parts with a wavelength &alpha; that is an integer fraction of the overall wavelength &lambda; will be reinforced. So the wavelengths we will see in our "spectrum of waveforms", will be those where &alpha; = &lambda; / n ( n=1,2,3...).

    So... there it is... .... What do you think ? Anything I need to elaborate ?
  12. Aug 21, 2003 #11
    I forgot.. there's a catch. The idea only works if the spectrum is temperature dependent. And according to current theory, it's not.

    If I'm not mistaken though, each of the hydrogen series (Lyman, Balmer, etc) are produced at different temperatures, which works to my favor. But I don't consider it very likely that 20th century physicist could have missed the fact that the lines moved when the temperature changed...

    .. but who knows ?
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