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Quick 'n' easy question about doppler effect

  1. Dec 4, 2004 #1
    Doppler effect revisited

    Am I right when I say that the doppler effect formula "f=f0((v+vo)/(v+vs))" is derivated from the classical theorem of speed addition and this is why the doppler effect for light and EM waves is different?
    Last edited: Dec 4, 2004
  2. jcsd
  3. Dec 4, 2004 #2


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    Because light does not behave as a classical particle, it has a constant speed for all observers. It is frequency and wavelength which undergo a Doppler shift in light.
  4. Dec 4, 2004 #3
    I'm a little confused here...

    I don't get how on this page: hyperphysics they get the first equation... I don't see the relativistic doppler effect written that way anywhere and I don't understand how they get ot that equation and how to expand it with maple...Anybody can help?
  5. Dec 4, 2004 #4

    Doc Al

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    That equation is just the relativistic Doppler formula rewritten in a form convenient for deriving the low speed approximation. They took something that usually appears in a form like:
    [tex]\sqrt{\frac{1 + x}{1 - x}}[/tex]
    And rewrote it like:
    [tex]\frac{\sqrt{1 - x^2}}{1 - x}[/tex]
    These expressions are equivalent.
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