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Doppler Shift

  1. Feb 24, 2005 #1
    uestion is How does the second order term in teh relativistic doppler shift (v/c)^2 compare to the total classical Doppler Shift for the observer receeding away from the source?

    now the doppler shift (relativistic) is [tex] \Delta f = |f_{0} (1-\sqrt{\frac{1-\beta}{1+\beta}})|[/tex]

    for the classical shift it is

    [tex] \Delta f = |f_{0} (1-\frac{v_{rel}}{v}-1)| = | f_{0} \frac{v_{rel}}{v}|[/tex]

    but i dont see a (v/c)^2 term anywhere? How am i supposed to do this??
    refers to part E opf this thread
    Last edited: Feb 24, 2005
  2. jcsd
  3. Feb 25, 2005 #2

    Doc Al

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    Staff: Mentor

    Do a binomial expansion of:
    [tex] \sqrt{\frac{1-\beta}{1+\beta}} = \frac{\sqrt{1-\beta^2}}{1+\beta}[/tex]
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