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Doppler shift

  1. Feb 8, 2004 #1
    relativistic doppler shift

    I'm trying to show that this equation for the doppler shift for light:
    f' = [√(1+(v/c))/√(1-(v/c))] * f
    reduces to
    &Delta;f/f = -v/c for v<<c

    So I expanded (1+v/c)^(1/2) = 1 + v/(2c)
    and (1-v/c)^(1/2) = 1 - v/(2c)
    dropping higher order terms on the assumption that they are negligible.
    Now I have
    f' = [(1+v/(2c))/(1-(v/(2c))] * f
    &Delta;f = f' - f = f * [(1+v/(2c))/(1-(v/(2c)) - 1]
    &Delta;f/f = (2c + v)/(2c - v)

    That doesn't look like -v/c.

    On the other hand, if I multiply the original equation by &radic;(1-(v/c))/&radic;(1-(v/c)) I get
    f' = [&radic;(1-(v^2/c^2))/(1-(v/c))] * f
    and then expand &radic;(1-(v^2/c^2)) = 1-v^2/(2c^2)
    f' = [(1-v^2/(2c^2))/(1-(v/c))] * f
    &Delta;f/f = (1-v^2/(2c^2))/(1-(v/c)) - 1
    = (2c^2 - v^2)/(2c^2 - 2cv) - 1
    = (2cv - v^2)/(2c^2 - 2cv)
    = v(2c - v)/(c(2c - 2v))

    Close but no cigar.

    Or, can I just claim that (2c-v)/(2c-2v) &sim; 1?

    (The minus sign can be attributed to how we define &Delta;f.)
    Last edited: Feb 8, 2004
  2. jcsd
  3. Feb 8, 2004 #2


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    Re: relativistic doppler shift

    Starting here:

    Can't you say:

    [tex] \frac{\Delta f}{f} = \frac{1+\frac{v}{2c}}{1-\frac{v}{2c}} - 1 [/tex]
    [tex] \frac{\Delta f}{f} = \frac{2\frac{v}{2c}}{1-\frac{v}{2c}} [/tex]

    and since v<<c, 1>>v/2c --> 1-v/2c ~ 1, and then you've proven it.

    Or can you expand again at that point, so that: 1/(1+x) ~ 1 - x (for x close to zero), then:

    [tex] \frac{\Delta f}{f} = \frac{v}{c}(1-\frac{v}{2c}) [/tex]

    drop the second order terms and you've got the answer.

    I'm not sure if that's all right, but that'd be my guess; hope it helps.
  4. Feb 8, 2004 #3
    I guess that's it. I'm always uncertain as to which terms I can legitimately "drop", but I don't see any other argument that works here.

    Thanks, James.
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