1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook