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Is there a maximum doppler shift?

  1. Mar 1, 2007 #1
    I was playing around with a thought experiment where an observer is moving towards a light source at a speed infinitesimally less than C and it seems like, for any given wavelength, the maximum possible Doppler shift is 1/2. for example near infrared light with a wavelength of 800 nm would be doppler shifted to 400 nm if the observer was traveling extremely close to C.

    Is that correct? Or are there relativistic effects that would make the shift greater (or less) than 1/2 the wavelength?
    Last edited: Mar 2, 2007
  2. jcsd
  3. Mar 2, 2007 #2
    Yes there is a limit to the Doppler shift: it is 100% .
  4. Mar 2, 2007 #3

    Which Doppler shift formula do you use? What do you mean by Doppler shift in the limits of your question?
  5. Mar 2, 2007 #4
    I wasn't using a formula, I was approaching it visually by making a sine wave in an Adobe Illustrator and moving the wave in one direction and the observer in the opposite direction at the same speed. In this model, for a 1 hz wave, the observer would see 2 wave cycles pass in one second. That's where I got the doppler shift of either a doubling of frequency or a halving of wavelngth.

    If you imagine that both the observer and wave are traveling at C this would seem to indicate an upper maximum to any shift for a given frequency. But of course this doesn't account for relativistic effects (I don't know what effect if any it would have)
  6. Mar 2, 2007 #5
    doppler shift

    1. Do you consider the classical or the relativistic one. How do you add the velocities?
    2. The observer will never move with C.
  7. Mar 2, 2007 #6
    Most likely I'm thinking in classical terms but the question is basically an effort to discover what, if any, relativistic effects occur.

    I don't really add anything, it's just an observation. If you start out with the leading edge of the observer (a square) touching the leading edge of a sine wave, move the square to the right by 1 wavelength and sine wave left by 1 wavelength, the leading edge of the box "observes" two cycles of the wave.

    Not really sure how to represent that as an equation.

    True, but as the observer approaches C, the number of observed cycles approaches 2 x frequency
    Last edited: Mar 2, 2007
  8. Mar 2, 2007 #7

    George Jones

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    Relativity does modify the standard first-year textbook result.

    For the situation you outlined, the correct relativistic expression is

    [tex]\lambda = \lambda_0 \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}}.[/tex]

    This reduces to the standard non-relativistic expression for speeds much smaller than [itex]c.[/itex]
  9. Mar 2, 2007 #8

    Who is v in the answer you propose taking into account the scenario proposed by the author of the thread.
  10. Mar 3, 2007 #9

    George Jones

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    v is the relative speed between the reference frame of the observer and the reference frame of the light source.
  11. Mar 3, 2007 #10
    if you put v=c in this formula, you have 0 wavelength. that is, approaching c, you would soon be killed, if not disintegrated, by radiation. creepy, since it is due to energy that you yourself spent to gain high speed... interesting form of suicide.
  12. Mar 4, 2007 #11
    If you picture a single wave (peak to peak) and imagine the light traveling to the right at C, and yourself traveling to the left at C, you will meet in the middle. To have a shift greater than 1/2 you would have to travel past the half point which means traveling faster than C.

    Of course this has nothing to do with relativity, it just explains why 1/2 seems to be the maximum shift.
  13. Mar 4, 2007 #12


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    Huh? There isn't any maximum shift, as other posters have explained.

    An emitted frequency of 1 can be redshifted downward by any multiplicative factor greater than zero.

    An emitted frequency of 1 can be blueshifted upward by a multiplicative factor that has no upper limit.
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