Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Validity of the Doppler shift formula?

  1. Feb 19, 2009 #1
    The formula which accounts for the Doppler shift is derived from the invarainace of the phase of a plane wave. [1] The plane wave is a mathematical construction associated with an infinite source os oscillation or with a very high source-receiver distance.
    The same formula is derived considering that the involved time intervals (emission and reception) are very small. [2].
    Both approaches lead to the same formula.
    Could we consider that the formula which accounts for the Doppler shift has its limitations holding correctly only in the case of very high source-observer distance or only in the case of very small periods (very high frequencies).
    Thanks in advance for your answer.
    [1] Bohm, The Special Theory of Relativity
    [2] Yuan Zhong Zhan, Special Relativity and its Experimental Foundations
     
  2. jcsd
  3. Feb 19, 2009 #2

    clem

    User Avatar
    Science Advisor

    "Could we consider that the formula which accounts for the Doppler shift has its limitations holding correctly only in the case of very high source-observer distance or only in the case of very small periods (very high frequencies)"?
    Why would you expect that since those assumptions are used nowhere in the derivation?
     
  4. Feb 20, 2009 #3
    Thank you for your question.
    I consider the non-colinear Doppler Effect when the direction in which the observer moves does not coincide with the direction in which the light signal propagates. French [1] derives the formula that acounts for it making the assumption that "the diatnce travelled by the source during one cycle of its transmitter signals is very much less than the distance from source to observer".
    Peres [2] and Yuan Zhong Zhang [3] present the formula that accounts for the Doppler shift involving infinitesimal (dt;dt') periods of emission and reception suggesting that the formulas they derive hld only in the case of very high frequencies.

    Do you know a derivation of the Doppler shift formula free oif such assumptions?
    Thanks for your answer.

    [1] A.P. French, Special Relativity (Nelson, 1968) pp.139-144
    [2] Asher Peres, "Relativistic telemetry" Am.J.Phys. 55 516 (1987
    [3] Yuan Zhong Zhang Special Relativity and its Experimental Foundation, (World Scientific, 1996) pp.43-45
     
  5. Feb 20, 2009 #4

    clem

    User Avatar
    Science Advisor

    The fact that the components [tex]k^\mu=[\omega,{\bf k}][/tex] form a four-vector can be used to find the effect of a Lorentz transformation on the frequency of a light ray. Consider a star moving with velocity v relative to an observer, while emitting light of angular frequency [tex]\omega[/tex] in the star's rest system. The Lorentz transformed wave-vector in the observer's rest system is given by
    [tex]\omega'=\gamma(\omega+{\bf v\cdot k})=\gamma\omega(1+v\cos\theta/c)[/tex].
    Why does French have to make his approximation?
    Simply using infinitesimals has nothing to do with the frequencies involved.
     
  6. Feb 20, 2009 #5

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    I haven't actually looked at any of the derivations you quoted. But I think this is connected with calculus.

    For colinear motion at constant velocity, the doppler factor remains constant everywhere, so there's no problem.

    For other motion, the doppler factor varies as the relative positions of transmitter and receiver change. To derive it rigorously you need to use calculus techniques, considering a limit as [itex]\delta t \rightarrow 0[/itex] or something equivalent. To derive it slightly less rigorously, you can consider small times, or high frequencies, or large distances, so that the change in geometry within one cycle or wavelength is negligible.

    This begs the question: how do you measure the frequency of a signal that does not have a constant frequency? There's more than one answer to that question. You can't just count the number of cycles in one second. In this case, a signal

    [tex]e^{j\omega t}[/tex]​

    is transmitted with a constant frequency [itex]\omega[/itex] and is received as

    [tex]e^{j k \omega T + \phi}[/tex]​

    at some "instantaneous frequency" [itex]k \omega[/itex], where the doppler factor k varies with the relative positions of transmitter and receiver (and [itex]\phi[/itex] is some constant phase offset). I suspect this approach might work without explicitly needing to use calculus. (There's a sort of calculus-related assumption in my reformulation above.)
    _____________

    I wrote the above before I saw Clem's post. His is probably the more sophisticated way to tackle this. (His k is not the same as my k.)
     
  7. Feb 20, 2009 #6
    Thank you for your answer which explains my questions. I have to think about the last part of your message.
    I think that what we measure in a Doppler shift experiment is the time interval between the reception of two positive or negative wave crests T(r) and reckon the frequency 1/T(r). The concept of "instantaneous frequency" is strange for me, and there are physicists who consider that we can not measure it. During the reception of two successive wave crests we have no enough information for finding out the period and during that time interval a change in the speed of the receiver or in the angle under which he receives the light signal a fact that is not taken into account when we derive the accredit Doppler shift formula.
    I also think that we measure is the wavelength and then we reckon the wave vector.
    Kind regards
     
  8. Feb 23, 2009 #7

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    The point is, if a wave's frequency is not constant, you can't accurately measure its frequency just by timing one cycle, or measuring one wavelength. You need a more sophisticated procedure (based on some sort of calculus technique) to get a meaningful answer. Measuring the duration of one cycle is an approximation, which gets better for high frequency signals, or at large distances where the frequency changes very slowly.

    Note that the doppler factor is a ratio of two frequencies (which does not depend on the emitter frequency) so we can consider the limit as the frequency tends to infinity, i.e. the cycle period tends to zero, or the wavelength tends to zero, so the high frequency approximation can be made rigorous.
     
  9. Mar 2, 2009 #8
    Thanks for your answer.
    Consider the phase of a plane electromagnetic wave at a point M(r,[tex]\theta[/tex]) at a time tr when detected from I
    Fr=fs(tr-r/c) (1)
    where fs represents the frequency of the electromagnetic oscillations at the origin O. Taking into account that the time derivative of Fr represents the frequency fr at point M we obtain
    fr=fs[1-(V/c)cos[tex]\theta[/tex]] (2)
    because dr/dtr=Vcos[tex]\theta[/tex] represents the radial compoent of the instantaneous velocity of an obserfver located at point M and in a state of rest in I'. So (2) would account for the classical Doppler shift.
    Please tell me which are the supplementary assumptions made deriving (2): "very high source observer distance" or "very high frequency"?
     
  10. Mar 4, 2009 #9

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    In this case, neither.

    As you are now defining frequency as [itex]f_r = dF_r/dt[/itex], instead of measuring a time period or wavelength between peaks, this is exact, not approximate, so no extra assumptions are required.

    (And this same techique adapts to both Newtonian and relativistic theory.)

    The above is a better definition of frequency than the one I attempted in an earlier post.
     
  11. Mar 5, 2009 #10
    Thanks
    Consider please the clocks K0 and K of the I frame located at its origin O and at a point M(r,[tex]\theta[/tex]) respectivelly. When K0 reads te a source S located at O emits a light signal along the direction [tex]\theta[/tex] synchronizing K to K0, K reading tr and so
    tr=te+r/c (1)
    Differentiating (1) side by side and taking into account that
    dr/dtr=Vcos[tex]\theta[/tex] (2)
    represents the radial component of the instantaneous velocity of an observer of I', instantly located in front of K (1) leads to
    1-(V/c)cos[tex]\theta[/tex]=dte/dtr (3)
    If dte and dtr represent the periods at which the successive light signals are emitted and received respectively, then (3) is the accreditted formula for the classical non-longitudinal Doppler shift.
    Does the derivation presented above suggest that (3) holds only in the case of "very small" periods, i.e. when the moving observer receives the successive signals being located at the same point in space (locality convention)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Validity of the Doppler shift formula?
Loading...