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Relativistic Doppler and preferred frame

  1. Sep 5, 2012 #1
    To my understanding the Relativistic Doppler shift is a product of the classical velocity shift and the dilation factor arising from that velocity.
    The classical effect can be red or blue depending on whether the emitter and receiver are receding or closing respectively.
    The dilation factor in both cases effects a red shift offset. SO in closing the net effect can be blue shifted but it will be blue shi8fted less than the classical value by the gamma factor.
    In the case of light emission I understand it as a decrease in electron resonance frequency due to time dilation.
    So in one frame A we can say that signals received from B are red shifted by the additional gamma factor because the emitting electrons were time dilated and therefore emitted photons of a lower frequency.
    But if we consider signals sent from A to B from frame A we cannot say the signals will be red shifted when received in B because the electron frequencies in B are relatively dilated.
    If this were the case the result would be a blue shift in addition to the classical effect not the expected red shift.
    As far as I can see it necessitates an assumption that the dilation occurs in the emitter frame but does not apply to the receiver.
    In effect a preferred frame.

    Is there something simple I am not getting here?
  2. jcsd
  3. Sep 5, 2012 #2


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    The time dilation happens to either observer if they express the other frames time in their coordinates. The situation is symmetric or reciprocal so neither frame is picked out.
  4. Sep 5, 2012 #3
    yes of course time dilation in general is symmetric and reciprocal.
    But in all cases, as regarded from any frame the clocks in the other frame are running relatively slower. Agreed??

    SO if we are in the emitter frame and apply this factor we then have the clocks and electron resonance frequencies dilated ( running slower ) in the other frame (the receiver)
    If this is the case then the received signals frequency would certainly increase relative to the emission frequency rather than decrease , wouldn't you agree?

    I am disregarding the classical Doppler factor and only talking about the dilation effect.
  5. Sep 6, 2012 #4


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    I think I understand your question, and I think it is a good question to ask. My answer is as follows:

    If you ignore time dilation, using classical analysis of a sequence of signals emitted by one observer and received by another, when they are separating at speed v, using c=1, v< 1, and assuming c=1 for both frames, you would get a proposed redshift of:

    (1+v) using the receiver frame

    1/(1-v) using the emitter frame

    However, in the receiver frame, the emitter is time dilated, and the pulses are slower by a further factor of γ giving: γ (1+v) = √(1+v)/(1-v).

    In the emitter frame, the receiver is time dilated, so they get the signals with 'less time' between receptions, as you propose, so the shift should be corrected to:

    (1/γ)/(1-v) = √(1+v)/(1-v)

    Thus, the shift is symmetric, and time dilation is symmetric, but if you factor into classical doppler and time dilation in each frame, the factoring is different.
  6. Sep 6, 2012 #5


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    An additional observation is that you may counter argue: but in pre-SR physics, doppler would end up the same in both frames, unlike the uncorrected classical computations I did above?! This was achieved in classical case by requiring (and expecting) that the signal speed could not be the same in both frames. Using the same speed in both frames, and correcting for time dilation, gives a new (SR) red shift formula, that is again symmetric in the two frames.
    Last edited: Sep 6, 2012
  7. Sep 8, 2012 #6
    well it seems clear now.
    It's kind of funny. At one time I did the calculations for a mutual exchange scenario, not using the Doppler formula , but doing direct calculations of signal paths based on relative velocities. Where and when, in the emitter frame, the signals would catch up to the receiver. So I had it right in front of me but failed to notice that the dilation factor at the receiver was actually effecting a blue shift of the signals.
    SO later when I read various places that the relativistic Doppler factor was , in all cases a red shift correction of the classical motion Doppler , I didn't question it and adopted it as fact. It is this premise that was at the root of my question.
    It is now clear that it is actually incorrect. It is only true in those frames in which the transmitter is in motion. In the rest frame of the transmitter the relativistic Doppler is actually a blue shift adjustment of the value resulting purely from relative motion.
    SO although the end result, the observed shift is perfectly symmetric and reciprocal , the relative effects of motion and time dilation producing that result , are not.
    They depend on which frame, emitter or receiver , that you choose as the basis for analysis.

    So thank you for your help in clarifying the source of the problem.

    ANother related question I have been meaning to post.

    I happened to be in the hyperphysics site awhile back and just out of idle curiosity
    entered some values into the Doppler calculator.

    + and - 0.8c Emission frequency 1 x 1014Hz

    the results were;

    1.3145 x 10 14 for closing and

    0.76 x 10 14 for separating.

    Try as I might I can't find the factor of 3 and .333 respectively in these results.

    What am i doing wrong???
  8. Sep 11, 2012 #7


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  9. Sep 11, 2012 #8


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    It appears that you were entering 0.8 into the first field which gives 0.2668c in the second field. If you enter 0.8 into the second field, you'll get what you want. Note that the "c" which applies after the second field actually appears under the first field so this can be rather confusing until you figure out how to work it.

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