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Transformation of Doppler shift

  1. Jun 13, 2012 #1
    Given the Doppler shift equation quoted here: http://en.wikipedia.org/wiki/Doppler_effect#General

    [tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) [/tex]
    where
    c is the velocity of waves in the medium;
    vr is the velocity of the receiver relative to the medium;
    positive if the receiver is moving towards the source;
    vs is the velocity of the source relative to the medium;
    positive if the source is moving away from the receiver.

    Does this equation predict that the redshift seen by the receiver is the same if the observer considers the receiver to be stationary and the source to be moving away at v or if the source is stationary and the receiver is moving away at v?

    Special relativity predicts that the redshift seen by the receiver will be the same whatever the speeds of the source and receiver are relative to the observer, as long as the speed of the source relative to the receiver is the same. i.e in SR, if the source is stationary and the reciever is moving away at v, then the redshift will be the same as when the receiver is stationary and the source is moving away at v.

    For example (using the classical Doppler shift) let us say the source is moving from the receiver at 0.6c and the receiver (and medium) is stationary with respect to the observer. Using units where the speed of the wave relative to the medium is c=1 we have:
    [tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 + 0}{1 + 0.6} \right) = 0.625 [/tex]
    OK, now we switch to the point of view of an observer that is comoving with the source, so now the receiver appears to be moving at vr = -0.6c.

    If we assume the c in the equation is the speed of the waves relative to the medium (irrespective of the motion of the observer relative to the medium) then the transformed result is:
    [tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1 - 0.6}{1 + 0} \right) = 0.4 [/tex]

    which is a different result for the red shift from the one obtained above.

    Now if we assume the c in the equation is the speed of the waves relative to the observer (i.e. c = speed of the wave relative to the medium plus the speed of the medium relative to the observer) then the transformed result is:

    [tex] \frac{f_r}{f_s} = \left(\frac{c + v_r}{c + v_s} \right) = \left(\frac{1.6 - 0.6}{1.6 + 0} \right) = 0.625 [/tex]
    which is the same as the original result.

    Which is the correct way to perform the calculation?
     
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2
    That formula is used for sound waves, not for light waves. The frequency of a sound wave measured by someone depends on his speed relative to the medium in which it propogates. Since the speed of light does not depend on your velocity relative to an ether, the frequency only depends on the relative velocity. Note that in the case of sound it depends on the absolute velocity.





    and all that isn't correct.
    If you want the classical doppler shift formula, just use the relativistic one without the time dilation effect. Either vr or vs will appear in the formula, both won't.
     
    Last edited: Jun 13, 2012
  4. Jun 13, 2012 #3
    OK, thanks for that. I see that when I copied and pasted the definition from Wikipedia, that I overlooked that the velocities were measured relative to the medium and not relative to the observer which is the way I am used to velocities being measured. Obviously that invalidates the rest of my post because the formula does not take the motion of the observer into account. What I am trying to find out is what physicists would have predicted for light before they found out there was no absolute medium. You seem to be saying that for sound, the Doppler shift does not only depend on the relative velocity. Does that mean that scientists would have assumed that the speed of the light medium could have been detected by measuring Doppler shift, before they learnt about relativity?

    This is another part I am curious about. It seems that if you replace vs with vrelative you get a version of the Doppler shift that is compatible with relativity and light transmission once you factor in time dilation, but it does not work the other way around. vr never appears in the relativistic equation.
     
    Last edited: Jun 14, 2012
  5. Jun 13, 2012 #4
    Not sure about that.

    You must be talking about the formula on the wiki page. vr does appear in the equation. There are two versions of the equation, one for the moving detector case(the relative frequencies as seen by the emitter) and the other one for the moving emitter case(seen by the detector). Check out equations 12 and 18 here.I haven't seen the derivation, I could imagine why the results are the way they are.
     
  6. Jun 14, 2012 #5
    Ok, I see that equation 12 is for when the receiver is considered to be moving and the source is stationary and equation 18 is for when the source is considered to be moving and the receiver is stationary.

    I thought you might be interested in this more general relativistic equation obtained by taking relativistic velocity addition into account, that can be used when both the source and the receiver are moving relative to the observer.

    [tex]f_r = f_s * \sqrt{\frac{(1+v_s)(1-v_r)}{(1-v_s)(1+v_r)}} [/tex]
     
  7. Jun 15, 2012 #6
    I derived this equation without taking the velocity addition into account. The ratio of the frequency measured by the receiver to the freq. measured by the observer is [tex]\frac{f_r}{f_o} = \sqrt{\frac{1-v_r}{1+v_r}}[/tex]. The ratio of the frequency measured by the source to the observer is [tex]\frac{f_s}{f_o} = \sqrt{\frac{1-v_s}{1+v_s}}[/tex]. Now divide them.
     
  8. Jun 15, 2012 #7
    Yuiop:
    "What I am trying to find out is what physicists would have predicted for light before they found out there was no absolute medium."

    Maybe you can tease out what u seek from this discussion:

    http://en.wikipedia.org/wiki/Aether_drag_hypothesis

    Seems like scientists had a lot of stuff going on trying to figure out what was dragging what!!
     
  9. Jun 15, 2012 #8
    I guess that works too. I just wanted to add that I should of mentioned that vr and vs are the the velocities relative to the observer and assumes (the source and receiver are going away from each other and that vr-vs>0) or (they are approaching each other and vr-vs<0). If they are approaching each other or vr-vs<0, (but not both) then all the signs in the equation should be reversed. Yep, a bit complicated.
     
    Last edited: Jun 15, 2012
  10. Jun 15, 2012 #9
    Thanks for the link. Sounds like they were as confused as I am :tongue:
     
  11. Jun 15, 2012 #10
    I think that's not the right way to do it. One needs to consider the direction of propogation of the wave and the direction of travel of the source and the receiver. If the source and the observer move in the direction of the wave then the mentioned formula is to be used. For any of them moving opposite to the direction of the wave, the corresponding sign of velocity should be reversed. *In the general formula, the velocity of each of them is multiplied by cosθ, where θ is the angle between the direction of light and the direction of motion of the source(or the receiver).
    EDIT:Sorry, the part starting from * is wrong.
     
    Last edited: Jun 15, 2012
  12. Jun 15, 2012 #11
    I tested your idea with some numerical examples and it does not work.
    Eg. try vs=-0.2 and vr=+0.6 with the signal going from left to right, because the source is to the left of the receiver. The source is going in the opposite direction to the wave so by your rule we should switch the sign, but then we get the wrong result.

    My rule, simplified a bit, is if either of the conditions:

    (1) The source and the receiver are moving towards each other.
    (2) vr-vs is negative.

    are true, but not both, then reverse all the signs, or swap vr with vs which amounts to the same thing. It would be nice to come up with a rule related to the direction of the wave, but I cannot see it at the moment.
     
    Last edited: Jun 15, 2012
  13. Jun 15, 2012 #12
    As the definitions already indicate, all classical Doppler velocities are calculated wrt the medium. And it matters if the source moves or if the receiver moves.

    Note that classical Doppler is also used for light - even in SR - for calculations with a single reference frame, just as also Maxwell's theory works fine for that. For such cases all velocities are calculated wrt the "stationary frame" as if it is in rest in a light medium.

    PS about the sign: that's a matter of convention, see for a different one (and also for an interesting discussion) http://mathpages.com/rr/s2-04/2-04.htm
     
    Last edited: Jun 15, 2012
  14. Jun 15, 2012 #13
    Ah thanks for that, I have finally come to that conclusion for classical Doppler shift. Thanks for the informative link too. I see, as you say, that they use a different convention for the sign of the source velocity but otherwise the equations are the same. They do not however tackle the thorny issue of how to change the signs depending upon whether the source and receiver are going apart or approaching each other and whether the source velocity is greater than the receiver velocity.

    P.S. The simplest sign rule I can come up with is: if the source is emitting in the positive x direction use the original equation, otherwise reverse all the signs.
     
    Last edited: Jun 15, 2012
  15. Jun 16, 2012 #14
    Let's see the case of the ratio of frequencies as seen by the emitter. The emitter sees waves passing through at a velocity 1, so its frequency is proportional to this speed. As seen by the emitter, the detector is moving away from him, so it is also moving in the direction of travel of light. The emitter sees that waves travel through the detector at speed 1-vr, where vr is the velocity of the receiver as seen by the emitter. Hence, the frequency measured by the detector must be proportional to 1-vr. I've neglected time dilation. The same reasoning applies to your general equation. If the emitter sees the receiver travelling towards him, he sees that the receiver moves opposite to the direction of travel of light, the receiver passes through 1+ vr meters of the wave per second. That's all you need to understand.
     
  16. Jun 17, 2012 #15
    I find a sign convention with all - instead of + easier. Then we have:
    [tex] \frac{f_r}{f_s} = \left(\frac{c - v_r}{c - v_s} \right) [/tex]
    Take all velocities positive if proceeding in positive direction (easiest if we choose the signal to propagate along increasing X).

    Check:

    vr= +c => fr=0 OK
    vs= +c => fr=∞ OK
    vr= c/2 and vs= c/2 => fr=fs OK

    and source velocity greater than receiver velocity:
    vr= c/4 and vs= c/2 => fr=1.5 fs OK
    vr= -c/4 and vs= -c/2 => fr=5/6 fs OK
     
    Last edited: Jun 17, 2012
  17. Jun 17, 2012 #16
    I agree.
     
  18. Jun 17, 2012 #17
    That works, but that is for the classical Doppler shift equation rather than the relativistic Doppler equation as posted in #5 or http://mathpages.com/rr/s2-04/2-04.htm
     
  19. Jun 18, 2012 #18
    Yes of course, from your OP this has been about basic or "classical" Doppler, which is also valid for SR within a single reference system. Adding a frame transformation (or two, if you like), one has to multiply with gamma, as usual. For the complicated case of c-sound and c-light mathpages provides a good discussion.

    The common relativistic Doppler equation transforms from one to the other's rest frame with only c-light which simplifies the resulting equation.

    Thus, starting from basic Doppler:
    [tex] \frac{f_r}{f_s} = \frac{c - v_r}{c - v_s} [/tex]
    Taking the rest frame of the source vs=0 =>
    [tex] \frac{f_r}{f_s} = \frac{c - v_r}{c} [/tex]
    For example vr=+0.5c => fr/fs=0.5

    Next transforming to the rest frame S' of the source:
    [tex] \frac{f'_r}{f_s} = \frac{c - v_r}{\sqrt{c^2 - v_r^2}} [/tex] so that
    [tex] \frac{f'_r}{f_s} = \sqrt{\frac{c - v_r}{c + v_r}} [/tex]
    For example vr=+0.5c => f'r/fs=0.577 = 0.5γ.

    I notice that the usual relativistic equation follows directly from my notation of basic Doppler, without any messing around with the signs.

    Then for both emitter and receiver moving, as already discussed in posts #5 and #6 that gives, with S'' the rest frame of the source:

    [tex] \frac{f'_r}{f''_s} = \sqrt{\frac{(c+v_s)(c-v_r)}{(c-v_s)(c+v_r)}} [/tex]
     
    Last edited: Jun 18, 2012
  20. Jun 18, 2012 #19
    EDIT: This post is a mess, skip it.
    If the wave and receiver and emitter all move along +x, use the formula mentioned in post #5. The same formula is to be used if all these velocities are reversed. If the wave moves along +x, the receiver moves along -x and the source moves along +x, reverse the signs of vr in the equation. This form of the equation is to be used also if the wave moves along -x, receiver along +x and source along -x. If the wave moves along +x, receiver along +x and source along -x, reverse the signs of vs in the equation in post #5. This form of the equation is to be used also if the wave moves along -x, receiver along -x and source along +x. All cases have been covered.
    The moral of the story is to simply reverse the signs of velocity of the one (the wave, receiver or emitter) which travels in the -x direction.
     
    Last edited: Jun 18, 2012
  21. Jun 18, 2012 #20
    My previous post is messy and wrong. This will work: Use the formula given by Harrylin in post #15 with his convention. For the one(the wave, receiver or emitter) that travels in the -x direction, reverse the sign of its velocity. Then bring in the two time dilation factors. The direction of velocity is immaterial in these factors.
     
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