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Relativistic transverse doppler shift

  1. Oct 16, 2013 #1
    Hi,
    I was looking up the formula for this on wikipedia and it said that the frequency shift is:

    [itex]\frac{\sqrt{1 - v^2/c^2}}{1+\frac{v}{c}\cos{\theta}}[/itex]

    In the case where the the emitter is directly above the observer when the photon arrives it says it simplifies to:
    [itex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/itex]

    Now, I can't get that to work out.
    In this special case we have that:
    [itex]\cos{\theta}=\frac{v}{c}[/itex]
    So the doppler shift becomes:
    [itex]\frac{\sqrt{1 - v^2/c^2}}{1+v^2/c^2}[/itex]
    which doesn't cancel the numerator because there's a minus sign missing.
    If you take the same situation but reverse the direction of v nothing changes because the new angle becomes [itex]\pi-\theta[/itex], so the sign of the cosine term cancels the minus sign from v, which is clearly wrong since it should go from a blueshift to a redshift. Is there supposed to be a +/- in front of the cosine term?
     
  2. jcsd
  3. Oct 16, 2013 #2

    Bill_K

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    Transverse means cos θ = 0. In this case the formula reduces to √(1 - v2/c2) = 1/γ. It's less than 1, so this is a redshift.
     
  4. Oct 16, 2013 #3
    Ok, call it something else if you want, I still mean the [itex]\cos{\theta}=v/c[/itex] case.
     
  5. Oct 16, 2013 #4

    stevendaryl

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    Here are the special cases:
    Case 1: [itex]\theta = 0[/itex]

    This corresponds to the case in which the sender is moving in a straight line away from the receiver. In this case, the Doppler shift formula reduces to:

    [itex]\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}[/itex]

    The frequency is red-shifted.

    Case 2: [itex]\theta = \pi[/itex]

    This corresponds to the case in which the sender is moving in a straight line toward the receiver. In this case, the Doppler shift formula reduces to:

    [itex]\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}[/itex]

    The frequency is blue-shifted.


    Case 3: [itex]\theta = \frac{\pi}{2}[/itex]

    This corresponds to the case in which the sender is moving in a straight line that is perpendicular to the line connecting sender and receiver.

    [itex]\sqrt{1-\frac{v^2}{c^2}}[/itex]

    The frequency is shifted by the usual time-dilation factor.

    The frequency is blue-shifted.
     
  6. Oct 16, 2013 #5

    Bill_K

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    There is an aberration effect, which causes the angle in the frames of the emitter and receiver to differ.

    cos θ' = (cos θ + v/c)/(1 + v/c cos θ)

    So cos θ' = 0 may be the case you're looking for.
     
  7. Oct 16, 2013 #6
    You can call it what you like but Wikipedia means the case when ##\cos{\theta} = 0##. In this special case there is no red shift according to the classical Doppler calculation, but in the relativistic case there is red shift purely due to time dilation of the emitter frequency.
     
  8. Oct 16, 2013 #7

    PeterDonis

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    I'm not so sure. Here's how the Wikipedia page defines the angle ##\theta_o## (note that the subscript is an "o", for "observed", not a zero):

    So the "purely transverse" case, where the only redshift is, as you say, due to the time dilation of the emitter frequency, is ##\theta_o = \pi / 2## by Wikipedia's definition, not ##\theta_o = 0##. Plugging ##\theta_o = \pi / 2## into the formula does indeed give ##f_o = f_s / \gamma##; plugging in ##\theta_o = 0## does not.

    The case the OP is asking about is the case where the light is *received* at the moment of closest approach. In the frame of the receiver, the distance ##L## covered by the light and the distance ##D## covered by the emitter, during the travel time ##T## of the light, will therefore be related by

    $$
    \frac{D}{L} = \frac{v T}{c T} = \frac{v}{c}
    $$

    The question is, how does this relate to the angle ##\theta_o##, according to the Wikipedia definition? At first sight, it does appear that the angle between the emitter direction and the observed direction of the light at reception should be ##\theta_o##, which would mean the OP is correct that ##\theta_o = v / c##.

    However, if you go down further on the page, it gives the aberration formula that Bill_K mentioned:

    which leads to a formula for the Doppler shift in terms of ##\theta_s##:

    For the case where the light is received at the point of closest approach, ##\theta_s = \pi / 2## and the formula reduces to ##f_o = \gamma f_s##, i.e., to a blueshift, just as the Wikipedia page says. But now, plug ##\theta_s = \pi / 2## into the aberration formula; it gives

    $$
    \cos{\theta_o} = - \frac{v}{c}
    $$

    Note the minus sign! So evidently, the way Wikipedia is defining angles, the distance ratio ##D / L = v / c## that I referred to above corresponds to an angle of ##\pi - \theta_o##; i.e., ##\cos{\left( \pi - \theta_o \right)} = v / c##, which obviously gives ##\cos{\theta_o} = - v / c##.

    It would be really nice if the Wikipedia page had diagrams showing these cases; you can kind of see what it means by the angles by looking at the diagram for accelerated motion further down the page.
     
  9. Oct 16, 2013 #8
    You are right. I was posting from a mobile phone, so reading the Wikipedia article was difficult.
     
  10. Oct 16, 2013 #9
    I am now sitting at a PC with a proper screen and a multi tasking operating system and I have changed my mind again. What I actually said is "Wikipedia means the case when ##\cos{\theta} = 0##" and not "Wikipedia means the case when ##\theta = 0##".
    OK, it seems you are referring to this Wikipedia section. Paraphrasing it says:

    When the objects are closest together the redshift is:

    [itex]\frac{1}{\gamma} = \sqrt{1-v^2/c^2}[/itex] and

    when the objects are closest together the blueshift is:

    [itex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/itex]

    which is not the same thing as saying the second expression is a simplification of the first.

    What Wikipedia actually says is "In the case when ##\theta_0 = \pi/2##" which is the same thing as saying the case when ##\cos(\theta_0) =0## the transverse Doppler effect

    [tex]\frac{f_o}{f_s} = \frac{\sqrt{1-v^2/c^2}}{\left(1+\frac{v \cos(\theta_0)}{c}\right)}[/tex]

    simplifies to:

    [tex]\frac{f_o}{f_s} = \sqrt{1-v^2/c^2}[/tex].

    i.e. the frequency recieved as measured in the rest frame of the receiver, is less than the frequency emmited as measured in the rest frame of the emitter, by a factor of gamma.
     
  11. Oct 16, 2013 #10

    PeterDonis

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    Ah, yes, I see you did. :redface: But you're still apparently missing something: the Wikipedia page talks about *two* cases, not just one.

    Yes, because the two expressions refer to two different scenarios. The first is when the light is emitted at the time of closest approach (time in the observer's frame); the second is when the light is received at the time of closest approach (in the observer's frame). Both expressions are simplifications of the general formula, but with different values for ##\cos{\theta_o}##: ##0## in the first case, ##- v / c## in the second. The first case is the one you were talking about; but the second case is the one the OP is talking about. The key thing the OP missed was the minus sign in front of ##v / c## in his case, which arises because of the (non-intuitive, IMO) way that the Wikipedia page defines the angle ##\theta_o##.
     
  12. Oct 17, 2013 #11
    This explains it, thank you!
     
  13. Oct 17, 2013 #12
    Although Wikipedia uses a subscript of zero for theta in one section and a subscript of "o" for observer in the next section, they are in fact the same quantity. It appears the two sections have been written by different people in an inconsistent way, which is a bit confusing. They are the same because the Doppler formula is the same for ##\theta_0## and ##\theta_o##.

    To try and clear up any remaining confusion, I have attached a diagram showing how ##\theta = \theta_0 = \theta_o## is defined. It is the angle ##\theta## = BCD rather than the intuitive angle ##\phi## = ACB, where as you say, ##\theta = (\pi-\phi)## and ##\cos(\theta) = -\cos(\phi)##.

    By way of example, when the source is on heading directly towards the receiver on a collision course, ##\theta = \pi## or 180 degrees and when the source is going directly away from the receiver ##\theta = 0##.

    Yes, this all seems correct.
     

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    Last edited: Oct 17, 2013
  14. Oct 17, 2013 #13

    PeterDonis

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    Yes, agreed. Thanks for the diagram--there should be one like it on the Wikipedia page.
     
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