When observer sees source in actual nearest position, there is a transverse Doppler effect due to time dilation. From the view point of observer, the source experience a shorter time by $1/\gamma$, thus the observer receives a lower frequency by $1/\gamma$. This situation is described in https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Receiver_sees_the_source_as_being_at_its_closest_point
My problem is, from the rest frame of source, the observer experience a dilated time and thus should receive a higher frequency $\gamma f$, which is contradicted to the argument above.
An answer in Transverse doppler effect in light argues that the time dilation in rest frame of source is cancelled by a classical doppler effect by $1/\gamma^2$, but where is this classical doppler effect from, since the relative speed of observer and source is perpendicular to their displacement from each other?

When the observer observes perpendicular light, in the source frame, the observer has already passed it by $\frac{v}{\sqrt{c^2-v^2}}L$. The classical Doppler effect comes from this.

  • $\begingroup$ My current guess is, the $1/\gamma^2$ factor comes from Lorentz transformation. For $t=0$, $t'=L/c$. However, when the next peak reaches, $t=\tau$, $t'=\gamma(\tau-vx/c^2)=\gamma(\tau-v^2\tau/c^2)=\tau/\gamma$. Time is not synchronized in both frames due to special relativity, which is what I neglected. $\endgroup$ Commented May 31, 2021 at 14:41
  • $\begingroup$ As always, the resolution of paradoxes in special relativity amounts to carefully analyzing what happens in each inertial frame. Note that there is light aberration in the rest frame of the observer (light is not travelling perpendicularly) and this extra longitudinal component is in fact blue-shifted off-sets the red shift coming from the time dilation. The result should match the simpler computation done in the rest frame of the source. Check en.wikipedia.org/wiki/Relativistic_Doppler_effect $\endgroup$ Commented May 31, 2021 at 14:45
  • $\begingroup$ Same problem exists when the observer moves straight towards source. Observer is time dilated according to source. Wasn't that the problem? So why not simplify the question. $\endgroup$
    – stuffu
    Commented Jun 2, 2021 at 16:33
  • $\begingroup$ @stuffu The situation with observer moving towards source is simpler. As is seen by source, the observer experience normal Doppler effect (1+v/c), and is also time diluted, raising the net effect to γ(1+v/c). As to observer, normal Doppler effect gives 1/(1-v/c) but diluted by 1/γ, giving the same result. The problem here is with traverse Doppler effect, witch is a pure relativistic effect by observer as he sees the light coming perpendicular to him. But by source, the observer is actually going away from it so the net redshift comes from a classical redshift plus a time dilation blueshift. $\endgroup$ Commented Jun 14, 2021 at 12:54

1 Answer 1


In my opinion, the Wikipedia explanation is incorrect or at least very confusing. They draw from reference (4) where a draft pdf is accessible (Chapter 11, David Morin). In that reference David Morin presents two different explanations, the first one is, in my opinion, confusing (unfortunately Wikipedia used it). However, the second explanation that he presents in section 11.8.2 as "REMARKS" is very clear. To understand you must consider that whether we have a blue shift or red shift is always judged from the point of view of the receiver. There a two cases

Case 1: the receiver is moving and the source is static (blue shift) $f_r=\gamma f_s$.

Case 2: the source is moving and the receiver is static (red shift) $f_r=f_s/\gamma$.


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