Doppler shift for an observer in circular motion

  • #1
BiGyElLoWhAt
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Say we have an observer in perfectly circular motion around a source, like a star.

Is it reasonable to apply the angle change formula ##cos \theta_o = \frac{cos \theta_s - \frac{v}{c}}{1-\frac{v}{c}cos \theta_s}## and then take the component of the motion parallel to the light wave in the observers frame and apply the doppler shift formula to it in order to obtain a doppler shift?
 

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  • #2
PeterDonis
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Doppler shift of what? Light coming from the star?
 
  • #3
BiGyElLoWhAt
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Yes.
 
  • #4
PeterDonis
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What answer do you get when you apply your method?
 
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  • #5
BiGyElLoWhAt
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Well, the velocity component parallel to light comes out to be ##-v^2/c## using a coordinate system such that the light from the source always points along the y axis (and thus the observer is moving along the x axis).
Plugging that into the doppler shift formula ##f_s/f_0 = \sqrt{\frac{1+\beta}{1-\beta}}## you get ##\frac{f_s}{f_0} = \frac{1+\frac{-v^2}{c^2}}{1-\frac{-v^2}{c^2}}## whiiichhh appears to be the transverse doppler shift. That's cool. Thanks.

*-x axis, because convention, +theta direction.
 
  • #6
BiGyElLoWhAt
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I suppose I should have just worked it out. I didn't realize that they led to the same thing.
 

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