# I Doppler shift for an observer in circular motion

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1. Apr 25, 2016

### BiGyElLoWhAt

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?

2. Apr 25, 2016

### Staff: Mentor

Doppler shift of what? Light coming from the star?

3. Apr 25, 2016

### BiGyElLoWhAt

Yes.

4. Apr 25, 2016

### Staff: Mentor

What answer do you get when you apply your method?

5. Apr 25, 2016

### BiGyElLoWhAt

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. Apr 25, 2016

### BiGyElLoWhAt

I suppose I should have just worked it out. I didn't realize that they led to the same thing.

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