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Doppler shift of a signal reflected in a mirror moving away from the observer. 
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#1
Mar1512, 09:51 PM

P: 3,967

I have rewritten this as as I accidently deleted my original post. I was wondering if the relativistic Doppler shift of a reflection from a mirror moving away from the observer was the same as the Newtonian equation in the special case that the mirror is orthogonal to the direction of motion.
I referred to equation (13) in this paper http://arxiv.org/ftp/physics/papers/0409/0409014.pdf and set theta to zero for this special case. I now think I have figured out the answer to my own question. The equation I gave in my my first post: [tex] f = f_0 \frac{12v/c+v^2/c^2}{1v^2/c^2} [/tex] Can be rearranged: [tex] f = f_0 \frac{(1v/c)(1v/c)}{(1v/c)(1+v/c)} [/tex] and simplified: [tex] f = f_0 \frac{(1v/c)}{(1+v/c)} [/tex] and this is exactly the same as the none relativistic equation for Doppler radar. Length contraction and time dilation is not involved in this special case of reflection in a moving mirror. 


#2
Mar1612, 04:15 AM

P: 939

Yes, this is kind of a trivial result though. You can just move to the coordinates of the mirror and argue that because of symmetries (light beam coming in 90 degrees wrt the mirror), outgoing angle must be same as incoming angle.



#3
Mar1612, 10:39 AM

P: 3,967




#4
Mar1712, 03:20 PM

P: 939

Doppler shift of a signal reflected in a mirror moving away from the observer.



#5
Mar1712, 04:03 PM

P: 3,967




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