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

1. Mar 15, 2012

### yuiop

I have re-written 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:

$$f = f_0 \frac{1-2v/c+v^2/c^2}{1-v^2/c^2}$$

Can be re-arranged:

$$f = f_0 \frac{(1-v/c)(1-v/c)}{(1-v/c)(1+v/c)}$$

and simplified:

$$f = f_0 \frac{(1-v/c)}{(1+v/c)}$$

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.

Last edited: Mar 15, 2012
2. Mar 16, 2012

### clamtrox

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. Mar 16, 2012

### yuiop

The question was not about angles but about red shift of a reflected signal. I chose the 90 degree angle wrt the mirror to eliminate complications due to relativistic aberration. I was inspired to ask the question because in another thread there was originally some doubt about whether time dilation or red shift are factors in the signal reflected from a mirror moving parallel to the observer. I am now fairly sure that they are not factors in that scenario.

4. Mar 17, 2012

### clamtrox

So more explicitly (if you want to think in terms of length contractions), you get a relativistic correction because the Lorentz transform from mirror coordinates to observer coordinates changes the angle of the mirror. If the mirror is perpendicular to the beam, you can just trivially transform to mirror coordinates and you have a moving observer observing a beam (bouncing off a stationary mirror, but you know what happens there already).

5. Mar 17, 2012

### yuiop

Thanks, yes it helpful to think of it in that context.