# Moving Reflector

1. Apr 4, 2012

### patricks

Hey everybody.

If I have a moving plane reflector with incident planar e&m waves, there is a Doppler shift in the reflected e&m wave, correct? So if the reflector is moving in the direction of the wave propagation, then the reflected waves are lower frequency then the incident waves. Therefore, the reflecting plane has absorbed energy from the wave, and heats up. Am I correct up to this point?

Now, if you goto a frame where the plane is at rest, the reflected waves have the same energy as the incident, and the plane does not heat up.

Can someone explain this to me from a relativity perspective?

2. Apr 4, 2012

### haruspex

When a photon is reflected, I don't see how heat can be generated straight off. Rather, there would be a change in motion of something. Indeed, by conservation of momentum, your initial observer will see the plane accelerate slightly.
First, let's suppose the plane is totally rigid, so all of the imparted energy goes into accelerating the plane as a unit. Your initial observer will measure the gain in kinetic energy as matching the loss in the reflected photon. An observer on the plane will see no change in either.
Now suppose there's some flexion in the plane, allowing some energy to be dissipated as heat within it. We can model this as small rigid reflector attached flexibly to the rest of the plane. An observer in that plane will see the recoil in the struck part and will now see the reflected photon as having lower frequency. The lost energy will correspond to the heat generated by the flexing.

3. Apr 4, 2012

### Redbelly98

Staff Emeritus
Another way of saying pretty much what haruspex said, but in more fundamental terms: When the mirror is moving, the e&m wave does work on the mirror, therefore the e&m wave has lost energy during the reflection.

4. Apr 4, 2012

### haruspex

I believe Patrick's issue was that an observer riding with the mirror would observe a heat gain but would not see the photon as having lost any energy.

5. Apr 4, 2012

### Redbelly98

Staff Emeritus
Oh, perhaps I was wrong to say my reasoning was paraphrasing yours, as I hadn't carefully read the second half of your post with the flexing mirror. I'd like to keep the discussion at a basic level: treat the mirror as a perfect reflector and a rigid body (no flexing).

My thinking is that no observer sees a heat gain; the loss in photon energy for the moving mirror case does not imply a heat gain. As you said, the energy change of the photon went into accelerating the mirror and increasing its kinetic energy -- if no force other than the photon acts on it.