Do parallel mirrors keep accelerating away from each other?

  • #1

Main Question or Discussion Point

Lets say you have two perfectly reflective parallel mirrors in a frictionless vacuum. You fire off a beam of light between them, so that it bounces between them continuously. Do they continuously accelerate away from each other due to radiation pressure? I think the answer must be no, but I can't figure out why.
 

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  • #2
Drakkith
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What makes you think they won't?
 
  • #3
Well I don't see where the energy would be coming from. They would both be gaining kinetic energy right?

I took a shower and thought about it a little more. I guess you could pose the same question with two paddles and a ping pong ball. As each paddle accelerates, its velocity relative to the ping pong ball is reduced. So each bounce is not as hard as the last and the velocity of the paddles approaches an upper limit.

With light, though, its velocity from the perspective of each mirror would always be the same (c), right? So it seems like they would just keep accelerating forever, but that can't be right. Can it? It seems very perpetual motiony.
 
  • #4
jbriggs444
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I took a shower and thought about it a little more. I guess you could pose the same question with two paddles and a ping pong ball. As each paddle accelerates, its velocity relative to the ping pong ball is reduced. So each bounce is not as hard as the last and the velocity of the paddles approaches an upper limit.
As light reflects from a mirror that is receding, what happens to the light's frequency and wavelength?
 
  • #5
Ok I think I got this. Thanks for pointing me in the right direction.

Since the mirrors are moving apart, the light would be redshifted. So the faster the mirrors go, the lower the frequency and longer the wavelength. The energy of a photon is related to its frequency by E = hf (where h is the Planck constant). And its mass is related to its energy by m = E/c^2. So its effectively loosing mass with each bounce, and thus loosing momentum. For a photon moving at c, the momentum p = mc = hf/c^2 * c = hf/c. So it does kinda work the same as the ping pong ball.

Light is weird.
 
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  • #7
ZapperZ
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Ok I think I got this. Thanks for pointing me in the right direction.

Since the mirrors are moving apart, the light would be redshifted. So the faster the mirrors go, the lower the frequency and longer the wavelength. The energy of a photon is related to its frequency by E = hf (where h is the Planck constant). And its mass is related to its energy by m = E/c^2. So its effectively loosing mass with each bounce, and thus loosing momentum. For a photon moving at c, the momentum p = mc = hf/c^2 * c = hf/c. So it does kinda work the same as the ping pong ball.

Light is weird.
While you get the idea, your explanation is incorrect since you are using a photon mass. https://www.physicsforums.com/threads/do-photons-have-mass.511175/ [Broken] on this. Photon has no rest mass and so, it cannot work the way you have described. It does, however, have momentum (and yes, a photon can have momentum but with no mass). It is this momentum that is changing in your situation, not its mass.

Zz.
 
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  • #8
anorlunda
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Nobody has mentioned an interesting aspect of this scenario. If a photon bounces off a mirror, it transfers momentum p=2hf/c. If it bounces off parallel mirrors a large number of times, the light is redshifted to zero. In that scenario, the total energy of the photon e=hf, is transfered to kinetic energy of the mirrors.

Think of a mirror light sail reflecting star light, as compared to a pair of parallel light sails. The single sail extracts only a tiny fraction of the light's energy. The pair extracts all of it in the form of kinetic energy (if none of the photons are absorbed).

I just thought that was interesting.
 
  • #9
While you get the idea, your explanation is incorrect since you are using a photon mass. https://www.physicsforums.com/threads/do-photons-have-mass.511175/ [Broken] on this. Photon has no rest mass and so, it cannot work the way you have described. It does, however, have momentum (and yes, a photon can have momentum but with no mass). It is this momentum that is changing in your situation, not its mass.

Zz.
Maybe I understand this conceptually but not mathematically. I did not think that its rest mass changes. The link says E = pc, so in the end you still get p = E/c = hf/c.

Nobody has mentioned an interesting aspect of this scenario. If a photon bounces off a mirror, it transfers momentum p=2hf/c. If it bounces off parallel mirrors a large number of times, the light is redshifted to zero. In that scenario, the total energy of the photon e=hf, is transfered to kinetic energy of the mirrors.

Think of a mirror light sail reflecting star light, as compared to a pair of parallel light sails. The single sail extracts only a tiny fraction of the light's energy. The pair extracts all of it in the form of kinetic energy (if none of the photons are absorbed).

I just thought that was interesting.
Neat. You could inject sunlight into the system by making one of them a one way mirror.
 
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  • #10
jbriggs444
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You could inject sunlight into the system by making one of them a one way mirror.
A perfectly reflecting (on one side) one-way mirror is a violation of the second law of thermodynamics.
 

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