Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If two perfect mirrors are placed facing one another and they are in proximity, and photons (don't ask me how) are traveling between them and toward one of them, what is to keep the radiation pressure from reaching incredible amounts?

I might be way off of base here because I am new to this field. I heard that the radiation pressure doubles on a mirror.

If this is the case, a laser beam focused on a mirror at any angle will cause the mirror to accelerate one of two ways-- no more.

This also violates energy conservation and thus isn't valid.

share|improve this question
What do you mean by 'cause the mirror to accelerate one of two ways'? –  zhermes Mar 23 '13 at 4:01
When a photon hits a mirror at an angle it bounces back at the opposite angle, so their is only two possible ways for the perfect mirror to travel (depending on which side was illuminated.) Or at least that's what it seems to me. –  Velox Mar 23 '13 at 14:51

1 Answer 1

up vote 5 down vote accepted

It takes twice the momentum to bounce a photon back from where it came, as it does to just absorb it. So the radiation pressure from reflecting a photon is double that of absorbing a photon. So, two opposing mirrors have simply twice as much radiation pressure as two absorbing surfaces. The pressure doesn't continue to grow.

share|improve this answer
In fact if the mirrors are allowed to accelerate then with each reflection the photons will be redshifted, eventually to the point where all of their energy and momentum is in the mirrors and there is no light left. –  Michael Brown Mar 23 '13 at 6:31
Ok but since they bounce back and forth tons of times, aren't they doing way more work than their energy allows them too? One strong blip of light would accelerate the mirrors to high velocity? The redshift effect will be incredibly minute until incredible speeds are achieved. –  Velox Mar 23 '13 at 14:49
@Velox "aren't they doing way more work than their energy" No. That's what Michael was saying. On each bounce they do a little work and *lose the same amount of energy (and therefor become a little lower in frequency). This (tiny!) effect is usually ignored because the mirrors' masses are so large compared to the photon energy that little energy is transferred, but you have set up a problem where it must not be ignored. –  dmckee Mar 23 '13 at 14:56
So, I think I will set it as an answer. Since it makes sense? :D –  Velox Mar 25 '13 at 14:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.