# Photons reflecting off mirrors, wheres the flaw?

#### Sakha

Hello.
Consider 2 spacecraft isolated in space, with a mirror on their back, facing each other. One light pulse is shot between them. The photons reflect off the first one, transferring a momentum off 2p, travels in the opposite directions, reflects off the second spacecraft, giving it 2p again. Repeat forever.
Where is the flaw in this reasoning? I think there has to be some redshift to maintain the Conservation of energy. What equation deals with this redshift?

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#### Dale

Mentor
Hello.
Consider 2 spacecraft isolated in space, with a mirror on their back, facing each other. One light pulse is shot between them. The photons reflect off the first one, transferring a momentum off 2p, travels in the opposite directions, reflects off the second spacecraft, giving it 2p again. Repeat forever.
Where is the flaw in this reasoning? I think there has to be some redshift to maintain the Conservation of energy. What equation deals with this redshift?
I think that conservation of momentum, de Broglie's relationship, and the Doppler effect will all show the same redshift.

#### Danger

Gold Member
Another thing to consider is that there is no such thing as a perfect mirror. There will be losses due to absorption and scattering at every instance of reflection. Within a very short period of time, the beam will vanish.

#### Andy Resnick

Hello.
Consider 2 spacecraft isolated in space, with a mirror on their back, facing each other. One light pulse is shot between them. The photons reflect off the first one, transferring a momentum off 2p, travels in the opposite directions, reflects off the second spacecraft, giving it 2p again. Repeat forever.
Where is the flaw in this reasoning? I think there has to be some redshift to maintain the Conservation of energy. What equation deals with this redshift?
The first flaw is not taking into account the recoil of the first ship caused by emitting the photon in the first place.

Reflection off a moving mirror has generally been considered in the context of interferometry, specifically gravitational wave sensing. Here's an example:

http://pra.aps.org/abstract/PRA/v51/i3/p2537_1

#### Sakha

So to resolve this scenario one needs complex math with Hamiltonians? Isn't there some simple math for it?
I was expecting a very straightforward solution.

#### Andy Resnick

You are coupling the momentum of ponderable matter with an electromagnetic field. Why do you think would be simple?

#### jnorman

excuse me, but no net momentum results from any photon which is absorbed and then re-emitted, right?

#### Andy Resnick

It depends- in this case, the absorption and emission events are not spherically symmetric; there is net momentum transfer.

#### Dale

Mentor
You are coupling the momentum of ponderable matter with an electromagnetic field. Why do you think would be simple?
I don't think it is so complicated. Conservation of momentum and energy together with de Broglie should do it. Obviously, that would be an approximation in the limit that the pulse is very brief.

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#### Sakha

I don't think it is so complicated. Conservation of momentum and energy together with de Broglie should do it. Obviously, that would be an approximation in the limit that the pulse is very brief.
When you say de Broglie you mean the 'simple' $$\lambda = \frac{h}{p}$$?

#### jnorman

andy - would that not mean that the energy of the emitted photon must be lower than the absorbed photon, due to the momentum gain?

Staff Emeritus
Isn't this a conservation of energy problem?

The photon has momentum p, so it has energy pc. At most, 100% of the energy can be transferred to the spaceship, so mv^2/2 = pc will give you the maximum velocity.

#### Dale

Mentor
When you say de Broglie you mean the 'simple' $$\lambda = \frac{h}{p}$$?
Yes, and I guess I forgot Planck's relationship between energy and frequency. I always think of both of those together as de Broglie's relationship and neglect to give Planck due credit.

#### Andy Resnick

andy - would that not mean that the energy of the emitted photon must be lower than the absorbed photon, due to the momentum gain?
That's a good question. I would naively agree, because E = pc and the transfer of momentum from the photon to the ship would imply a transfer of energy as well. But light can be confined in a high-Q cavity without suffering any frequency shift... Not sure where the relevant difference is between the two.

Edit: Ok, I think I got this part figured out: on resonance, there is no transfer of momentum from the field to the cavity; thus a broadband pulse injected into a high-Q cavity will evolve by transferring the momentum of off-resonant modes to the cavity; the on-resonant components will stably persist over time.

Allowing the mirrors to move means there is no stable resonant mode; field momentum is continuously transferred to the cavity mirrors at some rate equal to the rate of redshifting that occurs from the Doppler effect. Integrating this should give the same result that Vanadium 50 put up (except for a factor of 2; both ships move) in the limit t -> 00.

At least, that's what I came up with a little help from Mr. Noe... :)

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#### Andy Resnick

Isn't this a conservation of energy problem?

The photon has momentum p, so it has energy pc. At most, 100% of the energy can be transferred to the spaceship, so mv^2/2 = pc will give you the maximum velocity.
That is the result obtained in a limiting case. But consider light reflecting off a mirror that is free to move- in the macroscopic case, it's clear how to proceed: there's radiation pressure, leading to velocity of the mirror, and reflection off a moving mirror is doppler shifted. I found this: