Photons reflecting off mirrors, wheres the flaw?

In summary, this reasoning does not maintain the conservation of momentum. There is a loss of energy due to the recoil of the first ship and the mirrors moving. There is also no stable resonant mode, so field momentum is continuously transferred to the cavity mirrors.
  • #1
Sakha
297
0
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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  • #2
Sakha said:
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.
 
  • #3
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.
 
  • #4
Sakha said:
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
 
  • #5
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.
 
  • #6
You are coupling the momentum of ponderable matter with an electromagnetic field. Why do you think would be simple?
 
  • #7
excuse me, but no net momentum results from any photon which is absorbed and then re-emitted, right?
 
  • #8
It depends- in this case, the absorption and emission events are not spherically symmetric; there is net momentum transfer.
 
  • #9
Andy Resnick said:
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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  • #10
DaleSpam said:
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' [tex]\lambda = \frac{h}{p}[/tex]?
 
  • #11
andy - would that not mean that the energy of the emitted photon must be lower than the absorbed photon, due to the momentum gain?
 
  • #12
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.
 
  • #13
Sakha said:
When you say de Broglie you mean the 'simple' [tex]\lambda = \frac{h}{p}[/tex]?
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.
 
  • #14
jnorman said:
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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  • #15
Vanadium 50 said:
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:

http://www.google.com/url?sa=t&sour...sg=AFQjCNHrV7P1jSzdteHV4ZXJpDyznflcOA&cad=rja

But this again seems to be a macroscopic result: the mirror is set in motion, and light reflects off it. What happens if the field consists of a single photon? Photons are not localized in space, so I'm not sure how to proceed.
 
  • #16
Maybe when considering a single photon one would have to consider solid state physics and phonons? (I'm just thinking out loud)
 

1. What is the flaw in the concept of photons reflecting off mirrors?

The concept of photons reflecting off mirrors is based on the assumption that photons behave like particles. However, according to the theory of quantum mechanics, photons also exhibit wave-like properties. This means that they do not follow a straight trajectory when reflecting off a mirror, but rather exhibit diffraction and interference patterns.

2. Can photons actually bounce off a mirror?

Yes, photons can bounce off a mirror. When a photon hits a mirror, it is absorbed by the atoms on the surface and then re-emitted in a different direction. This process is known as reflection.

3. Do mirrors absorb photons?

Yes, mirrors do absorb photons. However, the absorption is only temporary as the photons are re-emitted in a different direction, resulting in the appearance of reflection.

4. Can photons pass through mirrors?

No, photons cannot pass through mirrors. Mirrors are made of materials that are opaque to photons, meaning that they cannot penetrate the surface and pass through it.

5. How does the angle of incidence affect the reflection of photons off a mirror?

The angle of incidence, which is the angle at which a photon hits the mirror, affects the angle at which it is reflected. According to the law of reflection, the angle of incidence is equal to the angle of reflection. This means that the direction of the reflected photon will be the same as the direction of the incident photon, but on the opposite side of the normal (an imaginary line perpendicular to the surface of the mirror).

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