# Reflection and radiation pressure.

1. Aug 8, 2007

### Repetit

Hey!

When a photon is reflected off an interface doesn't it deliver some of it's momentum to the object that it hits due to radiation pressure? If so, shouldn't the reflected light have a longer wavelength (smaller wavenumber) than the incident light? Is this effect present, but just so small that it is ignored?

Thanks

2. Aug 8, 2007

### Meir Achuz

You're quote is effective here. For a single photon, QM must be used, so the reflection and transmissiion coefficients represent probabilities.
The frequency of the photon is unchanged.

3. Aug 8, 2007

### cesiumfrog

Huh?

Reflection off a "mirror" is just a special case of Compton scattering (look it up on hyperphysicis). By assuming conservation of momentum and energy, you can simply derive the amount by which the reflected photon decreases in energy (increases in wavelength). Since those laws are so fundamental, the derivation is valid both for scattering from subatomic particles, reflection from macroscopic optics, and strong lensing by black holes. In the limit where the "mirror" is very heavy compared to the mass-energy of the photon (as in most practical cases and even more so if it is attached firmly to the ground), the fractional wavelength change is of course negligible.

4. Aug 8, 2007

### Meir Achuz

We have answered different questions, and I am not sure which one he asked.

5. Aug 9, 2007

### BobG

I would guess he's asking the question cesiumfrog answered correctly if only because he didn't ask it in the QM forum.

Except the light doesn't even have to be reflected. It still transfers momentum even if all of the light is absorbed - just half as much as if all the light were reflected by a perfect mirror. In fact, it transfers momentum even if the object is transparent - the momentum is just cancelled out when the light is emitted out the other side, leaving you with a net momentum change of zero.

The net change in energy (and wavelength) might be small, but it's effect builds up over time - especially if you start tossing in more interactions. If you light a candle and place it in a box lined with mirrors, how long will the light last after the oxygen runs out? It will never disappear completely, but every interaction that leaves some of the energy in the mirror absorbed as heat lowers the frequency and, unless you have a really big box, you're going to have a lot of interactions very quickly. The light drops out of the visible frequency range very quickly.

Last edited: Aug 9, 2007
6. Aug 9, 2007

### Cthugha

But this is just a minor effect. The main loss mechanisms are absorption of photons by the mirror material and photons escaping the box due to the transmission coefficient being pretty small, but not exactly 0.

If the effect you mentioned was that large, some modern lasers wouldn't work, especially microcavity lasers.

7. Aug 9, 2007

### cesiumfrog

Just to clarify something.. I agree that since light is very fast (and normally traverses the whole height of our atmosphere in a very short time) that if it is confined to remain in air (at standard pressure, such as one would use to burn a candle) then it would not take a long time (on our time scale) for the interactions (tending toward thermodynamic equilibrium) to reduce a visible photon to IR. (Of course, if the candle was in a mirror-box, most of the light would more likely be absorbed almost immediately by the opaque candle itself.)

But the size of the mirror-box is completely irrelevant to what you've described, and your mentioning it makes me wonder if you were really thinking of something more like a vacuum mirror-box containing nothing but photons. In this case, the reflection from the sides of the box will not alter colour of the photons (I don't mean "this is just a minor effect"). You can immediately prove this by conservation of energy (the rigid mirror-box is not accumulating kinetic energy). Unless the walls of the mirror-box move (i.e., unless the radiation pressure force is allowed to do work) then the effect can not occur (and even if photons only strike one wall at first then the effect will be cancelled since they reflect again next from the opposite wall).

8. May 23, 2010

### dan_b_

Insofar as wavelength shifts are concerned, they can be explained by the Doppler effect.
If a photon is reflected from a low mass mirror, during the interaction the mirror is in recoil and the reflected photon is red-shifted. In a closed box, if a photon is reflected from the left wall giving the box a small velocity to the left, the photon will be blue-shifted upon reflection of the approching mirror on the right wall. The time avearge of the momentum transfered to the wall will be zero, and the photon will not lose energy, provided it is not absorbed.

9. May 23, 2010

### manticmadman

I cannot speak for the smaller end of the world, but in our solar system the radiative force of the Sun(based on surface area of teh body) is precisely equal to the gravitational force between the two objects at this distance.

10. May 23, 2010

### cjameshuff

...only if you're an extremely lightweight solar sail.

Did you mean to say what you wrote? The ratio between the two is constant for any given body, both photon pressure and gravitational force falling off inversely to the square of distance, but the two are only precisely equal for perhaps some dust grains of a particular size, drifting out of the system due to the lack of anything to bind them here.

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