reflection of light off 4 90 degree mirrors

by curiouschris
Tags: degree, light, mirrors, reflection
 P: 110 Hi Part 1 If I shot a laser beam into a square box where each wall is a perfect mirror and the box is filled with a vacuum and devoid of any other material that could interfere with the light beam apart from the perfect mirrors. Q/ Would that light beam continue to bounce around forever? Part 2 Assuming I was able to measure the movement of the box. Q/ Would I see it 'jiggle'. In other words when the light beam hits each wall in succession would the box move in that direction. (solar sails use light so I assume this is exactly what would happen) Part 3 Would this jiggling continue forever, along with the original beam? CC
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P: 11,912
 Quote by curiouschris Hi Part 1 If I shot a laser beam into a square box where each wall is a perfect mirror and the box is filled with a vacuum and devoid of any other material that could interfere with the light beam apart from the perfect mirrors. Q/ Would that light beam continue to bounce around forever?
I'm going to assume you really mean a cubical box, and ask you: how does the light get inside a completely closed, perfectly reflecting enclosure in the first place.?
 P: 175 Redbelly, shooting it in and closing the hole very fast would probably do the trick chris, the photon would probably very quickly give all of it's energy to the box's walls and disappear. And the box would have to be very very light in order to see any motion, otherwise Heisenberg's Uncertainty Principle sets in...
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P: 5,183

reflection of light off 4 90 degree mirrors

 Quote by Crazy Tosser chris, the photon would probably very quickly give all of it's energy to the box's walls and disappear.
I thought he stipulated that the mirrors were perfectly reflecting.
P: 175
 Quote by cepheid I thought he stipulated that the mirrors were perfectly reflecting.
I don't think there's such a thing. But if we do assume the mirrors are perfectly reflecting, then the box will not move at all because the photon will not give any of it's energy away. That situation would in many ways contradict the Second law of Thermodynamics though...
 Mentor P: 5,183 You don't have to go to any great effort to convince me there's no such thing as a perfect mirror. Just to to be clear though...so the scenario is meaningless, then? In the same category as "what happens when an immovable object meets an unstoppable force?" I'm just curious...
 P: 175 Yes, it is a meaningless scenario :D And it sounds more like the Schrodinger's Cat Paradox... What happens to a cat that we know nothing about?
P: 2,050
 Quote by curiouschris Would this jiggling continue forever
No.

If the mirrors are fixed rigidly to a massive base, then there will be no jiggling and the light will continue forever basically. But if the mirrors are free (like a solar sail) then you can easily calculate how the jiggle is transferring energy to them and red-shifting the light, making the light less powerful each reflection (think of the jiggling dying down as the whole system thermalises).

But it isn't a meaningless scenario.
P: 175
 Quote by cesiumfrog No. If the mirrors are fixed rigidly to a massive base, then there will be no jiggling and the light will continue forever basically. But if the mirrors are free (like a solar sail) then you can easily calculate how the jiggle is transferring energy to them and red-shifting the light, making the light less powerful each reflection (think of the jiggling dying down as the whole system thermalises). But it isn't a meaningless scenario.
The mirror is composed of particles, man, and no matter what you do you cannot fix those in place... And I do assume that after a certain number of redshifts the photon will finally be completely absorbed by an electron.
P: 2,050
 Quote by cepheid [Is this a meaningless scenario, like] "what happens when an immovable object meets an unstoppable force?"
 Quote by Crazy Tosser Yes, [and also like Schroedinger's paradox:] What happens to a cat that we know nothing about? The mirror is composed of particles, man, and no matter what you do you cannot fix those in place... And I do assume that after a certain number of redshifts the photon will finally be completely absorbed by an electron.
The examples of meaningless scenarios were questions expected to be unanswerable in principle, unlike the OP's scenario.

I disagree with your method of employing quantum concepts to explain a classical physics phenomenon. This is certainly a situation in which classical loses should be expected to dominate. (Even if you wish to consider the mirror as a system of quantum mechanical particles, note that they need only be fixed on the scale of the wavelength and the wavelength may be chosen to be arbitrarily large.)
P: 110
 Quote by cesiumfrog No. If the mirrors are fixed rigidly to a massive base, then there will be no jiggling and the light will continue forever basically. But if the mirrors are free (like a solar sail) then you can easily calculate how the jiggle is transferring energy to them and red-shifting the light, making the light less powerful each reflection (think of the jiggling dying down as the whole system thermalises). But it isn't a meaningless scenario.
Thankyou cesiumfrog. Too many people are prepared to knock without really applying thought.

Most physics is worked up from ideal circumstances. As far as I know its the only way we can get an exact understanding of anything. We then slowly introduce real effects until the full problem is understood and all variations accounted for.
For example, teaching the physics of a bouncing ball is started without gravity etc... if we added gravity, air resistance, thermal losses etc etc students would quit, reeling from the enormity of the task.

Is that not true?

I can see what you mean by red shifting. It makes perfect sense, and opens a pandora's box :)

 The mirror is composed of particles, man, and no matter what you do you cannot fix those in place... And I do assume that after a certain number of redshifts the photon will finally be completely absorbed by an electron.
What absorption? as far as I am aware a perfect mirror absorbs nothing (its not perfect otherwise).

Can you explain how refraction works? it is analogous to reflection yet only involves a change of density, given the right angle of incidence and the right density of material does not refraction exactly mimic reflection? there is no absorption and re emission of photons with refraction why does there have to be with reflection?

CC
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 Quote by cesiumfrog No. If the mirrors are fixed rigidly to a massive base, then there will be no jiggling and the light will continue forever basically. But if the mirrors are free (like a solar sail) then you can easily calculate how the jiggle is transferring energy to them and red-shifting the light, making the light less powerful each reflection (think of the jiggling dying down as the whole system thermalises). But it isn't a meaningless scenario.
Actually (assuming the wave packet is shorter than the length of the box), there would be alternating red and blue shifts.

Imagine the box is stationary, with the wave pack moving to the right. As the wave reflects from the right-side wall, the box starts moving to the right (conservation of momentum) and the wave gets red-shifted as the box is moving away from it.

As the reflected wave moves toward the left-side wall, the wall is moving towards the wave, which will result in a blue-shift of the wave. Also, by conservation of momentum, the box will be stationary after the wave reflects from the left-side wall.

So if we imagine truly lossless reflections, the light does not continually lose energy. Rather, it exchanges energy back and forth with the box.
P: 110
 Quote by Redbelly98 Actually (assuming the wave packet is shorter than the length of the box), there would be alternating red and blue shifts. Imagine the box is stationary, with the wave pack moving to the right. As the wave reflects from the right-side wall, the box starts moving to the right (conservation of momentum) and the wave gets red-shifted as the box is moving away from it. As the reflected wave moves toward the left-side wall, the wall is moving towards the wave, which will result in a blue-shift of the wave. Also, by conservation of momentum, the box will be stationary after the wave reflects from the left-side wall. So if we imagine truly lossless reflections, the light does not continually lose energy. Rather, it exchanges energy back and forth with the box.

Around here thats crazy talk so I thought I better not point out the obvious ;)

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