What happens to light in a perfect reflective sphere?

Question

Let's say you have the ability to shine some light into a perfectly round sphere and the sphere's interior surface was perfectly smooth and reflective and there was no way for the light to escape.

If you could observe the inside of the sphere, what would you observe? A glow? And would temperature affect the outcome?

Seems silly, it's just something I've always thought about but never spent enough time (until now) to actually find an answer.

The act of observing would absorb light. Therefore you would see light and then it would dim. — Dale, Jul 18 '11 at 18:47
Let's say the sphere is a two-way mirror that doesn't absorb light. — doremi, Jul 18 '11 at 18:55
It doesn't work that way. If it is perfectly reflective then no energy gets out to be observed. Two way mirrors work because they are imperfect. — dmckee --- ex-moderator kitten, Jul 18 '11 at 19:01
It would weigh more due to the photons in it, per my previous question. But yes, photons are still massless. http://physics.stackexchange.com/q/10612/ Your question doesn't seem to be as well defined. — Alan Rominger, Jul 18 '11 at 19:03
The problem start with: How do you define 'a sphere' and how you measure it? Is it in motion? I suspect that this is not so simple. A similar question in PSE-trapping-a-lightray — Helder Velez, Jul 18 '11 at 22:11

score 9 · Accepted Answer · answered Jul 22 '11 at 06:54

OK, the inside of the sphere is perfectly-reflecting, and there's an ideal optical diode to let light in but keep it inside. As you keep the light turned on, the photon density in the sphere goes up and up, of course. It "looks" brighter and brighter, but you don't see that because the light can't escape. After turning the light off, it stays bright, the photons just keep bouncing around. If you "stick your head in" to look, you see a bright uniform glow that quickly dies away because your head and eyes are absorbing all the photons.

But do the photons bounce around forever? No!! Even a perfectly-reflective sphere will still interact with the light, because of radiation pressure. Each time a photon bounces off a wall, the wall gets kicked backwards, gaining energy at the expense of the photon (on average). Light can't produce a smooth force, only a series of kicks with shot noise statistics, because one photon hits the wall at a time. These kicks eventually heat up the walls, and cool down the photons. (From the photon's point of view, the photon frequency is going down because of Doppler-shifts during reflection off the moving walls.) Eventually everything equilibrates to a uniform temperature, hotter than the sphere started out. I don't know how long that would take. [In any realistic circumstance this radiation pressure effect can be ignored, because it is much less important than the "reflection is not 100% perfect" effect.]

i assume the optical diodes have a leaking rate which for big enough intensities, become equal to the input intensity, meaning you reach equilibrium and can't increase the density further, thats another mechanism of leakage — lurscher, Jul 22 '11 at 14:36
your assumption about photons leaking energy by reflection only happens with inelastic scattering, which means the photon needs to be actually absorbed by the atoms of the inner mirror (and then decay into a longer wavelength photon plus some kinetic energy). As long as it is being reflected no such thing will happen — lurscher, Jul 22 '11 at 14:38
@lurscher -- Yes, optical diodes cannot be perfect in reality, just as walls cannot be perfectly reflecting in reality. This is a silly hypothetical. Your statement about inelastic scattering is I think misleading: A photon reflecting off of a moving mirror will be redshifted if the mirror is moving away from the photon, or blueshifted if the mirror is moving towards the photon. I don't think it makes sense to call this "inelastic scattering", even though there is a change of frequency in some reference frame. At least it's not the usual "inelastic scattering" people think of. — Steve Byrnes, Jul 30 '11 at 00:43

score 2 · Answer 2 · answered Jul 18 '11 at 19:18

2

If you are observing the inside of the sphere, you are absorbing light to make the observation. The light would get dimmer and dimmer very quickly until you could see nothing at all.

answered Jul 18 '11 at 19:18

Dan

5,685

1

Would the sphere change in temperature? – doremi Jul 18 '11 at 19:31
That depends if the sphere emits radiation into the perfectly reflecting sphere. – Dale Jul 18 '11 at 20:26

Taiben · Answer 3 · 2014-09-01T09:02:11.787

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Just another perspective: Since the sphere is non-ergodic, your observation depends on your and the source locations inside the sphere. For ergodic shapes (ellipsoid, etc), you will see an evenly lit world.

edited Sep 01 '14 at 09:02

answered Sep 01 '14 at 08:46

Taiben

432

score -1 · Answer 4 · answered Jan 18 '14 at 02:55

As soon as the light shining in was turned off, the light in the sphere would disappear, not because observing depletes energy, it doesn't (but energy loss by the system is required for observation to occur). No one knows but this hypothesis that light beams persist when the light emitter is extinguished is just that. There is no supportive evidence for this theory, that I am aware of.

score -2 · Answer 5 · answered Mar 10 '16 at 08:52

The interaction of the light with any particles present (there can be no transfer of energy even light without matter) would lead to an increase of particle density and a vacuum until the sphere collapses in on itself. The idea that the light would just stop if the source was "switched off is nonsense. How we see the light from a long dead star?, Light continues it course to it's conclusion (absorbsion).

What happens to light in a perfect reflective sphere?

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