Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
The act of observing would absorb light. Therefore you would see light and then it would dim. –  JoeHobbit 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
2  
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 Jul 18 '11 at 19:01
1  
It would weigh more due to the photons in it, per my previous question. But yes, photons are still massless. physics.stackexchange.com/q/10612 Your question doesn't seem to be as well defined. –  AlanSE 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

4 Answers 4

up vote 3 down vote accepted

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.]

share|improve this answer
    
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 B Jul 30 '11 at 0:43

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
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. –  JoeHobbit Jul 18 '11 at 20:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.