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What happens to light in a perfect reflective sphere?

I was working on my toy ray tracer when I pondered on this:

Say we build a hollow sphere big enough to fit a person. The internal surface is perfect mirror, with no cracks or holes. We place an invisible observer with an invisible flashlight, just for the sake of argument, inside the sphere. The flashlight is turned on for, say, 1 second. What does the observer continue to see after the flashlight has been turned off, and why?

  1. it is pitch black, so he can't see anything.
  2. there is just as much light as when the flashlight was on, but it dims and eventually becomes pitch black?
  3. it isn't pitch black, and he can see.
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marked as duplicate by David Z Dec 5 '11 at 21:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

what is the significance of sphere in your question? would it matter if it was a cuboid room of mirrors?? Anyways, I would vote for option 2, since immediately after mirror is turned on, without the presence of any absorbing medium, light will not be absorbed, but will reflect back and forth the walls. – Vineet Menon Dec 5 '11 at 9:36
""We place an invisible observer with an invisible flashlight"" A invisible observer will not see anything at all, because that is a contradiction to invisibility. There were several questions like this already here. – Georg Dec 5 '11 at 9:54
Very related:… – Arnoques Dec 5 '11 at 11:03
@Georg The observer and the flashlight are invisible, just for the sake of it, so that they wouldn't absorb any of the energy. – Arlen Dec 5 '11 at 15:48
@Vineet Menon no real significance. We could use a cuboid, but I went with a sphere. – Arlen Dec 5 '11 at 15:53

Since there is nothing to disperse and diffract the light in the dome, and the curvature of sphere ,its concave nature (from the inside) will converge light and diverge depending the position of light wrt to the sphere.

and one might see an ray of light travelling in random directions, getting converged or diverged from time to time,

But to see light one will definitely have t absorb photons, which will be impossible for our invisible friend,

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I have a feeling that is shouldnt ve been downvoted – Tomarinator Dec 5 '11 at 15:28

(Note: This answer treats photons as if they were classical point particles; I am ignoring effects such as diffraction.)

An invisible observer is a contradiction, because something is invisible if it does not interact with photons, and if it does not interact with photons then it cannot observe them. An invisible flashlight is similarly impossible, but we can hypothetically assume we shine some light into the sphere from an opening and then slam the (perfectly mirrored) hatch shut, since light travels at a finite speed.

First, assuming a perfect reflector and nothing in the sphere (not even air, since that will interact slightly), then what we have is a volume full of electromagnetic radiation bouncing around. It contains a (nearly) fixed amount of energy and will continue to do so indefinitely (except for this effect).

If we introduce an observer, then the observer absorbs some of the light. If we assume the observer absorbs no light that it does not also observe (a bodiless perfect retina?), then it will observe an (exponentially) decreasing brightness. The photons have essentially random tracks through the sphere, so at any given moment some of them will be absorbed, and as the population of photons (amount of energy) reduces there will be fewer of them to be absorbed. (Some of the photons may be on paths which will never intersect the observer, but they can be just ignored.)

If we additionally grant your invisible flashlight — it doesn't actually have to be invisible, just to emit more light than it absorbs — then this simply adds to the photons in the sphere at some fixed rate. With both the flashlight and the observer, assuming that 1 second is sufficient time, then the amount of photons in the sphere will increase until the observer's absorption rate equals the flashlight's emission rate (or until the flashlight is turned off before then). When the flashlight is turned off, the photons will be absorbed as discussed above.

So, your option 2 is correct, but you have to think carefully about what kind of impossibilities your “ideal” objects are introducing.

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""(Note: This answer treats photons as if they were point particles; I am ignoring wave behavior."" Note, that photons do not have wave "behaviour" at all. You treat either light as a particle, or a wave. Photons with a wave behaviour are nonsense. – Georg Dec 5 '11 at 15:14
More nonsense than electrons with a wave behavior? This sounds pretty pedantic. – user2963 Dec 5 '11 at 15:20
@Georg What I meant is that I'm using a "infintesimal billiard ball" model with perfect collisions without considering such effects as refraction. I'm not sure how best to phrase that — I agree that "photons act like waves some of the time" is wrong. – Kevin Reid Dec 5 '11 at 15:21
@Kevin, that has nothing to do with diffraction, its fundamental: You may describe light as a photon, or as a wave, but never mix this ways of description. Its possible to describe diffraction of light in the particle model, although its rather inconvenient. – Georg Dec 5 '11 at 15:30
OK, I've specified classical particles. – Kevin Reid Dec 5 '11 at 15:44

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