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Given a solid whose interior is a hollow sphere with perfectly reflecting mirrors. A small hole is drilled in the sphere and a photon is sent in at some angle. Will it always eventually exit through the hole it entered?

Is there any arrangement of mirrors one can place inside the sphere such that the photon will never escape?

If not, is there a principle that explains it, or some way to prove it?

Qmechanic
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TROLLHUNTER
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  • If the photon could never escape, then it could never couple into the cavity. So a priori the answer is no. Ultra high Q cavities are used, especially in cavity QED. These cavities are actively stabilized and use "super mirrors" in a Fabry-Perot geometry. PMT's are used for detection. –  Aug 16 '13 at 02:11
  • Related: http://physics.stackexchange.com/q/13500/2451 and links therein. – Qmechanic Sep 29 '13 at 15:40

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For the first problem, yes. Because you shoot in the photon from the boundary of the sphere, the trajectory of the photon, using elementary geometry, will stay within a fixed plane (the plane is defined by the center of the sphere, the point of entry, and the initial direction of the photon). So you reduce the problem to an essentially two-dimensional problem. It is well known that the billiards trajectory in a circle is integrable, and the trajectory is either periodic or hits a dense set of points on the boundary. In the former case the photon will quickly exit the sphere, even in the limit as the size of the initial hole tends to zero. In the latter case for every non-zero sized hole, you can find some finite time $T$ such that the photon will be guaranteed to exit the hole after $T$. But as you shrink the size of the hole, $T$ increases unboundedly.

For the second problem, it is possible to set up trapping obstacles inside your sphere. For an example, see this question/answer on MathOverflow. However, in general you can only trap trajectories for a small set of initial angles; I am inclined to say that you cannot do so for all angles simultaneously, though I don't have a proof for that.

Community
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Willie Wong
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  • Will it be the "same" photon, though? Or will it be absorbed by the metal in the mirror and then another photon emitted? – Lagerbaer Apr 10 '11 at 05:07
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    @Lagerbaer: it depends on your definition of "same". For one thing, particles are indistinguishable. They are just fluctuations in the field. So, it's definitely the same in this sense. But if you consider Feynman diagram for reflection then it's obviously another photon that comes out from interaction. In this sense it's not the same. In any case, it's just terminology... – Marek Apr 10 '11 at 07:20
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    of course you can throw the photon in and close the hole before the half life of the photon-in-a-ball excitation expires, so you'll get statistically half of your balls enclosing photons – lurscher Apr 10 '11 at 14:33
  • I think the MathOverflow answer state that there is a solution for a trapped photon between surfaces, but no need for an 'entry hole' like we have here. If we have a hole, the beam cannot be trapped in a finite reflection cycle (it would hit the hole again) but must hit a new spot after every reflection. This however should be possible, if the locations in the alternating reflection sequence approach two limit points. – dronus Apr 08 '12 at 23:55
  • I think Willie's answer assumes the ideal case of no absorption and no diffraction. So if you're doing a computer simulation where these effects can be turned off, then it's fine. But this is not how a real cavity would behave. –  Aug 16 '13 at 20:15
  • @john: the user did say "photon" and "perfectly reflecting". The former (particle assumption) validates the lack of diffraction, the latter validates the lack of absorption. – Willie Wong Aug 19 '13 at 22:15
  • In that case Willie the question is devoid of any physical interest and is a purely mathematical question. What's it doing on physics.se? I'm serious, the system is impossible to realize because it violates fundamental laws of physics. What next, the physics of the tardis? –  Aug 19 '13 at 22:36
  • @John: geometric optics works well enough for us (and our ancestors) to build working eye glasses and telescopes, in spite of the fact that the thin lens approximation may not always be valid. Newton's original laws of motion runs counter to the "physical intuition" of Aristotles because in the real world there is friction and a ball doesn't go on rolling forever. Idealisation and abstraction are powerful aids to problem solving, and have long traditions in physics. – Willie Wong Aug 20 '13 at 00:07
  • @John: in addition, I find it rather surreal to have a professor of physics tell me that it is worthless to consider an idealised system that is impossible/impractical to realise in our physical world, and that older and approximate theories should be immediately discarded in their entirety just because they weakly violate a more newly described physical principle. But since I don't gain anything from defending this gedankenexperiment, I shan't argue further on this issue. – Willie Wong Aug 20 '13 at 00:14
  • You miss my point, or I didn't make it well. I'm not against an idealized model, I'm rejecting one that violates fundamental physics. I can for example tell you precisely the error associated with making an infinite wire approximation. Tell me how small you want epsilon and I'll tell you how long to make the wire. Your example, of geometrical optics shows what I mean. Tell me how good you want the answerto be and I'll tell you how large the frequency has to be. Without violating any physics. When you violate physics (in this case causality), you're left with nothing to hold on to. –  Aug 20 '13 at 01:42
  • The problem here is the formulation of the original question. It only has any meaning as a pure mathematics problem, which was explained above in terms of periodic orbits. If you want to discuss it as a physics problem it cannot violate fundamental laws of nature. Surely you see this. –  Aug 20 '13 at 01:49
  • @williewong as an example, I see no issue with the question that just came across se.physics about a 747 crashing into a diamond skyscraper. Do you see the difference? To a mathematician maybe these seem like similar abstractions. But I assure you they are not. –  Aug 20 '13 at 01:54
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A hollow sphere with perfectly reflecting mirrors is a particular case of a cavity resonator The others answers presume that the light ray (or a single photon if you prefer) travel along a plane that contains the centre of the sphere and the question set no constraint on this. Also the dimension of the sphere (radius) must be adequate to the wavelenght of the photon to trap it (total reflection). There are certain configurations of angles where the ray will return to the same initial point, and I suspect that in any situation it will always return to the origin. Probably it is already proved.
Is there any arrangement of mirrors one can place inside the sphere such that the photon will never escape?
Yes - The number of mirrors, size, the shape and orientation are important and some constraints on these must be previously known. The solution is trivial if we can change the mirrors configuration after the ray is inside or with a proper rotation of the sphere. I am not visualizing a solution if we can not change the configuration.

A cavity resonator is a hollow conductor blocked at both ends and along which an electromagnetic wave can be supported. It can be viewed as a waveguide short-circuited at both ends (see Microwave cavity).

physicists-trap-light-in-a-bottle and bottle microresonators
Even a perfect reflection can not avoid the Evanescent_wave so, there are no perfect mirrors. Some of the energy will evade.

In optics and acoustics, evanescent waves are formed when waves traveling in a medium undergo total internal reflection at its boundary because they strike it at an angle greater than the so-called critical angle. The physical explanation for the existence of the evanescent wave is that the electric and magnetic fields (or pressure gradients, in the case of acoustical waves) cannot be discontinuous at a boundary, as would be the case if there were no evanescent wave field. In quantum mechanics, the physical explanation is exactly analogous—the Schrödinger wave-function representing particle motion normal to the boundary cannot be discontinuous at the boundary.

In optics, evanescent-wave coupling is a process by which electromagnetic waves are transmitted from one medium to another by means of the evanescent, exponentially decaying electromagnetic field.

dielectric microsphere resonators

The whispering gallery modes (WGMs) of quartz microspheres are investigated for the purpose of strong coupling between single photons and atoms in cavity quantum electrodynamics (cavity QED). ...

In optics the treatment is done using only the wave properties of the light, and as a particle when it is absorbed by some atom/electron/...

EDIT add: The first sentence of this answer is not very interesting. In one situation the study uses 'geometric optics' because the wavelength is << sphere Radius and we must use EM pure treatment in a waveguide (the wavelength is in the order of transversal dimension of the waveguide).

Helder Velez
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  • If the light cannot escape, then by time reversal invariance it can never enter. Having built many cavities,I can tell you that the higher the Q of the cavity, the harder it is to couple into. –  Aug 16 '13 at 14:59
  • @John If the Wine cannot escape (of the sealed bottle), then by time reversal invariance it can never had entered. 'Travel to the past is impossible'. – Helder Velez Aug 17 '13 at 11:07
  • The jokes are much better on MO. Keep trying Helder. –  Aug 17 '13 at 15:22
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If you're asking a mathematical question: does a line starting at a point on the inside of a sphere, and reflecting off the sides of a sphere, ever come back to its starting point, then the answer is yes if and only if the angle the line makes with the tangent to the sphere at the starting point is 90/n degrees, for integral n > 0.

If you're asking a physical question: does a photon going through a finite-size hole and bouncing around inside ever come out, the answer is yes -- even if you close the hole up after you shoot the photon inside.

Walt Donovan
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A line entering a sphere will bounce around in the plane containing the original ray and a ray from the entering point through the center of the sphere. It will bounce around until it hits the entrance, then exit.

Your first example with theoretically perfect surfaces would work of course if you could perfectly close the entrance hole after you inject the photon, but I suspect this is not what you were looking for. Edit: noticed the other answers with the trapping arrangements so I won't try to answer that part here :)

BjornW
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Yes, it will always exit.

If you want to trap it - you will need to to be able to move mirror a little while photon is somewhere in the middle of your trap.

BarsMonster
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