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I was wondering and researching about light bending and reflection - I'm wondering if its possible to bend a light beam into a loop (orbit)? I know that light can be bent and reflected (even 100% reflection - TIR) is it possible to use a physical machine (array of lenses, optic cable etc) and feed it with a laser beam that will travel in a loop without losing any light?

Questions:

  • Can it be done? any suggestions or previous experiments?
  • If such case is possible (or not) does this light beam will be traveling through time?
  • What will happen if I broke the cycle after 1 hour?
  • do you mean putting a photon into orbit? – TanMath Dec 13 '14 at 01:39
  • yes a (group of photons) into an orbit by using a physical machine. – Shlomi Hassid Dec 13 '14 at 01:40
  • read this: http://www.universetoday.com/110682/can-light-orbit-a-black-hole/ – TanMath Dec 13 '14 at 01:42
  • I have seen that before and thats why i mentioned " a physical machine " photons follow the curve of space time which around a blackhole.... That doesn't answer my question I'm asking on manipulating them into an orbit and not losing them in the process. – Shlomi Hassid Dec 13 '14 at 01:48
  • Three mirrors in a triangle will do just fine, so will an optical fiber. – CuriousOne Dec 13 '14 at 01:49
  • Mirrors are reflecting 99.9 ( Perfect one 99.9998% ) of light that means that in a fraction I will loose my photons... – Shlomi Hassid Dec 13 '14 at 01:52
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    It's going to happen if you use a 'physical machine'. Nothing is guaranteed to interact with 100% of photons. – HDE 226868 Dec 13 '14 at 02:07
  • @HDE226868 - What? I'm sorry Im not sure what you are saying can you clarify? and TIR (Total internal reflection) is 100% - used in fiber optic cables. – Shlomi Hassid Dec 13 '14 at 02:13
  • Hmm. Let me try to explain it better. Well, as best I can put it, there is not a single case where one object is guaranteed to interact with another. Probability plays a huge role in the universe, especially on these scales, and it is never 100% probable that something will or will not happen (unless that something violates the laws of physics). Alternatively (and this may be the case here), the cables are not continuous, or perfectly smooth. There are going to be gaps, or impurities, and so the material will absorb photons - like AndreasH said. – HDE 226868 Dec 13 '14 at 02:17
  • So you are saying that according to this Andrea its not possible - and if the entire system is in a very low index material and the optic cable is curved in precision (not near the critical angle ) according to the "laws" it should be 100% reflected. – Shlomi Hassid Dec 13 '14 at 02:30
  • From what I can tell, not quite 100% should be reflected. In the vast majority of cases, such a technique will work, but in a tiny, tiny fraction of them, a photon or two will mess things up. – HDE 226868 Dec 13 '14 at 02:31
  • @ShlomiHassid: If you are losing 0.1% of your photons, then on average they make 1000 loops. Having said that, with modern mirrors and fiber one can do better than that, and one can always insert a laser amplifier... – CuriousOne Dec 13 '14 at 03:26
  • @CuriousOne, thank you for thinking about it, 1000 loops is not much even in huge scale when traveling at C - actually i think its possible not loosing any photon there are some new fiber cables that are extremely thin and are claimed to be 100% efficient - my question on this is actually understanding what will happen with such a device - will it bend the space time, will the photon trapped travel through time? let say i insert into this loop 8 photons will it be present in the past? – Shlomi Hassid Dec 13 '14 at 03:54
  • @ShlomiHassid: The number of loops is the number of loops at any speed of propagation. Like I said, you can make the system lossless with a fiber laser. There are plenty of such devices out there for different purposes. I have no idea where you are getting your ideas about "what will happen if" from, sounds like third rate scifi novels. Absolutely nothing special will happen when light goes around a loop. – CuriousOne Dec 13 '14 at 04:50

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In theory there is no in-principle reason why one cannot have a near to perfectly reflecting ring resonator as you wish, simply by making the ring of bigger and bigger radius, so that a repeated reflexion around the edge of a ring keeps the light confined to a circular path.

But you will always lose some light, even if the resonator's material is perfectly lossless. If it were possible for such a system to reach steady state, the phase fronts would be radial lines, all rotating at a constant angular velocity. All reflexions at interfaces, even TIR, penetrate into the reflecting medium a theoretically infinite distance. For example, in TIR, there is an exponetially dwindling with depth evanescent field beyond the reflecting interface, as I discuss in more detail in this Physics SE answer here.

At some critical radius, therefore, the phase fronts will "want" to move greater than $c$: what this means is that the light couples strongly to the radiation field and leaks out of the ring. If you're reflecting by TIR, you therefore always have a nonzero tunnelling to the radiation field.

In practice, though, such bending loss from resonant rings can easily be made so that the bending loss rate is less than one photon per the age of the Universe. The bending loss attenuation co-efficient decreases exponentially with bend radius: all of this is discussed in detail in Chapter 23 of Snyder and Love, "Optical Waveguide Theory". So in practice the limitation is going to be the intrinsic lossiness of your materials, not the bend.

In theory too you can use a nonlinear optical coupler to input light to the ring and then switch the coupler to shunt through light perfectly after the light has gone in (think of directing a train into a ring through a switch, and then switching the switch to keep the train in the ring.

None of this has any implications for time travel. If the light is confined to a dielectric, it isn't really light but a quantum superposition of excited matter states and free photons. You can think of the phenomenon as being repeated absorption and emission, and the delay as being simply owing to the absorber-emitters. It's simply like a game of "chinese whispers": if you open the switch an hour later, light pretty much the same as that which went in is going to come out. Nothing special has happenned. If the disturbance is moving at $c$, then there simply isn't any inertial frame which the light is at rest relative to. The time dilation is undefined and the light behaves just like any other light pulse, only that it has been diverted in a ring for a while. What you may be thinking of is experiments where unstable particles are put into rings and forced to travel near $c$: experimentally we do indeed observe that their lifetimes increase when travelling like this, and that the lifetime is extended exactly by the factor $\gamma$.