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Total internal reflection is used to send light signals long distances in fiber-optic cables. If the cable is shaped into a ring, the light will run in a circle until its intensity dwindles due to attenuation (optical ring resonators are examples of this). Light doesn't attenuate in a vacuum, however. If you were to create a ring-shaped vacuum chamber lined with a metamaterial with an index of refraction less than one, could you build up arbitrary light intensities without losses to attenuation?

Edit:

My question references materials with indices of refraction less than unity. I believe the speed which can be calculated using this index is the phase velocity of light which does not carry information and can exceed c. For example, to x-rays most solid materials are optically less dense than a vacuum and have a refractive index slightly less than one. You can get total external reflection this way.

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No. Even if you take on board somethinghere's correct answer and instead try to build the waveguide from wholly lossless, but physical, materials, the loss will still be there. The loss in a ring resonator is mostly bend loss, which arises even when the ring is made from perfectly lossless materials.

Bend loss arises because the waveguide is no longer translationally invariant (owing to the bend) so that the bound eigenmodes of the formerly straight waveguide are no longer eigenmodes but instead have a nonzero coupling co-efficient with the radiation field.

So the loss from the ring is radiated away from the ring. This is why a very tight ring bearing visible light glows.


Question from OP

At the risk of asking a turtles all the way down question, why does this coupling begin when the waveguide is curved? More specifically, why does the classical model of total reflection no longer apply?

This is an excellent question that exposes the tacit assumption I made in my answer above, i.e. that I was assuming a ring resonator where the EM field propagates power around the ring's circumference in a vibration often called a "whispering gallery mode". The answer above can be summarized as, "whispering gallery modes are never true modes owing to an unavoidable, everpresent coupling to the radiation field", and the answer still stands. Indeed, I believe this tacit assumption was also made in your answer, because I get the impression that you are talking about modes where energy moves around the ring's circumference.

However, why, as you ask, doesn't the classical notion of total internal reflexion work with bends? Let's anticipate and dispense with a sly answer before getting to the real one: one could say that the the usual model of TIR assumes Snell's law and thus plane waves, so that bends violate these assumptions. OK, but your question then becomes, "Why can't we derive a classical model for TIR for bends?".

The answer to this last one is that we can indeed derive such a model, and the ring structure does have lossless modes corresponding to the new, rotational cylindrical symmetry. If you write down Maxwell's equations in cylindrical co-ordinates, these lossless modes pop out (with Hankel function dependencies of the field polar components on the radial co-ordinate). These modes are akin to the lossless vibrations of a spherical cavity resonator. However, in these modes, there is no energy transfer around the ring's circumference. There is no nett power transfer at all: the device is a purely reactive device where energy shuttles to and fro outwards then back inwards across the ring, with no nett loss each cycle.

So there is a model of lossless TIR for bent waveguides, but the propagation directions are radial, not circumferential, so there is no transfer of energy around the ring as in a ring resonator. This last point is important in most laboratory ring resonators, which are excited by a coupled optical fiber with field propagating tangentially to the ring.

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  • $\begingroup$ So breaking down this answer without a background in Maxwell's equations, am I correct in understanding that not being translationally invariant means the waveguide (the vacuum tube lined with a metamaterial) curves as the light moves through it. The waveguide's eigenmode means the resonant electromagnetic frequency of the waveguide. Since it curves, the waveguide no longer has eigenmodes but instead has a coupling (loss) coefficient which represents energy transfered from the electromagnetic wave to the waveguide. Is this more or less correct? $\endgroup$ – Charles Vorbach Oct 17 '16 at 1:09
  • $\begingroup$ @CharlesVorbach exactly. Pretty impressive summary, even if you did have a background in Maxwell's equations. $\endgroup$ – WetSavannaAnimal Oct 17 '16 at 1:20
  • $\begingroup$ At the risk of asking a turtles all the way down question, why does this coupling begin when the waveguide is curved? More specifically, why does the classical model of total reflection no longer apply? $\endgroup$ – Charles Vorbach Oct 17 '16 at 1:33
  • $\begingroup$ @CharlesVorbach That's an excellent question. The answer is "it does", but the modes of vibration probably don't have the propagation properties that you are thinking of. See my edits to my answer. $\endgroup$ – WetSavannaAnimal Oct 17 '16 at 2:53
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This answer (as pointed out in a comment) turns out to be wrong since it is actually the phase velocity of light, which can travel faster than $c$ since it carries no information.

I am afraid such a system would be unphysical. If you look at the Wikipedia page for Refractive Index, you'll notice that the refractive index of a material is also a measure of light's relative velocity to vacuum when travelling through the material. So having a refractive index less than 1 would imply that light travels through the material at a velocity greater than $c$, which is not possible.

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  • $\begingroup$ Perhaps I am wrong, but I was under the impression that the refractive index references the phase velocity of light rather than the propagation speed of light. I believe the phase velocity carries no information and can exceed c. Am I wrong? $\endgroup$ – Charles Vorbach Oct 17 '16 at 1:28
  • $\begingroup$ It appears you are right. My bad. $\endgroup$ – somethinghere Oct 17 '16 at 16:34

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