Infinite reflection inside a glass loop

Question

If you had a loop made of completely transparent glass (or other material), in the shape of a donut; think atomic collider (but probably not needing to be so large :) ), and you introduced light from a "port" on the edge (that wouldn't allow light to escape if the photons are travelling in one direction), such as from a laser, would the photons continue to orbit the donut until the donut is broken, such that photons could be continuously added, and would there be a maximum "capacity" of photons in such an arrangement?

Answer 1

tl;dr: No, it doesn't circulate forever.

If light can be coupled into the ring, it will also couple out. To understand why this is the case in your particular proposal, I've drawn such a whispering gallery mode resonator with an exaggerated step for incoupling:

The light is coupled in very close to the surface, such that it is internally reflected under a very shallow angle. After 1 complete round-trip it passes the step. Therefore the ray which was initially very close to the surface finds itself quite a bit away from the surface, which in turn means it hits the surface under a steeper angle. The steepness of the internal reflection angle increases with every round-trip, so that after enough round-trips the internal reflection isn't total anymore. From then on, a part of the light is transmitted at each reflection until nothing is left.

Making the step smaller reduces the ability to couple light in, but it also reduces the increase in reflection angle accumulated over the roundtrips. Hence, the resonator can store the light for longer. In the limit of a vanishing step it is not possible anymore to couple light in or out – at least not from a free-space beam. It is still possible to approach an optical fiber close to the whispering gallery mode resonator and couple some light into the resonator using the evanescent waves around the fiber and the resonator mode, see for example in this post. Of course this coupling is also bidirectional; if light can go in, it can also go out. However, by positioning the fiber further away from the resonator one can arbitrarily reduce the coupling between the resonator and fiber modes and achieve relatively high storage times. Here is a paper, in which they trapped the light for $17 \, \text{µs}$, that is $8 \cdot 10^9$ optical cycles!

Besides coupling to the fiber there are other loss mechanisms: Roughness of the side walls acts like the step in your proposed design. Furthermore, there can be impurities in the transparent material which scatter some of the light into different directions, so that it eventually couples out. Besides these fabrication issues, there is a fundamental limit on the distance light can propagate in a medium: absorption. Every material with a refractive index $n > 1$ must also absorb light (see Kramers-Kronig relation).

Edit:
The question came up, if the situation is different when the light is guided in a fiber instead of a solid resonator. This is not the case, because the reason for the outcoupling is the increase in angle of incidence after passing the step. When people make the statement that "the reflection angles are limited" inside a fiber, they mean that light incident on the side walls under steeper angles couples out partially, such that it can't be guided over long distances inside the fiber. It does not mean that the fiber somehow violates the angle of incidence = angle of reflection law, making the angle of reflection shallower. The following sketch shows a ray in a circular fiber. Consistently keeping the reflection angle equal to the incidence angle, the ray hits the inner and outer side wall in a periodic pattern.