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Take an shell or bubble. The bubble is strong enough to maintain it's static sphere shape, except when a photon bounces off the inner surface.

A photon fires from inside the bubble.

The inner surface gets pelted by this photon, and the inner surface bounces it back similar to a trampoline bounce. The effect of the bounce on the inner surface pushes the photon back in the same velocity it arrived. The entire inside of the bubble is like this, but strong enough to stay it's static sphere shape. What happens? Does the photon ever escape?

Does the photon endlessly stay a photon in the bubble at this rate? CURIOUS!

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When "a photon" is reflected in the manner you say, this is actually an absorption followed by an emission of a new photon by the matter of the reflector. There are many classes of this absorption / emission of new photon process. What you are asking, therefore, is whether there is a lossless mechanism of this kind, where the re-emitted photon has exactly the same energy as the incident / absorbed one. Some of the classes are:

  1. Elastic interaction with an electron: total energy is conserved, some momentum / angular momentum may be transferred to the interacting medium, although this latter is negligible if the electron is an outer shell orbital electron so that the whole "bubble" is overwhelmingly massive compared with the electron;

  2. Fluorescence, where the emitted photon has a longer wavelength and some energy is tranferred to the fluorophore: this is not what you want;

Certain multilayer dielectric mirrors can be of very low loss: there can be many thousands of bounces before the photon is absorbed. So, you are talking about a spherical electromagnetic resonantor. The $Q$ or "quality factor" is $1/(1-p$, where $p$ is the probability that a photon will be re-emitted by the reflector. $Q$-factors of tens or hundreds of thousands can nowadays be made: see here for example.

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The basic fact to answer this question is that a photon, though an elementary particle, never changes velocity. It is $c$, the velocity of light. Its energy is equal to $h\times \nu$, and if it loses energy it will change frequency to a lower frequency.

In rough terms you are talking of total internal reflection. In reality there is always a probability that in "bouncing", (interacting with the molecules of the bubble) the photon will lose energy and be degraded. If it is a vacuum it will take a long time to lose enough energy to be indistinguishable from the vacuum background photons ( infrared from the black body radiation of the walls of the bubble). If in air it will also lose energy interacting with air molecules and again be indistinguishable from the background infrared radiation.

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