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First off, I confess I'm no physicist, but I have been asking people with a more extensive knowledge this one question, without a definitive answer so far.
Basically, I'm playing around with the idea of photons having mass as this part of the wiki page shows, it's not an entirely new concept... Especially the latest (mangled) sentence is of interest to me.

I have been reading up about light, the duality of it, photons and it being the only known gauge boson (massless thingamajig) and what have you. As a result, I'm even more confused, so I thought I'd ask my question here.

Setup:

Suppose we took two perfect mirrors, stretching out into infinity, perfectly parallel, facing each other. If I were to shoot a photon onto either one of these mirrors at a 45deg. angle, what would happen.

Hypotheses:

As far as I can know (or guess, or can imagine) either one of three things can happen:

  • The photon just bounces back and forth into infinity at that leisurely pace of $c$.
  • Calling on the particle part of light's duality: Every action causes an equal and opposite reaction. Upon colliding with the surface of the mirrors, energy is needed for the photon to change direction. As everything strides towards entropy, I'd assume there is some heat being released (photon's energy onto the mirror)?
    If that's the case, at some point the photon's electromagnetic "charge", ie energy-reserves should run out. What do I end up with? Slightly warmer mirrors and a massless, empty shell of a photon at the end? What is a photon that no longer has any energy anyway? Is that the famous dark-matter... or am I going all too scifi-crazy now? Because somewhere I did read that light, being massless, obviously has no rest-matter either, nor does it have an electric charge of its own. That causes me to think of a photon as some sort of carrier, an empty satchel and because it's not exactly huge, it can but contain a finite amount of energy (I think).
  • Last thing I can think of: because of my photon's bouncing, and my being at a terrible loss trying to grasp the formula's and theories about light's physical properties I've gotten the (perhaps silly) idea that the constant changing of the direction of propagation could affect the wavelength, essentially generating something more like gamma-rays. Again, I don't know what this entails for my mirror-setup, but when news breaks of an impending nuclear disaster, I don't think a mirror completely deflects gamma rays. In other words, I don't even think it unlikely if somebody told me that photon would just bugger off.

I hope someone can make sense of the bizarre meanders of a non-physicist's mind, but I would like to know the answer to a question I came up with about 10 years ago.

So far I've gotten the answers:

  • Oh, I'd have to check on that one.
  • Of course, they talk about the duality of light, but light is, essentially pure energy. they've developed this dual-character as a working model. Much like everything "'t is but a theory" (I particularly disliked this answer for some reason)
  • Do you know how they spot a black hole? (I replied: No) Because there is light, but none around it. All light is drawn to the black hole. (this was followed by an awkward silence, and a smug nod. Which met with a confused and monkey like gaze from my part)

Any more confusing ideas are always welcome.

Edit/recap:
Thanks to all of you for the info. In response to the comments, the kernel of the question is this: If I were able to follow the afore mentioned proton in this setup, what changes, if any, will I see along the line? Heat being generated? The photon "disintegrating" or dissipating, nothing (just endlessly bouncing back and forth...?

Reading the wiki on Total Internal Reflection, I noticed that this occurs with soundwaves, too. I immediately thought of that horrid screeching feedback noise you can get if you hold a mic to a speaker. I guess I sort of translated that phenomenon into the photon changing wavelengths.
Funny, but true: I remember as a child asking my father if you were able to create an infinite broadcast of sorts using two transmitters and two receivers playing a sound back and forth to them. In some way or another, I've always wondered about stuff like this as it turns out...

Mirror mass:
I suppose the mirrors would have to have infinite mass for them to stretch out into infinity. Though after some more checking, that complicates things considering $E = pc$. I've added that to my many light-related bookmarks, and I'll get back to you on that.

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    $\begingroup$ +1 because I think the infinite reflection question is a neat concept. Your question is all over the place though and I think you should edit it down to just the kernel of what you want to know. $\endgroup$
    – Brandon Enright
    May 2, 2013 at 23:52
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    $\begingroup$ Your infinite parallel mirrors setup is very similar to fiber optics and total internal reflection probably has a lot to say about the answer: en.wikipedia.org/wiki/Total_internal_reflection $\endgroup$
    – Brandon Enright
    May 2, 2013 at 23:53
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    $\begingroup$ See also: physics.stackexchange.com/questions/57766/… $\endgroup$
    – dmckee --- ex-moderator kitten
    May 3, 2013 at 0:03
  • $\begingroup$ Do the mirrors have infinite mass? $\endgroup$
    – joshphysics
    May 3, 2013 at 0:05
  • $\begingroup$ @dmckee: Thanks for the link, I now know I have about a thousand more wiki pages to read/decipher ;) $\endgroup$
    – Elias Van Ootegem
    May 3, 2013 at 10:21

2 Answers 2

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First, of course there's no perfect mirror. But let's assume there was one.

Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever.

But what if it does lose energy with each bounce? Well, your two mirrors form a cavity and if we appeal to the wave-aspect of light, only waves with wavelengths that "fit" into the cavity are allowed, so there'd be a minimum allowed wavelength, $\lambda_0$ with $\lambda_0 = L/2$ where $L$ is the distance between your mirrors and since energy and wavelength of a photon are intimately related, this means that the photon in your cavity has a minimum energy below which it cannot fall.

If you add the concept of heat / temperature / entropy to the mix, what you will get is that the walls (mirrors) are in thermal equilibrium with the photons in your cavity: Some of the energy is then stored in the walls and some in the photons. In fact, considering the situation of taking a cavity at some temperature and looking at the nature of the light that comes out of it (if you poke a tiny hole in it) is one of the phenomena that led to the discovery of quantum physics.

Some misconceptions: A photon has no "electromagnetic charge", it is a massless, chargeless particle. Now what if its energy "runs out"? Then it just ceases to exist. There is no photon without energy.

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  • $\begingroup$ qualify "ceases to exist" by maybe :it is so low in the ifra red spctrum that it is absorbed in a vibrational qm transition of a molecule. from momentum conservation it will be losing energy at each bounce, lowering wavelength, ulsess as Josh is asking the mirrors have infinite mass? $\endgroup$
    – anna v
    May 3, 2013 at 4:15
  • $\begingroup$ This might sound stupid, but I know a photon is massless and has no charge, but at the same time it's said to be "an elementary particle, the quantum of light and all other forms of electromagnetic radiation, and the force carrier for the electromagnetic force". So it has no charge, but carries energy... is energy even... what? who? how? *_- $\endgroup$
    – Elias Van Ootegem
    May 3, 2013 at 10:25
  • $\begingroup$ Also: if a photon ceases to exist if the energy "runs out", shouldn't you be left with an empty gauge boson particle? $\endgroup$
    – Elias Van Ootegem
    May 3, 2013 at 10:27
  • $\begingroup$ No, a photon cannot "run out" of energy independent of its other properties. Technically, if a photon "loses" energy, what really happens is that a photon of some initial energy $E_1$ is absorbed/destroyed and a new photon of some new energy $E_2$ is emitted. "Running out of energy" then just means that a new photon is never emitted. That can happen if a photon is absorbed by a crystal and the energy then re-emitted as lattice vibrations instead of a new photon. $\endgroup$
    – Lagerbaer
    May 3, 2013 at 15:10
  • $\begingroup$ @Lagerbaer: I'm sorry to be this thick, I got hung up on that gauge boson being matter, and matter should be conserved at all time... I have found out, now, that matter is not perfectly conserved, though I've also learned that -even though they're massless- photons still add mass. Anyway, I've got enough material out of this to study this matter (no pun intended) for a couple of days/weeks... meanwhile: Thanks for the info! $\endgroup$
    – Elias Van Ootegem
    May 3, 2013 at 16:30
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Pose the same question with slightly different parameters. Instead of parallel mirrors traveling to infinity, construct a sphere where the inside is reflective (read:mirror). Then pull a vacuum on the sphere so there is nothing to slow the photon. Then inject the photon at any angle and it will reflect forever at the speed of light and doing so allow the sphere to become incandescent (without the heat of course).

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  • $\begingroup$ Why didn't I think of a sphere... would've made things so much easier. So a perfect sphere, with a perfect reflective surface on the inside would have the photon bouncing around for ever. There is no loss of energy caused by the photon, effectively, colliding with the sphere. If we had a device that was able to measure the light inside the sphere, without breaking the necessarily closed system, we'd see a constant radiant flux, correct? $\endgroup$
    – Elias Van Ootegem
    Dec 30, 2016 at 17:38

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