# How are photons “consumed”?

I have very little background in physics, so I apologize if this question is painfully naive.

Consider the following thought experiment: an observer is in a closed room whose walls, floor, and ceiling are made entirely of mirrors, with a single light source in the middle of the room. When the light is on, the observer can see many copies of his reflection all over the place.

Suddenly, the light source turns itself off. Intuitively, I would expect the observer to "instantly" see darkness. However, I can't figure out why that is the case under the "particle" interpretation of light. There are obviously lots of photons already in the room from before. Furthermore, we know that they don't get "consumed" when they hit a wall, because otherwise the observer wouldn't see so many reflections of himself. Basically, when the light goes off, what happens to the photons already in the room?

I suspect the answer goes something like this: the photons in the room lose a little bit of energy every time they bounce off a mirror, but it's so minuscule that we can still see more reflections than our eye can resolve anyway. When the light goes off, however, it takes them a very small fraction of a second to bounce around the room enough times to diffuse completely, which our eye cannot detect.

Is that about right? If we had a theoretical "perfect reflector", would the light remain trapped in the room forever? If we had instruments that could measure such things very finely, would it take (slightly) longer for the light to go out in a room made of mirrors as opposed to a room made of, say, black cloth?

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this is an interesting thought experiment... – Timtam Aug 14 '11 at 1:11

When being reflected by a mirror, the photons do not lose "a tiny bit" of energy. Either they are reflected unchanged, or they are completely absorbed. A good mirror will reflect most of the photons, but will absorb a small fraction of them as well, say $0.1\%$ of them.

That is: Your photons don't lose energy over time; what happens is that the room loses photons over time: For each time a photon hits a wall, there is some probability $p$ that it will get absorbed ("consumed"). The chance that it doesn't get consumed after $N$ hits is $(1-p)^N$. Since the photons are very fast, they'll bounce off the walls very often in a short amount of time, so $N$ becomes really large really quick, and then $(1-p)^N$ becomes really small pretty fast, so after a short amount of time, all photons have been consumed with very high probability.

An important property of photons, that might not be entirely intuitive when coming from a "wave" background: The energy of an individual photon is determined entirely by the frequency of the light. Blue photons have higher energy than red photons. The intensity of light is determined not by the energy of your photons, but by their number.

If the photons would lose energy each time they bounce off a mirror, the reflections would change their color gradually, so that eventually blue light becomes red, then infrared etc. That doesn't happen: The mirrors don't change the color of the light. They only swallow some of the photons, i.e. they reduce the intensity.

With perfect mirrors, you could indeed expect to never lose any photons. But since anyone looking at the photons also absorbs then, the room would still get dark eventually. Unless there's nobody in there.

To be overly pedantic: Unless you keep the walls at zero temperature, you will always have some photons in the room as blackbody radiation. At "normal" temperatures, these photons are mostly in the infrared range, but if you make it really hot, the walls will start glowing.

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"When being reflected by a mirror, the photons do not lose "a tiny bit" of energy." I always assumed you'd have some amount of redshifting due to momentum transfer to the mirror? – TheSheepMan Aug 14 '11 at 2:30
That's a very insightful answer that has given me a lot to think about. Thanks for making it so accessible :) – Adrian Petrescu Aug 14 '11 at 3:43
@TheSheepMan: not that I know of. The photon can transfer momentum without losing any energy (i.e. without being redshifted) by simply changing direction. But that would make an interesting question to ask, if you can think of a reason why such redshifting might occur. – David Z Aug 14 '11 at 3:47
@David: sure, just looking at the photon, there's no reason to expect red-shifting. But transferring momentum to the mirror should also change the mirror's kinetic energy, and conservation of energy means the photons energy (and hence momentum) must change. At equilibrium the mirror is vibrating, of course, so the transfer of energy can be either sign. – wnoise Aug 14 '11 at 5:16
@wnoise: the momentum transfer could be compensated by a photon collision on the other side of the box, so that there's no net change in energy. But it is interesting to think about to what extent that actually happens. If anyone cares to discuss this further I think it'd be a good idea to move to the chat room. – David Z Aug 14 '11 at 5:40