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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?

Qmechanic
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Adrian Petrescu
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1 Answers1

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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.

Lagerbaer
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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 02:30
  • 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 03:43
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    @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 03:47
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    @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 05:16
  • "At equilibrium the mirror is vibrating, of course, so the transfer of energy can be either sign." - So would one expect a probability distribution for the frequency of individual photons ricocheting off the walls of the cage? Some sort of Gaussian peaked at the initial energy? – TheSheepMan Aug 14 '11 at 05:26
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    @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 05:40
  • Isn't that the idea behind Raman spectroscopy: A phonon hitting a mirror might create or destroy a phonon, and then energy conservation leads to a red/blueshift – Lagerbaer Aug 14 '11 at 17:19
  • "The energy of an individual photon is determined entirely by the frequency of the light. Blue photons have higher energy than red photons" The last sentence is only an approximation. A single photon does not have a unique frequency or wavelength, hence it does not have unique color. Quantities such as wavelength are sharply defined only for electromagnetic waves (the limit of so many photons). Photons have frequency distribution. – Revo Aug 17 '11 at 20:48
  • @David Can't photons change their energies by scattering off the mirrors like they do in Compton scattering? – Revo Aug 17 '11 at 20:50
  • @Revo: not if it's a perfect reflector. Also, re: 2 comments up, "blue photons have higher energy than red photons" is strictly true because the blue portion of the spectrum is completely located at higher frequencies than the red portion. – David Z Aug 17 '11 at 21:33
  • @Revo A photon is defined as an elementary excitation of the electromagnetic field and as such has one definite energy. – Lagerbaer Aug 17 '11 at 22:43
  • @David Correct, I was misunderstood. I only meant that one cannot assign a unique wavelength to photons unless we are in the classical limit. – Revo Aug 18 '11 at 00:07
  • @Lagerbaer If a photon has a definite energy, does it have definite wave length? – Revo Aug 18 '11 at 00:18
  • @Revo Yes. Which also makes sense if you think about boundary conditions in an electromagnetic cavity. – Lagerbaer Aug 18 '11 at 17:07
  • @Lagerbaer If a photon has a definite wavelength then there must be something I am missing here. Because one cannot define a unique wavelength to a single photon because wavelength is a property of electromagnetic waves. You can define wavelength for photons only operationally, namely, having bunch of them and plot a distribution (Reference: MIT series QM, French and Taylor, page 254 section 6.8 titled "statistical and classical properties of light"). If the distribution is well peaked around some number one can call that number the wavelength of the photons. – Revo Aug 18 '11 at 17:23
  • The Mössbauer effect shows precisely that absorbed and emitted photons do give up some energy to the recoil of individual atoms (and much less to the recoil of macroscopic structures). The problem should probably be analyzed in that way. But then again, I think that reflection in this sense is generally a collective (i.e. coherent) behavior. – dmckee --- ex-moderator kitten Sep 12 '13 at 03:17