Walk into one of those bathrooms in a fancy hotel, where there are mirrors on all the walls. Notice the multiple reflection of your face rapidly diminishes in intensity and it gets dark. This shows that the multiple reflections rapidly absorb.

A good mirror absorbs around 5%. Therefore after 20 bounces it is all gone.

If your box is one meter in size, 20 bounces implies a travel distance of 20 meters to get extinction. Divide that by the speed of light and your extinction time is 66 nanoseconds.

That the intensity goes down to zero virtually instantaneously after switching off says that the losses are very large for practically excellent mirrors. But even if the mirror losses would be small enough that enclosed photons had a life time of a few seconds, with increasing photon density, the photon injection channel would let photons also flow outwards. Equilibrium would ocurr if photon influx = photon outflux.

Can one stop photon outflux? Or at least reduce? Has it ever been noticed that saturation time is at least of a few seconds long?

And I think there are hardly many photon - photon interactions. So only significant losses you seem to suggest arises from the injection point. But then why photons do not keep reflecting in a closed box? I mean not trapped?

Walk into one of those bathrooms in a fancy hotel, where there are mirrors on all the walls. Notice the multiple reflection of your face rapidly diminishes in intensity and it gets dark. This shows that the multiple reflections rapidly absorb.

A good mirror absorbs around 5%. Therefore after 20 bounces it is all gone.

If your box is one meter in size, 20 bounces implies a travel distance of 20 meters to get extinction. Divide that by the speed of light and your extinction time is 66 nanoseconds.

Derek, so you need to slow down the photons to extremely slow speeds if at all you wish to trap. Or the dimensions of box need to be astronomical. That means the light from the sun to earth - well does that accumulate provided it gets bounded?

The nice book 'Exploring the quantum' by Haroche/Raimond rapports for superconducting mirrors for 20 GHz microwave photons 'very long Tc's [i.e. photon life times, my addition] up to a few tenth of a second' (p. 265). Photon loss by photon-photon interaction definitivly plays no role. With low tech mirrors, as Derek pointed out, you can't hold photons for longer than a few tenth of of a millisecond.

Dear RG team:

the chrology of contributions seems to be distorted here rather arbitrarily. Can you correct?

Dear Sir,

Your question puzzles me. How can the number of photons be steady in the box and turn to zero when you switch off the source if their number did not grow up when you turned on the source ?

I think that your question is relevant to the localisation of photons. The effect is based on Anderson's localisation in a disordered system where a photon always experiences two scattering pathways with a dephasing of Pi/2 that do interfere such as to slow down the photons.

See for example:

- Diederik S. Wiersma, Paolo Bartolini, Ad Lagendijk & Roberto Righini 1997). "Localization of light in a disordered medium". Nature 390 (6661): 671–673.

- M Störzer, P Gross, CM Aegerter, G Maret (2006). "Observation of the critical regime near Anderson localization of light". Phys. Rev. Lett. 96 (6)

I think there have also been experiments in Bose condensates, but could not find the reference.

Before I look for Roberto Righini's paper (my old friend from Florence) as suggested by Charles, I once again go back to Derek. Classically some 5% loss on each reflection can reduce successive intensities, but as for quantum photons what does it mean? If each photon suffers 5% loss each is gone. Because there is either a photon of some wavelength or none. These do not undergo inelastic reflections.

Dear V K Jindal,

you need to work through the basics. 5% loss, of course, means that for each photon there is a probability of 5% to get absorbed by the electrons in the mirror material (if it gets absorbed it is 100% destroyed, its energy and angular momentum is used to change the motion of the electrons). Further, there is 95% probability that an elastic collision happens at the mirror's surface.

VK:

It seem to me that what you're basically talking about is the Q (quality factor, or energy stored over energy dissipated) of a cavity. No real material is 100% reflective, so some intensity is lost with each reflection. Good dielectric mirrors that function in the visible spectrum can reflect more than 99% of the incident light, but that's still not 100%.

Cavities that have internal geometries commensurate with one half wavelength will show the highest Q values, even with lossy walls. In other words, the cavity is resonating with the excitation photons and energy storage will be maximized. Energy storage cannot go to infinity because as the energy density inside the cavity increases, so do the total losses. At some point, energy input equals energy loss and the energy storage is at its maximum.

Quantum mechanically, when a photon strikes a surface, it can either be reflected or absorbed. It cannot be partially reflected or partially absorbed. So, what does it mean to have a 99% reflectivity? In a quantum mechanical sense, reflectivity is a PROBABILITY that a photon will be reflected or not. Thus, a 99% reflectivity means that 99 out of one hundred photons striking the surface will rebound, 1% will be absorbed.

As the total number of photons increases inside the cavity (i.e.: the energy density is increasing), so also does the total number of photons absorbed increase. (e.g.: 1% of a bigger number is another bigger number). Eventually, the number of photons injected into the cavity will match the number absorbed, and energy storage (number of photons) reaches a maximum. Assuming you have a finite energy input of course.

Now, if as you suggest, we have a fictional "perfectly reflecting" material, then the energy density will build without limit even with a finite energy source. But this would be true only if the coupling mechanism of the photons into the cavity is also "perfectly unidirectional". That is, the coupling mechanism will dump photons into the cavity, but allow none to "leak" out backwards. Alas, there's no such thing as a perfectly reflecting surface, nor is there anything like a perfectly unidirectional coupler. And, of course no material can sustain infinite energy densities. Eventually, every material breaks down. Every atom in the periodic table has a finite ionization energy.

I hope this helps.

@Patrick,

your last sentence brings to mind the latin saying "si takuisses philosophus mansisses": we probably have in mind photons of (nearly) the same wavelenght. Choose this wavelenght long enough that the photon energy is lower than all (the finitely many) ionisation energies of elements. Then you may pump in as many photons as you like and no ionisation event will happen. Of course, radiation pressure and unharmonic effects will ultimately destroy the mirrors.

When you're at resonance, a node will occur at the walls and the field strength at the walls will be minimized. That's one reason the Q is highest there; resistive losses are minimized.

When you're off resonance, there are no nodes at the walls and field strength increases there, increasing losses. Q then effectively decreases. So, the amount of energy you can store in the cavity will go down when you move off resonance.

Ideally reflecting cavity walls may be considered theoretically, anyway. It seems, that with such walls, we do not need to discuss the temperature of the walls, and we may pump the cavity, i.e. with black body radiation spectrum with any energy density and any temperature (thought experiment). What should happened in case the walls of cavity run away from each other with high speed (up to c)? If size of cavity is very big, than two possibilities occur; Each photon increase its wave-length, decreasing the black body radiation temperature - which means each photon is extended through whole cavity (as a sound standing wave in a string getting longer). Or the photons far from the walls, keeps their spectrum, for longer time, due to the fact, that decrease of energy - increase of photon length, happens in the scattering from moving mirror. If walls move away with the speed c, than the spectrum temperature should not change at all, only an energy density. The time dependence of the temperature of microwave background radiation suggests the first answer, however, it seems to implicate that each background radiation photon quantum field is extended to whole universe (accordingly to quantum field theory?), and may be registered as a real photon at any place in the universe. In such case there is misleading to say that every place universe (i.e. our galaxy) is filled in with running everywhere photons which were send away some 13,4 10^9 years ago at a distance 13,4 10^9 light-years away.

@Charles

your answers puzzles me. When photons move in an optical cavity, where do you see disordered systems playing a role?

@Ulrich. There is no disorder in a cavity, I am writing about trying to keep photons in a trap which is the original question.

@Charles.

I think that you misinterpreted the question. I'm rather sure that the author of the question had nothing more sophisticated in mind than a mirror box with some means to bring photons in. Perhaps he tells us.

@Ulrich. You are probably right. Anyhow it remains the fact that if the number of photons diminishes when turning off the source of light it must grow when turning on. Don't you think?

@Charles. It must grow until the point where the losses equal the input, and this is after less than a millisecond unless one has exceptionally reflective mirrors. Right?

@Ulrich.Right, so the question is void then.

It is not possible to create an entrance aperture which is not also an exit aperture. See: http://labsphere.com/uploads/technical-guides/a-guide-to-integrating-sphere-theory-and-applications.pdf. Set the reflectivity to unity. The intensity does increase with time until it reaches steady state when the average intensity at the walls equals the intensity at the aperture.

@Patrick, we know this statistical interpretation of meaning of 99% reflectivity etc. taken to mean that 1% of photons will be absorbed. And when we talk of a single photon, we might interpret (and we do that way) in 100 reflections, you might consider the photon absorbed. And we still wonder how the photon carries the information that it has made around 99 reflections already and now it will be absorbed. It may be interpreted in a 'mean number' but the fact is this puzzles.

@V K Jindal,

what puzzles you here is not even the quantum nature of the process. The logic of Russian roulette is not different.

@Ulrich Mutze

Of course we are so used to all this. I thought may be someone will enlighten in simpler and convincing language..

VK:

No, that's not how statistics works. There's no memory or anything like that. Each time the photon strikes the wall, there's a non-unitary probability it will be reflected; that's all.

For example, when you flip a coin, there's a 50% probability it will land heads up. That doesn't mean if you flip the coin twice you get a 100% probability of getting a heads up. Probabilities don't add in that way.

Of course we know that. The question is why the statistics give us results for large numbers as if the events which are independent were correlated.

Dear V K,

For your information some Leighton has been stopped for one minute in a box for the first time a monté ago. See: http://link.aps.org/doi/10.1103/PhysRevLett.111.033601

Sort, some light has been stopped .... for the first time a month ago.

In a laser cavity (a one dimensional example of a photon trap) the gain medium injects photons and the intensity does build up. Mirrors can have losses much less than 1% and give a cavity Q of many hundreds. Eventually the intensity is high enough that the loss rate equals the source rate. Ideally one arranges this to occur when the gain of the lasing medium starts to drop and the cavity is then dumped in some manner to release the light.

Stopped light is not light that has been stopped. It is light that is generated in a stopped state. The phase velocity is high but the group velocity is zero. The medium must have significant dispersion to permit this.

Charles Hirlimann means trapped not stopped.

Stopped and trapped are a bit ambiguous. As I understand the situation the light starts off propagating through the crystal with normal phase and group velocities and then the dispersion of the crystal is modified with another laser beam which changes the dispersion relation so that the group velocity falls to zero. Whether you consider this to be trapping, stopping or whatever is down to semantics but it is clearly different from trapping in a hohlraum or cavity. Or maybe I have mis interpreted the articles on this.