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For what longest possible time it was possible to hold light in a closed volume with mirrored walls?

I would be most interested for results with empty volume but results with solid-state volume may be also interesting.

Qmechanic
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Anixx
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5 Answers5

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The set-up you are describing is essentially an optical cavity, and you are asking what is the longest lifetime which has been achieved in such a cavity.

In this paper (also described here), S. Kuhr et. al. describe a supraconducting cavity with a 130 ms lifetime. It is essentially 2 curved mirrors face to face. It works in microwaves (51 GHz), which has a long wavelength (6 mm), and this makes the manufacturing of mirror smooth at this scale much easier. This cavity is one of the key-element of this lab CQED (Cavity Quantum Electrodynamics) experiments.

I do not know if it is the best one, but I'm pretty sure it's more or less the state of the art in this domain.

Frédéric Grosshans
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    Wait, do the 2 curved mirrors manage to create stable paths for the ray? As in, given a slight deviation, it will smooth itself out and remain in the cavity 1000 reflections later. My intuition is that there are no optics that will do that. – Alan Rominger Feb 22 '12 at 15:27
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    Your intuition is wrong, pretty much anything other than a planar-planar cavity can be made stable. –  Feb 22 '12 at 15:36
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    Well not anything other than a planar-planar cavity, but yes, most optical cavities are indeed stable. Sometimes a cavity may be designed intentionally to be slightly unstable for some high power laser applications, but that is uncommon. Essentially any bulk medium laser will have a stable cavity with at least one of the mirrors curved. – Colin K Feb 22 '12 at 19:52
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    The stability conditions are described here http://en.wikipedia.org/wiki/Optical_cavity#Stability and computed here http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis#Resonator_stability . – Frédéric Grosshans Feb 23 '12 at 08:51
  • Superb book on the subject: Lasers by Anthony Siegman. – nibot Feb 27 '12 at 11:33
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    Just an update : the cavity described in this answer is a key element of the experiments of Serge Haroche (One of the two 2012 recipients of the Nobel prize in physics) – Frédéric Grosshans Jun 03 '13 at 11:24
  • What about white surfaces? Can they hold light so long? – Anixx Jan 18 '14 at 05:51
  • @FrédéricGrosshans how does that work given the laws of physics are time reversible? – John Dvorak Aug 22 '16 at 08:48
  • @JanDvorak : I don't see the link with time reversibility. Can you elaborate? – Frédéric Grosshans Aug 23 '16 at 11:02
  • @FrédéricGrosshans let's say you have a set of mirrors that causes a beam of light converge to an arbitrarily narrow beam. Now, the same set of mirrors should still converge the beam even if you reverse the beam (unless this is where I'm wrong?). But retracing any ray of light backwards should lead to its source, which is outside the region the beam is converged into. – John Dvorak Aug 23 '16 at 11:11
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    @JanDvorak: I see two problems with your reasoning 1. keeping the light in a cavity is about keeping the beam width constant, not about converging or diverging. 2. Converging = going from wide to narrow. If you reverse the beam, it goes from narrow to wide : it diverges – Frédéric Grosshans Aug 24 '16 at 18:10
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My answer is not much different from the other ones, but it is based on my personal experience. For several years I have been using Cavity ring-down spectrosopy(CRDS) - the method to measure very weak light absorption that is based on building such optical cavity and placing an absorbing sample inside. The light pulse will pass through the sample thousands of times, thus greatly enhancing its absorption. For highly reflective mirrors the formula given by Colin K can be approximated as $$ \tau_0=\frac{L}{c\left ( 1-R \right )} $$ State-of-the art mirrors for CRDS in the visible range can have reflectivity up to 99.999% (link) resulting in a decay time of $167 \mu s$ for $L=0.5m$. During this time the light travels 50000 km which means that it is reflected 100000 times.

PS: This experiment is done in vacuum (below 1 mbar). At atmospheric pressure the decay time is reduced to a couple of microseconds due to Rayleigh scattering (or absorption by dust particles).

gigacyan
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The lifetime of a photon in a resonant cavity is pretty trivial to compute, given the cavity length, internal losses, and mirror reflectivity. Switching momentarily to a wave description, we will let $L$ be the cavity length, $R_1$ and $R_2$ be the reflectivity of mirrors 1 and 3 respectively, and $T_i$ be the loss in the cavity medium. Clearly the intensity of a pulse of light in the cavity will exponentially decay, and the lifetime $\tau_c$ (defined by the $1 \over e$ threshold) can trivially be computed to be $$ \tau_c = -\frac{2 L}{c \ln{\left[R_1 R_2 (1-T_i)^2\right]}} $$

For a one meter cavity with no internal loss and 99% reflective mirrors, this gives a lifetime of roughly 330 ns.

There are much longer cavities, and the reflectivity of dielectric mirrors can have quite a few more "9"s tacked on. For example, the LIGO cavity is something on the order of a kilometer, and if we pretend that the mirrors are 99.999% reflective (that's three"9"s after the decimal place) 1 we get a lifetime of 0.333 seconds (wow).

Lifetime increases rapidly with mirror reflectivity once you get above 99%, so you'll see that if you repeat that calculation with $R=99.9999\%$, you get $\tau_c = 3.333$ seconds. That's an absurdly long time, but of course that fourth "9" after the decimal place is really starting to get unrealistic as well.

1: This is very much an order-of-magnitude guess. I'm not sure of the exact length of the LIGO cavity, and in fact the mirrors are not highly reflective because they are doing a trick called "Power Recycling," which ends up giving them a longer photon lifetime anyway. However, 99.999% is an impressive, although NOT an unrealistic number for a modern high quality dielectric mirror.

Colin K
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    The initial LIGO arm cavities were about 4 km in length (3995 m). The individual arm cavities had mirrors with power transmissivities T1=3e-2 and T2=10e-6, and a storage time of about 1/(85 Hz). The combined optical plant with power recycling had a storage time of about 1/(3 Hz). – nibot Feb 27 '12 at 12:54
  • What about perfect mirrors? (http://www.extremetech.com/computing/162322-mit-creates-the-first-perfect-mirror) – trans Oct 29 '15 at 23:42
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The photon lifetime in the LIGO interferometers is about 1 s. They use coupled cavities to get the storage time so high; the Michelson arms are composed of Fabry-Perot cavities and an additional mirror at the bright port of the Michelson (the power recycling mirror) forms a coupled cavity with them. The wavelength of the light is 1064 nm.

All of these answers are only referring to the statistical storage of photons in an optical cavity. You couldn't, for instance, select a single photon, store it for some time, and then use it later. This is the type of storage you would need for quantum computing with photons.

A schematic of the Advanced LIGO interferometers http://www.bssaonline.org/content/99/2B/980/F2.large.jpg

Chris Mueller
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    Clever answer! I'm sure that the OP doesn't have something like this in mind, but it's very interesting and moreover instantly seen to be plausible: 1s lifetime implies a cavity $Q$-factor of 37500, which is well within the advanced optical technologies used in LIGO. – Selene Routley May 23 '15 at 10:23
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This should be a comment but it will be too long.

edit: "I had said :I do not think anybody has designed and done such an experiment, so you could consider it a challenge and do it yourself." I was wrong in that, as an experiment is described by Frederic..

One can make some estimates: reflectivity will play a role. One can make highly reflective mirrors, but still some absorption will take place and the velocity of light is high. Even 10^-5 loss per hit will rapidly increase absorption in a second. Speed of light is 3*10^10cm/sec so in a 10cm box , thats a lot of hits per second.

In addition each hit by reversing the direction of the photon will transfer some momentum to the atoms/crystal lattice of the surface of the box and thus there will be a wavelength increase which will finally take it to infrared and invisibility. This needs numbers and calculations, but again, due to the high number of hits per second I expect that the time will be very small by which the initial photons will be completely absorbed by what is a perfect black box.

anna v
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    No, momentum transfer will not affect anything because if a photon transfers some momentum to the box at one wall, it will earn the same momentum when it would get reflected from the opposite wall. Total momentum transferred to the box will be zero (unlike pressure). Due to the same reason a gas in a box does not cool by itself. – Anixx Feb 22 '12 at 07:55
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    In cavity ringdown spectroscopy this principle is used for highly sensitive absorption measurements. The decay is exponential (so it is not strictly "light trapping") with characteristic time of about $100 \mu s$ for visible light. – gigacyan Feb 22 '12 at 09:08
  • @Anixx hmm.If lamda changes nu changes and this means energy is transferred to the box (E=h*nu). This cannot be given back from the opposite wall. Energy conservation is as strong as momentum conservation. – anna v Feb 22 '12 at 13:14
  • @gigacyan You could make an answer to the question then, if you have a link or so and explain how the measurement was done. – anna v Feb 22 '12 at 13:18
  • Lambda increases when reflecting from one wall and decreases when reflecting from the other. You forgot that momentum is a vector and not scalar value. – Anixx Feb 22 '12 at 14:03
  • @Anixx and you forget that energy is a scalar value. The energy lost on one wall cannot be gained from the other. – anna v Feb 22 '12 at 15:08
  • people seem to have a strange view of energy conservation, if the negative votes agree with Annix that the photon's energy is unaffected when its lamda increases. I would like to see a justification of that, because I firmly believe that energy conservation holds for photons too. When it bounces on the opposite wall it will change direction but lose another delta(E) and further increase in wavelength. – anna v Feb 22 '12 at 15:11
  • Hmm, are you sure that the wavelength changes? I have not thought this through completely but my intuition tells me that the wavelength should change if the mirror moves. The amplitude of the wave changes, but why it's frequency? – Alexander Feb 22 '12 at 15:22
  • anna v why do you think energy conservation means that the photon always gives energy to the wall? The energy is returned to the photon when it bounces back off the opposite wall. –  Feb 22 '12 at 15:44
  • @Alexander Think of Feynman diagrams, some will be elastic, then the wavelength will not change, some will be inelastic, then the wavelength will change , and it is the absorption part of 1_ .99999. in the collective wave mode. There are so many hits in a second that the wavelength will be getting larger and larger for those that correspond to the inelastic crossection. Inelastic means energy is lost to the atom/crystal/whatever. That cannot be gained back . – anna v Feb 22 '12 at 16:22
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    anna v please give an example of any real system in which this effect is not dwarfed by absorption. Or edit to make it clear you are talking about energy loss to the walls, not to macroscopic motion of the box. –  Feb 22 '12 at 16:37
  • @zephyr No. When it is an inelastic scatter the energy is lost, it is not a tennis game. The 1-.99999 number for absorption in the quote I made means that on average 1 in 10^5 scatters is inelastic , and there are another 10^4 scatters per second. – anna v Feb 22 '12 at 16:39