# Can light continue without a light source? [duplicate]

If you have 2 mirrors over for each other placed exactly so they face each other perfectly, and then use a laser light pen as the source into one of the mirrors so it bounces to the other mirror and back again:

1. Would the laser light line continue to exist between the 2 mirrors if the source of the laser stopped?
2. If so, how long could the laser light continue to be between the mirrors without a source?
3. Would it just continue existing between the mirrors using itself as the source or is this not possible for light?

Note: It doesn't have to be only 2 mirrors. It could be any amount of mirrors if that would change the outcome.

• Funnily enough, this is similar to how laser light is produced. Commented Nov 30, 2017 at 12:37
• Possible duplicate of Are infinite reflections between two mirrors really infinite? Commented Nov 30, 2017 at 12:50
• Why would it stop? Are you asking if the photons in transit cease to exist when the light is turned off? Commented Nov 30, 2017 at 15:57
• "If I turn off a ball launcher, do the already launched balls stop bouncing around?" Same question. Commented Nov 30, 2017 at 19:12

Yes, it would continue. But not forever, for two reasons. One is that no mirror is perfect, so a bit of light is lost at each bounce. The other is that no beam of light is perfectly parallel ("collimated"), so that the light spreads out over time, and light eventually falls outside the mirror.

Spherical mirrors will help, but they will not eliminate the problem at hand, which is diffraction. A perfectly collimated beam is not possible, but as a corollary to that it is also not possible to produce a beam with a finite cross section. Some portion of the beam always falls outside of the next mirror.

Using beams that are approximately Gaussian (perfect Gaussian beams are impossible) and spherical mirrors the amount of energy that misses the next mirror is small, but not zero.

The OP's question #3 is new. I don't quite know what you mean by "itself as the source". If you can somehow get the light going, it will continue for a while after you turn off the source. The decay, as others have pointed out is (approximately) exponential as roughly the same fraction is lost on each bounce. It's approximately exponential because the beam shape will change as bits are diffracted away, so the fraction lost changes slightly from bounce to bounce.

How long the light will persist depends entirely on the quality of the set up. The surface quality of the mirrors, their surface figure, the atmosphere, the materials used, the rigidity of the mounts ... It's probably possible to estimate the longest possible persistence time taking into account effects that others have mentioned in other answers: scattering, heating, momentum transfer, ... I don't know what the "theoretical" maximum would be, or the practical limit. A quick search on Fabry-Perot interferometers finds finesse values in the $10^6$ ball park, which would imply a persistence time for a 30 cm cavity of about 1 ms, but that's a very rough estimate.

• #2 can be solved with a spherical mirror. Commented Nov 30, 2017 at 7:44
• @SomeoneSomewhere the precision needed is probably impossibly high, and the curvature needs to self-adjust as the laser hits the same mirrors again. Commented Nov 30, 2017 at 14:20
• Detail question, if you don't mind: "No mirror is perfect", does it mean "we don't know how to make a perfect mirror", or "the laws of physics as we understand them mean a perfect mirror is not possible" ? Commented Nov 30, 2017 at 15:13
• @Cristol.GdM I'd assume the latter. You'd have surface roughness at the scale of individual atoms even in an otherwise as close to perfectly smooth surface. Commented Nov 30, 2017 at 16:46
• There's also the problem of the original light source. Typically, things that generate light can also absorb them, and so whatever generated the light and/or allowed it into a "perfect mirror" enclosure would constitute an imperfection. Commented Nov 30, 2017 at 17:51

would the laser light line continue to exist between the 2 mirrors if the source of the laser stopped.

As DJohnM commented,

light travel to the moon and back, even though the laser source is off for most of the round trip.

This happens because any light source emits photons. They are indivisible packages of energy and they are traveling as long as they not get absorbed by an obstacle or - more precise - by a subatomic particle.

Would it just continue existing between the mirrors using itself as the source or is this not possible for light?

As told above, once emitted, the photon is on it own and don’t anymore care about the source.

But where is another point about the mirrors. What garyp told you in his answer is about the technical imperfection. Beside this, any reflection process of photons is accompanied at least with the transfer of a momentum to the mirror. Every photon pushes the mirror a little bit and any photon - due to energy conservation - leaves the mirror with a lower frequency. So the light dies a infrared death and the mirror gains velocity, or the temperature raises.

• "... once emitted, the photon is on it own ... " On planetary scale, gravitation might be seen as relevant as so called "gravitational red shift" exists. I find this interesting as there are questions about free electrons and bound electrons. Maybe the photon does not care about the electron it was emitted from, but the phenomenon of gravitational redshift leads to think of the surrounding mass still exerting influence, and what is the difference between: free electron/atom-bound electron and atom-bound electron/its "embedding" gravitational mass and background. Commented Nov 13, 2022 at 14:34
• @PeterBernhard About the gravitational redshift some thoughts. Commented Nov 13, 2022 at 16:05

We recieve light from far away stars even they are dead when the signal reached to earth. So, the source is not there when we actualy see it (capture the light with space telescopes).

• While not a physics answer, your point is true and easy to understand. Commented Nov 30, 2017 at 10:55
• However, it is difficult to verify the stars are actually dead. Commented Nov 30, 2017 at 13:53
• @BrianDrummond if we observe it going nova, it must have died 10^n years ago. Commented Nov 30, 2017 at 14:31
• If a tree falls in the forest with nobody around to hear it... Commented Nov 30, 2017 at 15:27

A quick calculation which I heard of sometime ago (I don't remember where, I just remember the concept).

The only datum I found (ok, I looked only for a few minutes, so forgive me) on mirrors is on wikipedia. A dielectric mirror can reflect up to 99.999%. Pretty impressive, isn't it? Only 10 parts per million are lost every "bounce".

Now, how many "bounces" are needed so that the intensity is 1 millionth of the original source? Well, math says that it is about 1.38 million bounces.

Now, imagine the mirrors are 1 meter one from the other. The light will travel 1.38 million meters before degrading to 1/1000000 of its original intensity. How long does it take? Well, 4.6 ms.

So, with an almost perfect mirror and being able to detect until 1 millionth of the original light, will you see the light slowly dimming off? Well... No. A common camera acquiring at 25fps has one frame every 40 ms.

So even with an almost perfect mirror and an almost perfect detector, I'm afraid light is so damn fast..

• I really enjoyed this answer! It really drives home the points from some of the other answers. Commented Nov 30, 2017 at 22:39
• I wouldn't call that an "almost perfect detector" at all - that's like saying that a bicycle goes "almost light speed". Why not consider an UNcommon camera, with a frame rate of 4.4 trillion frames per second (fastest camera in history), or many tens of thousands of frames per second (many commercially available examples). I'll leave it to you to convert those to milliseconds. Commented Nov 30, 2017 at 22:45
• @Beanluc with "detector" I was not speaking about the speed of acquisition, but the light it can detect. After all, 1 millionth of the original evergy is a very low light to be detected. And moreover, the faster the shutter speed, the larger the light magnitude must be (did you notice that with high speed cameras they usually use very powerful illumination systems?). This means that if you have a faster camera the sensitivity is lower, so maybe you will reach the detection threshold much faster Commented Dec 1, 2017 at 9:33

Assuming ideal experimental conditions (using a perfect, or spherical mirror, for example) the laser light wouldn't reflect forever. This is because photons have momentum, $p = h/\lambda$, which means that each reflection will transfer momentum to the mirror, i.e.

\begin{align} \Delta p \, &= \, 0 \>\> \text{(conservation of momentum)} \\ \, &= \, \Delta p_{_{photon}} \, + \, \Delta p_{_{mirror}} \\ \, &= \, \big[ - p_{_{photon}} \, - \, ( + p_{_{photon}} ) \, \big] \, + \, \big[ \>\, p_{_{mirror}}^{^{final}} \, - \> ( \> p_{_{mirror}}^{^{initial}} = 0 \, ) \>\, \big] \\\\ \, &\implies p_{_{mirror}}^{^{final}} \, = \, 2 \, p_{_{photon}} \end{align}

This transfered momentum will heat up the mirror, after which the energy will be lost as thermal radiation, and the laser light will gradually decohere.

I believe that the calculation for the lifetime of the original laser photons might depend on variables such as the wavelength of the light, $\lambda$, the size, shape, and material of the mirror, and the temperature of the room. This might be helpful:

• How much of the thermal radiation is lost? I imagine most of it would return to the inside of the sphere where it would keep being reflected.
– Fax
Commented Nov 30, 2017 at 14:37
• If the mirror is perfect, then by definition it should reflect all the light energy incident on it. So a perfect mirror shouldn't absorb energy energy from photons. Am I wrong ? Commented Nov 30, 2017 at 14:55
• @123 It appears that colnegn is not referring to the energy gained by absorbing photons rather than reflecting them, which is what the ideal mirror helps us avoid. It appears colnegn is referring to the push against the mirror when the light bounces off of it, like what happens in a solar sail; different kind of energy gain. Commented Nov 30, 2017 at 15:31
• @colnegn Let's attach some ion drives or some other device to counter the change in velocity. Or perhaps the spherical mirror is also a mirror surface on the outside, and lasers on the outside are set up to detect and counter even the smallest changes in position or velocity to keep the sphere as motionless as possible. Then what? My assumption is that the sphere would still heat up from the velocity changes, and in fact twice as fast since there is now twice as much velocity change even if the sphere seems motionless. Could you touch on this in your answer? Commented Nov 30, 2017 at 15:36

With a perfect laser and perfect mirrors yes it would and this is provable from the conservation of energy.

If you turn on a one watt laser for 1 second it will use on joule of energy. That energy must exist after the light is turned off. If it has not been absorbed by the mirror in heat then it must still exist as light.

In reality as pointed out the light will scatter and be absorbed very quickly so this effect can't be observed.

well, light is both a radiation and a particle , in this case we'll consider it as a particle from the moment it got out of the source and started it's ongoing travel through space time , the particle of light will continue traveling until it hits the mirrors , and then there are two cases : -light will be 100% reflected by the mirror , if the mirror is considered as a perfect conductor -it'll be99% reflected by the not so perfect mirror and will vanish within 100 or so ..reflections so once a photon is thrown away by a source it'll continue it's journey forever until it reaches something , and no we can't consider a light wave a light source ,a light source is in brief something that radiates (emits energy ) energy to get rid of some "excess" in energy , while a light particle is just a finite quantity of energy , quantum, traveling through space-time .