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If you fired a laser at the perfect angle at a mirror (for either mirror setup), and then quickly moved a mirror to replace the laser, will the light oscillate between the mirrors, as shown in the picture?

The mirrors may absorb a small amount of light each time light "strikes" it, so will the light continue to oscillate between the mirrors until all of the energy is absorbed?

And in the second setup, if light does indeed reflect back and forth, doesn't that mean that the speed of the photons is zero at the point of impact (when it reverses it's velocity)?

enter image description here

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  • $\begingroup$ Distance/time is not the most fundamental way to define "the speed of a photon". Otherwise one can conclude the speed of a photon changes when it is about to be absorbed. The proper definition of the speed of a photon is done at the level of quantum field theory and it reduces to Distance/time under regular circumstances. $\endgroup$ – pathintegral Aug 16 '18 at 20:58
  • $\begingroup$ The laser itself is generally made up of a resonant cavity with a gain material inside. $\endgroup$ – The Photon Aug 17 '18 at 1:36
  • $\begingroup$ if you are interested in the behavior of light there is a series of MIT videos that are very instructive youtube.com/watch?v=zD6tTb74KdU $\endgroup$ – anna v Aug 17 '18 at 3:51
  • $\begingroup$ the speed of a photon never changes. As it reflects off a surface, the momentum changes (much like literally every other type of reflection/collision). However, even were there a point where the momentum reached zero, that does not imply the velocity is zero. You may be thinking that the momentum of a photon is $p=mv$, when it actually is $p=hf/c$. So, instead of velocity changing with momentum as in a massive object, it is frequency that changes with momentum for a photon $\endgroup$ – Jim Aug 17 '18 at 17:29
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I think it may be a bit misleading to always think of light as collection of photons. See the comments by John Rennie in this post, for example.

Regarding your mirror construction, yes the light can bound around when you build a cavity around, but the optical path length should be an integer multiple of the wavelength of light (the light is on resonance with the cavity). When off-resonance the cavity cannot support the coherence oscillation of light. Also, you drew flat mirrors, but in practice the mirrors would need some curvatures to compensate for the divergence of the propagating light. The first drawing will look similar to a bow-tie cavity and the second one to a Fabry-Perot cavity.

Regarding the possibility of "speed of the photon is zero", that is wrong. In microscopic sense, the speed of the photon is always $c$, although when it passes through an optically dense medium you may measure the group velocity of light to be less than $c$ due to refractive index (this is a macroscopic description, not microscopic).

Now let's look at the second drawing, for example. The light forms a standing wave, and at the surface of the mirror, the amplitude of the light is zero, so there is zero probability of detecting a photon right at the surface.

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  • $\begingroup$ You don't necessarily have to match the cavity dimension to n*wavelength or form a standing wave. For example, you could launch a pulse that's much shorter than the cavity length. $\endgroup$ – The Photon Aug 17 '18 at 1:30
  • $\begingroup$ @The Photon, right, but will the cavity sustain the pulse inside? $\endgroup$ – wcc Aug 17 '18 at 1:32
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    $\begingroup$ The only reason it won't is because no mirror is truly 100% reflective. (and diffractive losses) $\endgroup$ – The Photon Aug 17 '18 at 1:33
  • $\begingroup$ @The Photon, okay, but shouldn't the carrier frequency of the pulse be resonant with the cavity? Or could you lead me to a paper showing a long ring-down time of a pulse stored in a cavity? $\endgroup$ – wcc Aug 17 '18 at 1:37
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    $\begingroup$ Don't put spaces after @ or users are not notified. The proper address is: @ThePhoton $\endgroup$ – safesphere Aug 17 '18 at 4:23
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First, you cannot move a mirror, that has a rest mass faster then light to replace the source of light. but let's disregard that, and say that there is a photon bouncing between two perfect mirrors. This is called a photon clock.

Now photons are scattered elastically, that is called Rayleigh scattering froma mirror, that is the only way to keep the photons' energy and phases and build a mirror image. Now in the case of elastic scattering, the energy of the photon is kept, so in the case of a perfect mirror, this can go on forever. Of course there are no perfect mirrors, so after a while the photon will be absorbed or inelastically scattered and loses energy or gives all its energy to the absorbing atom's electron.

When the photon is scattered elastically off an atom of a mirror, it's wavefunction, that describes its probability desctribution for all of space, will change by changing the speed vector's direction of the photon. Everything else, all the other characteristics of the photon are kept, this is the only way to build a mirror image.

So your question, whether the photon is slowing down or not, it is not. Photons always travel at speed c in vacuum, when measured locally. When the scattering happens, the photon's speed vector's direction is changed, but the magnitude is always c. That means, that the speed of the photon is always c before, during, and after the scattering process, what changes is just the direction of the speed vector. (This is true for absorption re-emission too). If there would be an absorption, re-emission, the photon would seize to exist, until that moment, its speed is c, and after the emission, its speed is c instantaneously.

After the comments, it is important to note that whenever the photon bounces off inside the mirror, it exerts pressure (momentum) on the wall of the mirror.

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  • $\begingroup$ To sell an ideal model as the reality isn’t a good thing. The photon has an moment and every time it reaches the mirror and changes the direction, some part of this moment overgoes to the mirror. The photon loses energy and the bouncing photon is more and more redshifted. Even with photons a Perpetuum mobile is impossible. $\endgroup$ – HolgerFiedler Aug 17 '18 at 10:31

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