So in many applications like Optical Coherence Tomography, LIDAR, a mechanical scanning mirror is often used to reflect the laser to outside and also reflect the back scattered light to detector. Since light has a constant speed, when the light hits the mirror at scan angle A and goes out side, then gets back scattered by an object and goes back to the mirror, the mirror already moved to another angle. My understanding is that if the mirror scans extremely fast, or if the object is far far away, then the mirror will not be able to reflect the light back to the sensor.

On the other hand, assume I am going with the light, then time will freeze, in light's perspective, the mirror will be static between the two reflections, one going out, one going back to the detector.

I am sure I missed something here but could not figure out...

  • $\begingroup$ A classic use for your estimation skills. What angular difference occurs during the round-trip time? How does that compare to the angular size of the detector sub-system? Standard figure put into (possibly) more useful units: $c \approx 1 \,\mathrm{ft/ns} \approx 0.3\,\mathrm{m/ns}$. $\endgroup$ Jul 25, 2019 at 22:08
  • $\begingroup$ for instance if a lidar scans an object at 300m away, double pass means 2us of light traveling, and if this is a spinning lidar of 30 sweeps per second, which is 10800 degrees per second, so within 2us, the lidar will rotate about 0.02 degree... 0.02 degree is still small but i think it becomes noticeable $\endgroup$
    – shelper
    Jul 25, 2019 at 22:17
  • 1
    $\begingroup$ Whether or not that is significant depends on the layout of the guts of the machine. How big an angle is subtended by the detector system as seen from the mirror? (A side issue is that by using a segmented detector the designer can arrange the internals to use this angular offset as a cross-check on the delay.) Also there will be a range limit at which the return signal from most targets (i.e. things other than corner-cubes) won't be bright enough to register. $\endgroup$ Jul 25, 2019 at 22:26
  • $\begingroup$ Thanks for the comments, i agree with you, i just want to make sure that i understand the problem correctly. and i still don't understand from the light's perspective, is the mirror freezed or not... $\endgroup$
    – shelper
    Jul 25, 2019 at 22:39
  • 1
    $\begingroup$ @dmckee I wouldn't call this the twin paradox unless the question was whether the mirror experiences no time, as opposed to the light, that wasn't really the question, though it may be related. $\endgroup$
    – user234190
    Jul 25, 2019 at 23:35

2 Answers 2


From light's perspective, if it had one, it would see no distance or time, it would just arrive instantaneously. But since time is relative, it sees what you think of as the past and future as the same time. So in the lab frame, light can be sent from a source that misses an object if it doesn't make it there in time.

  • $\begingroup$ So from light's perspective it should always hit the detector 'cus the mirror didn't move, but from mirror's perspective, since it moved, light can miss the detector, although time relativity can be explained in different frame, but whether the detector sees the light or not should not be affected by the frame... $\endgroup$
    – shelper
    Jul 26, 2019 at 14:23
  • $\begingroup$ It doesn't really make sense to talk about light's frame of reference, nothing moves, no time occurs, it doesn't go anywhere, so the detector and the mirror are in the same place at the same time, etc. So really you should just be concerned with the lab frame of reference, light travels at a finite speed, so it takes time to get to a mirror, which it can miss if the mirror is far enough away. So you could see light leave a source and take time to reach a mirror that has moved during that time, but light will 'see' a mirror that has already moved as soon as it leaves the source, instantaneously $\endgroup$
    – user234190
    Jul 26, 2019 at 18:10
  • $\begingroup$ @shelper A lot of this discussion is just semantics, like a photon can't see anything of course, but one knows what one means by that. And you could switch the discussion to neutrinos moving close to the speed of light. So a neutrino would get to the mirror almost instantaneously, and it would see almost no distance covered. But when it reaches the mirror, it would see the mirror has already moved to where we saw it move during the time of the flight of the neutrino, because of relativity. $\endgroup$
    – user234190
    Jul 26, 2019 at 19:22

Photons are elementary particles, massless, they travel at speed c when measured locally in vacuum.

Now photons travel along lightlike worldlines, that is as long as the photon exists(between emission and absorption) it travels a spacetime distance of 0.

The emission and absorption are casually connected for the photon and these two events' spacetime distance is 0.

This is where you get confused.

Photons do not have a reference frame, you cannot say what it looks like from the photons view. In reality, you could say that the photon sees all this timescale between emission and absorption in one. It is because photons do not experience time as we do (who have rest mass).

Now from our frame (who have rest mass), this takes time, between emission and absorption, and in your case you can interact with the photon.

This interaction is called elastic scattering, this is what happens when a photon hits a mirror.

Now elastic scattering keeps the energy of the photon. Energy is conserved. It changes its momentum (vector), and yes, the photon exerts pressure on the mirror.

Photons travel at speed c in vacuum, when measured locally. It travels at speed c, then it changes angle, and travels at speed c.

Photons do not slow down, and speed up. They just change angle. Momentum is conserved.

Now where you get confused is how can the mirror rotate while the photon travels back?

You say time will freeze. It does not freeze. It is just that for the photon in spacetime emission and absorption are not separate, but connected by casuality.

This does not mean that we cannot interact with the photon. In our frame, the space and time distance the photon travels can have interactions with the photon. And in this frame, the mirror rotates.

The photon does not have a frame, but if you want to say what the photon sees, you shouldn't say time freezes. It is rather that the photon sees the whole timescale in one. That includes the rotation of the mirror, the elastic scattering.

  • 1
    $\begingroup$ This is just dancing around words. Words are used to convey meanings, if the meanings are understood, then they work. If you speak of what the photon sees, then from where does it see this? From what perspective, point of view, or frame of reference? If you say photons do not experience time, then that's what time freezing would mean. It's not like time freezing is something that can happen with particles with a rest mass. You can't have it both ways, if you want to describe what happens with the photon, it will always come back to describing what you are saying can't be described. $\endgroup$
    – user234190
    Jul 26, 2019 at 21:06
  • $\begingroup$ @user47014 "If you say photons do not experience time, then that's what time freezing would mean" I do not say that, what I said is that they do not experience time as we do (who have rest mass), they experience it differently. This could mean that they see the whole timescale in one. $\endgroup$ Jul 27, 2019 at 14:01
  • $\begingroup$ That's still not making clear the difference in the terms. $\endgroup$
    – user234190
    Jul 27, 2019 at 15:32
  • $\begingroup$ Nor from where the photon is experiencing time. $\endgroup$
    – user234190
    Jul 27, 2019 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.