Four points A1 (0; 0), A2 (1; 0; 0), B1 (0; 100; 0), B2 (1: 100; 0) are given. They are the corners of a long drawn rectangle. A light velocity c = 300,000,000 m/s is assumed. The axis unit is meter.

  1. A laser oriented along the line A1-B1 emits a light flash. It starts at point A1 at time t = 0 and arrives at the point B1 at time t = 100/c.
  2. If the laser moves at the constant velocity v = 0.01c in the direction of the line A1-A2 during the emission, the light flash would nevertheless have to arrive at the point B1 since the movement of the light is independent of the movement of the source .
  3. If the laser is permanent at point A1 and the target is at point B1 at the time of emission, but after the time t = 100/c at point B2 because it moves along the line B1-B2, the light flash miss the target. The target then has the coordinates (1; 100; 0; 100/c). The target is missed because it has moved out of the optical axis of the laser during the transit time of the light flash.
  4. This result must not change if the laser has also moved at the time of emission (see point 2). And it still must not change if the laser and the target have the same velocity vector: if the target moves out of the firing line, the flash will not hit even though the target is at rest relative to the source (The velocity vector of the source is no argument if light is self-propagating in space).

Is there a way out of the dilemma?

  • $\begingroup$ Your question is not clear. What dilemma? Also your coordinates are not clear,the " ; "and" :" are not standard coordinate notations. a drawing might help. $\endgroup$ – anna v Feb 8 '17 at 8:14

So, the laser points perpendicular to the line A1-A2 and is moving fast along the line A1-A2. Let's see what is the direction of laser beam and what is the velocity of each photon emitted be the laser.

At some moment let's see where are all the photons emitted by laser. The answer is: they are all located along the line perpendicular to A1-A2. Position of the line depends on time: the line is moving together with laser. It passes through A1 and B1.

But the velocity of each photon is not perpendicular to A1-A2! The speed of each photon is still $c$, but the direction of it's velocity is not perpendicular to A1-A2.

There is very little relativity-specific here. General picture would be the same if it was not the laser but a machine-gun firing bullets perpendicular to the gun's velocity. Relativity-specific here is only the fact that the speed of each photon is still $c$ and does not depend on fact if laser is moving or not. With machine-gun situation is different: it fires bullets with some speed, but the resulting speed of the bullets is bigger because the gun itself is moving.

  • $\begingroup$ As under point 1, the laser is at rest, all photons are aligned and move from A1 to B1. Why then, should the photons follow another line, if the laser is moving? This would mean that the motion of the light photon is depending on the motion of the source (like bullets of a gun), the contrary of what we learn in physics about light. The value of the speed isn't of interest here, only its direction. For most points on earth, the earth vector is changing daily from v to -v. Thus laser beams should show a daily recurrent aberation, this is not observed. $\endgroup$ – paule Feb 8 '17 at 13:15
  • $\begingroup$ Imaging a photon traveling in direction A1-B1. And imagine an observer, traveling fast in direction A1-A2. Now observer absorbs the photon. He can measure the speed of absorbed photon, he knows where the photon came from. And he would see that the photon was NOT flying parallel to A1-B1! Direction of photon speed depends on frame of reference. Now if the laser is moving, the photons it emits have velocity parallel to A1-B1 but only in the frame of reference of the laser! In staying-still frame of reference the direction will be different. $\endgroup$ – lesnik Feb 8 '17 at 13:51
  • $\begingroup$ Now to your example of a laser on Earth. Note, that observer is also on Earth. He would not observe any difference because he is in the same frame of reference as the laser. The laser is always staying still in his frame of reference, nothing changes. $\endgroup$ – lesnik Feb 8 '17 at 13:55
  • $\begingroup$ The picture of light you are using is almost perfect the one of a particle, like any other particle. The velocity of the source adds then to the velocity of the photon but only concerning its direction in space and not the value of the velocity. This is an interesting mixed picture between Newtonian light particles and special theory of relativity. But the latter is founded on the electrodynamic explanation of light, that regards light as a wave. No wave propagation can be influenced by the motion of the source, no change in direction too, if the second postulate of the SRT is true. $\endgroup$ – paule Feb 8 '17 at 15:08
  • $\begingroup$ @paule Motion of source does influence the waves emitted. It's called Doppler effect. But I am not ready to investigate how propagation of waves look like when we change system of reference. Too difficult for me. $\endgroup$ – lesnik Feb 8 '17 at 16:46

One small note. In both cases the laser pointer has to be directed not along the y-axis (A1 - B1). It has to be tilted a bit backward to direction of it's motion.

In the first case the laser pointer moves in the reference frame of the target, and due to aberration of light it takes some time for the photon to pass through the tube. So, the tube of laser pointer has to be tilted backward at oblique angle to direction of motion of the laser pointer. This angle does not depend of the lengths of the laser pointer. https://en.wikipedia.org/wiki/Aberration_of_light

In the second case the laser pointer is at rest, but it is tilted a bit backward in its frame. Thus, it launches a photon not "straight up", but at certain oblique angle (the same as in the first case) backward and both photon and B1 come the same point at the same moment.

Some visualization: https://www.youtube.com/watch?v=5-AAC4pemDI https://www.youtube.com/watch?v=hnphFr2Iai4

Please look carefully at the angles the source emits a photon in different configurations - whether it is at rest or in motion. In principle you can keep the laser pointer straight up in the both cases. But photon will hit B2 in both cases then.

  • $\begingroup$ Why should I tilt the laser at all? This would be a different experiment. If I understand the video right, a v-shaped tube next to the mirror would "guide" the photons to the mirror and back, if both, mirror and tube, are moving. Now the earth is, without any doubt, a moving system as we learn from astronomy but from the aberration of starlight too. This would mean that on earth one may use such a double-v-shaped tube in front of a mirror and the photons will be reflected anyway, if the mirror is parallel to the earth's vector. Seems to be strange. $\endgroup$ – paule Feb 8 '17 at 12:55
  • $\begingroup$ You can explain tilting of the laser two ways. 1) I am at rest and I just tilt the laser because I want to hit a target (episode 1), mirror is source. 2) I am in motion and I tilt a laser because of aberration. youtube.com/watch?v=5-AAC4pemDI (episode 2), the mirror is the source and the lamp is target. Or if you keep your laser straight up 1) I am in motion and keep the laser straight up and photon moves together with tube. But laser will come to the target at oblique angle then. If target has a tube this tube will be directed at oblique angle (episode 1), mirror is target. $\endgroup$ – Albert Feb 8 '17 at 13:33
  • $\begingroup$ I understand your critical approach, but it is not here. There is no controversy exactly in this experiment. $\endgroup$ – Albert Feb 8 '17 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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