The motion of a light flash emitted by a laser, which is at rest or moving 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.


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*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.

*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 .

*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.

*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?
 A: 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.
A: 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.
