0
$\begingroup$

According special relativity, the clock ticks slower on spaceship moving at relativistic speed because the light travels a longer diagonal distance with respect observer on ground.

If light is a wave then how it gains the velocity of spaceship? such that it travels in a diagonal.

can light make an interference pattern in this scenario?

enter image description here

Improve this question
$\endgroup$
3
  • $\begingroup$ Remember that an external observer sees this path. In the frame of the actual source, the light moves vertically. $\endgroup$
    – joseph h
    Commented Apr 24, 2023 at 5:39
  • $\begingroup$ The answers to this recent related question may be helpful: physics.stackexchange.com/q/759512/123208 $\endgroup$
    – PM 2Ring
    Commented Apr 24, 2023 at 5:41
  • $\begingroup$ On a general note... interference is the absence of self-interaction in a linear system (electrodynamics in vacuum at optical wavelengths is perfectly linear). Do electromagnetic fields in vacuum stay linear if viewed from a different coordinate system? Yes. $\endgroup$
    – FlatterMann
    Commented Apr 24, 2023 at 8:04

2 Answers 2

Reset to default
1
$\begingroup$

Ok, some corrections right?

The light never gain or lose velocity, its speed is always $c$ (in vacuum, anyway...).

So you are asking how can it travel in a diagonal, if it is a wave, right? I could ask the opposite to you, why couldn't it? What prevents the light to travel vertically in a referencial, and in a diagonal in another referencial? Why do you think it shouldn't be possible just because it is wave? If you can, you can imagine the light as photons, but it is not really necessary. Nothing prevents the wave propagation direction be different in differents referencial.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The particle can only gain or loose momentum if it has some mass. We know photon is a massless particle so how light gains momentum in direction of the spaceship? $\endgroup$
    – user140446
    Commented Apr 24, 2023 at 5:35
  • $\begingroup$ qph.cf2.quoracdn.net/… $\endgroup$
    – user140446
    Commented Apr 24, 2023 at 5:35
0
$\begingroup$

Your question should be closed as a duplicate. The answer has nothing to do with light, per se. The answer is that the direction of any motion is frame dependent. To see this, consider the following...

Suppose you stand at the origin of your horizontal x axis and shine a light vertically up. What that means is that the light moves further and further along your vertical y axis, but does not move at all along your x axis- the x coordinate of the light is always zero. That is the definition of vertical motion.

Now suppose that at the instant you shine the light, I happen to be walking past you at a meter per second. Let us consider the motion of the light in my frame. After a second, the light will be about 300,000 km above me, but not directly above me, since in my frame the light has an x coordinate of -1 meter. After two seconds, the light will be about 600,000 km above me, and now it will have drifted further behind me, with an x coordinate in my frame of -2 meters. And so on. With every second that passes, the light has an x coordinate in my frame that puts it an extra meter away from my vertical axis. So in my frame, as a person walking past you, the light is following a slightly angled path, not truly vertical one.

What I have depicted above is true of any kind of linear motion, be it the motion of light or of a bouncing ball. It has nothing to do with momentum etc.

Improve this answer
$\endgroup$

Your Answer

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