First of all, I saw few similar questions here but most of them address the question of why, when I want to address a question of HOW. Also one question is addressing the same problem but I find it's formulation rather complicated and this makes it difficult to understand the question and answers. When the problem can be described and visualized in much simple way:

Let's consider a spaceship floating under the Earth. There is a source of light and at time 0s the light will reach some point on Earth and the end of spaceship: enter image description here Then, after 1 second have passed on Earth, we will see the following picture: enter image description here

As we can see, relatively to the Earth the light have made a distance of 300,000 km in 1 second. But relatively to the spaceship the distance is half of that. When explaining how this can happen, usually they say that clock on the rocket will tick slower so speed calculated by folks on the spaceship is the same.

And it sounds believable until you imagine an other source of the light from an other end of the scene: enter image description here In this case by the same time the light from an other source have made a distance of 450,000 km. So if we use the same time as we did in previous case, the speed must be different.

So how can you imagine\describe the same speed for light from both sources and for both reference frames?


marked as duplicate by sammy gerbil, John Rennie special-relativity Aug 8 '16 at 9:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ There is no analogy to day-to-day experiences, because you have no experience with things that happen at sufficient speed to give you a referent. This simply is the way the world really works, and what you think of a making "sense" is an illusion that comes from only considering slow stuff at low precision. $\endgroup$ – dmckee Aug 7 '16 at 19:54
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    $\begingroup$ We get a lot of these questions, where you find a contradiction after applying length contraction and time dilation correctly. That's because there's a third major effect of relativity, called the relativity of simultaneity, which isn't covered in pop science books. You should look that term up for your answer. Basically, the clocks on the Earth and the rocket will disagree on when "0 s" is, and when the pulse of light was emitted. $\endgroup$ – knzhou Aug 7 '16 at 19:58
  • $\begingroup$ If the speed of light was not the same for all observers, regardless of their velocity, then Maxwell's Equations describing how electricity and magnetism interact would be in trouble, and experiments every day keep reassuring us that they are correct. emc2-explained.info/The-Constant-Speed-of-Light/#.V6eSpYpwbMI $\endgroup$ – user108787 Aug 7 '16 at 20:01
  • $\begingroup$ @sammygerbil, my question is how can you imagine\describe the phenomena using the thought experiment. Also there is no any answer there explaining how this may happen and how it may look. $\endgroup$ – Serhiy Aug 7 '16 at 20:17
  • $\begingroup$ @knzhou, when you say "the clocks on the Earth and the rocket will disagree" do you mean the bias? How this may affect the calculation of speed? $\endgroup$ – Serhiy Aug 7 '16 at 20:18

Its a question of simultaneity. In the earth reference frame the light sources emit simultaneously but in the frame of the rocket the light in front of the rocket will emit first so the ordering of events can become reversed according to its passengers.

  • $\begingroup$ Since in the Earth reference frame the light sources emit simultaneously does it mean, for observers from Earth the speed of the light relativele to the spaceship is not c? $\endgroup$ – Serhiy Aug 7 '16 at 20:22
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    $\begingroup$ @Serhiy Yes, of course. If I (on Earth) observe a spaceship go past at $0.9c$, and a light pulse in the same direction, I will measure their relative speeds as $0.1c$. $\endgroup$ – tfb Aug 7 '16 at 20:33
  • $\begingroup$ @tfb Ok, this is good :) then still it's unclear how guys on the spaceship can calculate it as c. The idea that their clock runs slower doesn't work, since we can put a light from the other end and the clock will have to run faster. The delay(shift) between clocks doesn't explain it neither, since there must be same delay between both points: when light reached one end of the rocket and other end. No matter how huge is this delay, the difference is probably constant. $\endgroup$ – Serhiy Aug 7 '16 at 20:39
  • $\begingroup$ Considering directly an arrangement with light sources at different locations always runs fast into an apparent wall because it makes it kinda difficult to justify relativity of simultaneity. Try considering first light beams emitted in opposite directions from a single location, like the middle of the rocket. The rocket crew will see the beams reaching the front and back of the rocket simultaneously, the Earth observers will not. And conversely. In other words, what one frame sees as simultaneous is no longer simultaneous in another frame. $\endgroup$ – udrv Aug 8 '16 at 4:27
  • $\begingroup$ Moreover, imagine a series of equidistant synchronized clocks along the rocket. While the rocket crew sees them beating time in unison, the Earth observers will see a different time on each one of them. And here lies the resolution: the times and time differences assigned by the Earth and by the rocket to the same two events are different, just as the distances are, while the ratio of distance/time remains the same in both frames. Calculating the whole thing does take some using to the idea though. $\endgroup$ – udrv Aug 8 '16 at 4:27

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