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Imagine a person going in a spacecraft moving away from the Earth, when he return from his journey he notice the time has passed slowly than for the habitants on the Earth. Noting this, we can conclude that the person that was moving is the spaceman, isn't it? Because when you move the time goes slowly than when you're not doing it so we could know when somenone moves and we couldn't have doubts if we move or not.

Tell me why I'm wrong please, I'm a bit confused. Thanks

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    $\begingroup$ en.wikipedia.org/wiki/Twin_paradox $\endgroup$ – PiKindOfGuy Feb 15 at 22:41
  • $\begingroup$ Because when you move the time goes slowly than when you're not I'm afraid you have really misunderstood time dilation here. I would suggest learning more about special relativity. Your doubts might then be cleared up. $\endgroup$ – Aaron Stevens Feb 16 at 3:25
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so we could know when somenone moves and we couldn't have doubts if we move or not.

This statement should be “ so we could know when somenone accelerates and we couldn't have doubts if we accelerate or not.”

In the twins paradox there is no ambiguity about which twin accelerated. The acceleration breaks the symmetry. Of course, with an accelerometer there is never any doubt about which twin accelerates anyway.

Note, that acceleration is not symmetric and relative in the way that velocity is.

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  • $\begingroup$ It should be noted that acceleration is not needed to explain the resolution of the twin paradox. Really it's just the change in inertial frames. So if at the "turn around" instead another "twin" moves past the traveling twin at the same velocity relative to Earth towards Earth and they synchronize clocks, the total time of the travelers will be shorter than what is recorded on Earth. $\endgroup$ – Aaron Stevens Feb 16 at 3:15
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    $\begingroup$ Yes, I am not proposing it as the solution to the twin’s paradox, but just a definite physical asymmetry. By the fact of the asymmetry the “doubts” are unfounded. $\endgroup$ – Dale Feb 16 at 3:19
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    $\begingroup$ Thanks Dale, I didn't remember that in my exemple one guy were accelerating $\endgroup$ – Sami Calvo Feb 18 at 7:09
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Take another example. Imagine that we have a huge spaceship landing platform, that is floating out in space. We also have three identical spaceships that have been out on a run for some time, but now all are heading back to the landing platform.

Spaceship "A" astronaut simply stops at the platform, and once done, those on the platform notice that the clocks on board Spaceship "A" are ticking at the normal speed.

The Spaceship "B" astronaut thinks he is top gun, and so he simply passes by the stationary Spaceship "A", and does so at a high velocity. Those on board the platform notice that the clocks on board Spaceship "B" are ticking at 1/2 the normal speed. Based upon that, they can calculate his velocity relative to the platform.

sqrt(1-(1/2)*(1/2)) * c = 0.866025440387 * c = 259,627.884 km/s

The Spaceship "C" astronaut thinks he is super top gun, and so he passes by the stationary Spaceship "A", and does so at an even higher velocity. Those on the platform notice that the clocks on board Spaceship "C" are ticking at 1/7 the normal speed. Based upon that, they can calculate his velocity relative to the platform.

sqrt(1-(1/7)*(1/7)) * c = 0.989743318611 * c = 296,717.582 km/s

However, the astronaut in Spaceship "B", sees things somewhat differently. As he passes by Spaceship "A", he sees the clocks on board Spaceship "A" as the clocks that are ticking at 1/2 speed compared to his own clocks.

On top of that, as Spaceship "C" passes by Spaceship "B", the astronaut on Spaceship "B" sees the clocks on board Spaceship "C" as clocks that are ticking at 1/2 speed compared to his own clocks.

And so taking all of these differing observations into account, it becomes obvious that these time dilations can not show exactly who's moving?

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  • $\begingroup$ Hey Sean, thank you for your answer but I still don't understand it very well. When the guy B approuches to A his clock goes slowly for A but how can be possible the clock A goes slowly too? I mean, the both spacecrafts, A and B, are deforming the space-time? $\endgroup$ – Sami Calvo Feb 18 at 7:00
  • $\begingroup$ If spaceship "B" had two clocks outside that are positioned with one at the front and one at the rear, and they appear to be synchronized to those on board ship "B", they would not be seen as synchronized by those on board ship "A". The clock at the rear would appear to be ahead of the clock at the front, and both would appear to be ticking at half speed. If ship "B" uses these two offset clocks to measure the ticking rate of a clock on spaceship "A", and do so while passing by, these time offsets are the key ingredient that make it appear as though ship "A's" clock is ticking at half speed. $\endgroup$ – Sean Feb 20 at 8:34

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