Einstein's relativity states that times goes slowly in a moving clock.That means if my friend moves at a speed of $v$ his time will go slowly. But I am also moving at a speed of $-v$ relative to him. So my clock should also move slowly too. Why doesn't that happen?


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    $\begingroup$ It happens, indeed. $\endgroup$ – Francesco Polizzi Aug 22 '17 at 16:39
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    $\begingroup$ It does. And when he turns round and comes back to you it will happen again. But the reason the clocks will show different times when you meet again is that one of you has accelerated to turn round (non-inertial frame) $\endgroup$ – Henry Aug 22 '17 at 16:41
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    $\begingroup$ It won't look different to you, only to him (because you are in motion relative to him, but not relative to yourself). $\endgroup$ – MPW Aug 22 '17 at 16:42
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    $\begingroup$ You have discovered the Twin Paradox. Perhaps the Wikipedia article will help you understand: en.wikipedia.org/wiki/Twin_paradox $\endgroup$ – Solomon Slow Aug 22 '17 at 20:12
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    $\begingroup$ See my answer therein : [How do I know which observer is running the time faster or slower?]{physics.stackexchange.com/questions/233649/…} $\endgroup$ – Frobenius Aug 22 '17 at 20:50

According to you:

At 1:00, your clock said 1:00, and his said 1:00.
At 2:00, your clock said 2:00, and his said 1:30.
At 3:00, your clock said 3:00 and his said 2:00.

According to your friend:

At 1:00, his clock said 1:00, and yours said 1:00
At 1:30, his clock said 1:30, and yours said 1:15
At 2:00, his clock said 2:00, and yours said 1:30

Notice that you disagree not just about the speed of each others' clocks, but about which events are simultaneous. You say that at the very moment when his clock said 1:30, yours said 2:00. He says that at the very moment when his clock said 1:30, your clock said 1:15. That's how you can each say that the other's clock is running slow.

  • $\begingroup$ "At 3:00, your clock said 3:00 and his said 2:00." No need to go fast to achieve particular effect, though; it happens once a year anyway. $\endgroup$ – a CVn Aug 23 '17 at 7:27

It helps to think of it like this.

Imagine a bouncing ball. In blatant disregard for the laws of physics, assume this ball will bounce forever up and down, between to points.

If you and a friend both have a magic ball, as your friend accelerates away from you, relative to you your friends ball will take longer to bounce between these two points. As it travels up and down, it also travels forward with the points it is bouncing between, and so the distance between the points, relative to you, is greater. It's going up, down, and forward.

To your friend, however, it is only moving up and down.

Assuming these magic balls are magically synched, the amount of time it took for your friends ball to complete the same number of bounces will have taken longer for you than it did for your friend, altho they will still be in synch.

Or, if that's too crazymouth for your sensibilities, look into Einstein's analogy of a train, a platform, and a magical synchronized double lightening strike.


Einsteinians systematically contradict special relativity when it comes to the traveling observer's time:

Brian Greene: "If you're moving relative to somebody else, time for you slows down." https://m.youtube.com/watch?v=QnmnLmwBmfE

Actually special relativity predicts the opposite:

If you're moving relative to somebody else, time for you SPEEDS UP.

You, the traveler, will discover this by checking the somebody else's (stationary) clocks against your (moving) clocks. Very few Einsteinians teach the correct prediction of special relativity:

David Morin, Introduction to Classical Mechanics With Problems and Solutions, Chapter 11, p. 14: "Twin A stays on the earth, while twin B flies quickly to a distant star and back. [...] For the entire outward and return parts of the trip, B does observe A's clock running slow..." http://www.people.fas.harvard.edu/~djmorin/chap11.pdf

"The situation is that a man sets off in a rocket travelling at high speed away from Earth, whilst his twin brother stays on Earth. [...] ...the twin in the spaceship considers himself to be the stationary twin, and therefore as he looks back towards Earth he sees his brother ageing more slowly than himself." http://topquark.hubpages.com/hub/Twin-Paradox

  • $\begingroup$ What if twin B never comes back.He stays in that distance star and observes twin A from there. $\endgroup$ – Theoretical Jan 21 '18 at 6:40

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