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?
-
2It happens, indeed. – Aug 22 '17 at 16:39
-
1It 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) – Henry Aug 22 '17 at 16:41
-
1It won't look different to you, only to him (because you are in motion relative to him, but not relative to yourself). – MPW Aug 22 '17 at 16:42
-
Everything depends on which reference frame you choose .Choose one reference frame,apply the equations,you will get the answer. – Soham Aug 22 '17 at 16:53
-
You both take a clock and measure the time. When he comes back you compare clocks and you'll observe that his clock went slower. Otherwise if your clock was slower it means it was you who were actually moving (faster). Because time is not absolute by itself, rather space and time are interconnected and affect each others. – Mihai B. Aug 22 '17 at 17:25
-
1You have discovered the Twin Paradox. Perhaps the Wikipedia article will help you understand: https://en.wikipedia.org/wiki/Twin_paradox – Solomon Slow Aug 22 '17 at 20:12
-
1See my answer therein : [How do I know which observer is running the time faster or slower?]{https://physics.stackexchange.com/questions/233649/how-do-i-know-which-observer-is-running-the-time-faster-or-slower/350359#350359} – Frobenius Aug 22 '17 at 20:50
-
The key is to understand that even though you both view the "others" clock(s) as the one(s) that is/are ticking slower, in actual fact only in one frame of reference are clocks ticking slower. Understanding exactly how all this occurs, is the key to fully understanding relativity. – Sean Aug 23 '17 at 23:44
-
@Mihai B. What if he doesn't come back and sees the other person from telescope or by any other mean.How will you clearify that???? – Theoretical Feb 07 '18 at 09:20
-
Rule #1: they both measure that the speed of light is c. They both have a clock which ticks with the same rate. They use lasers to signal each other, sending 2 pulses at a distance of 1 unit of time: (A) sends laser to (B) encoded his time; (B) receives and sends to (A) encoded his own time when the laser data arrived; (A) sends again to (B) his new time; (B) receives and sends to (A) his time. They now can calculate what velocity each has, via proper time calculus. Your confusion comes from red-blue shifting, because they will both observe the other one's laser red or blue shifted. – Mihai B. Feb 13 '18 at 10:51
-
relevant experiment: muon decay; high velocity muons coming from sun and arriving at earth surface https://en.wikipedia.org/wiki/Muon – Mihai B. Feb 13 '18 at 10:52
-
@Mihai B So you're saying that there are proper calculating methods. – Theoretical Feb 13 '18 at 13:05
3 Answers
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.

- 15,072
-
"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. – user Aug 23 '17 at 07:27
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.

- 9
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

- 316
-
What if twin B never comes back.He stays in that distance star and observes twin A from there. – Theoretical Jan 21 '18 at 06:40