# Why time slows only for moving object and not stationary observer? How the “stationary” & “moving” are decided? [duplicate]

This question already has an answer here:

Explanation for the questions in the title:
According to the the (special theory of) Relativity, if an observer is stationary and sees a fast moving object then time runs faster for the observer compared to the mover.

For example, persons 'A' & 'B' are somewhere far away in universe standing on a platform. There is nothing nearby for several light years. Hypothetically assume that 'A' is standing on the platform and 'B' boards the rocket and flies away with a significant % of light speed. Due to which the time runs 1.67 faster for 'A' compared to 'B'.
When 'B' returns to see 'A' after 10 years, 'A' has already passed 16.7 years. This is the premises of Relativity.

Now, my confusion starts here. Why only 'A' is considered "stationary" and 'B' as "moving"? Simulate the situation in other way for the same event. With respect to 'B', rocket is stationary and 'A' moves away with platform. And finally 'A' "returns" to see 'B'. In such case, 'B' should have grown older by 1.67 times.

But neither that happens nor both 'A' & 'B' age equally. It's just that 'B' remains younger.
On the funny note, the premises of relativity here should be who travels relative to other and not who burns the fuel! :-)
I have referred few questions in this forum, but couldn't get the answer.

## marked as duplicate by John Rennie, Norbert Schuch, Ali, Kyle Kanos, David Z♦Jan 12 '16 at 11:49

• This is the same case as that of the Twin paradox – Oswald Jan 12 '16 at 8:14
• @JohnRennie, When I read the wiki link provided by TheGhostOfPerdition, it indeed appears that my question is another form of "Twin Paradox". In your link, even after reading the accepted answer few times, I am not getting that why the stationary twin should age faster than the moving twin. May be because the explanation there relies on math equation alone. I would want someone to answer in a layman terms. Thanks. – iammilind Jan 12 '16 at 9:54
• Have a look at this , got ot from a link in the other paper,, not much math . scholarship.haverford.edu/cgi/… – anna v Jan 12 '16 at 13:26
• @annav, thanks for the link. It seems to discussing another question. According to the paper, if 2 twins are travelling in identical spaceships with same fuel & speed but separated by X0 distance, then after a while, their ages would differ. This actually creates 1 more question. :-) – iammilind Jan 12 '16 at 14:07
• It shows that the frame of reference is important. The lorenz transformation is in one direction, the direction the vector of velocity. This generates an asymmetry because it is not enough to say that they are at an X0 distance, the vector X is important too. – anna v Jan 12 '16 at 14:12

@CuriousOne posted this answer in the comments:

The premise of relativity is that the speed of light is the same for all observers. This has consequences, but it doesn't change time. All clocks still behave exactly the same for all observers traveling with their own clocks. It is only between observers that clocks are running at different rate. This change in relative clock times is one of the consequences of the constancy of the speed of light. So you always know what is at rest (the clock next to you) and what moves (the clock on the rocket). The astronaut has a resting clock next to him and you and your clock are moving for him.

• I agree that answers shouldn't be posted as comments, but I don't think this is an answer. Despite the title the question appears to be yet another about the twin paradox. – John Rennie Jan 12 '16 at 8:56
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Norbert Schuch Jan 13 '16 at 11:36
• @NorbertSchuch, I disagree, I think it's a fine answer. Please note the comment chain you're mentioning was deleted. – user1717828 Jan 13 '16 at 12:05
• @user1717828 This is an automated comment from the review queue. – Norbert Schuch Jan 13 '16 at 12:09

As the person B takes off in a rocket, both of them would see the other clock move at a slower rate, assuming the rocket to be moving at a constant speed, both are in an inertial frames of reference, but when B wants to return to A, B should make a turn somewhere, so a turn, is an acceleration, and accelerating frame of reference is non-inertial, and this 'turn' or the acceleration can be detected by B, and ofcourse by A, So by making a turn it is clear to both of them that it was B who was travelling and not A.

The same thing happens in the second case too, they both agree on who was actually travelling

• This argument is not sufficient for me. I thought about it while posting the question. I will counter-argue as following. Simulate the turn of 'B''s rocket in 2 ways: (1) The rocket din't actually make a turn, but was equipped to return in reverse with the same speed (2) Visualize the rocket frozen at the center and platform travelling farther and creating certain angle which appear to us as the rocket's turn. – iammilind Jan 12 '16 at 13:47
• @iammilind The reverse is an acceleration. – anna v Jan 12 '16 at 14:18
• @iammilind acceleration is a vector, it can change by changing the magnitude or the direction – Oswald Jan 12 '16 at 14:42
• @annav, TheGhostOfPerdition, I am not rejecting the acceleration. Just that, why do we give prominence to rocket while considering acceleration? For that matter, you can visualize the point-2 I listed. Think like the rocket is stand-still in that empty part of universe and the platform is accelerating around it. In other words, instead of rocket and platform, just consider 2 points. 1 point is moving and 1 is still; but which one! In nutshell, my main question is that what are the basis of categorizing "stationary" and "mover"; Why no one says that "platform" moved? – iammilind Jan 12 '16 at 14:49
• My intuition is that it is energy. It would take enormous energy to get the earth to move with -v as it appears at the rest frame of the rocket. I have to formulate this mathematically though. I am thinking about it. – anna v Jan 12 '16 at 14:52