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I recently bought a copy of Einstein's booklet Relativity, in which Einstein attempts to describe the basics of special relativity to laypersons. In Chapter 12, Einstein explains how the Lorentz transformations can be used to compute how the relative motion of frames of reference affects measuring rods and clocks.

Suppose I stand on a railway boarding platform, and my friend stands on a railway car which is moving past the platform. I have one frame of reference, and my friend has a second frame of reference which is in motion relative to the first. Each of us holds a meter stick and a stopwatch. At the moment he passes me, we both hold out our meter sticks and stopwatches to be compared. According to the Lorentz transforms, in my frame of reference I will observe his meter stick to be shorter than mine, and I will observe his watch to be ticking slower than mine. Similarly, in his frame of reference he will observe my meter stick to be shorter than his, and he will observe my stopwatch to be ticking slower than his own.

It seems odd to me that we would both observe our meter sticks to be shorter than the other person's, but I can accept this. What I can't reconcile is the stopwatches. Suppose my friend remains in motion for some time, but then his train loops back around to the station, stops, and we compare stopwatches. We now have the same frame of reference. Someone's watch must be ahead of the other person's, but whose?

Let me modify the experiment a bit. I have two stopwatches of identical construction which are synchronized. I set watch A on a table, and I put watch B on the edge of a merry-go-round and spin it very fast for a long time. Here the frames of reference are the table (k) and a point on the edge of the merry-go-round (k'). Because k' is moving relative to k, from perspective k B will tick slower. However, k is also moving with respect to k', so from perspective k' A is ticking slower. When I stop spinning the merry-go-round, which watch will be behind?

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  • $\begingroup$ The first example with the train coming back is the twin's paradox. The twin that does not accelerate (the one in the platform) is ahead en.wikipedia.org/wiki/Twin_paradox $\endgroup$ – user126422 Apr 16 '17 at 18:57
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Circular motion is accelerated so your friend on the train is not in an inertial frame. This introduces an asymmetry since your friends's frame has a non-zero proper acceleration and your frame does not. This is a variant of the twin paradox.

The time dilation in circular motion is discussed in my answers to Is gravitational time dilation different from other forms of time dilation? and Can a ultracentrifuge be used to test general relativity?.

The twin paradox is discussed in What is the proper way to explain the twin paradox?

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When two travellers that depart from each other at a certain time return to a common meeting point (anywhere), the one who has accelerated more will show less time passed on his watch. In the simplest case, where one stays stationary and the other goes at constant speed to a distant location, and returns at constant speed, the three worldline segments form a triangle in which all 3 legs are time-like. From the Einstein invariance formula, one can show that such a triangle disobeys the triangle inequality; the side going from the earliest event to the latest event (the worldline of the stationary guy) is longer than the sum of the other two sides. Just join the turnaround point to the point on the stationary guy's worldline, using a line that is a line of simultaneity for him (a line in space-time parallel to his space axis). Now apply the invariance formula, c*2t*2=x*2+c*2tau*2, to these two triangles and eliminate x. The sum of the two t's will be greater than the sum of the two taus. Another way to see this is to realize that a traveller's space axis rotates in the opposite direction to his time axis, so that events ahead of him are at an earlier time for him than for us, and events behind him (such as his starting point) are at a later time than for us. When he reverses direction , his x-axis rotates back towards ours and then beyond it, so that a large chunk of time at our location, but by his measure, suddenly disappears without his noticing it (unless he's watching us thru his telescope). He sees his starting point event as later than we do, as I said, and similarly he sees his arrival event as earlier than we do, because his x-axis is now tilted in the opposite direction. The time he measures for us jumped ahead at turnaround time because he changed the frame he measures it by. At his turnaround moment, he will see our time (by his measurement) suddenly jump from an early event in our time to a later event in our time. He does not record any of our time between those two events. But when he arrives back, he'll see we certainly recorded it! We'll be much older.

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  • $\begingroup$ If instead of a sudden change in direction you just decelerate and accelerate back, you will see the clock to speed up, not to jump, which is more physically realistic. $\endgroup$ – user126422 Apr 17 '17 at 4:31
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The twin paradox is an absurdity (either twin is younger than the other) unless, as Einstein explained in 1918, the turning-around acceleration of the traveling twin miraculously makes the stationary twin older:

http://topquark.hubpages.com/hub/Twin-Paradox "The Twin Paradox is a scenario that, at first glance, seems to make nonsense out of Einstein's theory of special relativity. 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. [...] What happens is that the twin on Earth, viewing himself as stationary and his brother as moving at high speed, sees his brother experiencing time dilation and thus ageing more slowly. At the same time, 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. Each sees the other as moving, and therefore as experiencing time dilation. But which brother is "correct" in the way he perceives the situation? Both are. Each sees the other as being younger than himself. How they were perceived by any onlooker would depend on which frame of reference the onlooker was in. It doesn't make sense to ask which brother is "really" older, because the answer depends on where you stand to ask the question! But what about when the brother in the spaceship returns to Earth? Surely the contradiction will be apparent then? Ah, but in order to return to Earth, the spaceship must slow down, stop moving, turn around and go back the other way. During those periods of deceleration and deceleration, it is not an inertial frame and therefore the normal rules of special relativity don't apply. When the twin in the spaceship turns around to make his journey home, the shift in his frame of reference causes his perception of his brother's age to change rapidly: he sees his brother getting suddenly older. This means that when the twins are finally reunited, the stay-at-home twin is the older of the two."

Of course, Einstein's 1918 explanation amounts to explaining the absurd in terms of the more absurd.

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    $\begingroup$ Special relativity is trivially consistent. $\endgroup$ – user12029 Apr 16 '17 at 21:43
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    $\begingroup$ This apparent paradox is very well studied, with Peter, Paul and Mary and all their cousins trying to find something inconsistent. The literature is clear. $\endgroup$ – ZeroTheHero Apr 16 '17 at 22:07
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    $\begingroup$ It is a shame to claim that something is absurd just because you had not dedicated enough time to understand it. Respect yourself a little more. The day you can understand a Minkowski diagram the absurdity will disappear. $\endgroup$ – user126422 Apr 17 '17 at 4:24
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    $\begingroup$ You are not qualified to answer this question as you clearly have no understanding whatever of Special Relativity. Downvoted. $\endgroup$ – m4r35n357 Apr 17 '17 at 9:20

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