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ImageTwin Paradox dictates that moving observer ages slowly or simply that moving clocks ticks slowly. Theoretically, it can be argued that both (moving and stationary) clocks will see each other to be ticking slowly compared to itself, however, experimentally it is obvious that only one of the clocks should be ticking slowly compared to other.

In this context, I assume that clocks represented by A and B in my following question form the moving system i.e. clock A and B are ticking slowly compared to P.

Imagine A and B co-moving observers separated by rest length L. They have their clocks synchronized using Einstein method. The direction of their movement is from A to B. (A is to the left for clarification).

There is another observer P which coincides with B at $(l=0, t=0)$. When A coincides with P, the time passed in P's clock will be $(L/v.\gamma)$. As per my argument above, the time shown by A should be $(L/v.\gamma^2)$ as seen by coinciding observer P.

But using Lorentz transformations, the event at A is given by $(l=-L, t=L/v)$ i.e. time passed in A is equal to time passed in P. Therefore, no time dilation as observer by B.

What is the resolution of this problem? What will be the time passed in A as seen by P when it reaches A?

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    $\begingroup$ Canonical post about the twin paradox: physics.stackexchange.com/q/2554/50583 $\endgroup$ – ACuriousMind Apr 10 '17 at 11:35
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    $\begingroup$ Also, you say "however, experimentally it is obvious that only one of the clocks should be ticking slowly compared to other" but that is precisely what is not obvious - compared to the other in what frame? $\endgroup$ – ACuriousMind Apr 10 '17 at 11:36
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    $\begingroup$ experimentally it is obvious that only one of the clocks should be ticking slowly compared to other - here. This isn't obvious, and in fact it isn't true. Each observer will measure the other one's clock to run slower. There is no contradiction in this. $\endgroup$ – Prof. Legolasov Apr 11 '17 at 4:59
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Apr 11 '17 at 10:41
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I'm a little hesitant to answer this because it's an easy exercise, especially if you take a moment to draw the right picture. But here it is anyway:

Here is the series of events according to $P$:

Time $T_0=-Lv/\sqrt{1-v^2}$: $A$ sets his clock to $0$.

Time $0$: $B$ and $P$ set their clocks to $0$.

Time $T_1=L\sqrt{1-v^2}/v$: $P$ meets $A$. At this time, $A$'s clock reads $A_1=L/v$, $P$'s clock reads $T_1=L\sqrt{1-v^2}/v$, and $B$'s clock reads $B_1=L(1-v^2)/v$.

So let's figure out how fast $B$'s and $A$'s clocks are moving (according to $P$, at the moment he meets $A$):

Since $B$ set his clock to $0$, time $T_1$ has passed, while B's clock has advanced by $B_1$. Therefore his clock is running at the rate $B_1/T_1=\sqrt{1-v^2}$. That's less than $1$ so $B$'s clock is slow.

Since $A$ set his clock to zero, time $X=T_1-T_0=L/(v\sqrt{1-v^2})$ has passed, while $A$'s clock has advanced by $A_1$. Therefore his clock is running at the rate $A_1/X= \sqrt{1-v^2}$. That's the same slow rate at which $B$'s clock runs.

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