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A rocket passes the earth and synchronises its clock with the earth. Years later, a rocket passes it going to the earth and synchronises its clock with the first rocket. When it reached the earth, will the clocks be synchronised?

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  • $\begingroup$ Well what do you know of time in special relativity? $\endgroup$ – Kyle Kanos Jul 29 '15 at 20:05
  • $\begingroup$ Well, what could you know or demand of the tick rates of various clocks at all if you didn't understand and use the relativistic methods for comparing them in the first place? $\endgroup$ – user12262 Jul 29 '15 at 20:50
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    $\begingroup$ This has the same resolution as the usual twin paradox. Really, it does. Add up the proper time along the various paths. $\endgroup$ – dmckee --- ex-moderator kitten Jul 30 '15 at 0:54
  • $\begingroup$ No it wont be... $\endgroup$ – Žarko Tomičić Jul 30 '15 at 8:00
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Let's say that the Earth is stationary, the two rockets have the same speed v and the refrence planes are: K for Eath, K' for rocket1 and K'' for rocket2.

We can separate the problem in to two parts.

1st part: The rocket goes from earth to a distance $L$(in K). enter image description here Because the time $t0$ is the same, at distance $L$ the time $t1'$ is equal to $t1'=γ(t1 - vx/c^2)$. Now at this exact same time the second rocket pass and gets time $t1'$ from the other rocket($t1'=t1''$)

2nd part:The other rocket goes from $L$ to earth. enter image description here

We want to find the relation betwen $t2''$ and $t2$.

Now for K: $L=v(t2-t1)$(1) and K'': $L''=-v(t2''-t1'')$(2) This hapens because in K'': $L''=γL$, the objerver in K'' things he is stationary and the earth approches him with speed $-v$

So: $L''=γL$ and $t1'=t1''$

$(1)/(2)==> 1/γ=v(t2-t1)/-v(t2''-t1'') t1''-t2''=γ(t2-t1) (t1-(vL/c^2))γ-t2''=γ(t2-t1)$

Solving the equation with $t1=L/v$, we get: $t2''=γ[(2/v^2 - 1/c^2)L -t2]$

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