What happens when a moving observer meets a stationary observer in special relativity? I understands how the time dilation work. The derivation of time dilation perfectly understandable for me. However the thought that two observers observes that time dilated for each other do puzzles me. For instance, observer A in his inertial frame of reference observed observer B in his own inertial frame of reference moving away at a certain velocity. Observer A observed that B's time is slowed (by whatever feasible measure), whereas B observed A's time slowed. What if B changes his direction of motion, heading back toward A. According to relativity, their time is still dilated.
Both observe one another having a slower time. What would happen when they come together?
 A: Yes. If B suddenly reversed direction at some point (or if we consider another frame C going in the opposite direction at the same speed and synchronized to B at the reversal point), his/her time would still appear time dilated to A. A and B would no longer be synchronized in time at their 2nd rendezvous.
The usual argument goes like this: Say A sees B reversing direction at time $t_A^0$ and position $x_A^0 = v t_A^0$. B's time at that moment is $t_B^0 = \gamma(t_A^0 - \frac{v}{c^2}x_A^0) = t_A^0/\gamma$ and he observes $A$ at position $x_B(A) = -v t_B^0 = -x_A^0/\gamma$. From the point of view of A, when A and B meet again in $x_A=0$, A's time is $t_A^0 + x_A^0/v = 2t_A^0$, while B's time is $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$. So A sees B as time dilated by a factor of $\gamma$.

But we still have a couple of problems: After B reverses direction, does s/he also see that A is time dilated relative to him? And how did A and B loose synchronization of their origins, even though they were synchronized in the beginning?

The answer to the first problem is again yes, B does observe A as time dilated. In order to prove this we need the answer to the 2nd problem, which is that B's reversal of direction implies a change in synchronization through a shift in coordinates, despite the fact that he does not change the coordinates of his location at the moment of reversal. The Lorentz transforms between A and B after B's reversal read
$$
\begin{eqnarray} x_A &=& \gamma (x_B - vt_B) + 2vt_A^0 \\
t_A &=& \gamma (t_B - \frac{v}{c^2}x_B)
\end{eqnarray}
$$
and
$$
\begin{eqnarray} x_B &=& \gamma (x_A + vt_A) - 2\gamma v t_A^0\\
t_B &=& \gamma (t_A + \frac{v}{c^2}x_A) - 2\gamma \left(\frac{v}{c}\right)^2t_A^0
\end{eqnarray}
$$
(All that is needed to derive these is to account for a shift in coordinate origins in the usual form of the Lorentz transforms. I am skipping the details, but we can easily verify that all coordinates verify their correct values for $t_A = t_A^0$.)
What happens as a consequence is that B sees A at a different time after his direction reversal: Right before the reversal B observed A at position $x_B(A) = - x_A^0/\gamma$, corresponding to A's time $t_A = \gamma(t_B^0 + \frac{v}{c^2}x_B(A)) = t_A^0/\gamma^2$. Right after the reversal the new Lorentz transforms show that B still observes A at position $x_B(A) = - x_A^0/\gamma$, but now this corresponds to A's time $\bar{t}_A = \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0$.
On the other hand, the 2nd rendezvous still takes place at time $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$ for B, and at $t_A^0 + x_A^0/v = 2t_A^0$ for A. So from B's point of view, s/he meets A again after a time $\Delta t_B = 2t_B^0 -t_B^0 = t_A^0/\gamma$, while A only observes a time lapse $\Delta t_A = 2t_A^0 - \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0 = t_A^0/\gamma^2 = \Delta t_B/\gamma$. In other words, B sees A undergoing time dilation, as expected.
