Clarification for the mutual time dilation for two observers in their inertial frame of reference

I am trying to understand the idea under the theory of special relativity recently. The time dilation predicted by special relativity baffles me. I used to think that I understand the idea underlies the theory, but when I thought about the mutual time dilation between two different observers in two different inertial frames, everything puzzles me once again.

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?

Extra question: I was following a worked example here(http://galileoandeinstein.physics.virginia.edu/lectures/time_dil.html). I am confused here "How, then, can Jill claim that Jack’s clocks C1, C2 are the ones that are running slow?" at the calculation of the relative time difference between C1 and C2 in the view of Jill. How come the Lv/c^2 ?

First regarding your extra question:

Both $C_1$ and $C_2$ are observed by Jill at the same instance of her time, say $ct'$. If their respective coordinates at that moment in Jack's frame are $(x_1, ct_1)$ and $(x_2, ct_2)$, then Jill knows from the Lorentz transforms that $t' = \gamma(t_1 - \frac{v}{c^2} x_1) = \gamma(t_2 - \frac{v}{c^2} x_2)$ ( $\gamma = \left(1-v^2/c^2\right)^{-\frac{1}{2}}$ being the time dilation factor), wherefrom $t_1 - t_2 = \frac{v}{c^2} (x_1 - x_2)$, etc.

As for the main question:

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.

• Thank you for all your incredible derivation for my questions. My brain is just obtuse facing the equations. It took me a while to go through all your explanation. I think I get it now, and thank you for all the help. – Johnson Zhou Dec 29 '15 at 13:13
• I wish I can handle the Lorentz transformation as handy as you can. Can you suggest some ways to strengthen my math and the physical implications systematically? (since I am a freshman in high school and exploring the glamorous realm of physics on my own) – Johnson Zhou Dec 29 '15 at 13:15
• Plus, how to determine quickly if it is to multiply γ or divide by γ. It seems confusing to me – Johnson Zhou Dec 29 '15 at 13:34
• You are very welcome. Please let me know if you'd like more detailed intermediate steps anywhere in the derivation. As to how to handle $\gamma$, the fast rules are: length contraction = division by $\gamma$; time dilation and Lorentz transformations = multiplication by $\gamma$. When in doubt always use the Lorentz transforms to see how events relate between different frames, taking into account both space and time coordinates. – udrv Dec 29 '15 at 21:34
• Use Lorentz transforms to find coordinates in one frame from coordinates in another frame, but also use with mixed coordinates: knowing space coordinate in frame 1 and time in frame 2, you can find time in frame 1 and space coordinate in frame 2. One other important thing to always keep in mind, because it easily gets forgotten or overlooked, is relativity of simultaneity: whatever is seen at the same time in one frame is not occurring at the same time in another frame. – udrv Dec 29 '15 at 21:34