# If someone flies away from me quickly, and then I catch up to them, will our clocks be in sync or out of sync?

At time 0, Alice and Bob are stationary in space with their clocks synced. Then Alice, for $$t=\text{1 day}$$ (measured by herself), flies in one direction at the speed $$v=0.99 c$$, and then stops. Due to length contraction, she'll've traveled a distance of $$D=\frac{v t}{\sqrt{1-(v/c)^2}}$$. Alice and Bob's clocks will be out of sync, and to Bob, it'll look like Alice was traveling for $$T = \frac{t}{\sqrt{1-(v/c)^2}}$$.

After the amount of time $$T$$ passes for Bob, he goes after Alice with speed $$v=0.99 c$$, reaches her after $$t=\text{1 day}$$ (as measured by Bob), then stops. My questions are

1. When Alice and Bob are together again, will their clocks be synced up?
2. If Bob had chased after Alice at a different speed, then when he reached her, would their clocks be synced up?
• Those "for $t=1$ day", measured by whom? What are their acceleration profiles? The answer will critically depend upon these things. Commented May 11, 2023 at 8:29
• Lorentz boosts do not commute. In general, the clocks will show different times. The twins paradox (it is not a paradox, we call it so because it is biologically weird) would not be a thing otherwise. You can resolve all your doubts by realizing that for these scenarios to happen, the frames have to stop being inertial, so every symmetry consideration based on universality etc breaks down, and the laws you have written are good for inertial frames Commented May 11, 2023 at 8:57
• It would probably be most clear if you would draw a spacetime diagram of the respective worldlines. Commented May 11, 2023 at 10:06
• Where are you stuck? Commented May 11, 2023 at 14:37

Alice's "first leg" of her trip through spacetime (where she travels for a specific time according to her clock at a given speed in some reference frame) is identical to Bob's "second leg". And Bob's "first leg" of his trip through spacetime (where he sits around and waits for a time $$T$$) is identical to Alice's "second leg". So the proper time elapsed on their clocks will be the same.