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

Question

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?

Answer 1

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.

As far as the second question goes: what you're asking is effectively "if an observer travels from spacetime event A to spacetime event B, do they necessarily see the same amount of time elapse on their clocks?" The answer is no; the amount of time elapsed on your clock depends on the path taken through spacetime. It happens that the two proper times are the same for the first two trajectories you describe, but in general they will not agree, and Bob's proper time might be bigger or smaller than Alice's depending on the trajectory he takes.