How does time dilation work for two people inertially floating in space but passing each other at near the speed of light relative to one another?

Question

Person A is floating in space, Person B zooms past them at near the speed of light. It's not possible for either of them to know which one is moving, they are simply moving relative to each other.

My first question was, which one of them is aging slower due to time dilation? The answer I found is that each one observes the other aging slower from their respective points of view.

However, if we assume both are moving inertially but end up crossing paths again due to curvature of spacetime. Will one of them have aged more than the other at their second meeting? More specifically, would they both measure the same amount of time passing from their initial meeting to their second meeting? If so, then how does time dilation work at all? Is it just an apparent slowing in time rather than actual?

Answer 1

If you consider time dilation strictly within special relativity, there is neither spacetime curvature nor acceleration. If B moves past A, then in the restframe of A, B is moving a distance $\Delta x_B$, between, say, two flags at rest relative to A, in a time $\Delta t_B$. B passing each of the flags is an event in spacetime. In the restframe of B, the flags are moving and passing B one after the other, at a temporal distance $\Delta\tilde t_B$, with a spatial distance of the two events (not of the flags as such) of $\Delta\tilde x_B=0$. This yields

$$(c\Delta t_B)^2-(\Delta x_B)^2=(c\Delta\tilde t_B)^2$$

meaning that $\Delta\tilde t_B$ ist smaller than $\Delta t_B$. So, one might be tempted to say, at least loosely speaking, that B is aging more slowly. But this way of loosely speaking is the source of confusion. A is moving relative to B, so just as well A should be aging more slowly, right? But note that the above setup is not symmetrical. We considered the time (and space) intervals between two events defined by B passing two flags at rest relative to A. The corresponding setup, with the roles of A and B switched, would have A passing two flags at rest relative to B.

The time intervals considered above (and the corresponding time intervals if the roles of A and B are switched) are the respective time intervals between the events defined by B passing the flags at rest relative to A (A passing the flags at rest relative to B) as observed by A and B in each case. These time intervals are really, physically different; this difference is not apparent or a delusion.

As a side note, as to the statement "observes the other aging slower": The actual visual impression of an observer is a more complicated issue, as the visual impression depends on the light from the observed object reaching the observer's eye at any given instant. This involves taking into account light travel time between object and observer.

Considering spacetime curvature: For an observer travelling on a trajectory $X^a(s)$, with $s$ a parameter of the trajectory and $0\leq a\leq3$, the proper time $\tau$ of the observer is obtained via

$$\left({d\tau\over ds}\right)^2=g_{ab}(s){dX^a\over ds}{dX^b\over ds}$$

with $g_{ab}$ the spacetime metric. Apply this to two trajectories, one for A, one for B, which are such that A and B meet twice along the trajectories. The proper times $\Delta\tau_A$ and $\Delta\tau_B$ between the two encounters will, in general, not be equal.

As a special case of the above, you could consider an inertial frame (laboratory) in which A is at rest. B is moving along a circle in this laboratory, repeatedly passing A. Evidently, this situation is not symmetrical between A and B. A will find B to age more slowly (each time B passes A, the two can compare clocks they have with them). B can attribute this difference to his motion relative to A, or to the "gravitational" time dilation caused by the centrifugal force B experiences.