In the Twin Paradox two twins initially at rest in the same reference frame are separated and take different journeys through spacetime. Eventually they are reunited. What is the condition for the reunion?

In Newtonian physics, they are reunited when they have the same spacetime coordinates. From a Special Relativity viewpoint they have different spacetime coordinates. In other words, how can the twins decide on a rendezvous point? It is OK to assume that one twin remains stationary.

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    $\begingroup$ The condition for reunion is that both twins have the same spacetime coordinates. This is true whether you are talking about Newtonian physics or relativistic physics. $\endgroup$ – David H Jan 21 '14 at 13:05
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    $\begingroup$ Each twin compares their own position with their twins position in their own coordinate system, and calculates the difference vector in their own coordinate system. The two twins may not agree what the difference vector is, but they will both agree that when the difference vector falls to zero they are at the same spacetime point. $\endgroup$ – John Rennie Jan 21 '14 at 15:08
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    $\begingroup$ The condition is that their worldlines cross. $\endgroup$ – WillO Jan 21 '14 at 16:00
  • $\begingroup$ I think your comment that they have "different spacetime coordinates" when they reunite might indicate some confusion. Say that at the moment they observe the other passing right next to themselves, twin A's clock shows 1000 seconds have passed since they departed, and twin B's clock shows 800 seconds have passed since they departed. Then the two events "twin A's clock reads 1000 s" and "twin B's clock reads 800 s" are both assigned the same spacetime coordinates as one another, regardless of what coordinate system we choose to use. $\endgroup$ – Hypnosifl Oct 5 '14 at 22:21

If I'm reading your question correctly, the problem you're describing is that, even if the twins decide to meet at the Starbucks on 5th and Main upon the astronaut twin's return, they won't be able to agree on a time since their respective clocks will be out of sync. Depending on the speed of the astronaut's travel, there could be several years difference in the clocks carried by the twins, so how does one do scheduling? I'm assuming there isn't a calendar app on their smartphones that does relativistic time zone corrections.

The most direct way would be for the twins to meet beforehand, with the astronaut's travel plans at hand (including exact travel speed), and work out what the resultant time difference would be. For a trip to Alpha Centauri and back at $0.9999c$, the Earth twin (Bob) would calculate that the return would happen 8.74 years (8 years and 270 days) after departure. The astronaut twin (Alice (these are fraternal twins)) would calculate that she would return in 0.124 years (45 days). Each would have to put very different appointments into their calendars.

This is actually difficult because the time difference varies a lot at such high speeds. If Alice made a mistake and only traveled at $0.999c$, the difference for the Bob on Earth would be 3 days later. The difference for Alice would be 100 days later. A similar situation using gravitational time dilation was depicted in the movie Interstellar on the ocean planet. A few minutes delay on the planet was experienced as a several years delay on the orbiting space craft.

About the only other way I can think of synchronizing their rendezvous is for Alice to signal the planet on its way back, perhaps by transmitting a radio beacon every time 10% of the distance between Alpha Centauri and Earth is crossed. That way, Bob can make time in his schedule with a few days notice (from Earth's perspective, these radio signals will not outpace the space ship by much). If the spaceship traveled more slowly, say at $0.5c$, this notice could increase to a year or more.

The fundamental problem is that it is impossible to synchronize clocks between observers traveling at different velocities.

Of course, the easiest solution would be for Alice to simply show up at Bob's house unannounced and drag him to the agreed upon Starbucks to show off pictures of her travels (which is exactly what my sister would do in this situation).


In the Twin Paradox two twins initially at rest in the same reference frame are separated

It is more usual (because conceptionally simpler) to consider "twin experiments" in which the two protagonists who were initially coincident to each other are separated. In this case, the requirement may even be dropped that any one, or both, of the twins were ever members of an inertial frame (i.e. at rest wrt. suitable other participants); initially, finally, or while having been separate from each other.

However, the given setup that the twins were "initially at rest (to each other)" without having met initially (i.e. without having initially been coincident, but separate from each other) can be considered as well by referring to Einstein's definition of simultaneity, for the two separate twins to agree on their respective "initial" indications.

are separated and take different journeys through spacetime. Eventually they are reunited.

Right, that's the usual setup prescription.

What is the condition for the reunion?

Well -- the notion of two (or more) distinguishable participants having been "united (a.k.a. "coincident"), or else "separated from each other", is pretty much considered self-evident (axiomatic) in RT. As Einstein put it:

"All our well-substantiated space-time propositions amount to the determination of space-time coincidences {such as} encounters between two or more {...} material points".

When being careful to distinguish "coincidence (instead of sequence) of distinct observations by any one participant" from "coincidence of several distinct participants (in the sense of a "reunion" or "meeting" or "having participated in the same coincidence event") then it may be said that

  • the two twins were "separate" from each other while they observed non-zero ping intervals between each other (i.e. for each signal indication each twin observed a corresponding "ping echo" from the other distinctly "after" having stated the signal); or conversely

  • the two twins had "met" (or "reunited") if for each signal indication each twin observed a corresponding "ping echo" from the other as the same indication (or in other words: "at once").

how can the twins decide on a rendezvous point?

The "rendezvous" event (in which the two twins participate, in coincidence) is simply characterized as the first event in which both twins participated after their respective "initial" indications (or, in the usual setup, after their common "initial" coincidence event).

Obviously, these considerations of physics and geometric relations don't involve or require any particular coordinate assignments to the twins and their individual indications, or even their representations as lines in $\mathbb R^n$.


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