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Suppose we have a person A and a person B.

Person B travels very close to speed of light and never returns. He's constant in speed. Then, we can say two things:

  1. B is younger than A.
  2. A is younger than B (since we can consider B's reference as inertial).

Who is correct between the two?


marked as duplicate by Qmechanic Jan 29 '18 at 13:27

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    $\begingroup$ possible duplicate of How is the classical twin paradox resolved? $\endgroup$ – Sklivvz Dec 24 '12 at 15:18
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    $\begingroup$ This is different from the classical twin paradox in that you don't have one of the twins turning around and returning to the starting point. $\endgroup$ – David Z Dec 24 '12 at 16:06

See the answers of this question: How is the classical twin paradox resolved?

The point is that both will never be able to compare their ages without experiencing acceleration. And, acceleration makes reference frame non-inertial for which physics isn't valid.

In case of negligible acceleration in orbit to compare ages, this paper addresses the issue:

The twin paradox in compact spaces
Authors: John D. Barrow, Janna Levin
Phys.Rev. A63 (2001) 044104

Abstract: Twins travelling at constant relative velocity will each see the other's time dilate leading to the apparent paradox that each twin believes the other ages more slowly. In a finite space, the twins can both be on inertial, periodic orbits so that they have the opportunity to compare their ages when their paths cross. As we show, they will agree on their respective ages and avoid the paradox. The resolution relies on the selection of a preferred frame singled out by the topology of the space.

  • $\begingroup$ Interesting reference, nice way to be able to compare clocks without leaving an inertial frame. $\endgroup$ – twistor59 Jun 17 '12 at 12:51
  • $\begingroup$ Will clocks diverge if a stationary massive object were present in one. Side of an orbit? $\endgroup$ – Argus Jul 8 '12 at 5:40

OK, let's restate the problem just a bit for the sake of clarity.

Two persons, a & b, observe that they are moving uniformly with respect to each other and that their relative speed is close to $c$.

Both a & b observe that the other ages relatively slowly.

Now, your question: which person is correct, i.e., which person is absolutely aging more slowly?

Answer: There is no absolute time in SR.

However, there is an invariant time (proper time) associated with each person and all observers agree on the elapsed proper time for each person.


As mentioned by Alfred Centauri, they both see each other aging more slowly. For example, consider the contrary. If one of them(A) sees the other twin(B) age fast, while the twin(B) sees A age slow, this implies there is a difference due to the frame of reference. As special relativity implies, there should be no preferred frame of reference. Hence, they both should see the same - age more slowly.

More explanation here


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