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I have read about twin paradox involving acceleration when turning around in a round trip. Due to accelaration, the travelling twin is concluded to be younger. But what about the case when there is no acceleration and only a straight trip is there ? Suppose a person X stands on earth . Another person Y who is of the same age , starts off a journey from a distant galaxy. He flies and comes to the Earth at a uniform velocity. According to X , Y's clock is slower and according to Y , X's clock is slower. X expects Y to be younger , when they meet on earth. While Y expects X to be younger. Who got younger actually when they met ?

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    $\begingroup$ To check who became younger you have to synchronize twice, thus it's necessary to make them meet twice, thus the need for acceleration. $\endgroup$ – Ruslan Dec 20 '16 at 18:00
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    $\begingroup$ It's the relativistic speeds reached that determine the time differential not the acceleration. The acceleration to reach a constant near light speed is incidental to the twin problem. $\endgroup$ – scrappedcola Dec 20 '16 at 18:03
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    $\begingroup$ @scrappedcola: If there's no acceleration, then there's no paradox, since the starting moment of the voyage can't be uniquely defined. See my answer below for further details. $\endgroup$ – Michael Seifert Dec 20 '16 at 18:32
  • $\begingroup$ @scrappedcola If there is no acceleration then in order for the twins to meet more than once then they must follow the same trajectory, as there is exactly one geodesic between any two events in flat spacetime. $\endgroup$ – tfb Dec 20 '16 at 21:19
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In your described situation, the two observers will not agree on "when" the voyage started, and so it is perfectly fine for them to disagree on how much time elapsed between the departure and the arrival. This can be seen most easily with a spacetime diagram:

enter image description here

In this diagram, the thick red line represents the worldline of your Person X, and the thick blue line represents the worldline of your person Y. Person Y departs from the distant galaxy at Event A, and arrives at Earth at Event B. The black dashed line represents a light pulse emitted from Event A towards Earth (drawn for reference only.) The coordinates $x$ and $t$ correspond to space and time as observed in the rest frame of the Earth.

Person X thinks that Event C is simultaneous with Event A. She then does the appropriate calculations, and concludes that the amount of proper time elapsed between Events A and B is less than the proper time elapsed between Events C and B. And she's right: $\tau_{AB} < \tau_{CB}$. Thus, she can conclude that "the amount of time elapsed on Person Y's clock between the moment he departed and the moment he arrived is less than the time elapsed on my clock."

Person Y, on the other hand, does not see Event C as simultaneous with Event A. Rather, he sees Event D as simultaneous with Event A.1 According to him, Event C is just an arbitrary moment that occurred some time before he left. He therefore does the same sorts of calculations that Person X did, and concludes that the proper time between Events A and B is greater than the proper time between Events D and B. He, too, is correct: $\tau_{AB} > \tau_{DB}$; and he can then say that "the amount of time elapsed on Person X's clock between the moment I departed and the moment I arrived is less than the time elapsed on my clock."

But there is no contradiction here, because the two observers mean different things by "the moment I/Person X departed". In fact, we have $\tau_{DB} < \tau_{AB} < \tau_{CB}$. The contradiction disappears once we carefully define our time intervals in terms of spacetime events, and realize that events that are simultaneous in one inertial reference frame are generally not simultaneous in another.


1 All the events that Person Y sees as simultaneous with Event A are indicated by the blue dotted line in the diagram, which is obtained by reflecting the worldline about the black dashed photon worldline.

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  • $\begingroup$ An absolutely great explanation !! Just one thing. If I got the Mirror reflection part right , isn't the line AD the world line of x w.r.t to Y $\endgroup$ – Shashaank Dec 20 '16 at 20:03

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