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The twin paradox in special relativity has been discussed over and over again. Send a twin on a spaceship out to someplace or another accelerating at Earth gravity, then have it go through a series of decelerations and accelerations -- all at 1g -- so it returns to Earth, where the other twin is waiting. And, depending on how long the trip was, the Earth-bound twin is, say, seventy years old while the spaceship twin has aged just a few years.

Within special relativity, it all makes sense. I have done the calculation etc. But what about general relativity? The fundamental observation of GR is, as I understand it, that all accelerating frames with a given acceleration are equivalent. (And I'm betting that my understanding of exactly what "equivalent" means is the answer to my question. But, proceeding ... ) So, the twin on the Earth experiences an acceleration of 1g, as does the twin on the spaceship. Why are their respective frames not equivalent, and they age differently? In a related question, what if we had triplets instead of twins and the third triplet spent the whole time weightless (ignoring health effects) in a space station orbiting the Earth? How would he age compared to his siblings?

Qmechanic
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bob.sacamento
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  • General relativity is an effect compound with that of special relativity. Despite the somewhat misleading names. i.e. if something is moving fast and accelerating, both effects need to be applied. – JMLCarter Oct 23 '17 at 20:17
  • Your related question is easier to answer. In a curved spacetime, we can consider all of the possible worldlines that start at Event A and end at Event B. An observer who is freely falling between these two events is said to be following a geodesic in the spacetime. One way to define a (timelike) geodesic in spacetime is that it is the path between two events that maximizes the amount of proper time between them. Thus, the orbiting triplet experiences the largest amount of time among the three of them, and thus ages the most. – Michael Seifert Oct 23 '17 at 20:31
  • I don't have a good answer for the main question, though. The role of observers in General Relativity is limited, to be honest, and there's not really an equivalence between them in the same way as there is in special relativity. I'll point you to this answer for now; if I think of a pithier way to explain it later, I'll be sure to post it. – Michael Seifert Oct 23 '17 at 20:33
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    The twin's world lines through the two events (parting and reuniting) are not 'equivalent'. In either the flat or curved spacetime version of the TP, the essential idea is that, through any two (time-like separated) events, there is one* word line that is extremal, i.e., accumulates the largest proper time lapse, and thus is a geodesic of the spacetime - the world line of an object in free-fall (no proper acceleration) . For two accelerated objects, it isn't enough that both have the same magnitude of (proper) acceleration to have the same elapsed time. – Alfred Centauri Oct 23 '17 at 21:40
  • As JMLCarter alludes to, the speed of the stay-at-home twin doesn't (ideally) change relative to Earth while the speed of the twin in the rocket does. – Alfred Centauri Oct 23 '17 at 21:45
  • @alfred-centauri OK, I figured someone would give an answer along those lines. It's just that, again, I thought that GR saw any deviation of a given rate from a geodesic as "equivalent". I guess maybe I'm taking "equivalent" too seriously? Care to elaborate? – bob.sacamento Oct 24 '17 at 14:19

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bob.sacamento asked: "But what about general relativity? "

In general relativity the twin that stayed on the planet can be younger than the twin travelling up and down again, thanks to the principle of maximized proper time.

bob.sacamento asked: "In a related question, what if we had triplets instead of twins and the third triplet spent the whole time weightless (ignoring health effects) in a space station orbiting the Earth? How would he age compared to his siblings?"

If we place one triplet (green) stationary at $\rm r_0=2 \ r_s$, launch another into orbit (red) and the third one (blue) is shot up with the required velocity to have him back at the same event the orbiting triplet returns, the blue triplet will have the longest proper time $(\tau=39.829 \rm \ GM/c^3)$, the one who stayed at home is in the middle $(\tau=8 \pi \surd 2=35.543 \rm \ GM/c^3)$ and the one in orbit will be the youngest $(\tau=8 \pi = 25.133 \rm \ GM/c^3)$, for the detailed calculation see here.

twin paradoxon general relativity

In this example the launch velocities required are $\rm v=c/ \surd 2$ for the red one in a circular orbit and $\rm v=0.574 \rm \ c$ for the blue one going up and down.

Yukterez
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According to The General Theory of Relativity, acceleration and gravitation are not "distinguishable from each other!" The "thought experiment in an elevator" is used to prove this! (As well as showing that light falls in a gravitational field!)

According to Special Relativity there is no "absolute speed", nor are there any "favored" frames of reference!

Postulating the Twin Paradox, is like taking a "walk out of Relativity", in order to postulate a twin as moving "close to the speed of light", when the other twin is "at rest on Earth".

(This is not "possible", to make such a statement, according to Special relativity, since all speeds are measured as being relative to something else!)

Anders Kouru
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