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What if we modify the experiment such that both twins have a spaceship and they both do exactly the same thing on time they agreed up front?

They accelerate for 1 day the opposite direction. Then they stop accelerating and drift away from each other for 10 years (in their local time) near c. Then they simultaneously decelerate and do the same thing in reverse: accelerate backwards, drift towards each other for some years then at the same time (by their pre-agreed timing) decelerate to meet again in the middle.

If they execute the above in a perfectly symmetrical manner then the acceleration phases will be identical for both of them and they should age the same way during accelerations so we can concentrate on the non-accelerating phases of their journey. And then the twin paradox is back again. From each others frame the other will be younger but they will meet in the end...

What is the resolution of this modified paradox?

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  • $\begingroup$ This post is a duplicate of the above. Sorry for that. $\endgroup$
    – xropi
    Nov 8, 2020 at 12:25

3 Answers 3

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The resolution is to calculate the spacetime interval; it is the same for both.

Age (proper time) is not frame dependent. They will both be younger than a third who stays at home. They did the same thing so they age the same as each other. There is only a "paradox" if you expect moving to (say) the left to be different to the right.

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  • $\begingroup$ But for the entire duration of the trip, each sees the other's time running slow, except for a time of zero duration when they are both at maximum separation. That's a paradox. $\endgroup$
    – JEB
    Nov 6, 2020 at 23:44
  • $\begingroup$ @JEB That's just usual time dilation with comparing a "moving" frame to a "stationary" one. $\endgroup$ Nov 7, 2020 at 4:21
  • $\begingroup$ I don't think so. Solving the problem in Minkowski space (by using proper time) doesn't explain the paradox, because the result isn't paradoxical in Minkowski space. That A can make be older (younger) than he by moving towards (away) from him, repeatedly and reversibly is basically the Andromeda Paradox, which remains perplexing. $\endgroup$
    – JEB
    Nov 7, 2020 at 15:52
  • $\begingroup$ @JEB what the two twins actually see is described in the doppler analysis here: math.ucr.edu/home//baez/physics/Relativity/SR/TwinParadox/…. The word "see" is abused atrocioulsy in most SR teaching! $\endgroup$
    – m4r35n357
    Nov 8, 2020 at 10:22
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This is a nice form of the twin paradox, (except for the initial and final accelerations: they only confuse things, so I'll ignore them). The paradox is that each twin sees the other aging slower than himself, yet they come back the same age.

For concreteness, we need to put some numbers to it, and fly with one twin, twin A.

Lets say they leave Earth with $\gamma=10$ ($v=0.995c$), and coast for $T'=10$ years (ship time) meaning they travel $10y\times v \times \gamma \approx 99.5 ly$ according to Earth, on each leg of the trip.

First Leg: In this time, twin A ages $T'=10$ years, but his lattice of co-moving observers with clocks and rulers see twin B flying by at $\gamma = 199$ ($u=0.999987c$), having aged a mere 18 days.

Begin Deceleration: When twin A begins his deceleration day, he sees Earth is $vT'=0.95\,$ly away, and twin B is only $uT'=9.99987\,$ly away.

As he decelerates, he finds that his co-moving observers are no good: they are in the past. He picks up new co-moving observers who keep moving forward in time as he slows, and thus the one at the same position and time as twin B becomes further away, and moves into what was Twin A's future.

$V_{Earth} = 0$: Twin A has stopped. He finds the twin B is his same age, not moving, and they are both $vT'\gamma=99.5\,$ly from Earth. This is but a moment. They all agree with Earth: while they have aged 10 years, the date is 100 years after they left.

Accelerate Towards Home: As Twin A accelerates home, he discovers that his lattice of co-moving observers over at Twin B is jumping into the Earth's future, fast, and as he reaches full coasting speed, twin B is 9 years and 347 days older than he is, not because he as aged fast, but because twin A's lattice of co-moving observers has moved forward in time. At this place and time, he is only 1/2 a light year from home in Earth's frame.

Coast Home Twin B ages 18 days, while Twin A ages 10 years, winding up at the same age. Earth ages another 100 years.

The point is that although Twin B's clock is always moving slower, when they are far apart there is no sure universal definition of simultaneity, and in the Earth's frame, A's definition of now at B is in the past on the way out, and in the future on the way home (and vice, versa of course).

It is the change in a distant "now" as velocity changes that accounts for the missing time. This is not time dilation, it is reversible (if one turns around again...time dilation doesn't go backwards, but changing your distant definition of "now" does).

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There are 2 ways to analise the problem.

  1. Choosing only one frame for the trip.

In this approach, the acceleration phases only complicates matter. If the frame of reference is: twin A spaceship going away, during all the first part of the jorney twin A is getting older than twin B.

When the return trip begins, that frame is now occupied by twin B, because he has exactly that speed and direction, what defines a frame. Now twin B is getting older than twin A.

By symmetry, both effects cancel, and they meet with the same age.

  1. Changing frames during the trip.

Here the acceleration phase is also not necessary. In the frame of reference of twin A, he ages more than twin B in the first part of the journey. Just before making a U-turn, the event: twin B is x years younger than me is simultaneous to the present age of twin A.

When the return trip begins, the simultaneity changes completely. Now, in his new frame of returning trip, Twin B is x years older than me is simultaneous with the present age of twin A. As the return trip goes on, twin B is getting younger gradually, until reach the same age when they meet.

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