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Two twin sisters synchronize their watches and simultaneously (from the earth frame) depart earth in different directions. Following a predetermined flight plan, each sister accelerates identically to 99.9%c and then returns home at the same time (again in the earth frame). The observers on earth see each twin as having aged identically, as the symmetry of the problem dictates. Each twin however should see the other as having aged... What gives?

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In a sense this is easier than the basic "paradox". It is sufficient to note that the difference of elapsed time observed by one twin must be equal to that of the other. To quantify it you use the same resolution as in the basic "paradox". – dmckee Jun 11 '12 at 17:38
For more on this see What is the proper way to explain the twin paradox? – John Rennie Mar 13 at 11:51

The resolution of this apparent paradox introduces no new ideas with respect to the canonical version of the paradox. As you point out, the journey of each twin is symmetric, and so each twin calculates that they will age the same amount by the time they return home.

If, instead of appealing to symmetry, you would like to do the calculation from the perspective of one twin reasoning about the passage of time of the other twin, the reasoning is as follows: During the outgoing phase of the journeys, each twin calculates that less time has elapsed on the other twin's clock than on their own. But during the turnaround, they calculate that the other twin ages rapidly to such an extent that, even accounting for the time dilation for the return journey, their ages will be the same when they meet up back at their starting point.

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We can simplify the problem by assuming that the twin sisters leave Earth already at 0.999c and when they return to the earth they are still doing 0.999c. That means the only acceleration takes place when the sisters reverse speed at the halfway point of their journey.

We can all compare watches on two points that everyone is in the same place i.e. when the sisters leave Earth we can all synchronise watches so they read the same, and when the sisters return we can all compare the readings on the watches. The fact the sisters are moving at this moment doesn't matter as long as both the sisters and we are in the same place.

So consider what sister A sees: working forward from the start point, sister A sees sister B moving at 0.9999995c (relativistic addition of 0.999c to 0.999c) so the time dilation is 1,000. So sister A sees sister B aging at 1/1000 of the normal rate. Likewise for sister B.

Now take the end point, at which the sisters are the same age, and follow the sister's trajectories back in time. Again sister A will see sister B aging at 1/1000 of the normal rate, and again likewise for sister B.

If we follow the trajectories forward from the beginning, and backwards from the end, then obviously at some point the trajectories must meet, and this means the sisters must have changed from the initial to the final inertial frame i.e. they must have accelerated to change inertial frames. Lets consider this from sister A's point of view.

During the acceleration sister A feels a force, but considers herself not to be moving. So from her perspective sister B accelerated to change her speed from 0.9999995c to -0.9999995c and even though sister B was aging at 1/1000 of the normal rate, during the acceleration sister B's time catches up with and overtakes sister A's time. During the inbound leg sister A sees sister B aging slowly again, but the age jump during the acceleration makes up the difference. Sister B sees exactly the same, i.e. she sees sister A age during the acceleration.

So that's why the sisters end up the same age. They do indeed see each other aging more slowly, but it's during the acceleration between the outbound and inbound legs that each others time catches up.

I haven't attempted to do the sums because to be honest I've long since forgotten it and I'd have to go away and look it up, but you can probably find it on the web. Contrary to popular belief you don't need general relativity as special relativity is perfectly capable of handling accelerations, as long as you're content to treat acceleration as absolute. My recollection of doing the calculation at college is that it was messy and actually not very illuminating.

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Good explanation +1. But you do need general relativity, to explain what happens in the acceleration phase. That is also why you elegantly skip that part. General relativity is also needed to explain the non-symetry of the normal twin paradox(one stays in the same inertial frame, and the other goes trough an acceleration at one point). – Hans-Peter E. Kristiansen Jun 11 '12 at 18:26
No you don't need GR--- the General Relativity is just ridiculous here. The proper answer is "draw the space-time diagram". – Ron Maimon Jun 14 '12 at 0:37
The problem with this answer is that the sister does not "see" the other sister age during the turn around, but the other sister "coordinate ages" because the coordinate now moves as the sister moves to a new reference frame. The whole thing is very simple if you look at the Euclidean geometry analog, as every other relativity paradox. – Ron Maimon Jun 14 '12 at 0:38
@Ron Maimon: the GR is not ridiculous. You can sit down and do the math from the perspective of one of the twins. During the acceleration due to the equivalence principle the twin sees a distance-dependent gravitational time dilation as causing the other twin to age. If you do the calculation you will see that you get the correct result. So your dismissal of this approach is unfair; it is a matter of personal/philosophical preference. – user1247 Jan 29 '13 at 19:45
@user1247: This is not exactly true--- it is only true if there is a uniform constant acceleration for one of the twins, and then it is true that you can view it either in the constant acceleration frame or in the unaccelerated frame, and then you do get the same answer, and it's equivalent. But the reason it's ridiculous anyway is because you are looking at flat space and answering "why is this broken line a different length than this straight line"? You wouldn't answer that by making a curvilinear coordinate system in Euclidean geometry, so you shouldn't do that in space-time either. – Ron Maimon Jan 31 '13 at 0:03

John Rennie's answer doesn't sound quite right. This problem can be restated with no acceleration. Suppose we have 4 sisters {A1,A2} and {B1,B2}. Sisters A1 and B1 leave the earth simultaneously and agree to keep traveling on a non accelerated trajectory indefinitely. At the same time (from the earth frame) sisters A2 and B2 start traveling towards earth. Some where in the middle Sister A2 passes A1 and B2 passes B1. Sisters A1 and B1 transmit their current wallclock time to their respective sisters. Here there is no acceleration, hence acceleration can not be part of the problem.

Or what if the universe is closed and they meet at the starting point after a finite time. Again, no acceleration.

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Closed universe is not relativistically invariant. As for the passing sisters, the "jump" comes from the changing frames, and is no more mysterious than when you rotate a frame that the y coordinate of something far away jumps. – Ron Maimon Jun 11 '12 at 22:48
Ok.. you don't like closed space... fair enough, how about flat space with periodic boundary conditions. That's invariant. Also, what jump are you referring to. I boosted no particles, I simply transmitted information between frames. And if both frames {A1,A2} are moving with the same absolute velocity (albeit in opposite directions) the time dilation should be constant from the perspective of other twins. There is no acceleration or boost. Something is screwy here. – ksh95 Jun 11 '12 at 23:01
Flat space with periodic boundary conditions is not invariant. When you boost, the boundaries become pasted with a time-jump. There is a preferred frame where the time-jump is zero, and this is the rest frame of the boundary conditions. There is nothing screwy--- draw the picture in space time and it will be obvious. Is it screwy that in a cylinder the path that wraps once around while going up is longer than the path that just went up? The cylinder is the Euclidean 2d periodic boundary in one of the dimensions. – Ron Maimon Jun 12 '12 at 2:47
+1 : The nice but wrong answer, together with @RonMaimon 's rebuttal helped me to understand the twin paradox better. – Frédéric Grosshans Jun 13 '12 at 17:28

A simple explanation appears to be that each twin are the same age after the journey because they traveled paths of equal length through spacetime. In other words their perspective watches measured the same amount of proper time.

In the canonical version of the twin paradox the difference in age of the earth bound twin, and the space-traveling twin can viewed as that the earth bound twin followed an essentially straight line through spacetime, and the space-traveling twin followed a curved path through spacetime. In Minkowski space, unlike Euclidean space, a curved path is shorter than a straight path.

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One poster commented that General Relativity was "needed to explain the non-symmetry of the normal twin paradox" since one twin accelerates. Special relativity can handle the case of accelerating objects (in this case a space-traveling twin). The key is to imagine the acceleration as a series of Lorentz transformations between inertial reference frames that differ infinitesimally in relative velocity. – PhysGrad Jun 13 '12 at 17:58
See, for example, "Spacetime Physics" by Taylor and Wheeler. This "solves" the twin paradox taking into account acceleration, entirely using SR. Indeed GR reduces to SR in the absence of gravity, and as gravity doesn't enter into the problem, there is no need for GR. – Peter Webb Mar 1 '15 at 17:50

I'd stick to simple for this one: Both twins accelerate and then decelerate equally, while the earth observer stays put with close to zero accumulated acceleration. So Minkowski's very first approach to answering this question -- "who is accelerating?" to paraphrase -- applies nicely. When everyone returns to the same location and velocity and compare clocks, the twins simply find they are the same (low) age, and that everyone on earth aged faster.

Incidentally, don't let discussions of "viewing" each other's ages through telescopes throw you off on this one. With enough data you can place brackets around permissible time deltas that way, but you can't give final answers because you don't know how much of the spatial distance between you may still end up converted into time distance. The shared location, shared frame finale in contrast gives answers that depend only on who spent more time accelerating.

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protected by Qmechanic Jan 29 '13 at 15:04

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