The so-called "triplets paradox" is a variation of the well known "twin paradox" of special relativity. They are not really paradoxes, but rather misinterpretations of the postulates of Einstein's relativity theory. Nevertheless, both can be quite puzzling to the uninitiated.

Let the triplets be Jim, Jack and Jill. Jim mans a space station, floating inertially somewhere in free space. His two siblings set off simultaneously from there, each to his/her own distant planet and back, using individual rocket propelled spacecraft. Apart from brief accelerations^[1] at the beginning, at turnaround and at their return, Jack and Jill also float inertially at a constant speed relative to the space station. They arrive back simultaneously, exactly two years later, by the space station clock.

As the space station measured things, Jill's target planet is 0.8 light years away and Jack's target planet 0.6 light years distant from the ship. So, Jill's round trip is 1.6 light years in length and Jack's 1.2 light years, both in space ship coordinates. We can easily work out the speeds of Jill and Jack relative to the spaceship: Jill must fly at 0.8c all the way there and back and Jack must fly at 0.6c, so they will turn around simultaneously, after exactly one year space station time.

The usual question is now: how much did Jack and Jill respectively age during the two "Jim-years" of their respective voyages?

The standard special relativity solution is simple - Jill's 'dilated' propertime interval is dτ_Jill = √[1-0.8²]dt = 0.6 x 2 = 1.2 years, while Jack's propertime interval is dτ_Jack = √[1-0.6²]dt = 0.8 x 2 = 1.6 years. Hence, Jack ages 0.4 years more than Jill, due to his slower speed relative to the station.

Now, the puzzle: flying in the same direction (the two planets are spatially lined up as shown above), there is a relative speed between Jack and Jill of around 0.4c on average.^[2] Jack should hence observe Jill to age slower than himself, obtaining her age increment as dτ_Jill = √[1-0.4²]dτ_Jack = 0.92 x 1.6 = 1.47 years. This gives a difference in aging between Jill and Jack of only 0.13 years, not the 0.4 years calculated in the standard solution above. Likewise, Jill should observe Jack to age slower than herself, obtaining his age increment as dτ_Jack = √[1-0.4²]dτ_Jill = 0.92 x 1.2 = 1.1 years. This gives a difference in aging between Jack and Jill of 0.1 years.

Huh!? This obviously cannot be, so where's the catch? It's in the figure to the left. Well, almost!

Although Jim's inertial frame has been chosen as the reference for speeds, we are free to choose any inertial frame as reference, provided we do our relativity correctly in that frame. In this Minkowski diagram, Jack's outbound frame has been chosen as reference, showing how the other inertial frames move relative to it. On his inbound flight (after accelerating and turning around), Jack himself moves at v=-0.88c relative to the chosen inertial frame. After her turnaround, Jill moves at v=-0.95c relative to this reference frame.

Note that the longer the spacetime path between two events on the Minkowski diagram, the less propertime it takes to move along that path. Jim's blue spacetime path is obviously the shortest and hence he records the longest time interval between the two events (departure and reunion). This non-intuitive behavior comes from the hyperbolic nature of the Minkowski diagram.^[3]

In case you want to check the numbers, the scale of a Minkowski worldline of an object moving spatially at speed v in the reference frame is given by:

Scale(moving) = √[(1+V²)/(1-V²)] x Scale(reference),

where 'scale' refers to linearly plotted length per geometric unit and V is implied to mean v/c, a decimal fraction of the speed of light.

The answer is now obvious: relative speed per se has nothing to do with age differences in relativity. Jack and Jill could have flown off in opposite directions at the same coordinate speeds, giving a much higher speed relative to each other, without influencing their relative ages at reunion. It is the length of the Minkowski spacetime path taken that is at the root of the issue. The longer a spacetime path between two given events, the shorter the elapsed time along that path. In such a case, "time movement is traded for spatial movement". Also, note that acceleration per se does not affect the rate of time either. It is the consequential change in spacetime path length that does it.

A simpler, but less useful answer to the puzzle is: you cannot use simple time dilation calculations (like dτ_Jill = √[1-0.4²]dτ_Jack) for time interval differences between two non-inertial frames,^[4] e.g. Jack and Jill's. At least one of them must be at rest in an inertial (non-accelerated) frame. That's why picking Jim's space station coordinates is the simplest for calculating time interval differences.

You are welcome to fire away with comments, questions (and/or soft tomatoes).

-J

[1] Assume that the acceleration times are negligible relative to the trip times. This can be achieved by very high accelerations, or by making the trips very long, e.g. i.e thousands of years, ignoring the mortality of the triplets...

[2] The relativistic subtraction of the two speeds is given by: V = (V_Jill - V_Jack)(1-V_Jill V_Jack) = 0.39, where V is implied to mean v/c, a decimal fraction of the speed of light. This is the case for most of the trip. Then there is short while (~0.05 years) where Jack has turned around and according to him, Jill is still going in the original direction, making their relative speed ~0.95c for that brief period. This gives an average relative speed of about 0.40c over the whole trip.

[3] You can read more on the hyperbolic nature of Minkowski spacetime diagrams on my website Relativity-4-Engineers, chapter 2, page 40, fig.2.7.

[4] It is possible to use any non-inertial frame as reference frame for any calculation that you wish, but it becomes much more complex to do so.