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# Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (http://www.relativity-4-engineers.com), in the eBook or in the snippets I post here.

Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).

A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"

Regards, Jorrie

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### The Triplets Puzzle

Posted March 29, 2009 11:00 PM by Jorrie

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 Jill = √[1-0.82]dt = 0.6 x 2 = 1.2 years, while Jack's propertime interval is 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 Jill = √[1-0.42]dτJack = 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 Jack = √[1-0.42]dτJill = 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 Jill = √[1-0.42]dτJack) 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 = (VJill - VJack)(1-VJill VJack) = 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.

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#1

### Re: The Triplets Puzzle

03/30/2009 1:01 AM

Jorrie,

Thanks for another of your wonderfully clear (by comparison to others I have tried to understand) explanation. I really enjoy these, but must admit that this old brain hurts!

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### Re: The Triplets Puzzle

03/30/2009 5:25 AM

It's worth remembering, that, if ever one ever travels at a significant fraction of c.

Long time no postings, Jorrie!

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### Re: The Triplets Puzzle

03/30/2009 5:53 AM

Hi PWSlack - "Long time no postings, Jorrie!"

No, no, not a long time - been away and back at hyper-velocity and it's only people who stayed put that think it was a long time!

-J

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### Re: The Triplets Puzzle

03/30/2009 11:31 AM

Jorrie,

I find footnote #2 interesting. With my first read it didn't occur to me that when moving to Jack's inertial frame that his relative velocity to Jill would be different than it was in Jim's inertial frame. When I looked at your footnote though, it became clear.

Thanks again for another thought provoking post.

Roger

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### Re: The Triplets Puzzle

03/31/2009 2:39 AM

I love this!! I really enjoy learning/reading about things that are just outside the periphery of my mental capability.... Where can I get more?

~Dan

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### Re: The Triplets Puzzle

04/01/2009 10:52 PM

Hi Dan, you asked: "Where can I get more?"

Well, one place is to look through this Blog. The index is here.

-J

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### Re: The Triplets Puzzle

04/01/2009 8:23 PM

Watch out, here it comes!

Just kidding.

Hi Jorrie,

Your write-up seemed clear, but let's see if I understood it. The "paradox" is that Jack's calculations and Jill's calculations didn't agree with each other or with Jim's calculations which were correct. The reason is that the math they used was not correct because they were accelerating. Your note 4 seems to say that Jack and Jill could have calculated correctly with more complicated math. Its complicated enough already! So, if Jack could see Jill (in real time) he would see her aging according to Jim's calculations. correct?

-S

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### Re: The Triplets Puzzle

04/01/2009 11:38 PM

Hi S: "Watch out, here it comes!"

I tried to duck, but relativistic tomatoes...

"So, if Jack could see Jill (in real time) he would see her aging according to Jim's calculations. correct?"

Yes, but only when they are finally together (or passing each other) again. While they are apart, Jack and Jill have no way of unequivocally observing each others aging. Indirectly, they can, but this opens up all sorts of interpretations and is usually very confusing.

The best is to assume that if Jill is younger than Jack at reunion, she steadily aged slower than Jack. I say this because it is accepted today that acceleration per say does not change inherent clock rates, unless we damage the clock. (I seem to recall that you are involved in QA - acceleration may surely provide accelerated aging om systems, in the engineering-sense!)

I know that this is not a very satisfactory answer and I can work through a specific calculation if you are curious - but it may create confusion!

-J

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### Re: The Triplets Puzzle

04/02/2009 7:13 PM

Hi Jorrie,

"...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."

But does average relative speed give the correct answer? I would doubt this as we talking about non-linear relationships here. You probably know the answer, but don't need to prove it to me.

-S

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### Re: The Triplets Puzzle

04/04/2009 6:58 AM

Hi S: "But does average relative speed give the correct answer?"

No, but the point illustrated was that it is invalid to just consider relative speed when non-inertial frames are considered. Even a detailed integration of time dilation, based on relative speed, then still gives a wrong answer. The correct answer can only be obtained by analysis in an inertial frame.

Another qualitative way of stating the result is: of all the possible observers that can be present at two separate spacetime events (like departure and reunion of the triplets), it is the inertial observer that will record the longest elapsed time and the observer with the largest cumulative velocity change that will record the shortest elapsed time. When plotted on an inertial Minkowski diagram, the latter happens to be the observer with the longest spacetime path.

-J

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