The "False Twin Paradox"

Jorrie

I have no problems with the resolution of the classical 'Twin Paradox', but the following "False Twin Paradox" does have me puzzled. (If it was discussed and explained elsewhere, please just point me there.)

Instead of the usual accelerating twin at the turnaround point, with the acceleration time taken to be insignificant for simplicity (like in this post by JesseM), we substitute the returning twin with a third inertial observer, called the "substitute away-twin", the red worldline in the attached diagram, showing a [itex]v=\pm 0.6c[/itex] scenario in semi-standard configuration.

At the flyby event of the away-twin and the substitute, the latter's clock is set to read the same date/time as the 'real' away-twin's. We can expect the substitute's calendar to read 2015 flyby event with the home-twin, whose calendar will read 2017, just as per the classical twin paradox. The argument is that the sum of the propertimes of the two away-twins between their respective events is less than the propertime of the home twin between the start and end events:

[tex]\tau'_{away} + \tau'_{away'} = \frac{\tau_{home}}{\gamma}[/tex]

My puzzle is twofold.

(i) Since the end results are the same, is this setup truly equivalent to the classical (accelerated) twins scenario with "instant acceleration" assumed?

(ii) Since we are here dealing with three purely inertial frames, being equivalent to each other, why can we decide that the sum of the proper times as measured by the two 'away-twins' is shorter than the proper time of the home twin? Does this not amount to some form of 'preferred frame', or at least deciding "who is doing the moving"?

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yuiop

Hi jorrie,

Normally we would say something like "the twin with longest path through spacetime experiences the shortest proper time" and this would be true from the point of view of any inertial observer. For your scenario we would probably say something like "the twin/s with the longest sum of paths through spacetime will experience the shortest sum of proper times". This does not amount to a preferred frame, because it is also true from the point of view of any inertial observer. Your scenario is also another way of showing that differential ageing in the twins paradox is not an issue of acceleration as there is no acceleration involved here. There are at least two other ways of demonstrating that acceleration is not the cause of the differential ageing that I can think of offhand.

Jorrie

Tx for the reply. I agree, but things get a bit more 'muddy' when we consider that by my scenario's definition, we can also say that because

[tex]
\tau'_{away} + \tau'_{away'} = \frac{\tau_{home}}{\gamma}
[/tex]

it means that

[tex]
\tau'_{away} = \tau'_{away'} = \frac{\tau_{home}}{2\gamma}
[/tex]

This has a hint of: the purely inertial away-twin's clock is "running slower" than the purely inertial home-twin's clock in an absolute way, i.e., when [itex]\tau'_{away} = 4 [/itex] then [itex]\tau_{home}=5[/itex].

We know that due to simultaneity issues, this is not acceptable thinking in SR, but it still raises the question: is it a good idea to portray the "false twin paradox" as equivalent to the classical twin paradox?

Hurkyl

Nope. It's simply the Minkowski triangle inequality -- for any triangle with time-like sides, the duration of one side is always longer than the sum of the durations of the other two sides.


IMO, no. The reason some people think the twin paradox is paradoxical no longer applies to the triplet paradox. But that fact has gone unnoticed by everyone I have seen argue that triplet paradox is paradoxical (presumably because they're too focused on invalidating the "one twin travels non-inertially" refutation).

IMO, mistakenly believing the triplet paradox itself is paradoxical requires some serious contortions of intuition. Pretty much the only way to find it paradoxical is to confusing it with the twin paradox .

michelcolman

Maybe not quite equivalent, but it's definitely a good example of a relativistic problem that makes you think, and therefore increases your understanding once you solved it. I never really understood relativity until one day I started reading about Twin Paradox, Ladder Paradox, Passing Trains, etc... Then it suddenly all made sense. Your False Twin Problem makes a nice addition to my list.

To solve this particular problem without drawing diagrams, consider the fact that, seen from the substitute away twin (SAT), the first away-twin (AT) is traveling at a very high speed (around 0.88c, the relativistic addition of 0.6c+0.6c) so his clock is inevitably much slower. In fact, earth's clock is ticking at a rate of 80% while the substitute-away-twin's is ticking at only 47%. This means he has "really" (as seen by SAT) been away a lot longer than his clock is indicating, more than twice as long (SAT's clock vs. AT's clock). During this time, earth's clock, even though it's only ticking at 80% during the entire experiment, has gotten plenty of time to get way ahead.

So, as seen by SAT, AT's trip took a lot longer than SAT's return trip. Which is only logical, since AT was only traveling at 0.28c relative to earth (=0.88c - 0.6c, in the same reference frame so simply a normal subtraction), less than half SAT's return speed of 0.6c relative to the earth. Of course the home twin will disagree...

I love twisting my head around these things :)

Al68

Another way it's equivalent is that like the standard version, the one way distance traveled by the "away" twin(s) is the proper distance/gamma. Ultimately each twin's proper time will equal his proper distance/velocity.

Regardless of acceleration, no acceleration, three frames, etc., greater proper distance equals greater proper time (assuming the same velocity).

Personally, I like to just consider a one-way trip and note that the ship's twin will have a greater elapsed time for a one-way trip. The rest will simply follow from that, regardless of the details of the scenario.

Jorrie

By 'proper distance', do you mean the distance between two events that are measured by an inertial frame in which they are observed to be simultaneous? (In analogy to 'proper time', measured in an inertial frame where the two events are co-located, both at least for the purely inertial case).

I do not quite agree that a one-way trip (without any acceleration) can tell you which 'twin' aged more, because the result will differ depending on who observes whom. AFAIK, you need a closed 'Minkowski triangle' to tell the difference in aging, as Hurkyl stated above.

Jorrie

By "triplet paradox", do you mean the type of "false twin paradox" (three frames, no acceleration) that I described? Or do you mean two of the triplets set off on non-inertial flights, with the other one staying inertial?

Hurkyl

Yes. IIRC, I've heard it called the triplet paradox; I had not heard the name "false twin paradox" before this thread.

Jorrie

OK, I've 'invented' the name "false twin paradox" as a title only - the idea was to find out to what degree it is deemed equivalent to the classical twin paradox (with 'instant acceleration').

Al68

I didn't say without acceleration, just without the return trip. We can just take the standard version except the ship just stops at the destination and stays there, at rest with earth. Ship's twin ages less. The only real difference is that we don't have the novelty of the twins' reunion.

Why would we need a closed triangle? As long as the twins end up at rest with each other, we have a common rest frame and know who's older at the end of the trip, even if the twins have to wait for their empirical evidence to arrive. Or for that matter, we could have a third observer waiting at the destination, at rest with earth, with a clock synched with earth's clock. Then the third observer's clock can be compared with the ship clock to show the ship twin aged less during the trip.

Hurkyl

There's something else you're missing -- the ability of the twins to directly compare their ages.

This version is more complicated, because in addition to space-time geometry, it depends on the choice and execution of some scheme for comparing remote clocks.


A closed loop is still required in order to make a comparison. In your example, the missing edge (and probably several more) comes from the procedure you used to "synchronize" the Earth and destination clocks.

Jorrie

I'm still interested in your definition of proper distance for the twins scenario.

Al68

Why would you say I'm missing it after I specifically mentioned it?It's not "missing", I just failed to spell it out. I assume that the standard SR convention will be used.

But these are issues of how the twins will know the results, not what the results are. The result is that less proper time will elapse for the ship twin. This would be true even if everyone lost their clocks.

Al68

The proper distance between two points or objects at rest with each other is the coordinate distance between them in their rest frame.

In the standard twins paradox, this proper distance is typically given as the distance between earth and a distant star or the midpoint of the turnaround.Yes. If we define events such as the turnaround and a specified earth clock reading, then the proper distance between those events is equal to the coordinate distance in earth's rest frame, if the turnaround is simultaneous with the specified earth clock reading in earth's rest frame.