(adsbygoogle = window.adsbygoogle || []).push({}); The "False Twin Paradox"

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 (https://www.physicsforums.com/showpost.php?p=2443751&postcount=31"), 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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# The False Twin Paradox

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