Re: [SR] Their proper times will necessarily be equal

Richard Hachel wrote:

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"If two different observers travel an identical path in equal observable times,

"Observable"?
WHO is doing the observing?  Presumably, an observer at earth, yes?

then their proper times will necessarily be equal.

This is not possible.

  Let's set up a Galilean frame of reference, in which a Galilean mobile moves from left to right on the x axis.
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  When it goes to O, in R, we click.
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  When it passes x, let's say 12 al, we click.
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  For example, we obtain To=12.915.
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  Or a speed of Vo=x/To=0.9291c
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  Tr=To.sqrt(1-Vo²/c²)=4.776

This is assuming a constant velocity of v = 0.9291c, yes.  The proper time
of the trip is 4.7746 years, using c = 299792458 m/s and one year equal to
3600*24*365.25 = 31557600 seconds. 

  Let us pose another body, but this time in uniformly accelerated motion,

WHO sees the "uniformly accelerated motion"?  The earth observer or the one
in the ship?

and whose speed will be specially chosen so that To=12.915.

Neither the earth observer nor the ship observer sees a "specially chosen"
speed.  Both will see the ship going faster and faster.  An ACCELERATION
must be specially chosen.

This means that in R, if they leave together, they arrive together (even if they do not have the same speed between them).

The equations are here (Equ. 6a):
https://en.wikipedia.org/wiki/Acceleration_(special_relativity)
I set the problem up using excel.  For the ships to arrive together at the
same time when viewed in the earth frame, the second ship must accelerate
at 10 m/sec^2.  The proper time (ship's time) will take 1.3632 years.  The
constant-v ship's time is 4.7746 years, so their proper times are NOT equal.