Yes, from the point any two observers in the same inertial frame 
synchronize clocks, their clocks will be synchronized in p-time BUT ONLY 
FROM THEN ON (we can't know if they were previously synchronized unless we 
know their acceleration histories). And only SO LONG AS they continue in 
the same inertial frame OR undergo symmetric accelerations. 

Same ages is just a way to ensure synchronized clocks at the birth event 
and make examples simpler. It has nothing to do with p-time synchrony per 

So in your next paragraph your and Jimbo's proper clocks ARE synchronized 
in p-time from then on under the conditions stated.

But I don't understand the rest of your example since you just stated that 
we are to ignore their PREVIOUS and SUBSEQUENT acceleration histories to 
preserve the synchronies but then you start giving an example with 
accelerations, which will obviously change their synchrony UNLESS they are 
symmetric. You seem to claim that the accelerations are symmetric but you 
keep describing them as stopping in different frames at different times 
which indicates they are NOT symmetric.

The only way to ensure the accelerations are symmetric is for both A and B 
to have the same proper accelerations at the same proper times AFTER they 
synchronize clocks. Are you doing that? If not you are not using MY method.

Also you seem to be switching from synchronized proper clocks which I 
assumed did NOT reflect actual ages to ACTUAL AGES which doesn't work.

I used actual ages synchronized at birth (twins) to avoid that kind of 



On Wednesday, March 5, 2014 4:23:54 PM UTC-5, jessem wrote:
> On Wed, Mar 5, 2014 at 2:42 PM, Edgar L. Owen <<javascript:>
> > wrote:
> Jesse,
> Yes, but respectfully, what I'm saying is that your example doesn't 
> represent my method OR results.
> In your example of A and B separated but moving at the same velocity and 
> direction, and C and D separated but moving at the same velocity and 
> direction, BUT the two PAIRS moving at different velocities, AND where B 
> and C happen to pass each other at the same point in spacetime here is my 
> result.
> Assuming the acceleration/gravitation histories of A and B are the same 
> and they are twins; AND the acceleration/gravitation histories of C and D 
> are the same and they are twins, then A(t1)=B(t1)=C(t2)=D(t2) which is 
> clearly transitive between all 4 parties.
> You earlier agreed that if two observers are at rest relative to each 
> other, then if they synchronize clocks in their rest frame, their clocks 
> will also be synchronized in p-time from then on. In your post at 
>  responded to
> ...

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