yogi said:
Triplet - in the triplet version - there is no acceleration - the outbound sibbling simply transfers his clock reading to an inbound third triplet - there is no acceleration felt at this transfer - so the traditional twin explanation often relied upon by relativists, including Einstein in his early papers, to wit: "that the turn-around twin feels the acceleration - therefore the experiment is not symmetrical" - is no longer applicable.
The acceleration in the Twin paradox is only needed to reconcile total elasped Earth time as measured by the stay at home twin with that as measured by the traveling twin.
In the triplet version, again all we need to do is reconcile what each triplet measures as the outcome.
Thusly:
Assume Triplet 1 sits at home.
Triplet 2 is outward bound at .866c relative to triplet 1
Triplet 3 is inward bound at .866c. relative to triplet 1
Triplets 2 and 3 pass each other 10ly from triplet 1 as measured by triplet 1
Thus from the view of triplet 1:
Triplet 2 takes 11.55 years to travel out to 10ly
Triplet 2's clock will only show 5.775 yrs at this point.
time from triplet 2 is transferred to triplet 3.
Triplet 3 takes 11.55 years to travel inward from 10ly, for a total elasped time of 23.1 years by triplet 1's clock.
Triplet 3's clock gains 5.775 yrs at this during this time, for a total elasped time of 11.5 years.
From triplet 2's view:
Due to length contraction, he travels out to 5ly until meeting triplet 3, at which time his clock reads 5ly/.866c = 5.775 yrs.
Triplet 1's clock would read 2.8875 yrs at this point.
time from triplet 2 is transferred to triplet 3
triplet 3 is moving at
\frac{.866c+.866c}{1+\frac{.866c^{2}}{c^{2}}}
or .9897c relative to triplet 2 as measured by triplets 2 & 3.
With triplet 1 receding at .866c and triplet 3 chasing at .9897c, it will take
\frac{5ly}{.9897c-.866c}= 40.41 yrs
by triplet 2's clock for triplet 3 to reach triplet 1
In which time triplet 1's clock will advance by 20.205 yrs. Add this to the 2.8875 yrs elasped previously and you get 23.1 years total elapsed time for triplet 1; The same time as measured by triplet one.
Triplet 3's clock will advance by
40.41yrs \sqrt{1-\frac{.989739c^{}2}{c^{2}}} = 5.775 yrs
The same time that triplet 1 sees this clock advance by. Add this to the 5.775 yrs transferred from clock 2, and you get a total elasped time of 11.55 yrs; again the same time as measured by triplet 1.
So far, triplets 1 & 2 both agree as to what the clocks read.
From the view of triplet 3:
He passes triplet 2 at 5 ly distant from triplet one and takes on the clock reading of 5.775 years. He takes 5.775 years to cover this distance a .866c, thus his clock reads 11.55 yrs upon reaching triplet 1; Again in agreement with triplets 1 & 2.
During this time he will see triplet 1's clock advance by 2.8875 yrs.
Now we must determine what time it is on triplet 1's clock according to triplet 3 at the time triplet 2 and triplet 3 pass each other.
We must assume that either triplet 1 sends out a signal to triplet 3 when triplet 2 leaves or triplet 3 is watching with a very powerful telescope and notes the time.
Since this info is moving away from triplet 1 at c as measured by triplet 1 and triplet 3 is moving in at .866c, triplet 3 will intercept this info when it is 10.72 lys from triplet 1, as measured from triplet.
From triplet 3's perpective this interception happens at 5.36 ly (again, this is due to length contraction.)
Having received this information, triplet 3 can now extrapolate back to when this info was sent. Since the info traveled at c wrt triplet 1, and triplet one is approaching at .866c, the time since emission(and triplet 1's clock reading 0) would be
\frac{5.36ly}{1c-.866c}= 39.99 yrs.
it will take an additional 0.4157 years by triplet 3's clock to intercept triplet 2
giving a total elapsed time at this point of 40.41 years since the initial launch of triplet 2. With time dilation factored in, triplet 1's clock would read 20.205 at the time of the passing of triplets 2 and 3, according to triplet 3. Adding this to the 2.8875 yrs that passes on triplet 1's clock while triplet 3 travels the remaining 5 ly, and you get a total elapsed time of 23.1 yrs on triplet 1's clock according to triplet 3. This again agrees what what triplets 1 & 2 measured.
Also, since triplet 2 is moving at .9897c relative to triplet 3 as measured by 3, applying time dilation for this velocity will show that triplet three will see triplet 2's clock advance by 5.775 yrs form time of leaving triplet 1 to intercept.
Thus all three are in agreement, and there is no paradox.