THE TWINS PARADOX
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The Story:
A vessel travels in space since several generations. The occupants see the stars passing, but they don't feel any movement. They feel themselves immobile, but they think that's impossible, they must have some speed, and they want to know it.
One day, they pass by the earth, and they decide to send a shuttle that will stay near the earth during 2 years, and will thereafter resume the journey at the same speed as the vessel. One of the systems will age less than the other, and by exchanging signals, they hope to find out which one, and so determine their speed.
But they fail.

The Explanation:
Whe have the earth E, the vessel V and the shuttle S.

1. E is motionless.
   We suppose that V and S are moving at 0.866c (c = the speed of light).
   --> their time is slowed to 1/2 (Lorentz's formula).

2. V sends S toward E with speed 0.
   S and V consider there age = 0 at that moment.
   --> S's time flows normaly, V's is still slowed.
   This goes on during 2 years (to S and E).
   --> S and V are at 1.732 L.Y.(Light Year) from each other. (1.732 = 0.866 x 2)
   --> V has aged 1 year, S 2 years.

3. S sends a signal SS toward V and leaves again with the same speed as V (0.866c).
   --> Its time is slowed again.
   At that moment V is 1 year old.
   SS propagates at speed c to V which is leaving at 0.866c.
   SS advances to V with relative speed of 0.134c (= c - 0.866c)
   The signal SS reaches V after 12.925 years (= 1.732 L.Y./ 0.134c)!
   During this time, V has aged 6.463 years (=12.925 / 2).
   --> V is 7.463 years old when he recieves the signal.

4. A few years after:
   When V is 10, he sends a signal SV to S (who is 11 years old).
   S is moving toward SV. Their relative speed is 1.866c (= c + 0.866c).
   SV reaches S after 0.928 years. (= 1.732 L.Y./ 1.866c)
   During this time, S has aged 0.464 years.
   S sends the signal back with the message that he was 11.464 years old when he recieved it.
   During the return of the signal, V continues his journey at speed 0.866c.
   SV approaches V at speed 0.134c (= c - 0.866c)
   The message reaches V after 12.925 years (= 1.732 A.L./ 0.134c).
   The two-way trip of the signal lasted 13.853 years (0.928 + 12.925).
   During this time V aged 6.927 years.
   He considers that S received the signal in the middle of the journey, thus after 3.464 years, when he, V, was 13.464 years old.

   But from the point of view of V, who considers himself immobile, S moved away from him (together with E), and stopped then.
   V considers that the time of S was slowed during his displacement.
   He thus finds logical that S was younger than he when he received the signal.
   But how far is S according to V, and what was his speed of remoteness?

   According to V:
   S received the signal SV after 3.464 years. He is thus 3.464 L.Y. away
   (So we see that lengths in the system of V equal the half of those of E and S!)
   S sent the signal SS when he left the earth (or, from the point of view of V, when S stopped).
   We saw that V was 7.463 years old when he received the signal SS.
   V considers that the signal took 3.464 years to arrive and that S sent it at the time 7.463 - 3.464 = 3.999.
   So S traveled 3.464 L.Y. in 3.999 years at speed 3.464 / 3.999 = 0.866c
   To V the time of S slowed down half during his displacement, that lasted 3.999 years for V, thus 2 years for S (S confirms this).
   When, at the start of the signal SV, V was 10 years old, S was 8.
   S received the signal SV when he was 8 + 3.464 = 11.464 years old.

5. What happens if S and V rejoin?
   Suppose that, when he is 20, V goes toward S at the same speed S went away.
   In the system E, S is 21 years old at this moment.
   Just like S in phase 2, V will stop during 2 years, while S, continuing his travel, ages one year.
   When they meet, they are both 22 years old.
   V considers that, when he was 20 years old, S was 18.
   He has, just as S in phase 3, aged 2 years during his displacement, that lasted 4 years for S.

   On the other hand, if S rejoins V, the calculation is more difficult.
   Suppose that S travels, as he did before, 2 years (for him) to go back to V.
   According to the transformation formulas, S has to move at speed 0.9897c, relative to E.
   S will join V after 14.016 years. (= 1.732 A.LS. / (0.9897C - 0.866C))
   If S left when he was 20 years old, V was 19 at this moment.
   At the time of their meeting, S is 22 years old and V is 26 (= 19 + 14.016 / 2).

   What is here the point of view of V?

   If S is 20 when he starts, V is 22 years old at this moment.
   For V, the return journey of S is symmetrical to the first part of the trip.
   So S will spend 4 years to come back, and will age of 2 years.
   S will join V at the age of 22 years; V will be 26 years old.

6. We could have made all the reasoning while considering V as immobile in phase 1 instead of E.
   In this case, it is the time and lengths of E that decrease half relative to V, but the result will be the same.

Conclusion:
Whatever is the initial speed of their system, the observers can consider theirselves immobile. Their measures will give coherent results. The only way to get a difference is to leave the system and to come back. In this case, the time of all system having moved in relation to the frame of reference will be slowed. All observers will arrive by their calculations to concordant results (with regard to the controllable facts). However, the justification of these results will be different if observers are in relative movement, but they won't be able to determine who is right. The version of all observers is as valid, since the fact to consider one or the other as immobile at the departure is all as arbitrary.

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