THE TWIN PARADOX
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Let's study this problem again with A(lbert) and B(runo).
A sends his brother in a rocket at 4/5 of the light speed. B travels uniformly during 15 years, then comes back at the same speed. The accelerations are instantaneous.
At that speed, the time slowing is 0.6.

During the journey, there is a symmetrical situation, that means, A et B both observe the same slowing of time by each other. Thus, B could expect that, at the end, A would have aged 0.6 times less than himself.
This, of course, does not happen.
In the Special Relativity Theory this is "explained" by the difficult notion of a "frame of reference jump" that would be done by B when he changes his direction.
Let's see what really happens.

During the all trip, A and B send signals to each other.
The first diagram gives A's point of vue.

B's displacement is represented by the blue line.
The journey lasts 30 years for A, but B has aged only 18 years. The turnaround takes place after 15 years for A, when B has aged 9 years.
The signals sent by A are red, those sent by B are green.
There are two time lines, one for A (30 years) and one for B (18 years).

After 1 year, A sends a signal that takes 4 years to reach B. But at that moment B has only aged 3 years!
The signal comes back to A at tA(9), 8 years after he sent it. A calculates that B covered a distance of 4 light-years in 5 years time, which corresponds to a speed of 4/5 c.


Also, the signal sent by B after 1 year (for him) arrives to A when this one has aged by 3 years and comes back to B after 8 years, at tB(9).
If B considers himself motionless, he also calculates that A moves at the speed of 4/5 c and ages 0.6 times less than himself.


The signals sent at tA(2) and tA(3) reach B at tB(6) and tB(9) and the signals sent at tB(2) and tB(3) reach A at tA(6) and tA(9). (see fig.1)


In the second part of the journey, the signal sent by B at tB(9) reaches A at tA(27) and comes back to B at tB(17).
To B, the signal reached A in the middle of the trip, that is at the moment tB(13).
The years tA(27) to tA(30) correspond, from his point of vue, to the period tB(13) to tB(18), thus A's time also slows by 0.6
However, the signals which were sent by B before the turnaround and received after it, that means the signals sent from tB(2) to tB(8), come back to him between tB(10) and tB(16). These signals last 8 years to cover the distance, just like the signal sent at tB(1).
So B observes that A stopped his displacement during this period.

On the other hand, the signals he receives after tB(9) show that A's time "runs" 3 times faster than his, which indicates that A moves toward him at the speed of 4/5c. B knows that the last signal he received before tB(9) was sent by A just before tA(3), which corresponds (from B's point of vue) to tB(5). If A turned around at that moment, he should join B at tB(10), which does not happen!

This diagram represents B's point of vue:
(The blue line shows A's displacement)

How should B interpret this contradictory facts?

Due to the acceleration, B knows he moved, between the sending and the reception, relatively to the source of the signals sent between tB(2) and tB(8).
That means he has no exact information about A's position from tB(9) on.
Only when, at tB(17), he receives the first signal that was sent after the acceleration, his observations again become trustable (and reciprocal to A's).

If B took only into account his observations made during the uniform phases and extrapolated his data to the "dubious" period, he would represent A's trip like this:
Here we see A's "jump in time" "observed" by B when he (A or B?) changes his "frame of reference".

This "jump in time" is not real!
Suppose that B's U-turn lasts 1 day, and that A sends a photo of himself and his clock every day.
All signals sent by A will arrive at B at regular intervals throughout the journey. At no time will B see a sudden advance in A's time. He must conclude that A's "jump in time" that he has calculated is only an illusion.

If B carefully measured the accelerations he felt at the start and at the turnaround, he could calculate himself his trajectory wrt A.

But after all, does A have the right to consider himself immobile wrt the source of the signals? A probably has also felt, in the past, some accelerations, and the all experience could as well have be done from the point of vue of a third observer, relatively to whom A is moving.

This diagram shows the same experience as seen by an observer C.
The speed of A wrt C is 1/5 c.

A and C don't measure the same speed for B (C even considers that B doesn't have the same speed in the return as in the outward journey).
But all the verifiable facts (i.e. the time indicated by the clocks at the departure and arrival of the signals) correspond perfectly.

A and C (and so every other observer moving uniformly during the experience) are equally allowed to consider themself immobile while making their measures.

One other observer we would mention is the one (D) who, coming from space, would join B at tB(9) and travel with him during the second part of his journey.
This diagram shows D's trajectory (in A's frame or reference):
As we can see, D has another perception than B of the place where A is at tB(9). To D, A is at that moment 7,2 light-years away.



This is D's point of vue:



In D's frame, 50 years pass between the moment B leaves A and when the two brothers meet (together with D).
A, who travels at 0,8c, will have aged 50 x 0,6 = 30 years. He needs 7,2 x 0,8 = 9 years to reach B from the moment on when B joins D.
B ages 9 years in the first part of his trip (he moves nearly as fast as light), and 9 years during the second part (just like D).


Bruno Van Rossum

june 2002


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