THE TWINS PARADOX
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
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, but B has aged only 18 years. The turnaround
takes place after 15 years, when B has aged 9 years.
The signals sent by A are red, those sent by B are green.
The signal sent by A at tA(1) takes 4 years to reach B. It arrives
to B at tA(5), which corresponds to tB(3), and comes back to A at tA(9).
A calculates that B covered 4 light-years in 5 years,
which corresponds to a speed of 4/5 c.
Also, the signal sent by B at tB(1) arrives to A at tA(3) and
back to B at tB(9).
If B considers himself motionless, he also calculates that A moves at the
speed of 4/5 c.
The signals tA(2) and tA(3) reach B at tB(6) and tB(9) and the
signals tB(2) and tB(3) reach A at tA(6) and tA(9).
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
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" (note the quotation marks).
If B carefully measured the accelerations he felt at the start by 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.
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