THE TWINS ALEX AND BRAD
Let Alex and Brad be twins (both in their thirties, located in the same position and both stationary relative to each other); Brad decides to leave with his spaceship covering a distance d (to then turn back, always moving at the same speed v), distance d is represented by one segment, segment AB. (initially Alex and Brad are positioned near point A and the distance d is the measure of the length to travel to get to a distant planet, the same value for both)
How long is the journey for Alex?
Alex will have to wait a time equal to 2*d/v to see his brother again. (we are simply considering a round trip)
The journey, “according to Brad”.
Compared to Brad, everything else moves with respect to him in the opposite direction at a speed v. (for Brad also the AB segment that was previously stationary is now in motion; it is also remembered that, while for Brad everything else is moving, for Alex only Brad is in motion and this is important to reach the conclusion)
As well indicated by the Lorentz transformations, the moving distance AB has a shorter length for Brad and measures d_1. (d_1 < d)
If we consider “Brad stationary”, at first he “waits” for the second extreme B of segment AB to meet him at speed v (this happens when a time interval equal to d_1/v passes, that is when Brad “arrives at his destination”), then Brad “waits” for the first extreme A of segment AB to meet him so that the journey ends. (and this still happens when a time equal to d_1/v passes, that is when Brad returns to Alex, Alex is the twin who has always stood stationary waiting for his brother near point A)
In conclusion, less time (equal to 2*d_1/v) has passed for Brad.
(d_1 < d, 2*d_1/v < 2*d/v)
To simplify the calculations, let’s imagine that the speed is equal to 0.866 c (about 90% of the speed of light, in this case the distance AB in motion according to Brad measures d_1 = 0.5*d; d_1 is half of d and also the elapsed time for Brad is half the elapsed time compared to Alex); again to simplify the numerical results we can also choose the value of d so that the elapsed time for Alex (2*d/v) is equal to 20 years. (in this case the travel time according to Brad is only 10 years)
When the twin brothers meet again, Alex is fifty and Brad is forty. (For Alex the trip lasted 20 years, while for Brad the trip only lasted 10 years, remember that before the trip both twins were in their thirties)
Conclusions
It is usually believed that Alex knows Special Relativity and expects Brad to be younger when he returns, but strangely instead (in my opinion) it is also believed that Alex’s clock should also slow down compared to Brad’s clock. When I wrote in my previous articles that in Brad’s frame there is no time difference I meant that there is obviously no time difference between Alex’s apparent motion and Brad’s motion in the Earth’s frame where distances are contracted.
Each of the twins (not only Brad, but also Alex if he does not know the Lorentz Transformations) may think (wrongly) that the other brother is the same age.
And, regardless of what each twin thinks of the other, for the first (Alex) a time equal to 2*d/v has passed and for the second (Brad) a shorter time equal to 2*d_1/v has passed, the time intervals are different and only Alex, if he knows the Lorentz Transformations, has the right to claim that his brother is younger. (Brad can’t do that)
However, the solution is the one that has been presented again now (and deepened in some of my previous articles in the case of non-constant speed, where it was necessary to resort to differential calculus):
the twin who has traveled, when he returns to his initial position, is younger than the brother who has remained waiting and the Lorentz Transformations allow us to verify this.
Massimiliano Dell’Aguzzo
