I propose to readers another article, where the characters may very well be two friends: Alpha and Beta. (and not necessarily twins!)
Beta decides to leave with his spaceship, first runs a reconnaissance lap increasing his speed and, just as Alfa recognizes him by raising his eyes to the sky, he is seen by the latter darting at a speed equal to v = 0.866*c.
(about 90% of the speed of light indicated by c)
Alpha knows that Beta is headed for a planet 17.32 light-years away and that the spaceship will not stop in the vicinity of the planet itself, continuing to travel in a uniform straight line (the two friends thus agreed before the departure of Beta , thus avoiding referring in the discussion to phases of acceleration and deceleration); in the example proposed it is in fact required that the spaceship cover the entire Earth-planet distance at a constant speed.
First, what does it mean that the distance is 17.32 light-years? It means that the light takes 17.32 years to cover it all. (0.32 years correspond to almost four months, actually 3 months and just over 25 days)
Beta’s spaceship is not as fast as light and will therefore take longer to travel that distance; inverting the simple formula v = s/t (because the speed is equal to the ratio between the distance traveled and the time taken), we obtain that t = s/v (t_Alfa = 17,32/0,866 = 20 years).
t_Alfa = 20 years is the travel time with respect to the Earth’s frame and Alpha, although not seeing his friend, knows that he will pass in the immediate vicinity of the planet when 20 years have elapsed for the clocks in the Earth’s frame.
We have so far analyzed the journey with respect to Alpha, the hourly equation of uniform rectilinear motion dictates that to travel a distance of 17.32 light-years at a constant speed of 0.866 * c takes 20 years. (and it can’t be otherwise!)
Now let’s consider the travel time measured with respect to the spaceship’s frame; if we hold the traveler friend still, it is not only the Earth that moves with Alpha, but the whole Universe does it (and every distance that apparently moves towards the spaceship with opposite speed is contracted by the Lorentz Transformations, that is minor); Beta is well aware of this also because she knows that the planets and stars are not ellipsoids (contracted spheres) so “accentuated”. At velocity v=0.866*c, when the reciprocal of the Lorentz factor is 0.5, the Earth-planet distance is 8.66 light-years for Beta (half of 17.32 light-years, where this last is the length to travel with respect to the frame of the Earth)
Still considering the hourly equation of uniform rectilinear motion, for Beta the travel time is t_Beta = 8.66/0.866 = 10 years. (half of t_Alfa, and even here it cannot be otherwise!)
If desired, one can imagine that a rigid rod with a length equal to the Earth-planet distance (and with a direction parallel to the Earth-planet distance itself) is attached to the tail of Beta’s spaceship. When the spaceship is in motion, the friend left on Earth sees the rod itself in motion (therefore contracted), and it is precisely when he sees the passage of the rod end in front of his eyes that he will know what the arrival time indicated by the Beta stopwatch.
When the auction ends, 10 years have passed for Alpha (in the Earth’s frame) and 5 years for Beta (in the spaceship’s frame), Beta has not yet reached the planet with respect to the Earth’s frame: the spaceship has covered half the effective Earth-Planet distance. (8.66 light-years)
Alpha and Beta are not the same age. (When ten years have passed for Alfa, only five have passed for Beta in the spaceship’s frame!)
The frequent mistake is to think that, if the spaceship is believed to be stationary, the “apparent travel” time of Alfa with respect to the spaceship’s frame must be slowed down and this is what creates the “false” paradox (as if there were a misleading symmetry that does not allow us to understand for which of the two observers the time elapsed is actually shorter); we still remember that for the spaceship the whole Universe is in motion (not only the Earth) and therefore, just as in Classical Physics, if the travel time with respect to Beta is 10 years, also the “apparent travel” time of the Earth-planet distance (until the planet “reaches” the spaceship) is 10 years. Sometimes, erroneously, considering Minkowski’s space-time graphs, the astronaut is considered to be stationary and an attempt is made to ideally trace the universe line of the Earth’s apparent motion with respect to the spaceship’s frame, in this case it is wrong do it because there is no time dilation (the time duration of the two motions is the same).
When I refer to the universe lines I mean the lines drawn in the Minkowski diagram.
However, it would be different if the astronaut remained stationary floating in Space and if only the Earth (for some strange reason) began to move. So yes, Alpha’s travel time would be less than travel time measured by Beta! If only the Earth moved (whatever the length of the distance), Beta would see only the Earth contract (while it is in motion, and not the whole Universe as seen above), and for this he would get older than his friend!
I conclude, stating that Lorentz Transformations are symmetrical and do not generate contradictions (you just need to pay attention), a body in motion in a reference frame will always be “younger” than the other observers who are stationary in the frame itself!