There are two twins on Earth; one of the two decides to leave (to a planet at distance d from Hearth and at high speed v with a spaceship), and on his return he will be younger than his brother left on Earth.

But if we consider the astronaut twin stationary, why can’t it be assumed that the opposite happens, that is, that the twin left on Earth is younger than the traveling brother?

Let us consider the symmetrical situation: all points of the Earth’s frame S (also the Earth and one of the twins) moving together at the same speed v opposite to that of the spaceship, covering the same distance d traveled by the spaceship in the Earth’s frame.

For the astronaut twin, the Earth-planet distance to travel is less than d. (so the travel time is shorter in the spaceship’s frame S_1). The planet and the Earth are the extremes of a length d, and this length is a moving distance in the spaceship’s frame S_1. (For the astronaut twin, the Earth-planet distance measures d/γ, at the beginning and end of the journey). It takes a time t_1 equal to (1/v)*(d/γ) for the planet to reach the spaceship.

In the spaceship’s frame S_1 the travel times of the twins are identical, and the twins are the same age.

When the Earth sees the spaceship at distance x (0 < x < d), the astronaut twin sees his brother at a distance x/γ, just as if the Earth was traveling a distance equal to x in the frame of the spaceship S_1 at a speed opposite to that of the spaceship (however, the Earth has not traveled any distance x in the spaceship’s frame S_1, it’s the Earth-planet distance moving by a distance x/γ in the frame of the spaceship). For the twin on Earth the travel time is therefore

t = (1/γ)*(x/v), while for the astronaut twin the travel time is

t_1 = (1/v)*(x/γ) = (1/γ)*(x/v).

If x = d then t = t_1 = (1/γ)*(d/v).

The two travel times are the same in the spaceship’s frame S_1! (it’s not t_1 = d/v)

Even for the planet it is like having traveled a distance d in the spaceship’s frame S_1 and its travel time is equal to that of the Earth, t = (1/γ)*(d/v).

But the time t in Earth’s frame S is determined by the arrival of the spaceship on the planet (t = d/v), and not from the travel time of the Earth in the spaceship’s frame S_1.

t = d/v and t_1 = (1/γ)*(d/v), t = γ*t_1!

Considering the round trip, t = 2*(d/v) and t_1 = (2/γ)*(d/v).

Even if we consider the astronaut twin to be stationary, the twin left on Earth is no younger than his astronaut brother!

If we consider the twin astronaut in motion, he must travel a distance d (moving in the Earth’s frame S) and, if we consider the Earth-planet distance in motion, it is as if the twin on Earth traveled a distance equal to d. (moving in the spaceship’s frame S_1)

The two travel times t and t_1 are equal only in the spaceship’s frame S_1!

The astronaut twin may therefore (wrongly) assume that the twin on Earth is the same age as him, but the twin on Earth is older. Actually, for the astronaut twin the Earth-planet distance is in motion (and the Earth and the planet belong to the Earth’s frame S). If we consider the Earth and the planet stationary, the Earth-planet distance is greater, and it is the spaceship that travels the entire distance d in the Earth’s frame S.

We can therefore state: in the frame of the spaceship S_1, the Earth travels the contracted Earth-planet distance d/γ and, as in Classical Physics, the spaceship travels the same (contracted) distance with respect to the Earth. Only in the Earth’s frame S (the Earth-planet distance has length d in Earth’s frame), the entire distance d is covered by the moving spaceship! And the astronaut twin, when he returns to Earth, will be younger than his twin brother who hasn’t traveled.

For interested readers, in my other article “MATHEMATICAL CALCULATIONS, APPLYING THE LORENTZ TRANSFORMATIONS”, the proof is more detailed.

Massimiliano Dell’Aguzzo

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