MATHEMATICAL INSIGHT

The Earth-star distance is d in the frame of the Earth, and imagine that the spaceship has a tail.

INITIAL SITUATION:

The nose of the spaceship and the Earth occupy the same position

(x = x_1 = 0) at the initial time. (t = t_1 = 0)

In the frame of the Earth the spaceship goes to the right, (x = v * t)

and in the spaceship frame the Earth goes to the left. (x_1 = - v * t_1)

If the tail length is d / gamma in the spaceship frame, I consider the two events A and B,

EVENT A:

in the frame of the Earth the spaceship reaches the star,

in the frame of the spaceship the star reaches the spaceship.

In the frame of the Earth, event A occurs at time t = d / v,

in the frame of the spaceship event A occurs at time t_1 = d / (gamma * v).

EVENT B:

in the frame of the Earth the end of the tail reaches the Earth,

in the frame of the spaceship the Earth reaches the end of the tail.

In the Earth frame, event B occurs at time t = d/(gamma*gamma*v),

in the frame of the spaceship event B occurs at time t_1 = d / (gamma * v).

Events A and B are simultaneous in the frame of the spaceship, but they are not simultaneous in the frame of the Earth.

In the ship frame the Earth’s clock slows down (and this is true), BUT events A and B are TWO DIFFERENT EVENTS.

If EVENT A occurs, the astronaut twin is younger than the Earth twin. (t_1 < t)

If EVENT B occurs, the Earth twin is younger than the astronaut twin. (t < t_1)

If we consider the spaceship in motion, the clock of the astronaut twin always slows down. (even before the turnaround) In the Earth frame the spaceship reaches the star at time t = d / v, and in the ship frame the star reaches the ship at time t_1 = d / (gamma * v): t_1 < t ! (and the reciprocal time dilation DOES NOT happen)

a) in the Earth frame the astronaut twin reaches the star at time t = d / v, and

b) in the ship frame, the star reaches the ship at time t_1 = d/(gamma*v).

The spaceship reaches a point X (the star for example), and the point X is the point where the turnaround takes place. Point X belongs to the frame of the Earth, the Earth can’t go back!

THE POINT X IS THE POINT WHERE THE TURNAROUND TAKES PLACE.

If the spaceship reaches point X, the spaceship also reaches all the infinite points of the Earth-X distance. And there are infinite events analogous to event A, the spaceship clock always slows down!

The two events A and B are simultaneous in the frame of the spaceship, but they are two different events.

IN THIS CASE THE SPACESHIP CLOCK ALWAYS SLOWS DOWN, ... AND THE TWIN PARADOX IS SOLVED!

Massimiliano Dell'Aguzzo