THE TWIN PARADOX IS NOT A PARADOX

I consider the Earth and a spaceship moving relative to each other at constant speed (as required in SR), and there are two scenarios:

SCENARIO 1

If you consider the Earth stationary, the spaceship moves from a starting point A to reach an end point B and the spaceship clock slows down relative to the Earth’s clock.

SCENARIO 2

If you consider the spaceship stationary, the Earth moves from the starting point A to reach an end point C (C is a point on the spaceship frame), and the Earth’s clock slows down relative to the spaceship clock. Imagine that the spaceship has a length, and the length of the spaceship is equal to d in the frame of the spaceship. In the ship frame, the Earth reaches the tail of the spaceship at time t_1 = d / v.

But the twin left on Earth “sees” the length of the spaceship contracted:

t = d / (gamma * v), t < t_1.

It’s difficult to visualize this, because we think of the spaceship moving in the frame of the Earth.

And if we think of the spaceship in motion, then t_1 < t. (SCENARIO 1)

If you consider SCENARIO 1 (the spaceship heading towards a star), then t_1 < t.

If you consider SCENARIO 2 (the Earth heading towards a point on the spaceship frame), then t < t_1.

BOTH SCENARIOS ARE CORRECT (and there is no contradiction), because the frame of the Earth and the frame of the spaceship are two different and independent frames. (the Earth and the spaceship keep moving away)

BUT WE KNOW … that in the frame of the Earth the spaceship reaches a star, and in this case SCENARIO 1 applies. Imagine that in the frame of the Earth the Earth-star distance is equal to d, the astronaut twin “sees” the Earth-star distance contracted: t = d / v and t_1 = d / (gamma * v),

t_1 < t.

In this case SCENARIO 1 applies, and the spaceship is “younger” than Earth. (BEFORE THE TURNAROUND)

IT’S SIMPLE! In the frame of the Earth the spaceship reaches the star and, in the frame of the spaceship, the star reaches the spaceship. The travel time is t = d / v, and the spaceship waits for the star.

t_1 = t / gamma, and t_1 < t.

I consider the two Lorentz Transformations:

a) x_1 = gamma * (x -v * t) ,

b) x = gamma * (x_1 + v * t_1) .

gamma is the Lorentz factor, and the other two Lorentz Transformations depend on a) and b).

SCENARIO 1 (x = v * t)

If x = v * t:

x_1 = 0

v * t = gamma * v * t_1

t = gamma * t_1

t_1 < t

SCENARIO 2 (x_1 = -v * t_1)

If x_1 =-v * t_1:

x = 0

-v * t_1 = -gamma * v *t

v * t_1 = gamma * v * t

t_1 = gamma * t

t < t_1

Both scenarios are valid (the frame of the Earth is moving relative to the spaceship frame), but in the Earth frame the spaceship reaches the star,

t = d / v and t_1 = d / (gamma * v), t_1 < t !

The twin paradox is solved, and the Lorentz Transformations are “ab omni naevo vindicatae”.

Massimiliano Dell’Aguzzo