There are two twins, twin A and twin B. Let’s imagine that the twin A has to reach (in a spaceship) a star at a distance d from the Earth. The star belongs to the frame of the Earth. Let’s also imagine that the spaceship has a tail of length d (in the frame of the spaceship), and that the astronaut twin is in the front of the spaceship.

At the initial time the positions of the twins coincide (x = x_1 = 0) and the twins are the same age.

What happens in the frame of the Earth?



Il gemello astronauta si muove a velocità costante v nel sistema di riferimento della Terra. (per dirigersi verso una stella che appartiene al sistema di riferimento terrestre, la distanza Terra-stella è quindi una lunghezza ben definita in tale sistema di riferimento)

Nel sistema di riferimento dell’astronave è la distanza Terra-stella a muoversi verso il gemello astronauta (per il gemello astronauta la distanza in movimento è contratta ed è minore della lunghezza propria Terra-stella) e, quando la stella raggiunge l’astronave, il gemello astronauta è più giovane del gemello rimasto sulla Terra.


To solve the twin paradox we consider the frame of the Earth (x, t), the frame of the spaceship (x_1, t_1), and the frame of the rocket R (x_2, t_2).

The Earth and the rocket R should not be confused, the rocket R is a rocket launched by the astronaut twin in the direction he sees the Earth moving away. (with a speed opposite to that of the spaceship)

The solution

The astronaut twin is in relative motion at speed v with respect to the frame of the Earth:

x_1 = gamma * (x -v*t)

x = gamma * (x_1…

There are two clocks A and B in motion relative to a given speed v.

In an empty space we have to choose:

1) The clock B moves in the frame of the clock A, B travels a distance d and the motion is a motion uniform rectilinear. (the time elapsed in the frame of A is t_A = d / v) In this case, in the frame of B, it is necessary to wait for a contracted distance. (the time elapsed in the frame of B is t_B = t_A / gamma, gamma is the Lorentz factor)



Ci sono due orologi A e B in moto relativo ad una data velocità v.

In uno spazio vuoto dobbiamo scegliere:

1) L’orologio B si muove nel sistema di riferimento di A percorrendo una distanza d, ed il moto è rettilineo uniforme. (il tempo trascorso nel sistema di riferimento di A è t_A= d /v) In questo caso, nel sistema di riferimento di B, occorre attendere una distanza contratta. (il tempo trascorso nel sistema di riferimento di B è t_B = t_A / gamma, con gamma ho indicato il fattore di Lorentz)


2) L’orologio A si muove nel sistema…


Initially, both twins do not move in the inertial frame of the Earth. (they are stationary like the Earth and like the star)

Then only one of the two leaves in the spaceship and heads towards the star but, before traveling at a constant speed, the spaceship must accelerate. (this is necessary)

In this time interval inside the spaceship there are the fictitious forces acting in the non-inertial frames.

The astronaut twin travels the Earth-star distance (a distance that does not change in the Earth's frame even during acceleration) while, in the frame of the spaceship…


Le jumeau astronaute se déplace à une vitesse v constante dans le référentiel terrestre. (pour se diriger vers une étoile qui appartient au référentiel terrestre, la distance Terre-étoile est donc une longueur bien définie dans ce référentiel)

Dans le référentiel du vaisseau spatial, c’est la distance Terre-étoile qui se déplace vers le jumeau astronaute (pour le jumeau astronaute, la distance de déplacement est contractée et est inférieure à la longueur Terre-étoile) et, lorsque l’étoile atteint le vaisseau spatial, le jumeau astronaute est plus jeune que le jumeau laissé sur Terre.



The astronaut twin moves at a constant speed v in the Earth’s frame. (to head towards a star that belongs to the Earth’s frame, the Earth-star distance is therefore a well-defined length in this frame)

In the frame of the spaceship, it’s the Earth-star distance that moves towards the astronaut twin (for the astronaut twin, the moving distance is contracted and is less than the Earth-star length) and, when the star reaches the spaceship, the astronaut twin is younger than the twin left on Earth.


In my article “MATHEMATICAL CALCULATIONS, APPLYING THE LORENTZ TRANSFORMATIONS” and in other precedents I wrote that, in the spaceship frame S_1, the astronaut twin can assume that he is the same age as his brother.

The distances of the frame S contract with respect to the frame S_1 when the frame S_1 is in motion, and the distances of the frame S_1 also contract (because the frame S is also in motion, with opposite speed). …

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…

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!

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