Let us consider two frames F (x, t) and F_1 (x_1, t_1), coinciding at the initial time t = t_1 = 0.

I consider the two Lorentz transformations:

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

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

With gamma I obviously indicated the Lorentz factor and I do not consider the other two Lorentz transformations because they depend on a) and b).

If the speed v is known, this is a system of two equations with four unknowns.

The two Lorentz transformations indicate that:

c) the frame F_1 moves with…


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?

In…


IL GEMELLO ASTRONAUTA E’ PIU’ GIOVANE

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.

SE LA STELLA RAGGIUNGE…


Let's imagine a spaceship moving in the frame of the Earth with uniform rectilinear motion at speed v, we know that the clock of the astronaut twin slows down compared to all the clocks of the frame of the Earth. (t_1 = t / gamma)

In this case it's not x_1 = - v * t_1 , the speed of the Earth in the frame of the spaceship is

- gamma * v. (The motion of the Earth is in advance of the uniform rectilinear motion x_1 = - v * t_1)

In both frames the distance between the twins…


-v * t_1 = -d (for any distance d and for t_1 < t) is a contradiction, because it means that the Earth DOES NOT MOVE with uniform rectilinear motion at speed -v.

Come to think of it, the Earth

at time t_1 = d / (v * gamma) reaches the second spaceship (x_1 = -d), the motion of the Earth in the frame of the spaceship is therefore represented by x_1 = - gamma * v * t_1.

The distance of the spaceship in the frame of the Earth is x = v * t, and in the frame…


Solving the CLOCK PARADOX means solving a system of two equations with four unknowns.

I consider the two Lorentz transformations:

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

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

With gamma I obviously indicated the Lorentz factor, and I do not consider the other two Lorentz transformations because they depend on a) and b).

In an empty space THE CLOCK PARADOX is not resolvable, it’s impossible to determine a single solution of a system of two equations with four unknowns.

BUT THE CLOCK PARADOX SOMETIMES “IS NOT A PARADOX”, consider…


If x = v * t (the spaceship moves with uniform rectilinear motion in the frame of the Earth), then:

1) in the frame of the spaceship there are gamma copies of the Earth-star distance contracted moving (with uniform rectilinear motion) at speed -v,

the times t = d / v and t_1 = d / (gamma * v) are correct!

2) in the frame of the Earth, there are gamma spaceships moving with uniform rectilinear motion at speed v, each spaceship has a tail and the length of each tail is equal to d (in the frame of the…


If the astronaut twin launches a rocket at the initial time (t = t_1 = 0 s) at a speed -v (in the direction in which the astronaut twin sees the Earth moving away), then the rocket is younger than the astronaut twin.

The clock of the astronaut twin slows down compared to the Earth's clock, and the rocket's clock slows down compared to the clock of the astronaut twin.

If we denote by (x_R, t_R) every event in the frame of the rocket, the two Lorentz transformations that describe the situation are:

x_1 = gamma * (x_R - v*…


Solving the twin paradox means solving the system of two equations:

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

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

With gamma I obviously indicated the Lorentz factor, I do not consider the other two Lorentz transformations because they depend on a) and b)

Usually we think of a single spaceship in the origin O_1 of the frame F_1, but the number of spaceships is infinite. Imagine a first spaceship positioned at the origin O_1 of the frame F_1, and a second spaceship positioned at point P (- d…


Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s)

You have to choose:

a) the frame F_1 moves with uniform rectilinear motion in the frame F at speed v. (x = v* t)

b) the frame F moves with uniform rectilinear motion in the frame F_1 at speed -v. (x_1 = -v * t_1)

THE CHOICE IS NECESSARY, a) AND b) ARE MUTUALLY EXCLUSIVE!

Imagine that the frame F_1 goes to the right (with respect to the…

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!

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