Let us consider two frames ** F (x, t)** and

*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

*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

There are two twins, ** twin A** and

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.

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

In both frames the distance between the twins…

** -v * t_1 = -d** (for any distance

Come to think of it, the Earth

at time** t_1 = d / (v * gamma) **reaches the second spaceship

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

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

Usually we think of a single spaceship in the origin ** O_1** of the frame

Consider two frames ** F** and

*You have to choose:*

** a)** the frame

** b)** the frame

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

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