# THE LORENTZ TRANSFORMATIONS AND FRAMES IN MOTION (THE TWIN PARADOX)

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

**, coinciding at the initial time**

*F_1 (x_1, t_1)*

*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

**and I do not consider the other two Lorentz transformations because they depend on**

*the Lorentz factor***and**

*a)***.**

*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

**moves with speed**

*F_1***with respect to the frame**

*v***, and travels a certain distance**

*F***in the frame**

*d*

*F.*** d)** the frame

**moves with speed**

*F***with respect to the frame**

*-v***, and travels a certain distance**

*F_1***in the frame**

*d_1*

*F_1.**c)** and **d)** are **MUTUALLY EXCLUSIVE!*

*What does** c) **mean?*

If the frame ** F_1** moves with speed

**with respect to the frame**

*v***then the contracted frame of**

*F,*

*F**(the distances between the points of the frame*

*F**are smaller due to*

*the Lorentz-FitzGerald contraction**)*moves with speed

**with respect to the frame**

*-v*

*F_1.*If *x = v * t, x_1 = 0, v * t = gamma * v * t_1, t_1 = t / gamma.*

The whole frame ** F_1** is younger than the whole frame

*F. (t_1 < t)*If the frame ** F_1** travels a distance

**while moving with respect to the frame**

*d*

*F***it is like solving:**

*(x = v * t = d, t = d / v, x_1 = 0)**d = gamma * v * t_1, t_1 = d / (v * gamma), t_1 < t.*

However, there is a frame ** F_2** that moves at

**speed with respect to the frame**

*-v***and the whole frame**

*F_1***is younger than the whole frame**

*F_2***.**

*F_1*

*(and I will elaborate on this concept later in this article)**What does **d)** mean?*

If the frame ** F** moves with speed

**with respect to the frame**

*-v***F_1,**then the contracted frame of

*F_1**(the distances between the points of the frame*

*F_1**are smaller due to*

*the Lorentz-FitzGerald contraction**)*moves with speed

**respect to the frame**

*v*

*F.*If *x_1 = -v * t_1, x = 0, -v * t_1 = -gamma * v * t, t = t_1 / gamma.*

The whole frame ** F** is younger than the whole frame

*F_1. (t < t_1)*If the frame ** F** travels a distance

**while moving with respect to the frame**

*d_1*

*F_1***it is like solving:**

*(x_1 = -v * t_1 = -d_1, t_1 = d_1 / v, x = 0)**-d_1 = -gamma * v * t, t = d_1 / (v * gamma), t < t_1.*

However, there is a frame ** F_3 **that moves at speed

**with respect to the frame**

*v***and the whole frame**

*F***is younger than the whole frame**

*F_3*

*F.*Let us now consider two twins, ** TWIN A** is positioned in the origin of the frame

**and**

*F_1***is positioned in the origin of the frame**

*TWIN B*

*F.*

*(THE WELL-KNOWN TWIN PARADOX)**WHAT IS THE SOLUTION?*

*The solution is: **(t_1 < t) AUT (t < t_1).*

*(TWIN A younger than TWIN B) AUT (TWIN B younger than TWIN A)*

** (t_1 < t)** and

**are not both true expressions,**

*(t < t_1)**if one of the two expressions is*

*true**the other is*

*false**:*

*1 OR 0 = 1*

*0 OR 1 = 1*

*(t_1 < t) OR (t < t_1) = 1*

*(t_1 < t) OR (t < t_1) is TRUE. (THERE IS NO CONTRADICTION!)*

*If we consider that the frame **F_1** travels a certain distance **d** while moving with uniform rectilinear motion in the frame **F**, then the whole frame **F_1** is younger than the whole frame **F**. On the other hand, if we consider that the frame **F** travels a certain distance **d_1** as it moves in the frame **F_1**, then the whole frame **F** is younger than the whole frame **F_1**. (there is no contradiction, it is not **t_1 < t** and **t < t_1**)*

In the best known version of the twin paradox, reference is made to one of the twins (for example ** TWIN A**) moving in a spaceship with uniform rectilinear motion in the frame of the Earth (the frame of

**).**

*TWIN B**(to reach a star, the Earth and the star belong to the frame of the Earth)*

In this case, ** the twin paradox vanishes** because the frame of

**moves with speed**

*TWIN A***with respect to the frame of**

*v*

*TWIN B:**TWIN A is younger than TWIN B,*

*even if TWIN A does not go back to Earth! (THE CLOCK PARADOX)*

*TWIN A** is being considered to travel a certain distance in the frame of the Earth. **(although the spaceship continues to travel after reaching the star)*

If ** TWIN A** travels a distance

**in the frame of**

*d***it is a mistake to think that**

*TWIN B***travels a distance**

*TWIN B***in the frame of**

*d / gamma*

*TWIN A.**t_1 = d / (v * gamma)**and** t = d / (v * gamma * gamma)**,* *ERROR!*

For ** TWIN B **the moving distances of the

**frame contract. Imagine that the spaceship has a tail, and the tail length is**

*TWIN A***in the frame of**

*d / gamma***.**

*TWIN A***The tail length of the spaceship is**

**in the frame of**

*d / (gamma * gamma)***.**

*TWIN B**TWIN B** does not travel the distance **d / gamma** in the frame of **TWIN A**, **TWIN A** travels a distance equal to **d / (gamma * gamma)** in the frame of **TWIN B** and the tail end of the spaceship reaches **TWIN B!*

If ** TWIN A **travels a distance

**in the frame of**

*d / (gamma * gamma)***, then:**

*TWIN B**t = d/(v*gamma*gamma) **and** t_1 = d/(v*gamma*gamma*gamma).*

The spaceship has not yet reached the star, the spaceship reaches the star at time *t = d / v.*

*The time** t** is **“the elapsed time”** in the Earth frame (the frame of **TWIN B**),*

*the time **t_1** is **“the elapsed time”** in the frame of the spaceship (the frame of **TWIN A**):*

*0 ≤ t ≤ d / v, 0 ≤ t_1 ≤ d / (v*gamma), t_1 = t / gamma.*

** TWIN A** is younger than

*TWIN B,**even if*

*TWIN A**does not return to Earth!*

*If applied correctly, **the Lorentz transformations** give us the solution:*

*THE LORENTZ TRANSFORMATIONS ARE “AB OMNI NAEVO VINDICATAE”.*

If the inertial frame of the spaceship moves with speed ** v** with respect to the inertial frame of the Earth

**, it makes no sense to consider the frame of the Earth in motion with respect to the frame of the spaceship.**

*(x = v * t)**( t_1 < t )**and*** (t < t_1)**

*is a contradiction as well explained in my article*

**,**

*“THE CONTRADICTION”*

*THE FRAME OF THE EARTH IS AT REST!**Is the twin paradox resolved in this case?*

As I said earlier, there is a frame** F_2** that moves at

**speed with respect to the frame of the spaceship (frame**

*-v***) and the whole frame**

*F_1***is younger than the whole frame**

*F_2*

*F_1.*If the astronaut twin ** (TWIN A)** launches a rocket

**at the initial time**

*(the rocket R)***and at the speed**

*t = t_1 = 0***in the direction in which**

*-v***sees the Earth moving away, then the entire frame of**

*TWIN A***is younger than the whole frame of the spaceship.**

*the rocket R*

*(t_R = t_1 / gamma)**If **TWIN B** hides in **the rocket R** with a probability of **50%** (before the ship’s departure), we will never know which of the twins is younger. **However, we do know that the entire frame of the spaceship is younger than the frame of the Earth. **(If **TWIN B** is not hiding in **the rocket R**, **TWIN A** is younger than **TWIN B**!)*

*Massimiliano Dell’Aguzzo*