# THE PARADOX OF WATCHES

There are two clocks ** A** and

**in motion relative to a given speed**

*B***.**

*v**In an empty space we have to choose:*

** 1)** The clock

**moves in the frame of the clock**

*B***,**

*A***travels a distance**

*B***and the motion is a motion uniform rectilinear. (the time elapsed in the frame of**

*d***is**

*A***) In this case, in the frame of**

*t_A = d / v***, it is necessary to wait for a contracted distance. (the time elapsed in the frame of**

*B***is**

*B***is the**

*t_B = t_A / gamma, gamma***)**

*Lorentz factor**AUT*

** 2)** The clock

**moves in the frame of the clock**

*A***,**

*B***travels a distance**

*A***and the motion is a motion uniform rectilinear. (the time elapsed in the frame of**

*d***is**

*B***) In this case, in the frame of**

*t_B = d / v***, it is necessary to wait for a contracted distance. (the time elapsed in the frame of**

*A***is**

*A***)**

*t_A = t_B / gamma**Many say*: if it is** 1)** clock

**also waits for a contracted distance**

*A***and, if it is**

*(t_A = t_B / gamma)***, clock**

*2)***also waits for a contracted distance.**

*B*

*(t_B = t_A / gamma)**THIS IS FOR ME A CONTRADICTION!*

If it is ** 1)** clock

**moves with an opposite speed compared to clock**

*A***but it is**

*B***(and not**

*t_B < t_A***) and, if it is**

*t_A < t_B***2)**, clock

**moves with an opposite speed with respect to clock**

*B***A**but it is

**and not**

*t_A < t_B. (*

*t_B < t_A)**If we do not initially make a choice it is:*

*(t_B < t_A) OR (t_A < t_B)*

**1 OR 0 = 1 (1 V 0 = 1)**

**0 OR 1 = 1 (0 V 1 = 1)**

** t_B < t_A** and

**cannot both be true, but**

*t_A < t_B***is a true proposition,**

*(t_B < t_A) OR (t_A < t_B)*

*(t_B < t_A) OR (t_A < t_B) = 1.**There is no contradiction!*

*In the** twin paradox **(with a twin left on Earth and an astronaut twin who reaches a star) there is no need to choose: it is the astronaut twin who travels the distance **d** in the frame of the Earth. (the star belongs to the frame of the Earth and the star is at a distance **d **from the Earth in the frame of the Earth) The astronaut twin does not start at a constant speed, on his journey he accelerates and decelerates but, if he reaches the star and comes back, **at the return of the spaceship to Earth he is younger than his brother.*

*INSIGHTS*

Two clocks ** A** and

**can both travel a same distance**

*B***in the frame of a third clock**

*d***(You can imagine two spaceships moving in the frame of the Earth, one spaceship travels to the left and the second spaceship travels to the right)**

*C.*If the velocity module is the same, the two watches are both younger than the third watch. ** (t_A < t_C, t_B < t_C, t_A = t_B)** It is wrong to think that clock

**travels a distance in the frame of clock**

*B***, or that clock**

*A***travels a distance in the frame of**

*A*

*B.**In this case we have chosen, both clocks **A** and **B** travel the same distance **d** in the frame of the clock **C**.*

*Massimiliano Dell’Aguzzo*