# THE TWIN PARADOX AND THE TWO INDEPENDENT UNIFORM LINEAR MOTIONS

*Consider a spaceship of length **L**, the nose of the spaceship and the twin on Earth occupy the same position at the initial times (**t = t’ = 0**) and the twins are in relative motion to each other at speed **v**.*

The nose of the spaceship moves in the frame of the Earth, and travels a distance ** L**. (to reach a star

**, suppose therefore that the Earth-star distance is**

*S***in the frame of the Earth)**

*L**T**: tail end of the spaceship*

*N**: nose of the spaceship (ship twin)*

*E**: Earth (earth twin)*

*S**: star **S*

*The ship twin is the nose of the spaceship, and the twin on Earth is the nose of a second spaceship, … the Earth-star spaceship. (Note that the distance **E-S** is **L **in the frame of the Earth, and also the distance **T-N** is **L** in the frame of the spaceship)*

If we consider the uniform linear motion of the spaceship in the frame of the Earth:

*1) **N** (ship twin) reaches the star **S** at time **t = L/v**,*

*2) the star **S** reaches **N** at time **t’ = L / (gamma*v)*

*If we consider the uniform linear motion of the Earth in the frame of the spaceship, then we need to consider something else: in this case it is necessary to consider the uniform linear motion of the star S in the frame of the spaceship.*

If we consider the uniform linear motion of the star ** S** in the frame of the spaceship:

*1)* *The star **S** reaches **N** at time **t’ = L/v*

*2)* *N**(ship twin) reaches the star **S **at time **t = L / (gamma*v)*

*If we consider the uniform linear motion of the Earth in the frame of the spaceship, the Earth-star distance does not contract.*

If we consider the unform linear motion of the spaceship in the frame of the Earth the Earth-star distance contracts in the frame of the spaceship, *but if we consider the uniform linear motion of the star in the frame of the spaceship … the distance L contracts in the frame of the Earth!*

However, we know that in the frame of the spaceship the star ** S** reaches the nose of the spaceship at time

**, the two times**

*t’ = L / (gamma*v)***and**

*t’ = L/v***cannot both be right.**

*t’ = L / (gamma*v)**And** t’ = L / (gamma*v)**, because we are considering the spaceship moving in the frame of the Earth!*

*So … the uniform linear motion of the spaceship in the frame of the Earth and the uniform linear motion of the Earth in the frame of the spaceship are two independent uniform linear motions:*

1) the uniform linear motion of the spaceship (in the frame of the Earth) does not depend on the uniform linear motion of the Earth (in the frame of the spaceship)

*AND*

2) the uniform linear motion of the Earth (in the frame of the spaceship) does not depend on the uniform linear motion of the spaceship. (in the frame of the Earth)

*Summarizing:*

*1) the clock of the spaceship slows down as the spaceship moves with uniform linear motion in the frame of the Earth*

*AND*

*2) the clock of the Earth (and the clock of the star) slow down while they move with uniform linear motion in the frame of the spaceship.*

If we consider the uniform linear motion of the twin on Earth in the frame of the spaceship the Earth-star distance does non contract *(… the distance **E-S** does not contract),** and if we consider the uniform linear motion of the ship twin in the frame of the Earth, the length of the spaceship does not contract. (… the distance T-N does not contract)*

*Obviously there is symmetry between the two scenarios!*

*The two distances are in relative motion to each other at speed **v**, and the two uniform linear motions are independent: **THE TWO DISTANCES OVERLAP!*

*The twins don’t know it, but the two distances overlap!*

*Massimiliano Dell’Aguzzo*