# THE TWIN PARADOX: FINAL CONCLUSIONS

Consider two frames ** F** and

**in relative motion to each other with speed**

*F_1***, the frames**

*v***and**

*F***are coincident at**

*F_1*

*the initial time*.*(t = t_1 = 0 s)**You have to choose:*

** a)** the frame

**moves**

*F_1***in the frame**

*with uniform rectilinear motion***at speed**

*F***.**

*v*

*(x = v* t)*** b)** the frame

**moves**

*F***in the frame**

*with uniform rectilinear motion***at speed**

*F_1***.**

*-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 frame

**) and imagine that the frame**

*F***goes to the left (with respect to the frame**

*F***).**

*F_1**Consider a point **(-d; 0)** of the frame **F_1** and a point** (d; 0) **of the frame **F**.*

*In the frame **F**, the point **P (-d; 0)** of the frame **F_1** reaches the origin **O** of the frame **F** at time** t = d/(gamma*v) **if** x_1 = -v * t_1**; in the frame **F_1**, at the same time **t_1 = d/(gamma*v)**, the point **Q (d; 0)** of the frame **F** reaches the origin **O_1** of the frame **F_1 **if** x = v * t**.*

*t = t_1 = d / (gamma * v)** for any value of **d**, whether it holds **a)**, or that it holds **b)**.*

*THE TWO SCENARIOS ARE SYMMETRICAL!*

Consider two frames ** F** and

**moving relative to each other at speed**

*F_1***:**

*v**in the frame **F_1**, the contracted distances of the frame **F** move with speed **-v** from right to left. (to the right of the origin **O_1**)*

*in the frame** F**, the contracted distances of the frame **F_1** move with speed **v** from left to right. (to the left of the origin **O**)*

*I am thinking only of the two distances **O-Q** and **O_1-P** moving relative to each other at speed **v**, and I don’t think about **x = v * t** or **x_1 = -v * t_1**.*

*(d/gamma) -v * t_1 = 0, and (-d /gamma) + v * t = 0.*

*THE LORENTZ TRANSFORMATIONS ARE SYMMETRIC!*

*If we choose **a)**, we know that **t_1 < t**.*

*Furthermore, the point **Q (d; 0)** reaches the origin **O_1** of the frame **F_1 **at time **t_1 = d / (gamma * v)**, and the origin **O** of the frame **F** also reaches the point **P (-d; 0)** at time **t_1 = d / (gamma * v)**.*

The contracted distances of the frame ** F** move with speed

**in the frame**

*-v***from right to left, the origin**

*F_1***so reaches the point**

*O***at time**

*P(-d; 0)***.**

*t_1 = d / (v * gamma)**If we choose a) (t_1 < t)**, it cannot be **v * t_1 = d. **(and this holds for any distance **d**)*

*The origin **O** of the frame **F **reaches the point **P (-d; 0) **of the frame **F_1*

*at time **t_1 = d / (gamma * v)**,*

*the origin **O** of the frame **F **is therefore **in advance of the uniform rectilinear motion** in the frame **F_1 (x_1 = -v * t_1)**.*

*The origin **O** moved with speed **-v * gamma** in the frame **F_1** (and **-v*gamma **is different from **-v, a)** and** b)** are mutually exclusive),*

*-v * gamma * t_1 = (-v*gamma*d) / (gamma*v) = -d**, for any value of **d**.*

*This is A CONTRADICTION, because the distance traveled by the origin O in the frame F_1 is d_1 = gamma * v * t_1 = d. In the frame F_1 the origin O is moving at speed -gamma * v (and not at speed -v),*

*and all clocks in frame F do not slow down compared to the clocks in frame F_1.*

*If we choose a), it’s not t < t_1 in the frame F_1.*

*If we choose **b)**, we know that* *t < t_1**.*

*Furthermore, the point **P(-d; 0)** reaches the origin **O** of the frame **F **at time **t = d / (gamma * v)**, and the origin **O_1** of the frame **F_1** also reaches the point **Q (d; 0)** at time **t = d / (gamma * v)**.*

If ** x_1 = -v * t_1**, then the contracted distances of the frame

**move with speed**

*F_1***in the frame**

*v***from left to right, the origin**

*F***so reaches the point**

*O_1*

*Q***at time**

*(d; 0)***.**

*t = d / (v * gamma)**If we choose b) (t < t_1)**, it cannot be **v * t = d. **(and this holds for any distance **d**)*

*The origin **O_1 **of the frame **F_1 **reaches the point **Q (d; 0) **of the frame **F*

*at time **t = d / (gamma * v)**,*

*the origin **O_1** of the frame **F_1 **is therefore **in advance of the uniform rectilinear motion** in the frame **F (x = v * t)**.*

*The origin **O_1 **moved with speed **v * gamma** in the frame **F** (and **v*gamma **is different from **v**,** a)** and** b)** are mutually exclusive!),*

*v * gamma * t = (v * gamma * d) / (gamma * v) = d**, for any value of **d**.*

*This is A CONTRADICTION, because the distance traveled by the origin O_1 in the frame F is d_1 = gamma * v * t = d. In the frame F the origin O_1 is moving at speed gamma * v (and not at speed v),*

*and all clocks in frame F_1 do not slow down compared to the clocks in frame F.*

*If we choose b), it’s not t_1 < t in the frame F.*

*(t_1 < t) and (t < t_1) are mutually exclusive!*

** Only one** of the two expressions

**or**

*(t_1 < t)***is**

*(t < t_1)*

*TRUE.**LET ‘S SOLVE THE TWIN PARADOX NOW**, we can think that frame **F** is the frame of the Earth, and that frame **F_1** is the frame of the spaceship. We know very well that the frame of the spaceship (at constant speed **v**, if the acceleration is **zero**) moves with uniform rectilinear motion in the frame of the Earth.*

*The choice **a)** is obligatory:* *the clock of the astronaut twin slows down compared to the Earth’s clock (and it can’t be otherwise)*

*The spaceship actually moves with uniform rectilinear motion between any two points of the Earth’s frame. The astronaut twin leaves from the Earth to reach a star, the Earth and the star belong to the frame of the Earth! (all the other points reached by the spaceship also belong to the frame of the Earth)*

*Consider** the Lorentz Transformations:*

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

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

The initial conditions cannot be both ** (x = 0)** and

**.**

*(x_1 = 0)**If **x_1 = 0** and **x = 0**, we get the **“useless solution”**: **x = x_1 = t = t_1 = 0**.*

The frame of the spaceship moves at speed ** v** (with uniform rectilinear motion) in the frame of the Earth, we know very well that

**.**

*x = v * t**(*

*and the Earth**, moving with respect to the spaceship,*

*DOES NOT MOVE**with the uniform rectilinear motion*

*x_1 = -v * t_1**)*

*If **x = v * t** **(x_1 = 0)** and **(x non-zero)**, with simple steps we obtain **t_1 < t**.*

*Consider the point **R (d / gamma, 0),** and think now that the point **R** belongs to the frame of the Earth. In the frame of the spaceship, the two events **A** and **B** representing*

*A:** the Earth reaches point **P (-d; 0)**,*

*B: **the point **R (d / gamma; 0) **reaches the spaceship*

*are not simultaneous!*

*The two events **A** and **B** are simultaneous in the frame of the Earth, but they are not simultaneous in the frame of the spaceship!*

*Let’s analyze event **A**: the Earth reaches point **P (-d; 0)*

In the frame of the Earth,

the point ** P** reaches the Earth at time

**and,**

*t = d / (gamma * v)*in the frame of the spaceship,

the Earth reaches the point ** P** at time

**.**

*t_1 = d / (gamma * v)*Point** P **represents the tail of the spaceship, imagine that the spaceship has a tail and let be the length of the tail

**in the frame of the spaceship (at the initial time**

*d***).**

*t = t_1 = 0 s*

*Important: Point P does not represent a second spaceship moving at speed v in the frame of the Earth.**In the frame of the Earth, if the point **P (-d; 0)** reaches the Earth, the spaceship also reaches the point **R (d/gamma; 0)**.*

In the frame of the spaceship, point ** R **reaches the spaceship at time

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

*When in the spaceship frame the elapsed time is*

*t_1 = d / (v * gamma * gamma)**,*

*in the frame of the Earth the elapsed time is **t = d / (v * gamma)**.*

*IMPORTANT!*

*If we think of the two distances **P-O_1** and **O-Q** moving relative to each other at speed **v**, then** t = t_1 = d / (gamma * v)** and the origin **O** (the Earth) reaches the point **P** at time **t_1 = d / (gamma * v)**.*

*But the uniform rectilinear motion **x = v * t** breaks the symmetry, in the frame of the Earth …*

*… the spaceship does not reach the star at all at time **t = d/(gamma*v)**, the spaceship reaches the star at time **t = d / v!*

Now consider the point ** W (-d / gamma, 0)** of the spaceship frame.

If we know that ** x = v * t**, at time

**the point**

*t_1 = d / (gamma * gamma * v)***reaches the origin**

*R***, and the origin**

*O_1***reaches the point**

*O***.**

*W**THIS IS IMPORTANT: in the frame of the spaceship the origin O (the Earth) reaches the point P at the time t_1 = d / (gamma * v), exactly when the point Q (the star) reaches the spaceship. Let us now consider in the point P (-d; 0) of the spaceship frame a second spaceship in motion with uniform rectilinear motion at speed v in the frame of the Earth (and not the tail of the spaceship as analyzed previously),*

*it’s t = d / v and t_1 = d / (v * gamma).*

*The whole frame of the spaceship moves with uniform rectilinear motion in the frame of the Earth, the number of spaceships is infinite!*

*THE UNIFORM RECTILINEAR MOTION x = v * t BREAKS THE SYMMETRY.*

*When in the frame of the Earth the elapsed time is **t = d / v**, in the spaceship frame the elapsed time is **t_1 = d / (v * gamma)**. If **x = v * t**, it makes no sense to imagine the tail of the spaceship moving with respect to the Earth.*

*If** t_1 < t** it cannot be** t_1 = t**, the point **P** and the origin **O_1** are two distinct spaceships, the point **P** does not belong to the tail of the spaceship positioned in the origin **O_1**,*

*THE ORIGIN O (THE EARTH) IS AT REST.*

We know that the spaceship moves with uniform rectilinear motion in the frame of the Earth, we know that ** x = v * t**.

*THE ASTRONAUT TWIN IS YOUNGER THAN EARTH, EVEN IF THE SPACESHIP DOES NOT GO BACK. (AND THE SPACESHIP CONTINUES TO TRAVEL AT CONSTANT SPEED)*

*The frame of the Earth is not a privileged frame, the spaceship is moving in the frame of the Earth with uniform rectilinear motion at speed v!*

The Earth and the star belong to the frame of the Earth, ** and the Earth-star distance has a well defined length in the frame of the Earth**, imagine that the Earth-star distance is

**in the frame of the Earth.**

*d**If the frame of the Earth moves at speed **-v** with respect to the frame of the spaceship with uniform rectilinear motion **x_1 = -v * t_1**, in the frame of the Earth it would be **x = gamma * v * t**, and the spaceship would reach the star in a time **t** less than **d / v**. **(THIS IS A CONTRADICTION!)*

*In the frame of the spaceship, the uniform rectilinear motion of the spaceship (in the frame of the Earth) does not allow the Earth at time t_1 = d / (v * gamma) to travel a distance equal to d in the frame of the spaceship with uniform rectilinear motion at speed -v.*

*(for any distance d)*

The uniform rectilinear motion of the spaceship ** x = v * t** in the frame of the Earth forces the Earth to move with a greater speed.

*(x_1 = -gamma * v * t_1)**In the frame of the spaceship, the position of the Earth relative to the spaceship is equal to the opposite position of the spaceship in the frame of the Earth.*

At time ** t**, in the frame of the Earth the position of the spaceship is

**.**

*x = v * t*At time ** t_1 = t / gamma**, in the frame of the spaceship the position of the Earth is

**.**

*x_1 = -gamma * v * t_1 = -(gamma * v * t) / gamma = -v * t**The Lorentz Transformations are “ab omni naevo vindicatae”.*

*Massimiliano Dell’Aguzzo*