THE TWIN PARADOX: FINAL CONCLUSIONS

Consider two frames F and F_1 in relative motion to each other with speed v, the frames F and F_1 are coincident at the initial time. (t = t_1 = 0 s)

You have to choose:

a) the frame F_1 moves with uniform rectilinear motion in the frame F at speed v. (x = v* t)

b) the frame F moves with uniform rectilinear motion in the frame F_1 at speed -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 F) and imagine that the frame F goes to the left (with respect to the frame 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 F_1 moving relative to each other at speed 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 -v in the frame F_1 from right to left, the origin O so reaches the point P(-d; 0) at time 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 F_1 move with speed v in the frame F from left to right, the origin O_1 so reaches the point Q (d; 0) at time 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 (t_1 < t) or (t < t_1) is 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 t = d / (gamma * v) and,

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 d in the frame of the spaceship (at the initial time 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 t_1 = d / (gamma * gamma * v) the point R reaches the origin O_1, and the origin O reaches the point 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 d in the frame of the Earth.

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

Condividiamo insieme la conoscenza!