Solving the twin paradox means solving the system of two equations:

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

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

With gamma I obviously indicated the Lorentz factor, I do not consider the other two Lorentz transformations because they depend on a) and b)

Usually we think of a single spaceship in the origin O_1 of the frame F_1, but the number of spaceships is infinite. Imagine a first spaceship positioned at the origin O_1 of the frame F_1, and a second spaceship positioned at point P (- d; 0) of the frame F_1. The position of the first spaceship obviously coincides with the position of the Earth (the origin O of the frame F) at the initial instant t = t_1 = 0.

In the frame F (the frame of the Earth) the distance between the two spaceships contracts, the distance between the origin O and the point Q also contracts in the frame F_1. (the frame of the spaceship) Initially, if we do not know the two frames F and F_1 , one should not think that one of the two frames moves with a uniform rectilinear motion in the other frame.

If we think about this, we are making a choice; we are choosing x = v * t, or we are choosing x_1 = - v * t_1.

Instead, it is necessary to think only of two lengths in relative motion with respect to each other. The first length has the extremes P and O_1, while the second has the extremes O and Q. If the two lengths are in relative motion to each other, it is important to know that at time t = d / (gamma * v) the point P coincides with the point O and, at time t_1 = d / (gamma * v), the point Q coincides with the point O_1. (this always happens, both if x = v * t and if x_1 = - v * t_1)

We can think of many spaceships, the first in the origin O_1, the second in the point P (-d ; 0), a third in the point X (-2d ; 0), a fourth in the point Y (-3d ; 0), and so on.

(The spaceships are points of the frame F_1) And we can then think of many planets (or many stars), the Earth in the origin O, a second planet in the point Q (d ; 0), a third planet in the point Z (2d ; 0), a fourth planet in the point W (3d ; 0) and so on. (The Earth and the other planets are points of the frame F)

In the frame of the Earth, the distance between two consecutive spaceships is d / gamma, even in the frame of the spaceship the distance between two consecutive planets is d / gamma.

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

(2d/gamma) – v*t_1 = 0 and (-2d /gamma) + v*t = 0

(3d / gamma) – v*t_1 = 0 and (-3d/gamma) + v*t = 0

(4d/gamma) – v*t_1 = 0 and (-4d/gamma) + v*t = 0

………………………………………………………………………………………

THIS IS THE TRUE SYMMETRY OF THE LORENTZ TRANSFORMATIONS, we don't know if x = v * t or if x_1 = - v * t_1. (You must choose one of the two, otherwise the system of two equations will not solve)

The distances are contracted because each frame measures the lengths of the other frame, but these distances have a well-defined length in each frame. (the length d)

Imagine the spaceship has a tail, the length of the tail is d in the frame of the spaceship.

At time t = d / (gamma * v), the end of the tail reaches the Earth in the frame of the Earth.

Meantime, the point Q (d; 0) of the frame F reaches the origin O_1 of the frame F_1 at time t_1 = d / (gamma * v).

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

(THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS!)

At the end of the tail of the spaceship is John, at time t = d / (gamma * v) the twin on Earth "sees" John and, in the frame of the spaceship, at time

t_1 = d / (gamma * v), John sees the Earth.

John is at a distance d from the nose of the spaceship, and we know that at the initial time (t = t_1 = 0 s) the nose of the spaceship and the Earth coincide. It is not necessary to know if x = v * t (or if x_1 = - v * t_1).

At the initial time (t = t_1 = 0 s), in the frame of the spaceship the Earth is at a distance d; the Earth is moving relative to the spaceship. For this reason, the Earth arrives at point P (John "sees" the Earth) at time t_1 = d / (gamma * v)

One can also imagine that John's position coincides with a point M of the frame of the Earth, the segment MO (O is the Earth, the origin of the Earth's frame!) is moving with speed - v with respect to John.

In an empty space the twin paradox is not resolved, it's impossible to determine a single solution of a system of two equations with four unknowns.

If we know that one of the twins is an astronaut, the paradox is solved. There is no need to choose, the choice x = v * t is obligatory.

If x = v *t, at time t = d / (gamma * v), the spaceship did not reach point Q. (a star, for example)

If x = v *t, at time t = d / (v * gamma), in the frame of the spaceship,

the time is t_1 = d / (v * gamma * gamma).

Imagine the point R (d / gamma; 0) of the frame of the Earth, and consider the two events A and B

A: Earth reaches point P.

B: point R reaches the spaceship.

The two events A and B are simultaneous in the frame of the Earth, the point P reaches the Earth at time t = d / (gamma * v), and the spaceship reaches the point R at time t = d / (gamma * v).

In the frame of the Earth, the length of the tail is contracted. If the spaceship travels a distance equal to d / gamma, the twin on Earth "sees" John.

The two events A and B are not simultaneous in the frame of the spaceship, point R reaches the spaceship at time t_1 = d / (gamma * gamma * v) and the Earth reaches point P at time t_1 = d / (gamma * v).

Now consider event C.

C: point Q reaches the spaceship.

Event C represents: the spaceship reaches point Q (in the frame of the Earth)

Event C represents: point Q reaches the spaceship (in the frame of the spaceship)

In the frame of the Earth the spaceship reaches the point Q at time t = d / v and, in the frame of the spaceship, the point Q reaches the spaceship at time t_1 = d / (gamma * v), exactly when the Earth reaches the point P.

THE CLOCK OF THE ASTRONAUT TWIN SLOWS DOWN, COMPARED TO THE EARTH'S CLOCK. (EVEN IF THE SPACESHIP CONTINUES TO TRAVEL)

THE CLOCK PARADOX IS ALSO SOLVED !

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)

If we know that one of the twins is an astronaut (and we choose x = v * t), the travel time is given by the correct application of uniform rectilinear motion.

The astronaut twin travels (at speed v) a distance d in the frame of the Earth and, in the frame of the Earth, the elapsed time t is necessarily t = d / v.

In the frame of the Earth, it does not matter that the tail of the spaceship arrives. (the tail of the spaceship has length d in the frame of the spaceship)

Instead, in the frame of the spaceship, the elapsed time t_1 depends on the arrival of the star. (and not from the point of arrival of the Earth, because the Earth does not move at speed - v with uniform rectilinear motion in the frame of the spaceship! )

In the frame of the spaceship it is as if "the dilated frame of the Earth" were in motion (a new frame, similar to that of the Earth, where every distance equal to d in the frame of the Earth, assumes a length equal to d * gamma)

In the frame of the spaceship the elapsed time t_1 is

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

v * t_1 = d / gamma.

If at time t_1 the Earth traveled a distance equal to d in the frame of the spaceship, then the Earth moved with speed - gamma * v, as if "the dilated frame of the Earth" had moved with speed -v.

If - v * t_1 = - d_1 / gamma, then d_1 = d * gamma.

Now imagine a spaceship with a tail in motion (the length of the tail of the spaceship is equal to d in the frame of the spaceship, the contracted tail reaches the Earth at time t = d / (gamma * v), it is impossible that the spaceship at the same time traveled a distance equal to d in the frame of the Earth! (CONTRADICTION!)

Let us now analyze a further contradiction.

In the frame of the spaceship, the star reaches the spaceship at the time

t_1 = d / (gamma * v), and the Earth travels a distance equal to d in the frame of the spaceship.

To the right of the spaceship the contracted Earth-star distance has moved, while to the left of the spaceship the dilated Earth-star distance has moved! (CONTRADICTION!)

In the frame of the spaceship the motion of the Earth is only "apparent !" (no need to add anything else!)

When in the frame of the spaceship, the star reaches the spaceship at time

t_1 = d / (v * gamma), the distance traveled by the Earth in the frame of the spaceship is only one:

d_1 = v * t_1, d_1 = d / gamma !

The spaceship moves with uniform rectilinear motion in the frame of the Earth, and the clock of the spaceship slows down compared to the clocks of the Earth frame. (even if the spaceship continues to travel with uniform rectilinear motion and does not return to Earth)

The uniform rectilinear motion of the spaceship in the frame of the Earth does not allow the Earth to travel a distance equal to d in the frame of the spaceship, at time t_1 = d / (v * gamma). (for any value of d)

In the frame of the Earth, at time t = d / (v * gamma), the spaceship travels a shorter distance, the right distance equal to d / gamma.

We cannot choose x_1 = - v * t_1, in this case THE TWIN PARADOX (THE CLOCK PARADOX) is solved!

On the other hand, if a frame F moves at speed - v with uniform rectilinear motion with respect to the frame F_1 (x_1 = - v * t_1), then all the clocks of the frame F slow down with respect to all the clocks of the frame F_1.

Only one of the two frames moves with uniform rectilinear motion at speed v in the other frame!

If x = v * t, then:

(c * t)^2 = (v * t)^2 + (c * t_1)^2 ,

PYTHAGORAS' THEOREM (LIGHT CLOCK)

If x_1 = - v * t_1, then:

(c * t_1)^2 = (v * t_1)^2 + (c * t)^2,

PYTHAGORAS' THEOREM (LIGHT CLOCK)

Massimiliano Dell'Aguzzo

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