THE LORENTZ TRANSFORMATIONS AND FRAMES IN MOTION (THE TWIN PARADOX)

Let us consider two frames F (x, t) and F_1 (x_1, t_1), coinciding at the initial time t = t_1 = 0.

I consider the two Lorentz transformations:

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

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

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

If the speed v is known, this is a system of two equations with four unknowns.

The two Lorentz transformations indicate that:

c) the frame F_1 moves with speed v with respect to the frame F, and travels a certain distance d in the frame F.

d) the frame F moves with speed -v with respect to the frame F_1, and travels a certain distance d_1 in the frame F_1.

c) and d) are MUTUALLY EXCLUSIVE!

What does c) mean?

If the frame F_1 moves with speed v with respect to the frame F, then the contracted frame of F (the distances between the points of the frame F are smaller due to the Lorentz-FitzGerald contraction) moves with speed -v with respect to the frame F_1.

If x = v * t, x_1 = 0, v * t = gamma * v * t_1, t_1 = t / gamma.

The whole frame F_1 is younger than the whole frame F. (t_1 < t)

If the frame F_1 travels a distance d while moving with respect to the frame F (x = v * t = d, t = d / v, x_1 = 0) it is like solving:

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

However, there is a frame F_2 that moves at -v speed with respect to the frame F_1 and the whole frame F_2 is younger than the whole frame F_1. (and I will elaborate on this concept later in this article)

What does d) mean?

If the frame F moves with speed -v with respect to the frame F_1, then the contracted frame of F_1 (the distances between the points of the frame F_1 are smaller due to the Lorentz-FitzGerald contraction) moves with speed v respect to the frame F.

If x_1 = -v * t_1, x = 0, -v * t_1 = -gamma * v * t, t = t_1 / gamma.

The whole frame F is younger than the whole frame F_1. (t < t_1)

If the frame F travels a distance d_1 while moving with respect to the frame F_1 (x_1 = -v * t_1 = -d_1, t_1 = d_1 / v, x = 0) it is like solving:

-d_1 = -gamma * v * t, t = d_1 / (v * gamma), t < t_1.

However, there is a frame F_3 that moves at speed v with respect to the frame F and the whole frame F_3 is younger than the whole frame F.

Let us now consider two twins, TWIN A is positioned in the origin of the frame F_1 and TWIN B is positioned in the origin of the frame F. (THE WELL-KNOWN TWIN PARADOX)

WHAT IS THE SOLUTION?

The solution is: (t_1 < t) AUT (t < t_1).

(TWIN A younger than TWIN B) AUT (TWIN B younger than TWIN A)

(t_1 < t) and (t < t_1) are not both true expressions, if one of the two expressions is true the other is false:

1 OR 0 = 1

0 OR 1 = 1

(t_1 < t) OR (t < t_1) = 1

(t_1 < t) OR (t < t_1) is TRUE. (THERE IS NO CONTRADICTION!)

If we consider that the frame F_1 travels a certain distance d while moving with uniform rectilinear motion in the frame F, then the whole frame F_1 is younger than the whole frame F. On the other hand, if we consider that the frame F travels a certain distance d_1 as it moves in the frame F_1, then the whole frame F is younger than the whole frame F_1. (there is no contradiction, it is not t_1 < t and t < t_1)

In the best known version of the twin paradox, reference is made to one of the twins (for example TWIN A) moving in a spaceship with uniform rectilinear motion in the frame of the Earth (the frame of TWIN B). (to reach a star, the Earth and the star belong to the frame of the Earth)

In this case, the twin paradox vanishes because the frame of TWIN A moves with speed v with respect to the frame of TWIN B:

TWIN A is younger than TWIN B,

even if TWIN A does not go back to Earth! (THE CLOCK PARADOX)

TWIN A is being considered to travel a certain distance in the frame of the Earth. (although the spaceship continues to travel after reaching the star)

If TWIN A travels a distance d in the frame of TWIN B it is a mistake to think that TWIN B travels a distance d / gamma in the frame of TWIN A.

t_1 = d / (v * gamma) and t = d / (v * gamma * gamma), ERROR!

For TWIN B the moving distances of the TWIN A frame contract. Imagine that the spaceship has a tail, and the tail length is d / gamma in the frame of TWIN A. The tail length of the spaceship is d / (gamma * gamma) in the frame of TWIN B.

TWIN B does not travel the distance d / gamma in the frame of TWIN A, TWIN A travels a distance equal to d / (gamma * gamma) in the frame of TWIN B and the tail end of the spaceship reaches TWIN B!

If TWIN A travels a distance d / (gamma * gamma) in the frame of TWIN B, then:

t = d/(v*gamma*gamma) and t_1 = d/(v*gamma*gamma*gamma).

The spaceship has not yet reached the star, the spaceship reaches the star at time t = d / v.

The time t is “the elapsed time” in the Earth frame (the frame of TWIN B),

the time t_1 is “the elapsed time” in the frame of the spaceship (the frame of TWIN A):

0 ≤ t ≤ d / v, 0 ≤ t_1 ≤ d / (v*gamma), t_1 = t / gamma.

TWIN A is younger than TWIN B, even if TWIN A does not return to Earth!

If applied correctly, the Lorentz transformations give us the solution:

THE LORENTZ TRANSFORMATIONS ARE “AB OMNI NAEVO VINDICATAE”.

If the inertial frame of the spaceship moves with speed v with respect to the inertial frame of the Earth (x = v * t), it makes no sense to consider the frame of the Earth in motion with respect to the frame of the spaceship.

( t_1 < t ) and (t < t_1) is a contradiction as well explained in my article “THE CONTRADICTION”, THE FRAME OF THE EARTH IS AT REST!

Is the twin paradox resolved in this case?

As I said earlier, there is a frame F_2 that moves at -v speed with respect to the frame of the spaceship (frame F_1) and the whole frame F_2 is younger than the whole frame F_1.

If the astronaut twin (TWIN A) launches a rocket (the rocket R) at the initial time t = t_1 = 0 and at the speed -v in the direction in which TWIN A sees the Earth moving away, then the entire frame of the rocket R is younger than the whole frame of the spaceship. (t_R = t_1 / gamma)

If TWIN B hides in the rocket R with a probability of 50% (before the ship’s departure), we will never know which of the twins is younger. However, we do know that the entire frame of the spaceship is younger than the frame of the Earth. (If TWIN B is not hiding in the rocket R, TWIN A is younger than TWIN B!)

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!