# THE CLOCK PARADOX AND THE LORENTZ TRANSFORMATIONS

Solving the CLOCK PARADOX means solving a system of two equations with four unknowns.

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).

In an empty space THE CLOCK PARADOX is not resolvable, it’s impossible to determine a single solution of a system of two equations with four unknowns.

BUT THE CLOCK PARADOX SOMETIMES “IS NOT A PARADOX”, consider for example the astronaut twin moving with speed v to reach a star. The Earth and the star both belong to the frame of the Earth, and let the Earth-star distance be equal to d in the frame of the Earth.

I represent every event in the frame of the Earth with the pair

position-time: (x, t), and

I represent every event in the frame of the spaceship with the pair

position-time: (x_1, t_1).

The initial position of the Earth (the origin O of the Earth’s frame) coincides with the initial position of the spaceship (the origin O_1 of the spaceship frame) at the initial time t = t_1 = 0 s. (x = x_1 = t = t_1 = 0)

Consider the event P (-d; 0) in the frame of the spaceship, the event 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: Event P does not represent a second spaceship moving at speed v in the frame of the Earth.

Also consider the event Q (d; 0) in the frame of the Earth, the event Q represents a star at a distance d from the Earth. (at the initial time t = t_1 = 0 s)

In the frame of the spaceship, at time t_1 = d / (gamma * v), the star reaches the spaceship; in the frame of the Earth, at time t = d / (gamma * v), the end tail of the spaceship reaches the Earth. 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 x = v * t, then the contracted distances of the Earth’s frame move with speed -v in the frame of the spaceship, and the Earth so reaches the end tail of the spaceship at time t_1 = d / (v * gamma).

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)

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.

Consider the Lorentz Transformations a) and b):

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

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

If x = -x_1, then t = t_1.

But we know that x = v * t (and not that x =-x_1),

THE EARTH’S FRAME IS AT REST.

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

Let’s consider again the system of two equations:

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

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

If x = v * t (x_1 = 0), then t_1 < t.

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

If x = v * t, THE SYMMETRY OF THE LORENTZ TRANSFORMATIONS IS BROKEN:

if we know that x = v * t , it cannot be -v * t_1 = -d. (for any distance d, because t_1 < t, t_1 = t / gamma)

The Earth is in advance of the uniform rectilinear motion x_1 = -v*t_1, because the Earth reaches the point P at time t_1 = d / (gamma * v). If x_1 = -v*t_1 and x_1 = -d, in the frame of the spaceship more than one “copy” of the contracted Earth-star distance “passes” the spaceship.

If gamma = 2 for example, then two “copies” of the contracted Earth-star distance “pass” the spaceship.

We know that in the frame of the spaceship (if gamma = 2) the star reaches the spaceship at time t_1 = d / (2 * v),

and at time t_1 = d / (2 * v) only “a copy” of the contracted Earth-star distance “passes” the spaceship. (If gamma = 2, v * t_1 = d / 2;

every contracted Earth-star distance is just equal to d / 2 !)

At time t_1 = d / (2 * v) two “copies” of the contracted Earth-star distance “do not pass” the spaceship, but the Earth reaches point P at time t_1 = d / (2 * v).

This is a contradiction, and it means that the Earth DOES NOT MOVE at speed -v with uniform rectilinear motion in the frame of the spaceship. In the frame of the spaceship the Earth at time t_1 = d / (v * gamma) reaches the point P (x_1 = -d), the motion of the Earth in the frame of the spaceship is so represented by

x_1 = -gamma * v * t_1.

At time t_1 = d / (gamma * v), the position of the Earth with respect to the spaceship is x_1 =-(gamma * v * d) / (gamma * v) = -d.

It is not possible that -v * t_1 = -d (for any distance d), and therefore it cannot be t < t_1!

Again the system of two equations:

a) x_1 = gamma * (x -v * t)
b) x = gamma * (x_1 + v * t_1)

If (x = v * t) and (x_1 = -v * t_1), we obtain “the useless solution”:

x = x_1 = t = t_1 = 0.

I indicate (x_1 = 0) with (c) , and

I indicate (x = 0) with (d).

(c) and (d) are MUTUALLY EXCLUSIVE!

If it is true (c), it is not true (d).

If it is true (d), it is not true (c).

(c): If x = v * t (x_1 = 0), then t_1 < t.

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

(d): If x_1 = -v * t_1 (x = 0), then t < t_1.

(- v * t_1 = -gamma * v * t, t = t_1 / gamma)

The spaceship actually moves with uniform rectilinear motion between any two points of the Earth’s frame (if the acceleration of the spaceship is zero). 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)

We cannot choose x_1 = -v * t_1, in this case THE TWIN PARADOX (THE CLOCK PARADOX) IS SOLVED! The choice x = v * t it’s obligatory: the clock of the astronaut twin slows down compared to the Earth’s clock. (even if the spaceship continues to travel and does not return to Earth)

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)

If x = v * t, we obtain:

(c * t)² = (v * t)² + (c * t_1)² ; THE ORIGIN O IS AT REST.

Instead, if x_1 = -v * t_1, we obtain:

(c * t_1)² = (v * t_1)² + (c * t)² ; THE ORIGIN O_1 IS AT REST.

In an empty space THE CLOCK PARADOX cannot be solved, it’s impossible to determine a single solution of a system of two equations with four unknowns; the solution is indeterminate: (t_1 < t) OR (t < t_1)

But, given two clocks, if we know that one of the two clocks (CLOCK A) moves with uniform rectilinear motion at speed v in the frame of the other clock (CLOCK B), THE CLOCK PARADOX IS SOLVED:

clock A slows down compared to clock B,

the Lorentz Transformations are “ab omni naevo vindicatae”.

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!