# MATHEMATICAL CALCULATIONS, APPLYING THE LORENTZ TRANSFORMATIONS

Let S be the frame of the Earth, S_1 the frame of the spaceship and d the Earth-planet distance in the Earth’s frame S (where the astronaut twin must arrive).

Let us consider the spaceship’s frame S_1 in motion at speed v. From the Lorentz transformation x = γ*(x_1 + v*t_1) we get x_1 = (x/γ) -v*t_1. The relationship just obtained considers the astronaut twin stationary and the Earth’s frame S (contracted) in motion with respect to the astronaut twin. (The speed of the entire frame S is opposite to that of the spaceship)

If x = d, at time t_1 = 0 we get x_1 = d/γ ;

if x = 0, at time t_1 = 0 we get x_1 = 0;

if x = -d, at time t_1 = 0 we get x_1 = -(d/γ).

x = -d is the abscissa of the point P diametrically opposite to the planet (with respect to the Earth).

In the frame S_1 the time in which the planet (x = d) reaches the spaceship is obtained by solving: (d/γ) -v*t_1 = 0.

t_1 = (1/γ)*(d/v).

In the spaceship’s frame S_1, at time t_1, the spaceship reaches the planet.

During this time interval the Earth (x = 0) reaches the point of abscissa x_1 so that x_1 = -v*t_1, x_1 = -(d/γ). At time t_1 (in the frame of the spaceship S_1), the Earth-planet distance continues to be d. (in the Earth’s frame S)

The Earth continues to be the origin of the frame S, x = γ*(x_1 + v*t_1) = γ*(-d/γ + d/γ ) = 0.

Note that, in the frame of spaceship S_1, the contracted Earth-planet distance d/γ has shifted all at time t_1 (No more time needed!). For the astronaut twin the moving distance is contracted, starting from the initial time t_1 = 0 up to time t_1 = (1/γ)*(d/v). In the spaceship’s frame S_1, the distance between the planet and the Earth is always at a distance d/γ, in fact (d/γ) -v*t_1 -(-v*t_1) = d/γ.

For the astronaut twin, at time t_1 = 0, the planet is at a distance d/γ (x_1 = d/γ) and this distance comes to meet him in a time t_1 = (1/γ)*(d/v).

If we wanted (x/γ) -v*t_1 = 0 at time t_1 = d/v, we would get x = γ*v*t_1, x = γ*d. (x > d), a point at a greater distance from the Earth, it is obviously not the position of the planet (x = d) in the Earth’s frame S.

In the Earth’s frame S the Earth-planet distance has distance d, in the spaceship’s frame S_1 the Earth-planet distance is moving and its length is d/γ. (the distance d of the frame S contracts, while the spaceship’s frame S_1 is in motion)

Even when the point P of abscissa -d reaches the Earth, in the spaceship’s frame S_1 the time is t_1 = (1/γ)*(d/v). In the point P of abscissa -d we can consider that there is a second spaceship, stationary with respect to the first spaceship. (which obviously has velocity v with respect to the Earth’s frame S). Also for the point P it is (d/γ) -v*t_1 = 0, the second spaceship reaches the Earth when the time is t_1 = (1/γ)*(d/v) in the spaceship’s frame S_1.

In the spaceship’s frame S_1, at time t_1, the second spaceship reaches the Earth.

Let’s now analyze what happens if all points of the Earth’s frame S travel a distance equal to d in the spaceship’s frame S_1. (with speed opposite to that of the spaceship)

In reality we analyze what happens if the Earth and the planet (together with the other points of the Earth’s frame S) move by a distance d/γ in the spaceship’s frame S_1. (the Earth-planet distance in motion contracts but, for the twin left on Earth, the planet is always at a distance d)

We now consider the frame of the Earth S in motion (with a speed opposite to that of the spaceship). From the Lorentz transformation x_1 = γ*(x -v*t) we get x = (x_1/γ) + v*t. The relationship just obtained considers the twin remained on the Earth stationary and the spaceship’s frame S_1 (contracted) in motion with respect to the twin remained on the Earth.

If x_1 = -d, at time t = 0 we get x = -(d/γ) ;

if x_1 = 0, at time t = 0 we get x = 0;

if x_1 = d, at time t = 0 we get x = d/γ.

The planet however continues to remain at a position x = d in the Earth’s frame S, the planet is stationary with respect to the Earth.

In the Earth’s frame S, at time t = (1/γ)*(d/v) the point diametrically opposite to the planet (x_1 = -d) reaches the Earth. In fact, solving -(d/γ) + v*t = 0, we obtain t = (1/γ)*(d/v). For the twin remaining on the Earth, at time t = (1/γ)*(d/v) a length equal to d/γ has passed all.

It is as if the Earth had reached the point P of abscissa -d at time t_1 = d/v (also in this case t = (1/γ)*(d/v)), the point P is now a point of the frame S_1. (as if the Earth had traveled the entire distance d in the spaceship’s frame)

In the spaceship’s frame S_1 a time equal to t_1 = d/v has not elapsed at all, but the Earth has reached the point P because the Earth moves in the spaceship’s frame S_1. (the distance d of the frame S_1 contracts, while the Earth’s frame S is in motion)

In the Earth’s frame S, at time t = (1/γ)*(d/v) the position of the spaceship (x_1 = 0) is x = v*t = d/γ. At time t = (1/γ)*(d/v) it’s v*t < d but, for the astronaut twin, the planet has reached the spaceship. (it was the planet that moved, in the spaceship’s frame S_1) If the Earth reaches point P at time t, the planet also reaches the spaceship (at time t). With respect to the planet, the position of the spaceship (at time t = 0) is -d/γ. At time t = (1/γ)*(d/v) also here -d/γ + v*t = 0. If a time t passes for the planet, the same time t passes for the twin on the Earth, the Earth and the planet are stationary in the Earth’s frame S!

It is as if the planet had reached the point O_1 of abscissa 0 at time t_1 = d/v (also in this case t = (1/γ)*(d/v)), the point O_1 is now the origin of the frame S_1, the spaceship. (as if the planet had traveled the entire distance d in the spaceship’s frame)

In the spaceship’s frame S_1 a time equal to t_1 = d/v has not elapsed at all, but the planet has reached the spaceship because the planet moves in the spaceship’s frame S_1. (also here the distance d of the frame S_1 contracts, while the Earth’s frame S is in motion)

At time t, the distance point P -spaceship continues to be d in the frame of spaceship S_1, the spaceship at time t = (1/γ)*(d/v) continues to be the origin of the frame S_1, x_1 = γ*(x -v*t) = γ*(d/γ -d/γ ) = 0.

For the twins, the same time flows in the spaceship’s frame S_1! (if all points of the Earth’s frame S travel a distance equal to d, with speed opposite to that of the spaceship)

For both twins (considered stationary) a distance equal to d/γ is moving at the same speed, the times are the same. And the distances d moving at speed v have length d/γ.

The Earth-planet distance d moving towards the astronaut twin is contracted and, also for the terrestrial twin, the distance d (whose extremes are the point P of the abscissa x_1 = -d and the spaceship x_1 = 0) moves contracted towards the Earth (and the planet). In the spaceship’s frame S_1, it’s t_1 = t = (1/γ)*(d/v).

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

If the frame S_1 moves through frame S, the frame S also moves through frame S_1 (times are the same!)

If all points of the Earth’s frame S travel a distance equal to d with speed opposite to that of the spaceship, the position of the spaceship at time t_1 = (1/γ)*(d/v) would coincide with that of the planet in the frame S_1. (from the Earth we do not see the astronaut twin on the planet, because it was the planet that reached the spaceship in the frame S_1, v*t = d/γ and v*t < d)

From the spaceship, at time t_1 = (1/γ)*(d/v), the astronaut twin sees his brother at a distance (d/γ); (the Earth, on the other hand, is at a greater distance d and the same time t = t_1 has passed for her)

In the frame S_1, the Earth and the planet are in motion at speed v (the whole frame S is in motion). If the astronaut twin wants to determine the time t in the Earth’s frame S and knows the Lorentz Transformations (even if he is wrong because the time elapsed is different for each of the two), he must consider that the frame S_1 is also in motion with respect to the frame S; in the spaceship’s frame S_1 it’s not t_1 = γ*t but t_1= t !

In the spaceship’s frame S_1 the time t = (1/γ)*(d/v) is correct, because the Earth -planet distance is contracted (d/γ) and v*t = v*(1/γ)*(d/v) = d/γ. (t = t_1 as in Classical Physics, but only in the frame S_1)

In the spaceship’s frame S_1 it’s not t_1 = γ*t, because the astronaut twin sees the entire frame S in motion. If he wants to determine the time t (wrongly, because the twin on Earth is older), he must consider that the whole frame S_1 is also in motion with respect to the frame S, the astronaut twin must not consider only the Earth (or only the planet) in motion, the astronaut twin must consider the frame S and the frame S_1 in relative motion to each other. Furthermore, for the astronaut twin, the time t_1 is determined by the arrival of the planet, it is not t_1 = d/v, and finally, for the astronaut twin the Earth does not travel a distance d moving in the frame S_1. At time t _1 > (1/γ)*(d/v) the Earth is at a distance d from the spaceship, but the Earth did not travel this distance: v*t_1 < d.

The astronaut twin may therefore (wrongly) assume that the twin on Earth is the same age as him, but the twin on Earth is older. Actually, for the astronaut twin the Earth-planet distance is in motion (and the Earth and the planet belong to the Earth’s frame S). If we consider the Earth and the planet stationary, the Earth-planet distance is greater, and it is the spaceship that travels the entire distance d in the Earth’s frame S. (without considering the point P diametrically opposite to the Earth)

It all depends on whether we are interested only in the arrival of the points (the points of the frame S, the time t and the time t_1 are determined in the frame S_1) or the arrival of the spaceship on the planet (in the frame S), the astronaut twin only cares that he arrives on the planet.

We know that the spaceship reaches the planet in the Earth’s frame S, the measure of time t in the Earth’s frame S is different. (because we know that the spaceship left the Earth to reach the planet)

t = d/v and t_1 = (1/γ)*(d/v). (the astronaut twin is younger than his brother, t > t_1)

Insights

The Earth-planet distance moved a length d/γ in the spaceship’s frame, and a time equal to t = (1/γ)*(d/v) has passed. When the Earth and the planet are moving within the frame S_1, it’s difficult to notice because we see only the origin of the spaceship’s frame S_1 (the spaceship). We only note that the length of the spaceship decreases, there are only the points of the Earth’s frame and the distances between them do not change. (even if all points of the Earth’s frame were moving with a speed opposite to that of the spaceship)

When the twin on Earth sees the spaceship moving, it is actually the entire spaceship’s frame S_1 that is in motion. (and the Earth’s frame S also moves in the opposite direction)

However, it’s possible to imagine a second spaceship positioned in the point P of abscissa x = -d (in the Earth’s frame S and x_1 = -d in the frame S_1 before departure, when t = t_1 < 0

and imagining that the two spaceships are connected by a rigid rod),

starting at the instant t = t_1 = 0 and at the same speed v of the other spaceship with one of the two twins. In the spaceship’s frame S_1, at time t_1 = (1/γ)*(d/v), the first spaceship reaches the planet.

What happens in the Earth’s frame? In the Earth’s frame S the first spaceship arrives on the planet at the time t = d/v, and the second spaceship also arrives on the Earth at the same time t = d/v. (?) This is true if the two spaceships are not joined together.

Since the two spaceships are joined together, the twin left on Earth sees a moving distance (and therefore contracted), the twin on Earth sees the second spaceship at a shorter distance d/γ, but in the frame S_1 the two spaceships remain at a distance d.

We know that x = (x_1/γ) + v*t and, only if x_1 = 0, then x = v*t.

If x_1 = -d, x = 0 at time t = (1/γ)*(d/v) and, come to think of it, it’s not a coincidence! In Earth’s frame S, at time t = (1/γ)*(d/v), the second spaceship reaches the Earth.

In the spaceship’s frame S_1, the twins are the same age. In both frames the equalities hold: (d/γ) -v*t_1 = 0 and -(d/γ) + v*t = 0. The example of the two spaceships represents just what happens when the frame S is in motion.

The distances of the frame S contract with respect to the frame S_1 when the frame S_1 is in motion, and the distances of the frame S_1 also contract. (because the frame S is also in motion, with opposite speed)

Massimiliano Dell’Aguzzo

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