THE LORENTZ TRANSFORMATIONS AND THE TEMPORAL EXPANSION

The Lorentz transformations are represented by a system of two equations with four unknowns (x, t, x_1, t_1):

x_1 = γ*(x -v*t),
x = γ*(x_1 + v*t_1).

I do not take the other two transformations into consideration because they are derived from these.

CASE A

If we consider the frame S_1 moving at speed v with respect to the frame S we have as initial conditions x_1 = 0 and x = d. (Note that the frame S is also in motion with respect S_1 with opposite speed)

We get t = d/v and t_1 = (1/γ)*(d/v). (t_1 < t)

As shown in my previous articles, the distance traveled by the whole frame S with respect to S_1 is d_1 less than d.

CASE B

If we consider the frame S moving at speed v with respect to the frame S_1 we have as initial conditions x = 0 and x_1 = -d_1. (Note that the frame S_1 is also in motion with respect S with opposite speed)

We get t_1 = d_1/v and t = (1/γ)*(d_1/v). (t < t_1)

Here too it can be shown that the distance traveled by the entire frame S_1 with respect to S is d less than d_1.

The situation just described is indeterminate. If we do not agree on which of the two frames (S or S_1) must travel a precise distance with respect to the other frame, everything is possible. We also need to know what frames they are!

The twin paradox

Let us consider the situation where S is the Earth’s frame and where S_1 is the frame of a spaceship. In the case of the twin paradox (the best known version) there is a twin who travels with his spaceship and then returns, so we know who is moving with respect to the Earth’s frame. (frame S)

Considering x_1 = 0 (the astronaut twin is stationary in his frame S_1) and x = d (a distance d has been traveled with respect to the Earth’s frame), we obtain necessarily t_1 < t.

Of the twins we know who remains stationary on the Earth and who moves along a given distance d (fixed distance with respect to the Earth’s frame), for us observers (within the Earth’s frame) the solution is well determined:

on his return, the astronaut twin is younger than his brother who remained on Earth to wait. (as also shown in my article “THE TWINS ALEX AND BRAD”)

Insights

Let us now consider the symmetrical situation: all points of the Earth’s frame S moving together at the same speed v opposite to that of the spaceship, covering the same distance d traveled by the spaceship in the Earth’s frame. (to then go back).

The set of points of the frame S (the set of all points of the Universe together with the Earth, the twin remained on the Earth itself and the planet) moves at the same speed within the frame of the astronaut (S_1, an empty copy of the fixed stars frame with only the spaceship inside at the origin O_1), and every distance between them contracts when observed in motion. (for the astronaut twin the distance between two points in the Universe is contracted, the distance between the Earth and the planet where the spaceship must arrive is also contracted)

And, as if he were traveling, the astronaut twin will arrive at his destination in less time (and then return); he does not care that the set of points of the frame S arrives (or rather each of them) in a position (and then comes back) at a certain distance as if measured in the frame S_1, the empty copy of the fixed stars frame. (the travel time of each point of the Universe that moves in the frame S_1 is equal to 2*d/v, considering the outward journey and the return journey, the same travel time as the spaceship in the Earth’s frame as analyzed above with conditions x = d and x_1 = 0)

Remember that d is the distance between the Earth and the planet where the astronaut must arrive. (d is the Earth-planet distance measured in the Earth’s frame)

If you think about it, even though the points of frame S still do not return to the initial position (where the spaceship is, from where they “started”), for the astronaut twin the travel time is not equal to 2*d/v.

The travel time for the astronaut twin is equal to (1/γ)*(2*d/v), and the same time passes for the terrestrial twin in motion. (also for him the distance d that comes towards him is contracting, he is moving in the frame S_1 and travels a distance d in the frame of the spaceship)

In this case, the apparent distance traveled by the spaceship in the round trip is also less than 2*d. If t = (1/γ)*(2*d/v), v*t < 2*d.

As already shown in my previous articles, in the frame S_1 there is no time difference between the twins.

……………………………………………………………………………..

This seems an equivalent situation to the previous one but it is not; in this case the distance traveled by the spaceship in the frame S does not coincide with the distance 2*d, this distance measures (1/γ)*(2*d) (a shorter distance). Initially it was said that the planet to be reached is at a distance d, length measured with respect to the Earth’s frame S.

Even if the speed of the set of points of the frame S becomes zero at the end of the outward journey and in the frame S the distance between the two origins of S and of S_1 has measure d, this does not mean that the value d represents the distance actually traveled by the spaceship: it can be deduced that the distance actually traveled by the spaceship x(t) < d.

In my article “THE INFINITESIMAL TRANSFORMATIONS OF LORENTZ” we consider a body (a spaceship) which, starting from the origin of the frame S, arrives at a point of the x axis of positive abscissa d with respect to the frame S. (even at non-constant speed)

It is shown that, if an observer in the frame S_1 measures the distance between the two origins of S and S_1 at the instant of time t_1, he obtains the value d_1m(t_1) = x(t)*γ^(-1) (t) but the distance x_1(t_1) traveled from the origin of S in the frame S_1 is less than d.

The relation x(t) < d can therefore be obtained by symmetry.

………………………………………………………………………………….

In this situation we consider the Earth the origin of the moving frame S and the destination the point P of abscissa -d, the point P is diametrically opposite to the planet (with respect to the Earth). For the planet, the destination is instead the position of the spaceship. Earth and planet move in a time interval shorter than d/v and, even if the spaceship respect to the Earth after the outward (or return) journey is at a distance d, the distance traveled by the spaceship in the Earth’s frame is shorter by d.

But, if it was the astronaut twin who left, the situation just described is obviously contradictory!

………………………………………………………………………………….

Furthermore, it must be said that if the set of points of the frame S (the set of all points in the Universe together with the Earth and the twin left on the Earth itself) traveled a distance equal to γ*d, the spaceship would travel a distance d in the Earth’s frame S but, in this case, the planet would overtake the spaceship at the end of his outward journey.

Conclusions

I repeat it once again, if we have x_1 = 0 and x = d as initial conditions we necessarily obtain t_1 < t.

This is the case in which, if the spaceship accelerates or decelerates with respect to the frame of the fixed stars, the well-known fictitious forces are generated inside it. (However, it is not necessary to introduce the Theory of General Relativity)

If you think about it, the initial condition x = d means that the spaceship on its outward journey moves a distance d with respect to the frame of the fixed stars.

The frame of the Earth coincides with the frame of the fixed stars. It is perfectly fine that the frame of the fixed stars is in motion relative to speed v with respect to the frame S_1 (an empty copy of the frame of the fixed stars), but the distance traveled by the spaceship in the frame S must be d and not a length less than d.

Before the departure of the astronaut twin, the distance between the Earth (with the other twin) and the planet to be reached is d (The distance that the spaceship must travel can’t be less).

If all the points of the Universe (together with the planet, the Earth and the twin on the Earth) were to move at a certain speed, the planet to be reached with respect to the Earth’s frame always remains at the same distance d, the spaceship is “obliged” to travel this same distance with respect to the Earth’s frame, not a shorter one!

When we correctly consider the frame S_1 (the spaceship) moving at speed v with respect to the frame S with initial conditions x_1 = 0 and x = d, it’s the distance d_1 traveled by the Earth in the frame S_1 that is less than d. The length d_1 coincides with the distance actually traveled by the spaceship (in the frame of the spaceship) and this length can be well determined by the twin astronaut himself.

For the traveling astronaut the distances are in fact contracted; the planets and the stars are not spheres but ellipsoids more and more flattened in the direction parallel to its motion, as the speed of the spaceship increases. (however he was on Earth before leaving and he knows their exact shape in the frame of the fixed stars)

The solution is therefore well determined: upon returning from his journey, the astronaut twin is younger than his brother who remained on Earth to wait. (and to verify this it is sufficient to know the Lorentz Transformations)

I finished, I think I’ve said everything. (for now, but there is more!)

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!

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