IL GEMELLO ASTRONAUTA E’ PIU’ GIOVANE

Il gemello astronauta si muove a velocità costante v nel sistema di riferimento della Terra. (per dirigersi verso una stella che appartiene al sistema di riferimento della Terra, la distanza Terra-stella è quindi una lunghezza ben definita nel sistema di riferimento terrestre)

Nel sistema di riferimento dell’astronave è la distanza Terra-stella a muoversi verso il gemello astronauta (per il gemello astronauta la distanza in movimento è contratta ed è minore della lunghezza propria Terra-stella) e, quando la stella raggiunge l’astronave, il gemello astronauta è più giovane del gemello rimasto sulla Terra.

Nel sistema di…


In my article “MATHEMATICAL CALCULATIONS, APPLYING THE LORENTZ TRANSFORMATIONS” and in other precedents I wrote that, in the spaceship frame S_1, the astronaut twin can assume that he is the same age as his brother.

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). …


There are two twins on Earth; one of the two decides to leave (to a planet at distance d from Hearth and at high speed v with a spaceship), and on his return he will be younger than his brother left on Earth.

But if we consider the astronaut twin stationary, why can’t it be assumed that the opposite happens, that is, that the twin left on Earth is younger than the traveling brother?

Let us consider the symmetrical situation: all points of the Earth’s frame S (also the Earth and one of the twins) moving together at the same…


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…


Le trasformazioni di Lorentz sono rappresentate da un sistema di due equazioni con quattro incognite (x, t, x_1, t_1):
x_1 = γ*(x -v*t),
x = γ*(x_1 + v*t_1).
Non prendo in considerazione le altre due trasformazioni perché derivano da queste.

CASO A
Se consideriamo il sistema di riferimento S_1 che si muove ad una velocità v rispetto al sistema di riferimento S, abbiamo come condizioni iniziali x_1 = 0 e x = d. (Notare che anche il sistema di riferimento S è in movimento rispetto a S_1 con velocità opposta)
Otteniamo t = d/v e t_1 = (1/γ)*(d/v). …


I repeat the resolution strategy that I showed in my previous article “THE LORENTZ’S TRANSFORMATIONS AND THE TEMPORAL EXPANSION” (sorry for having corrected it several times, but now there is its final version), trying to make it clearer.

This is the situation to be clarified

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…


Let Alex and Brad be twins (both in their thirties, located in the same position and both stationary relative to each other); Brad decides to leave with his spaceship covering a distance d (to then turn back, always moving at the same speed v), distance d is represented by one segment, segment AB. (initially Alex and Brad are positioned near point A and the distance d is the measure of the length to travel to get to a distant planet, the same value for both)

How long is the journey for Alex?

Alex will have to wait a time equal…


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)


Siano Alex e Brad due gemelli (entrambi trentenni, situati nella stessa posizione e tutti e due fermi l’uno rispetto all’altro); Brad decide di partire con la sua astronave percorrendo una distanza d (per poi tornare indietro), muovendosi sempre alla stessa velocità v.

La distanza d è rappresentata da un segmento, il segmento AB. (Inizialmente Alex e Brad sono posizionati in prossimità del punto A e la distanza d è la misura della lunghezza da percorrere per arrivare su un pianeta lontano, lo stesso valore per entrambi)

Quanto dura il viaggio per Alex?

Alex dovrà attendere un tempo pari a 2*d/v


Distances in motion

In the article it was shown that, if we denote by d_1m(t_1) the distance between the two origins of the frames S and S_1 measured at the instant of time t_1 by an observer in the frame S_1, it has the following value:
d_1m(t_1) = x(t)*γ^(-1) (t_1) or d_1m(t_1) = x (t)*γ^(-1) (t).

Regardless of the distance x_1(t_1) “apparently traveled” from the origin O of S with respect to the frame S_1, d_1m(t_1) is a different value that also depends on v_1(t_1).

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!

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