AN ALTERNATIVE RESOLUTION OF THE TWIN PARADOX

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). If there are two lengths that move towards each other (in relative motion at speed v), the elapsed times for both are necessarily and obviously the same.

The astronaut twin believes he is his own alter ego, the astronaut twin may mistakenly think that the twin on Earth is waiting for the arrival of a moving distance.

In the frame of the Earth S, the time t relative to the astronaut’s alter ego is the time of arrival of a moving distance and the time of arrival of the distance is shorter (d/γ < d). In this case, in the frame of the Earth S, the twin on Earth is the same age as the astronaut twin.

In my article “TEMPORAL DILATATION (IF THE SPEED IS CONSTANT)” I have already introduced the alter ego of the astronaut twin (when I gave the example of a spaceship in motion with a rigid rod of length equal to the Earth- planet distance attached to its tail). In that case, of course, the spaceship with the tail of length d is the alter ego of a moving spaceship (tailless).

In the frame of the Earth S, the tail end of the spaceship reaches the Earth at the time t = t_1 = (1/γ)*(d/v). This is the example that should make us reflect.

If the spaceship with the tail leaves with zero speed and arrives at zero speed (and if we are concerned only with the arrival of the end of the tail), the astronaut twin arrives on the planet at time t = (1/γ)*(d/v), t = t_1. On the other hand, if the spaceship with the tail travels at a constant speed v (and if we don’t care about the arrival of the end of the tail), the astronaut twin arrives on the planet at time t = d/v.

In the discussion of the twin paradox, the spaceship does not have a tail and the frame of the spaceship S_1 is empty with only the spaceship in the origin. We get then t > t_1.

The astronaut twin is mistaken in thinking that he is his own alter ego and, even if he sees the Earth in motion, it’s t > t_1. (he is always younger than the stationary observers of the frame of the Earth S)

If x = v*t then t = d/v (we know that x = d) and, considering the two Lorentz transformations x_1 = γ*(x -v*t) and x = γ*(x_1 + v*t_1) (from which the others are also obtained), we necessarily obtain x_1 = 0 and t_1 = (1/γ)*t.

By symmetry (the Lorentz transformations are symmetrical), we can imagine the alter ego of the Earth. If the alter ego of the Earth moves in the frame of spaceship S_1, he’s younger than the alter ego of the astronaut twin (t < t_1, in the frame of the spaceship S_1 the tail has no length d/γ, the tail has length d.). The Earth-planet distance is the tail of the “spaceship Earth”, for the Earth the treatment is reversed compared to what was said for the spaceship (where the astronaut twin is located). The alter ego of the Earth does not have a tail (of length equal to the Earth -planet distance), the alter ego of the Earth moves in the spaceship frame S_1 while the planet is motionless.

In this case x_1 = -v*t_1 and we therefore get t_1 = d/v (if x_1 = -d). Always considering the two Lorentz transformations x_1 = γ*(x -v*t) and x = γ*(x_1 + v*t_1), we necessarily obtain x = 0 and t = (1/γ)*t_1.

If the planet is in motion, the planet itself reaches the spaceship at time t_1 = (1/γ)*(d/v) in the frame S_1. To obtain t < t_1 only the Earth can be in motion, and not the whole Universe! In the discussion of the twin paradox it is not necessary to consider that the Earth travels a distance equal to d in the frame of the spaceship S_1, remember that the frame of the spaceship S_1 is empty. (there is only the spaceship in the origin O_1)

The astronaut twin can’t think that the terrestrial twin is younger, it can’t be t < t_1. The astronaut twin knows that in the frame of the Earth S the spaceship is heading in the opposite direction to the Earth, the spaceship must reach the planet. In the frame of the Earth only the spaceship is in motion (the frame of the spaceship is empty) and the Earth-planet distance is greater than observed in the spaceship, it is not the contracted distance d/γ. In the frame of the Earth S it is t > t_1 and it can’t be otherwise.

The twin on Earth does not wait for the arrival of contracted distances, the twin on Earth sees the uniform rectilinear motion of the spaceship.

(v*t = d, t = d/v)

Believe me, the Lorentz Transformations are “ab omni naevo vindicatae”. Although the Earth is in motion with respect to the astronaut twin, the time t (spent in the frame of the Earth S) is greater than the time t_1 (spent in the spaceship frame S_1) and the twin paradox vanishes.

Massimiliano Dell’Aguzzo

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