There are two clocks A and B in motion relative to a given speed v.

In an empty space we have to choose:

1) The clock B moves in the frame of the clock A, B travels a distance d and the motion is a motion uniform rectilinear. (the time elapsed in the frame of A is t_A = d / v) In this case, in the frame of B, it is necessary to wait for a contracted distance. (the time elapsed in the frame of B is t_B = t_A / gamma, gamma is the Lorentz factor)

AUT

2) The clock A moves in the frame of the clock B, A travels a distance d and the motion is a motion uniform rectilinear. (the time elapsed in the frame of B is t_B = d / v) In this case, in the frame of A, it is necessary to wait for a contracted distance. (the time elapsed in the frame of A is t_A = t_B / gamma)

Many say: if it is 1) clock A also waits for a contracted distance (t_A = t_B / gamma) and, if it is 2), clock B also waits for a contracted distance. (t_B = t_A / gamma)

THIS IS FOR ME A CONTRADICTION!

If it is 1) clock A moves with an opposite speed compared to clock B but it is t_B < t_A (and not t_A < t_B) and, if it is 2), clock B moves with an opposite speed with respect to clock A but it is t_A < t_B. (and not t_B < t_A)

If we do not initially make a choice it is:

(t_B < t_A) OR (t_A < t_B)

1 OR 0 = 1 (1 V 0 = 1)

0 OR 1 = 1 (0 V 1 = 1)

t_B < t_A and t_A < t_B cannot both be true, but (t_B < t_A) OR (t_A < t_B) is a true proposition, (t_B < t_A) OR (t_A < t_B) = 1.

In the twin paradox (with a twin left on Earth and an astronaut twin who reaches a star) there is no need to choose: it is the astronaut twin who travels the distance d in the frame of the Earth. (the star belongs to the frame of the Earth and the star is at a distance d from the Earth in the frame of the Earth) The astronaut twin does not start at a constant speed, on his journey he accelerates and decelerates but, if he reaches the star and comes back, at the return of the spaceship to Earth he is younger than his brother.

INSIGHTS

Two clocks A and B can both travel a same distance d in the frame of a third clock C. (You can imagine two spaceships moving in the frame of the Earth, one spaceship travels to the left and the second spaceship travels to the right)

If the velocity module is the same, the two watches are both younger than the third watch. (t_A < t_C, t_B < t_C, t_A = t_B) It is wrong to think that clock B travels a distance in the frame of clock A, or that clock A travels a distance in the frame of B.

In this case we have chosen, both clocks A and B travel the same distance d in the frame of the clock C.

Massimiliano Dell’Aguzzo

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