FURTHER CONSIDERATIONS

In an empty space, the twin paradox can’t be solved! The twins are TWIN A and TWIN B.

FIRST SOLUTION

We can consider the frame F_1 where TWIN B has zero speed. From x_1 = gamma * (x-v * t) and x = gamma * (x_1 + v * t_1), if x = v * t (x_1 = 0) we obtain v * t = gamma * v * t_1, we therefore obtain: t_1 = t / gamma. (TWIN A is younger than TWIN B)

In this case TWIN A travels a distance d_A = v * t (in TWIN B frame) and TWIN B travels a distance d_B = v * t_1. (in TWIN A frame) We have that d_A > d_B.

SECOND SOLUTION

There is another solution, we consider the frame F_2 where TWIN A has zero speed (TWIN B has an opposite speed) From x_1 = gamma * (x-v * t) and x = gamma * (x_1 + v * t_1), if x_1 = -v * t_1 (x = 0) we obtain -v * t_1 = -gamma * v * t, we therefore obtain: t = t_1 / gamma. (TWIN B is younger than TWIN A)

In this case TWIN B travels a distance d_B = v * t_1 (in TWIN A frame) and TWIN A travels a distance d_A = v * t (in TWIN B frame). We have that d_B > d_A.

THIRD SOLUTION

In an empty space, there is a frame F_3 where the two twins (TWIN A and TWIN B) travel at the same speed v_x. Just pay attention to the relativistic sum of the velocities and this velocity can be determined. In the frame F_3 the two twins are the same age (t_A = t_B) The two twins in the frame F_3 travel the same distance d_A = d_B = v_x * t_A = v_x * t_B. (d_A = d_B)

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In an empty space ALL THREE SOLUTIONS ARE RIGHT! (This is the TWIN PARADOX, and it can’t be solved) The twin paradox is indeterminate, there are infinite other solutions! All options are possible: t_1 < t, t < t_1 e t= t_1. In an empty space the velocity of each of the twins can go from 0 to v. (and the speed of the other twin can go from -v to 0) In an empty space both twins can move, or only one of them moves while the other is stationary! There is uncertainty, and the twin paradox can’t be resolved!

BUT, if we know that a ship (TWIN A) moves at a constant speed in the earth frame (the frame of TWIN B), then the FIRST SOLUTION IS NECESSARILY RIGHT! If the Earth-star distance is d (in TWIN B frame), t = d / v. (the time in the frame of TWIN B can’t be different, the Earth and the star belong to the same frame)

WE KNOW THAT THE SHIP REACHES THE STAR!

The Earth travels a shorter distance, the contracted Earth-star distance moves in the ship frame. (Lorentz-FitzGerald contraction) In earth frame the ship travels a greater distance, in ship frame the Earth travels a shorter distance! (d_A > d_B) In my opinion it is t_1 < t (only in an empty space t_1 < t or t < t_1) If we consider a ship moving in Earth frame I don’t believe in t_1 < t and t < t_1. (each of the twins can think what they want) It makes no sense that in the ship frame one thinks of the earth time, only one of the twins is younger. Even the earth twin may not know mathematics and may think that the Earth is younger. The two brothers may not know, but we do! In my opinion, neither of the twins has to think about the elapsed time in the brother frame.

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Imagine a ship with a tail, the length of the tail is d*gamma. In ship frame the time t_1 = (d * gamma )/v and, for earth twin, the time t is less, t = d/v. Earth twin is younger. But, in the ship frame, the star also moves.

Scenario 1

In ship frame, earth twin travels a distance d*gamma,

a) in ship frame t_1 = (d*gamma) / v,

b) in earth frame t = d/v.

In the frame of the ship, at time t_1 = (d*gamma)/v, the position of the ship is different from the position of the star.

Scenario 2

In earth frame, the ship travels a distance d,

c) in earth frame t = d/v,

d) in ship frame t_1 = d/(gamma *v).

I believe we have to choose , either scenario 1 or scenario 2. In both scenarios t = d/v.

SCENARIO 2 (THE RIGHT SCENARIO)

In the frame of the Earth the ship reaches the star and, in the frame of the ship, the star reaches the ship!

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The Earth moves relative to the ship, but it’s the ship that moves within the frame of the Earth. (The Earth and the star belong to the same frame) The contracted Earth-star distance moves in the ship frame, not only the Earth moves in the ship frame! If the Earth-star distance is d (in the earth frame), in the ship frame the time t_1 depends on the arrival of the star.

d/gamma -v * t_1 = 0,

d = gamma * v * t_1,

d = v * t. (t = gamma * t_1, t_1 < t)

It’s not -v * t_1 = -gamma * v * t, it’s not t < t_1!

The time t_1 depends on the arrival of the star, and not on the motion of the Earth!

Massimiliano Dell’Aguzzo