There are two twins, twin A and twin B. Let’s imagine that the twin A has to reach (in a spaceship) a star at a distance d from the Earth. The star belongs to the frame of the Earth. Let’s also imagine that the spaceship has a tail of length d (in the frame of the spaceship), and that the astronaut twin is in the front of the spaceship.

At the initial time the positions of the twins coincide (x = x_1 = 0) and the twins are the same age.

What happens in the frame of the Earth?

In the frame of the Earth, the end of the tail reaches the Earth in a shorter time. (a moving distance contracts) Assuming that the spaceship travels at constant speed v, the end of the tail arrives at time t_a = d / (v*gamma). (with gamma I obviously indicated the Lorentz factor)

At time t_a the spaceship did not reach the star, in the frame of the Earth the spaceship reaches the star at time t = d / v. (always assuming that the spaceship is traveling at a constant speed)

What happens in the frame of the spaceship?

In the frame of the spaceship, the star reaches the spaceship in a shorter time. (a moving distance contracts) If the spaceship travels at speed v in the frame of the Earth, the Earth and the star travel at an opposite speed to that of the spaceship. If the speed is constant (as previously analyzed), the arrival time of the star is t_1 = d / (v * gamma). At time t_1 the Earth does not reach the tail end of the spaceship, in the frame of the spaceship the Earth reaches the tail end at time t_b = d / v.

What interests us?

We said that the spaceship reaches the star in the frame of the Earth, so the elapsed time in the frame of the Earth is t = d / v and the elapsed time in the frame of the spaceship is t_1 = d / (v * gamma).

The spaceship does not travel at a constant speed but, for any speed value non-zero, the Earth-star distance contracts in the frame of the spaceship (the Lorentz-FitzGerald contraction!), while the length of the spaceship’s tail contracts for the twin left on Earth. (here too the Lorentz-FitzGerald contraction)

Just apply the Lorentz Transformations correctly, the astronaut twin is younger! The times t_a = d / (v * gamma) (in the frame of the Earth) and t_b = d / v (in the frame of the spaceship) are two “useless” times. In the frame of the spaceship at the time t_b = d / v the star has moved away from the spaceship, in the frame of the spaceship the star reaches the spaceship at the time t_1 = d / (v*gamma).

If the time in the frame of the spaceship is greater than t_1 the star has moved away from the spaceship, while in the frame of the Earth the spaceship at time t_a = d / (v * gamma) did not reach the star!

This is a contradiction, in the frame of the Earth the spaceship has not reached the star and, in the frame of the spaceship, the star has moved away from the spaceship. In the frame of the spaceship, the star reaches the spaceship at time t_1 = d / (v*gamma).

Further considerations

If the astronaut twin travels a distance d in the frame of the Earth (a), then the Earth travels a distance d / gamma in the frame of the spaceship. (b)

If the Earth travels a distance d in the frame of the spaceship (c), then the spaceship travels a distance d / gamma in the frame of the Earth (d).

If the Earth travels a distance d / gamma in the frame of the spaceship (e), then the spaceship travels a distance d / (gamma * gamma) in the frame of the Earth (f)

If the Earth travels a distance d*gamma in the frame of the spaceship, then the spaceship travels a distance d in the frame of the Earth. Also in this case, the star has moved away from the spaceship (in the frame of the spaceship).

If the spaceship reaches the star in the frame of the Earth it’s (a). If we consider (c)and (e) it does not hold (a). We said that (a) holds, where is the paradox?

The solution

If we denote the elapsed time in the frame of the Earth with t and if we denote the elapsed time in the frame of the spaceship with t_1, it’s t_1 < t ! (and not t < t_1!)

This is the solution, in the frame of the Earth at time t = d /v the spaceship reaches the star and, in the frame of the astronaut twin, the star reaches the spaceship at time t_1 = d / (v*gamma).

It has been said that the spaceship reaches the star in the frame of the Earth, the solution where the times are t = d / (v*gamma) and t_1 = d / v does not make sense. This solution makes sense if we consider a body that moves with uniform rectilinear motion in the frame of the spaceship. (for which the travel time in the frame of the spaceship is t_1 = d / v)

If the astronaut twin launches a rocket R in the direction he sees the Earth moving away, then the rocket R is younger than the astronaut twin.

The rocket R is younger than the astronaut twin (TWIN A), and the astronaut twin is younger than the twin left on Earth (TWIN B). The Rocket R and Earth can also travel at the same speed in the frame of the spaceship, but they are not the same age!

If the rocket R and the Earth travel at the same speed v (the speed opposite to that of the spaceship), indicating with t_1 the elapsed time in the frame of the spaceship and indicating with t_R the elapsed time in the frame of the rocket R, then it’s t_1 = d / v and t_R = d / (v*gamma) when the rocket R reaches the end of the spaceship tail.

I understand that doubts arise! (for Special Relativity it’s different)

I told you that the tail of the spaceship reaches the Earth at time t_a = d / (v * gamma), just when the time has the same value t_1 = d / (v * gamma) in which the star reaches the spaceship in the frame of the spaceship.

From x_1 = gamma * (x -v * t) and x = gamma * (x_1 + v * t_1)

(and these two Lorentz Transformations are enough), if x = 0 we get t < t_1:

x_1 = -gamma * v * t and x_1 = -v * t_1 (t = t_1 / gamma, t < t_1)

It seems strange, but this is true for the rocket R;

for the rocket R it’s x_1 =-v * t_1 and -v * t_1 = -gamma * v * t_R !

At time t = d / (v * gamma) no time t_1 = d / (v * gamma) has passed in the frame of the spaceship, because in the frame of the spaceship the elapsed time is t_1 = d / (v * gamma) only when in the frame of the Earth the elapsed time is t = d / v. Obviously, not even the time t_1 = d / v has passed in the frame of the spaceship.

But, even if a time equal to t_1 = d / v has not elapsed in the frame of the spaceship, we know that a length is in motion in the frame of the Earth. This length measures d in the frame of the spaceship and moves in the frame of the Earth (it is the tail of the spaceship, the tail of the spaceship has length d)

Even if the time t_1 is not t_1 = d / v and if we have a length equal to d (d = v * d/v), we consider: -d = -gamma * v * t (d = gamma * v * t)

d = gamma * v * t and therefore t = d / (gamma * v),

It is as if were -v * t_1 = -gamma * v *t, but it is not!

The Earth cares that the other end of the contracted length arrives (the tail end of the spaceship), it does not care that the Earth reaches that same end in the frame of the spaceship!

The time t depends on the arrival of the extreme of the contracted length, we are interested in the arrival of the end of the queue and it is therefore wrong to apply -v * t_1 = -gamma * v *t!

In the frame of the Earth we want the astronaut twin to reach the star and, in the frame of the spaceship, the star will reach the spaceship!

However, if the Earth moves in the frame of the spaceship, the star moves away from the spaceship in the frame of the astronaut twin (the star passes the spaceship) and the spaceship does not reach the star in the frame of the Earth. (It makes no sense at all to imagine the Earth moving relative to the spaceship, and this applies to any distance!)

This contracted distance d / gamma is the distance that moves towards a second spaceship moving at an opposite speed in the frame of the Earth. The contracted distance moves towards the second spaceship, but also towards the first. The second spaceship runs along the tail of the first spaceship, and the contracted tail moves in the frame of the second spaceship and in the frame of the Earth.

The tail of the contracted spaceship is in motion with respect to the Earth, but the Earth (the Earth is older than the astronaut twin) does not travel a distance d in the frame of the spaceship! It’s the rocket R that travels any distance in the frame of the spaceship.

If in the frame of the Earth the elapsed time is t = d / (gamma * v), in the frame of the spaceship the time t_1 is t_1 = d / (gamma * gamma * v).

When the time will be t_1 = d / v in the frame of the spaceship, then the distance traveled by the rocket R will be equal to d (v * t_1 = d) in the frame of the spaceship. It’s necessary to wait and, when the time will be t_1 = d / v in the frame of the spaceship, for the rocket R the time t_R will be

t_R = d / (gamma * v).

(it’s not t = t_1 / gamma in the frame of the spaceship!)

Consider the tail of the spaceship (of length equal to d in the frame of the spaceship), the tail is a moving length in the frame of the Earth.

It doesn’t matter that the Earth travels a distance equal to d in the frame of the spaceship, it matters that the spaceship travels a distance equal to d/gamma in the frame of the Earth, thus

t = d / (v * gamma) and t_1 = d / (v * gamma * gamma).

In the frame of the Earth the tail of the spaceship is contracted, the end of the tail reaches the twin left on Earth at time t = d / (v * gamma)!

The extremity of the tail arrives when the time is t = d / (v * gamma) in the frame of the Earth, but the astronaut twin (twin A) has not reached the star because the second end of the tail (where the astronaut twin is positioned) will reach the star at time t = d / v.

Also for the Earth it’s x_1 = -v * t_1 (only if we want to determine the position with respect to the spaceship), but the Earth and the rocket R must not be confused. In the frame of the spaceship the Earth-star distance contracts, the rocket R instead travels freely in the frame of the spaceship!

THE EARTH IS AT REST!