Consider two points A and B in relative motion with respect to each other at speed v, we have seen how to interpret the Lorentz transformations.
In this case, not knowing the initial conditions, the solution is:

(point A moves in the frame of point B and is younger than the point B)
or
(point B moves in the frame of point A and is younger than the point A)

The solution is indeterminate.

To be more precise, the conditions x = 2*d and x_1=0 (that I have set to solve the problem in the previous article) mean that one of the twins will be in motion with a certain speed compared to me. It seems strange, but we must consider that the two conditions x=2*d and x_1=0, once set, can also be valid in a mathematical sense for the twin who remains on Earth. (just as if I had blindfolded my eyes and I couldn’t see which of the twins will be in motion relative to me)

If we consider x=2*d and x_1=0, the solution is not yet well determined, but is represented by:

(one of the twins leaves with his spaceship and will be younger than his brother and me when he returns)

or

(only the Earth is moving in the frame of one twin stationary floating in space with me, the Earth is younger than me and the twin stationary)

I too should remain floating in space, otherwise if I left with the Earth (changing my frame) me and the twin left on Earth when “the Earth returns to the starting position” will be younger than the “astronaut” twin. Now I would be lucky, because one twin and I will be younger than the other, I would understand that it was the Earth that moved into the frame of a stationary twin (on Earth there was me and the younger twin!). But the observer must remain in his frame, the observer’s frame must not change!

I have more to be clear. In case it is only the Earth that moves I am the observer (blindfolded) and in this case I am together with the twin who with his spaceship has remained stationary to float in space. “If the astronaut twin stays stationary in space and if only the Earth (for some strange reason) starts moving”, you can imagine that the Earth reaches a point P in space (where there is a planet for example that I will call BETA).

Since the Earth is moving in the frame of the stationary twin, the Planet Beta-Earth distance is a fixed distance for me and for a stationary twin. For the Earth, however, this distance is contracted (due to Lorentz transformations) and that is why it will be younger than me.

This case (only the Earth is moving in the frame of one of the two stationary twins stood still floating in space with me) is impossible (at least it is very unlikely), so the solution becomes:

the twin who traveled with his spaceship in the Earth’s frame will be younger than his brother!

For example, when atmospheric muons decay they are younger than Earth and therefore travel a greater distance. The Earth does not move (only the Earth!) at the speed of muons while I am stationary with them. If that happened, the Earth would be younger than me and the muons. When in the experiments we notice that muons move relative to us, we can immediately discard the absurd hypothesis we have talked about, atmospheric muons are therefore younger than us! As we have seen on other occasions, even in the frame of muons the Earth is no younger than them!

As it was examined in THE TRANSFORMATIONS OF LORENTZ AB OMNI NAEVO VINDICATAE and in THE TWIN PARADOX (TRAVEL EXAMPLE) in the frame of the spaceship the astronaut twin is not older than those on Earth, there is no time difference.

In the frame of the Earth, atmospheric muons travel a distance d and do not come back (one-way trip at constant speed).
In the frame of the Earth the travel time is t=d/v, in the frame of the muons the travel times are both (for the Earth and for muons) t_1=(1/γ)*(d/v) (t_1 < t), if we think that the time t_1=d/v applies to the Earth we would make a mistake.
Considering the “second” Lorentz transformation x=γ*(x_1+v*t_1) with x_1=0, x=γ*v*(d/v), x = γ*d. (x > d, error!)
x > d means that the atmosferic muons in the Earth’s frame travel a greater distance but we know that x = d.
If we consider the outward (or return) journey of the spaceship, the reasoning is the same!

Another observer who is in another frame will measure something else and I don’t care because we are the ones who measure, the problem is indeterminate but we (in our frame) have solved it. (who has traveled is younger)

The frame of the Earth is not the privileged frame of reference, the objects that move in its frame are younger than the Earth itself.

We can therefore state that the two conditions x=2*d and x_1=0 mean that a twin leaves with his spaceship (one of the two) and the other stays on Earth with me (while the brother arrives at a distance d from the Earth, and then returns, d is a distance in the frame of the Earth).
(Blindfolded, I too am on Earth along with the twin who has not left).
Upon returning, I will take off the blindfold and who have traveled will be younger than his brother and than me.(it can’t be otherwise)
I don’t care about the measurements of other observers who are in other frames of reference. (I am on Earth!)

The Lorentz transformations are represented by a system of two equations in four unknowns, there are four Lorentz transformations, but two derive from the others.(we need to know some initial conditions)

I considered the case (one of the twins travels or the other travels).
If you wanted to, you might think that if they were traveling in two different spaceships at the same speed they would both be younger than me, or if they started at different speeds, the fastest would be the youngest. If neither of them left, the time spent for them would have been the same as for me.
In the examples I have now reported, however, there would still be mathematically to be discussed and I do not know if it makes sense because in truth one of them has to move with respect to me!

The results change according to the frame in which we make the measurements and this is so because the problem is indeterminate.
The problem is indeterminate, yes, but not impossible as we solve it brilliantly in the frame of the Earth! (the frame that interests us)

In the frame of the spaceship, the astronaut twin is not younger than the twin left on Earth, and the result obtained t_1 < t is valid even if the spaceship accelerates or decelerates on its journey, for an infinitesimal time dt in frame of the Earth it’s necessary to consider dt_1 = γ^(-1) (t) dt and then integrate.

dt_1 is less than dt, t_1 will also be less than t. (dt_1 < dt , t_1 < t)

Remember that the inverse of the Lorentz factor is less than 1, γ^(-1) (t) < 1!

It is also true, however, that if a spaceship accelerates in a reference frame, the objects inside it are subjected to apparent accelerations and therefore there is certainly a link with what I have said. But if we consider the motion of a point-like object of negligible mass?

Sometimes some other solution of “the twin paradox” is more difficult and not entirely convincing, to understand the one I have proposed you just need to know the Lorentz Transformations and solve a simple first degree equation.

Massimiliano Dell’Aguzzo

Condividiamo insieme la conoscenza!

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