THE TWIN PARADOX SOLVED BY A TERRESTRIAL OBSERVER

The indeterminate problem

Consider two points A and B in relative motion with respect to each other at speed v, Special Relativity shows us this “paradoxical solution”:

(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! (it does not have a single solution for all observers, but I am interested in a solution for terrestrial observers)

Thinking about it, in the classic case of the “twin paradox” (as we all know it but above all as I understand it) we can solve this paradoxical drawback.

The solution of a terrestrial observer

I am on Earth with two twins, one of the two leaves with his spaceship at speed v in the positive direction of the abscissas x and in the initial position x=0.

Let us consider the measure of the Earth’s time with t.

For one of the two with his spaceship its trajectory is described in time by:

x =v*t (uniform rectilinear motion in Earth’s frame),

we assume that the speed v is constant on the outward and return journeys

It is important to point out that x = v*t does not hold for the return journey, but is x = d -v*t. (However, the outward journey is symmetrical to the return journey)The other twin is waiting with me for his return.

The position of the astronaut twin in his frame is x_1=0 (he is always in the origin)

In fact, we know that “the best known Lorentz transformation” is: (all four are important…)

x_1=γ*(x-v*t) and confirm all:

x_1 = 0, γ *(x-v * t) = 0 but γ is different from 0, x -v*t = 0,

x = v*t.

The other Lorentz transformation symmetrical to the previous one is:

x = γ*(x_1+v*t_1)

but x_1=0 and therefore x=γ*v* t_1.

t_1 is the measure of time with respect to the astronaut twin while he is traveling.

We do not consider the other two Lorentz transformations because they derive from the first two, however it is not difficult to verify that,

for x_1 = 0, we obtain t =γ*t_1.

Let’s suppose that the round trip takes place at a constant speed (otherwise it is necessary to consider integrals that I learned about reading Sir Roger Penrose but which are found almost everywhere)

If the astronaut twin travels a distance 2*d (d on the outward and d on the return) the two previous equations become 2*d = v*t and 2*d =γ*v*t_1.

The distance v*t_1 represents the distance “traveled” from the Earth with respect to the spaceship’s frame and v*t_1 is less than 2*d (obviously it is also t_1 less than t , they are obtained in fact t = 2*d/v and t _1 = (1/γ)*(2*d/v).

If the astronaut twin’s journey takes place at high speed and for a few years, the effect of time dilation will be very evident.

Upon his return, The Astronaut Twin is younger than me and his brother. (t_1 < t)

The Lorentz transformations represent a system of two equations in four unknowns and to get the solution(a quatern) the initial conditions (2*d=v*t, x_1=0) helped us.

Conclusions

The starting problem is indeterminate, but we know that one of the twins stays on Earth with me and the other travels in his spaceship (we can solve the system of equations). Upon returning, the twin who traveled is younger than me and his brother (it can’t be otherwise). The astronaut twin can’t be older in Earth’s frame than me and his brother, let’s stop thinking about it!

Let’s examine the situation carefully

In the frame of the spaceship the distances move towards the spaceship itself at opposite speed, it is not only the Earth that moves. (but the entire Earth’s frame)

I clarify, in the frame of the Earth the spaceship moves from a starting point A to an end point B and in the spaceship’s frame it’s the distance AB that moves with opposite speed towards the spaceship.

If the astronaut twin arrives in the vicinity of the planet ALPHA which is at a distance d from the Earth (in Earth’s frame), the Earth-planet ALPHA distance will move in the frame of the spaceship, distances in motion are contracted (as imposed by the Lorentz Transformations) and therefore have a length less than their length measured in the Earth’s frame.

Let us now consider a new frame of reference which I call the contracted frame of the Earth (the frame of the Earth where distances are contracted). Even if the astronaut is stationary in his frame, he’s moving relative to this new frame.

Consider the two motions:

motion A: the contracted distances moving towards the spaceship,
motion B: the spaceship moving in the contracted frame of the Earth.

The two motions are evidently symmetrical to each other as in classical physics, (same distance traveled, same speed and same travel time), the astronaut travels a shorter distance (both on the outward journey and on the return journey) and the time for him is necessarily less than that measured on Earth. (t_1 <t)
In the frame of the spaceship, the Earth also travels the same contracted distance (v*t_1), in the spaceship’s frame there is no time difference between the Earth and the spaceship itself.
For “the second Lorentz transformation” x = γ*v*t_1 and that’s right because, come to think of it, v *t_1 = x/γ! (and v*t_1 < x)

In this case, the same time t_1 is measured for both the Earth and the spaceship, in the spaceship’s frame the Earth is no younger than the spaceship.

The two initial conditions (2*d = v*t and x_1=0) allowed us to resolve the problematic “TWIN PARADOX” in the frame of the Earth, another observer of a different frame will get different result, but I’m only interested in mine! The Lorentz Transformations in the frame of the Earth are “ab omni naevo vindicatae”. (avenged from every stain)

For those interested, in another article THE TWIN PARADOX (TRAVEL EXAMPLE) is examined an example of a journey by the astronaut twin where the speed is not constant. (avoiding however the resolution of integrals)

Massimiliano Dell’Aguzzo