The system of equations represented by the Lorentz transformations
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 x =v*t 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 trip, but it is x=d-v*t.(However, the outward journey is symmetrical to the return journey)
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 , with 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 will be younger than me and his brother who stayed on Earth waiting for him.
It’s clear that the Lorentz transformations represent a system of two equations in four unknowns and to have a single solution (a quatern) certain initial conditions we must know.
The solution of the system of equations is well determined in the Earth’s frame. When I refer to the quatern I mean a single solution (x, t, x_1, t_1) in which all four values are determined. I remind those who have recently begun to take an interest in Special Relativity that there are four Transformations of Lorentz, but from x_1 = γ*(x -v*t) and from x = γ*(x_1+v*t_1) the other two are obtained.
If someone asks me to solve the equation y -2x = 0 I can’t determine a single pair (x; y). I have to make a choice and say that if x = 5 then y = 10. We do not get a solution that holds in every frame but we are sure that the astronaut twin will be younger in the Earth’s frame when he returns.
For example, we are interested in understanding why atmospheric muons are younger than Earth (and in fact they travel a great distance before decaying). For muons the journey is one-way but the reasoning is the same.
The motion of the Earth in the spaceship’s frame and the distances traveled
I consider the motion of the Earth in the spaceship’s frame “an apparent motion” but it is the real movement in the frame of the spaceship. In my analysis, I consider a spaceship departing from Earth and returning after reaching a planet.
In the frame of the spaceship, the distance “traveled” by the Earth is not the same traveled by the spaceship in the Earth’s frame because the distance contracts, it is smaller for the Lorentz Transformations.
The distances traveled are different and that’s why there is no symmetry
To know who is younger you need to see who travels the greatest distance. If we consider the muons that are generated in the atmosphere, if we want to know who is younger(the Earth or the muons), we can say that they are the muons that are younger(and not the Earth) because the distance traveled by muons in the frame of the Earth is greater. If someone thought that the Earth was younger than muons they would be wrong because in the frame of muons the distance traveled by the Earth is less than the previous distance.
I indicate with γ the Lorentz factor, with t_1 the mean lifetime of the muons and with v their speed:
distance traveled a = v*t_1*γ (the distance traveled by muons in the Earth’s frame, on Earth the measurement of time is dilated t = γ*t_1)
distance traveled b = v*t_1 (the distance traveled by the Earth in the frame of the muons, for the muons the measure of time is t_1)
γ > 1, a > b !
The Earth’s frame is not the privileged one, the Earth’s frame is the one where the distance traveled by objects moving in it is greater than the distance traveled by the Earth.
Sometimes, erroneously, considering Minkowski’s space-time graphs, the astronaut is considered to be stationary and an attempt is made to ideally trace the universe line of the Earth’s apparent motion with respect to the spaceship’s frame, in this case it is wrong do it because there is no time dilation in the spaceship’s frame. (in the spaceship’s frame or in the frame of the muons the times are the same, both t_1) The universe lines are for me the lines drawn in the Minkoski diagram.
The speed of a moving object is not always constant (in reality it starts from a starting point to arrive at an end point at zero speed), in this case it is necessary to determine the distances actually traveled and for more information I recommend reading my other article “FURTHER INFORMATION CONCERNING LORENTZ TRANSFORMATIONS”.
“Not only the Earth is in motion?”
What do i mean when in previous articles i say that not only the Earth is in motion? This means that everything else is in motion. (except the spaceship) The frame of the spaceship is the frame in which the Earth’s frame is in motion.
Even if space was empty and there were only Earth, the spaceship and the planet ALPHA (planet ALPHA is where the astronaut is going)
we can consider two reference frames:
the frame of the spaceship,
the frame of the Earth.
They are two frames in relative motion to each other. If we consider the spaceship’s frame stationary, it is the entire Earth’s frame that moves at a speed opposite to that of the spaceship . But every reference frame has its points and between two points it is always possible to define a distance between them. Let A be the first point corresponding to the Earth, let B be the second point corresponding to the planet ALPHA and let C be the point corresponding to the spaceship. If we consider the spaceship stationary, even if the Universe were empty, there is a distance between points A and B and this distance contracts if it is moving (due to Lorentz transformations). For the twin remained on Earth the distance between points A and B is fixed. (because the astronaut twin travels to a planet ALPHA that is at a given distance from the Earth) For the twin on the spaceship the distance between points A and B is contracted (it is in motion!).
If we consider the twin stationary on the spaceship:
it is not spaceship (point C)that goes towards B, but it is B that goes towards spaceship (point C). Initially, points A and C coincide. In the end, points B and C coincide.
How can it be that the twin left on Earth is younger?
If the astronaut twin remained stationary floating in space and if only the Earth started moving (for some strange reason), then the Earth would be “younger”!
In this case the Earth reaches another planet (I call it BETA), where BETA is at some distance from the spaceship and the planet BETA is also stationary like the astronaut twin.
The center of Beta and the center of the Earth are two points, the distance Earth-planet BETA is contracted only in the Earth’s frame. (The distance Earth-planet BETA is in motion in Earth’s frame and is therefore shorter as the Lorentz Transformations tell us)
The Earth-planet BETA distance does not contract in the spaceship’s frame,it is not a moving distance and therefore the twin who is on Earth is “younger” than his brother!
The “paradoxical” (but impossible to realize) situation would be that the Earth moves (only it and not all the other points of the Universe, as I said before) in the frame of the twin floating in space.
So yes that on the return the twin left on Earth will be younger than his brother and me, I too remained floating in space (I’m the observer and I don’t change my frame). If I were instead on 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 as I said earlier, the observer’s frame must not change!
I add nothing more to show that the twin paradox is false in the frame of the Earth.
For example, muons that are generated in the atmosphere are able to travel greater distances because they are “younger” than anything that remains stationary on Earth.
(assuming that “atmospheric” muons are born and decay at a constant speed)
The opposite is not true (whoever remains stationary on Earth is not “younger” than muons in motion, not even in the frame of the muons), Lorentz transformations are sufficient to verify this.
I recommend reading THE TRANSFORMATIONS OF LORENTZ AB OMNI NAEVO VINDICATAE and THE TWIN PARADOX (TRAVEL EXAMPLE).
For me it is like considering Ockam’s razor, it is enough for me to have proof that Lorentz transformations mathematically represent the reality of a flat space-time in our reference frame. (privileged because it is the one from which we observe and study all physical phenomena)
The Lorentz transformations, never denied as the postulate relating to the constancy of the speed of light has always been verified, are self-sufficient in a “flat” space-time and, as I now believe, I retain them “ab omni naevo vindicatae”. (avenged from every stain) For me, even if the astronaut twin accelerates (or decelerates) dt_1 < dt and, if we integrate, also t_1 < t. (dt_1 = dt only if the speed is zero)
I will change my mind only when the second postulate of relativity is contradicted and, believe me, I will be among the first to do so because I am only interested in the outcome of the experiments.
If we consider an object in motion, it moves from a starting point to an end point in a frame of reference, I can’t find other words to define the meaning of the movement.
Now you can better understand how I think about the “twin paradox”, I want to think just like Newton wrote: “hypotheses non fingo” or better they were only helpful in the Earth’s frame the two postulates of Special Relativity and two initial conditions. (2*d=v*t and x_1=0)