# MY FINAL THOUGHTS ON THE TWIN PARADOX

I repeat the resolution strategy that I showed in my previous article ** “THE LORENTZ’S TRANSFORMATIONS AND THE TEMPORAL EXPANSION”** (sorry for having corrected it several times, but now there is its final version), trying to make it clearer.

*This is the situation to be clarified*

Let us now consider the symmetrical situation: all points of the Earth’s frame ** S** moving together at the same speed

**opposite to that of the spaceship, covering the same distance**

*v***traveled by the spaceship in the Earth’s frame. (to then go back).**

*d*The set of points of the frame** S** (the set of all points of the Universe together with the Earth, the twin remained on the Earth itself and the planet) moves at the same speed within the frame of the astronaut. (

**, an empty copy of the fixed stars frame with the spaceship inside it at the origin**

*S_1***)**

*O_1**Solution (analyzing only the outward journey)*

As for the travel time of the terrestrial twin, it’s ** t_a = (1/γ)*(d/v) **(now he’s in motion in the spaceship’s frame). For the terrestrial twin it is the distance

**that comes towards him and is therefore contracted.**

*d**(t_a is the travel time of the terrestrial twin)*

As for the astronaut twin, he sees the points in motion and that is why the Earth-planet distance does not measure ** d** but

**for the astronaut the elapsed time is equal to**

*d/γ,***the Earth-planet distance**

*t_b = (d/γ)/v = (1/γ)*(d/v) = t_a.*(**in motion contracts**

*d***)**

*(t_b is the travel time of the astronaut twin)*

*t_b = t_a;**this is why, if we consider the astronaut twin to be stationary, his brother who remained on Earth isn’t younger than him.*

*THERE IS NO TEMPORAL DIFFERENCE BETWEEN THE TWO TWINS (if we consider the astronaut twin stationary)*

*The travel time of the terrestrial twin is t_a = (1/γ)*(d/v) (the same time for the planet to reach the spaceship, in the frame S_1). For the astronaut twin, the time t_ c = d/v is not important, because the planet arrived earlier. (when a time equal to t_b has passed)*

It certainly is ** t_c = (d/v) > t_a = (1/γ)*(d/v)** (the same values assumed if we consider the spaceship in motion with respect to the Earth’s frame) and, if you think about it, it is not a coincidence.

*In my other articles I also wrote that, if the Earth (only our planet) started moving for some strange reason and then returned to where the stationary twin remained waiting (floating in space), then the twin remained on Earth (at the end of this strange journey) would be younger than his brother.*As I also wrote in my article ** “THE TWINS ALEX AND BRAD” **(regardless of what each twin thinks of the other), each of the two has his age.

*(time flows differently for the twins!)**Conclusions*

*To conclude, if the spaceship travels a distance **d** (to reach a planet for example), the elapsed time for the twin remaining on Earth is **d/v** and the elapsed time for the astronaut twin is less.*

*If the whole Universe (except the spaceship) travels a distance **d **(with opposite speed to that of the spaceship itself), the elapsed time for the twin remaining on Hearth is **(1/γ)*(d/v)** and the elapsed time for the twin astronaut is the same. (is being considered the journey in the frame of the spaceship, when the astronaut twin sees the planet coming)*

The two motions are not symmetrical because, if the whole Universe (except the spaceship) travels a distance ** d **(with respect to the spaceship’s frame), the spaceship travels a distance shorter than

**with respect to the Earth’s frame**

*d*

*S.**In the Earth’s frame, the astronaut twin must travel the distance d, otherwise he does not reach the planet!*

*Let’s avoid contradictions!*

The frame ** S_1** is an empty copy of the fixed stars frame (there is only the spaceship in the origin

**).**

*O_1*(The frame ** S_1** is therefore an empty copy of the Earth’s frame

**with the spaceship in the origin**

*S***)**

*O_1*The terrestrial twin and the astronaut twin therefore have *“the same ruler” *when measuring lengths.

We consider a length (firm with respect to the frame ** S**) that has a length

**in**

*d***;**

*S***the same length also has measure**

**in the frame**

*d***(if it’s not moving)**

*S_1.*We initially assumed that the Earth-planet distance was fixed and equal to ** d **in Earth’s frame,

**if the set of points in the Universe (together with the Earth and the planet) travels a distance**

**(where**

*d***is the same measure with respect to**

*d***and**

*S***),**

*S_1*

*it can be t = t_1 in the Earth’s frame!**v*t_1 = d/γ (that’s fine, the Earth-planet distance goes towards the astronaut twin and is contracted in the frame of the spaceship)*

*t_1 = (1/γ)*(d/v)** is the travel time in the frame of the spaceship.*

*v*t = d/γ (not acceptable, the Earth-planet distance in the Earth’s frame is always equal to d)*

*The travel time in the Earth’s frame is different, **t = d/v.*

*In the frame S the spaceship reaches the planet in a different (obviously greater) instant in time.*

*Each of the twins has their travel time, it is wrong for each of the two to care about the other.*

*What else could I add to convince you? I hope now everything is clearer, remembering what happens when the spaceship moves in the Earth’s frame S.*

*As I said, the problem is that the spaceship in this situation (in Earth’s frame) travels a distance less than the Earth-Planet distance. (Impossible! The spaceship is moving in the Earth’s frame)*

*I repeat, the planet to be reached is at a distance **d** with respect to the Earth’s frame, at the end of the journey (also considering the return) the astronaut twin is necessarily younger and it can’t be otherwise. **(t_1 < t **and** “THE TWIN PARADOX”, AS WE KNOW IT, DOES NOT EXIST)*

*If the astronaut twin is to arrive at a point that is at a distance **d** from the Earth (in the Earth’s frame **S**), this is the only solution. (even if there is nothing in that point)*

*As the well known twin paradox is formulated, it must be v*t = d!*

*Brief summary*

If we consider the spaceship in motion, it is ** t > t_1** and the spaceship travels a distance

**in the Earth’s frame.**

*d*If we consider the points of the frame

**to move with speed opposite to that of the spaceship, it is**

*S***in the frame of the spaceship.**

*t = t_1*Before the points of the frame ** S** departed, the distance between the astronaut and the planet is

**(the same distance between the Earth and the planet in the Earth’s frame, this is the distance that the spaceship must travel)**

*d.*Only if we consider the spaceship moving (only the spaceship) at the same speed towards the planet, the astronaut twin travel the right distance ** d.** (measured before departure), if we consider that all points of the Earth’s frame move with speed opposite to that of the spaceship it’s not the same thing.

*In this case the astronaut twin travels a shorter distance, but the planet is at a distance d in the frame S.**If t = t_1, both distances are equal to d/γ. (v*t = v*t_1)But we know that one of the two is d (the Earth-planet distance in the frame of the Earth), and this happens only if we consider the spaceship to move in the Earth’s frame. This is the exact interpretation. As I have shown in my previous articles, in this case both the spaceship and the Earth (in the spaceship’s frame) travel the same contracted distance d/γ.*

If it must be ** x = v*t = d**, then

*t = d/v.**(*

*t**can’t be a shorter time)*

Thus *x_1 = γ*(x -v*t) = γ*(d -v*t) = 0.*

Since we know that ** x = γ*(x_1 + v*t_1),** we obtain that

*d = γ*v*t_1.**Finally we derive **t_1 < t. (and it can’t be otherwise)*

*t_1 is equal to t only in the spaceship’s frame, because the moving distance between the Earth and the planet contracts. In the frame of the spaceship, the twins are certainly the same age but, at the end of the journey (when the twins meet again on Earth),**the astronaut twin will be younger! (if the destination is a planet or other that is at a distance d from the Earth in the frame of the Earth)*

*In the spaceship’s frame, the distance d in motion is perceived to be shorter and the distance traveled by the spaceship in the Earth’s frame will also be less than d. (the two distances are the same!)*

*In the frame of the spaceship, when the planet comes close to the spaceship, for the terrestrial twin the same time has passed (in the Earth’s frame). But, in the Earth’s frame, the spaceship did not reach the planet!*

…………………………………………………………………………………

*Instead,** the case is very different if the astronaut twin does not have a precise destination **d** in Earth’s frame **S** (now there is no planet to reach), it could be that the Earth moves away from the spaceship by a fixed distance **d_1** **(x_1 = -d_1,** the abscissa of the distance is negative with respect to the spaceship’s frame **S_1**).*

*The problem is therefore, as already said on other occasions, indeterminate.** It isn’t the fault of Special Relativity, **to solve a system of two equations in four unknowns we need to initially know at least two unknowns. **(No mathematician has ever been desperate for this!)*

*Dear readers, thank you for your attention, hoping to have shown mathematically everything I mentioned initially* *in my first article, **“THE TRANSFORMATIONS OF LORENTZ AB OMNI NAEVO VINDICATAE”. **I wish I was always clear. (I understand that the topic is not simple)*

*Massimiliano Dell’Aguzzo*