Re: Langevin's paradox again

Le 16/07/2024 à 13:37, "Paul.B.Andersen" a écrit :

 means that the direction of the acceleration (a vector) is
always toward Terrence, but since Stella at the first passing is
moving away from  Terrence at the speed 0.8c, her speed will
first be reduced (she's braking) and eventually reach zero,
and thereafter she will move towards Terrence at increasing speed.
It should not be hard to guess what the speed is when she passes Terrence the second time.
  Paul

I speak English very poorly, and sometimes I may misunderstand a question.
Thank you for the linguistic clarifications you have just provided.
As for the question, I've already answered it indirectly, but I'll do it more specifically.
We are therefore in the presence of a Stella which crosses the earth at a constant uniform speed of Vo=0.8c.
According to the criteria of Richard Verret and Richard Hachel, we therefore have Vr=(4/3)c.
At this precise moment, Stella transforms into a Bella, and sets up an acceleration system of approximately 10m/s², which we will assume to be precisely a=1 ly/y² acceleration towards the earth, which she wants to cross a second time.
In the Galilean reference frame of Stella (which has not yet transformed into Bella), we will have a distance for the earth which will gradually increase according to x=To.Vo. We will also have, for Stella 1 observing Stella 2 (now Bella), x=(c²/a).sqrt(1+a²To²/c).
When Bella (new Stella) crosses the earth, we will necessarily have x=x.
Let To.Vo=(c²/a).sqrt(1+a²To²/c).
This equation has two roots:
The first is To=0 and x=0.
This is the first crossing.
The second root gives: To=(40/9)ans.
Let x=32/9 al
This is the second observed crossing of the old Stella repository.
But this does not tell us the proper time of Bella (new Stella),
nor Terrence.
Note that so far, physicists agree with Doctor Hachel.
They will still be if we ask Terrence's own time between the two crossings.
Tr=To.sqrt(1-Vo²/c²)=24/9 years (or 8/3)
Where they will no longer be is when it is necessary to calculate Stella's own time (now Bella).
The fact that Bella accelerates from rest in Stella's frame of reference allows us to say:
“If the paths are equal, and the observable times equal, then the proper times will be equal”
Hence Tr=24/9 years for her too (or 8/3).
What physicists deny, but which I nevertheless confirm.
Note that if we have x and a, i.e. x=32/9 and a=1 we immediately have Tr for the Stella accelerated according to Tr=sqrt(2x/a) a very Newtonian formula, but which nevertheless applies here. Let Tr=sqrt[2*(32/9)/1]=24/9 years.
R.H.