Re: Langevin's paradox again

Le 15/07/2024 à 14:57, "Paul.B.Andersen" a écrit :

Den 14.07.2024 23:49, skrev Richard Hachel:

The scenario is:
 Terrence is inertial.
Stella passes Terrence with the speed 0.8c relative to Terrence.
At the instant when Stella is adjacent to Terrence they both set
their clocks to zero, and Stella starts her rocket engine so that
she accelerates at the constant acceleration c per year (≈ 0.97g)
towards Terrence.
Some time later, Stella will again pass Terrence at the speed 0.8c.
 The only question I want answered is:
What do Stella's clock and Terrence's clock show
at the instant when Stella passes Terence the second time?
 It's two invariant proper times, so they are "absolute".

 Well.
 I hadn't understood the question correctly, and I thought it was the question: Stella goes to 0.8c, and immediately Terrence sends Bella (a=1al/an²) to join her.
The question was when will Bella join Stella, and where?
The answer is quite simple, since we have x=Vo.To and x=(c²/a)[sqrt(1+To²a²/c²)-1]
Let for x=x, two root possibilities To=0 and To=40/9 years
And x=0 (the start) and x=32/9 ly.
It remains to be seen what the clean times will be for Stella and Bella.
I'm responding hastily, and I hope without any miscalculation.
Tr(Stella)=sqrt(To²-Et²)=sqrt(To²-x²/c²)=24/9 years
Other mode Tr(Stella)=To.sqrt(1-Vo²/c²)=24/9 years
Tr(Bella)=x/Vr=x.sqrt(1-Vo²/c²)/Vo=(32/9)*0.6/0.8=24/9 years
We notice that here, the proper times are equal.
But that's not what you're asking, your question is: "As Stella passes Vo=0.8c, it accelerates towards the earth (a=1 ly/y²)"
I ask you for a few moments, and I will answer you.
R.H.