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Subject: Re: Langevin's paradox again
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Date: Mon, 15 Jul 24 18:17:46 +0000
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From: Richard Hachel <r.hachel@wanadou.fr>
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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.