Path: ...!2.eu.feeder.erje.net!feeder.erje.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: "Paul.B.Andersen" <relativity@paulba.no>
Newsgroups: sci.physics.relativity
Subject: Re: Langevin's paradox again
Date: Tue, 16 Jul 2024 20:33:33 +0200
Organization: A noiseless patient Spider
Lines: 112
Message-ID: <v76e49$1cn1q$1@dont-email.me>
References: <FER4K03RCuXsBiIlfVNSgR0vilQ@jntp> <v710l9$7u9s$1@dont-email.me>
 <h2wfltG60Qx3kQd2ONdS3W4MJKs@jntp> <v71c86$a0te$1@dont-email.me>
 <fCidXYmBeggYm31OO1-5v64V0S8@jntp> <v736bn$mo0c$1@dont-email.me>
 <yX3K1wOobcIp1KzCuRuRL0Htcvg@jntp> <v73otr$ptku$1@dont-email.me>
 <cGLopKEv1S9qR2nNCvaF8IqKf-s@jntp> <v75m18$1828i$2@dont-email.me>
 <esHcGgnWa-kSH-yFGrczTDSFLzc@jntp>
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Date: Tue, 16 Jul 2024 20:28:25 +0200 (CEST)
Injection-Info: dont-email.me; posting-host="8154274a8b4f9e673e1f6a21fcfb5fc6";
	logging-data="1465402"; mail-complaints-to="abuse@eternal-september.org";	posting-account="U2FsdGVkX19nCI5y57t5X5aRTr9LLiub"
User-Agent: Mozilla Thunderbird
Cancel-Lock: sha1:ngS0a9T+WLtr8Sgr6HUaeMzuXHk=
In-Reply-To: <esHcGgnWa-kSH-yFGrczTDSFLzc@jntp>
Content-Language: en-GB
Bytes: 5497


Den 16.07.2024 15:25, skrev Richard Hachel:
 > Le 16/07/2024 à 13:37, "Paul.B.Andersen" a écrit :
 >>
 >> 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?
 >>
 >> "she accelerates at the constant acceleration c per year (≈ 0.97g)
 >>  towards Terrence."
 >> 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.
 >>
 >
 > I speak English very poorly, and sometimes I may misunderstand a 
question.

Use Google translator!

 > 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)

This is the same result as SR gives.
Terrence's proper time is τₜ = 8/3 years ≈ 2.6667 years
Well done!

 >
 > 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.

I think the right word is "claim", not "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.

So your theory predicts that Stella and Terrence ages equally.

According to SR Stella's proper time is τₛ ≈ 2.19722 years.

So "the travelling twin" ages less than the "stay at home twin".

The ageing of the twins in the "twin paradox" is experimentally verified 
to be as predicted by SR.

Your theory is falsified.

-- 
Paul

https://paulba.no/