Path: ...!news.nobody.at!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: "Paul.B.Andersen" Newsgroups: sci.physics.relativity Subject: Re: Langevin's paradox again Date: Sun, 14 Jul 2024 01:34:44 +0200 Organization: A noiseless patient Spider Lines: 90 Message-ID: References: <9oTvw4-YSIPb1dubtdBwcc_MeX8@jntp> <4O9Y8U3gtfBKakbkPS0LmREorbI@jntp> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 14 Jul 2024 01:29:41 +0200 (CEST) Injection-Info: dont-email.me; posting-host="0e0e6ccf56bfdd84e3a861a26db2692b"; logging-data="3993159"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX185FBJiIU+HMgRE0NXVwgF+" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:ari+KNevIuMHImFUY+ZO/DsPYCQ= Content-Language: en-GB In-Reply-To: Bytes: 4333 Den 12.07.2024 15:55, skrev Richard Hachel: > Le 12/07/2024 à 15:24, "Paul.B.Andersen" a écrit : >> Den 11.07.2024 20:29, skrev Richard Hachel: >>> >>> < snip whining and heavy breathing > >>> >>> >>> I repeat, during the U-turn, nothing happens at all on the TIME side. >> >> Except that TIME is passing? > > Yes. > For Stella : tau(stella)=24 hours  tau(Terrence)=40 hours > > For Terrence : tau(Terrence)=40 hours  tau(Stella)=24 hours Why do you write French? But we have Google translator. > These values ​​are insignificant and useless to calculate to understand > correctly the Langevin paradox, which has no connection with the > acceleration phases, none. Langevin's example was a constant speed out, instant turnaround, and constant speed back. Instant turnaround means infinite acceleration, which is impossible, and will be a mathematical singularity. ---- Your scenario is trivially simple. Measured in Terrence's inertial rest frame Stella is moving at the constant speed v = 0.8c, γ = 1/0.6. Since Terrence's proper time for the whole journey is τₜ = 70 hours, Stella's proper time must be τₛ = τₜ/γ = 42 hours. If the rate of Terrence's clock is constant fₜ, the rate of Stella's clock is constant fₛ = fₜ/γ (Stella's clock runs slow). As long as Stella's speed is constant, the shape of Stella's path is irrelevant, it can be circular, elliptic, partly straight and partly curved, or whatever shape you might like. (But no sharp corners, he path must be an analytic function.) Of course Stella must accelerate at some part of the journey in order to get back to Terrence, but when the speed is constant the acceleration is transversal and will have no effect on her proper time. > > Everything is played out during the simple Galilean phases. All. Galilean phase? If there were a "Galilean phase" Terrence's and Stella's clocks would always show the same. > > You must simply use the formula > Tapp=Tr.(1+cosµ.Vo/c)/sqrt(1-Vo²/c²) each time, > and for all measurements. This is the relativistic Doppler shift equation. We can use Doppler shift to calculate how much Terrence and Stella will age. See: https://paulba.no/pdf/TwinsByDoppler.pdf > The concordance is mathematical. > We absolutely don't need anything else. It is true that we don't need anything but Doppler shift to calculate how much Terrence and Stella will age. But when Stella is accelerating, the method will be rather cumbersome. So it is better to do it like this: https://paulba.no/pdf/TwinsByMetric.pdf > The rest is pure nonsense from relativistic physicists, who, > understanding that pouic, say anything (time-gap, watches that > panic, and other joys). > Their ridiculous space-time à la Minkowski plunged them 120 years > into deep theoretical darkness. > R.H. Whining again, Richard? -- Paul https://paulba.no/