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Path: ...!news.mixmin.net!weretis.net!feeder8.news.weretis.net!usenet.goja.nl.eu.org!pasdenom.info!from-devjntp Message-ID: <PPUqkCq0zWGMReLKBS2EOpI0shY@jntp> JNTP-Route: nemoweb.net JNTP-DataType: Article Subject: Re: The problem of relativistic synchronisation References: <m_uze6jFLkrMPuR4XaNmQntFPLY@jntp> <siZVeXFhx1b-RHNvgyKaFJEz2Sc@jntp> <vb28vm$1i5d6$2@dont-email.me> <2VJMHmUL3oTjzHTxkbHeeVgwp1A@jntp> <vb2tvf$1ls0b$1@dont-email.me> <vb4sas$2u11j$1@dont-email.me> <FwN11HvPTqkgQgrbwnddFC1OY98@jntp> <vb79g4$3cout$1@dont-email.me> <FUkaOI4ar_TcsN7KN74WFA8f3Yw@jntp> <vbactr$3v15v$1@dont-email.me> Newsgroups: sci.physics.relativity JNTP-HashClient: QX7svg0OG9bqGh9xAsdwDeRcO1c JNTP-ThreadID: 76aXiNOmFXLuhuEL0ulE9Z2aOsc JNTP-Uri: https://www.nemoweb.net/?DataID=PPUqkCq0zWGMReLKBS2EOpI0shY@jntp User-Agent: Nemo/1.0 JNTP-OriginServer: nemoweb.net Date: Wed, 04 Sep 24 23:02:24 +0000 Organization: Nemoweb JNTP-Browser: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/128.0.0.0 Safari/537.36 Injection-Info: nemoweb.net; posting-host="e8cbf2474b472b9bb79db3dccb6a856bc1d05409"; logging-data="2024-09-04T23:02:24Z/9010945"; posting-account="4@nemoweb.net"; mail-complaints-to="julien.arlandis@gmail.com" JNTP-ProtocolVersion: 0.21.1 JNTP-Server: PhpNemoServer/0.94.5 MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-JNTP-JsonNewsGateway: 0.96 From: Richard Hachel <r.hachel@wanadou.fr> Bytes: 2679 Lines: 33 Le 04/09/2024 à 21:37, "Paul.B.Andersen" a écrit : > Den 04.09.2024 02:32, skrev Richard Hachel: >> This is unavoidable and it is mathematical. >> Let tA'(e2)=tA'(e1)+(A'B'/Vapp') >> Let tA'(e2)= 0 + 3.10^8/(4/9)c >> tA'(2)=2.25 sec > > Nonsense. But no! I simply use a synchronization system based not on M but on the watches themselves, and, moreover, I involve universal anisochrony. Once well understood, my logic is quite simple, but the brain has been so clouded by the notion of universal and absolute present time for centuries, that it is difficult to convince those who read me. > At e2, tB' = d/v = 1.25 s, > and clock A' will simultaneously in K' show tA' = 1.25 s > > Understand this: > A and B are always simultaneously showing the same in K > A' and B' are always simultaneously showing the same in K' > > So how can you imagine that at event e2, when tB' = 1.25 s. > then clock A' should simultaneously show 2.25 s ? You need to add the anisochronous delay. There is one second between A and B in R, and one second between A' and B' in R'. R.H.