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Path: ...!weretis.net!feeder8.news.weretis.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: The problem of relativistic synchronisation Date: Wed, 4 Sep 2024 21:38:53 +0200 Organization: A noiseless patient Spider Lines: 120 Message-ID: <vbactr$3v15v$1@dont-email.me> References: <m_uze6jFLkrMPuR4XaNmQntFPLY@jntp> <NVcS6uZDkN8UGhkdIwwkCs4R7x8@jntp> <vb1mk4$1g551$1@dont-email.me> <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> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Wed, 04 Sep 2024 21:37:31 +0200 (CEST) Injection-Info: dont-email.me; posting-host="e34c553dc35c84cba304f9ef8a55d998"; logging-data="4162751"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19ymMg0APZAMjbv3R+NfbdU" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:uK8IkjyC+HRd0gU8xaY/Yn2quVg= Content-Language: en-GB In-Reply-To: <FUkaOI4ar_TcsN7KN74WFA8f3Yw@jntp> Bytes: 5191 Den 04.09.2024 02:32, skrev Richard Hachel: > Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit : >> Den 03.09.2024 12:03, skrev Richard Hachel: >>> >>> We have: >>> >>> tA(e1) = 0 >>> tA'(e1)= 0 >>> tA(e2) = 0.75 s >> >> OK. > >>> And what's missing: >>> tA'(e2) = 2.25 >>> >> >> This is nonsense, and demonstrates that you >> don't know what an event is. >> e2 is the event that clock A and clock B' are adjacent >> tA'(e2) is meaningless. > > It is obvious that tA'(e2) has a meaning for A'. > > It is the time at which in his frame of reference (A'), the event E2 > exists. > > Paul, Paul, you can't say it's meaningless. A little more consideration > for the other posters, and please a little more practical intelligence: > there is indeed a moment, where, for A, the event e2 exists in his frame > of reference, and if A' was synchronized at the start, there will be a > time, and only one time of e2 that will be written on his watch. > > We can easily, if we correctly master the notion of relativistic > simultaneity and the notions of relative chronotropies, > reveal what this time written on the clock A' thus synchronized during > e1 will be: tA'(e1)=0. e2 is short for "the event that clock A and clock B' are adjacent" At this event, tA = (d/v)⋅√(1−v²/c²) = 0.75 s and clock B will simultaneously in K show tB = 0.75 s At this event, tB' = d/v = 1.25 s, and clock A' will simultaneously in K' show tA' = 1.25 s > > I beg you not to say that it is absurd or meaningless. > > How to proceed? > > We KNOW that the travel time of A in A'B' At event e1 tA = 0, at event e2 tA = (d/v)⋅√(1−v²/c²) = 0.75 s so the travel time for A to go from e1 to e2 is = 0.75 s > will be equal to the distance > A'B' in R' divided by the apparent escape velocity of an object moving > at v=0.8c. And what will you escape from? A and B are moving with the speed v = 0.8c in K' <- A' and B' are moving with the speed v = 0.8c in K -> Nothing is moving with any other speed than v. There are no 'apparent speeds'. > 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. 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 ? > > Please note, I am talking about the exact time when A' > actually perceives e2 in direct-live, and which represents > the real time of things. Please note, I am talking about the exact time when Paul actually perceives that Richard snap his fingers in direct-live, and which represents the real time of things. > > > If we want to judge, and count based on synchronization > abstract, based on "the transverse speed of light", it is necessary > remove a second of anisochrony between A' and B'. Which gives a time > supposed (but false) of 1.25 sec. Well said! I have always admired your ability to clearly explain things. Of course we have to remove a second when we are talking about "the transverse speed of light". > > Please confirm that you have understood and that you validate, which > will allow us to go further and explain all the predictive values that > I have already given, but without explaining yet. I have understood that we have to remove a second based on "the transverse speed of light". Based on "the longitudinal speed of light" we would have to add a second, obviously. -- Paul https://paulba.no/