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Subject: Re: Relativistic synchronisation method
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From: Python <jpierre.messager@gmail.com>
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Le 17/12/2024 à 18:19, Richard Hachel  a écrit :
> Le 17/12/2024 à 17:42, Python a écrit :
>> 
>> Not at all. There is NO ambiguity. tA and t'A are measured by clock A (for 
>> events that happen there), tB is measured by clock B (for an event that happens 
>> there too).
>> 
>> And, obvisously, the fact that "clock X measure tX for an event happening there" 
>> is the same fact for everyone. Only fools of your kind can pretend that 1 + 1 = 3 
>> for some observers.
> 
> Will you stop being an idiot?
> 
> Le crétin, c'est toi.

You've, so far, proven the opposite. You are a idiot.

> Il y a des choses pour lesquelles il n'y a pas d'ambiguïtés ; et d'autres pour 
> lesquelles il peut y avoir des ambiguïtés. 
> 
> Tu dis que tA et tA' sont mesurés par l'horloge A, tu as raison.
> 
> Tu dis que tA'-tA=2AB/c et tu as raison.

I didn't say that. It can be the case or not. The POINT is that if it is 
the case, it is the case. Everyone agrees on that. And that if is not the 
case then everyone agrees on that.

Same for tB - tA = t'A - tB it is either true or false for everyone.

> Tu dis que tB est mesuré par l'horloge B (plus précisément tB(e2) en notation 
> Hachel), et tu as raison. 
> 
> Tu dis qu'il n'y a là aucune ambiguïté. 
> 
> Mais comme toujours tu oublies mon immense génie, et ta folie furieuse. 
> 
> 1. Je n'ai pas dit le contraire.
> 
> 2. Tu travestis ma pensée en me faisant dire ce que je n'ai pas dit. Je n'ai 
> jamais dit que ce que marquait une montre lors d'un événement conjoint était 
> relatif par changement d'observateur. 

It is your words : "but by omitting to say WHO takes the measurement of 
that".

There is no admission : an observer at A measures time shown on clock A 
for two events, an observer at B measures time shown on another clock at 
another event.

Clocks can show any values, nothing is supposed than they tick at the same 
rate.

I wrote an application to show this that can be understood by 7 years old 
children :

https://noedge.net/e/


>  Cela reviendrait à dire que si le train entre en gare de Marseille à 12h00 
> sur l'horloge de la gare,
> un observateur très lointain observe bien le train entrer en gare, mais voit 
> qu'il est 16h45 sur l'horloge de la gare.
> 
>  C'est absurde.

You missed the point that the procedure is not supposing that the clock 
are synchronized in any way, as my application illustrates. Then if the 
checking procedure involving ONLY tA, tB, t'A concludes that clocks A and 
B are NOT synchronized you can compute the offset to apply, at your 
choice, to A, to B or to both.

>  [snip remaining demented bullshit]