Re: The problem of relativistic synchronisation

W dniu 04.09.2024 o 21:38, Paul.B.Andersen pisze:

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:

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We have:
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tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

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OK.

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And what's missing:
tA'(e2) = 2.25
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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.

 

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It is obvious that tA'(e2) has a meaning for A'.
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It is the time at which in his frame of reference (A'), the event E2 exists.
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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.
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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

Fortunately, we have GPS now, so we can be
absolutely sure that your gedanken bullshit
has nothing in common with real clocks,
real simultaneity or real whatever.