Re: The problem of relativistic synchronisation

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.

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Slowly please.
Our readers need to understand correctly and be able to judge correctly.
We have determined, by simple synchronization procedure AA'
which we have called event e1, that tA(e1)=0, and that tA'(e1)=0.
We have then very easily concluded that tA(e2)=0.75.
So far, so good.
We won't come back to this.
Now, we are moving forward, we are looking, at the time of e2,
what time will be displayed on clock A' when B' crosses A.
It already seems that there is a difficulty here.
You say with certainty that tA'(e2)= ? ? ? , and I say with the same certainty that tA'(e2)=2.25.
One of us is necessarily wrong.
We must understand why.
R.H.