Re: The problem of relativistic synchronisation

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.