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:

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.

 

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.

 

 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

This is perfectly correct, and I am happy that we say the same thing, even if I give other values, because I speak in H (Hachel) synchronization and you in M ​​(Einstein) synchronization.
With your method of synchronization, that is to say (I beg you to understand what I am saying, because it is very important), you place yourself at the level of an observer M as I have defined it.
I remind you that each observer, even stationary, has his own hypercone of present time, and that only an imaginary observer placed OUTSIDE the 3D universe can synchronize the universe in his "absolute" fashion.
Everything then depends on the present moment considered FOR WHOM.
In Einstein synchronization, everything you have just said is true.
In HAchel synchronization (that is to say by putting oneself in the place not of M, but of A, or B, or A' or B', everything I say is also true.
My notation is however MORE true, because it describes exactly the time on the watch at the moment when, for it, the event occurs (live).
It is not a fantasy, on my part. If I do it, attracting contempt, hatred, misunderstandings, it is because it is necessary for the song.
tA(e1)=0
tA'(e1)=0
tA(e2)=0.75
tA'(e2)=2.25
You, you note in synchronization M.
I, I note in synchronization H, according to what the watch SEES for itself, on itself when the event occurs.
You, you note in synchronization M, according to what M perceives of the event (abstract absolute synchronization).
tM(e1)=0
tM(e2)=0.75
tM'(e1)=0
tM'(e2)=1.25
Note :
tM(e1)=tM'(e1).sqrt(1-Vo²/c²)
tM(e2)=tM'(e2).sqrt(1-Vo²/c²)
This is classic chronotropic dilation.
R.H.