Re: Sync two clocks

Liste des GroupesRevenir à physics 
Sujet : Re: Sync two clocks
De : r.hachel (at) *nospam* jesauspu.fr (Richard Hachel)
Groupes : sci.physics.relativity
Date : 16. Aug 2024, 14:24:33
Autres entêtes
Organisation : Nemoweb
Message-ID : <vPP1Z1BJfE1Dt7SYhCzEo7ZQWFI@jntp>
References : 1 2
User-Agent : Nemo/0.999a
Le 16/08/2024 à 12:47, Python a écrit :
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
 t_A is the time shown by clock A when a light signal is emitted;
 t_B is the time shown by clock B when the signal is received and re-emitted;
 t'_A is the time shown by clock A when the returned signal is received.
Here's the clown continuing.
If we look closely, what he says seems sensible.
This is why all of humanity is wrong about the theory of special relativity as taught.
Because the more we rub our eyes, the more sensible it seems.
Except that... breathe, blow...
That clock A notes a time t1 when the signal leaves A (we don't care about the value by the way,
it could be t1=0 or it could be t1=153), I'm willing.
We place B at 3.10^8m, and time is measured in seconds.
That clock A notes a time t2 when the signal returns, I'm willing again, and as Jean-Pierre
Messager says (sometimes he says intelligent things, although it's rare), I'm going to say that t2=2 or that t2=155.
How will Jean-Pierre achieve this prophetic feat?
Let's not get carried away, I know how to do it too.
Here's how I do it, breathe, blow.
t2 = t1 + 2AB/c
And this is so true that it applies to the entire universe, and all inertial frames of reference.
But unlike Jean-Pierre, Henri or Albert, I'll stop here.
Because this is where the vast ocean of relativistic science begins.
What time is it in B when B receives the information? I don't know at all,
and first of all, depending on how I synchronized B, it could be t=4532 or t=-12.
So I don't know at all.
Jean-Pierre is intelligent enough to understand that it is therefore necessary to first synchronize B with A,
to have something coherent, because saying that tA=0 tB=4532 and tA'=2 is always feasible, we are not lying, but it is very unhelpful.
Except that Jean-Pierre still has not understood Hachel's thinking, and he remains in the hypothesis of a flat present (the horizontal plane of the present time), as others remain in the hypothesis of the flat earth.
Nature is not made like that, that's not how it works.
So what time is it in B?
Jean-Pierre does not bother with embellishments: "We only have to artificially set tB=(t2-t1)/2 and thus, everything will be very simple and very practical".
Except that it is an artificial synchronization.
And except that it will not be true for A, nor for B.
It will only be true for M, a point placed at an equal distance from A and B, and the synchronization will be called M synchronization.
Because in the universe of A, this M synchronization is completely false, everything that is part of the "3D present time" of M is not part, and we are infinitely far from it, of the present time of A, and ditto for B.
Each chosen point, A, B, or M have the same 3D inertial frame, but they are not part of the same 4D frame, and each can only have its own (because of anisochrony, and the fourth component t).
The synchronization of Einstein, Poincaré, physicists, is therefore only an abstract synchronization,
which represents a point M, placed very far perpendicularly, in an imaginary fourth dimension,
and which apprehends all the points of the 3D universe at the same distance and at the same present moment of M.
It is obviously totally imaginary, but it is very useful.
For this point, indeed, we can say that tB=(t2-t1)/2 but it is a convention M.
For A as for B, it is absolutely impossible to synchronize these two watches between them FOR them.
As it is also impossible to synchronize A or B with the imaginary point M.
Always, always, always, there will remain a universal anisochrony.
And always, always, always, in the reality of things, if we have practiced a synchronization M:
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
I don't know if it will take Lengruche four years to understand that (a+b)(a-b)=a²-b²
but it is certain that in 30 years Ybmuche will still not have understood what I have just detailed here.
R.H.

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