Relativistic synchronization

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Sujet : Relativistic synchronization
De : r.hachel (at) *nospam* liscati.fr.invalid (Richard Hachel)
Groupes : sci.physics.relativity
Date : 16. Sep 2024, 15:54:49
Autres entêtes
Organisation : Nemoweb
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I have repeated a thousand times that it is impossible to "absolutely" synchronize two distant watches
due to the reality of an unavoidable physical phenomenon: universal anisochony.
Let's imagine an observer who wants to synchronize his watch with another watch placed very far away.
What is synchronizing?
Is it making sure that two watches have the same chronotropy? Not at all. If we synchronize two watches that cross at high speed, we note t'=t=0, and, at that moment, they are synchronized by definition. At the same present moment, in the same place, they momentarily mark the same time. They are synchronized, although they do not have the same reciprocal internal chronotropy, since according to the words of physicists who are right on this, each one paradoxically turns faster than the other.
Good. Let's say that A wants to synchronize his watch with B, located 3.10^8m away.
This is entirely possible FOR A.
It is enough that, in the reality of things, at the present moment common to A and B, FOR A,
A considers that B is marking the same time as him.
If the two watches are of equal design (they were made in Switzerland), they have the same chronotropy, and will remain permanently tuned TO A.
But a difficulty will arise. What tells me that the notion of universal simultaneity is not dependent on the place that performs the measurement?
Here, he should not even need to answer, as the truth should be obvious to anyone who has really understood the theory, and not in a vague and indistinct way.
Is the present time hyperplane of A the same as the present time hyperplane of B?
According to Hachel, we can answer that no. This is the notion of universal anisochrony.
When we synchronize B with A, FOR A, the synchronization seems perfect, but B looks at A with a rather particular astonishment. For B, the two watches are this time out of tune by Δt=2AB/c.
There can therefore be no absolute synchronization of type A.
What is the present time hyperplane of A concerns only A.
Once this is understood (but what a birth must be done in the foggy minds of men!)
we can then ask the question: "But what happens in the present time hyperplane of A' when it crosses A at high speed".
The answer is dramatically simple: the two present time hyperplanes are absolutely confused.
Everything that one perceives, live, is absolutely perceived by the other in the same universal present moment. This is what the magnificent Poincaré-Lorentz transformations say if we understand them correctly.
I restate them here in positive form and Doctor Hachel notation.
x'=(x+Vo.To)/sqrt(1-Vo²/c²)
y'=y
z'=z
To'=(To+x.Vo/c²)/sqrt(1-Vo²/c²)
t'=t
If we notice that t'=t=0 this means that at the moment of the crossing, the two universes are perfectly simultaneous.
We notice however that the x component poses a particular problem. We therefore have two identical present time hyperplanes, but deformed in x.
This refers to the aberration of the position of the stars. The star that becomes a superova explodes at the moment for the two observers, since the hyperplanes of simultaneity are the same for the whole universe, but not in the same place. There is spatial deformation of the position of objects in x.
This being well understood, let us return to Einstein's synchronization, which in his immense boasting,
Richard Hachel calls a synchronization of type M (as opposed to a synchronization of type A).
In Hachel, this synchronization is valid only in the sense that we convince ourselves that it is an imaginary, abstract synchronization, placed on the hyperplane of simultaneity of a point M located at an equal distance, in an imaginary fourth dimension, from all the universal points located in the chosen inertial frame of reference.
If we change the stationary frame of reference, we change the point M'.
How to synchronize on M and obtain a sort of absolute synchronization impossible in practice?
M sends an imaginary flash. It goes without saying that, for M, located at an equal distance from all points of the stationary universe considered, and whatever the speed of the flash for the outward journey and the speed of the response for the return journey (we can use Jean-Pierre Messager's pea cannon), all impacts and all responses will take place simultaneously.
M can then ask the entire universe (stationary relative to it) to start its clocks upon receipt of any flash or pea cannon shot.
We will then have a perfect synchronization, although imaginary, and this will allow us to label
in a useful way all the events of this universe.
All the more so since, this universe being stationary, no notion of different chronotropy will intervene.
The system will remain infinitely stable and us
R.H.

Date Sujet#  Auteur
16 Sep 24 * Relativistic synchronization3Richard Hachel
16 Sep 24 `* Re: Relativistic synchronization2Paul.B.Andersen
16 Sep 24  `- Re: Relativistic synchronization1Richard Hachel

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