Re: Relativistic synchronisation method

Le 01/01/2025 à 21:20, "Paul.B.Andersen" a écrit :

  Den 01.01.2025 18:55, skrev Richard Hachel:

Le 01/01/2025 à 12:35, "Paul.B.Andersen" a écrit :>

Den 31.12.2024 11:58, skrev Richard Hachel:

Le 31/12/2024 à 11:13, "Paul.B.Andersen" a écrit :

>
At home you set your clock to  UTC+1h.
You know the station clock shows UTC+1h.
You expect the clocks will be synchronous within a second
when you arrive at the station.
>

 

If the watches are well tuned, it is logical that when I find myself in the presence of the station clock, my watch will note the same time.
The opposite would also be absurd, since by definition they must be tuned.

 

No, what I say is not ridiculous at all.

 "Being well tuned" is obviously the same as "being synchronous",
so what you say is OK.
 But why do you use a different word than what is common in physics?
 

It is you who do not understand what I have been saying for several years now.

 I understand that you have called synchronised clocks for "well tuned"
for several years.
 

 I was already saying the same thing forty years ago, and I have never changed an inch.

 Why haven't you during those 40 years discovered that what you call
"well tuned" is what physicists call  "synchronised"?

Yes and no.
Words can sometimes take on different meanings.
What do you call the fact that two well-designed clocks (the one in the cupboard and the one on the mantelpiece) tick at the same speed?
The best term I can find, because it is important to be precise, even excellent, is: "they have the same chronotropy".
I can also say: "they are synchronous".
At the limit, they do not even mark the same time. But they are synchronous.
Now let's imagine two clocks that cross each other at very high speed (observable speed Vo = 0.8c). At the moment of their crossing,
they must set their clock to noon, no matter how they do it. Are they synchronized, yes, by definition, and they will always remain so because they have had a very clear and very well-defined synchronization process.
But will they be synchronous? No, even before they cross, at the moment they cross, and after they cross, they will not be synchronous. In the sense that neither their beat nor their chronotropy will be.
Their external beat will play like t'=t.(1+cosµVo/c)/sqrt(1-Vo²/c²).
Their internal chronotropy like To"=To/sqrt(1-Vo²/c²)
They will be synchronized, on a place at a time, but never synchronous.
R.H.