Re: Sync two clocks

Le 26/08/2024 à 08:13, Thomas Heger a écrit :

Am Sonntag000025, 25.08.2024 um 09:24 schrieb Python:

Le 24/08/2024 à 08:50, Thomas Heger a écrit :
...

But the necessary step was missing, that the remote station had to add the transit time to the received timing value.
>
No such such statement can be found in Einstein's paper, hence we are forced to beleive, that he didn't wanted to compensate the delay.

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THis is silly. You are not "forced" to believe such a idiotic thing.
>
Basic algebra shows that what you call "delay" is embedded in the
very definition of synchronization. This is something that you
are not "forced" to "believe", but forced to acknowledge.

 Where have you found the delay-corrections 'embedded' in Einstein's paper???

Einstein's method is actually based on the idea to take the perceived time from a hypothical clock at a point in the middle between A and B.

There is ZERO mention of a "middle point between A and B" in Einstein's
article.

This is why Einstein had no time-measure, which is valid throughout an entire coordinate system, but regarded time as dependent on the position of the clocks.

Definitely NOT. This is M.D. Lengrand/Hachel's mantra that you are
describing here.
This is quite amusing: as a crank you accuse Einstein to have in mind
what another crank, Hachel proposes and crank Hachel is accusing
Einstein to have not chosen what he proposed.

But 'position' means 'point within that particular coordinate system'. So Einstein excluded uniform time within a certain coordinate system.

Quite the opposite actually.

This is a strong indication, that he didn't plan to compensate delay.

This is baloney, something you made up so it cannot be a indication of
anything about Einstein's plan.

If you think otherwise, than you should quote the statement, which I had apparently overlooked.

You did, and you have no excuses as me and others have pointed out where
the delay appears numerous time.
Again :
In paragraph I.1 in Einstein's 1905 article you can read :
(*)   2AB/(t'_A - t_A) = c
(**)  t_B - t_A = t'_A - t_B
In a setup with two mutually  at rest clocks at position A and B in
a given system of reference. --> and <-- represents a light signal
emission/reception, all time values are recorded by both clocks at
time or receptions/re-emission:
Step 1:
A-->               B
t_A
Step 2:
A               -->B
                  t_B
A               <--B
Step 3:
A<--               B
t'_A
So if you read the only equations in paragraph I.1,
assuming clocks are synchronized (which is the point of
this paragraph: state what it MEANS to be synchronized):
(*)   2AB/(t'_A - t_A) = c
(**)  t_B - t_A = t'_A - t_B
t_B - t_A is a delay (between emission at A and reception at B)
t_'A - t_B is a delay (between emission at B and reception at A)
t'A - t_A is a delay (round trip time delay for a light signal
going from A to B bounced back to A)
 From (*) you can get : t'_A - t_A = 2AB/c so another way to
describe the same delay : twice the distance AB divided by c.
So there such a delay is present in paragraph I.1. THREE times
as a term in an equation and ONCE as a term you can obtain in
ONE step of basic algebra.
It is difficult to believe you've "overlooked" this and continue
to do so for YEARS.
At first it could have been a symptom of your complete inability
to understand a single sentence of the article (i.e. sheer stupidity),
since you've published your idiotic comments and got some clues from
numerous people here it is definitely a symptom of your dishonesty
Thomas.