Re: Sync two clocks

On Mon, 19 Aug 2024 23:15:28 +0000, Richard Hachel wrote:

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Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :

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Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
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We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
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In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time
is the same from B to A. (This follows from Einstein's definition
of simultaneity).
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At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
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At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
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In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
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Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
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When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
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A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
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Einstein:
  "The two clocks synchronise if  tB − tA = t'A − tB."
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Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
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That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So  tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
  δ = (n-m) + x seconds.
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After this correction, we have:
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  tB  − tA = (m - n) seconds +  δ     = x seconds
  t'A − tB = (n + 2x - m) seconds - δ = x seconds
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The clocks are now synchronised.
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Please explain what in the above you find impossible
to do in your lab.

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I have explained these things a hundred times.

And your explanation has been refuted millions of times :-))

It is impossible to synchronize two watches A and B located
in different places.

“There is no point in using the word 'impossible' to describe
something that has clearly happened.” – Douglas Adams

It is not a technical problem, it is not a problem of
intelligence, it is simply that it is as impossible as

finding a round square, or a natural number between 5

and 6.  Because nature is not made like that.

“There is no point in using the word 'impossible' to describe
something that has clearly happened.” – Douglas Adams

Anyone who wants to synchronize watches with each other does
not realize that he is looking for rabbit horns.

https://allthatsinteresting.com/jackalope

The only thing we can do is to find a kind of abstract,
imaginary synchronization procedure, called the Einstein
procedure where tAB is supposed to be equal to tBA

How can tAB not be equal to tBA for light is vacuum?

not only for the point M which is the only real origin of
the synchronization, but for everyone.

If tAB = tBA, then Paul's procedure works.  If tAB <> tBA,
then your point M method won't work either.

We then have a useful procedure, but false.

I set my watch to my wall clock months ago and it's still
in sync.  You're spouting baloney, Richard.

[Rest of misinformed baloney deleted]