Re: Sync two clocks

On 2024-08-19 23:15:28 +0000, Richard Hachel said:

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

Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
 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.
 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).
 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.
 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.
 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.
 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.
 When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
 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.
 Einstein:
"The two clocks synchronise if  tB − tA = t'A − tB."
 Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
 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.
  After this correction, we have:
 tB  − tA = (m - n) seconds +  δ     = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
 The clocks are now synchronised.
 Please explain what in the above you find impossible
to do in your lab.

 I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in different places.

So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
--
Mikko