Re: Sync two clocks

Den 16.08.2024 14:24, skrev Richard Hachel:

Le 16/08/2024 à 12:47, Python a écrit :

If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
>
t_A is the time shown by clock A when a light signal is emitted;
>
t_B is the time shown by clock B when the signal is received and re-emitted;
>
t'_A is the time shown by clock A when the returned signal is received.

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.

Here's the clown continuing.
 If we look closely, what he says seems sensible.
 This is why all of humanity is wrong about the theory of special relativity as taught.
 Because the more we rub our eyes, the more sensible it seems.
 Except that... breathe, blow...
 That clock A notes a time t1 when the signal leaves A (we don't care about the value by the way,
it could be t1=0 or it could be t1=153), I'm willing.

.. or n seconds.

 We place B at 3.10^8m, and time is measured in seconds.
 That clock A notes a time t2 when the signal returns, I'm willing again,

or tA' = n + 2x  (where x = 1 second in your case)

and as Jean-Pierre
Messager says (sometimes he says intelligent things, although it's rare), I'm going to say that t2=2 or that t2=155.
 How will Jean-Pierre achieve this prophetic feat?

All Jean-Pierre has to do is to note that tB = m seconds
when he see the flash.

 Let's not get carried away, I know how to do it too.
 Here's how I do it, breathe, blow.
 t2 = t1 + 2AB/c

No. The clocks are not synchronous yet. tB = m

 And this is so true that it applies to the entire universe, and all inertial frames of reference.
 But unlike Jean-Pierre, Henri or Albert, I'll stop here.
 Because this is where the vast ocean of relativistic science begins.
 What time is it in B when B receives the information? I don't know at all,
and first of all, depending on how I synchronized B, it could be t=4532 or t=-12.

Quite right. Hachel can't know what clock B will show.

 So I don't know at all.

But Jean-Pierre knows that tB = m seconds

 Jean-Pierre is intelligent enough to understand that it is therefore necessary to first synchronize B with A,
to have something coherent, because saying that tA=0 tB=4532 and tA'=2 is always feasible, we are not lying, but it is very unhelpful.

But since everything is happening in Hachel's lab,
Hachel will see that the computer at A read
tA = n and tA' = n + 2 seconds, and he can tell this to
the intelligent Jean-Pierre, who will understand that to make
his clock synchronous with the clock at A, he will have to
adjust his clock with  δ = (n-m) + 1 seconds
We will then have:
tB  − tA = t'A − tB = 1 second
The clocks are synchronised.

 Except that Jean-Pierre still has not understood Hachel's thinking, and he remains in the hypothesis of a flat present (the horizontal plane of the present time), as others remain in the hypothesis of the flat earth.
 Nature is not made like that, that's not how it works.
 So what time is it in B?
 Jean-Pierre does not bother with embellishments: "We only have to artificially set tB=(t2-t1)/2 and thus, everything will be very simple and very practical".

This is nonsense.
According to your definition above: t2 = t1 + 2AB/c
(t2-t1)/2 is the transit time AB/c (= 1 second)
Setting tB = AB/c would be meaningless.
You probably meant tB = (t2+t1)/2 which is the criterion
for synchronism.
But you can't set a clock which isn't synchronous to this.
Since it is necessary that the observations made at A are sent
to B (or told by Hachel), this must necessary take some time,
so if the the clock B was set to tB = n + 1 (AB/c) second it would lag
clock A with at least 1 second (AB/c).
It has to be adjusted by  δ = (n-m) + 1 seconds.
So this is why you thought each clocks would be
AB/c later that the other! :-D

 Except that it is an artificial synchronization.

"Artificial" as opposed to "Natural"? :-D

 And except that it will not be true for A, nor for B.

What should this mean?
Is it true for me that your clock shows GMT + 2h?
Is it true for you that my clock shows GMT + 2h?

 It will only be true for M, a point placed at an equal distance from A and B, and the synchronization will be called M synchronization.

You are babbling again.
All clocks in the world showing UTC are synchronous in the ECI
frame BY DEFINITION.
This is true. Period.

 Because in the universe of A, this M synchronization is completely false, everything that is part of the "3D present time" of M is not part, and we are infinitely far from it, of the present time of A, and ditto for B.
 Each chosen point, A, B, or M have the same 3D inertial frame, but they are not part of the same 4D frame, and each can only have its own (because of anisochrony, and the fourth component t).
 The synchronization of Einstein, Poincaré, physicists, is therefore only an abstract synchronization,
which represents a point M, placed very far perpendicularly, in an imaginary fourth dimension,
and which apprehends all the points of the 3D universe at the same distance and at the same present moment of M.

This is meaningless babble.
With the clocks synchronised as above,
we can _measure_ that the transit time from A to B
is L/c and the transit time from B to A is L/c.
(where L is the distance between A and B)
This is _only_ possible if clock A and clock B are synchronous
according to Einstein's definition.
And to measure the speed of an aeroplane flying from Oslo
to Paris as (t_Paris - t_Oslo)/distance, the clocks in
Oslo And Paris have to be synchronous. Remember?
(You kept fleeing, Richard. Chicken!)

 It is obviously totally imaginary, but it is very useful.

Words like "imaginary" and "abstract" are rather meaningless
in this context.

 For this point, indeed, we can say that tB=(t2-t1)/2 but it is a convention M.

It is a man made _definition_.
Einstein's _definition_ of simultaneity is based on symmetry.
The transit time is the same in both direction.
It is possible to define simultaneity in other ways, but
it would be very inconvenient if the speed of light were
not isotropic.

 For A as for B, it is absolutely impossible to synchronize these two watches between them FOR them.
 As it is also impossible to synchronize A or B with the imaginary point M.
 Always, always, always, there will remain a universal anisochrony.
 And always, always, always, in the reality of things, if we have practiced a synchronization M:
 FOR A:
tB-t1=2AB/c
t2-tB=0
 FOR B:
tB-t1=0
t2-tB=2AB/c

What an idiot!
tB is a time measured by B.
t1 and t2 are measured by A.
They are proper times and can't be
different for A and B.
If the clocks are synchronous:
tB = (t1+t2)/2
t2 = t1 + 2AB/c
For both!

 I don't know if it will take Lengruche four years to understand that (a+b)(a-b)=a²-b²
but it is certain that in 30 years Ybmuche will still not have understood what I have just detailed here.

If Ybmuche is a sane person, you are probably right. :-D
--
Paul
https://paulba.no/