Re: Sync two clocks

Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:

Den 20.08.2024 17:12, skrev Richard Hachel:

Le 20/08/2024 à 15:39, Python a écrit :
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Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.

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You are lying.
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I do not claim it "now". This is what I have always said for at least 40 years.
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Now, yes, obviously I assume it.
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The value (tA'-tA) = 2AB/c is the same not only for A and B, but also for all the stationary points of the inertial frame of reference of A and B.
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Better, if I change frame of reference it will remain true, by invariance of the transverse speed of light in any frame of reference.
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On the other hand the value tB-tA (go) will vary for most observers in R (where A and B are stationary), as will the value tA'-tB (return).
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But you cannot understand this, because 1. You are stupid and because 2. because you are tied up with relativistic thoughts all learned, but false.
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R.H.

 Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
 Nobody can have another opinion of what time YOU read of YOUR watch.

This is not, what 'invariant' means in the context of relativity.
Meant is, that time would not change, if you switch from one frame of reference to another.
But 'own time' (as 'proper time') is actually the only time you would have, because you cannot move in respect to yourself.
But other observers could and usually do.
Now the other observers movement would cause relative movement and that had an impect on what the other observer would regard as your time.
Therefore your time isn't invariant, if observed from somewhere else.

How is it possible to fail to understand this?
 If we have two stationary clocks in an inertial frame,
  and clock A shows tA = t1 when it emits light,
  and clock B shows tB = t1 + td when the light hits it,
  and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
 then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
   tB − tA = t'A − tB = td

If time is a local phenomenon, you cannot assume, that perceived delay (or 'transit time') would be independent of movement.
Let's take you as 'A' and some other observer 'B' in inertial motion moving away with c/2.
Then the remote observer would see your timing signals shifted down in frequency and would regard your clock as going slow.
Seen from your perspective the same would happen and you would see B's clocks run slow.
Therefore, 'time-stretching' is an apparent effect, caused by relative motion.
To avoid this effect, you would need to compensate delay altogether by measurements and by adding the measured delay to the received time in the timing signal.

The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
 

Whether this is actually true or not is irrelevant here, because you need to compensate the delay somehow. Otherwise you would get a 'mutual time stretching' effect, what cannot be a real physical effect, because it is visible only at the far side by the remote observer.
Since both of these observers are of equal rights, both could claim the other time reading invalid, hence both readings ARE invalid.
Therefore you must compensate the delay 'by hand'.
...
TH