Re: Sync two clocks

Le 23/08/2024 à 08:32, Thomas Heger a écrit :

Am Donnerstag000022, 22.08.2024 um 13:11 schrieb Python:
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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.
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Nobody can have another opinion of what time YOU read of YOUR watch.
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How is it possible to fail to understand this?
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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,
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then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
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  tB − tA = t'A − tB = td
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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.

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You introduced t_d or 'transit time' (aka 'delay'), while Einstein didn't use any of these terms.

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But he write down two equations that implies directly that a delay
is taken into account.

  The equation on page 3 COULD be interpeted as calculation of the delay.

There is NO other way. This is algebra dude...

But Einstein wrote, that would be the definition of the speed of light.

Nope. Again something you've made up.

I would miss the word 'delay' in this context (or something similar).
 Also the remainder of this paper does not mention delay neither.
 So: where have YOU found any use of delay or transit time in this paper?

Equations there implies t'_A = t_B - (AB)/c, (AB)/c is the light
propagation delay between A and B or B and A.

I 'combed' the text very carefully and could not find any statement, which eventually would match this discription.

Because you are very stupid.
But now that you've been shown where the delay is, you should retract
your "critics". Not doing so is called dishonesty.