Re: Sync two clocks

Am Freitag000023, 23.08.2024 um 10:51 schrieb Mikko:
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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.

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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.

 Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that
describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.

But this has nothing to do with synchronicity, but with a process to turn remote clocks to the same time value.
Time is not equal to what clocks say, because clocks are measuring devices, which measure time, but do not determine, what the measured quantity is.
The process to synchronize clocks require technical means, which are here light signals:
The clock at some point A emmits a timing signal, which a remote clock receives a little time later, because such signals have finite speed.
Now it should be obvious, that the remote clock had to compensate this delay, because otherwise it would not show the time of the master clock, but an asynchronous value.
For uncertain reasons Einstein had not mentioned this requirement at all, even if transit time per se was actually mentionend.
But the necessary step was missing, that the remote station had to add the transit time to the received timing value.
No such such statement can be found in Einstein's paper, hence we are forced to beleive, that he didn't wanted to compensate the delay.
TH