Re: Sync two clocks

Den 20.08.2024 17:12, skrev Richard Hachel:

Le 20/08/2024 à 15:39, Python a écrit :
 

Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.

 You are lying.
 I do not claim it "now". This is what I have always said for at least 40 years.
 Now, yes, obviously I assume it.
 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.
 Better, if I change frame of reference it will remain true, by invariance of the transverse speed of light in any frame of reference.
 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).
 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.
 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.
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
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.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
--
Paul
https://paulba.no/