Re: The problem of relativistic synchronisation

Am Donnerstag000005, 05.09.2024 um 14:25 schrieb Paul.B.Andersen:

Den 05.09.2024 01:23, skrev Richard Hachel:

>
Vo=0.8c
>
Vapp=Vo/(1+cosµ.Vo/c)
>
Vapp'=(4/9)c
>
Vapp"=4c
>
R.H.

 How confused is it possible to be? :-D
 You must know that this 'apparent speed' is a visual
observation (telescope).
  From whence did you get the idiotic idea that somebody
is visually observing any of the clocks in this paper?
 https://paulba.no/pdf/Mutual_time_dilation.pdf
 A  and B  are moving with the speed v = 0.8c in K'  <-
A' and B' are moving with the speed v = 0.8c in K   ->
 Nothing is moving with any other speed than v.
There are no 'apparent speeds'.
 Is this too hard for you to understand?

Any velocity is between an object and a point of reference.
Usually we have an observer (say 'A') at a certain spot (also called 'A'), who measures distances from his own position.
The measured object (say 'B') moves - say- away at a certain speed v.
But seen from B the point A moves away with the same speed, though into the opposite direction.
The relevant coordinate system can now be attatched to A or B, depending on where the observer is placed.
So: the coordinate system K is placed, that its center coincides with 'A', while the certer of K' coincides with B.
(the primed version of A and B make no sense, hence could be left away).
Now we have two systems K and K' which both receed from an imaginary point in the center (called 'M', for simpility), which is assumed to be not moving.
This would mean, that A would receed from M by v=0.8c and from B with 1,6 c, hence drops out of the realm of visiblity, because the image of A gets redshifted below the value 0 Hz, if seen from the remote side (here B).
But, nevertheless, both (A and B) could remain well and good, because who cares about distant observers, which you cannot see?
...
TH