Re: The problem of relativistic synchronisation

Den 04.09.2024 02:32, skrev Richard Hachel:

Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :

Den 03.09.2024 12:03, skrev Richard Hachel:

>
We have:
>
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

>
OK.

 

And what's missing:
tA'(e2) = 2.25
>

>
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
   tA'(e2) is meaningless.

 It is obvious that tA'(e2) has a meaning for A'.
 It is the time at which in his frame of reference (A'), the event E2 exists.
 Paul, Paul, you can't say it's meaningless. A little more consideration for the other posters, and please a little more practical intelligence: there is indeed a moment, where, for A, the event e2 exists in his frame of reference, and if A' was synchronized at the start, there will be a time, and only one time of e2 that will be written on his watch.
 We can easily, if we correctly master the notion of relativistic simultaneity and the notions of relative chronotropies,
reveal what this time written on the clock A' thus synchronized during e1 will be: tA'(e1)=0.

e2 is short for "the event that clock A and clock B' are adjacent"
At this event, tA  = (d/v)⋅√(1−v²/c²) = 0.75 s
and clock B will simultaneously in K show tB = 0.75 s
At this event,  tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s

 I beg you not to say that it is absurd or meaningless.
 How to proceed?
 We KNOW that the travel time of A in A'B'

At event e1 tA = 0, at event e2 tA = (d/v)⋅√(1−v²/c²) = 0.75 s
so the travel time for A to go from e1 to e2 is = 0.75 s

will be equal to the distance A'B' in R' divided by the apparent escape velocity of an object moving at v=0.8c.

And what will you escape from?
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'.

This is unavoidable and it is mathematical.
Let tA'(e2)=tA'(e1)+(A'B'/Vapp')
Let tA'(e2)= 0 + 3.10^8/(4/9)c
tA'(2)=2.25 sec

Nonsense.
At e2, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s
Understand this:
A and B are always simultaneously showing the same in K
A' and B' are always simultaneously showing the same in K'
So how can you imagine that at event e2, when tB' = 1.25 s.
then clock A' should simultaneously show 2.25 s ?

 Please note, I am talking about the exact time when A' actually perceives e2 in direct-live, and which represents the real time of things.

Please note, I am talking about the exact time when Paul
actually perceives that Richard snap his fingers in direct-live,
and which represents the real time of things.

  If we want to judge, and count based on synchronization abstract, based on "the transverse speed of light", it is necessary remove a second of anisochrony between A' and B'. Which gives a time supposed (but false) of 1.25 sec.

Well said!
I have always admired your ability to clearly explain things.
Of course we have to remove a second when we are talking
about "the transverse speed of light".

 Please confirm that you have understood and that you validate, which will allow us to go further and explain all the predictive values ​​that I have already given, but without explaining yet.

I have understood that we have to remove a second based on
"the transverse speed of light".
Based on "the longitudinal speed of light" we would have to
add a second, obviously.
--
Paul
https://paulba.no/