Re: Sync two clocks

On 2024-08-23 05:41:50 +0000, Thomas Heger said:

Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:

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.

 This is not, what 'invariant' means in the context of relativity.

Yes, it is.

Meant is, that time would not change, if you switch from one frame of reference to another.

No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is invariant.
--
Mikko