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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.physics.relativity
Subject: Re: Sync two clocks
Date: Fri, 23 Aug 2024 11:55:31 +0300
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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