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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.physics.relativity
Subject: Re: Sync two clocks
Date: Fri, 23 Aug 2024 11:51:42 +0300
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On 2024-08-22 07:02:47 +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.
>> 
>> 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.
> 
> You introduced t_d or 'transit time' (aka 'delay'), while Einstein 
> didn't use any of these terms.

Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that
describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.

-- 
Mikko