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From: Python <python@invalid.org>
Newsgroups: sci.physics.relativity
Subject: Re: Sync two clocks
Date: Thu, 22 Aug 2024 13:11:20 +0200
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Le 22/08/2024 à 09:02, Thomas Heger a écrit :
> 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.

But he write down two equations that implies directly that a delay
is taken into account.

> Therefore, you have read something, that should be there, but wasn't.

Paul has a functioning brain. You haven't.

> In fact I have spent a lot of time to verify, that 'delay' or anything 
> equaivalent was actually missing in Einstein's 1905 paper.

You'd spend a more valuable time trying to understand the meaning of
equations stated in part I.1.

> Now you have invented in your own mind something, what should be there 
> (but wasn't).

Then you would have discovered that it actually is there.

> To verify my statement yourself, you need to go carefully through the 
> paper and identify the statement, where you think, that Einstein had 
> delay (or anything equivalent) in mind.

Equations stated in part I.1. imply t'_A = t_B - (AB)/c

(AB)/c the exact delay you were looking for: distance between A and
B divided by celerity of light.

> But I was unsuccesful in this realm, because Einstein simply forgot delay.

He didn't. You missed it because you didn't understand a word of this
part.

Remember Thomas: it took you *years* to get that A and B are mutually at
rest! As an hypothetical teacher, if you were a student, I would sent
you back to kindergarten.

> That's why you can search as long as you like for 't_d' or 'delay' or 
> 'transit time', because they are not present.
> 
> Also no equation or any other statement can possibly be interpreted as 
> calculation of transit time.

They can :

Equations stated in part I.1. imply t'_A = t_B - (AB)/c

(AB)/c the exact delay you were looking for: distance between A and
B divided by celerity of light.

> It's simply not there!

It is there.