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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.physics.relativity
Subject: Re: Sync two clocks
Date: Sat, 24 Aug 2024 11:02:01 +0300
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On 2024-08-24 06:50:34 +0000, Thomas Heger said:

> Am Freitag000023, 23.08.2024 um 10:51 schrieb Mikko:
> ...
>>>>> 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.
> 
> But this has nothing to do with synchronicity, but with a process to 
> turn remote clocks to the same time value.

That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.

> Time is not equal to what clocks say, because clocks are measuring 
> devices, which measure time, but do not determine, what the measured 
> quantity is.

The intent is to adjust the clocks so that the value shown is the time,
or at least to find out the difference so that the time can be inferred
from the shown value.

> The process to synchronize clocks require technical means, which are 
> here light signals:
> 
> The clock at some point A emmits a timing signal, which a remote clock 
> receives a little time later, because such signals have finite speed.

And also a light signal to the opposite direction.

> Now it should be obvious, that the remote clock had to compensate this 
> delay, because otherwise it would not show the time of the master 
> clock, but an asynchronous value.

And in order to do that, the delay needs be known. Therefore the requirement
to adjust so that the apparent delay is the same in both directions.

> For uncertain reasons Einstein had not mentioned this requirement at 
> all, even if transit time per se was actually mentionend.

For obvious reason, Einstein required what he required.

-- 
Mikko