Re: ? ? ?

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Sujet : Re: ? ? ?
De : python (at) *nospam* invalid.org (Python)
Groupes : sci.physics.relativity
Date : 26. Apr 2024, 11:27:34
Autres entêtes
Organisation : CCCP
Message-ID : <v0fvin$3l14j$1@dont-email.me>
References : 1 2 3 4 5
User-Agent : Mozilla Thunderbird
Le 26/04/2024 à 09:11, Thomas Heger a écrit :
...
In the context at stake here, i.e. part I.1 of A.E. paper,
clocks at A and B are in mutual rest.
  Actually Einstein didn't say so, but didn't mention relative velocity neither.
He said so quite clearly:
"Let us take a system of co-ordinates ..." => a SINGLE system
"If a material point is at rest relatively to this system of co-
ordinates ..." all "points" mentioned later are at relatively
to this system, this obviously implies that they are mutually
at rest.

So, let's assume, that A and B are points in space and mutually at rest towards each other.
  So A and B are points at rest in respect to each other. Because point a is assumed to be at rest, too, the point B is also at rest.
 Now 'motion' or 'velocity' do not make sense, because everything mentionend is at rest in a stationary system.
 This is actually ok.
Good to hear. Why did it take you YEARS to understand this Thomas ?

But why then didn't Einstein calculate the delay of the light signals for the transit from A to B  and back?
 It should technically easy to send a signal from A to B, get it reflected there and measure the dealy, cut that in half and add this one-way delay to the time value imbedded in the time-coded signal, which A receives from B.
 This would eliminate the influence of the speed of light and would allow mutally equal synchronization between clocks at A and B.
 But this was not, what Einstein had done.
 Instead he had the strange idea, that the time value seen on the rmeote clock would be the time at the remote location.
Absolutely NOT, there is nothing of that kind in Einstein paper !
from t_B - t_A = t'_A - t_B
and (2AB)/(t'_A - t_A) = c
[i.e. t'A - t_A = (2AB)/c ! What is (2AB)/c if not - obviously -
such a delay (twice the delay actually) you stupidly complain that
it wouldn't have been taken into account ???]
 From these two simple equations you can deduce immediately :
t'_A = t_B + (AB)/c
i.e. time shown on clock A when receiving a signal sent by clock B when
clock B was showing t_B is t'_A = t_B + (AB)/c
(AB)/c is the delay : the time taken by light to travel from B to A.
If you cannot spot this at first read it means that your are not a
member of the expected audience of this article i.e. non-morons.
Which makes your attempt to "evaluate" it as a teacher quite pathetic.

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