Post by Ricardo JimenezPost by ***@gmail.comSeparating clocks that have been together synchronized
will create more distance for light to have to move through
creating a synchronization delay.
If, and only if, the clocks are considered to be in motion. (This is the
first statement that I read from you here that is remotely correct and not
just word salad. Congratulations.)
Post by Ricardo JimenezI'll try to explain again the difficulty I am having with the
synchronization procudure in the first 1905 Einstein relativity paper.
Einstein does NOT describe any “synchronization procedure” in that paper.
He describes what *would be* the case *if* two clocks *are* *synchronous*.
Have you even read it, or only read out-of-context claims about it at
Instagram University? *facepalm*
Post by Ricardo JimenezWhat is implied by the two equations he gives is that to synchronize
all clocks with the one at A, you send out a time signal to all of
them and they just need to set their clocks, at the instant they
receive that signal, to the recieived value plus the transport time
(distance/c). A doesn't do anything to his clock. Then when clock B
sends a time signal to A, at the instant A receives it, the value on
A's clock will be the same value as the received signal from B's
clock.
No, or only *in* the received signal, provided that B adjusts their clock
and submission so that it so. Typically, it will be a *different* value as
time has elapsed in the meantime (and a clock would show that), but
(Einstein assumes that) the *delay* will be the same:
t
^
: : :
t_A' + - : - - - .-
: :`. :
: : `. :
: : `. :
: : `.:
t_B + - :- - - - +
: : .':
: : .' :
: : .' :
: :.' :
t_A + - : - - - -:-
. . .
. . .
+---+--------+--> x
0 : :
:<--d--->:
: :
A B
And so Einstein argues (on page 148 of the official translation of his
papers) that *if*
t_B − t_A = t_A' − t_B,
*then* we must *conclude* that the two clocks A and B are synchronous.
[Notice that, contrary to your misconception earlier,
none of the clocks *need* to be in any origin. AISB.]
Post by Ricardo JimenezThus all clocks in space are synchronized to A's. However, B
still sees time signals from A equal to B's clock time minus
distance(A,B)/c.
Of course.
Post by Ricardo JimenezHowever, Einstein claims that clock synchronization is reflexive,
symmetric and transitive.
He does not claim that, he assumes it. And it actually (and trivially) is
so.
And regarding your “however”, which indicates your fallacy: Obviously, clock
synchronization does NOT mean that each clock receives the other’s time in
the signal at the time of arrival of the signal, but the other’s time when
that signal was submitted. Synchronization does NOT mean that the time *in*
the signal must be the same as the time when the signal is received. Given
that the signal is in this case a light signal, and travels at the speed of
light which is *finite*, such an understanding would be absurd; once the
clocks *are* synchronized, the signal MUST then contain a time value of the
past (of the receiving clock).
Synchronization means instead that both clocks would *show* the same time
*where they are* (which one could confirm if one would be able to travel
from one to the other instantly). Common sense suffices.
Or, if you (O) would be located exactly in the middle between the clocks
(and co-moving with them), you would *see* them showing the same time t_C –
a time in their (and your) *past*, though, because it takes light (the same)
time to travel to you (from each of them):
t
^
: : : :
+ - :- - - - + - - - -:- - present (moment of observation)
: : .':'. :
: : .' : '. :
: : .' : '. : past (of the moment of observation)
: :.' : '.:
t_C + - +- - - - : - - - -+
. . . .
. . . .
+---+--------+--------+---> x
0 : : :
:<--d/2->:<--d/2->:
: : :
A O B
Post by Ricardo JimenezDid he mean that you can redo clock synchronization with B as the
master clock so everything is synchronized with it?
No. Einstein does not even use the technical terms that you used (WTF are
your sources?), but explains quite clearly (in the official translation):
,-<https://einsteinpapers.press.princeton.edu/vol2-trans/157>
|
| […]
|
| We assume that it is possible for this definition of synchronism to be
| free of contradictions, and to be so for arbitrarily many points, and
| that the following relations are therefore generally valid:
|
| 1. If the clock in B is synchronous with the clock in A, then the clock in
| A is synchronous with the clock in B.
|
| 2. If the clock in A is synchronous with the clock in B as well as with
| the clock in C, then the clocks in B and C are also synchronous
| relative to each other.
That is so *trivial* and – again – “common sense” that usually it would not
even need to be mentioned. However, ISTM that he wanted to be very clear
about what one may assume and what not, as in the next section he shows that
certain assumptions that are thought to be obvious do not hold; namely that
those clocks would also be synchronous for someone who considers them to be
in motion.
PointedEars
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)