Discussion:
Clock Synchonization
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Ricardo Jimenez
2021-10-11 17:09:25 UTC
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AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
Maciej Wozniak
2021-10-11 17:39:26 UTC
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Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
You're missing a simple fact: apart of their brilliant gedankens
your idiot gurus have never synchronized any clocks and they
have no idea how to do it for real.
carl eto
2021-10-11 20:35:37 UTC
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GPS (Global Positioning System) is used to justify the measurement of the velocity of light but the GPS is produced by the intensity differences of the satellites radio signals not by the velocity of the radio signals since the electrons of the GPS system that are propagating at the velocity of 10^6 m/s cannot measure the time difference of radio waves propagating at the velocity of light.
mitchr...@gmail.com
2021-10-11 22:30:14 UTC
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Bring clocks together to align them and then separate them and they will be synchronous
but there is a light travel time delay in effect in their observations of one another.
Where light arrives together at the same time is called simultaneity.

Mitchell Raemsch
Thomas 'PointedEars' Lahn
2021-10-17 00:24:55 UTC
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Post by ***@gmail.com
Bring clocks together to align them and then separate them and they will be synchronous
No, because in order to separate them, one of them has to be *moved*
relative to the other. Then less proper time τ elapses for one of the
clocks than for the other as the spatial displacement for the "moved" clock
is larger (sometimes popular-scientifically phrased as “the faster you move
through space, the slower you move through time”).

Lorentz transformation:

t' = γ (t − v/c² x) = (t − v/c² x)/√(1 − v²/c²)
v ≠ 0, x ≠ 0 ⇒ t' ≠ t. ∎

Minkowski metric (equivalentl):

ds² = c²dt² − dx² − dy² − dz².

dy = dz ≔ 0:

ds² = c²dt² − dx².

Clock at relative rest in a frame (dx = 0), measures proper time τ:

ds² = c²dτ².

⇒ ds² = c²dτ² = c²dt² − dx²
⇔ c² (dτ/dt)² = c² − (dx/dt)² = c² − v²
⇔ (dτ/dt)² = 1 − v²/c²
⇔ dτ/dt = √(1 − v²/c²)
⇔ dτ = dt √(1 − v²/c²) = dt/γ

v ≠ 0 ⇒ dτ < dt. ∎


PointedEars
--
Q: What did the nuclear physicist post on the laboratory door
when he went camping?
A: 'Gone fission'.
(from: WolframAlpha)
Maciej Wozniak
2021-10-17 09:33:08 UTC
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Post by Thomas 'PointedEars' Lahn
Post by ***@gmail.com
Bring clocks together to align them and then separate them and they will be synchronous
No, because in order to separate them, one of them has to be *moved*
relative to the other. Then less proper time τ elapses for one of the
clocks than for the other as the spatial displacement for the "moved" clock
is larger (sometimes popular-scientifically phrased as “the faster you move
through space, the slower you move through time”).
In the meantime in the real world, however, GPS clocks keep
measuring t'=t, just like serious clocks always did.
Thomas 'PointedEars' Lahn
2021-10-17 00:25:23 UTC
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Post by carl eto
GPS (Global Positioning System) is used to justify the measurement of the
velocity of light
Not at all.


PointedEars
--
Q: Why is electricity so dangerous?
A: It doesn't conduct itself.

(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-10-11 23:22:29 UTC
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Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space.
It is not required that any of the clocks to be synchronized with each other
is located in an *arbitrarily* defined “space origin”.
Post by Ricardo Jimenez
If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin?
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).

But if you need the Lorentz transformation to see it, then consider that

t'' = γ(v') (t' − v'/c² x').

where t'' is the time that a second clock shows that is at rest relative to
a "moving" master clock that shows time t'. Then as the second clock is at
rest relative to the master clock, we have v' = 0 ⇒ γ(v') = 1, therefore

t'' = t'.

IOW, it does not matter where the second clock is (what x' is); as long as
it is *co-moving* with the master clock, it remains synchronized to the
latter *in that reference frame* (and it is NOT synchronized with the latter
in all other reference frames).
Post by Ricardo Jimenez
What am I missing?
The (Galilean) principle of relativity.


PointedEars
--
Q: What did the nuclear physicist order for lunch?
A: Fission chips.

(from: WolframAlpha)
Maciej Wozniak
2021-10-12 04:43:39 UTC
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That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
Ricardo Jimenez
2021-10-12 15:42:41 UTC
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On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant. The
procedure supposedly guarantees that any observer located at a chosen
space point, say the origin of 3 space, sees the time signal from the
clocks at other space points all equal to his own clock reading. I
asked a simple question amounting to whether or not this implies that
an observer located at some other space point will also see this and
all I have gotten back for an answer is BS. We are within the same
inertial coodinate system here.

There is a paper that seems to claim to give the answer:
https://aapt.scitation.org/doi/10.1119/1.13500
Macdonald - Clock synchronization, a universal light speed, and the
terrestrial red-shift experiment. Has anybody read it and can
comment?
Maciej Wozniak
2021-10-12 16:15:16 UTC
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Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant.
In the meantime in the real world, however, synchronizedf GPS
clocks keep measuring t'=t, just like all serious clocks always did.

Once again: you're asking people that have NEVER really
synchronized anything; they only have some brilliant
theoretical concepts about it.
Ricardo Jimenez
2021-10-12 16:26:44 UTC
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On Tue, 12 Oct 2021 09:15:16 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant.
In the meantime in the real world, however, synchronizedf GPS
clocks keep measuring t'=t, just like all serious clocks always did.
Once again: you're asking people that have NEVER really
synchronized anything; they only have some brilliant
theoretical concepts about it.
Have you read Galison's book? The procedures for synchronization of
clocks were being discussed, used by the railroads, and patented in
1905 when Einstein discussed them in his paper.
Maciej Wozniak
2021-10-12 17:54:59 UTC
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Post by Ricardo Jimenez
On Tue, 12 Oct 2021 09:15:16 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant.
In the meantime in the real world, however, synchronizedf GPS
clocks keep measuring t'=t, just like all serious clocks always did.
Once again: you're asking people that have NEVER really
synchronized anything; they only have some brilliant
theoretical concepts about it.
Have you read Galison's book? The procedures for synchronization of
clocks were being discussed, used by the railroads,
And - with what accuracy?

Grow on, poor fanatic. It was more than 100 years ago.
Outside your preciou physics - telecommunication and
its procedures have changed a bit during this time. Didn't
you notice? Because your idiot gurus surely didn't.
Thomas 'PointedEars' Lahn
2021-10-12 18:11:26 UTC
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Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
No, because they are in relative motion to that observer, to begin with.
They are also at a different distance from the center of mass than the
Earthbound observer. So they are certainly NOT in the same reference frame.
Post by Ricardo Jimenez
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant.
No, it does not. That is just pseudo-scientific word salad.
Post by Ricardo Jimenez
The procedure supposedly guarantees that any observer located at a chosen
space point, say the origin of 3 space, sees the time signal from the
clocks at other space points all equal to his own clock reading.
No, it does not. It only guarantees that the observer who is *at rest*
relative to those clocks says that they all show the same time.

Special relativity shows that those clocks are then not synchronized if they
are all moving relative to an observer, an effect called “the relativity of
simultaneity” (after the title of the section in Einstein’s 1905 paper).
Post by Ricardo Jimenez
I asked a simple question amounting to whether or not this implies that
an observer located at some other space point will also see this
No, you did not. That was NOT the content of your question in any valid
interpretation of English. Your question was:

“If you perform a translation of space, how do you show all clocks are
synchonized with the one at the new origin?”

Do you deny that?

I have answered that question: It follows from the principle of relativity,
and it can also be shown using the Lorentz transformation.

Do you deny that?
Post by Ricardo Jimenez
and all I have gotten back for an answer is BS.
*My* answer is not “BS”. My answer is correct. It is “BS” *to you* because
you appear to be ignorant about what you are asking about, and not willing
to understand the answer. And so you do not understand the answer, and
apparently even have asked the question wrong, or the wrong question.

It is your fault, not mine, if you do not like my correct answer.

And calling my well-formulated answer, into which I invested a considerable
amount of my precious free time in orderr to make it easily understandable
(to you), “BS”, is not an incentive for me to answer any more of your
questions.


PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-10-12 18:13:22 UTC
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Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
No, because they are in relative motion to that observer, to begin with.
They are also at a different distance from the center of mass than the
Earthbound observer. So they are certainly NOT in the same reference frame.
Post by Ricardo Jimenez
The point of clock synchonization is that it gives observational
meaning to the 3-space in Minkowski Space-Time, t=constant.
No, it does not. That is just pseudo-scientific word salad.
Post by Ricardo Jimenez
The procedure supposedly guarantees that any observer located at a chosen
space point, say the origin of 3 space, sees the time signal from the
clocks at other space points all equal to his own clock reading.
No, it does not. It only guarantees that the observer who is *at rest*
relative to those clocks says that they all show the same time.

Special relativity shows that those clocks are then not synchronized if they
are all moving relative to an observer, an effect called “the relativity of
simultaneity” (after the title of the section in Einstein’s 1905 paper).
Post by Ricardo Jimenez
I asked a simple question amounting to whether or not this implies that
an observer located at some other space point will also see this
No, you did not. That was NOT the content of your question in any valid
interpretation of English. Your question was:

“If you perform a translation of space, how do you show all clocks are
synchonized with the one at the new origin?”

Do you deny that?

I have answered that question: It follows from the principle of relativity,
and it can also be shown using the Lorentz transformation.

Do you deny that?
Post by Ricardo Jimenez
and all I have gotten back for an answer is BS.
*My* answer is not “BS”. My answer is correct. It is “BS” *to you* because
you appear to be ignorant about what you are asking about, and not willing
to understand the answer. And so you do not understand the answer, and
apparently even have asked the question wrong, or the wrong question.

It is your fault, not mine, if you do not like my correct answer.

And calling my well-formulated answer, into which I invested a considerable
amount of my precious free time in order to make it easily understandable
(to you), “BS”, is not an incentive for me to answer any more of your
questions.


PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)
Ricardo Jimenez
2021-10-12 18:27:24 UTC
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On Tue, 12 Oct 2021 20:13:22 +0200, Thomas 'PointedEars' Lahn
Post by Thomas 'PointedEars' Lahn
And calling my well-formulated answer, into which I invested a considerable
amount of my precious free time in order to make it easily understandable
(to you), “BS”, is not an incentive for me to answer any more of your
questions.
Yes, please don't spend any more time answering my questions because
all I get from you is handwaving appeals to the "relativity
principle", unintelligible word salad and unreadable equations because
Forte Agent shows many characters as ?. I explicitly said that all
clocks in question were at rest with respect to the observer at the
origin since they are in the same space coordinate system and you
ignored that. Where did you get the idea that you are a master of the
theory of relativity?
Thomas 'PointedEars' Lahn
2021-10-12 18:43:39 UTC
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Post by Ricardo Jimenez
On Tue, 12 Oct 2021 20:13:22 +0200, Thomas 'PointedEars' Lahn
Post by Thomas 'PointedEars' Lahn
And calling my well-formulated answer, into which I invested a
considerable amount of my precious free time in order to make it easily
understandable (to you), “BS”, is not an incentive for me to answer any
more of your questions.
Yes, please don't spend any more time answering my questions
As you wish. I have certainly better things to do than to teach ignorant
wannabes.
Post by Ricardo Jimenez
because all I get from you is handwaving appeals to the "relativity
principle",
It is not handwaving, it is a observable fact that there is no experiment by
which one can tell whether one is at rest or in uniform motion.
Post by Ricardo Jimenez
unintelligible word salad
It was not word salad.
Post by Ricardo Jimenez
and unreadable equations because Forte Agent shows many characters as ?.
My posting was correctly encoded. It is not my fault if you use broken
software or are unable to configure your software correctly.
Post by Ricardo Jimenez
I explicitly said that all clocks in question were at rest with respect to
the observer at the origin since they are in the same space coordinate
system and you ignored that.
No, I did not.
Post by Ricardo Jimenez
Where did you get the idea that you are a master of the theory of
relativity?
I am certainly not a master of the _theories_, but I know that there are
*two* of them, and I passed the “Mechanics Ⅰ with [Special] Relativity” exam
while studying Astrophysics. What about you?


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Maciej Wozniak
2021-10-12 18:49:29 UTC
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Post by Thomas 'PointedEars' Lahn
It is not handwaving, it is a observable fact that there is no experiment by
which one can tell whether one is at rest or in uniform motion.
Because it isn't and it never was a matter of your precious
experiments. It's assumptions - axioms, definitions, postulates -
that decide.
Maciej Wozniak
2021-10-12 18:44:57 UTC
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Post by Thomas 'PointedEars' Lahn
Post by Ricardo Jimenez
On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
Post by Maciej Wozniak
That is already implied by the principle of relativity. The so-called
"inertially moving" clocks are at rest relative to a co-moving observer,
so that observer MUST observe the same as if both the clocks and the
observer would be considered at rest (by another observer who considers
themselves to be at rest).
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
No, because they are in relative motion to that observer, to begin with.
They are also at a different distance from the center of mass than the
Earthbound observer. So they are certainly NOT in the same reference frame.
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
mitchr...@gmail.com
2021-10-12 18:49:01 UTC
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Separating clocks that have been together synchronized
will create more distance for light to have to move through
creating a synchronization delay.
Ricardo Jimenez
2021-10-12 20:30:47 UTC
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Post by ***@gmail.com
Separating clocks that have been together synchronized
will create more distance for light to have to move through
creating a synchronization delay.
I'll try to explain again the difficulty I am having with the
synchronization procudure in the first 1905 Einstein relativity paper.
What 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. Thus 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. However, Einstein claims that clock synchronization
is reflexive, symmetric and transitive.

Did he mean that you can redo clock synchronization with B as the
master clock so everything is synchronized with it?
Thomas 'PointedEars' Lahn
2021-10-13 00:01:34 UTC
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Post by Ricardo Jimenez
Post by ***@gmail.com
Separating 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 Jimenez
I'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 Jimenez
What 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 Jimenez
Thus 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 Jimenez
However, 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 Jimenez
Did 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)
Townes Olson
2021-10-13 00:02:31 UTC
Reply
Permalink
Post by Ricardo Jimenez
I'll try to explain again the difficulty I am having with the
synchronization procudure in the first 1905 Einstein relativity paper.
More than one synchronization procedure is described in that paper. You seem to be mixing them up.
Post by Ricardo Jimenez
What is implied by the two equations he gives...
The paper contains far more than two equations. You need to specify.
Post by Ricardo Jimenez
... is that to synchronize all clocks with the one at A, you send out a
time signal...
No, in the inertia-based synchronization there are no "time signals", there are simply pulses of light.
Post by Ricardo Jimenez
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).
No, that isn't the procedure for inertia-based synchronization described in the paper. There is no "received value" in this procedure. The B clock is set so that it is the average of the readings of A when a light pulse is emitted from A and when the reflection from B gets back to A. In other words, using the nomenclature of paragraph 1, we have tB = (tA+tA')/2. The values of tA and tA' are the emission and reception times at A, and the value of tB is the reflection time at B.
Post by Ricardo Jimenez
A doesn't do anything to his clock. Then when clock B sends a time signal to A...
B does not send a time signal, nor does A. All that's happening is that A is sending a pulse of light that reflect off B and goes back to A. Now, if you want to think about the engineering of how you communicate such that B knows how to adjust his clock so that tB = (tA+tA')/2 for each emission, reflection, and reception events, you can do that in infinitely many ways. The important point is to understand the defining aspect of inertial synchronization, which is that the speed of the signal is the same in both directions in terms of these coordinates.
Post by Ricardo Jimenez
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.
Huh? That's very confused. Look, if you want an explicit recipe for how they would go about achieving this synchronization, you could have A send is current reading tA by radio to B, and when B receives that signal he could immediately send his current reading "trialB" back to A, which A receives at time tA'. Now, A knows that B's clock should have been reading tB = (tA+tA')/2 at the reflection event, but it was actually reading "trialB", so A can now send a message to B telling him that his clock is off by delt = tB - trialB. So B can set his clock forward by delt and he will then be inertially synchronized with A.
Post by Ricardo Jimenez
Thus all clocks in space are synchronized to A's.
Well, any clock that is at rest in this frame can by synchronized with any other clock that is at rest in this frame using this procedure.
Post by Ricardo Jimenez
However, B still sees time signals from A equal to B's clock time minus distance(A,B)/c.
Well, obviously for any system of inertia-based coordinates x,t and any light signal going from x1,t1 to x2,t2 we have (x2-x1)/(t2-t1) = c.
Post by Ricardo Jimenez
However, Einstein claims that clock synchronization is reflexive, symmetric and transitive.
Right, the synchronization described above has all those properties.
Post by Ricardo Jimenez
Did he mean that you can redo clock synchronization with B as the
master clock so everything is synchronized with it?
There is no particular "master clock". The point is that inertia-based coordinates x,t are defined such that for any pulse of light going from x1,t1 to x2,t2 we have (x2-x1)/(t2-t1) = c. At each location we can initialize our clocks so that they read the t coordinate. A simpler way to synchronize clocks at the ends of a rod is just to put a light bulb at the mid-point of the rod and flash the bulb. Set the clocks to noon when the flashes arrive. They will be synchronized in the frame of the rod. You can do this for a whole array of clocks located at the nodes of a grid covering an entire region of space. So you can have all the clocks synchronized in the frame of the grid.
Thomas 'PointedEars' Lahn
2021-10-15 23:23:00 UTC
Reply
Permalink
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”

GPS Interface Specification (2021). “3.3.1.1 Frequency Plan.”
<https://www.gps.gov/technical/icwg/IS-GPS-200M.pdf>

Linked from <https://www.gps.gov/technical/icwg/>.


Which part of “offset” and “compensate” did you not get?


PointedEars
--
Q: Why is electricity so dangerous?
A: It doesn't conduct itself.

(from: WolframAlpha)
Michael Moroney
2021-10-16 00:28:15 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”
GPS Interface Specification (2021). “3.3.1.1 Frequency Plan.”
<https://www.gps.gov/technical/icwg/IS-GPS-200M.pdf>
Linked from <https://www.gps.gov/technical/icwg/>.
Which part of “offset” and “compensate” did you not get?
He doesn't "get" any of that. He's just some broken record player,
autisticly repeating "In the meantime in the real world, however,
synchronized GPS clocks keep measuring t'=t, just like all serious
clocks always did." <click> "In the meantime in the real world, however,
synchronized GPS clocks keep measuring t'=t, just like all serious
clocks always did." <click> "In the meantime in the real world, however,
synchronized GPS clocks keep measuring t'=t, just like all serious
clocks always did." <click> ...
Maciej Wozniak
2021-10-16 05:41:57 UTC
Reply
Permalink
Post by Michael Moroney
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”
GPS Interface Specification (2021). “3.3.1.1 Frequency Plan.”
<https://www.gps.gov/technical/icwg/IS-GPS-200M.pdf>
Linked from <https://www.gps.gov/technical/icwg/>.
Which part of “offset” and “compensate” did you not get?
He doesn't "get" any of that. He's just some broken record player,
autisticly repeating "In the meantime in the real world, however,
See, stupid Mike: I'm one of the best logicians the humanity
ever had and you're just a poor religious crank of an insane
ideology. You can't discuss against me, of course, all you can
do is barking, spitting and slandering. But you will do what
you can for The Shit.
Thomas 'PointedEars' Lahn
2021-10-17 00:38:10 UTC
Reply
Permalink
Post by Maciej Wozniak
See, stupid Mike: I'm one of the best logicians the humanity
ever had […]
(sic!)

And what does your psychiatrist say about that, NUTCASE?


PointedEars
--
Q: What did the nuclear physicist post on the laboratory door
when he went camping?
A: 'Gone fission'.
(from: WolframAlpha)
Maciej Wozniak
2021-10-17 09:35:41 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
See, stupid Mike: I'm one of the best logicians the humanity
ever had […]
(sic!)
And what does your psychiatrist say about that, NUTCASE?
See, Thomas: I'm one of the best logicians the humanity
ever had and you're just a poor religious crank of an insane
ideology. You can't discuss against me, of course, all you can
do is barking, spitting and waving your arms. But you will do
what you can for The Shit you love.
Thomas 'PointedEars' Lahn
2021-10-17 23:27:22 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
See, stupid Mike: I'm one of the best logicians the humanity
ever had […]
(sic!)
And what does your psychiatrist say about that, NUTCASE?
See, Thomas: I'm one of the best logicians the humanity
ever had
No; it does not take a psychiatrist to see that you are seriously mentally
ill, and are suffering from delusions of grandeur, instead. Get well soon.

<https://en.wikipedia.org/wiki/Grandiose_delusions>


PointedEars
--
I heard that entropy isn't what it used to be.

(from: WolframAlpha)
mitchr...@gmail.com
2021-10-17 23:30:35 UTC
Reply
Permalink
Clocks remain syncronized after they move apart
as long as the share the same Gamma for gravity and speed.

Mitchell Raemsch
Maciej Wozniak
2021-10-18 05:49:58 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
See, stupid Mike: I'm one of the best logicians the humanity
ever had […]
(sic!)
And what does your psychiatrist say about that, NUTCASE?
See, Thomas: I'm one of the best logicians the humanity
ever had
No; it does not take a psychiatrist to see that you are seriously mentally
ill, and are suffering from delusions of grandeur, instead.
See, Thomas: I'm one of the best logicians the humanity
ever had and you're just a poor religious crank of an insane
ideology. You can't discuss against me, of course, all you can
do is barking, spitting and waving your arms. But you will do
what you can for The Shit you love.

And in the meantime in the real world, GPS clocks will keep
measuring t'=t, just like all serious clocks always did.
Michael Moroney
2021-10-17 19:29:27 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Michael Moroney
Post by Thomas 'PointedEars' Lahn
Which part of “offset” and “compensate” did you not get?
He doesn't "get" any of that. He's just some broken record player,
autisticly repeating "In the meantime in the real world, however,
See, stupid Mike: I'm one of the best logicians the humanity
ever had
Hahahahaha!!!! The "best logician" is a broken record player that can
only repeat "In the meantime in the real world, however, GPS clocks keep
measuring t'=t, just like serious clocks always did." <click> "In the
meantime in the real world, however, GPS clocks keep measuring t'=t,
just like serious clocks always did." <click> ...

You can't discuss against me, of course,

Of course it's impossible to discuss anything with a broken record player.
Post by Maciej Wozniak
all you can
do is barking, spitting and slandering.
That's all you can do, when not in broken record player mode.
Maciej Wozniak
2021-10-18 05:41:29 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Michael Moroney
Post by Thomas 'PointedEars' Lahn
Which part of “offset” and “compensate” did you not get?
He doesn't "get" any of that. He's just some broken record player,
autisticly repeating "In the meantime in the real world, however,
See, stupid Mike: I'm one of the best logicians the humanity
ever had
Hahahahaha!!!! The "best logician" is a broken record player that can
only repeat "In the meantime in the real world, however, GPS clocks keep
measuring t'=t, just like serious clocks always did." <click> "In the
See, stupid Mike: I'm one of the best logicians the humanity
ever had and you're just a poor religious crank of an insane
ideology. You can't discuss against me, of course, all you can
do is barking, spitting and slandering. But you will do
what you can for The Shit you love.
Maciej Wozniak
2021-10-16 05:22:29 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”
Who cares what would maybe appear to an imagined
person? It's the measuremant result that matters, and
it is t'=t. Just like always.
Or maybe you're trying to convince me that the result
is invalid because the measurement equipment was
calibrated. Are you?
Michael Moroney
2021-10-17 19:39:52 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”
Who cares what would maybe appear to an imagined
person?
An observer doesn't have to be a physical person. In fact the usual
physics observer is an instrument of some sort. As to the satellites
themselves, it depends on whether the satellites use any process locally
which depends on correct *local* time (not adjusted signals sent to earth).
Post by Maciej Wozniak
It's the measuremant result that matters,
Which is why I said about any satellite-local process. Such as tracking
its own orbit or whatever.
Post by Maciej Wozniak
and it is t'=t. Just like always.
Not on the GPS. Unless you insist that 1/9192631770 = 1/9192631774.1.
Post by Maciej Wozniak
Or maybe you're trying to convince me that the result
is invalid because the measurement equipment was
calibrated. Are you?
As a crank with an idée fixe, educating you is a hopeless task. I know
that. I am here to watch crackpots and post to get reactions from them,
usually by posting corrections and facts that make them (you) whine.

Rarely something useful comes along, typically from the likes of Tom R.
Maciej Wozniak
2021-10-18 05:47:39 UTC
Reply
Permalink
Post by Michael Moroney
Post by Maciej Wozniak
Post by Thomas 'PointedEars' Lahn
Post by Maciej Wozniak
Nobody says they're in the same reference frame, poor halfbrain;
still, they're measuring t'=t, just like all serious clocks always did,
and anyone can observe that.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”
Who cares what would maybe appear to an imagined
person?
An observer doesn't have to be a physical person. In fact the usual
physics observer is an instrument of some sort.
And, of course Wise Gurus will tell us what appears to
it, right, stupid Mike?
Post by Michael Moroney
Post by Maciej Wozniak
It's the measuremant result that matters,
Which is why I said about any satellite-local process. Such as tracking
its own orbit or whatever.
Post by Maciej Wozniak
and it is t'=t. Just like always.
Not on the GPS. Unless you insist that 1/9192631770 = 1/9192631774.1.
As I've told you many times, these values are T, not t,
and your Shit is actually predicting them to be equal.
Post by Michael Moroney
Post by Maciej Wozniak
Or maybe you're trying to convince me that the result
is invalid because the measurement equipment was
calibrated. Are you?
As a crank with an idée fixe, educating you is a hopeless task. I know
that. I am here to watch crackpots and post to get reactions from them,
See, stupid Mike: I'm one of the best logicians the humanity
ever had, while you're just a fanatic idiot worshipping an
insane crazie. Of course, you can't discuss against me,
you can only spit, bark and slander. But you will do what
you can for The Shit you love.
Thomas 'PointedEars' Lahn
2021-10-15 23:24:16 UTC
Reply
Permalink
Post by Maciej Wozniak
In the meantime in the real world, however, an observer MUST
observe GPS clocks measuring t'=t, just like all serious clocks
always did.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”

GPS Interface Specification (2021). “3.3.1.1 Frequency Plan.”
<https://www.gps.gov/technical/icwg/IS-GPS-200M.pdf>

Linked from <https://www.gps.gov/technical/icwg/>.


Which part of “offset” and “compensate” did you not get?


PointedEars
--
Q: What did the nuclear physicist order for lunch?
A: Fission chips.

(from: WolframAlpha)
Julio Di Egidio
2021-10-12 19:58:38 UTC
Reply
Permalink
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space.
That's not it: to the letter that just says that a pulse of light emitted from the origin at some time t(0) gets to some other location at t(0) plus the trip time.

The most basic synchronisation of two clocks is with an observer mid-way (all at rest in the same inertial frame): both clocks send a pulse of light with each tick, they are synchronised when the receiver in the middle sees the two pulses of light coming at the same time. On top of that one can build other schemes.

Synchronisation is done on the simultaneity plane of some common inertial frame. Then the clocks can even part ways: they'll keep ticking the same (proper) time ad infinitum, i.e. as long as they keep working, just not anymore in a common simultaneity plane.
Post by Ricardo Jimenez
If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin?
If you just translate (coordinates!) to another frame, physically nothing is happening. Indeed, along the lines of what Thomas was saying, do translate coordinates then recompute the relevant quantities: you will see that nothing has happened.

Julio
Maciej Wozniak
2021-10-12 20:24:40 UTC
Reply
Permalink
Post by Julio Di Egidio
Synchronisation is done on the simultaneity plane of some common inertial frame.
A pity they exist only in your mystical tales. Still,
GPS staff have managed anyway.
Ricardo Jimenez
2021-10-12 20:46:51 UTC
Reply
Permalink
On Tue, 12 Oct 2021 12:58:38 -0700 (PDT), Julio Di Egidio
Post by Julio Di Egidio
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space.
That's not it: to the letter that just says that a pulse of light emitted from the origin at some time t(0) gets to some other location at t(0) plus the trip time.
The most basic synchronisation of two clocks is with an observer mid-way (all at rest in the same inertial frame): both clocks send a pulse of light with each tick, they are synchronised when the receiver in the middle sees the two pulses of light coming at the same time. On top of that one can build other schemes.
Synchronisation is done on the simultaneity plane of some common inertial frame. Then the clocks can even part ways: they'll keep ticking the same (proper) time ad infinitum, i.e. as long as they keep working, just not anymore in a common simultaneity plane.
Post by Ricardo Jimenez
If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin?
If you just translate (coordinates!) to another frame, physically nothing is happening. Indeed, along the lines of what Thomas was saying, do translate coordinates then recompute the relevant quantities: you will see that nothing has happened.
Julio
I don't understand how that works for more than two clocks. As I
pointed out in another post, if you want all clocks in space to be in
sync, you have to choose a master clock and synch all the others with
it. You can pick the origin in space as the position of the master
clock for all simultaneity 3 spaces. I don't know why they are
incorrectly called planes.
Julio Di Egidio
2021-10-12 22:49:50 UTC
Reply
Permalink
Post by Ricardo Jimenez
Post by Julio Di Egidio
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space.
That's not it: to the letter that just says that a pulse of light emitted from the origin at some time t(0) gets to some other location at t(0) plus the trip time.
The most basic synchronisation of two clocks is with an observer mid-way (all at rest in the same inertial frame): both clocks send a pulse of light with each tick, they are synchronised when the receiver in the middle sees the two pulses of light coming at the same time. On top of that one can build other schemes.
Synchronisation is done on the simultaneity plane of some common inertial frame. Then the clocks can even part ways: they'll keep ticking the same (proper) time ad infinitum, i.e. as long as they keep working, just not anymore in a common simultaneity plane.
Post by Ricardo Jimenez
If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin?
If you just translate (coordinates!) to another frame, physically nothing is happening. Indeed, along the lines of what Thomas was saying, do translate coordinates then recompute the relevant quantities: you will see that nothing has happened.
I don't understand how that works for more than two clocks.
Essentially, just doing it pairwise. More below.
Post by Ricardo Jimenez
As I
pointed out in another post, if you want all clocks in space to be in
sync, you have to choose a master clock and synch all the others with
it
No, you might be thinking a "master" clock as in "let's all (you) sync to my clock", but that's not the point: the choice of (space) origin is just as arbitrary as the choice of time origin. You can do it pairwise... Indeed, I don't know if e.g. in the article you have linked, say for reasons of "propagation of the syncing", they pick a "center" of the lattice or not: but, even if they do, that choice remains *arbitrary*, you pick the one you want.
Post by Ricardo Jimenez
You can pick the origin in space as the position of the master
clock for all simultaneity 3 spaces. I don't know why they are
incorrectly called planes.
The "plane of simultaneity" is a *hyper-plane*: namely, given a frame of reference, it is the whole of space at some point in time, namely a section (hence the "plane") for fixed time of the space-time manifold (in a first approximation). What you said is not even meaningful.

HTH,

Julio
Ricardo Jimenez
2021-10-13 00:12:30 UTC
Reply
Permalink
On Tue, 12 Oct 2021 15:49:50 -0700 (PDT), Julio Di Egidio
Post by Julio Di Egidio
Post by Ricardo Jimenez
Post by Julio Di Egidio
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space.
That's not it: to the letter that just says that a pulse of light emitted from the origin at some time t(0) gets to some other location at t(0) plus the trip time.
The most basic synchronisation of two clocks is with an observer mid-way (all at rest in the same inertial frame): both clocks send a pulse of light with each tick, they are synchronised when the receiver in the middle sees the two pulses of light coming at the same time. On top of that one can build other schemes.
Synchronisation is done on the simultaneity plane of some common inertial frame. Then the clocks can even part ways: they'll keep ticking the same (proper) time ad infinitum, i.e. as long as they keep working, just not anymore in a common simultaneity plane.
Post by Ricardo Jimenez
If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin?
If you just translate (coordinates!) to another frame, physically nothing is happening. Indeed, along the lines of what Thomas was saying, do translate coordinates then recompute the relevant quantities: you will see that nothing has happened.
I don't understand how that works for more than two clocks.
Essentially, just doing it pairwise. More below.
Post by Ricardo Jimenez
As I
pointed out in another post, if you want all clocks in space to be in
sync, you have to choose a master clock and synch all the others with
it
No, you might be thinking a "master" clock as in "let's all (you) sync to my clock", but that's not the point: the choice of (space) origin is just as arbitrary as the choice of time origin. You can do it pairwise... Indeed, I don't know if e.g. in the article you have linked, say for reasons of "propagation of the syncing", they pick a "center" of the lattice or not: but, even if they do, that choice remains *arbitrary*, you pick the one you want.
Post by Ricardo Jimenez
You can pick the origin in space as the position of the master
clock for all simultaneity 3 spaces. I don't know why they are
incorrectly called planes.
The "plane of simultaneity" is a *hyper-plane*: namely, given a frame of reference, it is the whole of space at some point in time, namely a section (hence the "plane") for fixed time of the space-time manifold (in a first approximation). What you said is not even meaningful.
HTH,
Julio
The meaning is that if all clocks in the same inirtial systen are
synched to the clock at a fixed point, say the origin, comparing the
time coordinates of events can be done by comparing the times on the
clock at the origin that signals from those events reach the origin.
Well at least I think so.

In the inaugural paper, Einstein doesn't mention the procedure for
synching two clocks by putting an observer in the middle. I got the
definition of synching with the clock at a fixed point by
simultaneously solving the two linear equations he gives. Defined
your way, simultaneity might lead to an equivalence relation among a
finite number of clocks which might be enough clocks to do relativity
theory. However, this seems to require some extra assumptions given
say, in the Wikipedia article
https://en.wikipedia.org/wiki/Einstein_synchronisation where
Macdonald's paper is mentioned.
Thomas 'PointedEars' Lahn
2021-10-13 11:44:57 UTC
Reply
Permalink
Post by Ricardo Jimenez
The meaning is that if all clocks in the same inirtial systen are
synched to the clock at a fixed point, say the origin, comparing the
time coordinates of events can be done by comparing the times on the
clock at the origin that signals from those events reach the origin.
No; instead, the idea is that the synchronization signal originates at the
master clock (in the middle).
Post by Ricardo Jimenez
Well at least I think so.
You are mistaken about that, too.
Post by Ricardo Jimenez
In the inaugural paper, Einstein doesn't mention the procedure for
synching two clocks by putting an observer in the middle.
He does not mention *any* synchronization *procedure* there, but describes
what would be the case *if* two clocks were _synchronous_. Thereby he
*defines* what he means when he says that they are _synchronous_.
Post by Ricardo Jimenez
I got the definition of synching with the clock at a fixed point by
simultaneously solving the two linear equations he gives.
He simply does not do that there. Get that /idée fixe/ out of your mind.
Post by Ricardo Jimenez
Defined your way, simultaneity might lead to an equivalence relation among
a finite number of clocks which might be enough clocks to do relativity
theory.
“Might”? *facepalm*

If clock A is synchronous with clock B, then B is synchronous with A. If B
is synchronous with A, and C is synchronous with B, then C is of course
synchronous with A, too: A and B show the *same* time. How can you even
begin to think that would not be so?
Post by Ricardo Jimenez
However, this seems to require some extra assumptions given
say, in the Wikipedia article
https://en.wikipedia.org/wiki/Einstein_synchronisation where
Macdonald's paper is mentioned.
Einstein synchronization is NOT what Einstein describes in the paper that
you are referring to.


You should take this online course:

<https://online.stanford.edu/courses/som-y0009-understanding-einstein-special-theory-relativity>

I can highly recommend it from first-hand experience: I took it (and passed
it early with full points) before I started studying Astrophysics at the
(another, real-life) university. It was one of the things that showed me
that I would be able to understand this, that I wanted to study and
understand it more deeply, and that I wanted to go to university again in
order to do that.

You are lucky: The next round (it is scheduled for 8 weeks, then it repeats)
is starting *today*.


PointedEars
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)
carl eto
2021-10-13 16:37:00 UTC
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You cannot measure the velocity of light using electrons that are propagating slower than the velocity of light
Odd Bodkin
2021-10-13 18:03:04 UTC
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Post by carl eto
You cannot measure the velocity of light using electrons that are
propagating slower than the velocity of light
Of course you can.

You can also measure the speed of a race car going 200 mph with just people
holding stopwatches, who are going no faster than walking pace.
--
Odd Bodkin -- maker of fine toys, tools, tables
carl eto
2021-10-13 18:07:06 UTC
Reply
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Post by Odd Bodkin
Post by carl eto
You cannot measure the velocity of light using electrons that are
propagating slower than the velocity of light
Of course you can.
You can also measure the speed of a race car going 200 mph with just people
holding stopwatches, who are going no faster than walking pace.
--
Odd Bodkin -- maker of fine toys, tools, tables
How?
Odd Bodkin
2021-10-13 19:53:13 UTC
Reply
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Post by Odd Bodkin
Post by carl eto
You cannot measure the velocity of light using electrons that are
propagating slower than the velocity of light
Of course you can.
You can also measure the speed of a race car going 200 mph with just people
holding stopwatches, who are going no faster than walking pace.
--
Odd Bodkin -- maker of fine toys, tools, tables
How?
Oh, come on. There are two people, Bob and Tom, standing at two locations
on the track. They both started their stopwatches when the race started.
The race car passes Bob and Bob clicks his stopwatch. The race car then
passes Tom and Tom clicks his stopwatch. Bob and Tom then casually walk to
a tent a few hundred feet away and compare their stopwatch times. The
difference in times goes in the denominator, the distance between where Bob
and Tom were standing goes in the numerator, and the quotient is the speed
of the race car.

This eludes you? What’s wrong with you? ….. Oh never mind, I remember now
what’s wrong with you.
--
Odd Bodkin -- maker of fine toys, tools, tables
carl eto
2021-10-13 20:36:04 UTC
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Permalink
You cannot use a stop watch to measure the velocity of light. You have to use an electrical circuit that signal is produced by propagating electrons. So when you attempt to measure the velocity of a radio wave the radio wave has already passed by. Plus, the satellite signal is continuous and GPS is formed by the intensity difference.
Odd Bodkin
2021-10-13 21:51:52 UTC
Reply
Permalink
Post by carl eto
You cannot use a stop watch to measure the velocity of light.
That’s true. But you can use a slow device to measure a fast signal. I just
gave you an example. Slow humans, fast car.
Post by carl eto
You have to use an electrical circuit that signal is produced by
propagating electrons. So when you attempt to measure the velocity of a
radio wave the radio wave has already passed by. Plus, the satellite
signal is continuous and GPS is formed by the intensity difference.
--
Odd Bodkin -- maker of fine toys, tools, tables
carl eto
2021-10-14 16:50:50 UTC
Reply
Permalink
Post by Odd Bodkin
Post by carl eto
You cannot use a stop watch to measure the velocity of light.
That’s true. But you can use a slow device to measure a fast signal. I just
gave you an example. Slow humans, fast car.
Post by carl eto
You have to use an electrical circuit that signal is produced by
propagating electrons. So when you attempt to measure the velocity of a
radio wave the radio wave has already passed by. Plus, the satellite
signal is continuous and GPS is formed by the intensity difference.
--
Odd Bodkin -- maker of fine toys, tools, tables
carl eto
2021-10-14 16:53:44 UTC
Reply
Permalink
Post by Odd Bodkin
Post by carl eto
You cannot use a stop watch to measure the velocity of light.
That’s true. But you can use a slow device to measure a fast signal. I just
gave you an example. Slow humans, fast car.
Post by carl eto
You have to use an electrical circuit that signal is produced by
propagating electrons. So when you attempt to measure the velocity of a
radio wave the radio wave has already passed by. Plus, the satellite
signal is continuous and GPS is formed by the intensity difference.
--
Odd Bodkin -- maker of fine toys, tools, tables
The question is how does the GPS function ---- by the time difference of the radio waves or the intensity of the radio waves since the time difference implies that the velocity of light is being measured.
mitchr...@gmail.com
2021-10-14 18:52:58 UTC
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Time across the universe has gone on since the initial
conditions of the BB. Gravity would be in the way
of beginning singularity expanding.

How did the singularity form first Hydrogen atoms
if its gravity didn't allow it to expand?

Mitchell Raemsch
rotchm
2021-10-13 18:41:22 UTC
Reply
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You are stroking the troll....
Odd Bodkin
2021-10-13 19:53:14 UTC
Reply
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Post by rotchm
You are stroking the troll....
I’m certainly paying a visit to the psych wing at the hospital, yes.
--
Odd Bodkin -- maker of fine toys, tools, tables
Kendale Gross
2021-10-14 14:21:48 UTC
Reply
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Post by Odd Bodkin
Post by rotchm
You are stroking the troll....
I’m certainly paying a visit to the psych wing at the hospital, yes.
certainly you just did, your "rotchm" is a known imbecile.
Michael Moroney
2021-10-12 23:26:22 UTC
Reply
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Post by Ricardo Jimenez
I don't understand how that works for more than two clocks. As I
pointed out in another post, if you want all clocks in space to be in
sync, you have to choose a master clock and synch all the others with
it.
If Clock A is synchronized with Clock B, and Clock B is synchronized
with Clock C, then Clock A is synchronized with Clock C.

From the 1905 SR paper, I believe.

Just synchronize each clock with an already synchronized clock until done.
JanPB
2021-10-14 14:59:20 UTC
Reply
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Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
This is actually not Einstein's sync process exactly.

The sync process is tB - tA = tA' - tB, where tA = clock reading at the
light pulse emission site, tB = clock reading at the mirror, tA' = clock
reading again at the emission site after the light ray returns.

It is not immediately obvious that this process is free of contradiction.
Einstein in his 1905 paper simply says we assume that it is - that's
because it's easy to demonstrate and he was writing for professional
physicists.

Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
requires an extra consideration to take care of (the other two are
trivial from the definition): one has to assume that given a single
clock at some location A and two mirrors at B and C (so that
a triangular light path from A to A is formed), the elapsed time
recorded by that single clock for light going A->B->C->A is equal
to the time with the light going the other way: A->C->B->A.

This follows from the presumed (silently) homogeneity of space and
it had been in fact verified experimentally by Fizeau (to acceptable
limits).

What you describe seems to be taken from Taylor & Wheeler's book
(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
and your question is valid, and it has the resolution I've described,
I think.

--
Jan
Ricardo Jimenez
2021-10-14 15:42:33 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
This is actually not Einstein's sync process exactly.
The sync process is tB - tA = tA' - tB, where tA = clock reading at the
light pulse emission site, tB = clock reading at the mirror, tA' = clock
reading again at the emission site after the light ray returns.
It is not immediately obvious that this process is free of contradiction.
Einstein in his 1905 paper simply says we assume that it is - that's
because it's easy to demonstrate and he was writing for professional
physicists.
Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
requires an extra consideration to take care of (the other two are
trivial from the definition): one has to assume that given a single
clock at some location A and two mirrors at B and C (so that
a triangular light path from A to A is formed), the elapsed time
recorded by that single clock for light going A->B->C->A is equal
to the time with the light going the other way: A->C->B->A.
This follows from the presumed (silently) homogeneity of space and
it had been in fact verified experimentally by Fizeau (to acceptable
limits).
What you describe seems to be taken from Taylor & Wheeler's book
(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
and your question is valid, and it has the resolution I've described,
I think.
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things. It is fairly terse and I am still trying to absorb it.
I am at first, taken aback, by the use of Einstein's definition of two
clocks being synchronized since it allows one to add any amount of
time to either clock's setting and they will still be synchronized
according to it. But I suppose you might want to call two clocks,
using the official time in two different time zones, synchronized.
JanPB
2021-10-14 19:22:12 UTC
Reply
Permalink
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
This is actually not Einstein's sync process exactly.
The sync process is tB - tA = tA' - tB, where tA = clock reading at the
light pulse emission site, tB = clock reading at the mirror, tA' = clock
reading again at the emission site after the light ray returns.
It is not immediately obvious that this process is free of contradiction.
Einstein in his 1905 paper simply says we assume that it is - that's
because it's easy to demonstrate and he was writing for professional
physicists.
Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive. Only the transitivity
requires an extra consideration to take care of (the other two are
trivial from the definition): one has to assume that given a single
clock at some location A and two mirrors at B and C (so that
a triangular light path from A to A is formed), the elapsed time
recorded by that single clock for light going A->B->C->A is equal
to the time with the light going the other way: A->C->B->A.
This follows from the presumed (silently) homogeneity of space and
it had been in fact verified experimentally by Fizeau (to acceptable
limits).
What you describe seems to be taken from Taylor & Wheeler's book
(I'm not a fan of this book BTW, I think pedagogically it's atrocious)
and your question is valid, and it has the resolution I've described,
I think.
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
Post by Ricardo Jimenez
I am at first, taken aback, by the use of Einstein's definition of two
clocks being synchronized since it allows one to add any amount of
time to either clock's setting and they will still be synchronized
according to it.
No, just look at the sync criterion again: tB - tA = tA' - tB.

If you add some nonzero amount of time, T, to clock B, the above
equation becomes:

(tB + T) - tA = tA' - (tB + T)

i.e.

tB - tA + T = tA' - tB - T

i.e.

T = -T

...which is plainly false. Adding T to clock A results in a similar
contradiction.

--
Jan
Ricardo Jimenez
2021-10-14 20:16:09 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
Post by JanPB
Post by Ricardo Jimenez
I am at first, taken aback, by the use of Einstein's definition of two
clocks being synchronized since it allows one to add any amount of
time to either clock's setting and they will still be synchronized
according to it.
No, just look at the sync criterion again: tB - tA = tA' - tB.
If you add some nonzero amount of time, T, to clock B, the above
(tB + T) - tA = tA' - (tB + T)
i.e.
tB - tA + T = tA' - tB - T
i.e.
T = -T
...which is plainly false. Adding T to clock A results in a similar
contradiction.
--
Jan
Oops, I should checked it before writing. :-)
Thomas 'PointedEars' Lahn
2021-10-14 22:47:17 UTC
Reply
Permalink
If only your references would be that precise. *facepalm*

Remove the *unncessary* attribution novel (it is all in the header fields),
and add the *necessary* precision to your references instead.
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3
Who is “he”, and the pages of which paper are you referring to?

You *really* need to learn how to cite your sources *properly* (including
*unambiguously*). “Einstein, p.115” or “there is a short paper on …” on
$WEBSITE simply does not cut it.


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Thomas 'PointedEars' Lahn
2021-10-14 22:47:45 UTC
Reply
Permalink
[Subject repaired]
If only your references would be that precise. *facepalm*

Remove the *unncessary* attribution novel (it is all in the header fields),
and add the *necessary* precision to your references instead.
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3
Who is “he”, and the pages of which paper are you referring to?

You *really* need to learn how to cite your sources *properly* (including
*unambiguously*). “Einstein, p.115” or “there is a short paper on …” on
$WEBSITE simply does not cut it.


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Thomas 'PointedEars' Lahn
2021-10-14 22:48:13 UTC
Reply
Permalink
[Subject repaired]
If only your references would be that precise. *facepalm*

Remove the *unnecessary* attribution novel (it is all in the header fields),
and add the *necessary* precision to your references instead.
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3
Who is “he”, and the pages of which paper are you referring to?

You *really* need to learn how to cite your sources *properly* (including
*unambiguously*). “Einstein, p.115” or “there is a short paper on …” on
$WEBSITE simply does not cut it.


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
JanPB
2021-10-15 07:47:53 UTC
Reply
Permalink
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
I don't see any mention of transitivity in either Einstein or Alan Macdonald's
paper you've referenced above.

--
Jan
Ricardo Jimenez
2021-10-15 16:28:21 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
I don't see any mention of transitivity in either Einstein or Alan Macdonald's
paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.

Macdonald: pg 2 - Clock synchronization etc.
We now show that (ii) is a necessary and sufficient condition that
clocks at nodes P and Q are synchronized with each other once they are
both synchronized with the clock at O.

They didn't use the word but that is what they were talking about.
carl eto
2021-10-15 17:22:04 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but
Let's assume that Lorentz invariant works, then What? Lorentz (1899) uses the invariant to reverse the negative result of Michelson experiment to justify the ether, composed of matter, and, Einstein (1917) uses the reversal of MMX to justify the ether yet vacuum proves the ether does not exist. Also, both Lorentz and Einstein use relativity (coordinate system transformation) to justify Maxwell theory but Maxwell equations are derived using Faraday induction effect that is not luminous. Using mumble jumble mathematical gymnastics cannot be used to justify the wave theory of light and modern theoretical physics (QM, QED, string theory, QFT,......etc. etc. that are based on the gauge since an electromagnetic wave (quantized or not) is based on a wave formed by the motion of an ether, composed of matter, where the physical structure (ether) has predencence before any abstract mathematical formulation.
JanPB
2021-10-15 21:09:02 UTC
Reply
Permalink
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
I you go to this webpage http://www.faculty.luther.edu/~macdonal/
there is a short paper on Clock Synchonization which it is claimed
fixes things.
He just states Einstein's definition I quoted above. He does not address
the transitivity requirement which Einstein left unsaid but is necessary
for the scheme to be consistent (meaning, the equation tB - tA = tA' - tB
never leads to a situation in which one clock is assigned two *different*
times).
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
I don't see any mention of transitivity in either Einstein or Alan Macdonald's
paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions (since he
knew the reader would be able to supply the proof without difficulty).
Post by Ricardo Jimenez
Macdonald: pg 2 - Clock synchronization etc.
We now show that (ii) is a necessary and sufficient condition that
clocks at nodes P and Q are synchronized with each other once they are
both synchronized with the clock at O.
They didn't use the word but that is what they were talking about.
OK.

--
Jan
Townes Olson
2021-10-15 23:17:27 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
Post by JanPB
I don't see any mention of transitivity in either Einstein or Alan Macdonald's
paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions...
Stachel's translation of Einstein's paper actually *does* say that, talking about "running",etc. A better translation in Einstein's collected papers says

"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."

This is better because Stachel's use of "running" can mislead people into thinking he is talking about rates.
Post by JanPB
he knew the reader would be able to supply the proof without difficulty.
Well, the reader certainly understands that the consistency of this all relies on the existence of a system of coordinates in terms of which light propagates isotropically in vacuum at speed c independent of the speed of the source. One of the shortcomings of the paper, in retrospect, is that most of the propositions are actually smuggled in via the definitions.
Thomas 'PointedEars' Lahn
2021-10-15 23:40:47 UTC
Reply
Permalink
Post by JanPB
Post by Ricardo Jimenez
Post by JanPB
I don't see any mention of transitivity in either Einstein or Alan
Macdonald's paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions...
Stachel's translation of Einstein's paper […]
… is not the official translation, therefore quite irrelevant.
The official translation can be found here instead:

<https://einsteinpapers.press.princeton.edu/vol2-trans/154>


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Townes Olson
2021-10-16 01:37:05 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Townes Olson
Post by JanPB
Post by Ricardo Jimenez
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions...
Stachel's translation of Einstein's paper actually *does* say that, talking about
"running", etc. A better translation in Einstein's collected papers says
"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
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."
This is better because Stachel's use of "running" can mislead people into thinking
he is talking about rates.
… is not the official translation, therefore quite irrelevant.
<https://einsteinpapers.press.princeton.edu/vol2-trans/154>
You are confused. The translation I quoted in my message was the one from the Einstein Collected Papers, and I explained why it is more clear than Stachel's translation. I accept your apology. Jan mistakenly thought Einstein didn't write that (and not just because of the slight translational variations). Also, it's dumb for you to talk about an "official" translation, in a sophomoric appeal to authority, rather than someone interested in the truth. If you do a straight literal translation from the original German, you can find justification for Stachel's choice of the word "run" rather than the Collected Paper's "is". Einstein used the German "lauft", which does indeed mean run. Here's a simplistic literal translation:

"We assume that this definition of synchronism is possible in a non-contradicting manner, and that for any number of points, so that the following relationships generally apply: 1. If the clock in B runs synchronously with the clock in A, then the clock in A runs synchronous with the clock in B. 2. If the clock in A runs synchronously with both the clock in B and the clock in C, the clocks in B and C also run synchronously relative to one another."

So, does this agree better with Stachel, or with CPAE? Miller gives another translation, in which he renders 2 as "If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other". So he avoids using either "runs" or "is".
Post by Thomas 'PointedEars' Lahn
he knew the reader would be able to supply the proof without difficulty.
Well, the reader certainly understands that the consistency of this all relies on the existence of a system of coordinates in terms of which light propagates isotropically in vacuum at speed c independent of the speed of the source. One of the shortcomings of the paper, in retrospect, is that most of the propositions are actually smuggled in via the definitions.
Thomas 'PointedEars' Lahn
2021-10-16 02:12:17 UTC
Reply
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Post by Townes Olson
Post by Thomas 'PointedEars' Lahn
Post by Townes Olson
Post by JanPB
Post by Ricardo Jimenez
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well
as with the clock at C, then the clocks at B and C also run
synchronously relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions...
Stachel's translation of Einstein's paper actually *does* say that,
talking about "running", etc. A better translation in Einstein's
collected papers says
"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
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."
This is better because Stachel's use of "running" can mislead people
into thinking he is talking about rates.
… is not the official translation, therefore quite irrelevant.
<https://einsteinpapers.press.princeton.edu/vol2-trans/154>
You are confused. The translation I quoted in my message was the one from
the Einstein Collected Papers, and I explained why it is more clear than
Stachel's translation.
I see.
Post by Townes Olson
I accept your apology.
LOL, the arrogance.
Post by Townes Olson
Jan mistakenly thought Einstein didn't write that (and not just because of
the slight translational variations).
OK.
Post by Townes Olson
Also, it's dumb for you to talk about an "official" translation, in a
sophomoric appeal to authority,
Bullshit. If you had read what you are quoting, you would know that it *is*
the official translation:

<https://einsteinpapers.press.princeton.edu/vol2-trans/11>
Post by Townes Olson
rather than someone interested in the truth. If you do a straight literal
translation from the original German, you can find justification for
Stachel's choice of the word "run" rather than the Collected Paper's "is".
Einstein used the German "lauft",
LOL, the arrogance again.

The word *would be* „läuft“, with an a-umlaut character, the declensed form
of the verb „laufen“ (to go, run, etc.). My native language is German
because I was born in Germany. And I have read the paper in the original
German, too. So I really do not need to be lectured by you about German.

Also, the translator of the papers into English, Dr. Anna Beck, is both a
native German speaker and trained in science. It is the peak of arrogance
on your part to assume that she understood German less well than you, or
Stachel who is not a native speaker either. There is a reason why in the
1989 an new, official translation was being published by Princeton
University Press.
Post by Townes Olson
which does indeed mean run.
Like many words in many languages, it has many meanings in German, you
wannabe.
Post by Townes Olson
Here's a simplistic literal translation: […]
LOL. If you had any clue about translations, you would know that such a
translation hardly reflects the original meaning.


PointedEars
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)
Townes Olson
2021-10-16 02:31:23 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Thomas 'PointedEars' Lahn
Post by Townes Olson
Post by JanPB
Post by Ricardo Jimenez
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well
as with the clock at C, then the clocks at B and C also run
synchronously relative to each other.
Yes, that's the point. But Einstein himself did not say this, he just
said we assume this definition is free of contradictions...
Stachel's translation of Einstein's paper actually *does* say that,
talking about "running", etc. A better translation in Einstein's
collected papers says
"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
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."
This is better because Stachel's use of "running" can mislead people
into thinking he is talking about rates.
… is not the official translation, therefore quite irrelevant.
<https://einsteinpapers.press.princeton.edu/vol2-trans/154>
You are confused. The translation I quoted in my message was the one from
the Einstein Collected Papers, and I explained why it is more clear than
Stachel's translation.
I see.
Good. I accept your apology.
Post by Thomas 'PointedEars' Lahn
Jan mistakenly thought Einstein didn't write that (and not just because of
the slight translational variations).
OK.
Good. I accept your apology.
Post by Thomas 'PointedEars' Lahn
Also, it's dumb for you to talk about an "official" translation, in a
sophomoric appeal to authority,
If you had read what you are quoting, you would know that it *is*
There is no such thing as an "official" translation. Sheesh.
Post by Thomas 'PointedEars' Lahn
rather than someone interested in the truth. If you do a straight literal
translation from the original German, you can find justification for
Stachel's choice of the word "run" rather than the Collected Paper's "is".
Einstein used the German "lauft",
LOL, the arrogance again.
You seem to be having a mental breakdown. There is nothing "arrogant" about stating the relevant fact that Einstein used the German word "lauft" (no, I'm not interested in rendering the umlaut in ASCII text... speaking of pomposity), which Stachel translated as "run" whereas Beck translated it as "is"... and Miller avoided it entirely.
Post by Thomas 'PointedEars' Lahn
Here's a simplistic literal translation: […]
LOL. If you had any clue about translations, you would know that such a
translation hardly reflects the original meaning.
Again, you seem to be suffering a severe mental breakdown. The four quoted translations are all virtually identical, differing only in the use of "run" (in two translations) versus "is" (in one translation), and neither (in one translation). For you to claim that one or more of those four nearly identical translations "hardly reflects the original meaning" is simply bizarre.
JanPB
2021-10-16 11:09:37 UTC
Reply
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Post by Townes Olson
Jan mistakenly thought Einstein didn't write that (and not just because of
the slight translational variations).
Yes, my bad. It wasn't the differences in translations, it was my forgetfulness.

--
Jan
Michael Moroney
2021-10-16 14:42:42 UTC
Reply
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Post by Townes Olson
Post by Thomas 'PointedEars' Lahn
LOL, the arrogance again.
You seem to be having a mental breakdown. There is nothing "arrogant" about stating the relevant fact that Einstein used the German word "lauft" (no, I'm not interested in rendering the umlaut in ASCII text... speaking of pomposity),
In German the umlaut changes the pronunciation and often the meaning.
There are some word pairs in German where two words are spelled the same
other than one having an umlaut while the other does not, and they have
completely different and unrelated meanings.

My high school German cannot remember any such word pairs, or whether
the umlauted and un-umlauted letters are considered different letters in
the alphabet or not.

The rest of your post is rather arrogant.
Townes Olson
2021-10-16 16:11:37 UTC
Reply
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In German...
Here is the original German sentence (E):
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.

As you can see, Einstein uses both läuft and laufen. Now, here are five widely available translations of this sentence into English, two of which render as "run/runs":

Translation 1 (G):
If the clock in A runs synchronously with both the clock in B and the clock in C, the clocks in B and C also run synchronously relative to one another.

Translation 2 (PJ):
If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other.

Translation 3 (S):
If the clock at A runs synchronously with the clock at B as well as with the clock at C, then the clocks at B and C also run synchronously relative to each other.

Translation 4 (B):
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.

Translation 5 (M):
If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other.

So, to be clear, which of these translations -- G, PJ, S, B, or M -- do you claim most accurately renders E into English? Mind you, I'm not asking which of them you think more clearly expresses what you think E *should* have said. I'm asking, purely linguistically, which of the translations is a more accurate and literal rendition of E.
The rest of your post is rather arrogant.
Can you explain what motivated you to type that sentence?
carl eto
2021-10-16 17:11:31 UTC
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Let's assume that Lorentz invariant works, then What? Lorentz (1899) uses the invariant to reverse the negative result of Michelson experiment to justify the ether, composed of matter, and, Einstein (1917) uses the reversal of MMX to justify the ether yet vacuum proves the ether does not exist. Also, both Lorentz and Einstein use relativity (coordinate system transformation) to justify Maxwell theory but Maxwell equations are derived using Faraday induction effect that is not luminous. Using mumble jumble mathematical gymnastics cannot be used to justify the wave theory of light and modern theoretical physics (QM, QED, string theory, QFT,......etc. etc. that are based on the gauge since an electromagnetic wave (quantized or not) is based on a wave formed by the motion of an ether, composed of matter, where the physical structure (ether) has precedence before any abstract mathematical formulation.
Odd Bodkin
2021-10-16 20:07:15 UTC
Reply
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Post by Townes Olson
Post by Michael Moroney
The rest of your post is rather arrogant.
Can you explain what motivated you to type that sentence?
I took it as an observation, one that does not require a defense or a
motivation.

You have received that observation from a number of people on this group,
and I suspect you therefore have received it in other contexts as well.

It may be that you are aware of this criticism but intend to do nothing
about it. It may also be that you are keenly aware of yourself having this
trait, but you’d prefer that it not be brought up in interactions. It may
also be that you are unaware of this criticism, having ignored it and
pushed it out of your mind every time it is raised to you.
--
Odd Bodkin -- maker of fine toys, tools, tables
JanPB
2021-10-17 07:11:35 UTC
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The normalization of Schrodinger
Schroedinger.
_Schrödinger_
Yes, that's why I said earlier "when the diacritics are not available".

--
Jan
Thomas 'PointedEars' Lahn
2021-10-17 20:34:13 UTC
Reply
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Post by JanPB
The normalization of Schrodinger
Schroedinger.
_Schrödinger_
Yes, that's why I said earlier "when the diacritics are not available".
Even if you do not have the „Ö“ key (and similar keys) on your keyboard, you
can switch to a different keyboard layout (in software), or use Compose
sequences, character map applications, or a spell checker to input the
umlaut characters (and other non-ASCII characters). If all else fails,
you can copy them from a file, the Web (e.g. Wikipedia), or Usenet.

And there is no *working* PC software that does not support displaying them.

So they are *always* available nowadays.


PointedEars
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)
Michael Moroney
2021-10-17 20:40:07 UTC
Reply
Permalink
On 10/17/2021 4:34 PM, Thomas 'PointedEars' Lahn wrote:
[]

Quick question: In German are letters with umlauts considered to be
completely different letters than plain letters with no umlaut? Would
sorting software and phone books sort "Mütter" to follow (or precede?)
"Mutter" always or are they considered identical? How about "ß"
compared to "ss", how does that work?
Tom Roberts
2021-10-17 21:19:09 UTC
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Post by Michael Moroney
Quick question: In German are letters with umlauts considered to be
completely different letters than plain letters with no umlaut?  Would
sorting software and phone books sort "Mütter" to follow (or precede?)
"Mutter" always or are they considered identical?  How about "ß"
compared to "ss", how does that work?
In my Langenscheidt "Tashenwörterbuch Englisch", purchased in Germany, ö
is alphabetized immediately after o, but when considering the rest of
the word the umlaut is ignored: Mutter, ..., mütterlich, ...,
Mutterliebe. Ditto for other vowels. ß is alphabetized as if it were ss.

Tom Roberts
Michael Moroney
2021-10-17 18:42:24 UTC
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Post by Townes Olson
In German...
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.
As you can see, Einstein uses both läuft and laufen.
Do you understand at all how verbs are conjugated in German to agree
with the subject? English has that concept but just barely. It's the
equivalent of writing "the clock runs" and "the clocks run" in English,
and saying "look, the author uses both 'run' and 'runs'". 'Laufen' is
somewhat an irregular verb in German by using 'läuft' for third person
singular.
See, you do seem to understand the difference between 'run' and 'runs'.
'Laufen' and 'läuft' are German equivalents, although German has more
than two forms.

<snip translations>

Nothing to do with your claim that insisting on using the umlaut in
'läuft' is "pompous". The usual (English) solution to German umlauts is
an additional 'e' after the letter without the umlaut, although I
dislike that.
Post by Townes Olson
The rest of your post is rather arrogant.
Can you explain what motivated you to type that sentence?
Several of your responses to Thomas. I don't have a dog in that fight,
so I'll leave you two to go at it, or not, and I will say no more.
Townes Olson
2021-10-17 20:06:02 UTC
Reply
Permalink
Post by Townes Olson
In German...
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.
As you can see, Einstein uses both läuft and laufen.
Do you understand...
My claim is that the sentence (in German) in Einstein's paper is as quoted above (do you deny this?), and that the sentence contains the words läuft and laufen (do you deny this?), and that -- as supported by verbatim quotes of 5 widely available English translations -- this has led some translators to use the English words "run" and "runs" in their translations (do you deny this?). Lastly, I claim that it is wrong to assert (as you and Thomas do) that any translation using the words "run" or "runs" is "not official" and "therefore quite irrelevant", and "hardly reflects the original meaning".
<snip translations>
Huh? This is the whole point of the discussion. Remember, the OP claimed that Einstein's 1905 paper included the sentence as translated by Stachel, and Jan (incorrectly) told the OP that this sentence was not in Einstein's paper. I posted a message to point out that the sentence (in German, of course) does appear in Einstein's paper, and I presumed that Jan had just forgotten this, and was not quibbling over the translation, although I noted in my very first message that one could quibble over the use of the word run/runs, since it confuses rates with initialization. Jan graciously acknowledged that, indeed, he had simply forgotten that the sentence was in Einstein's paper, and was not quibbling over the translation.

Now, as I said, one *could* challenge the translation, but the use of the verb "run" is arguably not very apt, and indeed some of the English translations avoid using "run". We may prefer these translations for external reasons, but that is different from saying it is strictly wrong to use the word run/runs in an English translation of that sentence (as you and Thomas claim). I posted the original German sentence and 5 widely available English translations, asked both you and Thomas to tell me which of those you think is the most accurate literal translation. Neither of you has answered... and neither of you ever will.
Nothing to do with your claim that insisting on using the umlaut in
'läuft' is "pompous".
You misunderstood. My comment was not a critique of the German language, it was a critique of Thomas' pompous diversion from the subject of how Einstein treated transitivity of simultaneity in his 1905 paper to an idiotic complaint that when noting that Einstein's paper uses the word lauft in a Usenet post I didn't bother to render the umlaut. The English language can be rightly criticized for lacking the resources to adequately express the monumental stupidity and pointlessness of that complaint... which you promptly endorsed.
Post by Townes Olson
The rest of your post is rather arrogant.
Can you explain what motivated you to type that sentence?
Several of your responses to Thomas.
Surely that's just begging the question. Can you point to something (anything) specific? I've summarized my claims above, and asked if you dispute *any* of them. Do you? If so, which ones? If not, then what exactly is your complaint?
Michael Moroney
2021-10-18 01:47:55 UTC
Reply
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Post by Townes Olson
Post by Townes Olson
In German...
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.
As you can see, Einstein uses both läuft and laufen.
Do you understand [at all how verbs are conjugated in German to agree with the subject?]...
My claim is that the sentence (in German) in Einstein's paper is as quoted above (do you deny this?), and that the sentence contains the words läuft and laufen (do you deny this?),
but it does not contain the non-word "lauft". While Thomas Lahn can be
extremely nitpicky, he is correct. If you don't know German, you should
have said so or said "sorry my keyboard doesn't have any 'ä' key" or "I
don't know the significance of those dots" or something similar.
Post by Townes Olson
and that -- as supported by verbatim quotes of 5 widely available English translations -- this has led some translators to use the English words "run" and "runs" in their translations (do you deny this?). Lastly, I claim that it is wrong to assert (as you and Thomas do) that any translation using the words "run" or "runs" is "not official" and "therefore quite irrelevant", and "hardly reflects the original meaning".
I will not comment on whether a translation using "run" or "runs" is
correct or not. Each language has their words and phrases which cannot
be word-for-word translated. And my German is too rusty.
Post by Townes Olson
<snip translations>
Huh? This is the whole point of the discussion.
I merely commented on your snarky reply to Thomas' comment.
Post by Townes Olson
Now, as I said, one *could* challenge the translation, but the use of the verb "run" is arguably not very apt, and indeed some of the English translations avoid using "run". We may prefer these translations for external reasons, but that is different from saying it is strictly wrong to use the word run/runs in an English translation of that sentence (as you and Thomas claim).
I made no such claim.
Post by Townes Olson
I posted the original German sentence and 5 widely available English translations, asked both you and Thomas to tell me which of those you think is the most accurate literal translation. Neither of you has answered... and neither of you ever will.
My high school German isn't good enough for me to make a meaningful
comment as to which translation is the most faithful to the original
German. So I won't.
Post by Townes Olson
Nothing to do with your claim that insisting on using the umlaut in
'läuft' is "pompous".
You misunderstood. My comment was not a critique of the German language, it was a critique of Thomas' pompous diversion from the subject of how Einstein treated transitivity of simultaneity in his 1905 paper to an idiotic complaint that when noting that Einstein's paper uses the word lauft in a Usenet post I didn't bother to render the umlaut. The English language can be rightly criticized for lacking the resources to adequately express the monumental stupidity and pointlessness of that complaint... which you promptly endorsed.
Umlauts are significant in German. They don't have them in order to try
to look cool like rock groups ("Blue Öyster Cult") often do.
Post by Townes Olson
Post by Townes Olson
The rest of your post is rather arrogant.
Can you explain what motivated you to type that sentence?
Several of your responses to Thomas.
Surely that's just begging the question. Can you point to something (anything) specific?
"I accept your apology." (at least 3 times) when Thomas wasn't
apologizing for anything.

"You seem to be having a mental breakdown."

Complaining that pointing out the lack of umlaut was "pomposity" as well
as stating essentially "I'm not going to bother with any umlaut".

"the monumental stupidity and pointlessness of that complaint [about
umlauts]".

Both of you appear to be rather hard headed. I bow out again so that
you two can fight over which translation is best/official or whether
umlauts matter. This may be quite amusing to watch.
Townes Olson
2021-10-18 03:35:20 UTC
Reply
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Post by Townes Olson
Post by Townes Olson
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.
My claim is that the sentence (in German) in Einstein's paper is as quoted above
(do you deny this?)...
Apparently you do not deny that... good.
Post by Townes Olson
and that the sentence contains the words läuft and laufen (do you deny this?),
[No] but it does not contain the non-word "lauft".
Again, Einstein's sentence -- and the words it contains -- are not in dispute, and are available online for anyone to view for themselves, so your maniacal fixation on whether someone referring to those word in a Usenet message includes the umlaut is toweringly stupid. The question at issue is whether John Stachel's translation of that sentence is "wrong", as Thomas claims (and you've endorsed his claim). The only difference in the Stachel translation (and the literal translation) is his use of the word run/runs, which obviously he got from the German words läuft and laufen. Do you deny this?
Post by Townes Olson
and that -- as supported by verbatim quotes of 5 widely available English
translations -- this has led some translators to use the English words "run"
and "runs" in their translations (do you deny this?).
Apparently you do not deny this.... good.
Lastly, I claim that it is wrong to assert (as you and Thomas do) that any translation
using the words "run" or "runs" is "not official" and "therefore quite irrelevant", and
"hardly reflects the original meaning".
I will not comment on whether a translation using "run" or "runs" is correct or not.
Huh? This is the entire subject of the discussion, i.e., the transitivity of synchronization as discussed in Einstein's 1905 paper. If you have no comment on this, then why are you commenting?
I merely commented on your snarky reply to Thomas' comment.
Hold on. My original post in this sub-discussion was nothing but a reply to Jan, pointing out that the sentence quoted by the OP (from a Stachel translation) actually *does* appear in Einstein's paper -- which Jan immediately agreed -- and I also noted that the Stachel translation (quoted by the OP, not by me) differed from the CPAE translation in its use of the word run/runs, which could be what was confusing the OP since it conflates rate with initialization. It might also have been why Jan didn't immediately recognize it. However, I also pointed out that there is some justification for the word run/runs in the actual German of Einstein's paper, so Stachel didn't fabricate it out of whole cloth. My message was completely unobjectionable and correct, as everyone (now) agrees.

At that point your esteemed ally Thomas jumped in to dismiss Stachel's translation as "not official" [whatever that means] and "therefore irrelevant", and to haughtily suggest that I am an ignoramous for not being aware of the "official" CPAE version. I pointed out that, in fact, the version I quoted in my original message *was* the CPAE version, which Thomas would have known if he hadn't just snipped and ignored my message. It quoted that version verbatim and compared and contrasted it with the Stachel version (and subsequently with several others). At this point your ally Thomas conceded that his haughty and supercilious allegations had been unfounded... but weirdly this didn't deter him from launching a completely pointless digression into the world of umlauts. Then you jumped in to endorse Thomas' position and add your insults to his.

In response I tried to direct the discussion back to the actual topic, by quoting verbatim the original German sentence, and then 5 widely available translations, and then asking which of those he and you think is the most literally accurate translation into English of what Einstein actually wrote. Neither he nor you ever responded.... and you never will. You've even admitted that you have no opinion on the matter. All you want to say is that you feel I have not treated your friend Thomas with the dignity and respect he deserves. Sheesh.

This shouldn't be so difficult for you. Even if you are totally ignorant of German, you know how to access Google, right? Just paste the German sentence in and see what you get. Your ally claims any translation that has the word run/runs is self-evidently inaccurate and doesn't come close to conveying the actual meaning. So, after this little exercise, do you still have confidence in his claim that John Stachel's translation was totally inaccurate?
Townes Olson
2021-10-18 12:53:43 UTC
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Post by Townes Olson
Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander.
This shouldn't be so difficult for you. Even if you are totally ignorant of German, you know how to access Google, right? Just paste the German sentence in and see what you get. Your ally claims any translation that has the word run/runs is self-evidently inaccurate and doesn't come close to conveying the actual meaning. So, after this little exercise, do you still have confidence in his claim that John Stachel's translation was totally inaccurate?
Here's another little test that you are fully capable of performing: Enter the CPAE English translation into the Google translator and see what you get in German. Since you are too lazy and disinterested, I'll spoon feed it to you:

"Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron ist, dann sind auch die Uhren in B und C relativ zueinander synchron."

Note that läuft/laufen do not appear, and instead we have "ist", as in "synchron ist" (is synchronous), just as one would expect. Now, one could argue that this is what Einstein *should* have said (and indeed I suggested this in my original post), but it is not what he actually said. He said "synchron läuft" (runs synchronously). So, in view of this, do you still claim (along with your pointy-eared friend) that John Stachel's translation (quoted by the OP) is blatantly inaccurate?
Michael Moroney
2021-10-17 18:59:11 UTC
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Post by Michael Moroney
In German the umlaut changes the pronunciation and often the meaning.
There are some word pairs in German where two words are spelled the same
other than one having an umlaut while the other does not, and they have
completely different and unrelated meanings.
"schon" (already) and "schön" (beautiful)

"Küchen" (kitchens [plural] and "Kuchen" (cake)

"schwül" (humid) and "schwul" (gay).

"Mutter" (mother) and Mütter" (mothers [plural]). A few other nouns form
plurals by adding an umlaut.
Thomas 'PointedEars' Lahn
2021-10-16 23:47:00 UTC
Reply
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Post by Townes Olson
(no, I'm not interested in rendering the umlaut in ASCII text... speaking
of pomposity),
You can add an "e" after the letter instead of the Umlaut. It's commonly
done in German when the diacritics are not available.
It is linguistically wrong, though; „oe“, for example, is NOT equivalent to
„ö“. Not only are there names where the original name did not have an
umlaut character, but pronounced as if it were, for example „Goethe“.
Also, it is more likely then that the word is mispronounced with a glottal
stop between the „o“-sound and the „e“-sound. Finally, there are a few
words where the glottal stop is actually correct; e.g. „Tetraeder” which
must be pronounced very differently from the (non-existing) word „Teträder“
(in fact, „Räder“ [“wheels”] is an existing word, and the reader trained to
read „ae“ is if it were an umlaut would be more inclined to assume that this
would be, or example, a compositum or a misspelling of another existing word
„Treträder“¹). It is just a bad idea.


PointedEars
___________
¹ <https://de.wiktionary.org/wiki/Tretrad>
--
I heard that entropy isn't what it used to be.

(from: WolframAlpha)
JanPB
2021-10-17 07:15:15 UTC
Reply
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Post by Thomas 'PointedEars' Lahn
Post by Townes Olson
(no, I'm not interested in rendering the umlaut in ASCII text... speaking
of pomposity),
You can add an "e" after the letter instead of the Umlaut. It's commonly
done in German when the diacritics are not available.
It is linguistically wrong, though; „oe“, for example, is NOT equivalent to
„ö“.
Yes, but it's used when the diacritics are not available.
Post by Thomas 'PointedEars' Lahn
Not only are there names where the original name did not have an
umlaut character, but pronounced as if it were, for example „Goethe“.
Or "Noether".
Post by Thomas 'PointedEars' Lahn
Also, it is more likely then that the word is mispronounced with a glottal
stop between the „o“-sound and the „e“-sound. Finally, there are a few
words where the glottal stop is actually correct; e.g. „Tetraeder” which
must be pronounced very differently from the (non-existing) word „Teträder“
(in fact, „Räder“ [“wheels”] is an existing word, and the reader trained to
read „ae“ is if it were an umlaut would be more inclined to assume that this
would be, or example, a compositum or a misspelling of another existing word
„Treträder“¹). It is just a bad idea.
Cool examples, thanks. Just came back from Switzerland where the German
acquires a whole new meaning :-)

--
Jan
Thomas 'PointedEars' Lahn
2021-10-16 23:53:42 UTC
Reply
Permalink
Post by Townes Olson
(no, I'm not interested in rendering the umlaut in ASCII text... speaking
of pomposity),
You can add an "e" after the letter instead of the Umlaut. It's commonly
done in German when the diacritics are not available.
It is NOT, and there is no reasonable circumstance in which the umlaut
characters are not available. Not even in Usenet: Windows-1252/ISO-5589-1
exists since time immemorial, and since RFC 5536 (November 2009) a working
newsreader must support Unicode as part of the requirement to support MIME.


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-10-16 23:54:14 UTC
Reply
Permalink
Post by Townes Olson
(no, I'm not interested in rendering the umlaut in ASCII text... speaking
of pomposity),
You can add an "e" after the letter instead of the Umlaut. It's commonly
done in German when the diacritics are not available.
It is NOT, and there is no reasonable circumstance in which the umlaut
characters are not available. Not even in Usenet: Windows-1252/ISO-8859-1
exists since time immemorial, and since RFC 5536 (November 2009) a working
newsreader must support Unicode as part of the requirement to support MIME.


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)
Townes Olson
2021-10-17 00:57:46 UTC
Reply
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The purpose ... is to provide “a careful, accurate translation..."
Here's the original German sentence:
"Wenn die Uhr in A sowohl mit der Uhr in B als auch mit der Uhr in C synchron läuft, so laufen auch die Uhren in B und C synchron relativ zueinander."

As you can see, Einstein uses both läuft and laufen. Now, here are 5 widely available translations of this sentence into English, two of which render as "run/runs":

Translation 1:
"If the clock in A runs synchronously with both the clock in B and the clock in C, the clocks in B and C also run synchronously relative to one another."

Translation 2:
"If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other."

Translation 3:
"If the clock at A runs synchronously with the clock at B as well as with the clock at C, then the clocks at B and C also run synchronously relative to each other."

Translation 4:
"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."

Translation 5:
"If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other."

So, to be clear, which of these translations do you claim most accurately renders the original German into English? Bear in mind, I'm not asking which of them you think more clearly expresses what you think Einstein *should* have said. I'm asking, purely linguistically, which of the translations is a more accurate and literal rendition of the German sentence.
Tom Roberts
2021-10-16 04:57:26 UTC
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Post by Thomas 'PointedEars' Lahn
<https://einsteinpapers.press.princeton.edu/vol2-trans/154>
[That's an English translation of the paper in question.]
Post by Thomas 'PointedEars' Lahn
If you had read what you are quoting, you would know that it *is*
<https://einsteinpapers.press.princeton.edu/vol2-trans/11>
[That's the introduction to the translations of volume 2.
Nowhere does it claim to be "Official".]

This is neither a legal nor a diplomatic document. There is no such
thing as an "official translation", because there is no organization to
grant such status. The only person or organization that could possibly
confer such status on a particular translation is Einstein himself, and
he did not do so. There are multiple translations of this paper, of
varying faithfulness to the original, and varying conformance to
standard English style for physics papers; different translations are
favored by different people.

The only version of this paper that has any claim to being "official" is
the original German. After all, that is what Einstein actually wrote.
Even so, "official" is the wrong word -- "original" is much more
appropriate.

Tom Roberts
Tom Roberts
2021-10-17 15:38:31 UTC
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Post by Thomas 'PointedEars' Lahn
If you had read what you are quoting, you would know that it
<https://einsteinpapers.press.princeton.edu/vol2-trans/11>
[That's the introduction to the translations of volume 2. Nowhere
does it claim to be "Official".]
Yet another person who can’t read.
No. YOU are reading more into this than is actually there. I repeat:
Nowhere does it claim to be "official". They say they are new
translations, not "official" ones. I suppose they could have used that
word, but that would strike scientists' ears as rather strange and
overly officious.
<https://einsteinpapers.press.princeton.edu/about>
There's nothing "official" there -- YOU are reading more into it than is
actually there.

Yes, Princeton University performed a large and serious project to
translate Einstein's papers into English, and it was supported by the
National Science Foundation (an agency of the U.S. government). But
there is nothing "official" about it, because as I said, there is no
person or organization who could confer such status on any particular
translation. The only person who could plausibly do that is Einstein
himself, and he did not do so.

Your German sensibilities may wish for an "official version", but in
this case that is a forlorn hope. Science does not work that way.

[AFAICT the only documents that have "official
translations" are legal and diplomatic documents.
That status is conferred by the original authors.]

Tom Roberts
Thomas 'PointedEars' Lahn
2021-10-15 23:04:50 UTC
Reply
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Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
I don't see any mention of transitivity in either Einstein or Alan
Macdonald's paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
This is a rewording of what Einstein writes in 1905 paper.
Which apparently you have *still* not read. Hopeless.

The triangles clearly exist merely in your fantasy.
Post by Ricardo Jimenez
Macdonald: pg 2 - Clock synchronization etc.
We now show that (ii) is a necessary and sufficient condition that
clocks at nodes P and Q are synchronized with each other once they are
both synchronized with the clock at O.
No triangles either there.
Post by Ricardo Jimenez
They didn't use the word but that is what they were talking about.
It is NOT what you claimed they would be talking about.


PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-10-15 23:05:13 UTC
Reply
Permalink
Post by Ricardo Jimenez
Post by JanPB
Post by Ricardo Jimenez
He claims to address transitivity on pgs 2 and 3 with the assumption
going around the triangle in either direction yields the same result.
And he references a laser experiment for justification of that
assumption. Is he wrong? Einstein does state transitivity holds but
doesn't offer a proof.
I don't see any mention of transitivity in either Einstein or Alan
Macdonald's paper you've referenced above.
Einstein: pg 127 in Stachel ed - Einstein's Miraculous Year
2. If the clock at A runs synchronously with the clock at B as well as
with the clock at C, then the clocks at B and C also run synchronously
relative to each other.
This is a rewording of what Einstein writes in 1905 paper.
Which apparently you have *still* not read. Hopeless.

The triangles clearly exist merely in your fantasy.
Post by Ricardo Jimenez
Macdonald: pg 2 - Clock synchronization etc.
We now show that (ii) is a necessary and sufficient condition that
clocks at nodes P and Q are synchronized with each other once they are
both synchronized with the clock at O.
No triangles either there.
Post by Ricardo Jimenez
They didn't use the word but that is what they were talking about.
It is NOT what you claimed they would be talking about.


PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)
Maciej Wozniak
2021-10-15 05:00:45 UTC
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Post by Ricardo Jimenez
I am at first, taken aback, by the use of Einstein's definition of
two clocks being synchronized since it allows one to add any amount
of time to either clock's setting and they will still be synchronized
according to it.
I have no idea where you got this notion. It is WRONG.
Note that Einstein described several different techniques for
synchronizing two clocks, in different papers over many years.
And note that in the meantime in the real world,
"impossibly" synchronized GPS clocks keep
measuring t'=t, just like all serious clocks always
did.
Thomas 'PointedEars' Lahn
2021-10-15 23:25:09 UTC
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Post by Maciej Wozniak
And note that in the meantime in the real world,
"impossibly" synchronized GPS clocks keep
measuring t'=t, just like all serious clocks always
did.
“[…] The SV [space vehicle; the ed.] carrier frequency and clock rates –
as they would appear to an observer located in the SV – are offset to
compensate for relativistic effects. […]”

GPS Interface Specification (2021). “3.3.1.1 Frequency Plan.”
<https://www.gps.gov/technical/icwg/IS-GPS-200M.pdf>

Linked from <https://www.gps.gov/technical/icwg/>.


Which part of “offset” and “compensate” did you not get?


PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)
Maciej Wozniak
2021-10-14 17:17:37 UTC
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Post by JanPB
Post by Ricardo Jimenez
AFAIK, the standard procedure to sync all clocks in a frame with the
one at the space origin is t(p) = t(0) + |p|/c where p is an arbitrary
point in space. If you perform a translation of space, how do you
show all clocks are synchonized with the one at the new origin? If
you try to evaluate t(p + q) where q is the translation vector you get
t(p + q) = t(q) + (|p + q| - |q|)/c which doesn't seem to do the
trick. What am I missing?
This is actually not Einstein's sync process exactly.
The sync process is tB - tA = tA' - tB, where tA = clock reading at the
light pulse emission site, tB = clock reading at the mirror, tA' = clock
reading again at the emission site after the light ray returns.
It is not immediately obvious that this process is free of contradiction.
Einstein in his 1905 paper simply says we assume that it is - that's
because it's easy to demonstrate and he was writing for professional
physicists.
Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive
No, it doesn't.
Thomas 'PointedEars' Lahn
2021-10-15 23:25:54 UTC
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Post by Maciej Wozniak
Post by JanPB
Anyway, to prove that there is no contradiction there, it suffices to
demonstrate that the "sync relation" between clocks is an equivalence
relation, i.e., reflexive, symmetric, and transitive
No, it doesn't.
How eloquent :->


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)
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