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?

On Monday, 11 October 2021 at 19:09:31 UTC+2, Ricardo Jimenez wrote:
> 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.

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.

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

Ricardo Jimenez wrote:

> 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”.

> 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).

> What am I missing?

The (Galilean) principle of relativity.

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

(from: WolframAlpha)

On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn wrote:

> 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.

On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
<maluwozniak@gmail.com> wrote:

>On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn wrote:
>
>> 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?

On Tuesday, 12 October 2021 at 17:42:47 UTC+2, Ricardo Jimenez wrote:
> On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
> <maluw...@gmail.com> wrote:
>
> >On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn wrote:
> >
> >> 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.

On Tue, 12 Oct 2021 09:15:16 -0700 (PDT), Maciej Wozniak
<maluwozniak@gmail.com> wrote:

>On Tuesday, 12 October 2021 at 17:42:47 UTC+2, Ricardo Jimenez wrote:
>> On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
>> <maluw...@gmail.com> wrote:
>>
>> >On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn wrote:
>> >
>> >> 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.

On Tuesday, 12 October 2021 at 18:26:49 UTC+2, Ricardo Jimenez wrote:
> On Tue, 12 Oct 2021 09:15:16 -0700 (PDT), Maciej Wozniak
> <maluw...@gmail.com> wrote:
>
> >On Tuesday, 12 October 2021 at 17:42:47 UTC+2, Ricardo Jimenez wrote:
> >> On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
> >> <maluw...@gmail.com> wrote:
> >>
> >> >On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn wrote:
> >> >
> >> >> 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.

Ricardo Jimenez wrote:

> On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
> <maluwozniak@gmail.com> wrote:
>> On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn
>> wrote:
>>> 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.
> 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.

> 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).

> 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?

> 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)

Ricardo Jimenez wrote:

> On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
> <maluwozniak@gmail.com> wrote:
>> On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn
>> wrote:
>>> 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.
> 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.

> 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).

> 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?

> 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)

On Tue, 12 Oct 2021 20:13:22 +0200, Thomas 'PointedEars' Lahn
<PointedEars@web.de> wrote:

>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?

Ricardo Jimenez wrote:

> On Tue, 12 Oct 2021 20:13:22 +0200, Thomas 'PointedEars' Lahn
> <PointedEars@web.de> wrote:
>>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.

> 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.

> unintelligible word salad

It was not word salad.

> 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.

> 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.

> 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

On Tuesday, 12 October 2021 at 20:13:25 UTC+2, Thomas 'PointedEars' Lahn wrote:
> Ricardo Jimenez wrote:
>
> > On Mon, 11 Oct 2021 21:43:39 -0700 (PDT), Maciej Wozniak
> > <maluw...@gmail.com> wrote:
> >> On Tuesday, 12 October 2021 at 01:22:32 UTC+2, Thomas 'PointedEars' Lahn
> >> wrote:
> >>> 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.

On Tuesday, 12 October 2021 at 20:43:42 UTC+2, Thomas 'PointedEars' Lahn wrote:

> 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.

On Monday, 11 October 2021 at 19:09:31 UTC+2, Ricardo Jimenez wrote:
> 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.

> 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

On Tuesday, 12 October 2021 at 21:58:39 UTC+2, ju...@diegidio.name wrote:

> 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.

On Tue, 12 Oct 2021 11:49:01 -0700 (PDT), "mitchr...@gmail.com"
<mitchrae3323@gmail.com> wrote:

>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?

On Tue, 12 Oct 2021 12:58:38 -0700 (PDT), Julio Di Egidio
<julio@diegidio.name> wrote:

>On Monday, 11 October 2021 at 19:09:31 UTC+2, Ricardo Jimenez wrote:
>> 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.
>
>> 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.

On Tuesday, 12 October 2021 at 22:46:57 UTC+2, Ricardo Jimenez wrote:
> On Tue, 12 Oct 2021 12:58:38 -0700 (PDT), Julio Di Egidio wrote:
> >On Monday, 11 October 2021 at 19:09:31 UTC+2, Ricardo Jimenez wrote:
> >> 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.
> >
> > > 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.

> 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.

> 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

On 10/12/2021 4:46 PM, Ricardo Jimenez wrote:

> 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.

Ricardo Jimenez wrote:

> On Tue, 12 Oct 2021 11:49:01 -0700 (PDT), "mitchr...@gmail.com"
> <mitchrae3323@gmail.com> wrote:
>>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.)

> 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*

> 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.]

> 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.

> 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

> 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)

On Tuesday, October 12, 2021 at 1:30:54 PM UTC-7, Ricardo Jimenez wrote:
> 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.

> What is implied by the two equations he gives...

The paper contains far more than two equations. You need to specify.

> ... 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.

> 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.

> 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.

> 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.

> 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.

> 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.

> However, Einstein claims that clock synchronization is reflexive, symmetric and transitive.

Right, the synchronization described above has all those properties.

> 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.

On Tue, 12 Oct 2021 15:49:50 -0700 (PDT), Julio Di Egidio
<julio@diegidio.name> wrote:

>On Tuesday, 12 October 2021 at 22:46:57 UTC+2, Ricardo Jimenez wrote:
>> On Tue, 12 Oct 2021 12:58:38 -0700 (PDT), Julio Di Egidio wrote:
>> >On Monday, 11 October 2021 at 19:09:31 UTC+2, Ricardo Jimenez wrote:
>> >> 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.
>> >
>> > > 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.
>
>> 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.
>
>> 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.