The synchronization of clocks and the relativity of motion

[AlMetis]

In his thought experiment on the relativity of simultaneity, Einstein synchronizes two clocks A and B (at either end of a rod) to the clock C in a stationary frame, while A and B (the rod) is moving relative to C. The equations Einstein finishes with demonstrate how observers moving with A and B, using the same method of synchronization between A and B will find A and B are not synchronized to each other, while an observer in C will find they are.

It follows from the same reasoning that when A and B are synchronized to each other and C, while at rest with C, setting A and B moving relative to C will similarly break the synchronization between all.

The principle of relativity stipulates the motion of A and B (the rod) relative to C is not a property of either AB or C, but a measure of the kinematics of each by the other. The reciprocity of these kinematics is a fundamental premise of the principle.

Once the synchronicity of A, B and C are broken by their relative motion, we can set C in the same motion as A and B setting all back at rest again, which will, according to the premise of the original experiment, set A and B and C synchronous once again.

We know this is not the case, and we know Einstein knew this was not the case when he wrote the equations accounting for the motion of the rod AB and attributing the lack of synchronization to this motion when A and B use the identical method that had been used at rest.

So it is not the motion of the rod AB relative to the clock C in the “stationary” system that breaks the synchronicity.

What is the motion relative to that it also prevents the synchronicity being regained by the same state that set it in the first place?

 

[PeterDonis]

In his thought experiment on the relativity of simultaneity, Einstein synchronizes two clocks A and B (at either end of a rod) to the clock C in a stationary frame, while A and B (the rod) is moving relative to C.

Which thought experiment are you referring to? Please give a reference.

 

[AlMetis]

Which thought experiment are you referring to? Please give a reference.

On The Electrodynamics of Moving Bodies
Thought experiment 2.
§ 2. On the Relativity of Lengths and Times

 

[PeterDonis]

On The Electrodynamics of Moving Bodies
Thought experiment 2.
§ 2. On the Relativity of Lengths and Times

Ok. I don't see where this source says any of the following:

Einstein synchronizes two clocks A and B (at either end of a rod) to the clock C in a stationary frame, while A and B (the rod) is moving relative to C.

The equations Einstein finishes with demonstrate how observers moving with A and B, using the same method of synchronization between A and B will find A and B are not synchronized to each other, while an observer in C will find they are.

Once the synchronicity of A, B and C are broken by their relative motion, we can set C in the same motion as A and B setting all back at rest again, which will, according to the premise of the original experiment, set A and B and C synchronous once again.

We know this is not the case, and we know Einstein knew this was not the case when he wrote the equations accounting for the motion of the rod AB and attributing the lack of synchronization to this motion when A and B use the identical method that had been used at rest.

 

[Sagittarius A-Star]

§ 2. On the Relativity of Lengths and Times

Be aware, that this chapter §2 from the 1905 paper contains a bad explanation of the relativity of simultaneity, because it uses "clocks" at A and B, that must tick faster than their proper time, to be synchronous with the "stationary" frame's coordinate-time.

A thought experiment with better didactic is used in Einstein's popular book from 1916, Section 9 - The Relativity of Simultaneity (train & embankment).
https://en.wikisource.org/wiki/Rela..._I#Section_9_-_The_Relativity_of_Simultaneity

 

[Dale]

Once the synchronicity of A, B and C are broken by their relative motion, we can set C in the same motion as A and B setting all back at rest again, which will, according to the premise of the original experiment, set A and B and C synchronous once again.

I have no idea what you are trying to say here. Synchronization isn’t something physical that happens to a clock or doesn’t. It is a convention. And different inertial frames disagree about the convention.

 

[vanhees71]

Indeed clock synchronization is a convention, but this convention is realized in the real world "all the time". It's, e.g., crucial to be able to have a very accurate positioning system like the GPS, where everything hinges on very precise and very precisely synchronized clocks, taking into account the full general-relativistic spacetime model! Relativity is not a mere play with mathematical concepts but it's real-world physics (and applied in engineering) too!

 

[Ibix]

What is the motion relative to that it also prevents the synchronicity being regained by the same state that set it in the first place?

There are two aspects to clock synchronisation. One is checking that the clocks run at the same rate and the other is checking that the clocks show the same time. The rate depends only on the current state of motion. In that case if you accelerate A and B and then decelerate them again so that they are once again at rest with respect to C then all the clocks will tick at the same rate.

However, whether the clocks show the same time depends on both how they were "zeroed" and the history of their states of motion since then. That's why A and B won't generally be synchronised after they are returned to rest - they remember their different histories. So they are not really in the same state they were in before they were disturbed.

 

[jbriggs444]

However, whether the clocks show the same time depends on both how they were "zeroed" and the history of their states of motion since then.

In addition to starting the two clocks "at the same time" using some process, we must worry about how the two clocks are stopped "at the same time". We must specify the process by which the clocks are stopped or compared.

i.e. whether two [remote] clocks show the same time at the same time depends both on the clocks and on the definition of "at the same time".

 

[AlMetis]

Clock synchronization and the persistence of it is a test of physical law. We can choose a convention just as we might choose for a ruler marked out in inches, or centimeters. But once chosen, what is one inch in our measure of the “physical” world does not change to one centimeter when we change ruler conventions.

When the “moving” observers use Einstein synchronization to measure the time of the two clocks A and B at either end of the rod that is in motion relative to the stationary system, they find the clocks are not synchronized “to each other”. They were synchronized to each other when at rest relative to the stationary system.

The principle of relativity lets us measure the motion of the stationary system and the motion of the rod. The laws are upheld in both frames as neither is a privileged frame with respect to the laws.

If I say the rod is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, one could reason the acceleration of the rod was the cause. This is the way out of the Twin paradox, but in this case we are measuring the “current” kinematics, not reasoning the history of frame changes.

If I say the stationary frame is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, the acceleration of the stationary frame is not a “physical” cause for the clocks at A and B to loose their synchronicity.

 

[Ibix]

If I say the stationary frame is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, the acceleration of the stationary frame is not a “physical” cause for the clocks at A and B to loose their synchronicity.

...and they would not do so. You can "set a frame in motion" all you want (although that only makes a restricted kind of sense, which may be your problem) but it will not change the synchronisation or otherwise of clocks. It might change your definition of what it means for clocks to be synchronised, but that is not a physical change, just you changing your mind.

 

[Sagittarius A-Star]

If I say the rod is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, one could reason the acceleration of the rod was the cause. This is the way out of the Twin paradox, but in this case we are measuring the “current” kinematics, not reasoning the history of frame changes.

If I say the stationary frame is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, the acceleration of the stationary frame is not a “physical” cause for the clocks at A and B to loose their synchronicity.

You must distinguish between proper acceleration and coordinate-acceleration. If you assume the "stationary" frame to be inertial, then you are speaking about a proper acceleration of the rod. In this case, you have in the accelerated rest-frame of the rod a pseudo-gravity and therefore also gravitational time-dilation. The clock at the front of the accelerated rod ticks faster than the clock at the rear-end of the rod, with reference to the accelerated frame.

 

[Nugatory]

This is the way out of the Twin paradox....

More like a dead end.... In the most common form of the paradox the time drift is independent of the amount of acceleration, and indeed it is possible to construct zero-acceleration variants of the paradox.

There must be some asymmetry in the histories of the two twins, and acceleration is often part of that asymmetry, but the idea that acceleration resolves the paradox is one of those things that has to be unlearned before one can properly understand relativity.

 

[Dale]

They were synchronized to each other when at rest relative to the stationary system.

This is not part of Einstein’s scenario. It is also not something that is either implied by or inferred from his specified scenario.

If I say the rod is set in motion, and the observers moving with the rod find the clocks are no longer synchronous, one could reason the acceleration of the rod was the cause

One could indeed reason this but such reasoning would be fallacious. First, the rod is described as having accelerated, but not the clocks. The clocks were never described as ever being at rest in the “stationary” system. Consequently they were also never described as being accelerated. All we are told is that:

1) they are placed at each end of the moving rod
2) they show time in the “stationary” frame

Taking standard clocks, attaching them to the rod before acceleration, synchronizing them, and then accelerating the rod with the attached clocks, would not lead to 2). So we cannot infer from 2) that they were accelerated.

Forget acceleration. It is not relevant for this scenario. The rod was accelerated in his description, but it was unnecessary. The clocks were never described as accelerated, and assuming acceleration does not produce the result he specified.

In order to synchronize the moving clocks in the stationary frame they must adjust their tick rate so that their display indicates more time than their actual proper time and (once already moving inertially and in position) they must adjust their display to match some “stationary” clock as it passes the stationary clock.

 

[Janus]

Be aware, that this chapter §2 from the 1905 paper contains a bad explanation of the relativity of simultaneity, because it uses "clocks" at A and B, that must tick faster than their proper time, to be synchronous with the "stationary" frame's coordinate-time.

A thought experiment with better didactic is used in Einstein's popular book from 1916, Section 9 - The Relativity of Simultaneity (train & embankment).
https://en.wikisource.org/wiki/Rela..._I#Section_9_-_The_Relativity_of_Simultaneity

So the thought experiment starts off assuming something like this:

But when you switch to the the frame in which the top row of clocks are at rest, you get this:

 

[PeterDonis]

1) they are placed at each end of the moving rod

We can't even say that. All we can say is that clock #1 is arranged to be co-located with one end of the rod when its clock reading is T, and clock #2 is arranged to be co-located with the other end of the rod when its clock reading is T. This justifies the statement that these two events occur at the same time T with respect to the stationary frame, so that the difference in their spatial coordinates in the stationary frame gives the length of the moving rod in the stationary frame.

But it is never asserted or implied that the two clocks are co-located with the two ends of the rod at any other events besides the ones described above.

 

[Sagittarius A-Star]

But it is never asserted or implied that the two clocks are co-located with the two ends of the rod at any other events besides the ones described above.

I think it is asserted. The German text says:

Wir denken uns ferner an den beiden Stabenden (A und B) Uhren angebracht, welche mit den Uhren des ruhenden Systems synchron sind, d. h. deren Angaben jeweilen der ,,Zeit des ruhenden Systems" an den Orten, an welchen sie sich gerade befinden, entsprechen; diese Uhren sind also ,,synchron im ruhenden System".

Wir denken uns ferner, daß sich bei jeder Uhr ein mit ihr bewegter Beobachter befinde, und daß diese Beobachter auf die beiden Uhren das im § 1 aufgestellte Kriterium für den synchronen Gang zweier Uhren anwenden.

Source (see PDF download link):
https://onlinelibrary.wiley.com/doi/10.1002/andp.19053221004

Translation by me:
"We also think of clocks attached to the two rod ends (A and B) which are synchronous with the clocks of the system at rest, that means, whose readings correspond occasionally to the "time of the system at rest" at the places where they just are. These clocks are therefore "synchronous in the system at rest".
We also imagine that at every clock there is an observer, moving with it, and that these observers apply to the two clocks the criterion set out in § 1 for the synchronous operation of two clocks."

Automatic translation:
https://www.deepl.com/de/translator...für den synchronen Gang zweier Uhren anwenden

 

[AlMetis]

These thought experiments have nothing to do with acceleration.
Forget I said “set in motion”. I should have been more clear and said “consider the other frame moving”.
This is about the relativity of motion and simultaneity in the "kinematics" of light.

Einstein defines his method of synchronization using the times of flight of a ray of light from A to B, and from B back to A, in order to find a common time for A and B that accounts for the time of flight between them.

“In accordance with definition the two clocks synchronize if t B − t A = t ′A − t B”

Assuming the one-way time of flight, as Einstein does in the original paper, his synchronization method finds a difference between the two legs when the rod is moving relative to the stationary system. The difference is a measure made by the observers moving with the rod and the difference is explicitly attributed to the motion of the rod relative to the stationary system. (Please see his concluding equations for confirmation, I have not learned LaTex yet.)

Is the rod moving or is the stationary system moving? (keep in mind “stationary” is the name he assigns to the system, not its state of motion)

Assuming the two-way time of flight is used, there is no difference found between the two legs, which means the clocks at A and B are synchronous, which contradicts Einstein’s equations and principle of the relativity of simultaneity.

We know the equations are empirically true, we know the principle works, so how is the principle of relativity ignored in order to find a difference in the times of flight from A to B and B to A, by claiming it is “explicitly” the motion of AB relative to the stationary system?Again, if I invoke the principle of relativity as the premise of the theory allows, indeed demands, AB is at rest, the stationary system is in motion, and the difference in the times of flight is attributed to the motion of the stationary system. A system that has nothing to do with the kinematics of light observed at and between A and B.

 

[Dale]

Forget I said “set in motion”.

Sure. But then your whole argument falls apart even more, since it was explicitly based on the idea that “the acceleration of the rod was the cause”

how is the principle of relativity ignored in order to find a difference in the times of flight from A to B and B to A, by claiming it is “explicitly” the motion of AB relative to the stationary system?

It isn’t ignored. The principle of relativity says that the laws of physics are symmetric. It does not require that boundary conditions or equipment setup be symmetric. The claim that the principle of relativity is ignored in this case is simply an incorrect claim.

At this point, if you wish to continue making the claim that the thought experiment in section 2 violates the principle of relativity then you must find a professional scientific reference that makes and supports this claim. PF is not the place to push such unsupported personal speculation

 

[Ibix]

The difference is a measure made by the observers moving with the rod and the difference is explicitly attributed to the motion of the rod relative to the stationary system.

Let me correct that for you: The difference is a measure made by the observers moving with the rod and using clocks synchronised in the stationary system, not in the rod's frame, and the difference is explicitly attributed to the motion of the rod relative to the stationary system.

As noted in the thread linked by @Sagittarius A-Star in #5, Einstein's explanation is not at all clear. Indeed, it implicitly involves clocks that are deliberately designed to tick at a wrong rate, since they move with the rod but keep time in the stationary system, but this is not noted in the text. Later presentations are much better. For example, a simpler approach would have been to equip the stationary system with a line of clocks and have the rod-riding observers get the pulse timings by checking the clock they are then passing.

This is not a failure of the principle of relativity. This is an unnecessarily complicated experimental setup that disguises the fact that it's using measures from a frame that is not the observer's rest frame.

 

[AlMetis]

Let me correct that for you: The difference is a measure made by the observers moving with the rod and using clocks synchronised in the stationary system, not in the rod's frame, and the difference is explicitly attributed to the motion of the rod relative to the stationary system.

The rod is already in motion relative to the stationary system when the observers at A and B apply the synchronization method outlined in “§ 1. Definition of Simultaneity” to the clocks at A and B.
In the process they find the one-way times of flight of a light ray between the ends of the rigid rod differs with the direction of flight (AB and BA) parallel to the motion of AB relative to the stationary system.The rod, the clocks A and B at each end of the rod, and the observers at each of the rod are all in the same rest frame.
Given speed of light is a “physical” constant, why do the observers measure a difference between the times of flight AB and BA?

In the manuscript referenced below, the previous synchronization of the clocks at A and B were with the clocks in the stationary system not with each other.
The point of that synchronization is to show the expectation of the transitive synchronization between clocks at rest (A=C and B=C therefore A=B), does not hold when A and B are in motion relative to C.https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

 

[Ibix]

The rod is already in motion relative to the stationary system when the observers at A and B apply the synchronization method outlined in “§ 1. Definition of Simultaneity” to the clocks at A and B.

Quoting from the translation linked in #5: By means of stationary clocks set up in the stationary system and synchronizing in accordance with § 1 (emphasis mine). These clocks are not Einstein synchronised in the rod's rest frame. If they were, no time difference would be detected.

I repeat - the description in this paper is over complicated and confusing, and this is what is causing you trouble. Much better explanations are available. Why not learn from them, and then come back and read Einstein's paper for historcal interest if you wish?

 

[Sagittarius A-Star]

The rod, the clocks A and B at each end of the rod, and the observers at each of the rod are all in the same rest frame.
Given speed of light is a “physical” constant, why do the observers measure a difference between the times of flight AB and BA?

They don't measure in their rest frame a difference between the times of flight AB and BA. They measure instead, that the two clocks are out-of-sync with reference to their rest frame, based on the definition, cited below.

We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A.

Source:
https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

The one-way-speed of light is generally not a “physical” constant. However, it is usually defined to be equal to the “physical” constant $c$. The two-way-speed of light is a “physical” constant.

 

[Dale]

The rod is already in motion relative to the stationary system when the observers at A and B apply the synchronization method outlined in “§ 1. Definition of Simultaneity” to the clocks at A and B.

The observers at A and B in section 2 do not use the method in section 1 to synchronize their clocks. They method they use to synchronize their clocks in section 2 is “their indications correspond at any instant to the ‘time of the stationary system’ at the places where they happen to be”.

The clocks at rest in the “stationary” system are synchronized using the method of section 1. The clocks A and B are synchronized by always matching the nearby “stationary” clock. It is shown that in the rod’s frame those A and B clocks are not synchronized by the convention in section 1. The conclusion being that the synchronization procedure in section 1 is frame-dependent. Clocks that are section 1 synchronized in the “stationary” frame are not section 1 synchronized in the “moving” frame. This is the relativity of simultaneity.

Given speed of light is a “physical” constant

The two way speed of light is a physical constant. The one way speed of light is a convention, specifically it is the synchronization convention.

why do the observers measure a difference between the times of flight AB and BA?

Because the clocks are not Einstein synchronized in that frame.

 

[Sagittarius A-Star]

In the manuscript referenced below, the previous synchronization of the clocks at A and B were with the clocks in the stationary system not with each other.

That's wrong. In the manuscript, the "clocks" at A and B are also in sync with each other - with reference to the "stationary frame".

I propose, that you learn this topic from chapter 1.3.1 of Morin's relativity book, according to the proposal of @Ibix in posting #22. The chapter 1 can be downloaded here:
https://scholar.harvard.edu/david-morin/special-relativity

 

[Motore]

I see you didn't take @Dale 's advice from the previous thread and I fear the conclusion won't be much different:

Dale said:

Take that time to formulate the question well and do not post a new thread until the beginning of next month.

When you do, please label all important events, worldlines, measures, and frames. Do not refer to any frame variant quantity (speed, distance, time) without explicitly mentioning the frame. And use LaTeX to write symbols.

https://www.physicsforums.com/threads/relativity-of-measures.1049680/post-6858890

 

[Ibix]

I see you didn't take @Dale 's advice from the previous thread and I fear the conclusion won't be much different:

To be fair to the OP, the problem here is that Einstein wasn't particularly clear about what was associated with what frame. Of course, you are correct that if OP wrote down what measures he thought were associated with which frame, he'd soon realise that "the clocks are synchronised in the rod rest frame" is inconsistent with "they show a difference in travel times" and hopefully realise for himself that he's misunderstood the (admittedly unclear) presentation.

 

[Vanadium 50]

To be fair to the OP, the problem here is that Einstein wasn't particularly clear

To be fair to Einstein, he was writing to an audience of people who with PhDs in physics, and got them in the second half of the 19th century. He's not the best starting point today. (There's also the fact that the world has had over a century to think about Einstein and to come up with clearer explanations)

 

[AlMetis]

That's wrong. In the manuscript, the "clocks" at A and B are also in sync with each other - with reference to the "stationary frame".

I propose, that you learn this topic from chapter 1.3.1 of Morin's relativity book, according to the proposal of @Ibix in posting #22. The chapter 1 can be downloaded here:
https://scholar.harvard.edu/david-morin/special-relativity

The manuscript says:
“…that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are therefore “synchronous in the stationary system.” (my emphasis)

In a less convoluted fashion, he is saying the clock at A is synchronized to the time of the stationary system and the clock at B is synchronized to the time of the stationary system, thus as far as any Newtonian in the stationary system is concerned, clock A and clock B are synchronized to each other., but they are not. This is what Newonian time (absolute) says is the physical, transitive property of time.
If we call the stationary system time C, then according to Newtonian time: If A=C and B=C then A=B
Einstein is showing that this is a false assumption. (when time is not absolute, but empirical evidence limited to c) He gets the observers at A and B to check this assumption via c:
“these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks.“ at which point they realize the clocks are not already synchronized.

I took your advice and read the section 1.3.1 of David Morin’s book.
(Special Relativity, For the Enthusiastic Beginner Copyright 2016, David Morin)

I have stated and illustrated my point in context of David’s thought experiment.
David begins with an assumption
“In A’s reference frame, the light hits the two receivers at the same time, “ This is an undefined measure, thus he is making an assertion/assumption based on the two-way convention that does not determine the one-way time of flight, it determines the one-way speed of light.

“If you want, you can think of A as being on a train, and B as standing on the ground. “
This is the model I have used in the diagrams below.
In order to demonstrate the one-way and two-way time of flight I have replaced David's detectors with mirrors at C and D.

David continues:
“With respect to B, the receivers (along with everything else in A’s frame) move to the right with speed v.”
Likewise, and as Dale said above, “The principle of relativity says that the laws of physics are symmetric. It does not require that boundary conditions or equipment setup be symmetric.”
To this end, (and rather than cluttering the diagram), when the “Platform” observer is moving relative to the “Train” observer, so too is everything that is not the Train and its contents.
I assume this satisfies the symmetry of the kinematics as per Dale’s comment.
Note the one-way Time of flights are not equal in Dia. 1, they are in Dia. 2
The reciprocal symmetry is in the two-way convention not the physical reality?

 

[Ibix]

The reciprocal symmetry is in the two-way convention not the physical reality?

The reciprocal symmetry is that you can do the same experiment starting from "In B’s reference frame, the light hits the two receivers at the same time" instead of A, as Morin has it. This is a different experiment (because "the same time" according to A and B are not the same), but apart from interchanging A/B and Train/Embankment the description and conclusions will be identical.

 

[Sagittarius A-Star]

In a less convoluted fashion, he is saying the clock at A is synchronized to the time of the stationary system and the clock at B is synchronized to the time of the stationary system, thus as far as any Newtonian in the stationary system is concerned, clock A and clock B are synchronized to each other., but they are not.
...
“these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks.“ at which point they realize the clocks are not already synchronized.

Such a statement about a frame-dependent quantity (time-offset of the "clocks") is meaningless if you don't mention the reference frame for it. See the related reminder of @Motore in #26.

The experiment of the observers at A and B, moving with the rod, does not rule-out, that "clock" A and "clock" B are synchronous to each other with reference to the "stationary frame".

 

[Dale]

as far as any Newtonian in the stationary system is concerned, clock A and clock B are synchronized to each other., but they are not

This is a mistake. They are indeed synchronized in the stationary system.

This is what Newonian time (absolute) says is the physical, transitive property of time.
If we call the stationary system time C, then according to Newtonian time: If A=C and B=C then A=B
Einstein is showing that this is a false assumption.

This is also a mistake. Synchronization is still transitive. What it is not, is frame-invariant. In a given reference frame if A is simultaneous with B and B is simultaneous with C then A is simultaneous with C. It is transitive.

If A and B are simultaneous in one frame then they may not be simultaneous in another frame. But that is invariance, not transitivity.

The reciprocal symmetry is in the two-way convention not the physical reality?

What are you asking here?

 

[AlMetis]

the light hits the two receivers at the same time" instead of A, as Morin has it.

Diagram 1.0 shows that does not happen.

The two-way convention calculates the one-way as a mean of each leg which sets the one-way time of flight the same for both observers. But as is shown in diagram 1.0 the one way time is not the same for both observers.

 

[AlMetis]

Such a statement about a frame-dependent quantity (time-offset of the "clocks") is meaningless if you don't mention the reference frame for it. See the related reminder of @Motore in #26.

You are asking me to define what Einstein did not.
I thought the context was understood as it comes from the same manuscript we have been talking about.
As one-way measures, there is no frame against which the equations are empirically true, it is a deduction from the constancy of light and the velocity of the rod. I suspect Einstein’s use of “we find that” was his way of saying it is a deduction.

“Taking into consideration the principle of the constancy of the velocity of light we find that

where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous. (my emphasis)

 

[AlMetis]

The experiment of the observers at A and B, moving with the rod, does not rule-out, that "clock" A and "clock" B are synchronous to each other with reference to the "stationary frame".

No it does not, but to get to the relativity of simultaneity, Einstein has to point out the synchronicity is not an “absolute”.