Einstein clock syncing with one way light emission absorber

[Alfredo Tifi]

Two spaceships with their engines shut off and identical radio receiver-amplifier-reemitting devices are in the empty space, very far from each other and from any celestial body. The lag time from absorbing to reemitting in the device is vary small compared to the return time of the signal (2t). If you want, this delay time can be determined by the constructing farm, by checking the two devices in the same laboratory. The two devices can be synchronyzed as clocks with the atomic clock held by A, in the following way.
First A send a "hello" radio signal to B and waits a 2t time for the received "hello" signal re-sent by B, and instantaneously (almost), with the same device, re-send the "hello" signal, taking note of the 2t time between the two emissions. Then A waits for another 2t' time of return from B. If 2t' is equal to 2t, then A and B are in relative rest to each other (this is their aim in traveling towards a very far galaxy). If 2t' is different from 2t A sends a radio-signal to turn on adjustment engines to reduce relative speed along the line between A and B. After some trial and errors 2t will remain constant. This means that A and B are in two different positions in the same inertial frame, and relatively motionless. Now A can continue to send hello signals every 2t pair seconds (0t, 2t, 4t...) while B re-send hello signals every 2t odd seconds (at 1t, 3t, 5t...). When A is perfectly confident they are both motionless send a radio signal "go-six" at an exact odd time (e.g. 5t). When B receives the signal at the unusual even time (i.e. 6t) he can a) set his one clock to 6t, b) start his engine at a prefixed power in the direction of the galaxy and, in the same instant, re-send the signal "go-six" to A. When A receives the t6 hello signal, he knows that B has turned the engine in that instant and does the same simultaneously. So the instantant t=6 is exactly the same instant for A and B. This is not conventional. Simultaneity makes sense in the same frame. Bea, that is Adam's wife, can send a kiss towards Adam's spaceship at any even time (t8, t10...) knowing her husband is kissing her in the same instant. So we can say "where is Mars and what's happening in Mars right in this moment?"
Moreover, if they know someway their distance, along the way, they can measure the famous one way light-speed and compare with the round-way one (c).

 

[Ibix]

You assume that the one-way speed of light is the same in both directions. You can choose to assume otherwise, as long as $2/c=1/c_++1/c_-$ with mo consequence except for the synchronisation. So your synchronisation is conventional.

 

[Alfredo Tifi]

You assume that the one-way speed of light is the same in both directions. You can choose to assume otherwise, as long as $2/c=1/c_++1/c_-$ with mo consequence except for the synchronisation. So your synchronisation is conventional.

I disagree. Radio signals are emitted both from A to B and from B to A and space between A and B is isotropous. The symmetry is such that nothing could justify a difference between c- and c+ in this case. There aren't "returning signals" in this case, but just re-emitted ones. No asymmetry between emitter and receiver.

 

[Ibix]

and space between A and B is isotropous.

That's an assumption, and therefore your approach is conventional. You need to prove it.

Hint: you can't. That's why simultaneity is conventional...

 

[Alfredo Tifi]

That's an assumption, and therefore your approach is conventional. You need to prove it.

Hint: you can't. That's why simultaneity is conventional...

I don't need to prove symmetry of space in a Universe where there are A and B only.
If I'll do, you should disprove the epicycles. Hint: you cant! Copernican system is conventional.

 

[Ibix]

I don't need to prove symmetry of space in a Universe where there are A and B only.

You do need to prove it if you want to claim you aren't assuming it.

If I'll do, you should disprove the epicycles. Hint: you cant! Copernican system is conventional.

True. What's your point? I'm not claiming anything foundational about physics. You, on the other hand, are claiming both that the isotropy of space is an assumption (which is what "I don't need to prove it" means) and that it is not an assumption (because that's what your "not conventional" claim requires).

 

[Alfredo Tifi]

... You, on the other hand, are claiming both that the isotropy of space is an assumption (which is what "I don't need to prove it" means) and that it is not an assumption (because that's what your "not conventional" claim requires).

If space is locally anisotropous respect to light propagation, and this is a local characteristic of space, then we have references in places or directions which remember us of absolute space or ether. If anisotropy of C+ speed is related to the emission trip, than this effect should be the same in the emission from B to A, so we would have a 2t equal to 2d/c+, with d = distance A-B. Being that -constant- A-B distance is not affected by traveling light, we could calculate a light speed equal to c+, as different from c. This result would be in contrast with the average velocity of a round trip, which is actually c (as experimentally determined).
So, anisotropy of space to respect light propagation must have some physical reality, not be an arbitrary matter of convention.
On the other side, isotropy can be taken as a Kantian apriori, in absence of evidences. Then we should image that special relativity has conventional bases. If these apriori are considered conventions, than all physics of all times is conventional, abstract, not referrable to reality. In contrast, I think that physics needs some honest assumption, as isotropy of propagation in space, to speak about reality.
In case, we could better pretend that special relativity has no basis at all because we don't have a way to decide if a reference system is inertial or not.

 

[Ibix]

If space is locally anisotropous respect to light propagation, and this is a local characteristic of space, then we have references in places or directions which remember us of absolute space or ether.

We only get absolute directions if the anisotropy is detectable, which you have yet to show. Your method simply assumes isotropy and is therefore obliged to be consistent with isotropy.

If anisotropy of C+ speed is related to the emission trip, than this effect should be the same in the emission from B to A, so we would have a 2t equal to 2d/c+, with d = distance A-B.

I cannot understand from this what experiment you are proposing. You seem to be saying that if the speed of light is $c_+$ in both directions then that is the speed we would measure. I have no idea how this relates to anisotropy.

The rest of your post is an attempt at a philosophical justification for assuming isotropy. That doesn't make it any less of an assumption. Unless you can provide a detailed description of an experiment that would give one outcome if the speed of light is isotropic and another outcome if it is not, then isotropy is simply a convention. Certainly, it's tempting to assume it and it makes the maths easier at no cost. But no experiment gives different results if you assume otherwise, so it is a matter of personal choice - not physics.

 

[sweet springs]

Hello

Bea, that is Adam's wife, can send a kiss towards Adam's spaceship at any even time (t8, t10...) knowing her husband is kissing her in the same instant.

They have been sharing the same inertial frame until ignition of the engines. After then gravity by acceleration makes difference.
Say Eve's spaceship accelerates toward Adam and Adam runs away, Adam's clock goes faster than Eve's. They observe they are separated further and further.

 

[Alfredo Tifi]

Hello
They have been sharing the same inertial frame until ignition of the engines. After then gravity by acceleration makes difference.
Say Eve's spaceship accelerates toward Adam and Adam runs away, Adam's clock goes faster than Eve's. They observe they are separated further and further.

Quite not. The have a way to ignite their engines at exactly the same instant of time in their inertial frame. Their engines and fuel flow, power and direction of motion along the A-B line and any other detail are equal. Their system will accelerate respect to another inertial frame, but their system will remain rigid, while their clocks will change (for apparent gravitational field) in the same way till they will judge from parallax they are approaching the galaxy fast enough. Whatever their common time pulses, they will decide to turn their engines off exactly in the same instant as for the ignition. Since then, they will maintain a constant velocity (referred to the galaxy) until they will approach enough to the galaxy to ride a free fall.

 

[Nugatory]

Their engines and fuel flow, power and direction of motion along the A-B line and any other detail are equal. Their system will accelerate respect to another inertial frame, but their system will remain rigid, while their clocks will change (for apparent gravitational field) in the same way

This can only be true in one inertial frame, and then only momentarily. Google for "Born rigid motion" and "Bell's spaceship paradox".

 

[sweet springs]

Thanks Nugatory.

The have a way to ignite their engines at exactly the same instant of time in their inertial frame.

Their kissing are simultaneous and the distance between them is kept constant in the inertial frame of reference at where they were still before ignition.
But in their frames of reference after ignition, things are as I mentioned in #9. Best.

 

[Alfredo Tifi]

Born rigid motion

thank you. Mathematically complex, but possible. About Born rigidity: "body can be brought in a Born rigid way from rest into any translational motion" ((wikipedia). This is exacltly what I pretend. Don't mean about rotatory motion. About Bell's paradox: " If the endpoints of a body are accelerated with constant proper acceleration in rectilinear direction so that the proper length remains constant, then the body satisfies Born rigidity". Problems arise if you try to keep a constant distance between the endpoints from the point of view of an external inertial frame, where, in that case, you should maintain a different acceleration profile which would violate Born-rigidity condition. So Born, through wikipedia, is telling us that the two spaceships will remain at constant distance even during the acceleration periods. So Adam and Bea, who won't care of those external observers, will keep flying straightaway towards their honeymoon galaxy, undisturbed.

 

[sweet springs]

Are you fine with https://en.wikipedia.org/wiki/Bell's_spaceship_paradox so Adam and Bea are happily kept tied with thread ?

 

[Alfredo Tifi]

Are you fine with https://en.wikipedia.org/wiki/Bell's_spaceship_paradox so Adam and Bea are happily kept tied with thread ?

Hello Sweet Springs. I'm compelled to admit that once A and B will start accelerating at the same t° in their (initially) inertial frame S, they will maintain exactly the same distance d° from the point of view of S, but they don't from the POV of their intantaneous inertial systems S', S''..., so their clocks won't keep synchronized anymore, because of relativity of simultaneity of S', S''... respect to S. So they will experiment a drift in distance (reciprocally increasing and observable from their non-inertial frame/s) and in time (observable from A as a second order drift of the rate of bunching signals respect to the atomic clock, but they should equally rely on the radio-signal pulses). These two drifts correspond to a constant tide effect in a constant apparent gravitational field with same acceleration a, that is felt in different times by the opposite parts of the body, so the effect is given by a constantly applied tide force Δa.
But if you remember that the two spaceships are at a constant distance d° from the point of view of the initial common inertial frame S, I argue that A will decide to have enough approaching velocity towards the galaxy and to broadcast a shut down signal at an even "internal" time t, received at an odd internal time t'+1 by B and the engines will be shut off at the different "internal" times t+1 and t'+1 (B before A). From that moment A and B re-enter into the same inertial system S^ (provided the third spaceship continues to see A at the same distance d° from each other). The retardation of A after B served to compensate the anticipation of B respect to A from the beginning of acceleration in the instantaneous inertial frames S', S'' etc. In case that wouldn't true they can reset their rest as at the beginning of their trip, and onece they know their distance is constant in time they will be in the same inertial frame and will reset their clocks to have the same time. When they will be approaching the galaxy, everything will be repeated by decelerating. They would as well decide to place side by side and feel the same (true) gravitational effects on time.

 

[Ibix]

If the rockets end up at relative rest then they must have stopped simultaneously as measured by the original frame. Their clocks will read the same time when they cut their engines, but they will not agree that they cut the engines at the same time. And, if they repeat the Einstein synchronisation procedure you outlined in your first post, they will find that their clocks are not synchronised, although observers at rest in the starting frame will say they are.

Are you still trying to maintain that there is some kind of absolute synchronisation here, or have you convinced yourself otherwise?

 

[Alfredo Tifi]

If the rockets end up at relative rest then they must have stopped simultaneously as measured by the original frame. Their clocks will read the same time when they cut their engines, but they will not agree that they cut the engines at the same time. And, if they repeat the Einstein synchronisation procedure you outlined in your first post, they will find that their clocks are not synchronised, although observers at rest in the starting frame will say they are.

Are you still trying to maintain that there is some kind of absolute synchronisation here, or have you convinced yourself otherwise?

 

[Alfredo Tifi]

Reassuming.
a) From the POV of the original frame they have never changed the A-B distances until they cut their engines;
b) due to relativistic non inertial net effects their clocks underwent a loss in synchronicity, so that they decided to cut their engines in different times from the POV of the original frame. That circumstance entailed a residual relative velocity, once acceleration became zero, that was observable from C and from A and B radars either, thanks to a non-constant rate in the returning signal. Since then, even C sees that the distance A-B is not anymore constant. Now we have three diffferent inertial frames.
c) the slower among A and B can adjust with rockets its velocity by trials and errors until the time interval between two subsequent signals remains constant. Then, once A and B are at rest in their new inertial frame S^, the clocks can be synchronized one more time. Now the beating pace of A and B is the same of the original (because A and B are again at rest in their inertial system, and their POV is what matters) but not if compared with C clock, because of Lorentz time dilatation, in case the signal coming from the original frame where C remained still reaches A & B.
But A & B don't care of C, and they are perfectly tuned and twinned with each other.

 

[Ibix]

Now the beating pace of A and B is the same of the original

A's pace is always constant from A's perspective. B's pace is always constant from B's perspective. C's pace is always constant from C's perspective. However, by your own admission, throughout most of this manoeuvre none of them agree that the others' clocks beat at the same pace as their own. Yet you claim that absolute synchronisation is possible.

And who is C anyway?

and their POV is what matters

More precisely, their POV is what you have chosen to care about.

but not if compared with C clock, because of Lorentz time dilatation

...and the relativity of simultaneity. If you don't want the relativity of simultaneity you need to replace the time Lorentz transform with $t'=\gamma t$. You're welcome to do that - it's just a change of coordinates - but the resulting velocity composition equation doesn't keep c invariant, so the speed of light is direction dependent in all but one frame.

But A & B don't care of C, and they are perfectly tuned and twinned with each other.

Unless they choose to use C's synchronisation convention. Or decide not to treat space as isotropic, just for fun.

 

[sweet springs]

Hi.

But A & B don't care of C, and they are perfectly tuned and twinned with each other.

Once they were closer in position but after cutting the engines off now separated more as ordinary and usual as a husband and his wife become.

 

[Alfredo Tifi]

Hi.

Once they were closer in position but after cutting the engines off now separated more as ordinary and usual as a husband and his wife become.

Ah ah! True, but sometimes useful :-) out of joke, it is counterintuitive that A and B are seen at the same distance from C (the third observer who remained still standing in the former inertial frame S as in A in Bell's "vertical arrangement" here: https://en.m.wikipedia.org/wiki/Bell's_spaceship_paradox) while they see each other farther (relented pace for the emission - return time of the radio signal). This could be understood imagining A and B observing from a distance a long rigid rule in their original system, having the same length of the A-B distance before turning their engines on. They would observe a Lorentz contraction in that rod. So, by inverting the point of view, the increased distance observed by A and B after their second clock synchronisation (which I'm NOT pretending to be identical with the former or with C's one) would be of a Lorentz factor longer than the initial distance dº independently by the acceleration story. So Adam's wife could well tollerate the increase of distance from Adam of a small reciprocal Lorentz factor, let's say, due to one tenth of light speed of relative velocity of their fast inertial frame S^ respect to previous S.

 

[Alfredo Tifi]

... Yet you claim that absolute synchronisation is possible.
[\QUOTE]
Sure! As I have already, explained, the second second synchronisation is like the first, but will lead to a different clocks' common pace respect to the previous synchronisation. A and B will recognize this reduced pace through the atomic clock of A, and they will justify this as due to the objective (from their POV) increase in distance observed after the acceleration period. They will be able to calculate their new distance d^ in their inertial system, as they could for dº in the former frame S.

And who is C anyway?[\QUOTE]

C is the third observer who remained still standing in the former inertial frame S, as "A" in Bell's "vertical arrangement" here: https://en.m.wikipedia.org/wiki/Bell's_spaceship_paradox

More precisely, their POV is what you have chosen to care about.[\QUOTE]

Yes, I wanted just to demonstrate that an "absolute now" makes sense from the point of view of any observer in an inertial frame, and that this is physics, not philosophy.

...and the relativity of simultaneity. If you don't want the relativity of simultaneity ...

Absolutely not. The relativity of simultaneity is physics, not convention, so I'm obligated to accept it. "it was concluded by Dewan & Beran and Bell, that relativistic stresses arise when all parts of an object are accelerated the same way with respect to an inertial frame, and that length contraction has real physical consequences" (same ref. by Wikipedia). But this does not affect the argument of REALITY of simultaneously in the same inertial frame and rationality of Einstein's method of syncing the clocks to permit such absolute now internal to the inertial frame.

 

[Ibix]

Yes, I wanted just to demonstrate that an "absolute now" makes sense from the point of view of any observer in an inertial frame, and that this is physics, not philosophy.

You have failed to do this because you are still assuming that the one way speed of light is the same as the two way speed. This is not an assumption that can be falsified because there is no experiment that you can do that will unambiguously have a different outcome depending on whether that assumption is correct or not. So it can only be a convenient (or not) choice.

It is possible that the assumption does reflect some underlying reality. But it is also possible that it does not, and there is no test that can confirm or deny that.

The relativity of simultaneity is physics, not convention, so I'm obligated to accept it.

No, it's a convention. You can make it go away by changing your simultaneity convention, at the cost of increasing the complexity of your maths. You merely end up changing your explanation for why the string breaks.

This should be obvious from the resolution of Bell's spaceships, where the accelerated and unaccelerated observers have different explanations for why the string breaks. Or the relativistic muons, where the muons and the ground observer have different explanations for how the muons reach the earth. Or the twin paradox, where the stay-at-home and traveling twins have different explanations for why the stay-at-home is older. How can there be two different explanations for one thing using absolute terms?

In fact, the Wikipedia link you quoted says exactly that in the next paragraph after the one you quoted:

However, Petkov (2009)[6] and Franklin (2009)[3]...denied the idea that those stresses are caused by length contraction in S. This is because, in their opinion, length contraction has no "physical reality", but is merely the result of a Lorentz transformation, i.e. a rotation in four-dimensional space which by itself can never cause any stress at all.

My bold. The same applies to simultaneity.

Let me put this another way. Can you find a single reputable reference saying it's possible to measure the one-way speed of light? Incidentally, PF rules require you to be able to do this for any claim you make. Do you really think that, if it were as easy as you claim, that no one would have done it in over a century?

 

[Alfredo Tifi]

You have failed to do this because you are still assuming that the one way speed of light is the same as the two way speed.

I have actually bypassed this problem through a double emission one way experiment. The radio signal send by A is received by B and instantaneously re-emitted by B. This is NOT a two way round trip of light: this kind of trip distinguishes outward and inward trips. There, the returning light has a direct causal relation with the emitting source which makes different the outbound and return trip, emission and reception of light. In the double emission round trip, provided A and B are at rest in their inertial frame, they are absolutely symmetric and there is no way to argue against the identity of "flight" time from A to B and B to A. Moreover, this experiment can be done. I don't have the necessary apparatus. If somebody would do it, and if they would found "c", conventionalists could as well continue to pretend that in a true (single emission) two way trip light speeds can be assumed as diverse in each half trip.

This should be obvious from the resolution of Bell's spaceships, where the accelerated and unaccelerated observers have different explanations for why the string breaks...

The physically absolute fact is that the string has got broken from both points of view. So, it is enough if one observer sees one cause: contraction, and a consequence, breakage. The opinion of Petkov and Franklin is more appealing, of course, because it is pre-relativistic (accelerations starts in different times)
Re-reading the Bell' paradox article for the fourth time I must admit I don't understand why the initial acceleration time of B and C is the same, but it is not the same during acceleration (cit. "the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity"). That is, why B don't see A (and viceversa) at constant distance while accelerating, as if they were connected by a rigid rod (as a single very long spaceship with equal synchronized engines in front and back). I would thank everybody can explain me that, because it is not really explained in the article.
The doubt behind my misunderstanding concerns this question: we know velocity is relative and summed with Lorentz transformations. But acceleration, to what is relative? Thank you in advance for your help.

 

[A.T.]

Re-reading the Bell' paradox article for the fourth time I must admit I don't understand why the initial acceleration time of B and C is the same, but it is not the same during acceleration (cit. "the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity").

How about reading the entire sentence?

"It also turns out that the rest length between the two has increased in the frames in which they are momentarily at rest (S′), because the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity"

So it's about different frames, not about initial vs. during acceleration.

 

[Alfredo Tifi]

How about reading the entire sentence?

"It also turns out that the rest length between the two has increased in the frames in which they are momentarily at rest (S′), because the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity"

So it's about different frames, not about initial vs. during acceleration.

I already did it, of course. Why 2 frameS? What is S'. They are initially at rest in the same system, S. Suppose they start moving at an infinitesimal equal velocity dV, having underwent a quasi-instantaneous acceleration, why should they acquire TWO different inertial systems? Why should they won't see each other at rest at constant distance anymore? Integrate many dV in many instants to get one tenth of light speed. When-why happens that they will be described by two different systems?

 

[A.T.]

Why should they acquire TWO different inertial systems?

What are talking about? There is just one S' mentioned.

 

[Ibix]

Can you please learn to use the quote feature? It's a / not a \ in the end tag.

I have actually bypassed this problem through a double emission one way experiment. The radio signal send by A is received by B and instantaneously re-emitted by B. This is NOT a two way round trip of light: this kind of trip distinguishes outward and inward trips.

Rubbish. Do you think a photon is keeping notes on whether it's been reflected or not? If it were, it would be trivially detectable.

What is not detectable is the situation where light moves at $c_+$ in the +x direction and $c_-$ in the -x direction. Un-reflected or reflected does not matter - only the direction. So your experiment is just Einstein synchronisation, which is conventional. You are just assuming that the one-way speeds are the same, as I have been trying to get you to understand for a week now.

The physically absolute fact is that the string has got broken from both points of view. So, it is enough if one observer sees one cause: contraction, and a consequence, breakage.

But if two observers see two different explanations for the break then they cannot be describing the reason for the break in absolute terms. So there can be no absolute synchronisation because the two observers disagree on whether simultaneity had anything to do with the breakage.

Re-reading the Bell' paradox article for the fourth time I must admit I don't understand why the initial acceleration time of B and C is the same, but it is not the same during acceleration (cit. "the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity").

The reason is simple: simultaneity is conventional. The equivalent of the Einstein simultaneity convention for accelerating observers is not the same as that for inertial observers, even those who are instantaneously traveling at the same speed. So the accelerating observers, assuming they choose an obvious definition of simultaneity, don't agree with any inertial frame on simultaneity.

 

[Alfredo Tifi]

Do you think a photon is keeping notes on whether it's been reflected or not? If it were, it would be trivially detectable.
... You are just assuming that the one-way speeds are the same...

Okay. The mirror can be considered as a second source, then the radio signal repetitor devise is only to synchronize clocks at very great distances. But Ockham razor's rule plus symmetry of space prevent me to consider different speeds for the +x and -x directions. Consider that which is the +x and which is -x direction in a straight line, respect to a central observer is really conventional. Light should take note of my decision to define a new direction +x' = -x to change speed properly before and after reflection.

... simultaneity is conventional. /QUOTE]

Disagreement in evaluating simultaneity of two events in different places in space-time is due to evaluation made from two or more inertial frames in relative motion one respect to each other. This assertion is not conventional, but subjected to experiment. It does not prevent us to speak of simultaneity in the same frame following any criterion. But Einstein's criterion is the only one which permits to all the standing observers to agree that if events E1 and E2 are evaluated as synchronous (with locally synchronised clocks) by observer O', then they would be evaluated as synchronous also by O'', even if light would "travel" to the left to reach O' from A & B and to the right to reach O''. (O' and O'' know their x distance from A and B, of course.)
Let's consider for example the one way light speed as "conventionally" infinite. This would permit O, from the center of coordinates, to say "All the events I see right now in different places are simultaneous" (everything we see is a one way signal). The problem is that in this case O', which is at rest on the very far +x place, cannot say the same for an event Z which is 1 millimeter close to him and for the event A which happened in x=0, the same place of O, and, at the same time, O'' who is very far from both O and O' and the events in his -x home, cannot receive simultaneously signals from A and Z. More simply: O' and O'' are symmetrically placed at +x and -x. O' is very close to the left of event place E' while O'' is very close to the left of event place E''.
so, the situation of x-axis is as follows:

-x <--- O''___E'' ____________long distance______________O'___E' --->+x

Suppose O'' sees simultaneously E' and E''. Obviously light has covered instantanelusly a long way (from E') and a short way (from (E'') coming from +x towards -x. Then, if we can conventionally choose a + infinite velocity in the one way also O' sees events E' and E'' simultaneously, but this would imply that light propagates instantaneously in both directions. In this case it would be troublesome to find c = 299792458 km/s in a two way round measure.
The only way to save coherence with the infinite one way light speed is to specify that this is the "first way" speed, the speed from any event to any observer, or emission speed, independently of who is to the right or to the left of each other. Eventually you may recover my wasted rubbish to save freedom of conventional choice, by considering c_{source to receiver} as possibly different from c_{mirror to receiver} independently by space directions +x or -x.

The equivalent of the Einstein simultaneity convention for accelerating observers is not the same as that for inertial observers, even those who are instantaneously traveling at the same speed.

Okay, I infer that is true, but can't figure the why, because I don't know Einstein's convention for syncing clocks between accelerating observers. In other words, what happens to the clocks of O' and O'' when they start accelerating at a very low acceleration a for both, initiating at exactly the same instant viewed from their common time (from the original frame O which remains unaccelerated)? The "same instant as viewed by O is also the same instant for O' and O'', even a few instants after they initiated to accelerate, because acceleration is very small and they are still almost at rest. Their frame is still "almost inertial". No matter of the time needed, starting with infinitesimal velocities, they can reach one tenth of light speed in a finite time. So, why will they get farther and farther? when this would happen?
Finally, I also reformulate a question for you: if an acceleration the same if evaluated from different inertial frames? Or, perhaps, is it the same only for frames having instantaneous velocities which aren't relativistic respect to the instantaneous accelerating frame? Is acceleration still calculated as dv/dt (that is, frame dependent)?

 

[Alfredo Tifi]

What are talking about? There is just one S' mentioned.

Thanks Advisor.
Reading the sentence "the rest length between the two has increased in the frames in which they are momentarily at rest (S′)" (underlined bold mine) I have to deduce that B and C are flying in two different frames, first because their distance is increasing (even though I don't understand why) and also because wikipedia uses the term "frames" (plural), not frame. The S' is referred to the situation in which they are already accelerating (if it were referred to the instant before accelerated motion they should have used the term "S", not S'). Then, to make sense of "momentarily at rest" while they are accelerating, I must pick an instantaneous velocity and image an inertial frame with that velocity (respect to S). So, again, the S' is plural because changing in time but also because it is supposed (but I don't understand why, yet) that B and C spaceships can't have the same velocity due to a different timing of acceleration from the point of view of each frame.
I sincerely hope you'll find some hidden flaw in my interpretation. That would help.

 

[Ibix]

Okay. The mirror can be considered as a second source, then the radio signal repetitor devise is only to synchronize clocks at very great distances. But Ockham razor's rule plus symmetry of space prevent me to consider different speeds for the +x and -x directions.

So we're agreed it's an assumption, since you cannot prove either Ockham's razor or spatial isotropy.

Consider that which is the +x and which is -x direction in a straight line, respect to a central observer is really conventional. Light should take note of my decision to define a new direction +x' = -x to change speed properly before and after reflection.

If this were the case I could turn a train around just by turning my head because its speed is initially +v so it must be +v after I turn around. No.

Physics doesn't care about coordinates. Your failure to grasp this is at the core of this whole thread. You changing your mind about which way is +x simply means that what you previously called $c_+$ you now call $c_-$ and vice versa. There is no trick to get around the fact that you can only measure the two-way speed of light without making some assumption about simultaneity.

 

[A.T.]

Reading the sentence "the rest length between the two has increased in the frames in which they are momentarily at rest (S′)" I have to deduce that B and C are flying in two different frames,

No, emphasis corrected. It's about multiple moments, not B and C.

 

[Alfredo Tifi]

No, emphasis corrected. It's about multiple moments, not B and C.

So in different times during acceleration B and C remain in the same frame S' at any moment. So, why aren't their clocks anymore synchronised, given they were synchronised before, in the inertial system S? Why they get farther? Why they don't accelerate in the "same instant"? How is treated simultaneity in an accelerated frame?

 

[A.T.]

So in different times during acceleration B and C remain in the same frame S' at any moment.

No, it's not the same S' for any moment during acceleration. If the acceleration confuses you, consider just one S', where they both end up at rest, after both engines shutdown, and they don't accelerate anymore:
- In S the engines start and stop simultaneously, so no change in distance.
- In S' the engines don't start and stop simultaneously, so the distance changes.

 

[Alfredo Tifi]

No, it's not the same S' for any moment during acceleration.

Obviously! I grasped the necessity that there are as many instantaneous S's as many velocities acquired respect to S due to acceleration (these velocities are the same from the POV of S in every S time).

... after both engines shutdown, and they don't accelerate anymore:
...- In S' the engines don't start and stop simultaneously, so the distance changes.

To imagine how this tenet works I think to how B and C evaluate (from S') their velocity respect to A, which remained in the original inertial frame S: they started accelerating with a small Δv in a small time Δt when their clocks were telling the same time t_ignition of A. So, t_ignition is the same for A, B, C. Then imagine that at t_ignition+Δt' (measured in B and C clocks) B and C soon shutted down their engines. Now we have only two inertial frames. One in which C and B are at rest in S' and view A (the other frame) moving at -Δv in the -x direction. They see a relented time pace in A (Δt < Δt') in the frame S (therefore they know their time is not synced with S time anymore), but they started viewing A moving left and relenting its time exactly in the same instant from their point of view.
So, firstly I can't grasp how could they start accelerating in different times in S' and see a change in their reciprocal distance and time. The entire scientific world agree on that, so I'm wrong, but... where is my flaw?
Secondly, given you and any relativistic scientist are right and relative motion at -Δv of A does not occur at same time if viewed by C and B, that is the same, B and C don't start changing velocity (accelerating) simultaneously, then I can't stand they are in the same instantaneous frame S'! How could two observers with different instantaneous velocities stand in the same frame in that moment? This is true only from the POV of A (where B and C do have the same velocity in any moment). For this reason I have considered, in the previous replies, that there are not only many frames S' for many moments, but a double series of S'_B and S'_C instantaneous frames. There is a reason which I am overlooking for which two frames B & C, which have the same velocity respect to A, are not the same inertial frame. The reason you maintain is that B and C acquired shifted times once they started accelerating at the same internal time. Direction of accelerated motion breaks the symmetry between B and C, provided B and C are not in the same place in the x direction. If their velocity were a constant Δv respect to S, they would stand at rest in the same frame S'.
Thirdly, if the relativistic stress experimented by an accelerated frame is, from the point of view of B and C (S') determined by asynchronicity in time of initial acceleration, then the story of linear increasing in velocity in time for S'_B would be the same of S'_C, but just shifted in time. This would give two parallel speed lines with a constant difference in velocity Δv' created at the beginning of acceleration because B and C didn't start simultaneously with acceleration a' (as judged from their pov). I suppose the constant Δv' could be calculated as a function of a' and of distance d° between B and C. I have no idea about how to calculate acceleration of S' a as viewed from A, given a' (acceleration in S'). Also the retardation of B's clock respect to C's clock remains constant during acceleration and can be calculated. The relativistic effect could be reverted shutting down the engine of the front spaceship (let's say B) before the back spaceship (C), in such way they will end at equal velocities, that is in the same inertial frame from their pov and from the pov of A. In this case the relativistic stress during acceleration could have no serious consequences provided there is no thread connecting B and C.

Thank you in advance for your help.