# I Einstein clock syncing with one way light emission absorber

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1. Jun 28, 2017

### 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 travelling 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).

2. Jun 28, 2017

### 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.

3. Jun 28, 2017

### Alfredo Tifi

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.

4. Jun 28, 2017

### Ibix

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...

Last edited: Jun 28, 2017
5. Jun 28, 2017

### Alfredo Tifi

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.

6. Jun 28, 2017

### Ibix

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

Last edited: Jun 28, 2017
7. Jun 28, 2017

### Alfredo Tifi

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 travelling 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 physic 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.

8. Jun 29, 2017

### Ibix

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.
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.

9. Jun 29, 2017

### sweet springs

Hello

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

10. Jun 29, 2017

### Alfredo Tifi

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.

11. Jun 29, 2017

### Staff: Mentor

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

12. Jun 29, 2017

### sweet springs

Thanks Nugatory.

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.

13. Jun 29, 2017

### Alfredo Tifi

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.

14. Jun 30, 2017

### sweet springs

15. Jul 1, 2017

### Alfredo Tifi

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.

16. Jul 1, 2017

### 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?

17. Jul 1, 2017

### Alfredo Tifi

18. Jul 1, 2017

### 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.

19. Jul 1, 2017

### Ibix

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?
More precisely, their POV is what you have chosen to care about.
...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.
Unless they choose to use C's synchronisation convention. Or decide not to treat space as isotropic, just for fun.

20. Jul 2, 2017

### sweet springs

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.