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On the Lighter Side => New Theories => Topic started by: mad aetherist on 14/10/2018 09:31:17

Title: Does the thread break?
Post by: mad aetherist on 14/10/2018 09:31:17
Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby.
X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.
(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.
(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

O is stationary & X & Y are going past at hi speed connected by the tight thread.
Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(B1) X & Y decelerate & there is no observer on O which is stationary.
(B2) X & Y decelerate & there is an observer on O which is stationary.
(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

Title: Re: Does the thread break?
Post by: Halc on 14/10/2018 14:10:46
Edit:  This answer of mine is wrong, corrected by the one below.
I am leaving it here for reference.

Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby.
X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.
(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.
Observers have no effect on what happens.  Perhaps you are thinking of quantum mechanics where observers arguably play a role.  So the first two situations are the same, and the string stretches or doesn’t, regardless of if anyone is watching.

If X and Y accelerate, the string stretches because force is being applied to each end to accelerate it, and the middle is going to sag a bit until the greater tension at the front and lower tension at the back can account for the acceleration of the mass of the string.  That’s Newton’s answer, not Einstein’s.

Einstein says that acceleration itself has no effect, but as speed picks up, the string maintains its tension at all times, but its length contracts relative to the original frame of O as the speed of the two ships X and Y reach relativistic speeds.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship.  That string will be in an identical state for the duration of the acceleration from the perspective of the ship, but relative to frame O, the ship and its string gets shorter as its speed becomes a significant fraction of light speed.  This shortening has no effect on the tension of the string, as it gets actually shorter, not just less stretched.
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(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.
...
(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.
If our observer is accelerating with X and Y, he sees zero change for the duration of the experiment. Frame O is not inertial in A3 or B3. There is no local way of distinguishing this from X, Y, and O all being stationary on a planet with gravity identical to the acceleration rate of the ships.
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O is stationary & X & Y are going past at hi speed connected by the tight thread.
Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(B1) X & Y decelerate & there is no observer on O which is stationary.
(B2) X & Y decelerate & there is an observer on O which is stationary.
This is just the reverse of above.  The thread has identical tension the whole time in any frame, but relative to inertial frame O, its length becomes less dilated (shortened) as the thread speed is reduced.

As for the question of if the thread breaks: No.  The tension is the same the whole time while accelerating, and even if it wasn't, it was described as an elastic thread, so one end could be tied to O as the ships drag the other end away, making it very long, but not broken.  If it breaks beyond a length L, then L was not specified.
Title: Re: Does the thread break?
Post by: Halc on 14/10/2018 16:18:01
2nd answer, not at all the same as the first answer.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship.
Little different, but different nonetheless.  That explanation almost works if X and Y are pretty close to each other, but in fact X and Y clocks will be dilated relative to each other.  Y will run faster (being less deep in the equivalent gravity well) and will measure the same speed change as having taken place in more time, thus less acceleration, which violates the 'same acceleration' stipulation of the OP.

Another way of looking at it is to imagine one ship that is 10 light years in length, with Y at the top and X at the bottom at the 'origin location'.  It accelerates in one day (as measured by O frame say) up to .866c.  In that time, X is less than half a light day (let's say 10 light-hours) from its starting point, and the length of the ship is dilated to only 5 light years in length now, so Y must be only 5 light years away + 10 light hours.
That's impossible.  It puts Y closer to the 'origin' than when it started out.  The answer I gave above should not have used the one-ship analogy.

For two ships accelerating like that, beginning simultaneously (in frame O) for one day, they both move ahead by that 10 light hours. The string is still 10 light years long in O's frame, but stretched now, since its new proper length is 20 light years.
In the new frame 'P' of X and Y (.866c relative to O) where both those ships will be stationary after the acceleration: well, first of all, they have to stop accelerating and coast for over 17 years to be relatively stationary again.  So they do that.  X and Y, initially moving at .866c relative to P, with X in front, are only 5 light years apart, so the string's dilated length is 5 light years.  Y accelerates to a halt first, then X over 17 years later. When they're done doing that, the string is stationary and 20 light years long now. It doesn't break because it is elastic, but if it wasn't elastic, yes it would break.
Title: Re: Does the thread break?
Post by: mad aetherist on 14/10/2018 22:42:24
I will digest all of that later & after some more replies give my own ideas.
I think that anyone can impose any extra conditions they like, but it might get very complicated.
I am happy to assume that the string & spaceships are massless & that there is no nearby mass etc in that part of the cosmos (ie true zero gravity) -- i prefer to imagine X & Y to be in line horizontally, both facing to left, with X ahead (both going rt to left), & with O sitting on the centerline of the picture (at least initially) & well below X & Y.
Title: Re: Does the thread break?
Post by: mad aetherist on 16/10/2018 00:14:33
Edit:  This answer of mine is wrong, corrected by the one below. I am leaving it here for reference.
Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby.
X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.
(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.
Observers have no effect on what happens.  Perhaps you are thinking of quantum mechanics where observers arguably play a role.  So the first two situations are the same, and the string stretches or doesn’t, regardless of if anyone is watching.  Comment. Ok, Einsteinians & aetherists would agree here.

If X and Y accelerate, the string stretches because force is being applied to each end to accelerate it, and the middle is going to sag a bit until the greater tension at the front and lower tension at the back can account for the acceleration of the mass of the string.  That’s Newton’s answer, not Einstein’s. Comment. Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.

Einstein says that acceleration itself has no effect, but as speed picks up, the string maintains its tension at all times, but its length contracts relative to the original frame of O as the speed of the two ships X and Y reach relativistic speeds. Comment. I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here). However observer X & observer Y would say that the distance tween X & Y doesnt change, & they would say that there is no stretching or slackening of the string in agreement with observer O but for a different reason.

Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian), i believe that an aetherwind blows throo the spaceships, & i believe that all objects shorten (or lengthen) depending on the change in the apparent aetherwind. Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary), plus i would need to know the magnitude & direction of the subsequent acceleration of X~Y. If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken. If the initial aetherwind was a tailwind then early on the string would slacken & then later the string would stretch.

That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens. The thing is that in accordance with aether theory all observers will see the actual length & size & shape of all objects at all times, the magnitude & direction of the aetherwind blowing throo the observer or blowing throo the object makes no difference. This is because any change in size or shape of the object is matched by the change in the size & shape of the observer's eye(s).

Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame). In the ARF the aetherwind is zero kmps, hencely there is no contraction, an object's length is its true length. Now the strange thing is that all observers will see an object's true length at all times, as long as the observer & the object have the same velocity (eg if both appear to be stationary out in space). But if there is a relative velocity tween observer & object then it is highly unlikely that the observer will be seeing the object's true length (because the distortion of the eyes doesnt match the distortion of the object).

There is little need to look at B123, the above aetheristic rules can be applied to tell whether the string stretches or slackens or no change (or whether the string does one thing for a while & then the other). 

In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.

The situation is little different from one long ship with X being the bottom and Y being the top, with a string between them running along the interior of the ship.  That string will be in an identical state for the duration of the acceleration from the perspective of the ship, but relative to frame O, the ship and its string gets shorter as its speed becomes a significant fraction of light speed.  This shortening has no effect on the tension of the string, as it gets actually shorter, not just less stretched.
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(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.
...
(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.
If our observer is accelerating with X and Y, he sees zero change for the duration of the experiment. Frame O is not inertial in A3 or B3. There is no local way of distinguishing this from X, Y, and O all being stationary on a planet with gravity identical to the acceleration rate of the ships. Comment.
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O is stationary & X & Y are going past at hi speed connected by the tight thread.
Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(B1) X & Y decelerate & there is no observer on O which is stationary.
(B2) X & Y decelerate & there is an observer on O which is stationary.
This is just the reverse of above.  The thread has identical tension the whole time in any frame, but relative to inertial frame O, its length becomes less dilated (shortened) as the thread speed is reduced. Comment.

As for the question of if the thread breaks: No.  The tension is the same the whole time while accelerating, and even if it wasn't, it was described as an elastic thread, so one end could be tied to O as the ships drag the other end away, making it very long, but not broken.  If it breaks beyond a length L, then L was not specified. Comment.
Title: Re: Does the thread break?
Post by: Halc on 16/10/2018 03:23:35
Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.
Fair enough, unless you start drawing conclusions from such impossibilities.  If I tug on a long nearly-massless string, the other end will not move right away, subject to the speed of sound.  Likewise the top of any rocket cannot move right away when the bottom starts to move.  The sting is going to stretch, massless or no.  We can ignore that, but keep it in mind.
Meanwhile, I consider the two ships to be points.  It just complicates things to give them length.  It's the point where the string is attached that matters, not the rest.

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I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here).
Well I did say that the post to which you are responding was wrong.  I assumed incorrectly that X and Y were opposite ends of a ship that had thrusters all along its length, not just at the bottom, with the string running the length inside.  Read my 2nd reply.  The string stretches, and would break if not elastic.

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Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian),
Gotcha.  I suppose that's why you're posting in this forum and not the main one.  I should have clued off from your user name when I first replied.

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Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary),
You said all objects were stationary at the beginning.  Isn't that definition enough for you?
The B examples had O stationary and X and Y moving.  You were quite clear about that, except in B3 where there really isn't anything that stands still.

Disclaimer: I read B3 wrong.  I didn't see that O was accelerating in the opposite direction.  I took it to work like A3 where they all went the same way.  My answer for B3 was pretty much identical to A3 for that reason.

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If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken.
It more depends on if X or Y starts accelerating first, or more specifically, in which frame they 'simultaneously' start their identical changes of velocity.  If you're going to have an absolute reference, you need to specify it in your question.  In the B examples, it seems to be the frame of X and Y after they finish.

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That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens.
No, that cannot be.  Observer O is not in the presence of either X or Y, so does not see any of their events as they happen, but only later when light reaches O from those events.  O can compute the length of the string, but does not directly observe it 'as it happens'.  O can do that same computation by simply consulting the flight schedule of X and Y and need not observe them at all.

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Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame).
I am familiar with the view.

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In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.
The relative velocity remains constant in that scenario.  You mention no acceleration.  It seems to be a description of a fly-by.  Perhaps I misread it and you mean something comes in, slowing to a stop, then returning the way it came, in which case, only one of X and Y can actually stop at O, so the other is never 'level with O', and the string will change tension with speed, breaking if it can stretch only so far.
If your aether theory doesn't predict that, then it is wrong.  But it does predict that.
Title: Re: Does the thread break?
Post by: mad aetherist on 16/10/2018 08:51:38
Ok, but i wasnt really looking for real-life effects, ie we would usually assume the string & spaceships to be massless.
Fair enough, unless you start drawing conclusions from such impossibilities.  If I tug on a long nearly-massless string, the other end will not move right away, subject to the speed of sound.  Likewise the top of any rocket cannot move right away when the bottom starts to move.  The sting is going to stretch, massless or no.  We can ignore that, but keep it in mind.  Comment. Ok.
Meanwhile, I consider the two ships to be points.  It just complicates things to give them length.  It's the point where the string is attached that matters, not the rest.  Comment. I think that the shortening of spaceships is critical (or can be), in the aetheric universe it magnifies any stretching or slackening of the string. All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction). But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy.
Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.
The additional problem in the Einsteinian universe being that nothing need be real, every relativity question is solved using Einstein's gamma which is a math-trick-model that is designed to give goodish answers, reality is not needed. The golden example of this is of course the Twin Paradox (i dont want to start a new argument here), which involves the gamma for ticking not length, but paradoxes involving length are no less serious, but are not as breathtaking (eg the spaceships~thread paradox).
The answer is that Einsteinian paradoxes are not real paradoxes, because Einsteinian relativity situations are not real, they are math-trick-models designed to give a quick & sometimes good answer. I am ok with math-models & math-trick-models, but in the modern precision era Einsteinian relativity is found to be missing the target, & must be replaced by a better model, which will involve aether.

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I think that Einstein would say that while the relative speed tween XY & O is increasing then observer O would see that the distance tween X & Y gradually shortens (because space contracts along that line)(& anything in that space contracts accordingly)(ie string shortens, & X & Y each shorten), but, i think that heshe would say that there is no stretching or slackening of the string, it maintains its tension at all times (so u & i agree here).
Well I did say that the post to which you are responding was wrong.  I assumed incorrectly that X and Y were opposite ends of a ship that had thrusters all along its length, not just at the bottom, with the string running the length inside.  Read my 2nd reply.  The string stretches, and would break if not elastic.  Comment. I will have a look. But offhand i am puzzled at why the string might want to break, it should not want to break (ie stretch) from the viewpoint of any & all Einsteinian observers. I will have a look.

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Me myself (ie an aetherist)(ie a neo-Lorentzian)(ie an anti-Einsteinian),
Gotcha.  I suppose that's why you're posting in this forum and not the main one.  I should have clued off from your user name when I first replied.  Comment. Yes my comments are likely to involve a dynamic aether, non-mainstream.

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Therefore before voicing an opinion re the fate of the string in A1 & A2 i would firstly need to know the speed & direction of the (initial) aetherwind relative to X~Y~O (when stationary),
You said all objects were stationary at the beginning.  Isn't that definition enough for you?  Comment. No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers, because the aetherwind will have a projection~value along the X~Y line, & that initial condition is critical, eg it might initially be a headwind or a tailwind or (unlikely) zero. Perhaps i could have said that at minimum i need to know the value of the projection of the aetherwind along the X~Y line -- but i think that that is an error -- i need to know the full value of the full wind (the explanation is not long but i will give it a miss for now).
The B examples had O stationary and X and Y moving.  You were quite clear about that, except in B3 where there really isn't anything that stands still.  Comment. Yes, in B3 O is initially still but as u say doesnt stay still.

Disclaimer: I read B3 wrong.  I didn't see that O was accelerating in the opposite direction.  Comment. Yes i think i saw that.   I took it to work like A3 where they all went the same way.  My answer for B3 was pretty much identical to A3 for that reason.  Comment.

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If the acceleration resulted in a stronger apparent aetherwind felt by X & Y measured along the line of X~Y then (a) X would shorten & Y would shorten, which would result in a wider gap tween X & Y, & (b) the string would shorten, which together with the widening of the gap would mean that the string would need to stretch (& might break). But if the acceleration resulted in a weaker aetherwind then X & Y would each lengthen, the gap tween would shorten, & the string would itself lengthen, & the string would slacken.
It more depends on if X or Y starts accelerating first, or more specifically, in which frame they 'simultaneously' start their identical changes of velocity.  If you're going to have an absolute reference, you need to specify it in your question.  In the B examples, it seems to be the frame of X and Y after they finish.  Comment. I like your comment re simultaneity, i hadnt thort of that. Simultaneity can be a problem in an Einsteinian universe, but it aint a major problem in the aether universe.  In the aether universe the present instant of time is universal, & there is no such thing as time dilation, it is merely a ticking dilation. Either way the best way of ignoring the problems of ticking & simultaneity etc is i think to simply say that for the purposes of observation the speed of light is infinite.

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That describes what must happen, but my original question is moreso re what do observers XYO see happen. Observer O is stationary throughout & hencely shehe will see the string stretching or slackening precisely as it happens.
No, that cannot be.  Observer O is not in the presence of either X or Y, so does not see any of their events as they happen, but only later when light reaches O from those events.  O can compute the length of the string, but does not directly observe it 'as it happens'.  O can do that same computation by simply consulting the flight schedule of X and Y and need not observe them at all.  Comment. Good points. This is solved by assuming that observational light travels at infinite speed.
We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).

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Aetherists believe that an objects proper true length & size & shape is the length & size & shape that it enjoys when in the absolute reference frame (sometimes called preferred frame).
I am familiar with the view.

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In a sense similar things can happen in the Einsteinian universe, eg if X & Y are at first getting closer, then become level with O, & then fly away, then the relative velocity relative to O will be decreasing then zero then increasing, but in all 3 cases the Einsteinian string will not change its Einsteinian tension.
The relative velocity remains constant in that scenario.  You mention no acceleration.  It seems to be a description of a fly-by.  Perhaps I misread it and you mean something comes in, slowing to a stop, then returning the way it came, in which case, only one of X and Y can actually stop at O, so the other is never 'level with O', and the string will change tension with speed, breaking if it can stretch only so far.  Comment. True, in the Einsteinian universe we would need to recognize that as the array of X~string~Y passes only one point of the array can be directly level or opposite O at any one instant, & i imagine that this can result in some complications when a half of the string is coming while a half is going (from O's view). However this doesnt make any difference in the aether universe.
If your aether theory doesn't predict that, then it is wrong.  But it does predict that.   Comment.
Title: Re: Does the thread break?
Post by: Halc on 16/10/2018 14:43:33
All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction).
Your theory has the aetherwind changing?  That’s not the neo-Lortentzian view.
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But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy.
The gap changes only in the dimension of movement.  So if X and Y are side-by side, string or not between them, the gap between them and the string length remains the same as they accelerate in parallel. The ships get shorter, but no less wide.  The string length is unaltered but it gets thinner in one dimension, squashing to sort of an elliptical cross section.  Both Einstein and Lorentz claim this, but apparently not you, so your view contradicts empirical physics.
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Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.
If anybody says any of that, they don’t know their physics.  If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.

What you say about a string not breaking is true where X and Y are different ends of the same ship, accelerating as one unit, not two separate units.  This is exactly equivalent to a string hanging in a tall building with Y at the top and X at the bottom, with gravity at Y equal to the acceleration of Y.  The string just sits there, not breaking.
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Yes my comments are likely to involve a dynamic aether, non-mainstream.
OK.  I sort of figured that out above where you talked about the wind changing.  Let me know how that works out for you.  Do fast moving things contract in length?  I was going with the Lortentz view on this, but you’ve got different ideas, so I have no clue as to what your rules are.  Length contraction would seem to be in direct contradiction with a changing etherwind.
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No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers
How is stationary defined if not relative to the aetherwind?  Minkowski spacetime has inertial reference frames, but the typical Lortentz interpretation says only one of those reference frames is preferred: the one where the aetherwind doesn’t blow.

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In the aether universe the present instant of time is universal
Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas.

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Good points. This is solved by assuming that observational light travels at infinite speed.
Now you’re living in another universe.  I can assume no such thing. You can disprove any bit of physics you like with this assumption that directly contradicts empirical measurements.

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We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).
Nonsense.  You’re making up fiction.  The whole experiment revolves around finite but constant light speed.  Nobody is observing rays.  Light reaches a detector and is measured but once.
Title: Re: Does the thread break?
Post by: mad aetherist on 16/10/2018 23:58:54
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All objects (spaceship)(string) shorten or lengthen equally (in any given direction)(if the aetherwind changes in that direction).
Your theory has the aetherwind changing?  That’s not the neo-Lorentzian view.
 Comment.  The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit.  For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day (MMXs are done horizontally). And the apparent wind will be different if u are driving at speed. I say apparent, but of course it is not apparent to u or me, but can be if we have the right sort of detector. This is all in accord with neo-Lorentz, apart from minor disagreement re exact numbers.
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But in the aetheric universe the gap always does the opposite, when objects shrink the gap widens (in any given direction), hencely its a double whammy.
The gap changes only in the dimension of movement.   [Yes]  So if X and Y are side-by side, string or not between them, the gap between them and the string length remains the same as they accelerate in parallel. The ships get shorter, but no less wide.  The string length is unaltered but it gets thinner in one dimension, squashing to sort of an elliptical cross section.  Both Einstein and Lorentz claim this, but apparently not you, so your view contradicts empirical physics.
 Comment.  No i said that the spaceships were in line (ie line astern)(not side by side).
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Meanwhile over in the Einsteinian universe Einsteinians cannot agree amongst themselves. Most of them say that a change in relative velocity contracts or expands the space in that direction (as seen by the stationary observer)(meaning all observers)(because all observers are stationary from their own point of view), & any object sitting or moving or accelerating in that space contracts or expands with that space. In which case the string never stretches or slackens & never breaks.
If anybody says any of that, they don’t know their physics.  If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.
 Comment. I was referring to the rezults of a poll at CERN – some said that the string doesn’t break, some said that it breaks (last i heard all of them were still getting paid).

What you say about a string not breaking is true where X and Y are different ends of the same ship, accelerating as one unit, not two separate units.  This is exactly equivalent to a string hanging in a tall building with Y at the top and X at the bottom, with gravity at Y equal to the acceleration of Y.  The string just sits there, not breaking.
 Comment.  In my string-thort-X X & Y are allways doing exactly the same thing (ie at the same time). I don’t know about that tall building stuff, my thort-X is out in space well away from mass etc.
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Yes my comments are likely to involve a dynamic aether, non-mainstream.
OK.  I sort of figured that out above where you talked about the wind changing.  Let me know how that works out for you.  Do fast moving things contract in length?  I was going with the Lortentz view on this, but you’ve got different ideas, so I have no clue as to what your rules are.  Length contraction would seem to be in direct contradiction with a changing etherwind.
 Comment.  Fast things might be contracted or might not be. That’s why i said I needed to know the exact details of the background aetherwind. If the fast thing is going against the background wind then it will be contracted, if with then it will be longer. If almost side-on to the wind (& slightly with)(say at 91 deg) then the length might be non-changed.
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No, i dont think that it is enough, i need to know the aetherwind blowing throo the stationary observers
How is stationary defined if not relative to the aetherwind?  Minkowski spacetime has inertial reference frames, but the typical Lortentz interpretation says only one of those reference frames is preferred: the one where the aetherwind doesn’t blow.
 Comment.  Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps. In my thort-X when i say that O is stationary i don’t mean that O's apparent wind is zero, i mean O is stationary in relation to something else, but it might be a good idea to consider O to be stationary in the absolute aether frame (it might make the thort-X simpler).

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In the aether universe the present instant of time is universal
Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas.
 Comment.  I don’t remember exactly what Lorentz thort (it was so long ago). Lorentz i think thort of time in his time-gamma-stuff as being a convenient math-trick, not real, useful briefly & then best ignored (like sex). Einstein elevated time to the forefront. Neo-Lorentzians reckon that there is no such thing as time, or if u like there is but it is the now, the present instant, & the present instant is universal. Thusly nL's do not believe in time dilation, we believe in ticking dilation.

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Good points. This is solved by assuming that observational light travels at infinite speed.
Now you’re living in another universe.  I can assume no such thing. You can disprove any bit of physics you like with this assumption that directly contradicts empirical measurements.
 Comment.  Yes i can say-assume that observational light travels at infinite speed in my string-thort-X. No, i cannot disprove any bit of physics i like with that assumption. And whats with this "contradicts empirical measurements" krapp.

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We see the same problem in Einstein's train & lightning thortX. Einstein assumes that (1) observational light travels at infinite speed (but doesn't say so) & (2) assumes that an observer can see a ray of light (this would need an assistant with a smoke machine).
Nonsense.  You’re making up fiction.  The whole experiment revolves around finite but constant light speed.  Nobody is observing rays.  Light reaches a detector and is measured but once.
 Comment.  Ok i had another look at Einstein's train-lightning-thort-X & u are correct. I was thinking of the modern youtube cartoons of this thort-X, where the platform observer is standing well away from the train. Einstein did indeed hav the 2 flashes & the 2 (midpoint) observers  more or less on the same alignment (or within inches) where their eyes were directly hit by the tips of the rays (no smoke or infinite speed needed). However there was no detector, they did observe rays with their eyes (not important).
In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldn’t work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed for O (not a problem).
Title: Re: Does the thread break?
Post by: Halc on 17/10/2018 01:04:47
Your theory has the aetherwind changing?  That’s not the neo-Lorentzian view.
 The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit.
That’s us changing velocity, not the aether.

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For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day
How the heck did he get a variance of 340 kmps if the variance of our velocity is only 61 kmps?  What exactly was being measured, since the aether is not detectable?  I find no references to this experiment.

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In which case the string never stretches or slackens & never breaks.
If anybody says any of that, they don’t know their physics.  If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.
I was referring to the rezults of a poll at CERN – some said that the string doesn’t break, some said that it breaks (last i heard all of them were still getting paid).
A poll was taken at CERN about your string question here?  And they answered that way?  Love to see a reference that confirms that.  I personally find that very hard to believe.

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I don’t know about that tall building stuff, my thort-X is out in space well away from mass etc.
More to the point: Two ships, not two ends of one long ship.  It makes a very big difference.

You say this:
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Fast things might be contracted or might not be.
and then this:
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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.
The first comment implies that something might be a ‘fast thing’ that is nevertheless not contracted because it is stationary in the preferred frame.  That doesn’t sound like a ‘fast thing’ to me.
It’s fine in the Einstein view where that thing is just fast is some other frame, but that’s not the typical language adopted by a self-proclaimed aetherist.

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In the aether universe the present instant of time is universal
Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas.
I don’t remember exactly what Lorentz thort (it was so long ago).
Lorentz envisioned 4D spacetime with a preferred frame but not a preferred moment.  Both Minkowski and Einstein borrowed from Lorentz’s work in this area in the formulation of their theories.  Spacetime has no ‘present instant’, but yes, I find that typical nL’s posit one anyway.

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Yes i can say-assume that observational light travels at infinite speed in my string-thort-X.
You are on your own then.  I cannot take your examples seriously with this assumption.
Is it really so hard to say that some event happens 1 light-hour away, so it gets seen 1 hour after it happens?  We’re both quite capable of subtracting off the signal travel time.

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In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldn’t work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed. That’s not a problem.
That works quite nicely as well, putting your observer way back, so you see all events in the order they happen in the frame of that observer.  Yes, you still need to select a frame for the observer, even if he’s well back.  One of them can be on another train running in parallel with the first.

Not sure what this distant observer expects to see.  Light signals from the two flashes going sideways to the observer?  That works.  He cannot see light move to some observer on the platform, but he can see the platform (or the train guy) signal some way when each flash is perceived.

Just guessing.  I’ve not seen your ‘string-thort-X’.  It isn’t going to predict anything different than relativity theory.
Title: Re: Does the thread break?
Post by: mad aetherist on 17/10/2018 03:41:51
Your theory has the aetherwind changing?  That’s not the neo-Lorentzian view.
 Comment. If u change velocity then the apparent aetherwind blowing throo u must change, the background wind can be considered constant.
 The background aetherwind blowing at 500 kmps south to north at 20 deg off Earth's spin-axis might not change much, but the apparent wind blowing throo u & me gradually changes during each day & year due to Earth's 0.4 kmps spin & 30 kmps orbit.
That’s us changing velocity, not the aether.
 Comment. Yes, the background aetherwind probly doesnt change much during a day.
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For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day
How the heck did he get a variance of 340 kmps if the variance of our velocity is only 61 kmps?  What exactly was being measured, since the aether is not detectable?  I find no references to this experiment.
 Comment. google -- Michelson-type interferometer operating at effects of first order with respect to v/c -- V V Demjanov -- 2010.
I think that the 61 kmps has only a small effect. The main difference arises due to the latitude, ie its effect on the horizontal angle of the 500 kmps. At the poles an MMX would show very little change in the horizontal component of the wind speed (praps 61 kmps), at the equator it would be at a max (praps 500 kmps), at the latitude of Obninsk (near Moscow) it was 340 kmps (on 22 June 1970).
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In which case the string never stretches or slackens & never breaks.
If anybody says any of that, they don’t know their physics.  If you want to call them Einsteinians, fine, but you seem to be talking about uneducated people.
I was referring to the rezults of a poll at CERN – some said that the string doesn’t break, some said that it breaks (last i heard all of them were still getting paid).
A poll was taken at CERN about your string question here?  And they answered that way?  Love to see a reference that confirms that.  I personally find that very hard to believe.
 Comment.  I will have a look for it.

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I don’t know about that tall building stuff, my thort-X is out in space well away from mass etc.
More to the point: Two ships, not two ends of one long ship.  It makes a very big difference.
You say this:
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Fast things might be contracted or might not be.
and then this:
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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.
The first comment implies that something might be a ‘fast thing’ that is nevertheless not contracted because it is stationary in the preferred frame.  That doesn’t sound like a ‘fast thing’ to me.
It’s fine in the Einstein view where that thing is just fast is some other frame, but that’s not the typical language adopted by a self-proclaimed aetherist.
 Comment. If the wind is a tailwind then a fast thing will be longer.

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In the aether universe the present instant of time is universal
Somehow I guessed that you would also adopt this divergence from Lorentz’s ideas.
I don’t remember exactly what Lorentz thort (it was so long ago).
Lorentz envisioned 4D spacetime with a preferred frame but not a preferred moment.  Both Minkowski and Einstein borrowed from Lorentz’s work in this area in the formulation of their theories.  Spacetime has no ‘present instant’, but yes, I find that typical nL’s posit one anyway.
 Comment.  I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.

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Yes i can say-assume that observational light travels at infinite speed in my string-thort-X.
You are on your own then.  I cannot take your examples seriously with this assumption.
Is it really so hard to say that some event happens 1 light-hour away, so it gets seen 1 hour after it happens?  We’re both quite capable of subtracting off the signal travel time.
 Comment.  Yes it is much too hard. Infinite observational speed of light fixes that. 

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In my string-thort-X i too can place my observer O within inches of the alignment of X~Y, but then the thort-X wouldn’t work proper, O needs to be well back (a very long way back), for my thort-X to make much sense. So, i need an infinite observational-light speed. That’s not a problem.
That works quite nicely as well, putting your observer way back, so you see all events in the order they happen in the frame of that observer.  Yes, you still need to select a frame for the observer, even if he’s well back.  One of them can be on another train running in parallel with the first.
 Comment. The Einsteinian universe needs to consider the frame of O, the aetheric universe doesnt need to. The aetheric universe can ignore O, it only needs to consider the aetherwind blowing throo XY, this will tell whether the string stretches or slackens. Then re what does O see, O sees exactly what happens, always, no matter what is O's speed or acceleration or position (constrained only by hisher ability to see instantaneously)(which is why i prefer an infinite observational light speed).

Not sure what this distant observer expects to see.  Light signals from the two flashes going sideways to the observer?  That works.  He cannot see light move to some observer on the platform, but he can see the platform (or the train guy) signal some way when each flash is perceived. Comment.   

Just guessing.  I’ve not seen your ‘string-thort-X’.  It isn’t going to predict anything different than relativity theory.
 Comment. My string-thort-X is merely the A123 & B123 question stuff. But i know that there are versions of a spaceship-string-spaceship thort-X out there on google & wiki. I must have a look to see what they reckon.
Title: Re: Does the thread break?
Post by: Halc on 17/10/2018 13:17:49
For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day
OK, the measurements are just done at different angles, with a device too cumbersome to just point different ways, so it uses Earth rotation to do that work.  I cannot figure what it is measuring.  It isn't speed relative to CMB since that cannot produce a value greater than about 370 km/sec (pointed at say Leo constellation), which is the speed typically promoted by aetherists.

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If the wind is a tailwind then a fast thing will be longer.
How can something that is stationary be a tail wind?

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I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.
I think Minkowski coined that term.  But Lorentz first envisioned it, and did the coordinate rotation mathematics for it, which is still called the Lorentz transformation.
Title: Re: Does the thread break?
Post by: mad aetherist on 17/10/2018 14:07:54
For example Demjanov's 1970 MMX in Obninsk showed that the horizontal projection of the wind was 140 kmps at one time of day & had a max of 480 kmps at another time of day
OK, the measurements are just done at different angles, with a device too cumbersome to just point different ways, so it uses Earth rotation to do that work.  I cannot figure what it is measuring.  It isn't speed relative to CMB since that cannot produce a value greater than about 370 km/sec (pointed at say Leo constellation), which is the speed typically promoted by aetherists.
 Comment. Demjanov used the smallest MMX ever, smaller than all of the silly little Year-1 MMX's done every year in Physics 101 at Colleges all over theusofa. His MMX was small because it was a twin media (air~carbondisulphide) MMX, which was 1000 times as sensitive as the old fashioned MMXs. It certainly wasnt one of those big MMXs that were frozen in place & needed the Earth's spin to work (i think there were 2 of these somewhere, Munera had one). Demjanov's MMX spun & gave a result for fringeshift in less than a minute -- the result changing during the course of a day. Why are we even arguing about Einstein -- what sort of proof is needed that aether & aetherwind exists.

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If the wind is a tailwind then a fast thing will be longer.
How can something that is stationary be a tail wind?
 Comment.  I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.

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I am dubious re Lorentz using the word spacetime before Einstein & Co. I will have a look.
I think Minkowski coined that term.  But Lorentz first envisioned it, and did the coordinate rotation mathematics for it, which is still called the Lorentz transformation.
 Comment.  Not forgetting that Lorentz & Michelson & Miller & Morley & Ives & Poincare & Silberstein all died believing in aether. And Einstein died believing in aether.
Title: Re: Does the thread break?
Post by: Halc on 17/10/2018 15:17:31
Quote from: Halc
How can something that is stationary be a tail wind?
Comment.  I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.
Yes you did:
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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.
That is pretty much an assertion that the aether is stationary relative to the absolute frame, consistent with the neo-Lorentzian view.  As such, it cannot be a tail wind.  If there is wind, it is because you are moving and the wind will come from your direction of movement: a head wind.

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Not forgetting that Lorentz & Michelson & Miller & Morley & Ives & Poincare & Silberstein all died believing in aether. And Einstein died believing in aether.
Not all of them called it that name, but arguably yes.  Can't bend or expand the fabric of spacetime if there is no fabric to bend.
Title: Re: Does the thread break?
Post by: David Cooper on 17/10/2018 20:41:16
Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby.
X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.
(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.
(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

I'm not interested in the observers as the thread either breaks for all of them or doesn't break for any of them.

As X and Y accelerate, they contract in length but stay the same distance apart, so the thread snaps.

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O is stationary & X & Y are going past at hi speed connected by the tight thread.
Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(B1) X & Y decelerate & there is no observer on O which is stationary.
(B2) X & Y decelerate & there is an observer on O which is stationary.
(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

As X and Y decelerate, they lose the length contraction that they started with, so the new thread goes slack. However, that only happens if their clocks are still synchronised for the first experiment. If they resynchronise their clocks for the frame in which they start this deceleration, the new thread will break as before because the leading ship will start its deceleration late.
Title: Re: Does the thread break?
Post by: mad aetherist on 18/10/2018 01:14:00
Quote from: Halc
How can something that is stationary be a tail wind?
Comment.  I didnt stipulate what the aethewind is or isnt, ie speed & direction, thusly it could be a tailwind or it could be a headwind. But stationary would be simpler for sure.
Yes you did:
 Comment.  No, not in the original questions. But the aetheric answer demands the knowledge of the initial
 background aetherwind (which at the same time tells u about the initial apparent wind)(which at the same time tells u about the change in the apparent wind during the events), & depending on the initial wind the aetheric answer can be (a) stretches, or (b) slackens, or (c) stretches then slackens, or (d) slackens then stretches -- but with the knowledge of the initial wind the answer in each of A123 & B123 will be one of these choices, & the answers for each of the 6 cases might be different, except that the answer for A1 will allways be the same as for A2, & B1 will allways be the same as B2 (unless u nominate different winds).
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Yes, if u are stationary in the preferred or absolute reference frame then the wind is zero kmps.
That is pretty much an assertion that the aether is stationary relative to the absolute frame, consistent with the neo-Lorentzian view.  As such, it cannot be a tail wind.  If there is wind, it is because you are moving and the wind will come from your direction of movement: a head wind.
 Comment. It can be a tailwind, depending on what u nominate, & as i said above the aetheric answer depends on the kmps & degrees nominated for the wind (whereas the Einsteinian answer probably doesnt need any such nomination).
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Not forgetting that Lorentz & Michelson & Miller & Morley & Ives & Poincare & Silberstein all died believing in aether. And Einstein died believing in aether.
Not all of them called it that name, but arguably yes.  Can't bend or expand the fabric of spacetime if there is no fabric to bend. Comment. I am ok with that.
Title: Re: Does the thread break?
Post by: mad aetherist on 18/10/2018 01:37:59
Some relativity question(s). I am especially interested in what Einstein would say, but there are other theories.

Spaceship X is connected to spaceship Y (line ahead) by a tight elastic thread & spaceship O is nearby.
X & Y & O are stationary. Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(A1) X & Y accelerate at the same rate & there is no observer on O which is stationary.
(A2) X & Y accelerate at the same rate & there is an observer on O which is stationary.
(A3) X & Y accelerate at the same rate & there is an observer on O which accelerates likewise.

I'm not interested in the observers as the thread either breaks for all of them or doesn't break for any of them.
 Comment. That is of course in an Einsteinian universe (& is likewise the the aetheric universe)(depending on the initial wind nominated)

As X and Y accelerate, they contract in length but stay the same distance apart, so the thread snaps.
 Comment. My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that.
If your answer refers to the aetheric universe then as i have said elsewhere u need to know the initial wind before answering.
But i suppose that u could say that your answer is based on what happens in the long run, ie after a long time -- but here we might get into trouble because in the long run the ultimate speed will be nearly c (on the other hand perhaps there aint a problem here)(i think there aint a problem).  But anyhow re such possible trouble -- in the aetheric universe it is possible for X~Y to be moving at almost c in one direction, & for O to be moving at almost c in the opposite direction, in which case the relative speed is almost 2c, however light from X~Y will eventually reach O after a long time (which makes my stipulation for an infinite speed of observational light even more necessary).

Re that there relative speed of almost 2c -- here i mean true relative speed, ie as observed by an observer in the absolute aether rest frame (where the aetherwind for the observer is 00 kmps). For our  observer O, who might have an aetherwind blowing throo himher, hisher eyes will be deformed (contracted) by the wind, & heshe will always see the true lengths & true speeds of X & string & Y. Likewise observers X & Y will usually have deformed eyes & they too will always see the true lengths & speeds of X & string & Y (so i don know why i needed to write this paragraph)(ignore it).

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O is stationary & X & Y are going past at hi speed connected by the tight thread.
Do observers on X & Y & O see the thread stretch or slacken or stay the same when.....
(B1) X & Y decelerate & there is no observer on O which is stationary.
(B2) X & Y decelerate & there is an observer on O which is stationary.
(B3) X & Y decelerate & there is an observer on O which accelerates at the same rate in the opposite direction.

As X and Y decelerate, they lose the length contraction that they started with, so the new thread goes slack. However, that only happens if their clocks are still synchronized for the first experiment. If they resynchronize their clocks for the frame in which they start this deceleration, the new thread will break as before because the leading ship will start its deceleration late.
 Comment. I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something. Especially as all of this happens in deep outer space a long way from mass (such mass might affect the clocks differentially if nearer to one).
How can the leading ship X start its acceleration late?

In the aetheric universe the clocks X & Y would too have to be synchronized at the start of the event, & here too i dont see how they would lose synchronicity if they both did exactly the same things thereafter. In B3 clock O would have to be synchronized with X & Y.
Title: Re: Does the thread break?
Post by: David Cooper on 18/10/2018 20:47:19
Note that in the original question you can find the wording "are stationary". That already implies that there is a fabric of space which they are stationary relative to, so it invites answers like mine which are based on Lorentz Ether Theory rather than SR.

My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that.

The correct Einsteinian answers match the correct LET answers, other than in the mechanism (because they accept contradictory accounts as all being equally correct, whereas LET recognises only one account as accurate due to its more rational rejection of contradiction).

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I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something.

No, you aren't missing anything - so long as they stick with the original synchronisation, every simultaneous manoeuvre that they make will be simultaneous in the original frame, so the thread will snap on acceleration, then the replacement one will go slack as they decelerate.

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How can the leading ship X start its acceleration late?

If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.
Title: Re: Does the thread break?
Post by: mad aetherist on 18/10/2018 22:49:06
Note that in the original question you can find the wording "are stationary". That already implies that there is a fabric of space which they are stationary relative to, so it invites answers like mine which are based on Lorentz Ether Theory rather than SR.
My Einsteinian answer is that the thread doesnt stretch or slacken or snap. But i think i noticed in wiki that the knee-jerk Einsteinian answer is that there is no change -- & wiki says that a more considered Einsteinian answer needs to account for simultaneity, in which case the thread snaps. I will have to look into that.
The correct Einsteinian answers match the correct LET answers, other than in the mechanism (because they accept contradictory accounts as all being equally correct, whereas LET recognises only one account as accurate due to its more rational rejection of contradiction).
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I dont understand how there can ever be a simultaneity~synchronization issue for X & Y in A123 & B123 (i mean assuming that they successfully synchronize at some time before)(& assuming that they do the same thing at the same time up untill the event)(& during the event). During the event they always do exactly the same thing (accelerate or decelerate) at exactly the same time. So why the problem. Am i missing something.
No, you aren't missing anything - so long as they stick with the original synchronisation, every simultaneous manoeuvre that they make will be simultaneous in the original frame, so the thread will snap on acceleration, then the replacement one will go slack as they decelerate.
 Comment. No i dont see the thread snapping. Unless u mean from the g force of acceleration.
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How can the leading ship X start its acceleration late?
If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.
 Comment. Wow, if thats the Einsteinian answer then Einsteinians are crazier than i ever thort. I dont see how X or Y can ever see the thread snap in the Einsteinian universe -- X & Y must always think themselves to be stationary relative to each other.
And i dont see how O might see the thread snap. I get it that Einsteinian SR says that if X & Y are simultaneous in their frame then they cannot be simultaneous in any other frame (this rule is wrong)(but i can follow their logic)(albeit flawed). Thusly O will see a difference in activity of X compared to Y, but the thread doesnt snap in the XY frame, therefore the thread doesnt snap, therefore O cannot see the thread snap (still sticking to Einstein's SR rules in an Einstein universe). Anything else is batshit crazy.
Title: Re: Does the thread break?
Post by: David Cooper on 18/10/2018 23:17:56
Comment. No i dont see the thread snapping. Unless u mean from the g force of acceleration.

Length contraction of the ships will cause the thread to break on acceleration. The ships maintain the same distance between the centres of mass of each, but because they both contract in length, the thread will snap. This happens both in LET (Lorentz Ether Theory) and in SR.

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If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.
Comment. Wow, if thats the Einsteinian answer then Einsteinians are crazier than i ever thort. I dont see how X or Y can ever see the thread snap in the Einsteinian universe -- X & Y must always think themselves to be stationary relative to each other.

They are stationary relative to each other, but you make a good point - as they accelerate, the lead one will appear to accelerate faster than the trailing one from the point of view of the people in those two ships because the apparent distance between them will increase. The leading one will also appear to stop accelerating sooner while the trailing one appears to continue accelerating until it's up to the same speed, and yet both of them accelerated at the same rate for the same length of time. When they decelerate, if they do this without resynchronising clocks, the trailing one will appear to start its deceleration first (from the point of view of the people in those two ships) and then the leading one will appear to decelerate more strongly, and yet again the reality is that they both started their deceleration at the same time and decelerated at the same rate. This is again an LET description, but it takes you straight to the correct SR description too - it's much harder to work out what would happen if you only know SR because it's so easy to misunderstand the rules in SR. Your idea that the ships should appear to remain stationary relative to each other under acceleration from the point of view of the occupants in SR is incorrect.

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And i dont see how O might see the thread snap.

The thread either snaps or it doesn't. If it snaps and O can't see it, O needs to buy a bigger telescope.

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I get it that Einsteinian SR says that if X & Y are simultaneous in their frame then they cannot be simultaneous in any other frame (this rule is wrong)(but i can follow their logic)(albeit flawed). Thusly O will see a difference in activity of X compared to Y, but the thread doesnt snap in the XY frame, therefore the thread doesnt snap, therefore O cannot see the thread snap (still sticking to Einstein's SR rules in an Einstein universe). Anything else is batshit crazy. [/color]

There is no question of the thread snapping for some observers and not for others - no theory of relativity allows that. The thread will always snap if the acceleration starts at the same time for leading and following ship with clocks synchronised for the frame in which they begin from at rest. If they synchronise the clocks in advance for the target frame in which they'll be at rest once they've stopped accelerating, then the thread will go slack instead. If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again. This applies equally for LET and SR.
Title: Re: Does the thread break?
Post by: mad aetherist on 18/10/2018 23:58:10
Comment. No i dont see the thread snapping. Unless u mean from the g force of acceleration.

Length contraction of the ships will cause the thread to break on acceleration. The ships maintain the same distance between the centres of mass of each, but because they both contract in length, the thread will snap. This happens both in LET (Lorentz Ether Theory) and in SR.

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If they resynchronise their clocks after accelerating and do so on the basis that they're now stationary, they'll end up with the leading one running behind the other (when observed by a stationary observer in the original frame). That's why the leading one will start decelerating after the trailing one, but they will see themselves as simultaneously beginning that deceleration. If they don't resynchronise their clocks though, the leading ship will not start the deceleration late, and to the people in the two ships it will appear as if the leading one starts its deceleration first.
Comment. Wow, if thats the Einsteinian answer then Einsteinians are crazier than i ever thort. I dont see how X or Y can ever see the thread snap in the Einsteinian universe -- X & Y must always think themselves to be stationary relative to each other.

They are stationary relative to each other, but you make a good point - as they accelerate, the lead one will appear to accelerate faster than the trailing one from the point of view of the people in those two ships because the apparent distance between them will increase. The leading one will also appear to stop accelerating sooner while the trailing one appears to continue accelerating until it's up to the same speed, and yet both of them accelerated at the same rate for the same length of time. When they decelerate, if they do this without resynchronising clocks, the trailing one will appear to start its deceleration first (from the point of view of the people in those two ships) and then the leading one will appear to decelerate more strongly, and yet again the reality is that they both started their deceleration at the same time and decelerated at the same rate. This is again an LET description, but it takes you straight to the correct SR description too - it's much harder to work out what would happen if you only know SR because it's so easy to misunderstand the rules in SR. Your idea that the ships should appear to remain stationary relative to each other under acceleration from the point of view of the occupants in SR is incorrect.

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And i dont see how O might see the thread snap.

The thread either snaps or it doesn't. If it snaps and O can't see it, O needs to buy a bigger telescope.

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I get it that Einsteinian SR says that if X & Y are simultaneous in their frame then they cannot be simultaneous in any other frame (this rule is wrong)(but i can follow their logic)(albeit flawed). Thusly O will see a difference in activity of X compared to Y, but the thread doesnt snap in the XY frame, therefore the thread doesnt snap, therefore O cannot see the thread snap (still sticking to Einstein's SR rules in an Einstein universe). Anything else is batshit crazy. [/color]

There is no question of the thread snapping for some observers and not for others - no theory of relativity allows that. The thread will always snap if the acceleration starts at the same time for leading and following ship with clocks synchronised for the frame in which they begin from at rest. If they synchronise the clocks in advance for the target frame in which they'll be at rest once they've stopped accelerating, then the thread will go slack instead. If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again. This applies equally for LET and SR.
Yep, i dont understand SR & GR. Re LET i agree that the thread might snap, & that the distance tween center of mass of X & Y doesnt change etc, all of that is common sense. But in SR & GR space along any one line must contract etc evenly & equally along every point along that line, to infinity. Hencely if X contracts then so does Y & so does the string & so does space, hencely the string doesnt snap.

In addition nothing in SR or GR is real, hencely there can be no contraction of any sort anywhere, therefore the string doesnt snap.

In addition observer O might see an apparent change in velocity or position or length of X & string & Y, if that is what the rule dictates, & this might require that the string snap & be seen to snap, but the real string is not bound to actually do what O is bound to see. O will not see the string snap, because it doesnt snap.

Einstein's gamma is merely a math-trick to ensure that the speed of light is isotropic in all frames. But the resulting changes in length ticking etc are not real. Hencely not being real this allows the lengths tickings etc to be different for different observers. If they can be different in that way then clearly they are not real changes. Which begs the question what use is SR & GR. The answer is that they give good numbers for some realworld happenings. But u shoodnt then peer into any & every little baby interim term etc & say that that baby is itself real & any associated baby number good just because the big adult end term is a goodish number.
Title: Re: Does the thread break?
Post by: Halc on 19/10/2018 00:11:46
If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again.
A nit correction: Other way around.  It could be tight at first, and gets slack as the two ships approach the frame in which their clocks are synchronized, and then get tight again at the end.

Yep, i dont understand SR & GR. Re LET i agree that the thread might snap, & that the distance tween center of mass of X & Y doesnt change etc, all of that is common sense.
Well since both theories make the same predictions, SR will also predict the thread breaking then.
Title: Re: Does the thread break?
Post by: mad aetherist on 19/10/2018 01:34:58
If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again.
A nit correction: Other way around.  It could be tight at first, and gets slack as the two ships approach the frame in which their clocks are synchronized, and then get tight again at the end.

Yep, i dont understand SR & GR. Re LET i agree that the thread might snap, & that the distance tween center of mass of X & Y doesnt change etc, all of that is common sense.
Well since both theories make the same predictions, SR will also predict the thread breaking then.
No i dont think that i ever said that both make the same predictions. My understanding is that Lorentz relativity says that only objects contract, space does not contract (string stretches or slackens depending on aetherwind). Whereas Einstein says that space contracts, & everything in that space contracts in sympathy with the space (string never stretches nor slackens).
Title: Re: Does the thread break?
Post by: Halc on 19/10/2018 02:11:32
No i dont think that i ever said that both make the same predictions.
You probably didn't say that, but I did.  If your theory is different than standard neo-Lorentz theory, then perhaps your personal theory makes different predictions than either of them.

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Whereas Einstein says that space contracts, & everything in that space contracts in sympathy with the space (string never stretches nor slackens).
You're misinterpreting Einstein.  He says something like that in context of different frames, but you're considering the distance between X and Y in O's frame, and the space between them never changes in that frame since they're doing identical things at the same time.

In the accelerating frame of X or Y, Y is moving sooner than X, increasing the distance between them and breaking the string.
Title: Re: Does the thread break?
Post by: mad aetherist on 19/10/2018 04:14:04
No i dont think that i ever said that both make the same predictions.
You probably didn't say that, but I did.  If your theory is different than standard neo-Lorentz theory, then perhaps your personal theory makes different predictions than either of them.

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Whereas Einstein says that space contracts, & everything in that space contracts in sympathy with the space (string never stretches nor slackens).
You're misinterpreting Einstein.  He says something like that in context of different frames, but you're considering the distance between X and Y in O's frame, and the space between them never changes in that frame since they're doing identical things at the same time.

In the accelerating frame of X or Y, Y is moving sooner than X, increasing the distance between them and breaking the string.
Thats the bit i dont understand, how one can start accelerating before the other (X is in front so if X starts first then the string might snap)(u said Y starts first).
Title: Re: Does the thread break?
Post by: Halc on 19/10/2018 04:34:28
Thats the bit i dont understand, how one can start accelerating before the other (X is in front so if X starts first then the string might snap)(u said Y starts first).
I guess the OP was vague about this.  I thought "Y(line ahead)" meant that Y was in front.
Reverse the labels in my comments if X is in front.  In any frame of either ship after it starts its acceleration, the ship in front moves first, breaking the string.
Title: Re: Does the thread break?
Post by: David Cooper on 19/10/2018 18:42:02
If they start with the clocks synchronised for a frame half way in between and start with a slack line with just the right amount of slack in it, it could become tight half way through and then go slack again.
A nit correction: Other way around.  It could be tight at first, and gets slack as the two ships approach the frame in which their clocks are synchronized, and then get tight again at the end.

You're right. I realised that as soon as I'd switched off my computer and considered booting it up again to correct it, but decided just to leave it there as an intelligence test. Usually I can correct things like that the next day before anyone picks up on it, but you're too quick witted.
Title: Re: Does the thread break?
Post by: David Cooper on 19/10/2018 19:33:06
Yep, i dont understand SR & GR. Re LET i agree that the thread might snap, & that the distance tween center of mass of X & Y doesnt change etc, all of that is common sense. But in SR & GR space along any one line must contract etc evenly & equally along every point along that line, to infinity. Hencely if X contracts then so does Y & so does the string & so does space, hencely the string doesnt snap.

It is odd, because even in SR it should be recognised that the leading ship has accelerated at exactly the same rate as the trailing one. The acceleration of the leading ship appears to be stronger because of the changing synchronisation issues - if they keep resynchronising clocks while they accelerate, they'll both calculate that the leading ship is accelerating more strongly than the other regardless of what their gauges are telling them about the actual power put down. If they both adjust the power to try to maintain the right output for the amount of "corrected" time that's passed, the leading ship will accelerate less and the trailing one will accelerate more, and the result of this will be length-contraction of the distance between their centres of mass to match the length-contraction of each ship. They should notice at the end of this process that the trailing ship has used up more fuel though. This is a weird issue, so it's easy to see why a lot of people in the SR camp might produce incorrect answers for it, but the ones who apply SR correctly will get the same answers as we do by using LET. While it initially looks as if this illustrates another broken aspect of SR, it's actually quite easy to explain away - when they are up to speed and the leading rocket appears to be further ahead despite using the same amount of fuel, retrospectively it will simply look as if it began its acceleration first and stopped accelerating first too - the original idea that the beginning of the acceleration looked simultaneous for both ships at the time will now be regarded as an illusion.

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Einstein's gamma is merely a math-trick to ensure that the speed of light is isotropic in all frames. But the resulting changes in length ticking etc are not real. Hencely not being real this allows the lengths tickings etc to be different for different observers. If they can be different in that way then clearly they are not real changes. Which begs the question what use is SR & GR. The answer is that they give good numbers for some realworld happenings. But u shoodnt then peer into any & every little baby interim term etc & say that that baby is itself real & any associated baby number good just because the big adult end term is a goodish number.

SR and GR are indeed just a mathematical abstraction - they turn time into a space dimension (in which every fundamental component of anything is of infinite length) and they lose actual time from the model altogether, so they're left with a static block universe in which nothing ever happens and in which the future was never generated from the past in order of causation, rendering all the apparent causation fake. It's all a fraud. The original version of SR wasn't 4D, but it generated contradictions, so Minkowski tried to fix it by creating the mathematical abstraction, and in the 4D versions, light doesn't even have a speed because it reduces all the paths that it follows to zero length. Whenever Einsteinists talk about the speed of light being c, they're mixing incompatible models.

The worst consequences of GR are found when they look at black holes. Look at the nonsense that comes out of there when they talk about wormholes and about things in black holes disappearing from the universe due to weird things happening to the space and "time" dimensions. In reality, nothing weird happens there at all - apparent time simply slows because the local speed of light slows to zero (relative to the black hole) and all functionality stops, but time itself ticks on completely unslowed while the content of space there freezes up - it's merely impossible for any clock to tick along with it. The laws of LET don't break in black holes - they tell us exactly what's going on inside them, and it's all rational. Nothing can cross the event horizon inwards, never mind out, because it can't go faster than light, so all the material that falls "into" a black hole actually gets stuck just outside the event horizon until the energy density there becomes sufficiently high to generate a new event horizon to the other side of it. There's also no singularity on the inside - its just a globe of space in which light can't move and everything else is frozen in place too, scattered all throughout that globe. Some warping of that globe is possible though, and this becomes considerable when two black holes merge. I speculated recently that the two black holes might not merge fully, maintaining a separation where their event horizons meet, but I realise now that the energy density will be so high that the gap in between will have the speed of light in it reduced to zero too, so it just becomes like any other part of the interior of the black hole. Many physicists have wasted their entire careers studying fantasy physics by applying GR to black holes, pumping out all sorts of nonsense that misleads the public in the process. GR and SR have both been disproved, but they're incapable of accepting that reality.
Title: Re: Does the thread break?
Post by: Halc on 19/10/2018 21:31:50
It is odd, because even in SR it should be recognised that the leading ship has accelerated at exactly the same rate as the trailing one.
They have.  If some observer is in one of the ships, he’d not be able to tell which one he was on by noting a less-than scheduled acceleration, or by how long the thrust lasted. This is as stipulated in the OP.  The two ships could be a line of 100 ships, each one (except the end ones) being both the leading and the trailing ship of the one on either side respectively.

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The acceleration of the leading ship appears to be stronger because of the changing synchronisation issues - if they keep resynchronising clocks while they accelerate, they'll both calculate that the leading ship is accelerating more strongly than the other regardless of what their gauges are telling them about the actual power put down.
They’d calculate that the lead ship is effectively further from the equivalent gravity well so the time in the lead ship is dilated less.  Time moves faster for the lead ship than the trailing one, so it gets its thrusting done sooner in the accelerating frame of one ship or the other.  Except at the beginning and the end (when the rear ship finishes), they are never both stationary in any frame, so there is no frame where they can obviously re-sync their clocks as they move.

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If they both adjust the power to try to maintain the right output for the amount of "corrected" time that's passed, the leading ship will accelerate less and the trailing one will accelerate more, and the result of this will be length-contraction of the distance between their centres of mass to match the length-contraction of each ship. They should notice at the end of this process that the trailing ship has used up more fuel though.
Same fuel consumption since they’ll both be going the same speed at the end, just like they do with no speed correction to keep the proper distance between the ships constant like that.


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Minkowski tried to fix it by creating the mathematical abstraction, and in the 4D versions, light doesn't even have a speed because it reduces all the paths that it follows to zero length.
aaand out comes the stuff you can’t push on the main forum….
Title: Re: Does the thread break?
Post by: David Cooper on 19/10/2018 23:12:14
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Minkowski tried to fix it by creating the mathematical abstraction, and in the 4D versions, light doesn't even have a speed because it reduces all the paths that it follows to zero length.
aaand out comes the stuff you can’t push on the main forum….

What makes you think such points can't be made on the main forum? It is a correct description of SR and GR which is beyond dispute. The issue of light reducing all paths to zero length came up here in the physics section several years ago, and when I proved the point, the moderator who was on the other side of the argument suddenly disappeared and was never heard of here again, despite being asked by a neutral participant to respond to what I had demonstrated. (He didn't just happen to pop off at that moment - I recently found that he's still very much alive and active on a maths forum.)

A correct understanding of Spacetime shows that every point in the universe is zero distance away from every other point for light. The faster an object moves, the shorter the distance it needs to travel through space to get from A to B, and the shorter the length of the trip in the time dimension will be too. When you get to objects moving at 0.99999999999c, they're already making the trip close to zero length and zero time, and there's no possibility of light travelling from A to B covering a longer distance or taking a longer time than such high speed objects to make that trip. It is futile to pretend that this is not a feature of the 4D models. You are logically forced to accept that that is how they are structured.
Title: Re: Does the thread break?
Post by: Halc on 19/10/2018 23:38:07
A correct understanding of Spacetime shows that every point in the universe is zero distance away from every other point for light.
It is what is approached (locally) as an object approaches light speed, yes.
But there is no 'for light'.  You make it sound like from the reference frame of light, and there is no such valid reference frame.  It was the reference to light not having a speed that was wrong.  Light moves locally at fixed speed in all frames, and there is thus no frame in which a photon is locally stationary.
Title: Re: Does the thread break?
Post by: mad aetherist on 20/10/2018 00:54:56
Yep, i dont understand SR & GR. Re LET i agree that the thread might snap, & that the distance tween center of mass of X & Y doesnt change etc, all of that is common sense. But in SR & GR space along any one line must contract etc evenly & equally along every point along that line, to infinity. Hencely if X contracts then so does Y & so does the string & so does space, hencely the string doesnt snap.
It is odd, because even in SR it should be recognised that the leading ship has accelerated at exactly the same rate as the trailing one. The acceleration of the leading ship appears to be stronger because of the changing synchronisation issues - if they keep resynchronising clocks while they accelerate, they'll both calculate that the leading ship is accelerating more strongly than the other regardless of what their gauges are telling them about the actual power put down. If they both adjust the power to try to maintain the right output for the amount of "corrected" time that's passed, the leading ship will accelerate less and the trailing one will accelerate more, and the result of this will be length-contraction of the distance between their centres of mass to match the length-contraction of each ship. They should notice at the end of this process that the trailing ship has used up more fuel though. This is a weird issue, so it's easy to see why a lot of people in the SR camp might produce incorrect answers for it, but the ones who apply SR correctly will get the same answers as we do by using LET. While it initially looks as if this illustrates another broken aspect of SR, it's actually quite easy to explain away - when they are up to speed and the leading rocket appears to be further ahead despite using the same amount of fuel, retrospectively it will simply look as if it began its acceleration first and stopped accelerating first too - the original idea that the beginning of the acceleration looked simultaneous for both ships at the time will now be regarded as an illusion.
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Einstein's gamma is merely a math-trick to ensure that the speed of light is isotropic in all frames. But the resulting changes in length ticking etc are not real. Hencely not being real this allows the lengths tickings etc to be different for different observers. If they can be different in that way then clearly they are not real changes. Which begs the question what use is SR & GR. The answer is that they give good numbers for some realworld happenings. But u shoodnt then peer into any & every little baby interim term etc & say that that baby is itself real & any associated baby number good just because the big adult end term is a goodish number.
SR and GR are indeed just a mathematical abstraction - they turn time into a space dimension (in which every fundamental component of anything is of infinite length) and they lose actual time from the model altogether, so they're left with a static block universe in which nothing ever happens and in which the future was never generated from the past in order of causation, rendering all the apparent causation fake. It's all a fraud. The original version of SR wasn't 4D, but it generated contradictions, so Minkowski tried to fix it by creating the mathematical abstraction, and in the 4D versions, light doesn't even have a speed because it reduces all the paths that it follows to zero length. Whenever Einsteinists talk about the speed of light being c, they're mixing incompatible models.

The worst consequences of GR are found when they look at black holes. Look at the nonsense that comes out of there when they talk about wormholes and about things in black holes disappearing from the universe due to weird things happening to the space and "time" dimensions. In reality, nothing weird happens there at all - apparent time simply slows because the local speed of light slows to zero (relative to the black hole) and all functionality stops, but time itself ticks on completely unslowed while the content of space there freezes up - it's merely impossible for any clock to tick along with it. The laws of LET don't break in black holes - they tell us exactly what's going on inside them, and it's all rational. Nothing can cross the event horizon inwards, never mind out, because it can't go faster than light, so all the material that falls "into" a black hole actually gets stuck just outside the event horizon until the energy density there becomes sufficiently high to generate a new event horizon to the other side of it. There's also no singularity on the inside - its just a globe of space in which light can't move and everything else is frozen in place too, scattered all throughout that globe. Some warping of that globe is possible though, and this becomes considerable when two black holes merge. I speculated recently that the two black holes might not merge fully, maintaining a separation where their event horizons meet, but I realise now that the energy density will be so high that the gap in between will have the speed of light in it reduced to zero too, so it just becomes like any other part of the interior of the black hole. Many physicists have wasted their entire careers studying fantasy physics by applying GR to black holes, pumping out all sorts of nonsense that misleads the public in the process. GR and SR have both been disproved, but they're incapable of accepting that reality.
I see that in the upwards accelerating elevator analogy Einsteinians say  that a clock near the ceiling will tick faster than a clock near the floor, thusly giving perfect equivalence (to an elevator stationed in a g field). The X-string-Y spaceship analogy is probly partly a horizontal rehash of the vertical elevator thort-X (talking bout the Einsteinian universe here). So, initially X & Y are stationary, then X starts acceleration (to the left in my world) at the same true time as Y, & during acceleration Einsteinians deem that X's clock ticks faster than Y's clock, & X (unknowingly) accelerates more than Y (because X uses fuel at a greater rate), & eventually the string must snap, & when it snaps then X & Y will both see it snap (albeit Y first, X later). I think that that is one (Einsteinian) answer. But i have some queries.

(1) X uses fuel at a greater rate than Y. But initially X uses fuel at the same  rate as Y. Then initially they must both accelerate equally (both accelerations are say 1.0). But somehow X's clock immediately starts to tick faster than Y's, even tho they are momentarily accelerating at the same rate. Somewhere in that first second X's clock's ticking jumps from 1.0 to 1.1, & X's fuel flow jumps from 1.0 to 1.1, & X's accel jumps from 1.0 to 1.1 (X & Y start with fuelflow-needles & acceleration-needles sitting on 1.0)(& all needles of course remain on 1.0 throughout).  Where exactly in that first second do these three jumps happen? (This refers to X, Y doesnt suffer any jumps)(X of course doesnt see or feel the jumps)(but X & Y will both see that something is wrong, ie they will see that the other ship is drifting)(& both will know that something is wrong when the string eventually snaps). So, where exactly in that first second do these three jumps happen?

(2) X & Y are stationary at 0.0000 sec, & both ships have the same length at 0.0000 sec. Both start accelerating, & initially both start to shorten, equally.  If the jumps happen at 0.0001 sec, then at that time spaceship X must start to shorten at a greater rate than Y.  Which happens first, shortening of X, or faster ticking in X. (2a) Do clocks tick faster because of shortening, or (2b) does shortening happen because the clocks tick faster. (2c) Or can both faster ticking & shortening happen together, at the same time. (3) Thems three questions apply to both X & to Y, but similar questions can be asked re the jump in ticking & shortening in X, u can call these three questions (3a) & (3b) & (3c).

(4) A slight variation. Spaceship X is identical to Y, but Y has moved its clock & thrusters to its nose (Y's thrusters still point backwards)(to my right)(whilst X's clock & thrusters are at the tail as built)(& point to my right)(ie point to Y).
Initially X & Y are stationary (as in (1)), then immediately both accelerate (as in (1)), etc (going to my left)(as in (1)).  Momentarily, both clocks tick at the same rate (as in (1)) etc. But, unlike in (1), will not Y be the ship that accelerates faster etc (4a) (in which case the string will slacken), & (4b) will Y need to fill out an accident report, ie will it eventually collide with X.
Title: Re: Does the thread break?
Post by: mad aetherist on 20/10/2018 00:59:50
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Minkowski tried to fix it by creating the mathematical abstraction, and in the 4D versions, light doesn't even have a speed because it reduces all the paths that it follows to zero length.
aaand out comes the stuff you can’t push on the main forum….

What makes you think such points can't be made on the main forum? It is a correct description of SR and GR which is beyond dispute. The issue of light reducing all paths to zero length came up here in the physics section several years ago, and when I proved the point, the moderator who was on the other side of the argument suddenly disappeared and was never heard of here again, despite being asked by a neutral participant to respond to what I had demonstrated. (He didn't just happen to pop off at that moment - I recently found that he's still very much alive and active on a maths forum.)

A correct understanding of Spacetime shows that every point in the universe is zero distance away from every other point for light. The faster an object moves, the shorter the distance it needs to travel through space to get from A to B, and the shorter the length of the trip in the time dimension will be too. When you get to objects moving at 0.99999999999c, they're already making the trip close to zero length and zero time, and there's no possibility of light travelling from A to B covering a longer distance or taking a longer time than such high speed objects to make that trip. It is futile to pretend that this is not a feature of the 4D models. You are logically forced to accept that that is how they are structured.
So, at high speeds trips can take almost zero time, in which case if c=c then it can be said that the trip distance is also almost zero. Interesting.
Title: Re: Does the thread break?
Post by: Halc on 20/10/2018 01:12:24
So, at high speeds trips can take almost zero time, in which case if c=c then it can be said that the trip distance is also almost zero.
In the frame where the speed is high, the trip takes a long time.  A trip cannot be taken at c, but it can get arbitrarily close.  The distance traveled is still a long way and takes a long time.

In the frame of the thing 'making the trip', it has negligible speed, so it just sits still for a short time.
Title: Re: Does the thread break?
Post by: Halc on 20/10/2018 01:31:42
I see that in the upwards accelerating elevator analogy Einsteinians say  that a clock near the ceiling will tick faster than a clock near the floor, thusly giving perfect equivalence (to an elevator stationed in a g field).
They would not word it that way, but essentially, yes.
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The X-string-Y spaceship analogy is probly partly a horizontal rehash of the vertical elevator thort-X (talking bout the Einsteinian universe here). So, initially X & Y are stationary, then X starts acceleration (to the left in my world) at the same true time as Y, & during acceleration Einsteinians deem that X's clock ticks faster than Y's clock
No.  Both ships are doing the same thing, so both clocks tick at the same rate in the frame in which they were inititally stationary.  If you’re talking about a different frame, it needs to be specified.  Absent that, I assume we’re using the original frame.
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& X (unknowingly) accelerates more than Y (because X uses fuel at a greater rate),
Nope.  Identical.
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& eventually the string must snap, & when it snaps then X & Y will both see it snap (albeit Y first, X later).
If it can’t stretch at all, it snaps right away.  If it lasts a while and breaks in the middle, who sees it first seems frame dependent, but yes, I think Y sees it first in O’s frame because Y is moving towards the break, meeting the light from it, while X is moving away and the light needs to go further to catch up.
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(1) X uses fuel at a greater rate than Y.
Same rate the whole time.
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But somehow X's clock immediately starts to tick faster than Y's
No
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So, where exactly in that first second do these three jumps happen?
There are never any ‘jumps’ no matter the frame you choose.  There are no discontinuities in this example.
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If the jumps happen at 0.0001 sec, then at that time spaceship X must start to shorten at a greater rate than Y.
There are no jumps.  Both ship lengths dilate down equally.
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(4) A slight variation. Spaceship X is identical to Y, but Y has moved its clock & thrusters to its nose (Y's thrusters still point backwards)(to my right)(whilst X's clock & thrusters are at the tail as built)(& point to my right)(ie point to Y).
There would be no difference in this scenario.  Ships still accelerate identically and the string breaks.  Where the engine is has no significant imact.
Title: Re: Does the thread break?
Post by: David Cooper on 20/10/2018 21:13:17
But there is no 'for light'.  You make it sound like from the reference frame of light, and there is no such valid reference frame.

The word "for" does not mean light is at rest in any frame of reference. There has to be some reality for light though, and we have more and more extreme frames reducing the distance that fast moving matter has to travel through the space dimensions and time dimension with these tending to zero. For light, the distance through space and time dimensions is always going to be zero because it can't be anything greater without making it possible for it to be overtaken by matter.

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It was the reference to light not having a speed that was wrong.  Light moves locally at fixed speed in all frames, and there is thus no frame in which a photon is locally stationary.

If it's covering all its trips in zero distance, it makes no sense to say it has a speed. It merely has an apparent speed in 4D models. When you say there is no frame in which it's stationary, that's only because the range of frames is infinite, never quite reaching light's equivalent of a frame, but light itself still has a kind of frame which merely isn't a member of the set of frames of reference. In reality though, the 4D models are merely a mathematical abstraction, so we're arguing about how many angels can dance on the head of a pin. For a real description of nature, you need to use LET instead.
Title: Re: Does the thread break?
Post by: David Cooper on 20/10/2018 21:50:00
I see that in the upwards accelerating elevator analogy Einsteinians say  that a clock near the ceiling will tick faster than a clock near the floor, thusly giving perfect equivalence (to an elevator stationed in a g field).

The phenomenon of relativity (I'm not talking about a theory of relativity here, but about nature itself) provides a lot of coincidences, and one of those coincidences is that the maths of acceleration matches up to the maths of gravitational pull, but they are not the same thing - it's a fun game using GR to interpret the scenario in question, but it is nothing more than a game, and it certainly doesn't provide any real understanding of what's going on.

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(1) X uses fuel at a greater rate than Y. But initially X uses fuel at the same  rate as Y. Then initially they must both accelerate equally (both accelerations are say 1.0). But somehow X's clock immediately starts to tick faster than Y's, even tho they are momentarily accelerating at the same rate. Somewhere in that first second X's clock's ticking jumps from 1.0 to 1.1, & X's fuel flow jumps...

What actually happens is that the change in synchronisation leads to them continually rewriting the history of the start times for the acceleration for the two ships because they always try to account for it using the frame in which they are currently at rest, and this repeatedly pushes forward the claimed start time for the leading ship's acceleration. An earlier start leads to the apparent gap between the two ships continually opening wider.

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Where exactly in that first second do these three jumps happen?

They're continual - an infinite series of tiny jumps, but they aren't really jumps because they're actually just recalculations about the history of events, rewriting the time gap between the two ships starting their accelerations. If you don't rewrite history in that way and just stick to a single frame, there is no difference between the two accelerations and no rewriting of the start times.

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(2a) Do clocks tick faster because of shortening, or (2b) does shortening happen because the clocks tick faster. (2c) Or can both faster ticking & shortening happen together, at the same time.

If you're continually changing the frame you're using to analyse events (as a person on either ship automatically does when looking with their own eyes at the events unfolding for them), then no change in ticking rate or ship lengths is seen at all. These changes are only measured if you stick to one frame for all your measurements. Choose one frame and you'll measure that the clocks are running slower and slower while the ships contract in length, but choose another frame and you can measure the clocks ticking faster instead while the ships lengthen.

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(4) A slight variation. Spaceship X is identical to Y, but Y has moved its clock & thrusters to its nose (Y's thrusters still point backwards)(to my right)(whilst X's clock & thrusters are at the tail as built)(& point to my right)(ie point to Y).

Even if you stick the rockets on Y on the end of a long rod that puts them ahead of X and if you also have a long rod putting X's rockets behind Y, the behaviour of the two ships won't change at all, just so long as the person pushing the start button is in the same place as before. The signal will take time to get to the rockets, and that delay will cancel out the positioning differences with the rockets. It's all about synchronisation.
Title: Re: Does the thread break?
Post by: mad aetherist on 21/10/2018 02:19:22
I see that in the upwards accelerating elevator analogy Einsteinians say  that a clock near the ceiling will tick faster than a clock near the floor, thusly giving perfect equivalence (to an elevator stationed in a g field).

The phenomenon of relativity (I'm not talking about a theory of relativity here, but about nature itself) provides a lot of coincidences, and one of those coincidences is that the maths of acceleration matches up to the maths of gravitational pull, but they are not the same thing - it's a fun game using GR to interpret the scenario in question, but it is nothing more than a game, and it certainly doesn't provide any real understanding of what's going on.

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(1) X uses fuel at a greater rate than Y. But initially X uses fuel at the same  rate as Y. Then initially they must both accelerate equally (both accelerations are say 1.0). But somehow X's clock immediately starts to tick faster than Y's, even tho they are momentarily accelerating at the same rate. Somewhere in that first second X's clock's ticking jumps from 1.0 to 1.1, & X's fuel flow jumps...

What actually happens is that the change in synchronisation leads to them continually rewriting the history of the start times for the acceleration for the two ships because they always try to account for it using the frame in which they are currently at rest, and this repeatedly pushes forward the claimed start time for the leading ship's acceleration. An earlier start leads to the apparent gap between the two ships continually opening wider.

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Where exactly in that first second do these three jumps happen?

They're continual - an infinite series of tiny jumps, but they aren't really jumps because they're actually just recalculations about the history of events, rewriting the time gap between the two ships starting their accelerations. If you don't rewrite history in that way and just stick to a single frame, there is no difference between the two accelerations and no rewriting of the start times.

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(2a) Do clocks tick faster because of shortening, or (2b) does shortening happen because the clocks tick faster. (2c) Or can both faster ticking & shortening happen together, at the same time.

If you're continually changing the frame you're using to analyse events (as a person on either ship automatically does when looking with their own eyes at the events unfolding for them), then no change in ticking rate or ship lengths is seen at all. These changes are only measured if you stick to one frame for all your measurements. Choose one frame and you'll measure that the clocks are running slower and slower while the ships contract in length, but choose another frame and you can measure the clocks ticking faster instead while the ships lengthen.

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(4) A slight variation. Spaceship X is identical to Y, but Y has moved its clock & thrusters to its nose (Y's thrusters still point backwards)(to my right)(whilst X's clock & thrusters are at the tail as built)(& point to my right)(ie point to Y).

Even if you stick the rockets on Y on the end of a long rod that puts them ahead of X and if you also have a long rod putting X's rockets behind Y, the behaviour of the two ships won't change at all, just so long as the person pushing the start button is in the same place as before. The signal will take time to get to the rockets, and that delay will cancel out the positioning differences with the rockets. It's all about synchronisation.
Yes i like that rocket on long rod stuff. But note that i said that the clocks & thrusters are both together on X & Y, & i also meant but didnt actually say the start button (& Captain) etc. So, if all of the essentials were on the ends of your long rods on X & Y such that X's were then behind Y's then i guess that the string would be affected in the opposite way (eg slacking instead of snapping).
Title: Re: Does the thread break?
Post by: mad aetherist on 21/10/2018 02:25:28
I see that in the upwards accelerating elevator analogy Einsteinians say  that a clock near the ceiling will tick faster than a clock near the floor, thusly giving perfect equivalence (to an elevator stationed in a g field).
They would not word it that way, but essentially, yes.
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The X-string-Y spaceship analogy is probly partly a horizontal rehash of the vertical elevator thort-X (talking bout the Einsteinian universe here). So, initially X & Y are stationary, then X starts acceleration (to the left in my world) at the same true time as Y, & during acceleration Einsteinians deem that X's clock ticks faster than Y's clock
No.  Both ships are doing the same thing, so both clocks tick at the same rate in the frame in which they were inititally stationary.  If you’re talking about a different frame, it needs to be specified.  Absent that, I assume we’re using the original frame.
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& X (unknowingly) accelerates more than Y (because X uses fuel at a greater rate),
Nope.  Identical.
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& eventually the string must snap, & when it snaps then X & Y will both see it snap (albeit Y first, X later).
If it can’t stretch at all, it snaps right away.  If it lasts a while and breaks in the middle, who sees it first seems frame dependent, but yes, I think Y sees it first in O’s frame because Y is moving towards the break, meeting the light from it, while X is moving away and the light needs to go further to catch up.
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(1) X uses fuel at a greater rate than Y.
Same rate the whole time.
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But somehow X's clock immediately starts to tick faster than Y's
No
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So, where exactly in that first second do these three jumps happen?
There are never any ‘jumps’ no matter the frame you choose.  There are no discontinuities in this example.
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If the jumps happen at 0.0001 sec, then at that time spaceship X must start to shorten at a greater rate than Y.
There are no jumps.  Both ship lengths dilate down equally.
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(4) A slight variation. Spaceship X is identical to Y, but Y has moved its clock & thrusters to its nose (Y's thrusters still point backwards)(to my right)(whilst X's clock & thrusters are at the tail as built)(& point to my right)(ie point to Y).
There would be no difference in this scenario.  Ships still accelerate identically and the string breaks.  Where the engine is has no significant impact.
Hmmmmm -- I must admit that i dont understand. But ok what about this -- X & Y are going along at speed to the left together with a new tight thread tween -- & then X & Y decelerate at the same time & rate etc -- what happens now?
Title: Re: Does the thread break?
Post by: Halc on 21/10/2018 04:57:48
I must admit that i dont understand. But ok what about this -- X & Y are going along at speed to the left together with a new tight thread tween -- & then X & Y decelerate at the same time & rate etc -- what happens now?
'At the same time' in which frame?  The answer very much depends on this.

Assuming you mean simultaneously in your observer O's frame in which X and Y are going along at speed, then the string will go slack.
Title: Re: Does the thread break?
Post by: Halc on 21/10/2018 05:05:31
There has to be some reality for light though, and we have more and more extreme frames reducing the distance that fast moving matter has to travel through the space dimensions and time dimension with these tending to zero.
I already pointed out that this is wrong.  A fast thing still needs to go through the full distance.  My twin going to some star 10 light years away is going to need at least 10 years to get there at any speed.  That his clock doesn't log that doesn't make the distance shorter.
If you look at it in his frame, he's not going anywhere, so it just takes say 1 year for the star to travel the one light year to him, and those events (star location at start and end) were always that close in that frame.

The only frame that actually reduce or otherwise alter lengths are accelerating frames.
Title: Re: Does the thread break?
Post by: mad aetherist on 21/10/2018 07:19:28
I must admit that i dont understand. But ok what about this -- X & Y are going along at speed to the left together with a new tight thread tween -- & then X & Y decelerate at the same time & rate etc -- what happens now?
'At the same time' in which frame?  The answer very much depends on this.

Assuming you mean simultaneously in your observer O's frame in which X and Y are going along at speed, then the string will go slack.
Ok, then what about in X's frame & in Y's frame?
Title: Re: Does the thread break?
Post by: Halc on 21/10/2018 15:12:10
Ok, then what about in X's frame & in Y's frame?[/quote]
That's your original scenario A.  The ships start accelerating in the frame in which they're initially at rest.  The string breaks.
Title: Re: Does the thread break?
Post by: mad aetherist on 21/10/2018 21:11:37
I must admit that i dont understand. But ok what about this -- X & Y are going along at speed to the left together with a new tight thread tween -- & then X & Y decelerate at the same time & rate etc -- what happens now?
'At the same time' in which frame?  The answer very much depends on this.
Assuming you mean simultaneously in your observer O's frame in which X and Y are going along at speed, then the string will go slack.
Ok, then what about in X's frame & in Y's frame?

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Ok, then what about in X's frame & in Y's frame?
That's your original scenario A.  The ships start accelerating in the frame in which they're initially at rest.  The string breaks.
No its scenario B2, X & Y are going along & then decelerate. There are 3 questions in B2, they can be called B2X B2Y & B2O, ie what do observers X & Y & O see, & u in effect answered B2O (what does O see)(O is stationary) & my question was in effect B2X & B2Y (what do X & Y see).
Title: Re: Does the thread break?
Post by: David Cooper on 21/10/2018 22:57:37
There has to be some reality for light though, and we have more and more extreme frames reducing the distance that fast moving matter has to travel through the space dimensions and time dimension with these tending to zero.
I already pointed out that this is wrong.

It isn't wrong - in the 4D geometry you can always consider a moving object to be stationary, thereby allowing it to make a journey from one point in Spacetime to another just by sitting still in the space dimensions and moving through the time dimension - such a path always exists for any object, and the smaller the number of ticks its clock makes between those two points, the shorter the length of that path will be through time. For an object moving a tiny fraction more slowly than light, it will reduce the distance of what appears to be a long trip to other observers to zero length and almost zero time. Light will reduce any similar trip to zero distance and zero time.

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A fast thing still needs to go through the full distance.

The distance is always zero in the space dimensions of 4D Spacetime for any travelling object - not just light. It is only the time dimension that the lengths vary for different objects.

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My twin going to some star 10 light years away is going to need at least 10 years to get there at any speed.  That his clock doesn't log that doesn't make the distance shorter.

He says that he isn't travelling any distance at all, but that you are, but he's wrong too. In 4D Spacetime, you are both travelling zero distance through space for your "trips".

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If you look at it in his frame, he's not going anywhere, so it just takes say 1 year for the star to travel the one light year to him, and those events (star location at start and end) were always that close in that frame.

The only frame that actually reduce or otherwise alter lengths are accelerating frames.

The important point is that these zero-length paths are always available to matter (except that they aren't zero through the time dimension, but they can get close to zero for that too), and they must also be available to light too (with the length through the time dimension actually reaching zero). And if that sounds bonkers, don't blame me. It's because SR is bonkers.
Title: Re: Does the thread break?
Post by: Halc on 22/10/2018 00:45:36
Ok, then what about in X's frame
Quote from: Halc
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Ok, then what about in X's frame & in Y's frame?
That's your original scenario A.  The ships start accelerating in the frame in which they're initially at rest.  The string breaks.
No its scenario B2, X & Y are going along & then decelerate.
You asked to do it in X and Y's frame.  In that frame, X and Y are stationary, not decelerating.  In that frame, X and Y start stationary, and accelerate to some speed, which is scenario A.

Anyway, I think I see what you mean.  Describe the situation from frame O, but X and Y start 'declerating' simultaneously as defined in their own frame, not in O's frame.  Then yes the string breaks because in O's frame, Y starts slowing down first, immediately breaking the string.
Title: Re: Does the thread break?
Post by: mad aetherist on 22/10/2018 01:39:30
Ok, then what about in X's frame
Quote from: Halc
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Ok, then what about in X's frame & in Y's frame?
That's your original scenario A.  The ships start accelerating in the frame in which they're initially at rest.  The string breaks.
No its scenario B2, X & Y are going along & then decelerate.
You asked to do it in X and Y's frame.  In that frame, X and Y are stationary, not decelerating.  In that frame, X and Y start stationary, and accelerate to some speed, which is scenario A.

Anyway, I think I see what you mean.  Describe the situation from frame O, but X and Y start 'declerating' simultaneously as defined in their own frame, not in O's frame.  Then yes the string breaks because in O's frame, Y starts slowing down first, immediately breaking the string.
So, in every scenario ship X & ship Y apart from any obvious mechanical strains must suffer a relativistic strain along their full lengths (but praps moreso near midlength) which is stretching or compressing ship X (& similarly ship Y), & this strain must result in a relativistic force-stress.
Or, there is a relativistic strain but not a realworld strain hencely no associated realworld force-stress (force).
And i suspect that that strain & force-stress depends on whether u are observing from XY or from O.
Title: Re: Does the thread break?
Post by: Halc on 22/10/2018 03:08:45
So, in every scenario ship X & ship Y apart from any obvious mechanical strains must suffer a relativistic strain along their full lengths (but praps moreso near midlength) which is stretching or compressing ship X (& similarly ship Y), & this strain must result in a relativistic force-stress.
A typical rocket is one piece and has all the thrust coming from the rear, which is going to naturally put compressive stress on the thing, and the relativistic shortening I suppose will relieve that a tiny bit.
A railroad train on the other hand is pulled from the front, and if it is long enough, no amount of strength will prevent the train from breaking due to the additional tension relativity adds.  The limit to the length of the train, as a function of the acceleration of the engine, is called the Rindler Horizon, which is the acceleration equivalent of the event horizon of a black hole.  So when you accelerate with your car from a stop sign, you define such a Rindler Horizon somewhere behind you where an object would need to move at light speed in order to stay stationary with you in your frame.  The harder you accelerate, the closer that horizon is.

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Or, there is a relativistic strain but not a realworld strain hencely no associated realworld force-stress (force).
It's quite real.  The string really breaks due to actual stress.

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And i suspect that that strain & force-stress depends on whether u are observing from XY or from O.
It has nothing to do with being observed or not, or by whom.
Title: Re: Does the thread break?
Post by: David Cooper on 22/10/2018 22:49:58
So, in every scenario ship X & ship Y apart from any obvious mechanical strains must suffer a relativistic strain along their full lengths (but praps moreso near midlength) which is stretching or compressing ship X (& similarly ship Y), & this strain must result in a relativistic force-stress.
Or, there is a relativistic strain but not a realworld strain hencely no associated realworld force-stress (force).
And i suspect that that strain & force-stress depends on whether u are observing from XY or from O.

We can't tell if a ship is accelerating or decelerating, so it may be contracting in length or extending (uncontracting). If we could measure the difference, we would break relativity and identify the absolute frame. Any delay in contracting or extending should produce a stretching or compression force. This will be a smaller effect than the compression or stretching of the object caused by the force being applied to change its speed, but we can ignore any such compression or stretching from the push or pull because it will be the same for both cases - the effect we're looking for will be much smaller, but it will either be a compression or a stretch. I suspect it will be cancelled out in some way though so that it can't be measured, and it's easy to see straight away that any ruler-like tool you use to measure the effect will be affected equally, so it will mis-read and hide the effect. But what about a light clock with mirrors at opposite ends of a space ship? Well, again, if you compare that with a smaller light clock, both will be compressed or stretched in the same way and the effect we're looking for will be masked.

Any attempt to measure the ship's length by an observer who isn't accelerating will always measure the ship as contracting if he thinks it's accelerating and extending if he thinks it's decelerating, so again the effect we're looking for is hidden.

What if we could just watch two atoms though and measure the distance between them visually? The atoms themselves should contract or extend, and the spacing between them needs to adjust accordingly, so if it's a deceleration, they should keep finding themselves too close together and push apart a bit more often than the opposite. The trouble with that is that they're already going to be moving around a lot and the effect we want to see is infinitesimal by comparison, but the question is whether it is in principle detectable or not. Also, if the acceleration is severe, the compression or stretch won't be so terribly small. What happens to a camera that's looking in from the side? The film/CCD isn't a single point, so it will be suffering from the same compression or stretch and the effect we're looking for will be masked. The shape of the atoms will also be stretched or compressed to the same extent, so any stretch or compression of the space between them will be matched by distortions of their own shape, and with the same distortions applying to the camera, there's no way to reveal the reality as to whether there's compression or stretch. It's a hopeless task - the phenomenon of relativity wins every time, hiding the truth from us.
Title: Re: Does the thread break?
Post by: David Cooper on 23/10/2018 23:01:05
I've come up with an idea for an experiment.

Imagine a rocket with a lot of mass aligned perpendicular to its direction of travel (to avoid it having any length). Sticking out ahead of it though, and behind, are four rods: two pointing ahead and two behind. On the ends of one of each pair of rods is a large mass, while the rods themselves weigh very little. The idea here is that when we accelerate/decelerate the rods, any stretch or compression caused by length contraction/extension will take different amounts of time to adjust the length of a rod with a large mass on the end of it compared with the rod next to it with no such mass at the end.

We're looking for a difference in length between the two rods of each pair, but we'll only get that during an acceleration/deceleration, and while there's a force applying, there will already be compression or stretch in play in the rods due to the acceleration itself (while any length contraction adjustment is a much smaller effect). The rods ahead of the ship will be compressed and the ones behind will be stretched. If we always apply the same rate of acceleration, the difference between the two rods in a pair will always be the same - the ones with the large mass at their end will be worse affected than their partner, the leading one being more compressed and the trailing one being more extended. We can make a mark on the rod that's longer than its partner during acceleration to show how far along it its partner reaches. Any effect from length-contraction adjustments should show up as variations in length of the shorter rod (only shorter during acceleration) away from that mark.

It may be worth naming the rods, so F means forward-pointing while R means rearward. H means heavy and L means light (though we're really talking about mass). We thus have four rods called: FH, FL, RH and RL. The compression under acceleration will make FH shorter than FL, and stretch will make RH longer than RL, so we can put marks on FL and RH to show how far the other rods will reach along them, and these marks can be called MF and MR.

At the moment when the ship makes a transition from deceleration to acceleration, there should be no extra compression or stretch from length contraction, so the ends of the shorter rods should exactly meet their mark. Here's the important point: this means that any variation at other times should allow us to identify the absolute frame (unless I've missed some important factor).

Case 1:-

If the rocket is accelerating and there is a change in the amount of length contraction needed for the current speed, that means a contraction has to be applied through the rods becoming slightly more stretched (RH and RL) or less compressed (FH and FL), and the ones with and without large masses at the end of them should take different lengths of time to adjust. The mass of the ship is much greater than the large masses at the ends of the rods, but because we also have rods pointing both ways, they have to change length without the ship end of any rods accelerating at a different rate from the ship - it's the ends with the masses on the end of them that must migrate.

Rod RH (already stretched by the acceleration) has a bit of extra effective stretch added to it by length contraction, as does RL, and it should take longer to adjust for this contraction than RL because rod RL has more work to do to haul the large mass in, so the end of RL should move away from mark MR, and it should do so in a direction taking it even further away from the end of RH.

[Rods FH and FL are more complex, so I'll leave them till later.]

Case 2:-

If the rocket is decelerating, the rods will have to extend rather than contracting, so the rods should become slightly more compressed (RH and RL) or less stretched (FH and FL), and the ones with and without large masses at the end should again take different lengths of time to adjust.

Rods FH and FL are easier to handle in case 2 because they will behave in a similar way to FH and FL did in case 1, the difference being that we're adding compressions rather than stretches. They are compressed by the acceleration and now have a bit of effective compression added by the length extension, so FH will take longer to respond and its end will fall short of mark MF on FL.

Rod RH (already stretched by the acceleration) should have a bit of effective compression added to it by length extension (or decontraction), as does RL, removing some of the stretch and allowing the rod to lengthen, but I'm not sure how it would react. Is it a hindrance as before, or is it now going to help extend the rod more quickly? If the latter, then it could hide the effect we're trying to see, but remember that it should still show up when the ship is momentarily stationary (moving from deceleration to acceleration), because at that point the ends of the shorter rods would line up with the marks. Either way then, we should have a method by which the absolute frame could be identified, unless there's a fault somewhere in the argument (which I fully expect to be the case, but if it turns out that there isn't, it would be a shame to miss the experiment that finds the aether by assuming that no such experiment can exist). This looks viable, but I've been here before several times with ideas for experiments that looked as if they could break relativity, and I haven't put a lot of time into attacking this one yet, so don't get excited. I've just posted it up front on the off chance that it might stand up, and if it does, the time and date stamp on it could be handy.

Demolitions invited.
Title: Re: Does the thread break?
Post by: mad aetherist on 24/10/2018 00:00:50
I've come up with an idea for an experiment.

Imagine a rocket with a lot of mass aligned perpendicular to its direction of travel (to avoid it having any length). Sticking out ahead of it though, and behind, are four rods: two pointing ahead and two behind. On the ends of one of each pair of rods is a large mass, while the rods themselves weigh very little. The idea here is that when we accelerate/decelerate the rods, any stretch or compression caused by length contraction/extension will take different amounts of time to adjust the length of a rod with a large mass on the end of it compared with the rod next to it with no such mass at the end.

We're looking for a difference in length between the two rods of each pair, but we'll only get that during an acceleration/deceleration, and while there's a force applying, there will already be compression or stretch in play in the rods due to the acceleration itself (while any length contraction adjustment is a much smaller effect). The rods ahead of the ship will be compressed and the ones behind will be stretched. If we always apply the same rate of acceleration, the difference between the two rods in a pair will always be the same - the ones with the large mass at their end will be worse affected than their partner, the leading one being more compressed and the trailing one being more extended. We can make a mark on the rod that's longer than its partner during acceleration to show how far along it its partner reaches. Any effect from length-contraction adjustments should show up as variations in length of the shorter rod (only shorter during acceleration) away from that mark.

It may be worth naming the rods, so F means forward-pointing while R means rearward. H means heavy and L means light (though we're really talking about mass). We thus have four rods called: FH, FL, RH and RL. The compression under acceleration will make FH shorter than FL, and stretch will make RH longer than RL, so we can put marks on FL and RH to show how far the other rods will reach along them, and these marks can be called MF and MR.

At the moment when the ship makes a transition from deceleration to acceleration, there should be no extra compression or stretch from length contraction, so the ends of the shorter rods should exactly meet their mark. Here's the important point: this means that any variation at other times should allow us to identify the absolute frame (unless I've missed some important factor).

Case 1:-

If the rocket is accelerating and there is a change in the amount of length contraction needed for the current speed, that means a contraction has to be applied through the rods becoming slightly more stretched (RH and RL) or less compressed (FH and FL), and the ones with and without large masses at the end of them should take different lengths of time to adjust. The mass of the ship is much greater than the large masses at the ends of the rods, but because we also have rods pointing both ways, they have to change length without the ship end of any rods accelerating at a different rate from the ship - it's the ends with the masses on the end of them that must migrate.

Rod RH (already stretched by the acceleration) has a bit of extra effective stretch added to it by length contraction, as does RL, and it should take longer to adjust for this contraction than RL because rod RL has more work to do to haul the large mass in, so the end of RL should move away from mark MR, and it should do so in a direction taking it even further away from the end of RH.

[Rods FH and FL are more complex, so I'll leave them till later.]

Case 2:-

If the rocket is decelerating, the rods will have to extend rather than contracting, so the rods should become slightly more compressed (RH and RL) or less stretched (FH and FL), and the ones with and without large masses at the end should again take different lengths of time to adjust.

Rods FH and FL are easier to handle in case 2 because they will behave in a similar way to FH and FL did in case 1, the difference being that we're adding compressions rather than stretches. They are compressed by the acceleration and now have a bit of effective compression added by the length extension, so FH will take longer to respond and its end will fall short of mark MF on FL.

Rod RH (already stretched by the acceleration) should have a bit of effective compression added to it by length extension (or decontraction), as does RL, removing some of the stretch and allowing the rod to lengthen, but I'm not sure how it would react. Is it a hindrance as before, or is it now going to help extend the rod more quickly? If the latter, then it could hide the effect we're trying to see, but remember that it should still show up when the ship is momentarily stationary (moving from deceleration to acceleration), because at that point the ends of the shorter rods would line up with the marks. Either way then, we should have a method by which the absolute frame could be identified, unless there's a fault somewhere in the argument (which I fully expect to be the case, but if it turns out that there isn't, it would be a shame to miss the experiment that finds the aether by assuming that no such experiment can exist). This looks viable, but I've been here before several times with ideas for experiments that looked as if they could break relativity, and I haven't put a lot of time into attacking this one yet, so don't get excited. I've just posted it up front on the off chance that it might stand up, and if it does, the time and date stamp on it could be handy.
Demolitions invited.
Are the rods connected to the ship?
Is it an Einsteinian universe?  Or an aether universe?
In an aether universe any object would contract-stretch on each side of its center of mass. An observer moving with the object would not see any change in shape, but might notice any slight movement of the center of mass. That movement would be made up of a real movement, plus a faux-movement -- the faux being due to the movement of the observer's eyes in relation to the observer's own center of mass.

In an Einstein universe there is no real contraction or stretching. And no apparent contraction or stretching, any such contraction or stretching being a math-trick model. Or, if u like, the contraction or stretching are apparent, but are due to a change in the measuring rod & measuring clock.
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 00:48:58
Are the rods connected to the ship?

The description says so.

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Is it an Einsteinian universe?  Or an aether universe?

I'm working to LET - makes most sense to work with a rational theory where you don't get tied up in unnecessary complexities that make it hard to see what you're doing. Point is though, this is a test that could be made in the real universe which would, if there's no fault in the idea, pin down the absolute frame. If that happened, LET would survive, but some disproven theories would be disproved through direct experimental observations without needing to apply any of that weird reasoning witchcraft from the mathematicians that physicists don't like.

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In an aether universe any object would contract-stretch on each side of its center of mass.

Exactly - the rods will contract or uncontract, pulling in towards the centre where all the mass of the space ship is sitting (distributed sideways so that we can avoid worrying about its contraction interfering).

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An observer moving with the object would not see any change in shape, but might notice any slight movement of the center of mass. That movement would be made up of a real movement, plus a faux-movement -- the faux being due to the movement of the observer's eyes in relation to the observer's own center of mass.

The only thing we need to observe is the location of the ends of the shorter rods relative to the marks on the longer ones (while the rods are only of different lengths while accelerating - when at rest, they're all the same length).

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In an Einstein universe there is no real contraction or stretching. And no apparent contraction or stretching, any such contraction or stretching being a math-trick model. Or, if u like, the contraction or stretching are apparent, but are due to a change in the measuring rod & measuring clock.

I'm not bothered about what SR or GR have to say on the matter - if this experiment identifies the absolute frame, it doesn't matter what they have to say on the matter as the symmetry will have been broken. That said though, if it's too hard to carry out the experiment due to the infinitesimal size of the effect we're looking for, it would still provide an interesting distinction between LET and SR/GR which the latter cannot possibly match. The big question though is, does LET actually predict this or have I made a mistake somewhere with the experiment and my predicted results?
Title: Re: Does the thread break?
Post by: Halc on 24/10/2018 00:59:24
I've come up with an idea for an experiment.

Imagine a rocket with a lot of mass aligned perpendicular to its direction of travel (to avoid it having any length). Sticking out ahead of it though, and behind, are four rods: two pointing ahead and two behind. On the ends of one of each pair of rods is a large mass, while the rods themselves weigh very little. The idea here is that when we accelerate/decelerate the rods, any stretch or compression caused by length contraction/extension will take different amounts of time to adjust the length of a rod with a large mass on the end of it compared with the rod next to it with no such mass at the end.

We're looking for a difference in length between the two rods of each pair, but we'll only get that during an acceleration/deceleration, and while there's a force applying, there will already be compression or stretch in play in the rods due to the acceleration itself (while any length contraction adjustment is a much smaller effect). The rods ahead of the ship will be compressed and the ones behind will be stretched. If we always apply the same rate of acceleration, the difference between the two rods in a pair will always be the same - the ones with the large mass at their end will be worse affected than their partner, the leading one being more compressed and the trailing one being more extended. We can make a mark on the rod that's longer than its partner during acceleration to show how far along it its partner reaches. Any effect from length-contraction adjustments should show up as variations in length of the shorter rod (only shorter during acceleration) away from that mark.

It may be worth naming the rods, so F means forward-pointing while R means rearward. H means heavy and L means light (though we're really talking about mass). We thus have four rods called: FH, FL, RH and RL. The compression under acceleration will make FH shorter than FL, and stretch will make RH longer than RL, so we can put marks on FL and RH to show how far the other rods will reach along them, and these marks can be called MF and MR.

At the moment when the ship makes a transition from deceleration to acceleration, there should be no extra compression or stretch from length contraction, so the ends of the shorter rods should exactly meet their mark.
I followed all that.

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Here's the important point: this means that any variation at other times should allow us to identify the absolute frame (unless I've missed some important factor).
What???  It allows us to identify the direction of acceleration.  Acceleration is absolute (sort of), so you’ve found a complicated way to measure that.  A simple plumb line would also work.

Still reading, but not sure what you expect here.

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Case 1:-

If the rocket is accelerating and there is a change in the amount of length contraction needed for the current speed, that means a contraction has to be applied through the rods becoming slightly more stretched (RH and RL) or less compressed (FH and FL), and the ones with and without large masses at the end of them should take different lengths of time to adjust. The mass of the ship is much greater than the large masses at the ends of the rods, but because we also have rods pointing both ways, they have to change length without the ship end of any rods accelerating at a different rate from the ship - it's the ends with the masses on the end of them that must migrate.
You’re saying that you’re making a change to the acceleration rate and it effects the strain on all 4 of the rods, taking some finite time to find a new equilibrium.
Not sure where relativistic contraction is involved.  Perhaps you need to identify the frame in which these measurements are going to be taken, because I don’t think you mean the frame of the ship.  Maybe you do.  Hard to tell.  There’s no relativistic contraction in ship frame, just static strain, or dynamic strain if the acceleration is not constant.


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Rod RH (already stretched by the acceleration) has a bit of extra effective stretch added to it by length contraction,
How does contraction add to stretch?  Wouldn’t they potentially cancel if they happen to be equal?  Sorry to interrupt mid-sentence...
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as does RL, and it should take longer to adjust for this contraction than RL because rod RL has more work to do to haul the large mass in, so the end of RL should move away from mark MR, and it should do so in a direction taking it even further away from the end of RH.
If you increase your acceleration, then yes.  Constant proper acceleration, going ever slower or faster in some other inertial frame, will not align RL with a different place on RH.


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Case 2:-

If the rocket is decelerating,
OK, definitely a different frame.  I know which one now.
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the rods will have to extend rather than contracting, so the rods should become slightly more compressed (RH and RL) or less stretched (FH and FL), and the ones with and without large masses at the end should again take different lengths of time to adjust.
OK, I think this makes a little sense if RH and RL are in the direction of acceleration, compressed strain.  You sort of have the ship backwards from the way I envision it.
For one, if something is moving, it is relativistically contracted, and as it slows in your frame, it will become less contracted, but never actually extended (longer than its proper length).  The rods RH and RL are compressed by strain, and the one endpoint stays at the mark on the partner rod.

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Rods FH and FL are easier to handle in case 2 because they will behave in a similar way to FH and FL did in case 1, the difference being that we're adding compressions rather than stretches. They are compressed by the acceleration and now have a bit of effective compression added by the length extension, so FH will take longer to respond and its end will fall short of mark MF on FL.
Sorry,  I just cannot parse this.  “effective compression added by the length extension” seems self contradictory.  I don’t know what you’re trying to convey with those words.  Extension does not add to compression, it would seem to relieve it.

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Rod RH (already stretched by the acceleration)
I thought we were decellerating.  You need to clarify which way the ship faces, and which way thrust is being applied.  It seems to switch from sentence to sentence, so I have a very hard time critiquing the experiment.

I got lost after that.  I will look again with some clarifications.  Not saying anything is wrong, just lacking clarity so far.
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 01:01:56
At the moment when the ship makes a transition from deceleration to acceleration, there should be no extra compression or stretch from length contraction, so the ends of the shorter rods should exactly meet their mark. Here's the important point: this means that any variation at other times should allow us to identify the absolute frame (unless I've missed some important factor).

There may be an error there if the speed of the rods isn't the same all the way along them due to changes in length as middle hits zero speed, and that will doubtless destroy the ability of the experiment to identify the absolute frame if case 1 and case 2 produce the same changes away from the marks (in the same direction rather than in opposite directions), so it's the last part of case 2 that needs to be resolved. Does the mass hinder or help the rod lengthen?

A reminder of that part:-

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Rod RH (already stretched by the acceleration) should have a bit of effective compression added to it by length extension (or decontraction), as does RL, removing some of the stretch and allowing the rod to lengthen, but I'm not sure how it would react. Is it a hindrance as before, or is it now going to help extend the rod more quickly?
Title: Re: Does the thread break?
Post by: mad aetherist on 24/10/2018 01:17:45
Are the rods connected to the ship?

The description says so.

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Is it an Einsteinian universe?  Or an aether universe?

I'm working to LET - makes most sense to work with a rational theory where you don't get tied up in unnecessary complexities that make it hard to see what you're doing. Point is though, this is a test that could be made in the real universe which would, if there's no fault in the idea, pin down the absolute frame. If that happened, LET would survive, but some disproven theories would be disproved through direct experimental observations without needing to apply any of that weird reasoning witchcraft from the mathematicians that physicists don't like.

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In an aether universe any object would contract-stretch on each side of its center of mass.

Exactly - the rods will contract or uncontract, pulling in towards the centre where all the mass of the space ship is sitting (distributed sideways so that we can avoid worrying about its contraction interfering).

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An observer moving with the object would not see any change in shape, but might notice any slight movement of the center of mass. That movement would be made up of a real movement, plus a faux-movement -- the faux being due to the movement of the observer's eyes in relation to the observer's own center of mass.

The only thing we need to observe is the location of the ends of the shorter rods relative to the marks on the longer ones (while the rods are only of different lengths while accelerating - when at rest, they're all the same length).

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In an Einstein universe there is no real contraction or stretching. And no apparent contraction or stretching, any such contraction or stretching being a math-trick model. Or, if u like, the contraction or stretching are apparent, but are due to a change in the measuring rod & measuring clock.

I'm not bothered about what SR or GR have to say on the matter - if this experiment identifies the absolute frame, it doesn't matter what they have to say on the matter as the symmetry will have been broken. That said though, if it's too hard to carry out the experiment due to the infinitesimal size of the effect we're looking for, it would still provide an interesting distinction between LET and SR/GR which the latter cannot possibly match. The big question though is, does LET actually predict this or have I made a mistake somewhere with the experiment and my predicted results?
U mention an absolute frame but u dont mention whether u expect an allowance for LC in accordance with relativity (gamma).
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 01:51:34
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Here's the important point: this means that any variation at other times should allow us to identify the absolute frame (unless I've missed some important factor).
What???  It allows us to identify the direction of acceleration.  Acceleration is absolute (sort of), so you’ve found a complicated way to measure that.  A simple plumb line would also work.

In an LET universe, when an object accelerates, it contracts in length regardless of whether it's being pulled or pushed up to speeds. If you take an elastic band and accelerate every atom of it up to 0.86c simultaneously in an instant, it will find itself to be stretched to twice its unstressed length, so it will shorten. The question is whether that stress could be detected in any way during an acceleration, and if it can, it would be different for a deceleration from 0.86c to zero, because that would lengthen it instead and lead to it going loose (or to a solid rod being compressed after the deceleration and needing to extend).

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You’re saying that you’re making a change to the acceleration rate and it effects the strain on all 4 of the rods, taking some finite time to find a new equilibrium.

The acceleration of the ship as a whole is constant, but the ends will either contract in or extend out, and I'm looking to see if that can be detected and if the two different things can be told apart.

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Not sure where relativistic contraction is involved.  Perhaps you need to identify the frame in which these measurements are going to be taken, because I don’t think you mean the frame of the ship.  Maybe you do.  Hard to tell.  There’s no relativistic contraction in ship frame, just static strain, or dynamic strain if the acceleration is not constant.

The measurements are made in the ship's frame of reference, but LET says that it is either contracting or extending (= uncontracting). Throw off your SR glasses for this and try to see it through LET. You may be the only other person here capable of thinking through this stuff properly, so I'd certainly value your help.

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Rod RH (already stretched by the acceleration) has a bit of extra effective stretch added to it by length contraction,
How does contraction add to stretch?  Wouldn’t they potentially cancel if they happen to be equal?  Sorry to interrupt mid-sentence...

If you have an piece of rubber a lightyear long which is capable of being stretched to twice its normal length without breaking, accelerating every part of it to 0.866c in a second would leave it in a stretched state due to length contraction. It will then take a good few years to contract, although it might break up in the attempt.

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as does RL, and it should take longer to adjust for this contraction than RL because rod RL has more work to do to haul the large mass in, so the end of RL should move away from mark MR, and it should do so in a direction taking it even further away from the end of RH.
If you increase your acceleration, then yes.  Constant proper acceleration, going ever slower or faster in some other inertial frame, will not align RL with a different place on RH.

The length contraction should accelerate the end of the rear rods more than the ship as a whole, and that's the effect we want to detect. If RL contracts more quickly than RH because of the large mass on the end of RH, they should contract at different rates and show up the length contraction that can't normally be detected, making it visible to observers in all frames.

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the rods will have to extend rather than contracting, so the rods should become slightly more compressed (RH and RL) or less stretched (FH and FL), and the ones with and without large masses at the end should again take different lengths of time to adjust.
OK, I think this makes a little sense if RH and RL are in the direction of acceleration, compressed strain.  You sort of have the ship backwards from the way I envision it.

Rods RH and RL are pointing rearward from the ship, while FH and FL point forwards. The people on the ship imagine that they're accelerating (in the forward direction) in both cases, but if they were moving to begin with, they may actually be decelerating. Case 1 and case 2 seem identical to the people in the ship, unless they see different behaviour in the rods regarding whether the ends of the shorter ones are level with the marks on the longer ones. In both cases, rods RH and RL are being stretched as they're being pulled along, whereas FH and FL are being pushed by the ship and are being compressed.

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For one, if something is moving, it is relativistically contracted, and as it slows in your frame, it will become less contracted, but never actually extended (longer than its proper length).

Indeed, which is why I also call the extension decontraction - it is an undoing of a contraction as the ship decelerates, slowing down relative to the absolute frame that the experiment is designed to try to identify.

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The rods RH and RL are compressed by strain, and the one endpoint stays at the mark on the partner rod.

It's FH and FL that are compressed by the acceleration of the ship as they're the ones pointing forwards. Remember that the F and R in the names stand for forward and rearward while the other letters refer to whether they're heavy or light (meaning that they either have a large mass attached to their far end or they have nothing there at all.)

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Rods FH and FL are easier to handle in case 2 because they will behave in a similar way to FH and FL did in case 1, the difference being that we're adding compressions rather than stretches. They are compressed by the acceleration and now have a bit of effective compression added by the length extension, so FH will take longer to respond and its end will fall short of mark MF on FL.
Sorry,  I just cannot parse this.  “effective compression added by the length extension” seems self contradictory.  I don’t know what you’re trying to convey with those words.  Extension does not add to compression, it would seem to relieve it.

If we slow something from 0.86c to zero in an instant, it will be compressed, and it will immediately extend as a result. When we decelerate, we get extension (loss of contraction), and that extension is driven by forces pushing the particles further apart to remove compression. FH and FL are compressed by the acceleration, but if it's actually a deceleration, there's extension to add to this, and that means they feel more compressed and push harder. FL reacts more quickly than FH because it doesn't have a great mass at the end to push forward.

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Rod RH (already stretched by the acceleration)
I thought we were decellerating.  You need to clarify which way the ship faces, and which way thrust is being applied.  It seems to switch from sentence to sentence, so I have a very hard time critiquing the experiment.

All the information needed was there, but it takes effort and time to take in what's going on in experiments of this kind, so it's easy to miss crucial details. RH faces rearward and is being pulled along, so it is stretched by that. The part that I'm having difficulty with though is working out what happens when a compression (from length extension [undoing a contraction]) is added to that stretched rod and its partner RL. What will happen to the end of RL and the mark on RH? If it moves the same way in case 2 as it does in case 1, then relativity survives as the absolute frame is not identified (as there would doubtless be something that hides the effect at zero speed too, as I mentioned in my previous post).
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 02:07:17
U mention an absolute frame but u dont mention whether u expect an allowance for LC in accordance with relativity (gamma).

I don't know what you're getting at. We're trying to measure the length contraction or the removal of contraction and using Lorentz Ether Theory as our guide. If the ship is accelerating, the rods will contract. If it's decelerating, they'll lengthen instead as the contraction is gradually removed. What I've tried to do is design an experiment where this change in length is slower for some rods than for others so that the difference will show up, but it won't show up if the visual result is the same for acceleration and deceleration. I was expecting to show that the effect was the same for both, but I haven't managed to prove to myself yet that it will be. If the ship slows to zero and then accelerates again, for a moment there will be no contraction or extension applying and it looks as if that might show up a difference unless it's hidden by the fact that not all parts of any of the rods are quite moving at the same speed as each other. I expect that will mask the effect and the experiment will be incapable of detecting the absolute frame, but the easiest place to push for an answer is probably to work out what will happen to the rearward-pointing rods in case 2 (of the forward-pointing ones in case 1). I'm just struggling to find the answer for that at the moment and probably need to think about it for a day or so to find the right way to handle it.
Title: Re: Does the thread break?
Post by: mad aetherist on 24/10/2018 02:19:08
U mention an absolute frame but u dont mention whether u expect an allowance for LC in accordance with relativity (gamma).

I don't know what you're getting at. We're trying to measure the length contraction or the removal of contraction and using Lorentz Ether Theory as our guide. If the ship is accelerating, the rods will contract. If it's decelerating, they'll lengthen instead as the contraction is gradually removed. What I've tried to do is design an experiment where this change in length is slower for some rods than for others so that the difference will show up, but it won't show up if the visual result is the same for acceleration and deceleration. I was expecting to show that the effect was the same for both, but I haven't managed to prove to myself yet that it will be. If the ship slows to zero and then accelerates again, for a moment there will be no contraction or extension applying and it looks as if that might show up a difference unless it's hidden by the fact that not all parts of any of the rods are quite moving at the same speed as each other. I expect that will mask the effect and the experiment will be incapable of detecting the absolute frame, but the easiest place to push for an answer is probably to work out what will happen to the rearward-pointing rods in case 2 (of the forward-pointing ones in case 1). I'm just struggling to find the answer for that at the moment and probably need to think about it for a day or so to find the right way to handle it.
Ok so gamma is needed for LC. And here gamma only applys to absolute speed, ie not to acceleration.

And we dont have to include some sort of silly Einsteinian LC related to gravitational potential (ie based on equivalence of GP to inertial acceleration). I believe in gravitational potential affecting LC due to nearness to mass (but not for GR reasons), but i dont believe in this LC being present when inertial acceleration is acting (unlike GR).
Title: Re: Does the thread break?
Post by: Halc on 24/10/2018 04:48:55
In an LET universe, when an object accelerates, it contracts in length regardless of whether it's being pulled or pushed up to speeds. If you take an elastic band and accelerate every atom of it up to 0.86c simultaneously in an instant, it will find itself to be stretched to twice its unstressed length, so it will shorten. The question is whether that stress could be detected in any way during an acceleration, and if it can, it would be different for a deceleration from 0.86c to zero, because that would lengthen it instead and lead to it going loose (or to a solid rod being compressed after the deceleration and needing to extend).
I will respond to the whole post in time, but I must point out that the two universes predict the same things in every way.  So if I decelerate an unstressed rubber band instantly, it will find itself too short and need to spring back to its full dimensions just like you describe.  This works in LET as well as Einstein’s view.  If you find the two different, you’ve probably got a mistake somewhere.  This one was pretty easy to spot.

So I agree with your description there, but you make it sound like SR would predict otherwise.
Title: Re: Does the thread break?
Post by: Halc on 24/10/2018 04:53:59
(ie based on equivalence of GP to inertial acceleration)
Assuming GP is gravitational potential, those two are not equivalent.  Acceleration is equivalent to gravitational force, not gravitational potential.  Dilation is a function of GP, which makes it equivalent to speed (not acceleration) in that regard.
Title: Re: Does the thread break?
Post by: Halc on 24/10/2018 15:06:55
OK, I’ll try to comment on the whole thing this time.

In an LET universe, when an object accelerates, it contracts in length regardless of whether it's being pulled or pushed up to speeds.
A pulled thing will stretch due to strain.  A long railroad train is usually one or two railroad-cars longer in length than when the same train is being pushed.
Speed, not acceleration, causes relativistic contraction, and that is true in both LET and elsewhere.

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If you take an elastic band and accelerate every atom of it up to 0.86c simultaneously in an instant, it will find itself to be stretched to twice its unstressed length, so it will shorten.
Yes, but that was neither pushed nor pulled.  The acceleration was applied everywhere, so all the strain is due to the instantaneous proper length expansion.

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The question is whether that stress could be detected in any way during an acceleration, and if it can, it would be different for a deceleration from 0.86c to zero, because that would lengthen it instead and lead to it going loose (or to a solid rod being compressed after the deceleration and needing to extend).
It can be detected, but you are pushing and pulling your rods, and the strain from doing that is going to completely overwhelm the miniscule strain from the gradual relativistic length changes. 

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The acceleration of the ship as a whole is constant, but the ends will either contract in or extend out, and I'm looking to see if that can be detected and if the two different things can be told apart.
From different frames, yes, they’ll be different.  You can get any data you want by selecting the frame that gives that data.  That seems to be what you’re doing here.  You never say in which frame these measurements are to be taken.  Ship frame?  The rods are static in that frame.

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The measurements are made in the ship's frame of reference,
Oh wow, I never got that.  Yes, they’d have to be if you’re trying to detect the aether.  Any other frame probably begs that you already know it.  But nothing is going to be different in ship frame between accelerating and decelerating.  Principle of relativity would be violated, and you’d win a Nobel prize.  Surely you’re aware of this.

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but LET says that it is either contracting or extending (= uncontracting). Throw off your SR glasses for this and try to see it through LET. You may be the only other person here capable of thinking through this stuff properly, so I'd certainly value your help.
I can do that.

First of all, your ship has to face in the way it is thrusting.  I don’t totally get that from your descriptions where I wasn’t sure in the deceleration case which way the ship is facing.  If it is facing forward during deceleration, then you presume to know which way is the aether, and you’re oriented your ship that way.  So it always faces in the direction of thrust just like the space shuttle does when it drops out of orbit.  I’d say it always accelerates forward, but you have a different definition of acceleration which really inhibits communication.  FL and FH are always compressed by strain, and RL and RH are always stretched by strain.  Both are contracted by speed, but that isn’t strain.

An automobile violates this.  It faces away from its thrust when decelerating.  The occupants holding plumb lines can very much tell the difference.  The automobile presumes to know the direction of the road-wind, and is thus no proof at all that the road is stationary.
The physics of your ship is much more like the asteroids video game and far less like the physics in star wars.  Yes, there is aether in asteroids, but detecting it is subtle (assuming the ship cannot see the screen edge) if you play a version with no friction.  I’ve seen some versions with friction, which of course totally different physics.


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Quote from: Halc
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Rod RH (already stretched by the acceleration) has a bit of extra effective stretch added to it by length contraction,
How does contraction add to stretch?  Wouldn’t they potentially cancel if they happen to be equal?  Sorry to interrupt mid-sentence...

If you have a piece of rubber a lightyear long which is capable of being stretched to twice its normal length without breaking, accelerating every part of it to 0.866c in a second would leave it in a stretched state due to length contraction. It will then take a good few years to contract, although it might break up in the attempt.
This analogy is inappropriate.  Your ship is in steady state, not accelerating from a stop to some speed in an instant.  The rubber rods are respectively pulled and pushed by that acceleration, and are not getting thrust applied to their parts all at once.  Your rods would not in any way behave like this piece of rubber you’re describing here.  They can be modelled as rubber, which is just a metal rod with more obvious strain resulting from a given stress.  I can even ignore the fact that a long rod will buckle if you push too hard on it.  The trailing rods have a finite length limit that is a function of acceleration.  A trailing rod cannot be longer than the distance to the Rindler horizon, even given infinite strength.   The rod in front can be as long as you want, so long as you don’t expect the ship to turn easily.

All that said, you have answered my question.  The trailing rods have additional strain from the fact that the ends of them need to accelerate harder than the rest of the ship.  That strain adds to (not cancels out) the strain put on them by the thrust of the ship.  Still, that strain is fixed and a function of acceleration, not a function of speed.  It isn’t going to change as your speed goes up and down.

This was the thing that confused me through my whole reply earlier, leading me to wonder if you’re facing the ship in the direction of travel rather than in the direction of thrust.  You are always facing thrustward, as a spacecraft should.  I am re-interpreting things in that light.

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The length contraction should accelerate the end of the rear rods more than the ship as a whole, and that's the effect we want to detect. If RL contracts more quickly than RH because of the large mass on the end of RH, they should contract at different rates and show up the length contraction that can't normally be detected, making it visible to observers in all frames.
How do RL and RH contract at different rates?  Under constant proper speed change, the two rods will be stretched to some fixed strain, and the two marks will always line up.  The weight on the end of one rod is why the rods are different lengths, yes, but the weight does not affect length contraction except that it accelerates a little harder, being a tad further behind.  The point where the two marks line up are always accelerating at the exact same rate as each other, but still more than the ship as a whole, as you point out.
This is going to be true whether the ship is accelerating or decelerating.

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Case 1 and case 2 seem identical to the people in the ship, unless they see different behaviour in the rods regarding whether the ends of the shorter ones are level with the marks on the longer ones.
Yes, that would be a difference.  But they’ll not see it.  Each shorter rod should always meet the mark on the longer rod in both cases. I am attempting to see how you might think that is not so.


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If we slow something from 0.86c to zero in an instant, it will be compressed, and it will immediately extend as a result.
This is a different case than the steady acceleration.
If you are coasting and go from 0.86c to 0 instantly, the rear rods will break off right at the ship and keep going at .86c.  The front rods will also continue to move for a while, compressing their length to almost nothing, before they spring back to their proper length.  A maximum strength front rod must still compress to about 0.28 its length before the end beings to decelerate.

Anyway, I realize you’re not proposing this, just trying to clarify the parts I could not parse.
I’ll have a second go at it.

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FL reacts more quickly than FH because it doesn't have a great mass at the end to push forward.
In the abrupt stop, the two react (begin to decelerate) at almost the same time.  FH actually begins first because it is closer to the ship, and it reacts harder (greater peak acceleration) because the rod is going to compress further than FL due to that weight that is going to have a shorter distance to achieve its deceleration.  Again, none of this is relevant to the steady thrust scenario.

I’m going to comment again on the original paragraph that I edited off.  You’ve clarified things, thanks.

This is the deceleration case, where you propose a difference in local observation.  What we locally observe is that the two marks always line up, both front and back.  Your job I suppose is to explain why that would not be the case during deceleration.

Rod RH (already stretched by the acceleration) should have a bit of effective compression added to it by length extension (or decontraction), as does RL, removing some of the stretch and allowing the rod to lengthen, but I'm not sure how it would react. Is it a hindrance as before, or is it now going to help extend the rod more quickly?
The ends of the R rods are moving faster than a decelerating ship, so they already have the momentum needed to extend the rods without the extension effect adding or subtracting to the stress and strain on the rods.  So it doesn’t have any additional compression effect on the rods.  It is not pushing the rods out, because the rod ends are already moving out and require no force adjustment to do their extending.  So it seems to be neither help nor hindrance.

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If the latter, then it could hide the effect we're trying to see, but remember that it should still show up when the ship is momentarily stationary (moving from deceleration to acceleration), because at that point the ends of the shorter rods would line up with the marks.
There should never be a time in any frame where the ends of the short rods don’t line up with the marks on the longer.  Anybody in any frame can look at that because they’re in each other’s presence.  See my remark below about the validity of this statement.

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Either way then, we should have a method by which the absolute frame could be identified, unless there's a fault somewhere in the argument (which I fully expect to be the case, but if it turns out that there isn't, it would be a shame to miss the experiment that finds the aether by assuming that no such experiment can exist).
My statement just above makes that very assumption, so take it as exactly that, just an assertion that comes from Galileo's time.  If we’re questioning that assumption, it would be begging to assume the principle.  Still, the comment about no compressive force resulting from deceleration is the one that applies here.  That’s why the rods don’t change alignment between acceleration and deceleration.
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 18:43:29
Thanks for giving it a go, Halc. I think I've found the answer (having had the unfair advantage of 24 extra hours to think about it), but before I can explain it, I need to make sure that you understand the experiment. From the point of view of the people in the ship, it's always moving in the same direction, but it may be moving backwards and decelerating relative to the absolute frame, or it may be moving forwards and accelerating relative to that frame (or it could be doing the first of those thing until it stops relative to the absolute frame, at which point it will start to do the second thing). While it's doing the first thing (decelerating), there should be length extension, and while it's doing the second, there will be contraction instead. For the rearward-pointing rods, extension should reduce the amount of acceleration acting on them, while contraction should increase it. If you increase the acceleration force, the rod with the large mass on the end will lengthen more than its partner, whereas if you decrease the force, it will shorten more than its partner, which means the end of the rod without the mass on its end will move relative to the mark.

The solution to this though is to recognise that while the force changes due to length contraction/extension and doesn't match up to the constancy of acceleration of the ship as a whole, the change in force is different for the different rods due to the difference in tension in each - a higher amount of tension in the rod with the large mass on its end leads to a greater adjustment force induced in it by the length contraction, so the two rods lengthen or contract in equal measure, completely hiding the effect from view in all cases.

I got there by replacing the rods with elastic to multiply the differences and make things easier to see clearly. If you imagine one piece of elastic with a weight on the end of it and the other without, the one with the weight may stretch out to twice the length under acceleration while the other one hardly stretches at all. Add in some length contraction and it's acting on atoms in the stretched elastic that are much more strongly trying to pull in, so the length contraction is acting on a lot more force and its effect is proportional to the amount of force it's acting on. In the unweighted elastic, there's hardly any force there for it to affect, so it has less input. The key thing is that it isn't an input of force, but something that adjusts the force that's already there, so it affects it in direct proportion to the amount of force that's active there. It is not an equal additional acceleration force in both rods, and the mistake I made at the start was in assuming that it would be an equal input of force to each.
Title: Re: Does the thread break?
Post by: David Cooper on 24/10/2018 19:03:04
This is just another example of how the absolute frame remains hidden from us - every experiment we try to design to reveal these physical differences for different speeds of travel through the fabric of space is masked in one way or another. It looks impossible to design any experiment that should be able to detect the absolute frame (or the aether wind).

Mad Aetherist frequently points to MMX experiments with gas in them providing different results from ones using a vacuum, but all the interactions between the light and gas should obey the same rules and mask the effect that Cahill claims has been detected. I doubt that anything unusual is going on there that could make a difference: the gas molecules and their atoms will length contract like anything else, and their rate of interaction with any photons they encounter will be slowed by their speed of movement through space. What mechanism could there be that would enable a gas-filled MMX to detect the aether that wouldn't also interfere with the speed of communications through air?
Title: Re: Does the thread break?
Post by: Halc on 25/10/2018 02:03:59
From the point of view of the people in the ship, it's always moving in the same direction,
It is always thrusting in the same direction.  They can’t detect motion at all, and for all they know (without  a window), they’re sitting in a building on a planet.  They detect a force, and that’s it.  The rods and stuff would all behave the same from their POV if it were uniform gravity and no speed whatsoever.

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For the rearward-pointing rods, extension should reduce the amount of acceleration acting on them, while contraction should increase it.
This is incorrect. Both increase the acceleration.  In both cases, the R rods experience more g force than the main ship (and the F rods less g force).  In the decelerating case, it is because the far ends of the R rods are already moving faster than the ship, but are slowing more than is the ship, to eventually match the ship’s speed when it reaches the absolute frame.

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If you increase the acceleration force, the rod with the large mass on the end will lengthen more than its partner, whereas if you decrease the force, it will shorten more than its partner, which means the end of the rod without the mass on its end will move relative to the mark.
That it will, but our example has constant acceleration of the ship, so the marks remain aligned forever.

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The solution to this though is to recognise that while the force changes due to length contraction/extension
No it doesn’t.  The force never alters.  The force is a function of acceleration, not of speed, so it never wavers for any given point on the ship or a rod.  It is different at one point than another, but always constant at a given point.

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I got there by replacing the rods with elastic to multiply the differences and make things easier to see clearly. If you imagine one piece of elastic with a weight on the end of it and the other without, the one with the weight may stretch out to twice the length under acceleration while the other one hardly stretches at all.
Yes, the effect is more noticeable with less stiff materials.  A real science experiment would probably still use some form of metal which has very predictable strain properties as opposed to elastic which tends to permanently deform after not much time under stress.  The weight scales for vegetables in the stores use metal springs for this reason, not elastic ones.  For the thought experiment, elastic works fine.
Title: Re: Does the thread break?
Post by: David Cooper on 25/10/2018 18:59:19
From the point of view of the people in the ship, it's always moving in the same direction,
It is always thrusting in the same direction.  They can’t detect motion at all, and for all they know (without  a window), they’re sitting in a building on a planet.  They detect a force, and that’s it.  The rods and stuff would all behave the same from their POV if it were uniform gravity and no speed whatsoever.

Coincidence that the two cases look the same. The physics is very different as no length contraction is involved in the gravity case. The ship has windows though in any case, and this is happening in deep space with the ship not tied to or resting on anything.

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For the rearward-pointing rods, extension should reduce the amount of acceleration acting on them, while contraction should increase it.
This is incorrect. Both increase the acceleration.

If contraction acts on the rearward-pointing rods, the ends are pulled in towards the ship. If extension acts on them, the ends are pushed out away from the ship. In the former case, that increases the acceleration of the ends, while in the latter case it decreases their acceleration (compared with the acceleration acting on the ship as a whole).

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In both cases, the R rods experience more g force than the main ship (and the F rods less g force).  In the decelerating case, it is because the far ends of the R rods are already moving faster than the ship, but are slowing more than is the ship, to eventually match the ship’s speed when it reaches the absolute frame.

If you're analysing events from the absolute frame and you start with the ship at rest, when the acceleration begins, we get a contraction which means that the ends of the rearward-pointing rods must be accelerating more. In the opposite case where the ship is moving backwards at a constant speed, then decelerates (while thinking it's accelerating forwards), the extension of the rearward-pointing rods must reduce the acceleration acting on their outermost ends.

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If you increase the acceleration force, the rod with the large mass on the end will lengthen more than its partner, whereas if you decrease the force, it will shorten more than its partner, which means the end of the rod without the mass on its end will move relative to the mark.
That it will, but our example has constant acceleration of the ship, so the marks remain aligned forever.

Only because the extra or reduced acceleration from the length extension/contraction is masked by it varying in strength according to the different amount of force running through the rods.

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The solution to this though is to recognise that while the force changes due to length contraction/extension
No it doesn’t.  The force never alters.  The force is a function of acceleration, not of speed, so it never wavers for any given point on the ship or a rod.  It is different at one point than another, but always constant at a given point.

There are two factors involved in producing the acceleration force - one is the constant acceleration of the ship and the other is length contraction/extension. The latter one will be visible to observers at rest in the absolute frame and they will see the truth of what is going on, but they will not know it to be the truth as they won't know which frame is the absolute frame.
Title: Re: Does the thread break?
Post by: Halc on 25/10/2018 22:42:44
They can’t detect motion at all, and for all they know (without  a window), they’re sitting in a building on a planet.  They detect a force, and that’s it.
Coincidence that the two cases look the same. The physics is very different as no length contraction is involved in the gravity case.
Of course there is, else you’d have a local test for dilation when light speed is measured at greater than c in a gravity well because clocks run slow down there but the lengths are unaltered.
The two cases look the same not by coincidence, but because the one was derived from the other.
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For the rearward-pointing rods, extension should reduce the amount of acceleration acting on them, while contraction should increase it.
You’re slipping up David.  I just realized you used the word acceleration correctly when discussing an instance where extending is going on.  That’s like cheating on your lover.
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Quote from: Halc
This is incorrect. Both increase the acceleration (edit: deceleration).
If contraction acts on the rearward-pointing rods, the ends are pulled in towards the ship. If extension acts on them, the ends are pushed out away from the ship.
Yes, but all that reduced deceleration was done when the ship started decelerating and the ends of the R rods were allowed to continue at a greater speed than the already decelerating ship.  The force (the push for extending that you’re talking about) is a function of the second derivative of speed, not the first derivative.  That means it is felt when the ship starts accelerating/decelerating, but during steady deceleration, that component is absent and all you have remaining is the R ends going faster trying to slow down (tension) to match the ship’s speed, which they will when it gets to the absolute frame.
The F rods initially decelerated to a speed less than the ship (momentarily reducing the steady state compression from a nonzero second derivative), but after that (steady state) spend their time decelerating less until the ship speed falls enough to match it.

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If you're analysing events from the absolute frame and you start with the ship at rest, when the acceleration begins, we get a contraction which means that the ends of the rearward-pointing rods must be accelerating more. In the opposite case where the ship is moving backwards at a constant speed, then decelerates (while thinking it's accelerating forwards), the extension of the rearward-pointing rods must reduce the acceleration acting on their outermost ends.
They’re going faster and have to stop at the same time as the ship, so they have to be decelerating more than the ship, not less.  It is no different than running the acceleration case in reverse.  The situation is totally time symmetric, so the R rods are going to have the same g force in both cases.

You asked for my opinion.  That’s it, and it is with my etherist hat on as best I can.

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There are two factors involved in producing the acceleration force - one is the constant acceleration of the ship and the other is length contraction/extension. The latter one will be visible to observers at rest in the absolute frame and they will see the truth of what is going on, but they will not know it to be the truth as they won't know which frame is the absolute frame.
Agree, but the truth is that the R ends are going faster than the ship, and are thus decelerating harder.  The observers in the rest frame will see this even if they don’t know they’re in that frame.
Title: Re: Does the thread break?
Post by: Halc on 26/10/2018 12:31:54
The physics is very different as no length contraction is involved in the gravity case.
Of course there is, else you’d have a local test for dilation when light speed is measured at greater than c in a gravity well because clocks run slow down there but the lengths are unaltered.
This reply I gave here is hasty.  I looked up this sort of thing and the answer is quite complicated.  Similar to the accelerating ship, the length will be contracted only in the direction of acceleration, meaning a ruler held vertical will be shorter than the same ruler outside a gravitational field.  It is actually longer if held horizontal.

The calculations involved are quite different for a free-falling observer in the gravity field vs a stationary one (the latter being the case for a planet or for an accelerating ship).
Title: Re: Does the thread break?
Post by: opportunity on 26/10/2018 12:59:41
These are one of the long questions based on the fundamentals used in the question. The question asks for a completeness of relativity that should already be implied in the a-priori definitions of time and space, right?
Title: Re: Does the thread break?
Post by: alancalverd on 26/10/2018 13:12:02
Relativity is very simple as long as you remember that it is relativity. If observers are not moving with respect to one another, there  can be no relativistic effect between them. If they are, then Einstein's equations seem to have been tested to an extraordinary degree of precision and found correct.
Title: Re: Does the thread break?
Post by: opportunity on 26/10/2018 13:21:27
Time and space are already relative without asking the allegiance of a theory that can't explain everything such as Einstein's.

sh1t, why do I have to say that?
Title: Re: Does the thread break?
Post by: David Cooper on 26/10/2018 21:29:42
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...The physics is very different as no length contraction is involved in the gravity case.
Of course there is, else you’d have a local test for dilation when light speed is measured at greater than c in a gravity well because clocks run slow down there but the lengths are unaltered.

I'm not talking about a detectable difference, but an underlying one. In the acceleration case, you get more and more length contraction the longer you go on accelerating. If you're just sitting on the surface of a planet, you get no change in length no matter how long you sit there.

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For the rearward-pointing rods, extension should reduce the amount of acceleration acting on them, while contraction should increase it.
You’re slipping up David.  I just realized you used the word acceleration correctly when discussing an instance where extending is going on.  That’s like cheating on your lover.[/quote]

It was right for one of them, and I assumed you could convert for the other if you wanted to.

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Quote from: Halc
This is incorrect. Both increase the acceleration (edit: deceleration).
If contraction acts on the rearward-pointing rods, the ends are pulled in towards the ship. If extension acts on them, the ends are pushed out away from the ship.
Yes, but all that reduced deceleration was done when the ship started decelerating and the ends of the R rods were allowed to continue at a greater speed than the already decelerating ship.

The reduced deceleration acts for as long as the ship is decelerating, and the rods continue to extend in length until the point where no contraction is acting on them, at which point they are at maximum distance from the centre of the ship. Once the deceleration turns becomes acceleration, They contract and move closer to the centre of the ship. This is not something that all happens at the instant that the deceleration/acceleration begins.

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The force (the push for extending that you’re talking about) is a function of the second derivative of speed, not the first derivative.  That means it is felt when the ship starts accelerating/decelerating, but during steady deceleration, that component is absent and all you have remaining is the R ends going faster trying to slow down (tension) to match the ship’s speed, which they will when it gets to the absolute frame.

That can't be right because when the deceleration of the ship become acceleration (when it is momentarily at rest in the absolute frame, the extension turns into contraction. There has to be an on-going reduction in deceleration up to that point and a new, then on-going increase in acceleration after that point - if that wasn't the case, it would just adjust to a fixed length for the rest of the acceleration.

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The F rods initially decelerated to a speed less than the ship (momentarily reducing the steady state compression from a nonzero second derivative), but after that (steady state) spend their time decelerating less until the ship speed falls enough to match it.

When the deceleration begins, they are compressed, but the length extension reduces their shortening all the way to the point when the ship stops in the absolute frame, and then they are more shortened after that as the contraction kicks in, so again this is not something that all happens at the start of the deceleration.

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They’re going faster and have to stop at the same time as the ship, so they have to be decelerating more than the ship, not less.

I wasn't discussing the ship stopping, but clearly if you switch the rockets off during acceleration, the ends of the trailing rods should (in the absence of the front rods) continue to accelerate the ship for a moment due to their faster movement. During a deceleration though, when the rockets are switched off, the end of the rods will be slower than the ship, so the ship will appear to be decelerated for a moment from the point of view of the people inside it. That would show up the absolute frame, so either relativity breaks or I've made another mistake somewhere.

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You asked for my opinion.  That’s it, and it is with my etherist hat on as best I can.

Thanks for going through it - you've just brought something else out that needs clearing up, and it's a much simpler thought experiment:-

We now have a ship with a single rod sticking out behind it and a large mass on the end. Under deceleration (which still feels like acceleration to the people in the ship), length extension (contraction being lost) leads to the mass moving away from the ship, whereas under acceleration the rod will be contracting instead, leading to the mass moving towards the ship, so when the rockets are turned off, what happens? Do the people in the ship measure a momentary deceleration of the ship in the former case? Of course, there's tension in the rod when the rockets are firing, so the mass will always pull the ship towards itself a bit when the rockets are switched off because that tension force has to be removed, but in one case there will be a tiny bit of extra force added to that, while in the other case it will subtract instead.

How can this effect be masked?

Edit: It may be a synchronisation issue, because those are also reversed for the two cases.
Title: Re: Does the thread break?
Post by: Halc on 27/10/2018 01:08:55
Of course there is, else you’d have a local test for dilation when light speed is measured at greater than c in a gravity well because clocks run slow down there but the lengths are unaltered.
I'm not talking about a detectable difference, but an underlying one. In the acceleration case, you get more and more length contraction the longer you go on accelerating. If you're just sitting on the surface of a planet, you get no change in length no matter how long you sit there.
And there’s me making the derrivative mistake.  Yes of course, it is not equivalent to length contraction over time, which is not a local effect.  There is dilation of sorts in a static field, just as there is static dilation on an accelerating ship.  I mentioned that in the prior post where I found myself unable to justify the statement you quoted here.  Anyway, standing on a planet (not in freefall), length is contracted a fixed amount in the vertical direction, and is actually a bit longer in the horizontal direction.  I imagine both these would be true locally on a ship, but the don’t necessarily translate to any kind of absolute change.
The effect might be part of why you cannot orbit near a black hole, where these effects are very significant.  It takes constant acceleration (power) to keep out of a black hole beyond a certain limit that is at a radius half again where the event horizon is.  I digress...

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The reduced deceleration acts for as long as the ship is decelerating, and the rods continue to extend in length until the point where no contraction is acting on them, at which point they are at maximum distance from the centre of the ship.
We can repeat all we want.  So I guess we agree to disagree on this point.
Tell me what the observer will see then.  I imagine you have a ship decelerating until it changes to acceleration, and there is some change (the marks suddenly line up differently?) and you have your empirical test for being stationary, at least in the dimension along which the ship is running.  You’d have to do it 3 times to get all 3 dimenstions.
The lines don’t budge if my argument is correct.  There is no observed change when the mode changes to acceleration.
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That can't be right because when the deceleration of the ship become acceleration (when it is momentarily at rest in the absolute frame, the extension turns into contraction.
Yes, it does, but that is no change to the derivative of velocity, only the velocity itself, moving in a straight unbroken line from negative into positive territory, with no discontinuity in the line.

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I wasn't discussing the ship stopping, but clearly if you switch the rockets off during acceleration,
Stopping = reaching speed zero, not shutting anything off.  The plan was to keep the engines on and go right into acceleration as you described it.  You can’t shut off at the absolute frame if you haven’t yet measured it, which was the point of this experiment.

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Under deceleration (which still feels like acceleration to the people in the ship),
Funny that.  Almost like Einstein was correct, eh?  Sorry.  My ether-hat fell off there for a second...
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so when the rockets are turned off, what happens? Do the people in the ship measure a momentary deceleration of the ship in the former case? Of course, there's tension in the rod when the rockets are firing, so the mass will always pull the ship towards itself a bit when the rockets are switched off because that tension force has to be removed, but in one case there will be a tiny bit of extra force added to that, while in the other case it will subtract instead.
The force due to the tension and strain will cause a brief deceleration of the main ship.  Newton says that.  It will cause the rod to vibrate actually, but it dies down.  So we account for that, and are left with the fact that the weight out there is going faster than the ship when we shut off the thrust.  It is moving in or moving away depending on if it was accelerating or decelerating, and the final speed will be a result of that momentum being transferred to the ship.  Yes, a difference between the two cases, but only if you make the computation in the known rest frame.  The people on the ship who are not at rest will feel no difference because in their frame, the weight at the end of the strained rod is stationary the whole time.  It isn’t a test for determining the rest frame if you need to make your measurements from it to detect it.
Title: Re: Does the thread break?
Post by: David Cooper on 27/10/2018 19:32:27
It was indeed a synchronisation issue. It's all to do with the time that it takes for the removal of the acceleration force to propagate to the weight at the end of the rod. In the case where the ship is decelerating and then switches off the rockets, the weight at that moment is moving more slowly than the ship, but it takes a long time for the removal of force to reach the end of the rod by the weight, so it's still accelerating the weight right up to that time. In this case, the effect of the length extension is cancelled out by the extra time the weight spends accelerating.

If the ship is accelerating, length contraction ensures that the weight is moving faster than the ship when the rockets are switched off, but this time the propagation of the loss of the acceleration force travels the length of the rod much faster, giving the weight less time to continue accelerating.