[Halc (OP)]

Proper length of an object is frame invariant.

This is debatable.  Proper length is of an object is, by definition the length in its own frame, but the figure can be computed in a different frame by integrating length/contraction over all simultaneous points in a given frame, which will give a different answer if the state (stretched or not say) of the object changes over time.  For the purpose of demanding that the proper length never changes, it is meaningful to compute said value in a frame of choice and it should not be different, but only because of this rigid property we're giving the object.  A non-rigid object could have a frame dependent proper length at a given time as defined by the chosen frame.
So if this computation yields a different proper length figure, we know we've violated the rules.

[Halc (OP)]

I did a quick one in my head (plus the numbers in post 88), using two waves instead of one.  The original wave was to 3135 km/sec which took 2 days for the wave and 4 days to move one light hour at that speed.  Total time is nearly 6 days.

So even if we allowed singularities, this solution is not optimal since we spend 4 days just coasting inertially, days which could be put to good use with some extra acceleration and deceleration, which would be perhaps another wave.  So of course more waves are better, especially since they also get smaller as you increase their number, and smaller waves have smaller singularities which gradually fade to nonexistence.

Lets bump the rear up to 2220 km/sec, at which speed the wave moves to the other end in only 1 day.  Then we immediately bump the rear again by a similar 2nd wave.  The time to move the light hour is now 2.8 days instead of 4, at least for the parts of the object moving at full speed, which none of them do for the whole distance.  The total time to move the object is now 4.8 days (each point is stopped for one day, half speed for 2 days, and full speed for 1.8 days).

Here I want to note that doing it in 2 waves reduces the full-speed coasting time to 1.8 days from 4.  Half the original 4 days were simultaneous in the original frame.  There are 2 full days where the entire object is moving at 3135 km/sec in the original frame, and 2 days that any particular point is dead stopped.  With the 2-wave method, there is 1 day of stopped and 1.8 days of full speed, and at no point in the original frame does the entire object move at that speed.  Some part of the object is always moving at half speed.  Both the stopped time and the 1.8 days of high speed is still a waste, time that could be used to milk a little more efficiency from the thing.  The optimal solution would reduce both stopped time and max-speed time to zero just like the 55 day solution.  For any given part of the object, acceleration will continue (at fixed or varying acceleration) until max speed is hit, at which point deceleration will immediately commence. 
I will try to write something to do this, and it will seemingly not need any parameters like the first function.  There will be no sweet spot for which I need to search.  It will just keep reducing the wave magnitude and use more and more of them until we cannot add more without overshooting our destination.  As wave magnitude decreases, the answer will converge to a valid solution.  When adding more waves makes no significant change to that final duration, we've got our answer.  At that point we can attack/defend the plan for being valid or not, and also search for further improvements to shave off even more time.

[guest4091]

DC#133;

All inertial frames measure light speed as constant because they assert a particular speed for it and adjust everything else to conform to that requirement.

[Observers have no control over the effects of td and lc, which modifies their measurements. The effects result from a constant independent speed of light. I.e. it's built into the physical behavior of the universe.]

Quote
This has to be as observed from outside. A moving object is affected by length contraction to the same degree as time dilation. That's why an observer moving with the object cannot measure any difference.

An observer on the ship will see the trip as taking no time due to the practically halted functionality. An observer stationary relative to the start and finish lines will see the ship travelling for an hour without contracting - it will appear to be an unstable length, but its functionality is frozen, as expected for an object moving at such high speed

.

[Length contraction occurs at high speeds.]

Quote
An observers world gets smaller the faster he moves in space. Near zero distance is perceived by the moving observer only.

In the 4D non-Euclidean geometry of STR and GTR, those zero-length distances exist. If we sent a ship at nearly c to M32 and back, it would return to us at a different location from the one it left, but the first leg of its trip would be shortened to next to zero length, and so would the second leg of its trip. The time it would pass through for the trip would be next to zero too. Light would make an equivalent trip in zero time of zero distance. That is a mathematical necessity of the model.

[A distance of zero is meaningless. SR is Euclidean geometry. GR is non- Euclidean geometry. Observer A's clock would record less time, as observed by E on the earth. A has a choice.
1. Assume an inertial frame and conclude the universe has contracted, thus M32 arrived early.
2. Assume he left E in a ship, knows SR, so concludes he is experiencing time dilation.]

Quote
Relativity defines the propagation speed of light, relative to space, not relative to an object!

Which Relativity? With LET, yes - it's the propagation speed of light relative to space. With STR and GTR, it's just a constant which represents the apparent propagation speed of light relative to space, but with those models the real speed of light is zero. If you don't want it to be zero, don't use a 4D model.

[SR or LET, there is no difference, especially since the coordinate transformations are equivalent! Both used Maxwell's equations as a basis for light propagation. The Lorentz version required a late correction (1905) by Poincare to maintain invariance.

[Bored chemist]

I just wondered something; how does an object know if it is "long"?

[Halc (OP)]

I just wondered something; how does an object know if it is "long"?

You get a small ruler (not moving relative to the segment being measured) and see how many of them fit from end to end.  That's its proper length. You can also just paint marks along the length of it, sort of like boats do to show draft (or draught), but painted marks become incorrect if the object experiences strain.  I'm not allowing any strain on my object, so that's not a problem for this topic.
Knowing the measurement doesn't tell you if it is "long" since nobody has provided a definition for that.  My example object is 100 light years in proper length, which seems "long" until you start talking about distances within and between galaxies.

[David Cooper]

I'm going to focus on a single point here which must not be lost in the noise.

With your method where you claim the separations are always comfortable

They're not merely comfortable.  They're exactly correct.  There is not a small wiggle margin I'm allowing.

They are not correct. Look at the atoms sitting at the back. Here's a diagram of them:-

O-------------------O-------------------O-------------------O-------------------O-------------------

That's them sitting at rest before the starting gun. They're going to move to the right when the gun fires. What do they look like the tiniest moment of time after the gun goes off? This:-

0-0--0---0----0-----

You have to teleport them to the left before they can start moving to the right.

[David Cooper]

Observers have no control over the effects of td and lc, which modifies their measurements. The effects result from a constant independent speed of light. I.e. it's built into the physical behavior of the universe.

If you are stationary or moving at constant speed and you assert that the speed of light relative to you from the north is c, if you then accelerate north to a different constant speed, the speed of light relative to you from the north is no longer c. If you want to claim it is now c relative to you, it can not have been c relative to you before. If you want it to be the same speed relative to you for both frames, you need to use a 4D Spacetime model in which the speed for light relative to you in both cases is zero. There is no valid alternative to these options in mathematics.

Length contraction occurs at high speeds.

Length contraction is not applied by magic. If you start with two ships sitting a mile apart in frame A with one directly ahead of the other and you accelerate them both identically to 0.866c (in the direction they're pointing), they will still be a mile apart in frame A after the acceleration. If you do the same thing with a single ship that's a mile long, that ship will have contracted to half a mile long (if the acceleration is gentle enough for the contraction to apply or if the acceleration was applied from the rear). In one case the contraction was applied, but in the other it was not. If I accelerate every atom of a metre long stick to a fraction under c in an instant, those atoms will still be sitting at the same separations after the acceleration - the contraction has not been applied, just as it wasn't with the two ships a mile apart. With the functionality of this stick effectively frozen, the contraction will be applied very slowly - so slowly that it will be impossible to measure it if the speed is close enough to c. For any finite length of time that you want to prevent the contraction from becoming visible, there is a valid speed less than c which will provide that.

A distance of zero is meaningless. SR is Euclidean geometry. GR is non- Euclidean geometry.

The same 4D model covers both. And the distance of zero is not meaningless - it means that there is a path available that has no separation between two points. Such paths exist between all Spacetime locations (although most of these require two zero-length steps to be combined).

Observer A's clock would record less time, as observed by E on the earth. A has a choice.
1. Assume an inertial frame and conclude the universe has contracted/, thus M32 arrived early.
2. Assume he left E in a ship, knows SR, so concludes he is experiencing time dilation.

If he chooses 2, he is using an absolute frame mechanism, so he's abandoned STR.

Which Relativity? With LET, yes - it's the propagation speed of light relative to space. With STR and GTR, it's just a constant which represents the apparent propagation speed of light relative to space, but with those models the real speed of light is zero. If you don't want it to be zero, don't use a 4D model.

SR or LET, there is no difference, especially since the coordinate transformations are equivalent! Both used Maxwell's equations as a basis for light propagation. The Lorentz version required a late correction (1905) by Poincare to maintain invariance.

There is a major difference. However, if you keep mixing models and imagine that you're doing SR while you're actually mixing LET and SR, then no wonder you're confused. In LET there is an absolute frame. In SR there is not, so you are banned from having time dilation - time cannot dilate for a stationary object, and all objects are stationary in SR.

Take the twins paradox. Twin A stays at home. Twin B goes away and back, recording less time passing than twin A. Did time dilate? Twin B is not moving according to SR during the first leg, so no - it could not dilate. Twin B is not moving according to SR during the second leg, so again no - it could not dilate. Twin A is also not moving according to SR, so again time did not dilate. You cannot have time dilation without an absolute frame mechanism, and you cannot have multiple absolute frames to switch between whenever it suits you because you are changing the speed of light relative to the content every time you change frame, breaking the rules of the universe where the speed of light is constant through space.

It is only by going to the 4D model that you can eliminate the contradictions, and you can only do that with a static block universe version of it, but even there the contradictions are still actually present - they are merely brushed under the carpet by refusing to address the generation phase of a block universe.

[Halc (OP)]

I'm going to focus on a single point here which must not be lost in the noise.

With your method where you claim the separations are always comfortable

They're not merely comfortable.  They're exactly correct.  There is not a small wiggle margin I'm allowing.

They are not correct. Look at the atoms sitting at the back. Here's a diagram of them:-

O-------------------O-------------------O-------------------O-------------------O-------------------

That's them sitting at rest before the starting gun. They're going to move to the right when the gun fires. What do they look like the tiniest moment of time after the gun goes off? This:-

0-0--0---0----0-----

You have to teleport them to the left before they can start moving to the right.

Not entirely clear picture.  If they move to the right, why does the picture show them having moved to the left?
Anyway, perhaps I see what you're trying to convey. The left-most one accelerates the most and closes the distance to the next atom the most.  So you'd expect them not to be the same distance apart since they're not moving at the same speed, and thus contracting differently along its length.  It seems you understand this since you've drawn it, but point it out like it is something wrong.

Oh....  You really are moving them to the left.  No, I'm not doing that. No acceleration takes place so fast that things have to move left to keep the thing the correct length. The left atom needs to move to the right to catch up with the ones to the right.  Nothing moves left.  You're contracting the thing before it begins to move.  Run the numbers into the acceleration formula I gave and you see that movement is always to the right, even if the picture looks like you 2nd one after the time it takes to move the leftmost atom a considerable distance to the right.

Here's a picture of the progression of the atoms you picture:

The seven lines at the bottom are the atoms, vertical at first (no speed), but each curving to the right (never left) as it picks up speed, getting closer to each other all the time at any horizontal line (simultaneous in the original frame) as the contraction takes effect.  Never mind the q= lines, which are irrelevant to what you're pointing out.  The q lines complete a coordinate system.

The picture assumes continuous acceleration, not reversing when some speed like 452km/sec is reached, so it is a fair depiction of the leftmost atoms of our meter-stick being moved a light-hour.

[esquire]

lets make this simple. when gamma rays burst are detected via multiple radio waves signals and the radio signals appear as mutliple repeating signal burst, at the same time from the same location in space, it can be constued that they at one time constitued a single burst from a single source. if this in fact is the case, you have the answer as to
" What limits does relativity put on acceleration of long objects?". a simple time measurement of a single radio burst from a larger packet of radio bursts, can be back functioned in conjunction with a calculated algorithm, that determines the time/length spread function between qamma rays and radio wave over distance in space. once the spread of how long it takes for a gamma wave to elongate into a radio wave, you have the basis of determining the distance, speed and length variables for an accelerated light object. by interpolating the spread length betwen gamma and radio waves, multiplied by a time period of a single radio burst in a mulitple radio burst packet,  you have an approximation of velocity, distance and length. simple.



.

[Halc (OP)]

lets make this simple. when gamma rays burst are detected via multiple radio waves signals and the radio signals appear as mutliple repeating signal burst, at the same time from the same location in space, it can be constued that they at one time constitued a single burst from a single source. if this in fact is the case, you have the answer as to
" What limits does relativity put on acceleration of long objects?"

A gamma ray burst is not a rigid object, nor does it accelerate from a stop nor come to a stop at its destination.

[David Cooper]

Anyway, perhaps I see what you're trying to convey. The left-most one accelerates the most and closes the distance to the next atom the most.  So you'd expect them not to be the same distance apart since they're not moving at the same speed, and thus contracting differently along its length.  It seems you understand this since you've drawn it, but point it out like it is something wrong.

But it does show something wrong. I had two options as to how to illustrate the problem, but I chose the one where the rear atom stays almost where it was before the starting gun while the rest move to their correct spacings from it for their newly acquired high speed. That requires them to be teleported to the left. The alternative way to illustrate the problem would have been to teleport the whole lot to the right, and that would be the better illustration because the object before the starting gun has every part of it stationary and no length contraction acting on it, whereas immediately after the gun there will be length contraction acting on it which means you have to jump the rear atoms forwards in an instant. How big a problem is this? Well, with your 100 lightyear long ship you're going to have significant length contraction acting on it even with the tail only moving at 452km/s. It's hard to work out the right contraction when different parts are moving at different speeds (and the front end doing 0), but if the whole thing was doing 452km/s the length contraction on the ship would be one light-hour. I'm going to make a guess that the contraction on the actual ship might be one light-minute rather than a light-hour, which would mean your rear atom is going to have to move that far before all the atoms are at comfortable separations, and given that it's moving at about 0.0015c, it's going to take 11 hours for it to settle to the right length, and throughout that time it is breaking your rules.

You'll have the same issue with the front end at the finish line if you want to halt that part of the ship at its destination: another 11 hours of breaking the rules.

[Halc (OP)]

But it does show something wrong. I had two options as to how to illustrate the problem, but I chose the one where the rear atom stays almost where it was before the starting gun while the rest move to their correct spacings from it for their newly acquired high speed.  That requires them to be teleported to the left.

If you look at the diagram, you see that nothing gets this 'newly acquired high speed' ahead of its allowed contraction.  The first atom accelerates far more than the next one, which is essentially still stationary at the first moment no matter how close we put the first atom to the dotted line.
Yes, if we accelerated the right any more than this maximum limit (such as you suggest in all your posts trying to hide the lack of contraction by reducing its duration to negligible time in one frame), the the atoms to the right would indeed need to teleport left to maintain correct spacing.  That's what keeps us from getting there any faster.

The alternative way to illustrate the problem would have been to teleport the whole lot to the right

No teleportation.  Each part of the object (be they atoms or light-year markers) follow the acceleration given by my specified formula c²/D, which is depicted by the dark lines in the picture.  Note that motion is contiguous (no teleportation) and always to the right, not the left.  Local contraction is always appropriate for the local speed.

How big a problem is this? Well, with your 100 lightyear long ship you're going to have significant length contraction acting on it even with the tail only moving at 452km/s.

Yes.  Exactly one light hour of contraction to be exact, which is the distance we wish to move.  It takes 55 days for the head (100 LY of D) to accelerate to that speed.  It takes more than 12 hours for the 1LY mark to get to that speed.  These hours are needed to prevent needing to move anything backwards to account for length contraction.

It's hard to work out the right contraction when different parts are moving at different speeds (and the front end doing 0),

The whole thing is doing 0 at first, but the tail has high acceleration so it doesn't stay that speed for long. All parts could just keep accelerating forever at that rate if we weren't worried about ever stopping the thing.

but if the whole thing was doing 452km/s the length contraction on the ship would be one light-hour.

In the original frame, the whole thing never goes full speed like that.  At mid-trip, the object has a total contraction of perhaps 3/4 light hour, but the speed varies along its length, going fastest in the middle.

I'm going to make a guess that the contraction on the actual ship might be one light-minute rather than a light-hour

In what frame?  I guessed 3/4 hour.  It isn't exact, but at such low speeds, it is really close to that.  At higher speeds, the 3/4 figure goes up, so it's hard to compute.  Calculus is your friend.
I notice you've not run any numbers demonstrating this fictional strain that you claim.  Show any segment of the object at some moment during the motion and the length of that segment will be correct for its speed.  If the two ends of the segment are moving at significantly different speeds, you will have to integrate the contraction over the length of the object, or just consider the contraction to be some figure that falls between the contraction for the two different speeds.

It seems I cannot describe a different way of moving the object when you cannot even see that the original slow way is a valid solution, if not optimal.  You claim the contraction will break it, but you've demonstrated no separation/length numbers that don't match.

[David Cooper]

If you look at the diagram, you see that nothing gets this 'newly acquired high speed' ahead of its allowed contraction.  The first atom accelerates far more than the next one, which is essentially still stationary at the first moment no matter how close we put the first atom to the dotted line.

You are now using the caterpillar method for perhaps the first 11 hours of the trip before you get to a point where you have a regular speed distribution in place along the ship. You can reduce that time by moving the atoms faster during this phase, and indeed you'll have to if they're to catch up with the places they should be in, so you'll have to decelerate them to 452km/s once the length is right so as to avoid damage with the ship absorbing all the excess force. Your method isn't as pure as you made it out to be.

How big a problem is this? Well, with your 100 lightyear long ship you're going to have significant length contraction acting on it even with the tail only moving at 452km/s.

Yes.  Exactly one light hour of contraction to be exact, which is the distance we wish to move.[/quote]

One light hour would be the contraction acting on it if the whole ship was moving at 452km/s,  but you only have one end of the ship doing that at a time, so the total contraction will be a lot less.

I notice you've not run any numbers demonstrating this fictional strain that you claim.  Show any segment of the object at some moment during the motion and the length of that segment will be correct for its speed.

I've already shown you that it's wrong at the start. The rear atom has to travel at 904km/s for perhaps 11 hours to get to where it should be, and when it gets there, the excess energy that it's carrying has to be taken off it to get it to the correct speed to conform to your description of how the ship behaves.

It seems I cannot describe a different way of moving the object when you cannot even see that the original slow way is a valid solution, if not optimal.  You claim the contraction will break it, but you've demonstrated no separation/length numbers that don't match.

It may be a valid solution, but the atoms at the back have to follow more complex rules than the ones they follow subsequently - it already depends on the caterpillar method.

[Halc (OP)]

If you look at the diagram, you see that nothing gets this 'newly acquired high speed' ahead of its allowed contraction.  The first atom accelerates far more than the next one, which is essentially still stationary at the first moment no matter how close we put the first atom to the dotted line.

You are now using the caterpillar method for perhaps the first 11 hours of the trip before you get to a point where you have a regular speed distribution in place along the ship.

No idea what you're talking about.  There is no wave, except I suppose for the point at which the ship moves at max speed, which moves south to north in frames where the object is mostly northbound, moves north to south in frames where the object is mostly southbound.  But if we're just talking about acceleration and not turning around, there is no wave at all.  c²/D does not involve a wave.  The new method I'm working on does.

One light hour would be the contraction acting on it if the whole ship was moving at 452km/s,  but you only have one end of the ship doing that at a time, so the total contraction will be a lot less.

Not a lot less.  3/4 I figure since halfway, the ship is moving at 3/4 of the max speed on average.
452 was chosen because in the ship frame, the universe contracts a light hour over 100 LY, so in that frame the tail at the beginning and the head at the finish line are nearly (arbitrarily close) simultaneous events.  Go any faster than that, and the universe contracts more than a light hour and we'll have overshot the destination before we can stop.  So it took me by surprise to find a method that allowed us to go a lot faster, but it does it by having no inertial frame in which the object is at rest but the universe is unreasonably contracted.

I notice you've not run any numbers demonstrating this fictional strain that you claim.  Show any segment of the object at some moment during the motion and the length of that segment will be correct for its speed.

I've already shown you that it's wrong at the start.

No, you're just asserting it. You've not shown anything. The acceleration of everything is always to the right.  No correction is made for inappropriate contraction, so show me that the contraction doesn't match the speed it is going.  Use numbers...

The rear atom has to travel at 904km/s for perhaps 11 hours to get to where it should be,

904 km/s for 11 hours is 36 million km (ish) or a couple light minutes.  Why does it need to be there in 11 hours?
The contraction will be a lot more than 36M km.  I think that was perhaps a guess on your part, but we have 27.6 days of the rear moving faster than the front to allow it to reduce the separation of the two, so it does it in 27.6 days, not 11 hours.  No need for anything to move at 904. 

It may be a valid solution, but the atoms at the back have to follow more complex rules than the ones they follow subsequently - it already depends on the caterpillar method.

Another thing I don't understand.  All atoms everywhere accelerate at a finite c²/D, and decelerate at c²/D' (where D and D' are distances to points in space arbitrarily close to the start event at the rear and beyond the finish event at the nose, respectively).  Since all atoms have finite acceleration and deceleration, they're all treated identically.  No special rules for any point.

[Halc (OP)]

Not a lot less.  3/4 I figure since halfway, the ship is moving at 3/4 of the max speed on average.

Bogus math on my part.  If the object is moving an average speed of 339, that yields an overall contraction somewhere around half an hour, not 3/4 of an hour.  The nose at the tail draw about that much closer in the original frame at mid-trip.

[Halc (OP)]

"If the rear acceleration takes 10 years (measured in local accelerating frame) to get up to say .866c, the front acceleration will take place for 10.866 years to get to that speed iff it ignites and ceases at the same time (object frame) as the rear acceleration."

.866c is essentially light, is it not?

Not sure what you mean.  Light is not a speed, even if it has a speed.
.866c is not essentially light speed since the acceleration can continue at the same g force for any amount of time past the 10 years and still not get to "essentially light speed", meaning anybody onboard might notice any difference without looking out of the window.

the object is a light year long, it must travel as a very intense frequency? does it not?

Objects like rocks don't travel at a frequency, intense or otherwise.  So no idea what you're talking about here.

[esquire]

"If the rear acceleration takes 10 years (measured in local accelerating frame) to get up to say .866c, the front acceleration will take place for 10.866 years to get to that speed iff it ignites and ceases at the same time (object frame) as the rear acceleration."

.866c is essentially light, is it not?

Not sure what you mean.  Light is not a speed, even if it has a speed.
.866c is not essentially light speed since the acceleration can continue at the same g force for any amount of time past the 10 years and still not get to "essentially light speed", meaning anybody onboard might notice any difference without looking out of the window.

the object is a light year long, it must travel as a very intense frequency? does it not?

Objects like rocks don't travel at a frequency, intense or otherwise.  So no idea what you're talking about here.

it seems obvious to me that anything traveling at or near the speed of light must adopt the parameters of light.
or else rock would be capable of traveling at the speed of light. nevermind!
 

[guest4091]

DC#146;

1. If you are stationary or moving at constant speed and you assert that the speed of light relative to you from the north is c, if you then accelerate north to a different constant speed, the speed of light relative to you from the north is no longer c. If you want to claim it is now c relative to you, it can not have been c relative to you before.
2. If you want it to be the same speed relative to you for both frames, you need to use a 4D Spacetime model in which the speed for light relative to you in both cases is zero. There is no valid alternative to these options in mathematics.

[1. That's as observed by a second party. In my inertial frame, light speed in space is always c, regardless of my speed. Light speed is independent of its source, i.e. it does not acquire the speed of the source, which differs from material objects.
2. "the speed for light relative to you in both cases is zero.", as observed by ?]

If you start with two ships sitting a mile apart in frame A with one directly ahead of the other and you accelerate them both identically to 0.866c (in the direction they're pointing), they will still be a mile apart in frame A after the acceleration.

[Agree for frame A. In the ship frames, they will measure their separation as .5 miles.]

If I accelerate every atom of a metre long stick to a fraction under c in an instant, those atoms will still be sitting at the same separations after the acceleration - the contraction has not been applied

[A mass cannot be accelerated instantly. Length contraction happens at light speed over microscopic distances (electron cloud), which will be faster than a transfer of energy between particles. The incremental energy transfer will require increasing transit times as it progresses  This is also the reason why a material object cannot be accelerated to light speed. It's NOT due to increasing mass.]

If he chooses 2, he is using an absolute frame mechanism, so he's abandoned STR.

[ His conclusion is based on 'he knows SR', and there is no absolute frame, which is the basis for the 'relativity principle'.]

The following quotes from the 1905 paper by the author of SR:

"Examples of this sort, together with the unsuccessful attempts to discover any motion of the earth relatively to the ``light medium,'' suggest that the phenomena of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute rest."

"The introduction of a ``luminiferous ether'' will prove to be superfluous inasmuch as the view here to be developed will not require an ``absolutely stationary space'' provided with special properties,"

[guest4091]

DC#146;
continued:

There is a major difference. However, if you keep mixing models and imagine that you're doing SR while you're actually mixing LET and SR, then no wonder you're confused. In LET there is an absolute frame. In SR there is not, so you are banned from having time dilation - time cannot dilate for a stationary object, and all objects are stationary in SR

[There is no mixing. It's common knowledge that LET hypothesizes a fixed ether frame that serves as a medium for light  Human thinking desires to interpret new things in terms of older established things (supposedly) understood. Government laws may ban certain activities, theories do not.
Time dilation doesn't require an ether or a fixed frame. The effect results from motion, which alters the distance light must move in any EM process. There is no difference. In SR, events don’t move, which is equivalent to a fixed medium.
So who is confused? Have you read any publications on SR?]

Take the twins paradox. Twin A stays at home. Twin B goes away and back, recording less time passing than twin A. Did time dilate? Twin B is not moving according to SR during the first leg, so no - it could not dilate. Twin B is not moving according to SR during the second leg, so again no - it could not dilate. Twin A is also not moving according to SR, so again time did not dilate. You cannot have time dilation without an absolute frame mechanism, and you cannot have multiple absolute frames to switch between whenever it suits you because you are changing the speed of light relative to the content every time you change frame, breaking the rules of the universe where the speed of light is constant through space

[There's one serious misinterpretation (red). SR does not state or imply the red portion. It states postulate 1 as: "the laws of physics are the same for all inertial (constant velocity) frames of reference". This translates to, any inertial frame may serve as a reference. A and B qualify as inertial frames, but still have to consider their relative motion in any measurement process.]
The purpose of the 'twin scenario' is to demonstrate that relative motion causes clocks (EM processes) to run slower relative to a reference clock.

From the 1905 paper, par.4:
"If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be second slow."

The graphic is another 'twin scenario'. It consists of 2 clocks, A and B, moving in separate spaceships to the right. The A path 0 to .4 at .4c and the B path 0 to.8 at .8c.
We examine it from before 1900. The trip time for each is 1.00. If B moved a greater distance than A, in the same time interval, B must have moved faster than A.
Now examine it after 1905, when theory predicts the faster moving clock runs slowest.
The revised conclusion is, the B clock indicates less time than the A clock. If the B clock changes direction to return to A via the 3rd short line, at .8c, its time is still less. All clocks are moving, and therefore all lose time. We only measure the difference. Where the circular arc crosses the vertical distance lines indicates the clock time. At reunion, At=.92 and Bt = .60.
No magic, just a little geometry.]

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[Halc (OP)]

If you start with two ships sitting a mile apart in frame A with one directly ahead of the other and you accelerate them both identically to 0.866c (in the direction they're pointing), they will still be a mile apart in frame A after the acceleration.

[Agree for frame A. In the ship frames, they will measure their separation as .5 miles.]

For frame A, the separation is a mile, just like in the beginning.  In the frame of the ships, the separation will be 2 miles. David knows his relativity at least this well.

If I accelerate every atom of a metre long stick to a fraction under c in an instant, those atoms will still be sitting at the same separations after the acceleration - the contraction has not been applied

[A mass cannot be accelerated instantly. Length contraction happens at light speed over microscopic distances (electron cloud), which will be faster than a transfer of energy between particles. The incremental energy transfer will require increasing transit times as it progresses  This is also the reason why a material object cannot be accelerated to light speed. It's NOT due to increasing mass.]

Well, I agree with the assessment of instant acceleration, but the rest is word salad.  If I accelerate every atom of a meter long stick to a fraction under c in a much shorter time than it take light to travel a meter, those atoms will still be sitting at the same separations after the acceleration.  The stick will still be a meter long, or 2 meters in the frame of the stick. This is the same concept as the first quote above.