[Halc (OP)]

I meant the eyes of the pilot tell him that the accelerations are the same, the same way as the eyes of drag racers tell the guys in the cars what the difference of accelerations is.

The drag racers are side by side, not end to end, and not moving at relativistic speeds, and are doing a non-local observation.

I don't think eyes can detect acceleration.  If I put a camera in a car pointing down at the back seat (not out the window), and somewhere in a minute of footage the car accelerates, I could not tell from watching the camera footage when that acceleration starts.  Acceleration is felt, and possibly visually detected by seeing something bend to the side or some other accelerometer, and it will be felt very different at one end of the ship than the other.

In my example of a long ship moving a short way (post 20), the people at either end of the ship die the strawberry jam death, and the ones in the middle die of boredom.  That would be something the pilot could see except the trip is shorter than the time needed for the image of the dead people at either end to reach the pilot in the middle.

If an array of particles all launch at the same time, with the same acceleration vector, their mutual relative velocities will remain zero so they will  remain at the same separation relative to one another. No relative motion = no relativistic effects.

This is all from the inertial frame perspective.  Their mutual relative velocities will remain zero in the original stopped inertial frame, yes.  There will be relativistic effect: time will pass for the particles more slowly but identically because they're moving.

If the initial array was a straight line, then an observer at the launch point will see the line contract in the radial direction, but not in the tangential direction, as they approach c relative to the launch point. Relative motion = relativistic effects.

No.  They're all accelerating identically, so no contraction.  See my very carefully worded example at the bottom of post 25 which describes exactly this kind of uniformly accelerating array.  For contraction to happen, the tail needs to accelerate more than does the front, bringing it closer to the front in the original inertial frame.

I encourage you to reword that example with the same acceleration to give the coordinates of the various parts of the ship after the acceleration completes.  Each piece accelerated identically for 1.3 years (ship time) or about 1.7 years in the inertial frame.  They all move exactly one light year (which is why those numbers were chosen), and that makes them all still the same distance apart.  No contraction of the space between them.  If you contract that space, some of the ships need to move far faster than light to get the ship length down in time.

Read and comment on the example (bottom of post 25) and stop waving it off because you know a different answer.  All I get is you holding your ears going "Lalala" when the contradiction is pointed out, or when I find several articles on the web (post 28) that support what I've been trying to say.

Describing in the frame of the array cannot be done because there is no one such frame.  Each of the elements is moving at a different speed because the ones at the front have been accelerating for a longer time than the ones in the rear.  Their clocks were in sync in the inertial frame, so they cannot be synced in any other (except you appear to also deny relativity of simultaneity).  That means they've accelerated for different amounts of time, and that's why they pull apart from each other in the frame of any of the pieces.

My thread isn't even about this.  I assumed one knows that different parts of the ship accelerate differently, and the question was if this limits a long ship from going places anytime quickly.

[Halc (OP)]

Thus the entire ship must accelerate as a single entity. There being no change in length, there can be no relative velocity or acceleration between the front and the back of the ship and thus no change in perceived clock rates between observers on the ship.

We need to pinpoint those rules.
There can be no change in proper length.  But there is very much going to be changes in relativistic length.
For it to have a proper length, it needs to be stationary in its own frame.  I fretted a lot about that one since it seems to be difficult to avoid, but decided it was a mandatory requirement, and that the solution to the problem lies exactly in that requirement.

Clearly my assessment of the rules was wrong.  It seems a brittle ship doesn't need to be stationary in any frame, and thus its proper length is only defined as the collective proper length of the parts.
In particular, a ship that is accelerating has parts that are necessarily moving at different velocities in other frames (like the original 'stopped' frame), and my description kept the ship stationary only in its own frame. If it can move at different speeds in that frame, it can do it in its own, and that means it might not even have a frame that is its own.

So David's post clued me into that.  You can have the bottom half of the ship going at .9c and the top half stopped.  That doesn't break the brittle ship unless any time is allowed to pass without moving the line between the fast stuff and the slow stuff.

So picture a causal diagram for the exact point/event E in the ship where this velocity difference exists.  Draw a big X with the event in the middle.
The stuff at the bottom is the past light cone, and nothing is moving in that cone.  As far as E has measured, the ship is stationary and has no reason to feel stress or strain.
The region above the X is the future light cone, and everything moves at .9c there.  As far as E will measure, the ship will always be uniformly fast and has no reason to feel stress or strain.  The stuff to the left and right are very different, but out of the causal cone.  The ship at E will need to accelerate to .9 instantly as the wave passes by.  It does not see the wave coming since it travels at faster than light, so the computer needs to know the plan, but we always said that.

Interestingly, the wave travels at just over light speed if the velocity change is near c, but far faster than light if the velocity change is low.  Because of this, the fastest way to get from A to B is to go slow.  I think my original plan is still the optimal one.  I could not think of a way to get the ship from A to B faster by allowing parts to move at different velocities like that.
I moved my ship one light hour in 55 days by keeping the speed down, but if we go at .999c and let the wave move up the ship, it takes 50 years (plus an hour).  I need to run the numbers for a wave resulting from an abrupt acceleration to 452 km/sec, which was the winning speed before.

[David Cooper]

Can you combine the two methods? Your way produces a faster time for the front of the ship (by setting it moving sooner), while mine produces a faster trip time for the back of the ship. Apply your method first to set a wave of acceleration through it much faster than the speed of light, then apply my method afterwards to increase the speed of the back end to nearly c with this wave running through the ship at a slower speed. The next thing to look at would be whether the trip time for the middle of the ship can be improved too.

[evan_au]

To drive the points home, my objects will be large and fast, but never with impossible properties like being massless, infinite rigidity, or with instantaneous acceleration

The ship at E will need to to accelerate to .9c instantly...

Hmmm... A few premises being violated there...

I don't think eyes can detect acceleration.

But eyes and ears can. And butts certainly can.

Have you ever been in an aeroplane, stationary at the end of a runway? The aisle looks perfectly level.
Then the plane accelerates down the runway. The aisle looks like it is tilting up towards the front of the plane. But looking out the window, you see that the plane has not rotated yet (ie it hasn't lifted the front wheel off the runway). Your ears can detect the acceleration, and they tell your brain that the direction of "up and down" has shifted, so your brain interprets the aisle as being tilted.

Even with your eyes closed, your butt feels the acceleration in your seat. It doesn't take much brains to detect acceleration.
But it does take external inputs to distinguish acceleration from the effects of gravity.

[alancalverd]

Halc: try reading about relativity (Einstein's book https://www.penguinrandomhouse.com/books/297702/relativity-by-albert-einstein/9780143039822 is probably the easiest to read because it was written by a genius rather than someone trying to bask in reflected glory) instead of making it up and getting it wrong.

[David Cooper]

Halc: try reading about relativity (Einstein's book https://www.penguinrandomhouse.com/books/297702/relativity-by-albert-einstein/9780143039822 is probably the easiest to read because it was written by a genius rather than someone trying to bask in reflected glory) instead of making it up and getting it wrong.

I don't think you should be making such a recommendation to someone who understands relativity much better than you do. Halc knows what he's talking about, but you keep repeating the same mistake.

If an array of particles all launch at the same time, with the same acceleration vector, their mutual relative velocities will remain zero so they will  remain at the same separation relative to one another. No relative motion = no relativistic effects.

They will indeed maintain the same separation as measured by the frame in which they were at rest to begin with, but they will also be seen to contract in length, so the gaps between them will open up and they would, if they were capable of seeing and thinking, see themselves moving further apart and imagine themselves to be accelerating at different rates with the leading ones accelerating more strongly than the ones following them.

[alancalverd]

Have you ever been in an aeroplane, stationary at the end of a runway? The aisle looks perfectly level.
Then the plane accelerates down the runway. The aisle looks like it is tilting up towards the front of the plane. But looking out the window, you see that the plane has not rotated yet (ie it hasn't lifted the front wheel off the runway). Your ears can detect the acceleration, and they tell your brain that the direction of "up and down" has shifted, so your brain interprets the aisle as being tilted.

Unusually, I beg to differ from my learned friend. We use a combination of tilt and optical illusion to create the sensation of linear acceleration in a simulator, but a real plane on a runway really does start to tilt very early in its trajectory. The semicircular canals are sensitive to rotation but not to linear acceleration, and the body's initial response to  backwards rotation is to try to maintain the head horizontal, so you will see the aisle rotate even without an external reference, because your head will rend to rotate forwards to compensate. Baroreceptors respond to pressure caused by linear acceleration, and horizontal acceleration is no different  in a plane, bus or train - you don't notice the aisle of a train rotating upwards as it leaves the station,do you?  The reason the simulator "works" is because a strong  visual input (the projected horizon doesn't move up or down the "windscreen") plus baroreceptor sensation overrides a weak rotation signal if the sim tilts slowly.

The initial sensation of takeoff can be very confusing. The front seat of a high performance training glider is supine when on the ground. Initial acceleration from a winch is very rapid, so you are being accelerated feet-first, which we normally interpret as falling, which is alarming. Next phase is to rotate forwards to get the tailwheel off the ground, so for a second or so you are "falling" and actually rotating towards a face-down position - very unpleasant! Problem is that the instructor sits in the back seat, almost upright, and just feels the expected pressure in his back, so can't understand why the passenger/student is confused.

[David Cooper]

Can my big ship get to its destination quicker because the engines fail to accelerate the occupants along with every other particle?

A frog has been levitated in the lab. It may be possible to levitate all the passengers with sufficient force to prevent acceleration forces flattening them. Probably better to develop technology to freeze them in a non-lethal manner though, but even then you need to guard against them snapping.

[Halc (OP)]

I have a ship that cannot take stress as I've described.  It is 100 light years in length, stopped in Frame F.  I want to move it forward by one light hour (about 1.08 billion km) in frame F.  How quickly (as measured in F) can I do that?  Obviously a tiny object can do it in about an hour at maximum speed.

We can't go at maximum speed.  If the tail T accelerates to c in an instant, the nose N requires 100 years to get up to that speed and by that time it has moved far further than a lousy light hour.  So it seems we want to instantly accelerate T just enough to get the nose of the ship to our destination (100 LY + 1 light hour) in the frame of T, and then instantly accelerate N by the opposite amount to drag the tail up the same distance.  Boom.  We've moved the desired distance.

So let's see, there are 876000 hours in a century, so we want to dilate the distance between the target of N and the current position of T by that factor (876000/875999) which happens at about 452 km/sec.  So I instantly accelerate T to 452 km/sec, and then slow down at a steady pace until I stop.  That's an average of 226 km/sec, so it takes over 55.3 days to move my ship that far.  It cannot be done faster.

I am going to redo this experiment with moving the same ship one light minute instead of a light hour.  That's about 18 million km.
Dilation factor needed is 1.000000019 which you get at 58.5 km/sec, or average speed half that.  The trip takes 7 days 3 hours.  I was wondering if it would take the same time, but no, it is apparently also a function of distance.

Next I will attempt using the caterpillar method to move that ship one light hour, the original scenario.  Using a hybrid method will be more complicated.

The ship at E will need to to accelerate to .9c instantly...

Hmmm... A few premises being violated there...

Fair enough.   As fast as we can then, and I gave my ships very powerful engines.  If instant acceleration can demonstrate some inconsistency, then it cannot be allowed.  Infinite rigid rods for instance can be used to signal distant places faster than light.  Unobtanium can be used to extract objects from black holes.

I don't think eyes can detect acceleration.

But eyes and ears can. And butts certainly can.

The ship is quiet of course.  If it accelerates everything perfectly, there's no vibration.  If it accelerates everything but the butts, the brittle chairs would probably shatter.  Think of it like the Enterprise hitting warp speed and not splatting everybody against the far wall, unless of course the director tells them all to pitch themselves to the right on his signal.
It's a thought experiment, not an exercise in engineering.  I can give the ship all the power I want and kill its occupants if that is convenient, but I can't violate rules like breaking light speed.

Have you ever been in an aeroplane, stationary at the end of a runway? The aisle looks perfectly level.
Then the plane accelerates down the runway. The aisle looks like it is tilting up towards the front of the plane.

That's how it feels, not looks.  Somebody watching a camera in the plane would not detect that tilt (and wouldn't detect it when the rotation actually happens), and would only notice Betty there getting pressed into her seat as she's playing the role of the accelerometer.

But looking out the window, you see that the plane has not rotated yet (ie it hasn't lifted the front wheel off the runway). Your ears can detect the acceleration, and they tell your brain that the direction of "up and down" has shifted, so your brain interprets the aisle as being tilted.

Yup.  Inertial cues, not visual ones, except for the window, which is a non-local measurement.

Even with your eyes closed, your butt feels the acceleration in your seat. It doesn't take much brains to detect acceleration.
But it does take external inputs to distinguish acceleration from the effects of gravity.

Agree.
Do any of these points detract from the main topic?  Can my big ship get to its destination quicker because the engines fail to accelerate the occupants along with every other particle?

Can you combine the two methods? Your way produces a faster time for the front of the ship (by setting it moving sooner), while mine produces a faster trip time for the back of the ship.

Did you run the numbers?  I don't think any part of the ship gets anywhere faster by your way.  It opened my eyes to an unnecessary condition I had been putting on the ship, but I don't think it helps win the race.  Not sure.

I admit that I have not yet run the numbers for a pure caterpillar movement at say 300 km/sec.  Mine moves at an average pace of 226 km/sec, with the middle achieving twice that halfway through the trip.  300 might be too slow, since it needs to add travel time (many days) to the wave propagate time (also days?).  Faster one is, the slower the other one is.  There must be an optimal point, but I've not plugged the numbers yet.

I ran the numbers on the simple case.  Had to write a program to do it:

Code: [Select]

#include <stdio.h>
#include <math.h>
#define SpD 86400.       // seconds per day
#define kmLH 1080000000. // km in a Light hour
int main(int ac, char **av)
{
  double factorD,        // lorentz factor (down)
         speedC,         // speed as fraction of c
         speedK,         // speed in km/sec
         speedK10, step, // range limit km/sec
         tDmov, tDwav;   // time in days
  sscanf(*++av, "%lf", &speedK);
  sscanf(*++av, "%lf", &speedK10);
  step = (speedK10-speedK)/10.0;
  speedK10 += step/3;
  while (speedK < speedK10)
  {
    speedC = speedK / 300000.;      // Compute speed as fraction of c
    factorD = sqrt(1 - speedC * speedC);   // Lorentz contraction
    tDmov = kmLH / speedK / SpD;    // Time to move 1 LH at that speed
    tDwav = 36500 * (1. - factorD); // Wave time: 36500 light day ship len
    printf("S %.2f T %.14f W %.14f = %.14f \n",
           speedK, tDmov, tDwav, tDmov+tDwav);
    speedK += step;
  }
  return 0;
}

One inputs the low and high range of speed to test in km/sec, and it gives travel time T for one light hour at that speed plus wave time W to get the caterpillar wave to traverse the length of the 100 light year ship.
Output is in days.

This first run gave me a general sense of where the sweet spot is:

Code: [Select]

> caterpillar 100 5100
S 100.00 T 125.00000000000000 W 0.00202777783209 = 125.00202777783210
S 600.00 T 20.83333333333333 W 0.07300007300048 = 20.90633340633381  < Close to my prior solution of 452 km/sec
S 1100.00 T 11.36363636363636 W 0.24536193580077 = 11.60899829943713
S 1600.00 T 7.81250000000000 W 0.51911480262024 = 8.33161480262024
S 2100.00 T 5.95238095238095 W 0.89426095483064 = 6.84664190721159
S 2600.00 T 4.80769230769231 W 1.37080351890573 = 6.17849582659804
S 3100.00 T 4.03225806451613 W 1.94874646653875 = 5.98100453105488
S 3600.00 T 3.47222222222222 W 2.62809461481267 = 6.10031683703489
S 4100.00 T 3.04878048780488 W 3.40885362640275 = 6.45763411420763
S 4600.00 T 2.71739130434783 W 4.29103000981162 = 7.00842131415945
S 5100.00 T 2.45098039215686 W 5.27463111963672 = 7.72561151179358
After narrowing it down, I got this final run:

Code: [Select]

> caterpillar 3135.26 3135.36
S 3135.26 T 3.98691017650849 W 1.99333063672691 = 5.98024081323540
S 3135.27 T 3.98689746018684 W 1.99334335266144 = 5.98024081284828
S 3135.28 T 3.98688474394631 W 1.99335606864054 = 5.98024081258685
S 3135.29 T 3.98687202778690 W 1.99336878465611 = 5.98024081244302
S 3135.30 T 3.98685931170861 W 1.99338150071221 = 5.98024081242082  <<< Sweet spot
S 3135.31 T 3.98684659571143 W 1.99339421680883 = 5.98024081252026
S 3135.32 T 3.98683387979536 W 1.99340693295003 = 5.98024081274539
S 3135.33 T 3.98682116396041 W 1.99341964912769 = 5.98024081308810
S 3135.34 T 3.98680844820657 W 1.99343236534588 = 5.98024081355245
S 3135.35 T 3.98679573253385 W 1.99344508160865 = 5.98024081414249
S 3135.36 T 3.98678301694223 W 1.99345779791194 = 5.98024081485416

It seems the caterpillar method can get the ship up to 3135 km/sec and do the trip in under 6 days, doing the movement in under 4 days and the wave taking just under 2 days to move up the 100 light-year ship.  That's like 11% of the 55 days I got via the other method.

Did I get all the math right?  In particular, I got a wave propagation speed of 1/LC where LC is the lorentz contraction factor of the speed of movement.  That speed is just over light speed if ship speed is close to c, but is far greater than c for slow speeds.  The first printout really shows this.

It works.  The hybrid method should probably shave something off that 4 day T figure, so we can play with that.  Complicated....

[David Cooper]

Are you moving a 100 lightyear long ship one lighthour? If so, the back end can in principle move one lighthour in a fraction over an hour (with astronomical energy costs). You want to get the front end moving as soon as possible, so you need a wave that propagates much more quickly than the speed of light through the length of the ship. I'm assuming you've found a way to find the ideal speed for that, but due to pressing work I haven't put the effort into trying to follow the maths to check it. Every other part of the ship should be able to make the trip faster than the front end does, and I'm envisaging sending out multiple waves from the back end to achieve that (an infinite number of them), each one taking particles to a higher speed than the previous one but taking longer to propagate. As soon as they reach their required destination, they stop. The nearer a particle is to the front of the ship, the lower the speed it will be doing when it reaches its destination and has to stop.

[Halc (OP)]

Are you moving a 100 lightyear long ship one lighthour? If so, the back end can in principle move one lighthour in a fraction over an hour (with astronomical energy costs).

Yes, 100 LY ship moved 1 LH.  Being energy efficient is not a goal of this topic.
That I can move part of it in an hour doesn't help me.  That data point is on my chart, and it takes a century to move the ship that way.
The rule is the entire ship must move, and come to a stop, in minimal time.  By stopping, there is less ambiguity over how long the procedure takes, but time in the inertial frame is the one I'm counting.  The ship in this scenario doesn't move fast enough to make much difference in what the ships clocks will read when we're done.

You want to get the front end moving as soon as possible, so you need a wave that propagates much more quickly than the speed of light through the length of the ship.

That's the hybrid approach, and the problem is hardly trivial.  Yes, get the front moving ASAP (I can do that instantly), and by symmetry, keep the back end moving until the end.  The optimum solution probably has no point-discontinuity of speed like the simple example I just ran.  I suspect an ugly curve and plenty of integration to do.

Every other part of the ship should be able to make the trip faster than the front end does,

By symmetry, this cannot be so.  The back end moves just like the front, but in reverse: The back goes fastest at first, but slows until the end, just like in my post 20 example.  The front gathers speed the whole time and slows much more quickly at the end.

and I'm envisaging sending out multiple waves from the back end to achieve that (an infinite number of them), each one taking particles to a higher speed than the previous one but taking longer to propagate.

Yes, a continuous curve, with no distinct 'waves'.
What we need is a formula for each section of the ship: Accelerate exactly like this.  Only the exact middle of the ship will have a symmetrical waveform to its speed profile.

As soon as they reach their required destination, they stop.

That should be all at the same time.  The entire ship moves the entire duration, which is not what the scenario above depicts, where all parts move for 4 days, and are stationary for 2 days (partly on either side of the moving period).

The nearer a particle is to the front of the ship, the lower the speed it will be doing when it reaches its destination and has to stop.

I came to the opposite conclusion: The front of the ship is moving the fastest at the end, same speed as the tail goes at the beginning.

Our bottleneck seems to be the middle.  I can yank the back up to speed fast, and the front can slow very fast, but the middle has to accelerate steady and carefully for both halves.  As far as the wave speed goes, the ends are the bottlenecks.  So I don't have an intuitive feel as for what our winning plan will look like, or how to go about proving there's no better plan.

[David Cooper]

The back end can go straight up to (an infinitesimal fraction less than) c, but the front end has to accelerate slowly to avoid breaking off the part behind it - you can accelerate a long chunk of ship behind it too at the same time, but length contraction will break it if the front end isn't the slowest in that acceleration, and if that long chunk of ship doesn't include the back end, it will break somewhere. The back end can slow from nearly c to zero in an instant, and the front can slow from whatever speed it's doing to zero in an instant too. If you want the fastest journey time for the ship, the back end will cover the distance in the shortest time and the front end will take the longest. You appear to want the back end to be as slow as the front end overall, but that's a slower average journey time. The distance travelled will limit the top speed of the front part of the ship, while its rate of acceleration is determined by the length of the ship. If you can work out that relationship, you then have a way to work out how quickly any in-between part of the ship can accelerate because any such point is equivalent to the front end of a shorter ship.

[Halc (OP)]

If you want the fastest journey time for the ship, the back end will cover the distance in the shortest time and the front end will take the longest. You appear to want the back end to be as slow as the front end overall, but that's a slower average journey time.

It may be that it doesn't matter, but if it matters, the solution will be symmetrical, meaning the front accelerates in a pattern that is the reverse of the rear of the ship.  If you play the video in reverse, it should look the same.  All scenarios attempted so far have been symmetrical.  If it is important to begin moving the nose ASAP (I think it is), then it is important to keep the tail moving right until the end.  I'm certain of this.
The limit is probably due to the endpoints, but not sure.  I'm thinking of a slow acceleration of the tail at first to get the 'wave' moving to midships much quicker than the uniform acceleration that was done in the output above.

The distance travelled will limit the top speed of the front part of the ship, while its rate of acceleration is determined by the length of the ship. If you can work out that relationship, you then have a way to work out how quickly any in-between part of the ship can accelerate because any such point is equivalent to the front end of a shorter ship.

As I said, 55 days if you do it that way.  That's exactly what was done in post 20.

Edit: That is not exactly what was done.  I only got each end up to 452 km/sec, not near lightspeed.

I think I see what you're saying.  Floor it from the tail, but stop the nose once it gets far enough, which is well before it is simultaneous with its tail.  The solution is asymmetrical, meaning the final speed of the tail doesn't matter.

- - -

Calculating the max acceleration of the nose of my ship turns out to be straightforward.  From the opening comments of the Rindler article:

whenever we talk about an inverse acceleration being equal to a distance, or vice versa, their product will equal c² in conventional units.

So c² is 9e16 (units of meters and seconds), and my ship is 100 light years long, or 9.46e17 meters giving us an acceleration of 0.09513 m/sec² which takes 78 days to move that light hour.

Did I do that correctly??  That's 320 km/sec at the end, slower than my 452, which I limited for contraction reasons.  Did I exceed some limit in post 20?

Why is the simple caterpillar method so much faster?

[David Cooper]

I realise now that my method can't combine with yours. If I accelerate the back end straight up to a fraction under c, it has to catch the particle ahead of it quickly in order for it to maintain connection with it at that speed, so the particle ahead can't accelerate until it's practically been caught as any earlier movement will open up a gap far greater than the length contraction requires for the two of them. That means a very long wait before the front end can start moving. To move the front end in the minimum amount of time, the back end has to accelerate more gently, though it can always accelerate more aggressively than any particle further forward. One limit on how quickly it can accelerate depends on how long it can "feel" stretched before it detaches from the particle ahead, and I don't know how that would be calculated.

[Halc (OP)]

It takes me about 100 days to find my stupid error

Calculating the max acceleration of the nose of my ship turns out to be straightforward.  From the opening comments of the Rindler article:

whenever we talk about an inverse acceleration being equal to a distance, or vice versa, their product will equal c² in conventional units.

So c² is 9e16 (units of meters and seconds), and my ship is 100 light years long, or 9.46e17 meters giving us an acceleration of 0.09513 m/sec² which takes 78 days to move that light hour.

Did I do that correctly??  That's 320 km/sec at the end, slower than my 452, which I limited for contraction reasons.  Did I exceed some limit in post 20?

I did not do that calcuation right.  Somehow I thought the 55 day figure was wrong, but no, this one is.  It takes just over 55 days to go one light hour at .09513 m/sec²

This is in agreement with the very different method of arriving at this figure from post 20

Off topic sort of, but...

I don't think eyes can detect acceleration.

But eyes and ears can. And butts certainly can.

The ship is quiet of course.  If it accelerates everything perfectly, there's no vibration.

Clearly I read the full comment incorrectly. It isn't about noise. Can inner ear detect actual acceleration?  Butt's do it the same way, but if it is detectable, then a lot of it will be unhealthy.  I'm thinking of the perfect acceleration couch like in Sci-Fi where the occupant is immersed in liquid including the lungs like they do with preemie babies.  How much yanking around can you then take without noticing, or without injury?

As I said, this is a thought experiment, and my very capable ship can accelerate each part of itself (including the occupants) as hard as it wants.  They'll feel nothing, else it wasn't engineered right.  I'm going to further explore scenarios with the ship and occupants staying intact (no stress or strain) despite the head moving at a vastly different speed than the feet.  I ran with David's idea and it showed that the ship can get to its destination a helluva lot faster than 55 days.  I want to improve on the 6 day trip time before I move on to pure stationary objects.

To drive the points home, my objects will be large and fast, but never with impossible properties like being massless, infinite rigidity, or with instantaneous acceleration

The ship at E will need to to accelerate to .9c instantly...

Hmmm... A few premises being violated there...

This comment prompted me to edit the OP with this:

Edit:  There seems to be nothing impossible about near instantaneous acceleration.  Many of my examples assume as a limit an acceleration to a desired speed in negligible time.  If this is found to violate finite light speed or some other law, kindly post details since it will effect my answers for minimum time to get a big thing somewhere.

Nobody has posted any blatant violations of physics resulting from instantaneous acceleration as opposed to arbitrarily high acceleration.  Since nobody seems up to the task, I will attempt to do it myself, and in the process, sink the caterpillar method of moving the ship.  Sorry David.

The post is rather long, so bear with me, since I think it is all relevant.

Summary so far:

We have a born-rigid ship that is 100 light years long, and we wish to move that ship a distance of 1 light-hour (north let's say), with it stopped at either end of the trip.  No solution that involves strain on the ship is allowed.

I presumed the ship had to be always stationary in its own frame along its entire length during the whole trip.  From this presumption, I computed a trip time of about 55 days in post 20.
David Cooper suggested what has become known as the caterpillar method of moving the ship.  The idea is to accelerate the tail quickly to some speed and propagate that acceleration to the front of the ship as the relativistic contraction allows.  The faster the speed chosen, the greater the contraction and the slower the 'wave' of acceleration moves to the front of the ship.  The wave always moves faster than light.  I found the optimal speed at 3135 km/sec which gives 4 days to move the light-hour and 2 days for the wave to propagate 100 light years for a total trip time of 6 days.  I intend to demonstrate that the method only works with infinite acceleration, and thus is a violation of my initial premise.  It is effectively like suggesting unobtanium.

Seeming inconsistencies with the method which are not in fact problematic:

The solution seems asymmetric.  The ship is stopped and to get it going, you must accelerate it from the tail at first and let the wave propagate to the front.  This allows the tail to move first and is consistent with the ship contracting in length as it moves at 3135 km/sec.  But then we accelerate it the other way and again this must be initiated at the tail, not the nose, allowing the ship to expand to its original length as it stops.  That's the asymmetry.  Acceleration in both directions requires the wave to move from south to north.  If I try to stop the ship by stopping the nose first, then the north end of the ship moves for much less time than the south end, and the ship stops much shorter than its original length, and it breaks.

So I tried to visualize the trip in the frame of the moving ship where it is stationary (and has undilated length) at the middle of the trip.  Surprise!   In that frame, the ship is moving south at first, and the acceleration starts at the north end (nose) of the ship and takes 2 days to reach the south end.  The wave moves the other way in that frame.  The solution is symmetrical after all.  This is not intuitive at first, but an interesting side effect of relativity of simultaneity.

The fatal discovery:

So what if I consider the situation in the frame halfway between the two frames described above?  Which way does the wave move then??  Turns out it doesn't.  The ship is partially contracted when 'parked', but moving south with the space station, and identically contracted when moving north at the same speed but opposite direction.  The solution only works because the ship instantaneously (in that frame) changes direction without ever changing speed.  If it took a millisecond to do this, then for that millisecond, it would be longer that it is before and after, and it would shatter.

Infinite acceleration cannot be allowed.

But why can't I accelerate over a minute to 3135 km/sec, still using the wave method?

This was the obvious solution for a while.  OK, so the trip takes a minute longer.  No problem.  But it doesn't work.
I tried to optimize the trip by doing several smaller waves.  Accelerate to 500 km/sec.  The smaller speed difference lets the wave move at a larger speed, and it takes less than 2 days to get to the front.  Then, a few minutes later, initiate another wave like that.  Send about 20 waves like that, each getting to the front in less than a day, and the ship now gets a total speed of 10000 km/sec and the time to move is reduced to 1.3 days.  Total less than 2 days, right?  Wrong!  It is not obvious until you look at it in the other frames and realize the same waves  move the opposite way and are thus arranged in the opposite order.  Any gradual acceleration results in compression of that acceleration until singularities occur.

Since finite acceleration can be integrated as a series of small step speed-changes, that finite acceleration would become infinite at some point along the line, so the limit of the acceleration is one where that singularity occurs forward of the front of the ship, and that's the solution posted in post 20.

[David Cooper]

We have a born-rigid ship that is 100 light years long, and we wish to move that ship a distance of 1 light-hour (north let's say), with it stopped at either end of the trip.  No solution that involves strain on the ship is allowed.

You have to allow some strain on it to accelerate it because of length contraction - there's no possible way to accelerate it at all without strain. What we have to avoid though is any move that leaves atoms sitting the wrong distance apart such that they are free to cause crumples or rips, but we are allowed to move an atom in such a way that the forces are very strong while we're accelerating it just so long as when we stop accelerating it it will be able to sit comfortably where we put it at the speed we've set it to. With the caterpillar method, we take the atom at the tail end and accelerate it towards the atom ahead of it, but we don't let it go until we've accelerated the next atom to the same speed, by which time the forces between them are back to comfortable levels (such that they will sit that distance apart naturally), and we delay the start of that second atom's acceleration so that they end up the right distance apart when we've let go of them both. The practicality of doing this is close to zero, but we're interested in an extreme case where the "ship" might just be a very long chain of carbon atoms with hydrogen's stuck along the sides - this molecule could be held sufficiently flat that all our external machinery for accelerating it can access every single atom and precisely control its acceleration. This will not work for any human passengers which can't be flattened to match, but perhaps we can invent some way of beaming forces through to individual atoms within complex structures where they're hidden deep beneath layers of many others - if we're allowed to do this, then we're still discussing the extreme limit of the potentially-theoretically-possible even if there are live humans on board.

I presumed the ship had to be always stationary in its own frame along its entire length during the whole trip.  From this presumption, I computed a trip time of about 55 days in post 37.

By the way, I never understood the full details of your method, so I just assumed that you know what you're doing with the maths (and continue to make that assumption). I wouldn't be surprised if you're the only person here who understands it. Perhaps it could be made easier to understand and discuss if the lightyears aspect was removed to cut it down to a better size. If we just work in units of d (distance), then the length of the ship can be 100d and the move can be d. We can then imagine a ship of 100 atoms in length all ending up one atom further along from where they started. Or we could imagine a ship of 10 atoms in length all ending up 1/10 of an atom further on by the finish. That would make it easier to discuss what happens in a way that can be visualised easily, and which could be simulated too with JavaScript to provide a moving diagram of the action governed by the relevant maths. The number of atoms used may affect the result, but we'll be able to see the relationship between different versions of the experiment as we go from 10 to 100 to 1000 atoms, and then we can calculate how it would work with a 100 lightyear ship with its atoms at normal atom spacing, but everyone would be able to gain a good understanding first from the simpler cases. (Time should be in units of t based on t=1 for light travelling the distance d.)

David Cooper suggested what has become known as the caterpillar method of moving the ship.  The idea is to accelerate the tail quickly to some speed and propagate that acceleration to the front of the ship as the relativistic contraction allows.  The faster the speed chosen, the greater the contraction and the slower the 'wave' of acceleration moves to the front of the ship.

That's right, because you can accelerate the particle ahead sooner as the space between them will be bigger at the target speed than for a higher speed, but you can then follow it up with another wave of acceleration to take the atoms to a higher speed, and you can do that for all possible speeds with a different wave for each. The wave with the slowest target speed will lead to the front atom being accelerated almost immediately, but it's so small a change in speed that it won't look as if that atom is moving at all.

The wave always moves faster than light.  I found the optimal speed at 3135 km/sec which gives 4 days to move the light-hour and 2 days for the wave to propagate 100 light years for a total trip time of 6 days.  I intend to demonstrate that the method only works with infinite acceleration, and thus is a violation of my initial premise.  It is effectively like suggesting unobtanium.

So long as you aren't accelerating any particles to c, the acceleration will be finite.

The fatal discovery:

So what if I consider the situation in the frame halfway between the two frames described above?  Which way does the wave move then??  Turns out it doesn't.  The ship is partially contracted when 'parked', but moving south with the space station, and identically contracted when moving north at the same speed but opposite direction.  The solution only works because the ship instantaneously (in that frame) changes direction without ever changing speed.  If it took a millisecond to do this, then for that millisecond, it would be longer that it is before and after, and it would shatter.

If you pick the frame half way in between, you start with the whole ship moving south and end up with the whole ship moving south at the end, but it will have spent some of its time moving north, and in this frame the whole ship will suddenly be moving north at the same time, then it will suddenly be moving south again some time later. The contraction will be the same for both directions of travel through this frame, so it's easy for the whole ship to make these accelerations simultaneously in this frame. The ship has changed direction twice, and it's also changed speed twice (if you consider the southward movement to be at -v and the northward movement to be at +v). If you're going to make the change in speed from one direction to the other take a millisecond, then you will have extreme forces applying during that moment trying to extend the ship, but these are no different from the forces that you have to handle on an atom-by-atom basis when looking at it from the other frames, because when you accelerate the end atom towards the one ahead of it, you are sending it towards an atom which is for most of that time applying a force to try to stop the accelerated atom from moving that way, and that opposing force will strengthen as they get closer together. That opposition of forces is only removed when you start accelerate the second atom up to the right speed for the two to sit comfortably together at their new separation distance. Clearly, if we're very precise in how we apply the acceleration to the first atom, it won't be pushed off course, and we can amplify our acceleration force too as it gets closer to the atom ahead of it. Some of the energy we're putting into the first atom will be transferred to the second atom, so that will affect the amount of force we have to apply directly to that atom later, but if we're accelerating the first atom to a speed close to c, the second atom doesn't have much time to respond to this before we start to move it anyway, so this becomes more of an issue the lower our target speed is. The cure for it is accuracy so that when high opposing forces exist, they are acting directly along the direction in which the ship is aligned - damage would only be done if the atoms get slightly out of alignment, because that would lead to strong sideways accelerations which would blow the ship apart, but we're already handling that in the original frame in the way we control the acceleration of our atoms (making sure they aren't going out of line and being deflected sideways), so we must be handling that in all other frames too.

But why can't I accelerate over a minute to 3135 km/sec, still using the wave method?

This was the obvious solution for a while.  OK, so the trip takes a minute longer.  No problem.  But it doesn't work.
I tried to optimize the trip by doing several smaller waves.  Accelerate to 500 km/sec.  The smaller speed difference lets the wave move at a larger speed, and it takes less than 2 days to get to the front.  Then, a few minutes later, initiate another wave like that.  Send about 20 waves like that, each getting to the front in less than a day, and the ship now gets a total speed of 10000 km/sec and the time to move is reduced to 1.3 days.  Total less than 2 days, right?  Wrong!  It is not obvious until you look at it in the other frames and realize the same waves  move the opposite way and are thus arranged in the opposite order.  Any gradual acceleration results in compression of that acceleration until singularities occur.

If something works in one frame, it has to be compatible with all other frames. If that wasn't the case, relativity would break. Consider just two waves. We have one where we accelerate atoms nearly to c, but we have a second wave where we accelerate them to 0.5c. The latter acceleration will propagate from atom to atom at a higher speed than the former, with both propagating at speeds higher than c, but these things are fully possible in the frame of reference in which the starting speed is zero. This must be compatible with the other frames that you're considering. In the frame moving at nearly c, we see what looks like a deceleration of the atoms from nearly -c to zero, and because we're uncontracting the ship from the point of view of this frame, we see the wave move from the front of the ship to the back. There' no problem there. Let's add our second wave in.This wave starts at the same time as the other and from the same location, but it reaches the other end first and then stops, so we have to wait for the first wave (the one described first) to reach the same finish line. When we view this from the other frame, the first wave really is the first wave in that it's seen to leave first, and the second wave is sent a little later. They both reach the finish (the tail of the ship) at the same time as each other), the first wave not quite stopping the atoms, but the second wave finishes that job.

Since finite acceleration can be integrated as a series of small step speed-changes, that finite acceleration would become infinite at some point along the line, so the limit of the acceleration is one where that singularity occurs forward of the front of the ship, and that's the solution posted in post 37.

I still think that the fastest way to move the front of the ship to its destination will involve the caterpillar method with an infinite number of waves moving at different speeds so as to maximise the acceleration of the leading atom, but I don't know how to handle the maths for combining all those waves.

[yor_on]

It depends on what one look at Halc.
'Light' has no acceleration
It has a 'release' and a 'absorption' sort of, but there is no acceleration involved in this.

Only conservation laws.

[yor_on]

And no, a engine that 'pushes' in all directions won't 'move'. You need to create a 'chamber' in where the 'action' acts in one direction to make it work.

[yor_on]

Ok, something being accelerated without a intrinsic engine? "   Is there a stress-free way of accelerating a long object? "
Yeah.

[Halc (OP)]

Thanks for looking at this David.

I'm arguing against your post not because it is wrong, but more as a critical review.  You have some strong points.  I feel like I'm on thin ice with some of my replies.

We have a born-rigid ship that is 100 light years long, and we wish to move that ship a distance of 1 light-hour (north let's say), with it stopped at either end of the trip.  No solution that involves strain on the ship is allowed.

You have to allow some strain on it to accelerate it because of length contraction - there's no possible way to accelerate it at all without strain.

I don't understand your point at all.  Of course its possible, at least mathematically, and this is a mathematical exercise after all, not an engineering one.  I'm not really concerned about accelerating individual atoms.  I consider the object to be a homogeneous rod, perhaps with length marks along it.

Nearest way to implement such a thing is the rod completely enclosed in a 100 light-year rail gun that can accelerate each portion at the appropriate rate.

What we have to avoid though is any move that leaves atoms sitting the wrong distance apart such that they are free to cause crumples or rips, but we are allowed to move an atom in such a way that the forces are very strong while we're accelerating it just so long as when we stop accelerating it it will be able to sit comfortably where we put it at the speed we've set it to.

'When we stop accelerating' is a frame dependent thing.  In my standard description (no wave), the object is stationary in its own frame for its entire length, but not so in any other frame, where different parts are moving at different speeds.  One cannot ever suddenly stop accelerating it 'simultaneously' in one of those other frames.
So for instance, the caterpillar method had the object moving at different speeds in any frame, as long as the wave existed.  Until the wave reaches the other end, the acceleration process cannot stop without shattering the object.

With the caterpillar method, we take the atom at the tail end and accelerate it towards the atom ahead of it, but we don't let it go until we've accelerated the next atom to the same speed, by which time the forces between them are back to comfortable levels

There never should be force between those two atoms.  No need to hold them until they're moving identically.  Well sort of...  If they're not moving identically, they obviously are being accelerated.  There is thus a force on each one, and that counts as being 'held' I suppose.
Two atoms can move at different speeds and not result in strain between them.  It takes time to build up strain, and we don't give it that time.
I sort of ran the proof of that in post 48.  There is zero strain on the object between any two points.  No strain means no stress.  The fault in the description was that it posited zero time for the acceleration, and that is unreasonable.  It can be done over a finite time to give finite acceleration, but the interval of time to do it shortens as the wave moves forward, and it shortens to zero before it gets to the other end of the object.  That forced singularity is why the method doesn't work.  You can make the acceleration gradual enough that the singularity doesn't happen before the other end of the object, but then you're back to a 55 day trip.

(such that they will sit that distance apart naturally), and we delay the start of that second atom's acceleration so that they end up the right distance apart when we've let go of them both.

Yes, we stop accelerating them once this is achieved.  But at no point is there stress between the two of them, even though one may be moving and the other not.  We're holding on not to keep them at an unnatural separation, but to apply the force needed for acceleration.
Interestingly, in the instant-acceleration scenario, that requires neither infinite force nor infinite power.  Both work out to finite numbers, at least in most frames of reference.  That wasn't obvious at first.

The practicality of doing this is close to zero,

As I said, this is a mathematical exercise and I'm not worried about practical issues.  My disproof is a mathematical one, not a demonstration that humans could not be kept alive if onboard.

I presumed the ship had to be always stationary in its own frame along its entire length during the whole trip.  From this presumption, I computed a trip time of about 55 days in post 20.

By the way, I never understood the full details of your method, so I just assumed that you know what you're doing with the maths (and continue to make that assumption).

The 55 day thing required 2 steps.

Step 1: Compute the exact speed of the ship.  There are two ways to do this.

1A: Use a lorentz conversion to compute the speed the universe would need to go to get it the 100 LY distance to contract by a light hour.  That gives me the speed the object needs to go (~452 km/sec).
1B: To confirm that, I did it a very different way in post 52, and made an arithmetic mistake, corrected in post 54.  Just compute c?/100 LY to get the maximum acceleration possible at the nose of the ship, and punch that number into a trip calculator with destination of 1 light-hour.  That latter part is what I get wrong since my trip calculator assumed I wanted to decelerate the 2nd half of the trip, but no, I want to accelerate the whole way.

Both methods yielded the same 452 km/sec

Step 2:
In post 20, I realize that the nose of the ship will accelerate at a constant rate (and the tail decelerate similarly), and hence move at an average speed of half the 452, so 226 km/sec.  So 55 days of that speed gets you one light hour.  Simple multiplication.

I wouldn't be surprised if you're the only person here who understands it.

There are a few who know their stuff better, but they've not posted much in this thread.  All I get are the ones that have a death-grip on their Newtonian assumptions, and it turns out that we've been making some of them ourselves.  I had to find my own errors since nobody else was pointing them out.

Perhaps it could be made easier to understand and discuss if the lightyears aspect was removed to cut it down to a better size. If we just work in units of d (distance), then the length of the ship can be 100d and the move can be d.

Well then I don't get a concrete answer.  It takes a long time to move my object 1 light-year, and yet I can move my meter-stick a centimeter in a moment.  I need real units, not abstract ones, and I need a long object so the relativity aspect is obvious.
I can move it a light-year if that makes it simpler for you.  The max speed would be 42082 km/sec. so half that is 21041, at which speed it takes 14¼ years to do the move.

We can then imagine a ship of 100 atoms in length all ending up one atom further along from where they started.

Yea, but who can relate to the bazillionth of a second it takes to do that?  The relativity isn't going to be apparent.

That would make it easier to discuss what happens in a way that can be visualised easily, and which could be simulated too with JavaScript to provide a moving diagram of the action governed by the relevant maths.

I find it easier to visualize on a large scale, where there are hours or days difference in times depending on frame of choice.  I can have length-marks on my object, which seems to serve the same purpose as your discreet atoms, but I found no need to refer to them to compute the times required by the various methods, or to demonstrate that the caterpillar method doesn't work at all.

David Cooper suggested what has become known as the caterpillar method of moving the ship.  The idea is to accelerate the tail quickly to some speed and propagate that acceleration to the front of the ship as the relativistic contraction allows.  The faster the speed chosen, the greater the contraction and the slower the 'wave' of acceleration moves to the front of the ship.

That's right, because you can accelerate the particle ahead sooner as the space between them will be bigger at the target speed than for a higher speed, but you can then follow it up with another wave of acceleration to take the atoms to a higher speed, and you can do that for all possible speeds with a different wave for each.

I twas trying to do that, finding a faster way by using multiple waves.  It doesn't work.  The waves catch up to each other, which wasn't at all obvious at first.

The fatal discovery:

So what if I consider the situation in the frame halfway between the two frames described above?  Which way does the wave move then??  Turns out it doesn't.  The ship is partially contracted when 'parked', but moving south with the space station, and identically contracted when moving north at the same speed but opposite direction.  The solution only works because the ship instantaneously (in that frame) changes direction without ever changing speed.  If it took a millisecond to do this, then for that millisecond, it would be longer that it is before and after, and it would shatter.

If you pick the frame half way in between, you start with the whole ship moving south and end up with the whole ship moving south at the end, but it will have spent some of its time moving north, and in this frame the whole ship will suddenly be moving north at the same time, then it will suddenly be moving south again some time later. The contraction will be the same for both directions of travel through this frame, so it's easy for the whole ship to make these accelerations simultaneously in this frame.

Here we differ.  In all other cases, the 'instant acceleration' was simply the limit of things as acceleration reached an arbitrary value, but in this case, the value with which we're concerned (length contraction) does not approach zero change as the time to make the acceleration decreases.  So I find it unreasonable to say that since it happens in zero time that one can get from one speed to another speed without hitting the speeds in between.
Avoiding the length contraction by doing the acceleration 'while God blinks' so to speak seems a cheat.  I'd accept the method if it worked with a finite (fraction of a second) acceleration that still propagated up the length in a wave.  I was trying to get the mathematics right on that scenario and was wondering why it didn't work.  Considering it in this middle-frame shows me why it doesn't work.

If you're going to make the change in speed from one direction to the other take a millisecond, then you will have extreme forces applying during that moment trying to extend the ship, but these are no different from the forces that you have to handle on an atom-by-atom basis when looking at it from the other frames

Again we differ.  These forces do not exist when doing it over a millisecond, atom by atom.  At no point are two atoms in a stressful arrangement.  The argument would indeed have merit if this temporary force existed, but it doesn't.  I showed that it doesn't.  Two atoms cannot exert a force during a pair of event separated in a space-like manner.
By the same argument, that force (stress/strain) admittedly cannot exist during an instantaneous direction-change.  I agree that far, but I find it to be cheating to use this singularity to get a value that is not approached by arbitrarily high acceleration.   5n/n is 5 for all values of n except zero, and it is valid to assign an arbitrary value (like 13) to the undefined value at n=0, but contradictions can be reached if you allow such mathematics at the singularities.  I can prove that 1=2 if you allow that.

because when you accelerate the end atom towards the one ahead of it, you are sending it towards an atom which is for most of that time applying a force to try to stop the accelerated atom from moving that way, and that opposing force will strengthen as they get closer together. That opposition of forces is only removed when you start accelerate the second atom up to the right speed for the two to sit comfortably together at their new separation distance.

Well, no.  It sort of works exactly because this doesn't happen.  Acceleration of an atom doesn't result in any force against its unaccelerated neighbor.  What does this is displacement, and displacement takes time.  For it to put a force on the neighbor would be to have a causal effect at greater than light speed, which cannot happen.
My argument against this method is not that there would be stress, but that it isn't a solution that is approached by arbitrarily high acceleration.  Only at the singularity does it work, and therein lies my protest.  The lack of approached limit is what I find unacceptable.

Your approach is also on the discreet atomic level rather than the homogeneous mathemeatical level, but if we model the 'ship' as a series of discreet points that are accelerated individually, then there really is no length of the object, just spacings between the atoms which are to be ignored for the duration that a force is applied to them.  If we allow that, even for a shorter duration than the speed of light between adjacent atoms, then the length of the object seems meaningless.  The rules are to be ignored while we briefly take tongs to each atom in turn and change its velocity.

How does QM handle this?  I swat the sun away, and its gravity still pulls Earth from where it used to be for 9 minutes.  Eventually the corrections are made by gravity waves that make the necessary adjustments to keeps the conservation laws happy.  Something similar must happen when we accelerate an atom to a new frame all outside the light cone of its neighbors. That must emit some sort of EM equivalent of graviton that is going to apply the stress and strain that the immediate speed of light prevented.  I'm no expert here.  I was approaching this mathematically, not from a quantum physics standpoint.

Some of the energy we're putting into the first atom will be transferred to the second atom

Due to QM effects perhaps, but otherwise no.  The EM field will aways consider adjacent atoms to be stationary relative to each other, but speed-of-light particles will be emitted by the acceleration, and I suppose those would put stress on our object.  There isn't supposed to be any, but at a QM level (as opposed to mathematically), I suppose it happens.

but if we're accelerating the first atom to a speed close to c, the second atom doesn't have much time to respond to this before we start to move it anyway

It will be accelerated itself long before the acceleration of its neighbor can possibly be noticed.  We're not doing it anywhere near c.  A few hundred or thousand km/sec.  1% of c at best.

But why can't I accelerate over a minute to 3135 km/sec, still using the wave method?

This was the obvious solution for a while.  OK, so the trip takes a minute longer.  No problem.  But it doesn't work.
I tried to optimize the trip by doing several smaller waves.  Accelerate to 500 km/sec.  The smaller speed difference lets the wave move at a larger speed, and it takes less than 2 days to get to the front.  Then, a few minutes later, initiate another wave like that.  Send about 20 waves like that, each getting to the front in less than a day, and the ship now gets a total speed of 10000 km/sec and the time to move is reduced to 1.3 days.  Total less than 2 days, right?  Wrong!  It is not obvious until you look at it in the other frames and realize the same waves  move the opposite way and are thus arranged in the opposite order.  Any gradual acceleration results in compression of that acceleration until singularities occur.

If something works in one frame, it has to be compatible with all other frames. If that wasn't the case, relativity would break.

It apparently doesn't work in any frame, but it isn't so intuitive in say the initial rest frame.  The one (middle) frame just made it real obvious why it didn't work, and yes, per relativity, the other frame thus must also not work.

Consider just two waves. We have one where we accelerate atoms nearly to c, but we have a second wave where we accelerate them to 0.5c. The latter acceleration will propagate from atom to atom at a higher speed than the former, with both propagating at speeds higher than c, but these things are fully possible in the frame of reference in which the starting speed is zero.

No, they're not possible in that (or any) frame.  It just wasn't initially obvious to either of us.  It became more apparent when I started to attempt optimizations and was running into so much trouble.

This must be compatible with the other frames that you're considering. In the frame moving at nearly c, we see what looks like a deceleration of the atoms from nearly -c to zero, and because we're uncontracting the ship from the point of view of this frame, we see the wave move from the front of the ship to the back. There's no problem there.

There is a problem there.
I think it stems mostly from the difference between acceleration and proper acceleration.  We have to assume constant proper acceleration else the motion will not be symmetric from one reference frame to the next.  But the actual acceleration (in a given frame) needs to be constant in order for the wave to progress at a uniform speed.  It works only with infinite acceleration because there is no difference between actual and proper acceleration, both being infinite and without duration.

I still think that the fastest way to move the front of the ship to its destination will involve the caterpillar method with an infinite number of waves moving at different speeds so as to maximise the acceleration of the leading atom, but I don't know how to handle the maths for combining all those waves.

Yes, that's one of the ideas I was trying and not getting to work.

In the frame moving at nearly c, we see what looks like a deceleration of the atoms from nearly -c to zero, and because we're uncontracting the ship from the point of view of this frame, we see the wave move from the front of the ship to the back. There' no problem there.

Let me describe the contradiction a different way.  Suppose we create a wave not instantly, but over a short period like a minute or an hour.  It doesn't matter if it is in a frame that considers it an increase or decrease in speed.  The point is that it is uniform acceleration for the duration of that minute or hour, and all points on the object will experience that uniform acceleration for that time.  But this cannot be since an accelerating object must accelerate harder at the tail than at the head, as we worked out in the beginning of this thread and in that other thread in New Theories you were involved in.  The ship will compress or pull apart if it has uniform acceleration, and a moving wave must produce that uniform acceleration, or the wave must change shape (lengthen or contract at least) as it moves, which violates the way we are envisioning it as just a contracted object leaving a normal length object behind it as it decelerates.  No such wave can traverse the object except a super-low acceleration one that changes slowly as it moves, slowly enough that the distortion of the wave doesn't result in a singularity before reaching the other end of the object.  It takes 55 days to do this.