Some Special Relativity exercises with rigid objects

I just don't get what I'm missing.
Since the elements of the object are not connected, they drop out of the problem related to starting and stopping at a doorway.
Since no acceleration is specified, it is unlimited.
Since a door can be represented by a single drone, the question simply dissolves to "How fast can one drone - with no explicit limit on its acceleration - take to travel one light hour?"

And the answer is: a time that approaches zero (given acceleration without bound).

There must still be some condition, obvious to you, going unstated. Otherwise, where does my solution go wrong?
 
Since the elements of the object are not connected, they drop out of the problem related to starting and stopping at a doorway.
Since you get along with the drone description, just imagine that every light hour there is one drone of a different color, and we must line up that drone with a mark along the path. No door to worry about.

Since no acceleration is specified, it is unlimited.
That's right. We place no limit on proper acceleration. 100 G is fine. So is a trillion G. At a mere 100 G, the time required would be measurably over 5.51 days. At 100000 G or a trillion G, the difference would not register at 3 digits of precision.

Since a door can be represented by a single drone, the question simply dissolves to "How fast can one drone - with no explicit limit on its acceleration - take to travel one light hour?"
It has to stay at the same proper distance from each of the other drones, so no, the problem certainly doesn't dissolve to a single drone case, which would be simply "it takes one hour".

And the answer is: a time that approaches zero (given acceleration without bound).
No, a single drone would take at least an hour to move one light hour, despite any level of proper acceleration. Physics also puts frame-dependent limits on any coordinate acceleration, but proper acceleration is neither frame dependent nor limited.

There must still be some condition, obvious to you, going unstated.
Yes, the physics of our universe. Surely you must know that no object can move faster than light despite any level of force or proper acceleration put on it.
 
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Sorry, I guess I was tired and in a hurry. Of course it approaches the limit of one hour, not zero.


So nevermind for the moment where your 5.5 days comes from, why is your answer anything other than "just over one hour"?
 
Exercise P:

Since Halc has been looking at this from the train's rest frame, I decided to look at it from the track's rest frame. For now, I am not using Halc's parameters of one light year and one light hour. I just want to run this simple example to see what it does.

I simply have 10 individual particles with no interaction between them, like 10 separate drones. The rear most drone instantaneously accelerates to my chosen velocity which happens to be v=0.866c. It could have been any velocity less than v=1.000c, but I just prefer to use gamma=2.000 while I work things out.

Then, at a pre-programmed time, the next drone accelerates the same way as the first. The pre-programmed time makes the proper distance between the drones remain the same as it was before they accelerated. We can see this in the diagram below, because the distance between them reduces by 1/gamma. This process repeats until all of the drones are moving at v=0.866c and they are the correct distance between each other.

6N2oIjf.png


Some interesting things to note:

1. This would not be possible if these were interactive particles being pushed by one force as one object, because the speed of the propagation of the accelerations is 1.000/0.577=1.732c which would be information moving faster than c. It is only possible in this special case because of the pre-programmed timing of the accelerations, which works totally fine for individual drones with synchronised clocks.

2. In this special case, the accelerometers on the drones all display the same proper acceleration, however they do not all do so at the same time. The first particle accelerates first, and then the next accelerates afterward, and so on. At the end of the exercise, where the drones are supposed to be at rest in the track frame again, this process would have to be repeated so that the length contraction incrementally disappears in the track frame, just as it incrementally occurred in the first place.
 
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Ned, do you understand why Halc gets any answer other than "just a little over an hour"?

I understand that it will take some time for all the drones to "compress" to the correct spacing as they gradually all achieve speed. I also understand it will take the same amount of time for them to "decompress" again at the end, in order to come to rest with the tracks. That makes the process take a different amount of time than "just a little over an hour" for his parameters. However, I have not yet calculated to see if it would match his 5.5 days per light year, or whatever it is that he has calculated.

If you study my diagram above, you should start to see the issue. I am not sure whether that is exactly the same issue that Halc is trying to illustrate or not.
 
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So nevermind for the moment where your 5.5 days comes from, why is your answer anything other than "just over one hour"?
Yet again, because of the requirement to maintain constant proper separation with all the other drones. That can't be done if they all move at nearly c.

Since Halc has been looking at this from the train's rest frame
Well, mostly frame S actually, the frame of the track, but yes, I expressed contraction of the track in the train frame (T), enough contraction to bring in one more light hour of track.
With the rotating examples, there seems to be but the one frame to work with.

I just want to run this simple example to see what it does.
I saw the picture. You're getting ahead of me. Yes, what you drew works.

Then, at a pre-programmed time, the next drone accelerates the same way as the first. The pre-programmed time makes the proper distance between the drones remain the same as it was before they accelerated. We can see this in the diagram below, because the distance between them reduces by 1/gamma. This process repeats until all of the drones are moving at v=0.866c and they are the correct distance between each other.
Exactly. If viewed as a material (and not a cloud of drones), there would be zero strain on your object. Each drone maintains proper separation from its neighbor, even if there is no proper separation defined between it and the drones further away.

This would not be possible if these were interactive particles connected together, because the speed of the propagation of the accelerations is 1.000/0.577=1.732c which would be information moving faster than c.
What you draw is indeed not possible just by local interaction. Clean rigid acceleration (where all particles maintain constant proper separation) is possible with force only at one location and only local interactions, but it is only possible if acceleration is continuous and constant. There would be no way to bring it to a halt relative to S. So I agree with you.

It is only possible in this special case because of the pre-programmed timing of the accelerations
Agree. I've said as much in the OP, in different words.

2. In this special case, the accelerometers on the drones all display the same proper acceleration, however they do not all do so at the same time.
Well, in the case you've drawn, each drone gets an impulse, which is a finite velocity change over zero time, which is technically infinite proper acceleration and then just coasting. None of your dots changes velocity except at that one moment. In my example (taking 5.5 days), each of the drones continuously accelerates one way or the other, until it stops at the end. There is never a period of coasting for any of them. There is in your example, and I was going to explore exactly that way of doing it, but I'm spending a lot of posts clarifying things, which is a good thing.

This process would then have to be reversed so that the length contraction incrementally disappears in the track frame, just as it incrementally occurred in the first place.
Exactly. After each drone moves say one light hour, a second impulse is applied that stops it. The whole object then moves not all at once, but in a wave very much like a caterpillar walking with a wave of movement that propagates from his butt to his head. You can measure the speed of your wave above. The points at which the impulse is taking place are separated by time 0.577 and each is a distance 1 apart, making the wave move at 1/0.577 = 1.732 c which is √3 c. At that speed, it would take about .577 years to traverse the 1-light-year object. What if you slowed your velocity down to say 0.1c? How fast does the wave propagate then?
 
Here is the diagram with worldlines drawn on top:

kNljHSG.png


I can see now why Rindler liked constant proper acceleration which would make nice hyperbolas, instead of straight lines.
 
However, I have not yet calculated to see if it would match his 5.5 days per light year, or whatever it is that he has calculated.
5.5 days for a single light-year object, which is different than 5.5 days per light year. For instance, to move a 10 LY object the same 1 lt-hr, we use the same relativity of simultaneity equation t'=Lv:

L=(87601/87600) * 3650 = 3650.0417 light days
and v = 0.00478 (the speed at which the Lorentz factor is 87601/87600)
t' = 3650.04 * 0.00478 = 17.44 days, which is around √10 times as long, not 10x

Ned, do you understand why Halc gets any answer other than "just a little over an hour"?
Suppose we just accelerated each drone instantly to say .99c. The length of a object with a 1-LY proper length moving that fast in S should be about 1/7 in S, but it's length in S is still 1 LY, seven times what it should be, which violates the stipulation that the drones all maintain a constant proper separation.
To maintain that constant proper separation, they take hyperbolic trajectories as described in articles about 'Rindler motion'. Those articles explain things better than I can.

You might find Øyvind Grøn's "Space geometry in rotating reference frames:A historical appraisal" useful.
That seems to concern actual rotation of an object, which cannot be done with Euclidean material, and all my examples involve Euclidean objects.
 
Yet again, because of the requirement to maintain constant proper separation with all the other drones. That can't be done if they all move at nearly c.
It can't be done at any speed, if you measure precisely enough.

At what speed do you decide length contraction is signficant enough? .9c? .1c? .01c? 100mph?


This violates Halc's requirement:

upload_2023-10-29_11-0-21.png

The object starts off with two units between drones and evolves to approx. one unit between drones. So it's not "maintaining constant proper separation with all the other drones".
 
This problem is poorly-defined. There are too many unspoken assumptions.
Trying to extract the assumptions is like pulling teeth. And we are on post 32 now.

The problem now has been reduced back to the non-rigidity issue I predicted. At no speed can a matrix of independently controlled drones maintain a fixed proper distance to an arbitrary level of precision. At any speed above zero, length contraction kicks in.

The onus is on the OP to properly define the conditions in a timely fashion, not upon the readers to beg for details.

Sry Halc.
 
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At what speed do you decide length contraction is signficant enough? .9c? .1c? .01c? 100mph?
That depends on the size of the object (or cloud of drones) and how far we have to move it. Example P needed 0.0151c, or about 4500 km/sec. Any faster than that and it overshoots its destination.


The object starts off with two units between drones and evolves to approx. one unit between drones.
So it's not "maintaining constant proper separation with all the other drones".
But my OP stipulation was that it maintain constant separation with its immediate neighbors, which Neddy's drawing does not violate.
It very much does violate Rindler motion since the object is not everywhere stationary in its own proper frame.

The solution works (does not cause any strain to the object), but is not optimal since it takes about 7 months instead of 5 days.

This problem is poorly-defined. There are too many unspoken assumptions.
Others seem to be able to follow, so apologies that you think there's still stuff unstated.


The problem now has been reduced back to the non-rigidity issue I predicted. At no speed can a matrix of independently controlled drones maintain a fixed proper distance to an arbitrary level of precision.[/QUOTE]This is a false assertion that has not been demonstrated. Drones undergoing Rindler motion must maintain fixed proper distance with all other drones.

At any speed above zero, length contraction kicks in.
Yes it does, which is why they don't accelerate equally. That's the whole point.
There are websites explaining all this better than I can. Read up on Rindler motion and then come back with what you find to be contradictory. Don't assert something false.

I wanted to discuss these problems with those that know the basics of rigid/Rindler/hyperbolic motion, mostly since there have been several threads opened by people (Fontenot in particular) that clearly do not know said basics. The first exercise is directly applicable to that. The others more indirectly.
 
That depends on the size of the object (or cloud of drones) and how far we have to move it.
No, it simply depends on your level of precision.

All non-pointlike objects, spread over any non-zero volume, and moving any distance will experience some form of length contraction, within and without themselves - if you measure carefully enough. There is no lower limit on speed to Lorentz contraction.

Since you allow for unlimited accleration, it follows that you must allow for unlimited precision - otherwise, such numbers as 5.5 days are arbitrary.

But my OP stipulation was that it maintain constant separation with its immediate neighbors, which Neddy's drawing does not violate.
It does. As I pointed out in Ned's own diagram. They start at two units apart and rapidly evolve to about one unit apart. The extended object shrinks, starting at one end.

This is a false assertion that has not been demonstrated.
It is the premise of SR. You are making assertions that appear to go against it - it is your conjecture on display here, so the onus is on you to demonstrate.

I wanted to discuss these problems with those that know the basics of rigid/Rindler/hyperbolic motion,
That would have been one of those things you could have referenced anytime in the last 33 posts when asked for clarification. But you did say you deliberately left out your work.

... mostly since there have been several threads opened by people (Fontenot in particular) that clearly do not know said basics.
You've started a thread half way through your thinking process - you confessed to deliberately withholding your logic - and won't be surprised if readers such as Fontenot stumble over the bits you mete out. I'd say you've got a self-fulfilling prophecy going here. This thread will evolve exactly as you expect it will, because you're contriving it to.

My cynic alarm is going off at your motivations. I give up jumping through your hoops.
 
This (Neddy's diagram) violates Halc's requirement:

The object starts off with two units between drones and evolves to approx. one unit between drones. So it's not "maintaining constant proper separation with all the other drones".

You are correct on this point. In my approach, even in the rest frame of the first drone to accelerate, the distance to the next drone changes. Before the acceleration it is 1 unit, immediately after the acceleration, the distance is 2 units, and then over time, as the first drone approaches the second, the distance reduces back to 1 unit. At the time, the second drone accelerates, and then two drones will remain 1 unit apart over time.

The changing distance would lead to "the string breaks," so my solution doesn't really work.

It can't be done at any speed, if you measure precisely enough.

There is an exact theoretical solution, which is the hyperbolic motion that is described by Rindler.

https://en.wikipedia.org/wiki/Rindler_coordinates

Note how the graph is similar to mine, but with hyperbolas instead of straight lines. This makes the length contraction exactly what it needs to be for the "string to not break" so-to-speak.
 
After thinking about this further, I realized this part of my post was incorrect:

You are correct on this point. In my approach, even in the rest frame of the first drone to accelerate, the distance to the next drone changes. Before the acceleration it is 1 unit, immediately after the acceleration, the distance is 2 units, and then over time, as the first drone approaches the second, the distance reduces back to 1 unit. At the time, the second drone accelerates, and then two drones will remain 1 unit apart over time.

What really happens, (I think), is the following. In the rest frame of the first drone to accelerate, before the acceleration the distance to the second drone is 1 unit, immediately after the acceleration, the distance to the second drone is still 1, because the second drone had already launched (time difference due to RoS). The first drone's frame would measure the distance to the launch pad of the second drone as length-contracted, so the launch pad would be nearer after the acceleration, but the second drone would have already launched, so it would already be 1 unit away from the first drone. And they would be stationary with respect to one another after that, until the next acceleration.

So maybe Halc is correct that my solution is valid. I don't know, but I tend to think the only true solution is the Rindler one.
 
All non-pointlike objects, spread over any non-zero volume, and moving any distance will experience some form of length contraction, within and without themselves - if you measure carefully enough. There is no lower limit on speed to Lorentz contraction.
This makes it sound like I ever said otherwise. Length contraction is exactly why it takes more time than an hour to move your object. You could do it in under a minute under Newtonian physics.

Since you allow for unlimited accleration, it follows that you must allow for unlimited precision, otherwise, such numbers as 5.5 days are arbitrary.
Non-sequitur. The 5.51 day limit is not a function of the ceiling on acceleration. It is a function of precision. If we kept it to 2 digits instead of 3, it would be only 5.5 days. 5.512 with 4 digits.

It does. As I pointed out in Ned's own diagram. They start at two units apart and rapidly evolve to about one unit apart.
From 2 to 1 unit apart in S, in which frame it is moving at 0.866c. The proper separation between the two adjacent points is unaltered. Do you even know what proper separation means?

It is the premise of SR.
There is no premise of SR that states: "At no speed can a matrix of independently controlled drones maintain a fixed proper distance to an arbitrary level of precision.". You're making up nonsense now.

You are making assertions that appear to go against it - it is your conjecture on display here, so the onus is on you to demonstrate.
I'm not making anything up. This is straight from established sites, anything that describes Rindler motion for instance. This was all worked out before I was born.

But you did say you deliberately left out your work.
I was hoping that somebody had the expertise here. I suspect they do, but nobody has bothered to post numbers before I do. I have since posted numbers, showing the work. It is a trivial relativity of simultaneity calculation. The latter two problems are far more complicated and I still have no numbers.

I give up jumping through your hoops.
Perhaps the topic could make quicker progress were that statement true.
 
In my approach, even in the rest frame of the first drone to accelerate, the distance to the next drone changes.
That's because 1) the drones do not exhibit Rindler motion, and 2) there are discreet gaps between them, not true of a mathematically solid object.
The second point can be resolved by simply noting that the proper separation between two objects is momentarily undefined since the neither drone is stationary in the rest frame of the other. It can also be resolved by making the object continuous instead of discreet.

The first point can be resolved by having all the dots accelerate per Rindler motion, which means each has a proper acceleration of 1/R where R is the distance to the Rindler horizon, which is the left dot if it has infinite proper acceleration. This is what yields the 5.51 day number.

Before the acceleration it is 1 unit, immediately after the acceleration, the distance is 2 units, and then over time, as the first drone approaches the second, the distance reduces back to 1 unit.
Agree. Proper separation is undefined during this short interval.

The changing distance would lead to "the string breaks," so my solution doesn't really work.
No it doesn't break because the string in continuous, not moving at all at first, and more and more of it moving (with instant acceleration) until the acceleration 'wave' reaches the next drone. This all assumes that the 'string' is self propelled (each bit of it), which of course just reduces the string to just another line of drones, which people tend to think of as discreet.

Note how the graph is similar to mine, but with hyperbolas instead of straight lines. This makes the length contraction exactly what it needs to be for the "string to not break" so-to-speak.
Yes. The page doesn't discuss reversing the procedure to bring the 'train' to a halt in S, but it's just a matter of starting from stationary in T and accelerating 0.0151c the other direction.

Note that in my example (with natural units of a light day), the acceleration of the right-most point is 1/365 light days/day² which is slightly below 1 g.


What really happens, (I think), is the following. In the rest frame of the first drone to accelerate, before the acceleration the distance to the second drone is 1 unit, immediately after the acceleration, the distance to the second drone is still 1, because the second drone had already launched (time difference due to RoS).
Let's compute it then. RoS computation: t' = Lv where L is 1 and v = 0.866. Here, t' is the time change at what you call the 'launchpad' of the 2nd drone. Pretty impressive for a point-size drone to get one of these.
Anyway, RoS time change at the pad is 0.866, and the drone did its acceleration at time 0.577, so indeed, the 2nd drone is already moving in that frame. All of them are as a matter of fact.
It still isn't Rindler (hyperbolic) motion, but what you say here works.

I tend to think the only true solution is the Rindler one.
Depends on what you call a 'true' solution. I did specify somewhat relaxed requirements in the OP precisely to allow a solution like the one you gave.
Did you compute how long it takes to move the light-year object this way?
 
Non-sequitur. The 5.51 day limit is not a function of the ceiling on acceleration. It is a function of precision. If we kept it to 2 digits instead of 3, it would be only 5.5 days. 5.512 with 4 digits.
Sorry, that's not what I meant. I expressed that poorly. I was trying to argue that a few decimals of precison in acceleration would make a big difference in time elapsed. - i.e. you might get 5 seconds or somesuch, rather than 5 days. )

But the point is moot.

The subject math (Rindler motion) does appear to out of my wheel house (something that was not obvious at the start).


So I'll step back fer realz this time.
 
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That seems to concern actual rotation of an object, which cannot be done with Euclidean material, and all my examples involve Euclidean objects.

Fig. 9. Part C is Euclidean material as lines from each of the emitted points to the camera reflect the distance traveled.
The result is shown in Fig. 9. Part C of the figure shows the “optical appearance” of a rolling ring, i.e. the positions of emission events where the emitted light from all the points arrives at a fixed point of time at the point of contact of the ring with the ground. In other words it is the position of the points when they emitted light that arrives at a camera on the ground just as the ring passes the camera.
 
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