Relativistic paradox involving torsion.

[Cleonis]

Some time ago someone introduced me to the following scenario.

Imagine you ship loads of oil rig pipes into space, and you assemble them to a length of many kilometers, and you start that long pipe spinning on its long axis. You add vibration damping, to get rid of any transverse, longitudinal and torsional vibration.

That long pipe can then be used as a mechanical device to establish simultaneity along the length of that pipe. Spaceships along the length can synchronize clocks by monitoring the instantaneous orientation of the long pipe.

Let there be two sets of spaceships, one set is co-moving with the long pipe, arranged with equal spacings along the length of the pipe. [added] (They are co-moving with the center-of-mass of the spinning pipe, but not "co-spinning".) [/added] The other set of spaceships is passing by the long pipe at relativistic speed, in the direction parallel to the pipe. Since this is a thought experiment we can extend the length of the pipe indefinately, so it's not in itself necessary to think about what happens at the ends.

Question:
As mapped in the coordinate system that is co-moving with the set of spaceships that is passing by, what is the shape of the spinning pipe?

The length of the pipe will be Lorentz-contracted, but here's the rub: if in that other frame the pipe would be simply straight it would violate relativity of simultaneity (as the pipe can be used to establish simultaneity).
The logical consequence is that as mapped in the coordinate system that is co-moving with the passing set of spaceships the long spinning pipe will be twisted, like the twist of a torsion bar under stress.

It's an illustration, of course, of the fact that (as in the example of Born rigidity) in special relativity perfectly rigid bodies do not exist. I don't quite fathom this long spinning pipe paradox yet. Accounting for the twist as mapped in the passing-by frame isn't straightforward. I'm interested in any comments and opinions.

Cleonis

 

[PeterDonis]

if in that other frame the pipe would be simply straight it would violate relativity of simultaneity (as the pipe can be used to establish simultaneity).

I haven't fully analyzed this setup yet, but one thing jumped out at me from your statement of the problem. The pipe can be used to establish a simultaneity (i.e., a time coordinate), but it will *not* be the simultaneity of an inertial frame! It can't be, because the observers co-moving with the pipe are rotating, and so experience acceleration; they are not in free fall.

You can still do analysis of these situations (similar to the way the rigid rotating disk is handled), but the analysis is more complicated than just the relativity of simultaneity of two inertial frames. So I'm not sure whether the pipe will actually have to look bent in the "rocket" frame; I think more analysis is required to see.

 

[Cleonis]

[...] the observers co-moving with the pipe are rotating, and so experience acceleration; they are not in free fall. [...]

I apologize for the ambiguity in my original statement of the setup. I have edited the original posting to address that.

Cleonis

 

[PeterDonis]

Your revision cleared up the scenario, but now I'm not sure what you mean by saying that having the pipe straight would violate the relativity of simultaneity. Can you elaborate?

 

[Cleonis]

Your revision cleared up the scenario, but now I'm not sure what you mean by saying that having the pipe straight would violate the relativity of simultaneity. Can you elaborate?

Well, that would involve explaining to you the very nature of the relativity of simultaneity that is implied by special relativity. I can do that, but that's not why I posted the paradox. I do teach, as you can read in my other postings, but didn't start this particular thread to do yet more teaching.

Anyway, here we go:
Imagine a straight line of external clocks, synchronized with Einstein synchronization procedure. A fleet of spaceships, arranged in a straight line (co-moving with the clocks) can maintain synchronized shipclocks for the whole fleet, either by monitoring the external clocks, or by setting up Einstein synchronization procedure among themselves. I will refer to this as 'the first fleet'.

Next imagine a second fleet of spaceships, moving at relativistic velocity, passing by the straight line of clocks. The ships of this passing-by fleet maintain among themselves synchronized fleet time with Einstein synchronization procedure.
- As mapped in the passing-by frame the distances between the external clocks will be Lorentz contracted.
- As mapped in the passing-by frame the external clocks will count time at a slower rate
- As mapped in the passing-by frame the external clocks will not be in synchronized state; each clock will be somewhat offset to the next. This implication is the one that is relevant in this setup.

Next step: the long spinning pipe.
For unconditional relativity of inertial motion the frame of the passing-by fleet must be just as servicable as the rest-frame for describing the events.

The crew of the second fleet takes time to assemble data from all of the ships of the fleet, building a comprehensive picture; a long spinning pipe is passing them by, and co-moving with that spinning pipe, (but not co-spinning) is the first fleet of spaceships.

Special relativity asserts that all synchronization procedures, be they based on light signals or mechanical devices, will give mutually consistent results. If mechanical device synchronization would would disagree with light signal synchronization you would not have the unconditional relativity of inertial motion that special relativity asserts.

As mapped in the second fleet frame, the clocks onboard the ships of the first fleet are all somewhat offset to the next. The procedure that the first fleet uses to keep synchronized fleet time is by monitoring the instantaneous orientation of the pipe. (NB none of the ships of the first fleet watches the entire length of the pipe, they monitor only the section of pipe _just adjacent_ to each spaceship. If they would also observe distant parts of the pipe they would have to take transmission delays into account. It suffices to monitor the _adjacent part_ of the pipe only.)
As mapped in second fleet frame, the synchronization procedure that the first fleet uses works fine for them, no discrepancy arises.

The demands:
- The synchronization procedure that the first fleet uses _must still work_ when it is mapped in the other frame.
- The results of any synchronization procedure that the first fleet uses _must be consistent_ with Einstein synchronization procedure.

The logical implication of those two demands is that the pipe cannot be straight in both frames. Something has to give.

Cleonis

 

[matheinste]

Hello cleonis.

The line on the rotating pole appearing at a certain orientation of the pole can be considered as the ticking of a clock in the frame in which the pole is at rest. So we can map the orientation of the pole to time intervals. Of course in the pole's frame the clocks are synched and can be represented by a straight line on the pole.

Now as viewed from the other frame the clocks are out of sync and so the ticks do not occur at the same in that frame. Mapping this back to the line on the pole the line will appear twisted.

If this analysis is correct then there is no problem. The marks on the pole appear straight in the poles frame and twisted in the other frame.

Matheinste.

 

[PeterDonis]

I understand how the relativity of simultaneity works. What I'm not sure about is what you mean by saying that, to observers in the second frame, the pipe does not look "straight". Do you mean one of the following, and if so, which?

(A) In the first frame, the pipe, along its length, is a straight line that is parallel to the line of spaceships; but in the second frame, the pipe, along its length, is not a straight line but is some kind of curve (so it is *not* parallel to the line of spaceships).

(B) The pipe has marks all along its length, which the observers in the first frame use to mark "ticks" of the "pipe clock" (each time a given observer's mark passes the window of his ship is a "tick"). The pipe itself is a straight line all along its length in both frames, but: In the first frame, the marks, when they pass the spaceships at each "tick" (which they all do simultaneously in this frame), all lie in a straight line that is exactly parallel to the line of spaceships; but in the second frame, the line of marks is twisted around the pipe so they don't all pass the ships at the same time.

I'm guessing you mean B (matheinste appears to agree), but some people, reading your description, might think A (I know I did at first).

 

[Cleonis]

[...] what [do] you mean by saying that, to observers in the second frame, the pipe does not look "straight".

Hello, PeterDonis,

Here is what I wrote about that in the original posting:

[...] as mapped in the coordinate system that is co-moving with the passing set of spaceships the long spinning pipe will be twisted, like the twist of a torsion bar under stress.

Also, the name that I chose for the thread might be a clue: 'Relativistic paradox involving torsion.'


On a more general note: I am keen to avoid ambiguities that arise from phrasings like: "to the observers of the second fleet the pipe appears to looks so-and-so". Essentially, that kind of ambiguity is introduced when transmission delays are invited to the story, and it can be eliminated.

The grid of spaceships is tightly packed. (As tight as we want, as this is a thought experiment.) The ships of both the first and the second fleet monitor only the part of the long spinning pipe that is _just adjacent_ to each spaceship, (hence no transmission delay), gather all the data from the individual ships, and assemble a comprehensive spacetime diagram.

The comprehensive spacetime diagram is free from ambiguity. It is the mapping of what is taking place as mapped in that inertial frame. The spacetime diagram is committed to a particular inertial frame, but not committed to any of the individual observers who are co-moving with that particular frame. You will never see me referring to observers, but always to: 'the events as mapped in the so-and-so frame.'

Cleonis

 

[PeterDonis]

So that would be B, then. In that case, I agree; I think B describes how the events in question would be mapped (to use your phraseology) in each frame. There's nothing physically inconsistent about it that I can see.

 

[Cleonis]

So that would be B, then. In that case, I agree; I think B describes how the events in question would be mapped (to use your phraseology) in each frame. There's nothing physically inconsistent about it that I can see.

(B) The pipe has marks all along its length, which the observers in the first frame use to mark "ticks" of the "pipe clock" (each time a given observer's mark passes the window of his ship is a "tick"). In the first frame, the marks, when they pass the spaceships at each "tick" (which they all do simultaneously in this frame), all lie in a straight line that is exactly parallel to the line of spaceships; but in the second frame, the line of marks is twisted around the pipe so they don't all pass the ships at the same time.

To emphasize a particular feature I stipulate that the cross section of the pipe is square to begin with. There's no need to apply any marking lines on the pipe, the corners serve as marks.

- As mapped in the coordinate system that is co-moving (but not co-spinning) with the long spinning pipe the pipe is untwisted.
- As mapped in the passing-by coordinate system the pipe is twisted; each corner traces a helix along the length of the pipe.

To twist a bar you have to apply a torque. Question: as mapped in the passing-by coordinate system: what is the source of that torque?
I'm only partially familiar with the right angled lever case, but I suspect that resolution of this case, the long spinning pipe, will be related to the resolution of the rigth angled lever case.

Cleonis

 

[JesseM]

To twist a bar you have to apply a torque. Question: as mapped in the passing-by coordinate system: what is the source of that torque?

In the frame where the bar has a nonzero linear velocity (the passing-by frame), whatever spun it up originally started applying torque to one end before the other (since in order to have it non-twisted in the comoving frame, the torque must have been applied simultaneously to all sections of the bar in the comoving frame--remember, there are no rigid objects in SR, you can't just apply a torque to one part of a bar and expect all parts to simultaneously start spinning in any frame)

 

[Cleonis]

In the frame where the bar has a nonzero linear velocity (the passing-by frame), whatever spun it up originally started applying torque to one end before the other (since in order to have it non-twisted in the comoving frame, the torque must have been applied simultaneously to all sections of the bar in the comoving frame--remember, there are no rigid objects in SR, you can't just apply a torque to one part of a bar and expect all parts to simultaneously start spinning in any frame)

I think what you have described applies for the case of a stack of non-connected disks, but not for the thought experiment of this thread, the spinning pipe (the pipe being used as a mechanical device to synchronize spatially separated clocks.)


Let me discuss the case of a stack of non-connected disks.

In the frame co-moving with the stack of disks, when the engineers want to have the stack spinning on its long axis, they must apply the same torque to each disk simultaneously. (To be more precise, exactly what procedure they use to synchronize the spinning disks is interexchangable, the relevant aspect is that they maintain a synchronized state of the disks, maintaining synchronizing timing markings on the disks.)

As mapped in an inertial coordinate system in which the stack-of disks has a velocity (along the axis that is co-axial with the stack's long axis), the torques on each of the individual disks have not been applied simultaneously, as per relativity of simultaneity. Thus it is predicted and explained why as mapped in the coordinate system in which the stack has linear velocity each disk is offset to the next disk at a slight angle, in such a way that relativity of simultaneity holds good.


The thought experiment of this thread is about a long pipe (which can be regarded as a stack of mechanically connected disks, of course). When a pipe is twisted like a torsion bar under torsional stress it will tend to straighten itself out again. To remain twisted a bar must be subjected to a sustained torque.

We have that in this thought experiment the long spinning pipe is non-twisted in the co-moving frame, but twisted in the frames in which it has a velocity. The question is: in the frames in which the pipe is twisted, what is the source of the torque that sustains the twist?

In the opening posting of this thread I speculated about a possible connection with Born rigidity effects. On closer inspection I think a comparison with Born rigidity effects doesn't hold. It's something different.

Cleonis

 

[matheinste]

We have that in this thought experiment the long spinning pipe is non-twisted in the co-moving frame, but twisted in the frames in which it has a velocity. The question is: in the frames in which the pipe is twisted, what is the source of the torque that sustains the twist?

Cleonis

As JesseM pointed out there is a problem with setting the pipe in motion as the torsion takes time to propogate along its length. If we consider the pipe as by some means earlier set in motion and stable along its length in its own frame then I think that the twist as observed from the frame moving relative to it is not any sort of mechanical effect, how can it be as no forces are involved. Is it not just a question of a progressive desynching of the "clocks" along its length as viewed by observers moving in parallel relative to it.

Matheinste.

 

[Cleonis]

As JesseM pointed out there is a problem with setting the pipe in motion as the torsion takes time to propogate along its length. [...] Is it not just a question of a progressive desynching of the "clocks" along its length as viewed by observers moving in parallel relative to it.
Matheinste.

As in the case of the JesseM posting, my assessment is that what you describe is applicable for another thought experiment than the long spinning pipe setup. What you describe applies for the case of a stack of non-connected disks. That is why in my previous message I included a discussion of the stack-of-disks setup.

Cleonis

 

[matheinste]

As in the case of the JesseM posting, my assessment is that what you describe is applicable for another thought experiment than the long spinning pipe setup. What you describe applies for the case of a stack of non-connected disks. That is why in my previous message I included a discussion of the stack-of-disks setup.

Cleonis

In principle if you ignore the start up phase and accept an already spinning pipe, non twisted in its own frame, I don't see any difference between non-conected discs and a solid pipe. Marks on the discs, aligned in the discs' frame, will appear non aligned as observed from the frame moving relative to it, due to loss of synchronization. And a line on the pole will appear to be twisted for the same reason.

Matheinste.

 

[Cleonis]

I realized that I need to be more precise as to what distinguishes the case of a stack of unconnected disks, and the solid pipe.

For instance, if the solid pipe is replaced by two co-axial pipes, tightly back-to-back, would that make a difference? No it wouldn't; there wouldn't be an offset angle between the two pipe ends that are back-to-back.

Another try: the stack of disks is not back-to-back disks, but disks that are very thin compared to the gaps between the disks. For instance, let the disks be 1 unit of length thick, and the gaps 99 units of length wide. Then rather than describing the state as a continuous twist over the entire length of some solid pipe, it is definable to describe the state as a concatenation of slight offset angles between the consecutive disks. The only reason the offset angle is definable lies in the presence of gaps between consecutive disks, gaps that are much wider than the thickness of the disks.

Well, that makes the distinction between the two setups much smaller than I thought earlier. It doen't change my overall assessment, though.

Cleonis

 

[matheinste]

I realized that I need to be more precise as to what distinguishes the case of a stack of unconnected disks, and the solid pipe.

For instance, if the solid pipe is replaced by two co-axial pipes, tightly back-to-back, would that make a difference? No it wouldn't; there wouldn't be an offset angle between the two pipe ends that are back-to-back.

Another try: the stack of disks is not back-to-back disks, but disks that are very thin compared to the gaps between the disks. For instance, let the disks be 1 unit of length thick, and the gaps 99 units of length wide. Then rather than describing the state as a continuous twist over the entire length of some solid pipe, it is definable to describe the state as a concatenation of slight offset angles between the consecutive disks. The only reason the offset angle is definable lies in the presence of gaps between consecutive disks, gaps that are much wider than the thickness of the disks.

Well, that makes the distinction between the two setups much smaller than I thought earlier. It doen't change my overall assessment, though.

Cleonis

I still can't see how any of these different arrangements alter the scenario conceptually.

Matheinste.

 

[Dale]

There is no additional stress in the rotating pipe in another frame. Remember, in the inertial frame where the center of mass of the pipe is at rest there is in fact a centripetal force which results in some measurable stress and strain in the pipe. Because the pipe is not at rest in that frame the 4-force has a timelike component. As you boost that 4-force into other frames you get the twisted forces and the twisted stresses and strains that you expect. This does not require the invention of any other forces, simply the boost of the existing force including its timelike component.

 

[DrGreg]

There is no additional stress in the rotating pipe in another frame. Remember, in the inertial frame where the center of mass of the pipe is at rest there is in fact a centripetal force which results in some measurable stress and strain in the pipe. Because the pipe is not at rest in that frame the 4-force has a timelike component. As you boost that 4-force into other frames you get the twisted forces and the twisted stresses and strains that you expect. This does not require the invention of any other forces, simply the boost of the existing force including its timelike component.

When I first read this, it seemed like a viable explanation. But then I actually thought what the 4-force components would be and realized it wasn't.

Let's say the pipe's axis is along the x-axis. Then, in the centre-of-mass frame, the radial 4-force acting on a particle within the pipe will lie entirely in the yz-plane, i.e. its t- and x-components are zero.

Now apply a Lorentz boost along the x-axis. This affects the t- and x-components only, which are both zero. So the the transformed components of 4-force are unchanged.

(By the way, this is one example where we need to consider all 4 dimensions of spacetime. You can't ignore one dimension, so it's almost impossible to visualise a spacetime diagram.)

 

[Dale]

Let's say the pipe's axis is along the x-axis. Then, in the centre-of-mass frame, the radial 4-force acting on a particle within the pipe will lie entirely in the yz-plane, i.e. its t- and x-components are zero.

No, only the x component is zero. The t component is non-zero.

Consider a differential element of the pipe. The t component is only zero in the momentarily co-moving inertial frame. The rest frame of the COM is not that frame. So when you boost along x the t component results in a non-zero component in the x' direction.

 

[DrGreg]

No, only the x component is zero. The t component is non-zero.

Consider a differential element of the pipe. The t component is only zero in the momentarily co-moving inertial frame. The rest frame of the COM is not that frame. So when you boost along x the t component results in a non-zero component in the x' direction.

Sorry, I don't follow that at all. In the centre-of-mass frame the 4-momentum of an element has a constant t-component (viz. energy), and when you differentiate that to get 4-force you get a zero component.

 

[Dale]

Hi DrGreg,

You are correct. Let's say we are considering the differential element that is on the y axis, so it is moving in the z direction, the net force is in the -y direction. So if we start in its momentarily co-moving inertial frame and then boost in the z direction we still wind up with a four force in the -y direction only, with no t component. I wasn't thinking through the first boost correctly and was incorrectly assuming that I knew that there was a t component.

I guess you have to consider the stress energy tensor rather than the four-force. The stresses that already exist in the rest frame of the center of mass must transform to a torsion in the other frame, but I cannot justify it. Now that I think about it, it makes sense that the four-force on each differential element must remain purely radial, otherwise you would get angular or linear acceleration.

 

[Cleonis]

(By the way, this is one example where we need to consider all 4 dimensions of spacetime. You can't ignore one dimension, so it's almost impossible to visualise a spacetime diagram.)

I guess you have to consider the stress energy tensor [...]. The stresses that already exist in the rest frame of the center of mass must transform to a torsion in the other frame, [...]

I am pleased to read the discussion between the two of you. You have taken the problem to a deeper level.

The long spinning pipe thought experiment is not as evocative as I hoped; as drGreg points out, all four dimensions need to be considered, making it rather inaccessible.

On a more philosophical note: at what point can a paradox be regarded as explained? For showing that relativistic physics implies the twist consideration of relativity of simultaneity suffices. And my impression is that some argued there's nothing more to the story than showing the logical implication. Or is the explanation only complete with a demonstration that the existent stresses in the center-of-mass frame transform to a torsion in the other frame?

I think it's interesting to see the relativistic paradoxes as challenges to the self-consistency of the theory. Then a complete discussion must not only demonstrate that the relativistic effect will occur, it must also show that out of the scenario no self-contradiction can be constructed.

As can be inferred from my postings, I'm not sufficiently familiar with dealing with stresses in relativistic physics. What I know is that internal stresses do need to be transformed.

Cleonis

 

[DrGreg]

I'm coming to the tentative conclusion that the alleged torsion here is just an illusion. I could be wrong.

Rotating bodies that are also moving along their axis of rotation naturally "twist". The geometry of spacetime forces it to happen and no extra torsion is required to explain it. Torsion is required only to distort the object from its natural shape, which in this case is twisted relative to the frame under consideration.

I think DaleSpam is right that an analysis using the energy-momentum-stress tensor might clear this up, but that is outside my area of expertise.

(This is probably irrelevant, but I see an analogy with de Broglie wavelength. The twisted pole has a "wavelength", i.e. the distance through which the twist is $2\pi$, which becomes infinite in the inertial rest frame of the centre of mass. And a frequency defined by the rotation rate (which also varies by frame).)

 

[Cleonis]

I'm coming to the tentative conclusion that the alleged torsion here is just an illusion. I could be wrong.

Since this is relativistic physics the qualification 'is just an illusion' is ambiguous.

I propose to specify by comparison, taking Lorentz contraction as yardstick. I submit: the torsion of the long spinning pipe in the other frame has the same level of physical reality as Lorentz contraction.

I'm aware that some people prefer to think of Lorentz contraction as 'just an illusion', and that others argue the only option is to regard Lorentz contraction as physical reality. Such is the range of interpretation that is in circulation. But even people who disagree as to where to place certain phenomena in the illusion/reality spectrum, they can agree that certain phenomena (in this case Lorentz contraction in the other frame and the torsion in the other frame) are in the same league.

Cleonis

 

[DrGreg]

I'm coming to the tentative conclusion that the alleged torsion here is just an illusion. I could be wrong.

Since this is relativistic physics the qualification 'is just an illusion' is ambiguous.

I propose to specify by comparison, taking Lorentz contraction as yardstick. I submit: the torsion of the long spinning pipe in the other frame has the same level of physical reality as Lorentz contraction.

I'm aware that some people prefer to think of Lorentz contraction as 'just an illusion', and that others argue the only option is to regard Lorentz contraction as physical reality. Such is the range of interpretation that is in circulation. But even people who disagree as to where to place certain phenomena in the illusion/reality spectrum, they can agree that certain phenomena (in this case Lorentz contraction in the other frame and the torsion in the other frame) are in the same league.

Cleonis

I phrased it badly, and got my torques and torsions confused. If "torsion" is just another name for "twist", then no it is not an illusion, there really is a twist. What I really meant to say was that to infer a torque from the twist would be illusory. I think?

 

[Dale]

On a more philosophical note: at what point can a paradox be regarded as explained? For showing that relativistic physics implies the twist consideration of relativity of simultaneity suffices. And my impression is that some argued there's nothing more to the story than showing the logical implication. Or is the explanation only complete with a demonstration that the existent stresses in the center-of-mass frame transform to a torsion in the other frame?

I think it's interesting to see the relativistic paradoxes as challenges to the self-consistency of the theory. Then a complete discussion must not only demonstrate that the relativistic effect will occur, it must also show that out of the scenario no self-contradiction can be constructed.

A paradox is a logical self contradiction. In other words, a paradox only exists if using a single theory and a single set of boundary conditions it is possible to derive two mutually exclusive results, e.g. of the form "X and not X". So the requirements for a real paradox are pretty clear cut and not open to a lot of interpretation.

In this case there never was a paradox established anyway. "Not twisted in frame A" and "twisted in frame B" are not mutually exclusive propositions. For relativity, there is no question about itself consistency. Since it is simply Minkowski geometry it is completely self consistent. These challenges can be fun to discuss and learn about but realistically they hardly rise to the level of a challenge to the self-consistency of the theory.

I propose to specify by comparison, taking Lorentz contraction as yardstick. I submit: the torsion of the long spinning pipe in the other frame has the same level of physical reality as Lorentz contraction.

I agree. They are both coordinate-dependent measurable effects.

 

[Dale]

Rotating bodies that are also moving along their axis of rotation naturally "twist". The geometry of spacetime forces it to happen and no extra torsion is required to explain it. Torsion is required only to distort the object from its natural shape, which in this case is twisted relative to the frame under consideration.

Another way to say it is that rotation is a twist in time. Since the Lorentz transforms map one frame's time into another frame's space then obviously the twist in time in one frame must show up as a twist in space in another frame.

(This is probably irrelevant, but I see an analogy with de Broglie wavelength. The twisted pole has a "wavelength", i.e. the distance through which the twist is $2\pi$, which becomes infinite in the inertial rest frame of the centre of mass. And a frequency defined by the rotation rate (which also varies by frame).)

I like the analogy because it also allows a simplification of the spacetime diagram back down to the convenient 1 space and 1 time dimension. Specifically, you can plot the "phase" of the rod across one dimension of space, and across time. This makes it pretty clear where the twist comes from.

 

[PeterDonis]

Hi DrGreg,

You are correct. Let's say we are considering the differential element that is on the y axis, so it is moving in the z direction, the net force is in the -y direction. So if we start in its momentarily co-moving inertial frame and then boost in the z direction we still wind up with a four force in the -y direction only, with no t component. I wasn't thinking through the first boost correctly and was incorrectly assuming that I knew that there was a t component.

I guess you have to consider the stress energy tensor rather than the four-force. The stresses that already exist in the rest frame of the center of mass must transform to a torsion in the other frame, but I cannot justify it. Now that I think about it, it makes sense that the four-force on each differential element must remain purely radial, otherwise you would get angular or linear acceleration.

If the net force on a differential element of the pipe is purely in the y-z plane, then there are no stresses between adjacent segments of the pipe ("adjacent" meaning "next to each other in the x-direction"). If that's the case, then there are no stresses to be "twisted"--the situation is that of the "adjacent spinning disks" that Cleonis described, and in the moving frame, all the stresses on each segment ("disk") will still be entirely in the y-z plane. (The adjacent disks will *look* "twisted" relative to one another, but someone with strain gauges in the moving frame will still measure zero strain between two adjacent disks.)

If, on the other hand, you assume that adjacent segments of the pipe *do* exert stresses on each other (i.e., the pipe "tries" to prevent itself from twisting), then those stresses have components in the x-direction. In other words, in this case the 4-force on each element of the pipe is *not* purely radial--there's a longitudinal (x-direction) component in the pipe CoM frame, and therefore the 4-force *will* acquire a timelike component (i.e., a twist) when boosted.

 

[Dale]

If the net force on a differential element of the pipe is purely in the y-z plane, then there are no stresses between adjacent segments of the pipe ("adjacent" meaning "next to each other in the x-direction").

I'm not sure about this. Stress and net force are pretty much independent of each other. For example, if the whole pipe were in compression then there would be stress between adjacent segments of the pipe despite the fact that the net force on a differential element remains in the y-z plane. I know this is the case in non-relativistic physics, but unfortunately I never took any relativistic statics courses!

 

[PeterDonis]

I'm not sure about this. Stress and net force are pretty much independent of each other. For example, if the whole pipe were in compression then there would be stress between adjacent segments of the pipe despite the fact that the net force on a differential element remains in the y-z plane. I know this is the case in non-relativistic physics, but unfortunately I never took any relativistic statics courses!

I agree that the net force on a differential element is in the y-z plane. However, that force is not the "torsion" that Cleonis' "paradox" is referring to. That torsion is the stress in the pipe that the boosted observer would see as the pipe "trying to untwist itself"; the source of that stress is the force exerted on each other by pipe elements which are adjacent to each other in the x-direction. *That* stress therefore has a component in the x-direction in the pipe's CoM frame, which acquires a timelike component in the boosted frame.

I did phrase it somewhat poorly in my previous post; if the pipe is in a stationary state (i.e., it has a constant angular velocity in its CoM frame), the stresses in the x-t plane (the "torsion" ones) do *not* have any net force associated with them; the stresses in opposing directions balance out, unlike the ones in the y-z plane (the plane in which the net force due to the pipe's rotation acts). However, the stresses in the x-t plane are still nonzero, and those are what look like "torsion" in the boosted frame.

 

[PeterDonis]

Warning: long post!
(EDIT: I've added another post correcting some errors in this one--please read both before responding!)

After thinking this over some more, I decided to bite the bullet and actually look at the stress-energy tensor of the pipe in the CoM rest frame and in the boosted frame.

Once again I found that Greg Egan has done most of the hard work, at his page on http://www.gregegan.net/SCIENCE/Rings/Rings.html" . I'm not going to go to the level of detail he does, but I recommend looking at his work as a check on what I'm going to say below about which components of the stress-energy tensor are nonzero.

Egan treats the case of a thin disk or hoop, but that will work fine for dealing with a thin slice of our pipe that's at a particular longitudinal coordinate. I'm going to use z instead of x for the longitudinal direction (the direction in which we're going to boost the frame), since the coordinates we'll end up using will be cylindrical, with r being the radial coordinate and $$\theta$$ being the angular coordinate. All we'll have to do is add one term to Egan's stress-energy tensor, to deal with stresses in the z-direction (which he doesn't include).

Egan's web page actually expresses the stress-energy tensor in a frame which is "co-rotating" with the disk or hoop, but he gives enough information to make it easy to re-express it in the non-rotating frame which is at rest with respect to the pipe's CoM. When we do that, we find that the only nonzero components are (I'm not writing down the detailed expressions that Egan does; we're only interested here in which components are nonzero, not what their specific values are):

$$T_{tt} = E$$
-- energy density

$$T_{t \theta} = \Phi$$
-- angular momentum

$$T_{\theta \theta} = p_{\theta}$$
$$T_{rr} = p_r$$
$$T_{zz} = p_z$$
-- pressure in each of the coordinate directions.

The absence of any other nonzero components in the pipe's CoM frame corresponds to our assumption that there is no torsion in that frame. (I should note, here, that until I took this detailed look I hadn't really thought specifically about which stress-energy tensor components would indicate torsion; some of the discussion in this thread has indicated that torsion might have a timelike component--i.e., one of its indices would be t--but that is *not* correct, as we will see.)

Now, what happens when we boost this tensor in the z-direction? Boosting a tensor is kind of like boosting a vector squared; each component gets two coefficients from the boost matrix. Since we're boosting along the z-direction, our boost matrix, which I'll call L, has the following nonzero components:

$$L^{t}_{t'} = L^{z}_{z'} = \gamma$$

$$L^{t}_{z'} = L^{z}_{t'} = - \beta \gamma$$

$$L^{r}_{r'} = L^{\theta}_{\theta'} = 1$$

We then transform the tensor components using the boost matrix as follows: for a generic tensor component with indices a', b' in the boosted frame, transformation matrix L, and indices m, n in the original frame, we have:

$$T_{a' b'} = L^{m}_{a'} L^{n}_{b'} T_{mn}$$

Note that we are using the Einstein summation convention, so the right-hand side of the above is a sum over all m, n. So the components of our transformed tensor will be (writing only the non-zero components of each sum):

$$T_{t' t'} = L^{t}_{t'}^2 T_{tt} + L^{z}_{t'}^2 T_{zz}$$

$$T_{t' z'} = L^{t}_{t'} L^{t}_{z'} T_{tt} + L^{z}_{t'} L^{z}_{z'} T_{zz}$$

$$T_{z' z'} = L^{t}_{z'}^2 T_{tt} + L^{z}_{z'}^2 T_{zz}$$

$$T_{t' \theta'} = L^{t}_{t'} L^{\theta}_{\theta'} T_{t \theta}$$

$$T_{r' r'} = L^{r}_{r'}^2 T_{rr}$$

$$T_{\theta' \theta'} = L^{\theta}_{\theta'}^2 T_{\theta \theta}$$

Substituting, we obtain:

$$T_{t' t'} = \gamma^2 \left( E + \beta^2 p_z \right)$$

$$T_{t' z'} = - \beta \gamma^2 \left( E + p_z \right)$$

$$T_{z' z'} = \gamma^2 \left( \beta^2 E + p_z \right)$$

$$T_{t' \theta'} = \gamma \Phi$$

$$T_{\theta' \theta'} = p_{\theta}$$

$$T_{r' r'} = p_r$$

Notice that the only other nonzero term that appears is the t-z term, which corresponds to the pipe's momentum in the z-direction in the boosted frame. It is negative because in this frame the pipe is moving backwards (i.e., in the negative z' direction). As I noted above, this term does *not* represent torsion; it's just linear momentum.

Notice again that there are *no* other nonzero terms; i.e., *there is no torsion*! Torsion (i.e., shear stress) would look like a term with two different spatial indices, such as r-z, r-$$\theta$$, or z-$$\theta$$. There are no such terms in the boosted tensor above.

What does this mean? It means that, although the pipe "looks twisted" in the boosted frame (in the sense that a line marked on the pipe, which is parallel to the z-axis in the CoM frame, appears twisted around the pipe in the boosted frame), there is *no* torsional stress in the pipe corresponding to the apparent "twist". So the twist is, in that sense, an "illusion".
(EDIT: No it isn't--I missed a nonzero term in the boosted tensor above. See my next post.)

 

[PeterDonis]

OOPS!

I missed a term! (Serves me right for not checking thoroughly enough.) Here's what I left out:

$$T_{z' \theta'} = L^{t}_{z'} L^{\theta}_{\theta'} T_{t \theta}$$

or, substituting,

$$T_{z' \theta'} = - \beta \gamma \Phi$$

So I was wrong in the above post: there *is* a torsional stress in the z-$$\theta$$ plane. The negative sign means that the stress is opposing the rotation of the pipe.

So what keeps this stress from slowing the pipe's rotation in the boosted frame? I think the answer is that the angular momentum, $$\Phi$$, is *larger* in the boosted frame (by the factor $$\gamma$$) than in the CoM frame. I think this extra angular momentum is what compensates for the torsional stress and keeps it from untwisting the pipe.

 

[Dale]

Hi Peter, thanks for the very nice derivation!

So what keeps this stress from slowing the pipe's rotation in the boosted frame?

Stress wouldn't slow the rotation anyway since stress is a pair of forces that balance out to produce no net force. AFAIK the only way to get a net force is for the stress to be non-constant, which is not the case here, it is constant in both z and theta.

 

[PeterDonis]

DaleSpam: Yes, you're right; there would have to be a gradient in the stress in the negative $$\theta$$ direction to slow down the spin, and there isn't; the only variation in any of the tensor components is radial.