Thin rotating disc under constant acceleration.

[yuiop]

Hm, ok, I'll have to think about this. The intuition behind my thinking that if the argument works for a 2-sphere, it should work for a ring, is this: suppose I have a way to spin up a 2-sphere about a single axis in a Born rigid manner (which, by their argument, should be possible). Then the restriction of this method to the "equatorial plane" of the 2-sphere should give a way to spin up a ring (the equator of the 2-sphere) in a Born rigid manner, since the motion of the equator should lie entirely in that plane. ...

I agree. In fig.1 of the RQF paper http://arxiv.org/abs/0810.0072 the sphere contracts in the horizontal plane but extends in the vertical direction to maintain the Born rigidity of the 2 surface as the sphere spins up. Any acceleration orthogonal to the equatorial plane that maintains the proper distance between adjacent lines of latitude would be canceled out at the equator. Therefore a ring on the great circle (equator) should be able to be spun up in a manner that maintains the proper circumference measurement (and a limited defintion of Born rigidity) as long as the radius is allowed to shrink in the appropriate manner with increasing angular velocity.

I like this argument intuitively, but it seems to lead, with only a little stretch, to the following argument for the ability to Born rigidly spin up zero thickness disc, which is well known to be impossible:

Consider a thick disc bounded by two plane discs orthogonally connected by cylinder section. Then, your argument seems to show that we should be able to Born rigidly spin up one of the bounding discs, because all its motion should be in one plane. There is something mysterious going on here.

Remember that the RQF version of rigidity allows "shape changing" from the point of view of an non rotating inertial external observer (O1) that remains at rest with respect to the COM of the object. In your example the discs at the top and bottom would have to be allowed to dish out of the plane, so that the radius increases while the circumference decreases from the point of view of O1. Internal measurements of the solid are ignored and only measurements between neighbouring points on the two surface have to maintain their mutual proper separations to qualify as a Quasi Rigid frame. Smoothly continuous surfaces with no sharp corners are more suited to RQF.

P.S. Also remember that the topic of this thread is for a disc with constant angular velocity, so considerations of maintaining Born rigidity of a disc as it spins up are slighly off topic, (but not entirely irrelevant).

 

[PeterDonis]

Consider a thick disc bounded by two plane discs orthogonally connected by cylinder section.

In other words, a 2-surface with topology S2, but in the shape of a cylinder (with top and bottom "caps") instead of a sphere, correct?

I think the difference between this case and the sphere case is that the equatorial plane of the sphere is placed symmetrically between the top and bottom hemispheres, so the spin-up process can't cause it to move out of plane--if it did, that would create an asymmetry between the top and bottom halves.

In the cylinder case, the top and bottom "caps" are not symmetrically placed, so there is no such corresponding constraint: I would guess, therefore, that any Born rigid spin-up of the cylinder as a whole would indeed involve the top and bottom "caps" moving out of plane as the cylinder changed shape under acceleration.

[Edit: I see yuiop managed to type faster than me on this one. ]

 

[PeterDonis]

the topic of this thread is for a disc with constant angular velocity, so considerations of maintaining Born rigidity of a disc as it spins up are slighly off topic, (but not entirely irrelevant).

Yes, the reason I brought up the spin-up case is that I am hoping it will give some insight into the linear acceleration of rotating disc case--i.e., that it is indeed not entirely irrelevant.

 

[PAllen]

Maybe a tack for getting a clearer understanding of the problem of uniformly accelerating a constantly rotating disk is to use the methods of the Epp et.al. paper to first ask what happens for rotating round 2-sphere uniformly accelerated in the direction of its rotation axis. What happens here could be big clue to the disc case.

 

[PAllen]

In other words, a 2-surface with topology S2, but in the shape of a cylinder (with top and bottom "caps") instead of a sphere, correct?

I think the difference between this case and the sphere case is that the equatorial plane of the sphere is placed symmetrically between the top and bottom hemispheres, so the spin-up process can't cause it to move out of plane--if it did, that would create an asymmetry between the top and bottom halves.

In the cylinder case, the top and bottom "caps" are not symmetrically placed, so there is no such corresponding constraint: I would guess, therefore, that any Born rigid spin-up of the cylinder as a whole would indeed involve the top and bottom "caps" moving out of plane as the cylinder changed shape under acceleration.

Remember that the RQF version of rigidity allows "shape changing" from the point of view of an non rotating inertial external observer (O1) that remains at rest with respect to the COM of the object. In your example the discs at the top and bottom would have to be allowed to dish out of the plane, so that the radius increases while the circumference decreases from the point of view of O1. Internal measurements of the solid are ignored and only measurements between neighbouring points on the two surface have to maintain their mutual proper separations to qualify as a Quasi Rigid frame. Smoothly continuous surfaces with no sharp corners are more suited to RQF.

I'm not sure I see how this fully resolves the mystery. Suppose we have the surface I describe (yes, Peter, your description is what I meant), and it is Born rigidly spun up - considering only points on the surface. So what if the a cap, viewed from the starting inertial frame changes shape and is not planar. By definition of RQF, it is supposedly still planar and unchanged in shape for the family of surface observers. Now look only at the top surface. How can it be Born rigid as part of a 2-sphere but not be born rigid when cut out, but otherwise undergoing identical motions? More generally, if a closed 2-surface is undergoing Born rigid motion, how is any simply connected subset of it not doing so?

 

[PeterDonis]

So what if the a cap, viewed from the starting inertial frame changes shape and is not planar. By definition of RQF, it is supposedly still planar and unchanged in shape for the family of surface observers.

I don't think it will be planar and unchanged in shape. It will look the same locally to the family of observers, but it won't look the same globally, because the observers are in relative motion when they weren't before.

Here's what I think the cylinder example is telling us: there *are* ways to spin up a disk (not a ring) in a Born rigid manner, but it won't stay a disk--it won't stay all in one plane. Instead, there are two possible solutions: one where the disk "bends up" out of plane and the other where the disk "bends down" out of plane. The two "caps" on the cylinder each realize one of these solutions.

Now look only at the top surface. How can it be Born rigid as part of a 2-sphere but not be born rigid when cut out, but otherwise undergoing identical motions?

I'm not sure I understand. There is no flat disk that undergoes Born rigid motion as part of a 2-sphere. The equator of the sphere does, but the equator is a ring, not a disk. Similarly, the ring at the junction between each "cap" of the cylinder and the side of the cylinder *can* undergo Born rigid spin-up while remaining all in one plane, just as the equator of the sphere can. But the *disk* (the ring plus its interior), which is a subset of the cylinder but not a subset of the sphere, cannot--it can only undergo a Born rigid spin-up by bending out of plane.

 

[PAllen]

I don't think it will be planar and unchanged in shape. It will look the same locally to the family of observers, but it won't look the same globally, because the observers are in relative motion when they weren't before.

Here's what I think the cylinder example is telling us: there *are* ways to spin up a disk (not a ring) in a Born rigid manner, but it won't stay a disk--it won't stay all in one plane. Instead, there are two possible solutions: one where the disk "bends up" out of plane and the other where the disk "bends down" out of plane. The two "caps" on the cylinder each realize one of these solutions.

Epp et. al. place great emphasis on the fact that shape changes observer from an inertial frame do not imply shape changes for the family of surface observers; that for the latter, shape is rigidly preserved. At least that is how I understand their claims.

In a sense, it doesn't matter much. You have result that a disc can be spun up born rigidly - it starts as a disc, and everywhere and when meets the Born rigid condition within the surface - which is all there is for disc with no thickness. They may be right, but this contradicts my understanding.

I'm not sure I understand. There is no flat disk that undergoes Born rigid motion as part of a 2-sphere. The equator of the sphere does, but the equator is a ring, not a disk. Similarly, the ring at the junction between each "cap" of the cylinder and the side of the cylinder *can* undergo Born rigid spin-up while remaining all in one plane, just as the equator of the sphere can. But the *disk* (the ring plus its interior), which is a subset of the cylinder but not a subset of the sphere, cannot--it can only undergo a Born rigid spin-up by bending out of plane.

I meant topological 2-sphere in the sense of Epp. et.al. As with their language, I would use round 2-sphere if I want to include the geometry. I should have made this clear.

 

[PeterDonis]

Epp et. al. place great emphasis on the fact that shape changes observer from an inertial frame do not imply shape changes for the family of surface observers; that for the latter, shape is rigidly preserved. At least that is how I understand their claims.

I think they mean locally, not globally; in other words, locally each observer sees his near neighbors in the same geometry, but the local geometries don't "fit together" into a global geometry the way they did when the object was moving inertially, prior to acceleration. So none of the observers sees the same *global* shape as they did before.

I meant topological 2-sphere in the sense of Epp. et.al.

Ah, ok.

 

[PAllen]

I think they mean locally, not globally; in other words, locally each observer sees his near neighbors in the same geometry, but the local geometries don't "fit together" into a global geometry the way they did when the object was moving inertially, prior to acceleration. So none of the observers sees the same *global* shape as they did before.

But then you still have Born rigid spin up of a (zero thickness) disc, with the sole caveat that it doesn't remain flat in an inertial frame. Born rigidity is a local criterion, so global shape change is irrelevant. Anyway, here is how Epp.et.al describe it in words (whether their math justifies this is complicated):

"the size
and shape, respectively, of the boundary of the finite spatial
volume—as seen by our observers, do not change with
time:"

 

[PeterDonis]

But then you still have Born rigid spin up of a (zero thickness) disc, with the sole caveat that it doesn't remain flat in an inertial frame.

Yes, but as I understand the original Ehrenfest paradox, it assumed that the disc *did* remain flat in an inertial frame; so the Epps et al. paper is not, strictly speaking, inconsistent with the statement that the Ehrenfest paradox requires that there is no way to Born rigidly spin up a disc--since the latter statement really includes the qualifier that the disk has to remain flat in an inertial frame.

"the size
and shape, respectively, of the boundary of the finite spatial
volume—as seen by our observers, do not change with
time:"

Hm, that seems to be a stronger statement than just looking the same locally.

 

[yuiop]

Epp et. al. place great emphasis on the fact that shape changes observer from an inertial frame do not imply shape changes for the family of surface observers; that for the latter, shape is rigidly preserved. At least that is how I understand their claims.

In a sense, it doesn't matter much. You have result that a disc can be spun up born rigidly - it starts as a disc, and everywhere and when meets the Born rigid condition within the surface - which is all there is for disc with no thickness. They may be right, but this contradicts my understanding. ...

This is how I am seeing it. Imagine a sheet of paper representing a plane with a grid of points printed on it. On the sheet of paper are a set of 2D observers that make measurements between neighbouring points on the paper, but their measuring rods are never allowed to leave the surface of the paper. If the sheet is rolled into a cylinder (in 3 space) the 2D observers on the sheet of paper see no change in distance between the reference points on the sheet and so conclude that no change of shape (or size) has ocurred. It is this limited (short sighted) version of Born rigidity (the 2D view) that Epp et. al. are using.

 

[yuiop]

Yes, but as I understand the original Ehrenfest paradox, it assumed that the disc *did* remain flat in an inertial frame; so the Epps et al. paper is not, strictly speaking, inconsistent with the statement that the Ehrenfest paradox requires that there is no way to Born rigidly spin up a disc--since the latter statement really includes the qualifier that the disk has to remain flat in an inertial frame. ...

I think we all agree that the conclusion of the Ehrenfest paradox, that there is no way to spin up a disc in a strict Born rigid manner while the plane of the disc remains flat, is perfectly valid and straightforward. The only way to avoid that hard truth is to relax the defintion of rigidity or remove the flatness constraint. However, I am not convinced that things are so clear cut when it comes to a spinning disc with constant angular velocity and constant linear acceleration along its spin axis.

Rotation is notoriously tricky in relativity. I think one thing that would go a long way towards resolving the prime issure raised by this thread would be to deeply analyse the linear analogue as described by pervect. Does a rod sliding along the floor of an Einstein elevator undergo length contraction over and above the classic SR velocity related length contraction due to orthogonal acceleration?

Consider this scenario. We have an observer (A1) that remains at rest with the floor of an Einstein lift that is infinitely wide and initially unaccelerated. A train travels along the floor at a constant velocity v and has length L. A disc with its rotation axis orthogonal to the floor rotates at a constant angular velocity w and has radius R and circumference P. Now the lift accelerates upwards (orthogonal to the floor). A1again measures v,L,w,R and P and finds nothing has changed. Acceleration has changed nothing. A second observer (A2) is on the train and a third observer (A3) is riding on the rotating disc and they also find that their measurements of distance and velocity, before and after the acceleration are unchanged. That is my intuitive guess of what will be observed, but it difficult to prove. All the observers will off course notice the proper acceleration of the plane, but other measurements that remain entirely within the plane will be constant.

 

[PAllen]

I now wonder if non-rigourous presentations of Herglotz-Noether have me (others?) fooled. Epp et.al. make passing reference to Herglotz-Noether depending on a 3-parameter congruence (that is a body with volume). Earlier in this thread I noted the following paper giving a modern, rigorous proof:

http://arxiv.org/abs/0802.4345

Indeed, when I read section 3.3, I note a critical point: the vector field defining the congruence must be defined on an open subset of Minkowski space. A 2x1-surface is not an open subset of Minkowski space. Combined with Epp. et.al. it now seems to me that:

- any simply connected 2-surface can be Born rigidly accelerated and rotated, with 6 degrees of freedom. This is also true of any curve (which is also not an open subset of Minkowski space).

Statements to the contrary are probably based on adding additional constraints based on appearance in an inertial frame - which have nothing to do with the Born rigid condition (as a local geometric constraint).

 

[PeterDonis]

the vector field defining the congruence must be defined on an open subset of Minkowski space.

I see this statement in the paper, but it confuses me a little, because the intuitive picture of a solid object is that it occupies a *closed* subset of spacetime, because the object includes its boundary as well as its interior. For example, the world-tube of a 2-sphere plus interior in flat spacetime will be a closed subset of Minkowski spacetime. Taken strictly, the requirement of an open subset would seem to imply that only the interior of an object could satisfy the conditions; we could never include the object's surface.

It may be that what they really meant to say was "an open subset plus its boundary". This would exclude cases like a 2-sphere without its interior which are discussed in Epp et al., while including cases like ordinary solid objects. It would also exclude the case of a flat disk of zero thickness, since the "zero thickness" part would prevent the disk's world tube from being an open subset of spacetime.

 

[PAllen]

I see this statement in the paper, but it confuses me a little, because the intuitive picture of a solid object is that it occupies a *closed* subset of spacetime, because the object includes its boundary as well as its interior. For example, the world-tube of a 2-sphere plus interior in flat spacetime will be a closed subset of Minkowski spacetime. Taken strictly, the requirement of an open subset would seem to imply that only the interior of an object could satisfy the conditions; we could never include the object's surface.

It may be that what they really meant to say was "an open subset plus its boundary". This would exclude cases like a 2-sphere without its interior which are discussed in Epp et al., while including cases like ordinary solid objects. It would also exclude the case of a flat disk of zero thickness, since the "zero thickness" part would prevent the disk's world tube from being an open subset of spacetime.

I think this is typical mathematical physics. Without loss of generality, and with fewer words in the propositions, you can subtract boundary from an object. Sort of why 1 is not considered a prime.

 

[WannabeNewton]

Not really; a manifold with boundary is a different beast from a manifold. There are different constructions and theorems for manifolds with boundaries.

 

[PAllen]

Not really; a manifold with boundary is a different beast from a manifold. There are different constructions and theorems for manifolds with boundaries.

But a physical object is not a mathematical object. Can you tell the difference between a physical object with purported boundary, and the same object with the purported boundary removed? You might as well pick the mathematical model of a physical object that is simpler to work with. If ever there were a difference in physical prediction for an object with or without a boundary, that would be something!

 

[yuiop]

Now, when you try to apply accelerations "at the same time" on a ring in a Born rigid manner, using the Einstein convention, as you work your way pairwise along the ring you find you can't accelerate all parts of the ring "at the same time" :-(, because it's impossible to Einstein-synchronize all the clocks in a rotating ring. ...

Yes it is impossible to Einstein-synchronize all the clocks in a rotating ring, but there is another way to synchronise the clocks. A signal is sent from the centre of the ring and all the clocks on the ring are started when they receive the signal. This is the most natural way to synchronise clocks for a rotational frame. This is not Einstein synchronisation and one of the side effects is that the speed of light is not isotropic when measured by observers on the ring when the ring is rotating. It may be that the Hertglotz-Noether theorem assumes Einstein synchronisation and isotropic speed of light (as do most other relativity theorems) and that may be why that theorem does not appear to cope with rotation very well. A lot depends on the assumptions made. One disadvantage of using Einstein synchronisation in an accelerating reference frame is that the clocks get out of sync with a change in velocity, which by definition is continuously happening in an accelerating reference frame. However, if radar measurements are used, then the synchronisation method is irrelevant as radar measurements only use a single clock.

Now going back to the idea of shrinking the radius of the ring to maintain the proper circumference of the ring, it occurred to me that shrinking the radius tends to increase the tangential velocity (to maintain angular momentum) which in turn tends to increase the radar measurement of the circumference by observers on the ring. Just to make sure that the two factors do not cancel each other out, I carried out an example calculation.

Let us say we have a ring with radius r1=1, mass m1=1 and tangential velocity v1 =0.6 using units where the speed of light c=1. The proper circumference (C1) of this ring is 2*pi*r/sqrt(1-v1^2) = 7.853982. The angular momentum of the ring is L1 = m*r1*v1/sqrt(1-v1^2) = 0.75 and the angular velocity is W1 = v1/r1 = 0.6.

Now if we reduce the radius to 0.75 and increase the tangential velocity to 0.8, the angular velocity increases from 0.6 to 1.06666, the angular momentum increases from 0.75 to 1 and the proper circumference remains constant. Therefore we can maintain the circumference measurement made by observers on the ring while spinning up the ring. Of course this is not the whole story, as I am just considering snapshots at different angular velocities, rather than measurements made under continuous angular acceleration.

 

[PeterDonis]

Without loss of generality, and with fewer words in the propositions, you can subtract boundary from an object.

But can you? I.e., do all of the theorems proved in the paper you cited, still hold if we add back the boundary? For example, if the assumption of an open set translates to an assumption that every point the theorems apply to must have an open neighborhood that lies within the body, then points on the boundary would violate that assumption.

Also, physically speaking, points on the boundary of the object differ from points in the interior in a key respect: they are less constrained in their motion by the presence of neighboring parts of the object. The assumption of an open set in the proofs appears to disregard that physical difference.

 

[PAllen]

But can you? I.e., do all of the theorems proved in the paper you cited, still hold if we add back the boundary? For example, if the assumption of an open set translates to an assumption that every point the theorems apply to must have an open neighborhood that lies within the body, then points on the boundary would violate that assumption.

Also, physically speaking, points on the boundary of the object differ from points in the interior in a key respect: they are less constrained in their motion by the presence of neighboring parts of the object. The assumption of an open set in the proofs appears to disregard that physical difference.

Note, the paper referencing the open set is the paper claiming a modern, rigorous proof of Herglotz-Noether. Very few books give a proof (none of my SR books have a proof). If ever a mathematical model of reality gave a different prediction depending on whether an object with volume was considered to contain or not contain its mathematical boundary, this would be an amazing result. Looking at this case: consider the object as having a boundary. Enclose it in a slightly larger object. Now, what was the boundary is interior. Now remove the boundary of the slightly larger object. I think the physical assumption in the proof is that no one would believe a theory that gave a different answer for these two cases. Note also, that while the boundary-less object does have the property that every point has a neighborhood within the object, it is also true that for every size ball, there are points of the object for which said ball would include both interior and exterior points.

Physically, a boundary layer has no relation to mathematical boundary. One experimentally measures whether such layers are 1 molecule thick, 2 molecules, consist really of more than one micro-layer, etc. How many atoms (or even electrons) are removed from an object purported to have mathematical boundary, when you remove the mathematical boundary? Or, for perfect classical objects, how many grams are removed?

I remain convinced that this is all red herring. For example, I could propose one additional (possibly not necessary) assumption to make the proof accommodate bodies including their boundary: We assume that a mathematical boundary of an object with volume has no physical significance.

The main point remains: a modern proof of Herglotz-Noether does, indeed, assume a 3-parameter famliy of congruences under the guise of its open set definition; this is consistent with what Epp. et. al. claim about Herglotz-Noether; and they (Epp. et.al.) claim to have established that for simply connected closed 2-surfaces, Born rigid motions is possible with all 6 degrees of freedom expected from Newtonian physics. I further argue that their logic establishes a case they were not interested - that a simply connected 2-surface that is not necessarily closed has the same generality of Born rigid motion (they were only interested in closed 2-surfaces because they want to enclose a volume).

 

[PeterDonis]

Looking at this case: consider the object as having a boundary. Enclose it in a slightly larger object. Now, what was the boundary is interior. Now remove the boundary of the slightly larger object. I think the physical assumption in the proof is that no one would believe a theory that gave a different answer for these two cases.

Why not? As I said in my last post, the boundary has a physical meaning: the forces on atoms at the boundary are different than the forces on atoms at the interior. So I would expect a physical theory that took that difference in forces into account to make different predictions from one that didn't.

Note also, that while the boundary-less object does have the property that every point has a neighborhood within the object, it is also true that for every size ball, there are points of the object for which said ball would include both interior and exterior points.

Yes, but the size of ball that is relevant is the size of ball within which the interactions between atoms are significant. In a typical ordinary object, I believe that size is probably a few atom diameters at most; as I understand it, almost all internal forces in ordinary objects are between adjacent atoms. (I say "ordinary objects" because there are more exotic states of matter, such as Bose-Einstein condensates, where longer-range interactions are important.) So the "boundary" of the object would be determined by which atoms had a neighborhood of that size that included exterior points.

Physically, a boundary layer has no relation to mathematical boundary.

I agree that, physically, the boundary is not zero thickness; but that doesn't necessarily mean the mathematical boundary can't be a reasonable idealized model of the physical boundary. I don't really know enough about how manifolds with boundary differ, mathematically, from manifolds without boundary to know whether the differences are reasonable models for the physical aspects of boundaries.

 

[PAllen]

Why not? As I said in my last post, the boundary has a physical meaning: the forces on atoms at the boundary are different than the forces on atoms at the interior. So I would expect a physical theory that took that difference in forces into account to make different predictions from one that didn't.

I guess we agree to disagree. To me, atoms on a boundary or fluid surface layer are completely unrelated to mathematical boundaries, and a theory of them requires modeling thickness. For example, it must model that surface / interior separation breaks down for droplets below a certain size. Nonzero thickness is critical for such a model.

Yes, but the size of ball that is relevant is the size of ball within which the interactions between atoms are significant. In a typical ordinary object, I believe that size is probably a few atom diameters at most; as I understand it, almost all internal forces in ordinary objects are between adjacent atoms. (I say "ordinary objects" because there are more exotic states of matter, such as Bose-Einstein condensates, where longer-range interactions are important.) So the "boundary" of the object would be determined by which atoms had a neighborhood of that size that included exterior points.

One atoms worth of ball encompasses more volume than the mathematical boundary of a galaxy.

I agree that, physically, the boundary is not zero thickness; but that doesn't necessarily mean the mathematical boundary can't be a reasonable idealized model of the physical boundary. I don't really know enough about how manifolds with boundary differ, mathematically, from manifolds without boundary to know whether the differences are reasonable models for the physical aspects of boundaries.

For reasons stated above, I doubt a mathematical boundary would have properties relevant to modeling a physical boundary. Anyway, this whole question amounts to questioning a proof Herglotz-Noether in a way not relevant to the surprising conclusions of Epp et. al. What would be relevant is if someone found a claimed proof of Herglotz that claimed to cover 2-surfaces of congruences. Otherwise, you are just critiquing what you believe is an unwarranted assumption in a standard proof of Herglotz.

 

[PeterDonis]

To me, atoms on a boundary or fluid surface layer are completely unrelated to mathematical boundaries, and a theory of them requires modeling thickness.

This may be true; as I said, I don't know enough about the mathematics of manifolds with boundaries vs. manifolds without boundaries to know whether the math has features that reasonably model at least some portion of the physics.

What would be relevant is if someone found a claimed proof of Herglotz that claimed to cover 2-surfaces of congruences.

But if zero thickness mathematical boundaries have no physical relevance, then neither do 2-surfaces of zero thickness. So the Epps et al. results don't appear to me to have any physical relevance if we insist that zero thickness surfaces are physically irrelevant.

Otherwise, you are just critiquing what you believe is an unwarranted assumption in a standard proof of Herglotz.

I'm trying to understand what that assumption entails, because if zero thickness surfaces are not physically relevant, then the Herglotz-Noether theorem should apply to any actual object, even if it doesn't apply to zero thickness surfaces.

 

[PAllen]

But if zero thickness mathematical boundaries have no physical relevance, then neither do 2-surfaces of zero thickness. So the Epps et al. results don't appear to me to have any physical relevance if we insist that zero thickness surfaces are physically irrelevant.

.

The generalization of the Epps results to non-closed simply connected 2-surfaces are pure math of no physical significance. The only relevance is to conflicting claims in the literature about Born rigid motion of such pure mathematical objects.

For closed 2-surfaces, I think there is significance in that more general 'near rigid motion' is defined by allowing the body itself to deviate from Born rigidity, but only as constrained by the Born rigid motion of the mathematical 2-surface.

While the Epps results are mathematically interesting, I rather prefer (physically) the approach of Lhosa et. al. to generalizing Born rigid motion because it applies a constraint and says specific things about the whole 3-parameter congruence.

 

[PAllen]

I'm trying to understand what that assumption entails, because if zero thickness surfaces are not physically relevant, then the Herglotz-Noether theorem should apply to any actual object, even if it doesn't apply to zero thickness surfaces.

I believe Herglotz-Noether does apply to any actual object: you cannot, even in principle, Born rigidly move (by magic force application) any actual object except within the constraints of Herglotz-Noether. A spinning disc cannot be abstrcted to zero thickness.

One insight on this is that you can say that if you want to apply Herglotz to a surface, you add the requirement that the surface must move in a way compatible with being contained in a bounding volume. This rules out some of the non-local dependencies required for the Epps result.

 

[PeterDonis]

I believe Herglotz-Noether does apply to any actual object: you cannot, even in principle, Born rigidly move (by magic force application) any actual object except within the constraints of Herglotz-Noether.

But, to bring this back to the OP of the thread, that implies that a spinning disc that is linearly accelerated along its spin axis can't realize a Born rigid motion (since this motion is not one of the ones allowed by H-N). Which implies that a spinning disc sitting at a constant altitude in a gravitational field can't realize a Born rigid motion.

The two alternative approaches to try to avoid the worst consequences of the above appear to be:

(1) Epps et al.: the surface of the accelerated spinning disc can move Born rigidly, but the interior can't. The question here is, what is the motion of the interior? Is it non-rigid only because the interior has to undergo some kind of periodic non-rigid fluctuation? Or is it non-rigid because interior stresses will gradually build up, non-periodically, until the disc tears itself apart? As far as I can tell, this paper does not address that question at all.

(2) Lhosa et al.: the disc can't move exactly Born rigidly, but it can realize some kind of "approximately rigid" motion. Basically, their paper uses "strain rate" to characterize the rigidity of the motion: Born rigidity means a strain rate of 0. This raises the same question as before: for the actual motion of an accelerated spinning disc, is the strain rate nonzero because it's some periodic function that stays reasonably bounded? Or is it nonzero because strain gradually builds up, non-periodically, until the disc tears itself apart? I can't tell from this paper what the authors' answer to this question is.

I'm also not sure what *I* think the answer to the above question is. On the one hand, looking at particular congruences that might describe an accelerated rotating disc appears to show that these congruences must have nonzero shear, which tends to point at the "disc will eventually tear itself apart" conclusion. On the other hand, actual experience seems to show that rotating discs can sit at a constant altitude in a gravitational field indefinitely, which tends to point at the "strains remain reasonably bounded" conclusion. But if that's the case, then we must not be looking at the right congruence to describe the disc's motion--so what *is* the right congruence?

 

[PAllen]

Now we're getting to physically interesting questions. I think we should focus on seeing if any bounds on strain over time can be deduced for the Lhosa et. al. construction, if not in general, at least for the uniformly accelerating spinning disc.

 

[PAllen]

Llosa et.al. state that the strain rate is of order rotation squared (without grown of this bound) for FN congruences for the case where Born rigid motion is impossible. Then they perturb this, reducing the order of strain rate, to produce their near rigid congruence. It seems to me that there is no indication of anything but a 'small' bounded strain rate for any given rotation. Only if the rotation increases without bound will the strain increase without bound.

 

[PeterDonis]

I think we should focus on seeing if any bounds on strain over time can be deduced for the Lhosa et. al. construction, if not in general, at least for the uniformly accelerating spinning disc.

That's what I've been trying to get from the Lhosa et al. paper, but I'm not sure if I understand their notation well enough.

A quick back of the envelope calculation, if I'm understanding them correctly, might look like this:

(1) On p. 15, they say that for "small" rotations (which basically means the radius of the object is small enough compared to the angular velocity that the rim of the object moves much slower than the speed of light, relative to the center of mass frame), the strain rate is proportional to the square of the angular velocity. It looks like they mean this to be in appropriate dimensionless units (because the angular velocity itself is supposed to be a "small parameter" here); so really the strain rate should be proportional to the velocity of the object's rim, roughly speaking, in dimensionless units (i.e., v/c).

So for a typical object, say a 1 meter radius disc rotating at 3 radian/sec (to make the numbers come out nicely), we have $v / c = \omega r / c = 10^{-8}$. So the strain rate should be of order $v^2 / c^2 = 10^{-16}$.

(2) The tensile strength of the strongest materials we know of is in the range of 10 - 100 GPa; take the higher of the two. This equates to an energy density of $10^{11} J/m^3$ (since GPa *is* $J/m^3$). The dimensionless strain in the material, at the point of failure, is just the tension divided by the energy density of its rest mass; the densest known material (osmium) has a density of $22 g/cm^2$, or $2 x 10^4 kg/m^3$, or about $10^{20} J/m^3$. So the dimensionless strain at the failure point will be of order $10^{11}/10^{20} = 10^{-9}$.

(3) At a strain rate of $10^{-16}$, it will therefore take approximately $10^7$ "units" to reach failure. It's not entirely clear to me what the "units" of the rate are supposed to be here, but the "naive" obvious choice would be seconds, which would equate to the object reaching failure point in about 1/3 of a year, or 4 months, if the strain rate were constant (i.e., non-periodic). But I don't see any indication in the paper of whether they think the strain rate is in fact non-periodic.

 

[PeterDonis]

It seems to me that there is no indication of anything but a 'small' bounded strain rate for any given rotation. Only if the rotation increases without bound will the strain increase without bound.

A bounded strain rate does not necessarily imply a bounded strain; if the strain rate is non-periodic, it can still build up over time, if the time is sufficiently long. (Of course this also depends on what a "sufficiently long" time is; as I noted in my previous post, that's one of the things I haven't really gotten a handle on from their paper.)

 

[yuiop]

Here are my reasons (from a number of different perspectives) for believing that it should be possible to maintain the Born rigidity of a disc that is accelerating along its spin axis.

1) It is implied by "the clock postulate". See http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html. The extended version suggests that not only clock rates, but also length contraction are independent of acceleration.

2) It is implied by the Schwarzschild metric. Objects moving horizontally at height h have length contraction that is simply a function of the velocity relative to a stationary observer also at height h and is independent of h or the gravitational acceleration at h.

3) Nature. No observations have ever been made of objects undergoing Herglotz-Noether time dependent rigidity failure, despite the strong possibility that some objects have been rotating at high speeds in gravitational fields for millions or billions of years. Maybe there is time yet. ("The Herglotz-Noether time bomb"is ticking.)

4) Analysis of the coordinates given by pervect in an old thread https://www.physicsforums.com/showpost.php?p=4466113&postcount=25 show that length contraction of an object moving horizontally in an upward accelerating Einstein elevator are independent of time or acceleration.

5)Initial analysis of of a different set of coordinates given by pervect in a new thread https://www.physicsforums.com/showthread.php?t=709033 show that the spatial measurents of the 2+1 dimensional disc that is both rotating and linearly accelerating are independent of the linear acceleration.

It seems that the often made assertion, that a thin disc cannot rotate in Born rigid manner when accelerated along its spin axis are an incorrect conclusion or interpretation of the Herglotz-Noether theorem. Some of the previous poster's in this thread seem to suggesting that maybe the Herglotz-Noether theorem does not cover the case of an infinitesimally thin rotating disc, depending on how boundaries are defined.