Thin rotating disc under constant acceleration.

[WannabeNewton]

Thanks Peter, that's great! But is there a way to make more explicit the need for a tangential acceleration of the ring when the ring is linearly (and uniformly) accelerated along the axis perpendicular to the plane of rotation? Also what is the nature of this tangential acceleration e.g. is it constant?

 

[PeterDonis]

But is there a way to make more explicit the need for a tangential acceleration of the ring when the ring is linearly (and uniformly) accelerated along the axis perpendicular to the plane of rotation?

As the ring linearly accelerates, if no tangential acceleration is applied (i.e., if the angular velocity of the ring remains constant, as seen by an observer at the ring's center of mass), the locally measured distance between neighboring pieces of the ring will change, because their motion has a component in the same direction as the linear acceleration, and that component changes as a result of the linear acceleration. (This is the same thing that gives rise to the "out of plane" Thomas precession that yuiop referred to.)

In other words, the ratio of the (constant) tangential velocity of the ring (due to its constant angular velocity as seen by a comoving observer) to the total linear velocity of a given piece of the ring, *changes* as a result of the linear acceleration, if no tangential acceleration is present. But for a Born rigid motion, that ratio must remain constant; so tangential acceleration has to be applied to change the ring's tangential velocity "in step" with the change in its velocity due to the linear acceleration.

Also what is the nature of this tangential acceleration e.g. is it constant?

I haven't done the math, but I think it will have to vary with time if tangential Born rigidity is maintained (i.e., if each "ring" at a given radius perpendicular to the linear acceleration maintains Born rigidity).

 

[PeterDonis]

I think it will have to vary with time if tangential Born rigidity is maintained (i.e., if each "ring" at a given radius perpendicular to the linear acceleration maintains Born rigidity).

Actually, thinking it over some more, this may not be right; the tangential acceleration has to vary with radius in the disk, but I'm not sure it has to vary with time.

 

[yuiop]

But *how* is Born rigid motion realized for this object? First consider the case of a rotating ring whose CoM is moving inertially, but which has to be "spun up" from the non-rotating state. Can this be done in a Born rigid manner? Yes, it can (remember this is a *ring*, not a disk!). All we need to do is adjust the acceleration profile of each little piece of the ring so that everything stays "in step" as it spins up. From the standpoint of the inertial frame in which the ring is initially at rest, what we're doing is a sort of circular analogue of the Rindler congruence; but instead of the acceleration varying in space (with the x coordinate), it varies in *time*--we *change* the acceleration of the disk as it spins up, in a way that just compensates for the length contraction and time dilation of the disk relative to the global inertial frame, so that the distance between neighboring pieces of the ring, as measured by observers riding along with the pieces, remains constant.

I agree that Born rigid spin up of a ring is possible, but surely you meant centripetal acceleration towards the cetre of the ring is required rather than tangential acceleration. Just to be sure we are on the same page I am using these definitions:

Here is how I think Born rigidity of a ring is maintained during spin up. Hopefully we can agree that for a ring Born rigidity is acheived if the proper circuference (measured by an accelerated observer riding on the ring) remains constant and we ignire measurements of the radius.
First, consider the point of view of a non rotating inertial observer $O_1$ that remains at rest with the centre of the ring. Initially the ring is at rest and the circumference $C_1$ is $2*\pi*R_1$ as per the Euclidean expectation. The ring is then spun up by applying acceleration tangential to the ring in the plane of the ring as per the above diagram, until the tangential velocity $v_1$ of the ring is a constant non zerovnalue. According to observer $O_1$ the radius and the circumference of the ring remains constant, whatever the angular velocity of the ring. (Assume just the right centripetal acceleration is applied to ensure the radius remains contant). Now according to an accelerated observer on the ring ($O_2$) the circumference ($C_2$) varies with the tangential velocity of the ring according to $C_2 = 2*\pi*R_1*\sqrt{1-v_1^2/c^2}$. In order to retain Born rigidity, the radius will have to shrink (as measured by $O_1$) and this can be acheived by applying additional centripetal acceleration over and above that required to maintain constant radius as measured by $O_1$.

Note that changing thr radius changes the tangential velocity for a given angualar velocity so will have to bear in mind what type of velocity we are reffering to.

The case of a rotating ring that is linearly accelerated works similarly; the only difference is that now, we add an acceleration component perpendicular to the ring plane

This would be along the rings rotation axis or the z axis as labelled in the drawing in my last post in this thread, which is consistent with the axes convention used by Pervect in the metric he introduced earlier.

(in the above case, the acceleration was entirely within the ring plane). But we can still keep the ring's motion Born rigid, provided we are allowed to *adjust* the ring's acceleration--i.e., we need to be able to add an arbitrary *tangential* acceleration, just as we would if we were "spinning up" the ring, in order to compensate for the changing length contraction and time dilation of the ring, due to linear acceleration, relative to a global inertial frame. (Note that we have not yet discussed, in this thread, the fact I just mentioned, that Born rigid motion of the ring requires this tangential acceleration.)

You have lost me a bit here. If in a given inertial reference frame (S) a rod of proper length L is moving along its own length in the x direction at velocity $v_x$, then its coordinate length is $L*\sqrt{1-v_x^2/c^2}$. Now if the rod is accelerated in the z direction, orthogonal to its length, then its coordinate length does NOT change as measured by the observer at rest in S and its proper length does not change either. However, the velocity in the x direction does slow down to $v_x*\sqrt{1-v_z^2/c^2}$ as measured in S.

 

[PeterDonis]

I am using these definitions

These are what I was using as well, no disconnect here.

for a ring Born rigidity is acheived if the proper circuference (measured by an accelerated observer riding on the ring) remains constant and we ignire measurements of the radius.

Yes, agreed. In fact the radius will have to change, which is an important point. See below.

First, consider the point of view of a non rotating inertial observer $O_1$ that remains at rest with the centre of the ring. Initially the ring is at rest and the circumference $C_1$ is $2*\pi*R_1$ as per the Euclidean expectation. The ring is then spun up by applying acceleration tangential to the ring in the plane of the ring

Yes. But this is tangential acceleration, and will remain tangential acceleration as seen by an observer moving with the ring; so it can't be that there is no tangential acceleration, as you appear to be claiming. But perhaps I'm misunderstanding exactly what you're claiming. See further comments below.

According to observer $O_1$ the radius and the circumference of the ring remains constant, whatever the angular velocity of the ring.

I was confused on reading this at first, because it's obvious that, for the ring's circumference to remain the same, as seen by an observer moving with the ring, as the ring spins up, the ring's radius, as seen by $O_1$, must decrease. Then I saw that you say just this later on:

In order to retain Born rigidity, the radius will have to shrink (as measured by $O_1$) and this can be acheived by applying additional centripetal acceleration over and above that required to maintain constant radius as measured by $O_1$.

I think this is right: while the ring is spinning up, the centripetal acceleration seen by a small piece of the ring will be larger than it is when the ring's angular velocity is constant. But that doesn't take away the tangential acceleration during the spin-up. (Of course there is no tangential acceleration once the spin-up is complete and the ring's angular velocity is constant.)

It does, however, mean that my later disclaimer about the tangential acceleration was probably correct: the tangential acceleration doesn't have to change with time to keep the spin-up Born rigid, since the additional centripetal acceleration is what's compensating for the increasing length contraction of the ring, as seen by $O_1$. So if that's what you were referring to when you said there should be centripetal acceleration instead of tangential acceleration, I think I agree. (I still haven't worked through the math in detail, though; there may be other things waiting to bite. )

Note that changing thr radius changes the tangential velocity for a given angualar velocity so will have to bear in mind what type of velocity we are reffering to.

Yes, good point.

This would be along the rings rotation axis or the z axis as labelled in the drawing in my last post in this thread, which is consistent with the axes convention used by Pervect in the metric he introduced earlier.

Yes.

You have lost me a bit here.

Well, in view of the above, I was a little lost myself, since when I wrote that post I was still thinking in terms of a changing tangential acceleration being required to maintain Born rigidity in the spin-up case. So now we need to re-think the linear acceleration + rotation case in the light of the above. I'll have to ponder that some more.

 

[yuiop]

Hi Pervect, I looked at your derivation https://www.physicsforums.com/showpost.php?p=4445483&postcount=9 of the coordinates in the accelerating frame in the relativistic weight thread and everything seems in order. I think the transformation you gave from the accelerating frame to the inertial frame https://www.physicsforums.com/showpost.php?p=4466113&postcount=25 would be useful to this thread, if we had the reverse transformation, i.e. (t,x,z) -> (T,X,Z) but I was unable to derive the reverse myself. Is it possible?

 

[PAllen]

Two papers have been noted in this thread that propose generalizations of Born Rigid motion to situations where it is impossible per Herglotz-Noether, by relaxing the Born rigid conditions in some minimal way:

http://arxiv.org/abs/0810.0072
http://arxiv.org/abs/1103.4475

The approach of these two papers is completely different.

Epp et.al. derive that for closed two surface bounding a spatial volume, a 2-parameter congruence of timelike world lines representing the boundary are able to meet the full Born rigid condition. The problem arises trying to extend this to the interior, which is impossible. The interior (except the special cases covered by Herglotz-Noether) must be allowed to deviate from Born rigid motion (to flow and compress as needed). This suggests (to me) that what makes a disk of zero thickness impossible to rotate and accelerated rigidly is the absence of an interior to 'compensate' for the boundary. Alternatively, the the boundary and the interior are the same. I would sum this conclusion as: rigid motion of a shape is generally possible, but not rigid motion of body as a whole.

Llosa et.al. instead keep a strictly local definition of (extended) rigidity that extends throughout a volume. They find one inspired by Born conditions, but weaker, that is always possible, but non-unique. To make it nearly unique they show how to find an instance of the congruence that is a perturbation from Fermi-Normal coordinates (with rotation), for a chosen world line and rotation. This automatically leads to Born rigid motion whenever FN coordinates don't need to be perturbed to be one of their extended congruence class. Otherwise, this process leads to a nearly unique choice among their congruence class. In sum: a notion of near rigid motion of a body as a whole is possible (by relaxing Born conditions and using Fermi-Normal coordinates as a disambiguating 'framework'.)

 

[pervect]

.

But *how* is Born rigid motion realized for this object? First consider the case of a rotating ring whose CoM is moving inertially, but which has to be "spun up" from the non-rotating state. Can this be done in a Born rigid manner? Yes, it can (remember this is a *ring*, not a disk!).

I hate to be the bearer of bad news, but I recall an argument (I believe by Gron, but I don't recall the details that you can't change the rotation rate of a rotating ring in a Born rigid manner.

I'll paraphrase the argument very loosely from memory.

Since I am doing this from memory, it's possible my recollection is flawed, (of course if I spotted any flaw I wouldn't post the argument!).

Let's start with a simpler case, just a small bar.

If you have a small bar, and you want to accelerate that small bar in a Born rigid manner, you need to apply an acceleration at the front of the bar and the rear of the bar 'at the same time".

If you apply the accelerations at different times, the bar doesn't maintain it's rigidity, the front moves first (or the rear moves first) and the length of the bar changes.

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.

You can make the amount of stretching as small as you like by making the change in rotation slow enough, but it's never truly "Born Rigid".

 

[PeterDonis]

I hate to be the bearer of bad news, but I recall an argument (I believe by Gron, but I don't recall the details that you can't change the rotation rate of a rotating ring in a Born rigid manner.

One of the papers that PAllen linked to just now argues that you can--not in so many words, but it argues that a closed 2-surface can undergo a Born rigid motion with arbitrary acceleration. That implies that a ring, which is a closed 1-surface, should also be able to undergo a Born rigid motion with arbitrary acceleration.

If you have a small bar, and you want to accelerate that small bar in a Born rigid manner, you need to apply an acceleration at the front of the bar and the rear of the bar 'at the same time".

I don't think this is the correct general way of characterizing the constraint. One reason is that the acceleration varies in space as well as in time, so just saying that the two ends have to be accelerated "at the same time" isn't sufficient. Also, the characterization should be invariant, and "at the same time" isn't. Also, the characterization should be local.

Here's how I think the Born rigidity condition could be characterized to meet the above conditions: in the local inertial frame of a particular small segment of the bar, at a particular instant of that segment's proper time, the proper acceleration of the bar varies in space according to the Rindler formula ($a = 1 / x$, possibly with some constants thrown in)--i.e., in just the right way to keep all the local distances the same. In the case of the bar, the local inertial frames all happen to "line up" globally in a simple way, so all of these local conditions can be "stitched together" into a single global Rindler chart.

However, I don't think the latter condition is required for Born rigid motion, and of course it isn't satisfied in the case of the ring. But I think the former condition can still be satisfied for a ring being spun up, *if*, as yuiop pointed out, there is an extra centripetal acceleration imposed (so that the radius of the ring gradually shrinks as it spins up). The extra centripetal acceleration corrects for the fact that the tangential accelerations applied to neighboring pieces of the ring are not exactly in the same direction, so that the end result, when viewed in the LIF of any individual piece of the ring, shows the proper accelerations and velocities of neighboring pieces varying in space in just the right way to keep all the local distances the same.

I haven't done the math to compute this, so it's just a heuristic argument, but if the paper PAllen linked to is correct, something of this sort must be going on.

 

[PAllen]

Actually, I think Pervect may be right here. My non-rigorous argument is as follows. The Epp et.al. paper I linked (on rigid motion of a closed 2-surface) has a strong reliance on the genus of the 2 surface being zero. Thus, despite the over-broad abstract, general, quasilocal rigid motion (as they've defined it) is not possible for a teacup. Within 3 or 4 space, I think a ring is more appropriately considered the zero volume limit of a torus than a dimensional reduction of a 2-sphere to a 1-sphere.

Anyway, a full argument needs to mathematical - either using the techniques of one or another proof of Herglotz-Noether, or setting up the differential system of Epp et.al. and seeing if it is not over-constrained and has solutions.

 

[PeterDonis]

The Epp et.al. paper I linked (on rigid motion of a closed 2-surface) has a strong reliance on the genus of the 2 surface being zero.

Hm, I'll have to read through the paper again, I didn't see that restriction when I read through it before.

Within 3 or 4 space, I think a ring is more appropriately considered the zero volume limit of a torus than a dimensional reduction of a 2-sphere to a 1-sphere.

I think it depends on exactly how the genus of the surface comes into the argument. For example, does their argument still work if it is restricted to a great circle on a 2-sphere?

 

[PAllen]

The section where they explicitly rely on genus of the 2-surface is page 6, as follows"

"The first equation
tells us that F (the radial perturbation) is determined
by the vector field fi (the tangential perturbation),
and the second tells us that fi must be a conformal
Killing vector (CKV) field on the unit round sphere.
It is well known that any two-surface with the topology
S2 admits precisely six CKVs (compared with two
CKVs for a torus, and zero CKVs for any closed surface of
higher genus[18]), and as generators of infinitesimal diffeomorphisms
they form a representation of the Lorentz
algebra.[12]"

[edit: This is how they ensure enough degrees of freedom for the motion of the boundary. So, I would think the thing to do is try to recast their argument in terms of a closed curve rather than a 2-surface, and see what happens. Obviously, all closed curves are topologically identical. ]

 

[PeterDonis]

This is how they ensure enough degrees of freedom for the motion of the boundary. So, I would think the thing to do is try to recast their argument in terms of a closed curve rather than a 2-surface, and see what happens. Obviously, all closed curves are topologically identical.

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. But if this is right, there should be a way to formulate it in terms of the degrees of freedom argument; that will take some more consideration.

 

[PAllen]

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. But if this is right, there should be a way to formulate it in terms of the degrees of freedom argument; that will take some more consideration.

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.

 

[PeterDonis]

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.

No, the argument doesn't carry over to a disk, because the point of just having a ring--and the point of only having a bounding 2-sphere instead of a 2-sphere plus its interior--is to allow the surface (2-sphere or ring) to adjust itself radially as well as tangentially. In other words, the absence of the interior--more precisely, the absence of a requirement to maintain constant proper distance to any neighbors in the interior direction--adds crucial degrees of freedom.

 

[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).