Thin rotating disc under constant acceleration.

[PeterDonis]

The MTW discussion of Fermi coordinates encompassing rotation is section 13.6.

Ah, yes, this does discuss what they call the "proper reference frame of an accelerated observer", where the "accelerated" can arise from any arbitrary combination of linear acceleration and rotation. But I don't see them using the term "Fermi coordinates" here (though that's really an issue of terminology, not physics). They do make it clear that the spatial basis vectors are Fermi-Walker transported only in the case of zero rotation.

The paper shows that Fermi coordinates for rotation around inertial world line describe Born rigid motion.

Yes, but not for rotation around a linearly accelerated worldline.

 

[PAllen]

Ah, yes, this does discuss what they call the "proper reference frame of an accelerated observer", where the "accelerated" can arise from any arbitrary combination of linear acceleration and rotation. But I don't see them using the term "Fermi coordinates" here (though that's really an issue of terminology, not physics). They do make it clear that the spatial basis vectors are Fermi-Walker transported only in the case of zero rotation.

Yes, but many later writers (as exemplified by a couple of my references - I have seen many others over the years) call this simply Fermi-Normal coordinates or generalized Fermi coordinates. Anyway, it is just terminology.

Yes, but not for rotation around a linearly accelerated worldline.

I never said otherwise. Here is the exact quote of mine you initially questioned:

"1) For cases where Born rigid motion is possible (e.g. uniform acceleration; rotation without acceleration), is this FN-coordinate body a Born rigid body?"

Then, after I wrote that, I found I found the paper which confirmed that this is true: whenever FN coordinates (including rotation) are applied to scenario where Born rigid motion is possible, the congruence of constant spatial coordinate positions describe Born rigid motion.

The paper also showed the authors entertained my other suggestion: FN (extended, if you will) congruences are as close as possible to rigid motion for other cases. Unfortunately, they showed this was not so - you can do better with a more complex construction (without, of course, fully achieving Born rigidity; there is no dispute with Herglotz-Noether)

 

[pervect]

I've been busy recently, but had a few thoughts on the topic.

It's helpful to consider the case of a block sliding along the floor of an Einstein's elevator before tackling the rotating case. There is some discussion by me in another thread https://www.physicsforums.com/showthread.php?t=701257, the discussion includes a derivation of the trajectory of the moving block in an inertial frame, and a convenient set of coordinate transforms and an associated metric.

It's a bit sketchy, but the supporting math is there, though the "convenient" coordinates aren't Fermi-normal coordinates. One could get Fermi-normal coordinates at the expense of making the metric time dependent, but I'll leave that to the interested reader (my personal guess is that the approach I took will be easier to manage and understand).

I'll summarize the results a bit less mathematically.

In Fig 1, we see the usual picture of the block sliding along the elevator floor. The solid and hatched block represents the position of the block at two different times.

A gyroscope attached to the block will rotate, because of effects similar to Thomas precession, whose axis is in the a (cross) v plane, cross being the vector cross product. The plane of rotation is the a^v plane, i.e. it is the plane formed by the accleration vector and the the velocity vector. It's not particularly obvious why the block is rotating viewing this diagram.

In Fig 2, we see the block in an instantaneously co-moving inertial frame. The elevator floor now appears curved, due to the relativity of simultaneity. The block is stationary, and the floor slides in the other direction, without rotation. The block itself rotates so that it's bottom of the block is always parallel to the curved floor - this is again due to Thomas precession, it's not caused by any torque from the floor. It's a bit more obvious in this diagram why the block must rotate.

The main effect of the motion of the block as far as dynamics go will be an increase in its weight.

If we consider the rotating disk case now, I believe the following should be true (I haven't done separate detailed calcuations)

1) We will have the same increase in weight at the edge of the disk relative to the center of the disk. This could be regarded as stresses on the disk (though the stresses will be immediately translated to the floor).

2) There will be additional Thomas precession due to the centripetal acceleration and the velocity of the disk (one might argue that this is in fact the true meaning of Thomas precession). This will induce some stress on the disk as well, though I find it difficult to vizulaize the details. However, these stresses would be present in a rotating disk that was NOT in an Einstein's elevator, so I don't think these stresses are of interest to the Born rigidity question.

I don't see, at the moment, a source for any other stresses than 1) and 2).

 

[yuiop]

I've been busy recently, but had a few thoughts on the topic.

All your inputs are gratefully received and worth any wait

It's helpful to consider the case of a block sliding along the floor of an Einstein's elevator before tackling the rotating case. There is some discussion by me in another thread https://www.physicsforums.com/showthread.php?t=701257, the discussion includes a derivation of the trajectory of the moving block in an inertial frame, and a convenient set of coordinate transforms and an associated metric.

This is a useful simplification that may assist in resolving the question raised in this thread. Your metric is interesting although I have not fully grasped it yet. Can you confirm that the the proper length of the block along the x-axis remains constant or are their issues with rotation that make this unclear?

A gyroscope attached to the block will rotate, because of effects similar to Thomas precession, whose axis is in the a (cross) v plane, cross being the vector cross product. The plane of rotation is the a^v plane, i.e. it is the plane formed by the accleration vector and the the velocity vector. It's not particularly obvious why the block is rotating viewing this diagram.

This is exactly the precession I was trying to describe in the last paragraph of post #9. I have attached a diagram to show the various vectors you describe as applied to the rotating disc. It is easy to see from the diagram that the Thomas rotation effect of opposing sides of the disc counteract each other and there is no net rotation of the disc, although there may still be additional internal stresses induced within the disc as a result.

1) We will have the same increase in weight at the edge of the disk relative to the center of the disk. This could be regarded as stresses on the disk (though the stresses will be immediately translated to the floor).

Agree. With a properly supported disc on suitable bearings, this should have no effect on the geometry or Born rigidity of the disc.

2) There will be additional Thomas precession due to the centripetal acceleration and the velocity of the disk (one might argue that this is in fact the true meaning of Thomas precession). This will induce some stress on the disk as well, though I find it difficult to vizulaize the details. However, these stresses would be present in a rotating disk that was NOT in an Einstein's elevator, so I don't think these stresses are of interest to the Born rigidity question.

I don't see, at the moment, a source for any other stresses than 1) and 2).

Again I agree. It is difficult to see any intuitive reason why a disc with constant angular velocity, at constant altitude with constant proper linear acceleration should have a breakdown of Born rigidity.

 

[PeterDonis]

I don't see, at the moment, a source for any other stresses than 1) and 2).

As I've stressed before (pun intended ), the key issue isn't stress, it's shear--not shear stress, but shear in the congruence of worldlines describing the disk. I think your picture makes it clearer where the shear comes from in an accelerated, rotating disk. I'll approach this in two stages.

(1) First consider a rotating, accelerated *ring*--i.e., a circular object with negligible thickness in the direction of its linear acceleration, *and* negligible annular thickness (i.e., radial dimension perpendicular to its linear acceleration). As the papers linked to earlier in this thread show, Born rigid motion *is* possible for this object.

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.

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 (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.)

(2) Now consider the case of a rotating, linearly accelerated *disk*. We can think of the disk as a set of concentric rings with different diameters. But that at once shows us the key difficulty with trying to make the disk's motion as a whole Born rigid: the tangential acceleration component that we have to add in, per the above, in order to keep each individual ring's motion Born rigid, is *different* for the different rings! This is because the magnitude of the required tangential acceleration varies with the radius of the ring.

But this change in tangential acceleration from ring to ring must produce shear between the rings--in other words, it's impossible to both keep each individual ring's motion Born rigid, *and* to maintain the rings stationary relative to each other. It's the same problem encountered in trying to spin up a disk (instead of a ring) from a non-rotating to a rotating state in a Born rigid manner: each ring within the disk requires a *different* tangential acceleration profile to keep it Born rigid within itself as it is spun up.

I think this is the intuitive picture I was looking for earlier; thinking over pervect's scenario helped to break it loose.

 

[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.