Thin rotating disc under constant acceleration.

[PeterDonis]

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

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

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

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

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

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

 

[PAllen]

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

 

[PAllen]

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

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

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

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

 

[PAllen]

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

 

[PAllen]

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

 

[PAllen]

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

 

[PeterDonis]

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

 

[yuiop]

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

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

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

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

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

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

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