Ehrenfest's Paradox: Resolving the Kinematical Solution

[center o bass]

I'm reading about Ehrenfest's paradox where one considers a rotating disk. If one let r' be the radius of the disc in an inertial frame and r be the radius of the disc when it is at rest. Then the periphery must be Lorentz' contracted such that $2\pi r' < 2\pi r$, but since the radial line is perpendicular to the direction of motion of the disk it is not Lorent'z contracted so that $r'=r$. Thus the paradox.

The supposed kinematical solution is to consider the disk being accelerated up to a given angular velocity. By an analysis involving the relativity of simultaneity it is found that it is inconsistent to require that the disk be accelerated up to the angular velocity and additionally require it being done while keeping the periphery 'Born rigid'. I.e. accelerated in such a way that the rest length of each line element along the periphery remains constant.

This is claimed to resolve the paradox and my question is why it does that.
More specifically; why does one need to assume a 'Born rigid' acceleration of the disk in order to accept the paradox?

 

[PeterDonis]

I'm reading about Ehrenfest's paradox where one considers a rotating disk. If one let r' be the radius of the disc in an inertial frame and r be the radius of the disc when it is at rest. Then the periphery must be Lorentz' contracted such that $2\pi r' < 2\pi r$, but since the radial line is perpendicular to the direction of motion of the disk it is not Lorent'z contracted so that $r'=r$. Thus the paradox.

But what, exactly, is the paradox? Is it that the actual, physical circumference of the disk is not equal to 2 pi times its actual, physical radius? That's an interesting fact, certainly, but why is it a paradox? There's nothing self-contradictory about it that I can see; it just shows that the spatial geometry of the rotating disk is not Euclidean. (Note that there are issues with even defining what "the spatial geometry of the rotating disk" means; we'll see if we end up opening that can of worms in this thread, as it has been in previous ones on this topic. )

The supposed kinematical solution is to consider the disk being accelerated up to a given angular velocity. By an analysis involving the relativity of simultaneity it is found that it is inconsistent to require that the disk be accelerated up to the angular velocity and additionally require it being done while keeping the periphery 'Born rigid'. I.e. accelerated in such a way that the rest length of each line element along the periphery remains constant.

This is claimed to resolve the paradox and my question is why it does that.

Because it shows how the spatial geometry of the rotating disk can change as it is spun up.

More specifically; why does one need to assume a 'Born rigid' acceleration of the disk in order to accept the paradox?

Because that amounts to assuming that the spatial geometry of the disk (again, with caveats as to how that actually is defined) remains unchanged as it spins up. In other words, it amounts to assuming that either (a) there are no stresses induced in the disk as it spins up, or (b) whatever stresses are induced in the disk as it spins up don't deform the disk. When put this way, of course, the resolution is obvious: the act of spinning up the disk does induce stresses in the disk, and the stresses do deform the disk, changing its spatial geometry.

 

[Ibix]

In a Newtonian world you can simply spin a disc faster and faster. There's a limit to a real disc, though. It'll fly apart under its centrifugal stresses at some speed.

The Special Relativistic case adds stresses due to the different Lorentz contractions experienced by the different parts of the disc. The disc will deform and eventually fly apart under these forces.

There is only a paradox if you have an ideal material that can neither deform nor disintegrate. If you can find such a material, I'd like a suit of armour...

 

[center o bass]

Because that amounts to assuming that the spatial geometry of the disk (again, with caveats as to how that actually is defined) remains unchanged as it spins up. In other words, it amounts to assuming that either (a) there are no stresses induced in the disk as it spins up, or (b) whatever stresses are induced in the disk as it spins up don't deform the disk. When put this way, of course, the resolution is obvious: the act of spinning up the disk does induce stresses in the disk, and the stresses do deform the disk, changing its spatial geometry.

Ah, so in order to even write that $2\pi r' < 2 \pi r$ one are implicitly assuming that the spatial geometry of the disk have not changed after it was accelerated up to a given angular velocity. Otherwise it would not be generally true that the circumference were $2\pi$ times the radius.

Thus, if I understand it correctly, the kinematical resolution is actually stating that the spatial geometry of the disk must be deformed under the acceleration process? And relativity is also telling us what that geometry must be at a given angular velocity?

Namely a physical circumference of

$$ \frac{2\pi r}{\sqrt{1 - \frac{\omega^2 r^2}{c^2}}}$$

I'm not sure if I've totally missed the point when I say that I'm a bit surprised that relativity gives a requirement on what shape the disk deform into without an input theory of how the parts of the disk are 'bound together' (elasticity etc.)

 

[Bill_K]

I'm not sure if I've totally missed the point when I say that I'm a bit surprised that relativity gives a requirement on what shape the disk deform into without an input theory of how the parts of the disk are 'bound together' (elasticity etc.)

The input is Born Rigidity. That is we assume the disk does not stretch, consequently its elastic properties do not enter.

 

[center o bass]

The input is Born Rigidity. That is we assume the disk does not stretch, consequently its elastic properties do not enter.

Where does that enter? What parts of the disk are required to be Born rigid? As I understand it, it was inconsistent to require elements on the periphery be Born rigid under the acceleration.

Btw, just to make sure: Even though the spatial geometry of the disk is non-Euclidean, it does not mean that spacetime is curved, right? One can't change the geometry of the space by a coordinate transformation.

 

[PeterDonis]

The input is Born Rigidity. That is we assume the disk does not stretch, consequently its elastic properties do not enter.

But isn't that the "input" that creates the paradox? In other words, the point of the paradox, as I understand it, is that the disk cannot be spun up without deforming, so the assumption of Born rigidity cannot be true for an actual disk. Put another way, for the case of angular acceleration (unlike the case of linear acceleration), it is impossible to induce Born rigid motion even with a "conspiracy of forces" applied to each individual point of the disk.

 

[PeterDonis]

Even though the spatial geometry of the disk is non-Euclidean, it does not mean that spacetime is curved, right? One can't change the geometry of the space by a coordinate transformation.

Correct. It is assumed in this scenario that the disk has negligible mass and therefore negligible gravity, so spacetime as a whole is flat. As you say, that's true regardless of which coordinates we adopt.

 

[center o bass]

But isn't that the "input" that creates the paradox? In other words, the point of the paradox, as I understand it, is that the disk cannot be spun up without deforming, so the assumption of Born rigidity cannot be true for an actual disk. Put another way, for the case of angular acceleration (unlike the case of linear acceleration), it is impossible to induce Born rigid motion even with a "conspiracy of forces" applied to each individual point of the disk.

That was my current understanding as well.. :)

 

[Bill_K]

But isn't that the "input" that creates the paradox? In other words, the point of the paradox, as I understand it, is that the disk cannot be spun up without deforming, so the assumption of Born rigidity cannot be true for an actual disk.

Yes, exactly.

 

[pervect]

There are a ton of papers out there on the Erhenfest paradox. I've found http://arxiv.org/abs/gr-qc/9805089 to be useful.

A key insight here is the following:

As a consequence, in any reference frame for which the vortex tensor differs from zero, the concept of ”the whole physical space at a given instant” turns out to be conventional, in the sense that it is lacking an operational meaning because of the impossibility of a symmetrical and transitive synchronization procedure at
large.

So you can't Einstein synchronize all the clocks on a rotating disk - or a rotating globe. There are systems that by convention synchronize clocks on a rotating disk (or globe), but they aren't Einstein synchronized.

If you can't agree on a concept of "physical space", it becomes obvious that it's confusing to talk about it's circumference.

I hope this somewhat different perspective helps. But at eny event, with a little googling, you should be able to find a TON of papers on the rotating disk paradox, enough to give you much more specific questions than the ones you've been asking.

 

[Ben Niehoff]

Btw, just to make sure: Even though the spatial geometry of the disk is non-Euclidean, it does not mean that spacetime is curved, right? One can't change the geometry of the space by a coordinate transformation.

Correct. It is assumed in this scenario that the disk has negligible mass and therefore negligible gravity, so spacetime as a whole is flat. As you say, that's true regardless of which coordinates we adopt.

I disagree. The disk is spinning at relativistic speeds. Its angular momentum is not negligible; it contributes to the energy-momentum tensor, and thus to the local curvature of spacetime. This is precisely why the geometry of the disk becomes non-Euclidean.

Rotation is an absolute motion and not merely a coordinate transformation. Every point on the disk experiences proper acceleration.

 

[PeterDonis]

I disagree. The disk is spinning at relativistic speeds. Its angular momentum is not negligible; it contributes to the energy-momentum tensor, and thus to the local curvature of spacetime.

The angular *velocity* of the disk is certainly not negligible; but does that automatically imply that its angular momentum is not negligible? What if we make the disk out of test particles, whose mass is so small that the disk's effective moment of inertia is small enough to make the angular momentum negligible even if the angular velocity is not?

Also, if you are right and it is impossible for the disk's angular momentum to be non-negligible, wouldn't that mean that it's impossible to realize the Ehrenfest paradox in flat spacetime? It seems to me that a lot of the literature on the paradox uses flat spacetime to model it.

This is precisely why the geometry of the disk becomes non-Euclidean.

I'm not sure I agree with this, in the sense that there is certainly a congruence of timelike worldlines in flat spacetime, the Langevin congruence, for which Einstein clock synchronization can't be used to define surfaces of simultaneity, as pervect pointed out. The "spatial geometry" associated with such a congruence can only be defined as a quotient space, not as a foliation of identical spacelike slices; and this spatial geometry will be non-Euclidean. So it's certainly possible to have an object in flat spacetime with non-Euclidean spatial geometry.

Rotation is an absolute motion and not merely a coordinate transformation. Every point on the disk experiences proper acceleration.

Yes, agreed; this is true regardless of how the angular momentum issue comes out. The proper acceleration vector associated with the Langevin congruence is nonzero.

 

[Ben Niehoff]

Hmm...you're probably right.

 

[center o bass]

But isn't that the "input" that creates the paradox? In other words, the point of the paradox, as I understand it, is that the disk cannot be spun up without deforming, so the assumption of Born rigidity cannot be true for an actual disk. Put another way, for the case of angular acceleration (unlike the case of linear acceleration), it is impossible to induce Born rigid motion even with a "conspiracy of forces" applied to each individual point of the disk.

Yes, exactly.

According to an argument given by Øyvind Grøn it seems like it's precisely the fact that standard measuring rods remain Born rigid, while the disk can not, that makes the geometry non-Euclidean. As the disk is accelerated there are space for more standard measuring rods and hence the circumference around the periphery will appear to be longer.

From that point of view it does not seems like the resulting geometry of the disk is a consequence of Born rigidity of the disk.The argument is given here at page 35 under 'A rotating disk with angular acceleration':
http://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CDUQFjAB&url=http%3A%2F%2Fareeweb.polito.it%2Fricerca%2Frelgrav%2Fsolciclos%2Fgron_d.pdf&ei=ddWdUauVIYXl4QS834FY&usg=AFQjCNFhAQw918KWnBS7B2Xl-3NOECKsrg&sig2=SsZ0Z0-A01kbTT22mM0iqA&bvm=bv.46865395,d.bGE&cad=rja

 

[center o bass]

Correct. It is assumed in this scenario that the disk has negligible mass and therefore negligible gravity, so spacetime as a whole is flat. As you say, that's true regardless of which coordinates we adopt.

Hmm...you're probably right.

According to Nathaniel I. Reden there has not yet been found a definition of time for which the time curvature of the Riemann tensor for the spacetime geometry of the rotating frame disappear.

https://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CEYQFjAC&url=https%3A%2F%2Fwww.amherst.edu%2Fmedia%2Fview%2F10267%2Foriginal%2Freden05.pdf&ei=Nu6dUbDtA8S24AS3hoCoCQ&usg=AFQjCNG6x91ToivTx2HJK1UhQmhj41Dpsg&sig2=9DAQIwiovmKpA5u_rO-OwA&bvm=bv.46865395,d.bGE&cad=rja

 

[pervect]

According to Nathaniel I. Reden there has not yet been found a definition of time for which the time curvature of the Riemann tensor for the spacetime geometry of the rotating frame disappear.

https://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CEYQFjAC&url=https%3A%2F%2Fwww.amherst.edu%2Fmedia%2Fview%2F10267%2Foriginal%2Freden05.pdf&ei=Nu6dUbDtA8S24AS3hoCoCQ&usg=AFQjCNG6x91ToivTx2HJK1UhQmhj41Dpsg&sig2=9DAQIwiovmKpA5u_rO-OwA&bvm=bv.46865395,d.bGE&cad=rja

Where does he say that exactly?

At the moment, I'm thinking that if the components of any tensor vanish in one coordinate system, they must vanish in all coordinate systems.

And in a non-rotating coordinate system with no matter, the Riemann is zero.

So if you have an empty rotating frame with no matter, it must also have a zero Riemann.

 

[Mentz114]

According to Nathaniel I. Reden there has not yet been found a definition of time for which the time curvature of the Riemann tensor for the spacetime geometry of the rotating frame disappear.

In the cited paper Reden seems to be working in 3 dimensions. It is true that the 3 dimensional sub-manifold of the rotating frame is curved. But it is not spacetime curvature.

 

[center o bass]

Where does he say that exactly?

At the moment, I'm thinking that if the components of any tensor vanish in one coordinate system, they must vanish in all coordinate systems.

And in a non-rotating coordinate system with no matter, the Riemann is zero.

So if you have an empty rotating frame with no matter, it must also have a zero Riemann.

At page 25 under summary he states:

'Moreover, none of the three results in a Riemann curvature tensor with all
components zero, so we have not solved the problems mentioned in section 1.7.'

I would think that some of the components being non-vanishing implies a non-vanishing Riemann tensor and thus a non vanishing curvature.

 

[center o bass]

In the cited paper Reden seems to be working in 3 dimensions. It is true that the 3 dimensional sub-manifold of the rotating frame is curved. But it is not spacetime curvature.

It is indeed working in three dimensions in section one, but in section two he includes the time coordinate and still gets non vanishing Riemann tensor components.

 

[Mentz114]

I've looked a section 2 in Reden's paper. None of the metrics he writes has zero Einstein tensor and they are *not* physical solutions of the EFE ( in my estimation ). So they don't tell us anything useful.

The Langevin frame is the correct treatment and shows that only a 3D submanifold is curved, not the global spacetime.

For rotating disks in GR we need something like the Neugebauer-Meinel disks*, see for instance arXiv:gr-qc/9703077v1 26 Mar and references therein.

*Neugebauer & Meinel, Phys. Rev. Lett. 75 (1995) 3046

 

[PeterDonis]

The Langevin frame is the correct treatment and shows that only a 3D submanifold is curved, not the global spacetime.

Can you clarify what you mean by "Langevin frame"? My understanding is that there is *no* 3D submanifold of the spacetime as a whole that corresponds to "the spatial geometry of the rotating disk". This is because of the failure of Einstein clock synchronization for the worldlines of the Langevin congruence. The "spatial geometry" of the rotating disk is the quotient space of the Langevin congruence over proper time for Langevin observers (I think I've stated that correctly--the idea is that each point of the quotient space corresponds to a particular worldline in the Langevin congruence). This quotient space is not a submanifold of the spacetime, is it?

 

[Mentz114]

Can you clarify what you mean by "Langevin frame"? My understanding is that there is *no* 3D submanifold of the spacetime as a whole that corresponds to "the spatial geometry of the rotating disk". This is because of the failure of Einstein clock synchronization for the worldlines of the Langevin congruence. The "spatial geometry" of the rotating disk is the quotient space of the Langevin congruence over proper time for Langevin observers (I think I've stated that correctly--the idea is that each point of the quotient space corresponds to a particular worldline in the Langevin congruence). This quotient space is not a submanifold of the spacetime, is it?

In answer to the question I've highlighted I can only quote the Wiki article http://en.wikipedia.org/wiki/Born_coordinates

That is, we can consider the quotient space of Minkowski spacetime (or rather, the region 0 < R < 1/ω) by the Langevin congruence, which is a three-dimensional topological manifold. Even better, we can place a Riemannian metric on this quotient manifold, turning it into a three dimensional Riemannian manifold, in such a way that the metric has a simple operational significance.

which shows I've mis-read it. The quotient manifold is not a submanifold of the global spacetime.

 

[WannabeNewton]

The wiki paragraph quoted is very incomplete as far as I can tell. Why is the quotient space a topological manifold? The quotient is a very ill-behaved operation in the category of topological spaces and in general does not take a Hausdorff space to a Hausdorff space or a locally euclidean space to a locally euclidean space etc. Moreover, why is it a smooth manifold? The article assumes, on top of the topological structure that it doesn't prove it has (and, as far as I can tell, fails to give reference to something that does), that the quotient also has a smooth structure because it freely uses the theorem that any smooth manifold can be given a Riemannian metric.

I checked the Landau & Lifgarbagez reference as well in my copy of classical theory of fields, and they don't prove it either.

EDIT: Never-mind, I found an exercise in one of my manifolds texts that deals with the issue of quotient manifolds obtained from vector fields :)

 

[Mentz114]

In view of the posts above, can we say that the circumference problem is explained by the non-Euclidean nature of the 3D space transported along a worldline of the Langevin congruence ?

 

[pervect]

Deleting duplicate, saving attachment...

 

[pervect]

Here is my very quick interpretation of the "Rotating Disk" with some "visual aids" that describes one picture of the non-euclidean space.

First look at the wiki diagram of the Langevian syncrhonization.



The red line indicates the worldline of one observer on the disk, say "point 0" in that Reden paper.

The blue line is a curve that you get by Lorentz synchronizing observers in each of the local comoving inertial frames at the edge of the disk - it's orthogonal to the congruence of red lines representing the worldlines at various points around the disk.

Let's call the lower interesection of the red and blue lines point "A" and the upper intersection of the red and blue lines point "B".

we see that "A" and "B" occur at different times. THis is what's causing the confusion.
Tartaglia mentioned a key important insight - that our "spatial" surface is a quotient space.

The wiki defintion of this is:

In topology and related areas of mathematics, a quotient space (also called an identification space) is, intuitively speaking, the result of identifying or "gluing together" certain points of a given space. The points to be identified are specified by an equivalence relation. This is commonly done in order to construct new spaces from given ones.

How do we construct this space? How do we represent this space on our diagram?

Well, we know it contains the winding outer blue line. That's represents the circumference of our space.

If we imagine a slightly smaller cylinder, of smaller radius, it has a winding blue helical line around it, which is also in our space.

Finally we come to the center, which is just a point.

Where is the point located on our diagram? In the center, obviously, but where on the z axis - at what point in time?

I believe the answer is conventional, so I chose the "obvious" location, midway on the z axis between A and B.

As much as we might like to use Einstein clock synchronization throughout - we can't. Hence, the need for a non-conventional radial clock synchronization.

When we put it all together, we visualize something like the screw thread, below, but oriented vertically to match the first diagram.

This is our "space".

If we smash it flat, we get somethign like the attachment below. It's a circle, with a bit of an extra piece glued onto it.

Point A on our diagram represents point A on the Langevian cylinder, the lower intersection of the red and blue lines. Point B on our diagram represents point B on the Langevian culinder, the upper intersection of the red and blue lines.

Now, we "glue" together (in the sense of our definition of a quotient space) the dotted line to the dashed line on our diagram. And that's our space.

In this particular representation, our space is flat everywhere (we made it out of flat paper), it's the "gluing operation", the identification of points, that makes the space non-Euclidian.

 

[PeterDonis]

The blue line is a curve that you get by Lorentz synchronizing observers in each of the local comoving inertial frames at the edge of the disk - it's orthogonal to the congruence of red lines representing the worldlines at various points around the disk.

Yes, and as has been pointed out already in this thread (I can't remember if it was you that did so), the fact that the blue line is not closed indicates the failure of Einstein clock synchronization. However, I think the consequence of this is more drastic than you are saying. See below.

we see that "A" and "B" occur at different times.

Different times according to the underlying Minkowski coordinates, yes. But the whole point of the failure of Einstein clock synchronization is that, since both A and B lie on the same blue line, they both occur "at the same time" according to Einstein clock synchronization. *That* is why Einstein clock synchronization fails: it ends up assigning the same time to timelike separated events.

But more than that, it means the blue line cannot be used to define the quotient space. See below.

How do we represent this space on our diagram?

You can't; it's not a submanifold of the spacetime, as I posted before.

Well, we know it contains the winding outer blue line. That's represents the circumference of our space.

No, it doesn't; it can't, because, as I said above, the blue line contains timelike separated events. Same comment for the smaller and smaller cylinders as you go towards the center.

Finally we come to the center, which is just a point.

Where is the point located on our diagram? In the center, obviously, but where on the z axis - at what point in time?

At no point in time; as noted, the quotient space is not a submanifold of the spacetime. See below.

I believe the answer is conventional

I don't think so; I think the answer is that the quotient space is not a submanifold of the spacetime, so points in the quotient space cannot be identified with points in the spacetime. Instead, points in the quotient space are identified with *worldlines* in the Langevin congruence. For example, the point at the center of the quotient space is identified with the entire worldline of the center of the disk.

As much as we might like to use Einstein clock synchronization throughout - we can't. Hence, the need for a non-conventional radial clock synchronization.

This can be done, but I don't think it's the same as defining the quotient space; it's a separate question.

In this particular representation, our space is flat everywhere (we made it out of flat paper), it's the "gluing operation", the identification of points, that makes the space non-Euclidian.

I don't think this is right either. Since points in the quotient space represent worldlines in the Langevin congruence, in order to define the geometry of the quotient space, we have to figure out what the appropriate metric is on the quotient space, which requires defining a notion of "distance between adjacent worldlines" that doesn't depend on which point on any given worldline we pick. As I understand it, the way this is done is as described on the Wiki page on Born coordinates, in the section "Radar distance in the small" (this is the section that the quote that Mentz114 gave earlier came from).

 

[WannabeNewton]

Indeed the quotient space by itself is simply a topological space (the quotient space being the space obtained by taking the subset of Minkowski space-time covered by the Langevin congruence and identifying all points along a given worldline in the congruence). Recall that these points are an equivalence class of events in the original subset of Minkowski space-time. The reason we can do this is because in the Born chart, the Langevin congruence given by the integral curves of the 4-velocity field $e_0 = \frac{1}{\sqrt{1 - \omega^2 r^2}}\partial_{t}$ which follow an orbit of the time-like killing vector field $\partial_{t}$ so the Langevin observers are stationary with respect to the Born chart. This means that we can identify all the events with the same spatial location but different times, as represented in the Born chart, and simply represent the worldline of a given observer in the congruence by a single point in the quotient space. However as Peter notes, the quotient manifold has no notion of geometry until a Riemannian metric is introduced, such as the Langevin-Landau-Lifgarbagez metric described in the article.

Note the part of the article that talks about how Frobenius' theorem fails for the 4-vector field of the Langevin congruence (i.e. $(e_{0})_{[\mu}\nabla_{\nu}(e_{0})_{\alpha]}\neq 0$) which means there is no one-parameter family of orthogonal space-like hypersurfaces associated with the 4-velocity field of the Langevin congruence. As the article notes, this lends to troubles in defining the "spatial geometry" of the rotating disk.

 

[center o bass]

Just to take up a point at least I did not get entirely clear; Bill_K mentioned that the shape the disc 'bends' into (the spatial geometry) of the disk arises due to the 'input' that the disc be Born rigid. But then I read in W.Rindler - Relativity that

" A body moving rigidly cannot start to rotate, since circumferences of circles described by points of the body would have to shrink, while their radii would have to remain constant, which is impossible. In general, therefore, the motion of one point of a rigidly moving body determines that of all others"

With a footnote on the end stating

"For this reason, a ‘rigidly moving’ disk that is set to rotate would have to bend."

And to me this seems to indicate that Rindler claims that if the rotating disk starts to bend it's no longer Born rigid. I.e. only non-rotating objects can remain Born rigid. But then if the requirement of Born rigidity of the disk is not possible; what then determines the geometry of the rotating disk? Are we not discussing the geometry of a rotating, Born rigid disk?

 

[PeterDonis]

this seems to indicate that Rindler claims that if the rotating disk starts to bend it's no longer Born rigid.

More precisely, it means that it is impossible for the *acceleration* of the disk to be a Born rigid motion. It is perfectly possible for rotation of the disk *at constant angular velocity* to be Born rigid.

I.e. only non-rotating objects can remain Born rigid.

No, that's a stronger claim, which is false. See above. Only angular acceleration cannot be Born rigid.

what then determines the geometry of the rotating disk?

See my #28 in response to pervect. The gist of it is, if by "geometry of the rotating disk" you mean the geometry of a spacelike slice of "constant time" for the disk, there is no such thing, because it's impossible to do Einstein clock synchronization on a rotating disk. There is a quotient space that can be interpreted as "the geometry of the rotating disk", but it does not correspond to any spacelike slice of the spacetime.

 

[center o bass]

More precisely, it means that it is impossible for the *acceleration* of the disk to be a Born rigid motion. It is perfectly possible for rotation of the disk *at constant angular velocity* to be Born rigid.
No, that's a stronger claim, which is false. See above. Only angular acceleration cannot be Born rigid.

Alright, but if it does not rotate to begin with it can not go into a rotation while remaining Born rigid throughout the motion.

This would imply, I suppose, that for any object which is required to be Born rigid throughout the motion with the initial condition that it was not rotating, the motion would have to be a translational motion?

But thanks. This became a lot clearer to me. We are studying a Born rigid disk, rotating at constant angular velocity which is fine :)

 

[PeterDonis]

Alright, but if it does not rotate to begin with it can not go into a rotation while remaining Born rigid throughout the motion.

Correct.

This would imply, I suppose, that for any object which is required to be Born rigid throughout the motion with the initial condition that it was not rotating, the motion would have to be a translational motion?

Yes; linear acceleration can be done in a Born rigid manner (though it requires very precisely specified forces to be applied at *every* point of the object, hence is not going to happen in any practical sense).

 

[pervect]

Yes, and as has been pointed out already in this thread (I can't remember if it was you that did so), the fact that the blue line is not closed indicates the failure of Einstein clock synchronization. However, I think the consequence of this is more drastic than you are saying. See below.

But more than that, it means the blue line cannot be used to define the quotient space. See below.

I'm probably confused about what constitutes a "quotient space" then, as I don't have any math books that cover them formally, and the GR textbooks and papers I have only seem to refer to them when this particular question (rotating frames) arise.

I'm pretty sure a quotient space doesn't have to be a manifold though, and I remain uncertain about exactly which axioms that a quotient space is supposed to satisfy.

You can't; it's not a submanifold of the spacetime, as I posted before.

Even wiki says that a quotient space derived from a manifold isn't necessarily a manifold - so I'm not sure this shows that my construction "isn't a quotient space". It does follow the rather "informal" construction that wiki suggests.

Actually, what I don't like about my construction, now that I look at it, is that it doesn't measure radial distances.

Are you suggesting perhaps that the more usual interpretation of "quotient space" is to identify timelike wordlines in the timelike congruence with "points" in a space?

 

[WannabeNewton]

I'm pretty sure a quotient space doesn't have to be a manifold though, and I remain uncertain about exactly which axioms that a quotient space is supposed to satisfy.

You are correct that in general a quotient space doesn't have to be a manifold. There are no axioms per say that a quotient space satisfies. Here is the definition of a quotient space: let $X$ be a topological space, $Y$ be any set, and $q:X\rightarrow Y$ be a surjective map. We define a topology on $Y$ by saying that $U\subseteq Y$ is open iff $q^{-1}(U)\subseteq X$ is open. We call this the quotient topology on $Y$ and $q$ the quotient map. The most common application is as follows: let $\sim$ be an equivalence relation on a topological space $X$ and let $q:X\rightarrow X/\sim$ be the map sending each $p\in X$ to its equivalence class $[p]$. Then, the set $X / \sim$ combined with the quotient topology induced by $q$ is called the quotient space of $X$. So as you can see, the only real thing we need to define a quotient space is an equivalence relation, there are no "axioms" in the usual sense.

In the case of the Langevin congruence and the rotating disk, we take the region $U$ of Minkowski space-time defined in the wiki article, which is filled completely by the Langevin congruence, and we define an equivalence relation by stating that if two events lie on the worldline of a given Langevin observer then they are equivalent. So here we have a well defined equivalence relation, and with this we can get the associated quotient space. In this quotient space, a single point is actually the worldline of a Langevin observer in the Langevin congruence.