Ehrenfest's Paradox: Resolving the Kinematical Solution

[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.

 

[DrGreg]

For reference some previous threads which included discussion of quotient spaces and rotating disks:

• Large rotating disk, particularly post #17 onwards
  
• analogy between centrifugal force and gravity

 

[DrGreg]

Coincidentally, this has just cropped up in another thread:

When expressed in cylindrical polar coordinates, these are known as Born coordinates
ds^2 = -\left( c^2 - \omega^2 \, r^2 \right) \, dt^2 + 2 \, \omega \, r^2 \, dt \, d\phi + dz^2 + dr^2 + r^2 \, d\phi^2

so I might as well point out that if you "complete the square" on this formula you get<br /> ds^2 = -\left( c^2 - \omega^2 \, r^2 \right) \left( dt - \frac{\omega r^2}{c^2 - \omega^2 r^2} d\phi \right)^2 + dz^2 + dr^2 + \frac{r^2}{c^2 - \omega^2 r^2} \, d\phi^2<br />The expression<br /> dz^2 + dr^2 + \frac{r^2}{c^2 - \omega^2 r^2} \, d\phi^2<br />represents the non-flat metric of the quotient space. The same technique applies to any stationary spacetime.

Ref: Rindler (2006), Relativity: Special, General and Cosmological, 2nd ed. p. 198