# Tall rotating cylinder near a black hole

1. Aug 12, 2013

### yuiop

Imagine we have a very tall vertical cylinder like a very elongated telegraph pole, that is rotating at 200 rpm about its long axis on near perfect bearings. Initially the cylinder is sufficiently far from a black hole, that differences in gravitational time dilation between the top and bottom of the cylinder are negligible for our purposes. Let us say there is observer that is stationary with respect to the black hole and located at the bottom of the cylinder. The cylinder is slowly lowered past this observer so that the top of cylinder becomes adjacent to this observer and lower part is very close to the black hole event horizon. Does anyone agree that the observer will see the part of the cylinder that is local to him, slow down significantly, (say by about 90%) and that it will speed back up to 200 rpm when the cylinder is raised again? Will an observer that accompanies the lower part of the cylinder always see that part rotate at approximately 200 rpm? If so, what does this say about conservation of angular momentum?

P.S. Assume the cylinder is sufficiently rigid that it does not get twisted into knots by differences in speed between the top and bottom and that the lowering process is sufficiently slow for the rotation speeds to even out along the length of the cylinder.

Last edited: Aug 12, 2013
2. Aug 12, 2013

### Staff: Mentor

Sounds extremely complicated to analyze. I don't have an intuitive feel for it.

3. Aug 12, 2013

### Staff: Mentor

I remember a similar question coming up quite some time ago, and I don't think it ever really got resolved. I think the reason is that, if the cylinder is long enough that gravitational time dilation can vary significantly along its length if it's close enough to the hole (which basically means the cylinder's proper length is of the same order of magnitude as the hole's mass, in "natural" units where G = c = 1), the curvature of the spacetime will interact with the cylinder's motion and it's no longer clear what, if anything, will be "held constant" or "conserved".

For example, suppose your hypothesis is true and the stationary observer sees the local piece of the cylinder in his vicinity rotate more and more slowly as the cylinder is lowered, and more and more rapidly as the cylinder is raised again. Does this mean angular momentum conservation is violated? Not necessarily, for two reasons:

(1) When evaluating the cylinder's angular momentum, there should be a factor that depends on spacetime curvature--for example, if you try to integrate the angular momentum of each small piece of the cylinder at a given radial coordinate, over the range of radial coordinates it occupies, the integrand should have a factor that depends on the "redshift factor" $\sqrt{ 1 - 2M / r }$. So changing the range of $r$ coordinates that the cylinder occupies will change its angular momentum, even if its angular velocity remains constant.

(2) As the cylinder is lowered or raised, the change in spacetime curvature might change its angular velocity, because of higher-order gravitational effects that aren't captured in the simple model of "changing gravitational time dilation", which implicitly assumes that the test object being lowered or raised has no internal properties other than its rest mass. The cylinder's angular momentum, since it's due to spin and not to being in orbit about the hole (your scenario can be realized with the cylinder moving purely radially), is an additional internal property.

4. Aug 12, 2013

### yuiop

Hi Dale and Peter. It seems to me that if "nothing" happened to the cylinder and it continued to spin at the same rate as observed from the top and with no twisting along its length, then it conceivable that we could end up with a situation where the rim velocity of the part of the cylinder nearest the black hole could exceed the speed of light as measured locally. Obviously that should not happen, so it seems reasonable to conclude that the cylinder must slow down along its entire length, although it may appear to be constant to an observer that remains adjacent to the lower part of the cylinder. I only assume that because I assume angular momentum is conserved in some form, maybe only locally, but then again that is obviously not happening form the point of view of the observer near the top. Maybe the conservation happens in an averaging sort of way along the length of the cylinder or it is conserved when we take (gravitational) relativistic mass into account. After all conservation of linear momentum in flat spacetime requires relativistic mass to be taken into account (if I recall correctly). Anyway, I am not looking for exact quantitative answers here (yet :P), but just qualitative gut instinct. Does the cylinder slow down (or speed up on the way back up up) .... or not?

Oh.. almost forgot... momentum and energy are conserved as a pair in relativity, rather than independently. Might need to take that into account.

Last edited: Aug 12, 2013
5. Aug 12, 2013

### Staff: Mentor

I don't think that there is any way that you can reasonably assume "nothing" happened to the cylinder. If it is being lowered slowly then it can be used to do work so energy can be extracted from it, and it certainly will undergo tidal stresses.

I just don't have a qualitative gut instinct. It could be that it will slow down, it could be that stresses become infinite. I don't know.

6. Aug 12, 2013

### Staff: Mentor

Does this mean, as observed by an observer who moves along with the top of the cylinder, or as observed by an observer at infinity? I.e., are you assuming that the cylinder's angular velocity is constant with respect to an observer comoving with the cylinder's top (meaning that, with respect to infinity, the angular velocity will change as the "redshift factor" $\sqrt{1 - 2M / r}$ changes), or are you assuming that the cylinder's angular velocity is constant with respect to infinity, i.e., with respect to Schwarzschild coordinate time $t$?

This makes it seem to me like you are assuming that the cylinder's angular velocity $\omega$ is constant with respect to infinity; then the velocity of the cylinder's rim at the bottom of the cylinder, as measured locally, will be $v = \omega R / \sqrt{1 - 2M / r}$, where $R$ is the cylinder's radius, as measured locally, and $r$ is the (global) radial coordinate of the bottom of the cylinder.

Again, it appears that you mean "slow down with respect to infinity", i.e., its angular velocity $\omega$ with respect to infinity has to decrease to keep the locally measured $v$ less than 1, given the formula above. This seems reasonable to me, but it doesn't say much, because it doesn't tell you what constraint to impose. One obvious constraint, as you note, would be to hold the locally measured $v$ constant at the bottom of the cylinder; but I don't see any reason why that would *have* to be true.

I don't think "gut instinct" is of any value here. This situation is too different from the kind of situations our gut instinct can cope with. The only way to know would be to actually model it mathematically, and I'm not sure how to do that.

7. Aug 12, 2013

### timmdeeg

Locally measured, the speed of light is c. In another frame of reference, that of "the far away observer" any speed close to the event horizon, including that of the light and that of the rotation of the cylinder slows down.

But how about the Lense-Thirring effect? I am not sure in this case but suspect that the rotating cylinder would cause the black hole to begin to rotate.

Oh, PeterDonis has already answered in the meantime.

Last edited: Aug 12, 2013
8. Aug 12, 2013

### Staff: Mentor

That is only present for objects with nonzero *orbital* angular momentum; it's not present for purely radial motion.

(It also requires the black hole itself to be rotating, but that brings up another issue--see below.)

I think it's even weirder than that. If the cylinder's angular momentum (which is solely due to its spin about its own axis) is large enough to significantly affect spacetime curvature, then the spacetime as a whole could *not* be Schwarzschild spacetime; it would already have to include an extra effect in the metric due to the cylinder's angular momentum.

Put another way, an observer much farther away from the hole than the top of the cylinder would see the "central body" in the spacetime as already rotating--more precisely, as already having nonzero angular momentum. The observer would not be able to tell for sure whether that angular momentum was due to the hole or to the cylinder; the two together would look like one rotating "central body".

9. Aug 12, 2013

### yuiop

OK, I was deliberately vague because I was I not sure myself. I think we can agree that a slow down in angular velocity must be seen by an observer at infinity. In a more simplified scenario, we could imagine lowering a small flat rotating cylinder, which is effectively a clock and it will almost certainly slow down in accord with the gravitational time dilation factor. I an trying to get at the more interesting physical situation of two such flywheel clocks separated by a great height but connected by a rigid rod so that they are forced to spin at the same rate, but of course we have define what we mean by the same rate and who measures it and where it is measured. So, having given it some thought and assuming we all agree that the observer at infinity will see a slow down, I would like to define "nothing happens" as no change in rotational velocity observed by the observer that remains static relative to the gravitational field and remains local to some part of the cylinder as it lowered past him. I think that is the more interesting measurement as it more "physical" than the coordinate calculationss made by the observer at infinity.

In this case there is a situation where v>c is possible, but I hope we can all agree that constant angular velocity measured at infinity will not happen.

Upon reflection, I agree. If it helps any, it seems that the angular momentum per unit mass is constant for a free falling particle as it spirals in towards a black hole, as long as the angular velocity is measured in terms of the proper time of the infalling particle. See http://www.fourmilab.ch/gravitation/orbits/

I also seem to recall that the Lagrangian formulation can be useful in determining conserved quantities for free falling particles in General Relativity, but I am not sure if that is relevant to this situation.

So to conclude, does a local static observer see a change in the angular velocity of the part of the cylinder nearest him as the cylinder is lowered past him? Intuitively I think yes.

I am also guessing that an observer that remains local to the bottom of the cylinder as it lowered, will see the bottom speed up slightly and an observer that remains local to the top of the cylinder will see the the top slow down slightly as the cylinder is lowered.

Last edited: Aug 12, 2013
10. Aug 12, 2013

### yuiop

Please assume for the purposes of this thread, that the black hole is sufficiently massive, that the mass and angular momentum of the cylinder is insignificant and that Schwarzschild spacetime applies with reasonable accuracy.

11. Aug 12, 2013

### WannabeNewton

The problem is you're talking about the rotation of an extended "rigid" body in curved space-time. There are a plethora of mathematical subtelties involved. You can't use your intuitions from rotation of point particle congruences.

12. Aug 12, 2013

### yuiop

OK, let's try a slight variation and see it sheds any insight on the situation. Imagine there are large clocks located at the top and bottom of the rotating cylinder. The top observer sees the lower clock ticking ten times slower than his local clock (and the bottom observer sees the top clock ticking ten times faster than his lower clock). Now strong motors are attached to the top and bottom of the cylinder and run in such a way that that locally the cylinder is forced to spin at 200 rpm as measured by their respective local clocks. The top observer sees the bottom of the cylinder as rotating ten times slower than the top of the cylinder. Will the cylinder be twisted and eventually snap?

Last edited: Aug 12, 2013
13. Aug 12, 2013

### Staff: Mentor

I agree, but in order to justify this I would start with an even simpler situation, an accelerated rotating disk in flat spacetime. Suppose we have a disk that is rotating in the y-z plane in flat spacetime, and is accelerated linearly in the x direction. Since the acceleration is orthogonal to the plane of rotation, it should apply no torque to the disk (we assume the acceleration is applied at the disk's center of mass), so it should not change the angular velocity of the disk, as measured by an observer comoving with the disk. Therefore, an observer at rest in a global inertial frame, who starts out at rest with respect to the disk, will see the disk's angular velocity decrease as it accelerates away from him, according to the time dilation factor due to relative velocity.

Yes, this is an issue, and I would investigate this by considering the simpler case of two rotating disks in flat spacetime which are accelerated such that the proper distance between them, as seen by an observer comoving with either disk, remains constant. That means each disk is following a "Rindler observer" worldline: if both disks are accelerating in the positive x direction (and rotating in the y-z plane), then the disk with the smaller x coordinate will have the larger acceleration, and its "rate of time flow" will be slower, as seen by an observer comoving with either disk, than the "rate of time flow" of the disk with the larger x coordinate.

Let's assume first that the two disks are not physically connected, so they are each free to rotate; and let's also assume that each disk's angular velocity is the same, as seen by an observer comoving with that particular disk (i.e., the disks are identical as seen locally). How will their respective angular velocities compare, as seen by an observer comoving with either disk? It seems evident that the lower observer (the one with the smaller x coordinate) will see the upper disk rotating faster than his own, and the upper observer (the one with the larger x coordinate) will see the lower disk rotating slower than his own.

If this is correct, then if the two disks are connected by a fixed rigid rod, so that they are forced to spin at the same rate, as seen by an observer comoving with either disk, then the rod will experience a torsional stress, since the upper end of the rod will be "trying" to spin faster and the lower end of the rod will be "trying" to spin slower. The magnitude of the stress should depend on the difference in the "rate of time flow" between the disks, which in turn will depend on the proper distance between the disks (which will also be the proper length of the rod).

This is orbital angular momentum, not spin angular momentum; orbital angular momentum is zero for purely radial trajectories, so the fact that it's a constant of the motion doesn't help here. (Also, it's only a constant for free-fall motion, as you note, and we're talking about accelerated motion.)

Let's see how this looks in the flat spacetime case I described above. Suppose we have an observer who, at some instant of his proper time, is co-located with the lower disk, but whose proper acceleration is slightly greater than that of the lower disk, so he is slowly moving "up" the rod from the lower disk to the upper disk. (We'll assume that we can make this observer's velocity relative to the disks constant; we're not really concerned with this observer's exact acceleration or velocity profile.) What will this observer see?

By hypothesis, both disks have the same angular velocity as seen by observers comoving with either disk (and the rod connecting them is experiencing a torsional stress, as above). But that does not mean that an observer comoving with the upper disk will see both disks spinning with the *same* angular velocity as an observer comoving with the lower disk; that can't be, since their rates of time flow differ. The lower disk observer should see both disks (and the rod) spinning with a *larger* angular velocity than the upper disk observer. Therefore, the third observer, who moves with a constant speed from the lower disk to the upper disk, should see the angular velocity of both disks (and the rod) relative to him *decrease*.

There are still some differences between this scenario and the curved spacetime scenario, so I'm not fully convinced that the argument I just gave carries over. See below.

I'm not sure how to get this out of the flat spacetime model above; in that model, the observers comoving with each disk see the angular velocity of the whole assembly as remaining constant (though the lower disk observer sees a larger constant angular velocity than the upper one, as above). But the way I modeled the "lowering" of the assembly isn't quite analogous to the curved spacetime case: in my flat spacetime model, the proper acceleration of each disk is constant, but in the curved spacetime "lowering" case, each disk's proper acceleration would be slowly increasing.

The problem with trying to extend the flat spacetime model to cover a slowly increasing proper acceleration of each disk is that it's not clear what to hold constant. What "remains the same" as the assembly is slowly lowered? The curved spacetime case has the same problem; what are you holding constant that allows you to guess the observations you guessed in the above quote?

14. Aug 12, 2013

### WannabeNewton

Just two questions Peter:

What exactly do you mean by "rate of time flow"?

Could you explain this part? Relative to the observer comoving with the upper disk, the lower disk is rotating slower than his own so when we affix the rod he sees the upper disk rotate slower as it has to match the rotation of the lower disk. To the observer comoving with the lower disk, the upper disk is rotating faster than his own so when we affix the rod he would also see the upper disk rotate slower as the rod is trying to grind down the upper disk so that it corotates with the lower disk. Now why is the final angular velocity of the corotating system as seen by the lower observer necessarily larger (as opposed to lower) than that seen by the upper observer? I guess if you explain what is meant by "rate of time flow" being slower for the lower observer and faster for the upper observer, this will become clearer.

Also, are we imaging that the observers are comoving with respect to their disks while at the center of their respective disks?

Thanks!

15. Aug 12, 2013

### Staff: Mentor

There's not really a standard term for what I mean by this, but the idea is simply that, if two Rindler observers exchange light signals, and time the intervals between them, the "lower" observer will see less elapsed proper time between two successive signals than the "upper" one will. "Rate of time flow" is just the rate at which a given observer's proper time advances between successive signals.

I wasn't talking about how the observed angular velocity changes when you hook up the rod; I was just talking about what final angular velocity will be observed by each observer, given that the angular velocity of every part of the assembly (lower disk, rod, upper disk) must be the same as measured by each observer.

Suppose the two observers are exchanging light signals, and counting the number of rotations of the assembly (by hypothesis, each observer sees the entire assembly rotating with the same angular velocity) between two successive signals. Each observer must see the same number of rotations between two light signals, because the assembly is rotating as a unit; but the lower observer sees less of his own proper time elapse between two successive signals than the upper one does. So the lower observer must see a *larger* angular velocity for the assembly: the same number of rotations in a smaller amount of proper time. In other words, since proper time is in the denominator of angular velocity, angular velocity varies inversely with "rate of time flow" as I defined it above.

Yes.

16. Aug 12, 2013

### WannabeNewton

Gotcha! Thanks Peter.

17. Aug 13, 2013

### yuiop

Nice idea to use acceleration in flat spacetime to try and analyse this. Thanks. I think we are in broad agreement that the static observer will probably see a local slow down of the cylinder as it is lowered past him. I think we also agree that if local observers at the top and bottom of the cylinder see the same local rotational rate then the cylinder will be under significant torsional stress and something has to give. I would now like to introduce a 3rd scenario that might clarify one aspect of the topic.

Scenario 3:
In this variation, the rotating cylinder is held at a constant distance from the black hole. The bottom of the cylinder is driven by a motor that keep the bottom part of the cylinder rotating at a constant 200 rpm as measured locally. The top of the cylinder is free to rotate in very efficient bearings and is "rigidly" connected to the bottom of the cylinder. Assuming the gravitational time dilation factor is ten times greater at the bottom of the cylinder than at the top, will the observer local to the top of the cylinder see the top part rotating at 20 rpm as measured locally and will the top observer also observe the bottom of the cylinder to be rotating at 20 rpm?

18. Aug 13, 2013

### Staff: Mentor

The top observer should certainly see the entire cylinder rotating at the same angular velocity (and so should the bottom observer). I think my previous analysis implies that the ratio of angular velocities seen by the two observers (top and bottom) should be the ratio of their time dilation factors, in which case the answer would, in principle, be yes, the top observer would see the entire cylinder rotating at 20 rpm.

However, there is at least one possible wrinkle: the situation as you state it might be impossible to set up because of the torsional stress in the cylinder; the magnitude of that stress goes up with the cylinder's proper length, and if it gets too large, it violates the energy conditions (roughly speaking, the stress can't be greater than the cylinder's energy density). I haven't tried to calculate whether the conditions as you state them would violate that limit.

Another possible wrinkle is tidal gravity, which is not present in flat spacetime, and might affect the correctness of my flat-spacetime analysis when it is transferred to the curved spacetime case. I don't think radial tidal gravity is an issue, since the effect of that is to change the relationship between proper acceleration and position along the cylinder; but my analysis didn't make any specific assumptions about that relationship, it only assumed that the acceleration was orthogonal to the plane of the cylinder's rotation.

Tangential tidal gravity might be an issue, since there's no analog to that in the flat-spacetime case at all. I *think* its effects will be negligible as long as the *width* of the cylinder (i.e., its size in the plane of its rotation) is very small compared to the mass of the hole (again, in units where G = c = 1). But I haven't done any calculations to make sure.

19. Aug 13, 2013

### Staff: Mentor

Strictly speaking, my flat spacetime analysis doesn't show that. What it shows is that, if we have a cylinder "hovering" at constant altitude, an observer who slowly moves up the cylinder will see its angular velocity, relative to him, decrease.

I understand that you feel a strong intuitive "tug" towards extending that conclusion to cover the case where the observer is "hovering" at constant altitude and the cylinder is slowly being lowered past him. But I'm not sure that intuition is valid, for the reasons I gave at the end of post #13.

20. Aug 15, 2013

### DrGreg

I googled for "Born rigid motion" and discovered the Herglotz-Noether theorem which implies that in special relativity the centre of mass of a Born-rigid rotating object can move only inertially.

So in GR in regions where tidal forces are negligible you couldn't have a rigid rotating object held at a constant altitude, it could only be in free-fall to avoid distortion. Presumably tidal forces could make things worse rather than better.

21. Aug 15, 2013

### Staff: Mentor

Oy--evidently I was right when I said gut instinct isn't a reliable guide here.

I found this paper by Giulini which discusses the theorem towards the end:

http://arxiv.org/abs/0802.4345

The key result is his Theorem 18, which he labels as "Noether & Herglotz, part 1", and which says that any rotational rigid motion, i.e., any rigid motion with nonzero vorticity, must be a Killing motion, i.e., every point within the object must be following an orbit of a Killing vector field. Just to quickly summarize my understanding of what this means:

A rigidly rotating object whose center of mass moves inertially satisfies the condition. The KVF is (in the global inertial frame where the object's CoM is at rest) $\partial_t + \omega \partial_{\phi}$, where $\omega$ is the (constant) angular velocity. This is a KVF since it is a linear combination with constant coefficients of $\partial_t$ and $\partial_{\phi}$, both of which are KVFs.

A linearly accelerating object which is *not* rotating also satisfies the condition. The KVF here is (in a global inertial frame) $cosh ( a \tau ) \partial_t + sinh ( a \tau ) \partial_x$, where $a$ is the (constant) acceleration and $\tau$ is the proper time of the accelerating object. This is not manifestly a KVF in the coordinates given, but one can see that it is by plugging into Killing's equation, or, alternatively, by transforming to Rindler coordinates, where the KVF is just $\partial_T$ ($T$ is the Rindler coordinate time), and the metric is independent of $T$ so $\partial_T$ is manifestly a KVF. (I've often seen this KVF called the "boost" KVF.)

However, a linearly accelerating object which is also rotating can't satisfy the condition, because the worldlines of each of its points would be orbits of a vector field which, at best, would be a linear combination of KVFs (some mixture of the "boost" KVF with $\partial_{\phi}$) with *non* constant coefficients. Such a vector field is not a KVF.

I'm still cogitating over how to visualize this physically, but one thing strikes me: another implication of the theorem is the observation (originally due to Ehrenfest, as the paper notes) that a disk (whose CoM is here assumed to be moving inertially) cannot be rigidly *spun up* from non-rotating to rotating motion. (Mathematically, this is because the disk's points, again, can't be following orbits of a KVF, since at best the worldlines would be orbits of a linear combination of $\partial_t$ and $\partial_{\phi}$ with non-constant coefficients.) The physical manifestation of this is that the spatial geometry of the rotating disk is flat (i.e., Euclidean), while the "spatial geometry" (there are caveats to this usage, which is why I put it in quotes) of the rotating disk is curved (i.e., non-Euclidean).

It might be that linearly accelerating a rotating object would act on it, physically, much as "spinning up" a non-rotating object does: it would be like constantly trying to induce a "change in geometry" in the object. Another way of looking at this might be to consider the reason why the "spatial geometry" of the rotating disk has to be carefully defined: there is no global notion of simultaneity shared by all points on the rotating disk. There can't be, since they are all in relative motion. A linearly accelerated object that is not rotating *can* have a globally shared notion of simultaneity (the one embodied in Rindler coordinates); but points at different "heights" along the object will experience different rates of time flow (i.e., different rates of change of their proper time relative to Rindler coordinate time, which defines the common simultaneity convention). The theorem seems to be saying, basically, that you can't mix these two things (lack of global simultaneity and differences in the rate of time flow due to acceleration) and still have a rigid motion.

Sorry for the lengthy post, I wanted to go ahead and get down my thoughts after a quick read-through of the paper.

22. Aug 15, 2013

### WannabeNewton

So basically, for our purposes here, we cannot take a previously Born rigidly rotating object in SR and then uniformly accelerate it in some direction while maintaining Born rigidity of the rotation? And this is because as per the theorem cited, any motion that is a Born rigid rotation must be a killing motion, whereas supposedly a motion combining a previous Born rigid rotation and uniform acceleration could not be a killing motion hence won't result in another Born rigid rotation?

Also, since $\mathcal{L}_{u}h_{ab} = 2\sigma_{ab} + \frac{2}{3}\theta h_{ab}$, and any Born rigid motion is defined as $\mathcal{L}_{u}h_{ab} = 0$, this implies the motion is Born rigid iff $\sigma_{ab} = \theta = 0$ where $\sigma_{ab},\theta$ are the shear and expansion of the congruence respectively. Is there a way to visualize the failure of a motion combining both Born rigid rotation and uniform acceleration to itself be a Born rigid rotation in terms of the presence of shear and/or expansion (in SR that is)?

Translating this over to GR, if we consider any extended object in Schwarzschild space-time formed by a fluid with non-vanishing vorticity such that each fluid particle remains at a constant altitude (so that the entire object is at constant altitude), the fluid would have to have a non-vanishing expansion and/or shear so as to fail to have Born rigid rotation?

Last edited: Aug 15, 2013
23. Aug 15, 2013

### Staff: Mentor

More precisely, *something* about the motion must fail to be Born rigid. The theorem, as I read it, doesn't specify that the rotation is what has to fail to be rigid; it could be that the linear acceleration fails to be rigid. Or it could be a combination of both.

I think the "supposedly" is pretty easy to see, although I have not worked through it in detail. Try writing the frame field describing the accelerated, rotating object; I think you'll find that there's no way to write it as a linear combination of Killing vectors with constant coefficients.

Yes, that's the best invariant way to describe Born rigid motion, and it's the one the paper uses.

This is what I'm still cogitating over; but your way of describing it in terms of a nonzero shear or expansion having to appear somewhere sounds like a good way to look at it. It might be that the combination of linear acceleration plus rotation forces either the linear acceleration to develop nonzero expansion or the rotation to develop nonzero shear (or some combination thereof). But I'm still cogitating.

Yes, that's what the theorem appears to me to imply.

24. Aug 15, 2013

### yuiop

Suppose we have a flywheel that remains on the accelerating worldline at x=0.4 in the Rindler chart below and another that remains at x=1.0 and we adjust the rotation speeds of these flywheels so that all the accelerating observers agree they are rotating at the same speed as each (although they will disagree on what that speed is). Once adjusted and assuming perfect bearings there is no reason for the flywheels to slow down or speed up as long as they held at constant altitude. I think you will agree that the dimensions or mutual orientations of the flywheels as measured by the observers on the flywheels will remain constant for all time. Does that not satisfy the requirements of rigidity? Now if we connect the two flywheels by a rod, we have an approximation of the cylinder in the OP when initially stationary in a gravitational field.

This seems to contradict the conclusion that we cannot have a "rigid rotating object held at a constant altitude". However, I guess it depends on whether by Born rigid rotation we really mean Born rigid angular acceleration and exclude the trivial case of constant angular velocity.
We note that in real life that objects can be spun up from non rotating to rotating motion even with the complication of having a gravitational field present. However, I agree that that Born rigidity is not maintained during the spin up, but that is not important to the question in the OP because I was referring to a slowly rotating cylinder of say 200 rpm where Special relativistic effects are negligible because I wanted to focus on the GR effects. This lack of Born rigidity does not make the scenario physically impossible, but just more difficult to analyse mathematically. I know I mentioned the term rigid on several occasions but I meant in the everyday sense of fairly rigid and not infinitely rigid or strictly Born rigid. The lack of Born rigidity means that stresses will be set up in the cylinder and as long as the stresses do not increase without bound to destruction over time, I do not see this as a significant problem. It is worth noting that a thin ring can be spun be spun up while maintaining a form of Born rigidity (i.e constant proper circumference) as long as the radius is allowed to shrink as the ring speeds up. One simplification of the problem would be to consider a thin walled tube (essentially a stack of rings) instead of a solid cylinder.

When considering just the circumference it is possible to have a notion of simultaneity on the spinning ring using a reference clock at the rotational centre, instead of using classic Einstein clock synchronisation method. The points on the disc are not in relative motion from the point of view of an observer at rest on the disc.

25. Aug 15, 2013

### yuiop

I think the Special Relativistic effects at 200 rpm are negligable compared to the effects of gravitational time dilation factors of ten or more, so I would like to put aside the Born rigidity issues for a moment (although I find them interesting). To further simply things I would like to consider a stack of discs on top of each other on perfect bearings that are allowed to rotate independently. They are set spinning in such a way that they all appear to spin at 20 rpm as measured by the top observer (A) and so appear to rotate as a single cylinder. A second observer (B) at the bottom of the stack measures the discs to be rotating at 200 rpm and this observer remains at constant altitude. The stack is slowly lowered past the bottom observer and each disc that goes past him appears to rotating slower than the previous one and when the top disc is alongside him he measures that disc to be rotating at 20 rpm. The lowest disc way below him now appears to be rotating at say 1 rpm (this is just a rough figure and the additional complication of gravitational length contraction would have to be taken into account to figure out just how far the cylinder extends downwards). I think it would be safe to assume that if a clutch mechanism was now engaged so that independent discs rotated as a single cylinder, the final averaged out rotation rate of the cylinder would be less than 20 rpm and greater than 1 rpm as measured by observer B. So lets say this final figure is 10 rpm. With the discs still locked together, this observer should see the nearest part of the effective cylinder increase from 10 rpm to 200 rpm as the cylinder is raised back up to its original position. Now factor in all the fine details.