Locally non-rotating observers

[WannabeNewton]

Hey guys. I have a question about two (possibly ostensibly) different definitions of a locally non-rotating observer that I have come across in my texts.

The first is specifically for stationary, axisymmetric space-times in which we have a canonical global time function $t$ associated with the time-like killing vector field. We define a locally non-rotating observer to be one who follows an orbit of $\nabla^{a}t$ i.e. his 4-velocity is given by $\xi^{a} = (-\nabla^{b}t\nabla_{b}t)^{-1/2}\nabla^{a}t = \alpha \nabla^{a}t$. Such an observer can be deemed as locally non-rotating because his angular momentum $L = \xi^{a}\psi_{a} = \alpha g_{\mu\nu}g^{\mu\gamma}\nabla_{\gamma}t\delta^{\nu}_{\varphi} = \alpha \delta^{\gamma}_{\nu}\delta^{t}_{\gamma}\delta^{\nu}_{\varphi} = \alpha \delta^{t}_{\varphi} = 0$ where $\psi^{a}$ is the axial killing vector field; these observers are also called ZAMOs for this reason. It might also be worth noting that the time-like congruence defined by the family of ZAMOs has vanishing twist $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = \alpha\epsilon^{ab[cd]}\nabla_{b}t\nabla_{(c}\nabla_{d)}t + \alpha^{3}\epsilon^{a[b|c|d]}\nabla_{(b}t \nabla_{d)}t\nabla^{e}t\nabla_{c}\nabla_{e}t = 0$.The second definition I have seen is the much more general notion of Fermi-Walker transport. That is, if we choose an initial Lorentz frame $\{\xi^{a}, u^{a},v^{a},w^{a}\}$ and the spatial basis vectors evolve according to $\xi^{b}\nabla_{b}u^{a} = \xi^{a}u_{b}a^{b}$, $\xi^{b}\nabla_{b}v^{a} = \xi^{a}v_{b}a^{b}$, and $\xi^{b}\nabla_{b}w^{a} = \xi^{a}w_{b}a^{b}$, where $a^{b} = \xi^{a}\nabla_{a}\xi^{b}$ is the 4-acceleration, then the observer is said to be locally non-rotating.

My question is, to what extent are these two definitions equivalent (both mathematically and physically) whenever they can both be applied? In other words, in what sense is the qualifier 'rotation' being used in each case?

Let me elucidate my question a little bit. I know that for asymptotically flat axisymmetric space-times, there must exist a fixed rotation axis on which $\psi^{a}$ vanishes. Then the first definition tells us that the ZAMOs have no orbital rotation about this fixed rotation axis (for example no orbital rotation about a Kerr black hole); because these observers are at rest with respect to the $t = \text{const.}$ hypersurfaces, these observers are as close to stationary hovering observers as we can get in a space-time with a rotating source. We know however that such observers have an instrinsic angular velocity $\omega$; if we imagine a ZAMO holding a small sphere with frictionless prongs sticking out with beads through the prongs then a ZAMO should be able to notice his intrinsic angular velocity $\omega$ by seeing that the beads are thrown outwards along the prongs at any given instant. Is this correct?

Now, on the other hand, the second definition of local non-rotation (using Fermi-Walker transport) seems to be saying sort of the opposite. That is, it seems to be telling us which observers can carry such spheres and never see the beads get thrown outwards i.e. the orientation of the sphere will remain constant (the orientation will be represented by the spatial basis vectors of a Lorentz frame) so these are the observers who have no intrinsic angular velocity i.e. no self-rotation. This is why the two definitions confused me because they seem to be talking about two totally different kinds of non-rotation.

Thank you in advance.

 

[atyy]

I'm not sure if this is relevant, but Malament states that there is a unique concept of local rotation in GR, but not of orbital rotation http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf (p219, section 3.2)

 

[WannabeNewton]

He seems to be saying the problem is for extended bodies (e.g. a one dimensional ring) as opposed to particles/observers (which is what I was talking about above). However his definition of local non-rotation has managed to confuse me even more because he seems to define it only for a time-like congruence (a family of non-intersecting observers) with 4-velocity field $\xi^{a}$ and says that it is non-rotating if the twist $\omega^{a} = \epsilon_{abcd}\xi^{b}\nabla^{c}\xi^{d} = 0$ which is equivalent to saying that $\xi^{a}$ is hypersurface orthogonal i.e. $\xi_{[a}\nabla_{b}\xi_{c]} = 0$ meaning that the worldlines of individual observers in the congruence don't twist about one another. So this is a statement about rotation of neighboring worldlines in the congruence relative to a given reference worldline in the congruence; I can't immediately see if this fits in at all physically with either of the two definitions above.

Perhaps I have misunderstood what it means for the locally non-rotating observers as per the first definition above (the ZAMOS) to have zero angular momentum (i.e. that maybe it is not related to orbital rotation) but all the texts I've come across make it seem like the vanishing angular momentum means that the ZAMOs have no orbital rotation about some axis, that they manage to hover in place while having an induced intrinsic angular velocity that causes them to rotate in the sense that it can be detected using test gyroscopes held by the ZAMOs. The Fermi-Walker definition on the other hand seems to be talking about observers who can't detect any rotation in the local test gyroscope sense (so no intrinsic angular velocity).

 

[Mentz114]

Hey guys. I have a question about two (possibly ostensibly) different definitions of a locally non-rotating observer that I have come across in my texts.
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Now, on the other hand, the second definition of local non-rotation (using Fermi-Walker transport) seems to be saying sort of the opposite. That is, it seems to be telling us which observers can carry such spheres and never see the beads get thrown outwards i.e. the orientation of the sphere will remain constant (the orientation will be represented by the spatial basis vectors of a Lorentz frame) so these are the observers who have no intrinsic angular velocity i.e. no self-rotation. This is why the two definitions confused me because they seem to be talking about two totally different kinds of non-rotation.

My understanding is that the first case, the twist of the congruence ( AKA vorticity) is a statement about nearby trajectories. If the shear and expansion are zero, but twist is non-zero, the rotation is rigid. The trajectories do not overtake or lag behind each other. ( Stephani, page 178).

The second case seems to be something that happens to the vector space being transported, and your suggestion that a 'spin detector' would register seems rerasonable,

I have to say, I've always found the first kind rotation a bit puzzling myself. Still do, in fact.

[Edit] We posted at the same time, so now I'll read the above.

 

[PeterDonis]

Perhaps I have misunderstood what it means for the locally non-rotating observers as per the first definition above (the ZAMOS) to have zero angular momentum (i.e. that maybe it is not related to orbital rotation)

Correct; it isn't. In a spacetime like Kerr spacetime, which is stationary and axisymmetric but not static, ZAMOs have zero angular momentum but nonzero angular velocity.

but all the texts I've come across make it seem like the vanishing angular momentum means that the ZAMOs have no orbital rotation about some axis, that they manage to hover in place

Can you give some examples? I've never seen anything in a textbook that conveyed this to me.

Consider: if the spacetime is stationary but not static, then the timelike KVF is not hypersurface orthogonal; that means the orbits of the timelike KVF are different from the orbits of observers who are orthogonal to the surfaces of constant time.

The latter observers are the ZAMOs; I think it's fairly straightforward to show that zero angular momentum is equivalent to having a 4-velocity that's orthogonal to surfaces of constant time. (In fact, your OP more or less does so.)

But the *former* observers, the ones whose worldlines are orbits of the timelike KVF, are the ones who are "hovering in place", since the orbits of the timelike KVF define what it means to "hover in place". So the ZAMOs in a stationary but nonstatic spacetime cannot be hovering in place.

 

[PeterDonis]

If the shear and expansion are zero, but twist is non-zero, the rotation is rigid. The trajectories do not overtake or lag behind each other. ( Stephani, page 178).

Just as a note, since I brought up Kerr spacetime: the ZAMO congruence in Kerr spacetime has nonzero shear, because the angular velocity that corresponds to zero angular momentum varies with the radial coordinate. So when trying to use that congruence to think about twist, you have to be sure to factor out the nonzero shear to avoid conflating the two effects.

 

[WannabeNewton]

I'll compile the references I read them in from which I interpreted things in the above manner but in the meantime, what is confusing me is the nature of the induced angular velocity of the ZAMOs. I thought the induced angular velocity was an intrinsic one (a self-rotation kind of thing), not one that involved rotation about some other body/axis. This article seems to imply that: http://www34.homepage.villanova.edu/robert.jantzen/research/articles/mg9-2002-GEMrot.pdf (2nd paragraph of the first page).

I also found a passage in this text: http://books.google.com/books?id=n0...epage&q=locally nonrotating observers&f=false but I'm having a hard time understanding what kind of rotation they are referring to when they say "Now we introduce into the external space of a rotating black hole another reference frame which does not rotate in the sense given above" because they talk about things being rotated around a black hole as well as local rotation of frames (in the gyroscope sense) relative to a Fermi-Walker transported frame (which they call local Lorentz frames).

 

[PeterDonis]

My question is, to what extent are these two definitions equivalent (both mathematically and physically) whenever they can both be applied? In other words, in what sense is the qualifier 'rotation' being used in each case?

As I understand it, the first definition depends on the second, because the definitions of the elements of the kinematic decomposition--expansion, shear, and twist (or vorticity)--implicitly assume that the spatial basis vectors of all observers are Fermi-Walker transported along their worldlines; the physical interpretation of the meaning of expansion, shear, and twist depends on the Fermi-Walker transport assumption.

Put another way, if we consider a particular timelike congruence, then an observer following any particular worldline in the congruence has, in general, two choices for how to maintain the orientation of his spatial basis vectors: he can choose to Fermi-Walker transport them, regardless of how that might change their orientation relative to neighboring observers in the same congruence; or he can choose to keep them oriented in a certain way relative to neighboring observers in the congruence, regardless of how that might change their orientation relative to Fermi-Walker transport.

For example, consider two congruences in Kerr spacetime: the ZAMO congruence, which you defined in your OP, and the "hovering" congruence, which I implicitly defined in a previous post: it is the congruence of integral curves of the timelike KVF. Both congruences have zero expansion, so we can leave that out of the discussion. The ZAMO congruence has zero twist but nonzero shear; whereas the hovering congruence has zero shear but nonzero twist.

Now, for each congruence, consider a formation of five observers: a "fiducial" observer F, at the "center" of the formation; L, the "leading" observer, who is at the same $r$ as F but slightly greater $\phi$; T, the "trailing" observer, at the same $r$ as F but slightly smaller $\phi$; I, the "inner" observer, at the same $\phi$ as F but slightly smaller $r$; and O, the "outer" observer, at the same $\phi$ as F but slightly larger $r$. At some particular event on F's worldline, he orients his spatial basis vectors so that one points directly at L, one points directly at T, one points directly at I, and one points directly at O.

Suppose that F uses gyroscopes to maintain the orientation of his spatial basis vectors; this amounts to Fermi-Walker transporting them. How will his basis vectors behave, relative to the directions he observes L, T, I, and O to be, as all the observers continue along their worldlines, for each congruence given (ZAMO and hovering)?

First, a note: neither of these congruences is a geodesic congruence, so all of these observers will have nonzero proper acceleration. That means we need to be careful to factor out any effects of proper acceleration in both cases.

Take the hovering congruence first since its behavior is easier to see because the angular velocity of all the observers is zero. In this case, F's basis vectors will appear to rotate in the opposite sense to the rotation of the Kerr black hole, relative to the directions of L, T, I, and O (nonzero twist), but the rate of rotation will be the same in all directions (zero shear). Because of the zero angular velocity, rotation relative to L, T, I, and O is equivalent to rotation relative to infinity (or, if you like, the "distant stars"), so F's basis vectors will appear to rotate relative to those as well (whereas L, T, I, and O will stay in fixed positions, as seen by F, against the background of the distant stars).

Now for the ZAMO congruence. In this case, F's basis vectors will *not* appear to rotate relative to the directions of L and T (zero twist). However, F's basis vector pointing inward (originally towards I) will gradually fall behind I, while F's basis vector pointing outward (originally towards O) will gradually move ahead of O (nonzero shear). I don't call this latter behavior "rotation" because it can be attributed entirely to the difference in angular velocity between F, I, and O, due to them all being at slightly different radial coordinates; if that difference is factored out, F's basis vectors do not rotate relative to I and O either (zero twist).

What about the distant stars in the ZAMO case? Well, since the ZAMO congruence has nonzero angular velocity, obviously F's basis vectors will rotate relative to the distant stars, and the rate of their rotation will be equal to F's angular velocity about the hole. But L and T will also be moving against the background of the distant stars, as seen by F, so they stay with F's basis vectors pointing in those directions; and I and O will also be moving against the background of the distant stars, as seen by F, though slightly differently than L and T because of the difference in angular velocity.

 

[WannabeNewton]

Basically, my question comes down to: if the ZAMOs have an angular velocity $\omega$ in the sense that they rotate about some fixed rotation axis at a constant $r$ and $\theta$ then in what sense are they locally non-rotating? I ask because the Fermi-Walker definition of locally non-rotating seems to be talking about the kind of non-rotation that I was picturing in the sense of the sphere apparatus I talked about before; so Fermi-Walker transport tells us which frames are locally non-rotating in a sense analogous to a non-rotating frame in Newtonian mechanics i.e. that the spatial basis vectors of the frame have a fixed orientation. But I have no clue in what sense the first definition (the one regarding obervers following orbits of $\nabla^{a}t$) defines a local non-rotation.

EDIT: I replied right when you replied Peter so I apologize, let me now read your latest reply.

 

[TrickyDicky]

He seems to be saying the problem is for extended bodies (e.g. a one dimensional ring) as opposed to particles/observers (which is what I was talking about above).

My understanding is that observers are by definition extended (contrary to what the name might suggest) so not much to do with (point)particles. OTOH I'm not sure in what sense can one determine whether a particle is rotating or non-rotating if not with reference to something, in this sense many "non-rotation" definitions will disagree to some extent depending on the set up.

 

[PeterDonis]

I thought the induced angular velocity was an intrinsic one (a self-rotation kind of thing), not one that involved rotation about some other body/axis. This article seems to imply that: http://www34.homepage.villanova.edu/robert.jantzen/research/articles/mg9-2002-GEMrot.pdf (2nd paragraph of the first page).

Yes, they appear to be using "intrinsic rotation" to mean "rotation about the black hole", which is a very confusing way to put it.

I also found a passage in this text: http://books.google.com/books?id=n0...epage&q=locally nonrotating observers&f=false but I'm having a hard time understanding what kind of rotation they are referring to when they say "Now we introduce into the external space of a rotating black hole another reference frame which does not rotate in the sense given above" because they talk about things being rotated around a black hole as well as local rotation of frames (in the gyroscope sense) relative to a Fermi-Walker transported frame (which they call local Lorentz frames).

Yes, again they are using words in a very confusing way, but it's evident, since they say the "locally non-rotating" congruence is everywhere orthogonal to the surfaces of constant time, that they are talking about the ZAMO congruence. IMO this is a good illustration of why not to think in terms of "frames" and instead to focus on the physical invariants, like orthogonality (or lack thereof).

 

[PeterDonis]

we need to be careful to factor out any effects of proper acceleration in both cases.

Actually, we can do more than that: we can use the spatial direction of the proper acceleration as another check on the behavior of F's spatial vectors. In the case of the hovering congruence, F's spatial basis vectors will rotate with respect to the direction of his proper acceleration. In the case of the ZAMO congruence, they will not. (This, btw, also gives us a way to show that the apparent change in the directions of I and O, relative to F's spatial basis vectors, is entirely due to their different angular velocities.)

 

[WannabeNewton]

Now, for each congruence, consider a formation of five observers: a "fiducial" observer F, at the "center" of the formation; L, the "leading" observer, who is at the same $r$ as F but slightly greater $\phi$; T, the "trailing" observer, at the same $r$ as F but slightly smaller $\phi$; I, the "inner" observer, at the same $\phi$ as F but slightly smaller $r$; and O, the "outer" observer, at the same $\phi$ as F but slightly larger $r$. At some particular event on F's worldline, he orients his spatial basis vectors so that one points directly at L, one points directly at T, one points directly at I, and one points directly at O.

Ok so I'm picturing a diamond like shape with F at the center of the diamond and the other 4 observers at the endpoints of the diamond. What I don't get is, if he only has 3 spatial basis vectors at his disposal then how can he have each one pointing to a different observer in the group of 4? Usually what I see in such scenarios are connecting vectors attached to the 4 observers from F.

Suppose that F uses gyroscopes to maintain the orientation of his spatial basis vectors; this amounts to Fermi-Walker transporting them. How will his basis vectors behave, relative to the directions he observes L, T, I, and O to be, as all the observers continue along their worldlines, for each congruence given (ZAMO and hovering)?

So just to clarify, is the local non-rotation as per the first definition literally equivalent to the statement that the congruence defined by the ZAMOs has zero twist i.e. that at any given instant those 4 observers are not rotating around F himself, relative to F? But even though the individual observers in the congruence don't rotate about each other, they still rotate about some fixed rotation axis with some angular velocity, relative to an observer at infinity (or distant fixed stars), that depends on $r$ and $\theta$? What does it physically mean then for the ZAMOs to have vanishing angular momentum?

Take the hovering congruence first since its behavior is easier to see because the angular velocity of all the observers is zero. In this case, F's basis vectors will appear to rotate in the opposite sense to the rotation of the Kerr black hole, relative to the directions of L, T, I, and O (nonzero twist), but the rate of rotation will be the same in all directions (zero shear). Because of the zero angular velocity, rotation relative to L, T, I, and O is equivalent to rotation relative to infinity (or, if you like, the "distant stars"), so F's basis vectors will appear to rotate relative to those as well (whereas L, T, I, and O will stay in fixed positions, as seen by F, against the background of the distant stars).

So in this case can I imagine the endpoints of the diamond as rotating about F but they maintain their distances from F (rigid rotation)? I just don't get the part about rotation relative to the distant stars (or an observer at infinity) because I don't get how F's basis vectors can rotate relative to them if this observer has zero space-time induced angular velocity i.e. he is not rotating around the black hole.

What about the distant stars in the ZAMO case? Well, since the ZAMO congruence has nonzero angular velocity, obviously F's basis vectors will rotate relative to the distant stars, and the rate of their rotation will be equal to F's angular velocity about the hole. But L and T will also be moving against the background of the distant stars, as seen by F, so they stay with F's basis vectors pointing in those directions; and I and O will also be moving against the background of the distant stars, as seen by F, though slightly differently than L and T because of the difference in angular velocity.

If I was F, why would I would see L and T not stay beside me in a constant manner while I see the distant stars rotate around me, at a rate equal to the angular velocity of F around the black hole, as opposed to seeing L and T moving as well? Shouldn't I only see I and O moving (O lagging behind and I going forward) while the stars rotate around me?

Finally, I guess my trouble is in picturing the local non-rotation defined by Fermi-Walker transport because I imagined it as non-rotation in the usual sense (like if I'm standing on a skating rink but not spinning around in place) but it seems to be that vanishing twist of the 4 observers around F means that F is not spinning around in the skating rink sense, relative to an observer at infinity so I'm confused by that.

Thanks Peter.

 

[PeterDonis]

Ok so I'm picturing a diamond like shape with F at the center of the diamond and the other 4 observers at the endpoints of the diamond.

Yes.

What I don't get is, if he only has 3 spatial basis vectors at his disposal then how can he have each one pointing to a different observer in the group of 4?

The term "basis vectors" is a bit sloppy here, yes. Actually I'm only using two of the 3 basis vectors, plus the same 2 with a sign flip. To use all 3 basis vectors, you would add two more nearby observers, U ("up") and D ("down"). I'll leave the actual behavior of F's 3rd basis vector (and its sign-flipped counterpart) relative to U and D in the two cases as an exercise for the reader.

So just to clarify, is the local non-rotation as per the first definition literally equivalent to the statement that the congruence defined by the ZAMOs has zero twist i.e. that at any given instant those 4 observers are not rotating around F himself, relative to F?

As far as I can tell, yes.

But even though the individual observers in the [ZAMO] congruence don't rotate about each other, they still rotate about some fixed rotation axis with some angular velocity, relative to an observer at infinity (or distant fixed stars), that depends on $r$ and $\theta$?

Yes. I was only considering the $r$ dependence here (I implicitly assumed that all the observers were in the equatorial plane, which is one reason why I didn't add the U and D observers).

So in this case [the hovering congruence] can I imagine the endpoints of the diamond as rotating about F but they maintain their distances from F (rigid rotation)?

Yes; since there is zero shear and zero expansion, the congruence is rigid, as Mentz114 said.

I just don't get the part about rotation relative to the distant stars (or an observer at infinity) because I don't get how F's basis vectors can rotate relative to them if this observer has zero space-time induced angular velocity i.e. he is not rotating around the black hole.

Because the hole itself is rotating. The hovering congruence around a Schwarzschild hole has zero twist, so F's basis vectors would not rotate at all in that case.

If I was F [in the ZAMO congruence], why would I would see L and T not stay beside me in a constant manner

You do, if you mean relative to F's basis vectors. F's basis vectors rotate relative to the distant stars, and so do the observed directions of L and T, so L and T stay exactly in step with F's basis vectors.

 

[WannabeNewton]

Ok I'm just having a bit of trouble picturing the non-rotation/rotation as defined by twist, the rotation due to the space-time induced angular velocity, and the non-rotation as defined by Fermi-Walker transport.

For the ZAMO congruence, if I was F then I wouldn't see any of my neighboring ZAMOs rotate around me (zero twist), I would see the ZAMOs who are nearby on the same circle as me stay exactly in place relative to me, see the ZAMOs directly in front of and behind me fall behind and go forward due to shear (but not rotate around me), and finally I will see the distant stars rotate around me at a rate equal to my angular velocity around the black hole (by the way I was talking about the angular velocity as measured at infinity). For the hovering congruence, shouldn't I see my neighboring observers rotate around me due to non-vanishing twist (but maintain constant distance due to vanishing shear) but see the distant stars fixed relative to me (due to vanishing angular velocity around the black hole relative to infinity)?

Finally, how would I picture Fermi-Walker transport with regards to non-rotation? Because it uses gyroscopes and is an absolute measure of non-rotation (and not relative to anything else unlike the above) I don't really know how to picture it in the above sense.

Thanks Peter.

 

[PeterDonis]

For the ZAMO congruence, if I was F then I wouldn't see any of my neighboring ZAMOs rotate around me (zero twist), I would see the ZAMOs who are nearby on the same circle as me stay exactly in place relative to me, see the ZAMOs directly in front of and behind me fall behind and go forward due to shear (but not rotate around me), and finally I will see the distant stars rotate around me at a rate equal to my angular velocity around the black hole (by the way I was talking about the angular velocity as measured at infinity).

All ok, and I agree on the definition of angular velocity, that's the one I was using too.

For the hovering congruence, shouldn't I see my neighboring observers rotate around me due to non-vanishing twist (but maintain constant distance due to vanishing shear)

Relative to your basis vectors, yes. *Not* relative to the distant stars. See below.

but see the distant stars fixed relative to me (due to vanishing angular velocity around the black hole relative to infinity)?

Not relative to your basis vectors. Remember the whole point of the example is that "nonzero angular velocity" and "no rotation of basis vectors relative to infinity" are uncoupled, because the spacetime is stationary but not static, so the orbits of the timelike KVF (which defines what "nonzero angular velocity" means) are not orthogonal to the surfaces of constant time (which defines what "no rotation of basis vectors relative to infinity" means). Yes, this is highly counterintuitive, because our intuitions are Newtonian, and there is no Newtonian analogue to the non-orthogonality that appears here.

Finally, how would I picture Fermi-Walker transport with regards to non-rotation? Because it uses gyroscopes and is an absolute measure of non-rotation (and not relative to anything else unlike the above) I don't really know how to picture it in the above sense.

For visualization I more or less work it backwards--I first visualize things globally, relative to infinity, and then make adjustments to come up with how things will look locally.

 

[WannabeNewton]

Wait so if I was F (in the hovering congruence) then would I see the other 4 observers rotate relative to my basis vectors in such a way so that they are fixed against the background stars (so that if one of the 4 observers happened to coincide with a fixed star at a given instant then at the next instant he would still coincide with it after having rotated around me)? I just don't get how F can be hovering in the sense that he stays put in one place if he sees the stars rotating around him; wouldn't that mean he must be rotating around the black hole and not in one place (if viewed from infinity)? And where exactly does the statement that F's neighbors are fixed relative to the background stars as they all rotate about F come from?

The fact that F, in the ZAMO congruence, sees his neighbors not rotate around him but sees the distant stars at infinity rotate around him (meaning F has non-vanishing angular velocity relative to infinity but none of F's neighbors have any angular velocity relative to F?) but that F, in the hovering congruence, sees both his neighbors and the stars at infinity rotate around him confuses me because I don't see in what sense F is hovering.

Thanks again for your time Peter.

 

[PeterDonis]

Wait so if I was F (in the hovering congruence) then would I see the other 4 observers rotate relative to my basis vectors in such a way so that they are fixed against the background stars (so that if one of the 4 observers happened to coincide with a fixed star at a given instant then at the next instant he would still coincide with it after having rotated around me)?

Yes. That should be obvious from the fact that all of the observers in the congruence are following orbits of the timelike KVF, which means their spatial locations relative to infinity are fixed.

I just don't get how F can be hovering in the sense that he stays put in one place if he sees the stars rotating around him; wouldn't that mean he must be rotating around the black hole and not in one place (if viewed from infinity)?

In a static spacetime, yes. But this spacetime is not static. Once again, I realize this is highly counterintuitive, but that's part of what "stationary but not static" *means*.

And where exactly does the statement that F's neighbors are fixed relative to the background stars as they all rotate about F come from?

From the fact that they are following orbits of the timelike KVF. See above.

I don't see in what sense F is hovering.

He's following an orbit of the timelike KVF. Once again, the key is that orbits of the timelike KVF are not orthogonal to surfaces of constant time. Take a step back and think carefully about what that means. Your intuitions are set up to assume that it can't even happen--that surfaces of constant time are always orthogonal to your 4-velocity.

In fact, we can sharpen that even more by asking: what do the surfaces of constant time look like in F's local inertial frame, in the hovering case? It should be obvious, from the non-orthogonality, that F's surfaces of constant time, in his LIF, are *not* parallel to the global surfaces of constant coordinate time! Now: which way do F's surfaces of constant time "tilt", relative to the global surfaces of constant time? (For a ZAMO, they don't tilt at all; they *are* parallel, even though the ZAMO has non-zero angular velocity. You might want to think about that too.)

 

[WannabeNewton]

Ok I think I can understand it better if I can see the physical difference between the angular momentum $L$ and the angular velocity $\omega$ relative to infinity in a non-static but stationary space-time. For nonstatic but stationary space-times, static observers and ZAMOs don't coincide so if I am static relative to infinity then I will still have a component of 4-velocity along $\psi^{a}$ so my angular momentum won't vanish but I can't see this as being the reason for why I would see the distant stars rotate around me because the ZAMOs have vanishing angular momentum but they also see the distant stars rotate around them (due to their induced angular velocity around the black hole relative to infinity).

 

[PeterDonis]

if I am static relative to infinity then I will still have a component of 4-velocity along $\psi^{a}$

If I'm understanding your notation correctly, this is wrong. The 4-velocity of a static observer is proportional to the timelike KVF; it does not have a component in the direction of the axial KVF, which is what I think you are using $\psi^{a}$ to denote.

so my angular momentum won't vanish

Correct, it won't (despite the fact about the 4-velocity that I just gave), and nonvanishing angular momentum is a better thing to look at if you're trying to understand why the hovering observer's basis vectors rotate with respect to infinity. Question: what is the *sign* of the hovering observer's angular momentum? Is is positive or negative? And how does that relate to the sign of the twist of hovering congruence?

 

[WannabeNewton]

I think what I meant to say was that $\xi^{a}$ and $\psi^{a}$ are not orthogonal due to the presence of the cross terms in the coordinate chart adapted to the KVFs (if they were orthogonal then the angular momentum would vanish because $L = u^{a}\psi_{a} = \alpha \xi^{a}\psi_{a}$). And $L = \alpha g_{t\varphi} = \alpha (\frac{4mcra\sin^2{\theta}}{\rho^{2}})$ where $\rho^{2} = r^{2} + a^{2}\cos^{2}{\theta}$. So the angular momentum should have the same sign as $a$ (the rotation of the black hole). I'm not sure what the sign of the twist is since it looks rather painful to compute but is there a relation between the sign of the twist to the sign of the angular momentum? I would think that the neighboring static observers should rotate around me in the same direction as my angular momentum if they are to remain fixed relative to the distant stars but I'm not really sure what *kind* of rotation angular momentum represents here. Is it also a kind of rotation around the black hole?

 

[WannabeNewton]

It might help if I knew what the observers at infinity were seeing of a single ZAMO and a single static observer (ignoring twist for now). Would I just see a ZAMO as orbiting the rotating black hole with a given angular velocity $\omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ (or rotating star or what have you)? And would I see a static observer in front of the rotating black hole/star as literally staying in one place, much like a static observer in Schwarzschild space-time just sits in one place in front of a star? This latter observation is what I'm not sure about. If the static observer has a non-vanishing angular momentum (although I'm still unsure about how this angular momentum will look visually from infinity) then how can he be sitting idly by in front of the rotating black hole/star as seen by an observer at infinity?

Wald says that "The closest analog to a family of static observers outside the black hole in the charged Kerr geometry are the 'locally nonrotating observers' " so I don't really get in what intuitive/physical sense the observers who follow an orbit of $\xi^{a}$ in kerr space-time are actually static if Wald says the closest thing we have to static observers are the observers who follow orbits of $\nabla^{a} t$ (bottom of page 319). This just goes back to my question above i.e. that I can't imagine how the so called static observers in the kerr geometry are at the same spatial position with respect to an observer at infinity but still have an angular momentum.

 

[PeterDonis]

So the angular momentum should have the same sign as $a$ (the rotation of the black hole).

No; the angular momentum of a hovering observer has the *opposite* sign to $a$. You have to be *very* careful with signs when doing this computation.

I'm not sure what the sign of the twist is since it looks rather painful to compute but is there a relation between the sign of the twist to the sign of the angular momentum?

Yes; the sign of the twist is negative for the hovering congruence, just as the sign of the hovering observer's angular momentum is. These two facts are connected. I'll post separately about the details of the computation; I don't think it's quite as painful as you fear , but it does require some care.

I would think that the neighboring static observers should rotate around me in the same direction as my angular momentum

No, they should rotate around you in the *opposite* sense to your angular momentum. Think of the simple case of rotating in flat spacetime, while observers nearby remain fixed relative to the distant stars. Which way will they appear to rotate relative to you?

I'm not really sure what *kind* of rotation angular momentum represents here. Is it also a kind of rotation around the black hole?

It's rotation relative to the ZAMO congruence; i.e., it's rotation relative to the family of worldlines that is orthogonal to the global surfaces of constant time.

 

[WannabeNewton]

No; the angular momentum of a hovering observer has the *opposite* sign to $a$. You have to be *very* careful with signs when doing this computation.

Isn't $L = \alpha g_{t\phi}$ and for the kerr metric in the usual coordinates isn't $g_{t\phi}$ equal to some positive terms times $a$?

Yes; the sign of the twist is negative for the hovering congruence, just as the sign of the hovering observer's angular momentum is. These two facts are connected. I'll post separately about the details of the computation; I don't think it's quite as painful as you fear , but it does require some care.

No, they should rotate around you in the *opposite* sense to your angular momentum. Think of the simple case of rotating in flat spacetime, while observers nearby remain fixed relative to the distant stars. Which way will they appear to rotate relative to you?

Ok so I will see the stars rotate around me in some direction and in order for the nearby observers in the congruence to stay fixed relative to the stars, they have to rotate around me in the same direction as the stars and since I see the stars rotate in a direction opposite to that of my own angular momentum, the rotation of the nearby observers around me is opposite to that of my own angular momentum? But then how is the sign of the angular momentum the same as that of the twist (both negative) if they have opposite senses of rotation?

It's rotation relative to the ZAMO congruence; i.e., it's rotation relative to the family of worldlines that is orthogonal to the global surfaces of constant time.

So the $L$ of a static observer is rotation about/around a nearby ZAMO? All these different rotations are making me get woozy :p I guess what I don't get is, if the rotation due to angular momentum of a given static observer is about a nearby ZAMO, then why would the observer see the distant stars rotate around him as opposed to just seeing a ZAMO rotate around him and the distant stars fixed relative to him?

Sorry if this is going in circles Peter (no pun intended xP) but it's just really really hard for me to picture this. I'm just having trouble understanding in what visual/physical sense these guys are static (as opposed to the mathematical definition which I'm fine with) if they see the distant stars rotating around them while using the distant stars as a measure of being "at rest"/static in the kerr space-time. In other words, I don't get how my neighboring static observers can be fixed relative to the distant stars and yet I see the distant stars rotating around me if we are all static observers and are all meant to be at rest relative to the fixed stars (or to an observer at infinity if that makes it easier).

Thanks Peter.

 

[PeterDonis]

Isn't $L = \alpha g_{t\phi}$ and for the kerr metric in the usual coordinates isn't $g_{t\phi}$ equal to some positive terms times $a$?

The sign of $g_{t \phi}$ depends on the metric sign convention you're using; the key is that it has the same sign as $g_{tt}$. However, the crucial sign in $L$ is the sign of $\xi^a \psi_a$. For the hovering observer, that sign is negative, because the component of $\xi^a$ in the tangential direction is negative.

Ok so I will see the stars rotate around me in some direction and in order for the nearby observers in the congruence to stay fixed relative to the stars, they have to rotate around me in the same direction as the stars and since I see the stars rotate in a direction opposite to that of my own angular momentum, the rotation of the nearby observers around me is opposite to that of my own angular momentum?

Yes.

But then how is the sign of the angular momentum the same as that of the twist (both negative) if they have opposite senses of rotation?

They don't. The twist is your rotation relative to the nearby observers (i.e., relative to adjacent worldlines in the same congruence as your worldline), not their rotation relative to you. As you have just established, your rotation relative to the nearby observers is the same as your rotation relative to infinity for this congruence.

So the $L$ of a static observer is rotation about/around a nearby ZAMO?

Not about the ZAMO itself, no. (It's hard to express these things in English, which might be part of why it's hard to visualize them.) The hovering observer's basis vectors are rotating relative to the basis vectors of a ZAMO that is passing him; that's the nonzero twist. The hovering observer's basis vectors are also rotating relative to infinity; that's the angular momentum.

why would the observer see the distant stars rotate around him as opposed to just seeing a ZAMO rotate around him and the distant stars fixed relative to him?

The ZAMO isn't rotating around him; the ZAMO's basis vectors are rotating relative to his basis vectors (as are the distant stars).

I'm just having trouble understanding in what visual/physical sense these guys are static

An observer at infinity would see them hovering, with zero angular velocity about the hole. An observer at infinity would see a ZAMO as rotating about the hole, with positive angular velocity (i.e., rotating around the hole in the same sense as the hole's own rotation).

if they see the distant stars rotating around them while using the distant stars as a measure of being "at rest"/static in the kerr space-time.

They see the distant stars rotating relative to spatial basis vectors stabilized by gyroscopes. You have to be careful to distinguish that from other senses of "rotation". If you look at the 4-velocities of the static observers, they have only one component, the $t$ component (in the global coordinates adapted to the timelike KVF); i.e., they have zero angular velocity. But their spatial basis vectors don't stay at fixed directions relative to the distant stars as they travel along their worldlines.

It seems like you're having trouble imagining how an observer can have zero angular velocity but still have his basis vectors rotating relative to the distant stars. Once again, I think it might help to think carefully about what the non-orthogonality of the timelike KVF to the surfaces of constant coordinate time means; that's what breaks the link you're used to having between angular velocity and rotation of the gyro-stabilized basis vectors relative to the distant stars.

 

[WannabeNewton]

The sign of $g_{t \phi}$ depends on the metric sign convention you're using; the key is that it has the same sign as $g_{tt}$. However, the crucial sign in $L$ is the sign of $\xi^a \psi_a$. For the hovering observer, that sign is negative, because the component of $\xi^a$ in the tangential direction is negative.

Oops I was looking at a source that used the (+ - - -) convention and didn't realize it while I was using the opposite convention in my head. Yeah ok so it's the opposite direction. Sorry about that!

They don't. The twist is your rotation relative to the nearby observers (i.e., relative to adjacent worldlines in the same congruence as your worldline), not their rotation relative to you. As you have just established, your rotation relative to the nearby observers is the same as your rotation relative to infinity for this congruence.

Oh ok wow all this time I thought that twist was the measure of how nearby observers rotate around me because it seems like all the texts I've read explain it this way i.e. if I imagine a small sphere of observers around me then they will rotate around me as per the twist. I guess this is just easier to visualize than me rotating relative to the nearby observers (I'm not even sure how I would visualize that since there are a multitude of nearby observers).

So since the observers rotate in an opposite sense around me to my angular momentum, the twist itself will be in the same sense as the angular momentum and hence the same sign since the actual sign of the twist is opposite to that of the rotation of the nearby observers around me?

Not about the ZAMO itself, no. (It's hard to express these things in English, which might be part of why it's hard to visualize them.) The hovering observer's basis vectors are rotating relative to the basis vectors of a ZAMO that is passing him; that's the nonzero twist. The hovering observer's basis vectors are also rotating relative to infinity; that's the angular momentum.

Ok I just wanted to separate the following two questions:

(1) I thought the non-zero twist meant the hovering observer's connecting vectors to nearby hovering observers are rotating? Are you saying they are rotating relative to the connecting vectors setup by a passing ZAMO because a ZAMO defines local non-rotation in the sense of vanishing twist? If we imagine the diamond setup from before for 5 static observers (so that there is a connecting vector from the center to each endpoint of the diamond) and then have the diamond setup for a setup of 5 passing ZAMOs, then when the two coincide we can imagine the connecting vectors to each endpoint of the diamond for the static observers as rotating at that instant relative to the respective connecting vectors to each endpoint of the diamond for the ZAMO observers (i.e. the connecting vector from the static fiducial observer to the leading static observer rotates relative to the connecting vector from the ZAMO fiducial observer to the leading ZAMO observer at the instant they coincide)?

Sorry if this is convoluted but this is the only way I can relate it to the fact that the ZAMOs define the local non-rotation in terms of vanishing twist. It doesn't seem the same to me as saying the spatial basis vectors of a single hovering observer himself rotates relative to the spatial basis vectors of a single ZAMO observer without any reference to neighboring observers in either case; it doesn't seem to relate to twist in any way.

(2) So from the perspective of an observer at infinity, the static observer is hovering in place (so no orbital rotation around the black hole) but self-rotating sort of like if I watched you as you sat on a chair that was rotating in place (aren't those chairs just fun :p)? Here when you say the hovering observer's basis vectors are rotating relative to infinity, do you mean the connecting vectors from the observer to his nearby hovering neighbors (like the diamond) or just the spatial basis vectors carried by the observer himself (with no relation to neighboring observers)?

Thanks Peter. Hopefully I'm getting closer to getting the physical intuition for this and can then just tie it to the math involved. Sorry if this is dragging on for too long (again no pun intended xP).

 

[PeterDonis]

So since the observers rotate in an opposite sense around me to my angular momentum, the twist itself will be in the same sense as the angular momentum and hence the same sign since the actual sign of the twist is opposite to that of the rotation of the nearby observers around me?

Actually, I think I got this part backwards; you were right, the observers rotate around you in the same sense as your angular momentum. But the twist also has the same sign as your angular momentum, because it *is* the way the nearby observers rotate around you. I was confusing the basis vectors with the connecting vectors in the congruence. Sorry about the mixup. See further comments below.

(1) I thought the non-zero twist meant the hovering observer's connecting vectors to nearby hovering observers are rotating?

Relative to his basis vectors, yes. Although I was phrasing it the other way around, as a rotation of the observer's basis vectors relative to the vectors connecting the observer to nearby observers in the same congruence. But this is another place where I think I mis-stated things, which shows how hard it is to keep all this stuff straight when you're trying to visualize it. See further comments below.

If we imagine the diamond setup from before for 5 static observers (so that there is a connecting vector from the center to each endpoint of the diamond) and then have the diamond setup for a setup of 5 passing ZAMOs, then when the two coincide we can imagine the connecting vectors to each endpoint of the diamond for the static observers as rotating at that instant relative to the respective connecting vectors to each endpoint of the diamond for the ZAMO observers (i.e. the connecting vector from the static fiducial observer to the leading static observer rotates relative to the connecting vector from the ZAMO fiducial observer to the leading ZAMO observer at the instant they coincide)?

Yes. The connecting vectors for the hovering diamond will be rotating "backwards" (i.e., in the opposite sense to the hole's rotation) relative to the connecting vectors for the ZAMO diamond.

It doesn't seem the same to me as saying the spatial basis vectors of a single hovering observer himself rotates relative to the spatial basis vectors of a single ZAMO observer without any reference to neighboring observers in either case; it doesn't seem to relate to twist in any way.

The twist is a property of a congruence of worldlines, not a single worldline, yes. But the standard for a vector being "unchanged" as it travels along a worldline is Fermi-Walker transport. Another way of saying that the twist of the ZAMO congruence vanishes is that the connecting vectors between neighboring members of the congruence stay exactly in line with each individual member's Fermi-Walker transported basis vectors. And another way of saying that the twist of the hovering congruence does not vanish is that the connecting vectors between neighboring members do *not* stay in line with each individual member's Fermi-Walker transported basis vectors.

(2) So from the perspective of an observer at infinity, the static observer is hovering in place (so no orbital rotation around the black hole) but self-rotating sort of like if I watched you as you sat on a chair that was rotating in place (aren't those chairs just fun :p)?

Yes.

Here when you say the hovering observer's basis vectors are rotating relative to infinity, do you mean the connecting vectors from the observer to his nearby hovering neighbors (like the diamond) or just the spatial basis vectors carried by the observer himself (with no relation to neighboring observers)?

Here's where I might have gotten mixed up before. Let me re-state things briefly for both congruences. All basis vectors are assumed to be Fermi-Walker transported (i.e., gyro-stabilized).

For the ZAMO congruence, the connecting vectors in the diamond *do* rotate relative to infinity, in the same sense as the hole's rotation. They do *not* rotate relative to the fiducial observer's basis vectors; this means the twist of the congruence vanishes. The rate of rotation of both sets of vectors (connecting vectors and basis vectors), relative to infinity, is the same as the angular velocity of the fiducial ZAMO about the hole.

For the hovering congruence, the connecting vectors in the diamond do *not* rotate relative to infinity. They rotate backwards (i.e., opposite to the sense of the hole's rotation) relative to the fiducial observer's basis vectors; this means the twist of the congruence is negative. That means (I think--see below) that the fiducial observer's basis vectors *do* rotate relative to infinity, in the *same* sense as the hole's rotation (I think I got this the wrong way around in a previous post).

The last part is particularly counterintuitive, since the hovering observer's angular momentum about the hole is negative, but his Fermi-Walker transported basis vectors are rotating, relative to infinity, in the *positive* direction. (Intuitively, this is because the basis vectors, or rather the gyroscopes that point in the basis directions, are being "frame dragged" by the hole's rotation.) However, I think the resolution to this is that the fiducial observer's basis vectors are indeed rotating that way relative to infinity, but at a slower rate than the ZAMO's basis vectors. So relative to a ZAMO's basis vectors, the hovering observer's *basis vectors* are rotating backwards--in addition to the connecting vectors of the hovering diamond rotating backwards relative to the connecting vectors of the ZAMO diamond. (And the two relative rates of rotation will be different--the backwards rate of the hovering diamond, relative to the ZAMO diamond, will be larger than the backwards rate of the hovering basis vectors, relative to the ZAMO basis vectors. This should be evident from the above.)

Hopefully this didn't tangle things up too much further.

 

[WannabeNewton]

The twist is a property of a congruence of worldlines, not a single worldline, yes. But the standard for a vector being "unchanged" as it travels along a worldline is Fermi-Walker transport. Another way of saying that the twist of the ZAMO congruence vanishes is that the connecting vectors between neighboring members of the congruence stay exactly in line with each individual member's Fermi-Walker transported basis vectors. And another way of saying that the twist of the hovering congruence does not vanish is that the connecting vectors between neighboring members do *not* stay in line with each individual member's Fermi-Walker transported basis vectors.

So for example, if the fiducial observer in the ZAMO diamond attached a connecting vector to the observer directly in front and initially aligned this connecting vector with one of his spatial basis vectors (parallel or anti-parallel) and Fermi-Transported his spatial basis vectors, then the connecting vector will remain aligned to said spatial basis vector at the next instant? On the other hand for the hovering diamond, if the fiducial observer attaches a connecting vector to the front observer and initially aligns it with a spatial basis vector (again parallel or anti-parallel as needed) then at the next instant this connecting vector will have rotated relative to said spatial basis vector?

For the ZAMO congruence, the connecting vectors in the diamond *do* rotate relative to infinity, in the same sense as the hole's rotation. They do *not* rotate relative to the fiducial observer's basis vectors; this means the twist of the congruence vanishes. The rate of rotation of both sets of vectors (connecting vectors and basis vectors), relative to infinity, is the same as the angular velocity of the fiducial ZAMO about the hole.

For the hovering congruence, the connecting vectors in the diamond do *not* rotate relative to infinity. They rotate backwards (i.e., opposite to the sense of the hole's rotation) relative to the fiducial observer's basis vectors; this means the twist of the congruence is negative. That means (I think--see below) that the fiducial observer's basis vectors *do* rotate relative to infinity, in the *same* sense as the hole's rotation (I think I got this the wrong way around in a previous post).

Ok this is the crucial part so I want to make sure I'm getting it physically, because I think it's the different kinds of rotations involved that confused me before. Let me start with the second scenario. So the connecting vectors to the nearby hovering observers from a given hovering observer (call him $O$) don't rotate relative to infinity because the nearby observers are at rest relative to infinity since they are static correct? But they do rotate relative to $O$'s spatial basis vectors, giving rise to a non-zero twist. Now I want to make sure that I understand what kind of rotation this is: the rotation of the connecting vectors, from $O$ to nearby hovering observers, relative to $O$'s spatial basis vectors (i.e. rotating about $O$) is nothing more than $O$ hovering in place but self-rotating (like me spinning around in place) as seen from infinity? If I then went to a different hovering observer and attached a new set of connecting vectors and did the same analysis, this observer will too be self-rotating in place as seen from infinity? So basically if I view things from infinity will I basically see every hovering observer as staying put in place but self-rotating (again in the sense of me spinning around in place)?

On the other hand, for the ZAMO congruence if I take an observer $O$ and attach connecting vectors to nearby observers then these will not be rotating relative to $O$'s spatial basis vectors but the spatial basis vectors (hence the connecting vectors) will be rotating relative to infinity but not in the above sense. That is, if I view things from infinity then I will not see any of the ZAMO observers have any self-rotation (the connecting vectors don't rotate relative to the spatial basis vectors of the ZAMO observers) but I will see the spatial basis vectors rotate about the black hole meaning the ZAMOs have an angular velocity about the black hole.

So in the first case (static observers) we have self-rotation relative to infinity but no rotation about the black hole relative to infinity whereas in the second case (ZAMOs) we have no self-rotation relative to infinity but we do have rotation about the black hole relative to infinity. I hope I got that correct because its the presence of these different kinds of rotation that really confused me.

The last part is particularly counterintuitive, since the hovering observer's angular momentum about the hole is negative...

I'm still confused by how we would visualize the angular momentum relative to infinity because from what you just said it makes it seem like the observer is orbiting the black hole ("angular momentum about the hole") but relative to infinity these observers should be hovering in place so I'm not sure how to interpret that. But I'll leave that till after I make sure I got the above stuff understood physically.

Thank you Peter.

 

[PeterDonis]

if the fiducial observer in the ZAMO diamond attached a connecting vector to the observer directly in front and initially aligned this connecting vector with one of his spatial basis vectors (parallel or anti-parallel) and Fermi-Transported his spatial basis vectors, then the connecting vector will remain aligned to said spatial basis vector at the next instant?

Yes.

On the other hand for the hovering diamond, if the fiducial observer attaches a connecting vector to the front observer and initially aligns it with a spatial basis vector (again parallel or anti-parallel as needed) then at the next instant this connecting vector will have rotated relative to said spatial basis vector?

Yes.

So the connecting vectors to the nearby hovering observers from a given hovering observer (call him $O$) don't rotate relative to infinity because the nearby observers are at rest relative to infinity since they are static correct?

Yes.

But they do rotate relative to $O$'s spatial basis vectors, giving rise to a non-zero twist.

Yes.

Now I want to make sure that I understand what kind of rotation this is: the rotation of the connecting vectors, from $O$ to nearby hovering observers, relative to $O$'s spatial basis vectors (i.e. rotating about $O$) is nothing more than $O$ hovering in place but self-rotating (like me spinning around in place) as seen from infinity?

Yes. An observer at infinity would see $O$'s spatial basis vectors spinning around in place.

If I then went to a different hovering observer and attached a new set of connecting vectors and did the same analysis, this observer will too be self-rotating in place as seen from infinity? So basically if I view things from infinity will I basically see every hovering observer as staying put in place but self-rotating (again in the sense of me spinning around in place)?

Yes, but the rate of rotation will depend on the hovering observer's $r$ and $\theta$ coordinates.

for the ZAMO congruence if I take an observer $O$ and attach connecting vectors to nearby observers then these will not be rotating relative to $O$'s spatial basis vectors but the spatial basis vectors (hence the connecting vectors) will be rotating relative to infinity

Yes.

but not in the above sense.

I think so; the key is that the rate of rotation relative to infinity of the ZAMO's spatial basis vectors is exactly equal to the ZAMO's angular velocity around the hole, so the rotation of the basis vectors can be entirely accounted for by the ZAMO's rotation around the hole; there is no "extra" spinning over and above that.

That is, if I view things from infinity then I will not see any of the ZAMO observers have any self-rotation (the connecting vectors don't rotate relative to the spatial basis vectors of the ZAMO observers) but I will see the spatial basis vectors rotate about the black hole meaning the ZAMOs have an angular velocity about the black hole.

I think this is the same as what I just said above, so yes.

So in the first case (static observers) we have self-rotation relative to infinity but no rotation about the black hole relative to infinity whereas in the second case (ZAMOs) we have no self-rotation relative to infinity but we do have rotation about the black hole relative to infinity.

Again, I think this is the same as what I said above, so yes.

I'm still confused by how we would visualize the angular momentum relative to infinity because from what you just said it makes it seem like the observer is orbiting the black hole ("angular momentum about the hole") but relative to infinity these observers should be hovering in place so I'm not sure how to interpret that.

Once again, the non-orthogonality of the timelike KVF and the surfaces of constant time uncouples angular momentum relative to infinity from angular velocity relative to infinity.

Something that might help is to consider the Langevin congruence in flat spacetime, i.e., the congruence of observers "on a rotating disk", all rotating about the same axis with the same angular velocity. This congruence is described, for example, here:

http://en.wikipedia.org/wiki/Born_coordinates

Note that in this case (flat spacetime), the twist of the congruence depends on the angular velocity, so it's zero when the angular velocity is zero. In Kerr spacetime, the frame dragging due to the hole's rotation shifts the zero angular momentum point: it is no longer zero angular velocity, but ZAMO angular velocity. So the twist of congruences in Kerr spacetime, as well as the angular momentum of observers following them, depends on the angular velocity *relative to a ZAMO*, rather than the angular velocity relative to zero.

The other element that changes in Kerr spacetime is that there can be shear present, and for the types of congruences we're considering, the shear is a function of the angular velocity relative to zero (*not* relative to a ZAMO). So the ZAMO congruence has zero twist but nonzero shear, while the hovering congruence (which has negative angular velocity relative to a ZAMO) has zero shear but nonzero (negative) twist.

 

[WannabeNewton]

Ok but if zero angular momentum is defined by the ZAMOs and angular momentum is measured relative to them, then what do we mean by angular momentum "about the black hole" relative to infinity? And similarly if the hovering observers define zero angular velocity and angular velocity should be measured relative to them, then what do we mean by angular velocity "about the black hole" relative to infinity?

 

[PeterDonis]

Ok but if zero angular momentum is defined by the ZAMOs and angular momentum is measured relative to them, then what do we mean by angular momentum "about the black hole" relative to infinity?

Angular momentum isn't "defined" by the ZAMOs; it's defined as a parameter of the motion for a test object. A test object with 4-momentum $p^a$ has angular momentum $L = g_{ab} \psi^a p^b$, where $\psi^a$ is the axial KVF (same notation as you were using). It happens that ZAMOs are the particular observers for which that formula for $L$ gives zero, but that doesn't mean angular momentum is "defined" by the ZAMOs; it's defined by the properties of the spacetime (and the 4-momentum of the observer, of course).

The term "relative to infinity" might not be a very good one, particularly with respect to angular momentum, since its behavior here doesn't match our intuitions (an observer with zero angular velocity has nonzero angular momentum, and vice versa). I don't know that there is any simple measurement that observers at infinity can make that gives an object's angular momentum, as there is for angular velocity (see below). My intent was simply to say something like the above.

And similarly if the hovering observers define zero angular velocity and angular velocity should be measured relative to them, then what do we mean by angular velocity "about the black hole" relative to infinity?

Again, the hovering observers don't define zero angular velocity; the timelike KVF does, since it defines what it means to be "at rest" spatially. The hovering observers are just the ones who are following orbits of the timelike KVF. Here the term "relative to infinity" makes more sense since an observer at infinity can actually time the motion of objects and measure their angular velocity directly.

(The only complication here is that, of course, the result observers at infinity get will be different, because of gravitational time dilation, than the result an observer at finite radius with nonzero angular velocity, such as a ZAMO, will get when he times the passages directly overhead of some object at infinity. But that's easy to take into account if you need to.)

 

[WannabeNewton]

Alrighty then. Just one more question: is there a way to show mathematically that the static observers will be rotating in place as seen from infinity? Physically it can be argued as we did above but what about mathematically? The reason I ask is, we can calculate the twist of the static congruence but this will tell us how a swarm of static observers rotates around a given static observer in the congruence so how would we then translate this mathematically to an observation at infinity which would conclude that the given observer is rotating in place? I guess what I'm asking is, how would we mathematically codify the static observer rotating in place as seen from infinity in contrast with the ZAMO having an angular velocity representing him going around the black hole as seen from infinity.

Thank you so much for all the help Peter!

 

[PeterDonis]

is there a way to show mathematically that the static observers will be rotating in place as seen from infinity?

The "in place" part is easy, since the static observer follows an orbit of the timelike KVF by definition.

For the "rotating" part, the brute force way is to compute Fermi-Walker transport for the spatial basis vectors, given some starting orientation (for example, assume that one basis vector points radially outward at some starting event, so it only has an $r$ component, and see if Fermi-Walker transporting it along the worldline adds any component besides an $r$ component), using the metric and connection coefficients and the formula for the components of the 4-velocity of the hovering observer. MTW goes into how to do that at some point, I don't have my copy handy to check where. I think it's somewhere in Wald as well, but I don't know how much detail he goes into about the actual computation.

There might well be a slicker way to do it, possibly using vierbeins and tetrads, but I'm not familiar enough with those to give any useful info.

 

[WannabeNewton]

You know that's actually something else that's been confusing me ever since I read this particular section in Geroch's notes: http://postimg.org/image/fqhd7g665/

As per the above, aren't we prematurely assuming that the observer is not rotating (in the self-rotation sense) if we assume his basis vectors are Fermi-Walker transported? Don't we measure rotation (again in the self-rotation sense) by seeing if the basis vectors fail to be Fermi-Walker transported?

 

[PeterDonis]

As per the above, aren't we prematurely assuming that the observer is not rotating (in the self-rotation sense) if we assume his basis vectors are Fermi-Walker transported?

We're not so much assuming this as using it as a definition: we *define* "zero self-rotation" to mean Fermi-Walker transporting the spatial basis vectors. We do this because it makes sense: Fermi-Walker transport basically corresponds, physically, to stabilizing the vectors with gyroscopes.