# Locally non-rotating observers

1. Jul 21, 2013

### 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.

2. Jul 21, 2013

### atyy

3. Jul 22, 2013

### 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).

Last edited: Jul 22, 2013
4. Jul 22, 2013

### Mentz114

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.

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

5. Jul 22, 2013

### Staff: Mentor

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.

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.

6. Jul 22, 2013

### Staff: Mentor

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.

7. Jul 22, 2013

### 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).

8. Jul 22, 2013

### Staff: Mentor

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.

9. Jul 22, 2013

### 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.

Last edited: Jul 22, 2013
10. Jul 22, 2013

### TrickyDicky

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.

11. Jul 22, 2013

### Staff: Mentor

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

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).

12. Jul 22, 2013

### Staff: Mentor

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.)

13. Jul 22, 2013

### WannabeNewton

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.

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?

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.

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.

Last edited: Jul 22, 2013
14. Jul 22, 2013

### Staff: Mentor

Yes.

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.

As far as I can tell, yes.

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).

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

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.

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.

Last edited: Jul 22, 2013
15. Jul 22, 2013

### 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.

16. Jul 22, 2013

### Staff: Mentor

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

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

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.

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.

17. Jul 22, 2013

### 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.

18. Jul 22, 2013

### Staff: Mentor

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.

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*.

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

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.)

19. Jul 22, 2013

### 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).

20. Jul 22, 2013

### Staff: Mentor

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.

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?

21. Jul 22, 2013

### 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?

Last edited: Jul 22, 2013
22. Jul 22, 2013

### 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.

Last edited: Jul 22, 2013
23. Jul 22, 2013

### Staff: Mentor

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.

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?

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.

24. Jul 22, 2013

### WannabeNewton

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 $a$?

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?

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.

Last edited: Jul 22, 2013
25. Jul 22, 2013

### Staff: Mentor

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.

Yes.

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.

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.

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

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).

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.