Practical measurements of rotation in the Kerr metric

[PeterDonis]

when you say "Ok, thinking about this some more, there is one obvious example of a free-fall object released by a static observer where the angular velocity of the free-fall object *has* to change fairly quickly", with respect to whom/what are we measuring the angular velocity of the freely falling object to be changing fairly quickly?

With respect to infinity; but there are certainly subtleties involved. To list a couple:

(1) Inside the static limit, there are no "static observers", so there's no obvious way to relate angular velocity with respect to infinity to angular velocity with respect to a local static observer. (To put this another way, inside the static limit the KVF labeled by $\partial / \partial t$ is spacelike--it's null *at* the static limit--so it becomes problematic to interpret "angular velocity" using it as a standard.)

(2) If timelike geodesics have a changing $d \phi / dt$ in Boyer-Lindquist coordinates, then null geodesics should too, so it becomes problematic to try and use light signals from a freely falling object to an observer at infinity to measure angular velocity with respect to infinity.

 

[PeterDonis]

Is there a physical reason for why a static observer would measure a freely falling particle as having constant angular velocity in the local Lorentz frame of the static observer?

If the angular velocity with respect to infinity of the freely falling particle is in fact changing, then the way the local inertial frames "line up" must also be changing from one radius to another. In other words, I think the solution to the apparent "paradox" here is going to boil down to *something* in the local inertial frames of static observers having to "change direction" with a change in radius, even though the "acceleration due to gravity" does *not* change direction. But I'm still having trouble imagining what that something could be.

 

[WannabeNewton]

Is it possible that the intrinsic spin of a ZAM particle about its own axis (relative to the distant stars) gives it an angular momentum that cancels out the angular momentum of the particle due to its orbital velocity around the gravitational body? (The two rotations are in opposite directions).

ZAMO frames don't have an intrinsic spin by the way. As noted earlier, the time-like congruence formed by the family of ZAMOs, described by the 4-velocity field $u^{\mu} = \alpha \nabla^{\mu}t$, is locally non-rotating in the sense that the intrinsic angular velocity (also called the twist 4-vector or the "absolute" rotation) vanishes identically:

$\omega^{\mu} = \epsilon^{\mu\nu\gamma\sigma}u_{\nu}\nabla_{\gamma}u_{\sigma} = \alpha^2\epsilon^{\mu\nu[\gamma\sigma]}\nabla_{\nu}t\nabla_{[\gamma}\nabla_{\sigma]}t - \alpha \epsilon^{\mu\gamma[\nu\sigma]}\nabla_{[\nu}t\nabla_{\sigma]}t\nabla_{\gamma}\alpha = 0$

This means that the space-like hypersurfaces $t = \text{const.}$ form global simultaneity slices for the ZAMOs i.e. Einstein synchrony ($dt = 0$) can be established amongst the ZAMOS. If ZAMO frames had an intrinsic spin then Einstein synchrony ($dt = 0$) could not be established amongst the ZAMOs.

In fact it is the static frames in Kerr space-time that possesses an intrinsic spin. The $dt = 0$ Einstein synchrony cannot be established amongst the family of static observers because the 4-velocity field of the family of static observers is parallel to the vector field $\xi^{\mu} = \delta^{\mu}_t$ and it can be shown that $\omega^{\mu} = \epsilon^{\mu\nu\gamma\sigma}\xi_{\nu}\nabla_{\gamma}\xi_{\sigma} \neq 0$ i.e. the integral curves of $\xi^{\mu}$ twist around one another. In fact one can prove the general result that static frames in stationary space-times posses no intrinsic spin if and only if $\xi^{\mu}$ is twist-free.

In other words zero angular momentum does not have to mean zero angular velocity around the gravitational body.

The converse is true as well: zero angular velocity does not imply zero angular momentum. The reason is again because $\nabla^{\mu}t$ is not parallel to $\xi^{\mu}$ in Kerr space-time, unlike in Schwarzschild space-time.

 

[yuiop]

According to this text (see p.62), in the local Lorentz frame of a static observer, a freely falling object has only a polar and radial component to gravitational acceleration as measured in said local Lorentz frame: http://books.google.com/books?id=n0...#v=onepage&q=static frame kerr metric&f=false
which is basically the same result as in that paper linked in the previous page of this thread.

Unfortunately the google preview does not allow me to see the pages 62 and 63 :(
Also, I cannot see the diagrams in this paper linked by peter. Any alternative link?

How is a static observer defined here? Is it an observer with $d\phi/dt=0$ in the Kerr metric or a ZAMO? There would be no azimuthal acceleration in the ZAMO reference frame.

Is there a physical reason for why a static observer would measure a freely falling particle as having constant angular velocity in the local Lorentz frame of the static observer?

Consider an object with an initial angular momentum of zero at infinity. If the trajectory of the free falling particle remained purely radial then it would be acquiring non-zero angular momentum, even thought it had zero angular velocity in Kerr coordinates. (i.e. there would be a violation of conservation of angular momentum.) This is because as mentioned earlier, a 'stationary' ring in the Kerr metric with $d\phi/dt=0$ has a positive Sagnac result. The paper by Malament that you linked earlier on page 225, just below figure 3.2.4 states: "Imagine further that we keep track of whether the pulses arrive back at Q simultaneously, using, for example, an interferometer). It turns out that this will be the case—i.e., they will arrive back simultaneously—if(and only if) the ring has zero angular momentum".

The $d\phi/dt=0$ state of the ring does not qualify as a zero angular momentum state because of the positive Sagnac result. For the free falling particle to maintain zero angular momentum its angular velocity at any height r has to be $\Omega(r) = -g_{t\phi}/g_{\phi\phi}$, ie that of a ZAMO.

P.S. I think we cross posted. I think you said basically the same thing in the last line of your last post.

 

[PeterDonis]

This means that the space-like hypersurfaces $t = \text{const.}$ form global simultaneity slices for the ZAMOs i.e. Einstein synchrony ($dt = 0$) can be established amongst the ZAMOS.

I'm not sure this is quite true. The ZAMO congruence has zero vorticity, as you note, but it has nonzero shear, so it is not a rigid congruence. I'm not sure you can do Einstein clock synchronization with a non-rigid congruence, since different members of the congruence are in relative motion.

If ZAMO frames had an intrinsic spin then Einstein synchrony ($dt = 0$) could not be established amongst the ZAMOs.

I think this is true, but I'm not sure it quite has the intuitive connotations one might expect. See further comments below.

In fact it is the static frames in Kerr space-time that possesses an intrinsic spin.

Yes, this congruence has nonzero vorticity, but that may not imply quite what you think. See below.

$dt = 0$ Einstein synchrony cannot be established amongst the family of static observers

Since the observers are all following orbits of the timelike KVF, there is an obvious clock synchronization that can be used between them, although it is not, strictly speaking, equivalent to Einstein clock synchronization, since static observers at different spatial coordinates will have different clock rates. But they will all agree on whether a given pair of events are simultaneous or not. In other words, the surfaces of constant $t$ *are* simultaneity surfaces for this congruence.

What the nonzero twist of the static congruence means is that, as you say, neighboring members of the congruence will be twisting around each other. But twisting relative to what? Relative to gyroscopes that each member of the congruence carries, i.e., relative to a set of spatial basis vectors that are Fermi-Walker transported along their worldlines. But that, in itself, has nothing to do with simultaneity.

 

[WannabeNewton]

Unfortunately the google preview does not allow me to see the pages 62 and 63 :(

Weird, pages 62 and 63 show up in the google preview for me; maybe it's a UK vs US thing? I'll try to find an alternative link.

Also, I cannot see the diagrams in this paper linked by peter. Any alternative link?

The contents of that paper are more or less the same as those of sections 3.2 and 3.3 of the GR notes by Malament that I linked you to, as far as I can tell.

How is a static observer defined here? Is it an observer with $d\phi/dt=0$ in the Kerr metric or a ZAMO?

A static observer is formally defined as one whose 4-velocity is parallel to the time-like killing field $\xi^{\mu} = \delta^{\mu}_{t}$. Such an observer will indeed have $\frac{d\phi}{dt} = 0$ everywhere along his/her worldline. A ZAMO in Kerr space-time does not correspond to a static observer in Kerr space-time for the reasons mentioned in the post directly above yours.

...
The $d\phi/dt=0$ state of the ring does not qualify as a zero angular momentum state because of the positive Sagnac result. For the free falling particle to maintain zero angular momentum its angular velocity at any height r as to be $\Omega(r) = -g_{t\phi}/g_{\phi\phi}$, ie that of a ZAMO.

I agree with all of this. The non-intuitive aspect of all of this, as Peter noted several times prior in the thread, is that in the local Lorentz frame of a static observer, the acceleration due to gravity of a freely falling particle passing by the static observer only has radial and azimuthal components so it's angular velocity with respect to this static observer would (within the confines of the static observer's local Lorentz frame) be constant; on the other hand we know that with respect to an observer at infinity the angular velocity of the freely falling particle increases as the particle falls towards the black hole. So we're basically trying to figure out why physically things play out this in the local Lorentz frames of static observers.

 

[WannabeNewton]

I'm not sure this is quite true. The ZAMO congruence has zero vorticity, as you note, but it has nonzero shear, so it is not a rigid congruence. I'm not sure you can do Einstein clock synchronization with a non-rigid congruence, since different members of the congruence are in relative motion.

As far as I know, the $dt =0$ Einstein synchrony amongst observers in a time-like congruence only requires that the congruence be hypersurface orthogonal to the $t = \text{const.}$ foliation of space-time, which is equivalent to it being twist-free. Does the lack of rigidity of a twist-free time-like congruence really prevent global simultaneity from being established amongst the observers in the congruence?

In other words, the surfaces of constant $t$ *are* simultaneity surfaces for this congruence.

I don't think that's true. If you have "Gravitation and Inertia"-Wheeler handy, check out pp.100-102. In particular note that only if the space-time is static do the $t = \text{const.}$ slices form global simultaneity slices for static observers. A space-time being static is equivalent to $\xi^{\mu}$ being twist-free. See also the very last paragraph of p.62 of the text I linked earlier: http://books.google.com/books?id=n0...#v=onepage&q=static frame kerr metric&f=false

EDIT: if the preview doesn't work for you I can take a screenshot of the relevant paragraph

 

[PeterDonis]

Does the lack of rigidity of a twist-free time-like congruence really prevent global simultaneity from being established amongst the observers in the congruence?

Yes, because, as I said before, the nonzero shear means neighboring members of the ZAMO congruence are in relative motion. You can't do Einstein clock synchronization between observers in relative motion; their local simultaneity surfaces obviously don't match up. You need a rigid congruence.

I think the fact that the ZAMO congruence is orthogonal everywhere to surfaces of constant $t$ is somewhat misleading in this connection, because orthogonality depends on the metric, and the metric is a function of $r$ (and $\theta$, of course, if we're not in the equatorial plane--but just being a function of $r$ is enough for here). So a ZAMO at one value of $r$ has simultaneity surfaces that are *not* parallel to the simultaneity surfaces of a ZAMO at a different value of $r$, even though both of their worldlines are orthogonal, locally, to surfaces of constant $t$ at their respective values of $r$.

Another way of putting this might be that the ZAMO congruence is not a Killing congruence; its worldlines, all taken together, are not integral curves of a single KVF. If we just look at the members of the congruence at a single value of $r$, then that "sub-congruence" is a Killing congruence; it's a linear combination of $\partial / \partial t$ and $\partial / \partial \phi$ with constant coefficients. But the coefficients are only constant because $r$ is constant; looking at the whole congruence, over the entire range of $r$, the congruence is a linear combination of $\partial / \partial t$ and $\partial / \partial \phi$ with coefficients that vary with $r$, so it isn't a Killing congruence.

I don't think that's true.

Wait, you're right, if the timelike KVF isn't hypersurface orthogonal then the paths of radial geodesics won't behave the way they need to for the clock synchronization to work the way it does in a static spacetime. So zero vorticity is a necessary, but not, I think, a sufficient (see above) condition for global simultaneity in this sense being possible--it looks like you need a rigid, twist-free congruence, i.e., the expansion, shear, and vorticity *all* need to be zero.

(I say "in this sense" because a family of observers can always *adopt* a standard of simultaneity that does not "match up" with their natural local surfaces of simultaneity. For example, in our everyday life on the surface of the Earth we all use the same simultaneity convention even though many of us are in relative motion purely due to the Earth's rotation. But that convention is not the "natural" one for any of them.)

Another way of putting the requirement, to build on what I said above about Killing congruences, might be that for a global standard of simultaneity, you need to have a Killing congruence, *and* the Killing congruence needs to be hypersurface orthogonal. The latter condition ensures zero vorticity: I *think* the former condition ensures zero expansion and shear, but I'm not sure.

 

[WannabeNewton]

According to the following text ("General Relativity for Mathematicians"-Sachs and Wu), the ZAMO frames given by the congruence $u^{\mu} = \alpha \nabla^{\mu} t$ where $\alpha$ is as usual the normalization scalar field, are Einstein synchronizable because $u^{\mu}$ is parallel to the gradient of a scalar field $\nabla^{\mu} t$; a congruence being twist-free i.e. $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ is equivalent to local Einstein synchrony: http://postimg.org/image/5qb0mj25v/

Note the book uses $\omega = -hdt$ to represent $u^{\mu} = \alpha \nabla^{\mu} t$ and $\omega \wedge d\omega =0$ to represent $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$.

There is no requirement of rigidity.

 

[yuiop]

Weird, pages 62 and 63 show up in the google preview for me; maybe it's a UK vs US thing? I'll try to find an alternative link.

If I was cynical, I would suspect the preview algorithm is more likely to block a page if a lot of users try to access it at about the same time.

The contents of that paper are more or less the same as those of sections 3.2 and 3.3 of the GR notes by Malament that I linked you to, as far as I can tell.

I so wanted to see a picture of lazy Susan and her bucket with attached telescope.

... on the other hand we know that with respect to an observer at infinity the angular velocity of the freely falling particle increases as the particle falls towards the black hole. So we're basically trying to figure out why physically things play out this in the local Lorentz frames of static observers.

Is there a conflict in how an observer at infinity and a local static observer define vertical? For example if we define vertical as the direction parallel to a plumb bob, then this definition of vertical would not be parallel to a radial line according to an observer at infinity in the Kerr metric.

 

[PeterDonis]

According to the following text ("General Relativity for Mathematicians"-Sachs and Wu), the ZAMO frames given by the congruence $u^{\mu} = \alpha \nabla^{\mu} t$ where $\alpha$ is as usual the normalization scalar field, are Einstein synchronizable because $u^{\mu}$ is parallel to the gradient of a scalar field $\nabla^{\mu} t$; a congruence being twist-free i.e. $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ is equivalent to local Einstein synchrony: http://postimg.org/image/5qb0mj25v/

Are you sure their definition of "synchronizable" is equivalent to Einstein clock synchronization? One of the key requirements of Einstein clock synchronization is that clocks that have been once synchronized *stay* synchronized. I don't see how you could meet that requirement with clocks in relative motion.

To put this another way, are you sure that the definition of synchronization you are really thinking of in this thread is Einstein synchronization? Or is it something else?

 

[PeterDonis]

if we define vertical as the direction parallel to a plumb bob, then this definition of vertical would not be parallel to a radial line according to an observer at infinity in the Kerr metric.

Why not?

 

[WannabeNewton]

Or is it something else?

I was thinking of synchronization by means of radar in the following manner:

http://postimg.org/image/anfuzj6b7/

http://postimg.org/image/ls4ltco0r/

I had thought that radar synchronization by means of the $1/2$ averaging scheme was called Einstein synchronization in this very general context. Is that inaccurate terminology? If so I apologize.

 

[yuiop]

Yes, because, as I said before, the nonzero shear means neighboring members of the ZAMO congruence are in relative motion. You can't do Einstein clock synchronization between observers in relative motion; their local simultaneity surfaces obviously don't match up. You need a rigid congruence.

There should be no problem or conflict with a bunch of ZAM observers in the equatorial plane using Einstein synchronisation as long as all the ZAMO's in the group share the same radial coordinate. A co-moving ring that connects such a group would have angular velocity $d\phi/dt= -g_{t\phi}/g_{\phi\phi}$ and light signals sent clockwise and anticlockwise around the ring arrive back simultaneously, so unlike the case of a rotating disc in flat space time, it is possible for observers all the way around the ring to use Einstein clock synchronisation without there being a conflict when synchronising the last clock with the first clock. That is my initial thoughts. Might be wrong...

 

[yuiop]

For example if we define vertical as the direction parallel to a plumb bob, then this definition of vertical would not be parallel to a radial line according to an observer at infinity in the Kerr metric.

Why not?

If there was azimuthal acceleration as well as regular radial acceleration due to gravity, the plumb bob would be accelerated downwards and sideways, so if the plumb bob is used by an observer as a local reference for vertical, he will not detect any azimuthal acceleration. However, I am not sure if that is how a local observer defines vertical and probably by such a definition he would not detect any polar acceleration either. Any thoughts? Just trying to get to the bottom of this puzzle.

 

[WannabeNewton]

...and light signals sent clockwise and anticlockwise around the ring arrive back simultaneously, so unlike the case of a rotating disc in flat space time, it is possible for observers all the way around the ring to use Einstein clock synchronisation without there being a conflict when synchronising the last clock with the first clock. That is my initial thoughts. Might be wrong...

No this is definitely true. So if we take a Sagnac ring at some fixed $r$ value that has zero angular momentum then there is no Sagnac effect for light signals circulating in opposite senses around the ring as you saw in Malament's GR notes (take a look at Proposition 3.2.2). This means that if we start synchronizing clocks laid out around the ring by starting from some initial clock in the ring then we will remain synchronized to the initial clock after making a complete circuit around the ring by passing from clock to clock; as you noted this would fail for a rotating ring in flat space-time but no so in Kerr space-time because we can have zero angular momentum rotating Sagnac rings in kerr space-time.

Peter wasn't objecting to this in any way. He was commenting about Sagnac rings located at different $r$ values if you look again at his post #43.

I think I may just be using the term Einstein synchrony incorrectly in the context of radar synchronization as defined in post #48.

 

[yuiop]

Peter wasn't objecting to this in any way. He was commenting about Sagnac rings located at different $r$ values if you look again at his post #43.

So he did. Should of read it more carefully. Sorry Peter!

If I may, I would like to briefly return to the issue of gyroscope measurements of rotation.

In the diagram below, the sketch on the left is how a gyroscope attached to a ring that satisfies the ZAM criteria in the Kerr metric, would behave if it simultaneously satisfies the CIR or GTR criteria. Labels T1..T5 indicate successive positions and successive time intervals.

The sketch on the right of the above diagram, is how a gyroscope behaves when attached to a rotating ring around a Schwarzschild black hole. The gyroscope ≈generally≈ continues to point ≈roughly≈ in the direction of the distant stars and and rotates relative to a local point on the rotating ring. The gyroscope does not return exactly to its original location with its original orientation due to geodetic precession or the de Sitter effect and this is hinted at by the dashed line at position T5.

Now consider a Schwarzschild BH that captures a single atom with non zero angular momentum so that it technically becomes a Kerr BH. Would the gyroscope suddenly stop behaving as depicted in the right hand sketch (generally pointing at the distant stars) and start behaving as depicted by the left hand sketch (remaining stationary relative to a local part of the ring) on account of a single atom added to the black hole? This seems unlikely, so it also seems unlikely that a ring that satisfies the ZAM criterion for non rotation in the Kerr metric would also satisfy the CIR or GRT criteria for non rotation in the Kerr metric. As far as I can tell, there would be just be an additional (but lesser?) Lense Thirring precession on top of the de Sitter precession.

 

[WannabeNewton]

Firstly, I agree with both your sketches (once the caveat with the de-Sitter precession is added in as you noted) with regards to the compass of inertia criterion (and the equivalent gyroscope tangency criterion), at least upon a first glance. If I again call the 4-velocity field of the time-like congruence describing the ring by $u^{\mu}$ then we will have $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ as proven earlier in the thread; based on Proposition 3.2.1 in the notes I linked earlier, this is equivalent to the condition that $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\sigma}\psi^{\sigma}) = 0$ i.e. that the tangent field $\psi^{\mu}$ to the ring is Fermi-Walker transported along the worldline of a gyroscope mounted on the ring. The only way for this to be possible, based on the geometrical picture of the tangent field $\psi^{\mu}$ to the ring, is for the gyroscope to continuously change its spin axis relative to a distant fixed star of the asymptotic Minkowski frame in exactly the manner which you have drawn in your left diagram.

In the Schwarzschild case the ring would not satisfy the compass of inertia criterion because in this case $u_{[\gamma}\nabla_{\mu}u_{\nu]} \neq 0$ i.e. it would not be non-rotating according to a local compass of inertia. Consequently $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\sigma}\psi^{\sigma}) \neq 0$ along the worldline of a gyroscope mounted on the ring and hence the angle between $\psi^{\mu}$ and the spin axis of the gyroscope must be continuously changing. If there were no de-Sitter precession then the spin axis would thus have to be pointed towards a distant fixed star throughout the trajectory as in your right diagram but there is a de-Sitter precession which obstructs that, as you noted. Regardless, $\psi^{\mu}$ must rotate relative to the spin axis of the gyroscope in the Schwarzschild case unlike in the Kerr case for the aforementioned reasons.

So again at first glance I agree with your diagrams themselves. However:

This seems unlikely, so it also seems unlikely that a ring that satisfies the ZAM criterion for non rotation in the Kerr metric would also satisfy the CIR or GRT criteria for non rotation in the Kerr metric.

(1) If I read your intentions correctly, you want to describe a perturbation of the Schwarzschild metric about a small angular momentum $\epsilon J$ of the self-gravitating source where $\epsilon << 1$ yes? Did I read your description correctly? I'm trying to simplify your description a bit so that I can describe it mathematically because discontinuous perturbative jumps to zeroth order solutions are not easy to describe mathematically whereas continuous one-parameter perturbations of zeroth order solutions are easy to describe.

If we do perturb the Schwarzschild solution about a small angular momentum $J$ (where I have absorbed the expansion parameter $\epsilon$ into $J$), then the new solution will be $ds^{2} = -(1 - \frac{2M}{r})dt^{2} + (1 - \frac{2M}{r})^{-1}dr^2 + r^2d\Omega^2 - 4\frac{J}{r}\sin^{2}\theta dtd\phi + O(J^2)$ where I have dropped all terms of 2nd order and higher in $J$. The presence of the cross term $g_{t\phi} = -2\frac{J}{r}\sin^{2}\theta$ will lead to the exact same frame dragging phenomena as in Kerr space-time so that in particular the compass of inertia criterion plays out as in your left diagram.

(2) Why would this somehow prevent a ring with zero angular momentum in Kerr space-time from also being non-rotating according to a local compass of inertia? As already shown previously in the thread, the 4-velocity field $u^{\mu}$ of the ring satisfies both $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ and $L = \psi^{\mu}u_{\mu} = 0$ in the allowed regions of Kerr space-time. This by definition means that the ring has zero angular momentum (i.e. no Sagnac effect for counter-propagating light signals around the ring) and that the ring is non-rotating according to a local compass of inertia.

Sorry for the long post but I just wanted to make sure that I myself understood what you were trying to describe in the first place. Cheers.

 

[PeterDonis]

I had thought that radar synchronization by means of the $1/2$ averaging scheme was called Einstein synchronization in this very general context. Is that inaccurate terminology?

No, but as it's described in the pages you showed, it doesn't apply to the ZAMO congruence in Kerr spacetime, because the book pages say the congruence must be geodesic, and the ZAMO congruence is not geodesic.

To further check the definition given in the book pages, I would recommend trying to apply it in the following cases (I'll give how I think it will come out in each case in parentheses):

(1) The obvious "canonical" case: two inertial observers in Minkowski spacetime, at rest relative to each other. (Obviously these will turn out to be synchronized, and the global time function $t$ will just be the $t$ coordinate of their common global inertial frame.)

(2) Two inertial observers in Minkowski spacetime, *not* at rest relative to each other. (These will turn out *not* to be synchronized, because the "consistency condition" will not hold. Note that the worldlines of these observers do not form a congruence, because they will intersect; but I think this is still a good test case to show how the consistency condition fails for observers in relative motion.)

(3) The Rindler congruence in Minkowski spacetime. (This will turn out to be synchronized, I think--however, there is a wrinkle here which I don't see addressed in the book pages. Two different Rindler observers have different "rates of proper time flow": if we pick two pairs of events on the two observers' worldlines, each of which are simultaneous by the synchronization criterion, and use these to define two successive surfaces of simultaneity for the congruence, the observer who is "higher up"--i.e., feels less proper acceleration--will experience more elapsed proper time between the two surfaces of simultaneity. As I read the synchronization in the book, this violates it; the condition as given appears to require that all observers experience the *same* elapsed proper time between two given surfaces of simultaneity.)

(4) The Bell congruence in Minkowski spacetime. (This will turn out *not* to be synchronized--no common surfaces of simultaneity can be defined, even if we leave out the issue of different elapsed proper time that came up for the Rindler congruence above--because different members of the congruence are in relative motion: the congruence has nonzero expansion.)

(5) The congruence of static observers in Schwarzschild spacetime. (This will turn out to be synchronized, but with the same wrinkle about differences in elapsed proper time that arose with the Rindler congruence.)

(6) The congruence of Painleve observers in Schwarzschild spacetime. (This will turn out *not* to be synchronized, even though there are a set of surfaces--the surfaces of constant Painleve coordinate time--which are orthogonal to all the Painleve worldlines. Synchronization will fail because, once again, the "consistency condition" will fail to hold. Note that this congruence has nonzero expansion, unlike the static congruence.)

Note that #3, #4, and #5 are all non-geodesic congruences, so as I read the book definition, once again, it isn't even applicable to them; but I think it's still instructive to imagine trying to do "radar synchronization" with them and seeing how it comes out.

 

[PeterDonis]

Now consider a Schwarzschild BH that captures a single atom with non zero angular momentum so that it technically becomes a Kerr BH. Would the gyroscope suddenly stop behaving as depicted in the right hand sketch (generally pointing at the distant stars) and start behaving as depicted by the left hand sketch (remaining stationary relative to a local part of the ring) on account of a single atom added to the black hole?

Yes, but bear in mind that your sketches only show the spatial part of the two cases. If you draw them in spacetime, so that you can take into account the angular velocity of a ZAMO in the Kerr case, you will see that the two cases (Schwarzschild BH -> Kerr BH with very, very small angular momentum) are very close, because the angular velocity required for zero angular momentum in the Kerr case is so small. So the worldline of the rotating observer with his gyro will look like a very, very stretched out helix, whose tangent vectors are almost vertical: and adding the very, very small angular momentum to the BH will only need to change the behavior of the gyroscope by a very, very small amount to switch it from the right hand sketch behavior to the left hand sketch behavior.

 

[WannabeNewton]

No, but as it's described in the pages you showed, it doesn't apply to the ZAMO congruence in Kerr spacetime, because the book pages say the congruence must be geodesic, and the ZAMO congruence is not geodesic.

Take a look at the exercise in the second image where one must generalize the radar synchronization method to non-geodesic congruences.

Thanks for the detailed post I'll read it in just a bit.

 

[PeterDonis]

Take a look at the exercise in the second image where one must generalize the radar synchronization method to non-geodesic congruences.

Do you mean where it talks about $a_{\gamma}$ being nonzero for some members of the congruence? That's not talking about non-geodesic worldlines: $a_{\gamma}$ isn't proper acceleration, it's just a clock offset (which I admit is lousy notation). In other words, $a_{\gamma}$ being nonzero just means that the members of the congruence weren't lucky enough to all have clocks that started out with compatible "zero" points, so some of them will have to adjust their clock settings by a constant in order to be synchronized.

 

[WannabeNewton]

Do you mean where it talks about $a_{\gamma}$ being nonzero for some members of the congruence? That's not talking about proper acceleration: $a_{\gamma}$ isn't proper acceleration, it's just a clock offset (which I admit is lousy notation). In other words, $a_{\gamma}$ being nonzero just means that the members of the congruence weren't lucky enough to all have clocks that started out with compatible "zero" points, so some of them will have to adjust their clock settings by a constant in order to be synchronized.

Exercise 5.3.1 at the bottom of the second image where it says to generalize the situation to the case wherein the congruence is synchronizable but not proper time synchronizable. If you recall from this image: http://postimg.org/image/5qb0mj25v/ synchronizable means that the 4-velocity field of the congruence is parallel to the gradient of a scalar field such as in the case of the ZAMO congruence.

 

[PeterDonis]

Exercise 5.3.1 at the bottom of the second image where it says to generalize the situation to the case wherein the congruence is synchronizable but not proper time synchronizable.

Ok, I'm getting a bit confused trying to keep all their definitions straight, particularly which are special cases of which, and that may be what's causing me problems. Let me try to enumerate the conditions:

Locally proper time synchronizable: In one place it says this is equivalent to the congruence being geodesic and irrotational; in another place it says this is equivalent to $d \omega = 0$. I think I sort of see how these conditions are equivalent; assuming they are, then this condition can't apply to the ZAMO congruence because the latter is non-geodesic.

Proper time synchronizable: This is equivalent to there being a scalar function $t$ such that $\omega = - dt$ (stated in two places in slightly different notation). I agree after re-reading that there doesn't seem to be a geodesic requirement here, but the ZAMO congruence still doesn't meet this requirement because of the radial dependence; see below. (As the text notes, $\omega = - dt$ ensures that $d \omega = 0$, so proper time synchronizable implies locally proper time synchronizable: but the converse is *not* true. So this is a more restrictive condition than the first.)

Locally synchronizable: This is equivalent to $\omega \wedge d \omega = 0$. See below for discussion of whether the ZAMO congruence meets this condition.

Synchronizable: This is equivalent to there being scalar functions $h$ and $t$ such that $\omega = - h dt$. The text says that this ensures that $\omega \wedge d\omega = 0$, meaning that synchronizable implies locally synchronizable (but the converse is not true); I think I see how this works although I probably need to brush up on wedge products to fully check it for myself . The ZAMO congruence meets this condition, with $h$ being a function of $r$ and $\theta$ (which is why we can't just rescale the time coordinate to make $\omega = - dt$, so the congruence is *not* proper time synchronizable) that I won't write down here; which means it also meets the above condition assuming that the implication just noted holds.

So to briefly recap, the chain of implications runs like this:

proper time synchronizable -> locally proper time synchronizable

proper time synchronizable -> synchronizable

locally proper time synchronizable -> locally synchronizable

synchronizable -> locally synchronizable

(If we drew this out as a diagram, it would form a diamond-shaped pattern.)

Now, looking at the exercise you mention, I see an important comment in it (bolded in the quote below):

Generalize the above discussion to the case that $Z$ is synchronizable but not proper time synchronizable by assuming one autocrat $\gamma$ and other observers who regard the consistency condition...as more important than insisting on their own proper time.

In other words, the "synchronized" clocks for all of the "other observers" will *not* run at the same rate as their proper time clocks do. That is what was bothering me about using the term "Einstein clock synchronization" with observers in relative motion: I've always understood Einstein clock synchronization to be a way of matching up spatially separated clocks whose proper time "runs at the same rate", which is of course a very restrictive condition (it basically means proper time synchronizable). If that requirement is dropped, so all we need are "common surfaces of simultaneity" without requiring that every observer's proper time ticks off the same amount between two such surfaces, then I agree that a much wider range of congruences can meet that requirement. (For the ZAMO congruence, the "autocrat" would be an observer at infinity, whose proper time is the same as Boyer-Lindquist coordinate time, which would be the "synchronized" time for the entire congruence--but would not match the proper time of any members of the congruence not at infinity.)

And just to briefly review the other congruences I mentioned in a previous post:

* A congruence of inertial observers in Minkowski spacetime obviously meets all four conditions.

* The Rindler congruence is synchronizable but not proper time synchronizable (the scalar $h$ is a function of the Rindler spatial coordinate $x$). Same for the congruence of static observers in Schwarzschild spacetime.

* The Bell congruence does not meet any of the four conditions, as far as I can tell.

* The Painleve congruence is the interesting one: by the definitions given, it is proper time synchronizable (with the scalar $t$ being Painleve coordinate time), which means it meets all four conditions. However, it has nonzero expansion, which violates my intuition that only a rigid congruence can be proper time synchronizable (to phrase it in the proper terminology given all of the above). I'll need to consider this one some more.

 

[WannabeNewton]

Locally proper time synchronizable: In one place it says this is equivalent to the congruence being geodesic and irrotational; in another place it says this is equivalent to $d \omega = 0$.

Yeah I wish the text was available online so I apologize for the inconvenience with regards to the notation (which I myself am not used to as well) and the dispersion of terms. To put the above in notation we're more comfortable with, I'll use the 4-velocity field $u^{\mu}$ again instead of the associated 1-form $\omega$ that Sachs and Wu use.

Then $u^{\mu}$ is locally proper time synchronizable if and only if $\nabla_{[\mu}u_{\nu]} = 0$. Recall that for a general time-like congruence, we can decompose $\nabla_{\mu}u_{\nu}$ into the rotation tensor, expansion tensor, and 4-acceleration as $\nabla_{\mu}u_{\nu} = \omega_{\mu\nu}+ \theta_{\mu\nu}- u_{\mu}u^{\gamma}\nabla_{\gamma}u_{\nu}$.

Now $\nabla_{[\mu}u_{\nu]} = \omega_{\mu\nu}- u_{[\mu}u^{\gamma}\nabla_{|\gamma|}u_{\nu]}$; if $u^{\mu}$ is geodesic and irrotational then clearly $\nabla_{[\mu}u_{\nu]} = 0$ and conversely if $\nabla_{[\mu}u_{\nu]} = 0$ then note that $0=u^{\mu}\nabla_{[\mu}u_{\nu]} = u^{\gamma}\nabla_{\gamma}u_{\nu} + u_{\nu}u^{\gamma}u^{\mu}\nabla_{\gamma}u_{\mu} = u^{\gamma}\nabla_{\gamma}u_{\nu} = 0$ which consequently implies $\omega_{\mu\nu}$ so $u^{\mu}$ is geodesic and irrotational.

Locally synchronizable: This is equivalent to $\omega \wedge d \omega = 0$. See below for discussion of whether the ZAMO congruence meets this condition.

Synchronizable: This is equivalent to there being scalar functions $h$ and $t$ such that $\omega = - h dt$. The text says that this ensures that $\omega \wedge d\omega = 0$, meaning that synchronizable implies locally synchronizable (but the converse is not true); I think I see how this works although I probably need to brush up on wedge products to fully check it for myself . The ZAMO congruence meets this condition, with $h$ being a function of $r$ and $\theta$ (which is why we can't just rescale the time coordinate to make $\omega = - dt$, so the congruence is *not* proper time synchronizable) that I won't write down here; which means it also meets the above condition assuming that the implication just noted holds.

Yes I agree with everything you said. As for synchronizability implying local synchronizability, using our usual notation we know that if $u^{\mu} = \alpha \nabla^{\mu} t$ for a scalar field $\alpha$ then $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ as shown earlier in the thread. The condition $\omega \wedge d\omega =0$ is the exact same as $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$ if we let $\omega_{\mu} = u_{\mu}$. See Wald pp.428-429.

In other words, the "synchronized" clocks for all of the "other observers" will *not* run at the same rate as their proper time clocks do. That is what was bothering me about using the term "Einstein clock synchronization" with observers in relative motion: I've always understood Einstein clock synchronization to be a way of matching up spatially separated clocks whose proper time "runs at the same rate", which is of course a very restrictive condition (it basically means proper time synchronizable). If that requirement is dropped, so all we need are "common surfaces of simultaneity" without requiring that every observer's proper time ticks off the same amount between two such surfaces, then I agree that a much wider range of congruences can meet that requirement.

Alright so I was in fact using the term Einstein synchrony incorrectly then, sorry again about that. For some reason I had it in my head that if a synchronization scheme used the usual $1/2$ averaging method for clock times then it was Einstein synchrony but as you pointed out there is the additional constraint that the local clock times (proper times) must tick at equal rates upon synchronization in order for us to have the canonical form of Einstein synchrony. But at least we've agreed now on the fact that using the radar synchronization method, observers in twist-free congruences can locally synchronize their clocks and observers in congruences whose tangent fields are parallel to gradients of scalar fields can globally synchronize their clocks.

(For the ZAMO congruence, the "autocrat" would be an observer at infinity, whose proper time is the same as Boyer-Lindquist coordinate time, which would be the "synchronized" time for the entire congruence--but would not match the proper time of any members of the congruence not at infinity.)

That's how I understood the exercise as well when applied to the ZAMO congruence.

If you have time, I would appreciate it if you could take a look at page 104 of "Gravitation and Inertia"-Wheeler et. al wherein the authors describe simultaneity and clock synchronization for arbitrary space-times, and compare/contrast with what we have discussed thus far in this thread. Thanks again!

 

[yuiop]

Sorry for the long post but I just wanted to make sure that I myself understood what you were trying to describe in the first place. Cheers.

I think you interpreted what I was getting at correctly.

So the worldline of the rotating observer with his gyro will look like a very, very stretched out helix, whose tangent vectors are almost vertical: and adding the very, very small angular momentum to the BH will only need to change the behavior of the gyroscope by a very, very small amount to switch it from the right hand sketch behavior to the left hand sketch behavior.

After giving WBN's post some thought, I came to a similar conclusion that the time factor and the slow rotation involved makes the apparent 'step change' in the behaviour of the gyroscopes after a small perturbation, less dramatic, so I will concede I have no strong counter arguments to WBN's position in post #53.

Back to the question of azimuthal acceleration or lack of. On page 66 of this book, equation (3.3.37) states that the azimuthal acceleration is zero, but this is from the point of view of a ZAMO observer, so this is to be expected. I am still unable to see pages 62 and 63 of the same book, that gives the acceleration according to a static local observer with $(d\phi=d\theta=dr=0)$ in the Kerr metric. Any chance of someone forwarding those 2 pages to me?

 

[WannabeNewton]

Any chance of someone forwarding those 2 pages to me?

Here you go bud

http://postimg.org/gallery/8e3o2wjk/791e59dc/

 

[WannabeNewton]

Peter, you might be interested in sections 5 and 6 of this paper: http://arxiv.org/pdf/0708.0170v1.pdf

It initially states exactly what Sachs and Wu state i.e. that observers belonging to a time-like congruence $\xi^{\mu}$ can (locally) synchronize their clocks using radar if and only if $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} = 0$ but uses notation that both you and I are well accustomed to; note however the caveat that the clocks must be infinitesimally separated (top of p.17).

After that it introduces Proposition 4 (also p.17) which in some sense generalizes the radar synchronization method above to clocks that are not infinitesimally separated but close enough so that radar can be utilized. The added restriction is that now $\xi^{\mu}$ has to satisfy both $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} = 0$ and $\nabla^{(\mu}\xi^{\nu)} = 0$ i.e. it must be both irrotational and rigid.

 

[yuiop]

Here you go bud

http://postimg.org/gallery/8e3o2wjk/791e59dc/

Thanks WBN! I appreciate you taking the time to upload the pages. It seems we are still stuck with the paradox Peter mentioned earlier. Why does the coordinate angular velocity of the free falling particle in the equatorial plane, increase if there is no angular acceleration due to gravity?

I think I may have to resort back to the idea of the connection between spin angular momentum and orbital angular momentum that I mentioned earlier. Normally these quantities are conserved separately. Spin angular momentum of the particle is conserved if there are no external torques acting on it, but there is an external torque acting on the particle, altering its intrinsic spin in the Kerr metric (if its orbital velocity is not that of a ZAMO), so the particle can not be considered as an isolated system. A particle with initial zero spin and zero orbital velocity in the Kerr metric, will generally have its spin increased and to conserve total angular momentum the orbital velocity has to increase in the opposite direction to conserve a total angular momentum of zero. Basically, in the Kerr metric, there appears to be spin-orbit coupling and these quantities are not individually conserved.

Normally spin-orbit coupling is associated with atoms and electrons but it can apply in large systems. Consider a rotor arm that has a fixed rotation axis at one end and a flywheel attached to the other end of the rotor via an electric motor, such that the rotation axes of the rotor and the flywheel are parallel and orthogonal to the rotor arm. Initially with the motor switched off, the total angular momentum is zero. When the motor is energised, it spins up the flywheel, but conservation of angular momentum dictates that the rotor arm must rotate in the opposite sense so that the flywheel starts orbiting the fixed axis of the rotor arm. The coupling of the spin and orbital motion in this system, conserves total angular momentum, but the spin and orbital angular momenta are not individually conserved. The coupling between the rotor and the flywheel is due to the electromagnetic field inside the energised motor. In the Kerr metric, the coupling is due to the gravitational field.

An example of gravitational spin-orbit coupling is the Earth, Moon system. Tidal friction is causing the Earth intrinsic spin to gradually slow down. To conserve the total angular momentum of the system, the orbital angular momentum of the Moon has to increase and this is reflected in the gradual increase of the orbital radius of the Moon around the Earth. Here again, spin and orbital angular momentum are not individually conserved in a coupled system.

 

[PeterDonis]

Why does the coordinate angular velocity of the free falling particle in the equatorial plane, increase if there is no angular acceleration due to gravity?

I think the answer is that there *is* angular acceleration due to gravity, for objects that have nonzero radial velocity. In other words, my earlier claim that the acceleration due to gravity is the same for a static observer as for an object freely falling radially past him, was wrong.

Suppose we have an object which is only moving radially, i.e., its 4-velocity (in Boyer-Lindquist coordinates) only has components $(u^t, u^r)$. The proper acceleration of such an object is given by

$$
a^b = u^a \nabla_a u^b = u^a \left( \partial_a u^b + \Gamma^b{}_{ac} u^c \right)
$$

Expanding this out, we have

$$
a^b = u^t \left( \partial_t u^b + \Gamma^b{}_{tc} u^c \right) + u^r \left( \partial_r u^b + \Gamma^b{}_{rc} u^c \right)
$$

We are interested in whether the $b$ index can be anything other than $t$ or $r$. The partial derivative terms can't contribute any terms like that (since $u^b$ only has $t$ and $r$ components), but the connection coefficient terms can--in Kerr spacetime, not Schwarzschild spacetime. In Schwarzschild spacetime, the only connection coefficients with $tt$, $tr$, or $rr$ as lower indexes (which is required to match the only nonzero $u^c$ components) have $t$ or $r$ as upper indexes. But in Kerr spacetime, there is at least one that doesn't (assuming that I've done the Maxima inputs correctly): $\Gamma^{\phi}{}_{tr} = \Gamma^{\phi}{}_{rt}$. This will contribute a nonzero $a^{\phi}$ component, but only if $u^r$ is nonzero. So a static observer will not see any angular acceleration due to gravity, but an observer free-falling radially will.

 

[WannabeNewton]

According to the following paper, $\Gamma^{\phi}_{tr}$ is non-vanishing: http://articles.adsabs.harvard.edu//full/1999MNRAS.308..863S/0000874.000.html (see p.874)

That being said, for a freely falling particle we have $u^{\gamma}\nabla_{\gamma}u^{\mu} = 0$ so that in particular $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2} = -\Gamma ^{\phi}_{\mu\nu}u^{\mu}u^{\nu}$.

If $u^{\mu}|_{\tau = 0} = (u^t_0, u^r_0,0,0)$ then $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} = -2\Gamma ^{\phi}_{r t}u^{t}_0 u^{r}_0$.

Hence $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} \neq 0$ and thus $\frac{\mathrm{d}^2 \phi}{\mathrm{d} t^2}|_{\tau = 0} = (u^t_0)^{-2}\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} + u^{\phi}_{0}\frac{\mathrm{d} ^{2}\tau}{\mathrm{d} t^2}|_{\tau = 0} = (u^t_0)^{-2}\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0}\neq 0$.

So even if we have $\Omega_0 = 0$ for a freely falling particle, $\dot{\Omega}_0 \neq 0$ which will start increasing the freely falling particle's angular velocity relative to infinity.

Also, I don't think there's any problem with the acceleration of the freely falling particle as it passes by the origin of the local Lorentz frame of a static observer. If a freely falling particle passes by the origin of the Fermi-Normal coordinates of a static observer, then at that event the static observer will measure the acceleration due to gravity of the freely falling particle to be $\vec{g} = -\vec{a} - 2\vec{\Omega}\times \vec{v} + 2(\vec{a}\cdot \vec{v})\vec{v}$ (see exercise 13.14 in MTW). Here $\vec{a}$ is the static observer's acceleration, $\vec{v}$ is the 3-velocity of the freely falling particle, and $\vec{\Omega}$ is the spin of the static observer's local Lorentz frame; all of these quantities are of course relative to the Fermi-Normal coordinates of the static observer.

So $\vec{g}$ isn't just $-\vec{a}$ (inertial acceleration) but also $-2\vec{\Omega}\times \vec{v}$ (Coriolis acceleration) and $2(\vec{a}\cdot \vec{v})\vec{v}$ (relativistic correction to inertial acceleration).

Now as we already know the congruence of static observers in Kerr space-time has a 4-velocity field that's parallel to the time-like killing field $\xi^{\mu}$, which itself has a non-vanishing twist (vorticity) given by $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta}$. According to the following paper, the vorticity has non-vanishing radial and polar components in the local Lorentz frame of a given static observer in the congruence: http://arxiv.org/pdf/1210.6127.pdf (see eq.(35) in p.4). Furthermore, according to eq.(2.161) in p.54 of Straumann's GR text, the spin of the static observer's local Lorentz frame is given by $\Omega^{\mu} = -\frac{1}{2}(\xi^{\gamma}\xi_{\gamma})^{-1}\epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta}$.

So in particular the spin should have a polar component and if we have a freely falling particle pass through the origin of the static observer's local Lorentz frame such that it only has a radial 3-velocity at that instant then the cross product of its radial 3-velocity with the polar component of the spin of the static observer's local Lorentz frame should yield an azimuthal component to the Coriolis acceleration that contributes to $\vec{g}$ at that instant.

 

[PeterDonis]

According to the following paper, $\Gamma^{\phi}_{tr}$ is non-vanishing: http://articles.adsabs.harvard.edu//full/1999MNRAS.308..863S/0000874.000.html (see p.874)

Good, then I did the Maxima inputs right.

That being said, for a freely falling particle we have $u^{\gamma}\nabla_{\gamma}u^{\mu} = 0$ so that in particular $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2} = -\Gamma ^{\phi}_{\mu\nu}u^{\mu}u^{\nu}$.

If $u^{\mu}|_{\tau = 0} = (u^t_0, u^r_0,0,0)$ then $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} = -2\Gamma ^{\phi}_{r t}u^{t}_0 u^{r}_0$.

Hence $\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} \neq 0$ and thus $\frac{\mathrm{d}^2 \phi}{\mathrm{d} t^2}|_{\tau = 0} = (u^t_0)^{-2}\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0} + u^{\phi}_{0}\frac{\mathrm{d} ^{2}\tau}{\mathrm{d} t^2}|_{\tau = 0} = (u^t_0)^{-2}\frac{\mathrm{d} ^{2}\phi}{\mathrm{d} \tau^2}|_{\tau = 0}\neq 0$.

So even if we have $\Omega_0 = 0$ for a freely falling particle, $\dot{\Omega}_0 \neq 0$ which will start increasing the freely falling particle's angular velocity relative to infinity.

Yes, but note that you need to have $u^r_0 \neq 0$ in order to have $\dot{\Omega} \neq 0$. So an object that is released by a static observer in free fall, so that it initially has $u^r = 0$, will have $\dot{\Omega} = 0$ initially. But as it builds up radial velocity, it will start to have $\dot{\Omega} \neq 0$.

So $\vec{g}$ isn't just $-\vec{a}$ (inertial acceleration) but also $-2\vec{\Omega}\times \vec{v}$ (Coriolis acceleration) and $2(\vec{a}\cdot \vec{v})\vec{v}$ (relativistic correction to inertial acceleration).

Yes, I agree.

 

[WannabeNewton]

So I guess then that the black holes text from the google preview linked earlier in the thread forgot to take into account the Coriolis acceleration when calculating the acceleration due to gravity on the freely falling particle in the local Lorentz frame of a static observer? Because it claimed that the acceleration due to gravity on the freely falling particle in such a static frame would just be $\vec{g} = -\vec{a}$ but this would neglect the Coriolis acceleration, and hence the azimuthal component of acceleration due to gravity, on the particle due to the spinning of the static frame and we know that a static frame in Kerr space-time must be spinning due to frame dragging so the Coriolis acceleration can't be eliminated.

 

[Mentz114]

If there was azimuthal acceleration as well as regular radial acceleration due to gravity, the plumb bob would be accelerated downwards and sideways, so if the plumb bob is used by an observer as a local reference for vertical, he will not detect any azimuthal acceleration. However, I am not sure if that is how a local observer defines vertical and probably by such a definition he would not detect any polar acceleration either. Any thoughts? Just trying to get to the bottom of this puzzle.

Going back a bit to this question. Some time ago I worked out ( with Maxima) the kinematics for a static ( hovering ) frame in Boyer-Lundqvist coordinates.

acceleration has components in the $r$ and $\theta$-directions. The $r$ term contains $\cos(\theta)$ with values

$a_r = \frac{m\,r-{a}^{2}}{{r}^{2}\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}}$ with $\cos(\theta)=0$ ( azimuthal) and $a_r = \frac{m\,\left( r-a\right) \,\left( r+a\right) }{{\left( {r}^{2}+{a}^{2}\right) }^{2}\,\left( {r}^{2}-2\,m\,r+{a}^{2}\right) }$ with $\cos(\theta)=1$ ( polar).

In the $\theta$-direction $a_\theta = -\frac{{a}^{2}\,\cos\left( \theta\right) \,\sin\left( \theta\right) }{{\left( {a}^{2}\,{\cos\left( \theta\right) }^{2}+{r}^{2}\right) }^{{3}/{2}}}$

If these are correct, the polar and azimuthal bovering frames need only an $r$-acceleration, but in btween there is also a $\theta$-acceleration required.

There is vorticity ( spin around the $r$-axis) of $-\frac{a\,\sqrt{{r}^{2}-2\,m\,r+{a}^{2}}\,\cos\left( \theta\right) }{{\left( {a}^{2}\,{\cos\left( \theta\right) }^{2}+{r}^{2}\right) }^{{3}/{2}}}$ which is greatest at the pole and falls to zero in the azimuthal plane.

 

[George Jones]

So I guess then that the black holes text from the google preview linked earlier in the thread forgot to take into account the Coriolis acceleration when calculating the acceleration due to gravity on the freely falling particle in the local Lorentz frame of a static observer?

I haven't been following this thread at all, but this is really hard to believe. Frolov and Novikov! Is it possible that the book calculates something different?