Practical measurements of rotation in the Kerr metric

[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?

 

[PeterDonis]

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.

This doesn't look right. The vorticity should be around the $z$ axis (or at least there should be a component around that axis--i.e., the axis perpendicular to the $r-\phi$ plane), and it should not be zero in the equatorial (i.e., $\theta = \pi / 2$) plane.

 

[George Jones]

I have now read the passage in the book. The book says

Let us look at the forces acting in this frame due to the presence of a rotating black hole.

This means "At all p, compute the relative acceleration between a static observer at p and a momentarily comoving ($\vec{v}=0$) freely falling observer at p."

So, as shown by WannabeNewton and PeterDonis, the book is correct.

 

[WannabeNewton]

This means "At all p, compute the relative acceleration between a static observer at p and a momentarily comoving ($\vec{v}=0$) freely falling observer at p."

Ah ok, all is well then

Merry Christmas!

 

[Mentz114]

This doesn't look right. The vorticity should be around the $z$ axis (or at least there should be a component around that axis--i.e., the axis perpendicular to the $r-\phi$ plane), and it should not be zero in the equatorial (i.e., $\theta = \pi / 2$) plane.

Agreed. I've checked for gross errors but baffled by this.. The comoving frame basis I'm using mixes $t$ and $\phi$ so I'm not sure what 'comoving' means here.

 

[yuiop]

Hope you all had a pleasant Christmas!

It seems that the velocity dependent Coriolis force explanation given by Peter and WBN, resolves the paradoxical issues mentioned earlier. Equations 4.24 and 4.22 of this text gives an expression for the angular velocity $d\phi/d\tau$ (assuming the affine parameter is the proper time of the particle) for a free falling particle, in terms of $M, \alpha, L, E$ and r. All these parameters are constants of motion except for r which can be expressed as a function of proper time $r(\tau)$. If $d r(\tau)/d\tau =0$ then $r(\tau)$ is a constant then there is no time dependent parameter left in the expression for $d\phi/d\tau$, so $d^2 \phi/ d\tau^2$ must be zero under those conditions. This agrees with the velocity dependent nature of the 'fictitious' Coriolis force.

Something that is still puzzling me is the section on frame dragging in this Wikipedia article on the Kerr metric.

Qualitatively, frame-dragging can be viewed as the gravitational analog of electromagnetic induction. An "ice skater", in orbit over the equator and rotationally at rest with respect to the stars, extends her arms. The arm extended toward the black hole will be torqued spinward. The arm extended away from the black hole will be torqued anti-spinward. She will therefore be rotationally sped up, in a counter-rotating sense to the black hole. This is the opposite of what happens in everyday experience. If she is already rotating at a certain speed when she extends her arms, inertial effects and frame-dragging effects will balance and her spin will not change. Due to the Principle of Equivalence gravitational effects are locally indistinguishable from inertial effects, so this rotation rate, at which when she extends her arms nothing happens, is her local reference for non-rotation. This frame is rotating with respect to the fixed stars and counter-rotating with respect to the black hole. A useful metaphor is a planetary gear system with the black hole being the sun gear, the ice skater being a planetary gear and the outside universe being the ring gear. This can be also be interpreted through Mach's principle.

Is this qualitative description correct? It says nothing about changing orbital radius so presumably includes perfectly circular orbits. It seems to imply that any orbiting object with a given (possibly constant) orbital velocity and radius, has an associated spin rate and if the object does not have this critical spin rate it will be subjected to a spin torque until the spin reaches the associated critical value.

 

[WannabeNewton]

I find that the quoted qualitative description is poorly worded.

To summarize, frame dragging, at the simplest level, can manifest itself in two ways:

In the case of a ZAMO frame, frame dragging causes the ZAMO frame to have a non-zero angular velocity relative to infinity. However the ZAMO frame has no intrinsic spin again because the ZAMO congruence has a 4-velocity field $\eta^{\mu} = \alpha\nabla^{\mu}t$ that has identically vanishing vorticity: $\eta^{[\gamma}\nabla^{\mu}\eta^{\nu]} = 0$. In other words the ZAMO frame is locally non-rotating: it sets the local standard of non-rotation just like an inertial frame in Minkowski space-time sets the standard of non-rotation. Mutually orthogonal gyroscopes at rest in the ZAMO frame don't precess and hence the ZAMO frame itself doesn't precess relative to asymptotic Minkowski frames since asymptotic Minkowski frames are also locally non-rotating. But the ZAMO frames do, as mentioned, have an orbital angular velocity relative to asymptotic Minkowski frames.

On the other hand, we have for a static frame a manifest intrinsic spin due to frame dragging; this is again because the static congruence has a 4-velocity field $\xi^{\mu} = \beta (\frac{\partial}{\partial t})^{\mu}$ that has non-zero vorticity: $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} \neq 0$. The static frame will obviously not have an orbital angular velocity relative to asymptotic Minkowski frames but it will rotate relative to a local ZAMO frame, that is mutually orthogonal gyroscopes at rest in the static frame will precess and hence the static frame itself will precess relative to asymptotic Minkowski frames.

EDIT: if the relationship between $\eta^{[\gamma}\nabla^{\mu}\eta^{\nu]} = 0$ and a lack of gyroscopic precession isn't clear then see section II.C of the following paper: http://arxiv.org/pdf/1210.6127v4.pdf

 

[Mentz114]

Mistake in paper ?

I have a query about the paper referred to earlier ( http://arxiv.org/pdf/1210.6127v4.pdf). It has a section on the Kerr metric at the top of the second column on page 4. I think the frame basis and cobasis vectors given are wrong, in that they do not give the $d\phi^2$ term in the metric above. This is easy to see because the contributions to the $d\phi^2$ term come from $e^\hat{t}$ and $e^\hat{\phi}$, i.e. $\left( \frac{2\,a\,m\,r\,{\sin\left( \theta\right) }^{2}}{\sqrt{\Sigma}\,\sqrt{\Sigma-2\,m\,r}} \right)^2 + \left(\frac{\sin\left( \theta\right) \,\sqrt{\Delta}\,\sqrt{\Sigma}}{\sqrt{\Sigma-2\,m\,r}} \right)^2 = \frac{{\sin\left( \theta\right) }^{2}\,\left( \Delta\,{\Sigma}^{2}+4\,{a}^{2}\,{m}^{2}\,{r}^{2}\,{\sin\left( \theta\right) }^{2}\right) }{\Sigma\,\left( \Sigma-2\,m\,r\right) }$.

Have I made a mistake, or missed something ?

 

[yuiop]

... The static frame will obviously not have an orbital angular velocity relative to asymptotic Minkowski frames but it will rotate relative to a local ZAMO frame, that is mutually orthogonal gyroscopes at rest in the static frame will precess and hence the static frame itself will precess relative to asymptotic Minkowski frames.

Thanks for the explanation WBN. I am curious if the axis of gyroscopes in the static frame rotates at the same rate as the axis of a gyroscope in the ZAMO frame with equal radius as seen by the very distant observer? To clarify, consider a ZAMO observer that orbits a Kerr BH once per year as viewed by the observer at 'infinity'. The distant observer sees the axis of the ZAMO gyroscope rotate once per year while maintaining a constant angle wrt the instantaneous radial axis of the ZAMO observer. Will the axis of a gyroscope held by a static observer at the same radius also rotate once per year as observed by the distant observer, or is not as simple as that?

Now back to the Wikpedia ice skater...

... An "ice skater", in orbit over the equator and rotationally at rest with respect to the stars, extends her arms. The arm extended toward the black hole will be torqued spinward. The arm extended away from the black hole will be torqued anti-spinward. She will therefore be rotationally sped up, in a counter-rotating sense to the black hole. This is the opposite of what happens in everyday experience. ..

I think I 'get' this now. While the orbiting ice skater is at rest wrt the distant stars, she does not have zero angular momentum. When she extends her arms, conservation of angular momentum requires that her spin angular velocity slows down, but to the distant observer, this looks like an increase in spin angular velocity relative to the distant stars. In fact, this idea provides a practical alternative to a gyroscope to determine angular spin momentum (or lack of). If extending your arms causes no change in your angular spin velocity, then you have no spin momentum (around your vertical axis), even if you appear to be rotating relative to the distant stars or relative to a radial vector pointing at the gravitational source or whatever other reference point you choose.

 

[WannabeNewton]

Will the axis of a gyroscope held by a static observer at the same radius also rotate once per year as observed by the distant observer, or is not as simple as that?

I hope I'm understanding your use of the term "rotation" correctly here in that "rotation of the ZAMO gyroscope" refers to the orbital rotation of the gyroscope around the central Kerr BH relative to infinity, since it has no spin rotation (as you noted its angle is constant wrt the instantaneous spatial axes of a ZAMO frame), whereas "rotation of the static gyroscope" refers to the spin rotation of the gyroscope relative to infinity.

If so, it isn't as simple as what you stated but it can still be described mathematically at the surface level. Consider the entire family of static observers in Kerr space-time; the congruence of worldlines of these observers has, as noted earlier, the 4-velocity field $\xi^{\mu} = \frac{1}{\sqrt{-g_{tt}}}\delta^{\mu}_{t} = \gamma \delta^{\mu}_t$.

Consider an observer $O$ in this family. We attach to $O$ a frame, which, by definition, is a choice of orthonormal basis wherein the time-like basis vector is simply the 4-velocity of $O$. The particular frame we attach to $O$ is given by $\{\xi^{\mu}, \eta^{\mu}_1, \eta^{\mu}_2, \eta^{\mu}_3 \}$ where $\eta^{\mu}_{i}$ are spatial basis vectors setup in a special way: $O$ has three infinitesimally neighboring static observers separated from him along the directions $\eta^{\mu}_1,\eta^{\mu}_2,\eta^{\mu}_3$ of this frame and each of $\eta^{\mu}_1,\eta^{\mu}_2,\eta^{\mu}_3$ points from him to the respective neighbors. In order to force the $\eta^{\mu}_{i}$ to remain locked to the respective neighbors everywhere along his worldline, $O$ uses what's called Lie transport: $\mathcal{L}_{\xi}\eta^{\mu}_{i} = \xi^{\gamma}\nabla_{\gamma}\eta^{\mu}_{i} - \eta^{\gamma}_{i}\nabla_{\gamma}\xi^{\mu} = 0$. We have thus defined a frame for $O$; let's call this a static frame.

*Of course there's a physical reasoning behind this choice of frame and it relates to the way in which we aim to measure the rotation of the frame. Imagine we are in flat space-time and we have a rotating disk. Two superimposed observers $O$ and $O'$ sit at the center of the disk; $O$ belongs to a family of observers, the rest of whom are situated at all other points on the disk. Imagine that relative to $O'$ all these other observers have instantaneous tangential velocities $\vec{v} = \vec{\omega}\times \vec{r}$ where $\vec{\omega}$ is the angular velocity of the disk and $\vec{r}$ the position vector to a given observer, whereas relative to $O$ all these other observers are stationary on the disk. Then clearly $O$ is spinning in place relative to $O'$.

Now if we attach to $O$ the frame $\{\xi^{\mu}, \eta^{\mu}_1, \eta^{\mu}_2, \eta^{\mu}_3 \}$ from above then the Lie transport condition $\mathcal{L}_{\xi}\eta^{\mu}_{i} = 0$ that locked the spatial basis vectors to neighboring observers is the exact same thing as having the observers sitting at all other points on the disk remain stationary relative to $O$. On the other hand $O'$ has an inertial frame attached to him given by some $\{e^{\mu}_0, e^{\mu}_1,e^{\mu}_2,e^{\mu}_3 \}$ and because this is an inertial frame, the $e^{\mu}_{i}$ are physically realized by torque-free gyroscopes. In other words this frame constitutes what we call a non-rotating frame. Above we said that the observers sitting at all other points on the disk have tangential velocities relative to $O'$; this is exactly the same thing as saying that the spatial basis vectors $\eta^{\mu}_{i}$ of $O$'s frame rotate relative to the gyroscopes $e^{\mu}_{i}$. This measures the rotation of $O$'s frame relative to the non-rotating frame of $O'$ by means of torque-free gyroscopes.*

Coming back to the congruence of static observers in Kerr space-time, we measure the rotation of a static frame almost identically to what was described above. The main difference now is that we must work with locally non-rotating frames so we superimpose a locally non-rotating observer on our chosen static observer $O$ (meaning they share the same worldline) and the rotation of the static frame attached to $O$ (more precisely, the rotation of the $\eta^{\mu}_{i}$) is, at each event, measured relative to the torque-free gyroscopes of the locally non-rotating frame attached to the locally non-rotating observer superimposed on $O$. We define the vorticity 4-vector $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta}$ as the rotation of the $\eta^{\mu}_{i}$ (and hence of the static frame). This is just a relativistic generalization of the vorticity 3-vector from fluid mechanics. It's magnitude is simply $\omega = \sqrt{\omega^{\mu}\omega_{\mu}}$ and the units of $\omega$ are $[\omega] = \frac{1}{s}$.

The periodicity of $\omega$ is with respect to a clock carried by a locally non-rotating observer sharing the same worldline as our static observer $O$ so in order to get $\omega_{\infty}$ we simply tack on the "gamma factor" $\gamma = \frac{dt}{d\tau}$ where $d\tau$ is the proper time along the shared worldline of $O$ and a locally non-rotating observer. Doing this we get $\omega_{\infty} = \gamma^{-1}\omega$.

Letting $\Omega$ be the angular velocity of a ZAMO observer about the central Kerr BH, we can then compute $\frac{\omega_{\infty}}{\Omega}$ in order to compare the two rotation rates. I haven't explicitly done the calculation yet but just looking at the form of $\omega^{\mu}$ from the paper I linked earlier, a priori I can't see any reason why we would get $\frac{\omega_{\infty}}{\Omega} = 1$.

Sorry for the incredibly long post but I just wanted to describe, once and for all, what we are really measuring when it comes to the rotation of static frames.

 

[PeterDonis]

Letting $\Omega$ be the angular velocity of a ZAMO observer about the central Kerr BH, we can then compute $\frac{\omega_{\infty}}{\Omega}$ in order to compare the two rotation rates. I haven't explicitly done the calculation yet but just looking at the form of $\omega^{\mu}$ from the paper I linked earlier, a priori I can't see any reason why we would get $\frac{\omega_{\infty}}{\Omega} = 1$.

If $\Omega$ is also measured at infinity (i.e., using time at infinity), then you will have $\omega_{\infty} / \Omega_{\infty} = 1$. Think of the distant observer looking down on a static observer directly below him (radially), and a family of ZAMOs all at the same radius as the static observer (this is all in the "equatorial plane" so the family of ZAMOs occupies the "orbital ring" in the equatorial plane at that radius). At some time $t = 0$ the distant observer marks which ZAMO is just passing the static observer; he chooses the time so that the static observer's gyroscope also points directly at him at the same instant. Then he waits until the same ZAMO comes around to pass the static observer again.

Observe, first, that the ZAMO's gyroscope must also be pointing directly at the distant observer at $t = 0$. (That has to be the case because the ZAMO's gyroscope always points directly radially outward.) Observe next that the ZAMO's gyroscope and the static observer's gyroscope will always be parallel (because they must both be rotating, relative to infinity, at the same rate, since they are at the same radius). That means the static observer's gyro will once again point at the distant observer (i.e., will have completed one rotation) at the same time the ZAMO's does; but that will be precisely when the ZAMO passes the static observer again, as seen by the observer at infinity.

The part that may be counterintuitive is that the static observer and the ZAMO do *not* measure the same angular velocity of rotation; $\Omega$ as measured by the ZAMO (by watching successive passages overhead of an object at infinity) is *not* the same as $\omega$ as measured by the static observer (by the method you describe). That's because they are in relative motion, so their respective $\gamma$ factors relative to infinity are not the same.

 

[WannabeNewton]

(because they must both be rotating, relative to infinity, at the same rate, since they are at the same radius)

I don't understand why the two rotation rates relative to infinity must be the same, even if the ZAMO and the static observer are at the same radius. Wouldn't that require $\omega_{\infty} = \Omega_{\infty}$? We have $\Omega_{\infty} = \frac{2Mar}{(r^2 + a^2)^2 - \Delta a^2 sin^2\theta}$. As for $\omega_{\infty} = \gamma^{-1}\sqrt{\omega^{\mu}\omega_{\mu}}$ we have to compute the vorticity $\omega^{\mu}$ of the congruence of static observers with 4-velocity field $\xi^{\mu} =\gamma\delta^{\mu}_{t}$. If the expressions for $\omega^{\mu}$ in the following paper are correct, then I don't see why $\omega_{\infty} = \Omega_{\infty}$ would have to hold just by inspection (I'm not saying it won't but I feel like crying just at the sight of the computation that would be required to verify this ): http://arxiv.org/pdf/1210.6127v4.pdf (p.4)

 

[PeterDonis]

Wouldn't that require $\omega_{\infty} = \Omega_{\infty}$?

Yes; but I haven't actually done the computation, I just gave a physical argument that seems convincing to me. I don't relish the computation either.

 

[WannabeNewton]

Yes; but I haven't actually done the computation, I just gave a physical argument that seems convincing to me. I don't relish the computation either.

Sorry, I think I'm not understanding the physical argument well enough then. Could you explain again why $\omega_{\infty} = \Omega_{\infty}$ should hold physically, at least when restricted to a single radius?

 

[yuiop]

Sorry, I think I'm not understanding the physical argument well enough then. Could you explain again why $\omega_{\infty} = \Omega_{\infty}$ should hold physically, at least when restricted to a single radius?

I am beginning to suspect that $\omega_{\infty} \ne \Omega_{\infty}$ by considering what happens in the Schwarzschild case due to geodetic and/or the Lense-Thirring precession. In the Schwarzschild metric a non orbiting gyroscope at constant altitude continues to point at a given distant star indefinitely, while an orbiting gyroscope does not, or at least that is what the Gravity probe B experiment is claimed to demonstrate. It seems clear that in the Schwarzschild metric, the equivalent of $\omega_{\infty} = \Omega_{\infty}$ does not hold, so there in strong reason to think it should automatically hold in the Kerr metric. In fact the Lense-Thirring effect is the precession of an orbiting gyroscope caused by the rotation of the gravitational body, so the equations for that should be directly relevant to the Kerr metric.

 

[PeterDonis]

I am beginning to suspect that $\omega_{\infty} \ne \Omega_{\infty}$ by considering what happens in the Schwarzschild case

Unless I'm misunderstanding the terminology, $\omega$ and $\Omega$ are both zero in Schwarzschild spacetime. A ZAMO in Schwarzschild spacetime has zero angular velocity relative to infinity, so ZAMO = static observer, and a static observer's vorticity is zero (more precisely, the vorticity of the static congruence is zero), so his gyroscopes stay pointing in the same direction relative to infinity; they don't precess. The only reason the question arises at all in Kerr spacetime is that a ZAMO is *not* the same as a static observer at finite $r$.

due to geodetic and/or the Lense-Thirring precession.

These are only nonzero for objects with nonzero angular velocity in Schwarzschild spacetime, which, as above, are neither static (of course) nor ZAMOs (since ZAMOs are static, as above).

 

[PeterDonis]

Could you explain again why $\omega_{\infty} = \Omega_{\infty}$ should hold physically, at least when restricted to a single radius?

The simplest way to put it is that, given a ZAMO and a static observer at the same $r$, viewed from infinity:

(1) The ZAMO's gyroscope rotates with the same angular velocity as the ZAMO himself revolves around the hole (because the ZAMO's gyroscope always points directly radially outward);

(2) The static observer's gyroscope rotates with the same angular velocity as the ZAMO's gyroscope (because they are both gyroscopes at the same radius and therefore subject to the same "frame dragging" effect).

(3) #1 and #2 together imply that the static observer's gyroscope rotates (which is what I understand $\omega_{\infty}$ to mean) with the same angular velocity as the ZAMO revolves around the hole (which is what I understand $\Omega_{\infty}$ to mean).

 

[WannabeNewton]

(1) The ZAMO's gyroscope rotates with the same angular velocity as the ZAMO himself revolves around the hole (because the ZAMO's gyroscope always points directly radially outward);

Gyroscopic rotation in the above context refers to the orbital rotation of the gyroscope around the BH correct? If so then I agree.

(2) The static observer's gyroscope rotates with the same angular velocity as the ZAMO's gyroscope (because they are both gyroscopes at the same radius and therefore subject to the same "frame dragging" effect).

Given a static observer $O$ in the congruence of static observers in Kerr space-time, $\omega = \sqrt{\omega^{\mu}\omega_{\mu}}$ represents the rotation of a set of spatial basis vectors Lie transported along the worldline of $O$, meaning they are locked to neighboring static observers, relative to a set of spatial basis vectors Fermi-Walker transported along the worldline of $O$, which can be physically realized as torque-free gyroscopes carried by a locally non-rotating observer sharing the same worldline as $O$. By inserting $\gamma$ appropriately we get $\omega_{\infty}$. I apologize if I explained that poorly before. So when we say "the static observer's gyroscopes" are we referring to the Lie transported spatial basis vectors (which precess relative to infinity) or colocated Fermi-Walker transported spatial basis vectors?

 

[PeterDonis]

Gyroscopic rotation in the above context refers to the orbital rotation of the gyroscope around the BH correct?

Strictly speaking, it refers to the rotation of the direction the gyroscope is pointing, relative to infinity. That is the same as the rate of orbital rotation about the hole for a ZAMO, so in that particular case, yes, it refers to the rate of orbital rotation as well. But for a non-ZAMO observer the two are not the same.

Given a static observer $O$ in the congruence of static observers in Kerr space-time, $\omega = \sqrt{\omega^{\mu}\omega_{\mu}}$ represents the rotation of a set of spatial basis vectors Lie transported along the worldline of $O$, meaning they are locked to neighboring static observers, relative to a set of spatial basis vectors Fermi-Walker transported along the worldline of $O$, which can be physically realized as torque-free gyroscopes carried by a locally non-rotating observer sharing the same worldline as $O$. By inserting $\gamma$ appropriately we get $\omega_{\infty}$.

Ok, then it may just be a matter of signs; $\omega_{\infty}$ as you've defined it is a *backwards* rotation (i.e., in the opposite sense to the rotation of the hole), because it's the rotation of the Lie transported basis vectors (which always point in the same direction relative to infinity) relative to the Fermi-Walker transported basis vectors (whose direction relative to infinity rotates, in the same sense as the hole rotates). I was thinking of $\omega_{\infty}$ as the angular velocity relative to infinity of the gyroscopes carried by the static observers, which has the same magnitude, but opposite sign, to the $\omega_{\infty}$ that you've defined. We're really concerned here with the magnitude, not the sign, so I don't think it really matters which definition we adopt.

So when we say "the static observer's gyroscopes" are we referring to the Lie transported spatial basis vectors (which precess relative to infinity) or colocated Fermi-Walker transported spatial basis vectors?

The latter; see above.

 

[WannabeNewton]

Ah ok, I see now; I was confusing the two. Thanks!

 

[PeterDonis]

(2) The static observer's gyroscope rotates with the same angular velocity as the ZAMO's gyroscope (because they are both gyroscopes at the same radius and therefore subject to the same "frame dragging" effect).

Well, it appears that this, however intuitively plausible it seemed, was wrong. When I compute the vorticity for the static congruence, I get

$$
\omega = \frac{M a}{r^3 \left( 1 - 2M / r \right)}
$$

which agrees with the result given in this paper (the one WBN linked to in an earlier post). This gives (in the equatorial plane)

$$
\omega_{\infty} = \gamma^{-1} \omega = \frac{M a}{r^3 \sqrt{1 - 2M / r}}
$$

We have $\Omega_{\infty} = - g_{t \phi} / g_{\phi \phi}$, which gives (in the equatorial plane, and neglecting signs since we're only concerned with the magnitudes)

$$
\Omega_{\infty} = \frac{2 M a}{r \left( r^2 + a^2 \right)}
$$

Clearly these are not the same; in fact we can take their ratio easily:

$$
\frac{\omega_{\infty}}{\Omega_{\infty}} = \frac{r^2 + a^2}{2 r^2 \sqrt{1 - 2M / r}} = \frac{1}{2} \left( 1 + \frac{a^2}{r^2} \right) \sqrt{\frac{r}{r - 2M}}
$$

This goes to 1/2 (from above) for large $r$, and there will be *some* $r$ where it becomes 1; it grows without bound as $r \rightarrow 2M$. This is even more counterintuitive than the fact that they're not the same: it's saying that, as $r$ decreases, there comes a point where gyroscopes carried by static observers are rotating, relative to infinity, *faster* than gyroscopes carried by ZAMOs (which equates to rotating faster than the ZAMOs themselves are orbiting around the hole).

I think what is actually going on here is geodetic precession, which means I was also too quick to dismiss yuiop's earlier comment about that (sorry yuiop!). If we think of geodetic precession as acting on the angular momentum of an object, then of course a static observer in Kerr spacetime will see a nonzero precession, even though a static observer in Schwarzschild spacetime does not.

 

[WannabeNewton]

Peter, let me try to summarize what we've talked about thus far and you can tell me if it's sound:

Consider first the congruence of static observers with 4-velocity $\xi^{\mu} = \gamma \delta^{\mu}_{t}$ and choose a reference observer $O$ in the congruence with worldline $\lambda$. Along $\lambda$ we attach two separate frames:

The first frame is what we want to call the "static frame" or "Copernican frame" $S$ and it consists of the orthonormal basis $\{\xi^{\mu}, \eta^{\mu}_1,\eta^{\mu}_2,\eta^{\mu}_3\}$ wherein the $\eta^{\mu}_{i}$ are locked to infinitesimally neighboring static observers by means of Lie transport. $S$ therefore does not rotate relative to the distant stars (hence the name "Copernican frame") because the spatial axes $\eta^{\mu}_{i}$ of $S$ are locked to neighboring static observers and since the static observers are fixed relative to the distant stars, the spatial axes of $S$ must also be fixed relative to the distant stars i.e. they do not rotate relative to the distant stars. We can think of $S$ as the natural frame of $O$.

The second frame is what we want to call the "locally non-rotating frame" $S'$ and it consists of the orthonormal basis $\{\xi^{\mu}, e^{\mu}_1,e^{\mu}_2,e^{\mu}_3\}$ wherein the $e^{\mu}_{i}$ are physically realized by gyroscopes in torque-free motion by Fermi-transporting them along $\lambda$. We can think of $S'$ as the natural frame of a locally non-rotating observer $O'$ with the same worldline $\lambda$. Therefore $O$ and $O'$ are described by the same worldline $\lambda$ with the only difference being that $O$ has the static frame $S$ whereas $O'$ has the locally non-rotating frame $S'$.

As we know the vorticity 4-vector $\omega^{\mu}$ describes exactly the failure of the spatial axes of the static frame $S$ to be Fermi-transported along $\lambda$. In other words $\omega^{\mu}$ measures the rotation of the spatial axes of $S$ relative to the torque-free (Fermi-transported) gyroscopes attached to the comoving locally non-rotating frame $S'$.

And finally, since the spatial axes of $S$ are fixed relative to the distant stars but rotate relative to the gyroscopes of $S'$ with angular velocity $\omega^{\mu}$, the gyroscopes of $S'$ must rotate relative to the distant stars with angular velocity $\omega^{\mu}_{\infty} = -\gamma^{-1}\omega^{\mu}$.

Would you agree with all of the above?

 

[WannabeNewton]

If we think of geodetic precession as acting on the angular momentum of an object, then of course a static observer in Kerr spacetime will see a nonzero precession, even though a static observer in Schwarzschild spacetime does not.

In linearized gravity the geodetic precession of the spin axis of a torque-free (Fermi-transported) gyroscope relative to the distant stars arises from the coupling of the 3-velocity of the gyroscope relative to a background global inertial frame (i.e. the distant fixed stars) to the gradient of the gravitational potential (i.e. the gravitational acceleration); it explicitly takes the form $\frac{3}{2}\vec{v}\times \vec{\nabla}\varphi$. In Schwarzschild space-time the geodetic precession of a torque-free gyroscope relative to the distant stars arises similarly from the coupling of the orbital angular velocity of the gyroscope relative to the distant stars to the connection coefficients (which we know correspond in a sense to the gravitational acceleration).

A gyroscope Fermi-transported along the worldline of a static observer in Kerr space-time has by definition no 3-velocity relative to the distant stars so why would it experience a geodetic precession? It should only experience a Lense-Thirring precession, which arises solely from the coupling to space-time geometry that no object can avoid.

 

[yuiop]

Unless I'm misunderstanding the terminology, $\omega$ and $\Omega$ are both zero in Schwarzschild spacetime. A ZAMO in Schwarzschild spacetime has zero angular velocity relative to infinity, so ZAMO = static observer, and a static observer's vorticity is zero (more precisely, the vorticity of the static congruence is zero), so his gyroscopes stay pointing in the same direction relative to infinity; they don't precess. The only reason the question arises at all in Kerr spacetime is that a ZAMO is *not* the same as a static observer at finite $r$...

I agree with all your comments here, but maybe I should clear up my intended meaning when I said "the equivalent of $\omega_{\infty} = \Omega_{\infty}$" in the Schwarzschild metric. In order to have a meaning for $\omega_{\infty}$ that is equally useful in both metrics, we need to adjust the definition of $\omega_{\infty}$ slightly. $\Omega_{\infty}$ remains unchanged and is the angular spin velocity of a ZAMO as measured at infinity. As you note, the ZAMO in the SM is the static observer. The ZAMO in both metrics has zero angular velocity relative to the instantaneous radial vector pointing at the gravitational source and zero angular spin momentum relative to a gyroscope. This two conditions effectively define a ZAMO in a general way. The ZAMO in both metrics is not moving on a geodesic. To make $\omega_{\infty}$ applicable to both metrics in a general sense, $\omega_{\infty}$ means the angular spin velocity of an observer at the same radius, that has an arbitrary orbital velocity not equal to the ZAMO orbital velocity. Expressed like this we can say that $\omega_{\infty} \ne \Omega_{\infty}$ is true in both metrics. I apologise for changing the meaning of $\omega_{\infty}$ to make it more generally applicable in both metrics, (so the static observer in the Kerr metric and the orbiting observer in the Schwarzschild metric are just special cases of $\omega_{\infty}$), without making it clear. (My bad).

P.S. I would like to thank you for coming up with a quantitative solution. That was unexpected!

P.P.S It seems that $\omega_{\infty} = \Omega_{\infty}$ is what we would expect if geodetic and L-T precession did not exist, so maybe those effects are defined more generally, relative to the ZAMO frame, rather than the distant stars.

 

[WannabeNewton]

To make $\omega_{\infty}$ applicable to both metrics in a general sense, $\omega_{\infty}$ means the angular spin velocity of an observer at the same radius, that has an arbitrary orbital velocity not equal to the ZAMO orbital velocity.

Keep in mind that the physical interpretation of the vorticity $\omega^{\mu}$ of a time-like congruence as the rotation, relative to the torque-free gyroscopes in a locally non-rotating frame, of a frame whose spatial axes are Lie transported along a given worldline in this congruence only works if the congruence is rigid.

This is expected since we thought of the Lie transported spatial axes as orthonormal spatial basis vectors locked (or anchored) to neighboring observers in the congruence. If the congruence is rigid, as is the congruence of static observers, then the spatial distances between neighboring observers in the congruence remain constant in their instantaneous rest frames and so we have a well-defined orthonormal basis. But if the congruence is not rigid then the spatial distances between neighboring observers in the congruence will be changing in their instantaneous rest frames and so the lengths of the vectors carried by a given observer in the congruence that are locked to neighboring observers will be changing over time (as read by a clock carried by the given observer)-we don't have a well-defined orthonormal basis anymore-the lengths of spatial basis vectors can't be changing! The fact that the "coordinate lattice" defined by the observers in the congruence fails to be a "rigid coordinate lattice" robs $\omega^{\mu}$ of its direct physical meaning in terms of gyroscopic precession.

The reason I mention this is that neither the congruence of circularly orbiting observers in Schwarzschild space-time nor the ZAMO congruence in Kerr space-time is rigid.

*Rigid here means Born-Rigid.

 

[yuiop]

... The reason I mention this is that while the congruence of circularly orbiting observers in Schwarzschild space-time is rigid, the ZAMO congruence in Kerr space-time is not...

Could you clarify. As I understand it, the congruence of circularly orbiting observers in Schwarzschild space-time is only Born rigid if we choose observers all with the same orbital angular velocity as measured at infinity. If the observers are in natural circular geodesic orbits, they will not have equal orbital angular velocities, so no Born rigidity in this case unless we specify equal radius. Likewise, I assume a ZAMO congruence in the Kerr metric will only be Born rigid iff they have equal orbital radii. By contrast, a congruence of ZAMO observers in the SM (i.e. static) with arbitrary radii, can be Born rigid. Would you agree with these statements?

P.S. I see you edited your comment after I quoted it, so we might be a cross purposes here. Still, it won't hurt to clarify.

 

[WannabeNewton]

You are entirely correct yuiop. I was confusing the properties of circularly orbiting observers in Schwarzschild space-time with those of observers sitting on a rotating disk in flat space-time. I actually edited my post in order to correct this while you were replying, I'm really sorry about that!

A congruence with a 4-velocity field of the form $u^{\mu} = \gamma (\delta^{\mu}_t + \omega \delta^{\mu}_{\phi})$ where $\omega$ is constant along the worldlines of the congruence will be Born-rigid if and only if $\omega$ is identically constant in space-time (however there are no additional constraints placed on $\gamma$). That is, $u^{\mu}\nabla_{\mu}\omega = 0$ is assumed (meaning the angular velocity relative to infinity is constant along each worldline in the congruence) but only if $\nabla_{\mu}\omega = 0$ (meaning $\omega$ is identically constant in space-time) do we have a Born-rigid congruence. This happens for the congruence of observers sitting on a rotating disk in flat space-time but not so for the congruence of circularly orbiting observers in Schwarzschild space-time.

 

[PeterDonis]

Would you agree with all of the above?

Yes, all looks right to me.

 

[yuiop]

... This happens for the congruence of observers sitting on a rotating disk in flat space-time but not so for the congruence of circularly orbiting observers in Schwarzschild space-time.

It occurred to me that I am using the word "orbiting" to mean circular motion around a gravitational body at an arbitrary velocity and not necessarily at the correct orbital velocity for geodesic motion. Perhaps that is an abuse of the term? Is there a word for a "non geodesic orbit" or does orbiting always mean following a geodesic path?

Also, Wikipedia makes this intriguing statement:

The term geodetic effect has two slightly different meanings as the moving body may be spinning or non-spinning. Non-spinning bodies move in geodesics, whereas spinning bodies move in slightly different orbits.[3] (Rindler Page 254)

Any thoughts? Do they really mean spinning objects cannot follow a geodesic?

 

[PeterDonis]

To make $\omega_{\infty}$ applicable to both metrics in a general sense, $\omega_{\infty}$ means the angular spin velocity of an observer at the same radius, that has an arbitrary orbital velocity not equal to the ZAMO orbital velocity. Expressed like this we can say that $\omega_{\infty} \ne \Omega_{\infty}$ is true in both metrics.

Just to clarify, this won't always be true as you've defined it; it will only "almost always" be true. In the Schwarzschild case, $\Omega_{\infty} = 0$ by definition, and a static observer has $\omega_{\infty} = 0$, but an observer with any nonzero angular velocity does not. In the Kerr case, $\Omega_{\infty}$ depends on $r$ (and also on $\theta$ outside the equatorial plane, but I'll restrict to the equatorial plane in what follows for simplicity), but for any given $r$, there will be *some* angular velocity for which $\omega_{\infty} = \Omega_{\infty}$. (At the particular radius where the formula I posted before gives $\omega_{\infty} / \Omega_{\infty} = 1$, that angular velocity will be zero--the static observer's gyroscope will precess, relative to infinity, at the same rate as the ZAMO's gyroscope. But at other values of $r$, there will, I think, be some nonzero angular velocity for which $\omega_{\infty} = \Omega_{\infty}$.)

 

[WannabeNewton]

Perhaps that is an abuse of the term?

No it's a perfectly fine and standard use of terminology. See here for more: http://en.wikipedia.org/wiki/Frame_...agihara_observers_in_the_Schwarzschild_vacuum

Any thoughts? Do they really mean spinning objects cannot follow a geodesic?

I checked out the page in Rindler that the wiki quote referenced. The wiki quote actually inaccurately paraphrased what Rindler was saying. Rindler's exact words were "freely falling gyroscope" and a gyroscope has no intrinsic spin by definition so here Rindler is specifically talking about a non-spinning object following a geodesic. However there is no reason why we can't have a spinning object following a geodesic. Just take an inertial observer in Minkowski space-time and have the observer spin in place; the now spinning observer still follows a geodesic in Minkowski space-time (a straight line) but has an intrinsic spin.