Practical measurements of rotation in the Kerr metric

[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.

 

[PeterDonis]

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).

I haven't looked at the derivation of this, but does the coupling arise directly to angular velocity, or indirectly via orbital angular momentum? If it's the latter, that makes a difference in Kerr spacetime, even though it ends up working out the same in Schwarzschild spacetime (since zero orbital angular momentum equals zero orbital angular velocity in the latter but not the former).

Lense-Thirring precession, which arises solely from the coupling to space-time geometry that no object can avoid.

This is true, and the proper interpretation of the results I posted should include L-T precession as well. I just think it's still possible that they should include geodetic precession for a static observer--but it depends, of course, on the details of how geodetic precession is derived, as above.

 

[WannabeNewton]

I haven't looked at the derivation of this, but does the coupling arise directly to angular velocity, or indirectly via orbital angular momentum? If it's the latter, that makes a difference in Kerr spacetime, even though it ends up working out the same in Schwarzschild spacetime (since zero orbital angular momentum equals zero orbital angular velocity in the latter but not the former).

Do you have "Gravitation and Inertia"-Wheeler and Ciufolini nearby, by any chance? If so, check out section 3.4.3.

See also here: http://books.google.com/books?id=Vc...ic precession schwarzschild spacetime&f=false

Yes, all looks right to me.

Great, thanks!

 

[PeterDonis]

Do you have "Gravitation and Inertia"-Wheeler and Ciufolini nearby, by any chance? If so, check out section 3.4.3.

Not handy right now, but I'll take a look when I get a chance (assuming I can find my copy ).

See also here: http://books.google.com/books?id=Vc...ic precession schwarzschild spacetime&f=false

This gives the transport equations for the gyroscope in terms of the connection coefficients. In Schwarzschild spacetime there is no coupling to the connection coefficients involving $\phi$ unless the angular velocity is nonzero. That may not be true in Kerr spacetime; the "cross term" in the metric (in Boyer-Lindquist coordinates) adds some connection coefficients involving $\phi$ that aren't there in Schwarzschild spacetime. I would want to do a similar computation to this one using the Kerr connection coefficients to see what difference that makes.

 

[yuiop]

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.

Are there any conditions in either metric where a small test object with intrinsic spin will follow a different geodesic path to a similar small test object with no intrinsic spin, if dropped from the same location in a vacuum?

 

[WannabeNewton]

Are there any conditions in either metric where a small test object with intrinsic spin will follow a different geodesic path to a similar small test object with no intrinsic spin, if dropped from the same location in a vacuum?

Let me clarify what I said before and in doing so hopefully answer your question. In order for a worldline $\gamma$ to be a geodesic, all we require is that its 4-velocity $u^{\mu}$ satisfy $u^{\gamma}\nabla_{\gamma}u^{\mu} = 0$. There is no constraint whatsoever on the kind of spatial axes an observer following $\gamma$ must carry; we can have the observer carry a non-spinning set of spatial axes along $\gamma$ or a spinning set of spatial axes along $\gamma$ but this won't change the fact that the observer is still following a geodesic $\gamma$. So if "spinning object" simply refers to an observer carrying a spinning frame then the above holds.

However it seems to me that what you are talking about (and perhaps what the wiki article was talking about as well) is the deviation of a small but finitely sized spinning object (where small means the characteristic size is much smaller than the curvature scales) from geodesic motion. This is governed by the Papapetrou equation: http://en.wikipedia.org/wiki/Mathisson–Papapetrou–Dixon_equations