Practical measurements of rotation in the Kerr metric

[yuiop]

In another thread WannabeNewton mentioned:

... The mounted gyroscope can yield a positive result for rotation of the planet but at the same exact time the Sagnac effect can yield a negative result for rotation of the planet and vice versa.

and gave this reference:

... See section 3.2 of the following notes: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf

Until WBN mentioned it, I had never given any thought to the difference between these methods of measuring rotation, so I would like to explore those ideas further here, particularly in relation to the Kerr metric.

Consider a *large* stationary thin ring in the equatorial plane of a Kerr black hole, centred on the rotation axis of the KBH. 'Stationary' here is defined as $dr=d\phi=d\theta=0$ as measured in standard B-L Kerr coordinate system as defined in Eq1 here. The ring is hollow with a reflective inner surface such that a light signal can be sent all the way around the inside of ring in either direction and so can be used as a large Sagnac device. A quick calculation reveals that if the gravitational body has non zero angular momentum, then the time for light to travel in opposite directions inside the ring is unequal and the Sagnac device indicates the 'stationary' ring is rotating. The coordinate angular velocity of the light inside the ring can be described by an equation of the form $d\phi/dt = a \pm \sqrt{b}$ where a and b are functions of radius r. If there is a radius where the 'stationary' ring indicates non rotation, then the value of b is zero for that radius. The outermost solution turns out to be $r=GM + \sqrt{GM-\alpha^2}$ where $\alpha$ is the angular momentum per unit mass of the gravitational body. (This equation is equally valid for any Sagnac ring centred on the rotation axis of the gravitational body and lying in a plane parallel to the equatorial plane.) This radius also happens to be the outer event horizon or 'static limit' where no object can maintain constant radius. Therefore there is no location outside the event horizon of a Kerr black hole, where a 'stationary' Sagac ring indicates zero rotation according to the Sagnac effect. (Additionally no ring can be rotationally stationary (i.e $d\phi = 0$) in the Kerr metric, within the ergosphere of the KBH.).

If 3 axis gyroscopes are attached to a large 'stationary' Sagnac ring, would they indicate any rotation at all? How is this calculated?

Now consider a large equatorial Sagnac ring that is rotating with angular velocity:

$$\Omega = - \frac{g_{t \phi}}{g_{\phi \phi}}$$

This is the angular velocity (at that radius) at which an inertial reference frame is said to 'dragged' by the rotating black hole. Would the large Sagnac device still record non zero rotation?

For reference, define 'distant stars' as being stationary with respect to the Kerr metric and so far away that they have negligible effect on measurements local to the KBH. What would the small gyroscopes attached to the ring indicate? As I understand it, when a small box containing a 3 axis gyroscope indicates zero rotation, the box will be rotating relative to a local part of the ring and rotating on the opposite sense to the rotation of the black hole and rotating relative to distant stars. This in turn implies that no global notion of "zero rotation" can be defined using local gyroscopes. Is this correct? Also, as I understand it, a particle in free fall and will have its angular velocity $(d\phi/dt)$ increased or decreased until it matches $\Omega = - g_{t \phi}/g_{\phi\phi}$. Now if the natural orbital angular velocity of a particle in a circular orbit with radius r is not equal to $\Omega = - g_{t \phi}/g_{\phi \phi}$ then it seems to follow that there is no such thing as a natural stable circular orbit outside a KBH except maybe at a certain critical radius. Does that make any sense?

Sorry for all the questions. Basically I would like to know under what conditions (preferably with an equation) would gyroscopes attached to a large ring indicate zero rotation and when would a large Sagnac ring indicate zero rotation?

 

[PeterDonis]

Basically I would like to know under what conditions (preferably with an equation) would gyroscopes attached to a large ring indicate zero rotation and when would a large Sagnac ring indicate zero rotation?

For Kerr spacetime, the two criteria are the same, and the "zero rotation" condition is that, with respect to an observer at infinity, a ring at radial coordinate $r$ is rotating around the hole, in the same sense as the hole's rotation, with angular velocity

$$
\Omega (r) = - g_{t \phi} / g_{\phi \phi} = \frac{2 M \alpha}{r^3 + \alpha^2 \left( r + 2M \right)}
$$

(Note that the plane of rotation is the hole's "equatorial plane", i.e., $\theta = \pi / 2$. Also note, btw, that the standard coordinate labeling for Boyer-Lindquist coordinates has $\phi$ as the angular coordinate corresponding to the rotational Killing vector field associated with the hole, whereas you appear to be using $\theta$ to label that coordinate.) So the Sagnac ring that is being "dragged along with the hole" will show zero rotation by both criteria. (This state of rotation is often called the "ZAMO" condition, for "Zero Angular Momentum Observer", because the angular momentum of an object rotating with this angular velocity about the hole is zero.)

 

[PeterDonis]

As a side note, an interesting effect of a ring rotating with angular velocity $\Omega = - g_{t \theta}/g_{\theta \theta}$ is that a clock attached to the rotating ring will show the same elapsed time as a clock on a stationary ring with the same radius, each time they pass each other. This seems weird, so I hope I have calculated that right.

I'm not sure about this; I'll have to check when I get a chance.

Also, as I understand it, a particle in free fall and will have its angular velocity $(d\theta/dt)$ increased or decreased until it matches $\Omega = - g_{t \theta}/g_{\theta \theta}$.

No, it won't; the relationship between angular velocity and angular momentum is affected by the value of $\Omega$ at the particle's orbital radius (see my previous post), but the "dragging" due to the hole's rotation doesn't exert any force that will change the particle's angular velocity or force it to be a certain value.

What is true, however, is that the closer the particle's orbit is to the hole's horizon, the smaller the range of possible angular velocities is, and that range is always "centered" on $\Omega (r)$. Inside the ergosphere (i.e., at radial coordinates smaller than the static limit), the possible range of angular velocities no longer includes zero; i.e., it is impossible to remain stationary with respect to infinity. But there is still *some* allowed range; the angular velocity is only forced to be $\Omega (r)$ in the limiting case of an "orbit" exactly at the hole's horizon (which isn't actually physically possible, of course).

 

[yuiop]

... Also note, btw, that the standard coordinate labeling for Boyer-Lindquist coordinates has $\phi$ as the angular coordinate corresponding to the rotational Killing vector field associated with the hole, whereas you appear to be using $\theta$ to label that coordinate.)

Sorry, I meant to use the same notation as in the Wikipedia link. I have now corrected the notation in the OP.

I'm not sure about this; I'll have to check when I get a chance.

I have deleted that part. I have checked it again and there was an error in that calculation.

I will respond to your other responses when I have digested them ;)

 

[PeterDonis]

Sorry, I meant to use the same notation as in the Wikipedia link. I have now corrected the notation in the OP.

Don't forget to correct the subscripts in the metric coefficient expressions as well.

 

[PeterDonis]

Also, a good quick reference for Kerr spacetime is this paper by Matt Visser:

http://arxiv.org/abs/0706.0622

And this article by Malament is a good discussion of "no rotation" conditions:

http://arxiv.org/pdf/gr-qc/0011094‎

(Note, however, that Malament's CIA condition is *not* the same as the "gyroscope" condition in the OP here.)

 

[yuiop]

... So the Sagnac ring that is being "dragged along with the hole" will show zero rotation by both criteria. (This state of rotation is often called the "ZAMO" condition, for "Zero Angular Momentum Observer", because the angular momentum of an object rotating with this angular velocity about the hole is zero.)

Section 3.2 of the paper linked by WBN compares the Zero Angular Momentum (ZAM) and Compass of Inertia on the Ring (CIR) criteria and seems to conclude in proposition 3.3.4. (No-Go Theorem) that they are not the same in the Kerr metric, but I am not sure if I have read it right as the paper is a bit too technical for me.

If a gyroscope is attached to the ring as in Figure 3.2.3 of that paper, it seems inevitable that it will precess relative to the tangent of the ring when the ring is rotating. To a first approximation the gyroscope will be pointing to the 'distant stars' and on top of that there will be Thomas precession plus whatever the rotating KBH adds to the equation. Additionally the Wikipedia article mentions that:

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. ... 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 implies that when the 'ice scater' is not rotating locally (when compared to a local gyroscope or when using WBN's beaded porcupine device*), she is rotating relative to the gravitational body, the distant stars and the ring.

*

... The observer can also carry with him an apparatus consisting of a sphere with beaded prongs attached isotropically. Clearly if the observer had non-zero spin then the beads would be thrown outwards due to centrifugal forces; if the observer had vanishing spin then the beads would remain in place. ...

 

[WannabeNewton]

Hey yuiop! Peter already gave a very detailed reply so let met just add one overarching point regarding one of your questions, that of differentiating the Sagnac effect test from the gyroscope test in a general stationary axisymmetric space-time:

As you already noted in your post, non-rotation of the ring as per the Sagnac effect is equivalent to the ring having zero angular momentum. On the other hand, if we mounted a gyroscope on the ring then non-rotation of the ring as per the gyroscope means that the tangent field to the ring is Fermi-Walker transported along the worldline of the gyroscope; this is equivalent to the time-like congruence formed by the worldlines of the points on the ring having zero vorticity i.e. said time-like congruence must be irrotational.

The paper I linked you gives Godel space-time as an example wherein the Sagnac effect test and gyroscope test fail to be equivalent for rings in different regions of Godel space-time.

For Kerr space-time, as Peter noted, they will be equivalent in well-behaved regions. This is because the ring would be described by the time-like congruence $u^{\mu} = \alpha \nabla^{\mu}t$ where $t$ is the time-coordinate in the coordinate system used above and $\alpha$ is the normalization factor. The angular momentum is defined as $L = \psi_{\mu}u^{\mu}$ where $\psi^{\mu}$ is the axial killing field. We have $L = \alpha g_{\mu\nu}\psi^{\mu}g^{\nu\gamma}\nabla_{\gamma}t = \alpha \delta^{\gamma}_{\mu}\psi^{\mu}\nabla_{\gamma}t = \alpha\delta^{t}_{\phi} = 0$. Furthermore since this time-like congruence is the gradient of a scalar field, it is trivially irrotational i.e. $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$. Hence the Sagnac effect test and gyroscope test would be equivalent in the well-behaved regions.

A ring at constant coordinate radius $R$ will then have the angular velocity $\Omega = \frac{u^{\phi}}{u^{t}} = \frac{g^{\phi \phi}}{g^{tt}} = -\frac{g_{t\phi}}{g_{\phi\phi}} = \frac{2MRa}{(R^2 + a^2)^2 - a^2\Delta \sin^2\theta}$ relative to a stationary observer at spatial infinity (see Schutz pp. 313-314). As Peter noted, this frame dragging effect can equally characterize the state of non-rotation.

EDIT: And to clarify, $\Omega$ is in the same sense as the rotation of the black hole, which is given by $J = Ma$.

 

[WannabeNewton]

Section 3.2 of the paper linked by WBN compares the Zero Angular Momentum (ZAM) and Compass of Inertia on the Ring (CIR) criteria and seems to conclude in proposition 3.3.4. (No-Go Theorem) that they are not the same in the Kerr metric, but I am not sure if I have read it right as the paper is a bit too technical for me.

The No-Go Theorem basically states that any general notion of non-rotation in stationary axisymemtric space-times obeying the three specific conditions given in the paper cannot exist in Kerr space-time. It's a different discussion from that of the previous section wherein the equivalence of the ZAMO and compass of inertia tests are discussed; there the counter-example given is Godel space-time not Kerr space-time.

 

[yuiop]

...No, it won't; the relationship between angular velocity and angular momentum is affected by the value of $\Omega$ at the particle's orbital radius (see my previous post), but the "dragging" due to the hole's rotation doesn't exert any force that will change the particle's angular velocity or force it to be a certain value.

As I understand it, if a particle with initially zero angular momentum is dropped from infinity, it will not free fall along a radial path, but will acquire a horizontal component with $d\phi/dt \ne 0$ (well before reaching the ergosphere). This implies that a rocket hovering at r with $dr=d\phi=d\theta=0$ will not only require vertical thrust in the radial direction to maintain $dr/dt=0$, but will also require a small horizontal thrust in order to maintain $d\phi/dt =0$.

 

[WannabeNewton]

Oh before I forget, I really recommend purchasing the following book if you want detailed discussions of rotation, gyroscopic precession by means of frame dragging, de-Sitter effect etc. in the context of GR: https://www.amazon.com/dp/0691033234/?tag=pfamazon01-20

As you can see there are used copies for like 7 bucks so it's totally worth it. It's one of my most favorite GR texts.

 

[WannabeNewton]

As I understand it, if a particle with initially zero angular momentum is dropped from infinity, it will not free fall along a radial path, but will acquire a horizontal component with $d\phi/dt \ne 0$ (well before reaching the ergosphere). This implies that a rocket hovering at r with $dr=d\phi=d\theta=0$ will not only require vertical thrust in the radial direction to maintain $dr/dt=0$, but will also require a small horizontal thrust in order to maintain $d\phi/dt =0$.

This is true. The static observer so mentioned has a 4-velocity $u^{\mu} = \alpha \xi^{\mu}$ where $\xi^{\mu}$ is the time-like killing field and $\alpha$ is the normalization factor. If one computes $a^{\mu} = u^{\gamma}\nabla_{\gamma}u^{\mu}$ then one finds that not only is there a radial component to the acceleration (which is the vertical rocket thrust you mentioned) but also a polar component to the acceleration (which is the horizontal rocket thrust you mentioned). See p.14 of the following: http://arxiv.org/pdf/0704.0986v1.pdf

 

[yuiop]

Oh before I forget, I really recommend purchasing the following book if you want detailed discussions of rotation, gyroscopic precession by means of frame dragging, de-Sitter effect etc. in the context of GR: https://www.amazon.com/dp/0691033234/?tag=pfamazon01-20

As you can see there are used copies for like 7 bucks so it's totally worth it. It's one of my most favorite GR texts.

If Santa doesn't deliver, I'll treat myself

 

[PeterDonis]

This is true.

No, it isn't. Read the paper you linked to carefully: the components of the proper acceleration required to remain static are in the $r$ and $\theta$ directions; there is no component in the $\phi$ direction.

The $\theta$ component is only nonzero outside the "equatorial plane", so if we adopted cylindrical coordinates the only component would be radial--meaning radial in the cylindrical sense, away from the axis of the cylinder. That is, the $\theta$ component is really a coordinate artifact; it's there because we are trying to use spherical coordinates in a spacetime that has only axial symmetry.

[STRIKE]So a freely falling observer will fall inward in the "radial" direction in the axial sense--i.e., radially inward towards the axis.[/STRIKE] [Edit: not sure the previous statement is true, see follow-up in post #18.] He will *not* acquire any component in the tangential (i.e., $\phi$) direction.

 

[PeterDonis]

As I understand it, if a particle with initially zero angular momentum is dropped from infinity, it will not free fall along a radial path, but will acquire a horizontal component with $d\phi/dt \ne 0$ (well before reaching the ergosphere).

This is not correct; as I said in my reply to WN, there is a nonzero $\theta$ component (if the particle is outside the equatorial plane), but no $\phi$ component. As I noted there, what this is really saying is that the particle free-falls radially inward in the axial rather than spherical sense, i.e., towards the axis of rotation of the hole.

The effect that the "dragging" of spacetime by the hole has is that as the particle free-falls, it will acquire a spin, in the opposite sense to the rotation of the hole. The paper WN linked to notes this on pp. 14-15, though they don't put it in quite those terms; but that's what the comments about the "gravitational Larmor's theorem" mean.

 

[WannabeNewton]

No, it isn't. Read the paper you linked to carefully: the components of the proper acceleration required to remain static are in the $r$ and $\theta$ directions; there is no component in the $\phi$ direction.

Yes I said there is an acceleration in the polar direction, not the azimuthal direction and if I recall yuiop was using the convention that $\phi$ was polar.

 

[PeterDonis]

Yes I said there is an acceleration in the polar direction, not the azimuthal direction and if I recall yuiop was using the convention that $\phi$ was polar.

Not after I pointed out that that isn't the usual convention. See posts #2, #4, and #5.

 

[PeterDonis]

So a freely falling observer will fall inward in the "radial" direction in the axial sense--i.e., radially inward towards the axis.

On looking again at the paper WN linked to, I'm not sure this is true, except possibly in the initial stages of free-fall for a object that starts from rest far enough away from the axis. An obvious case where it can't be true is an object that is free-falling inward *on* the axis: it certainly won't stay put, so free-fall in general can't be purely in the "radial" direction in the axial (as opposed to spherical) sense.

 

[WannabeNewton]

So a freely falling observer will fall inward in the "radial" direction in the axial sense--i.e., radially inward towards the axis. He will *not* acquire any component in the tangential (i.e., $\phi$) direction.

I was hoping you could clarify this point a bit more. We have (using the regular coordinate conventions for the polar and azimuthal angles) $p^{\phi} = g^{\phi t}p_{t} + g^{\phi\phi}p_{\phi}$ and $p^{t} = g^{t t}p_{t} + g^{\phi t}p_{\phi}$. If we drop in a particle with $p_{\phi} = \psi^{\mu}p_{\mu} = L = 0$ from infinity then $\omega = \frac{d\phi}{dt}= \frac{p^{\phi}}{p^{t}} = \frac{g^{t\phi}}{g^{\phi\phi}}$ so the observer has an angular velocity $\omega$ relative to an observer at infinity when dropped with zero angular momentum $L = 0$ from infinity. Could you explain how to interpret this properly then? Thanks!

Not after I pointed out that that isn't the usual convention. See posts #2, #4, and #5.

Ah ok, missed that sorry!

 

[PeterDonis]

If we drop in a particle with $p_{\phi} = \psi^{\mu}p_{\mu} = L = 0$ from infinity then $\omega = \frac{d\phi}{dt}= \frac{p^{\phi}}{p^{t}} = \frac{g^{t\phi}}{g^{\phi\phi}}$ so the observer has an angular velocity $\omega$ relative to an observer at infinity when dropped with zero angular momentum $L = 0$ from infinity. Could you explain how to interpret this properly then? Thanks!

At infinity, $\Omega = 0$, so the particle dropped from infinity starts out with zero angular velocity relative to infinity. I.e., at infinity, zero angular velocity is the same state of motion as zero angular momentum. Since there is no $\phi$ component induced in his motion by free-fall, his angular velocity relative to infinity remains zero all the way in. But since zero angular velocity (note that this is *orbital* angular velocity, $d \phi / dt$, to be distinguished from *intrinsic* angular velocity, i.e., spin) is *not* the same as zero angular momentum at a finite radius, the free-falling particle will, as I noted before, acquire a nonzero angular momentum, i.e., a spin, in the opposite sense to the hole's rotation, induced by "frame dragging".

(More precisely, this is what will happen for free fall in the "equatorial plane"; I'm not sure the induced angular momentum will always have the same sign. For example, a particle free-falling inward along the axis of rotation will, if I am reading sources like MTW correctly, acquire angular momentum in the *same* sense as the hole's rotation.)

 

[WannabeNewton]

At infinity, $\Omega = 0$, so the particle dropped from infinity starts out with zero angular velocity relative to infinity. I.e., at infinity, zero angular velocity is the same state of motion as zero angular momentum.

I agree that initially, that is at infinity, $\Omega= 0$ but as the particle nears finite radii won't $\Omega$ start increasing in the same sense as the hole's rotation?

But since zero angular velocity is *not* the same as zero angular momentum at a finite radius, the free-falling particle will, as I noted before, acquire a nonzero angular momentum, i.e., a spin, in the opposite sense to the hole's rotation, induced by "frame dragging".

I'm a bit confused by this. If we started off with $L = 0$ at an initial event at infinity for the freely falling particle then won't $L = 0$ at every event on the particle's worldline since $L$ is a conserved quantity?

 

[PeterDonis]

I agree that initially, that is at infinity, $\Omega= 0$ but as the particle nears finite radii won't $\Omega$ start increasing in the same sense as the hole's rotation?

Yes, $\Omega$ does, but there is nothing forcing the angular velocity of the free-falling observer to always be equal to $\Omega$. The only constraint is that, if the free-falling observer starts out with zero angular momentum, then its initial angular velocity must be equal to $\Omega$ at the radius where it starts. So if it starts at infinity, its initial angular velocity will be equal to $\Omega$ at infinity, i.e., zero.

I'm a bit confused by this. If we started off with $L = 0$ at an initial event at infinity for the freely falling particle then won't $L = 0$ at every event on the particle's worldline since $L$ is a conserved quantity?

I agree there's something weird here; but the paper you linked to makes clear that there is no $\phi$ component induced in free-fall motion. I'll have to think about this some more.

 

[WannabeNewton]

Yes, $\Omega$ does, but there is nothing forcing the angular velocity of the free-falling observer to always be equal to $\Omega$. The only constraint is that, if the free-falling observer starts out with zero angular momentum, then its initial angular velocity must be equal to $\Omega$ at the radius where it starts. So if it starts at infinity, its initial angular velocity will be equal to $\Omega$ at infinity, i.e., zero.

Ok but if we agree with the usual result that $L = \psi^{\mu}p_{\mu}$ is a conserved quantity in Kerr space-time, so that if $L = 0$ initially on the particle's trajectory then $L = 0$ for all proper time along the particle's worldline, then won't $\Omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ always equal the angular velocity $\frac{d\phi}{dt}$ of said freely falling particle as per post #19? In other words, $\Omega$ will only fail to be the angular velocity of the freely falling particle if the particle goes from initially having vanishing angular momentum to having a non-zero angular momentum along its trajectory right? But this is what doesn't make sense to me since as noted $L$ is a conserved quantity along this trajectory.

 

[PeterDonis]

Ok but if we agree with the usual result that $L = \psi^{\mu}p_{\mu}$ is a conserved quantity in Kerr space-time, so that if $L = 0$ initially on the particle's trajectory then $L = 0$ for all proper time along the particle's worldline, then won't $\Omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ always equal the angular velocity $\frac{d\phi}{dt}$ of said freely falling particle as per post #19?

That seems plausible, but it's not consistent with the paper you linked to, which, as I noted, makes clear that the only components of free-fall motion that change are the $r$ and $\theta$ components; the $\phi$ component does not change. So we must be missing something; I'm just not sure what.

 

[WannabeNewton]

Thanks. I'll have to reread the paper more carefully in a bit but at a cursory glance of the paragraph at the very bottom of p.14, I think the difference in form between the angular velocity of the freely falling observer at the very bottom of p.14 of the paper and the angular velocity of the freely falling observer who starts at infinity with zero angular momentum is in the initial conditions of the free fall motion.

We were talking about an observer freely falling from infinity with an initial angular momentum $L = 0$. On the other hand, if I read the bottom of p.14 correctly, the paragraph is instead talking about an observer who is initially static at some fixed spatial coordinate using a rocket to hover in place and then at some instant switches off the rocket thrusters so as to go into free fall; the initial conditions of the free fall motion in this case would be the state of the static observer at the very instant the rocket thrusters are switched off. In particular, the initial angular momentum would not be zero since the static observer has non-zero angular momentum and I think this is what changes the form of the angular velocity between the two aforementioned cases.

 

[PeterDonis]

We were talking about an observer freely falling from infinity with an initial angular momentum $L = 0$. On the other hand, if I read the bottom of p.14 correctly, the paragraph is instead talking about an observer who is initially static at some fixed spatial coordinate using a rocket to hover in place and then at some instant switches off the rocket thrusters so as to go into free fall; the initial conditions of the free fall motion in this case would be the state of the static observer at the very instant the rocket thrusters are switched off. In particular, the initial angular momentum would not be zero since the static observer has non-zero angular momentum and I think this is what changes the form of the angular velocity between the two aforementioned cases.

Yes, the initial conditions are different, but the "acceleration due to gravity" should be the same. Consider the simpler case of Schwarzschild spacetime: suppose we have a static observer at radius $r$ who releases a rock from rest relative to him at the same instant that another rock is falling radially inward past him with nonzero radial velocity $v$ relative to him. The coordinate acceleration of both rocks, relative to him, will be the same, even though they have different velocities relative to him.

For another reference which appears to support the $\phi$ component of radial free-fall motion not changing in Kerr spacetime, see section III of Hamilton's classic paper on the "river model" of black holes, which talks about how the model extends to Kerr spacetime:

http://arxiv.org/abs/gr-qc/0411060

 

[WannabeNewton]

Yes, the initial conditions are different, but the "acceleration due to gravity" should be the same. Consider the simpler case of Schwarzschild spacetime: suppose we have a static observer at radius $r$ who releases a rock from rest relative to him at the same instant that another rock is falling radially inward past him with nonzero radial velocity $v$ relative to him. The coordinate acceleration of both rocks, relative to him, will be the same, even though they have different velocities relative to him.

Right I agree that the acceleration due to gravity won't depend on the initial conditions but won't the angular velocity $\frac{d\phi}{dt}$ depend on the initial choice of $L$? In particular, we have $g_{t\phi}p^{t} + g_{\phi\phi}p^{\phi} = g_{t\phi}p^{t}(1 + \frac{g_{\phi\phi}}{g_{t\phi}}\frac{\mathrm{d} \phi}{\mathrm{d} t}) = L$. If we arrange for the initial conditions to have $L = 0$ then we get back $\frac{\mathrm{d} \phi}{\mathrm{d} t} = -\frac{g_{t\phi}}{g_{\phi\phi}}$. On the other hand if we choose for example $L = g_{t\phi}p^{t}$ for the initial conditions then we get $\frac{\mathrm{d} \phi}{\mathrm{d} t} = 0$; for a static observer we have, at any instant, $L = g_{t\phi}p^{t}$.

For another reference which appears to support the $\phi$ component of radial free-fall motion not changing in Kerr spacetime, see section III of Hamilton's classic paper on the "river model" of black holes, which talks about how the model extends to Kerr spacetime:

http://arxiv.org/abs/gr-qc/0411060

I'll take a look, thanks!

 

[PeterDonis]

I agree that the acceleration due to gravity won't depend on the initial conditions

Ok, good.

won't the angular velocity $\frac{d\phi}{dt}$ depend on the initial choice of $L$?

The *initial* angular velocity will, yes. What I'm trying to understand is how, if at all, the angular velocity changes as the radius changes along a free-fall trajectory.

if we choose for example $L = g_{t\phi}p^{t}$ for the initial conditions then we get $\frac{\mathrm{d} \phi}{\mathrm{d} t} = 0$; for a static observer we have, at any instant, $L = g_{t\phi}p^{t}$.

Yes, but what about a free-fall object released by the static observer? If its $L$ remains constant, then its angular velocity will have to change as it falls, since $g_{t \phi}$ will change along its trajectory. But if the "acceleration due to gravity" everywhere has no $\phi$ component, how can the angular velocity change?

 

[PeterDonis]

For another reference which appears to support the $\phi$ component of radial free-fall motion not changing in Kerr spacetime

Hm, looking at this some more, I'm not sure it does support this. At the start of section III it does say that the "river" velocity has no azimuthal component; the effect of "frame dragging" by the hole's rotation is to add a *twist* (spatial rotation) to the river, and it says that gradients in the river flow velocity (which are purely radial) induce a Lorentz boost in "fish" swimming in the river, whereas changes in the twist (which are the only things that are azimuthal) induce a spatial rotation in the "fish". But later on it says that objects that free-fall in from rest at infinity with zero angular momentum have constant $\phi_{ff}$, which is *not* the same as constant $\phi$.

If I'm reading this right, then there is a significant difference in the "river" model between the Schwarzschild and Kerr cases: in the Schwarzschild case, the "flow lines" of the river are the trajectories of observers that free-fall in from rest at infinity with zero angular momentum, whereas in the Kerr case they are not (because the flow lines have constant $\phi$, whereas the free-fall trajectories have constant $\phi_{ff}$. So it looks like I still need to give this more thought.

 

[yuiop]

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

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

 

[yuiop]

Right I agree that the acceleration due to gravity won't depend on the initial conditions ...!

Coordinate acceleration due to gravity does depend on initial conditions. In Schwarzschild coordinates a free falling particle that was released from infinity will be de-accelerating near the event horizon (in coordinate terms), while a particle released from rest near the event horizon, will be accelerating.

 

[PeterDonis]

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

Intrinsic spin should certainly contribute to total angular momentum, but the constant of the motion is not total angular momentum, it's orbital angular momentum--more precisely, it's the component of orbital angular momentum that is parallel to the axial KVF. So I'm not sure that intrinsic spin can play a part in the question we're considering.

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

That's already true in Kerr spacetime simply because the axial KVF and the timelike KVF are not orthogonal, so an object that is static (i.e., following an orbit of the timelike KVF) has a nonzero momentum component that's parallel to the axial KVF. A ZAMO is an observer whose 4-momentum has an extra component in the axial direction that's just right to make its total 4-momentum orthogonal to the axial KVF. Intrinsic spin doesn't play a part in it.

 

[PeterDonis]

Coordinate acceleration due to gravity does depend on initial conditions.

That depends on the coordinates. The coordinates I was referring to were the coordinates in the local inertial frame of a static observer, not the global Schwarzschild coordinates. I should have made that clearer in my post.

 

[PeterDonis]

...what about a free-fall object released by the static observer? If its $L$ remains constant, then its angular velocity will have to change as it falls, since $g_{t \phi}$ will change along its trajectory. But if the "acceleration due to gravity" everywhere has no $\phi$ component, how can the angular velocity change?

Ok, thinking about this some more, there is one obvious example of a free-fall object released by a static observer where the angular velocity of the free-fall object *has* to change fairly quickly. Consider a static observer "hovering" just outside the static limit, who releases an object in free-fall. The object's initial angular velocity will be zero, but it will have to change quickly, because the object will soon be below the static limit, where there are no timelike worldlines with zero angular velocity. So it's certainly not possible for a free-falling worldline (or any worldline) to have zero angular velocity "all the way down", so to speak.

The question then becomes, what is it that gives the free-falling object its nonzero angular velocity? And is whatever it is present everywhere, or just below the static limit? (I now suspect that the effect *is* present everywhere, contrary to what I thought earlier, but I'm still not sure how to formulate a "local" description of what's going on.)

 

[WannabeNewton]

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

which is basically the same result as in that paper linked in the previous page of this thread. Is there a physical reason for why a static observer would measure a freely falling particle as having constant angular velocity in the local Lorentz frame of the static observer?

To ask a slightly different question, when you say "Ok, thinking about this some more, there is one obvious example of a free-fall object released by a static observer where the angular velocity of the free-fall object *has* to change fairly quickly", with respect to whom/what are we measuring the angular velocity of the freely falling object to be changing fairly quickly? With respect to a static observer within the gravitational field? Because that would be at arms with the above. Or is it with respect to the Boyer–Lindquist coordinate chart on Kerr space-time i.e. with respect to a static observer at infinity?