Practical measurements of rotation in the Kerr metric

[PeterDonis]

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

Just realized that this isn't quite right; I missed a factor of $\left( 1 + 2M / r \right)$ multiplying $a^2$ in $\Omega_{\infty}$. The correct formula for the ratio is

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

This doesn't change any of the key conclusions, but I wanted to correct it for the record.

 

[yuiop]

Just realized that this isn't quite right; I missed a factor of $\left( 1 + 2M / r \right)$ multiplying $a^2$ in $\Omega_{\infty}$...

Yes, I was going to point out that the equation for $\Omega_{\infty} = - g_{t \phi} / g_{\phi \phi}$ should have been:

$\Omega_{\infty} = \frac{2 M a}{r^3 + a^2 (r +2 M )}$

but as you correctly noted it does not change your key conclusions and $\omega_{\infty}/\Omega_{\infty}$ at large r remains 1/2. I only noticed because I was trying to get that pesky 1/2 to go away, but it remains despite the correction. I wonder if it is somehow connected to the "Thomas half" that pops up in relation to electron spin orbit? At very large r we are essentially in flat space and Thomas precession becomes more relevant than de Sitter or Lense-Thirring precession.

Another observation is that another way of expressing equal rotation rates for $\omega_{\infty}$ and $\Omega_{\infty}$ is to require $\omega_{\infty}-\Omega_{\infty}=0$. Expressed like this $\omega_{\infty}=\Omega_{\infty}$ when $r \rightarrow \infty$. This is slightly puzzling, but nevertheless it still remains true that above a critical radius (near the photon orbit), $\omega_{\infty}<\Omega_{\infty}$ and below the critical radius $\omega_{\infty}>\Omega_{\infty}$. At the critical radius $\omega_{\infty}=\Omega_{\infty}$ in the Kerr metric, so you were right to correct my earlier statement that $\omega_{\infty}\ne \Omega_{\infty}$ is always true.

Now for another puzzle. Equation (52) of the paper linked by WBN gives the angular spin velocity $(\Omega)$ as measured at infinity in the Schwarzschild metric of an un-torqued orbiting gyroscope axis as:

$\Omega= \frac{3M-r}{2M-r+\omega^2 r^3}$,

where $\omega$ in this case is the orbital velocity of the gyroscope. For an object in a geodesic orbit of constant radius, the angular orbital velocity of the object as measured at infinity in the SM is:

$\omega = \sqrt{\frac{M}{r^3}}$

Substituting this expression into the equation above it gives $\Omega = \omega$. (this argument is outlined in the paper). However it implies that an object in a geodesic orbit in the Schwarzschild metric does not rotate relative to the instantaneous radial vector and returns to its original position and orientation once per orbit. Where has the geodetic and L-T precession gone? It also implies that a gyroscope in the Gravity Probe B spacecraft precessed a full rotation approximately every 46 minutes relative to the spacecraft telescope fixed on the guide star, which I am pretty sure was not the case.

Is this discrepancy at all related to the Papapetrou equations that WBN mentioned (Thanks WBN ;), which indicate a spinning object does not follow a normal geodesic path? How big is the deviation according to this effect? This article implies it can be quite significant as indicated in the diagram below:

The solid path is the path of a spinning particle and the dashed in-falling path is that of a spin-less particle, both starting with the same velocity and location.

 

[PeterDonis]

$\omega_{\infty}/\Omega_{\infty}$ at large r remains 1/2. I only noticed because I was trying to get that pesky 1/2 to go away, but it remains despite the correction. I wonder if it is somehow connected to the "Thomas half" that pops up in relation to electron spin orbit? At very large r we are essentially in flat space and Thomas precession becomes more relevant than de Sitter or Lense-Thirring precession.

Yes, that factor of 1/2 remains, but remember that both $\omega_{\infty}$ and $\Omega_{\infty}$ go to zero as $r \rightarrow \infty$. So their ratio going to 1/2 at large $r$ really just says that they don't go to zero at exactly the same "rate", so to speak. Given this, I don't think Thomas precession is the reason the ratio is 1/2, since Thomas precession requires nonzero angular velocity, and both angular velocities are going to zero at large $r$.

 

[yuiop]

Any thoughts on the problem I mentioned with equation (52) in my previous post?

 

[PeterDonis]

Any thoughts on the problem I mentioned with equation (52) in my previous post?

Wait for it...

$\Omega= \frac{3M-r}{2M-r+\omega^2 r^3}$

There should be an extra factor of $\omega$ on the RHS, correct?

Where has the geodetic and L-T precession gone?

Write the formula this way:

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

where $\gamma^2 = 1 / \left( 1 - 2M / r - \omega^2 r^2 \right)$ is just the [Edit: square of the] "time dilation factor" of the orbiting object relative to infinity (the $2M / r$ is gravitational time dilation and the $\omega^2 r^2$ is from the nonzero velocity $v$ relative to a static observer at $r$). The factor of $\gamma^2$ is the Thomas precession (note that the corresponding formula in flat spacetime is $\Omega = \gamma^2 \omega$), and the factor of $\left( 1 - 3M / r \right)$ is the de Sitter precession. It just so happens that, for a geodesic orbit, $\gamma^2 = 1 / \left( 1 - 3M / r \right)$, so the increase in $\Omega$ due to Thomas precession exactly cancels the decrease in $\Omega$ due to de Sitter precession, leaving, as you note, $\Omega = \omega$. (Note that there is no Lense-Thirring precession because we are in Schwarzschild spacetime, not Kerr spacetime.)

Note also what $\Omega = \omega$ means. It means that the Lie transported spatial basis vectors carried by the orbiting object (i.e., the ones that always point at neighboring objects orbiting with the same angular velocity--the easiest one to think about is the one that always points radially outward) are rotating in the "forward" sense (the same sense as the object is orbiting) with respect to Fermi-Walker transported basis vectors (i.e., ones oriented by gyroscopes) with angular velocity $\omega$, i.e., the same angular velocity as the object is orbiting. In other words, this is just like the Newtonian case, where a gyroscope that starts out pointed at some particular object at infinity stays pointed at that object forever (assuming it is not subjected to any torque). But again, as above, this is not because there is no Thomas precession or de Sitter precession; it's because they just happen to cancel each other out for a geodesic orbit.

Another interesting thing to ponder: what happens at $r = 3M$? The above formula makes it clear that $\Omega = 0$, i.e., gyroscopes carried by an orbiting object now point in the same direction as Lie transported basis vectors, i.e., the gyroscopes precess, relative to infinity, at the same rate that the object is orbiting (so one of them always points radially outward). Of course "orbiting" here means the more general sense of maintaining the same $r$ with constant (not necessarily geodesic) angular velocity, since the only geodesic orbits at $r = 3M$ are the photon orbits.

For $r < 3M$, it gets weirder: $\Omega$ is now *negative*, meaning that gyroscopes carried by an "orbiting" object precess with a *larger* angular velocity than the object itself orbits. This is related to what is sometimes called "centrifugal force reversal" near a black hole, which I have posted about on my PF blog:

https://www.physicsforums.com/blog.php?b=4327

 

[WannabeNewton]

Another interesting thing to ponder: what happens at $r = 3M$? The above formula makes it clear that $\Omega = 0$, i.e., gyroscopes carried by an orbiting object now point in the same direction as Lie transported basis vectors, i.e., the gyroscopes precess, relative to infinity, at the same rate that the object is orbiting (so one of them always points radially outward).

Another interesting aspect of $r = 3M$ and gyroscopes: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf (pp. 233-235 of the notes).

 

[WannabeNewton]

Morning. I just have two additional comments on top of Peter's detailed reply:

...gives the angular spin velocity $(\Omega)$ as measured at infinity in the Schwarzschild metric of an un-torqued orbiting gyroscope axis as:

The gyroscopic precession that the article writes down is the precession relative to the distant stars. In deriving it (and in deriving the geodetic precession) we assume that the gyroscope is Fermi-transported along its worldline which is equivalent to the gyroscope being torque-free, as you noted. Therefore:

Is this discrepancy at all related to the Papapetrou equations that WBN mentioned (Thanks WBN ;), which indicate a spinning object does not follow a normal geodesic path? How big is the deviation according to this effect? This article implies it can be quite significant as indicated in the diagram below:

This is true but keep in mind that this is for spinning objects of small but non-zero size (so such an object will have an intrinsic moment of inertia and mass quadrupole moment e.g. that of an ellipsoid whose semi-major axis is much smaller than the radius of curvature of space-time). If we assume, for simplicitly, that the center of mass of such an object follows a geodesic then the gravitational torques exerted on the spinning object will be directly proportional to the Riemann curvature tensor (which is second order) and the mass quadrupole moment of the object. If we have a small sphere then there will be no gravitational tidal toques exerted on the object because it has a vanishing mass quadrupole moment and so a torque-free gyroscope whose center of mass follows a geodesic can be modeled by a small sphere.

 

[WannabeNewton]

Just to add a bit more detail. Consider a small spinning body of mass density $\rho$ in some space-time. Again small means that the characteristic size of the object is much smaller than the space-time curvature scales so that $R_{\mu\nu\alpha\beta}$ is constant across the body. Assume the center of mass of the body follows a geodesic with 4-velocity $u^{\mu}$ and let $\xi^{\mu}$ be the separation vector from the center of mass to a mass element $\rho d^3x$ on the body.

Now go to the local inertial frame of the center of mass; the local inertial frame has the associated coordinate system $\{x^{\mu}\}$. The center of mass follows a geodesic so the acceleration of the mass element $\rho d^3 x$ relative to the center of mass is just given by the equation of geodesic deviation $a^{i} = -R_{\mu\nu \delta}{}{}^{i}\xi^{\nu}u^{\mu}u^{\delta} = -R_{0 j 0}{}{}^{i}x^{j}$ where I used the fact that $u^{\mu} = \delta^{\mu}_0$ and that $\xi^0 = 0, \xi^i = x^i$ in the center of mass's local inertial frame. Therefore the mass element $\rho d^3 x$ feels a tidal force $-(\rho R_{0 j 0}{}{}^{i}x^{j})d^3 x$ relative to the center of mass and thus a tidal torque $-(\epsilon _{ijk}R_{0 l 0}{}{}^{j}x^k x^{l})\rho d^3 x$ (since $\vec{\tau} = \vec{r}\times \vec{F}$).

If we attach a spin vector $S^{\mu}$ to the center of mass then $\frac{\mathrm{d} S_{i}}{\mathrm{d} t} = -\epsilon _{ijk}R_{0 l 0}{}{}^{j} \int \rho x^k x^{l} d^3 x$. Now note that $\delta^{kl}\epsilon _{ijk}R_{0 l 0}=\epsilon _{ijk}R_{0}{}{}^{ k}{}{} _{0}{}{}^{j} = \epsilon _{i[jk]}R_{0}{}{}^{ [k}{}{} _{0}{}{}^{j]} = 0$ since $R_{0}{}{}^{ k}{}{} _{0}{}{}^{j} = R_{0}{}{}^{ j}{}{} _{0}{}{}^{k}$. Therefore we can add the term $\int -\frac{1}{3}\rho r^2 \delta^{kl} d^{3}x$ to our original $\int \rho x^k x^{l} d^3 x$ without changing $\frac{\mathrm{d} S_{i}}{\mathrm{d} t}$. Note that $Q^{kl} \equiv \int \rho(x^{k}x^l - \frac{1}{3}r^2 \delta^{kl})$ is just the mass quadrupole moment of the spinning body hence we can write $\frac{\mathrm{d} S_{i}}{\mathrm{d} t}$ covariantly as $u^{\mu}\nabla_{\mu}S^{\rho} = \epsilon^{\rho \beta \alpha \mu} R^{\nu}{}{}_{\sigma \alpha\lambda}u_{\mu}u^{\sigma}u^{\lambda}Q_{\beta\nu}$.

This gives us the tidal torque experienced by a small spinning body in a gravitational field. If we have a uniform gravitational field or a small spherical body then clearly the tidal torque vanishes.

 

[yuiop]

Wait for it...

I am just as bad with Christmas presents ... hehe

There should be an extra factor of $\omega$ on the RHS, correct?

Oops, yes! It should have been $\Omega= \frac{\omega(3M-r)}{2M-r+\omega^2 r^3}$

Write the formula this way:

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

where $\gamma^2 = 1 / \left( 1 - 2M / r - \omega^2 r^2 \right)$ is just the "time dilation factor" of the orbiting object relative to infinity (the $2M / r$ is gravitational time dilation and the $\omega^2 r^2$ is from the nonzero velocity $v$ relative to a static observer at $r$).

The time dilation factor bit makes sense. Not entirely sure what $\Omega$ is measured relative to. Here it seems to that $\Omega$ is the precession rate of a gyroscope axis measured relative to the instantaneous radial vector while in post #90, $\Omega$ seems to be relative to the distant stars.

The factor of $\gamma^2$ is the Thomas precession (note that the corresponding formula in flat spacetime is $\Omega = \gamma^2 \omega$), and the factor of $\left( 1 - 3M / r \right)$ is the de Sitter precession. It just so happens that, for a geodesic orbit, $\gamma^2 = 1 / \left( 1 - 3M / r \right)$, so the increase in $\Omega$ due to Thomas precession exactly cancels the decrease in $\Omega$ due to de Sitter precession, leaving, as you note, $\Omega = \omega$. (Note that there is no Lense-Thirring precession because we are in Schwarzschild spacetime, not Kerr spacetime.)

This is all nice and clear. The only issue is that some texts insist that the precession in a gravitational field cannot be broken down into a Thomas precession component because that only applies in flat space, but I think that is an interpretational issue. Your description certainly has a nice logical feel to it.

Note also what $\Omega = \omega$ means. It means that the Lie transported spatial basis vectors carried by the orbiting object (i.e., the ones that always point at neighboring objects orbiting with the same angular velocity--the easiest one to think about is the one that always points radially outward) are rotating in the "forward" sense (the same sense as the object is orbiting) with respect to Fermi-Walker transported basis vectors (i.e., ones oriented by gyroscopes) with angular velocity $\omega$, i.e., the same angular velocity as the object is orbiting.

OK, so by this definition, the geodesically orbiting gyroscope axis (FWTB vector) is precessing relative to the LTSB vector at a rate of $-\Omega$ and in the geodesic orbit case is equal in magnitude to the orbital velocity $\omega$ in the Schwarzschild metric. Here $\Omega$ is relative to the instantaneous radial vector rather than the distant stars.

In other words, this is just like the Newtonian case, where a gyroscope that starts out pointed at some particular object at infinity stays pointed at that object forever (assuming it is not subjected to any torque).

This is the problem I am having. The GBP experiment measuring how much the un-torqued gyroscopes in the perfectly geodesically orbiting spacecraft , precessed relative to a distant star and they expected (and measured) that the gyroscopes would not remain pointing at the distant guide star forever.

But again, as above, this is not because there is no Thomas precession or de Sitter precession; it's because they just happen to cancel each other out for a geodesic orbit.

This suggests that the GPB experiment should only have detected L-T precession due to the rotation of the Earth and no geodetic effect due to orbiting (because of the self cancelling).

For $r < 3M$, it gets weirder: $\Omega$ is now *negative*, meaning that gyroscopes carried by an "orbiting" object precess with a *larger* angular velocity than the object itself orbits. This is related to what is sometimes called "centrifugal force reversal" near a black hole, which I have posted about on my PF blog:

https://www.physicsforums.com/blog.php?b=4327

Nice blog entry. Very clear. It was nice to see that that with the apropriate transformation and change of sign convention, that your equation for the centrifugal force agrees with the one I gave in the old thread that was "inconclusive" due to excessive bickering. (Last equation of this post).

 

[yuiop]

The gyroscopic precession that the article writes down is the precession relative to the distant stars. In deriving it (and in deriving the geodetic precession) we assume that the gyroscope is Fermi-transported along its worldline which is equivalent to the gyroscope being torque-free, as you noted.

From Peter's post, it seems that the gyroscopic precession $\Omega$ is relative to the instantaneous LTSB vector rather than the distant stars. I am not 100% sure. See my comments in previous post.

... This gives us the tidal torque experienced by a small spinning body in a gravitational field. If we have a uniform gravitational field or a small spherical body then clearly the tidal torque vanishes.

Ah OK thanks. So for small test bodies, we can generally treat the Papapetrou effect on a small spinning body as negligible.

 

[WannabeNewton]

From Peter's post, it seems that the gyroscopic precession $\Omega$ is relative to the instantaneous LTSB vector rather than the distant stars.

Allow me to correct myself. Yes $\Omega$ measures the precession of torque-free gyroscopes carried along the worldline of a circularly orbiting observer in Schwarzschild space-time relative to Lie transported spatial basis vectors carried along said observer's worldline. Geodetic precession on the other hand is measured relative to the distant stars.

Precession relative to the stars and precession relative to the Lie transported spatial basis vectors coincide for the static observers in Kerr space-time which might be why I made the error above, since we've been talking about them for so long now

 

[WannabeNewton]

This is the problem I am having. The GBP experiment measuring how much the un-torqued gyroscopes in the perfectly geodesically orbiting spacecraft , precessed relative to a distant star and they expected (and measured) that the gyroscopes would not remain pointing at the distant guide star forever.

What gravity probe B measures is the geodetic precession, see here: http://physics.umd.edu/lecdem/services/refs_scanned_WIP/3%20-%20Vinit%27s%20LECDEM/D401/2/AJP001248.pdf

In the PPN (post parametrized Newtonian) formalism (of which Schwarzschild space-time is a special case), one can separate the Thomas Precession from the Geodetic precession and separate Lense-Thirring precession from both of these.

For a detailed derivation of gyroscopic precession (relative to the distant stars) in the PPN formalism, in which the three precession effects are separated, see section 3.4.3. of "Gravitation and Inertia"-Wheeler and Ciufolini.

 

[PeterDonis]

What gravity probe B measures is the geodetic precession, see here: http://physics.umd.edu/lecdem/services/refs_scanned_WIP/3%20-%20Vinit%27s%20LECDEM/D401/2/AJP001248.pdf

According to the Introduction of that paper, it measures both, which is consistent with what I've read in other sources about Gravity Probe B. Later on in the paper, when they derive the formula for geodetic precession, the derivation also includes the Thomas Precession term, so that is included as well.

Not entirely sure what $\Omega$ is measured relative to. Here it seems to that $\Omega$ is the precession rate of a gyroscope axis measured relative to the instantaneous radial vector

Yes, that's correct; sorry for switching notation in mid-stream. There should be more forms of the Greek letter omega.

The only issue is that some texts insist that the precession in a gravitational field cannot be broken down into a Thomas precession component because that only applies in flat space, but I think that is an interpretational issue.

As the paper WBN linked to makes clear (equations 46 and 47), both the geodetic and the Thomas precession have the same general form of some constant times $\vec{v} \times \vec{a}$, where $\vec{v}$ is the orbital velocity and $\vec{a}$ is the "acceleration due to gravity" (the paper writes it as $\nabla \phi_G$, the gradient of the Newtonian potential, which amounts to the same thing). So I think it is indeed an interpretational issue.

(Btw, when it comes to interpretational issues, normally I tend to come down on the side of *not* depending on analogies that go "here's what happens in curved spacetime, and here's what would have happened if spacetime were flat", since the "flat background" is unobservable. But in this case, conceptually, I think it helps--at any rate it helps me--to look first at the flat spacetime formula for Thomas precession, to get the $\gamma^2$ factor, and then look at how the formula changes in curved spacetime to see the geodetic effect due to the central mass. But ultimately that's more pedagogy than physics.)

OK, so by this definition, the geodesically orbiting gyroscope axis (FWTB vector) is precessing relative to the LTSB vector at a rate of $-\Omega$ and in the geodesic orbit case is equal in magnitude to the orbital velocity $\omega$ in the Schwarzschild metric. Here $\Omega$ is relative to the instantaneous radial vector rather than the distant stars.

Yes.

This is the problem I am having. The GBP experiment measuring how much the un-torqued gyroscopes in the perfectly geodesically orbiting spacecraft , precessed relative to a distant star and they expected (and measured) that the gyroscopes would not remain pointing at the distant guide star forever.

This suggests that the GPB experiment should only have detected L-T precession due to the rotation of the Earth and no geodetic effect due to orbiting (because of the self cancelling).

Well, the paper's result for the geodetic precession doesn't match the one we have been using, which came from the other paper WBN linked to, so something is going on. This paper's result is (equation 4 in the paper, and I'm switching units so that $G = 1$)

$$
\Omega = \frac{3}{2r} \left( \frac{M}{r} \right)^{\frac{3}{2}}
$$

which can be rewritten, using $\omega = \sqrt{M / r^3}$, as

$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$

However, if we look at the details of the derivation (leading up to equation 48 in the paper), we see that the formula as I have just written it is actually generally applicable to any angular velocity, not just a geodesic orbit, at least in the weak field approximation (we substitute $\omega r$ for $v$ in the general formulas in the paper and use $\nabla \phi = M / r^2$ in the Newtonian limit). So this latter formula should be compared with the general formula from the other paper,

$$
\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}
$$

I'm not sure how to resolve this discrepancy.

Nice blog entry. Very clear. It was nice to see that that with the apropriate transformation and change of sign convention, that your equation for the centrifugal force agrees with the one I gave in the old thread that was "inconclusive" due to excessive bickering. (Last equation of this post).

Thanks! Feel free to link to the blog entry in future threads if this comes up.

 

[WannabeNewton]

According to the Introduction of that paper, it measures both, which is consistent with what I've read in other sources about Gravity Probe B. Later on in the paper, when they derive the formula for geodetic precession, the derivation also includes the Thomas Precession term, so that is included as well.

Perhaps I missed it but I don't see in the introduction any mention of Thomas precession. They only mention the Lense-Thirring precession and the geodetic precession (the term proportional to $v \times \nabla \phi_g$). The Thomas precession is a term proportional to $v \times a$.

EDIT: the following might be interesting:

http://postimg.org/image/k3ny4zptx/
http://postimg.org/image/wh0sbwfid/

$$
\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}
$$

But that's the magnitude of the vorticity $\omega^{\mu}$ which measures the rotation of separation vectors Lie transported along a reference worldline in a congruence relative to Fermi-transported spatial basis vectors (gyroscopes) along this reference worldline. This doesn't necessarily equal (up to a sign) the gyroscopic precession relative to the distant stars. It only does in the case of static observers because Lie transport locks separation vectors to the distant stars.

 

[PeterDonis]

Perhaps I missed it but I don't see in the introduction any mention of Thomas precession.

Equation 47 in the paper.

The Thomas precession is a term proportional to $v \times a$.

And $a = \nabla \phi$. At least, that's how the paper you linked to earlier appears to be obtaining equation 47. But this usage of $a$ appears to differ from other sources; see below.

Also, for comparison with the following, this paper says that the net precession of $- (3/2) \vec{v} \times \nabla \phi$ is due to two sources: a geodetic precession of $- 2 \vec{v} \times \nabla \phi$, and a "Thomas precession" of $(1/2) \vec{v} \times \nabla \phi$. (The sign convention is because they are defining the precession of gyroscopes relative to Lie transported basis vectors, rather than the reverse; but the key is that the two effects are of opposite sign, unlike the excerpt discussed below.)

EDIT: the following might be interesting:

http://postimg.org/image/k3ny4zptx/
http://postimg.org/image/wh0sbwfid/

This excerpt breaks things up somewhat differently, and appears to use different nomenclature. What it calls "Thomas precession" is indeed driven by proper acceleration, not coordinate acceleration, and is therefore zero for a geodesic orbit. But it breaks up the total precession of $(3/2) \vec{v} \times \nabla \phi$ differently than the paper you linked to previously; it says that $(1/2) \vec{v} \times \nabla \phi$ is due to "precession in the gravitomagnetic field" while $\vec{v} \times \nabla \phi$ is due to "the curvature of space". So there are definitely some interpretational variations here.

But that's the magnitude of the vorticity $\omega^{\mu}$ which measures the rotation of separation vectors Lie transported along a reference worldline in a congruence relative to Fermi-transported spatial basis vectors (gyroscopes) along this reference worldline. This doesn't necessarily equal (up to a sign) the gyroscopic precession relative to the distant stars.

This is true, but the difference is just a correction factor of $\gamma = \sqrt{1 / \left( 1 - 2M / r - \omega^2 r^2 \right)}$ multiplying equation 4 in the paper you linked to, which doesn't resolve the discrepancy.

 

[WannabeNewton]

Equation 47 in the paper.

So there are definitely some interpretational variations here.

Thanks! The flurry of differences regarding the decomposition of gyroscopic precession is making my head spin

This is true, but the difference is just a correction factor of $\gamma = \sqrt{1 / \left( 1 - 2M / r - \omega^2 r^2 \right)}$ multiplying equation 4 in the paper you linked to, which doesn't resolve the discrepancy.

Ah ok I see now what you were originally referring to. I'm burnt out right now but let me see what I can come up with (of course knowing you, you'll probably figure it out before I do )

 

[yuiop]

...
I'm not sure how to resolve this discrepancy.

Since we are looking for an error, Mentz mentioned their might be one in the original paper back in post #77 which I have quoted below:

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]

which can be rewritten, using $\omega = \sqrt{M / r^3}$, as

$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$

However, if we look at the details of the derivation (leading up to equation 48 in the paper), we see that the formula as I have just written it is actually generally applicable to any angular velocity, not just a geodesic orbit, at least in the weak field approximation (we substitute $\omega r$ for $v$ in the general formulas in the paper and use $\nabla \phi = M / r^2$ in the Newtonian limit). So this latter formula should be compared with the general formula from the other paper,

$$
\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}
$$

I'm not sure how to resolve this discrepancy...

As mentioned before the equation $\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}$ implies a geodesically orbiting gyroscope remains locked on the distant stars which is obviously not correct.

It turns out that this is the expression for the rotation of a gyroscope axis relative to the LTSB vectors in terms of the proper time of the orbiting gyroscope. This is effectively the rate at which a gyroscope must rotate relative to the LTSB vectors (in the opposite sense to the orbital angular velocity) in order to remain fixed fixed on the distant stars, but because this is in proper time, it fails to do so when corrected for coordinate time.

When we apply the correction gamma factor $\gamma = \sqrt{1 / \left( 1 - 2M / r - \omega^2 r^2 \right)}$ (as mentioned by WBN) the correct expression becomes:

$\Omega = \omega \frac{1 - 3M / r}{\sqrt{1 - 2M / r - \omega^2 r^2}}$

so that now both $\Omega$ and $\omega$ are coordinate angular velocities measured by an observer at infinity.

For a geodesic orbit, $\omega^2 = M/r^3$ and when substituted into the above equation we get:

$\Omega = \omega \sqrt{1 - 3M / r}$.

What we want is the precession rate relative to the distant stars rather than the LTSB vectors , so we must subtract the above expression from the orbital velocity to obtain:

$\Omega_{stars} = \omega \left(1-\sqrt{1 - 3M / r} \right)$

When we carry out a series expansion of $\Omega_{stars}/\omega$ we get:

$\Omega_{stars}/\omega = (3 M)/(2 r)+(9 M^2)/(8 r^2)+(27 M^3)/(16 r^3)+(405 M^4)/(128 r^4) ...+O(M^7)$

which makes it clear that the equation in the Gravity Probe B paper:

$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$

is just an approximation of $\Omega_{stars}$ as defined above.

 

[PeterDonis]

It turns out that this is the expression for the rotation of a gyroscope axis relative to the LTSB vectors in terms of the proper time of the orbiting gyroscope. This is effectively the rate at which a gyroscope must rotate relative to the LTSB vectors (in the opposite sense to the orbital angular velocity) in order to remain fixed fixed on the distant stars, but because this is in proper time, it fails to do so when corrected for coordinate time.

Ah, of course. So really the formula (for an arbitrary orbit, not necessarily a geodesic one) should read, using our earlier notation,

$$
\Omega = \gamma^2 \omega_{\infty} \left( 1 - \frac{3M}{r} \right)
$$

where $\Omega$, without the subscript $\infty$, is relative to the proper time of the orbiting object. To convert this to a frequency relative to infinity, we apply the correction factor of $\gamma^{-1}$ to obtain

$$
\Omega_{\infty} = \gamma \omega_{\infty} \left( 1 - \frac{3M}{r} \right)
$$

For an orbiting object, $\gamma = 1 / \sqrt{1 - 3M / r}$, and the rest follows. Cool!

Btw, this means the notation in the paper WBN linked to (which is apparently fairly common since a paper by Rindler is referenced that gives the same result) is rather misleading; by stating the result for an orbiting object as $\Omega = \omega$ it invites the incorrect interpretation that I made. Using the $\infty$ subscript makes it clear that $\Omega = \omega_{\infty}$ is comparing apples and oranges; the "apples to apples" result is $\Omega_{\infty} = \gamma^{-1} \omega_{\infty}$.

 

[Mentz114]

Since we are looking for an error, Mentz mentioned their might be one in the original paper back in post #77 which I have quoted below:

I corresponded with the second author (梁為傑) and we concluded that the frame and coframe bases are correctly written in the paper. I disagree with the components of the vorticity vector, but get the same value for $\Omega=\left( \omega^\mu \omega_\mu \right)^{(1/2)}$ as they do.

 

[WannabeNewton]

which makes it clear that the equation in the Gravity Probe B paper:
...
is just an approximation of $\Omega_{stars}$ as defined above.

Awesome thanks! That clears up everything.

 

[WannabeNewton]

I thought I should bring something up that I completely missed at the inception of this thread. Consider the congruence of ZAMOs in Kerr space-time with 4-velocity field $u$; the most natural frame field for this congruence is: $$u = e_{t} = e^{-\nu}(\partial_t + \omega \partial_{\phi})\\ e_{r} = e^{-\mu}\partial_{r}\\ e_{\theta} = e^{-\lambda}\partial_{\theta}\\ e_{\phi} = e^{-\psi}\partial_{\phi}$$ where $e^{2\nu} = \frac{\rho^2 \Delta}{\Sigma^2}, e^{2\mu} = \frac{\rho^2}{\Delta}, e^{2\psi} = \frac{\Sigma^2}{\rho^2}\sin ^2\theta, e^{2\lambda} = \rho^2, \omega = \frac{2Ma r}{\Sigma^2}$ in Boyer–Lindquist coordinates.

For future convenience, let $\eta = \partial_t + \omega \partial_{\phi}$ and let $\psi^{\mu} = (\partial_{\phi})^{\mu}$. We know that $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ i.e. that the vorticity satisfies $\omega^{\mu} = \epsilon^{\mu\nu\gamma\delta}\eta_{\nu}\nabla_{\gamma}\eta_{\delta} = 0$. However keep in mind that $\nabla_{(\mu}\eta_{\nu)} \neq 0$, that is, the congruence of ZAMOs is not rigid. This is because the angular velocity $\omega$ is not constant across the entire congruence but rather only along the worldlines of individual ZAMOs.

Note that $\mathcal{L}_{u}\psi^{\mu} = 0$ and $\psi_{\mu}a^{\mu} = \psi_{\mu}u^{\mu} = 0$ where $a^{\mu}$ is the 4-acceleration of the ZAMOs.

Therefore $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\delta}\psi^{\delta}) = u^{\gamma}\nabla_{\gamma}\psi^{\mu} = F_{u}\psi^{\mu}$ where $h_{\mu\nu}$ is the spatial metric relative to the ZAMOs and $F_{u}\psi^{\mu}$ is the Fermi-Walker derivative along the worldline of an individual ZAMO. Note also that $F_{u}e^{\mu}_{\phi} = e^{-\psi}F_{u}\psi^{\mu} + \psi^{\mu}(u^{\nu}\partial_{\nu}e^{-\psi}) = e^{-\psi}F_{u}\psi^{\mu}$.

Recall from Malament's text that $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\delta}\psi^{\delta}) = 0$ if and only if $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ (see p.223: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf). If we apply this proposition to the congruence of ZAMOs and use the results from the previous paragraph then $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ would imply that $F_{u}e_{\phi} = 0$.

In other words, a gyroscope carried by a ZAMO observer would not precess relative to the $e_{\phi}$ axis of the natural frame we've attached to this ZAMO. But from exercise 33.4 in MTW, such a gyroscope precesses relative to the $e_{r}$ axis of this frame with an angular velocity $\Omega^{r} \propto \partial_{\theta}\omega / \rho$. If the gyroscope precesses relative to $e_r$ then it also has to precess relative to $e_{\phi}$.

There's no contradiction of course because the proposition in Malament's text required $\eta$ to be a Killing field i.e. it required the congruence described by $\eta$ to be rigid everywhere and not just when restricted to a single ring (see p.221 and p.223). The fact that the congruence of ZAMOs is not rigid (as mentioned before) lends to a gyroscopic precession in the natural ZAMO frame defined above. From exercise 33.4 in MTW we in fact see that the gyroscopic precession in this frame exists only because $\omega$ is not constant everywhere across the congruence which is exactly what prevents the congruence of ZAMOs from being rigid.

I just thought I should make note of that because I totally missed it the first time I read Malament's text.

 

[PeterDonis]

from exercise 33.4 in MTW, such a gyroscope precesses relative to the $e_{r}$ axis of this frame with an angular velocity $\Omega^{r} \propto \partial_{\theta}\omega / \rho$.

But note that in the equatorial plane, $\theta = \pi / 2$, we have $\partial_{\theta} \omega = 0$, so this precession goes to zero in the equatorial plane, which is where we've been doing most of our analysis in this thread. It is definitely worth remembering, though, that Kerr spacetime is only axisymmetric, so an analysis restricted to the equatorial plane leaves out significant physics.

 

[WannabeNewton]

I do have a related comment though. On p.221, Malament states the following: "We want to think of the ring as being in a state of rigid rotation, i.e., rotation with the distance between points on the ring remaining constant. So we are further led to restrict attention to just those congruences of timelike curves on $\mathcal{R}$ that are invariant under all isometries generated by $\tilde{t}^a$. Equivalently (moving from the curves themselves to their tangent fields), we are led to consider future-directed timelike vector fields on $\mathcal{R}$ of the form $(\tilde{t}^a+ k \phi^a)$, where $k$ is a number."

Here $\mathcal{R}$ represents the world tube of the ring. However I have a slight problem with this definition of rigid rotation of the ring. The definition starts out by saying that the congruence of time-like curves forming $\mathcal{R}$ must be invariant under all isometries generated by the time-like killing field $\tilde{t}^a$ which is fine but note that even the ring we obtain by restricting the congruence of ZAMOs in Kerr space-time to a single radius $R$ satisfies this condition because the tangent field to such a ring would be $\eta^a = \tilde{t}^a + \omega \phi^a$, where $\frac{2MaR}{\Sigma^2}$, hence $\mathcal{L}_{\tilde{t}}\eta^{a} = \mathcal{L}_{\tilde{t}}\tilde{t}^a + \omega \mathcal{L}_{\tilde{t}}\phi^a + \phi^a \tilde{t}^a \partial_{a}\omega =0$.

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically. However the definition quoted above then subsequently says that the ring satisfies rigid rotation if its tangent field is of the form $\eta^a = \tilde{t}^a + k \phi^a$ where $k$ is constant everywhere in space-time*. This would disqualify the ring in the previous paragraph from being rigid because $\omega$ is not constant everywhere in space-time, only on the ring itself. However we know for a fact that the ring from the previous paragraph satisfies Born rigidity. Isn't the definition quoted above a much stronger condition than simply requiring that the ring exhibit rigid rotation? By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

*It isn't clear from the wording whether $k$ is meant to be constant just on $\mathcal{R}$ or everywhere in space-time beccause all that's stated is "$k$ is a number". However it is made evident later on, albeit indirectly, that $k$ is constant everywhere in space-time because Malament writes the zero vorticity condition $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ as $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ (p.222) which clearly assumes that $k$ is constant everywhere in space-time otherwise there would be a term $\tilde{t}_{[a}\phi_b \nabla_{b]} k$.

 

[PeterDonis]

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically.

It isn't for the ZAMO congruence, correct. But you could pick another congruence where $\omega$ *was* constant everywhere (subject to some limitations--see below).

By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

Yes. More precisely, if you have a congruence that includes worldlines with different $r$ and/or $\theta$, the congruence will only be rigid if the "rings" at different $r$ and $\theta$ are all rigidly rotating with respect to each other, which means that the angular velocity $\omega$ (which is set by the constant $k$ in the 4-velocity field of the congruence) must be the same for all "rings". This in turn implies that at most one of these "rings" can be a ring of ZAMOs, i.e., only at at most one $(r, \theta)$ in the congruence will $\omega$, the angular velocity of the entire congruence, be equal to the ZAMO angular velocity $\Omega ( r, \theta )$, which of course, as the notation shows, varies with $r$ and $\theta$. (Actually, technically, there could be two such values of $(r, \theta)$ if the congruence was symmetric about the equatorial plane, since the ZAMO angular velocity is a function of $\sin^2 \theta$ and $\sin \theta = \sin \left( \pi - \theta \right)$.)

Of course such a rigid congruence will also be limited in spatial extent, since for fixed $\omega$ there will be a "boundary" in the $( r, \theta )$ plane at which the tangential velocity reaches the speed of light, so the congruence can't extend further outward than this boundary.

 

[yuiop]

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically.

$\omega$ can be constant everywhere (within boundaries) if we do not insist that all the "other" rings also be ZAMO's, but I am not sure that has been specified. I think that is basically what Peter is saying.

However the definition quoted above then subsequently says that the ring satisfies rigid rotation if its tangent field is of the form $\eta^a = \tilde{t}^a + k \phi^a$ where $k$ is constant everywhere in space-time*. This would disqualify the ring in the previous paragraph from being rigid because $\omega$ is not constant everywhere in space-time, only on the ring itself. However we know for a fact that the ring from the previous paragraph satisfies Born rigidity. Isn't the definition quoted above a much stronger condition than simply requiring that the ring exhibit rigid rotation? By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

Condition (3.2.1) on page 222 is a condition that must hold if the ring is to qualify as non rotating by the CIR criterion, so it does not explicitly state that this is a condition that must hold on $\mathcal{R}$ for the ring to qualify as having rigid rotation. I think the qualifier "must hold on $\mathcal{R}$" may be significant. See below.

*It isn't clear from the wording whether $k$ is meant to be constant just on $\mathcal{R}$ or everywhere in space-time because all that's stated is "$k$ is a number". However it is made evident later on, albeit indirectly, that $k$ is constant everywhere in space-time because Malament writes the zero vorticity condition $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ as $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ (p.222) which clearly assumes that $k$ is constant everywhere in space-time otherwise there would be a term $\tilde{t}_{[a}\phi_b \nabla_{b]} k$.

Page 231 makes it clear that k is a function of r (it gives an example in the Godel spacetime) so for k to be a constant, r must equal $\mathcal{R}$. This is also implied in the qualifier to condition (3.2.1) that it "must hold on $\mathcal{R}$".

Page 237 states that if we have two concentric rings that have Born rigid rotation with respect to each other, then if one ring qualifies as non-rotating by a given criteria, then the other ring can fail the non-rotating test using the same criteria. This is certainly true in the equatorial plane in the Kerr metric.

P.S. I hope I am not speaking out of turn

 

[WannabeNewton]

This in turn implies that at most one of these "rings" can be a ring of ZAMOs, i.e., only at at most one $(r, \theta)$ in the congruence will $\omega$, the angular velocity of the entire congruence, be equal to the ZAMO angular velocity $\Omega ( r, \theta )$, which of course, as the notation shows, varies with $r$ and $\theta$.

Ah right so let's say we have a ring of ZAMOs in Kerr space-time with some angular velocity $\omega_0 = \omega|_{r_0,\theta_0}$. We then consider the timelike congruence with tangent field $\eta^a = \tilde{t}^a + \omega_0 \phi^a$. This is of course not the congruence of ZAMOs but the ring generated by $\eta^a$ at $(r_0,\theta_0)$ will match the respective ring of ZAMOs. Then all the properties we derive for $\eta^a$ will also apply to the respective ring of ZAMOs simply by restricting $\eta^a$ to $(r_0,\theta_0)$.

In particular if we find that $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ then we know Fermi-transport of the spatial direction $\hat{\varphi}^{a} = \frac{h^{a}{}{}_{m}\phi^n}{\left \| h^{a}{}{}_{m}\phi^n\right \|}$ of $\phi^a$ holds: $F_{\hat{\eta}}\hat{\varphi}^a = h^{a}{}{}_{b}\hat{\eta}^m \nabla_m \hat{\varphi}^b = 0$. This then implies that there is no gyroscopic precession along the worldline of any ZAMO (relative to the natural ZAMO frame) if the ZAMO belongs to the ring of ZAMOs at $(r_0,\theta_0)$.

If we let $\xi^a = \tilde{t}^a + \omega \phi^a$ represent the tangent field to the ZAMO congruence, where $\omega = \frac{2M a r}{\Sigma^2}$ is now the non-constant angular velocity of the ZAMO congruence, then we know that $\xi_{[a}\nabla_b \xi_{c]} = 0$. However this does not (necessarily) imply that $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ right? Otherwise all rings of ZAMOs in Kerr space-time would satisfy the property that gyroscopes mounted on the rings don't precess relative to the natural ZAMO frame which isn't true according to MTW.

In fact $\xi_{[a}\nabla_b \xi_{c]} = 0$ implies that $$\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + \omega\tilde{t}_{[a}\nabla_{b}\phi_{c]} + \omega\phi_{[a}\nabla_{b}\tilde{t}_{c]} + \omega^2 \phi_{[a}\nabla_{b}\phi_{c]} + \tilde{t}_{[a}\phi_b \nabla_{c]} \omega = 0$$ whereas $$\eta_{[a}\nabla_{b}\eta_{c]} = \tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + \omega_0\tilde{t}_{[a}\nabla_{b}\phi_{c]} + \omega_0\phi_{[a}\nabla_{b}\tilde{t}_{c]} + \omega_0^2 \phi_{[a}\nabla_{b}\phi_{c]}$$
So $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ if and only if $(\tilde{t}_{[a}\phi_b \nabla_{c]} \omega )|_{r_0,\theta_0} = 0$. I don't know what ring of ZAMOs in Kerr space-time actually satisfies this.

 

[WannabeNewton]

$\omega$ can be constant everywhere (within boundaries) if we do not insist iff we insist that all the "other" rings also be ZAMO's...

Sure but I was talking about the ZAMO congruence in particular.

Condition (3.2.1) on page 222 is a condition that must hold if the ring is to qualify as non rotating by the CIR criterion, so it does not explicitly state that this is a condition that must hold on R for the ring to qualify as having rigid rotation.

That condition can't hold unless the congruence is rigid. See below.

Page 231 makes it clear that k if a function of r (it gives an example in the Godel spacetime) so for k to be a constant, r must equal $\mathcal{R}$. This is also implied in the qualifier to condition (3.2.1) that it "must hold on $\mathcal{R}$".

Given a congruence with tangent field $\eta^a = \tilde{t}^a + k \phi^a$, $k$ cannot vary across this congruence; this is what it means for the entire congruence to be rigid. The condition $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ can clearly only hold if the congruence is rigid otherwise there would be terms involving derivatives of $k$. So for each choice of $\eta^a$ there is a choice of $k$ that is constant across the entire congruence associated with $\eta^a$. What the author shows in Godel space-time is that there are different such choices of $k$ for different values of $r$ which result in the associated $\eta^a$ being non-rotating according to the CIR criterion.

There may be some overlap here with post #132.

 

[yuiop]

Sure but I was talking about the ZAMO congruence in particular.

You seem to be talking about a congruence of ZAM observers with different r, whereas the paper is (possibly) restricting itself to a congruence of ZAM observers with equal r that constitute the congruence of smooth timelike curves on the two dimensional submanifold $\mathcal{R}$ or what they call the striated orbit cylinder.

That condition can't hold unless the congruence is rigid. See below.

A congruence of ZAMO's with unequal r in the Kerr metric, cannot have rigid rotation, so there would be no way for a ring to qualify as non rotating by the CIR criteria if we insist on this stricter definition.

Sometimes it is useful to restrict ourselves to a ring of constant r. It seems to me that we can have a definition of Born rigid motion for a ring by reference only to events on the ring. While we can have a definition of Born rigid rotation for a disc if the disc has constant angular rotation, it is not possible to have Born rigid motion for a disc that has angular acceleration. A ring on the other hand can have Born rigid motion, even when the ring is undergoing angular acceleration, so consideration of a ring is less restrictive.

Just my 2 cents.

 

[WannabeNewton]

...whereas the paper is (possibly) restricting itself to a congruence of ZAM observers with equal r that constitute the congruence of smooth timelike curves on the two dimensional submanifold $\mathcal{R}$ or what they call the striated orbit cylinder.
...
Sometimes it is useful to restrict ourselves to a ring of constant r. It seems to me that we can have a definition of Born rigid motion for a ring by reference only to events on the ring. While we can have a definition of Born rigid rotation for a disc if the disc has constant angular rotation, it is not possible to have Born rigid motion for a disc that has angular acceleration. A ring on the other hand can have Born rigid motion, even when the ring is undergoing angular acceleration, so consideration of a ring is less restrictive.

I think you're forgetting one small but important detail: it isn't enough to just have $\eta^a$ on $\mathcal{R}$ itself. If we only had $\eta^a$ on the cylinder then $\eta_{[a}\nabla_b \eta_{c]} = 0$ would have no meaning because we need to know the behavior of $\eta^a$ in a neighborhood of the cylinder in order to take derivatives of the form $\eta_{[a}\nabla_b \eta_{c]} = 0$. It isn't enough to just have $\eta^a$ on the cylinder itself.

For this reason, if a ring has angular velocity $\omega_0$ we simply consider the vector field $\eta^a = \tilde{t}^a + \omega_0 \phi^a$ on the entirety of space-time so that we can take derivatives off of the cylinder, ergo the impetus for my qualms in post #129 and the posts following it.

 

[yuiop]

So far in this thread we have been a bit vague about the direction of precession relative to the direction of orbital velocity or relative to the direction of rotation of the gravitational body in the Kerr metric.

Could someone confirm that if what we want is the rotation $\Omega_{gyr}$ in proper time, of a gyroscope axis relative to the LTSB vectors transported along with gyroscope, then we should use the opposite sign to that indicated in this paper.

If that is the case then:

In the Kerr metric $\Omega_{gyr} = \frac{-ma}{r^3(1-2m/r)}$.

In the Schwarzschild metric $\Omega_{gyr} = \frac{-\omega(1-3m/r)}{(1-2m/r-\omega^2r^2)}$

In the Minkowski metric $\Omega_{gyr} = \frac{-\omega}{(1-\omega^2r^2)}$.

In general for large r and small $\omega$, the $\Omega_{gyr}$ is in the opposite sense to the orbital angular velocity $\omega$ or to the angular velocity $a$ of the gravitational body. This should guarantee that in asymptotically flat spacetime in the Newtonian regime, an untorqed gyroscope continues to point more or less at the same distant stars. Does that seem right?

 

[yuiop]

I think you're forgetting one small but important detail: it isn't enough to just have $\eta^a$ on $\mathcal{R}$ itself. If we only had $\eta^a$ on the cylinder then $\eta_{[a}\nabla_b \eta_{c]} = 0$ would have no meaning because we need to know the behavior of $\eta^a$ in a neighborhood of the cylinder in order to take derivatives of the form $\eta_{[a}\nabla_b \eta_{c]} = 0$. It isn't enough to just have $\eta^a$ on the cylinder itself.

I will bow to your vastly greater knowledge on this topic

 

[WannabeNewton]

Does that seem right?

Keep in mind that $\Omega$ as defined in the paper is only the magnitude of the gyroscopic precession relative to a connecting vector. The direction of rotation is contained in the vorticity vector $\omega^{\mu}$ but yes if we want the gyroscopic precession relative to a connecting vector then we need to look at $-\omega^{\mu}$ because $\omega^{\mu}$ itself is the rotation of the connecting vector relative to local torque-free gyroscopes.

I will bow to your vastly greater knowledge on this topic

Sorry I wasn't trying to disagree with you or anything. I agree with you on all your points. I was just trying to clarify things for myself. Thanks for the help :)

EDIT: In other words, your statement in the following quote made it clear to me that the author's definition of rigid rotation for a single ring by use of a vector field $\eta^a = \tilde{t}^a + k\phi^a$ spanning all of space-time has no loss of generality in it so as long as we stick to talking about a single ring because we can simply construct a different $\eta^a$ for each ring we wish to analyze:

$\omega$ can be constant everywhere (within boundaries) if we do not insist that all the "other" rings also be ZAMO's, but I am not sure that has been specified. I think that is basically what Peter is saying.

However such a mode of analysis would have us be cautious for even though the ZAMO congruence is itself vorticity free, it doesn't mean all the rings of ZAMOs satisfy the compass of inertia criterion because each such ring corresponds to a different $\eta^a$. That's what I was trying to say in post #132.

 

[yuiop]

Keep in mind that $\Omega$ as defined in the paper is only the magnitude of the gyroscopic precession relative to a connecting vector. The direction of rotation is contained in the vorticity vector $\omega^{\mu}$ but yes if we want the gyroscopic precession relative to a connecting vector then we need to look at $-\omega^{\mu}$ because $\omega^{\mu}$ itself is the rotation of the connecting vector relative to local torque-free gyroscopes.

Thanks for the confirmation!

Sorry I wasn't trying to disagree with you or anything. I agree with you on all your points. I was just trying to clarify things for myself. Thanks for the help :)

No probs. I assume in your final conclusion is similar to Peter's position, that for a set of concentric rings that have rigid angular motion, "that at most one of these "rings" can be a ring of ZAMOs", (in the Kerr metric).

 

[yuiop]

In the Kerr metric $\Omega_{gyr} = \frac{-ma}{r^3(1-2m/r)}$.

In the Schwarzschild metric $\Omega_{gyr} = \frac{-\omega(1-3m/r)}{(1-2m/r-\omega^2r^2)}$

In the Minkowski metric $\Omega_{gyr} = \frac{-\omega}{(1-\omega^2r^2)}$.

Assuming the above is correct, then:

In the Minkowski metric $\Omega_{gyr} = -\gamma\omega$,

where $\gamma$ is the time dilation factor. This is the purely kinematic Thomas precession. (Strictly speaking it is the Wigner precession, as Thomas precession is the accumulation of Wigner precession over a complete orbit.)

In the Schwarzschild metric $\Omega_{gyr} = -\gamma\omega + 3\gamma m/r$.

Expressed like this, the total precession is a combination of Thomas precession and another effect (geodetic?) that operates in the opposite direction. I think this is something that Peter was referring to in a earlier post.

Basically I am trying to break down the components of the total precession. The precession in the Kerr metric is a bit more complex. (That is for later.)