Practical measurements of rotation in the Kerr metric

[WannabeNewton]

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

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

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

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

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

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

Would you agree with all of the above?

 

[WannabeNewton]

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

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

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

 

[yuiop]

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

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

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

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

 

[WannabeNewton]

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

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

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

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

*Rigid here means Born-Rigid.

 

[yuiop]

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

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

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

 

[WannabeNewton]

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

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

 

[PeterDonis]

Would you agree with all of the above?

Yes, all looks right to me.

 

[yuiop]

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

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

Also, Wikipedia makes this intriguing statement:

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

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

 

[PeterDonis]

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

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

 

[WannabeNewton]

Perhaps that is an abuse of the term?

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

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

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

 

[PeterDonis]

In Schwarzschild space-time the geodetic precession of a torque-free gyroscope relative to the distant stars arises similarly from the coupling of the orbital angular velocity of the gyroscope relative to the distant stars to the connection coefficients (which we know correspond in a sense to the gravitational acceleration).

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

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

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

 

[WannabeNewton]

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

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

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

Yes, all looks right to me.

Great, thanks!

 

[PeterDonis]

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

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

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

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

 

[yuiop]

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

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

 

[WannabeNewton]

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

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

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

 

[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.