Practical measurements of rotation in the Kerr metric

[PeterDonis]

In other words, this definition of ring non-rotation seems to require more physics than just the physics of the given ring that we wish to analyze (the properties of space-time itself included in the physics of the given ring of course).

I think this is correct; although Malament never explicitly states it, I think it's implicit in his definition of a "striated orbit cylinder" that the coefficient $k$ that appears in the 4-velocity field is constant. (One piece of indirect evidence for this is that he wants the ring to be in a state of "rigid" rotation; but for the striated orbit cylinder congruence to be rigid, $k$ must be constant. Even though technically the value of $k$ off of the cylinder is never "sampled" by any member of the congruence, it still is required, as you note, to evaluate derivatives.)

In the later discussion of the "relative rotation criterion" for different rings, Malament does allow for the possibility of different rings having different values of $k$; rings of ZAMOs at different values of $r$ could be treated using the method he uses there. But that method would treat rings of ZAMOs at different $r$ as belonging to different congruences (different striated orbit cylinders). I don't think his methods would work if you tried to treat all ZAMO rings (at different values of $r$) as members of the same congruence.

 

[WannabeNewton]

Thanks Peter!

 

[yuiop]

Can I conclude from all the above, that a ZAMO ring in the Kerr metric is not rotating by the ZAM criterion (Sagnac effect) but generally speaking does not qualify as not rotating by the CIR criterion? In other words, a gyroscope that is fixed by a suitable gimble to a Kerr ZAMO ring (so that it remains untorqued) so that it is initially tangential to the ZAMO ring, will not remain tangential to the the ZAMO ring, as a general rule. Furthermore, it seems other than satisfying the ZAM criterion, there is nothing special about gyrosopes attached to a ZAMO ring in the Kerr metric, such as remaining pointed at a fixed point at infinity or any other readily identifiable reference point.

 

[WannabeNewton]

Yes.

By the way, I just remembered the existence of another paper of Malament's in which the relative angular velocity of two neighboring observers is discussed in much more detail. Note that Malament starts out by using telescopes and null geodesics (light beams) to define a unit direction vector representing the orientation of the telescope between the two neighboring observers and then relates the angular velocity of the telescope (i.e. the relative angular velocity of the two neighboring observers) to the "cross product" of the unit direction vector with its Fermi-derivative, which physically is just the rotation of the telescope relative to a local compass of inertia (mutually orthogonal gyroscope axes). This is a general relationship that holds for all space-times.

Only then does Malament specializes to two neighboring observers following orbits of a time-like Killing field with the aim of relating the angular velocity of the telescope to the vorticity of the time-like Killing field plus a term involving the unit direction vector (which is, importantly, automatically Lie transported by the Killing field). So while the Fermi-derivative definition of relative angular velocity holds true in general cases, it would seem based on the calculations in Straumann's text, Malament's text, and the following paper of Malament's that the strict relationship between the telescope angular velocity and vorticity only holds for time-like Killing fields (which are trivially Born rigid vector fields).

It's a very instructive read albeit the TeX is very awkward: http://philsci-archive.pitt.edu/117/1/RelOrbitalRotation.pdf

 

[yuiop]

It's a very instructive read albeit the TeX is very awkward: http://philsci-archive.pitt.edu/117/1/RelOrbitalRotation.pdf

Thanks for the link. it is fairly lengthy, but it looks interesting so I will digest at my leisure.

I have another question. While Rindler refers to a rigid rotating grid as a reference for his gyroscope equations, other authors seem to calculate the vorticity relative to a (not necessarily rigid) congruence of observers with identical angular velocity. Almost universely, the precession of a gyroscope held by a static observer in the Kerr metric is given as:

$\frac{ma}{r^3(1-2m/r)}$

It is tempting to assume this special case is relative to a rigid grid that is not rotating relative to the distant stars. However, if the precession is relative to a congruence of observers with identical angular momentum, then this assumption is not safe. This is because static observers in the Kerr metric, do not have zero angular momentum, and static observers at different altitudes do not have the same angular momentum. I am having difficulty getting Rindler's equation to agree with other sources so I am wondering if this is due to the differences mentioned above. Unfortunately the rotation of the congruence of equal angular momentum observers is difficult to quantify.

P.S> I sent a PM ;)

 

[yuiop]

This paper http://arxiv.org/pdf/1304.6936v1.pdf probably answers the question in my last post. The paper focuses exclusively on an exact solution for the Lense Thirring precession in various metrics, relative to a reference frame that is stationary with respect to an observer at infinity. Equation (20) when specialised to the equatorial plane is identical to the equation given in the last post. This suggests that when Rindler's equation for the precession is extended to the general case, it should give $\frac{ma}{r^3(1-2m/r)}$ for precession of a gyroscope held by stationary observer in the Kerr metric, when the orbital angular velocity is zero.

 

[WannabeNewton]

This is because static observers in the Kerr metric, do not have zero angular momentum, and static observers at different altitudes do not have the same angular momentum.

I'm not seeing what the issue is. The static observers all have zero angular velocity, that's all that matters when calculating the gyroscopic precession via the vorticity. Angular momentum plays no role here (in fact the angular momentum is more directly related to the Sagnac effect).

Could you pinpoint exactly what in Rindler's paper is troubling you?

 

[yuiop]

I'm not seeing what the issue is. The static observers all have zero angular velocity, that's all that matters when calculating the gyroscopic precession via the vorticity. Angular momentum plays no role here (in fact the angular momentum is more directly related to the Sagnac effect).

I guess I was just distracted by earlier claims that it was the congruence of observers with equal momentum that was significant to vorticity and gyroscope precession. I suspect that angular momentum does play a role here. The fact that observers with zero angular velocity in the Kerr metric do not have equal angular momentum, is directly related to why gyroscopes precess relative to stationary observers in the Kerr metric and not relative to stationary observers in the Schwarzschild metric. The gyroscopes are acting like the paddle wheel that measures the local angular momentum of a fluid vortex. Perhaps I am just clutching at straws trying to make sense of a few things in the Rindler paper. It would certainly help if you could provide some technical assistance with the question I raised in post 56 of the parallel thread.

Could you pinpoint exactly what in Rindler's paper is troubling you?

I think there a few typos in the paper that were throwing me off balance. Perhaps you could check them out. He states in equation (37) that the precession is approximated by $3 \pi m /r$ and that due to the positive sign, the precession is prograde and the Thomas precession is in the opposite sense to the Minkowski case. If you carry out the series approximation, it seems he has the sign wrong for (37). It is easy to see that $-( \gamma \omega-\omega)$ is negative and retrograde. Perhaps this is the source of that myth.

In equation (16) he gives the geodesic orbital angular velocity as:

$\omega = \frac{1}{a\pm \sqrt{r^3/m}}$

If we assume a negative value for $\omega$ represents a retrograde orbit then for fixed r and m, an increasing value of ($a$) results in the retrograde orbital speed increasing which is the opposite of what you would expect. For the prograde orbit an increasing value of ($a$) results in a slowing down of the prograde orbit which again is the opposite of expectations. I suspect the formula should actually be:

$\omega = \frac{-1}{a\pm \sqrt{r^3/m}}$

It might seem a small typo, but when $\omega$ is inserted into the precession equations it does not result in a simple reversal of the sign of the precession. With this correction, the magnitude of precession of the gyroscope $\sqrt{m/r^3}$ in the rest frame of the rotating lattice is no longer the same in both the Schwarzschild and Kerr cases as he claims.

Lastly he states below equation (44) that the precession of gyroscope held by a stationary observer in the Kerr metric is:

$-ma \Delta t \sqrt{1-2m/r}$

when most other sources give the result as:

$\frac{-ma\Delta t}{r^3 \sqrt{1-2m/r}}$

 

[WannabeNewton]

It would certainly help if you could provide some technical assistance with the question I raised in post 56 of the parallel thread.

I think Peter would be able to help you much more efficiently than I for that because he has access to computational tensor algebra software that I don't unfortunately. Sorry about that.

I think there a few typos in the paper that were throwing me off balance. Perhaps you could check them out.

I have class right now but I will definitely scrutinize the calculations in just a couple of hours. Cheers.

 

[Bill_K]

This suggests that when Rindler's equation for the precession is extended to the general case, it should give $\frac{ma}{r^3(1-2m/r)}$ for precession of a gyroscope held by stationary observer in the Kerr metric, when the orbital angular velocity is zero.

This agrees with the result in my blog post,

Ω = γ²ω[r - 3M(1 - aω)]/r + γ²Ma(1 - aω)²/r³
γ² = (1 - (r² + a²)ω² - 2M(1 - aω⁾²/r)^-1

in the nonrotating case, ω = 0.

 

[WannabeNewton]

The fact that observers with zero angular velocity in the Kerr metric do not have equal angular momentum, is directly related to why gyroscopes precess relative to stationary observers in the Kerr metric and not relative to stationary observers in the Schwarzschild metric.

From here on out I'm going to drop the qualifier "natural" when I say "rest frame" since it's understood from context. The reason gyroscopes precess relative to the rest frame of static observers in Kerr space-time is the non-vanishing vorticity of the time-like Killing field $\xi^{\mu}$ in Kerr space-time; in Schwarzschild space-time the vorticity of the time-like Killing field vanishes and so gyroscopes don't precess relative to the rest frames of static observers in this space-time. After all, the formula for the gyroscopic precession is given solely in terms of the vorticity of the time-like Killing field.

More specifically it's given by $\Omega^{\mu}_{\text{gyro}} = -\frac{1}{2}(\xi_{\mu}\xi^{\mu})^{-1}\epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} =-\frac{1}{2}(\xi_{\mu}\xi^{\mu})^{-1}\omega^{\mu}$. In Schwarzschild space-time $\omega^{\mu} = 0$ whereas in Kerr space-time $\omega^{\mu} \neq 0$. On the other hand, the angular momentum (about the rotation axis) of observers following orbits of $\xi^{\mu}$ is given by $L = (-\xi_{\nu}\xi^{\nu})^{-1/2}\psi_{\mu}\xi^{\mu}$ where $\psi^{\mu}$ is the axial Killing field; coincidentally $L = 0$ in Schwarzschild space-time whereas in Kerr space-time $L \neq 0$...or is it really a coincidence?

First off, in Kerr space-time we can easily find examples of time-like Killing fields of the form $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$, wherein the angular velocity $\omega$ is constant, such that $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\eta_{\nu}\nabla_{\alpha}\eta_{\beta} \neq 0$ but $L = (-\eta_{\nu}\eta^{\nu})^{-1/2}\psi_{\mu}\eta^{\mu} = 0$ so there is definitely no general relationship between angular momentum and vorticity for non-static time-like Killing fields. As such, don't rely on your Newtonian intuitions when it comes to vorticity and angular momentum of non-static time-like Killing fields as they will surely deceive you in Kerr space-time.

With that in mind, let's come back to static observers but in a more general setting. Given any stationary axisymmetric space-time, $L = 0 \Leftrightarrow \omega^{\mu} = 0$ for static observers and this is very easy to see; as always the static observers by definition follow orbits of the time-like Killing field $\xi^{\mu}$ and as before $L = (-\xi^{\nu}\xi_{\nu})^{-1/2}\xi_{\mu}\psi^{\mu}$.

If we choose the coordinate system adapted to the axial and stationary symmetries (e.g. BL coordinates in Kerr space-time) then $L = (-g_{00})^{-1/2}g_{0\phi}$ and $\xi_{\mu} = g_{00}\delta^{0}_{\mu} + g_{0\phi}\delta^{\phi}_{\mu}$ so if $L = 0$ then $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = g_{00}\delta^{0}_{[[\gamma}\delta^{0}_{\nu]}\nabla_{\mu]}g_{00} = 0$ so $\omega^{\mu} = 0$.

Conversely, if $\omega^{\mu} = 0$ then $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ so there exists a one-parameter family of space-like hypersurfaces $\Sigma_{(t)}$ orthogonal to $\xi^{\mu}$ that foliate space-time. Furthermore $\mathcal{L}_{\xi}\psi^{\mu} = 0$ so combining these two we can choose coordinates $(t,x^1,x^2,\phi)$ for space-time adapted to $\xi^{\mu}$ and $\psi^{\mu}$. Now in these coordinates $\xi^{\mu} = \gamma\nabla^{\mu}t$ since $\xi^{\mu}$ is orthogonal to $\Sigma_{(t)}$ hence $\psi_{\mu}\xi^{\mu} = \gamma g_{\phi \mu}\nabla^{\mu}t = \gamma \delta^{\mu}_{\phi}\nabla_{\mu}t = 0$ therefore $L = 0$.

So at least for static observers in stationary axisymmetric space-times, $L = 0 \Leftrightarrow \omega^{\mu} = 0$ i.e. vorticity is equivalent to angular momentum (about the rotation axis) and therefore your intuition holds true.

 

[yuiop]

This agrees with the result in my blog post,

Ω = γ²ω[r - 3M(1 - aω)]/r + γ²Ma(1 - aω)²/r³
γ² = (1 - (r² + a²)ω² - 2M(1 - aω⁾²/r)^-1

in the nonrotating case, ω = 0.

I am happy to report that Bill's equation also agrees with the result by MTW for the ZAMO case, the geodesic orbit case by Rindler and is algebraically equivalent to the general equatorial equation (35) by Vishveshwara, when the gamma factor is fixed as mentioned in post 57 of the parallel thread.

With some tutoring on notation from WBN, this is how I obtained the result for the ZAMO case from the MTW precession equation:

$\Omega = \frac{1}{2}\sqrt{\frac{g_{\phi\phi}}{|g_{t\phi} - \omega^2 g_{\phi\phi}|}} \left[\frac{\omega_{,\theta}}{p}\epsilon_{\hat r}-\frac{\Delta^{1/2}\omega_{,r}}{p}\epsilon_{\hat \theta}\right]$

where $\Delta = (r^2-2mr+a^2)$. By considering only ZAMOs in the equatorial plane we can set $\omega_{,\theta}=0$,$p=r$ and after inserting the ZAMO orbital velocity $\omega^2 = (-g_{t\phi}/ g_{\phi\phi})^2$ I obtain after some simplification:

$\Omega = \frac{1}{2}\frac{g_{\phi\phi}}{\Delta^{1/2}} \left[-\frac{\Delta^{1/2}\omega_{,r}}{r}\epsilon_{\hat \theta}\right]$

Interpreting $\omega_{,r}$ as $\frac{\partial{}}{\partial{r}}(-\frac{g_{t\phi}}{ g_{\phi\phi}})$ and after carrying out the derivative the result is:

$\Omega = \frac{1}{2}\frac{g_{\phi\phi}}{\Delta^{1/2}} \left[-\frac{\Delta^{1/2}}{r}\left(\frac{-2ma(a^2+3r^2)}{r^2 (g_{\phi\phi})^2}\right)\epsilon_{\hat \theta}\right]$

$\Omega = \frac{ma(a^2+3r^2)}{r^3 (r^2+a^2+2ma^2/r)}\epsilon_{\hat \theta}$

where $\epsilon_{\hat \theta}$ is the unit vector indicating the axis of the resulting precession is orthogonal to the equatorial plane.

The same result (except for the sign) is obtained by inserting the ZAMO velocity into the general equatorial equation (35) by Vishveshwara or Bill's equation. The general equation also agrees with the result of $\sqrt{m/r^3}$ given by Rindler when we enter the geodesic orbital angular velocity $\omega = 1/(a\pm\sqrt{r^3}{m})$ also given by Rindler. None of these results are particularly easy to obtain from the general formulas and required the use of Maple and some manual simplification.

Bill's general equation which is slightly simpler can be expressed as:

$\Omega = \omega\gamma^2 - \omega\frac{3m}{r}(1-a\omega)\gamma^2 +\frac{ma}{r^3}(1-a\omega)^2\gamma^2$

where the first term is the kinematic Thomas precession, the second term is the geodetic precession due the curvature of spacetime around a gravitational body and the third term is the Lense-Thirring precession due the angular velocity (a) of a rotating gravitational body.

$\Omega$ in all the above formulas is the rotation rate of a rigid rotating lattice relative to a set of gyroscopes as measured by an observer at rest with (and co-spinning with) the lattice, so the precession rate of the gyroscopes is actually $-Omega$. The rotation rate $\omega$ of the rigid lattice is measured by the coordinate observer at infinity in the respective metric. The rotating lattice is equivalent to a congruence of observers all with identical angular velocity, (not momentum). To obtain the precession rate of the gyroscopes as measured by the observer at infinity we correct $\Omega$ by the time dilation factor and subtract the result from $\omega$ so that:

$\Omega_{\infty} = \omega -\omega\gamma + \omega\frac{3m}{r}(1-a\omega)\gamma -\frac{ma}{r^3}(1-a\omega)^2\gamma$

Note that setting a=0 gives the expected Schwarzschild precession and setting m=0 gives the expected Minkowski precession. This strongly suggests that the Thomas precession given by the second term is valid in a gravitational field and has the same sign as in flat space, contrary to some claims in the literature.

I have attached a maple worksheet that compares Bill's and Vishveshwara's precession formulas for the ZAMO case.

The hardest part is keeping track of the signs to determine the direction of precession and a lot of the literature is vague on this issue using $\pm$ signs or $\Omega^2$ to duck the issue. The NASA handouts for the direction for the Lense-Thirring precession in the GBP are full of contradictions on the direction. At the moment on balance, based on what is commonaly assumed for the Schwarzschild case, I am going with the L-T precession being retrograde. Also the MTW equation gives the opposite sign for the precession compared to the other equations. If anyone can shed some light on the direction of precession, I would be grateful.

 

[PeterDonis]

The hardest part is keeping track of the signs to determine the direction of precession

My understanding is that, if we are talking about the precession of gyroscopes relative to an observer at infinity, Thomas precession is retrograde, de Sitter (geodetic) precession is prograde, and Lense-Thirring precession is retrograde.

 

[WannabeNewton]

This strongly suggests that the Thomas precession given by the second term is valid in a gravitational field and has the same sign as in flat space, contrary to some claims in the literature.

Could you provide an example of such a contrarian? I don't see why anyone would think that Thomas precession is completely absent in a gravitational field. What they might have said is that it would be hard to interpret such a term as Thomas precession in the same sense as in flat space-time because we don't have a fixed background global inertial frame to apply consecutive Lorentz boosts from.

If anyone can shed some light on the direction of precession, I would be grateful.

Thomas Precession = retrograde
Geodetic Precession = prograde
Lense-Thirring Precession = retrograde

The retrograde nature of Thomas you already have an intuition for. The geodetic precession is just a spin-orbital coupling so we would expect it to be prograde. What may not be immediatly intuitive is the retrograde nature of the Lense-Thirring precession. Ohanian has a nice description of it's physical origin (see attachment). As a side note, Ohanian's book is one of those GR books that you just have to buy.

 

[yuiop]

Could you provide an example of such a contrarian? I don't see why anyone would think that Thomas precession is completely absent in a gravitational field. What they might have said is that it would be hard to interpret such a term as Thomas precession in the same sense as in flat space-time because we don't have a fixed background global inertial frame to apply consecutive Lorentz boosts from.

This is a quote from the Rindler precession paper:

... while one-third is essentially due to Thomas precession; however, the latter is now in the forward rather than the retrograde sense, for it is now the frame of the field that Thomas precesses around the gyroscope, which itself is free, i.e. unaccelerated.

These two image links

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

that you posted a while ago, state:

The second term is purely a special relativistic effect (Thomas precession) ... The second term will contribute whenever there is a non gravitational source of acceleration but will vanish for a gyroscope that is in free fall around a massive body.

This is from Wikipedia:

One can attempt to break down the de Sitter precession into a kinematic effect called Thomas precession combined with a geometric effect caused by gravitationally curved spacetime. At least one author[6] does describe it this way, but others state that "The Thomas precession comes into play for a gyroscope on the surface of the Earth ..., but not for a gyroscope in a freely moving satellite."[7] An objection to the former interpretation is that the Thomas precession required has the wrong sign.

[6] Rindler, Page 234
[7] Misner, Thorne, and Wheeler, Gravitation, p. 1118

What may not be immediatly intuitive is the retrograde nature of the Lense-Thirring precession. Ohanian has a nice description of it's physical origin (see attachment).

Ohanian's description for Lense-Thirring precession is very much like the paddle wheel in a fluid vortex analogy.

Did you get a chance to check out the the possible typos in the Rindler paper in post 158 here https://www.physicsforums.com/showthread.php?t=729416&page=9 ?

 

[Mordred]

not sure if this will help you all but I noticed this article contained a lot of the metrics you all are referring to thought it may help. Some applications of the ZAMO frame are applied in this article as well. Its 97 pages though lol.

Foundations of Black Hole Accretion Disk Theory
Marek A. Abramowicz, P. Chris Fragile
(Submitted on 28 Apr 2011 (v1), last revised 21 Jan 2013 (this version, v3))

This review covers the main aspects of black hole accretion disk theory. We begin with the view that one of the main goals of the theory is to better understand the nature of black holes themselves. In this light we discuss how accretion disks might reveal some of the unique signatures of strong gravity: the event horizon, the innermost stable circular orbit, and the ergosphere. We then review, from a first-principles perspective, the physical processes at play in accretion disks. ...

http://arxiv.org/abs/1104.5499

edit: I found the article a highly useful source of accretion disk dynamics, covers each aspect in it in some nice detail. Also includes the ZAVO metrics as well

 

[WannabeNewton]

I'll have to take a closer look at what the sources are saying regarding Thomas precession in gravitational fields because there seems to be contention surrounding it and a lot of it seems to result solely from calculations.

Ohanian's description for Lense-Thirring precession is very much like the paddle wheel in a fluid vortex analogy.

Well, we should be a little careful here. If you place a paddle wheel in a fluid vortex, there are in general torques applied to different paddles from the fluid velocity field and the wheel starts to rotate. As such it gains an intrinsic spin angular momentum. This is easy to quantify of course because it's simply give by the spin 4-vector $\vec{S}$ (which is by definition the Hodge dual of the spin 2-form) and the net torque on the paddle wheel relative to a given observer is simply $\frac{d \vec{S}}{d\tau}$. Operationally we can measure $\vec{S}$ by placing at the center of the paddle wheel a compass of inertia. Now imagine the paddle wheel is infinitesimally small so that each paddle is described by a single respective world-line. If the paddle wheel is to remain rigid, the world-lines better be integral curves of a time-like Killing field $\vec{\eta} = \vec{\xi}+ \omega \vec{\psi}$. Then the rotation/spin angular momentum $\vec{S}$ of the paddles relative to the compass of inertia at the center of the paddle wheel is exactly what $\vec{\omega} = \vec{\nabla}\times_{\eta} \vec{\eta}$ measures.

But this is obviously not the same thing as the orbital angular momentum $L$ which measures something entirely different and is in general very weakly related to $\vec{\omega}$, unlike $\vec{S}$ which is very strongly related to $\vec{\omega}$ (however as shown in post #161, for static observers $L$ and $\vec{\omega}$ are strongly related).

Did you get a chance to check out the the possible typos in the Rindler paper in post 158 here https://www.physicsforums.com/showthread.php?t=729416&page=9 ?

I'm doing that now. Looking through the calculations should also help shed light on the Thomas precession "issue" mentioned above.

 

[PeterDonis]

I'll have to take a closer look at what the sources are saying regarding Thomas precession in gravitational fields because there seems to be contention surrounding it and a lot of it seems to result solely from calculations.

Some of it may be more an issue of terminology than anything else. Consider the actual formulas for Minkowski, Schwarzschild, and Kerr spacetimes, with each individual term split out (these are precession rates of gyroscopes relative to a static observer at infinity, so I've adjusted the signs appropriately, divided by $\gamma$, and and subtracted out $\omega$ from the formulas in Bill_K's blog post):

$$
\Omega_{Minkowski} = - \left( \gamma - 1 \right) \omega
$$

$$
\Omega_{Schwarzschild} = - \left( \gamma - 1 \right) \omega + \gamma \omega \frac{3M}{r}
$$

$$
\Omega_{Kerr} = - \left( \gamma - 1 \right) \omega + \gamma \omega \frac{3M}{r} \left( 1 - a \omega \right) - \gamma \frac{M a}{r^3} \left( 1 - a \omega \right)^2
$$

As you can see, each successive formula introduces a new term; the standard nomenclature calls the first term (the one that appears in all three) "Thomas precession", the second term (which appears in the last two and has an extra factor in the third) "de Sitter precession" or "geodetic precession", and the third term (which appears only in the last formula) "Lense-Thirring precession" (I've also seen "Schiff precession", I believe that's what the Rindler paper calls it). With the signs as above, positive terms are prograde and negative terms are retrograde; some sources flip the signs.

Mathematically, this is all nice and neat, and invites a nice and neat terminology to match, which I've just given above. Physically, however, there are some issues if we try to get specific with our interpretation of Thomas precession. Thomas precession is standardly interpreted as being due to the non-commutativity of Lorentz boosts. But if that's the case, surely the sign of the precession should depend on the sign of the boost, i.e., on the sign of the proper acceleration. In the Minkowski case, the sign of the proper acceleration is always negative (i.e., radially inward), and the sign of the Thomas precession term in the formulas above is likewise always negative. But in the Schwarzschild and Kerr cases, the sign of the boost depends on $\omega$; it is positive for $\omega = 0$, becomes zero at the value of $\omega$ corresponding to a geodesic orbit, and is negative for larger values of $\omega$. (We'll leave out the other effects that come into play when the radius gets small enough.) Yet the term in the formula for Schwarzschild and Kerr is identical to the Minkowski one*. What gives?

To me, this just means you need to be careful with physical interpretation; it may not match your intuitions, and there may be different ways of trying to resolve any mismatch between the math and your intuitions. (For example, some sources talk about Thomas precession being due to "coordinate boosts" being non-commutative--i.e., the orbits in all three cases above are curved "inward" with respect to the asymptotic Lorentz frame at infinity, and by the "same amount", so of course the Thomas precession term will look the same in all three cases. But as the scare-quotes make clear, this is a hand-waving argument and doesn't really make sense in terms of physical invariants.)

* - Actually this isn't quite true, because of the definition of $\gamma$; as used here, it is the total "time dilation" factor relative to infinity, so in the Schwarzschild and Kerr cases it includes the effect of altitude as well as motion; e.g., in the Schwarzschild case $\gamma = 1 / \sqrt{1 - 2M / r - \omega^2 r^2}$. But this doesn't affect the sign of the precession; it only adjusts its magnitude.

 

[WannabeNewton]

But in the Schwarzschild and Kerr cases, the sign of the boost depends on $\omega$; it is positive for $\omega = 0$, becomes zero at the value of $\omega$ corresponding to a geodesic orbit, and is negative for larger values of $\omega$.

Here's where I think the various ambiguities come from. Ohanian's derivation of the geodetic precession in the weak field limit for example uses a frame fixed to the center of the Earth (ignoring its rotation) and considers a gyroscope in circular free fall orbit around the Earth. Then we apply consecutive Lorentz boosts from the fixed Earth frame to the continuous one-parameter family of local inertial frames of the gyroscope. Here $\omega$ corresponding to a geodesic orbit doesn't have a vanishing boost velocity but rather has a boost velocity corresponding to the tangential velocity due to the centripetal acceleration. A static gyroscope OTOH would have a vanishing boost velocity because it's at rest with respect to the fixed Earth frame. Other sources, such as the ones that claim Thomas precession vanishes for the circular geodesic orbits because the acceleration vanishes, are considering boosts with respect to the local inertial frames themselves. This is where I think the canonical interpretation of Thomas precession breaks down because there is no single fixed local inertial frame but rather a continuous one-parameter family of them so there is no background frame to boost from at each instant of orbit proper time to the instantaneously comoving local inertial frame if we're using the local inertial frames themselves as the standard for what to boost from-if we do conform to the latter then it doesn't make any sense to interpret it as Thomas precession in the same sense as in flat space-time wherein the local inertial frames and fixed static "Earth" frames / asymptotic Lorentz frames all coincide.

 

[yuiop]

(I've also seen "Schiff precession", I believe that's what the Rindler paper calls it).

Rindler seems to use the term "Schiff precession" specifically for precession in a geodesic orbit with no proper acceleration acting on the gyroscopes.

But in the Schwarzschild and Kerr cases, the sign of the boost depends on $\omega$; it is positive for $\omega = 0$, becomes zero at the value of $\omega$ corresponding to a geodesic orbit, and is negative for larger values of $\omega$. (We'll leave out the other effects that come into play when the radius gets small enough.) Yet the term in the formula for Schwarzschild and Kerr is identical to the Minkowski one*. What gives?

Is this not an argument that proper acceleration has nothing to do with Thomas precession. I think WBN said something similar in an earlier post.

Another interpretation is that a gyroscope can somehow sense centrifugal acceleration even when it is canceled out by the gravitational acceleration when in a geodesic orbit. Mathematically the proper acceleration acting on the gyroscope is zero. Is it possible that since no gyroscope is a mathematical point particle, it is sensitive to tiny tidal differences, so that even in a perfect geodesic orbit, extremities of the gyroscope can detect the slight differences between the centrifugal force and the gravitational force that only exactly cancel out at the centre of mass of the object?

To me, this just means you need to be careful with physical interpretation; it may not match your intuitions, and there may be different ways of trying to resolve any mismatch between the math and your intuitions. (For example, some sources talk about Thomas precession being due to "coordinate boosts" being non-commutative--i.e., the orbits in all three cases above are curved "inward" with respect to the asymptotic Lorentz frame at infinity, and by the "same amount", so of course the Thomas precession term will look the same in all three cases. But as the scare-quotes make clear, this is a hand-waving argument and doesn't really make sense in terms of physical invariants.)

Surely, the fact that spacetime is locally Minkowskian in a gravitational field (even if only on an infinitesimal scale) means that Lorentz type boosts are still valid locally even in a gravitational field? Certainly, WBN seemed to indicate that he preferred the non-commutative nature of Lorentz boosts as the explanation for Thomas precession, rather than the acceleration explanation. In fact that was the main topic of the other precession thread.

 

[yuiop]

The retrograde nature of Thomas you already have an intuition for. The geodetic precession is just a spin-orbital coupling so we would expect it to be prograde. ...

Kip Thorne gives a very easy to visualise demonstration of the geodetic precession effect, about a quarter way into this brief video clip.

https://einstein.stanford.edu/MISSION/mission1.html

If the demonstration is correct, it makes it very clear that geodetic precession is prograde, as you say.

 

[WannabeNewton]

Is it possible that since no gyroscope is a mathematical point particle, it is sensitive to tiny tidal differences, so that even in a perfect geodesic orbit, extremities of the gyroscope can detect the slight differences between the centrifugal force and the gravitational force that only exactly cancel out at the centre of mass of the object?

Well the only centrifugal force on the gyroscope would come from its intrinsic spin, which of course we use to define a spin axis for the gyroscope that gets Fermi-transported along a chosen world-line. But the gyroscope is assumed to be very small so the bulging effects will be entirely negligible and we can simply treat the gyroscope as a sphere (recall that the centrifugal force is first order in the radius of the object). We need spherical gyroscopes because it is a spherical shape that allows the gyroscope to be torque-free. If it weren't spherical then it would experience gravitational tidal torques and the usual equations for precession wouldn't hold because those assume the gyroscope is being Fermi-transported along the chosen world-line.

 

[yuiop]

Well the only centrifugal force on the gyroscope would come from its intrinsic spin, which of course we use to define a spin axis for the gyroscope that gets Fermi-transported along a chosen world-line. But the gyroscope is assumed to be very small so the bulging effects will be entirely negligible and we can simply treat the gyroscope as a sphere (recall that the centrifugal force is first order in the radius of the object). We need spherical gyroscopes because it is a spherical shape that allows the gyroscope to be torque-free. If it weren't spherical then it would experience gravitational tidal torques and the usual equations for precession wouldn't hold because those assume the gyroscope is being Fermi-transported along the chosen world-line.

Even a perfectly spherical object in geodesic orbit is subject to tidal forces which is the cause of the Roche limit for orbiting bodies, which causes them to break up in extreme circumstances. The only way to completely eliminate this effect is to have a point particle and I imagine by its very nature, a gyroscope cannot be a point particle.

http://en.wikipedia.org/wiki/Roche_limit

P.S. I am not suggesting that the tidal forces cause a direct torque on the gyroscope. I am just suggesting that somehow the geodesically orbiting gyroscope is aware of the centrifugal forces, in the same way scientists with sensitive enough instruments in a large enough geodesically orbiting lab, could determine they are not moving inertially in flat space, without looking out the window.

 

[WannabeNewton]

Even a perfectly spherical object in geodesic orbit is subject to tidal forces which is the cause of the Roche limit for orbiting bodies, which causes them to break up in extreme circumstances. The only way to completely eliminate this effect is to have a point particle and I imagine by its very nature, a gyroscope cannot be a point particle.

http://en.wikipedia.org/wiki/Roche_limit

Again, that is first order in the radius of the gyroscope. The radius of the gyroscope is assumed to be negligibly small so the effect won't matter. We don't need to completely eliminate it, we just need to make it higher order than the order of the effect we're interested in (the gyroscopic precession) and the assumption of a very small gyroscope does exactly that. Also bear in mind that the entire game of using gyroscopes is just for operational purposes and physical realization. Mathematically there is no need to ever consider a gyroscope-all we need is a spin vector to Fermi-transport and the precession is simply the precession of the spin vector.

 

[PeterDonis]

Is this not an argument that proper acceleration has nothing to do with Thomas precession.

I think it's an argument that the term in the mathematical formulas that is usually called "Thomas precession" can't be explained by proper acceleration, yes. However, I don't think that vindicates the "usual" explanation in terms of non-commutative Lorentz boosts either. See further comments below.

Surely, the fact that spacetime is locally Minkowskian in a gravitational field (even if only on an infinitesimal scale) means that Lorentz type boosts are still valid locally even in a gravitational field?

Yes, definitely; but relative to local inertial frames, the boosts at different events along a geodesic orbit in curved spacetime will "line up" differently than the boosts at different events along a non-geodesic orbit in flat spacetime, even if the "shape" of both orbits relative to infinity is the same. See further comments below.

Certainly, WBN seemed to indicate that he preferred the non-commutative nature of Lorentz boosts as the explanation for Thomas precession, rather than the acceleration explanation.

But the "boosts" in question are not "real" boosts--"real" boosts are always associated with proper acceleration. Put another way, "real" boosts are boosts relative to the local inertial frame at a given event, not relative to some asymptotic global "inertial" frame that isn't really inertial in a curved spacetime. The "boosts" whose non-commutativity is supposed to explain the term in the formulas that is called "Thomas precession", whose sign is always the same, are relative to an asymptotical global "inertial" frame, not relative to any actual local inertial frame; if the boosts were relative to local inertial frames, they would look different in curved spacetime than in flat spacetime, because the local inertial frames are different--the difference in proper acceleration is one manifestation of this. But boosts relative to the asymptotic global "inertial" frame have no physical meaning that I can see; they're just coordinate artifacts. So I don't see how these "boosts" can be used to explain a real, frame-independent physical effect.

(WBN was saying the same thing as the last part--he was just arguing that the "boosts" relative to the asymptotic global "inertial" frame can be used to explain why the term in the formulas that is called "Thomas precession" looks the same in flat spacetime as in curved spacetime. I'm still not sure I buy that because those boosts aren't "real", as I argued above.)

 

[WannabeNewton]

(WBN was saying the same thing as the last part--he was just arguing that the "boosts" relative to the asymptotic global "inertial" frame can be used to explain why the term in the formulas that is called "Thomas precession" looks the same in flat spacetime as in curved spacetime. I'm still not sure I buy that because those boosts aren't "real", as I argued above.)

Actually I'm entirely in agreement with you because the boosts that Ohanian applies in his derivation from the frame fixed to the center of the Earth (which is at rest with respect to the asymptotic Lorentz frame) to an instantaneously comoving local inertial frame aren't really boosts in the strict mathematical sense since in curved space-time boosts only make sense locally. However the derivation in Ohanian was in the weak field limit so to a good approximation we can perform such boosts from the asymptotic Lorentz frame to an instantaneously comoving local inertial frame since we only work to first order in the metric perturbation. In a more general setting it would of course be rather suspect.

 

[PeterDonis]

In a more general setting it would of course be rather suspect.

Yes, and the formulas I wrote down are valid generally, not just in the weak field limit.

 

[WannabeNewton]

http://physics.stackexchange.com/qu...ct-into-thomas-precession-and-spatial-curvatu

 

[WannabeNewton]

Hey guys I have a question that cropped up when reading a paper by Ashtekar on rotation in Kerr space-time. So we know that if we have a time-like vector field $\xi^{\mu}$ and $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ then we can synchronize the clocks following orbits of $\xi^{\mu}$ such that the resulting simultaneity surfaces are everywhere orthogonal to $\xi^{\mu}$ i.e. we can Einstein synchronize the clocks modulo gravitational time dilation effects and the likes. As far as I know this result applies to any and all time-like vector fields i.e. there are no requirements that it be a Killing field etc.

But this is for clocks of an entire vector field. Say we only cared about synchronizing a specific subset of the clocks in such a manner. For example say we have a single ZAMO ring in Kerr space-time and wish only to synchronize clocks at rest on the ZAMO ring. Denote by $\mathcal{R}$ the submanifold in space-time describing the ZAMO ring and let $\xi^{\mu}|_{\mathcal{R}}$ be its tangent field. Then is there a way to apply the result from above to $\xi^{\mu}|_{\mathcal{R}}$ and from it deduce that clocks at rest on the ZAMO ring are Einstein synchronizable? In other words if $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]}|_{\mathcal{R}} = 0$ can I say the clocks on the ZAMO ring are Einstein synchronizable in the sense defined above?

The reason I ask is, as we've talked about earlier, there is more than one way to extend $\xi^{\mu}|_{\mathcal{R}}$ to a neighborhood of $\mathcal{R}$ before computing $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]}|_{\mathcal{R}}$. We can extend it to $\xi^{\mu} = t^{\mu} + \omega \psi^{\mu}$, where $t^{\mu},\psi^{\mu}$ are the time-like and rotational Killing fields and $\omega := \omega(r,\theta)$ or we can extend it to $\tilde{\xi}^{\mu} = t^{\mu} + \omega_0 \psi^{\mu}$ where $\omega_0 := \omega(r_0,\theta_0)$ throughout space-time. As we've seen, $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ but $\tilde{\xi}_{[\gamma}\nabla_{\mu}\tilde{\xi}_{\nu]}\neq 0$ so if we applied the result aforementioned then one extension would yield synchronizable clocks on the ZAMO ring whereas the other extension would yield non-synchronizable clocks on the same ZAMO ring. But we've changed nothing about the kinematics of the ZAMO ring itself-whether or not the clocks on it are synchronizable should depend only on the behavior of $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]}|_{\mathcal{R}}$ and not on how we extend it which has nothing to do with the ZAMO ring itself. So would it be fair to say that the above result cannot be so easily restricted to subsets of a given space-time filling family of clocks?

 

[PeterDonis]

As far as I know this result applies to any and all time-like vector fields i.e. there are no requirements that it be a Killing field etc.

I think this is technically correct, but it has some weird physical implications. For example, the congruence of "comoving" observers in FRW spacetime meets the requirements, since it is everywhere timelike and irrotational; and the conclusion appears to hold since surfaces of constant FRW coordinate time are everywhere orthogonal to the congruence. But if you think about how you would "Einstein synchronize" comoving clocks, by sending round-trip light signals between them, while this would indeed give you the surfaces of constant FRW coordinate time as your surfaces of simultaneity, the time between successive round-trip light signals would not be constant; in the usual case of an expanding FRW spacetime, the round-trip light travel time between two comoving observers would be increasing. "Einstein synchronization" seems to be a weird term to describe this, since that term is usually taken to imply a constant round-trip light travel time. In other words, "Einstein synchronization" is usually taken to imply a rigid congruence, but the "comoving" congruence in FRW spacetime is of course not rigid, since it has nonzero expansion.

In other words if $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]}|_{\mathcal{R}} = 0$ can I say the clocks on the ZAMO ring are Einstein synchronizable in the sense defined above?

I'm not sure the derivatives you need to take will be well-defined if you restrict to a single ZAMO ring. The physical interpretation of the vorticity requires Lie transported vectors pointing to neighboring members of the congruence; but if you restrict to a single ZAMO ring, there are no "neighboring" members in any direction off the ring, so how can you take the appropriate derivatives in those directions?

If your response to the above is that we just restrict all derivatives to the two-dimensional subspace of the chosen ZAMO ring, consider that, in order to Einstein synchronize two clocks on the same ZAMO ring, they have to exchange light signals, and those light signals will not be restricted to that ZAMO ring (unless it happens to be at the same radius as a photon orbit). So any physical interpretation of the theorem in question will implicitly have to rely on some extension of the congruence off the chosen ZAMO ring.

whether or not the clocks on it are synchronizable should depend only on the behavior of $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]}|_{\mathcal{R}}$ and not on how we extend it which has nothing to do with the ZAMO ring itself.

I'm not sure this is true; see above.