# Intuitive understanding of Thomas precession

1. Jan 20, 2014

### yuiop

There are various explanations for Thomas precession in the literature and I would like to come to a deeper understanding of the cause for this precession, so that maybe I can understand better whether Thomas rotation is applicable to the precession of a gyroscope orbiting around a gravitational body.

This paper gives an explanation in terms of an aircraft following a circular path. The argument is that if the circle is approximated by a polygon, then the angle for each corner of the polygon as measured by an observer on the aircraft is larger than the angle measured by an observer at rest with the polygon. One problem with this explanation is that it implies the rate of precession of the rod is constant (for constant angular velocity of the aircraft). This is not necessarily the case (See below). Another question is why should a gyroscope behave as if it is measuring angles of a path that is at rest in another reference frame?

The paper then converts the equation for the Thomas precession into a form that suggests that the precession is a result of centrifugal force. The problem with this interpretation is that an gyroscope in a perfect geodesic orbit around a gravitational body, experiences no proper forces, so no Thomas precession should occur. Then again, some authors would argue that Thomas precession does not occur in the gravitational orbit and this interpretation would support that view. To quote Wikipedia:

Another point of view is that Thomas precession is simply the result of the rotation that occurs when two consecutive non parallel Lorentz boosts are carried out. This is the interpretation outlined in this mathpages article. A consequence of this interpretation is that the precession rate for a given gyroscope is not constant for a constant velocity orbit. The diagram below illustrates this effect.

In the diagram two orthogonal gyroscopes (initially represented by the blue and red vectors) start at the 3 O'clock position and progress around the circle in an anticlockwise fashion. It can be seen that from the point of view of an inertial observer at rest with the centre of the circle, that the blue and red gyroscopes do not remain at right angles with respect to each other and do not precess by the same amount per complete 'orbit' unless by chance the angular velocity has certain critical values. The equation normally given for Thomas precession is just an average figure for many orbits. (Mathpages explains this in quite a lot of detail).

Finally page 235 of this paper by Malament gives a intuitive description of the precession of gyroscope being the result of the gyroscope following a path that deviates from the path of a photon that initially travels in the same direction. This has the nice quality that it applies in both Special and General relativity.

Which, if any, of the above explanations best describe what is really happening in Thomas precession?

P.S. Is aberration a factor? If an observer travelling in a circle points a telescope at a distant star and compares the direction the telescope is pointing relative to a gyroscope, will the angle of the telescope oscillate due to the periodically changing aberration factor, according to an inertial observer at the centre of the circle?

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2. Jan 20, 2014

### WannabeNewton

The Thomas precession is a purely kinematic effect resulting from the non-commutativity of Lorentz boosts. In Galilean relativity the generators of boosts all commute with one another but in special relativity the generators of boosts yield an overall rotation generator (about an axis orthogonal to the axes of the boosts).

Imagine we have a torque-free gyroscope in some arbitrary trajectory; denote by $\vec{u}$ the 4-velocity of the gyroscope. It has a spin axis, represented by a spin vector $\vec{S}$ that is Fermi transported along its world line (which is basically what it means for the gyroscope to be torque free). The spin vector is defined by the condition $\vec{u}\cdot \vec{S} = 0$; Fermi transport of its spin axis is defined by $\frac{d \vec{S}}{d\tau} = \vec{u}(\vec{S}\cdot \vec{a})$. So relative to an observer comoving with the gyroscope, the spin axis is maintained. As you know, the rest frame of the gyroscope is effectively a smooth one-parameter family of momentarily comoving inertial frames (MCIFs). Therefore if we have a background global inertial frame fixed to the distant stars we can express $\vec{S}$ relative to the distant stars by performing a one-parameter family of consecutive Lorentz boosts corresponding to the one-parameter family of MCIFs describing the rest frame of the gyroscope. But in general this will involve consecutive Lorentz boosts in different directions; the commutator of the generators of boosts in different directions will, as noted above, yield an overall spatial rotation (the Thomas rotation) and this will manifest itself as the precession of the gyroscope relative to the distant stars, as can be easily derived by making use of the equations above.

So out of the three sources you listed, I would say mathpages has the proper explanation of the Thomas precession. When we introduce a gravitational field, there will be a gyroscopic precession due to the acceleration and velocity (Thomas precession) and a gyroscopic precession due to the gravitational field and velocity (geodetic precession). The gyroscopic precession due purely to the gravitational field is the Lense-Thirring precession.

I don't see what relevance stellar aberration has to the origin of the Thomas precession. What is true of course is that if we have a set of three mutually orthogonal telescopes carried by an observer who keeps them fixed to the distant stars and a gyroscope carried by said observer, the gyroscope will precess relative to these telescopes because the gyroscope is maintaining the direction of its spin axis in the observer's rest frame.

Last edited: Jan 20, 2014
3. Jan 21, 2014

### WannabeNewton

By the way, you have to be careful in your interpretation of that intuitive description because it is meant to relate directly to the contents of proposition 3.2.5. which says that in a static space-time (such as Schwarzschild space-time), an axisymmetric rigid ring is non-rotating in the sense that $\eta_{[a}\nabla_b \eta_{c]} =0$ for all possible angular velocities if and only if light can circulate the ring whilst freely falling. If you recall that $\eta_{[a}\nabla_b \eta_{c]} =0\Leftrightarrow F_{\hat{\eta}}(h^{a}{}{}_{b}\phi^b) = h^{a}{}{}_{b}\hat{\eta}^c \nabla_c (h^b{}{}_{d}\phi^{d}) = 0$ then this proposition basically states that in static space-times a gyroscope fixed to the ring with its spin axis initially aligned in the direction of $\phi^a$ remains aligned in the direction of $\phi^a$ if light can circulate the ring in free fall for all possible angular velocities of the ring. So this proposition isn't telling you why gyroscopic precession does or does not occur, it simply tells you how to detect it in very specialized systems.

4. Jan 21, 2014

### Bill_K

This is certainly false! All I can figure is that this is an artifact of replacing Thomas precession, which is a continuous process, by a discrete set of Lorentz transformations.

5. Jan 21, 2014

### yuiop

Mathpages gives this differential equation for the instantaneous change in precession angle $(d\theta)$ relative to the instantaneous change in orbital tangential vector $(\phi)$:

$d\theta = -d\phi \left( 1 + \frac{v^2}{1-v^2} \cos{(\phi)}^2 \right)$

where v is the instantaneous tangential orbital velocity and $\phi$ is the instantaneous angle of the gyroscope relative to the orbital tangential vector. This as you say is based on discrete (but infinitesimal) set of Lorentz transformation, which Mathpages then integrates to obtain a expression for larger rotation angles. The important aspect of the original differential equation is that the istantaneous precession rate depends on the angle of the gyroscope relative to the orbit tangential vector. When the gyroscope is at right angles to the instantaneous tangential velocity such that $\phi = \pi/2$ and $\cos(\phi/2) = 0$ there is no precession of the gyroscope relative to the distant stars because $d\theta = -d\phi$. Despite this, at the next instant, the gyroscope is no longer exactly perpendicular to the orbital tangential velocity and starts precessing, reaching a maximum precession rate when the gyroscope is parallel to the instantaneous tangential velocity.

If Thomas precession 'really' is a 'continuous process' so that the precession rate is constant given constant orbital velocity, then the cause of the precession cannot be due to Lorentz rotatation caused by succesive non parallel Lorentz boosts. The point of this thread is to try and determine what really is the cause. For example, centrifugal force is independent of the gyroscope angle and remains constant for constant angular velocity, so this cause would result in precession being a continuous (constant) process rather than an oscillating process. On the other hand, if the gyroscope is a perfect sphere, then the precession due to Lorentz boosts, would effectively be an average of a set of gyroscopes at various angles. Perhaps this is one reason the gravity Probe B experiment used nearly perfectly spherical gyroscopes.

It might be worth noting that even though the gyroscopes in the diagram I uploaded do not remain at right angles to each other (from the POV of the central inertial observer), that they almost certainly do remain at right angles in the rest frame of the orbiting satellite.

6. Jan 21, 2014

### WannabeNewton

You can't (analytically) integrate something discrete. Integration requires a continuous label e.g. a continuous one-parameter family of MCIFs and associated Lorentz boosts.

Why? That's exactly the cause. It's a simple calculation (the one on mathpages is overly cumbersome-standard textbooks will have much more elegant derivations of the Thomas precession) that shows the link between the non-commutativity of the Lorentz boost generators (Thomas rotation) and the Thomas precession.

EDIT: Bill has an awesome blog related to the above with explicit calculations done in the elegant way: https://www.physicsforums.com/blog.php?b=4448 [Broken]

Last edited by a moderator: May 6, 2017
7. Jan 21, 2014

### Bill_K

I'd give a failing grade to any discussion of Thomas Precession that relied on a series of discrete Lorentz transformations. This would be like trying to do classical mechanics "intuitively" by replacing a force and the continuous acceleration that results from it by a series of discrete impulses. This is (literally) why calculus was invented!

Unfortunately, elementary treatments of special relativity typically omit any discussion of acceleration or rotation, often leaving the impression that these are difficult topics, and even that they somehow lie beyond the scope of SR.

Understanding Thomas Precession relies first on an understanding of how the spin of an accelerating particle evolves under the absence of torque. The answer is Fermi-Walker transport. In order to remain spacelike in the particle's rest frame, the spin vector must remain orthogonal to v, and therefore cannot stay constant, it must evolve somehow. Fermi-Walker transport is the minimal change that meets this condition, namely a simple boost in the plane of v and a. The equation is DS/Dτ = (v ∧a)S.

It relies secondly on an understanding of Lorentz rotations, to the extent that we can pick out the spatial part of the F-W rotation in a particular inertial frame. Given the unit vector t in the time direction, one can form the quantity Wμ = ½ εμνστvνaσtτ. This quantity is a vector which is purely spacelike in the rest frame, with constant magnitude assuming v ∧a has constant magnitude, and pointing along the rotation axis.

Last edited: Jan 21, 2014
8. Jan 21, 2014

### WannabeNewton

I just realized that this explanation might not have been lucid enough. If we consider, for example, the family of observers in circular orbit about the origin of some global inertial frame with constant angular velocity $\vec{\Omega} = \Omega \vec{e}_{z}$ then we can represent their tangent field by $\vec{\eta} = \gamma (\vec{e}_{t} + \Omega r \vec{e}_{\phi})$ where $r$ is the radius of their circular orbit from the origin of the global inertial frame and $\gamma = (1 - \Omega^2 r^2)^{-1/2}$. The 1-form $\eta$ associated with $\vec{\eta}$ is then given by $\eta = \gamma(-e^t + \Omega r e^{\phi})$.

If we compute the vorticity of this family of observers we find that $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\eta_{\nu}\partial_{\alpha}\eta_{\beta} \\= -\gamma \Omega \epsilon^{\mu t r\phi}\partial_{r}(r \gamma)+\gamma \Omega r\epsilon^{\mu t r \phi}\partial_{r}\gamma \\= \gamma^2 \Omega\epsilon^{ tr\phi \mu}$.

so $\vec{\omega} = \gamma^2 \Omega \vec{e}_z$. Recall also that $\eta_{[\gamma}\partial_{\mu}\eta_{\nu]} \neq 0\Leftrightarrow \vec{\omega}\neq 0$. Since $\vec{\eta}$ is a Killing congruence, we can apply proposition 3.2.1. from the notes you linked, which states that $F_{\vec{\eta}}\vec{e}'_{\phi} = (\nabla_{\vec{\eta}}\vec{e}'_{\phi})_{\perp} \neq 0 \Leftrightarrow \eta_{[\gamma}\partial_{\mu}\eta_{\nu]} \neq 0$, to conclude that a gyroscope fixed to the ring will precess relative to the spatial basis vector $\vec{e}'_{\phi}$ that always points tangential to the circular orbit in the natural rest frame of the observer in this family comoving with the gyroscope.

More precisely, if we choose a reference observer from this family then recall (from previous threads) that $\vec{\omega}$ represents the angular velocity of rotation of neighboring observers relative to the compass of inertia (mutually orthonormal gyroscopes) carried by the reference observer, with respect to the reference observer's proper time. In other words if we attach the spatial basis vectors $\{\vec{e}'_r,\vec{e}'_{\phi}\}$ to the reference observer, wherein $\vec{e}'_r$ always points towards the origin, then $-\vec{\omega}$ represents the precession of the compass of inertia relative to $\vec{e}'_{\phi}$. Relative to the global inertial frame, $\vec{e}'_{\phi}$ itself precesses per orbit by $2\pi$ simply due to its rotation around the origin.

Hence $\Delta \phi = 2\pi[ 1 - (1 -r^2 \Omega^2)^{-1/2}]\approx -\pi r^2 \Omega^2 + O(v^4)$ represents the net precession of the compass of inertia relative to the global inertial frame. This is the usual Thomas Precession for circular orbits in the approximation of small orbital velocities.

All of this basically follows from proposition 3.2.1. of the notes you linked. However, it doesn't explain why this precession occurs. It simply relates the precession to the non-vanishing of the vorticity and in the process provides a very simple way of calculating the precession rate (for the special case of circular orbits that is!). The why of the precession is, as already stated, the non-commutativity of consecutive Lorentz boosts in different directions.

9. Jan 22, 2014

### yuiop

Are you saying the explanation and mathematical calculations given by mathages are incorrect?

Are you saying the precession of the gyroscope should be constant over time, given constant angular velocity (and therefore constant centrifugal acceleration) irrespective of the gyroscopes orientation relative to its instantaneous tangential velocity vector?

I took WBN's advice and took a look at your blog and I have a couple of questions. In flat space you derived:

d2a/dτ2 = -γ4ω2 a

You then said that this is a harmonic oscillator equation. If we represent the above equation as

d2a/dτ2 = -k a

where k represents a constant then the well known solution for such an equation is:

a(t) = A cos(√(k)t)

where A is a constant representing the amplitude of the oscillation. You then conclude that √(k) is the precession rate and that it must be constant. However, the nature of harmonic motion is that there is a quantity that oscillates over time in a non constant fashion. It is clear that if √(k) and A are constant that the value of a is not a constant. What is this varying quantity? We have defined the angular 'orbital' velocity and centrifugal acceleration to be constant, so the only possible time varying oscillating quantity is the precession rate. This suggests that the formula for the precession Ω should actually be:

Ω(τ) = A cos(γ2ωτ)

You also stated:
Shouldn't that be more quickly?

In the section where you address precession in the Schwarzschild metric there appears to be a mistake in your derivation. You stated:
I get that to come out as:

d2Sr/dτ2 = -γ4ω2(r - 3M)2/r2

Since Sr does not appear on the right and since everything on the right is a constant, harmonic motion does not apply here, since by definition the acceleration of the harmonic motion is a function of position and there there is no position variable (or any other variables) on the right for a constant orbit. This might suggest that the non constant periodic precession motion that appears in flat spacetime does not appear in the gravitational case. This seems to concur with MTW's claim that Thomas precession does not occur in the gravitational case. On the other hand, the error might be earlier in your derivation and the Sr was missed from one of the derivatives.

For the Kerr metric you give the precession equation as:

Ω = γ2ω[r - 3M(1 - aω)]/r + γ2Ma(1 - aω)2/r3

This has certain nice qualities such as when setting a=0 the equation becomes:

Ω = γ2ω(1 - 3M/r + γ2)

which is the precession in the Schwarzschild metric. Setting a=0 and M=0 gives:

Ω = γ2ω

which is the precession in flat space.

You also give γ2 = [1 - (r2 + a22 - 2M(1 - aω)2/r]-1 and when we set ω=0 in the precession equation for the Kerr case using that definition of γ, we get:

Ω = Ma/[r3(1-2M/r)]

which we know is correct for a static observer in that metric from previous threads.

So far your equation for the Kerr precession passes all the basic checks with flying colours, but the problem comes when solving:

Ω = γ2ω[r - 3M(1 - aω)]/r + γ2Ma(1 - aω)2/r3 =0 for ω,

because this should yield the ZAMO angular velocity, but that does not seem to be the case. I am no mathematician and I might well be mistaken. Perhaps someone could check that for me?

10. Jan 22, 2014

### WannabeNewton

I haven't taken a look at your other questions yet (sorry!) but I can quickly answer this one. Intuitively, since the congruence of ZAMOs has vanishing vorticity, you'd expect the precession of a gyroscope (or compass of inertia) carried by a ZAMO observer to vanish right? Not so because the ZAMO congruence is not rigid. In the natural rest frame of a ZAMO, the gyroscope (or compass of inertia) precesses relative to $e'_{\phi}$ (as defined in the previous post) and hence relative to the distant stars after correcting for gravitational/kinematical time dilation and the extra factor of $2\pi$ coming from the precession of $e'_{\phi}$ itself relative to the distant stars due to the circular orbit of the ZAMO. Again this precession comes about because of the choice to use the natural rest frame of the ZAMO in which the lack of rigidity manifests itself as precession relative to $e'_{\phi}$.

(Recall the discussion from page 8 of your other thread on measurements of rotation in Kerr space-time)

11. Jan 23, 2014

### yuiop

In the other thread we sort of assumed that the gyroscopes would not precess relative to the $e'_{\phi}$ vector of an object with ZAM orbital velocity. according to any observer. Peter gave the precession rate as being equal to the orbital velocity $-g_{t\phi}/g_{\phi\phi}$. However, I have tried a couple of formulas and none give a zero precession spin when the orbit velocity is that of a ZAMO so I am starting to doubt our previous assumption. Malament gives the example of ring in Godel spacetime that is not rotating by the ZAM criterion but is rotating by the CIR criterion. Therefore they do not have to be equal by definition. He does not address this issue directly in relation to the Kerr metric, where we have assumed the two criteria agree, but I would like to see a stronger proof of that.

I am not sure we fully resolved that issue. One of the last things Peter said was:
It does not really make sense to use concentric ZAMO rings (with different r and different angular velocity) as a reference frame. If a line tangential to a fixed point on given a given ring represent the horizontal axis of the frame (ie $e'_{\phi}$) and another line drawn orthogonally to a point on a larger ring represents the vertical axis, (ie $e'_{r}$. then after a period of time, the point on the larger ring could be on the opposite side of the gravitational body such that it crosses over $e'_{\phi}$ and at some points the initially orthogonal axes of the frame could be parallel to each other (in the rest frame of the observer) as they cross over making for a very poor set of reference axes.

Having read some of the MTW stuff you sent to me, I am totally confused as to what they consider the natural (orthonormal) rest frame of a non-rotating observer to be, so maybe we could try and clear that up here. I think it would help me a lot if we could refer to a diagram like this:

In the diagram the rectangles represent consecutive positions of an orbiting lab. It is perfectly square in its own rest frame but appears rectangular due to length contraction as a result of its orbital motion. The lab is anchored to the centre of the orbit such that $e'_{r}$ always points directly away from the centre of orbit and $e_{\phi}$ is always tangential to the orbital path. If the local $e_{/phi}$ and $e_{r}$ principle axes of the lab were made of steel and welded to each other such that they remain perpendicular to each other, would these axes represent the orthonormal rest frame of a locally non rotating observer, iff the orbital velocity of the lab is that of a ZAMO? If so, (staying with the assumption of ZAMO velocity) would gyroscopes in the lab precess relative to these axes or not? Would their behaviour relative to the non principle diagonal axes be any different? If the lab frame as I have presented it does not represent the orthonormal rest frame of a non rotating observer, what should such a frame be called and what is a orthonormal locally non rotating frame at rest with? Is the lab frame stationary with respect to Lie Transported Basis vectors or do those vectors always point at a stationary point at infinity? Finally, for completeness, of the lab was rotating at some velocity such that un-torqued gyroscopes are precessing in the lab, could two such gyroscopes that are orthogonal to each other represent Fermi Walker basis vectors? In other words, are un-torqued gyroscopes always at rest with FW basis vectors that are transported along with the gyroscopes, by definition?

The non principle diagonal axes demonstrate how the precession rate can be constant in one reference frame and not in another. In the rest frame of the lab, the diagonal axes are at 45 degrees. If there is a precessing gyroscope that rotates from $e_{r}$ to $e_{\phi}$ in 20 seconds in the lab rest frame, then it passes the intermediate diagonal axis in half that time (10 seconds) assuming constant precession in the lab. To an observer at the centre of the orbit, the gyroscope takes longer to get to the 45 degree position from $e'_{r}$ than it takes to get from the 45 degree position to $e'_{\phi}$ so the rotation is not constant but oscillates, getting faster as the gyroscope gets near the tangential position and slowing down again as it gets orthogonal to the tangential position. This pretty much agrees with drawing in the OP. The oscillating precession comes about as a result of the length contraction distortion of the lab fame in the external frame, so the two frames disagree on the position of the 45 degree axis relative to the principle orthogonal axes of the lab.

Another interesting observation (which I might of mentioned in a previous thread) is that the time dilation of a clock carried by a ZAMO relative to infinity is identical to the time dilation of a static observer relative to infinity in the Kerr metric. This seems quite surprising so maybe I have miscalculated. (They are both equal to $\sqrt{1-2m/r}$). A clock with any other orbital velocity does not have this property.

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12. Jan 23, 2014

### Staff: Mentor

To be clear, this is the precession rate of a gyroscope carried by a ZAMO, relative to a static observer at infinity. This means that an observer at infinity will see a gyroscope carried by the ZAMO precessing (rotating) with the same angular velocity as the ZAMO revolves around the hole. So, for example, a gyroscope that starts out pointing directly radially outward from the ZAMO will always point directly radially outward from the ZAMO; relative to the static observer at infinity, the direction the gyro is pointing will rotate because the ZAMO is revolving around the hole.

As WBN pointed out, the ZAMO congruence has nonzero shear, so a gyroscope carried by a ZAMO that points directly radially outward will *not* remain pointing directly at the same neighboring member of the ZAMO congruence. A ZAMO slightly further outward, radially, will "fall behind" the direction in which the first ZAMO's gyroscope is pointing.

Right, because the ZAMO congruence is not rigid.

Yes.

No, they wouldn't.

No; if the gyro isn't precessing relative to the principle axes, it isn't precessing relative to any axes.

I'm not sure about the more general question that is implicit here (and which you go into in more detail later in your post); I'm going to have to think about that some more.

I'm not sure I would use the term "by definition", because the fact that gyroscopes physically realize FW transported basis vectors is not a definition, it's a physical observation. But it's true, yes.

13. Jan 23, 2014

### WannabeNewton

We didn't assume that actually. See below.

It's not enough to just have an observer. You also need a frame with a choice of spatial axes for the gyroscope to precess relative to. Letting $\xi^{\mu} = \gamma \nabla^{\mu} t$ be the tangent field to the ZAMO congruence, because $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ there exists some choice of frame for a given ZAMO such that a gyroscope at rest in this frame does not precess relative to the spatial axes of the frame but the natural frame of the ZAMO (the one with the $e_{\phi}$ and $e_{r}$ spatial basis vectors as defined in MTW exercise 33.4) is not such a frame.

We agreed on page 8 of that thread that Malament's definition of non-rotation of a single axisymmetric rigid ring as per the compass of inertia on the ring only applies if the ring has a tangent field of the form $\eta^{\mu} = \tilde{t}^{\mu} + \omega_0 \phi^{\mu}$ where $\omega_0$ has to be constant everywhere in space-time and correspond to a ZAMO angular velocity. So in order for the ring to be non-rotating according to this criterion one would have to find a value of $\omega_0$ for which $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ holds. Keep in mind this is not the same thing as saying that the vorticity of the ZAMO congruence itself vanishes because Malament's definition of non-rotation as per the compass of inertia on the ring requires an everywhere rigid congruence extended to all of space-time from the congruence of world lines defined by the single axisymmetric rigid ring hence this will not be the same as the ZAMO congruence. In other words $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ does NOT necessarily imply that $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ yet it is the latter of the two that we want, for a particular value of $\omega_0$, in order to conclude that there exists at least one ring of ZAMOs in Kerr space-time non-rotating according to the compass of inertia on the ring.

I'll reply to the latter half of your post in a separately reply...EDIT: And I see Peter replied to most of it already :)

Last edited: Jan 23, 2014
14. Jan 23, 2014

### WannabeNewton

See the following: http://postimg.org/image/en60x7i6t/ [Broken], http://postimg.org/image/q3cfv8wd1/ [Broken]

What do you mean by stationary? The answer to your latter question is no and the answer to your former question is also no if you mean "do the spatial basis vectors of the above frame agree with a set of connecting vectors Lie transported along the world line of the chosen reference observer".

Last edited by a moderator: May 6, 2017
15. Jan 24, 2014

### yuiop

That is what I initially assumed, but this page link that WBN kindly sent me suggests that while groscopes do not precess relative to the principle axes, they do precess relative to other axes. http://postimg.org/image/q3cfv8wd1/ [Broken] To be honest, it is hard to understand exactly what they are saying. Some related pages for background:

http://postimg.org/image/4sevqze8l/
http://postimg.org/image/en60x7i6t/ [Broken]

Using the equation for the precession of a gyroscope with ZAMO angular velocity, the resut if I have interpreted the equation correctly is that the precession rate of a gyroscope relative to the lab axis is always ma/r^3. Setting w=0 yields precession of zero, because the angular velocity of ZAMO can only be zero, if the angular momentum of the black hole is also zero.

My calculation can be seen here.

I worked out the result of ma/r^3 on paper because it was too cumbersome for the software, and then asked what the orbital velocity would be to yield that precession rate as a reverse engineered confirmation.

I have interpreted the MTW equation as:

$\sqrt{\frac{g_{\phi\phi}}{|g_{tt}-\omega^2 g_{\phi\phi}|}} \left(-\frac{\omega \sqrt{\Delta}}{p}\right)$

I have restricted everything to the equatorial plane to keep things simple, so:

$p=r^2$ , $\Delta = (r^2+a^2-2mr)$ and $\omega_{,\theta} = 0$.

Bill's equation (from his blog) for the precession of a ZAMO also yields a non zero result, although his approach is different and the results cannot be directly related to the MTW equation. Bill's equation suggests there is *some* angular orbital velocity where the spin of the gyroscope is zero, but it is not the ZAMO orbital velocity.

I cannot find any straight forward calculation that directly demonstrates a ZAMO has zero spin in the Kerr metric.

Last edited by a moderator: May 6, 2017
16. Jan 24, 2014

### yuiop

LOL. I formulated the question in a form that hoped to solicit a yes or no answer. The links you sent are the pages I said were totally confusing me -->

I think that answers that question satisfactorally. Is there any simple description of Lie transported vectors in relation to the lab frame or FW transported vectors or anything else?

Last edited by a moderator: May 6, 2017
17. Jan 24, 2014

### yuiop

Question (B) on this page http://postimg.org/image/4sevqze8l/ makes the point that a 'locally non rotating' observer is not inertial. The orbital velocity of a ZAMo is less than the angular velocity required for geodesic motion so this observer experiences proper acceleration. Question (c) on the same page implies that these accelerations acting on the centre of mass of the gyroscope (so as not to cause torque) cause the gyroscope to precess according to the given equation. Does this imply that proper acceleration acting on the centre of mass of a gyroscope will always cause precession (even in flat space) or does this condition only apply to a ZAMO reference frame?

Now consider a rod that undergoes a single boost from rest, in a direction not parallel or orthogonal to the rod. This will cause a Lorentz rotation of the rod from the point of view of an observer in the original rest frame of the rod. I assume a gyroscope with its axis parallel to the rod that is boosted in the same way will remain parallel to the rod. If this is true, it implies that a gyroscope that is accelerated relative to momentarily co-moving inertial reference frame, will precess if the acceleration is not exactly parallel or orthogonal to the gyroscopes spin axis. This in turn implies that a gyroscope that is held by a static observer in Schwarzschild spacetime will precess, if it is not exactly horizontal or vertical. Is this momentum free precession such that the gyroscope will settle down in the horizontal position or will it oscillate back and forth like an un-damped pendulum? Any ideas?

18. Jan 24, 2014

### WannabeNewton

Nowhere in (c) does it say the proper acceleration is the cause of the precession and rightfully so because that isn't the cause of the precession. Here's the simplest way I can explain why it happens; again let $\xi^{\mu}$ be the tangent field to the ZAMO congruence. Imagine a reference ZAMO $O$ who, at some initial proper time on his clock, attaches a connecting vector $\lambda^{\mu}$ to another ZAMO $O'$ separated by an infinitesimal amount in the radial direction at this instant. In order to make sure $\lambda^{\mu}$ remains locked to $O'$, we have $O$ Lie transport $\lambda^{\mu}$ along $\xi^{\mu}$: $\mathcal{L}_{\xi}\lambda^{\mu} = 0$. Furthermore $O$ Fermi-transports a set of spatial axes along his world line, and these spatial axes are physically represented by gyroscopes; in particular, at this instant $O$ has a gyroscope that is colinear with $\lambda^{\mu}$.

Now what does it mean for $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$? Well it means that $\lambda^{\mu}$ does not precess relative to the aforementioned gyroscopes hence the gyroscope colinear with $\lambda^{\mu}$ will remain colinear with it everywhere along the world line of $O$. But $\lambda^{\mu}$ points from $O$ to the infinitesimally radially separated observer $O'$ who (without loss of generality) has a larger angular velocity than that of $O$. On the other hand the unit vector $e_{r}$ is not locked to $O'$ (or any observer for that matter) i.e. $\mathcal{L}_{\xi}e_r \neq 0$ so the gyroscope colinear with $\lambda^{\mu}$ will precess relative to $e_r$. Can you picture that?

If by "Lorentz rotation" you mean "Thomas rotation" then this only occurs after two consecutive Lorentz boosts in different directions and the resulting precession itself is relative to a single background global inertial frame in flat space-time; you on the other hand are talking about something relative to transient momentarily comoving local inertial frames in curved space-time so I have no idea what it is you're talking about. In other words, you're not talking about a gyroscopic precession relative to a single frame, such as the local inertial frame fixed to the distant stars, due to the Thomas rotation. And relative to the local inertial frame that's fixed to the distant stars, the static observers are at rest anyways (that's what it means for them to be static). I'm not at all sure what you're describing.

Last edited: Jan 24, 2014
19. Jan 24, 2014

### Staff: Mentor

Does it? The vorticity is zero, yes, but the change in $\lambda^{\mu}$ relative to gyroscopes depends on the shear as well as the vorticity, right? And the ZAMO congruence has nonzero shear, so $\lambda^{\mu}$ will change direction relative to gyroscopes even thought the vorticity is zero. At least, I think that's correct.

20. Jan 24, 2014

### WannabeNewton

AFAIK, the rotation of the connecting vector $\lambda^{\mu}$ relative to local gyroscopes is characterized entirely by the vorticity alone.

See for example the following from "Gravitation and Inertia"-Wheeler and Ciufolini: http://postimg.org/image/jrueziypt/ [Broken]

And also section II.C (p. 3) of the following paper: http://dspace.rri.res.in/bitstream/2289/1285/1/1993 PhyRevD V48 p5706.pdf keeping in mind that, as per the paper's definitions, a quasi-Killing trajectory is a linear combination of Killing fields but with non-constant coefficients (e.g. tangent field to the ZAMO congruence) whereas a Killing trajectory is a linear combination of Killing fields with constant coefficients (such as the congruence of observers atop a rotating disk).

That's not to say however that the shear doesn't affect the rotation of $\lambda^{\mu}$ (and hence of the local gyroscopes) relative to the radial unit vector $e_r$ since $\lambda^{\mu}$ is locked onto an observer in a neighboring orbit whereas $e_r$ is not (it's just the good old radial unit vector) so $\lambda^{\mu}$ will get dragged along the orbit of the neighboring observer due to the difference in angular velocities between the reference observer and neighboring observer while $e_r$ will not. Is there a flaw in this description?

Last edited by a moderator: May 6, 2017
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