Locally non-rotating observers

[PeterDonis]

Does Fermi-Walker transport conceptually involve a congruence, since Fermi-Walker differentiation seems to be defined for a vector field, but not a single vector?

I'm not sure. I can come up with a way of defining Fermi-Walker transport that doesn't require a congruence, only a single worldline; but I'm not sure if it covers all the cases in which a Fermi-Walker derivative might come up.

Let me briefly describe the definition I just referred to, since it's also relevant to the monkey wrench I just threw in my last post. Take a very simple example, linear acceleration in flat Minkowski spacetime. Suppose I have a worldline with constant proper acceleration in the $x$ direction, so we can just look at the $t, x$ plane. At some event O on this worldline, the worldline's instantaneous 4-velocity in a chosen global inertial frame is $(u^t, u^x) = (1, 0)$. An infinitesimal interval later, at event P, the worldline has accelerated and now has 4-velocity $(\gamma, \gamma v)$.

The question is, how do we Fermi-Walker transport the spatial basis vector in the $x$ direction along this infinitesimal interval? Well, we could do this: first, parallel transport the "zweibein" of orthonormal basis vectors in the $t, x$ directions from event O to event P. Since this is a global inertial frame in flat Minkowski spacetime, parallel transport just leaves all vector components unchanged. So the parallel transported zweibein vectors will be $(1, 0), (0, 1)$ at P, just as they were at event O.

Now find a Lorentz transformation that takes the parallel transported 4-velocity (the $t$ vector in the zweibein) at P to the actual 4-velocity at P. Obviously this is just a Lorentz boost in the $x$ direction with velocity $v$, which takes the vector $(1, 0)$ to the vector $(\gamma, \gamma v)$.

Then apply the same Lorentz transformation to *all* the vectors that were parallel transported from O to P; so we boost the vector $(0, 1)$ in the $x$ direction with velocity $v$ to obtain the new spatial basis vector $(\gamma v, \gamma)$ at P. This will be the Fermi-Walker transported spatial basis vector at P; i.e., the Fermi-Walker transported zweibein at P will be $(\gamma, \gamma v), (\gamma v, \gamma)$. (In the fully general case, we would apply the Lorentz transformation to all four basis vectors that were parallel transported from O to P; here this is a no-op for the basis vectors in the $y$ and $z$ directions, so including them makes no real difference.)

I believe this definition can be generalized to all cases of Fermi-Walker transport along a single worldline--you just allow *any* arbitrary Lorentz transformation at event P, which covers all possible non-inertial worldlines--and obviously it does not require a congruence; it only requires knowledge of the 4-velocity at each event on the worldline itself. (The case of a geodesic worldline, where Fermi-Walker transport is the same as parallel transport, is just the case where the Lorentz transformation you apply to the transported vectors is the identity.)

 

[WannabeNewton]

If we attach rods from a given Langevin observer to neighboring Langevin observers then why would they rotate relative to infinity with $\omega$, which is the angular speed at which each Langevin observer is rotating around the center of the disk? Why does the rotation $\Omega$ of the rods about the given Langevin observer due to the twist not affect what is seen at infinity? The $\omega$ part comes from the fact that all the Langevin observers rotate around the center of the disk with that speed so an observer at infinity will see the rods being dragged around the disk at the same rate right but why will $\Omega$ not affect what is seen at infinity?

 

[PeterDonis]

If we attach rods from a given Langevin observer to neighboring Langevin observers then why would they rotate relative to infinity with $\omega$, which is the angular speed at which each Langevin observer is rotating around the center of the disk?

Because when the Langevin observer has made one complete rotation around the center of the disk, the rods have also made one complete rotation around the center of the disk. For example, pick a distant star in a particular spatial direction from the center of the disk, say the $x$ direction in the global inertial frame in which the center of the disk is at rest. Suppose that one connecting rod, the one directed radially outward, points directly at that star when the Langevin observer is just seeing it overhead (i.e., when he is just crossing the $x$ axis). Then each time the Langevin observer sees that star directly overhead (i.e., each time he crosses the $x$ axis), the connecting rod that was pointed at that star when he started with the star directly overhead will be pointed directly at that star again--and it won't be pointed at that star anywhere else in the course of his rotation.

Another way of putting this is that the connecting rods are, in effect, rigidly attached to the disk, so they must rotate, relative to infinity, with the same angular velocity as the disk.

Why does the rotation $\Omega$ of the rods about the given Langevin observer due to the twist not affect what is seen at infinity?

Because $\Omega$ is defined relative to Fermi-Walker transport, not relative to infinity. As above, you can derive how an observer at infinity will see the connecting rods rotate without even knowing $\Omega$.

 

[WannabeNewton]

Ok so if we attached a rod radially outward as you said from an observer $O(R_0)$ to another one $O(R_1)$ then the reason why an observer at infinity will see the rod rotate at exactly $\omega$ around the disk is because both $O(R_0)$ and $O(R_1)$ are rotating around the disk at the same angular speed $\omega$ relative to infinity so the observer at infinity will never see the rod rotate due to $O(R_1)$ overtaking $O(R_0)$ or vice-versa and will only see it rotate due to the equal rotation rate $\omega$ of said two observers around the disk relative to infinity? So is this in contrast with what happens in Kerr space-time for the ZAMOs because observers at different radii have different angular velocities around the black hole and so a rod radially attached initially from one observer to another will not only rotate around the black hole relative to infinity due to the overall orbital rotation but also rotate because the inner observer is overtaking the outer observer?

As a side note, when we say "the twist causes nearby observers to rotate about the given reference observer from the perspective of the reference observer" do we really always mean "the twist causes the nearby observers to rotate relative to the Fermi-Walker transported spatial basis of the reference observer"? I ask because we discussed this earlier in the thread and it makes physical sense to me but some of my texts (e.g. Malament's text and Wald's text) just say "rotating about" and don't make any mention of Fermi-Walker transport.

Thanks!

 

[PeterDonis]

the reason why an observer at infinity will see the rod rotate at exactly $\omega$ around the disk is because both $O(R_0)$ and $O(R_1)$ are rotating around the disk at the same angular speed $\omega$ relative to infinity

Yes.

so the observer at infinity will never see the rod rotate due to $O(R_1)$ overtaking $O(R_0)$ or vice-versa and will only see it rotate due to the equal rotation rate $\omega$ of said two observers around the disk relative to infinity?

Yes.

So is this in contrast with what happens in Kerr space-time for the ZAMOs because observers at different radii have different angular velocities around the black hole

Yes, this is a difference, but there are some caveats to how we interpret it. See below.

so a rod radially attached initially from one observer to another will not only rotate around the black hole relative to infinity due to the overall orbital rotation but also rotate because the inner observer is overtaking the outer observer?

Actually that is shear, not twist. To factor out the shear, just consider the leading and trailing observers in the ZAMO congruence, relative to the fiducial observer; i.e., the observers who are at the same $r$ but slightly larger and smaller $\phi$. The connecting rods to these observers will rotate, relative to infinity, at exactly the ZAMO angular velocity for that $r$.

when we say "the twist causes nearby observers to rotate about the given reference observer from the perspective of the reference observer" do we really always mean "the twist causes the nearby observers to rotate relative to the Fermi-Walker transported spatial basis of the reference observer"?

That's my understanding, yes.

 

[WannabeNewton]

I guess it was easier for me to picture the twist of the static observers in the Kerr space-time because I could just think about how a given static observer would be "self-rotating" relative to the stars and will also see nearby static observers fixed to the stars being spun around him and in this sense result in a non-vanishing twist. Here, the Langevin observers aren't "self-rotating" relative to infinity in any way, they are all just spinning around the cylindrical walls so if I were to put myself in the rest frame of a given Langevin observer I can't really imagine why I would see nearby Langevin observers rotate around me (i.e. rotate relative to my Fermi transported spatial basis vectors). Even if the observers at different radii were rotating at different angular velocities, this would only result in a shear (much like for the ZAMO congruence) and not a twist but regardless here we have no shear at all and just a twist but I can't really picture this twist. If I imagine myself in the frame of a given Langevin observer, I just picture all the other nearby Langevin observers as being at rest with me.

 

[PeterDonis]

I were to put myself in the rest frame of a given Langevin observer I can't really imagine why I would see nearby Langevin observers rotate around me (i.e. rotate relative to my Fermi transported spatial basis vectors).

It's hard for me to imagine too, which is why I am looking for alternate ways to think about Fermi-Walker transport, such as what I described a few posts back (parallel transport the basis vectors, then Lorentz transform them so the 4-velocity matches what it actually is at the new event).

 

[WannabeNewton]

Peter do you happen to have the text "Gravitation and Inertia" by Wheeler et al?

 

[PeterDonis]

Peter do you happen to have the text "Gravitation and Inertia" by Wheeler et al?

Yes. Why?

 

[atyy]

It's hard for me to imagine too, which is why I am looking for alternate ways to think about Fermi-Walker transport, such as what I described a few posts back (parallel transport the basis vectors, then Lorentz transform them so the 4-velocity matches what it actually is at the new event).

Thanks for that explanation a few posts back. It does seem that Fermi-Walker transport of a vector along a curve is possible without defining a congruence.

 

[WannabeNewton]

Ok because I was reading pages 237-238 where they talks about rotation of a congruence with respect to Fermi-Walker transport much like you did before and I just had two questions that cropped up:

On page 238 they say "This expression explicitly shows that...give the rate of change with time of the separation...between neighboring particles, that is, the motions and the rate of change of the dimensions of an infinitesimal volume element of the fluid relative to a local comoving Fermi frame". Are we supposed to take the local comoving Fermi frame to be a local fluid element that Fermi transports its spatial axes?

On the same page, slightly below, they say "In particular, for an infinitesimal spherical surface described by the fluid particles with $\delta x^{i}\delta x_{i} = \delta r^{2}$ in the Fermi frame...and $\omega_{ik}$ its angular velocity relative to the Fermi axes, that is, relative to local gyroscopes". Is the Fermi frame supposed to be attached to a local fluid element that is at the center of the spherical surface defined above by the other fluid elements? In particular, in the image at the bottom of the page, the picture depicting the rotation is showing two dots (fluid elements) on the surface of the sphere that are rotating relative to the spatial axes of a Fermi frame attached to a fluid element at the center of that sphere (with the spatial axes representing gyros) correct? I ask because I didn't know what the dot at the top of that rotating sphere was representing.

 

[TrickyDicky]

Hey guys. I have a question about two (possibly ostensibly) different definitions of a locally non-rotating observer that I have come across in my texts.

The first is specifically for stationary, axisymmetric space-times in which we have a canonical global time function $t$ associated with the time-like killing vector field. We define a locally non-rotating observer to be one who follows an orbit of $\nabla^{a}t$ i.e. his 4-velocity is given by $\xi^{a} = (-\nabla^{b}t\nabla_{b}t)^{-1/2}\nabla^{a}t = \alpha \nabla^{a}t$. Such an observer can be deemed as locally non-rotating because his angular momentum $L = \xi^{a}\psi_{a} = \alpha g_{\mu\nu}g^{\mu\gamma}\nabla_{\gamma}t\delta^{\nu}_{\varphi} = \alpha \delta^{\gamma}_{\nu}\delta^{t}_{\gamma}\delta^{\nu}_{\varphi} = \alpha \delta^{t}_{\varphi} = 0$ where $\psi^{a}$ is the axial killing vector field; these observers are also called ZAMOs for this reason. It might also be worth noting that the time-like congruence defined by the family of ZAMOs has vanishing twist $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = \alpha\epsilon^{ab[cd]}\nabla_{b}t\nabla_{(c}\nabla_{d)}t + \alpha^{3}\epsilon^{a[b|c|d]}\nabla_{(b}t \nabla_{d)}t\nabla^{e}t\nabla_{c}\nabla_{e}t = 0$.


The second definition I have seen is the much more general notion of Fermi-Walker transport. That is, if we choose an initial Lorentz frame $\{\xi^{a}, u^{a},v^{a},w^{a}\}$ and the spatial basis vectors evolve according to $\xi^{b}\nabla_{b}u^{a} = \xi^{a}u_{b}a^{b}$, $\xi^{b}\nabla_{b}v^{a} = \xi^{a}v_{b}a^{b}$, and $\xi^{b}\nabla_{b}w^{a} = \xi^{a}w_{b}a^{b}$, where $a^{b} = \xi^{a}\nabla_{a}\xi^{b}$ is the 4-acceleration, then the observer is said to be locally non-rotating.

My question is, to what extent are these two definitions equivalent (both mathematically and physically) whenever they can both be applied? In other words, in what sense is the qualifier 'rotation' being used in each case?

Let me elucidate my question a little bit. I know that for asymptotically flat axisymmetric space-times, there must exist a fixed rotation axis on which $\psi^{a}$ vanishes. Then the first definition tells us that the ZAMOs have no orbital rotation about this fixed rotation axis (for example no orbital rotation about a Kerr black hole); because these observers are at rest with respect to the $t = \text{const.}$ hypersurfaces, these observers are as close to stationary hovering observers as we can get in a space-time with a rotating source. We know however that such observers have an instrinsic angular velocity $\omega$; if we imagine a ZAMO holding a small sphere with frictionless prongs sticking out with beads through the prongs then a ZAMO should be able to notice his intrinsic angular velocity $\omega$ by seeing that the beads are thrown outwards along the prongs at any given instant. Is this correct?

Now, on the other hand, the second definition of local non-rotation (using Fermi-Walker transport) seems to be saying sort of the opposite. That is, it seems to be telling us which observers can carry such spheres and never see the beads get thrown outwards i.e. the orientation of the sphere will remain constant (the orientation will be represented by the spatial basis vectors of a Lorentz frame) so these are the observers who have no intrinsic angular velocity i.e. no self-rotation. This is why the two definitions confused me because they seem to be talking about two totally different kinds of non-rotation.

Thank you in advance.

WN, first sorry about my remark regarding the definition of observers I posted above, I was in a hurry and didn't pay much attention, second thanks for this discussion you and Peter are having, very instructive.
Now, if I may ask something related to the OP: why would you want to reconcile definitions about local (non)rotation that are built on different solutions of the EFE, therefore different geometries? The first definition refers to the Kerr geometry with axisymmetry, but no spherical symmetry, and the second that is valid for the more general spherical symmetry. In the first case total angular momentum is not even conserved since the spacetime is not spherically symmetric about any of its points, while it is in the second.

Actually it is a bit of a puzzle for me how exactly the expected physical consequences of the different and sometimes mutually incompatible GR solutions are chosen and considered to be mainstream physics or not in the absence of empirical confirmation.
For instance the Kerr stationary axisymmetric solution that gives rise to your first definition and that it seems logical that it should apply to rotating objects such as planets and stars, produces the prediciton of the frame-dragging effect, but this geometry seems to be not compatible with the more general spherical symmetry (that on the other hand is observationally quite confirmed and theoretically expected globally) of the Schwarzschild solution that predicts the geodetic effect, in as much as the effect demands stationarity of the spacetime(but non-staticity).

 

[WannabeNewton]

The first definition applies to any axially symmetric, stationary space-time so not just the Kerr space-time. You don't need spherical symmetry for angular momentum to be conserved, you just need axial symmetry. If axial symmetry is available then there exists an axial killing field $\psi^{a}$ and the quantity $u_{a}\psi^{a}$ is conserved when $u^{a}$ is a geodesic 4-velocity as usual i.e. $u^{a}\nabla_{a}(\psi_{b}u^{b}) = \psi_{b}u^{a}\nabla_{a}u^{b} + u^{(a}u^{b)}\nabla_{[a}\psi_{b]} = 0$. We define the angular momentum as $L = u^{a}\psi_{a}$.

The Fermi-Walker condition can be applied in any space-time whatsoever; there are no restrictions there. I wasn't exactly looking for the two definitions to be reconciled per say but rather wanted to see what their differences were i.e. what kind of non-rotation does Fermi-Walker refer to and what kind of non-rotation does following an orbit of $\nabla^{a} t$ refer to. Peter helped reconcile these differences for me in the thread (thanks again Peter ).

The latter refers to a congruence of observers, all following orbits of $\nabla^{a} t$, with vanishing twist which is equivalent to hypersurface orthogonality i.e. $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0\Leftrightarrow \xi_{[a}\nabla_{b}\xi_{c]} = 0$ for any tangent field $\xi^{a}$. A given observer in the congruence will not see locally separated observers rotate around him as his proper time passes so it is local non-rotation in that sense. Furthermore since $\nabla^{a}t$ happens to be everywhere orthogonal to $\psi^{a}$, the angular momentum of these observers also vanishes so that's an added bonus.

The former condition, that of Fermi-Walker transport, is more akin (locally) to non-rotating frames in SR in the sense that locally we can imagine the Fermi frame as an observer with 3 mutually perpendicular gyroscopes enclosed in an elevator in SR with the gyroscopes yielding null results for rotation of the elevator in the SR sense.

 

[TrickyDicky]

The first definition applies to any axially symmetric, stationary space-time so not just the Kerr space-time.

Sure, i was using it as an example in reference to the second part of my post.

You don't need spherical symmetry for angular momentum to be conserved, you just need axial symmetry.

The component of angular momentum $L = u^{a}\psi_{a}$ of a particle along the rotation axis is conserved with axial symmetry but not the total angular momentum of the particle.

The Fermi-Walker condition can be applied in any space-time whatsoever; there are no restrictions there. I wasn't exactly looking for the two definitions to be reconciled per say but rather wanted to see what their differences were i.e. what kind of non-rotation does Fermi-Walker refer to and what kind of non-rotation does following an orbit of $\nabla^{a} t$ refer to.

Ok, this answers my first question.

 

[WannabeNewton]

The component of angular momentum $L = u^{a}\psi_{a}$ of a particle along the rotation axis is conserved with axial symmetry but not the total angular momentum of the particle.

What does "total angular momentum" even mean when only $L$ is available? It's all there is.

 

[PeterDonis]

Ok because I was reading pages 237-238 where they talks about rotation of a congruence with respect to Fermi-Walker transport much like you did before and I just had two questions that cropped up

Oh, you meant you actually wanted me to dig out my copy and look at it? I don't have it handy right now but I think I remember it well enough to give some useful comments.

On page 238 they say "This expression explicitly shows that...give the rate of change with time of the separation...between neighboring particles, that is, the motions and the rate of change of the dimensions of an infinitesimal volume element of the fluid relative to a local comoving Fermi frame". Are we supposed to take the local comoving Fermi frame to be a local fluid element that Fermi transports its spatial axes?

I believe that's correct, yes.

On the same page, slightly below, they say "In particular, for an infinitesimal spherical surface described by the fluid particles with $\delta x^{i}\delta x_{i} = \delta r^{2}$ in the Fermi frame...and $\omega_{ik}$ its angular velocity relative to the Fermi axes, that is, relative to local gyroscopes". Is the Fermi frame supposed to be attached to a local fluid element that is at the center of the spherical surface defined above by the other fluid elements?

IIRC, yes, that's right.

 

[PeterDonis]

the Kerr stationary axisymmetric solution that gives rise to your first definition and that it seems logical that it should apply to rotating objects such as planets and stars, produces the prediciton of the frame-dragging effect

It seems logical, yes, but AFAIK nobody has actually proven that the vacuum region exterior to a rotating object (not a black hole) such as a planet or star actually is the Kerr geometry. The corresponding proof in the spherically symmetric case requires Birkhoff's theorem, and there is no analogue of Birkhoff's theorem for the non-spherical but axisymmetric case.

(Also, AFAIK in practice the full Kerr geometry is not used in a lot of cases; a sort of perturbation approach is used where the leading order Kerr terms are added on to an underlying Schwarzschild or PPN-type solution. See below.)

but this geometry seems to be not compatible with the more general spherical symmetry (that on the other hand is observationally quite confirmed and theoretically expected globally)

No, it isn't observationally confirmed; it's an approximation that works well in many cases but is well understood to be an approximation. Even disregarding all the other asymmetries of a rotating planet like the Earth, it's an oblate spheroid, not a sphere.

of the Schwarzschild solution that predicts the geodetic effect

The Schwarzschild solution does not predict the geodetic effect (by which I assume you mean frame dragging). You need to add something to it to represent the rotation of the central body. As I understand it, the way this is done in practice (for things like analyzing Gravity Probe B data) is to start with the Schwarzschild metric and add a perturbation to it to represent the expected frame dragging effects. The form of the perturbation is obtained by expanding the Kerr metric around $a = 0$ (i.e., by assuming that $a << M$).

 

[WannabeNewton]

Oh, you meant you actually wanted me to dig out my copy and look at it?

At least it's not as heavy as MTW xD :)

Thanks for the confirmations! Much appreciated. I just wanted to make sure I was interpreting the text correctly because "Gravitation and Inertia" is the only text I own wherein the term "rotate about" is actually defined rigorously in terms of Fermi frames, in the way you described earlier in the thread, so I wanted to make sure I wasn't fudging anything up in my head. All my other texts just say "rotate about" and leave it at that unfortunately.

Speaking of Wheeler, they go into detail on the theoretical and experimental aspects of the geodetic and Lense-Thirring effects in section 3.4.3.

 

[TrickyDicky]

The Schwarzschild solution does not predict the geodetic effect (by which I assume you mean frame dragging).

No, the geodetic effect is not frame dragging, look up de Sitter precession.

 

[TrickyDicky]

What does "total angular momentum" even mean when only $L$ is available? It's all there is.

Exactly, that is all there is in an axisymmetric space, to be compared with the angular momentum available that is conserved in classical static spherically symmetric scenario( of SR and Schwarzschild metrics for instance or of classical mechanics for that matter).

 

[PeterDonis]

No, the geodetic effect is not frame dragging, look up de Sitter precession.

Ah, you're right, sorry for the mixup. But I'm still not sure I see the point of your comment; de Sitter precession will still be present in Kerr spacetime, because the Kerr metric contains the Schwarzschild metric terms (the Schwarzschild metric is just the Kerr metric with $a = 0$). So all the Schwarzschild effects are still there, even if the metric is Kerr.

 

[PeterDonis]

Exactly, that is all there is in an axisymmetric space, to be compared with the angular momentum available that is conserved in classical static spherically symmetric scenario

A spherically symmetric spacetime *is* axisymmetric; it has an axial KVF. (It just also has two additional spacelike KVFs.) So a spherically symmetric spacetime has all the properties of an axisymmetric spacetime, plus some extra ones due to the additional symmetries.

 

[TrickyDicky]

No, it isn't observationally confirmed; it's an approximation that works well in many cases but is well understood to be an approximation.

I wrote globally. Current cosmology expects isotropy not to be just an approximation globally.
Locally anyone can trivially see it is an approximation.

 

[WannabeNewton]

Exactly...

Oh ok, I guess we were just saying the same thing in different ways then. pew pew pew...pew >.> <.<

 

[TrickyDicky]

A spherically symmetric spacetime *is* axisymmetric; it has an axial KVF. (It just also has two additional spacelike KVFs.) So a spherically symmetric spacetime has all the properties of an axisymmetric spacetime, plus some extra ones due to the additional symmetries.

Sure, but not the other way around which is the case I'm addressing.

 

[PeterDonis]

I wrote globally. Current cosmology expects isotropy not to be just an approximation globally.

Really? Do you have a reference for this? My understanding is that it's an approximation "all the way up"; we use models that are globally isotropic but the models are not claimed to be exact.

 

[PeterDonis]

Sure, but not the other way around which is the case I'm addressing.

I agree that an axisymmetric spacetime is not spherically symmetric; that's obvious. But I don't understand what physical predictions you are saying require exact spherical symmetry, as opposed to just axisymmetry. The geodetic effect is present in an axisymmetric spacetime; you don't need exact spherical symmetry to predict it. I agree that there is no conserved "total angular momentum" in an axisymmetric spacetime; there's only angular momentum about the symmetry axis. But I don't see what physical predictions we are actually making that that invalidates.

 

[TrickyDicky]

Ah, you're right, sorry for the mixup. But I'm still not sure I see the point of your comment; de Sitter precession will still be present in Kerr spacetime, because the Kerr metric contains the Schwarzschild metric terms (the Schwarzschild metric is just the Kerr metric with $a = 0$). So all the Schwarzschild effects are still there, even if the metric is Kerr.

I'm precisely arguing that I fail to see how the Kerr geometry being stationary and explicitly non-static and non-spherically symmetric(the frame dragging effect rests critically on this) can be compatible with effects that are related to spherical symmetry.
Note again that the general spacetime here is Schwarzschild, that is both spherically symmetric and therefore axisymmetric and both static and stationary and shows the geodetic effect but no frame drsgging, while the
Kerr geometry is axisymmetric but not
spherically symmetric and stationary but not static and needs this geometrical configuration to have frame dragging.
In mainstream relativity it is expected that we should observe both effects, but it seems to me that logically we should observe either one or the other.
Where is my misunderstanding?

 

[WannabeNewton]

I'm not sure I get what you are asking. A lack of spherical symmetry doesn't imply geodetic precession can't exist; Kerr space-time can exhibit both geodetic precession and Lense-Thirring precession. Are you asking how an experiment can distinguish between the two?

 

[PeterDonis]

I'm precisely arguing that I fail to see how the Kerr geometry being stationary and explicitly non-static and non-spherically symmetric(the frame dragging effect rests critically on this) can be compatible with effects that are related to spherical symmetry.

Can you give an example of an effect that is "related to spherical symmetry" but *not* "related" to axisymmetry? The geodetic effect is "related" to both; see below. As I said before, nobody is claiming that the universe is globally exactly spherically symmetric, so that's not a good example either.

In mainstream relativity it is expected that we should observe both effects

Yes, because the Earth is rotating, so the vacuum geometry around it is *not* Schwarzschild.

but it seems to me that logically we should observe either one or the other.

If the Earth were *exactly* spherically symmetric (i.e., not rotating, plus made of some idealized material that formed an exactly symmetric sphere), then we would expect to observe only de Sitter precession (geodetic effect), not Lense-Thirring precession (frame dragging), yes. But I can't think of any conditions under which we would expect to observe frame dragging but *not* the geodetic effect. In Kerr spacetime both effects are predicted. Perhaps that's where your misunderstanding lies.

 

[TrickyDicky]

I'm not sure I get what you are asking. A lack of spherical symmetry doesn't imply geodetic precession can't exist; Kerr space-time can exhibit both geodetic precession and Lense-Thirring precession.

Well, my underdtanding is that in the Kerr spacetime the rotating effects are mixed in a way that is not meaningful to separate them in two different effects.

 

[WannabeNewton]

Peter, I have another question about frame dragging. In Wald's text (problem 7.3b) he states that the induced $\dot{\phi}$ around the central rotating mass for observers who follow orbits of $\nabla^{a} t$ in stationary, axisymmetric space-times is frame dragging. On the other hand, we've said that observers who follow orbits of the time-like KVF will spin in place, which Wheeler seems to call frame dragging as per part (c) of the following diagram: http://postimg.org/image/z69htcfdz/ because he depicts the small balls as spinning in place due to the spinning of the big sphere if I'm reading the diagram correctly. Also in problem 4.3c of Wald, one shows that for an observer at rest at the center of a thin slowly rotating shell (in the weak field regime) who parallel transports along his geodesic worldline (with 4-velocity $u^{a}$) a vector $S^{a}$ with $S^{a}u_{a} = 0$ (i.e. $S^{a}$ is a purely spatial vector), the inertial components $\vec{S}$ of $S^{a}$ precess as per $\frac{d\vec{S}}{dt} = \vec{\Omega} \times \vec{S}$ in a background global inertial frame. So if we represent this purely spatial vector by a gyroscope held by the observer then an observer at infinity will see the gyroscope (and the aforementioned observer holding it) spin around in place at the center of the shell much like an observer at infinity would see a static observer in Kerr space-time spin around in place correct? This is more akin to Wheeler's picture but Wald also calls this frame dragging.

Are they both considered forms of frame dragging? Intuitively the former (being pulled into orbital motion around the central rotating mass) seems like a kind of "dragging" to me since you're being dragged around the central rotating mass. The central rotating mass causing other objects to spin in place, such as the static observers in Kerr space-time, doesn't really seem like a kind of "dragging" to me.

EDIT: All of the above should be relative to the distant stars, if I'm not mistaken.

 

[PeterDonis]

Are they both considered forms of frame dragging?

AFAIK yes, the term "frame dragging" can be applied to both of these effects. However, as you have seen, usages can differ, so I wouldn't hang my hat on any particular usage of the term as being "standard". This is another illustration of why you have to look at the math to be sure you know what's being discussed; English, or any other natural language, just isn't precise enough.

 

[WannabeNewton]

Cool thanks! Also, for the thin rotating shell scenario mentioned above (rotating about the z-axis of a background global inertial coordinate system), if I have a congruence of observers at rest inside the shell (at rest as in static) then I find that the twist comes out to $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} = 2\partial_{[y}h_{|0|x]}\delta^{\mu z}= \frac{8}{3}\frac{M}{R}\omega \delta^{\mu z}$ where $h_{\mu\nu}$ is the perturbed metric inside the shell and $\omega$, $M$, and $R$ are the magnitude of the angular velocity, mass, and radius of the shell respectively. So if we imagine the purely spatial vector $S^{\mu}$ from above (the one that is parallel transported along the worldline of the observer at rest at the center of the shell) as a gyroscope held by the central observer in the congruence then this gyroscope is Fermi-Walker transported along this observer's worldline (since Fermi-Walker transport is equivalent to parallel transport for geodesics) and hence we can think of $\omega^{\mu}$ as representing the rotation of nearby static observers inside the shell relative to this gyroscope? So the central observer in the congruence will see nearby static observers rotate around him whilst being fixed to the distant stars, much like we imagined before with the static observers in Kerr space-time?

On the other hand the quantity $\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4}{3}\frac{M}{R}\vec{\omega}\times \vec{S}$ represents the rate at which an observer at infinity sees the aforementioned gyroscope spin in place (here $\vec{\omega}$ is the angular velocity of the shell)?

 

[WannabeNewton]

Well, my underdtanding is that in the Kerr spacetime the rotating effects are mixed in a way that is not meaningful to separate them in two different effects.

Do you know where I can read more on this? I can only seem to find things on the PPN formalism in my texts and elsewhere.

 

[TrickyDicky]

Do you know where I can read more on this? I can only seem to find things on the PPN formalism in my texts and elsewhere.

My dear mate WN, PF is the main place for me thanks to guys like you, Peter and a few others.
Let me try and explain where my reasoning(hopefully not flawed) comes from to deduce that one might not have a well defined and separated de Sitter effect, like the one calculated using the Schwarzschild mettric, added to the rotational frame dragging in the axisymmetric stationary solution.
In the static case the hypersurface orthogonality greatly simplifies obtaining the geodetic effect, the curvature of the spacetime for each time instant can be assigned to the spatial part of the metric which makes easier to visualize and compute how a gyroscope that is Fermi-Walker transported nevertheless experiences certain precession due to curvature(like any vector that is parallel transported in the presence of curvature).

Now, back to the stationary and axisymmetric case, the presence of cross terms with \phi will clearly give us a rotational frame dragging effect around a self-rotating source, but now it is much more difficult to separate from this effect a clear cut gyroscopic effect equivalent to the above described precession due to the lack of hypersurface orthogonality that prevents us from the neat foliation of space and time that the static case permits.

When this two effects were derived by de Sitter et al. and Lense,Thirring et al. separately around the same time !916-18, they didn't even have the Kerr geometry, and subsequently it seems textbooks just take the two effects and simply claim they just must both appear, adding the formula obtained from the Schwarzschild metric and the one obtained from lowest order approximation of axisymmetric solutions.

 

[PeterDonis]

In the static case the hypersurface orthogonality greatly simplifies obtaining the geodetic effect

I agree with this, but note that "greatly simplifies" is not the same thing as "makes possible".

(Thanks for the kudos, btw. )

it seems textbooks just take the two effects and simply claim they just must both appear, adding the formula obtained from the Schwarzschild metric and the one obtained from lowest order approximation of axisymmetric solutions.

I haven't gotten that impression from the textbooks I'm familiar with, but I'll take a look at MTW and Wald when I get a chance to refresh my memory of how they treat this.

That said, you do realize that the lowest order approximation of the axisymmetric solutions *is* the Schwarzschild metric, right? As I said before, if you take the Kerr metric and treat $a$ (the angular momentum per unit mass) as a small parameter (more precisely, you rewrite the line element in $a/M$ and treat that as a small parameter), the zeroth order term in the perturbation expansion, where $a = 0$, is the Schwarzschild metric. So I don't see any issue.

 

[WannabeNewton]

Wald doesn't do anything on the geodetic effect I'm afraid and the only things related to frame dragging that he does are the thin shell thing I talked about before and the ZAMOs :(

Btw Peter, was my understanding in post #84 ok?

 

[PeterDonis]

Btw Peter, was my understanding in post #84 ok?

I want to look at that section of Wald in more detail before responding.

 

[TrickyDicky]

I agree with this, but note that "greatly simplifies" is not the same thing as "makes possible".

(Thanks for the kudos, btw. )
I haven't gotten that impression from the textbooks I'm familiar with, but I'll take a look at MTW and Wald when I get a chance to refresh my memory of how they treat this.

That said, you do realize that the lowest order approximation of the axisymmetric solutions *is* the Schwarzschild metric, right? As I said before, if you take the Kerr metric and treat $a$ (the angular momentum per unit mass) as a small parameter (more precisely, you rewrite the line element in $a/M$ and treat that as a small parameter), the zeroth order term in the perturbation expansion, where $a = 0$, is the Schwarzschild metric. So I don't see any issue.

Yes, but I expressed myself confusingly there, I was actually trying to say that like WN alluded to(PPN formalism), all the derivations of the formula for the frame-dragging I've seen (certainly those initially done by Einstein himself around 1917 -see "The meaning of relativity" pages107-109- , and I guess by Thirring and Lense), are based simply on the linearization of the EFE in the weak field applied to a self-rotating source.

 

[PeterDonis]

all the derivations of the formula for the frame-dragging I've seen (certainly those initially done by Einstein himself around 1917 -see "The meaning of relativity" pages107-109- , and I guess by Thirring and Lense), are based simply on the linearization of the EFE in the weak field applied to a self-rotating source.

The same comment would apply here; the weak field linearized approximation to a self-rotating source should have leading order terms which are just the weak field linearized approximation to a non-rotating source. At least, that's the way it seems to me, but I'll need to look at MTW's discussion (since WannabeNewton has already checked Wald and found little to chew on) to see if there are subtleties I'm missing.

 

[PeterDonis]

this gyroscope is Fermi-Walker transported along this observer's worldline (since Fermi-Walker transport is equivalent to parallel transport for geodesics) and hence we can think of $\omega^{\mu}$ as representing the rotation of nearby static observers inside the shell relative to this gyroscope?

Yes, this looks right. But note that $\omega^{\mu}$ will be the rate of rotation of the nearby static observers relative to the gyroscope, with respect to the central observer's proper time, *not* with respect to coordinate time. I left that part out of our previous discussion, and I think it makes a difference. See below.

So the central observer in the congruence will see nearby static observers rotate around him whilst being fixed to the distant stars, much like we imagined before with the static observers in Kerr space-time?

Yes.

On the other hand the quantity $\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4}{3}\frac{M}{R}\vec{\omega}\times \vec{S}$ represents the rate at which an observer at infinity sees the aforementioned gyroscope spin in place (here $\vec{\omega}$ is the angular velocity of the shell)?

Yes, but as your notation makes clear, this rate is relative to coordinate time, *not* the central observer's proper time, which is why it's a different number than $\omega^{\mu}$, even though both are referring to the same thing (well, opposite signs of the same thing).

The reason I bring this up is that, as I mentioned above, I ignored the difference between coordinate time and proper time in our previous discussion, and putting it back in makes a difference. For example, consider the ZAMO and hovering congruences in Kerr spacetime. We talked about comparing two sets of basis vectors for each type of observer: a set of Fermi-Walker transported vectors (gyroscopes), and a set of vectors fixed by connecting rods to neighboring members of the same congruence. Consider what we said about them:

For the ZAMO congruence, the two sets of vectors remain aligned (assuming that they start out aligned); that means both of them rotate at the same rate relative to an observer at infinity. We said this rate is "the same" as the angular velocity of the ZAMO about the black hole, but actually it is different depending on who is observing it: the ZAMO will get a different actual number than the observer at infinity, because of time dilation (here a combination of the "redshift factor" from the $g_{tt}$ metric coefficient, and the SR time dilation due to nonzero tangential velocity).

For the hovering congruence, the two sets of vectors will *not* remain aligned (even if they start out aligned); they rotate relative to each other. But again, because of time dilation (here just the $g_{tt}$ "redshift factor"), the rate at which the hovering observer sees the neighboring static observers rotate around him, compared to his gyroscopes (which is a rate with respect to his proper time), will *not* be the same as the rate at which an observer at infinity sees the hovering observer's gyroscopes spinning in place (which is a rate with respect to coordinate time).

 

[WannabeNewton]

Ah, right twist is based on the proper time read by the observer in the congruence we're using as a reference whereas the other is the coordinate time as read by the observer at infinity so gravitational and kinematical time dilation will cause discrepancies. Thanks for pointing that out Peter, and thanks for clarifying the scenario with the static congruence inside the thin rotating shell. I was just using that scenario because it resembled the static observer scenario in Kerr space-time except it was somewhat easier to picture.

Thanks again!

 

[WannabeNewton]

Peter sorry to bother you again but I had a question that I meant to ask before but forgot to. If we take locally non-rotating to mean that the twist $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0$ for a time-like congruence $\xi^{a}$ then we know this implies that $\xi^{a} = \alpha \nabla^{a}\beta$ for smooth functions $\alpha,\beta$ but $L = \xi^{a}\psi_{a} = \alpha \psi_{a}\nabla^{a}\beta \neq 0$ in general so there is no equivalence between locally non-rotating in the sense of vanishing twist and zero angular momentum for an arbitrary twist free congruence. So is it just a coincidence that for the $\nabla^{a} t$ congruence we have local non-rotation in the sense that $\omega^{a} = \alpha \epsilon^{abcd}\nabla_{b} t \nabla_{c}(\alpha \nabla_{d}t) = 0$ as well as vanishing angular momentum $L = \alpha\psi_{a}\nabla^{a}t = 0$ or is there a physical connection between the two?

I ask because in Malament's GR text, local non-rotation is simply defined as vanishing twist of a congruence (so that the worldlines of the observers in the congruence are unfurled) and no mention of angular momentum is made whereas in Wald's text local non-rotation is specifically defined as being an observer in the $\nabla^{a} t$ congruence in which case $L = 0$ follows suit; I don't really get why Malament defines it in a much more general manner whereas Wald sticks to that specific case. I can tell intuitively that defining local non-rotation as $\omega^{a} = 0$ gives us a congruence which is hypersurface orthogonal which is quite important physically whereas I can't immediately see the relevance/use of the fact that for the specific case of the hypersurface orthogonal congruence given by $\nabla^{a} t$, we have $L = 0$ as well.

 

[TrickyDicky]

The same comment would apply here; the weak field linearized approximation to a self-rotating source should have leading order terms which are just the weak field linearized approximation to a non-rotating source. At least, that's the way it seems to me, but I'll need to look at MTW's discussion (since WannabeNewton has already checked Wald and found little to chew on) to see if there are subtleties I'm missing.

Well yes, when I say linearization I'm rather referring to the post-Newtonian expansion or the lowest order approximation from the Newtonian limit so it goes beyond linearization proper, sorry about the confusion. For instance one of the parameters of the parametrized post-Newtonian formalism is frame draging per angular momentum unit =Δ1.
The Lense-Thirring frame-dragging effect formula that was used in textbooks before the mathematical derivation from the Kerr geometry was available in the sixties is the one based in the gravitomagnetic field of the earth, that is basically the post-Newtonian approximation above mentioned, the weak field, low speed approximation of the GR equations of motion in the presence of angular momentum.

 

[PeterDonis]

Peter sorry to bother you again

No problem, I'm not bothered. I can talk about this stuff indefinitely.

there is no equivalence between locally non-rotating in the sense of vanishing twist and zero angular momentum for an arbitrary twist free congruence.

See below.

So is it just a coincidence that for the $\nabla^{a} t$ congruence we have local non-rotation in the sense that $\omega^{a} = \alpha \epsilon^{abcd}\nabla_{b} t \nabla_{c}(\alpha \nabla_{d}t) = 0$ as well as vanishing angular momentum $L = \alpha\psi_{a}\nabla^{a}t = 0$ or is there a physical connection between the two?

I think there might be a connection; see below.

I can tell intuitively that defining local non-rotation as $\omega^{a} = 0$ gives us a congruence which is hypersurface orthogonal which is quite important physically whereas I can't immediately see the relevance/use of the fact that for the specific case of the hypersurface orthogonal congruence given by $\nabla^{a} t$, we have $L = 0$ as well.

What other hypersurface orthogonal timelike congruences are there? Can you find one that has a nonzero $L$? I suspect there aren't any, which means that any twist free vector field $\alpha \nabla^a \beta$ must have $\beta = k t$ with $k$ constant. Which really means we set the coordinate chart up the right way for $t$ to have that property.

 

[TrickyDicky]

I looked at MTW and it is actually quite well explained in the PPN formalism in pages 1117-9. So if we expect both effects(de sitter precession and frame-dragging) to show up we must at the same time ignore the cross terms d\phi/dt (de Sitter effect) as it is done in Fermi-Walker transport, and not ignore them (frame dragging). Hmmm...well, one can say it is just an approximation after all, but if we refer to the actual GR solutions we must at the same time rely for the last effect on a geometry that explicitly rules out staticity (Kerr), and on a static geometry that that demands the cross terms to vanish for computation of the de Sitter effect. I don't get the logic behind this.

 

[Bill_K]

I don't get the logic behind this.

I have a blog which discusses precession in all its aspects: Thomas precession in flat space, de Sitter precession in Schwarschild, and Lense-Thirring in Kerr. The combined precession rate contains terms which can easily be identified with each of these effects.

 

[WannabeNewton]

What other hypersurface orthogonal timelike congruences are there? Can you find one that has a nonzero $L$? I suspect there aren't any, which means that any twist free vector field $\alpha \nabla^a \beta$ must have $\beta = k t$ with $k$ constant.

Well like in the Schwarzschild space-time, $\nabla^{a}t$ is everywhere time-like for $r > 2M$ since $g_{\mu\nu}\nabla^{\mu}t\nabla^{\nu}t = -(1 - \frac{2M}{r})^{-1}$ which is less than zero for $r > 2M$. Similarly, the function $f(t,r) = t + \frac{1}{2}[2M\ln (r - 2M) + r]$ gives us the vector field $\xi^{\mu} = \nabla^{\mu}f = g^{\mu t} + \frac{1}{2}g_{rr}g^{\mu r}$ and we have that $g_{\mu\nu}\xi^{\mu}\xi^{\nu} = g^{tt} + \frac{1}{4}g_{rr} = -\frac{3}{4}(1 - \frac{2M}{r})^{-1}$ which, much like the norm of $\nabla^{a} t$, is less than zero for $r > 2M$ hence everywhere time-like for $r > 2M$ just like $\nabla^{a} t$. $\xi^{\mu}$ is of course twist free but $\xi^{a} \neq \alpha \nabla^{a}t$ for some smooth function $\alpha$ however $L = \psi_{a}\xi^{a} = 0$ still holds so it would be possible to have twist free time-like vector fields in the space-time that are not colinear with $\nabla^{a} t$ but still have vanishing angular momentum if we included e.g. a radial component.

However if we want a twist free vector field that is time-like and has $L \neq 0$ then it would need to be non-orthogonal to $\psi^{a}$. I can't immediately find such a vector field in the Schwarzschild space-time (the computation got quite messy) nor can I prove one doesn't exist (what I tried to do was assume there existed a vector field $\xi^{a}$ such that $\xi^{a}\psi_{a}\neq 0$, which in the Schwarzschild space-time comes down to $\xi^{a}$ having a component along $\psi^{a}$, and then showing that $\xi^{a}$ can't satisfy both $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0$ and $\xi_{a}\xi^{a} < 0$ but it got insanely messy really fast and I can't immediately think of an easier/more elegant approach that would hold not just for Schwarzschild space-time but also in full generality for any stationary, axisymmetric space-time).

Anyways, the main reason I asked was in the specific case that we do have both vanishing twist (so local non-rotation) and vanishing angular momentum, is there a way to picture the vanishing angular momentum in terms of the vanishing twist? So in the case of the $\nabla^{a} t$ congruence in Kerr space-time, is there a way to visualize vanishing angular momentum for the observers in the congruence in terms of the fact that the congruence is twist free? I just can't really visualize vanishing angular momentum on its own so that's why I asked the question in the first place, to see if there was a way to physically connect them (since they are both true for the $\nabla^{a} t$ congruence) because unlike vanishing twist and angular velocity (as observed at infinity) which I can visualize, I can't seem to visualize vanishing angular momentum.

Thanks Peter!

 

[TrickyDicky]

Kerr space-time can exhibit both geodetic precession and Lense-Thirring precession. Are you asking how an experiment can distinguish between the two?

I can't think of any conditions under which we would expect to observe frame dragging but *not* the geodetic effect. In Kerr spacetime both effects are predicted.

Ok, but let's see an example of what I mean. Since it is easy to agree that the most general notion of locally non-rotating is the one given by Fermi-Walker transport that is valid for any GR solution by virtue of the orthonormal vectors of the frame field, let's stick to this definition and apply it to the actual physical gyroscopes of the satellite Gravity probe B, my questions:

-are the gyroscopes locally non-rotating in the sense of the Fermi-Walker transport (so that they can only precess due to effects of the central mass/energy but not due to rotation of the source) or not?

-Can they be both locally rotating and non-rotating at the same time?

-If they are locally non-rotating in the Fermi-Walker sense (gyroscopically stabilized), how can they measure the rotating effects of frame-dragging?

-If they are not Fermi-Walker stabilized how can they measure the full de Sitter precession?