Locally non-rotating observers

[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.