Locally non-rotating observers

[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?

 

[Bill_K]

-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?

They're locally nonrotating, which by definition means Fermi-Walker transported. If you read the blog I referred to, which discusses all three effects together, you won't need to ask silly questions like this.

 

[TrickyDicky]

They're locally nonrotating, which by definition means Fermi-Walker transported. If you read the blog I referred to, which discusses all three effects together, you won't need to ask silly questions like this.

Bill, by my experience silly questions are usually the best in science.
I wasn't able to elucidate this by reading your blog, that's why I asked. And you didn't answer my third question, can you?

 

[Bill_K]

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

Ok, here's a description of Gravity Probe B. Measurement comes from comparison of the locally nonrotating frame as determined by the onboard gyroscopes versus the "fixed stars". It's the relative motion that counts - whether you regard the gyroscopes as fixed (F-W frame) and the stars as precessing, or that GR (frame dragging) causes the gyroscopes themselves to precess doesn't matter.

 

[TrickyDicky]

Measurement comes from comparison of the locally nonrotating frame as determined by the onboard gyroscopes versus the "fixed stars".

Right, this gives us the de Sitter precession but again, locally non-rotating according to Fermi-Walker transport by definition (according to Wikipedia) "defines a reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame." How can it measure frame-dragging due to rotation of the frame?

It's the relative motion that counts - whether you regard the gyroscopes as fixed (F-W frame) and the stars as precessing, or that GR (frame dragging) causes the gyroscopes to precess doesn't matter.

Well, this is fine and doesn't affect the issue at hand, but I have heard you say a few times that rotation is absolute and independent of the distant stars. Now you are Machian suddenly?

 

[Bill_K]

Right, this gives us the de Sitter precession but again, locally non-rotating according to Fermi-Walker transport by definition (according to Wikipedia) "defines a reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame." How can it measure frame-dragging due to rotation of the frame?

The point my blog is supposed to illustrate, even skipping the mathematics, is that all three precessions are aspects of one phenomenon. You can illustrate it using a path in flat space, or in Schwarzschild, or in Kerr. You can point to terms in the result and call them de Sitter or whatever, but the distinction is merely circumstantial.

Well, this is fine and doesn't affect the issue at hand, but I have heard you say a few times that rotation is absolute and independent of the distant stars. Now you are Machian suddenly?

The local nonrotating frame depends only on the particle's world line. I only meant to say that Gravity Probe B measures the difference between the two frames.