Mach's principle, GR and the nature of space

[epovo]

I have been puzzled by this for years, so I would welcome some enlightenment.
It seems that Einstein was enamored with Mach's principle while searching for GR, but in the end GR does not seem compatible with it - or rather has nothing to say about it.
What I mean is that the proverbial spinning skater's arms are pulled out because rotation is absolute even for GR even if the skater is far away from any star. Where as Mach conjectured that it is the presence of the rest of the universe (however far it may be) that confers some meaning to the notion of rotation and therefore produces the pull of their arms, it seems that this idea is now abandoned. I tend to think that Mach's principle came from the notion that empty space is nothingness. This is not the prevalent idea now. Space seems to be a bubbly thing after all, and filled with the Higgs field. Is this the thing against which rotation happens? What is the current thinking about this?
Thank you.

 

[WannabeNewton]

Some of the ideas in Mach's principle are compatible with GR but most aren't. The idea that matter "content" everywhere in the universe affects the local laws of physics is at the heart of GR and probably can't be made more evident than from the effects of frame dragging, the Lense Thirring Precession, and the definition of "locally non-rotating" observers in GR. However as noted there are many aspects of Mach's principle which are not compatible with GR so GR is for the most part non-Machian.

This is one of my most favorite papers on the topic and in the field of gravitation as a whole: http://blogs.epfl.ch/document/19038

 

[Bill_K]

I have been puzzled by this for years, so I would welcome some enlightenment.
It seems that Einstein was enamored with Mach's principle while searching for GR, but in the end GR does not seem compatible with it - or rather has nothing to say about it.
What I mean is that the proverbial spinning skater's arms are pulled out because rotation is absolute even for GR even if the skater is far away from any star. Where as Mach conjectured that it is the presence of the rest of the universe (however far it may be) that confers some meaning to the notion of rotation and therefore produces the pull of their arms, it seems that this idea is now abandoned. I tend to think that Mach's principle came from the notion that empty space is nothingness. This is not the prevalent idea now. Space seems to be a bubbly thing after all, and filled with the Higgs field. Is this the thing against which rotation happens? What is the current thinking about this?
Thank you.

epovo, I can't speak for the current thinking, only my own opinion, and I largely agree with you, only even moreso. Mach's principle was a vague pre-Einsteinian notion, best forgotten! Whatever intuitive appeal it once had, Mach's principle is inconsistent with much of physics, and certainly inconsistent with General Relativity, which gives an absolute definition for the local nonrotating rest frame that's independent of what the "distant stars" are doing.

 

[WannabeNewton]

Bill, while I agree the so mentioned rotation is absolute, how is the definition of locally non-rotating not affected by the matter content in space-time? The family of locally non-rotating observers external to a static spherically symmetric mass is not the same as the family of locally non-rotating observers external to a rotating spherically symmetric mass.

 

[PeterDonis]

General Relativity, which gives an absolute definition for the local nonrotating rest frame that's independent of what the "distant stars" are doing.

Does it? In order to specify such a frame, you have to know the metric. In order to know the metric, you have to solve the EFE. In order to solve the EFE, you have to know the stress-energy tensor--or else you have to at least have an ansatz for one. Or, I suppose, you could work backwards, starting with an ansatz for the metric, computing its Einstein tensor, and multiplying that by $8 \pi$ and calling it the stress-energy tensor of your solution. (Which, of course, raises the question of how physically reasonable the resulting SET is, but I don't think we need to open that can of worms here. )

But in any case, you always have *some* specification of the matter-energy content of the spacetime, the SET, associated with your metric, and therefore associated with your definition of locally nonrotating frames. Which, to me, says that GR does *not* give an absolute definition of local nonrotating frames that is independent of what the "distant stars" are doing, because what the distant stars are doing is part of the overall SET of the spacetime, and therefore part of the overall solution from which your definition of local nonrotating frames is derived.

Does this mean that Mach's Principle *is* part of GR after all? IMO it depends on what you think "Mach's Principle" says. I don't really think it's a very important question, since the answer to it doesn't affect any predictions of GR.

 

[PeterDonis]

The family of locally non-rotating observers external to a static spherically symmetric mass is not the same as the family of locally non-rotating observers external to a rotating spherically symmetric mass.

Even more than that, the definition of locally non-rotating observers in an asymptotically flat spacetime (which both of your examples are) is different from that in a spacetime that isn't, such as an FRW spacetime. Asymptotic flatness itself is an assumption about the global matter-energy content of the spacetime--namely, that there isn't any other than the isolated body in the center, whether it's rotating or not.

 

[Bill_K]

how is the definition of locally non-rotating not affected by the matter content in space-time?

Not what Mach said. Here's what Mach said:

“It does not matter whether we think of the Earth rotating around its axis, or if we imagine a static Earth with the celestial bodies rotating around it.”

The Kerr metric does show that matter is affected by other matter, but you have to be careful what conclusions you draw from that. Orbits around Kerr exhibit "frame dragging", which means for example that even the orbit with angular momentum zero still "goes around" the central body. Closer to the issue is the gyroscopic precession of a Fermi-Walker transported particle (I have a blog post on this) And it happens not just for Kerr - not even just for Schwarzschild - it's a general feature of GR. So to say that rotation of the Kerr object is responsible ignores the other causes.

In place of Kerr, there's another GR solution by Thirring more relevant to the Mach issue. In linearized theory one can calculate the field inside a large, slowly rotating shell, mass M, radius R, angular velocity ω, and find there's inertial dragging inside: Ω ~ Mω/R. But notice what I said: inside a rotating shell. The rotation of the shell is absolute and can be detected and measured. So again this is all very interesting, masses do have an effect on other masses, but it's not what Mach said.

 

[Bill_K]

Even more than that, the definition of locally non-rotating observers in an asymptotically flat spacetime (which both of your examples are) is different from that in a spacetime that isn't, such as an FRW spacetime. Asymptotic flatness itself is an assumption about the global matter-energy content of the spacetime--namely, that there isn't any other than the isolated body in the center, whether it's rotating or not.

I'm not sure I see the point. FRW itself is nonrotating, and won't affect the locally nonrotating observers. There are rotating cosmologies, generalizations of FRW in fact, in which the neighbors of every "galaxy" rotate about it, but they're not needed for the observations.

 

[WannabeNewton]

Hi Bill, thanks for the response. Do you remember this problem you helped me with some while back: https://www.physicsforums.com/showthread.php?t=675475?

It was based on this problem from Wald: http://postimg.org/image/6m116zzl9/
In particular that very last sentence. Would you say you agree or disagree with it? This is how I intepreted Mach's principle with relation to what I said above.

 

[Bill_K]

Yeah, this is the Thirring solution. I'd only disagree where he says, "in a manner in accord with Mach's principle." Mach does not deserve credit every time something rotates!

In particular, note that for this solution the precession rate is much smaller than the rotation rate of the sphere, Ω << ω. Mach's Principle would have us believe that Ω = ω.

 

[WannabeNewton]

Right so I'm wondering in what way Wald is saying this effect is in accordance with Mach's principle? It just seems like a very loose association based on the idea that the central observer precesses due to the spinning of the thin shell which changes which family of observers are locally non-rotating. In other words, when saying such effects are "in a manner in accord with Mach's principle", how seriously is Wald interpreting the statements of Mach and is such a loose interpretation even valid? As you say it cannot be true word for word since the effect is not of the same order as the rotation of the shell itself!

P.S. he does it again in a related (albeit much easier) problem later on in the text: http://s18.postimg.org/bb2fqvf09/mach_2.png

 

[PeterDonis]

FRW itself is nonrotating, and won't affect the locally nonrotating observers.

Sure it does. It makes them nonrotating in the same sense that "FRW itself is nonrotating". If the metric were not FRW but something different, the definition of locally nonrotating observers would also be different.

There are rotating cosmologies, generalizations of FRW in fact, in which the neighbors of every "galaxy" rotate about it, but they're not needed for the observations.

Meaning, they don't match observations. Agreed. But that just proves my point: if the metric were different, observations, including the observations that determine our definition of "locally nonrotating observers" would be different.

Once again, I'm not saying this means Mach's Principle is correct; I think that depends on what you mean by "Mach's Principle". I'm just saying that it makes no sense to me to say that the definition of anything that is determined by the metric, including the definition of "locally nonrotating observers", is "absolute", independent of the rest of the matter in the universe, since the metric is not absolute--different solutions of the EFE lead to different metrics.

 

[WannabeNewton]

But in what sense is the world "absolute" being used here? If we go with Wald's definition that locally non-rotating observers in an axisymmetric stationary space-time are ones who follow orbits of $\nabla^{a} t$ (in the coordinates adapted to the killing vector fields) then all observers at a given event will unambiguously agree on who is locally non-rotating and who isn't because they can measure their own angular momentum $L = u^{a}\psi_{a}$ (where $\psi^{a}$ is the axial killing vector field) and if $L = 0$ then they know they have $u^{a} = \nabla^{a}t/(-\nabla^{b}t\nabla_{b}t)^{1/2}$ and if $L \neq 0$ then they know they aren't following such an orbit. However it is certainly true that the matter content affects the local standard of non-rotation since $\nabla^{a}t$ depends on the metric which itself depends on the matter content so different matter content will yield different families of locally non-rotating observers (ZAMOs in this case). For example in Schwarzschild space-time the ZAMOs will necessarily be the static observers who follow an orbit of the time-like killing vector field $\xi^{a}$ whereas in Kerr space-time the ZAMOs and the static observers will not be the same.

I'm assuming Wald is loosely interpreting Mach's principle to mean the second part regarding the matter content (based on the way he states it in the two exercises I linked above) whereas Bill's quote from Mach is saying something much different (at least from the way I interpret it) and something that isn't really in accord with the fact that observers can unambiguously agree on who is locally non-rotating, within the framework of GR.

 

[pervect]

My position isn't quite as extreme as Bill K's, but it's close.

I'm sure Mach's principle was inspirational, but I haven't seen much actually useful come directly from it - by useful, I mean experimental predictions based on the principle itself. I'll agree that historically it was important in inspiring GR (but that's not the standard by which I"m attempting to judge it, I'm attempting to find specific experimental predictions from the principle itself).

On the other hand, I haven't looked carefully, either!

The topic is apparently interesting enough that Julian Babour had a conference about it, which inspired a book. Mach's Principle: From Newton's Bucket to Quantum Gravity (Einstein Studies). Which, I might add, I haven't read.

Without a better grounding in the literature on this particular topic, I can't say more.

I suppose I can say a bit about my opinion on its usefulness in introductory GR courses. That opinion would be that it is not a good topic for introductory courses.

 

[yenchin]

See also https://www.physicsforums.com/showthread.php?t=505896

 

[Dale]

Once again, I'm not saying this means Mach's Principle is correct; I think that depends on what you mean by "Mach's Principle".

That is my big problem with Mach's principle: what does it mean experimentally? You talk about the equivalence principle or the principle of relativity and you can get pretty concrete testable statements, but Mach's principle usually reduces to untestable statements about how an "otherwise empty" universe would behave.

As a result, I don't usually care too much about Mach's principle. It isn't sufficiently well-defined to determine if it is correct or not.

 

[Bill_K]

As a result, I don't usually care too much about Mach's principle. It isn't sufficiently well-defined to determine if it is correct or not.

Here's another quote:

"When a body moves relatively to the fixed stars, centrifugal forces are produced; when it moves relatively to some different body, and not relatively to the fixed stars, no centrifugal forces are produced. I have no objection to calling the first rotation "absolute" rotation, if it be remembered that nothing is. meant by such a designation except relative rotation with respect to the fixed stars."

Nothing vague about that! The local nonrotating frame is the frame in which the distant stars do not rotate. Period, end of sentence.

And thus both the Thomas precession and the Lense-Thirring effect already violate Mach's Principle.

 

[Dale]

In which case Mach's principle is experimentally falsified, and I care about it even less!

I still think that is vague, mostly because there are no fixed stars. They all move relative to each other. But that is the most concrete statement I have seen. I think the "fixed stars" ambiguity can be resolved.

 

[WannabeNewton]

The local nonrotating frame is the frame in which the distant stars do not rotate. Period, end of sentence.

And thus both the Thomas precession and the Lense-Thirring effect already violate Mach's Principle.

Why does the second sentence follow from the first necessarily? The non-vanishing of the induced coordinate angular velocity of the locally non-rotating observer which would make the distant stars look like they were rotating about him/her from his/her viewpoint, according to Mach?

 

[PeterDonis]

And thus both the Thomas precession and the Lense-Thirring effect already violate Mach's Principle.

For extra fun, in at least one book by prominent relativists (Cuifolini and Wheeler's Gravitation and Inertia), exactly the opposite claim is made: effects like Thomas precession and Lense-Thirring *support* Mach's principle, because they show that the rotation of nearby matter *does* have an effect on inertia. Obviously they are using a different definition of "Mach's Principle" than Mach himself appeared to use!

 

[WannabeNewton]

All these different interpretations of Mach's principle are making my head hurt T_T. Yes it seems Cuifolini and Wheeler are interpreting it in the same way Wald is interpreting it because he says the exact same thing in his text.

 

[epovo]

I think we should remember that when Mach wrote empty space was nothingness. The same is true for GR. Things have changed. Space has been found to be something and presumably something that does not interfere with objects moving through it uniformly, but which interacts with stuff accelerating wrt to it. Can that be the origin of inertia? Isn't the Higgs field something like this?

 

[Nabeshin]

For extra fun, in at least one book by prominent relativists (Cuifolini and Wheeler's Gravitation and Inertia), exactly the opposite claim is made: effects like Thomas precession and Lense-Thirring *support* Mach's principle, because they show that the rotation of nearby matter *does* have an effect on inertia. Obviously they are using a different definition of "Mach's Principle" than Mach himself appeared to use!

It can best be understood thusly: Much in the sense of rectilinear inertia, objects do not 'want' to rotate with respect to one another. The bulk of the mass of the universe being made up of the 'fixed stars' (distant cosmic objects), the locally inertial frames are the ones which don't rotate with respect to these (since the fixed stars are the primary determiners of inertia).

However, a local object like a rotating body does get a small say in the matter, since it would 'like' for the local frames not to be rotating with respect to it. This contribution can be interpreted as the Lense-Thirring effect, and it's this interpretation which some find consistent with Mach.

(Note: Of historical interest, it's not clear exactly what Mach meant by his original quotations. A thorough reading seems to suggest he is suggesting nothing like the modern Mach's principle of Einstein or others, but rather that one cannot and should not speculate about circumstances beyond the realm of current experimental science. He believed science's goal was to provide as compact as possible an explanation for all known observational results, not really to predict new ones. In this sense, he did not believe in atoms because they were not an 'established observational result', although they had tremendous predictive power.)

 

[johne1618]

As far as I understand it Einstein described gravitation as an inertial effect, using the equivence principle, whereas Mach was seeking to describe inertia as a gravitational effect.

They were attacking the same problem from opposite points of view so perhaps that is why it is hard to derive Mach's principle from an Einsteinian starting point.

 

[Comradez]

Here's a thought-experiment that really has me stumped:

Imagine an Earth/moon system in isolation from the rest of the universe. No other matter or energy exists.

From the moon's vantage point, would it not be reasonable to conclude that the moon is stationary and not spinning, and that the Earth is stationarily spinning some distance apart?

From the Earth's vantage point, would it not be reasonable to conclude that the Earth is both stationary and not spinning, but rather, the moon is in orbit and is tidally rotating with one side always facing Earth?

How would one distinguish which case was actually occurring? Would coriolis effects still occur on the Earth without the reference of the fixed stars? Would they still occur on the moon? If so in both cases, then what are they each simultaneously rotating relative to?

 

[Nabeshin]

Here's a thought-experiment that really has me stumped:

Imagine an Earth/moon system in isolation from the rest of the universe. No other matter or energy exists.

From the moon's vantage point, would it not be reasonable to conclude that the moon is stationary and not spinning, and that the Earth is stationarily spinning some distance apart?

From the Earth's vantage point, would it not be reasonable to conclude that the Earth is both stationary and not spinning, but rather, the moon is in orbit and is tidally rotating with one side always facing Earth?

How would one distinguish which case was actually occurring? Would coriolis effects still occur on the Earth without the reference of the fixed stars? Would they still occur on the moon? If so in both cases, then what are they each simultaneously rotating relative to?

A Machian claims that if there's truly nothing else in the universe, the situation is equivalent to the two bodies simply being at rest. No rotation, Coriolis, or other effects occur.

 

[PeterDonis]

From the moon's vantage point, would it not be reasonable to conclude that the moon is stationary and not spinning, and that the Earth is stationarily spinning some distance apart?

It depends on what you mean by "stationary". You could certainly set up a coordinate chart on this spacetime in which the Moon was at rest and the Earth was not, but coordinates don't necessarily represent the physics.

From the Earth's vantage point, would it not be reasonable to conclude that the Earth is both stationary and not spinning, but rather, the moon is in orbit and is tidally rotating with one side always facing Earth?

You could certainly set up a different coordinate chart in which the Earth was at rest and the Moon was not, but again, coordinates don't necessarily represent the physics.

How would one distinguish which case was actually occurring?

It depends on what you mean by "actually". Both coordinate charts that I described above would be valid charts and could, in principle, be used to analyze the system. But there would also be lots of other charts that could be so used.

OTOH, one could look at physical invariants of the spacetime instead of coordinates to see what is "actually" going on; if we do that, then neither of the cases you describe is what is "actually" occurring. If we neglect the emission of gravitational waves by the system, the spacetime as a whole is stationary--it has a timelike Killing vector field, which is an invariant geometric feature. But neither the Earth's worldline nor the Moon's worldline will be an integral curve of that timelike KVF, so neither one will be "stationary" in this sense. What will be "stationary" in this sense is the center of mass of the Earth-Moon system; indeed, the natural coordinates adapted to the system's timelike KVF are usually called "center of mass coordinates".

Would coriolis effects still occur on the Earth without the reference of the fixed stars? Would they still occur on the moon?

Yes; at least, yes if we require that the spacetime as a whole is a solution to the Einstein Field Equation. The only way I know of to construct such a solution is to make the spacetime asymptotically flat; see below for what that implies.

If so in both cases, then what are they each simultaneously rotating relative to?

The asymptotically flat background that is required, as above, for the spacetime as a whole to be a solution of the EFE. In other words, it is not really possible to have a spacetime that is a solution to the EFE but that only has the Earth and Moon in it, and "nothing else". There also has to be a boundary condition, which is what the asymptotic flatness is. (Actually, one could also impose other boundary conditions, but asymptotic flatness is the simplest one, and there are already enough complications in this scenario without me bringing in new ones. )

 

[PeterDonis]

A Machian claims that if there's truly nothing else in the universe, the situation is equivalent to the two bodies simply being at rest. No rotation, Coriolis, or other effects occur.

But such a situation would not be a solution of the Einstein Field Equation: there is no solution of the EFE that has the Earth and Moon both remaining at rest relative to each other for all time, without moving. At least, not if the Earth or the Moon has nonzero mass.

 

[WannabeNewton]

If the space-time is indeed asymptotically flat, and on top of that axisymmetric (has axial KVF $\psi^{a}$) we have an absolute invariant of the space-time that corresponds to (in most cases) the total angular momentum of the space-time, given by $J = \frac{1}{16\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\psi^{d}$ where $S$ is a topological 2-sphere taken in the vacuum region; this is akin to the Komar mass. It is, as mentioned, an absolute quantity and can certainly distinguish truly isolated static configurations from truly isolated non-static ones (static meaning the space-time has a hypersurface orthogonal time-like KVF) without reference.

One can also define such an invariant corresponding to total angular momentum for non-axisymmetric (but still asymptotically flat) space-times but it is much less trivial.

 

[Mentz114]

If the space-time is indeed asymptotically flat, and on top of that axisymmetric (has axial KVF $\psi^{a}$) we have an absolute invariant of the space-time that corresponds to (in most cases) the total angular momentum of the space-time, given by $J = \frac{1}{16\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\psi^{d}$ where $S$ is a topological 2-sphere taken in the vacuum region; this is akin to the Komar mass.
...

Sorry if this is a dumb question, does the anti-symmetrization here $\epsilon_{abcd}\nabla^{c}\psi^{d}$ mean that the covariant derivative can be replaced with partial derivatives ?

Presumably the Minkowski space-time in the cylindrical chart has the necessary KVFs, but zero ang. momentum ?

 

[PeterDonis]

on top of that axisymmetric (has axial KVF $\psi^{a}$)

This wouldn't actually be true of the isolated Earth-Moon case, or indeed of any case involving two isolated bodies orbiting their mutual center of mass. You would need something like a rotating fluid.

 

[WannabeNewton]

Sorry if this is a dumb question, does the anti-symmetrization here $\epsilon_{abcd}\nabla^{c}\psi^{d}$ mean that the covariant derivative can be replaced with partial derivatives ?

Not in general, no. When you evaluate the expression in a coordinate basis what you end up calculating is $\epsilon_{\mu\nu\alpha\beta}g^{\alpha \sigma}g^{\beta \gamma}\nabla_{\sigma}\psi_{\gamma}$; the christoffel symbols are certainly important in the calculation. For example the Komar mass is defined in an extremely similar way as $M = -\frac{1}{8\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d}$. In the coordinates adapted to the time-like KVF $\xi^{a}$, its coordinate representation is given by $\xi^{\mu} = \delta^{\mu}_{t}$ hence $\nabla^{\nu}\xi^{\mu} = g^{\nu\alpha}\Gamma _{\alpha t}^{\mu}$. If you had on the other hand just used partials then $\partial^{\nu}\xi^{\mu} = g^{\nu\alpha}\partial_{\alpha}\delta^{\mu}_{t} = 0$ which is of course not a general result.

Presumably the Minkowski space-time in the cylindrical chart has the necessary KVFs, but zero ang. momentum ?

Yes the so defined angular momentum is coordinate independent as you can see (one can show that Riemann and Lebsgue integrals of differential forms over smooth manifolds are coordinate independent). In fact one can show using Stoke's theorem that $J = \frac{1}{16\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\psi^{d} = -\int _{\Sigma}T_{ab}n^{a}\psi^{b}dV$ where $\Sigma$ is a space-like hypersurface chosen so that $\psi^{a}$ is tangent to $\Sigma$, $S = \partial \Sigma$, and $n^{a}$ is the unit outward normal to $\Sigma$. Now for any static axisymmetric space-time we have that the time-like KVF $\xi^{a}$ is normal to some family of space-like hypersurfaces $\Sigma$ and that $\xi^{a}\psi_{a} = 0$ hence $\psi^{a}$ lies tangent to such $\Sigma$.

Choosing $n^{a} = \alpha \xi^{a}$ as our outward normal then, in the coordinates adapted to the KVFs we have $T_{\mu\nu}\xi^{a}\psi^{b} = T_{0\phi}$. Now we know that in these coordinates, $g_{0\phi} = 0$ and that $R_{0\phi} = 0$ hence by Einstein's equations $T_{0\phi} = 0$ thus $J = 0$. This is a covariant result so it holds for all coordinate systems. This includes Minkowski space-time and Schwarzschild space-time.

On the other hand, for stationary (but not necessarily static!) axisymmetric space-times, $J \neq 0$ in general. For example $J = Ma$ for the Kerr space-time of a completely isolated rotating spherically symmetric body; again $J$ is an unambiguous quantity.

 

[Mentz114]

Thanks for a complete answer, which will take me some time to verify. Right now, I don't believe you

 

[Nabeshin]

But such a situation would not be a solution of the Einstein Field Equation: there is no solution of the EFE that has the Earth and Moon both remaining at rest relative to each other for all time, without moving. At least, not if the Earth or the Moon has nonzero mass.

Yes, GR is decidedly non-Machian from this point of view.

Edit: In a fully Machian theory, the two bodies would likely just fall together at some (rapid) rate under their mutual gravitational pull. So they're not truly stationary. Regardless, GR doesn't describe this (the rate would be much much bigger than a GW-merger timescale).

 

[Nabeshin]

If the space-time is indeed asymptotically flat, and on top of that axisymmetric (has axial KVF $\psi^{a}$) we have an absolute invariant of the space-time that corresponds to (in most cases) the total angular momentum of the space-time, given by $J = \frac{1}{16\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\psi^{d}$ where $S$ is a topological 2-sphere taken in the vacuum region; this is akin to the Komar mass. It is, as mentioned, an absolute quantity and can certainly distinguish truly isolated static configurations from truly isolated non-static ones (static meaning the space-time has a hypersurface orthogonal time-like KVF) without reference.

One can also define such an invariant corresponding to total angular momentum for non-axisymmetric (but still asymptotically flat) space-times but it is much less trivial.

Do you know how this relates to the calculation of the Chern-Pontryagin invariant? I was given the impression the CP invariant also tells you about the 'spin' in a spacetime (region). However, I don't entirely understand it outside the trivial case of kerr, and know it (oddly enough) produces zero for the godel universe.