Kerr metric and rotating stars

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1. Jan 22, 2015

luinthoron

I have recently come across the notion that Kerr metric describes the spacetime outside a rotating black hole but not outside a rotating (electrically neutral) star. Unlike Schwarzschild metric, which works both for non-rotating spherically symetric black hole without charge as well as any other object with the same properties. Where does the difference come from? I feel it has something to do with moment of inertia of the spinning star's material but I'm not sure. I am looking forward to your replies.

2. Jan 22, 2015

jambaugh

I believe this "notion" is wrong. Keeping in mind that GR is a local theory the metric outside the event horizon of a spinning black hole has now way of knowing if what is inside (an arbitrary spherical boundary) is a spinning star or spinning black hole. All it "knows" is the neighboring stress-energy and that doesn't have an "eventually its a black hole" flag encoded in it. It only propagates information about energy-momentum flux. And spinning black holes and spinning stars have the same source effect on this.

3. Jan 22, 2015

Matterwave

The OP's statement is actually partially correct. The metric outside of a uniform, spinning object, is not necessarily the Kerr metric. This is due to the fact that there is no analogue of Birkhoff's theorem (which says basically that all spherically symmetric vacuum solutions of the Einstein Field Equations will be the Schwarzschild metric) for the Kerr metric. This is in addition to the fact that, unlike in the Schwarzschild case, nobody has yet found a compatible "inner Kerr metric" that describes a uniform spinning body and which matches smoothly with the Kerr metric at the boundary. However, I don't think there is a result that guarantees that the Kerr metric is NOT the metric outside of a uniform, spinning body either. Indeed, any metric outside of a uniform, spinning body must be approximately Kerr in nature. This is my understanding of the problem, feel free to correct me if I'm wrong. :)

4. Jan 22, 2015

Staff: Mentor

I don't have a reference handy (will try to look for one when I have more time), but I think it has been shown mathematically that the metric outside of a uniform spinning body must be asymptotically Kerr, i.e., it approaches the Kerr metric more and more closely as you get further and further away from the body.

Also, we know experimentally (from Gravity Probe B, for example) that outside a slowly rotating, weakly gravitating body (the Earth, in the case of Gravity Probe B), the metric is Schwarzschild plus the small correction terms characteristic of the Kerr metric (for example, frame dragging), and nothing else to the accuracy we have done experiments thus far, i.e, as far as we can tell the metric outside the Earth is what we would expect if the Kerr metric is the correct one for the vacuum region outside a uniformly rotating body, at least with slow rotation and weak gravity.

5. Jan 22, 2015

PAllen

I found the following which explores expected differences between vacuum outside compact astronomical bodies and the Kerr vacuum. The generalization of Kerr they use is the Hartle-Thorne metric. They define a dimensionless factor (quadropole moment * mass / angular momentum squared) which describes Kerr deviation (it is 1, in the units they use, for the Kerr vacuum). The Kerr factor can be substantially larger than 1 for some neutron stars.

http://arxiv.org/abs/1301.5925

Last edited: Jan 22, 2015
6. Jan 22, 2015

Staff: Mentor

Cool paper! Just a couple of comments: first, as I understand it, the deviations from the Kerr metric described in this paper for neutron stars are because they are not perfectly axisymmetric, the way a Kerr black hole is. So this paper doesn't directly address the theoretical question of whether the metric in the vacuum region outside a perfectly axisymmetric rotating body would be the Kerr metric.

Second comment: I don't think this metric is a generalization of the Kerr metric; I think it's the Schwarzschild metric plus extra terms due to (slow) rotation, but only up to a certain order of approximation, i.e., it's sort of in between the Schwarzschild metric (no rotation) and the Kerr metric (rotation included to all orders of approximation).

7. Jan 22, 2015

Matterwave

Yes, I think that's right. :)

8. Jan 22, 2015

PAllen

I disagree. There are no terms in the metric that depend on angular coordinates. This means there are no deviations from axial symmetry.

9. Jan 22, 2015

pervect

Staff Emeritus
I agree with PAllen - as far as I know, the quadrupole moment of, for example, the rotating Earth is due to the equatorial bulge, and does not require any deviations from axial symmetry.

I believe the quadrupole moment is typically called "J2", but I wasn't able to find a good source to refresh my recollections.

10. Jan 22, 2015

Staff: Mentor

I think what you mean is that there are none that depend on $\phi$, correct? There is a $\sin^2 \theta$ in one of the terms. and the $P_2$ factors depend on $\cos \theta$ (see below).

Yes, I see, since everything is a function of $r$ and $\theta$ only, there is an axial Killing vector field, $\partial / \partial \phi$, in addition to the timelike one.

Last edited: Jan 22, 2015
11. Jan 22, 2015

PAllen

Yes, sorry, I meant the axial angular coordinate.

The remaining, interesting, theoretical question is would there be any deviation from Kerr for a perfectly spherical rotating body, under any circumstances? Only theoretical, because any real body could not be that rigid - neutron stars are incredibly rigid by normal standards, but still have a bulge. In any case, the upshot of this paper is:

- for most practical purposes, Kerr is a good approximation
- the first order deviation from Kerr is due to quadrupole moment of a rotating body being different (in ratio with other parameters) as compared to a BH.

12. Jan 23, 2015

Staff: Mentor

The body would not be spherical, correct? In the perfectly idealized case, the body's surface would be an equipotential surface, which is an oblate spheroid, not a sphere. The Kerr metric would not be consistent with a perfectly spherical shape. See further comments below.

The rigidity of any body is limited by relativity; the sound speed can't be greater than the speed of light. So I don't think it's possible to have a body that's so rigid it resists centrifugal force and remains spherical when rotating.

(Of course, a body with a small enough mass can "resist" the forces that are trying to shape its surface into an equipotential surface, but that's true whether it's rotating or not. If we're talking about whether a rotating body can be spherical, we're presumably talking about a body with a large enough mass to force it into a spherical shape. But if its mass is large enough to do that, it's also presumably large enough to force the body's surface to be an equipotential surface, so if the body is rotating, it will have a bulge due to the rotation.)

13. Jan 23, 2015

PAllen

I see, and that is consistent with the finding that (for any equation of state considered by the paper), the Kerr factor of 1 is approached as compactness increases to BH. Thus, a Kerr solution incorporates a particular quadrupole moment (in relation to other parameters) representing this common limit of maximal compactness. For less compact bodies, the equation of state plays a role, and there are always small deviations from Kerr metric outside the body.

14. Jan 23, 2015

jambaugh

My apologies, I had not considered the multi-polar deviations from Kerr.

The question I would add to the mix is whether the equilibrium shape of a rotating star (discounting such things as internal convection currents) would most nearly conform to the source shape for a Kerr solution? Or deviate in a qualitatively substantial way?
Is there a specific set of values for say, viscosity and compressibility that would make this happen? Hmmm....

15. Jan 23, 2015

PAllen

For a star I can't point to specific calculations (I'm sure they exist, but I haven't searched). However, for neutron stars, the paper I cited above shows that the less compact (compactness being defined as the ratio of radius to BH radius for the same mass) the neutron star (correspondingly, the less massive it is within the allowed range of masses that neutron stars can occur in), the more its external vacuum solution deviates from Kerr. This was true for any of half a dozen candidate equations of state (and there was very little difference which was used). They also considered possible equations of state for hypothetical strange stars, finding different deviations as a function of compactness - but still the same qualitative statement. [I find it interesting that their results motivate BH has no hair via a very different route than the normal proofs - for any equation of state they considered, the deviation from Kerr vanishes as BH radius is approached].

Thus, I would guess that a star might have larger deviations from Kerr than a neutron star, since it is so much less compact.

Last edited: Jan 23, 2015
16. Jan 23, 2015

Staff: Mentor

Yes, that's how I understand it.

17. Jan 24, 2015

stevebd1

Thought this paper might be of interest as it features a Kerr interior metric (page 3)-

Rigid Rotation and the Kerr Metric
http://arxiv.org/abs/1404.5297v2

Abstract. The Einstein field equations have no known and acceptable interior solution that can be matched to an exterior Kerr field. In particular, there are no interior solutions that could represent objects like the Earth or other rigidly rotating astronomical bodies. It is shown here that there exist closed surfaces upon which the frame-dragging angular velocity and the red-shift factor for the Kerr metric are constant. These surfaces could serve as a boundary between rigidly rotating sources for the Kerr metric and the Kerr external field.​

18. Jan 24, 2015

PAllen

Unfortunately, as noted on p.4, the Burghardt interior metric is not acceptable as model of rotating, macroscopic body because it is singular. What you need, and what this paper does not provide, is an interior that is non-singular and satisfies the dominant energy condition (this last is required for the body to behave as expected for a classical body; the dominant energy condition is necessary and sufficient for the field equations to predict that a small body follows a timelike geodesic). Instead, picking up on an observation by Thorne, that while any realistic body (a star, a planet, not just a collapsing body) will deviate from Kerr near the body, there may be measure zero set of conditions for a body to actually match Kerr at the surface. This paper shows that it is possible to find a surface with the right properties for this to be so - but they do not exhibit any acceptable interior solution that achieves matching at such surfaces. Thus, the question of whether this measure zero set of conditions is empty or not, remains unsolved.

This paper also clarifies that since it is extra 'degrees of freedom' in how multipole moments relate to mass and angular momentum for real bodies that leads to near surface Kerr deviations for rotating bodies, you have a direct explanation for asymptotic approach to Kerr at a distance - multipole effects approach zero (quickly) with distance from a body.

[Edit: This paper also suggests my guess in post #15 is wrong. That for stars and planets (rather than neutron stars and hypothetical strange stars), the Kerr deviation should be vanishingly small. This is suggested by the table on p. 14, and calculations around it.]

Last edited: Jan 24, 2015