5
$\begingroup$

I hope this question is not mistreated as a duplicate, because that is not my intention.

The Kerr metric suffices to describe the exterior solution to an axisymmetric spacetime mass distribution of mass $M$, but fails to be matched with an interior solution in a covariant form (several works have been done for a Kerr interior solution, however there is still no covariant line element for matching the solution in the boundary of the horizon event, without using scalar fields or any other formalism rather than the Einstein field equations).

For matching the interior solution of an object, like a star, rotating (with axisymmetrical symmetry) there is a not so famous solution, known as the Hartle-Thorne metric, which can be found in a handful of articles made by Bardeen, Hartle & Thorne in the late 60s (early 70s):

$$ds^2 = -e^{2\nu} c^2 dt^2 - 2f(r) \sin^2\theta \,dt d\phi + e^{2\lambda} dr^2 + e^{2\alpha}r^2(d\theta^2+\sin^2\theta d\phi^2),$$

for $\nu, \lambda, \alpha$ are all functions depending on $r$ and $\theta$ (Legendre polynomials $P_2(\cos\theta)$ if we stay at the first order in the expansion). However, this line element is slightly more complicated than Kerr's, and yet Kerr is not a party everybody wants to attend to.

So my question is, how far from reality I can treat the outside part of a star with Kerr? And how can I confront how much the Hartle-Thorne treatment deviates from Kerr's? How can I expand in power series the Hartle-Thorne solution to model, for example, the event horizon in that space time; so that it improves or even matches Kerr's horizon event?

$\endgroup$
5
  • 1
    $\begingroup$ "The Kerr metric suffices to describe the exterior solution to an axisymmetric spacetime mass distribution of mass M". Does it? $\endgroup$
    – ProfRob
    Dec 14, 2023 at 8:48
  • 3
    $\begingroup$ I think the gist of the question is "we know the metric outside rotating black holes, whose mass is compact enough to give them an event horizon, but what about a rotating star, which lacks one?" $\endgroup$
    – J.G.
    Dec 14, 2023 at 8:59
  • 1
    $\begingroup$ @ProfRob at least many people uses it for this endeavor, but my point saying that is that it seems to be sufficient, but maybe not the best (not necessary, if you want to be careful with the language). And actually Hartle Thorne is reduced after some choice of quadrupole to Kerr's metric, but my question goes beyond. $\endgroup$
    – omivela17
    Dec 14, 2023 at 9:36
  • $\begingroup$ @J.G. You got it. But Hartle Thorne, or at least the line element I wrote above, should be able to reproduce a horizon event similar to Kerr's (Schwarzschild's horizon event matches an external solution to the spherical symmetric line element which is both internal and external), photons surfaces, trapped surfaces for photons and all that. At least, I want to know how a star would look like and until where I can "compactify" one, after adding rotation. $\endgroup$
    – omivela17
    Dec 14, 2023 at 9:42
  • 1
    $\begingroup$ Have a look at section 2.8 of this living review. $\endgroup$
    – A.V.S.
    Dec 14, 2023 at 12:04

1 Answer 1

3
$\begingroup$

I suspect that exactly what you want was published by Frutos-Alfaro (2019) in the form of a perturbative approach to the Kerr metric that includes terms due to a mass quadrupole, spin octopole and mass hexadecapole. They refer to this as an improved Hartle-Thorne metric, since that metric only included a mass quadrupole.

Possibly even more useful is the work by Pappas (2017), which similarly provides a perturbative expansion beyond the Kerr metric that includes the mass quadrupole, spin octopole and mass hexadecapole but then also compares the metric to the results of numerical relativity calculations and the "vanilla" Hartle-Thorne metric (see their Fig.2).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.