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Conformal compactification of Kerr spacetime

+ 5 like - 0 dislike
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I'm looking for a book/paper where the conformal compactification of Kerr spacetime is calculated. I've seen plenty of reference for the Minkowski, but none (explicitly calculated) for Kerr.

Thank you.

PS: prior to writing this post, I was already referred to "Large scale structure of spacetime" by Hawking and Ellis, but there they only discuss the maximal extension of Kerr (and other solutions), not conformal compactification calculation.

EDIT on what I meant by "compactification for Minkowski": The Minkowski metric is

$ds^{2}=-dt^{2}+dr^{2}+r^{2}d\Omega^{2}$

where $-\infty<t<\infty$ and $0\leq r<\infty$. (We now want to get the coordinates with finite ranges)

First go to the double-null coordinates $u=t-r$ and $v=t+r$, and then change to $T=\arctan u+\arctan v$, $R=\arctan v-\arctan u$. We then obtain

$ds^{2}=\omega^{-2}(T,R)(-dT^{2}+dR^{2}+\sin^{2}Rd\Omega^{2})$

with $\omega(T,R)=2\cos U \cos V=\cos T+\cos R$.

Therefore, the original metric is related to the new one by (explicitly given) conformal transf. $\omega^{2}$ as

$d\tilde{s}^{2}=\omega^{2}(T,R)ds^{2}=-dT^{2}+dR^{2}+\sin^{2}R d\Omega^{2}$

where $R,T$ have the finite ranges $0\leq R<\pi$ and $R-\pi <T <\pi -R$.

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
asked Oct 12, 2015 in Theoretical Physics by GregVoit (115 points) [ no revision ]
retagged Oct 17, 2015
Most voted comments show all comments
What do you mean by "calculate the conformal compactification"? Let $(M,g)$ be any asymptotically flat space time and let $(M,\Omega^2 g)$ be a conformal compactification, then for any function $f:M\to \mathbb{R}$ that is bounded both above and below, $(M,f\Omega^2 g)$ is another conformal compactification. Can you edit to tell us what you think is the answer for Minkowski, and what sort of information you want to get from such a computation?

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Willie Wong
@WillieWong I added more information to the question now.

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
@IgorKhavkine yes, I've seen the diagram in their book. That was also why I decided to ask the question here because I couldn't follow (at least without external help) the "methods similar to R-N solution". I thought maybe someone here knew where I could find the double-null form of the Kerr metric and then see what the conformal factor is.

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
Hmm, I guess the literature is really encouraging you to go through the exercise yourself. :-) Still doesn't have the explicit answer, but these notes on Black Holes by Harvey Reall contain some more detail about how the Reissner-Nordstrom case works.

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Igor Khavkine
ok, thank you @IgorKhavkine

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
Most recent comments show all comments
This is quite different from the situation in Reissner-Nordström where, thanks to the spherical symmetry of the solution, one can rather easily capture the entire (four-dimensional) spacetime by means of a Penrose diagram in which every point represents a two-sphere. So one might wonder how much Figure 28 in Hawking and Ellis' book actually tells us about the global conformal structure of the Kerr spacetime. If you are somewhat flexible about what you'd like your (conformal) diagram of Kerr to do then consider consulting Section 3.7 in Piotr Chruściel's notes...

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Umberto Lupo
... where "projection diagrams" are defined and discussed (see also this paper). The remarks towards the end of page 146 there are particularly relevant

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Umberto Lupo

2 Answers

+ 2 like - 0 dislike

See Section 8 of Pretorius and Israel, "Quasi-spherical light cones of the Kerr geometry". http://arxiv.org/abs/gr-qc/9803080 The paper contains quite a bit more of course. But I want to point out that in the expression you find there, $R^2$ is fairly reasonable as a quantity in the usual (say, Boyer-Lindquist) coordinates. But the function $r_*$ is quite difficult to get a handle on.

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Willie Wong
answered Oct 13, 2015 by Willie Wong (570 points) [ no revision ]
+ 2 like - 0 dislike

The full and slightly tedious 4D Kruskalization of Kerr  is given in Boyer & Lindquist (1967). Many lectures/lecture notes on the topic (such as these) give only a Kruskalization along the axis because that allows one to map all the accessible regions without having to care about the extra $\theta$-dependence and the $g_{t \phi}$ component of the metric.

The point of the axis-Kruskalization is, as in the case of Reissner-Nordström/Schwarzschild, to introduce 

  1.  a tortoise radial coordinate $r^*$ which makes $g_{r^*r^*}=-g_{tt}$,
  2. a set of natural "outgoing" and "ingoing" null coordinates $u,v=r^* \pm t$ and regularizing them by a light-cone reparametrization.

After Kruskalization, the compactification of coordinates is trivially done via an $\arctan$ of the light-like coordinates.


Note however, that these "canonical" compactifications are degenerate at null-infinity, i.e., the tangent space is destroyed at the boundary of the diagrams. One of the results is that they do not reduce to any conformal compactification of the Minkowski space-time at null infinities even though they should.

A satisfying conformal compactification of the Schwarzschild space-time (along with new regions beyond null infinity!) has been published only a year ago by Haláček & Ledvinka. I don't believe such "good" compactifications are known explicitly for the Kerr space-time. This may be only because nobody really needed them so far, the methods of Haláček & Ledvinka seem straightforward, the construction is just going to be rather toilsome for Kerr.

answered Oct 18, 2015 by Void (1,505 points) [ no revision ]

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