Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,354 answers , 22,792 comments
1,470 users with positive rep
820 active unimported users
More ...

  Conformal compactification of Kerr spacetime

+ 5 like - 0 dislike
3183 views

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
@IgorKhavkine and what about $h$? From you expression, it's what is "whatever else left" apart from the double null term, right? Do you assume any restrictions on $h$ or what it should look like?

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
and by the way, what about the case if $h=0$? would it work? The "attached" boundary is per definition $\Omega=0$ which should be a null hypersurface...

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
Just checking in here to point out a small fact which might shed some light on the matter. The diagrams in Figure 28 in Sec 5.6 of Hawking and Ellis' book mentioned by @IgorKhavkine are not (by the authors' own admission as one can see from the figure's caption) intended to capture the entire Kerr solution in the same way as the conformal diagrams of the R-N solution. What is being shown there, and in the corresponding figure in Reall's notes, is just a conformal diagram for the axis of symmetry of Kerr, which happens to be a truly two-dimensional, totally geodesic, submanifold!

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user Umberto Lupo
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
Most recent comments show all comments
ok, thank you @IgorKhavkine

This post imported from StackExchange MathOverflow at 2015-10-17 20:45 (UTC), posted by SE-user GregVoit
@IgorKhavkine question about your previous comment: you say one should bring the metric to the double null form. But why one would want that? For the "conf. compactification procedure" one needs a conformal factor, i.e. a positive function $\Omega^{2}$ such that $g_{new}=\Omega^{2}g_{old}$. How do double null coordinates help to find this $\Omega$?

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

3 Answers

+ 3 like - 0 dislike

The conformal completion of the Kerr space-time including a detailed discussion of the structure at $i^o$ can be found in Appendix  of A. Ashtekar and R.O. Hansen's paper J. Math. Phys. 19 (7) 1542 (1978)

answered Jan 21, 2019 by Abhay Ashtekar [ no revision ]
+ 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 (580 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 (1,645 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...