There are uniqueness theorems that classify Black holes according to its mass, angular momentum and charge. One of the theorem is Carter-Robinson theorem which has many assumptions and then it says axis-symmetric and stationary black holes are kerr black holes which depends only on two parameters. One of the assumption is that "there is no closed causal curve in the domain of outer communication". Please explain why this assumption is necessary or where can i find this reasoning?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
One of the assumption is that "there is no closed causal curve in the domain of outer communication".
Note, that e.g. in Robinson, 1975 the assumption is the absence of timelike curve which is a weaker requirement (causal curve can also be null) and it is easier to see the reasoning for this.
… why this assumption is necessary ?
Because if there is a closed timelike curve in the domain of outer communication (basically, the outside of past and event horizons) of a stationary spacetime then this domain is one vicious set, i.e. a set where there are both future and past directed timelike curves between any two events of the set. (This term “vicious set”, due to Carter, generalizes the notion of CTC, which could be called vicious loop).
Let us start with a closed timelike curve, disconnect it at some point and “wiggle” its start and end points a little. As a result, among those close curves there would necessarily be curves going back in time (as measured by clock of asymptotic observer). And due to stationary nature of spacetime we can connect this curve to its copy but translated back in time obtaining a spiral going backwards in time. So, there could be no meaningful global chronology in such spacetime, since an experimenter starting anywhere, could approach the region with CTCs, enter the trajectory going backwards in time, stay on this trajectory for large enough number of loops and afterwards fly to an arbitrary event in the (coordinate) past.
Examples of spacetimes with such property are overextreme Kerr (and overextreme Kerr–Newman) solutions. These spacetimes do not have horizons so the entire spacetime is a vicious set. In underextreme Kerr ($a<M$) the vicious set is the region $r\le r_{-}$ ($r_{-}$ is the inner horizon radius) so outside of the black hole there is normal causality.
… where can i find this reasoning ?
The original paper discussing causality in Kerr family of solutions as well as general criteria for vicious sets in stationary axisymmetric spacetimes is Carter, 1968.
-
$\begingroup$ can we say that we are avoiding causal closed curves so we can avoid time travel concept ??? but I don't see any relation with this uniqueness theorem !! $\endgroup$ Commented May 2 at 18:30
-
$\begingroup$ Not merely avoiding time travel as such but requiring well posedness of the usual physics outside bh (formulated e.g. as various Cauchy problems). $\endgroup$– A.V.S.Commented May 2 at 18:57
-
$\begingroup$ when you say "outside black hole", do you mean "outside event horizon of Black Hole ? $\endgroup$ Commented May 3 at 7:59
-
$\begingroup$ Yes. Or we can call this “domain of outer communication”. $\endgroup$– A.V.S.Commented May 3 at 11:22
-
$\begingroup$ Just one more question. Israel didn't assume this condition in his Israel's uniqueness theorem and the reason I found is because this theorem is about static Black Holes. Can you explain why there are no closed causal curves in static spacetime ?? $\endgroup$ Commented May 4 at 8:42