You cannot cross an Einstein Rosen Bridge because it needs faster than light speed to cross it. But the same is not true for a Kerr wormhole, it lies beyond the Cauchy Horizon and you do not need FTL speed to cross it. So, how do you cross the wormhole without falling into the ring singularity? Do you just 'drive' your rocket through the Cauchy horizon on the other side?
Asked
Active
Viewed 242 times
0
-
Related: For the Kerr metric, does taking the maximal analytic extension commute with taking the zero-spin limit?. The question is different, but the content is relevant, especially the picture in A.V.S.'s answer. – Chiral Anomaly Oct 05 '21 at 12:41
1 Answers
4
The short answer is: you do nothing.
It is surprisingly hard to hit the ring singularity in Kerr. The vast majority of geodesics (test particle trajectories) plunging into a Kerr black hole miss the singularity a pass back across the inner and outer horizon to exit the black hole (in a new asymptotically flat universe). Only geodesics that are in the equatorial plane of the Kerr geometry (and a small class if geodesics almost perfectly aligned with the symmetry axis) can ever reach $r=0$.
This means that all you have to do to make this voyage is jump in a Kerr black hole making sure you are off the equatorial plane, and the rest will take care of itself.
However, there is an important disclaimer: The interior Kerr solution is notoriously unstable at the Cauchy horizon. Essential any gravitational perturbation entering the black hole will tend to blow up here. The above is assuming an perfect pristine Kerr geometry and ignores any gravitational perturbations (e.g. those caused by you, the observer). One should therefore not take this too seriously. In practical real situations the interior is likely to be very different (in ways that we do not fully understand at this time).
TimRias
- 11,610
-
Thanks. So it simply means we just need to hit the Cauchy Horizon once again? – Nayeem1 Oct 05 '21 at 18:38