How do we know from the Kerr Metric that it is unstable near the Cauchy Horizon? What causes this unstability?

  • $\begingroup$ I think that is an open question. $\endgroup$
    – MBN
    Commented Oct 7, 2021 at 8:29
  • 1
    $\begingroup$ I assume you're referring to the mass inflation instability, as described in Poisson (1990), A look inside black holes. That paper isn't specifically about rotating black holes, but it says "These features are very general and do not depen on our specialized, spherical model; our conclusions should then be valid for the case of an uncharged and rotating black hole." $\endgroup$ Commented Oct 8, 2021 at 21:51
  • $\begingroup$ It may not be unstable see recent paper cited here wired.com/story/… $\endgroup$
    – bkocsis
    Commented Jun 18, 2023 at 11:36

1 Answer 1


That is a very good question. I will give you a rough argument of why we believe this to be true based on Bob and Alice. You can find a more detailed discussion at e.g. https://www.jstor.org/stable/79484.

Imagine Bob and Alice start together at $i$. Now, Bob has suicidical tendencies and decides to jump into a Kerr black hole, whereas Alice is more stable and decides to drift away to timelike infinity ($i^+$). In the penrose diagram below we represent the path of Bob with green and Alice by red.


Now, Alice decides to periodically send light signals to Bob (blue lines below). Alice lives forever as you need infinite proper time to reach $i^ +$, however Bob quickly reaches the Cauchy horizon $\mathcal{CH}^ +_R$. Nevertheless, as we can see in the picture, the light signals from Alice will all eventually reach Bob. Although Alice sends them at a fixed frequency, to receive infinite signals in a finite proper time, Bob will receive signals with an increasing frequency, that goes to infinity right before the Cauchy horizon. We call this an infinite blue-shift. In rough terms, this blue-shift translates to a spacetime instability at the Cauchy horizon, so although Kerr possesses a Cauchy horizon, we expect generic small perturbations to blow up at the interior horizon.

Further question for the reader, why does this argument fail for asymptotically de-Sitter spacetimes? In fact, recent research showed that the Cauchy horizon of near-extremal Reissner–Nordström de-Sitter is stable with respect to generic perturbations.

  • $\begingroup$ So what eventually happens? What does this instability cause? $\endgroup$ Commented Nov 30, 2021 at 12:55
  • $\begingroup$ The Cauchy horizon becomes a null singularity $\endgroup$ Commented Dec 1, 2021 at 13:08
  • $\begingroup$ Is this an assumption or are there analytical calculations or numetical simulations showing this? $\endgroup$
    – bkocsis
    Commented Jun 18, 2023 at 11:34

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