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

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  • $\begingroup$ I think that is an open question. $\endgroup$
    – MBN
    Commented Oct 7, 2021 at 8:29
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    $\begingroup$ 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

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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.

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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.

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  • $\begingroup$ 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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