I have noticed that in the literature on causality in general relativity one sees apparent counterexamples to the cosmic censorship hypothesis (somehow you have models for gravitational collapse which assume spherical symmetry and things like this so that naked singularities can in fact arise). Hawking conceded that these were counterexamples, but then re-instated the hypothesis because these examples are in some sense unrealistic or unphysical.

I was wondering if the Penrose conjecture is also likely to have 'unphysical' or 'unrealistic' violations (so somehow make some special assumptions and then cook up a black hole spacetime which violates the Penrose inequality), or whether the conjecture is that one can simply never create a counterexample at all to the inequality?

Edit: I am aware of the counterexample of Carrasco and Mars to a stronger version of the conjecture. In that paper they find slices of the Kruskal spacetime for which the outermost generalized apparent horizon has area strictly greater than $16 \pi M^2$, and so this is not a counterexample to the true Penrose inequality as far as I am aware.

Jarosław Kopiński has mentioned to me in private communication that there is in fact already a counterexample to the Penrose inequality with 'apparent horizon':

and so that it is not that surprising that one can construct counterexamples when the inner boundary is even more general.

  • $\begingroup$ Whoever gave this a negative vote care to explain why or just going to do a hit-and-run? $\endgroup$ – Hollis Williams Oct 27 at 22:49

For time-reversally symmetric initial data the Penrose conjecture states the Riemannian Penrose inequality, which has been proven in full generality by H.L. Bray (arXiv:math/9911173). Without time reversal symmetry there exist counter-examples, see A counter-example to a recent version of the Penrose conjecture.

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    $\begingroup$ No, that is a counter-example to a more generalised version of the Penrose conjecture which uses generalized horizons as opposed to the usual horizons for black hole spacetimes. I am talking about the original Penrose inequality. $\endgroup$ – Hollis Williams Oct 20 at 14:38

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