15
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

6
$\begingroup$

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.

$\endgroup$
1
  • 5
    $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2020 at 14:38
4
$\begingroup$

Having thought about this more and discussed it with others, the answer seems to be that there are likely no counterexamples to the Penrose inequality, even if one allows for unphysical violations.

For recent numerical evidence of this, a paper of Kulczycki and Malec investigates the Penrose inequality and other modifications, finding counterexamples to the modified versions but no violations of the original inequality.

$\endgroup$
1
  • $\begingroup$ Would you be willing to paraphrase the content of the conversations that led you to this conclusion? Is there an underlying heuristic? $\endgroup$
    – Leo Moos
    Commented Mar 30, 2021 at 9:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .