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  Geometric meaning of the black hole horizon

+ 2 like - 0 dislike

It is widely accepted that the singularity of the Schwarzschild metric at the event horizon is purely an artifact of the coordinates and no physical singularity exists at the horizon. However, as Karlhede had shown in 1982, the Karlhede's scalar $R^{ijkl;m}R_{ijkl;m}$ (the square of the covariant derivative of the Riemann tensor) changes sign at the Schwarzschild horizon and therefore, in principle, a freely falling observer can detect the moment of crossing the horizon by local measurements (see http://arxiv.org/abs/1404.1845 , Karlhede's invariant and the black hole firewall proposal, by J. W. Moffat and V. T. Toth).

What are geometric meanings of the Karlhede's scalar and black hole horizon (if the latter indeed can be defined in an invariant way)?

This post imported from StackExchange MathOverflow at 2015-03-30 11:45 (UTC), posted by SE-user Zurab Silagadze
asked Sep 1, 2014 in Resources and References by Zurab Silagadze (255 points) [ no revision ]
retagged Mar 30, 2015
Any arguments on the geometry of black hole spacetimes based entirely on the Schwarzschild example should be taken with a heavy grain of salt: it simply has too many symmetries. That said: for asymptotically flat space-times the even horizon certainly can be defined invariantly, as the boundary of $\mathcal{J}^-(\mathscr{I}^+)$. The apparent horizon, on the other hand, is much more difficult to define. The people studying dynamical horizons often use MOTS, but there are some problems with that even in Schwarzschild.

This post imported from StackExchange MathOverflow at 2015-03-30 11:45 (UTC), posted by SE-user Willie Wong
If you look at the arXiv pre-print you cited (page 2, near eq. 18), for something as simple as Kerr, Karlhede's invariant changes sign on the ergosphere, and not the event/apparent horizon. To me this is evidence enough that while the invariant may have some physical meaning, that its vanishing coincides with the event horizon in Schwarzschild is exactly that: a coincidence.

This post imported from StackExchange MathOverflow at 2015-03-30 11:45 (UTC), posted by SE-user Willie Wong
Probably this is indeed a coincidence. However it was shown in arxiv.org/abs/gr-qc/9808055 that the Karlhede's invariant vanishes at the horizon for any Schwarzschild like space-time and it was conjectured that it vanishes at regular horizons in any static axially symmetric space-time. It is true, however, that the horizon is not the only place where the Karlhede's invariant vanishes.

This post imported from StackExchange MathOverflow at 2015-03-30 11:45 (UTC), posted by SE-user Zurab Silagadze
As far as I can tell their notion of Schwarzschild-like means the union of Schwarzschild, Reissner-Nordstrom, and Sch-dS; all are spherically symmetric and static. In other words, where I said Schwarzschild in my first comment you can equivalently read "Schwarzschild-like", using this notion.

This post imported from StackExchange MathOverflow at 2015-03-30 11:45 (UTC), posted by SE-user Willie Wong

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