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Geometric meaning of the black hole horizon

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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 2015-11-16 07:38 (UTC), posted by SE-user Zurab Silagadze
asked Sep 1, 2014 in Theoretical Physics by Zurab Silagadze (255 points) [ no revision ]
retagged Nov 16, 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 2015-11-16 07:38 (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 2015-11-16 07:38 (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 2015-11-16 07:38 (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 2015-11-16 07:38 (UTC), posted by SE-user Willie Wong
The fundamental reason that this has to be a coincidence is that event horizons are observer-dependent. A good example is an observer with constant proper acceleration in Minkowski space. Such an observer sees an event horizon, which clearly can't be connected to any intrinsic property of the spacetime, since the Riemann tensor vanishes identically. The horizon that we customarily talk about in the case of the Schwarzschild spacetime just happens to be the one seen by an observer at null infinity.

This post imported from StackExchange MathOverflow at 2015-11-16 07:38 (UTC), posted by SE-user Ben Crowell

1 Answer

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There is no local definition of a horizon. Horizons are everywhere, at least if you choose the right set of observers. One of those choices are e.g. Rindler coordinates corresponding to a particular set of uniformly accelerating observers in Minkowski space-time. In Schwarzschild coordinates you see a set of uniformly accelerating observers at $r=const.$ which would have to accelerate infinitely to stay at the horizon. Up to topology, you see the very same effect in Rindler coordinates. I.e. the significance of black-hole horizon follows, to a certain extent, from a globally topological aspect of the space-time.

There are of course various quasi-local definitions of a horizon which however 1) do not all give the same horizons in non-stationary space-times, and 2) do not always give the global "possible to escape to infinity" horizon in non-stationary space-times. In fact, if you have ever had a taste of the theory of dynamical systems, it would seem absurd that a rich dynamical theory allowing for chaos would unambiguously store the fate of the whole slice of the phase space (i.e. geodesics with all possible velocities at the given point) in one simple number.

The problem is that even the curvature invariants in fact correspond to certain "semi-local" quantitites, i.e. quantities not detectable on the tangent bundle $\sim \Delta x$ but in second-order differences $\sim \Delta x^2$. By this title, the Karlhede scalar corresponds to a "semi-semi-local" quantity measurable only through $\sim \Delta x^3$ effects.

This suggests that the scalar could be related to some of the quasi-local horizon definitions. I.e. the fact that the Karlhede scalar vanishes at the edge of the ergoregion in Kerr might be correlated with the fact that the ergoregion is the trapping surface for meridional-plane geodesics (which are transversal to the Killing vectors).

To understand curvature invariants, it is sometimes useful to introduce a 3+1 Bel decomposition with respect to some observers (I also recommend this paper by Cherubini et al. for a crash-course and application). In the case of the Karlhede scalar it would probably yield that it is some kind of combination of "slap-force gradients" and "local spatial-curvature variation" with some kind of sign difference between the two. In general, the zero of the Karlhede scalar will simply mean that the "slap force gradients" and "local spatial curvature variations" cancel each other out "by chance".

Of course, only direct investigation can reveal further insight.

answered Nov 16, 2015 by Void (1,505 points) [ no revision ]

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