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  Can a static spacetime have a non-Killing horizon?

+ 2 like - 0 dislike
A stationary spacetime can have a non-Killing horizon, i.e., the horizon does not have a constant surface gravity. Can a static spacetime have a non-Killing horizon? If yes, give an example; if not, why?
asked Nov 23, 2014 in Theoretical Physics by renphysics (30 points) [ no revision ]

2 Answers

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I think that by its definition a Killing horizon requires the existence of a global asymptotic time-like Killing vector field. A static spacetime by its definition has such a Killing vector field $k_t$ (plus an additional condition that $k_t$ must be irrotational ). I do not see why a static space-time would have a non-Killing horizon if the all the required symmetries are actually preserved. Remember that for a Killing vector $\xi$ and two vector fields $X,Y$ we have $$\mathcal{L}_{\xi}(\nabla_X Y) = \nabla_{\mathcal{L}_{\xi}X}Y + \nabla_X (\mathcal{L}_{\xi}Y)$$ and for the case of non-Killing horizons some extra terms should arise from the Lie derivative of the Christoffel symbols (I do not see how this could arise in classical Einstein gravity). I think you can find such examples of non-Killing horizons if you consider Horava-Lifschitz gravity but I cannot help you further into this. But if you manage to find a counter-example let us know.
answered Nov 26, 2014 by conformal_gk (3,625 points) [ no revision ]
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I do not understand well what you mean by Killing horizon here. A Killing horizon in a 4-dimensional spacetime is, by definition,  a null 3-surface made of integral lines of a Killing field defined in a neighbourhood of that surface, which  becomes null exactly on that surface. As is known, the surface gravity turns out to be  always constant along each integral line (which can be proved to be geodesic).

I guess that actually you are wondering if the surface gravity must be constant also changing the integral curve of the Killing field while remaining on the horizon.

It is possible to prove that, whenever a spacetime admitting a Killing horizon satisfies Einstein equations and the dominant energy condition is verified, the surface gravity must be constant on
the whole horizon.

However a result, originally obtained by Carter, states that the surface gravity of a Killing horizon must be constant if (i) the Killing field is static or (ii) there is an additional Killing
field and the two Killing fields are 2-surface orthogonal.

All that can be found in Wald's textbook on general relativity. Now I cannot check the precise page, but I am sure on that.

answered Nov 26, 2014 by Valter Moretti (2,085 points) [ revision history ]
edited Nov 27, 2014 by Valter Moretti
I think he means non-Killing horizons.
What is the definition of non-Killing horizon? A null 3-surface such that...?
The non-Killing horizon I have in mind is in the sense of http://arxiv.org/abs/1212.4498
Sorry I was not able to find, in that paper, a definition of horizon or non-Killing horizon, it seems to me that these notions are simply used assuming that the reader is familiar with them. Could you, please, point out the exact point in the paper where the notion of non-Killing horizon is introduced, or mentioning another paper where it is discussed with more details? I am familiar with the notion of Killing horizon, bifurcate Killing horizon, event horizon (which is a global structure), but I do not know what a non-Killing horizon is.

What I cannot understand is how the spacetime determines a non-Killing horizon. It cannot be a null 3-surface simply, beacuse there are infinitely many such surfaces in every spacetime, none more interesting of the remaining  ones. Killing horizons are instead fixed by the structure of  Killing isometries. However, if Killing isometries do not enter the game by definition, how you pick out a preferred null surface? One could use conformal Killing vectors or other geometrical structures I guess, but I do not understand well.

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