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In the Schwarzschild spacetime with metric in standard Schwarzschild coordinates

$$ds^2=\rho(r)dt^2-\rho(r)^{-1}dr^2-r^2d\Omega^2,\quad \rho(r)=1-\dfrac{2GM}{r},$$

we have a coordinate singularity at $r = 2GM$. This is the event horizon, and is a surface of the spacetime manifold $M$, which, although here defined by coordinates, is certainly independent of any coordinate system.

Now in a more "conceptual language" if two observers start far away from the black hole, and one falls towards the central singularity, as soon as he passes $r = 2GM$, in other words, as soon as it crosses the horizon, the one that stayed will completely loose contact with him. A physicist I know used this to define a "causal horizon".

Now, this is an imprecise language. How can we characterize a causal horizon like this in general?

I believe the way would be on the definition of reference frames as timelike unit vector fields - namely, congruence of observers. In that case, we could say that $\mathscr{H}$ acts as a causal horizon for the reference frame $Z$ when:

  1. The causal horizon $\mathscr{H}$ splits $M$ into two regions $M_{\text{in}}$ and $M_{\text{out}}$ with $\mathscr{H}$ acting as a boundary between both.

  2. The reference frame $Z$ is supported in either one of the regions. In other words, its observers are located inside one of the regions.

This seems to be on the way, but I believe something more is required, related to the causality structure. Probably we should require $M_{\text{in}}$ and $M_{\text{out}}$ to be causally disconnected in the sense that for every $p\in M_{\text{in}}$ we have $J^{\pm}(p)\cap M_{\text{out}}=\emptyset$ and for every $p\in M_{\text{out}}$ we have $J^{\pm}(p)\cap M_{\text{in}}=\emptyset$.

Is that it? Is my approach correc to what should be a causal horizon in the sense of the "conceptual" example of Schwarzschild coordinates I described?

Is there a standard approach to this, different than this one?

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The general object used to define horizons is the irreducible past set, or $\mathrm{IP}$. An irreducible past set is :

  1. A past set, that is, if $A$ is a past set, then $I^-(A) = A$
  2. Irreducible, which means that $A \neq \varnothing$ and $A$ cannot be the union of two past sets. In other words, if $A = A_1 \cup A_2$ and $A_1, A_2$ are past sets, then $A = A_1 = A_2$.

It can be shown that every irreducible past set is the chronological past of a timelike curve,

$$\mathrm{IP}(\gamma) = \bigcup_{p \in \gamma} I^-(p)$$

The horizon of an observer is then the boundary of the irreducible past set associated to its curve.

$$\mathrm{Hor}(\gamma) = \partial \mathrm{IP}(\gamma)$$

To consider the more general horizons of an entire spacetime rather than a specific observer, you consider the $\mathrm{IP}$ of specific ideal points, called terminal irreducible past sets, or $\mathrm{TIP}$. They are irreducible past sets that cannot be written as $I^-(p)$, and can generally be considered to be the chronological past of ideal points of the spacetime, for instance points at infinity or singularities.

The horizon of black holes is generally done by considering the $\mathrm{TIP}$ associated with future null infinity, $\mathscr{I}^+$, which are the null points "at infinity" (this is where null curves end up if they do not crash into a black hole). The specific definition can be found for instance in Wald.

The horizon of a black hole is then the boundary of the union of $\mathrm{TIP}$s of those points at infinity.

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  • $\begingroup$ Thanks ! So for example, for the observers in the Rindler frame (uniformly accelerated in Minkowski), this would yield the boundary of the rindler wedge, and for a observer associated to the Schwarzschild coordoninates this would yield the usual event horizon? I'll have a look at Wald's book for details, is it in the black holes chapter? $\endgroup$ – user1620696 Jun 13 '18 at 11:24
  • $\begingroup$ Yes, although beware that for Schwarzschild observer, there may be more than one horizon : an observer of unbounded acceleration will have the Rindler type of horizon, as can be seen by taking the past of a point at null infinity. Any observer of finite acceleration of infinite proper time will have the Schwarzschild horizon, though. The specific thing to look for in Wald here is asymptotic flatness, since most often such boundaries are defined with respect to this. $\endgroup$ – Slereah Jun 13 '18 at 13:09

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