Why can apparent horizon be computed based on its local geometry? In the paper titled Black Holes, Geometric Flows, and the Penrose Inequality in General Relativity by Hubert L. Bray, has been written:

Another very interesting and natural phenomenon in general relativity is the existence of black holes. Instead of thinking of black holes as singularities in a spacetime, we will think of black holes in terms of their horizons. For example, suppose we are exploring the universe in a spacecraft capable of traveling any speed less than the speed of light. If we are investigating a black hole, we want to make sure that we do not get too close and get trapped by the “gravitational forces” of the black hole. We can imagine a “sphere of no return” beyond which escape from the black hole is impossible. It is called the event horizon of a black hole. However, one limitation of the notion of an event horizon is the difficulty of determining its location. One way is to let daredevil spacecraft see how close they can get to the black hole and still escape from it eventually. The problem with this approach (besides the cost in spacecraft) is that it is hard to know when to stop waiting for a daredevil spacecraft to return. Even if it has been fifty years, it could be that this particular daredevil was not trapped by the black hole but got so close that it will take one thousand or more years to return. Thus, to define the location of an event horizon even mathematically, we need to know the entire evolution of the spacetime. Hence, event horizons cannot be computed based only on the local geometry of the spacetime. This problem is solved (at least for the mathematician) by the notion of apparent horizons of black holes. Given a surface in a spacetime, suppose that it emits an outward shell of light. If the surface area of this shell of light is decreasing everywhere on the surface, then this is called a trapped surface. The outermost boundary of these trapped surfaces is called the apparent horizon of the black hole. Apparent horizons can be computed based on their local geometry, and an apparent horizon always implies the existence of an event horizon outside of it.

Question: Bray has argued that event horizons cannot be computed based only on the local geometry of the spacetime because we cannot define the location of an event horizon. On the other hand we know that apparent horizon is always inside (or coincide) the event horizon. How can we define the location of apparent horizon? When spacecraft cannot define the location of event horizon, how can it define that of apparent horizon? Unlike the event horizon, why can apparent horizon be computed based on its local geometry?

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    $\begingroup$ Related. Apparent horizons are related to a local observer (an analogy, in special relativity would be, for instance, an accelerated observer, which may have a fluctuating horizon, if its acceleration is not constant) , while events horizons are "global", while this means depending on the whole set of observers at spatial infinity. $\endgroup$ – Trimok Jan 1 '14 at 13:14
  • $\begingroup$ Apparent horizons depend on sections of the spacetime in question. Example: it is possible to slice the Schwarzschild geometry such that there is no apparent horizon, ever, though there is an event horizon. $\endgroup$ – user28355 Jan 14 '14 at 1:32

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