This is a non-obvious theorem, because due to time dilation, the distant observer may have a hard time judging whether infalling matter has really fallen through the horizon "now."Īnother way of stating the result of the theorem is that we actually have two independent Killing vectors at the horizon: the null one and another one that constitutes the extension of the timelike one that exists at infinity.
The Hawking rigidity theorem says that if the spacetime is a vacuum spacetime and is analytic and regular, then if a distant observer says it's in equilibrium (stationarity), then a nearby observer does as well (horizon is Killing). A light cone in Minkowski space is a Killing horizon.) (Not all Killing horizons are event horizons. A black hole in equilibrium is one whose event horizon is also a Killing horizon. The way we get around this is by the definition of a Killing horizon.
You can't visit the horizon once and then visit it again later to see if it's changed. We would like to say that somehow this observer doesn't see the horizon "changing over time," but the horizon is a null surface, so this doesn't quite make sense. However, this is hard, because if we poke the black hole with a stick, we lose our stick. To an observer who intends to poke and prod the black hole from nearby, we want some other definition.
To a distant observer, this is defined by whether the spacetime is stationary, i.e., has a timelike Killing vector "near infinity." We want to talk about the notion that gravitational collapse is complete, and a black hole has reached an equilibrium. 323-340, and looking at this talk, and this SE question. The following is just my quick and dirty understanding after skimming Hawking and Ellis pp.
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