From: jthorn@astro.indiana-zebra.edu   
      
   As background, we're assuming classical general relativity holds,   
   and we're considering the spherically-symmetric collapse of a dust   
   cloud (in an asymptotically-flat spacetime) to form a black hole.   
   The dust cloud is of total mass $M$, and is initially at rest and   
   contained in a finite region.   
      
   As Gregor Scholten noted, one convenient way to describe the collapse   
   is using ingoing Eddington-Finkelstein coordinates $(t,r)$, defined as   
   follows:   
   * $r$ is the usual areal radial coordinate $r$, defined such that   
     on any t=constant slice, an r=constant surface is a 2-sphere   
     (respecting the spherical symmetry) of area $4\pi r^2$.   
   * $t$ is the ingoing Eddington-Finkelstein time coordinate, defined by   
     requiring that $t+r$ be an ingoing null coordinate.  We can choose   
     the zero point of $t$ arbitrarily; it's convenient to make our initial   
     slice be $t=0$.   
   We can also introduce the usual $(\theta,\phi)$ angular coordinates   
   on t=constant r=constant 2-spheres.   
      
   In these coordinates the *ingoing* speed of light is $dr/dt = -1$.   
      
   Given the assumption that the dust cloud was initially at rest with   
   a finite radius, the dust cloud will then collapse as we move forward   
   in time.  That is, the dust cloud occupies the region   
     $r \le L(t)$   
   for some function suitable real-valued function $L$.   
      
   We are given that $L(0) > 0$ [the dust cloud initially occupies a   
   finite region] and that $dL/dt(t=0) = 0$ [the dust cloud is initially   
   at rest].  Because gravity is attractive, $L$ will decrease with time;   
   since $r = L(t)$ is the trajectory of the outermost dust particles we   
   we know that $dL/dt \ge -1$ (since -1 is the ingoing speed of light,   
   and the dust can't move as fast as that).   
      
   Let's denote by $t=Z$ the time at which $L$ reaches zero.   
   At that time all the dust has reached the origin, so a singularity   
   will certainly have formed there (it will probably have actually formed   
   earlier, but I'll get to that later).  In the context of classical   
   general relativity it's not meaningful to talk about the dust once   
   it's all hit a singularity, so we won't bother defining $L(t)$ for   
   $t > Z$.   
      
   *Outside* the dust cloud (i.e., for $r >= $L(t)$) there is no dust,   
   i.e., the spacetime is vacuum, so by Birkhoff's theorem this part   
   of the spacetime is isometric to part of the Schwarzschild black   
   hole spacetime.  That is, for $r \ge L(t)$ the spacetime metric is   
   the Schwarzschild (black hole) metric, which we are denoting by   
   $S(t,r)$.  Notice that there are never any dust particles in this   
   region!   
      
   	[   
   	In our $(t,r,\theta,phi)$ coordinates $S(t,r)$ is the   
   	line element   
   	$ds^2 = -(1 - 2M/r) dt^2 + 4M/r dt dr + (1 + 2M/r) dr^2   
   		+ r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2$   
   	]   
      
   Until a singularity forms, *inside* the dust cloud (i.e., for   
   $0 \le t < Z$ and $0 \le r \le L(t)$) the metric has some other   
   non-vacuum form which we are are denoting $R(t,r)$.   
      
           [Note that the "interior Schwarzschild solution",   
           a.k.a. the "Schwarzschild star" has essentially   
           nothing to do with $R(t,r)$: as I noted earlier in   
           this thread, the interior Schwarzschild solution is   
           a solution of Einstein's equations describing a star   
           made out of fluid-with-pressure, whereas dust has   
           zero pressure.]   
      
   That is, in general the spacetime metric is   
            { R(t,r)   if $0 \le t < Z$ and $0 \le r \le L(t)$   
     ds^2 = {   
            { S(t,r)   otherwise   
      
   The metric must be continuous across the outer boundary of the dust   
   cloud, so   
     $S(t, r=L(t)) = R(t, r=L(t))$  for all $0 \le t < Z$   
      
   In (crude, but still useful) ASCII-art, we thus have   
      
           t                        O marks the dust cloud   
          /|\   
           *                        * marks the singularity   
           *   
           *   
      t=Z  *   
           OO   
           OOOOO   
           OOOOOOOO   
           OOOOOOOOOOO   
           OOOOOOOOOOOOO   
           OOOOOOOOOOOOOOO   
           OOOOOOOOOOOOOOOOO   
           OOOOOOOOOOOOOOOOOOO   
           OOOOOOOOOOOOOOOOOOOO   
           OOOOOOOOOOOOOOOOOOOOO   
           OOOOOOOOOOOOOOOOOOOOOO   
      t=0  OOOOOOOOOOOOOOOOOOOOOO----> r   
      
   (This is essentially the Misner, Thorne, & Wheeler figure 32.1c to   
   which I referred earlier in this thread.)   
      
   [In my diagram the "*" points mark a singularity.  These points aren't   
   actually in our differential manifold, and the metric isn't well-defined   
   there.]   
      
   Now let's consider black hole formation.  Recall that the black hole   
   region of spacetime is defined as the set of events in spacetime from   
   which *no* future-pointing null geodesic can reach future null infinity   
   (informally, large $r$ far in the future).  That is, an event $(t,r)$   
   is *not* in the BH region if and only if (some of the light from) a   
   light flash emitted at $(t,r)$ (eventually) *does* reach future null   
   infinity.   
      
   The event horizon (EH) is defined to be the boundary of the BH region.   
   It's easy to see that the EH is a null surface in spacetime.   
      
   The BH region is defined to be a (4-dimensional) set of events in   
   spacetime, and the EH -- the boundary of the BH region -- is thus   
   a 3-dimensional set of events in spacetime.  However, it's often   
   convenient to speak of these in terms of their intersection with a   
   given t=constant slice, i.e., we will often refer to the BH as a   
   (possibly-empty) 3-dimensional set of events in a given slice,   
   and to the EH as a (possibly-empty) 2-dimensional set of events   
   in a given slice.   
      
   Depending on how compact the dust cloud is at $t=0$, the BH may   
   or may not be present in the $t=0$ slice.  The latter case is   
   pedagogically more interesting, so I'll focus on it.   
      
   So, let's assume that there is no BH (& hence no EH) at $t=0$.   
   [In spacetime terms, we are assuming that the BH and EH do not   
   intersect the $t=0$ slice.]  That is, we're assuming that (some of   
   the light from) a light flash emitted from the event $(t,r) = (0,0)$   
   will eventually reach future null infinity.   
      
   Now consider a sequence of light flashes emitted from $r=0$ at   
   successively later times.  As the dust cloud collapses and becomes   
   more compact, there will eventually be a (first) time $t=X$ such   
   that *none* of the light from a light flash emitted at the event   
   $(t,r) = (X,0)$ will ever reach future null infinity.  By definition,   
   this event is in the BH, and since we are taking $X$ to be the   
   *first* such time, this event is on the EH.   
      
   Note that at $t=X$ there need not be any singularity present, nor   
   will $L(t)$ have reached zero.  In fact, there need not be be anything   
   "interesting" in the dynamics at this time -- the property that   
      "none of the light from a light flash emitted at the event   
       $(t,r) = (X,0)$ will (ever) reach future null infinity"   
   is a *global* property of spacetime, with no local symptoms.   
   At the time t=X, most of the dust may still be outside $r=2M$.   
      
   At $t > X$ the EH will expand outwards at the (local) speed of light,   
   eventually becoming $r = 2M$ once all the dust has fallen through the   
      
   [continued in next message]   
      
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    * Origin: you cannot sedate... all the things you hate (1:229/2)