From: g.scholten@gmx.de   
      
   Sabbir Rahman wrote:   
      
   >>> At the   
   >>> moment that the black hole forms at t=t0   
   >>   
   >> Where t is the time coordinate of some coordinate system in which the   
   >> black hole forms within finite time, i.e. not Schwarzschild coordinates,   
   >> but e.g. Eddington-Finkelstein coordinates.   
   >   
   > The choice of coordinate system is not important   
      
   It is important in that way that it must allow for indicating a time   
   t=t0 when the black hole forms. This excludes e.g. Schwarzschild   
   coordinates, in which t0 would be in the infinite future.   
      
      
   >>> say, it is true that the   
   >>> boundary conditions require that R(r=2M,t=t0)=S(r=2M,t=t0).   
   >>   
   >> You mean because at the surface of the collapsing celestial body, the   
   >> metric R(r,t) must match the external Schwarzschild metric (outside the   
   >> body), and since this is known to match the internal Schwarzschild   
   >> metric S(r,t) there, it must match internal Schwarzschild metric S(r,t),   
   >> too?   
   >   
   > The metric outside the collapsing matter matches the exterior   
   > Schwarzschild metric up until the formation of the black hole.   
      
   You mean because you think that as soon as the radial coordinate R of   
   the dust cloud's surface becomes < rs, the metric in the region R < r <   
   rs would be the interior Schwarzschild metric? That's wrong. The metric   
   in the region r > R is the exterior Schwarzschild metric everywhere,   
   even after R has become < rs. So, in the region R < r < rs, the metric   
   is the exterior Schwarzschild metric, like in the region r > rs.   
      
      
   > After   
   > the formation of the black hole, according to the standard picture,   
   > the metric at the boundary of the collapsing matter should match   
   > the interior Schwarzschild metric.   
      
   Correct, and at this boundary (r = R), the interior Schwarzschild metric   
   matches the exterior Schwarzschild metric. Once again, after the black   
   hole has formed, i.e. R has become < rs, the metric is:   
      
   - exterior Schwarzschild metric for r > R, no matter if R < r < rs or r > rs   
      
   - exterior Schwarzschild metric = interior Schwarzschild metric = R(r,t)   
   for r = R (all three metrics match there)   
      
   - R(r,t)  for r < R   
      
      
   > This is a consequence of Birkhoff's theorem.   
   > I am claiming that this cannot be physically correct.   
      
   And you are wrong. It can be correct.   
      
      
   >>> However   
   >>> in general it will obviously not be the case that R(r,t)=S(r,t) for   
   >>> r<2m or t>t0 as R(r,t) is some non-black hole metric and S(r,t) is   
   >>> the Schwarzschild interior metric.   
   >>   
   >> What do you mean with R(r,t) being some non-black hole metric? R(r,t) is   
   >> the metric inside the surface of the collapsing celestial body, even   
   >> after the black hole has formed (and the collaps has not yet reached the   
   >> state of singularity at r = 0).   
   >>   
   >> Of course, R(r,t) will not match S(r,t) inside the surface, they only   
   >> match at the surface.   
   >   
   > R(r,t) is the metric inside the collapsing matter as discussed   
   > above. It can be a generic spherically symmetric metric depending   
   > upon the properties and distribution of the matter in the interior.   
   > For example, the collapsing matter could in principle consist of a   
   > large number of massive rockets which are free to thrust around at   
   > will into whatever configuration they choose. (Obviously the thrusting   
   > should be in the radial direction and a function of only the radius   
   > for spherical symmetry to be maintained, though in general they can   
   > form non-spherical distributions, and can even in principle delay   
   > the collapse indefinitely).   
      
   My question was due to what reason you mean that R(r,t) is some   
   non-black hole metric?   
      
      
   >>> Now, in the standard picture, because the outer shell of matter is   
   >>> part of the Schwarzschild interior, and _also_ part of the dust   
   >>> cloud, it is known from Birkhoff's theorem that it must end up in   
   >>> a singularity, and moreover because there can only be one metric   
   >>> on the interior submanifold, it must sweep all of the dust particles   
   >>> inside the cloud with it into the singularity.   
   >>>   
   >>> The problem is, however, that the particles in the dust cloud live   
   >>> in a manifold with metric R(r,t), and the particles in the boundary   
   >>> live in a manifold with metric S(r,t).   
   >>   
   >> No, that's wrong. The particles live in a manifold with a metric that is   
   >> R(r,t) inside a region r <= R, and the exterior Schwarzschild metric   
   >> outside that region, i.e. for r > R, where R ist the radial coordinate   
   >> of the surface of the dust cloud (aka celestial body).   
   >>   
   >> There is no manifold with metric S(r,t). Near the surface r = R, the   
   >> metric R(r,t) approximately matches the metric S(r,t) that *would be*   
   >> the metric inside the region r <= R if the dust cloud wat not   
   >> collapsing, but this not in any way mean that there would be a manifold   
   >> with metric S(r,t).   
   >   
   > Actually, yes there is. Any matter which happens to cross the black   
   > holes event horizon after formation of the black hole will cross   
   > the EH and enter a manifold which has the metric S(r,t) of the   
   > Schwarzschild interior. Again this follows from Birkhoff's theorem.   
      
   No, it does not follow from Birkhoff's theorem. Birkhoff's theorem tells   
   us that for r > R (R = radial coordinate of the surface of the dust   
   cloud), the exterior Schwarzschild metric applies. From this follows   
   that the metric R(r,t) inside the region r < R must match the exterior   
   Schwarzschild metric at the surface r = R. Since it is well-known that   
   the interior Schwarzschild metric S(r,t) matches the exterior   
   Schwarzschild metric there, this yields that the metric R(r,t) must   
   match the interior Schwarzschild metric there, at r = R.   
      
   This is what follows from Birkhoff's theorem. Birkhoff's theorem does   
   *not* tell us that particles from outside the dust cloud would enter a   
   manifold with the interior Schwarzschild metric S(r,t). Assume the black   
   hole has already formed, i.e. R < rs. Then the following happens: a   
   particle crosses the event horizont and enters the region R < r < rs,   
   where the exterior Schwarzschild metric applies like outside the event   
   horizon. A little later, the particle reaches the surface r = R of the   
   dust cloud. There, the metric is equal to (a) exterior Schwarzschild   
   metric, (b) interior Schwarzschild metric and (c) metric R(r,t). After   
   crossing the surface, the particle enters the region 0 < r < R, where   
   the metric R(r,t) applies.   
      
   You seem to think that in the region R < r < rs, the interior   
   Schwarzschild metric S(r,t) would apply. But that is wrong. It is the   
   exterior Schwarzschild metric that applies there, like in the region r >   
   rs. Birkhoff's theorem does not in any way reject this.   
      
      
   > The point that I am making is that the metric inside the collapsing   
   > matter and this Schwarzschild interior metric are two different   
   > metrics corresponding to two different manifolds. Both of them are   
   > present   
      
      
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