From: g.scholten@gmx.de   
      
   Another remark:   
      
   Sabbir Rahman wrote:   
      
   > At t0, the outermost shell only experiences M-dm, where dm is the   
   > infinitesimal mass of that shell. If you think carefully about it,   
   > _none_ of that infinitesimal shell experiences the full mass, and so   
   > actually _none_ of it is inside its Schwarzschild radius, and _none_ of   
   > it collapses unavoidably. Any further collapse is in principle   
   > avoidable. [That is not to say however that any matter outside that   
   > infinitesimal shell that subsequently crosses R=2M will not unavoidably   
   > collapse into the singularity. Indeed it must, by Birkhoff's theorem].   
      
   Maybe, your intended argument is the following:   
      
   The outermost shell of the matter configuration at R = 2GM experiences   
   the gravity of the mass M - dm only, whereas the infalling particle,   
   when it is currently at r = R = 2GM, experiences the gravity of the full   
   mass M. So that the outermost shell at R = 2GM is just able to resist   
   against gravity, but the infalling particle at R = 2GM is not.   
      
   This argument is wrong, though. Both, the outermost shell and the   
   infalling particles at R = 2GM experience the gravity of the mass M -   
   dm. So, both are in the same way just able to resist against the   
   gravitational forces. And this is not in contradiction to Birkhoff's   
   theorem:   
      
   The usual formulation of Birkhoff's theorem is that at r = R, the metric   
   is equal to S_M(R,t) which is the Schwarzschild metric for the full mass   
   M. But this usual formulation does not take the strict distinction   
   between M and M - dm into account that you make. By taking this   
   distinction into account, Birkhoff's theorem has to be re-formulated to:   
      
   - at r = R, the metric is the Schwarzschild metric S_{M-dm}(R,t) for the   
   mass M - dm   
      
   - at R + dr, i.e. infinitesimally outside the boundary of the matter   
   configuration, the metric is the Schwarzschild metric S_M(R+dr,t) for   
   the full mass M   
      
   So, both, the outermost shell and the infalling particle at r = R = 2GM   
   are on the edge of being able to resist against the gravitational forces.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)