From: intuitionist1@gmail.com   
      
   On Tuesday, June 19, 2018 at 10:34:48 AM UTC+3, Gregor Scholten wrote:   
   > 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:   
      
   No, they don't. The outermost shell experiences mass M-dm and does not   
   experience a black hole metric and can escape to infinity. The infalling   
   particle experiences the mass M, and falls into the black hole.   
      
   You are making the same mistake that everyone has made and are _assuming_   
   that there is only a single manifold (which is understandable, admittedly)   
   and that once the black hole forms, everything inside r=2GM is by default   
   inside the black hole and cannot escape.   
      
   There is actually no _physical_ reason why the matter inside r=2GM cannot   
   escape;- it does not know about the black hole and follows the geodesics of   
   R(r,t), not S_M(r,t). You seem now to be clutching at straws here - even to   
   the extent of trying to reformulate Birkhoff's theorem, to try to save the   
   standard picture whereby the interior is somehow swept into the black hole   
   which a priori it is completely oblivious to.   
      
   > 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   
      
   No, this is at r=R-dr, not r=R, which is _not_ inside the black hole. At   
   r=R, the metric is S_M(R,t), and this is the black hole metric.   
      
   > - 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   
      
   No, at R+dr the metric is S_{M+dm}(R+dr,t), which is part of the black hole   
   metric, just outside the event horizon.   
      
   > 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.   
      
   So what you have done here is physically moved the outermost shell from   
   where it actually is at r=2GM-dr (and where it does not experience the   
   black hole metric), to r=2GM where it is not (but where it would experience   
   the black hole metric, thus supporting your incorrect picture of what   
   happens - even though it requires superimposing the outermost shell on top   
   of the infalling matter, which does not make sense). This seems to be just   
   wishful thinking on your part.   
      
   If this argument had been correct, then it would basically imply that the   
   outermost shell must be swept up with the metric S_M(R,t) i.e. the   
   Schwarzschild black hole metric for the full mass M, and by induction, it   
   would sweep up all of the interior matter with it, which would also   
   experience the same metric for the full mass M. This is precisely the   
   standard picture, and it is incorrect, because it would require either that   
   an infinitesimal amount of mass has to make a finite change of the metric   
   from R(r,t) -> S_M(r,t) as it sweeps inwards, which is impossible, OR the   
   outer infinitesimal shell becomes of finite mass (so that it can effect   
   that change of metric) - but this would mean infinite mass density in that   
   outer shell, implying the formulation of a singularity there.   
      
   But here I am just repeating my earlier explanation as to why the standard   
   picture of collapse must be wrong. As I said, you seem now to be just   
   clutching at straws to try to uphold a misconception which was mistakenly   
   established several decades ago. To make progress we need to correct that   
   misconception and move on.   
      
   [[Mod. note -- Perhaps we should wind up this thread soon.  It doesn't   
   look like any of the main participants are likely to change their minds   
   at this point.   
   -- jt]]   
      
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