From: g.scholten@gmx.de   
      
   Sabbir Rahman wrote:   
      
   >>> After a short time, a black hole will be formed as soon as the   
   >>> entirety of the first collapsing dust shell has crossed the   
   >>> Schwarzschild radius at r=2M. Note that the entire rigid lattice   
   >>> in the interior remains fixed, so that the metric R(r,t) in the   
   >>> interior basically remains fixed throughout that interior. The   
   >>> lattice will be totally oblivious to the fact that a black hole has   
   >>> just formed.   
   >>   
   >> This is impossible. As soon as the black hole has formed, the rigid   
   >> lattice which is now inside the black hole starts to collaps. To resist   
   >> against the gravitational compression and remain fixed, its atoms had to   
   >> travel on spacelike wordlines, i.e. to exceed the speed of light in a   
   >> local frame.   
   >   
   > There is no 'gravitational compression' (=="abrupt change of   
   > metric") -  the rigid frame does not experience anything different than   
   > it did before the black hole formed.   
      
   And since it is well-known that the "rigid" structure is unavoidably   
   compressed by the gravitation as soon as the black hole has formed, this   
   implies that the "rigid" structure is already compressed before the   
   black hole forms. More precisely, according to GR, the repulsive forces   
   between the atoms of the "rigid" structure, that make the structure   
   rigid and supports it against gravitational collaps, unavoidably fail to   
   support as soon as the radial coordinate of the structure's surface is   
   lower than 3/2 rs = 3GM, where M is the mass of the structure.   
      
   So, your scenario, where the interatomic forces support the "rigid"   
   structur against collaps in a state where the radial coordinate of the   
   structure's surface is roughly higher than the Schwarzschild radius rs =   
   2GM of the structure, is completely unrealistic.   
      
   A scenario that is at least nearly realistic is one where the radial   
   coordinate of the strcture's surface is exact r = 3/2 rs = 3GM (M = mass   
   of the structure), so that the repulsive interatomic forces are just   
   able to balance gravity, i.e. there is an unstable equilibrium of   
   repulsive interatomic forces and gravity. Now imagine a dust shell of   
   little mass m << M that falls onto the surface of the (unstably) rigid   
   structure. Then there a two possible outcomes:   
      
   (1) We take into account that the atoms of the dust shell interact with   
   the atoms of the structure. Then the atoms of the surface of the   
   structure a pushed a little bit inwards by the atoms of the dust shell.   
   By this, the unstable equilibrium of repulsive interatomic forces and   
   gravity is destroyed, gravity becomes dominant and makes the outermost   
   shell of structure atoms collaps. In addition the push on the outermost   
   shell of structure atoms spreads as a sound wave inside the structure,   
   destroying the equlibrium for all inner atoms of the structure, causing   
   the structure as whole to collaps.   
      
   (2) We assume that the atoms of the dust shell do not interact with the   
   atoms of the structure. Then the dust shell falls through the outermost   
   shell of structure atoms. As soon as this has happened, the structure   
   atoms of the outermost shell feel the little additional gravity of the   
   dust atoms. By this, the unstable equilibrium of repulsive interatomic   
   forces and gravity is destroyed, gravity becomes dominant and makes the   
   outermost shell of structure atoms collaps. The resulting inward   
   movement of the outermost structure atoms causes a sound wave inside the   
   structure that is spreading inwards and destroys the equlibrium for all   
   inner atoms of the structure, causing the structure as whole to collaps.   
      
   In both cases, the interatomic forces that supported the structure   
   against gravitational collaps before are responsible for the sound wave   
   to form that disturbs the equlilibrium and let gravity become domimant.   
      
   Now let us consider the evolution of the metric in both cases:   
      
   (a) Before the dust shell encounters (falls onto) the surface of the   
   structure at r = 3GM, the metric at the the surface of the structure is   
      
   S_M(3GM,t)   
      
   where S_M is the Schwarzschild metric for the mass M of the structure.   
   Let R > 3GM be the radial coordinate of the surface of the dust shell,   
   the metric at the surface of the dust shell is   
      
   S_{M+m}(R,t)   
      
   where S_{M+m} is the Schwarzschild metric for the combined mass M+m of   
   the structure and the dust shell.   
      
   (b) When as the dust shell hast just encountered the the surface of the   
   structure at r = 3GM, the metric at the the surface of the structure is   
      
   S_{M+m}(3GM,t)   
      
   because the gravitational field there is no longer caused by the mass M   
   of the structure alone, but additionally by the mass m of the dust   
   shell. So, the metric at r=3GM changes by the finite difference   
      
   S_{M+m}(3GM,t) - S_M(3GM,t)   
      
   due to the finite additional mass m. We could as well assume an   
   infinitesimal mass dm for the dust shell, then the change in the metric   
   at r=3GM would be the infinitesimal difference   
      
   S_{M+dm}(3GM,t) - S_M(3GM,t)   
      
   Regarding the change in the metric at the surface of the structure at   
   r=3GM, there is no difference between the cases (1) and(2).   
      
      
   >> In addition, after the first dust shell has entered the region r < 2M,   
   >> the metric R(r,t) in that region is no longer determined by the matter   
   >> from the lattice alone, but also influenced by the matter of the dust   
   >> shell. The infinitesimal mass contribution dm of the dust shell yields a   
   >> infinitesimal change in the metric R(r,t).   
   >   
   > You ask me why I think the dust particles sweep the inner shells along   
   > with  them - and yet here you are insisting that they be swept inwards   
   > by that very same infinitesimal dust shell. I actualy agree with you -   
   > the infinitesimal dust shell can only make an infinitesimal change in   
   > the metric. So R(r,t) basically does not change in any noticeable way.   
   > For the entire rigid framework however to be somehow forced to collapse   
   > into a singularity would require an enormous change in the metric   
      
   No. As we have seen above, your scenario is completely unrealistic. In   
   the more realistic scenario where the radial coordinate of the surface   
   of the rigid structure is at r = 3GM, it turns out that there is highly   
   instable equlibrium of repulsive interatomic forces and gravity. Already   
   a small disturbance is sufficient to destroy the equlibrium and make   
   gravity become dominant, so that the whole structure collapses.   
      
      
   >>> After another short time, the second shell of dust will cross the   
   >>> Schwarzschild surface. From Birkhoff's theorem, this shell of dust   
   >>> will follow the geodesics of the Schwarzschild interior solution -   
   >>> with metric S(r,t) -   
   >>   
   >> Birkhoff's theorem tells us that at the outermost shell, the metric has   
   >> to equal the external Schwarzschild metric, and by this, the interior   
   >> Schwarzschild metric. However, you are not considerung the outermost   
      
   [continued in next message]   
      
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