From: g.scholten@gmx.de   
      
   Sabbir Rahman wrote:   
      
   >> I just read again the chapters 32.4 and 32.5 of MTW, and in fact, they   
   >> write what I already explained: all the particles inside the collapsing   
   >> dust cloud fall simultanously towards the point r = 0, there is no   
   >> "encountering" of the outermost shell of particles and the more inner   
   >> shells.   
   >>   
   >> In 32.4, they consider a dust cloud with zero pressure and argue that   
   >> the metric inside the cloud equals the Friedmann metric of a contracting=   
   >   
   >> universe. And like there is no encountering of shells of galaxies in a   
   >> contracting universe, there is no encountering of shells of dust particle=   
   > s.   
   >>   
   >> In 32.5, they consider a collapsing star with pressure gradient. They   
   >> write that the surface of the star is no longer free-falling due to the   
   >> forces caused by the pressure gradient, but that qualitatively seen the   
   >> process of collaps is like in the case of zero pressure.   
   >>   
   >> So, I am really wondering what makes you think that the standard picture   
   >> would be that the outer shell of dust particles encounters the more   
   >> inner shells and sweep them along with it?   
   >   
   > The solutions given in in MTW are incorrect, but the reason why is,   
   > admittedly, not at all obvious.   
      
   Obvios or not, please explain why you think that the solutions given in   
   in MTW are incorrect. In the following, you do not do that, you just   
   consider a different picture.   
      
   And you still did not answer my questions what makes you think that the   
   standard picture would be that the outer shell of dust particles   
   encounters the more inner shells and sweep them along with it?   
      
      
   > I think with hindsight that rather than considering a freely falling   
   > dust cloud, it would be somewhat easier to consider an (admittedly   
   > rather artificial) mass distribution which happens to have almost   
   > the same mass density as the dust cloud throughout the interior but   
   > which is NOT allowed to collapse by keeping the mass fixed in place   
   > through mechanical means of some kind.   
   >   
   > To be more concrete, let us consider an approximately spherically   
   > symmetric, rigid, solid (e.g. metallic), spherical lattice of mass   
   > M-dm (i.e. just less than the mass M of the dustcloud we have been   
   > considering) and radius r = 2M-dr (i.e. slightly less than the   
   > radius of the initial dust cloud).   
   >   
   > Then this rigid structure will have approximately the same average   
   > density as the dust cloud we started with, albeit with very slight   
   > less mass and of slightly smaller radius. Let us assume that it   
   > does not have enough mass within any radius to form a black hole.   
   >   
   > Now, at some distance outside this rigid structure let there be an   
   > infalling thick shell of dust of mass precisely dm just outside   
   > r=2M (the thickness should be finite but not too large).   
   >   
   > Just outside that shell, let there be a large number of concentric   
   > infalling shells of dust each of tiny (or infinitesimal) mass. I   
   > hope that this scenario is clear.   
   >   
   > Clearly this is what is going to happen:   
   >   
   > 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.   
      
   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).   
      
      
   > 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   
   shell here, but the the second shell seen from the innermost one.   
   Therefore, Birkhoff's theorem does not tell us in any way how the metric   
   has to be at this second shell.   
      
      
   > Now, in the standard picture of gravitational collapse, each component   
   > of the rigid lattice structure   
      
   You in fact want to tell us that the standard picture of gravitational   
   collapse does in any way refer to rigid lattice structures like the ones   
   you are describing? Once again my question: from which article or book   
   do you take your information about the standard picture of gravitational   
   collapse?   
      
      
   > which lies inside the black hole,   
   > must be swept up by the first infalling dust shell of mass dm as   
   > it passes that component and eventually end up in the Schwarzchild   
   > singularity. This means that the rigid metric R(r,t) must turn into   
   > the interior Schwarzschild solution S(r,t) as the thick dust shell   
   > (and specifically the outermost infinitesimal shell of that thick   
   > shell) sweeps by, taking the rigid lattice with it as it heads   
   > towards the Schwarzschild singularity.   
      
   You mean you want to consider a dust shell of inifinitesimal mass dm,   
   but finite thickniss Delta_r, implying a infinitesimal small dust   
   density? That makes little sense. We should rather assume a dust shell   
   of finite density, with infinitesimal mass dm and infinitesimal   
   thickness dr.   
      
   So, at the time t0, when the dust shell has the radial coordinate 2M +   
   dr so that the black hole has not yet formed, the metric at 2M + dr is   
   the interior Schwarzschild metric S(2M + dr, t0). The metric at 2M - dr,   
   i.e. at the boundary of the rigid structe, is   
      
   R(2M - dr, t0) = S(2M - dr, t0) + delta_g(2M - dr, t0)   
      
   It differs from the interior Schwarzschild metric S(2M - dr, t0) by a   
   infinitesimal difference delta_g(2M - dr, t0).   
      
   A little later, at time t0 + dt, the dust shell has the radial   
   coordinate 2M - dr. The black hole has formed and the dust shell   
   encounters the boundary of the rigid structure. The metric at 2M - dr,   
   the boundary of the rigid structe, is   
      
   R(2M - dr, t0 + dt) = S(2M - dr, t0 + dt)   
      
   then. It has changed by the infinitesimal difference delta_g(2M - dr,   
   t0). This is caused by the infinitesimal change in the matter   
   distribution that happens by the dust shell fall from 2M + dr to 2M - dr.   
      
      
   > There are two possibilities:   
   >   
   > 1) On the one hand, if this _does_ happen, then you need to explain   
   > the physical mechanism by which the infinitesimal shell of mass at   
      
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