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*To*: starship-design@lists.uoregon.edu*Subject*: Re: starship-design: Blackhole*From*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Date*: Thu, 21 Aug 1997 21:13:10 +0100*Reply-To*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Sender*: owner-starship-design

Steve replied to me: >My understanding (which is blissfully free of any messy equations :-) is >that the rate of virtual particle creation is proportional to the rate >of change of gravitational force near the black hole's event horizon. >Since the force of gravity at r is proportional to 1/r^2, then the >derivative of that is proportional to 1/r^3. Yup. >So although the surface >area of the black hole is proportional to r^2, the rate of evaporation >is proportional to r^2/r^3 or 1/r. Eh, yup. >Furthermore the Schwarzchild radius >for a "classical" black hole is directly proportional to the black hole >mass. So despite the shrinking surface area the rate of evaporation >increases asymptotically as the black hole shrinks and the black hole >finishes evaporating with an impressive bang. Yep, you convinced me. Timothy P.S. A little addition (which partially relates to the problem): F=GMm/r^2 r=2GM/c^2 Fsurface=GMm/4G^2M^2*c^4=mc^4/(4GM) Asurface=4pi r^2=4pi (4 G^2 M^2/c^4)=16 pi G^2 M^2/c^4 Fsurface/Asurface=c^8 m/(64 pi G^3 M^3) The surfaceforce per surface area increases with the 3th power when the mass decreases! (NB: The total surface area decreases with the 3th power too, and since we're sitting on the surface, the distance to the center decreases with the first power at the same time)

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