Widdekind Posted August 12, 2010 Posted August 12, 2010 The Schwarzschild Radius of an electron is ~10-57 m, whist that of ("bare", "undressed") u & d quarks is about 10x larger. Never-the-less, a pure point particle has zero radius, and, so, R = 0 < RS. Wouldn't, therefore, demanding that all particles be treated as classical pure point particles imply, that all matter is made up of mini-Black-Holes ? And, such mini-Black-Holes would evaporate by Hawking Radiation in ~10-98s. How could matter, made up of mini-Black-Holes, which evaporate in less than a Planck Time, make up (stable) matter ? (Doesn't the "smeared out" distribution of matter, in the Wave Functions of quantum theory, 'neatly circumvent such issues? To compress an electron, down to its Schwarzschild Radius, would require giving it a relativistic energy, via the Heisenberg Uncertainty Principle, of 300 gigatons mass equivalent.)
ajb Posted August 12, 2010 Posted August 12, 2010 Treating objects as point particles as an idealisation would not imply that particles are black holes. In reality quantum mechanics gives an effective size to particles and this prevents the formation of black holes in general. Try putting the Compton wavelength of a massive particle equal to the Schwarzschild radius. What is the mass required to form a black hole? Is this anything near the mass of particles in the standard model?
Widdekind Posted August 13, 2010 Author Posted August 13, 2010 Wouldn't that (basically) be the Planck Mass ? [math]\frac{2 G m}{c^2} = R_S = \lambda_C = \frac{\hbar}{m c}[/math] [math]m = \sqrt{\frac{\hbar c}{2 G}} \approx 10^{-8} kg[/math] Could, then, we attribute some "substantial reality", to the Wave Functions, of quantum objects -- treating, for example, [math]\rho_m = | \Psi |^2 = < \Psi^{*} | \Psi >[/math] as, effectively, a "microscopic mass density", of said quantum object* ? What about "Bohmian" Hidden Variables interpretations of QM, which treat quantum objects as classical point particles (albeit "guided", non-locally & instantaneously, by "pilot waves") ? Could one exclude such classical conceptions, of quantum objects, on the grounds that, even if their pilot waves "smear out" our macroscopic knowledge of those point particles, the particles themselves would still actually & really be "micro-Black-Holes", which would radiate away in under a Planck Time ?? * Along like lines, one could call [math]\rho_p = < \Psi^{*} | c \hat{ p } | \Psi >[/math] the (relativistic) momentum density; and, [math]\rho_E = < \Psi^{*} | \hat{ E } | \Psi >[/math] the (relativistic) energy density (?).
ajb Posted August 14, 2010 Posted August 14, 2010 It is indeed the Planck mass. I think the idea of a quantum particles mass density has been discussed before on this forum. Have a search of the forums.
Widdekind Posted August 19, 2010 Author Posted August 19, 2010 (thanks for the reply) It seems, in a sense, that quantum spatial dispersion acts like a "pair of snowshoes", for quantum objects, so that they don't "poke holes in spacetime", and "fall down into the snow".
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