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Re: M Physics 21: SR's E= SQRT(p^2c^2+ m^2c^4) is an ill-defined function mathematically & physically, with proof ***GR's Curved Spacetime in SR's Black Hole - 'X' Particle***Posted by Coppernicus on May 09, 2003 at 12:07:08: In Reply to: M Physics 21: SR's E= SQRT(p^2c^2+ m^2c^4) is an ill-defined function mathematically & physically, with proof ***GR's Curved Spacetime in SR's Black Hole - 'X' Particle*** posted by kx21 on May 07, 2003 at 16:49:12: Kx21, RE ur: 2) Massless Particle: E = pc (p > 0 & m = 0) 3) Massive Particle: E = mc^2 (p = 0 & m > 0) Does this mean that pc =(not)= mc^2? Does it mean that Massless Particles conversion into Massive Particles is missing something? And if it is missing something, then what is it? My way to get around this conundrum was to convert the equation into E/c^2 = m, then modify the resulting mass so that it is incomplete by the amount of g (5.9e-39), to read as E/c^2 = (m-g), with the resulting mass m set at A=1. The product this equation has a result just shy of =1, so that E/c^2 = [1-(5.9e-39)], which is almost A=1, but not quite. Does that help in any way what you are trying to achieve? I also had converted E = pc into (if at v=c then p=h/w) pc = hc/w, where w is EM wave lambda and h is Planck's constant, so that E/c^2 becomes hc/w/c^2, which works out to h/cw. Rewriting E/c^2 = (m-g) then becomes h/cw = (m-g), or more formally h/cw + g = m, which is set to m=1. Playing around with these numbers gives interesting results, one of which is that for lower w, you get higher g, and if g=1, you get total max gravity. Using G^2=gc^2 as a conversion formula, at g=1, G^2=c^2, which taking square root becomes G = c, so light speed is nulled by max gravity. Now, if either a Massless Particle or Massive Particle find themselves in the G = c zone (blackhole), then they both dissolve into pure energy. Is this cool or what? :) C.
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