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|    sci.physics.research    |    Current physics research. (Moderated)    |    17,516 messages    |
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|    Message 16,437 of 17,516    |
|    toadastronomer@gmail.com to toadast...@gmail.com    |
|    Re: gamma factor in special relativity    |
|    11 Feb 19 14:14:50    |
      [[Mod. note -- I apologise for the long delay in posting this article,       which was received by by s.p.r moderation system on 2019-01-28.       -- jt]]              On Friday, August 24, 2018 at 10:40:51 AM UTC-4, toadast...@gmail.com wrote:       [[Mod. note -- 88 excessively-quoted lines snipped here. -- jt]]       > 24-AUG-2018       >       >       > Gamma-limit Conjecture Briefly:       >       > The coupling, between the energy-momentum of a mass m, and constant       > scalar curvature S, of relevant 3-sphere, may become effective if       > a) the mass can be confined to within twice its Compton wavelength,       > and b) the particle's velocity satisfies the condition gamma = 1/2(S),       > where gamma = (((1 - v/c)^2)^1/2)^-1.       >       > According to Friedrich, a solution of the Einstein-Dirac equation is       > a field psi, satisfying the equations,       >       > Ric-(1/2)S*g = T_psi ; D(psi) = lambda(psi),       >       > where Ric is Ricci tensor, S = Tr(Ric) is scalar curvature, g is       > metric tensor, T_psi is energy-momentum tensor, D is Dirac operator       > and lambda is WK-number. Specifically, for the two metrics with WK-       > spinors on S^3, contained in the space N^3(K,L,M), where K = M = 1       > and L = 1/4S, I find       >       > psi = S / lambda ; lambda = S*psi^-1; S = lambda*psi = D(psi).       >       > Now, with gamma-limit defined as (gamma - 1) = (1/gamma), we put an       > electron with mass m, in a square potential with dimension       > d = 2(h/mc) and make the electron velocity 78.6151385% light speed.       > The electron's kinetic energy is E_n = (1/2)mv^2 and we then find the       > total energy is E = E_nS. . . At the gamma limit, the electron's       > energy is the product of the kinetic energy and the scalar curvature.       >       > In the inelastic scattering model, an incident photon gives the electron       > in the box its kick, the incident photon's frequency is       >       > nu_i = (E/h)*(1/2), with h Planck's constant (eV sec).       >       > If the diameter of the potential is made about 38% smaller, the incident       > photon frequency has to be corrected by another factor of S/2,       >       > nu_i = (E/h)*L; L = 1/4S.       >       > Just beyond this scale and gamma diverges. To within a factor of 4       > the input energy is E_i = 2(E). So maybe pair production to account       > for the gamma divergence. Dunno.       >       >       > Cheers,       > mark jonathan horn              28-JAN-2019                     I noticed, in a recent review of notes, that the mass term I defined was       NOT that of the electron, but rather the extrapolated       photocharge mass from an inelastic scattering model, found       to account for experimental data at optical energies from a       camera's output. Pretty sure any mass works at the gamma-limit,       but haven't done the calculations.              As a point of clarification then, the photocharge mass is:              m = (m* - m_e),              where effective mass m* = m_e(1 + eta^2)^1/2;              m_e is electron mass and eta is electric field strength parameter[1]              eta = e( |
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