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Planck Probablity Cloud

  1. Oct 27, 2005 #1
    Planck Probability Cloud...


    I am inquiring if there is anything incorrect with the conceptualism regarding a Planck Singularity as existing within a probability cloud?

    [tex]P(r) dr = | \psi |^2 dV[/tex]
    [tex]dV = 4 \pi r^2 dr[/tex]
    [tex]P(r) dr = 4 \pi r^2 | \psi |^2 dr[/tex]
    [tex]P(r) = 4 \pi r^2 | \psi |^2[/tex]
    [tex]r_p = \sqrt{\frac{\hbar G}{c^3}}[/tex]
    [tex]| \psi_{1s} |^2 = \left( \frac{1}{\pi r_p^3} \right) e^{- \frac{2r}{r_p}}[/tex]
    [tex]P_{1s}(r) = \left( \frac{4 r^2}{r_p^3} \right) e^{-\frac{2r}{r_p}} = 4 \left( \frac{c^3}{\hbar G} \right)^{\frac{3}{2}} r^2 e^{-\frac{2r}{r_p}}[/tex]
    Planck Singularity radial probability density:
    [tex]\boxed{P_{1s}(r) = 4 \left( \frac{c^3}{\hbar G} \right)^{\frac{3}{2}} r^2 e^{-\frac{2r}{r_p}}}[/tex]
    [tex]M_p = \sqrt{\frac{\hbar c}{G}}[/tex]
    [tex]\rho = \frac{M_p}{dV} = \frac{M_p}{4 \pi r_p^2 dr} = \frac{}{4 \pi} \sqrt{ \frac{c^7}{\hbar G^3}} \frac{}{dr}[/tex]
    Planck Singularity average probability cloud density:
    [tex]\boxed{\rho_{1s} = \frac{}{4 \pi} \sqrt{ \frac{c^7}{\hbar G^3}} \frac{}{dr}}[/tex]
     

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    Last edited: Oct 27, 2005
  2. jcsd
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