1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    Attached Files:

    Last edited: Oct 27, 2005
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted