# Planck Probablity Cloud

1. Oct 27, 2005

### Orion1

Planck Probability Cloud...

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

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

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Last edited: Oct 27, 2005