# [ExI] Fwd: Logical Proof of a Multiverse

Mon Jan 24 03:33:20 UTC 2022

Error: proving that "our causal cell or Hubble volume does not contain the
information storage capacity to compute the electron positions of a single
gold (Au) atom" does not directly prove that "either the multiverse exists
and is so mind bogglingly big as to be practically infinite OR the inner
working of atoms is governed by a universal wave function that is running
on a computer that is causally disconnected from and vastly larger than our
observable universe".

On Sun, Jan 23, 2022 at 1:20 PM Stuart LaForge via extropy-chat <
extropy-chat at lists.extropy.org> wrote:

>
>
> ----- Forwarded message from Stuart LaForge <stuart.laforge at gmail.com>
> -----
>     Date: Sun, 23 Jan 2022 13:10:19 -0800
>     From: Stuart LaForge <stuart.laforge at gmail.com>
> Subject: Fwd: Logical Proof of a Multiverse
>       To: Stuart LaForge <avant at sollegro.com>
>
> ---------- Forwarded message ---------
> From: Stuart LaForge <stuart.laforge at gmail.com>
> Date: Sun, Jan 23, 2022 at 1:00 PM
> Subject: Logical Proof of a Multiverse
>
>
> Claim: Our causal cell or Hubble volume does not contain the information
> storage capacity to compute the electron positions of a single gold (Au)
> atom.
>
> Proof: Let [image: S_i] represent the Shannon information capacity of our
> Hubble volume as calculated by generally accepted parameters for the Hubble
> volume applied to the formula for the Bekenstein bound as follows.
>
> The equation for the Bekenstein bound is
>
> [image: \mathbb{S}_i \leq \frac{2\pi c M R} {\hbar ln2}]
>
> The mass of the Hubble volume (including dark components) can be adequately
> represented by the product of the critical density [image: \rho_c] and the
> spherical Hubble volume [image: V_H] as follows where H is the Hubble
> parameter (constant).
>
> [image: \rho_c = \frac{3 {H}^2}{8 \pi G}]
>
> [image: V_H = \frac{4 \pi}{3} R_H^3]
>
> Here [image: R_H = \frac{c}{H}] is the Hubble Radius. [image: M_H = \rho_c
> V_H = \frac{c^3}{2 H G}] is the Hubble mass.
>
> Substituting the Hubble mass and the Hubble radius into the Bekenstein
> bound formula gives us the maximum total information content of our Hubble
> volume:
>
> [image: S_H \leq \frac{\pi c^5}{ln2 H^2 G \hbar} \approx 3.25 \times
> 10^{122} bits]
>
> Remember our Hubble volume is the set of all space-time that is subluminal
> and timelike relative to us; that is to say our causal cell is by one
> definition our causally bounded universe.
>
> Now let's try to compute the electron configuration of a gold atom. Gold
> (AU) has an atomic number 79 with 79 protons and 79 electrons. Now let us
> ignore the protons and simply try to compute the electron distribution
> across the atom. Since the gold atom has a radius of 144 picometers (pm)
> and a diameter of 288 pm then we could envisage a 3-dimensional
> approximation of a gold atom in a 288 x 288 x 288 cubic grid with a total
> of $288^3 = 23,887,872 voxels that any one or more of the 79 electrons > could occupy at any one time according to the multiparticle time-dependent > Schrodinger's equation. > > [image: \Psi(\vec{r}_1, \vec{r}_2,...\vec{r}_{79};t)] > > Since in order to determine a probability for each of the 79 electrons to > be in each of the 23,887,872 voxels, we would have to integrate over three > terms dx,dy, and dz for each of the 79 electrons to determine the > probabilities of each of the [image: 288^{3 \times 79} = 23887872^{79} > \approx 7.52 \times 10^{582}] possible states. > > Since each of the [image: 7.52 \times 10^{582}] states will have to be > represented by a scalar probability value, and each such scalar value must > exceed 1 bit of Shannon entropy, and [image: 7.52 \times 10^{582} >> 3.25 > \times 10^{122}], the claim is proved. > > Q.E.D. > > So logically either the multiverse exists and is so mind bogglingly big as > to be practically infinite OR the inner working of atoms is governed by a > universal wave function that is running on a computer that is causally > disconnected from and vastly larger than our observable universe. In either > case, there is much more to reality than meets the eye. > > Stuart LaForge > > > > > > > > > > > > > > > < > https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail&utm_term=icon > > > Virus-free. > www.avast.com > < > https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail&utm_term=link > > > > <#m_-6820246920155310988_m_1258230283594942653_m_8553758471747849954_DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2> > > ----- End forwarded message ----- > > > > > ---------- Forwarded message ---------- > From: Stuart LaForge <stuart.laforge at gmail.com> > To: Stuart LaForge <avant at sollegro.com> > Cc: > Bcc: > Date: Sun, 23 Jan 2022 13:10:19 -0800 > Subject: Fwd: Logical Proof of a Multiverse > > > ---------- Forwarded message --------- > From: Stuart LaForge <stuart.laforge at gmail.com> > Date: Sun, Jan 23, 2022 at 1:00 PM > Subject: Logical Proof of a Multiverse > To: <extropolis at googlegroups.com> > > > Claim: Our causal cell or Hubble volume does not contain the information > storage capacity to compute the electron positions of a single gold (Au) > atom. > > Proof: Let [image: S_i] represent the Shannon information capacity of our > Hubble volume as calculated by generally accepted parameters for the Hubble > volume applied to the formula for the Bekenstein bound as follows. > > The equation for the Bekenstein bound is > > [image: \mathbb{S}_i \leq \frac{2\pi c M R} {\hbar ln2}] > > The mass of the Hubble volume (including dark components) can be > adequately represented by the product of the critical density [image: > \rho_c] and the spherical Hubble volume [image: V_H] as follows where H > is the Hubble parameter (constant). > > [image: \rho_c = \frac{3 {H}^2}{8 \pi G}] > > [image: V_H = \frac{4 \pi}{3} R_H^3] > > Here [image: R_H = \frac{c}{H}] is the Hubble Radius. [image: M_H = > \rho_c V_H = \frac{c^3}{2 H G}] is the Hubble mass. > > Substituting the Hubble mass and the Hubble radius into the Bekenstein > bound formula gives us the maximum total information content of our Hubble > volume: > > [image: S_H \leq \frac{\pi c^5}{ln2 H^2 G \hbar} \approx 3.25 \times > 10^{122} bits] > > Remember our Hubble volume is the set of all space-time that is subluminal > and timelike relative to us; that is to say our causal cell is by one > definition our causally bounded universe. > > Now let's try to compute the electron configuration of a gold atom. Gold > (AU) has an atomic number 79 with 79 protons and 79 electrons. Now let us > ignore the protons and simply try to compute the electron distribution > across the atom. Since the gold atom has a radius of 144 picometers (pm) > and a diameter of 288 pm then we could envisage a 3-dimensional > approximation of a gold atom in a 288 x 288 x 288 cubic grid with a total > of$288^3 = 23,887,872 voxels that any one or more of the 79 electrons
> could occupy at any one time according to the multiparticle time-dependent
> Schrodinger's equation.
>
> [image: \Psi(\vec{r}_1, \vec{r}_2,...\vec{r}_{79};t)]
>
> Since in order to determine a probability for each of the 79 electrons to
> be in each of the 23,887,872 voxels, we would have to integrate over three
> terms dx,dy, and dz for each of the 79 electrons to determine the
> probabilities of each of the [image: 288^{3 \times 79} = 23887872^{79}
> \approx 7.52 \times 10^{582}] possible states.
>
> Since each of the [image: 7.52 \times 10^{582}] states will have to be
> represented by a scalar probability value, and each such scalar value must
> exceed 1 bit of Shannon entropy, and [image: 7.52 \times 10^{582} >> 3.25
> \times 10^{122}], the claim is proved.
>
> Q.E.D.
>
> So logically either the multiverse exists and is so mind bogglingly big as
> to be practically infinite OR the inner working of atoms is governed by a
> universal wave function that is running on a computer that is causally
> disconnected from and vastly larger than our observable universe. In either
> case, there is much more to reality than meets the eye.
>
> Stuart LaForge
>
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> www.avast.com