I get it now:

pradius = UnitConvert[codata["ProtonRMSChargeRadius"],"PlanckLength"]
= (5.206\[PlusMinus]0.012)*10^19Subscript[l, P]
pvolume=(4/3) Pi pradius^3
= (5.91\[PlusMinus]0.04)*10^59Subsuperscript[l, P, 3]
h2pvolume=codata["HubbleVolume"]/pvolume
= (1.024\[PlusMinus]0.020)*10^123
hsurface=UnitConvert[4 Pi codata["HubbleLength"]^2,"PlanckArea"]
= (8.99\[PlusMinus]0.11)*10^122Subsuperscript[l, P, 2]
RelativeError[QuantityMagnitude[h2pvolume],QuantityMagnitude[hsurface]]
= -0.122\[PlusMinus]0.023

As Dirac-style "Large Number Coincidences" go, a -12±2% relative error is
quite remarkable since Dirac was intrigued by coincidences with orders of
magnitude errors!

However, get a load of this:

CH4=2^(2^(2^(2^2-1)-1)-1)-1
= 170141183460469231731687303715884105727
protonAlphaG=(codata["PlanckMass"]/codata["ProtonMass"])^2
= (1.69315\[PlusMinus]0.00004)*10^38
RelativeError[protonAlphaG,CH4]
= 0.004880\[PlusMinus]0.000022

0.5±0.002% relative error!

Explain that.


On Sun, Mar 31, 2024 at 9:45 PM Matt Mahoney <mattmahone...@gmail.com>
wrote:

> On Sun, Mar 31, 2024, 9:46 PM James Bowery <jabow...@gmail.com> wrote:
>
>> Proton radius is about 5.2e19 Plank Lengths
>>
>
> The Hubble radius is 13.8e9 light-years = 8.09e60 Planck lengths. So
> 3.77e123 protons could be packed inside this sphere with surface area
> 8.22e122 Planck areas.
>
> The significance of the Planck area is it bounds the entropy within to A/4
> nats, or 2.95e122 bits. This makes a bit the size of 12.7 protons, or about
> a carbon nucleus. https://en.wikipedia.org/wiki/Bekenstein_bound
>
> 12.7 is about 4 x pi. It is a remarkable coincidence to derive properties
> of particles from only G, h, c, and the age of the universe.
>
>>
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