Mark Rohrbaugh's formula, that I used to calculate the proton radius to a higher degree of precision than QED or current measurements, results in a slightly higher relative error with respect to the Hubble Surface prediction, but that could be accounted for by the 11% tolerance in the Hubble Surface calculation derived from the Hubble Radius, or the 2% tolerance in the Hubble Volume calculation taken in ratio with the proton volume calculated from the proton radius:
pradiusRohrbaugh=(8.41235641\[NegativeVeryThinSpace]\[NegativeVeryThinSpace]\[NegativeVeryThinSpace](35\[NegativeThinSpace]\[PlusMinus]\[NegativeThinSpace]26\[NegativeVeryThinSpace])*10^-16)m pradiusRohrbaughPL=UnitConvert[pradiusRohrbaugh,"PlanckLength"] pvolumeRohrbaugh=(4/3) Pi pradiusRohrbaughPL^3 h2pvolumeRohrbaugh=codata["HubbleVolume"]/pvolumeRohrbaugh RelativeError[QuantityMagnitude[h2pvolumeRohrbaugh],QuantityMagnitude[hsurface]] (8.41235641\[NegativeVeryThinSpace]\[NegativeVeryThinSpace]\[NegativeVeryThinSpace](35\[NegativeThinSpace]\[PlusMinus]\[NegativeThinSpace]26\[NegativeVeryThinSpace])*10^-16)m (5.20484478\[NegativeVeryThinSpace]\[NegativeVeryThinSpace]\[NegativeVeryThinSpace](84\[NegativeThinSpace]\[PlusMinus]\[NegativeThinSpace]16\[NegativeVeryThinSpace])*10^19)Subscript[l, P] (5.90625180\[NegativeVeryThinSpace]\[NegativeVeryThinSpace]\[NegativeVeryThinSpace](6\[NegativeThinSpace]\[PlusMinus]\[NegativeThinSpace]5\[NegativeVeryThinSpace])*10^59)Subsuperscript[l, P, 3] = (1.025\[PlusMinus]0.019)*10^123 = -0.123\[PlusMinus]0.022 On Tue, Apr 2, 2024 at 9:16 AM James Bowery <[email protected]> wrote: > 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 <[email protected]> > wrote: > >> On Sun, Mar 31, 2024, 9:46 PM James Bowery <[email protected]> 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. >> >>> >>> *Artificial General Intelligence List <https://agi.topicbox.com/latest>* >> / AGI / see discussions <https://agi.topicbox.com/groups/agi> + >> participants <https://agi.topicbox.com/groups/agi/members> + >> delivery options <https://agi.topicbox.com/groups/agi/subscription> >> Permalink >> <https://agi.topicbox.com/groups/agi/Teaac2c1a9c4f4ce3-Me023643f4fef1483cfab3ad6> >> ------------------------------------------ Artificial General Intelligence List: AGI Permalink: https://agi.topicbox.com/groups/agi/Teaac2c1a9c4f4ce3-M17fccdbdbf49f194fe6532ef Delivery options: https://agi.topicbox.com/groups/agi/subscription
