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Fundamental J. Modern Physics, Vol. 2, Issue 1, 2011, Pages 23-72 Published online at http://www.frdint.com/ FRACTAL PHYSICS THEORY - NUCLEONS AND THE STRONG FORCE LEONARD J. MALINOWSKI BASF Dispersions and Resins Monaca, Pennsylvania USA Abstract This fourth article, in a series of five, intends to demonstrate the value of applying Fractal Physics Theory to further understanding of nucleon structure and nucleon interactions. An ideal neutron is assumed, as a limiting case, to be composed of 100% subquantum scale Hydrogen atoms. With this assumption and identifying the pre-solar system mass with the mass of a cosmic scale neutron, it appears possible to derive the masses of all the nucleons. The strong nuclear force is discussed by introducing subquantum scale fusion, lilliputian scale electromagnetic forces, and lilliputian scale gravity. 1. Introduction The first three articles of the Fractal Physics Theory series provide essential background information [1, 2, 3]. The binding energy of a nucleus is the energy released when individual protons and neutrons are brought together to form a nucleus. This article proposes the strong nuclear force is a combination of internucleon interactions and intra-nucleon interactions (Figure 1). The inter-nucleon interactions involve lilliputian scale (ls) gravitational attraction, the reducing effects of the intervening neutron’s ls-dielectric on proton-proton Coulomb repulsion and a variety of ls-electromagnetic attractions induced by a proton on adjacent neutrons, two of which are examined. First, if subquantum scale (sqs) conduction metals exist Keywords and phrases : fractals, lilliputian gravity, nucleons, strong force. Received April 4, 2011 © 2011 Fundamental Research and Development International 24 LEONARD J. MALINOWSKI in the neutron, then a net ls-Coulombic attraction will occur between a neutron and a proton. The electric field of the proton will attract any sqs-conduction electrons present while repelling their associated sqs-conduction lattice ions. Second, the electric field of the proton will induce sqs-electric dipoles in any non polar sqs-atoms comprising the neutron creating a short range attractive force. The intra-nucleon interactions involve sqs-fusion of the nucleon’s composite sqsatoms, which results in increasing the sqs-nuclear bonds of the sqs-atomic nuclei composing the nucleons. Idealized ls-chemical compositions of representative nuclei are calculated considering the inter-nucleon energies and basic fusion equations with atomic masses from initial idealized ls-chemical compositions of neutrons and protons. Figure 1. Diagram of fractal strong nuclear force. This article begins with idealized neutron beta decay and idealized ls-chemical compositions of the neutron, proton, and electron, followed by a brief discussion on the proposed origin of the muon. Next, the increased strength of gravity at the lilliputian scale is presented. Finally the lilliputian scale Coulombic interactions are examined with ls-chemical compositions resulting from sqs-fusion for a few representative nuclei. 2. Fractal Proton, Neutron and Electron Chemical Compositions 2.1. Cosmic scale neutron beta decay Most stars appear to be composed primarily of hydrogen. In order to examine nuclear lilliputian scale chemical compositions, an idealized cosmic scale free neutron is initially assumed to be composed of 100% Hydrogen. Consider the free neutron beta decay: n → p + + e − + antiν e FRACTAL PHYSICS THEORY - NUCLEONS AND … 25 The mass difference ∆m between reactant and products must equal the energy liberated by the reaction (Table 1). The mass lost in cosmic scale (cs) neutron beta decay will be modeled by the mass lost from fusion reactions using the isotopes in Table 2. Let the cosmic scale neutron be composed of 100% solid phase Hydrogen atoms of atomic weight 1.007940754 amu. Allow thermonuclear fusion to initiate, perhaps by continued cosmic ray bombardment of the dark star’s surface [2]. Let the cosmic scale electron be composed of the stable thermonuclear fusion end product Iron of mass 55.84514562 amu. Table 1. Summary of idealized compositions of fractal neutrons, protons, and electrons [1] Table 2. Elements assumed composing idealized cs-neutrons, cs-protons, and cselectrons [4] A H He Fe Mass (amu) Abundance (%) Mass (amu) 1 1.007825032 99.9885 1.007709132 2 2.014101778 0.0115 0.000231622 3 3.016029319 0.000134 0.000004041 4 4.002603254 99.999866 4.002597891 54 53.9396105 5.845 3.15277023 56 55.9349375 91.754 51.32254255 57 56.935394 2.119 1.20646100 58 57.9332756 0.282 0.16337184 Composition (amu) 1.007940754 4.002601932 55.84514562 2.2. Cosmic scale electron The mass of a cs-electron and the mass of iron are used to calculate the number of iron atoms composing the cs-electron: Moles of Iron: (1.083591301 × 1030 g ) 55.84514562 = 1.940350032 × 10 28 moles, 26 LEONARD J. MALINOWSKI Atoms of Iron: (1.940350032 × 10 28 moles ) ( 6.0221415 × 10 23 ) = 1.168506245 × 1052 atoms. The decaying cosmic scale neutron must fuse Hydrogen to form the cs-electron’s Iron: (1.168506245 × 1052 ) (56) = 6.543634972 × 1053 Hydrogen atoms fused. 2.3. Cosmic scale proton Mass of Hydrogen used to form the Iron of the cs-electron: ( 6.543634972 × 1053 atoms of H ) (1.007940754 ) ( 6.0221415 × 1023 ) (1000) = 1.095224409 × 10 27 kg of H consumed by the cs-neutron in forging the cselectron. Remaining cs-neutron mass after accounting for the cs-electron: 1.992381602 × 1030 kg − 1.095224409 × 10 27 kg = 1.991286378 × 1030 kg. Additional mass that must be radiated by fusion reactions to obtain the cs-proton mass: 1.991286378 × 1030 kg − 1.989639050 × 1030 kg = 1.647328000 × 10 27 kg. Let the 1.647328 × 10 27 kg be consumed by the following proton - proton chain: 3H 2 → H 2 + 4 He 2 1. p + H + e − → D + ( e − + e + ) + ν e , 2. p + H + e − → D + ( e − + e + ) + ν e , 3. D + H →3He 2 , 4. D + H →3He 2 , 5. 3 He 2 + 3He 2 → 4 He 2 + H 2 . The proton - proton chain results in combining four hydrogen atoms into a helium atom and releasing an amount of energy equivalent to the mass difference: FRACTAL PHYSICS THEORY - NUCLEONS AND … Initial Hydrogen : Final Helium : 27 (4) (1.007940754 amu ) = 4.031763016 amu, 4.002601932 amu = 0.029161084 amu, Mass difference, ∆m : 4.842311318 × 10 −29 kg = 4.352052374 × 10 −12 J = 27.163376148 MeV. The # of Helium atoms needed to be fused from Hydrogen to consume the remaining mass difference between the decaying cs-neutron and the final cs-proton (1.647328000 × 10 27 kg ): 1.647328000 × 1027 kg 4.842311318 × 10−29 kg = 3.401945666 × 1055 Helium atoms required. The cs-proton contains 3.401945666 × 10 55 Helium atoms with the rest of its mass attributed to Hydrogen (Table 3). Mass of Helium in the cs-proton: ( 3.401945666 × 1055 ) ( 4.002601932 ) ( 0.001) ( 6.0221415 × 1023 ) = 2.261095043 × 10 29 kg. Table 3. Chemical composition of idealized cosmic scale proton Element Hydrogen # of Atoms Helium 3.401945666 ×1055 1.053655626 × 10 Mass (kg) 57 30 % Mass 88.635652 2.261095043 ×10 29 11.364348 1.98963905× 1030 100.000000 1.763529545 × 10 3. Cosmic Scale Proton Central Pressure and Temperature Estimate 3.1. The pressure and temperature at the Sun’s center Pc = 19G ( Sun ,s Mass )2 ( Sun ,s radius )4 , G = 6.6742 × 10−11 Nm 2 kg 2 , M s = 1.99 × 1030 kg, (1) 28 LEONARD J. MALINOWSKI Rs = 6.96 × 108 m, Pc = 2.14 × 1016 Pa . Pc = nc kTc , (2) mc = mean particle mass at the Sun’s center is averaged over ions and electrons, nc = central density ( mc ) = 150g cm3 1.5 × 10−24 g = 1.0 × 1032 m −3 , k = 1.3806505 × 10 −23 J K , Tc = central temperature = 15 × 106 K. 3.2. The pressure and temperature at a cosmic scale proton’s center Let the cosmic scale beta decay of a cs-neutron end with mostly Helium formed in the core of the cs-proton. Central pressure, Pc = 19G (1.98963905 × 1030 kg )2 ( 3.788566 × 108 m )4 = 2.4366995 × 1017 Pa . Let the central Helium atoms have their two electrons per atom exist in their lsplasma phase. The translational kinetic energies of the two electrons have been divided amongst the electron’s composite sqs-Iron atoms in random directions. Thus, an ideal sqs-iron plasma gas is formed, contracted in size until the pressure of the delocalized electrons equal the ambient central pressure. PV = nRT = 24.587387 eV + 54.417760 eV = 1.265801923 × 10 −17 J, (3) Volume = 5.194739535 × 10 −35 m3 , radius = 2.314684 × 10 −12 m. Compare this Helium radius to the Helium shell radius of ~ 1 × 10 −12 m in the Radon atom. Helium atomic mass = ( 4.002601932 amu ) (1.66053886 × 10−27 kg amu ) = 6.646476049 × 10−27 kg. FRACTAL PHYSICS THEORY - NUCLEONS AND … 29 The Helium atomic density at the core of the cs-proton ρ = 1.279462811 × 108 kg m3 . Pc = nc kTc , mc = 6.646476049 × 10 −27 kg 3 = 2.215492016 × 10−27 kg, nc = (1.279462811 × 108 kg m 3 ) mc = 5.775072993 × 1034 m −3 , k = 1.3806505 × 10 −23 J K , Tc = 305600 K. The cosmic scale proton’s central temperature is 305600 K. The temperature decreases until 2.725 K at the cs-proton’s surface. Cs-proton’s must be in thermodynamic equilibrium with the microwave background radiation. 4. Proposed Origin of the Muon Cosmic scale beta decay of a cs-neutron ends with mostly Helium formed in the core of the cs-proton, based on the atomic densities in Table 4. Table 4. Densities of Hydrogen and Helium atoms [4] Element Radius (meters) Mass (amu) Density ( kg m 3 ) Hydrogen of H 2 3.7072 × 10 −11 1.007940754 7842.6 Helium 2.7339 × 10 −11 4.002601932 77650 When cosmic rays ( ~ 90% protons) strike Earth’s upper atmosphere (mainly Nitrogen and Oxygen nuclei) many pions are released that quickly decay into muons. The muon’s mass is 1.88353140 × 10−28 kg [4], which is 11.261% of the proton’s mass. The 1.90082550 × 10−28 kg of sqs-Helium roughly calculated to exist in a proton is extremely close, 100.9%, to the muon’s mass. It is proposed that the collision of a high energy proton and a Nitrogen or Oxygen nucleus are releasing coated sqs-Helium cores from protons, the pions. Consider the high energy head on collision of two protons (Figure 2). Tremendous ls-compression shock waves start at the proton-proton contact point and travel inwards and rebound off the ls-Helium 30 LEONARD J. MALINOWSKI cores. The net effect must be rapid sqs-fusion from sqs- H 2 to higher sqs-masses. The freed sqs-Helium core plus this extra sqs-fusion mass “coating” constitutes the charged pion mass. The charged pion then sheds its sqs-fusion mass layer and the sqs-Helium fuses to sqs-Iron, the muon. Helium 4 has zero spin and pions have zero spin. Figure 2. Protons colliding create pions. Let all the available Helium in the core of an idealized cs-proton fuse to Iron: ( 3.401945666 × 1055 He atoms ) 14 = 2.42996119 × 1054 Iron atoms. Mass of fused Iron: ( 2.42996119 × 1054 Iron atoms ) ( 55.84514562 ) ( 0.001) 6.0221415 × 10 23 = 2.253377 × 10 29 kg. Cosmic scale muon mass = (1.88353140 × 10 −28 kg ) (1.189533197 × 1057 ) = 2.240523 × 10 29 kg. This mass of fused Iron equals 100.6% of the cs-muon’s mass. 5. Fractal Gravity 5.1. Lilliputian scale gravity Modern Physics and Fractal Physics Theory (FPT) diverge significantly on this point. FPT calculates the gravitational forces between quantum scale objects as measured in the human scale to be vastly stronger than gravitational forces between these same quantum scale objects calculated by Modern Physics. FPT proposes that the gravitational constant G, in not scale invariant. To accurately depict gravitational forces between nucleons a larger gravitational constant is required [1]. The gravitational force, the Coulombic force and the magnetic force FRACTAL PHYSICS THEORY - NUCLEONS AND … 31 have equivalent scaling fractals (¥) . ¥ Fgravity = ¥ FCoulomb = ¥ FMagnetic = ¥ Mass ¥Length = (1.189533 × 1057 ) ( 3.788566 × 10 23 ) = 3.139797 × 1033 or more succinctly, ¥ Fg = ¥ FC = ¥ FM = ¥ M ¥ L = 3.140 × 1033. (4) Consider the equation, F = ma = Gm1m2 r 2 . (5) Rewritten, from the Principle of Scalativity [1], ¥ F = ¥ M ¥a = ¥G ¥ M 2 ¥ L2 , (6) where ¥a = ¥ L ¥ t 2 = ¥ L ¥ L2 = ¥ L−1. Then by substitution ¥ M ¥ a = ¥ M ¥ L−1 = ¥G ¥ M 2 ¥ L2 . Solving for ¥G = ¥ L ¥ M = ¥ Fg −1 = 3.184919 × 10 −34. (7) Therefore the product of the gravitational force scaling fractal and the gravitational constant scaling fractal are scale invariant, ¥ Fg ¥G = 1. (8) Consider the equation, F = ma = kC q1 q2 r 2 . (9) Rewritten, from the Principle of Scalativity, ¥ F = ¥ M ¥a = ¥ kC ¥ Q 2 ¥ L2 , (10) where ¥ a = ¥ L−1 and ¥Q = ¥ L1 2 ¥ M 1 2 . Then by substitution ¥ M ¥ a = ¥ M ¥ L−1 = ¥ kC ¥Q 2 ¥ L2 = ¥ kC ( ¥ L1 2 ¥ M 1 2 ) 2 ¥ L2 = ¥ kC ¥ L ¥ M ¥ L2 . Solving for ¥ k C = 1. Therefore the Coulomb constant is scale invariant. 5.2. Bohr Hydrogen atom [4] Proton mass, m p : 1.67262171 × 10−27 kg, Proton charge: 1.60217653 × 10−19 C, (11) 32 LEONARD J. MALINOWSKI Electron mass, me : 9.1093826 × 10−31 kg, Electron charge: − 1.60217653 × 10−19 C, Bohr radius: 5.2946541 × 10 −11 m, calculated with the reduced mass of the system, kC : 8987551788 Nm 2 C 2 , [G ]1, 0 : 6.6742 × 10 −11 m 3 ( s 2 kg ) , [G ]−1, 0 : 2.0956 × 1023 m3 ( s 2 kg ), where ¥G = [G ]1, 0 [G ]−1, 0 . FPT agrees with the Modern Physics calculation of the Coulombic force: FC = kC q 2 r 2 = 8.23 × 10 −8 N attraction between the proton and electron. Modern Physics calculation of the gravitational force: Fg = [G ]1, 0 m p me r 2 = 3.63 × 10−47 N attraction between the proton and electron. FPT calculation of the gravitational force: Fg = [G ]−1, 0 m p me r 2 = 1.14 × 10−13 N attraction between the proton and electron. The force of gravity does matter on the quantum scale, but for the Hydrogen atom it is still almost a million times weaker than Coulomb’s force. For heavier atoms such as Xenon, ls-gravity may be part of Van der Waals forces. 5.3. Dineutron Cosmic scale neutron radius: Volume Radius = (1.992381602 × 1030 kg ) ( 7842.6 kg m 3 ) = 2.540461 × 10 26 m 3 , = 3.9289 × 108 m. Neutron radius, rn = ( 3.9289 × 108 m ) ( ¥Length ) = 1.037 × 10 −15 m, Neutron mass, mn = 1.67492728 × 10 − 27 kg, [G ]−1, 0 = 2.0956 × 10 23 m 3 s 2 kg. FRACTAL PHYSICS THEORY - NUCLEONS AND … 33 Two neutrons side by side as in a nucleus will experience a gravitational potential energy: U g = −[G ]−1, 0 ( mn )2 ( 2rn ) = −2.835 × 10 −16 J = −1769 eV . (12) This binding energy should allow thermal neutrons to alter each other’s trajectories. However, due to the conservation of angular momentum, even thermal neutrons that may start to orbit will accelerate as the distance between them shortens. Soon their velocities will exceed their escape velocities. Escape velocity from the neutron, Vescape = ( [G ]−1, 0 mn rn )1 2 = [ ( 2.0956 × 10 23 m 3 ( s 2 kg ) * (1.6749 × 10 − 27 kg ) (1.037 × 10 −15 m ) ]1 2 = 582 km s . Two neutrons each with v = 2.2km s , orbiting each other at a radius r have an initial angular momentum: L i = 2(1.6749 × 10 −27 kg ) ( 2200 m s )r = ( 7.36956 × 10−24 kgm s )r. (13) Their final angular momentum L f = 2mn ( 582000 m s ) (1.037 × 10−15 m ) = 2.0217 × 10−36 kgm 2 s . From L i = L f , the initial farthest radius of orbit leading to a ls-gravitational bound dineutron r = 2.74 × 10 −13 m. 6. Proton-neutron Attraction, Charge-conduction Electrons 6.1. Cosmic scale neutron conduction electrons If a cs-neutron is composed of metal atoms, some valence electrons are free to wander about the rigid lattice structure formed by the positively charged ion cores. A cs-proton’s positive charge will attract the cs-neutron’s conduction electrons to the cs-neutron’s surface adjacent the cs-proton. A typical conductor contributes one conduction electron per atom, so for a dense metal, there are ~ 1 × 10 23 conduction electrons per cm 3 . Half a cs-neutron’s surface area will be directed towards an adjacent cs-proton. To simplify the following calculations, the cs-neutron’s radius 34 LEONARD J. MALINOWSKI will be considered the same as the cs-proton’s radius of 3.788566 × 108 m. Cosmic scale neutron radius : 3.788566 × 108 m, 1 Cosmic scale neutron' s surface area : 2 9.018402 × 1017 m 2 , The radius of a 1cm3 spherical volume is : 0.62035 cm, A radius 0.62035 cm makes a circular area : 1.209 cm 2 . From geometry, the maximum # of conduction e-s available on half a cs-neutron’s surface: ( 9.018402 × 1017 m 2 1.209 × 10−4 m 2 ) × (1 × 1023 e −s cm3 ) = 7.459 × 10 44 electrons potentially available. Assume the cs-proton’s charge pulls an equal but opposite amount of conduction electron charge (from the 7 × 10 44 electrons available) to half the adjacent csneutron’s surface, that is, 2.122881278 × 10 40 electrons [1]. With these electrons spread evenly over half the cs-neutron’s surface area, the electron surface # density is: e − # σ = ( 2.122881278 × 10 40 e −s ) ( 9.018402 × 1017 m 2 ) = 2.353944 × 1022 e −s m 2 . (14) Is such a high electron surface density realistic? Yes. Consider that the 1s 2 orbital in Radon ( Z = 86) contains 2 electrons and has an average diameter ~ 0.02Å [5]. Surface area of radon’s 1s 2 orbital shell: 1.2566 × 10 −23 m 2 . e − # σ = ( 2 electrons 1.2566 × 10 −23 m 2 ) = 1.6 × 10 23 electrons m 2 . 6.2. Coulomb binding energy between a cs-proton and a cs-neutron’s conduction electrons The cs-proton’s charge is 3.401230560 × 10 21 C. Assume the cs-neutron’s surface conduction electrons are spread out evenly over half the cs-neutron’s surface with e − # σ = 2.353944 × 10 22 e − s m 2 . The electric field E of the cs-proton varies FRACTAL PHYSICS THEORY - NUCLEONS AND … 35 over the cs-neutron’s surface so an approximation is made by calculating the E at the center of five slab areas. In Figure 3 below, the five blue circles drawn beyond the csproton’s surface represent five electric equipotential surfaces all centered on the origin of the axes, with radii as listed. These five blue circles slice the cs-neutron into five slabs, but only part of the fifth slab is required to account for 1 2 the csneutron’s surface. Define Slab 1 as the area contained within the intersection of the cs-proton’s equipotential equation with r = 5R 4 and the cs-neutron’s surface equation. Define Slab 2 as the area contained within three equations; the cs-proton’s equipotential equations with r = 5R 4 , and r = 6R 4 , and the cs-neutron’s surface equation. Continue to define Slabs 3 and 4, as areas bounded by the appropriate three equations, adjacent to and right of the preceding slab. Finally, define Slab 5 as the area contained within the cs-proton’s equipotential equation with r = 8R 4 , the vertical line through the cs-neutron’s center, and the cs-neutron’s surface equation. The areas of the first 4 slabs of the cs-neutron in Figure 3 are calculated below by integrating the curves through the appropriate limits. For example, Slab 1 is divided into two parts. Part I integrates the cs-neutron’s surface equation (positive Yaxis) from x1 = R to x2 = intersection of cs-neutron’s surface equation and csproton’s potential equation with r = 5R 4 . This area is multiplied by 2 (from symmetry) to obtain the total area of Slab 1, Part I. Part two integrates the cs-proton’s potential equation with r = 5R 4 (positive Y-axis) from x2 to x3 = 5R 4 . This area is multiplied by 2 (from symmetry) to obtain the total area of Slab 1, Part II. The straight forward but tedious calculations are reproduced as End Notes to this article. Tables 5.a and 5.b calculate the Coulombic binding energy between a csneutron’s conduction electrons and a cs-proton. A significant Coulomb binding energy ( − 1.7 × 10 44 J ) is expected to occur between a cosmic scale proton and a csneutron that are adjacent in a cs-nucleon if the cs-neutron is composed entirely of conducting metals. This binding energy in the cosmic scale as measured in the human scale is − 1.7229 × 10 44 J. Using the energy scaling fractal ¥ E = 1.189533 × 1057 , a fractally self-similar binding energy of 0.904 MeV is expected to occur between a proton and a neutron that are adjacent in a nucleon if the neutron is composed entirely of subquantum scale conducting metals. 36 LEONARD J. MALINOWSKI Table 5.a. Cosmic scale neutron slab values (areas calculated as End Notes to this article) (1) # of conduction e − s = (surface area) ( e − # σ = 2.353944 × 10 22 e − s m 2 ) (2) Qn Charge (C) = ( # of conduction e − s Slab )( − 1.60217653 × 10 −19 C ) (3) Slab #5 = 0.5πR 2 − ( ∑ Slabs 1 through 4) R 2 = 0.167729886 R 2 Table 5.b. Cosmic scale neutron conduction electrons and cs-proton coulomb binding energy, U C Slab Radius (1) R(m) Qn Charge (C )( 2) Binding Energy ( J )(3) 1 9 RP 8 4.2621368 × 108 − 3.7364590 × 10 20 − 2.6798489 × 10 43 2 11RP 8 5.2092783 × 108 − 7.0347353 × 10 20 − 4.1280760 × 10 43 3 13RP 8 6.1564198 × 108 − 9.1205964 × 10 20 − 4.5286889 × 10 43 4 15 RP 8 7.1035613 × 108 − 1.0488676 × 10 21 − 4.5135891 × 10 43 5 17 RP 8 8.0507028 × 108 − 3.6318396 × 10 20 − 1.3790193 × 10 43 − 3.4012306 × 10 21 − 1.7229222 × 10 44 ∑ (1) R p = 3.788566 × 108 m; the electric field E of the proton is calculated at the midpoint of the surface area slab (2) Qn Charge (C) from Table 5.a, column 6 (3) U C = kC Q p Qn R; kC = 8987551788 Nm 2 C 2 ; Q p = 3.401230560 × 10 21 C FRACTAL PHYSICS THEORY - NUCLEONS AND … 37 Figure 3. Cs-proton-neutron within a larger cs-nucleon, in 2-dimensions. 7. Proton-neutron Attraction, Charge-induced Electric Dipoles [6] For the Deuteron, sqs-conduction metals are not expected to be present in significant quantities. However, an applied electric field, such as that from a nuclear proton, will distort the electronic structures of any sqs-non metals present in the nuclear neutron. Fractal Physics Theory proposes that the proton’s electron field induces sqs-electric dipole moments in the non polar sqs-atoms of adjacent neutrons and tries to align these sqs-dipoles in the electric field’s direction. Two charges separated by a distance r as in Figure 4, constitute an electric dipole with magnitude µ = qr , directed from the negative charge to the positive charge. Dipole moments are measured in debye units, where 1 debye, D = 3.336 × 10 −30 Cm. Figure 4. Electric dipole or dipole moment, µ = qr. Polar solvents have two notable energetic effects when dissolving ionic solids: 1. One end of a dipole may be electrostatically attracted to an ion of opposite charge. 2. Polar solvent molecules reduce the strength of the Coulombic interactions between the ions in solution. The potential energy in vacuum between two ions: U = q1q2 ( 4πε 0 r ) . (15) In a solvent of relative permittivity ε r (also called the dielectric constant, a unit less and temperature dependent quantity) the potential energy between two ions: 38 LEONARD J. MALINOWSKI U = q1q2 ( 4πε r ) , where ε r = ε ε 0 . (16) For example water has a relative permittivity ε r = 78 at 25°C, thus the long range interionic Coulombic interaction energy is reduced by 78 from its magnitude in vacuum. The strong nuclear force is very short range, operating only over a few Fermi, but it appears to have the same strength between two adjacent neutrons, two adjacent protons or a neutron and a proton. The neutron carries no net charge, yet has a magnetic dipole moment µ = −1.91µ N , suggesting it has some sort of charged substructure. Consider the successes of the liquid drop model of the nucleus. FPT proposed that an ideal free neutron is composed of ~ 1.19 × 10 57 sqs-Hydrogen atoms. A large compact frozen solid collection of Hydrogen is dielectric in the presence of electric fields. The neutron permittivity, in many nuclei, is a major component of the strong nuclear force. The neutrons of a nucleus in between protons play the significant polar solvent role of reducing the Coulombic repulsive forces between the protons. The net effect of the proton’s electric field on the neutron results in the neutron’s polarization. The polarization P of the dielectric equals the average dipole moment per unit volume. P = µ ave Volume ( units : Cm m 3 = C m 2 ). (17) The polarization P also equals the surface charge density on opposite ends of a polarized bulk medium. P = σ = surface charge/(surface area on opposite ends of bulk medium). Non polar molecules may acquire a dipole moment in an electric field as long as the electric field is not too strong: µ Induced = αE* , (18) where α is the polarizability of the molecule, units = C 2 m N . E* is the electric field at the location of a molecule inside a bulk medium. This molecule experiences the applied E, and an additional E arising from the induced charges on the bulk medium’s opposing surfaces: E* = E + P ( 3ε 0 ) . (19) FRACTAL PHYSICS THEORY - NUCLEONS AND … 39 When E is too strong, the induced moment depends also on βE* 2 , where β = hyperpolarizability. A term often used is the polarizability volume α , = αkC = α ( 4πε 0 ) , units = 1Å 3 = 10 −24 cm 3 . The potential energy U of an electric dipole µ1 and a charge q 2 : U = − µ1q2 ( 4πεr 2 ). (20) This potential energy decreases more rapidly ( 1 r 2 ) than the potential energy between two charges ( 1 r ) . The deuteron is used to illustrate (Figure 5) how the strong nuclear force may arise involving the attraction between a proton and the induced sqs-electric dipoles of a neutron (assuming no sqs-conduction electrons are present). A free neutron’s magnetic dipole moment µ is −1.913 µ N . A deuteron’s magnetic dipole moment is + 0.857 µ N . The neutron in the deuteron does not experience a homogeneous electric field E from the proton, but rather a fanned out conic E because of the proton’s spherical shape. An effect of this non homogeneous E results in an apparent increase of the neutron’s magnetic dipole moment to − 1.936 µ N . The proton contributes its same magnetic dipole moment of + 2.793 µ N to the deuteron’s magnetic dipole moment. Figure 5. Deuteron. 8. Cosmic Scale Deuteron ( d + ) 8.1. Neutron and proton fusion The following nuclear reaction is observed in the human scale: n + p+ → d+ + γ 40 LEONARD J. MALINOWSKI 1.67492728 × 10 −27 kg Neutron mass : Proton mass : 1.67262171 × 10 − 27 kg 3.34754899 × 10 − 27 kg Reactant mass : Deuteron mass : 3.34358335 × 10 −27 kg [4] Reactant mass – (deuteron mass) = 3.96564 × 10 −30 kg = 3.5641395 × 10 −13 J = − B.E. of the deuteron and also the energy of the emitted gamma photon. Gamma photon frequency = ( 3.5641395 × 10 −13 J ( 6.6260693 × 10 −34 Js ) = 5.3789650 × 10 20 Hz. Gamma photon wavelength = ( 299792458 m s ) ( 5.3789650 × 10 20 Hz ) = 5.5734227 × 10 −13 m. It will take a time = ( 5.5734227 × 10 −13 m ) ( 299792458 m s ) = 1.859 × 10 −21 seconds for this photon to be emitted. An antineutrino is not observed so sqscontrolled thermonuclear fusion is not expected in this reaction. A self-similar cosmic scale reaction observed from the human scale: cs - n + cs - p + → cs - d + + cs - γ Cs - neutron mass : 1.99238160 × 1030 kg Cs - proton mass : 1.98963905 × 1030 kg Reactant mass : 3.98202065 × 1030 kg Cs-deuteron mass: ( 3.34358335 × 10 −27 kg ) ( ¥Mass = 1.189533197 × 1057 ) = 3.97730339 × 1030 kg. Reactant mass – (cs-deuteron mass) = 4.717260 × 10 27 kg = 4.239662 × 10 44 J = − B.E. of the cs-deuteron and also the energy of the emitted cs-gamma photon. FRACTAL PHYSICS THEORY - NUCLEONS AND … 41 Cs-gamma photon frequency = ( 4.239662 × 10 44 J ) ( [ h ]1, 0 = 2.986120904 × 10 47 Js ) = 1.4197891 × 10 −3 Hz. Cs-gamma photon wavelength = ( 299792458 m s ) (1.4197891 × 10 −3 Hz ) = 2.1115281 × 1011 m. It will take a time = ( 2.1115281 × 1011 m ( 299792458 m s ) = 11.7 minutes for this cs-photon to be emitted. A cs-antineutrino is not observed so controlled thermonuclear fusion is not expected in this reaction. A stellar thermonuclear explosion could take place in 11.7 minutes. The cosmic scale gamma photon energy will be approximated to result from the release of gravitational and Coulombic binding energies. 8.2. Cs-deuteron inter-nucleon gravitational binding energy Let the cs-neutron in the cs-deuteron carry half the cs-deuteron’s mass and be composed of 100% Hydrogen molecules. The mass of the cs-neutron in the csdeuteron is then = 1.9886517 × 1030 kg. U g = −( 6.6742 × 10 −11 Nm 2 kg 2 ) (1.9886517 × 1030 kg )2 ( 2*3.788566 × 108 m ) = −3.4834679 × 10 41 J. 8.3. Cs-deuteron inter-nucleon Coulombic binding energy If an H 2 molecule has mass = 2.0158815 amu = 3.3474496 × 10 −27 kg, then the cs-neutron in the cs-deuteron is composed of 5.9407965 × 1056 H 2 molecules. Let the radius of the cs-neutron and the cs-proton in the cs-deuteron both = 3.788566 × 108 m, with each having a volume = 2.2777874 × 10 26 m 3 . The csneutron is assumed to have a constant H 2 # density = 2.6081436 × 1030 H 2 m 3 . The proton’s electric field E is not constant through the neutron’s interior. The following slab integration technique approximates the E at various volumes of the neutron. In Figure 6, the eight blue circles drawn beyond the cs-proton’s surface represent 8 electric equipotential surfaces all centered on the origin of the axes, with radii as listed. These 8 blue circles slice the cs-neutron into 8 slabs. Define Slab 1 as the area contained within the intersection of the cs-proton’s equipotential equation 42 LEONARD J. MALINOWSKI with r = 5 R 4 and the cs-neutron’s surface equation. Define Slab 2 as the area contained within three equations; the cs-proton’s equipotential equations with r = 5 R 4 , and r = 6 R 4 , and the cs-neutron’s surface equation. Continue to define Slabs 3 through 7, sequentially, as areas bounded by the appropriate 3 equations, adjacent to and right of the preceding slab. Finally, define Slab 8 as the area contained beyond the cs-proton’s equipotential equation with r = 11 R 4 and within the cs-neutron’s surface equation. The areas of each of the 8 slabs of the cs-neutron in Figure 6 are calculated below by integrating the curves through the appropriate limits. Slab 1 is divided into two parts. Part I integrates the cs-neutron’s surface equation (positive Y-axis) from x1 = R to x2 = intersection of cs-neutron’s surface equation and proton’s potential equation with r = 5 R 4 . This area is multiplied by 2 (from symmetry) to obtain the total area of Slab 1, Part I. Part two integrates the cs-proton’s potential equation with r = 5 R 4 (positive Y-axis) from x2 to x3 = 5 R 4 . This area is multiplied by 2 (from symmetry) to obtain the total area of Slab 1, Part II. Calculations are reproduced as End Notes to this article. Figure 6. Deuteron in 2-dimensions. Tables 6.a and 6.b calculate the Coulombic binding energy in a cs-deuteron occurring between the cs-proton and the induced electric dipoles in the cs-neutron. The electric field inside the cs-neutron E* is underestimated to equal the cs-proton’s electric field E. The Hydrogen molecule polarizability volume α , = 0.819 × 10 −30 m 3 [4]. H 2 α = α , kC = ( 0.819 × 10 −30 m 3 ) ( 8.987551787 × 109 ) = 9.112604 × 10 −41 C 2 m N . FRACTAL PHYSICS THEORY - NUCLEONS AND … 43 The relative permittivity (dielectric constant) of liquid molecular hydrogen [4]: ε r = 1.2792, at temperature = 13.52 K and pressure = 0.1 MPa, ε r = 1.24827, at temperature = 15 K and pressure = 0.1 MPa, ε r = 1.26198, at temperature = 20 K and pressure = 10 MPa, at the extremely high pressures within a cs-neutron, the relative permittivity of H 2 is hereby estimated as ε r = 1.4. Table 6.a. Cosmic scale neutron slab values for cs-proton induced electric dipoles ( ε r = 1 .4 ) (1) # of H 2 molecules/Slab = (Slab volume)( H 2 # density = 2.6081436 × 1030 H 2 m 3 ) , Slab volume = (Table 6.a, column 3 cell) ( 2.2777874 × 10 26 m 3 ) 4.188790204 . (2) Radius for the electric field E of the proton is calculated at the midpoint of the neutron’s slab area, R = 3.788566 × 108 m. (3) Q p = cs-proton charge = 3.401230560 × 10 21 C; ε = ( ε 0 = 8.854187817 × 10 −12 C 2 Nm 2 ) ( ε r = 1.4 ). 44 LEONARD J. MALINOWSKI Table 6.b. Cs-neutron slab energies for cs-proton induced electric dipoles ( ε r = 1 .4 ) 1 molecule of H 2 1 molecule of H 2 ∑ of H 2 Slab µinduced = αE(Cm )(1) U = −µ i E ( J ) U = −µ i E ( J )(2) 1 1.0953103 × 10 −26 − 1.3165333 × 10 −12 − 4.2960635 × 10 43 2 7.3322427 × 10 −27 − 5.8997167 × 10 −13 − 3.6245785 × 10 43 3 5.2497122 × 10 −27 − 3.0243252 × 10 −13 − 2.4089641 × 10 43 4 3.9431172 × 10 −27 − 1.7062272 × 10 −13 − 1.5629178 × 10 43 5 3.0699009 × 10 −27 − 1.0342040 × 10 −13 − 1.0068847 × 10 43 6 2.4576215 × 10 −27 − 6.6280763 × 10 −14 − 6.3530633 × 10 42 7 2.0117945 × 10 −27 − 4.4414494 × 10−14 − 3.7430069 × 10 42 8 1.6771292 × 10 −27 − 3.0866725 × 10 −14 − 1.5826645 × 10 42 − 1.4067282 × 10 44 (1) E from Table 6.a, column 7 cell; α = 9.112604 × 10 −41 C 2 m N , (2) ∑ of H 2 Energy = (Table 6.a, column 4 cell)(Table 6.b, column 3 cell). 8.4. Summary of cs-deuteron binding energies Cs -deuteron inter - nucleon Coulombic B.E. = −1.4067282 × 10 44 J Cs -deuteron inter - nucleon gravitational B.E. = −3.4834679 × 10 41 J Total cs -deuteron inter - nucleon binding energy Total cs -deuteron binding energy Total cs -deuteron inter - nucleon B.E. Cs -deuteron intra - nucleon fusion B.E. = −1.410212 × 10 44 J = −4.239662 × 10 44 J = −( −1.410212 × 10 44 J ) = −2.829450 × 10 44 J When four Hydrogen atoms fuse into a Helium atom 4.352052374 × 10 −12 J is released. To radiate the remaining energy 2.829450 × 10 44 J, 6.5014153 × 10 55 Helium atoms must be fused from 2.6005661 × 1056 Hydrogen atoms. FRACTAL PHYSICS THEORY - NUCLEONS AND … 45 Table 7 lists the cs-deuteron idealized chemical composition. Table 7. Chemical composition of idealized cosmic scale deuteron Element Hydrogen (1.007940754) Helium (4.002601932) # of atoms 1.983049354 × 10 9.9033610 × 10 55 Mass (kg) 57 3.319078872 × 10 30 % Mass 83.450 6.582245182 × 10 29 16.550 3.97730339 × 10 30 100.000 9. Cosmic Scale Deuteron - Deuteron Collision Consider the reaction: d + + d + → t + (1.01 MeV ) + p + ( 3.02 MeV ). The deuteron-deuteron collisions illustrated in Figure 7 reproduce the two major branches observed; formation of the triton 50% of the time and formation of the helion 50% of the time. Reaction mass difference: 2( cs -d + ) = 2( 3.97730339 × 1030 kg ) − ( cs - t ) = − 5.95641633 × 1030 kg ) − ( cs -p + ) = − 1.98963905 × 1030 kg ) + Total ∆m = 8.551 × 10 27 kg = 7.685 × 10 44 J; the kinetic energy that the cs-triton and cs-proton share. There are no cs-neutrinos or cs-gamma photons included in the cs-triton fusion reaction, yet a lot of mass is converted into the kinetic energy of the cs-triton and csproton. It is reasonable to propose that the two colliding cs-deuterons trigger a thermonuclear explosion which launches a cs-proton in one direction and recoils the cs-triton in the other. The ejected explosion gaseous-plasma material remains bound to the product cosmic scale nuclei, eventually cooling and settling on them. 46 LEONARD J. MALINOWSKI Figure 7. d + + d + collisions (blue proton and brown neutron). 10. Cosmic Scale Triton (t+) 10.1. Cs-triton total binding energy 2(Cs - neutron) = 2(1.99238160 × 1030 kg Cs - proton = 1.98963905 × 1030 kg Reactant mass = 5.97440225 × 1030 kg Cs-triton mass: 5.95641633 × 1030 kg, Reactant mass − (cs- triton mass) = 1.798592 × 10 28 kg = 1.616493874 × 10 45 J = − B.E. of the cs-triton. Figure 8 depicts the proposed induced dielectrics of the triton. Figure 8. Triton, E field lines, S = − 1 2 + 1 2 + 1 2 = 1 2 ; µ = 2.122 + 2.793 −1.936 = 2.979; Red fill: negative dielectric; Blue fill: positive dielectric. FRACTAL PHYSICS THEORY - NUCLEONS AND … 47 10.2. Cs-triton inter-nucleon gravitational binding energy Approximate each nucleon mass in the triton by: ( t + mass 3 = 1.98547211 × 1030 kg. Let the cs-proton’s radius = the cs-neutron’s radius = r = 3.788566 × 108 m and the distance between the two cs-neutron centers be 4r. Gravity potential U = U n1- p + U n1− n 2 + U n 2 − p , where U n1− p = U n 2 − p , U g = −2Gm 2 (2r ) − Gm 2 (4r ) = −6.944675 × 10 41 J + −1.736169 × 10 41 J = −8.680844 × 10 41 J. (21) 10.3. Cs-triton inter-nucleon Coulombic binding energy The cs-proton would induce electric dipoles in both cs-neutrons providing a total charge-dipole Coulomb energy in the cs-triton double that calculated for the csdeuteron in Table 6.b. U C = 2( − 1.4067282 × 10 44 J ) = −2.8134564 × 10 44 J. 10.4. Summary of cs-triton binding energies Total cs-triton binding energy: − 1.616493874 × 10 45 J Total cs-triton inter-nucleon B.E.: − ( − 2.8221372 × 10 44 J ) Cs-triton intra-nucleon fusion B.E.: − 1.3342802 × 10 45 J When four Hydrogen atoms fuse into a Helium atom, 4.352052374 × 10 −12 J is released. To radiate the remaining energy 1.3342802 × 10 45 J, 3.0658643 × 10 56 Helium atoms must be fused from 1.2263457 × 10 57 Hydrogen atoms. Table 8 lists the cs-triton idealized chemical composition. Table 8. Chemical composition of idealized cosmic scale triton Element Hydrogen amu 1.007940754 # of atoms Mass (kg) 2.2062094 × 1057 3.6925874 × 1030 % Mass 61.993 Helium 4.002601932 3.4060589 × 1056 2.2638289 × 1030 38.007 5.95641633 × 10 30 100.000 48 LEONARD J. MALINOWSKI 11. Cosmic Scale Triton Beta Decay Consider the reaction: t + →3 He 2 + + e − + antineutrino. Cosmic scale beta decay is viewed in the human scale as a star radiating while fusing Hydrogen to Helium to Iron. Cs - triton = 5.95641633 × 1030 kg Cs - helion = − 5.955293438 × 1030 kg Cs -electron = − 1.083591301 × 10 27 kg Mass difference : 3.9300699 × 10 25 kg = 3.532170675 × 10 42 J This cs-electron requires the fusion of 1.168506245 × 1052 Iron atoms from the cstriton’s Hydrogen. Hydrogen 56*1.007940754 amu: 56.44468222 amu Iron 1*55.84514562 amu: –55.84514562 amu 0.59953660 amu = 8.94759153 5 × 10 −11 J Fusing a cs-electron’s Iron from Hydrogen releases 1.045531659 × 10 42 J. This leaves 3.532170675 × 10 42 J − 1.045531659 × 10 42 J = 2.486639 × 10 42 J for product kinetic energies and additional cs-antineutrino energy. 12. Cosmic scale helion ( 3 He 2 + ) 12.1. Cs-helion total binding energy 2(Cs - proton) = 2(1.98963905 × 1030 kg ) Cs - neutron = 1.99238160 × 1030 kg Reactant mass = 5.97165970 × 1030 kg Cs-helion mass: 5.955293438 × 1030 kg. Reactant mass – (cs-helion mass) = 1.6366262 × 10 28 kg = 1.470926273 × 10 45 J = −B.E. of the cs-helion. 12.2. Cs-helion inter-nucleon gravitational binding energy Approximate each nucleon mass in the helion by: ( 3 He 2 + mass ) 3 = 1.985097813 × 1030 kg. FRACTAL PHYSICS THEORY - NUCLEONS AND … 49 Let the cs-proton’s radius = the cs-neutron’s radius r = 3.788566 × 108 m and the distance between the two cs-proton centers be 4r. Gravity potential U = U p1− n + U p1− p 2 + U n − p 2 , where U p1− n = U n − p 2 , U g = −2Gm 2 (2r ) − Gm 2 (4r ) = −6.942057 × 10 41 J + −1.735514 × 10 41 J = −8.677571 × 10 41 J. (22) 12.3. Cs-helion inter-nucleon Coulombic binding energy The cs-protons induce electric dipoles on both sides of the cs-neutron providing a total charge-dipole Coulomb energy in the cs-helion double that calculated for the cs-deuteron in Table 6.b. U C = 2( − 1.4067282 × 10 44 J ) = −2.8134564 × 10 44 J. The cs-helion inter-nucleon cs-neutron dielectric effect on cs-proton-proton Coulombic repulsion is estimated: U C = Q p1Q p 2 ( 4πε 0 ε r* 4r ) = kC ( 3.4012306 × 10 21 C ) 2 (1.4* 4* 3.788566 × 108 m ) = 4.90061592 × 10 43 J. (23) 12.4. Total cs-helion inter-nucleon binding energy Cosmic scale helion inter-nucleon gravitational binding energy: − 8.677571 × 1041 J Cosmic scale helion inter-nucleon Coulombic binding energy: − 2.8134564 × 1044 J Cosmic scale helion inter-nucleon Coulombic repulsion energy: 4.90061592 × 1043 J − 2.3320724 × 1044 J 12.5. Summary of cs-helion binding energies Total cs-helion binding energy: − 1.470926273 × 10 45 J Total cs-helion inter-nucleon B.E.: − ( − 2.3320724 × 10 44 J ) Cs-helion intra-nucleon fusion B.E.: − 1.2377190 × 10 45 J When four Hydrogen atoms fuse into a Helium atom 4.352052374 × 10 −12 J is released. To radiate the remaining energy 1.2377190 × 10 45 J, 2.8439892 × 10 56 50 LEONARD J. MALINOWSKI Helium atoms must be fused from 1.1375957 × 10 57 Hydrogen atoms. Table 9 lists the cs-helion idealized chemical composition. Table 9. Chemical composition of idealized cosmic scale helion Element amu Hydrogen 1.007940754 # of atoms Helium 4.002601932 Mass (kg) 2.1585531 × 10 % Mass 57 3.6128238 × 10 30 60.666 56 2.3424696 × 1030 39.334 5.955293438 × 1030 100.000 3.5243783 × 10 13. Cosmic Scale Alpha Particle ( 4 He 2 + ) 13.1. Cs-alpha particle total binding energy Arbitrarily, the assumption is made here that the cosmic scale alpha particle is not composed of any conduction atoms. Alpha particle mass: 6.6446565 × 10 −27 kg [4] 2(Cs - proton) = 2(1.98963905 × 1030 kg ) 2(Cs - neutron) = 2(1.99238160 × 1030 kg ) Reactant mass = 7.96404130 × 1030 kg Cs - 4 He 2 + mass : 7.904039489 × 1030 kg. Reactant mass – (cs- α particle mass) = 6.0001811 × 10 28 kg = 5.392693837 × 10 45 J = −B.E. of the cs-alpha particle. Figure 9 depicts the alpha particle configuration used to estimate its binding energy. Figure 9. Alpha particle. 13.2. Cs-alpha particle inter-nucleon gravitational binding energy Approximate each nucleon mass in the alpha particle by: 0.25(alpha mass) = 1.976009872 × 1030 kg. Let the cs-proton’s radius = the cs-neutron’s radius r = FRACTAL PHYSICS THEORY - NUCLEONS AND … 51 3.788566 × 108 m. Based on Figure 9, the distance between the two cs-proton centers is 3.5r. Gravity potential U = U p1− n1 + U p1− n 2 + U p1− p 2 + U p 2 − n1 + U p 2 − n 2 + U n1− n 2 , where U p1− n1 = U p1− n 2 = U p 2 − n1 = U p 2 − n 2 = U n1− n 2 . Therefore U g = −5Gm 2 (2r ) + −Gm 2 (3.5r ). (24) U g = −1.719659993 × 10 42 J + −1.965325707 × 10 41 J = −1.916192564 × 10 42 J. 13.3. Cs-alpha particle inter-nucleon Coulombic binding energy Based on Figure 9, each cs-proton would induce dipoles and attract half of each cs-neutron towards it. A reasonable estimate sums the first 4 slabs from the csdeuteron calculation from Table 6.b, which totals − 1.1892524 × 10 44 J, reproduced in Table 10 below. The total estimate U C = 4* ( − 1.1892524 × 10 44 J ) = −4.7570096 × 10 44 J. Table 10. First 4 slabs from Table 6.b cs-deuteron induced electric dipoles ( ε r = 1 .4 ) Slab µinducd = αE(Cm ) U = −µ1E (Joules) U = −µ1E (Joules) 1 1.0953103 × 10 −26 − 1.3165333 × 10 −12 − 4.2960635 × 10 43 2 7.3322427 × 10 −27 − 5.8997167 × 10 −13 − 3.6245785 × 10 43 3 5.2497122 × 10 −27 − 3.0243252 × 10 −13 − 2.4089641 × 10 43 4 3.9431172 × 10 −27 − 1.7062272 × 10 −13 − 1.5629178 × 10 43 − 1.1892524 × 10 44 The Cs - 4 He 2 + inter-nucleon cs-neutron dielectric effect on cs-proton-proton Coulombic repulsion is estimated: U C = Q p1Q p 2 ( 4πε 0 ε r*3.5r ) = ( k C ) ( 3.4012306 × 10 21 C )2 (1.4*3.5*3.788566 × 108 m ) = 5.600703909 × 10 43 J. (25) 52 LEONARD J. MALINOWSKI 13.4. Total cs- 4 He 2 + inter-nucleon binding energy Cosmic scale 4 He 2 + inter - nucleon gravitational binding energy : Cosmic scale 4 He 2 + inter - nucleon Coulombic binding energy : − 1.916192564 × 10 42 J − 4.7570096 × 10 44 J Cosmic scale 4 He 2 + inter - nucleon Coulombic repulsion energy : 5.600703909 × 10 43 J − 4.216101135 × 10 44 J 13.5. Cs- 4 He 2 + intra-nucleon fusion binding energy Total cs- 4 He 2 + binding energy: − 5.392693837 × 10 45 J Total cs- 4 He 2+ inter-nucleon binding energy: − ( − 4.216101135 × 10 44 J ) − 4.971083724 × 10 45 J When four Hydrogen atoms fuse into a Helium atom 4.352052374 × 10 −12 J is released. Available Hydrogen to fuse: 2(cs -protons) = 2.107311252 × 1057 Hydrogen atoms 2(cs- neutrons) = 2.380775632 × 1057 Hydrogen atoms 4.488086884 × 1057 Hydrogen atoms Hydrogen from the 4 cs-nucleons can forge 1.122021721 × 1057 Helium atoms releasing 4.883097295 × 10 45 J. Total required cs- 4 He 2+ intra - nucleon fusion B.E.: Available from fusing all Hydrogen to Helium B.E. : − 4.971083724 × 10 45 J − ( − 4.883097295 × 10 45 J ) − 8.7986429 × 10 43 J At this point consider the cs-Helium 4 nucleus as composed of 100% Helium atoms. 2(cs- protons) = 6.803891332 × 1055 Helium atoms Fused Helium = 1.122021721 × 1057 Helium atoms 1.190060634 × 1057 Helium atoms Let 3 Helium atoms fuse to form one Carbon 12 atom. Reactants: 3(4.002601932 amu) =12.0078058 amu Product: =12.0000000 amu Energy released: =0.0078058 amu = 1.1649516 × 10−12 J FRACTAL PHYSICS THEORY - NUCLEONS AND … 53 To radiate the remaining energy 8.7986429 × 10 43 J, 7.5527970 × 10 55 Carbon 12 atoms must fuse from 2.2658391 × 1056 Helium atoms. Table 11 lists the cs-alpha particle idealized chemical composition. Table 11. Chemical composition of idealized cs-alpha particle Element # of atoms Mass (kg) % Mass Helium 9.6277093 × 1056 6.399033936 × 1030 80.959 7.5527970 × 1055 1.505005553 × 1030 19.041 7.904039489 × 1030 100.000 (4.002601932) Carbon (12.000000) At the high pressures inside a cs-alpha particle, the 19% Carbon is in the diamond crystal form. 14. Cosmic Scale Carbon 12 Nucleus ( 12 C 6 + ) 14.1. Cs-Carbon 12 nucleus total binding energy Carbon diamond Type I has a dielectric constant ε r = 5.87 at 300 K [4]. Arbitrarily, the assumption is made here that the cosmic scale Carbon 12 nucleus is composed primarily of Carbon with 10% Magnesium. To obtain the Carbon 12 nuclear mass the masses of 6 electrons must be subtracted, while the mass equivalent of the 6 ionization energies must be added (Table 12). Carbon 12 atomic mass : Six electron masses : 12*1.66053886 × 10 −27 kg = 6* 9.1093826 × 10 − 31 kg = 1.992646632 × 10 −26 kg − 5.46562956 × 10 − 30 kg 1.83633 × 10 − 33 kg Six I.E. mass equivalents : 1.992100253 × 10 − 26 kg Carbon 12 nuclear mass : Cosmic scale 12 C 6 + mass: 2.369669383 × 1031 kg 6(cs-proton) = 6(1.98963905 × 1030 kg ) 6(cs -neutron) = 6(1.99238160 × 1030 kg ) Reactant mass : 2.38921239 × 1031 kg Reactant mass − ( Cs -12 C 6 + mass ) = 1.9543007 × 10 29 kg 54 LEONARD J. MALINOWSKI = 1.7564379 × 10 46 J = −B.E. of the Cs -12 C 6 + . Table 12. Carbon atom ionization energies [4] Electron # Energy (eV) 1 11.2603 2 24.3833 3 47.8878 4 64.4939 5 392.087 6 489.9933 1030.1056 eV = 1.8363299 × 10−33 kg 14.2. Cosmic scale 12 C 6 + inter-nucleon gravitational binding energy The total gravitational binding energy of a cosmic scale 12 C 6 + mass is: U total g = −G ( 2.369669383 × 1031 kg )2 R = −3.6007382 × 10 43 J, (26) G = 6.6742 × 10 −11 Nm 2 kg 2 , R = 1.2 fm(12)1 3 ¥ L = 1.0408381 × 109 m. Only the gravitational binding energy of the 12 cs-nucleons, not the entire mass, is desired. The average gravitational binding energy of 1 cs-nucleon of the cs-Carbon 12 nucleus: U nucleon g = −G ( 2.369669383 × 1031 kg 12 )2 3.788566 × 108 m = −6.8696938 × 10 41 J. U total g = − 3.6007382 × 10 43 J −12(U nucleon g ) = − 12( − 6.8696938 × 1041 J ) − 2.7763749 × 10 43 J, cosmic (27) scale 12 6 + C inter-nucleon gravitational binding energy 14.3. Cosmic scale 12 C 6 + inter-nucleon cs-neutron dielectric effect on csproton-proton Coulombic repulsion Figures 10.a, 10.b, 10.c are used to calculate the inter-nucleon Coulombic energies. The six cosmic scale protons in the cs-Carbon 12 nucleus will have the following 15 interactions: FRACTAL PHYSICS THEORY - NUCLEONS AND … 55 U p1− p 2 + U p1− p 3 + U p1− p 4 + U p1− p5 + U p1− p 6 + U p 2 − p 3 + U p 2 − p 4 + U p 2 − p 5 +U p 2 − p 6 + U p3 − p 4 + U p3 − p5 + U p3 − p 6 + U p 4 − p5 + U p 4 − p 6 + U p 5 − p 6 . Figure 10.a. Carbon 12 nuclear structure, top-down view. Let the top layer be placed directly above the bottom layer with protons 4-6 above the 3 outer neutrons of the bottom layer. Figure 10.b. Carbon 12 nuclear structure, side view. From Figure 10.a symmetry these 15 interactions reduce to 4 calculations: A. In these four interactions, the intervening space is the cs-neutron, U p1− p 2 = U p1− p3 = U p 2 − p 3 = U p 4 − p 6 U C = 4*K C Q p 2 ( ε r x ) = 4* (1.335772653 × 10 43 J ) = 5.343090612 × 10 43 J, (28) x = 3.5* ( 3.788566 × 108 m ) , Q p = 3.401230560 × 10 21 C, ε r = 5.87, K C = 8987551788 Nm 2 C 2 . B. In these two interactions, half the space is vacuum and half the space is the csneutron, U p 4 − p 5 = U p 5 − p 6 . 56 LEONARD J. MALINOWSKI U C = KC ( 0.5Q p )2 [ ( ε r = 1 ) (3.5)( 3.788566 × 108 m ) ] = 1.960246368 × 1043 J U C = KC ( 0.5Q p )2 [ ( ε r = 5.87 ) (3.5)( 3.788566 × 108 m ) ] = 3.339431632 × 1042 J ∑ = 2.294189531 × 1043 J U C = 2*K C Q p 2 ( ε r x ) = 2* ( 2.294189531 × 10 43 J ) = 4.588379062 × 10 43 J. (29) C. Refer to Figure 10.b, to determine x = (8)1 2 R, U p1− p 4 = U p1− p 6 = U p 2 − p 4 = U p 2 − p5 = U p3− p5 = U p3− p 6 . U C = 6*K C Q p 2 ( ε r x ) = 6*K C Q p 2 (1* 8*R ) = 5.821634699 × 10 44 J. (30) D. Refer to Figure 10.c, to determine x = (20)1 2 R, U p1− p 5 = U p 2 − p 6 = U p3− p 4 . U C = 3*K C Q p 2 ( ε r x ) = 3*K C Q p 2 ( 5.87* 20*R ) = 3.136222377 × 10 43 J. (31) Total Coulombic repulsion in cs-Carbon 12 nuclei ∑ A - D = 7.128403904 × 10 44 J. Figure 10.c. Carbon 12 nuclear structure, side view. 14.4. Cosmic scale 12 C 6 + inter-nucleon Coulombic binding energy From Figure 10.a, six protons have 18 direct contact surfaces with six neutrons and 6 more distant contacts with six neutrons. Let the low density Magnesium metal be at the cs-neutron surfaces to provide cs-neutron conduction electron bonding with the cs-proton. The binding energy − 1.7229222 × 10 44 J per bond is taken from Table 5.b for 18 bonds. The same binding energy per bond for 6 bonds is reduced by 1 ( 8 − 1) by replacing R p in Table 5.b by r = R p ( 8 − 1 ) , the effect from the gap between these 6 protons and 6 neutrons (Figure 10.c). FRACTAL PHYSICS THEORY - NUCLEONS AND … U C = (18)* ( − 1.7229222 × 10 44 J ) = −3.1012600 × 10 45 J U C = (6 )* ( − 1.7229222 × 10 44 J ( 8 − 1 ) = −5.6537846 × 10 44 J Total Coulombic binding energy in cs-Carbon 12 nuclei = −3.6666385 × 10 45 J 57 14.5. Total cosmic scale 12 C 6 + inter-nucleon binding energy Cosmic scale 12 C 5 + inter -nucleon gravitational binding energy : − 2.7763749 × 10 43 J Cosmic scale 12 C 6 + inter -nucleon Coulombic binding energy : − 3.6666385 × 10 45 J Cosmic scale 12 C 5 + inter -nucleon Coulombic repulsion energy : 7.1284039 × 10 44 J − 2.9815619 × 10 45 J 14.6. Cosmic scale 12 C 6 + intra-nucleon fusion binding energy Total cs -12 C 6 + binding energy : − 1.7564379 × 10 46 J Total cs -12 C 6 + inter - nucleon B.E.: − ( − 2.9815619 × 10 45 J ) Cs -12 C 6 + intra - nucleon fusion B.E.: − 1.4582817 × 10 46 J Let 24 Hydrogen atoms fuse to form one Magnesium 24 atom. Reactants : 24(1.007940754 amu ) = 24.1905781 amu Product : Energy released : A Magnesium mass of = 23.9850417 amu = 0.2055364 amu = 3.0674620 × 10 −11 J 2.3696694 × 10 30 kg contains 5.9497434 × 10 55 Magnesium 24 atoms, which requires the fusion of 1.4279384 × 10 57 Hydrogen atoms, releasing 1.8250612 × 10 45 J. Cs -12 C 6 + intra -nucleon fusion B.E.: Fusing Hydrogen to Magnesium : Remaining energy to account for : − 1.4582817 × 10 46 J − ( − 1.8250612 × 10 45 J ) − 1.2757755 × 10 46 J Let 12 Hydrogen atoms fuse to form one Carbon 12 atom. Reactants : 12(1.007940754 amu ) = 12.0952890 amu Product : Energy released : = 12.0000000 amu = 0.0952890 amu = 1.4221101 × 10 −11 J To radiate the remaining energy 1.2757755 × 10 46 J, 8.9710037 × 10 56 Carbon 58 LEONARD J. MALINOWSKI 12 atoms must be fused. Table 13 lists the cs-Carbon 12 nucleus idealized chemical composition. Table 13. Chemical composition of idealized cosmic scale 12 C 6 + Element Hydrogen amu 1.007940754 # of atoms Mass (kg) Helium 4.002601932 2.0411674 × 1056 1.3566570 × 1030 5.72 Carbon 12.0000000 8.9710037 × 1056 1.7876040 × 1031 75.44 Magnesium 23.9850417 55 30 10.00 1.2512975 × 10 57 5.9497434 × 10 2.0943276 × 10 2.3696694 × 10 30 2.3696694 × 1031 % Mass 8.84 100.00 This trend of fusing larger, more energetically stable nuclei to account for the binding energies of more energetically stable cosmic scale nuclei is expected to continue. As a final example, the cosmic scale nuclear binding energies of one of the most stable nuclei, Iron 56, will be examined. 15. Cosmic scale Iron 56 nucleus ( 56 Fe 26 + ) 15.1. Cs-Iron 56 nuclear binding energy To obtain the Iron 56 nuclear mass the masses of 26 electrons must be subtracted, while the mass equivalent of the 26 ionization energies must be added (Table 14). Iron 56 atomic mass : 26 electron masses : 55.9349375 amu = 26* 9.1093826 × 10 − 31 kg = 9.288213735 × 10 −26 kg − 2.3684395 × 10 − 29 kg 6.16882 × 10 − 32 kg 26 I.E. mass equivalents : 9.285851464 × 10 − 26 kg Iron 56 nuclear mass : Cosmic scale 56 Fe 26 + mass: 1.104582858 × 10 32 kg 26(cs-proton) = 26(1.98963905 × 1030 kg ) 30(cs-neutron) = 30(1.99238160 × 1030 kg ) Reactant mass : 1.115020633 × 10 32 kg Reactant mass − (cs-56 Fe 26 + mass) = 1.0437775 × 1030 kg = 9.381004336 × 10 46 J = −B.E. of the cs- 56 Fe 26 + . FRACTAL PHYSICS THEORY - NUCLEONS AND … 59 Table 14. Iron atom ionization energies [4] e− # Energy (eV) e− # Energy (eV) e− # Energy (eV) 1 7.9024 10 262.1 19 1456 2 16.1877 11 290.2 20 1582 3 30.652 12 330.8 21 1689 4 54.8 13 361.0 22 1799 5 75.0 14 392.2 23 1950 6 99.1 15 457 24 2023 7 124.98 16 489.256 25 8828 8 151.06 17 1266 26 9277.69 9 233.6 18 1358 ∑ of 26 I.E. = 34604.528 eV = 6.16882 × 10 −32 kg 15.2. Cosmic scale 56 Fe 26 + inter-nucleon gravitational binding energy The total gravitational binding energy of a cosmic scale 56 Fe 26 + mass is: U total = −G (1.104582858 × 1032 kg ) 2 R = −4.6817732 × 10 44 J, g G = 6.6742 × 10 −11 Nm 2 kg 2 , R = 1.2fm (56)1 3 ¥ L = 1.739343849 × 109 m. (32) Only the gravitational binding energy of the 56 cs-nucleons, not the entire mass, is desired. The average gravitational binding energy of 1 cs-nucleon of the cs-Iron 56 nucleus: U nucleon g = −G(1.104582858 × 1032 kg 56 )2 3.788566 × 108 m = −6.8540127 × 10 41 J. − 4.6817732 × 10 44 J U total g = −56(U nucleon (33) g )= − 56( − 6.8540127 × 10 41 J ) − 4.2979485 × 10 44 J, cosmic scale 56 Fe 26 + inter- nucleon gravitational binding energy. 15.3. Cosmic scale 56 Fe 26 + intra-nucleon fusion binding energy The Iron 56 nucleus is one of the most energetically stable nuclei. Presume the 60 LEONARD J. MALINOWSKI cosmic scale Iron 56 nucleus is composed entirely of Iron 56 atoms. Let 56 Hydrogen atoms fuse to form one Iron 56 atom. Reactants : 56(1.007 940 754 amu ) = 56.444 6822 amu Product : = 55.934 9375 amu = 0.509 7447 amu = 7.607 521147 × 10 −11 J Energy released : Available Hydrogen to fuse: 26( cs -proton ) = 26(1.053 655 626 × 10 57 Hydrogen atoms ) 30( cs - neutron ) = 30(1.190 387 816 × 10 57 Hydrogen atoms ) Total Hydrogen available: = 6.310 668 076 × 10 58 Hydrogen atoms All available Hydrogen can forge 1.126 905 014 × 10 57 Iron 56 atoms, releasing 8.572 9537 × 10 46 J. Let 14 Helium atoms fuse to form one Iron 56 atom. Reactants : 14( 4.002 601 932 amu ) Product : Energy released : = 56.036 4270 amu = 55.934 9375 amu = 0.101 4895 amu = 1.514 647 465 × 10 −11 J Available Helium to fuse: 26(cs- protons) = 26( 3.401 945 666 × 10 55 ) = 8.845 058 732 × 10 56 Helium atoms. All available Helium can forge 6.317 899 094 × 10 55 Iron 56 atoms, releasing 9.569 3898 × 10 44 J. Total Cs- 56 Fe 26 + intra-nucleon fusion binding energy: − 8.6686476 × 10 46 J. 15.4. Cosmic scale 56 Fe 26 + inter-nucleon Coulombic binding energy Cosmic scale56 Fe 26+ binding energy : − 9.381 004336 × 10 46 J 56 26+ Cosmic scale Fe inter - nucleon gravitational binding energy : − ( − 4.2979485 × 10 44 J ) Cosmic scale 56 Fe 26+ intra - nucleon fusion binding energy : Cosmic scale 56 Fe 26+ inter - nucleon Coulombic binding energy : − ( − 8.6686476 × 10 46 J ) − 6.6937725 × 10 45 J If the conducting metals of the cs-Iron 56 nucleons are effective at shielding the csproton-proton repulsion, then there can be − 6.693 7725 × 10 45 J − 1.722 9222 × 10 44 J = 39 cs-proton-neutron Coulomb bonds in the cs-Iron 56 nucleus. FRACTAL PHYSICS THEORY - NUCLEONS AND … 61 16. Conclusion With two postulates, that the neutron is composed of 100% subquantum scale Hydrogen atoms and that the mass of the pre-solar system is the mass of a cs-neutron, Fractal Physics Theory appears able to calculate the fractal chemical compositions and binding energies of all the nuclei without postulating the existence of a unique “Strong Nuclear Force”. Most striking is the fact that the mass loss due to fusing all available sqs-Hydrogen and sqs-Helium from 56 separate nucleons into sqs-Iron 56 must occur to obtain the mass of the Iron 56 nucleus. No further sqs-fusion is possible. The Iron 56 nucleus is one of the most stable nuclei because it is composed entirely of sqs-Iron 56 atoms. Tables 15 and 16 summarize ideal values discussed in this article. Table 15. Idealized subquantum scale chemical compositions *Symbol [O]− 3, 0 refers to an object located in the subquantum scale as observed from the human scale. Table 16. Fractal strong nuclear force Particle Lilliputian Scale % Gravity Lilliputian Scale % Coulombic Subquantum Scale % Fusion d+ 0.08 33.18 66.74 + 0.05 17.41 82.54 t 3 He 2 + 0.06 15.80 84.14 4 2+ 0.04 7.78 92.18 12 6 + 0.16 16.81 83.03 56 0.46 7.13 92.41 He C Fe 26 + 62 LEONARD J. MALINOWSKI 17. End Notes - Calculation of the Areas of the 5 Slabs of the cs-neutron in Figure 3, and 8 Slabs in Figure 6 Slab 1, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (5 R 4 )2 , centered on the origin (0, 0 ). Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0 ) , with r = R. The two equations are solved simultaneously to determine their intersection point: 25 R 2 16 = 4 Rx − 3R 2 , x2 = 73R 64 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x2 + y2 = R2 , y = ( R 2 − x 2 )1 2 . ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ). [7] Translated limits, x1 = − R to x2 = −55R 64 , a = R. ∫ydx @ x2 = 0.5( − 55 R 64 )(( R 2 − (− 55 R 64) 2 )1 2 + 0.5R 2 sin −1 ( − 55 64) = −0.219718933R 2 + −0.517023074 R 2 = −0.736742007R 2 , ∫ydx @ x1 = −0.5 R( R 2 − (− R )2 )1 2 + 0.5 R 2 sin −1 (− 1) = 0 + −0.785398163R 2 = −0.785398163R 2 , ∫ydx @ x2 x1 = 0.048656156R 2 . Slab 1, Part I area = ( 0.048656156 R 2 ) ( 2) = 0.097312312 R 2 . Slab 1, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (5R 4)2 , FRACTAL PHYSICS THEORY - NUCLEONS AND … 63 centered on the origin (0, 0) . ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ). [7] Integration limits, x2 = 73R 64 to x3 = 5R 4, a = 5R 4 . x3 ∫ ydx @ ∫ ydx @ x2 = 0 + 0.5(5R 4 )2 sin −1 (1) = 1.227184630R 2 , = 0.5(73R 64) (( 5 R 4 )2 − (73R 64 )2 )1 2 + 0.5(5 R 4 )2 sin −1 (73 64 1.25) = 0.291626947R 2 + 0.897933066R 2 = 1.189560013R 2 , ∫ ydx @ x3 x2 = 0.037624617R 2 . Slab 1, Part II area = ( 0.037624617R 2 ) (2) = 0.075249234R 2 . Slab 1, Part I + Part II: 0.172561546R 2 . Slabs 1 + 2, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (6 R 4 )2 , centered on the origin (0, 0) . Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0) , with r = R. The two equations are solved simultaneously to determine their intersection point: 36 R 2 16 = 4 Rx − 3R 2 , x2 = 21R 16 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x2 + y2 = R2 , ( R 2 − x 2 )1 2 . ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = − 11R 16 , a = R. y= 64 LEONARD J. MALINOWSKI ∫ ydx @ x2 = 0.5(− 11R 16) ( R 2 − ( − 11R 16 )2 )1 2 + 0.5 R 2 sin −1 (− 11 16) = −0.249625879R 2 + −0.379020382R 2 = −0.628646261R 2 , ∫ ydx @ ∫ ydx @ x1 x2 = 0 + 0.5R 2 sin −1 (− 1) = −0.785398163R 2 , x1 = 0.156751902R 2 . Slabs 1 + 2, Part I area = ( 0.156751902R 2 ) ( 2) = 0.313503804R 2 . Slabs 1 + 2, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (6 R 4 )2 , centered on the origin (0, 0). ∫ (a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 21R 16 to x3 = 1.5 R, a = 1.5 R. ∫ ydx @ ∫ ydx @ x3 x2 = 0 + 0.5(1.5R )2 sin −1 (1) = 1.767145868R 2 , = 0.5(21R 16) ((1.5R )2 − ( 21R 16)2 )1 2 + 0.5(1.5R )2 sin −1 (21 16 1.5) = 0.476558497R 2 + 1.198615294R 2 = 1.675173791R 2 , ∫ ydx @ x3 x2 = 0.091972077R 2 . Slab 2, Part II area = ( 0.091972077 R 2 )(2 ) = 0.183944154R 2 . Slabs 1 + 2, Part I + Part II : subtract Slab 1: Slab 2 : 0.497 447 958R 2 − 0.172 561 546 R 2 0.324 886 412 R 2 Slabs 1 + 2 + 3, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (7 R 4 )2 , centered on the origin (0, 0) . FRACTAL PHYSICS THEORY - NUCLEONS AND … 65 Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0) , with r = R. The two equations are solved simultaneously to determine their intersection point: 49 R 2 16 = 4 Rx − 3R 2 , x2 = 97 R 64 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x 2 + y 2 = R 2 , y = ( R 2 − x 2 )1 2 . ∫ (a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = − 31R 64 , a = R, ∫ ydx @ x2 = 0.5(− 31R 64 ) (( R 2 − (− 31R 64 )2 )1 2 + 0.5 R 2 sin −1 (− 31 64 ) = −0.211880272R 2 + −0.252824313R 2 = −0.464704585R 2 , ∫ ydx @ ∫ ydx@ x1 x2 = 0 + 0.5R 2 sin −1 ( − 1) = −0.785398163R 2 , x1 = 0.320693578R 2 . Slabs 1 + 2 + 3, Part I area = ( 0.320693578R 2 ) (2 ) = 0.641387156R 2 . Slabs 1 + 2 + 3, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (7 R 4 )2 , centered on the origin (0, 0). ∫ (a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 97 R 64 to x3 = 7 R 4 , a = 7R 4 , ∫ ydx @ ∫ ydx @ x3 x2 = 0 + 0.5(7 R 4) 2 sin −1 (1) = 2.405281875R 2 , = 0.5(97 R 64 ) ((7 R 4 )2 − (97 R 64 )2 )1 2 + 0.5(7 R 4 )2 sin −1 (97 64 1.75) 66 LEONARD J. MALINOWSKI = 0.662980207R 2 + 1.603662212R 2 = 2.266642419R 2 , ∫ ydx @ x3 x2 = 0.138639456R 2 . Slabs 1 + 2 + 3, Part II area = ( 0.138639456R 2 ) ( 2) = 0.277278912R 2 . Slabs 1 through 3, Part I + Part II : subtract Slabs 1 + 2 : Slab 3: 0.918666068 R 2 − 0.497447958 R 2 0.421 218110R 2 Slabs 1 through 4, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = ( 2 R )2 , centered on the origin (0, 0). Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on (2 R, 0) , with r = R. The two equations are solved simultaneously to determine their intersection point: 4 R 2 = 4 Rx − 3R 2 , x2 = 7 R 4 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x 2 + y 2 = R 2 , y = ( R 2 − x 2 )1 2 , ∫ (a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = − R 4, a = R, ∫ ydx @ x2 = 0.5(− R 4 ) = ( R 2 − (− R 4 )2 )1 2 + 0.5 R 2 sin −1 (− 1 4 ) = −0.121030729R 2 + −0.126340127R 2 = −0.247370856R 2 , ∫ ydx @ ∫ ydx @ x1 x2 = 0 + 0.5 R 2 sin −1 (− 1) = −0.785398163R 2 , x1 = 0.538027307R 2 . Slabs 1 through 4, Part I area = ( 0.538 027 307 R 2 ) (2) = 1.076 054 614 R 2 . FRACTAL PHYSICS THEORY - NUCLEONS AND … 67 Slabs 1 through 4, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = ( 2 R )2 , centered on the origin (0, 0). ∫ (a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 7 R 4 to x3 = 2 R, a = 2R, ∫ ydx @ x3 ∫ ydx @ x2 = 0 + 0.5(2 R )2 sin −1 (1) = πR 2 , = 0.5(7 R 4) (( 2 R )2 − (7 R 4)2 )1 2 + 0.5(2 R )2 sin −1 (7 4 2) = 0.847215107R 2 + 2.130871633R 2 = 2.978086740R 2 , ∫ ydx @ x3 x2 = 0.163505913R 2 . Slabs 1 through 4, Part II area = ( 0.163 505 913R 2 ) (2) = 0.327 011 827 R 2 . Slabs 1 through 4, Part I + Part II : subtract Slabs 1 through 3: Slab 4 : 1.403 066 441R 2 − 0.918 666 068R 2 0.484 400 373R 2 Slabs 1 through 5, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (9 R 4)2 , centered on the origin (0, 0). Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on (2R, 0) , with r = R. The two equations are solved simultaneously to determine their intersection point: 81R 2 16 = 4 Rx − 3R 2 , x2 = 129 R 64 . To simplify the integration the neutron’s surface equation is translated to center on the origin. The integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: ( R 2 − x 2 )1 2 . x2 + y2 = R2 , y= 68 LEONARD J. MALINOWSKI ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = R 64, a = R, ∫ydx @ x2 = 0.5( R 64) ( R 2 − ( R 64)2 )1 2 + 0.5R 2 sin −1 (1 64) = 0.007811546R 2 + 0.007812817R 2 = 0.015624363R 2 , ∫ydx @ ∫ydx @ x1 x2 = 0 + 0.5R 2 sin −1 ( − 1) = −0.785398163R 2 , x1 = 0.801022526R 2 . Slabs 1 through 5, Part I area = ( 0.801022526R 2 ) (2 ) = 1.602045052R 2 . Slabs 1 through 5, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (9 R 4)2 , centered on the origin (0, 0). ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 129 R 64 to x3 = 9 R 4, a = 9 R 4 , ∫ydx @ ∫ ydx @ x3 x2 = 0 + 0.5(9 R 4)2 sin −1 (1) = 3.976078202R 2 , = 0.5(129 R 64 )((9 R 4 )2 − (129 R 64 ) 2 )1 2 + 0.5(9 R 4 )2 sin −1 (129 64 2.25) = 1.007689469R 2 + 2.81045422R 2 = 3.818143689R 2 , ∫ydx @ x3 x2 = 0.157934513R 2 . Slabs 1 through 5, Part II area = ( 0.157 934 513R 2 ) (2) = 0.315 869 026 R 2 . Slabs 1 through 5, Part I + Part II : subtract Slabs 1 through 4 : Slab 5 : 1.917 914 078R 2 − 1.403 066 441R 2 0.514 847 637 R 2 FRACTAL PHYSICS THEORY - NUCLEONS AND … 69 Slabs 1 through 6, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (5R 2)2 , centered on the origin (0, 0) . Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0) , with r = R. The two equations are solved simultaneously to determine their intersection point: 25R 2 4 = 4 Rx − 3R 2 , x2 = 37 R 16 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x 2 + y 2 = R 2 , y = ( R 2 − x 2 )1 2 , ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = 5R 16, a = R , ∫ydx @ x2 = 0.5(5R 16) ( R 2 − (5R 16)2 )1 2 + 0.5 R 2 sin −1 (5 16) = 0.148424649R 2 + 0.158911852R 2 = 0.307336501R 2 , ∫ydx @ ∫ydx @ x1 x2 = 0 + 0.5R 2 sin −1 ( − 1) = −0.785398163R 2 , x1 = 1.092734664R 2 . Slabs 1 through 6, Part I area = (1.092734664R 2 ) ( 2) = 2.185469328R 2 . Slabs 1 through 6, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (5R 2)2 , centered on the origin (0, 0) . ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 37 R 16 to x3 = 5R 2 , a = 5R 2 , 70 LEONARD J. MALINOWSKI ∫ydx @ x3 ∫ ydx @ = 0 + 0.5( 5R 2 ) 2 sin −1 (1) = 4.908738521R 2 , = 0.5(37 R 16 ) ((5 R 2 )2 − (37 R 16 )2 )1 2 + 0.5(5 R 2 )2 sin −1 (37 R 16 2.5) x2 = 1.098342410R 2 + 3.690736231R 2 = 4.789078641R 2 , ∫ydx @ x3 x2 = 0.119659880R 2 . Slabs 1 through 6, Part II area = ( 0.119659880R 2 ) (2 ) = 0.239319760R 2 . Slabs 1 through 6, Part I + Part II : subtract Slabs 1 through 5 : Slab 6 : 2.424 789 088R 2 − 1.917 914 078R 2 0.506 875 010 R 2 Slabs 1 through 7, Part I Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (11R 4 )2 , centered on the origin (0, 0) . Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0) with r = R. The two equations are solved simultaneously to determine their intersection point: 121R 2 16 = 4 Rx − 3R 2 , x2 = 169 R 64 . To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x 2 + y 2 = R 2 , y = ( R 2 − x 2 )1 2 , ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = 41R 64 , a = R, ∫ ydx @ x2 = 0.5(41R 64 ) ( R 2 − (41R 64 )2 )1 2 + 0.5 R 2 sin −1 (41 64) = 0.245953201R 2 + 0.347655973R 2 = 0.593609174R 2 , FRACTAL PHYSICS THEORY - NUCLEONS AND … ∫ydx @ ∫ydx @ x1 x2 71 = 0 + 0.5R 2 sin −1 ( − 1) = −0.785398163R 2 , x1 = 1.379007337 R 2 . Slabs 1 through 7, Part I area = (1.379007337R 2 ) ( 2) = 2.758014674R 2 . Slabs 1 through 7, Part II Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (11R 4 )2 , centered on the origin (0, 0) . ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Integration limits, x2 = 169R 64 to x3 = 11R 4, a = 11R 4 , ∫ydx @ ∫ydx @ x3 x2 = 0 + 0.5(11R 4)2 sin −1 (1) = 5.939573611R 2 , = 0.5(169 R 64) ((11R 4)2 − (169 R 64)2 )1 2 + 0.5(11R 4)2 sin −1 (169 64 2.75) = 1.013 807 100 R 2 + 4.869 550 615R 2 = 5.883 357 715R 2 , ∫ydx @ x3 x2 = 0.056215896R 2 . Slabs 1 through 7, Part II area = ( 0.056215826R 2 ) ( 2) = 0.112431792R 2 . Slabs 1 through 7, Part I + Part II : subtract Slabs 1 through 6 : Slab 7 : 2.870 446 466 R 2 − 2.424 789 088R 2 0.445 657 378R 2 Slabs 1 through 8 Equation of the proton’s electric equipotential surface: x 2 + y 2 = r 2 = (3R )2 , centered on the origin (0, 0) . Equation of the neutron’s surface: x 2 + y 2 = 4 Rx − 3R 2 , centered on ( 2R, 0) with r = R. 72 LEONARD J. MALINOWSKI To simplify the integration the neutron’s surface equation is translated to center on the origin. Integration limits, x1 = R and x2 = 3R are also translated by the same amount along the X-axis by subtracting 2R. Translated neutron surface equation: x 2 + y 2 = R 2 , y = ( R 2 − x 2 )1 2 , ∫( a 2 − x 2 )1 2 dx = 0.5 x( a 2 − x 2 )1 2 + 0.5a 2 sin −1 ( x a ) , Translated limits, x1 = − R to x2 = R, a = R, ∫ydx @ x2 x1 = 0 + 0.5R 2 sin −1 (1) − [ 0 + 0.5R 2 sin −1 ( − 1) ] = 0.25π + 0.25π = 0.5π. Slabs 1 through 8 = 0.5π( 2) : Subtract slabs 1 through 7 : Slab 8 : π R2 − 2.870 446 466 R 2 0.271146 187 R 2 References [1] L. J. Malinowski, Fractal Physics Theory - Foundation, Fundamental J. Modern Physics 1(2) (2011), 133-168. [2] L. J. Malinowski, Fractal Physics Theory - Cosmic Scale Nuclear Explosion Cosmology, Fundamental J. Modern Physics 1(2) (2011), 169-195. [3] L. J. Malinowski, Fractal Physics Theory - Electrons, Photons, Wave-Particles, and Atomic Capacitors, Fundamental J. Modern Physics 1(2) (2011), 197-221. [4] D. R. Lide, editor, Handbook of Chemistry and Physics, CRC Press, Boca Raton, FL, 2006. [5] Linus Pauling, General Chemistry, Dover Publications, Inc., New York, 1970. [6] P. W. Atkins, Physical Chemistry, 3rd ed., Copyright © 1986 Oxford University Press. [7] John B. Fraleigh, Calculus with Analytic Geometry, Addison-Wesley Publishing Company, Inc., Philippines, 1980.