Download fractal physics theory - nucleons and the strong force

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
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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,
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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)
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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
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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)
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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.