Ground Rules
Compare Electron Energy Levels to Proton and Neutron Energy Levels >
The Nuclear Gravitation Field solutions to Schrodinger Wave Equation differ from Schrodinger Wave Equation solutions to the Nuclear Gravitation Electric Field. Newton’s Law of Gravity assumes Masses M_{1} and M_{2} are spherical. Stars, Planets, and Moons are, typically, spherical, therefore the 1/r^{2} Gravity Field Potential Function attracting mass M_{2} to mass M_{1} of the equation is valid.
The Nuclear Gravitation Field Gravity Field Potential Function will be dependent upon the shape of the Nucleus as “seen” by the Proton or Neutron of interest next to Nucleus.
Elements with Electron Magic Numbers are in Group 18 at the right. Elements with Proton Magic Numbers are outlined in Red. Reference: http://www.webelements.com/index.html
Electrons fill the electron energy levels starting from left to right along each row and by rows from top to bottom. Hydrogen (H), with a 1s^{1} electron configuration, and Helium (He), with a 1s^{2} electron configuration, are placed at the top of the Periodic chart on both sides for the Periodic Table of the Elements because the first electron energy level consists only of an “s Orbital” and Hydrogen can take on the characteristic of either an Alkali Metal and as a Halogen and Helium is a Noble Gas.
Reference: http://www.webelements.com/index.html Nuclei with the number of Protons and/or Neutrons less than 50 typically will have a classical shape that deviates from a near perfect spherical shape. If the classical shape of the Nucleus is not spherical, then the Nuclear Gravitation Field would not be a 1/r^{2} potential function within the Schrodinger Wave Equation defining the Proton or Neutron Total Energy and Nuclear Energy Level fills for Protons and Neutrons would be expected to deviate from energy level fills for Electrons as observed. Therefore, either the Heisenberg Uncertainty Principle does not drive how Proton and Neutron Energy Levels are filled or the method of fill of the nuclear energy levels cannot confirm the Strong Nuclear Force being equivalent to Gravity.
NNuclei with the number of Protons and/or Neutrons greater than or equal to 50 will have a classical shape that matches a near perfect sphere. Therefore, if the Strong Nuclear Force and Gravity are the same, then the Solutions for the Schrodinger Wave Equation with a Nuclear Gravitation Field proportional to a 1/r^{2} function should result in the applicable Proton and Neutron Energy Levels filling identically to the filling of the Electron Energy Levels for the applicable energy levels.
The fill of Neutrons for the Seventh Energy Level at a change of 44 deviates from the Electrons and Protons for the Seventh Energy Level at a change of 32. The fill of Neutrons for the Eighth Energy Level at a change of 58 deviates from the projected fill of Electrons and Protons for the Eighth Energy Level at a change of 50. The Neutron Energy Level fill deviations is suspected to be the result of the strong accumulated Coulombic Repulsion Force tending to tear the nucleus apart. The need for additional Neutrons in the Nucleus is required to raise the Strong Nuclear Force to hold the Nucleus together. Note that for the heavier elements, the Neutron to Proton ratio rises from a 1 to 1 ratio for Light Nuclei to a 3 to 2 ratio for Heavy Nuclei. For stable Nuclei of the Heavier Elements, Neutrons fill the next higher energy level than the Protons fill. For example, the Nucleus for Lead208 (_{82}Pb^{208}), 82 Protons fill Six Energy Levels and 126 Neutrons fill Seven Energy Levels. All currently known Elements beyond Element 83, Bismuth209, (_{83}Bi^{209}), are observed to be radioactive – not stable. Table of the Nuclides Z = Number of Protons = Vertical Axis N = Number of Neutrons = Horizontal Axis Reference: http://atom.kaeri.re.kr When the number of Protons and Neutrons in the Nucleus each number 50 or greater, the classical shape of the Nucleus is a near perfect sphere. Proton Energy Levels Six through Eight fill in identical manner to Electron Energy Levels Six through Eight. Neutron Energy Level Six fills in identical manner to Electron Energy Level Six. The deviation for Neutron Energy Levels above Level Six can be accounted for by the need to raise the Nuclear Gravitation Field intensity to overcome the rising Coulombic Repulsion Force tending to break the Nucleus apart. It is safe to conclude that when the classical shape of the Nucleus is a near perfect sphere, then the Nuclear Gravitation Field Potential Function is proportional to 1/r^{2} and consistent with Newton’s Law of Gravity. Nuclear Gravitation Field Required to Overcome Nuclear Electric Field Repulsion Let’s assume we wanted to fuse a Proton (Hydrogen1 Nucleus) to a Tritium (Helium3) Nucleus. Let’s determine what the minimum Nuclear Gravitation Field would be required to overcome the Nuclear Electric Field repulsive force present at the surface of the Tritium Nucleus. The Nuclear Electric Field repulsive force must be determined for a Proton next to a Tritium Nucleus. We will assume the Proton is placed next to the Proton in the Tritium Nucleus. The radius, r, of each Proton is equal to 1.2 x 10^{15} meter, therefore, the distance between the center of one Proton and the center of the other Proton is twice the radius of each Proton or 2.4 x 10^{15} meter. Only the Protons sense a repulsive force because the Neutrons in the Tritium Nucleus are neutral in charge. The equation for Force between electrically charged particles is as follows: q_{1} is the charge of the Proton in the Tritium Nucleus; q_{2} is the charge of the Proton placed next to the Tritium Nucleus to establish the repulsive force, F; ∈_{0} is the permittivity of the Electric Field in free space; and r is the distance between the centers of each of the Protons. We also know that the classical physics calculation for force acting on a body, in this case the Tritium Nucleus acting upon the Proton, is as follows: Where F is Force, m_{p} is the mass of a proton, and a is the acceleration of the Proton. Using both equations, above, the acceleration sensed by the Proton being repelled by the Tritium Nucleus can be determined by solving for acceleration, a, as follows: Divide the acceleration field, a, by the acceleration of gravity on Earth, 9.81 meters/sec^{2}, to obtain the acceleration field in g’s. The value of acceleration, a, represents the repulsive electric field established by the Tritium Nucleus on the Proton of interest. In order to overcome this repulsive field, the minimum required attractive acceleration field required by the Strong Nuclear Force to overcome the Electric Field repulsion of the proton from the Tritium Nucleus must be at least 2.441 x 10^{27}g. In order to determine whether or not the Nuclear Gravitation Field at the surface of the nucleus has an intensity great enough to result in observable General Relativistic effects, one must compare the Nuclear Gravitation Field at the surface of the nucleus to the gravitational field in the vicinity of the Sun’s surface and in the vicinity of the surface of a Neutron Star. Gravitational fields in the vicinity of stars are more than intense enough for significant General Relativistic effects to be observed. The Neutron Star was selected as one of the cases to study because the density of the nucleus, which is made up of protons and neutrons, is very close to the density of a Neutron Star. A Neutron Star typically contains the mass approximately that of our Sun, however, the matter is concentrated into a spherical volume with a diameter of about 10 miles or 16 kilometers. The radius of a Neutron star is about 5 miles or 8 kilometers. The following figure illustrates the relative size of the Earth to the size of a White Dwarf star and a Neutron star. Comparative Sizes of the Earth, Reference: “The Life and Death of Stars,” by Donald A. Cooke, Page 131, Figure 8.12 Let’s determine the gravitational field at the surface of a Neutron Star. The Neutron Star is assumed to contain the same mass as the Sun, therefore, the mass of the Neutron Star, “M_{Neutron Star},” is equal to 1.99×10^{30} kg. The Neutron Star’s radius, which is defined as “R_{Neutron Star},” is about 5 miles equal to about 8 km or 8.0×10 a_{Neutron Star} = 2.07×10^{12}Nkg^{1} = 2.07×10^{12} meters/second^{2} To determine the “gforce” at the Neutron Star’s surface, the gravitational field of the Neutron Star must be normalized relative to Earth’s gravitational field in the same manner used to calculate the gforce at the Sun’s surface. Earth’s gravitational field is 1g. The ratio of the acceleration of gravity on the Neutron Star’s surface to the acceleration of gravity on the Earth’s surface represents the gforce on the Neutron Star’s surface. The gforce on the Neutron Star’s surface is calculated as follows: Albert Einstein’s General Relativity Theory was confirmed in 1919 when a total eclipse of the Sun occurred. With the Sun being blocked out by the Moon, a star directly behind the Sun could be seen on either side of the Sun due to the acceleration of light around the Sun. The gravitational field of the Sun at 27.8g can warp or bend SpaceTime. The gravitational field of a Neutron Star, at 2.10×10^{11}g, has an intensity of nearly a billion times greater than the Sun’s gravitational field. The Neutron Star’s gravitational field will result in substantial “SpaceTime Compression.” The minimum gravitational field intensity at the surface of the Tritium Nucleus was calculated to be General Relativity and SpaceTime Compression SpaceTime Compression: What is SpaceTime Compression and how is it related to Special Relativity and General Relativity? SpaceTime Compression is the relativistic effect of reducing the measured distance and light traveltime between two points in space as a result of the presence of either:
One way to quantitatively observe the SpaceTime compression effect is to look at the blue right triangle within a quarter circle of unity radius (radius, r = 1) displayed in the figure, below. The hypotenuse of the triangle is the side of the blue triangle starting from the center of the circle moving diagonally upward and to the right with a length unity, or 1, and equal to the radius of the quarter circle. The hypotenuse of the blue right triangle represents the velocity of light, c, as a fraction of the velocity of light, c, or c/c = 1. Trigonometric Identities and Relationship to Relativity The vertical side of the blue right triangle is the ratio of the velocity of the spacecraft to that of the velocity of light, or v/c and is the “opposite side” from the angle formed by the hypotenuse of the blue right triangle, or the radius of the quarter circle, and the horizontal leg of the blue right triangle. That angle is represented by the Greek letter θ (Theta). The angle θ spans from 0° to 90°. The length of the vertical side of the blue right triangle is equal to sinθ, therefore, v/c = sinθ. The horizontal side of the blue right triangle represents the amount of SpaceTime Compression that reduces the distance between two points in SpaceTime (length contraction) and reduces the time it takes light to travel between the two points (time dilation) based on the velocity of the spacecraft relative to the velocity of light. The length of the horizontal side of the blue right triangle is equal to cosθ. Using the Pythagorean Theorem, we can solve for cosθ. sin^{2}θ + cos^{2}θ = 1 v/c = sinθ Substitute v/c for sinθ, then solve for cosθ
(v/c)^{2} + cos^{2}θ = 1
cos^{2}θ = 1 – (v/c)^{2} cosθ = [1 – (v/c)^{2}]^{1/2} 1/cosθ = secθ = 1/[1 – (v/c)^{2}]^{1/2} = “SpaceTime Compression Factor” The “SpaceTime Compression Factor” is the multiplicative inverse of the value of length of the horizontal side of the blue right triangle equal to 1/cosθ, or secθ, and is designated by the Greek Letter γ (gamma). The length contraction and time dilation can be determined by solving for the length of the horizontal side of the triangle using the Pythagorean Theorem. Therefore, the distance between the Earth and the star of interest 10 lightyears away as measured by the observer in the spacecraft moving at a velocity, v, is defined as follows: d = d_{0} [1 – (v/c)^{2}]^{1/2} Where d_{0} is the measured distance to the star of interest in “Normal SpaceTime” or “Uncompressed SpaceTime” as measured by the observer on the Earth and d is the “Compressed SpaceTime” distance (or length contracted distance) as measured by the observer in the spacecraft moving at a velocity, v. The reduction of time for light to travel the distance between the star of interest and spacecraft in the vicinity of Earth is affected in the same manner by “SpaceTime Compression.” Einstein noted the equivalence of space and time, hence the term SpaceTime is used. The relationship of space and time is as follows: d = ct d represents distance, c represents the velocity of light, and t represents elapsed time. Substituting ct for d and ct_{0} for d_{0}, the distance compression equation can become a time dilation equation. ct = ct_{0} [1 – (v/c)^{2}] Therefore: t = t_{0} [1 – (v/c)^{2}]^{1/2} The following table provides the values for length contraction and “SpaceTime Compression Factor” as a function of velocity relative to the velocity of light, c = 299,750 km/sec = 186,300 miles/sec.
As indicated in the table, above, measured velocities do not contribute significantly to the “SpaceTime Compression” effect unless the measured velocity is a significant fraction of the the velocity of light, c. At a measured velocity of 0.995c, the “SpaceTime Compression Factor” is just above 10 and at a measured velocity of 0.999c, the “SpaceTime Compression Factor” is just under 22.4. Table “Velocity and SpaceTime Compression” introduces the concept of effective velocity. When the spacecraft is traveling at a measured velocity of 0.707c, the effective velocity, v_{eff}, of the spacecraft is 1.000c or the speed of light, c. Although the spacecraft only has a measured velocity as 0.707c, the length contraction along the line of travel is reduced to 0.707 (or 70.7%) of the original distance (which represents a “SpaceTime Compression Factor” equal to 1.414), therefore, the time to travel the uncompressed distance (which is a known quantity) is equal to the time it would take light to travel the uncompressed distance. The evaluation, above, discusses SpaceTime Compression as a function of a constant relativistic velocity, therefore is an evaluation of a inertial reference frame. General Relativity includes the evaluation of accelerated reference frames. Gravity fields establish accelerated reference frames because gravity accelerates light, electric fields, and magnetic fields. Therefore, gravity fields generate the Compressed SpaceTime due to the acceleration of light, electric fields, and magnetic fields. Since the speed of light will always be measured as propagating at a constant speed of 2.9975 x 10^{8} meters/sec, the distance traveled is reduced by the SpaceTime Compression Factor. Light, Electric Fields, and Magnetic Fields propagate based upon Compressed SpaceTime. General Relativity – Accelerated Reference Frames and Relation to Nuclear Gravitation Field We determined that the minimum Nuclear Gravitation Field acceleration has to be at least 1.403 x 10^{28}g, therefore, the effects of General Relativity must be considered. Gravity fields propagate based upon Uncompressed SpaceTime. In order to “see” or “measure” the Nuclear Gravitation Field propagating outward from the Nucleus omnidirectional in spherical symmetry dropping in intensity 1/r^{2} consistent with Newton’s Law of Gravity, we would have to measure the field intensity in Uncompressed SpaceTime. However, we live in the Compressed SpaceTime reference frame, therefore, we see events in Compressed SpaceTime. As previously discussed, gravity fields generate the Compressed SpaceTime due to the acceleration of light, electric fields, and magnetic fields. Light, Electric Fields, and Magnetic Fields propagate based upon Compressed SpaceTime. Table “Accelerated Reference Frame SpaceTime Compression Due to Gravity Field,” below, determines the Uncompressed SpaceTime acceleration of light as a function of the intensity of various gravity fields, determines the reduction in the distance traveled by light as observed in Compressed SpaceTime, and the resulting SpaceTime Compression Factors. Let’s assume the gravitational field next to the nucleus of the atom was equal to 2.9975 x 10^{8} meters/sec^{2}. If light was subjected to this acceleration field, in one second the speed of light would be doubled to 5.9950 x 10^{8} meters/sec equal to the speed of light in Uncompressed SpaceTime. Since the speed of light in free space is invariant with respect to the reference frame of the observer, the speed of light remains at 2.9975 x 10^{8} meters/sec. Therefore, the distance traveled by light in Compressed SpaceTime must be reduced to half the Uncompressed SpaceTime distance as indicated by the first entry highlighted in red in Table, “Accelerated Reference Frame SpaceTime Compression Due to Gravity Field,” below.
From Table “Accelerated Reference Frame SpaceTime Compression Due to Gravity Field,” we find that the SpaceTime Compressed distance that light travels is essentially a zero distance (at five significant digits} for any gravity field acceleration field greater than or equal to The Nuclear Gravitation Field next to the Tritium Nucleus rivals the field near a Neutron Star or Black Hole. The Nuclear Gravitation Field intensity drops about 27.5 decades just outside the Nucleus before any gravity field can be “seen” or “measured” propagating outward from the Nucleus because we observe the Nuclear Gravitation Field in a Compressed SpaceTime reference frame. The SpaceTime Compression occurring next to the Nucleus is so significant that if we could view the Atom in Uncompressed Space Time, the Electron Cloud would be one meter away from the Nucleus. Since the atom is at least 30,000 (3.0 x 10^{4}) times larger than the nucleus, the quantized Nuclear Gravitation Field intensity drops an additional 9 decades (4.5 x 2 decades) as it passes through the electron cloud of the atom to the outside of the atom due to Nuclear Gravitation Field intensity dropping 1/r^{2}. The final quantized Nuclear Gravitation Field intensity is on the order of 1.0 x 10^{8}g or less, extremely feeble. This value appears to be too large to be gravity at this value. One must realize that we are observing the Nuclear Gravitation Field intensity which is Quantized Gravity because the field is associated with a specific energy level spectrum. Nuclear Gravitation Field and The Nuclear Gravitation Field is stronger than the Nuclear Electric Field at the Nuclear Surface in order to hold the nucleons in the Nucleus together. Nuclear Gravitation Field intensity outside the Electron cloud is less than 1.0 x 10^{35} the intensity of Nuclear Electric Field outside the Electron cloud. SpaceTime Compression is directly proportional to the intensity of the Nuclear Gravitation Field. Quantized gravity will be on the order of 1.0 x 10^{8} to 1.0 x 10^{9} times greater than the average gravity measured outside the electron cloud of the Atom. Therefore, gravity leaving the electron cloud will be 1.0 x 10^{16}g or less. Quantized Gravity is analogous to the Classical Physics and Quantum Mechanical evaluation of Light and how the PhotoElectric Effect occurs. Quantized Light and PhotoElectric Effect Analogous to Quantized Gravity – Liberating Outer Electron from Sodium Atom: In order to compare Classical Physics to Quantum Mechanics, the energy of light shining on a surface must be assumed to be a continuous distribution. In other words, light energy is assumed neither to be discrete nor quantized. Based upon that assumption, the amount of energy available to be absorbed by an electron can be determined. That calculated value will then be compared to the results of Millikan’s “Photoelectric Effect” experiments. For this calculation, a 100 watt (Joules/second) orange light source with a wavelength of 6000 Angstroms is directed onto a square plate of Sodium 0.1 meter by 0.1 meter. The surface area of the square Sodium plate is 0.01 meter^{2}. It is assumed that all the light emitting from the orange light source is directed onto the Sodium plate. The atomic radius of the neutral Sodium atom is 2.23 Angstroms which is equal to 2.23×10^{10} meter. The diameter of the Sodium atom is twice the radius or 4.46 Angstroms equal to 4.46×10^{10} meter. It now must be determined how many Sodium atoms can fill the square surface of the Sodium plate assuming only the top layer of Sodium atoms (one Sodium atom deep). Although the Sodium atoms are spheres, this calculation will assume that they are square. The side of the “square Sodium atom” has the same length as the diameter of the spherical atom. A spherical atom of Sodium will fit into each of the theoretical “square Sodium atoms” that make up the top layer of Sodium atoms on the square plate. Therefore, each Sodium atom will take up the following surface area on the plate: Area of Sodium Atom = (4.46×10^{10} meter)×(4.46×10^{10} meter) = 1.989×10^{19} meter^{2} Number of Sodium Atoms on Surface of Plate = Area of Plate divided by Area of Sodium Atom Number of Sodium Atoms on Surface of Plate = (0.01 meter^{2})/(1.989×10^{19} meter^{2}) Number of Sodium Atoms on Surface of Plate = 5.027×10^{16} Sodium Atoms In one second, the Sodium plate surface absorbs 100 Joules of energy. 1 electron volt (eV) is equal to 1.6022×10^{19} Joules. The next step is to calculate the amount of energy imparted to one Sodium atom in eV assuming a continuous even distribution of light energy across the Sodium plate. The intent here is to perform a comparison of the values of the classical electron absorption energy to the Quantum Mechanical electron absorbed energy as provided in Figure “Sodium Plate PhotoElectric Effect Results,” below. PhotoElectric Effect on Sodium Plate Sodium Plate PhotoElectric Effect Results Reference: http://hyperphysics.phyastr.gsu.edu/hbase/mod2.html Energy Imparted to 1 Sodium Atom (E_{Na}) = Total Energy Imparted to Plate divided by Number of Sodium Atoms = 5.027 x 10^{16} Naatoms Each Sodium atom is receiving 1.242×10^{4} eV of energy each second. The light is only illuminating one side of the Sodium atom because it is coming from one direction, therefore, as the spherical Sodium atom is considered, half the surface area of the Sodium atom is illuminated by the light. The total surface area of a spherical Sodium Atom is calculated as follows: The illuminated portion of the sphere of the Sodium atom is equal to half the value calculated, above, or 3.124×10^{19} meter^{2}. In actuality, from a classical point of view, the size of the Sodium Atom is not important or required to determine how much light energy the electron will receive from the from the light source based on classical physics. The size and exposed surface area of the electron is all that is required to complete this calculation. In this calculation it is assumed that the density of an electron is the same as the density of a proton or neutron. The mass of a proton, neutron, or electron is proportional to the cube of its radius or its diameter. The surface area of either the proton, the neutron, or the electron is proportional to the cube root of its volume squared. The surface area of the electron, then, should be proportional to the surface area of a proton or neutron by the ratio of its mass to the mass of a proton or neutron to the 2/3 power. The diameter of a proton or neutron is about 1.0×10^{15} meter. The radius of a proton or neutron is equal to half the diameter or about 0.5×10^{15} meter. The total surface area of either a proton or neutron is calculated below: Since the light is shining from one direction, the light only illuminates half of the surface area of either a proton or neutron. Therefore, the illuminated surface area of a proton or neutron is equal to 1.571×10^{30} meter^{2}. The electron mass is only 1/1840 that of the proton or neutron. Therefore, the surface area of an electron will be equal to the surface area of a proton or neutron multiplied by the cube root of 1/1840 squared. The surface area of an electron can be calculated based upon the surface area of a proton or neutron (nucleon) as follows: A_{surfaceelectron} = 2.092×10^{32 }meter^{2} Since the light is shining from one direction, the light only illuminates half of the surface area of the electron. Therefore, the illuminated surface area of the electron is equal to 1.046×10^{32} meter^{2}. The calculated amount of energy by the classical physics illumination from the light source received by the Sodium’s electron is as follows: The Classical Physics analysis predicts the electron only receives 4.163×10^{10} eV of light energy per second. The amount of energy required to liberate an electron from the Sodium atom is on the order of 0.5 eV. The Classical Physics analysis result indicates that it is impossible for the photoelectric effect to ever take place. Quantum Mechanics predicts that the electron can absorb energies on the order of 0.5 eV or greater and can be liberated from the Sodium atom because the incoming light energy propagates in discrete packets or quanta of energy rather than as a continuous distribution of energy. The vast difference in magnitude of the energy that the electron would absorb based upon Classical Physics to the amount of the energy the electron will absorb by Quantum Mechanics is extremely important. It is quite reasonable to assume that this significant relative difference in magnitude of field intensity can also apply to the intensity of the Nuclear Gravitation Field. The Nuclear Gravitation Field would be much more intense if it was a discrete function rather than a continuous function. Figure “Sodium Plate PhotoElectric Effect Results,” above, states the electron Kinetic Energy is about 0.5 eV when it absorbs light at a wavelength of 6000 Angstroms. The electron must absorb a minimum amount of “Ionization Energy” to remove it from the Sodium atom before it obtains any Kinetic Energy. To be conservative, this calculation assumes the Ionization Energy of the electron in the 3s orbital of the Sodium atom to be equal to zero. The ratio of the quantized energy absorbed by the electron versus the classical calculated energy absorbed by the electron is as follows: The ratio of the amount of energy absorbed by the electron via Quantized Light, assuming the principles of Quantum Mechanics, versus the amount of energy absorbed by the electron, assuming Classical Physics, is on the order of 1.201×10^{9} times greater or over a billion times greater. The expectation is that Quantized Gravity would behave similar to Quantized Light, therefore, Quantized Gravity could be expected to have an intensity of 1.0 x 10^{8} to 1.0 x 10^{9} times greater than the average gravity measured outside the electron cloud of the atom. Strong Nuclear Force Properties Provide Case for Equivalence to Gravity Several properties of the Strong Nuclear Force are observed specifically because the Strong Nuclear Force is Gravity. The virtual vanishing of the Strong Nuclear Force just outside the Nuclear Surface is the primary indicator the Strong Nuclear Force is Gravity because of the intense SpaceTime Compression taking place. The addition of Neutrons to the Nucleus to boost the Strong Nuclear Force intensity to maintain its intensity above the Nuclear Electric Field intensity within the Nucleus and hold the Nucleus together is directly related to the General Relativistic effect of Gravity. If the Strong Nuclear Force had nothing to do with Gravity, no such accelerated field would be produced within the Nucleus affecting the propagation of Light, Electric Fields, or Magnetic Fields and SpaceTime Compression would be nonexistent. In that case, Protons, alone, would always remain sufficient to overcome the Coulombic Repulsion generated by the Nuclear Electric Field because the Strong Nuclear Force intensity per Proton will always remain above that of the Nuclear Electric Field intensity per Proton. Without SpaceTime Compression occurring within the Nucleus, the Nucleus would remain stable with any number of Protons from one to infinity. The observed stablity curve for Nuclei require about a 1 to 1 Neutron to Proton ratio for light Nuclei and require about a 3 to 2 Neutron to Proton ratio for heavy Nuclei. The Strong Nuclear Force (Gravity) propagates based upon Uncompressed SpaceTime within and outside the Nucleus and the Nuclear Electric Field propagates based upon Compressed SpaceTime within and outside the Nucleus. Nuclear Gravitation and Electric Fields Within the Nucleus Let’ assume the Nucleus has a constant homogeneous mass density and constant homogeneous charge density for simplicity of calculations. . Let’s determine the Strong Nuclear Force acceleration profile without SpaceTime Compression and with SpaceTime Compression within the Nucleus. Let’s determine the acceleration field for the Nuclear Electric Field within the Nucleus. R represents the radius of the Nucleus. r represents the variable radial position of a Proton being evaluated between the Nuclear Center and the outer radius, R, of the Nucleus, in order to determine the acceleration field profiles. Determination of Nuclear Gravitational Field acceleration, g_{IN}, as a function of internal distance from the Center of the Nucleus, r. Mass, M, is equal to constant mass density times Volume, ρ_{Mass} x V_{IN}, as a function of r, radial distance from Center of Nucleus. Therefore, the gravitational acceleration inside the Nucleus, g_{IN, is proportional to r.} The mass contributing to g_{IN} is only the mass from Center of Nucleus to position r inside the Nucleus. However, the g_{IN} calculated previously with linear rise relative to radial distance from the Center of Nucleus, r, is not correct because it assumes classical physics. The Gravity field internal to the Nucleus is extremely intense and the effects of General Relativity must be considered. g_{IN} must be evaluated with the effects of SpaceTime Compression occurring. Gravity propagates based upon Uncompressed SpaceTime. Light, Electric Fields, and Magnetic Fields propagate based upon Compressed SpaceTime. We observe the Nucleus of the Atom in our Compressed SpaceTime Reference Frame. g_{IN} must be reevaluated to see how it will behave in Compressed SpaceTime. The Mass of the Nucleus as a function of density and distance from the Center of the Nucleus is based upon the Compressed SpaceTime radial distance, r_{CST}, as indicated by the equation below: The Gravity Field inside the Nucleus, g_{IN}, must be calculated based upon Uncompressed SpaceTime radial distance from the Center of the Nucleus, r_{UST} as indicated by the equation below: Substituting the calculation value for M_{IN}, the resulting equation for calculating g_{IN} is as follows: r_{UST} rises faster than r_{CST} and the rate of rise of r_{UST} goes up because the Gravity field intensity rises as a function of r_{UST} resulting in rising Compressed SpaceTime. g_{IN} will no longer rise linearly, it will tend to level off as r_{UST} and r_{CST} rise. The figure, below, indicates the behavior of g_{INNSTC}, Nuclear Internal Gravity Field  No SpaceTime Compression present, and g_{INCST}, Nuclear Internal Gravity Field  With SpaceTime Compression present: Determination of Nuclear Electric Field acceleration, a_{EFIN}, as a function of internal distance from the Center of the Nucleus, r. The charge distribution inside the Nucleus contributing to a_{EFIN} is only the charge distribution from Center of Nucleus to position r inside the Nucleus. Therefore, the Nuclear Electric Field acceleration field inside the Nucleus, a_{EFIN} is proportional to r. The charge distribution inside the Nucleus contributing to a_{EFIN} is only the charge distribution from Center of Nucleus to position r inside the Nucleus. Figure “Nuclear Field Profiles Within the Nucleus,” below, provides the field profiles as a function of distance r from the Center of the Nucleus to the Nuclear Surface. Nuclear Gravitation Field Profiles Within the Nucleus NFNSTC = Strong Nuclear Force  No SpaceTime Compression The profiles of the Acceleration Fields listed, above, within the Nucleus as a function of Atomic Mass are provided in Figure, “Nuclear Gravitation Field and Nuclear Electric Field at Nuclear Surface as Function of Atomic Mass,” below. Nuclear Gravitation Field and Nuclear Electric Field at Nuclear Surface as Function of Atomic Mass NFNSTC = Strong Nuclear Force  No SpaceTime Compression Dropoff of SNFWSTC results in the apparent “saturation” of the Strong Nuclear Force and has a profile appearance similar to Binding Energy per Nucleon curve. Binding Energy Per Nucleon Nuclear Gravitation Field and Configuration of Lead208 and Bismuth209
Reference: http://atom.kaeri.re.kr/ton/nuc11.html The Lead208 isotope (_{82}Pb^{208}) is a “double magic” Nucleus containing 82 Protons and 126 Neutrons. For Lead208, 82 Protons fill Six Energy Levels and 126 Neutrons fill Seven Energy Levels. The Nuclear Gravitation Field for Lead208 is relatively strong and SpaceTime Compression next to the Nuclear Surface is very significant compared to average stable nuclei. The Lead Nucleus is also near the apparent limit where the Strong Nuclear Force is able to overcome the Coulombic Repulsion Force and remain a stable nucleus. The Bismuth209 isotope (_{83}Bi^{209}) has 83 Protons and 126 Neutrons. For Bismuth209, 82 of the 83 Protons fill Six Energy Levels and 126 Neutrons fill Seven Energy Levels. The 83^{rd} Proton is the lone Proton in Seventh Energy Level. For all the currently known stable isotopes of Elements, the Bismuth209 nuclear configuration is unique. There are no other stable isotopes of Elements on Earth with a similar configuration. Element 83, Bismuth209, is the last known stable isotope listed on the Periodic Table of the Elements. All identified isotopes of Elements beyond Bismuth are radioactive indicating the Coulombic Repulsion Force has become significant enough to affect the Strong Nuclear Force ability to hold those nuclei together. Radioactive decay occurs to ultimately change the Nuclei to a stable Nucleus. The lone proton in the Seventh Energy Level of the Bismuth Nucleus is “loosely” held to the Nucleus resulting in the Nuclear Gravitation Field for Bismuth209 being significantly weaker than the Nuclear Gravitation Field for Lead208. The weaker Nuclear Gravitation Field for Bismuth209 results in the Nuclear Gravitation Field undergoing a significantly lesser amount of SpaceTime Compression. Therefore, the gravity field outside the electron cloud for Bismuth will be greater than what would be determined by Bismuth209 having an atomic mass of 208.980399 amu (atomic mass units) and Lead208 having an atomic mass of 207.976652 amu. Note that the difference in mass between Bismuth209 and Lead208 is 1.003747 amu. The atomic mass of the Hydrogen1 isotope is 1.007825 amu, therefore, the mass defect to bind the 83rd proton to the Bismuth209 Nucleus is 0.004078 amu. Cavendish Experiments can be performed to demonstrate that Bismuth209 has a stronger gravitational field beyond that associated with its mass than the gravitational field of Lead208. The Cavendish Experiment was used to determine G in Newton’s Law of Gravity equation:
Perform the Cavendish Experiment with Lead208 to measure its Gravitation Constant G_{Pb}. Perform the Cavendish Experiment with Bismuth209 to measure its Gravitation Constant G_{Bi}. If the result of performing these Cavendish Experiments results in determining that the value for G_{Bi} is greater than the value for G_{Pb}, then this result will provide compelling evidence the Strong Nuclear Force and Gravity are one and the same. The Universal Gravitation Constant is not Universal but specific to every isotope of every Element. It is related to the Binding Energy Per Nucleon and does not vary much for stable isotopes of Elements with the exception of Lead208 and Bismuth209. Let’s first look at Newton’s Law of Gravity and the “Universal Gravitation Constant.” The following passage was extracted from “Physics, Parts I and II,” by David Halliday and Robert Resnick, pages 348 to 349. This passage discusses Lord Cavendish's Experiment designed to measure the Universal Gravitation Constant: To determine the value of G it is necessary to measure the force of attraction between two known masses. The first accurate measurement was made by Lord Cavendish in 1798. Significant improvements were made by Poynting and Boys in the nineteenth century. The present accepted value of G is 6.6726x10^{11} Newtonmeter^{2}/kg^{2}, accurate to about 0.0005x10^{11} Newtonmeter^{2}/kg^{2}. In the British Engineering System this value is 3.436x10^{8} lbft^{2}/slug^{2}. The constant G can be determined by the maximum deflection method illustrated in the Figure, “Cavendish Experiment,” below. Two small balls, each of mass m, are attached to the ends of a light rod. This rigid “dumbbell” is suspended, with its axis horizontal, by a fine vertical fiber. Two large balls each of mass M are placed near the ends of the dumbbell on opposite sides. When the large masses are in the positions A and B, the small masses are attracted, by the Law of Gravity, and a torque is exerted on the dumbbell rotating it counterclockwise, as viewed from above. When the large masses are in the positions A' and B', the dumbbell rotates clockwise. The fiber opposes these torques as it is twisted. The angle through which the fiber is twisted when the balls are moved from one position to the other is measured by observing the deflection of a beam of light reflected from the small mirror attached to it. If the values of each mass, the distances of the masses from one another, and the torsional constant of the fiber are known, then G can be calculated from the measured angle of twist. The force of attraction is very small so that the fiber must have an extremely small torsion constant if a detectable twist in the fiber is to be measured. The masses in the Cavendish balance of displayed in Figure, “Cavendish Experiment,” below, are, of course, not particles but extended objects. Since each of the masses are uniform spheres, they act gravitationally as though all their mass were concentrated at their centers. Because G is so small, the gravitational forces between bodies on the Earth’s surface are extremely small and can be neglected for ordinary purposes. The Cavendish Experiment
In the figure, below, a comparison of two Nuclear Gravitation Fields propagating outward from the Nuclear surface, Nuclear Gravitation Field 1 has a greater intensity than Nuclear Gravitation Field 2 at the Nuclear surface. When the effect of SpaceTime Compression is considered, Nuclear Gravitation Field 1 undergoes more SpaceTime Compression than Nuclear Gravitation Field 2. Therefore, Nuclear Gravitation Field 2 leaving the electron cloud of the atom will have a greater intensity than Nuclear Gravitation Field 1 outside the electron cloud of the atom. As previously noted, the Nuclear Gravitation Field for Bismuth209 should be greater than Nuclear Gravitation Field for Lead208 because it is significantly weaker at Nuclear surface. Comparison of Nuclear Gravitation Fields Conclusion Compelling evidence that the Strong Nuclear Force and Gravity are one and the same is provided below: Strong Nuclear Force = Gravity HYPERLINKS: Index and Direct Hyperlinks to the Other Web Pages on this Website:
