Chapter III:In order to evaluate whether or not the “Strong Nuclear Force” and Gravity are one and the same force, one must look at the strong forces associated with the nucleons in the nucleus from a classical physics point of view and then compare the results with those of Quantum Mechanics. The two forces of interest are the Electrostatic (Coulombic) Repulsion of the protons in the nucleus and the “Strong Nuclear Force” (of attraction) that holds the protons and neutrons in the nucleus together. Protons, each having the same positive charge, are repelled from each other by the electric field established between them. This repulsion force is known as the Electrostatic Repulsion force. The protons and neutrons are attracted to each other by an apparently extremely strong and short ranged “Strong Nuclear Force.” The “Strong Nuclear Force” appears to be much stronger than the Electrostatic Repulsion Force of the protons because the protons and neutrons that make up the nuclei of all atoms remain bound to one another. The majority of the Elements that make up the “Periodic Table of the Elements” (see Table 71) are stable elements. In general, nuclear stability exists when the nucleus has the proper proportion of neutrons to protons. For the lighter elements up to about the Element Iron (Fe), the most stable nuclear configurations typically consist of a ratio of one proton to one neutron. For the heavier elements beyond Iron, the most stable nuclear configuration typically consists of a ratio of two protons to three neutrons. The nuclear stability curve is displayed in Figure 72, “Table of the Nuclides,” in Chapter VII. The Electrostatic Repulsion Force The general equation for the Electrostatic Repulsion Force between two charged objects that are point sources or spherical in shape is represented by the following relationship: Where “F” represents the Electrostatic Force, “q_{1}” and “q_{2}” are the charged particles of interest, “1/4πε_{0}” represents a constant of proportionality to relate charge and distance to force, and “r” represents the distance between the center of charge of the two charged particles. If both objects are positive charged or if both objects are negative charged, then the Electrostatic Force will be repulsive. If one object is positive charged and the other object is negative charged, then the Electrostatic Force is attractive. In the more specific case where the Electrostatic Repulsion of two protons is considered, the variables q_{1} and q_{2} represent the electrostatic charge of each of the two protons. The charge on each proton is 1.6022×10^{19} Coulomb. One Coulomb of charge flowing through a conductor past a given point in one second equals one Ampere (or 1 Amp) of current. No nuclei exist in nature containing only two protons. Therefore, I will assume that the two positive charges that are repelling each other are two Deuterium nuclei, heavy Hydrogen, or deuterons, for this analysis. Figure 31: Hydrogen Isotopes: Reference: http://nautilus.fis.uc.pt/st2.5/scenese/elem/e00100.html Deuterium is an isotope of Hydrogen containing one proton and one neutron (_{1}H^{2})in the nucleus. When two Deuterium nuclei come in contact with each other, they can, theoretically, fuse together to form a Helium nucleus (_{2}He^{4}). This will not change the analysis of the Electrostatic Repulsion Force because the Electrostatic Repulsion Force is a function of electric charge and the distance between the electric charges. The Electrostatic Repulsion Force is not a function of the mass of the nuclei. In this case, the electrostatic charge of each Deuterium nucleus is identical to the charge of each proton. The constant, “ε_{0},” in the denominator of the Electrostatic Force Equation is the electric field permittivity constant: ε_{0} = 8.8542×10^{12} Farad/meter (F/m) = 8.8542×10^{12} Coulomb^{2}/Newtonmeter^{2}(C^{2}/Nm^{2}) The Newton (N) is a System Internationale (SI) unit for Force in the MKS (MeterKilogramSecond) system. 1 N is equal to 1 kilogrammeter/second^{2} (kgm/sec^{2}). A 1 kilogram mass on Earth sea level would experience a gravitational force of 9.8 N for comparison. Earth’s gravitational field, or acceleration, at sea level is 32.2 feet/sec^{2} equal to 9.8 meters/sec^{2} or 1g of acceleration. There is a relationship between the electric field permittivity constant in free space,“ε_{0},” and the magnetic permeability constant in a vacuum, “μ_{0},” related to speed of the propagation of light. A photon, or quanta of light energy, contains both an electric field wave and a magnetic field wave that propagate with the wave of light. The electric field and the magnetic field are perpendicular to each other and each are perpendicular to the direction of propagation of the photon. The speed of light in a vacuum, “c,” is equal to 2.99729×10^{8} meters/second and is related to ε_{0} and μ_{0} as follows: It should also be noted that on the microscopic, or atomic level, that mass, charge, and energy are not continuous functions but discrete values. The smallest possible electric charge is that which is associated with a proton or electron. Light energy exists in discrete quanta of energy known as “photons.” Light also has a wave characteristic because it has wavelength and frequency. Hence, energy, in the form of light, or electromagnetic radiation, has a dual particle and wave characteristic. Protons, electrons, and neutrons, which represent matter, have discrete rest masses. Matter also has a dual particle and wave characteristic. After completing the classical physics evaluation of Electrostatic Forces and Gravitational Forces, the evaluation of Quantum Mechanics will reveal a more accurate representation of energy and mass of light and matter, including charged and neutral particles, that make up the atom. The Schrodinger Wave Equation, which will be discussed later, is used to define the characteristics, propagation, and energy levels of the electrons around the nucleus and can also be used to define the energy levels of the protons and neutrons within the nucleus. Wavelengths of light, electrostatic fields, gravitational fields, electrons, protons, and neutrons will all play a significant part as to how my theory, the “Nuclear Gravitation Field Theory,” demonstrates how Gravity is the force that holds the nucleus together. I will calculate the total Electrostatic Force of Repulsion that is present when two Deuterium nuclei are a distance of 1 Angstrom from one another. A distance of 1 Angstrom is equal to 1.000×10^{10} meter. Using the Electrostatic Force Equation, as provided above, and assuming the radius of the atom, r = 1.000×10^{10} meter, the Electrostatic Force of Repulsion between the two Deuterium nuclei is equal to: Electrostatic (Coulombic) Repulsion Force: Newton’s Law of Gravity  The Attractive Force of Masses In accordance with Newton’s Law of Gravity, the gravitational force between two objects, which is an attractive force, is inversely proportional to the square of the distance between the center of gravities of each of the two objects. One will note that the form for the equation for gravitational force is very similar to the Electrostatic Force Equation. Newton’s Law of Gravity states: In this case, “F” represents the Force of Gravity, “G” represents Newton's Universal Gravitation Constant, “M_{1}” and “M_{2}” represent the masses of interest that are attracted to one another, and “r” represents the distance between the center of mass of each of the two masses. The masses are assumed to be either point sources of mass or spheres. In this calculation I will assume the two Deuterium nuclei are a distance of 1 Angstrom, or 1.000×10^{10} meter, which is the same distance between the Deuterium nuclei assumed in the previous Electrostatic Repulsion calculation. G is equal to 6.6726×10^{11} meter^{2}/kgsec^{2}. The rest mass of a Deuterium nucleus, or deuteron, is equal to 2.0135532 Atomic Mass Units (AMU) equal to 3.3436×10^{27} kilogram. The Gravitational Attraction Force between the two Deuterium nuclei is calculated below: Gravitational Attraction Force: In addition, I will calculate the Gravitational Attraction Force of the two Deuterium nuclei at an assumed distance where the Deuterium nuclei are close enough to one another for Nuclear Fusion to take place. The diameter of the atom, as mentioned earlier, is about 1.0×10^{10} meter. A small nucleus, such as a Deuterium nucleus, has an average diameter of 1.26×10^{15} meter. Using the value of the average diameter of a Deuterium nucleus, I will assume that the distance between the center of gravities of each of the Deuterium nuclei is 1.26×10^{15} meter. The mass of the Deuterium nucleus is 3.3436×10^{27} kilogram. Using Newton’s Law of Gravity the Gravitational Attraction Force between the two Deuterium nuclei is calculated below: Newton’s Law of Gravity: Gravitational Attraction Force: The Gravitational Attraction Force for the two Deuterium nuclei at the point where fusion can take place, a distance of 1.26×10^{15} meter, is calculated to be about 10^{10} times greater than when the two Deuterium Nuclei were a distance of 1.000 Angstrom equal to 1.000×10^{10} meter apart. That result is expected considering Newton’s Law of Gravity states the Gravitational Attraction Force is inversely proportional to the square of the distance between the two Deuterium nuclei. Comparison of Electrostatic Repulsion and Gravitational Attraction When comparing the Electrostatic Repulsion Force of the two Deuterium nuclei to that of the Gravitational Attraction Force of the two Deuterium nuclei, the result is the Electrostatic Repulsion Force is 3.0928×10^{35} times greater than the Gravitational Attraction Force. From a purely classical physics point of view, it would be concluded that the “Strong Nuclear Force” holding the protons and neutrons together in the nucleus couldn’t possibly be the same as Gravity. The “relatively feeble” Gravitational Attraction Force would not be able to overcome the Electrostatic Repulsion Force of the protons. It must be concluded that the “Strong Nuclear Force” must be a completely different force than Gravity ... OR ... Something else must be considered if Gravity can be, in fact, the same as the “Strong Nuclear Force.” It is a known fact that the electrons orbiting the nucleus of atoms have discrete, or quantized, energy levels. Likewise, it is a known fact that the protons and neutrons in the nucleus have discrete, or quantized, energy levels. Therefore, to perform an acceptable evaluation to determine if Gravity and the “Strong Nuclear Force” are one and the same, an evaluation must be performed analyzing Newton’s Law of Gravity, Electrostatic Repulsion, and the “Strong Nuclear Force,” incorporating the principles of Quantum Mechanics and Einstein’s General Relativity Theory as applicable to the atom and nucleus. Index and Direct Links to Other Chapters of Nuclear Gravitation Field Theory Nuclear Gravitation Field Theory
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