Author: Liang2a
Date: 07-31-12 07:43
LiTE: Liang’s Theory of Everything.
Part 4.4 What are forces. Sepcon forces in nucleus.
Physicists say nuclei are made of protons and neutrons. I say they are made of protons, electrons and sepcons.
Consider a square sheet of paper with a proton at each of its corners. Imagine a sharp pencil is poked through the center of the paper and an electron is attached to each end. A sepcons will form between each proton and each electron for a total of 8 sepcons. Each proton will have two sepcons and two protons (in adjacent corners) locking it into a straight line oscillation while each electron will have 4 sepcons locking it into a straight line oscillation.
I will discuss what happens to one of the four protons. The other three protons are symmetrical. Imagine a Cartesian plane. Place an electron at X=2^0.5 (square root of 2 or 1.4142) and Y=0. Place another electron at X=-2^0.5 and Y=0. Place a proton at X=0 and Y=2^0.5.
Draw a circle with radius 1.4142 or square root of 2 with center at (1.4142, 0). Draw another circle with radius 1.4142 with center at (0, 1.4142). Imagine the two circles to be two spheres. The intersection of these two spheres forms the sepcon disc for the right side electron. Call this right side sepcon. Now draw another circle with radius 2 and centered a (1.4142, 0) and imagine it to be a sphere. The space between this sphere and the right side sepcon is the repulsive region of the right side sepcon. Call this repulsive region R1(r). Draw another circle with radius 2.8284 and centered at (1.4142, 0). Imagine the circle to be a sphere. The space between the sphere with radius 2 and centered at (1.4142, 0) and the sphere with radius 2.8242 and centered at (1.4142, 0) is the attractive region of right side sepcon, A1(r).
Draw a circle with radius 1.4142 or square root of 2 with center at (-1.4142, 0). Draw another circle with radius 1.4142 with center at (0, 1.4142). Imagine the two circles to be two spheres. The intersection of these two spheres forms the sepcon disc for the left side electron. Call this left side sepcon. Now draw another circle with radius 2 and centered a (-1.4142, 0) and imagine it to be a sphere. The space between this sphere and the right side sepcon is the repulsive region of the left side sepcon. Call this repulsive region R1(l). Draw another circle with radius 2.8284 and centered at (-1.4142, 0). Imagine the circle to be a sphere. The space between the sphere with radius 2 and centered at (-1.4142, 0) and the sphere with radius 2.8242 and centered at (-1.4142, 0) is the attractive region of the left side sepcon, A1(l).
http://i58.photobucket.com/albums/g253/Liang1a/alphaparticle2jpg.jpg
As mentioned above, the proton is located on the Y-axis at (0, 1.4142). Apply a force in the downward direction. As the proton travels down the Y-axis it is moving through a volume of space overlapped by the repulsive region of the right sepcon and the repulsive region of the left sepcon. The strength of the two forces are equal. The lateral component of the forces cancel each other out while the vertical component of the forces combine to push the proton upward along the Y-axis. The proton will eventually stop and then be pushed upward along the Y-axis. When the proton rises above y=1.4142 it enters a volume of space overlapped by the attractive region of the right sepcon and the attractive region of the left sepcon. Again the strength of the two forces are equal. The lateral components of the forces cancel each other out while the downward components of the forces combine to pull the proton down along the Y-axis. Eventually the proton stops and begins to be pulled downward along the Y-axis.
In the absence of outside forces, the proton would oscillate up and down along the Y-axis forever. But consider what happens when a force is applied to move the proton to the left. Imagine the proton is located at (0, 1.4142) and a force is applied to it to make it move to the left in the negative X direction. It enters a volume of space overlapped by the repulsive region of the left sepcon, R1(l), and the attractive region of the right sepcon, A1(r). The repulsive force of the R1(1) would push the proton upward and to the right in the positive X direction while the A1(r) would pull the proton downward and to the right in the positive X direction. The combined force would pull the proton to the right in the positive X direction and back toward the Y-axis. The same result is obtained if the proton is pushed to the right side of the Y-axis. So the proton ends up centered on the Y-axis and oscillate up and down along the Y-axis.
If the proton is located above (0, 1.4142) and pushed to the left in the negative X direction as it is moving downward then it would enter a volume of space overlapped by the attractive region of the left sepcon, A1(l), and the attractive region of the right sepcon, A1(r). The proton would move downward until it reaches the volume of space overlapped by the repulsive region of the left sepcon, R1(l), and the attractive region of the right sepcon, A1(r). The result is as described above with the proton pushed and pulled back toward to the Y-axis. So any deviation from the Y-axis to the left or right will be corrected and the proton ends up back along the Y-axis.
Now, let me remind you that there are two protons located at the two adjacent corners of the square paper as described above. These are located respectively at (0, 0, 1.4142) and (0, 0, -1.4142). To distinguish among the three protons, let’s call the proton along the Y-axis proton 1 and the proton located at (0, 0, 1.4142) proton 2; and the third proton located at (0, 0, -1.4142) proton 3.
Proton 2 and 3 would exert a repulsive force on proton 1 along their respective line of sight with proton 1. When proton 1 is moving along the Y-axis the two repulsive forces are equal and cancel each other out in the lateral direction. This means that if proton 1 moves toward proton 2, then the repulsive force of proton 2 becomes greater than the repulsive force of proton 3. The result is proton 1 will be pushed back toward the Y-axis and oscillate up and down along the Y-axis.
Each of the two electrons has associated with it 4 sepcons that keep it centered on the X-axis and oscillating along the X-axis for the same reason that the two sepcons and the two adjacent protons associated with each of the 4 protons keep their associated proton centered along the Y-axis or Z-axis respectively.
From the above explanation, it is clear that the system will correct and stabilize itself so that the proton will oscillate up and down along the Y-axis while the electrons will oscillate along the X-axis forever. Furthermore, other protons will be repulsed by the two positive electric charges at greater distances and cannot approach close enough to upset the system. Electrons will be stopped at a certain specified distance by forming sepcons with the remaining two positive charges. More about this later. Suffice it to say here that electrons too cannot approach close enough to upset the system unless they are moving very fast. More about this later also.
If the above combination of 4 protons:8 sepcons:2 electrons breaks off from a larger nucleus it is called an alpha particle by the physicists who describe it as a combination of 2 protons and 2 neutrons. It is also called a helium nucleus.
If a combination of 1 proton:1 sepcon:1 electron breaks off from the above description of a 4 proton:8 sepcons:2 electron system or helium nucleus, then it is already a neutron. As explained before in a prior part, a neutron tends to spin uncontrollably and ultimately breaks apart.
If a 4 protons:8 sepcons:2 electrons system or helium nucleus breaks up into two equal parts then each part will consist of 2 protons:2 sepcons:1 electron. Such a combination is called a deuterium nucleus by the physicists who describe it as a proton and a neutron.
Since each proton of a deuterium has only one sepcon associated with it, it tends to spin like the proton in a neutron. But unlike the neutron, when the two deuterium protons spin above the horizon of the sepcons and come into sight of each other, they exert a force against each other along the line of sight. Such a force will slow the protons down and push them back away from each other. Ultimately, the protons will stay on the opposite side of the electron and spin within a narrow angle around a straight line that passes through the electron. So, a deuterium nucleus is more stable than a neutron but less stable than a helium nucleus.
It is also much easier to fuse two deuterium nuclei than to fuse two helium nuclei. Two deuterium nuclei might be fused according to the following scenario.
Imagine a Cartesian plane. Locate an electron, call it the left electron, at (x=-1.4142, y=0). Locate a proton, call it proton 1, at (x=-1.4142, y=2). Locate another proton, call it proton 2, at (x=-1.4142, y=-2). Imagine a sepcon between each of the two protons and the electron respectively. Let another electron, call it the right electron, be located on the X-axis beyond x=1.4142 with two protons associated by two respective sepcons. Call these protons proton 3 and proton 4. Let the right electron approach from the positive X direction along the x-axis with proton 3 and proton 4 in tow. When the right electron approaches x=1.4142 it will come under the attractive forces from proton 1 and proton 2. In response, the right electron will begin to send out attractive forces to the protons 1 and 2 which will move toward the Y-axis. When proton 1 reaches (x=0, y=1.4142) and proton 2 reaches (x=0, y=-1.4142) and the right electron reaches (x=1.4142, y=0), a sepcon will come into being between the right electron and proton 1 while another sepcon will come into being between the right electron and proton 2. Then proton 3 and proton 4 will swing around along the plane parallel to the X-axis and Z-axis and y=0, being constrained by the repulsive forces of proton 1 and proton 2. When proton 3 reaches the location (0, 0, 1.4142) a sepcon will come into being between it and the left electron. Similarly when proton 4 reaches the location (0, 0, -1.4142) a sepcon will come into being between it and the left electron. At this time a stable system of 4 protons:8 sepcons:2 electrons or helium nucleus is formed by the fusion of the two 2 protons:2 sepcons:1 electron systems or deuterium nuclei.
Again, I remind you that the forces are in the form of messenger units projected from and as a subset of the sepcon matter units and carrying data from the sepcon matter units to the protons and electrons to allow them to update their stored data with respect to their direction of travel and how many steps to take during each mobile period as explained in previous parts.
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