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The proton and neutron mathematically



The proton and neutron mathematically

When you are inter­ested in physics you must read “Unbe­liev­able”!

The pro­ton can be con­sid­ered to emerge out of fusion of n+1 positrons and n elec­trons. The for­mula for the total energy of the pro­ton becomes:

Rp=Re/(2n+1)

Adjudge to the pro­ton the radius Rp:

When the energy of the pro­ton is equally divided over mag­netic and elec­tro­sta­tic energy the radius Rp is cal­cu­lated with the mass Mp of the proton:

The radius is cal­cu­lated at Rp=1.535.10^-18 meter. This radius is a pos­si­ble indi­ca­tion for the radius of the point-​volume. A pro­ton, in real­ity emerges out of a neu­tron that ejects an elec­tron. Reversed, a neu­tron might emerge from the fusion of a pro­ton and elec­tron. The equa­tions for pro­ton en elec­tron are:

When both par­ti­cles fuse, the neu­tron emerges. Although the neu­tron is not charged, this does not mean the elec­tro­sta­tic energy of both par­ti­cles is lost. The neu­tron has, like the pro­ton and elec­tron, two degrees of free­dom to store energy. The neu­tron lacks the poten­tial to store elec­tro­sta­tic energy. Instead it has the elec­tro­mag­netic prop­erty to oscil­late. When we fuse the pro­ton and elec­tron we get the formula:

The first part of this equa­tion presents the mag­netic spin energy of the neu­tron. The sec­ond part the oscil­la­tion energy. Assum­ing the energy is equally divided over both degrees of free­dom we cal­cu­late Rn=1.533.10^-18 meter.

When an observer is sit­u­ated in the neu­tron the observer is at rest with the ether under control/​influence of the neu­tron. The observer spins with the neu­tron and can only observe the oscil­la­tion (sec­ond part Wn). When both charges +Qe and -Qe, in oscil­la­tion, over­lap com­pletely, the poten­tial energy of the oscil­la­tion is nil and the magnetic/​kinetic energy of the oscil­la­tion peaks. When the dis­tance between both charges is max­i­mal all oscil­la­tion energy is potential.

The elec­tro­sta­tic force between the two charges is cal­cu­lated with the dif­fer­en­tial of the sec­ond part of equa­tion Wn. The elec­tro­sta­tic force between both charges is:

Fn=-4.91.107 New­ton

The pos­i­tive and neg­a­tive point-​volume present each half of the energy/​mass of the neu­tron. Both point-​volumes have the force Fn (Fn=Fm) work­ing. When we cal­cu­late the accel­er­a­tion of each par­ti­cle with a=2Fn/Mn, the clas­si­cal way, we find that both par­ti­cles move exact with the speed of light when they pass each other.

a= 2Fn/​Mn S=1/2Rn T=(2S/a)^(1/2) T=(Rn/a)^(1/2) T=5.11.10^-27 time

S is the dis­tance each point-​volume trav­els before they pass each other. The dif­fer­ence between the cal­cu­lated speed of pas­sage V=aT and c is so small that the dif­fer­ence is explained by the mar­gin errors of the mass of the neu­tron. The cal­cu­la­tion fits exactly. The coin­ci­dence would be extreme, when the derived force accel­er­ates the par­ti­cles accord­ing to clas­si­cal mechan­ics exactly to the speed of light c, when the pro­posed rela­tion does not exist. The neu­tron is in a sim­ple way explained by means of the ether and clas­si­cal mechanics.

When cal­cu­lat­ing the accel­er­a­tion of both point-​volumes in the neu­tron we stated that Fn was work­ing on both sides. So we cal­cu­lated the accel­er­a­tion with a force equal to 2Fn. In the above we only accounted for de elec­tro­sta­tic force/​energy and there­fore we only incor­po­rated half the energy of the neu­tron. We neglected the mag­netic spin energy/​force of the neutron.

Now we assume the observer is sit­u­ated out­side the neu­tron at rest with the sur­round­ing ether. The observer sees now a neu­tral par­ti­cle. The oscil­la­tion is too fast to observe. The spin of the neu­tron induces a cur­rent in the sur­round­ing ether. The induced spin cur­rent in the ether sur­round­ing the neu­tron is not in bal­ance with the ether out­side the neu­tron. The mag­netic force, the rejec­tion by the sur­round­ing ether of this cur­rent, is derived by dif­fer­en­ti­a­tion of the first part of the equa­tion for the energyWn of the neutron.

The mag­netic force is equal to the elec­tro­sta­tic force (Fm=Fn). Cal­cu­lat­ing the oscil­lat­ing with 2Fn is jus­ti­fied because the total force is the sum of the elec­tro­sta­tic force Fn plus the mag­netic force Fm.

When the neu­tron oscil­lates over a dis­tance Rn it has to “push” away the sur­round­ing ether. The total energy of the neu­tron can also be stated as: Wn=Rn(Fm+Fn).

Next chap­ter: The pho­ton and the con­stant of Planck (h)

When you are interested in physics you must read “Unbelievable“!

The proton can be considered to emerge out of fusion of n+1 positrons and n electrons. The formula for the total energy of the proton becomes:

Rp=Re/(2n+1)

Adjudge to the proton the radius Rp:

When the energy of the proton is equally divided over magnetic and electrostatic energy the radius Rp is calculated with the mass Mp of the proton:

The radius is calculated at Rp=1.535.10^-18 meter. This radius is a possible indication for the radius of the point-volume. A proton, in reality emerges out of a neutron that ejects an electron. Reversed, a neutron might emerge from the fusion of a proton and electron. The equations for proton en electron are:

When both particles fuse, the neutron emerges. Although the neutron is not charged, this does not mean the electrostatic energy of both particles is lost. The neutron has, like the proton and electron, two degrees of freedom to store energy. The neutron lacks the potential to store electrostatic energy. Instead it has the electromagnetic property to oscillate. When we fuse the proton and electron we get the formula:

The first part of this equation presents the magnetic spin energy of the neutron. The second part the oscillation energy. Assuming the energy is equally divided over both degrees of freedom we calculate Rn=1.533.10^-18 meter.

When an observer is situated in the neutron the observer is at rest with the ether under control/influence of the neutron. The observer spins with the neutron and can only observe the oscillation (second part Wn). When both charges +Qe and –Qe, in oscillation, overlap completely, the potential energy of the oscillation is nil and the magnetic/kinetic energy of the oscillation peaks. When the distance between both charges is maximal all oscillation energy is potential.

The electrostatic force between the two charges is calculated with the differential of the second part of equation Wn. The electrostatic force between both charges is:

Fn=-4.91.10^7 Newton

The positive and negative point-volume present each half of the energy/mass of the neutron. Both point-volumes have the force Fn (Fn=Fm) working. When we calculate the acceleration of each particle with a=2Fn/Mn, the classical way, we find that both particles move exact with the speed of light when they pass each other.

a= 2Fn/Mn     S=1/2Rn    T=(2S/a)^(1/2)      T=(Rn/a)^(1/2)     T=5.11.10^-27 time 

is the distance each point-volume travels before they pass each other. The difference between the calculated speed of passage V=aT and c is so small that the difference is explained by the margin errors of the mass of the neutron. The calculation fits exactly. The coincidence would be extreme, when the derived force accelerates the particles according to classical mechanics exactly to the speed of light c, when the proposed relation does not exist. The neutron is in a simple way explained by means of the ether and classical mechanics.

When calculating the acceleration of both point-volumes in the neutron we stated that Fn was working on both sides. So we calculated the acceleration with a force equal to 2Fn. In the above we only accounted for de electrostatic force/energy and therefore we only incorporated half the energy of the neutron. We neglected the magnetic spin energy/force of the neutron.

Now we assume the observer is situated outside the neutron at rest with the surrounding ether. The observer sees now a neutral particle. The oscillation is too fast to observe. The spin of the neutron induces a current in the surrounding ether. The induced spin current in the ether surrounding the neutron is not in balance with the ether outside the neutron. The magnetic force, the rejection by the surrounding ether of this current, is derived by differentiation of the first part of the equation for the energyWn of the neutron.

The magnetic force is equal to the electrostatic force (Fm=Fn). Calculating the oscillating with 2Fn is justified because the total force is the sum of the electrostatic force Fn plus the magnetic force Fm.

When the neutron oscillates over a distance Rn it has to “push” away the surrounding ether. The total energy of the neutron can also be stated as: Wn=Rn(Fm+Fn).

Next chapter: The photon and the constant of Planck (h)

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