In order to quickly estimate the conversion of electromagnetism into rotational motion of a sphere of mass, I will use a distance equivalency methodology. Using the hydrogen atom and a single electron, I will use Bohr's values in order to calculate the amount of rotation equivalent to the electron's attraction to the proton due to the electromagnetic force.
This will exemplify how the electromagnetic force can be converted to rotational mass.
The electron absorbs and emits photons. The properties of the electron change when this happens. In this thought experiment, the electron is hypothesized to be sphere. I can calculate the relationship of rotational mass between a single massive sphere and the universe. The photon's frequency is directly related to the electron's rotation. The frequency vector can be related to the cross product of the primary rotation and secondary rotation vectors. The cross product vector also represents the average rotation of all the points of mass inside the spherical volume of the electron.
The absorption of a photon increases the energy of the electron. The emission of a photon decreases the energy of the electron. I can convert the photon (electromagnetic force) to motion using an arbitrarily defined total rotation vector (cross product vector), the emitted photon's frequency (energy), the gravitational constant and the mass of the electron. The energy in the photon is the rotational motion required to make the electron jump from the n=2 orbital to the n=6 orbital of a hydrogen atom.
I will calculate the force of attraction of the electron and its distance from the positive charge at both the n=2 and n=6 orbitals. I will use this information to find the equivalent velocity required to increase the gravitational attraction of the electron to empty space (total mass of the universe). I will find the velocity by dividing the displacement in metres by the time in seconds. I want to convert this total displacement to rotational motion, relating the gravitational acceleration to the rotational motion of a sphere. I divide the total velocity by the frequency of the emitted photon in order to define the circumference of a distance that is repeated multiple times thoughout the event.
I start by calculating the force on the electron at two different orbitals: the orbitals n=2 and n=6.
n=2 orbital, 2.11670884268e-10 metres from the proton
3.4 electronvolts of energy, equivalent to 5.44740051072e-19 Joules
J divided by orbital radius gives me the force on the electron = 2.57352376524e-9 Newtons
n=6 orbital, 1.90563795841e-9 metres from the proton
0.377777777778 electronvolts of energy, equivalent to 6.05266723414e-20 Joules
J divided by orbital radius gives me the force on the electron = 3.17618947892e-11 Newtons
I now have to convert this force to gravitational acceleration. The velocity required to increase the gravitational attraction of the electron to empty space (total mass of the universe) is the rotational displacement of the electron's mass over time. Using the equation for the gravitational field, I will find the positions equivalent to the increase in force due to the electromagnetic force. In this circumstance, I will use the gravitational constant, the electron's mass and the quantities of force calculated above to solve for the distance. The difference between the two distances over time is the average velocity of the electron necessary to increase gravitational force by an amount equivalent to the electromagnetic force.
Gravitational constant = 6.67408e-11
Electron's mass = 9.10938356e-31 kg
n=2 orbital force = 2.57352376524e-9 Newtons
n=6 orbital force = 3.17618947892e-11 Newtons
By isolating the d-variable in the gravitational field equation I find the following distances from the positive charge:
n=2 orbital equivalent position = 1.53700792691e-16 m
n=6 orbital equivalent position = 1.38352495515e-15 m
The difference between the two positions is equal to 1.22982416246e-15 metres travelled over 2.17790719213e-16 seconds, resulting in a velocity of 5.64681620458 metres per second. The seconds are calculated using the reduced Planck constant divided by the electronvolts.
This means that the mass of the electron needs to rotate at an average of 5.64681620458 metres per second in order to generate the same force as the electromagnetic force.
Finally, I can relate this velocity to the cross product vector and the frequency of the light emitted in order for the electron to fall back down to the n=2 orbital from the n=6 orbital. The wavelength of this light is 410 nm and its frequency is 7.30676958566e14 Hertz (cycles per second). A cycle is equivalent to the rotational motion travelled by the mass of the electron. I can divide the average velocity vector by the frequency to obtain a circumference of 7.72819799281e-15 metres.
By dividing this number by 2π, I obtain a radius of 1.22998091175e-15 metres. This is consistent with the classical electron radius of 2.8179403227e-15 metres. Keep in mind these are approximate results. I assume a sine of one in the cross product, meaning I assume there is a 90 degree angle between the primary rotation vector and the secondary rotation vector.
This estimation demonstrates that a photon starts forming at the boundary of the electron's classical radius and free space when the acceleration of an electron's rotational motion is larger than the acceleration in free space. The excess rotation is strong enough to accelerate surrounding particles (gravity) and penetrate free space. Therefore, the sum of all excess rotational motion propagating through the permeability of free space can be defined as a photon (Standard Model).
The classical electron radius is consistent with results of experimentation.
Properly estimating the volume of a sphere of mass requires isolating all motion vectors from gravitational acceleration vectors. The sum of all motion contributes to the total mass of a particle. With this example, I can isolate the mass due to the electromagnetic force. I will need to do the same for the other forces in order to solve for the volume of a sphere of mass. In this example, I am converting the electromagnetic force into motion using the kilogram which can be broken down into rotational mass due to the strong, weak and electromagnetic forces as well as the linear mass due to gravity. In the end, the constant speed of light will prove that all massive particles are spheres with identical volumes. Their differing properties are the result of the total motion imparted onto them by the Big Bang.
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