Spring Theory

I propose that there is no separation. They are infinitesimally close together (on top of each other so that the opposing magnetic fields align. Maybe I should go back to the absurd assumption about the photon. It is described at the quantum level with a spin of 1 and its electric field pointing along that spin as in Figure 1.1: Two oppositely charged plates have an electric field pointing from one to the other: The flawed assumption is that the photon can be modeled with a directional charge in a ribbon form: The argument that there is no charge, just an electric field, can be equated to there is a directional charge with the net effect zero. When the photon becomes polarized, the charges end up with peaks in the popular wave description and the opposing magnetic fields are presented: I propose the only way to naturally polarize a single photon is if it circles itself into a double orbit" After the collapse, the photon is now polarized and the electric charge points align: And the opposing magnetic fields align: And the dipole is formed. Defining charge as compression/decompression of space on wave fronts results in the deterministic model for particles. The inner charge is shielded to present a charged particle. Some very interesting properties fall out from these assumptions including the structure of the proton using three photons in orbit. Also a physical description of the wave function.

The next step is to determine the speed of the electron orbit, which is what I used in the above calculations. If we have (2) 1/2 ħ photon dipoles in orbit, then the total angular momentum would be about ħ: [latex] v_c =\frac{\hbar}{r_em_e} [/latex] = 4.108235899773E+10 m/s Plugging in the generally accepted values in this equation yields a rather high speed of the photon of about 137 times the speed of light. This normally leads to pause, and calls for the fact that nothing (especially something with mass) can move or spin faster than light. Another way to look at the concept is not that nothing travels faster than the speed of light, but that nothing travels faster than a photon. The photon, though, has variable speeds through space, especially trapped in ring orbits to form particles. The complete ring particle structure still has a speed of light limit though. Analyzing this calculated speed, which I will label the “Dirac velocity”, with the speed of light in a vacuum in a ratio is an interesting discovery: [latex] \frac{c}{v_c}= [/latex] 0.0072973525696658 = α = 0.0072973525693(11) This is close to the fine structure constant, within the standard uncertainty. An explanation for this high velocity is the positive charge on the inside of the electron. High compression of space means high photon velocity as show in Figure 3.12:

No, every electron has a positive charge inside it. If the dipole is a negative radius outside, then the inside radius will be positive and could be one mechanism (hand waving) to hold the photon pair together - gravity. If you take each dipole as half the mass and a positive radius from one dipole inside the electron to the other dipole, then gravity be the attraction for the two positive radius points inside the electron. The centrifugal force experienced at the dipoles is: [latex] F = ma = \frac{\frac{m_e} {2} v^2}{r_e} = 2.7342903649E+05 N [/latex] The gravitational force required to keep the photons in orbit at the classic electron radius would be: [latex] F = \frac{Gm_1 m_2 }{r^2} = \frac{G \frac{m_e} {2} \frac{m_e} {2} }{r^2} = \frac{G m_e^2} {4 r^2} [/latex] Since the radius is modeled as a pulse curvature inside the electron, these could potentially get right next to each other. The radius at which the gravitational force equals the centrifugal force: [latex] r = \sqrt { \frac {Gm_e }{4F} } [/latex] = 7.11605989070E-39 m The result is smaller than the Planck length. Would this be a problem? I'm interested. What experiments?

Yes, my Lagrangian is semi complete because I have chosen a photon propagating along the path of the z axis. It does however produce all the known properties of the photon when you derive the equations of motion. Hang in there, gentlemen. I didn't do any of this with motivated reasoning, I just started with these assumptions and everything just fell out of it. I'm confident you guys will have an aha moment when it all clicks. You're combining a black hole and a particle. I'm saying two photons can be retained in an orbit but the black hole is the genesis. I'm also not sold on the virtual particle thing. This may be a side effect of the dipole structure of perceived particles. So black holes may never evaporate. The dipole was formed when the photons magnetic field intersected in orbit per my previous graphic of the deterministic electron. The electric fields also intersected. There is a Princeton paper that describes the destructive interference that would happen with EM waves that results in an amplification of the electric fields: http://kirkmcd.princeton.edu/examples/destructive.pdf The classical radius is the best example we have as it has not been disproved in experiment.

So if you model the radius of the electron with a negative radius pulse at each dipole, a potential candidate for the radius formula would be: [latex] f(x)= r^2-\dfrac{\left(\frac{k}{\cos^2\left(x\right)+k^2}+r\right)^2}{2} [/latex] A graph of this with the radius at 1 looks like: If you plug this formula into an integral calculator you get: [latex] \int_{0}^{2\pi} f(x) = {\pi}r^2 [/latex] So the area is the same as the area of a circle. This makes the overall curvature of the electron and mass based on its classical radius. At the dipoles you have a negative radius pulse (negative curvature) which is what presents negative charge. For a positron you will have positive pulses at the dipoles pulsing positive curvature which presents positive charge. Now we have a mechanism behind charge. When a positive curvature (positive charge) meets a negative curvature (negative charge) they reinforce to create an attraction effect. When a negative curvature meets a negative curvature they oppose to create a repulsive effect: The overall curvature of the electron is still positive per the integral above, but the charge is represented by negative curvature pulses. The electron is spinning so Maxwell's equations are approximations as a point source.

I'm assuming you believe that black holes exist. Then you should see that it can bend space so much that a photon will curve its path around it. This is the photon sphere. Now make the black hole dense enough so photon sphere has the radius of an electron. In a radio, a wire receives the electromagnetic wave. The wave causes current to flow one direction and then the other as it oscillates. I'm not changing the mass of the electron. You can take those equations an add in a positive contribution and then add in a negative contribution. The result is you get the same equation. The Langragian of the photon that I derived is as follows (cylindrical coordinates): [latex] \mathcal{L} = \dfrac{\hbar \dot z^2}{ r_c^2 \dot \theta} {} + {2\hbar \dot \theta} - \dfrac {\hbar \dot z}{ r_c} [/latex] There is also a kinetic energy term for the charge in the form: [latex] KE_{charge} = \frac{1}{2} m_e \dot{r^2} [/latex] You would have one momentum term for positive charge (positive mass) and another term for negative charge (negative mass) which would cancel each other out. And now we get to what charge is. It is the radius momentum pulse at each electric field peak of the photon. You can model the radius as the classical radius but with a Dirac delta function like pulse at the dipole. Negative charge pulses a negative mass (negative radius) and positive charge pulses a positive mass (positive radius).

Each dipole has a spin of 1/2 which is the only interactive part of the electron. Any experiment will hit one of the dipoles and measure half spin. Two total dipoles makes a total spin of 1 (about). The same logic that you cannot have an electric field without a charge. If the photon has an electric field then what is the source, magnetic field? If it has a magnetic field then is it positive or negative? The answer is it has both. Between (2) charged plates, with the positive on the bottom and the negative on the top, the electric field points up. If you say the photon has an electric field pointing is some direction, then there has to be a positive side and a negative side like my ribbon graphic. The non-charge is net zero charge. The absurd premise here is that the photon has an oscillating charge so the net effect is no charge.

This speculation requires that you accept the possibility of the absurd premise.

Photons have magnetic fields. If a photon wraps around itself its magnetic fields that point in opposite directions will also align. Black holes have quasar events. I propose that these are matter being emitted from the black hole, not incoming matter being ejected. According to GR (light cone diagram shows this), anything that enters the black hole is destined to end up at the singularity. A spinning black hole has a singularity ring instead of a singularity point. So another prediction is that there are only spinning black holes but this may be already generally accepted. Anything that enters the event horizon travels to the center. What I propose is the matter breaks down into its photon components to bend around the singularity ring. Enough photons collect to collapse and create degeneracy pressure. This is the genesis of the quasar event that streams matter and photons from the center along the axis of rotation.

No, an electron has positive charge at its center which is really compressed space. A black hole can create electrons. The positive and negative indicate current flowing. It flows one way then another. All you would need is a black hole with a Schwarzschild radius of at least the radius of the electron. The Schwarzschild radius is: [latex] r_s = \frac{2GM}{c^2} [/latex] This means its mass would need to be: [latex] M = \frac{ r_ec^2 }{2G} [/latex] = 1.8973060E+12 kg The mass of a photon with a planck length wave length: [latex] m = \frac{ h}{\lambda_{pl} c} [/latex] = 1.367493897E-07 kg So the number of Planck length photons occupying the same space required to make a black hole would be 1.3874329E+19.

I was getting there, but it's hard to comprehend without these fundamentals. The nonsense comment is not very nice. The short answer is photons are put into orbit by bending space. The most likely candidate to bend space in this manner would be a rotating black hole with the particle created at the center. Once the opposing magnetic fields align you have a mechanism to keep the photons bound in the orbiting state. Each photon affects its pair by bending space to create a gradient in velocity such that the angular velocity is constant. The formed particles are expelled out of the black hole through quasar events which fall around the galaxy as the spiral arms as mostly protons and electrons to make hydrogen.

The Mass of the Electron: As derived from previous de Broglie’s momentum formulas for the photon, the mass formula is expressed as a function of velocity and curvature radius: [latex] mass =\dfrac{\hbar}{r_cv_c} = \dfrac{h}{2 \pi r_cv_c} = \dfrac{h}{\lambda _{db} v_c} [/latex] Where rc is the curvature radius and vc is the velocity along that curvature. The familiar ħ is the reduced plank’s constant (h divided by 2π). This designation of mass of the photon was the fundamental reason why the photon has momentum, but the idea of open curvature is why it has evasive relative mass to other objects. It is often stated in physics that photons have no mass and any object with no mass travels at the speed of light. As the photon collapses into a double orbit, its end point joins with its beginning point to create closed curvature. The mass now presents itself to all relative objects. From all radial directions, an object “feels” a positive radius. Since the radius of the electron is the curvature radius, we apply the photon mass equation and obtain the mass of each of the photons that make up the electron to be: [latex] mass_{\gamma} =\dfrac{h}{2\lambda_{db} v_c} [/latex] Where the de Broglie wavelength is the circumference of the electron and vc is the velocity along that circumference. Multiplying by 2 photons gives a total electron mass equation as follows: [latex] mass_e=\frac{h}{\lambda_{db} v_c} [/latex] or [latex] =\frac{\hbar}{r_e v_c} [/latex] Where re is the classical radius of the electron and vc is the velocity along that radius. The Radius of the Electron: The classical radius is usually defined as the radius of the charge of the electron. Of course the electron is still considered a point particle generally in physics, but the radius is useful to characterize its atomic interactions. The equation to determine the classical radius is derived from the electrical energy equation: [latex] U_e = m_ec^2 = \frac{q_e^2}{4\pi \epsilon_0 r_e} [/latex] => [latex] r_e = \dfrac{q_e^2}{4\pi \epsilon_0 m_e c^2} = 2.81794032620E-15 [/latex] Where qe is the charge of the electron me is the electron mass, c is the speed of light, and ε0 is the permittivity of free space. This should not be confused with permeability of free space, μ0, similarly named for some reason.

Please don't equate sharing an experience as preaching. I'm presenting a possible theory. I appreciate the time you all have put in responding. Part of the speculation is that charge does have a direction at the quantum level and the scalar field is just an approximation or average value of the charge in all directions. Each dipole on the electron points out along a straight line, but the average as it spins and moves around appears as a point source. I'm speculating that an existence of an electric field means there must be some positive or negative convention. Same with the magnetic field. It points one way, then points the other way. If there is a positive magnetic field, then there must also be a negative one to balance out. Same with electric fields but the charge source is in the photon that is also directional to create the directional electric field. The de Broglie Circumference of the Electron: Louis de Broglie won a Nobel prize because he postulated that the wave properties of light also apply to particles. His famous equation designates a wavelength to every particle: [latex] \lambda =\frac{h}{p} = \frac{h}{mv} [/latex] Where λ is the de Broglie wavelength, h is Planck’s constant, m is the mass of the particle and v is the velocity of the particle which is related to its momentum, p. As discussed previously, de Broglie’s hypothesis was confirmed when electrons had the same interference pattern as light when sent through the double slit experiment. If we apply this equation to the electron ring model, substitute the mass formula derived previously, then the de Broglie wavelength would represent the circumference of the electron: [latex] \lambda_{db} = \frac{h}{mv} = \frac{h}{\frac{\hbar}{r_cv_c} v} = \frac{h}{\dfrac{h/2π}{r_c} } =2\pi r_c = circumference [/latex] Instead of this calling this de Broglie wavelength, it would be more appropriate to call it the de Broglie circumference. The photon pairs travel along this circumference in an orbiting ring structure. The total path length of each photon is actually twice the de Broglie circumference.

Well, that's the way I learned how electromagnetic waves work in physics class. We measured them with an oscilloscope. They oscillate from positive to negative. You cannot have an electric field without it being positive or negative or zero or somewhere in between. You'lll have to take that as a definition then. A scalar is a value at any point in space (like temperature). What I'm saying is that space is a scalar field with a value of the speed of light at any point in space. When I've been discussing angular momentum so far, I've been using the magnitude, not the vector (cross product). By modeling the photon as a curvature spiral, you don't need a vector description yet. If you collapse the spiral modeled photon with opposite positive and negative fields you get closed curvature and a mass presented. The results in a reduction in radius a final angular momentum value of 1/2 ħ as shown in figure 3.7. Likewise if we took and opposite rotating photon and collapsed it to double ring orbit, it would result in a negative 1/2 ħ shown in figure 3.8. We have to be careful of convention because just as in decay, the photons will eject in opposite directions, creation results from the photons incoming from opposite directions. The negative and positive aspects of angular momentum are designated with respect to the photon itself. From a global position, the photon coming in from the right will also collapse into a positive 1/2 h bar angular momentum, This means to obey the quantum laws for a total angular momentum of ħ, the particle would require (2) collapsed photons with opposite spins as shown in figure 3.9. Note if the photons did not originally have opposite angular momentum states, when they collapsed to form a particle, the dipoles would be rotating in opposite directions. The resulting binary photon ring orbital system has the inside electric charge completely shielded so the it would exhibit only a negative electric charge as shown in figure 3.10. Now you have a deterministic electron.

What kind of electric field is not associated with a charge convention? Or magnetic field for that matter.