The Deterministic Ring Theory of Particles

[Spring Theory]

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.

Edited January 6 by Spring Theory

[swansont]

1 hour ago, Spring Theory said:

So if you model the radius of the electron with a negative radius pulse at each dipole,

What dipole is associated with an electron and how is this relevant?

1 hour ago, Spring Theory said:

 

a potential candidate for the radius formula would be:

f(x)=r2−(kcos2(x)+k2+r)22

A graph of this with the radius at 1 looks like:

 

If you plug this formula into an integral calculator you get:

∫2π0f(x)=πr2

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.

Is there any experimental evidence that the electron has a radius given by the classical radius?

1 hour ago, Spring Theory said:

Now we have a mechanism behind charge.

A hand-wave is not a mechanism

 

[Spring Theory]

On 1/6/2024 at 3:09 PM, joigus said:

You don't derive a Lagrangian. You either already know the problem well, and then the Lagrangian is pretty much prescribed, or you must postulate a Lagrangian based on symmetry principles (example: the standard-model Lagrangian when it was postulated) because you don't know the dynamics precisely.

Your Lagrangian seems to suggest a singled out direction of space, so it could hardly be fundamental, as it violates rotation symmetry.

Your Lagrangian, I'm afraid, cannot explain known properties of electrons, like interference, or the Bohm-Aharonov effect, or spin, or electron-electron scattering, electron-photon scattering etc. All of those can be accounted for by field theory. So why change? Just because it's intellectually pleasing to you?

Your "theory" is one of many pet theories that lead nowhere useful, as far as it seems.

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. 

On 1/6/2024 at 2:48 PM, swansont said:

And I asked you for this calculation. You got ~2x10^12 kg. Clearly, that can’t be correct. Do you have a better calculation?

What keeps this black hole from evaporating?

 

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. 

On 1/6/2024 at 5:12 PM, swansont said:

What dipole is associated with an electron and how is this relevant?

Is there any experimental evidence that the electron has a radius given by the classical radius?

A hand-wave is not a mechanism

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.

Edited January 10 by Spring Theory

[joigus]

1 hour ago, Spring Theory said:

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. 

I'll probably hang in somewhere else. I already had my aha moment, but in a different direction than what you suggest.

One more tip: Lose the smugness.

Edited January 10 by joigus
minor correction

[swansont]

1 hour ago, Spring Theory said:

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

But why would they stay in orbit if there is no longer a black hole? They tend to go in straight lines.

1 hour ago, Spring Theory said:

The classical radius is the best example we have as it has not been disproved in experiment.

LOL no. There have been experiments that yield a much smaller value

[Spring Theory]

On 12/30/2023 at 2:14 PM, Bufofrog said:

Every electron has a black hole at it's center?

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?

 

On 1/10/2024 at 9:53 AM, swansont said:

LOL no. There have been experiments that yield a much smaller value

I'm interested. What experiments?

Edited January 13 by Spring Theory

[swansont]

18 minutes ago, Spring Theory said:

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.

But without the BH, there is no appreciable gravity. Certainly not enough to do what you claim.

And: a dipole? What would the electric dipole moment be?

[Spring Theory]

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:

Edited January 13 by Spring Theory

[swansont]

2 minutes ago, Spring Theory said:

The next step is to determine the speed of the electron orbit, which is what I used in the above calculations.

!

Moderator Note

The next step needs to be addressing the many problems that have been pointed out, rather than building on top of a flawed foundation

 

[Spring Theory]

1 hour ago, swansont said:

And: a dipole? What would the electric dipole moment be?

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.

Edited January 13 by Spring Theory

[swansont]

1 hour ago, Spring Theory said:

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. 

Directional charge? Charge is a scalar.

!

Moderator Note

Piling nonsense on top of nonsense, and repeating assertions instead of addressing issues. A hand-wave is not a model.

We’re done here. Don’t bring this up again.