# The Deterministic Ring Theory of Particles

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25 minutes ago, Spring Theory said:

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

Do you understand the difference between a vector and a scalar ?

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19 hours ago, swansont said:

Direction convention ≠ charge convention

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.

19 hours ago, studiot said:

Do you understand the difference between a vector and a scalar ?

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.

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39 minutes ago, Spring Theory said:

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.

The positive and negative on the scope, or the graph, does not indicate charge.

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1 hour ago, Spring Theory said:

Well, that's the way I learned how electromagnetic waves work in physics class.

etc

So you are using this to preach that you know everything there is to know about these things instead of listening to see what others might know. ?

I asked if you know the difference between a scalar and a vector to try to help you you understand what swnasont and I are both saying, but from different viewpoints.

The positive and negative convention used in an electric field refers to the direction part of the electric field vector, it does not refer to the magnitude.

Charge on the other hand does refer to the magnitude, since charge is a scalar and has no direction.

Edited by studiot
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22 hours ago, studiot said:

So you are using this to preach that you know everything there is to know about these things instead of listening to see what others might know. ?

I asked if you know the difference between a scalar and a vector to try to help you you understand what swnasont and I are both saying, but from different viewpoints.

The positive and negative convention used in an electric field refers to the direction part of the electric field vector, it does not refer to the magnitude.

Charge on the other hand does refer to the magnitude, since charge is a scalar and has no direction.

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.

22 hours ago, swansont said:

The positive and negative on the scope, or the graph, does not indicate charge.

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:

$\lambda =\frac{h}{p} = \frac{h}{mv}$

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:

$\lambda_{db} = \frac{h}{mv} = \frac{h}{\frac{\hbar}{r_cv_c} v} = \frac{h}{\dfrac{h/2π}{r_c} } =2\pi r_c = circumference$

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.

Edited by Spring Theory
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14 minutes ago, Spring Theory said:

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.

Positive or negative convention? Do you understand what is meant by convention?

It points one way in some places and the other way in others.  But there’s no charge.

Quote

Part of the speculation is that charge does have a direction at the quantum level

Then you need to provide a mathematical model of what this means (not just drawings), and evidence to support it

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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:

$mass =\dfrac{\hbar}{r_cv_c} = \dfrac{h}{2 \pi r_cv_c} = \dfrac{h}{\lambda _{db} v_c}$

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:

$mass_{\gamma} =\dfrac{h}{2\lambda_{db} v_c}$

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:

$mass_e=\frac{h}{\lambda_{db} v_c}$ or $=\frac{\hbar}{r_e v_c}$

Where re is the classical radius of the electron and vc is the velocity along that radius.

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:

$U_e = m_ec^2 = \frac{q_e^2}{4\pi \epsilon_0 r_e}$ => $r_e = \dfrac{q_e^2}{4\pi \epsilon_0 m_e c^2} = 2.81794032620E-15$

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.

Edited by Spring Theory
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1 hour ago, Spring Theory said:

The photon pairs travel along this circumference in an orbiting ring structure.

!

Moderator Note

You’ve been asked several times why a photon would orbit. Instead of forging ahead with more nonsense, you need to address points that have been raised. If you don’t, this thread will be closed.

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4 hours ago, swansont said:
!

Moderator Note

You’ve been asked several times why a photon would orbit. Instead of forging ahead with more nonsense, you need to address points that have been raised. If you don’t, this thread will be closed.

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.

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35 minutes ago, Spring Theory said:

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.

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

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1 hour ago, Spring Theory said:

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.

This isn’t a small detail to gloss over. What is the mass of a black hole GR requires to do this, and how does it compare with your prediction?

1 hour ago, Spring Theory said:

Once the opposing magnetic fields align you have a mechanism to keep the photons bound in the orbiting state.

You’ll have to quantify this as well.And show that photons feel attraction from magnetic fields.

1 hour ago, Spring Theory said:

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.

Electrons have quasar events?

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1 hour ago, Bufofrog said:

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

Sorry, I clearly misunderstood what you were saying.

I think you misspoke when you said this:

2 hours ago, Spring Theory said:

a rotating black hole with the particle created at the center.

I don't believe a particle can exist in the center of a black hole, but for sure a particle inside the event horizon will never leave the black hole

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Posted (edited)
On 12/30/2023 at 2:14 PM, Bufofrog said:

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

No, an electron has positive charge at its center which is really compressed space. A black hole can create electrons.

On 12/29/2023 at 9:42 AM, swansont said:

The positive and negative on the scope, or the graph, does not indicate charge.

The positive and negative indicate current flowing. It flows one way then another.

On 12/30/2023 at 3:40 PM, swansont said:

This isn’t a small detail to gloss over. What is the mass of a black hole GR requires to do this, and how does it compare with your prediction?

All you would need is a black hole with a Schwarzschild radius of at least the radius of the electron. The Schwarzschild radius is:

$r_s = \frac{2GM}{c^2}$

This means its mass would need to be:

$M = \frac{ r_ec^2 }{2G}$ = 1.8973060E+12 kg

The mass of a photon with a planck length wave length:

$m = \frac{ h}{\lambda_{pl} c}$ =  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.

Edited by Spring Theory
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27 minutes ago, Spring Theory said:

No, an electron has positive charge at its center which is really compressed space. A black hole can create electrons.

Sorry, but this is just looks like unevidenced, illogical arm waving.

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Posted (edited)
On 12/30/2023 at 3:40 PM, swansont said:

You’ll have to quantify this as well.And show that photons feel attraction from magnetic fields.

Photons have magnetic fields. If a photon wraps around itself its magnetic fields that point in opposite directions will also align.

On 12/30/2023 at 3:40 PM, swansont said:

Electrons have quasar events?

Black holes have quasar events. I propose that these are matter being emitted from the black hole, not incoming matter being ejected.

On 12/30/2023 at 4:11 PM, Bufofrog said:

I don't believe a particle can exist in the center of a black hole, but for sure a particle inside the event horizon will never leave the black hole

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.

Edited by Spring Theory
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23 minutes ago, Spring Theory said:

If a photon wraps around itself its magnetic fields that point in opposite directions will also align.

-1.  You have been polite and respectful but you ignore members criticism and continue to double down on things like "photon wraps around itself" which is absurd.

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24 minutes ago, Spring Theory said:

What I propose is the matter breaks down into its photon components to bend around the singularity ring.

To add to your "to do" list: How do you make spin 1/2 from a sum of spin 1 "components"? Are those infinite sums?

I'm still waiting for your explanation of charge from non-charge, which we don't seem to be getting any closer to.

Nonsense is that which makes no sense, which so far seems quite appropriate.

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1 hour ago, Spring Theory said:

No, an electron has positive charge at its center which is really compressed space. A black hole can create electrons.

You went from photons bending space to black holes, so if that’s not it, what bends the space for these photons that purportedly comprise an electron?

1 hour ago, Spring Theory said:

The positive and negative indicate current flowing. It flows one way then another.

No current is required to have a varying electric and magnetic field. Why would current flow one way and then the other, in free space?

1 hour ago, Spring Theory said:

All you would need is a black hole with a Schwarzschild radius of at least the radius of the electron. The Schwarzschild radius is:

rs=2GMc2

This means its mass would need to be:

M=rec22G = 1.8973060E+12 kg

The mass of a photon with a planck length wave length:

m=hλplc 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.

Is there any evidence that an electron has a mass of 10^12 kg?

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4 hours ago, Bufofrog said:

-1.  You have been polite and respectful but you ignore members criticism and continue to double down on things like "photon wraps around itself" which is absurd.

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

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3 minutes ago, Spring Theory said:

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

That is an insurmountable problem for your idea I'm afraid.

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Posted (edited)
4 hours ago, joigus said:

To add to your "to do" list: How do you make spin 1/2 from a sum of spin 1 "components"? Are those infinite sums?

I'm still waiting for your explanation of charge from non-charge, which we don't seem to be getting any closer to.

Nonsense is that which makes no sense, which so far seems quite appropriate.

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.

Edited by Spring Theory
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1 minute ago, Spring Theory said:

The same logic that you cannot have an electric field without a charge.

We've already told you there is no such logic. Take Maxwell's equations. Choose both the charge-density and current-density terms to be identically zero. You get to what's known as vacuum solutions of classical EM. Those are known as electromagnetic waves. They are vacuum solutions (correspond to zero charge).

You are blissfully ignorant of basic physics, and a conversation of any kind is impossible.

Vacuum = sourceless = no charge

Study harder!

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Posted (edited)
3 hours ago, swansont said:

You went from photons bending space to black holes, so if that’s not it, what bends the space for these photons that purportedly comprise an electron?

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.

3 hours ago, swansont said:

No current is required to have a varying electric and magnetic field. Why would current flow one way and then the other, in free space?

In a radio, a wire receives the electromagnetic wave. The wave causes current to flow one direction and then the other as it oscillates.

3 hours ago, swansont said:

Is there any evidence that an electron has a mass of 10^12 kg?

I'm not changing the mass of the electron.

24 minutes ago, joigus said:

Study harder!

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):

$\mathcal{L} = \dfrac{\hbar \dot z^2}{ r_c^2 \dot \theta} {} + {2\hbar \dot \theta} - \dfrac {\hbar \dot z}{ r_c}$

There is also a kinetic energy term for the charge in the form:

$KE_{charge} = \frac{1}{2} m_e \dot{r^2}$

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).

Edited by Spring Theory
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33 minutes ago, Spring Theory said:

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.

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?

33 minutes ago, Spring Theory said:

In a radio, a wire receives the electromagnetic wave. The wave causes current to flow one direction and then the other as it oscillates.

33 minutes ago, Spring Theory said:

I'm not changing the mass of the electron.

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32 minutes ago, Spring Theory said:

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.

Adding and subtracting the same thing doesn't change the solutions. And they are divergence-free, namely: chargeless.

32 minutes ago, Spring Theory said:

The Langragian of the photon that I derived is as follows:

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.

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