# Photon Sinewave Travel

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## Photon Sinewave Travel

Didn't Miles suggest in his Superposition paper, in which he linked to someone's video clip of a photon's x-spin, that photons travel in sine waves? But it seems that would only apply to photons that have only an x-spin. Wouldn't the smallest photons have no spin at all, not even a-spin (axial spin)? So it seems to me the non-spinning photons would have straight-line motion; a-spin photons would also have straight-line motion; x-spin photons would have sinewave motion; and photons with stacked spins above the x-spin would have gyrating motion. So it looks like only a small percentage would have sinewave motion. Right?

I'm familiar with conventional diagrams of EM radiation, where photons are thought to have sinewave electric field motion perpendicular to sinewave magnetic field motion, each out of phase with the other by 90 degrees, I think. Is that where Miles figured science got the sinewave EM motion from? But aren't those diagrams just a representation of the strength of the E-field and M-field, rather than a representation of physical photon motion?

I'm familiar with conventional diagrams of EM radiation, where photons are thought to have sinewave electric field motion perpendicular to sinewave magnetic field motion, each out of phase with the other by 90 degrees, I think. Is that where Miles figured science got the sinewave EM motion from? But aren't those diagrams just a representation of the strength of the E-field and M-field, rather than a representation of physical photon motion?

**LloydK**- Posts : 448

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## Re: Photon Sinewave Travel

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I believe the path of the stacked-spin B-photon, gyrating motion you’ve described nicely is a consensus viewpoint. The only correction, x-spins form centroids and not sine waves. The boys at the lab don't have a colliding B-photons sim yet, I've been holding my breath for a while now.

Miles said that mainstream understanding of E/M waves were based on the mathematics for sound waves - I won't go beyond that. I was taught that electromagnetic waves were comprised of those cycling orthogonal electric and magnetic components and not in any way indicative of a physical motion. In my education and experience with standard physics, there was never any serious suggestion that the E/M wave represented an individual photon.

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I believe the path of the stacked-spin B-photon, gyrating motion you’ve described nicely is a consensus viewpoint. The only correction, x-spins form centroids and not sine waves. The boys at the lab don't have a colliding B-photons sim yet, I've been holding my breath for a while now.

Miles said that mainstream understanding of E/M waves were based on the mathematics for sound waves - I won't go beyond that. I was taught that electromagnetic waves were comprised of those cycling orthogonal electric and magnetic components and not in any way indicative of a physical motion. In my education and experience with standard physics, there was never any serious suggestion that the E/M wave represented an individual photon.

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## Re: Photon Sinewave Travel

His paper on The Wavelength and Frequency of Light are Reversed provides more info on how the sine wave is found through mainstream science. In essence, they are not measuring the actual motion of the particle, they are measuring gaps in a diffraction grating and then assigning that to a sine wave. Before that, they just assumed that light worked just like sound and gave it a frequency and a wavelength.

Photons

So the very first photon would have a nice sine wave motion but no other photon would. SpinSim shows this if you give the particle a linear velocity. The actual wave is nothing like a sine wave unless you generalise it, a lot. The more spin levels you add, the more it would need to be generalised to approximate a sine wave. But that is beside the point because the mainstream are not talking about the actual motion of the photon.

Photons

**always**spin. If it is not spinning then it is not a photon, it is a BPhoton. I am not certain, but I would probably allow the BPhoton to have an axial spin without calling it a photon. So, at least as far as I see it, a photon starts at the first X spin and works its way up to the electron.So the very first photon would have a nice sine wave motion but no other photon would. SpinSim shows this if you give the particle a linear velocity. The actual wave is nothing like a sine wave unless you generalise it, a lot. The more spin levels you add, the more it would need to be generalised to approximate a sine wave. But that is beside the point because the mainstream are not talking about the actual motion of the photon.

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## Re: Photon Sinewave Travel

Airman, what do you mean by centroid? I've dealt with centroids in my professional work but that referred to the averaged center of all points in a polygon, which may not always actually be inside of that polygon and that causes all sorts of pain.

X spins create perfect spirals and if you compress that down to 2D, looking from the side, then you get a sine wave. They are the only spin that creates such a nice motion.

X spins create perfect spirals and if you compress that down to 2D, looking from the side, then you get a sine wave. They are the only spin that creates such a nice motion.

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## Re: Photon Sinewave Travel

Sorry, I was thinking cycloids.

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## Re: Photon Sinewave Travel

While thinking about this topic, I have realised than I can create a magnetism app that shows a stream of spinning photons. Maybe put some electrons in there to show how they are pushed along by the current. Another version could have two streams, side-by-side, to show the interaction of the magnetic fields. I'll see what I can do when I get the chance.

Jared, you might be able to get a working model before I can. We just need some spheres moving along in the same direction and each sphere is spinning about an axis that is parallel to that direction. Use the right-hand-rule to find the right spin direction for the photons. We don't need stacked spins for this, they would just get in the way.

And just like that, we have the beginnings of a model for electricity! All this time and we just stumble upon it.

Oh, Airman, you might want to take a breath or buy some scuba tanks, it could be a while before we have a collision model. I can't figure out how to incorporate all of the spin levels in a collision. But I think I could create a collision model for this magnetism app. Allow two streams to collide at their edges at show the forces or something. That might help me see how to get all of the spins in there.

Jared, you might be able to get a working model before I can. We just need some spheres moving along in the same direction and each sphere is spinning about an axis that is parallel to that direction. Use the right-hand-rule to find the right spin direction for the photons. We don't need stacked spins for this, they would just get in the way.

And just like that, we have the beginnings of a model for electricity! All this time and we just stumble upon it.

Oh, Airman, you might want to take a breath or buy some scuba tanks, it could be a while before we have a collision model. I can't figure out how to incorporate all of the spin levels in a collision. But I think I could create a collision model for this magnetism app. Allow two streams to collide at their edges at show the forces or something. That might help me see how to get all of the spins in there.

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## Re: Photon Sinewave Travel

So are the diffraction grating gaps actually due to the cycloid motion of x-spins? It seems like I remember seeing light diffraction through slits in a Physics class in the late 60s. They made darker and lighter concentric partial circles that looked like a top view of a horizontal sine wave, as on a pond. It's interesting that light was first thought to be like sound. Steven Rado seemed to think that too in his Aethrokinematics CD.Nevyn said: His paper on The Wavelength and Frequency of Light are Reversed provides more info on how the sine wave is found through mainstream science. In essence, they are not measuring the actual motion of the particle, they are measuring gaps in a diffraction grating and then assigning that to a sine wave. Before that, they just assumed that light worked just like sound and gave it a frequency and a wavelength.

Aren't B-photons infrared? I thought that's what Miles determined was the lowest level photon, as in the Cosmic Microwave Background. I don't see why you distinguish B-photons from other photons. I think the B-photons go everywhere just like all the other photons do.Photons always spin. If it is not spinning then it is not a photon, it is a BPhoton. I am not certain, but I would probably allow the BPhoton to have an axial spin without calling it a photon. So, at least as far as I see it, a photon starts at the first X spin and works its way up to the electron.

I may make a note of your idea to simulate magnetism under Projects. Oh, good, you already did.

**LloydK**- Posts : 448

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## Re: Photon Sinewave Travel

I think they are actually measuring the interaction of light with the charge fields of the diffraction grating. Look into Miles paper on the two slit experiment.

Light and sound are completely different. Sound is a compression and decompression of a medium. It takes many different particles or atoms or molecules to form a sound. The density is what creates the wave for sound. It is the density of the medium that is waving. Light can create a wave with only one photon. It does not need a medium and it does not involve density, only motion. It is the spins that create the wave for each and every photon. That is why we can measure a single photon and still place it on the EM spectrum but you can't measure a single particle/atom/molecule and get sound out of it.

I distinguish photons from BPhotons because they are not the same thing. A BPhoton forms a photon, but it is not one itself. The BPhoton is just a rigid sphere. It is that which makes everything else. It can only do two things: move and collide. It can move in two ways, simultaneously: linear and axial spin; three ways if you include expansion. While it can also stack spins, once it does so we call it a photon. That is because a key characteristic of a photon is its wave motion. The BPhoton does not create a wave motion until it gains its first X spin.

The BPhoton is no where near the infrared. Miles stated that the average or most common photon was infrared and therefore the charge field is mostly in the infrared. But there are plenty of photons smaller than an infrared photon. Have a look at the EM spectrum and you will see photons on both sides of infrared. The BPhoton is smaller than all of them, it isn't even on that spectrum.

Light and sound are completely different. Sound is a compression and decompression of a medium. It takes many different particles or atoms or molecules to form a sound. The density is what creates the wave for sound. It is the density of the medium that is waving. Light can create a wave with only one photon. It does not need a medium and it does not involve density, only motion. It is the spins that create the wave for each and every photon. That is why we can measure a single photon and still place it on the EM spectrum but you can't measure a single particle/atom/molecule and get sound out of it.

I distinguish photons from BPhotons because they are not the same thing. A BPhoton forms a photon, but it is not one itself. The BPhoton is just a rigid sphere. It is that which makes everything else. It can only do two things: move and collide. It can move in two ways, simultaneously: linear and axial spin; three ways if you include expansion. While it can also stack spins, once it does so we call it a photon. That is because a key characteristic of a photon is its wave motion. The BPhoton does not create a wave motion until it gains its first X spin.

The BPhoton is no where near the infrared. Miles stated that the average or most common photon was infrared and therefore the charge field is mostly in the infrared. But there are plenty of photons smaller than an infrared photon. Have a look at the EM spectrum and you will see photons on both sides of infrared. The BPhoton is smaller than all of them, it isn't even on that spectrum.

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## Re: Photon Sinewave Travel

Well then, how many stacked spins are there from B-photon to Electron? Aren't there supposed to be 6? And if there are 6 spin levels, how is the spectrum a frequency/wavelength continuum, instead of discrete steps? I haven't understood that as yet.

A B-photon is a rigid sphere of what? Is it solid or hollow? Did Miles calculate its diameter?

Did Miles state that B-photons are not photons? If so, do you know where? I thought the B just meant bombarding photons.

A B-photon is a rigid sphere of what? Is it solid or hollow? Did Miles calculate its diameter?

Did Miles state that B-photons are not photons? If so, do you know where? I thought the B just meant bombarding photons.

**LloydK**- Posts : 448

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## Re: Photon Sinewave Travel

There is some confusion around how many levels are required to go from a BPhoton to an electron. There are certainly more than 6, at the very least I have calculated 14 spin levels for an electron and that is going by the specified radii of BPhoton and electron. However, Miles has stated that the BPhoton is

Let's put this into a table and see where it goes:

Therefore, there are roughly 34 spin levels in between a BPhoton and a proton. The numbers don't work out perfectly and this could be because the mainstream include the charge emission in the size of the proton.

I also assumed 4 spin levels to a spin set so the higher level axial spins are in there and they are assumed to be doubling the radius. Obviously, I am against this sort of interpretation but I wanted to calculate it as Miles has done and leave my own ideas out of it.

The EM spectrum is not exactly photons. Some points on the spectrum are specific photons but others are mixtures of different photons. Suppose we had some photons with 10 spin levels, say, and some other photons with 9 spin levels, then we would measure both at the same time and our machines would probably average them into 9.5 spin levels (or the equivalent of that in frequency/wavelength). See Miles' papers on color (rainbow papers) to find out more about this as there are actually only 2 photons that create color and the rest are mixtures of them.

With respect to spin levels being continuous or discrete, they are both but at different times. In order to

**G**times smaller than a proton, and the proton is supposed to be only 4 spin levels above the electron.**G**is 6.67408 x 10^-11, if we take the inverse of that then the proton is around 15 billion times larger than the BPhoton. If we divide that number by 2, 4 times, then we get around 936 million for the electron. That is, the electron is 936 million times larger than a BPhoton. If we do that again, we get about 58 million. Again and we get 3.6 million and we are 12 spin levels below the proton.Let's put this into a table and see where it goes:

Spin level below proton | Size compared to BPhoton |

0 | 1.49833 x 10^10 |

4 | 9.36458 x 10^8 |

8 | 5.85286 x 10^7 |

12 | 3.65804 x 10^6 |

16 | 2.28627 x 10^5 |

20 | 1.42892 x 10^4 |

24 | 8.93076 x 10^2 |

28 | 5.58172 x 10 |

32 | 3.48858 |

36 | 0.21803 |

Therefore, there are roughly 34 spin levels in between a BPhoton and a proton. The numbers don't work out perfectly and this could be because the mainstream include the charge emission in the size of the proton.

I also assumed 4 spin levels to a spin set so the higher level axial spins are in there and they are assumed to be doubling the radius. Obviously, I am against this sort of interpretation but I wanted to calculate it as Miles has done and leave my own ideas out of it.

The EM spectrum is not exactly photons. Some points on the spectrum are specific photons but others are mixtures of different photons. Suppose we had some photons with 10 spin levels, say, and some other photons with 9 spin levels, then we would measure both at the same time and our machines would probably average them into 9.5 spin levels (or the equivalent of that in frequency/wavelength). See Miles' papers on color (rainbow papers) to find out more about this as there are actually only 2 photons that create color and the rest are mixtures of them.

With respect to spin levels being continuous or discrete, they are both but at different times. In order to

**gain**a spin level, the jump is discrete and that is because it takes**c**worth of force to create that new spin level. However, when**losing**a spin level, it can be continuous and that is because any force, in the right dimension with respect to that spin level, can slow it down. I imagine that once a spin level gets below a certain speed, it will unravel. This may be why neutrons decay. They might have a top level spin that does not work well with the ambient field (here on Earth, at least) and the ambient field slowly slows down that top level spin.**Nevyn**- Admin
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## Re: Photon Sinewave Travel

LloydK wrote:A B-photon is a rigid sphere of what? Is it solid or hollow? Did Miles calculate its diameter?

Did Miles state that B-photons are not photons? If so, do you know where? I thought the B just meant bombarding photons.

I covered the sizes above but didn't comment on what the BPhoton is.

We don't really know what it is and we assume that it is rigid and indestructible. Rigid means that it does not deform under force. It remains a sphere no matter what happens to it. Indestructible means that it won't break and therefore it does not matter if it is solid or hollow because we can't break it apart to find out. I have a vague memory of Miles stating that nothing can be indestructible, but I don't see what use a broken BPhoton is, so I assume it is indestructible.

Yes, Miles has stated that BPhotons are not photons because photons require a wave motion. I don't remember what paper that is in but I would assume it is in the early papers on stacked spins.

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## Re: Photon Sinewave Travel

Here are the B-Photon citations I've found in various papers (sorry for not posting more lately..been reading along though):

**I have renamed the virtual or messenger photon as the B-photon, since it is neither virtual nor a messenger. It is real, mechanical, and works by bombardment.

...

They are traveling c. Newton said you needed an acceleration to create a force, but you don’t. If a car going a constant velocity hits you, it still creates a force. This is all that is happening with B-photon emission. The B-photons emitted by one plate are bombarding the molecules in the other plate, and this creates a force. Because one field is an acceleration and one is a velocity, they don’t change at the same rate. When we move the plates closer, we are making both the distance and the time smaller, right? Well, as we do that, the acceleration of the gravity field begins to overwhelm the velocity of the E/M field. Over smaller intervals, an acceleration works more like a velocity. It has less time to vary, so its “average” strength or size over the interval increases. Think of it this way: the acceleration of gravity is constant, so it creates a constant force. Which means that this force will be the same over a large interval or a small. If we look at a tiny time interval, gravity must still create its force in that time. This must mean that, compared to a constant velocity, its strength must appear to increase.

....

Now, Newton’s equation works because it gives us two correct field mechanisms. In the first part of this rewritten equation, we have the E/M field part of the unified field. The term GD

The second term V

(4π r

And you can now see why I labelled the distance as d instead of r. We have both in these equations, so we have to keep them separate.

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I am trying to explain mass in a non-technical and conversational manner here, but let me pause for a moment to give you links to papers where I address each facet of this complex question with a bit more rigor. In my latest paper on charge, I show that all [spinning] quanta are emitting the foundational E/M field. So it is not just the acceleration of the radius that is creating the mass or the elasticity or the ponderability; it is this bombarding or exclusionary field. This emitted field is a field of real particles, so the force that creates mass is ultimately a force of bombardment. This emitted field (which up to now has been called the charge field, and has been mediated by the virtual "messenger photon") I now call the foundational E/M field, or the B-photon field, and it has a real mass itself. This mass of the field is in fact the charge. In another recent paper, I have even found the radius of the B-photon. In my Unified Field paper I show how this field fits into Newton's gravitational equation. Since it has mass, it must do so, and it does so in a quite simple way, without any difficult math. In short, the mass variables in Newton's equation can be written as a volume times a density. The volume we give to the gravitational field; the density we give to the B-photon field. G then acts as a transform between to the two fields. In this way, mass actually becomes a compound effect. Without the emission of the foundational E/M field, we could not calculate a mass. The force of exclusion is created by this emitted field, and mass is more directly an attribute of the emitted field than it is of the gravitating particle.

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If we apply the spins to photons or electrons, we can explain the mysteries of quantization and superposition very easily and quickly and transparently. But if we apply them to hadrons, we can begin to explain the composition of nuclei, without any need of quarks. This is because the number of the degrees of freedom we achieve with such an analysis matches the degrees we achieve with quark theory [see just below], and the symmetry also matches current quark theory. Stacked spins replace the current quark quantum numbers, as well as explaining mechanically many things that have so far gone unexplained. Chirality and symmetry are given a physical basis, among many other things.

So these are the fundamental changes we have so far:

1) The mediating photons of charge are not virtual or “messenger.” There is no attraction. All charge is repulsive. Charge is mediated by B-photons, by straight bombardment. Not all particles must have charge, since not all particles must emit B-photons. Among particles with charge, this charge is a function of surface area. Therefore if we give the proton a charge of 1, the electron no longer has a charge of -1. It has a charge of about 1/1836 or .000545. The B-photon is also not mass-less and is not point-like. It has a calculable mass and radius, both of which are about G (6.67 x 10-11) times the mass and radius of the proton. That is, the B-photon is eight million times smaller than the electron.

2) Gravity is a measurable force at the quantum level. It has been hidden in the mis-defined and mis-applied charge field. The total E/M or charge field we now measure at the quantum level is not a single field, but a unified field. Once we un-unify it, or separate it, we find the gravitational field as a major player once more. As at the macro-level, we find that gravity is strictly a function of radius. Mathematically, it may be treated as a straight expansion of the sphere of any given particle. Along with the new positive charge field, this explains relative attractions, orbital distances, and unification.

3) We have four possible spins on every quantum. A quantum may have all these spins, or only some of them. A quantum that loses an outer spin will seem to change from one quantum to another.

4) No strong force. No weak force. No gluons. No bosons.

Here is a list of possible spin states of a baryon.

....

This means that the difference between particles and anti-particles is not a difference in the particles themselves, or in the size of the emission. The difference is in the combination of stacked spins. Certain combinations give us an emission field of B-photons that are spinning clockwise. Other combinations give us B-photons that are spinning counter-clockwise. When these two fields meet, they must cancel each other’s spins, and therefore each other’s repulsive energies. The particles emitting will therefore be unprotected: their fields are gone. They are prone to collision.

The neutron should also be prone to collision, since its B-photon field cannot get beyond its z-spin. But it is not as vulnerable to collision as some particles, since it retains its spins regardless. These spins are some protection, since they must be destroyed before the particle is “naked”. The z-spin has a lot of energy, whether B-photons are passing through it or not. This is the mechanical reason it survives for a short time outside the nucleus.

This is also the reason the neutron has more mass: it doesn’t lose the energy of emission. Since energy is mass, we may deduce that the mass equivalence of the emission of a baryon is 2.3 x 10-30 kg. The neutron traps this emission; the proton emits it.

Finally, this also explains the slight mass difference between the neutron and anti-neutron, in a direct mechanical way. The emission is trapped by both particles, making them neutral; but in the anti-neutron, the trapped emission does not cancel its own energy precisely. A clockwise spin will cancel a counter-clockwise spin, when two particles of equal mass meet head-on. But if two particles of equal mass meet head-on, and each has a clockwise spin, the spin is not canceled. No, it is doubled. Which means that B-photons trapped in anti-neutrons cannot cancel out completely. Only their kinetic energy, or energy from forward motion, cancels. But the spin energy of the emission remains. Depending on which of the four anti-neutrons we are talking about, this spin energy can either augment or tamp down the spin energy of the particle. So the anti-neutron can weigh slightly more or less than the neutron.

....

The energy failure in beta decay is explained very simply by the energy difference between the incoming positron and the outgoing electron. Both have the same mass and the same size charge, you will say. True, but charge is no longer simply plus or minus, as I showed above. Even the charge has spin. What I mean is that the same analysis I applied to the neutron and proton must be applied to the electron and positron. The electron and positron are also emitting a B-photon field. It is about 1,836 times less strong than the field emitted by the proton, but it exists and it is repulsive. Yes, both the electron and the positron are repulsing all other particles. In that sense they both have positive charges. Look above: both the proton and anti-proton have positive charges, according to my analysis. Both are emitting and therefore repulsing, by straight bombardment. Therefore their opposite natures are not really a function of charge. They are a function of spin: the spin of the B-photons that are being emitted. Well, the same thing applies to the electron and positron. Both are emitting and therefore both are repulsing, but the B-photons of the positron are upside-down, relative to the B-photons of the electron.

Let that sink in fully, and then go back to the beta decay. The neutron becomes a proton, by switching its z-spin. This switch allows the B-photon emission of the baryon to get through. So, before the positron arrived, we had no emission from the big particle. The positron hits it, and it begins emitting. But the total B-photon field—existing after this interaction—is made up of both particles, both the baryon and the lepton. The positron’s B-photons were upside-down, relative to the proton, but the electron’s B-photons are upside-up. The positron would have subtracted from the field energy, but the electron adds to it. And so the missing energy is not in an invisible neutrino, it is in the total B-photon field. The charge of the field has appeared to go up, because the charge itself has spin. The B-photons are spinning, and we have to monitor not only the number of photons but also the spin of the photons.

This explains the experiment of 1956, since an augmented B-photon field will act exactly like a theoretical field of neutrinos. It will travel and it will carry energy. This augmented B-photon field will have exactly the right energy and spin to facilitate anti-beta decay.

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Neutrino "Oscillation"

I showed at the end of my QCD paper that neutrino oscillation is actually a change in the B-photon or emission field, not the reincarnation of one particle to another. But here I give the muon neutrino to an electron triplet. Which is it? Both. In neutrino oscillation we go from one unseen flux to another. We call the first flux an electron neutrino and the second a muon neutrino. But in reality, the first flux is a flux of the B-photon field, with no neutrino present. The second is a flux of the electron field, with electrons colliding with free neutrons and losing their axial spins. Again, there is no neutrino present, but there is a new particle present after the second event: an electron triplet. In each stripping, a neutron is stripped of its z-spin and multiple electrons are stripped of their axial spins. It would appear that it takes three electrons to strip each neutron. The axial spin has a much lower energy than the z-spin, and only by adding their linear energy to the equation can the moving electrons successfully attack a neutron. I will have more to say about this in other papers.

------

From a previous paper, we know that the radius of the B-photon is G times less than the radius of the proton. This gives us a photon radius of 2.74 x 10-24m. The z-spin is 8 times the radius, so we should find a basic wavelength of 2.2 x 10-23m. Obviously, we don’t find photons with a wavelength that small. Why? Simply because the wavelength we measure has been stretched out by the velocity of the photon. The photon would be measured to have a wavelength of 2.2 x 10-23m only if it were at rest.

You will say, “Even if we accept that the photon is spinning, how can the z-spin be stretched? The spin would give us a spin radius, which is just a length. A length cannot be stretched by motion, unless you are proposing some kind of relativity here.”

I am not proposing relativity as the solution here. The answer to your question is that a spin is not just a radius, and is not just a length. A spin is a motion: a motion that takes time. Even if the photon were spinning at velocity c, one rotation must take some real time. We know that the linear velocity of light is not infinite, so we must assume the speed of spin is also not infinite. If it is not infinite, it must take time. If it takes time, then it will be stretched by the linear motion. While the surface of the photon is spinning, the photon as a whole is moving some linear distance x.

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In my paper on QCD, I show that the neutrino field actually IS this foundational B-photon field, at least in the case of beta decay.

**114a. Zero-point Energy and the Casimir Effect****I have renamed the virtual or messenger photon as the B-photon, since it is neither virtual nor a messenger. It is real, mechanical, and works by bombardment.

...

They are traveling c. Newton said you needed an acceleration to create a force, but you don’t. If a car going a constant velocity hits you, it still creates a force. This is all that is happening with B-photon emission. The B-photons emitted by one plate are bombarding the molecules in the other plate, and this creates a force. Because one field is an acceleration and one is a velocity, they don’t change at the same rate. When we move the plates closer, we are making both the distance and the time smaller, right? Well, as we do that, the acceleration of the gravity field begins to overwhelm the velocity of the E/M field. Over smaller intervals, an acceleration works more like a velocity. It has less time to vary, so its “average” strength or size over the interval increases. Think of it this way: the acceleration of gravity is constant, so it creates a constant force. Which means that this force will be the same over a large interval or a small. If we look at a tiny time interval, gravity must still create its force in that time. This must mean that, compared to a constant velocity, its strength must appear to increase.

....

Now, Newton’s equation works because it gives us two correct field mechanisms. In the first part of this rewritten equation, we have the E/M field part of the unified field. The term GD

^{2}/d^{2}gives us the force of the bombarding field. D is the molecular or macro-density of the field and G transforms that density into a B-photon density.** The B-photons cause the actual force, by contact, so that is the mechanical density that we are seeking. The d^{2}then just tells us how that field diminishes due to the spherical field. That d^{2}comes right out of the surface area equation, as I have shown elsewhere. SA = 4πr^{2}, which means that a field expanding spherically, or traveling out from a source spherically must diminish as the inverse square of the radius.The second term V

^{2}gives us the gravitational field strength, separated from the E/M field. I have shown that the solo gravity field, once split from the E/M field, must vary with radius and radius alone. Of course we can rewrite that volume as(4π r

^{3}/3)^{2}And you can now see why I labelled the distance as d instead of r. We have both in these equations, so we have to keep them separate.

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**155. The Third Wave - Part VII**I am trying to explain mass in a non-technical and conversational manner here, but let me pause for a moment to give you links to papers where I address each facet of this complex question with a bit more rigor. In my latest paper on charge, I show that all [spinning] quanta are emitting the foundational E/M field. So it is not just the acceleration of the radius that is creating the mass or the elasticity or the ponderability; it is this bombarding or exclusionary field. This emitted field is a field of real particles, so the force that creates mass is ultimately a force of bombardment. This emitted field (which up to now has been called the charge field, and has been mediated by the virtual "messenger photon") I now call the foundational E/M field, or the B-photon field, and it has a real mass itself. This mass of the field is in fact the charge. In another recent paper, I have even found the radius of the B-photon. In my Unified Field paper I show how this field fits into Newton's gravitational equation. Since it has mass, it must do so, and it does so in a quite simple way, without any difficult math. In short, the mass variables in Newton's equation can be written as a volume times a density. The volume we give to the gravitational field; the density we give to the B-photon field. G then acts as a transform between to the two fields. In this way, mass actually becomes a compound effect. Without the emission of the foundational E/M field, we could not calculate a mass. The force of exclusion is created by this emitted field, and mass is more directly an attribute of the emitted field than it is of the gravitating particle.

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**213a. A Reworking of Quantum Chromodynamics**If we apply the spins to photons or electrons, we can explain the mysteries of quantization and superposition very easily and quickly and transparently. But if we apply them to hadrons, we can begin to explain the composition of nuclei, without any need of quarks. This is because the number of the degrees of freedom we achieve with such an analysis matches the degrees we achieve with quark theory [see just below], and the symmetry also matches current quark theory. Stacked spins replace the current quark quantum numbers, as well as explaining mechanically many things that have so far gone unexplained. Chirality and symmetry are given a physical basis, among many other things.

So these are the fundamental changes we have so far:

1) The mediating photons of charge are not virtual or “messenger.” There is no attraction. All charge is repulsive. Charge is mediated by B-photons, by straight bombardment. Not all particles must have charge, since not all particles must emit B-photons. Among particles with charge, this charge is a function of surface area. Therefore if we give the proton a charge of 1, the electron no longer has a charge of -1. It has a charge of about 1/1836 or .000545. The B-photon is also not mass-less and is not point-like. It has a calculable mass and radius, both of which are about G (6.67 x 10-11) times the mass and radius of the proton. That is, the B-photon is eight million times smaller than the electron.

2) Gravity is a measurable force at the quantum level. It has been hidden in the mis-defined and mis-applied charge field. The total E/M or charge field we now measure at the quantum level is not a single field, but a unified field. Once we un-unify it, or separate it, we find the gravitational field as a major player once more. As at the macro-level, we find that gravity is strictly a function of radius. Mathematically, it may be treated as a straight expansion of the sphere of any given particle. Along with the new positive charge field, this explains relative attractions, orbital distances, and unification.

3) We have four possible spins on every quantum. A quantum may have all these spins, or only some of them. A quantum that loses an outer spin will seem to change from one quantum to another.

4) No strong force. No weak force. No gluons. No bosons.

Here is a list of possible spin states of a baryon.

....

This means that the difference between particles and anti-particles is not a difference in the particles themselves, or in the size of the emission. The difference is in the combination of stacked spins. Certain combinations give us an emission field of B-photons that are spinning clockwise. Other combinations give us B-photons that are spinning counter-clockwise. When these two fields meet, they must cancel each other’s spins, and therefore each other’s repulsive energies. The particles emitting will therefore be unprotected: their fields are gone. They are prone to collision.

The neutron should also be prone to collision, since its B-photon field cannot get beyond its z-spin. But it is not as vulnerable to collision as some particles, since it retains its spins regardless. These spins are some protection, since they must be destroyed before the particle is “naked”. The z-spin has a lot of energy, whether B-photons are passing through it or not. This is the mechanical reason it survives for a short time outside the nucleus.

This is also the reason the neutron has more mass: it doesn’t lose the energy of emission. Since energy is mass, we may deduce that the mass equivalence of the emission of a baryon is 2.3 x 10-30 kg. The neutron traps this emission; the proton emits it.

Finally, this also explains the slight mass difference between the neutron and anti-neutron, in a direct mechanical way. The emission is trapped by both particles, making them neutral; but in the anti-neutron, the trapped emission does not cancel its own energy precisely. A clockwise spin will cancel a counter-clockwise spin, when two particles of equal mass meet head-on. But if two particles of equal mass meet head-on, and each has a clockwise spin, the spin is not canceled. No, it is doubled. Which means that B-photons trapped in anti-neutrons cannot cancel out completely. Only their kinetic energy, or energy from forward motion, cancels. But the spin energy of the emission remains. Depending on which of the four anti-neutrons we are talking about, this spin energy can either augment or tamp down the spin energy of the particle. So the anti-neutron can weigh slightly more or less than the neutron.

....

The energy failure in beta decay is explained very simply by the energy difference between the incoming positron and the outgoing electron. Both have the same mass and the same size charge, you will say. True, but charge is no longer simply plus or minus, as I showed above. Even the charge has spin. What I mean is that the same analysis I applied to the neutron and proton must be applied to the electron and positron. The electron and positron are also emitting a B-photon field. It is about 1,836 times less strong than the field emitted by the proton, but it exists and it is repulsive. Yes, both the electron and the positron are repulsing all other particles. In that sense they both have positive charges. Look above: both the proton and anti-proton have positive charges, according to my analysis. Both are emitting and therefore repulsing, by straight bombardment. Therefore their opposite natures are not really a function of charge. They are a function of spin: the spin of the B-photons that are being emitted. Well, the same thing applies to the electron and positron. Both are emitting and therefore both are repulsing, but the B-photons of the positron are upside-down, relative to the B-photons of the electron.

Let that sink in fully, and then go back to the beta decay. The neutron becomes a proton, by switching its z-spin. This switch allows the B-photon emission of the baryon to get through. So, before the positron arrived, we had no emission from the big particle. The positron hits it, and it begins emitting. But the total B-photon field—existing after this interaction—is made up of both particles, both the baryon and the lepton. The positron’s B-photons were upside-down, relative to the proton, but the electron’s B-photons are upside-up. The positron would have subtracted from the field energy, but the electron adds to it. And so the missing energy is not in an invisible neutrino, it is in the total B-photon field. The charge of the field has appeared to go up, because the charge itself has spin. The B-photons are spinning, and we have to monitor not only the number of photons but also the spin of the photons.

This explains the experiment of 1956, since an augmented B-photon field will act exactly like a theoretical field of neutrinos. It will travel and it will carry energy. This augmented B-photon field will have exactly the right energy and spin to facilitate anti-beta decay.

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**253. Explaining Mesons without Quarks**Neutrino "Oscillation"

I showed at the end of my QCD paper that neutrino oscillation is actually a change in the B-photon or emission field, not the reincarnation of one particle to another. But here I give the muon neutrino to an electron triplet. Which is it? Both. In neutrino oscillation we go from one unseen flux to another. We call the first flux an electron neutrino and the second a muon neutrino. But in reality, the first flux is a flux of the B-photon field, with no neutrino present. The second is a flux of the electron field, with electrons colliding with free neutrons and losing their axial spins. Again, there is no neutrino present, but there is a new particle present after the second event: an electron triplet. In each stripping, a neutron is stripped of its z-spin and multiple electrons are stripped of their axial spins. It would appear that it takes three electrons to strip each neutron. The axial spin has a much lower energy than the z-spin, and only by adding their linear energy to the equation can the moving electrons successfully attack a neutron. I will have more to say about this in other papers.

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**258. HOW DO PHOTONS TRAVEL?**From a previous paper, we know that the radius of the B-photon is G times less than the radius of the proton. This gives us a photon radius of 2.74 x 10-24m. The z-spin is 8 times the radius, so we should find a basic wavelength of 2.2 x 10-23m. Obviously, we don’t find photons with a wavelength that small. Why? Simply because the wavelength we measure has been stretched out by the velocity of the photon. The photon would be measured to have a wavelength of 2.2 x 10-23m only if it were at rest.

You will say, “Even if we accept that the photon is spinning, how can the z-spin be stretched? The spin would give us a spin radius, which is just a length. A length cannot be stretched by motion, unless you are proposing some kind of relativity here.”

I am not proposing relativity as the solution here. The answer to your question is that a spin is not just a radius, and is not just a length. A spin is a motion: a motion that takes time. Even if the photon were spinning at velocity c, one rotation must take some real time. We know that the linear velocity of light is not infinite, so we must assume the speed of spin is also not infinite. If it is not infinite, it must take time. If it takes time, then it will be stretched by the linear motion. While the surface of the photon is spinning, the photon as a whole is moving some linear distance x.

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**276a. A Preliminary Study of the Pyramid as an Electrical Structure**In my paper on QCD, I show that the neutrino field actually IS this foundational B-photon field, at least in the case of beta decay.

**Cr6**- Admin
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## Re: Photon Sinewave Travel

It's really quite interesting, but also confusing. We're in uncharted territory here, paddling away as best we can. It's disheartening for me to have to rethink my stacked spin models - I JUST made it to the electron level, at the Z3 spin, finally! The video isn't ready yet, however. And it's evidently not the electron level. So I need to keep going. It's quite tedious in my Maya software, since it's designed for animation physics and special FX. But I'm making good progress.

Regarding the photon's wavelength, it will only be a sine at that first spin, I agree. But that spin would be far too small for these types of detectors anyway, it seems like? And viewed from any orthogonal, every stacking produces a pattern over a certain dt. It may be a very long dt, relative to the photon's size, but when viewed collectively through these shitty detectors from our past I have no problem visualizing waveforms. They just wouldn't necessarily be very clean, but then again the detectors are very far from clean devices as well. I have many unfinished animations where, upon stacking some spins, I then send the photon off at velocity and observe its trail from the side or top. Granted, my models were flawed. But even Nevyn's less-flawed (thank you, sir) models produce patterns. And at light speed, these patterns certainly appear to form waves. Groups of photons at varying energies and spin states will make these waves even rounder, smoother.

All that said, I'm not opposed to retiming and rethinking my entire spin model. But if we discard the axial spins (aside from the first, obviously), where does that take us?

I have to concur with you here. While the additional axials would change the motion, they wouldn't change the radius. If we trust your gut on this one, where would that put us in spin-stackings?

I'm having a tough time visualizing how a spun-up photon would spin axially alone upon a collision. It seems like there's only one single point on its path and one single vector on an incoming collider that might cause this? It seems... unlikely. I can handle the end-over spins pretty well, mentally, but the additional axials seem next to impossible.

Lloyd, I'll work on one of those videos to get it to a presentable state, so you can visualize what I mean by them always forming a cyclical "wave", on a long enough timeline. Accuracy aside, the concept should be pretty easy to visualize.

But I believe I had the frequency and wavelength confused.

At higher spins, the patterns get more complex, but they could still be called patterns.

Regarding the photon's wavelength, it will only be a sine at that first spin, I agree. But that spin would be far too small for these types of detectors anyway, it seems like? And viewed from any orthogonal, every stacking produces a pattern over a certain dt. It may be a very long dt, relative to the photon's size, but when viewed collectively through these shitty detectors from our past I have no problem visualizing waveforms. They just wouldn't necessarily be very clean, but then again the detectors are very far from clean devices as well. I have many unfinished animations where, upon stacking some spins, I then send the photon off at velocity and observe its trail from the side or top. Granted, my models were flawed. But even Nevyn's less-flawed (thank you, sir) models produce patterns. And at light speed, these patterns certainly appear to form waves. Groups of photons at varying energies and spin states will make these waves even rounder, smoother.

All that said, I'm not opposed to retiming and rethinking my entire spin model. But if we discard the axial spins (aside from the first, obviously), where does that take us?

Nevyn wrote:I also assumed 4 spin levels to a spin set so the higher level axial spins are in there and they are assumed to be doubling the radius. Obviously, I am against this sort of interpretation but I wanted to calculate it as Miles has done and leave my own ideas out of it.

I have to concur with you here. While the additional axials would change the motion, they wouldn't change the radius. If we trust your gut on this one, where would that put us in spin-stackings?

I'm having a tough time visualizing how a spun-up photon would spin axially alone upon a collision. It seems like there's only one single point on its path and one single vector on an incoming collider that might cause this? It seems... unlikely. I can handle the end-over spins pretty well, mentally, but the additional axials seem next to impossible.

Lloyd, I'll work on one of those videos to get it to a presentable state, so you can visualize what I mean by them always forming a cyclical "wave", on a long enough timeline. Accuracy aside, the concept should be pretty easy to visualize.

But I believe I had the frequency and wavelength confused.

At higher spins, the patterns get more complex, but they could still be called patterns.

**Jared Magneson**- Posts : 330

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## Photon WAVE Motion and Not a Sine Wave

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From

Photon motion is a cycloidal spin wave. Each individual photon creates wave motion by end-over-end spins.

A sine wave is a field wave describing the response of a fluid medium; sine waves refer to equal and opposite oscillating charge or pressure polarities based in an electromagnetic fluid medium aether. Miles has said that one may consider his charge field an aether; however, Miles’ aether does not behave as a fluid.

I may be loyal to a fault; someone must be a stick in the mud.

* twicebittendesigns turns up empty - That address doesn’t exist …

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**Jared wrote. Regarding the photon's wavelength, it will only be a sine at that first spin.**From

**The Wavelength and Frequency of Light are Reversed****Sine wave**? Incorrect. Please refer to Chris Wheeler’s photon wave motion diagram,*.Photon motion is a cycloidal spin wave. Each individual photon creates wave motion by end-over-end spins.

A sine wave is a field wave describing the response of a fluid medium; sine waves refer to equal and opposite oscillating charge or pressure polarities based in an electromagnetic fluid medium aether. Miles has said that one may consider his charge field an aether; however, Miles’ aether does not behave as a fluid.

**A spins?**Miles has said that axial spins can occur on top of any x, y, or z spin, just not on existing A spins. I don’t believe we’ve proven otherwise, nor have we (as far as I can tell) modeled any recycling A spins. As I’ve previously said, I believe the A spin is what a fourth end-over-end spin degenerates into, therefore there is a radius doubling. The A spin allows us to model electrons or protons as spheres with a tangential velocity of c. In my opinion, we should not show any levels below the A unless the top level spin is lost.**Wavelength or frequency**. I believe the particle's top level spin motion radius is the same as wavelength. Frequency is how often the photon’s highest spin repeats each second.I may be loyal to a fault; someone must be a stick in the mud.

* twicebittendesigns turns up empty - That address doesn’t exist …

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Last edited by LongtimeAirman on Tue Apr 18, 2017 2:47 pm; edited 3 times in total (Reason for editing : Typos)

**LongtimeAirman**- Admin
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## Re: Photon Sinewave Travel

I should have clarified - not a sine wave. A simple wave is what I should have said.

As for axial spins, it's easy to show (I mean as easy as everything else is in Maya, which is not at all/massively tedious/almost impossible to work with) but what's not easy is to see any actual difference, beyond say A2 or A3. It doesn't increase the radius at all, from a volume of influence/path point of view. It may however increase mass (ponderability), which is what we've kinda been tossing around in these threads.

I really would like to get to a point where I can import Nevyn's code in some fashion, instead of doing everything manually on my end. So we can make and see these kinds of changes rapidly, instead of it taking me days to time the animations in Maya. It's actually the VoI spheres that are my biggest headache right now.

As for axial spins, it's easy to show (I mean as easy as everything else is in Maya, which is not at all/massively tedious/almost impossible to work with) but what's not easy is to see any actual difference, beyond say A2 or A3. It doesn't increase the radius at all, from a volume of influence/path point of view. It may however increase mass (ponderability), which is what we've kinda been tossing around in these threads.

I really would like to get to a point where I can import Nevyn's code in some fashion, instead of doing everything manually on my end. So we can make and see these kinds of changes rapidly, instead of it taking me days to time the animations in Maya. It's actually the VoI spheres that are my biggest headache right now.

**Jared Magneson**- Posts : 330

Join date : 2016-10-11

## Re: Photon Sinewave Travel

I'm not convinced that that image from The Wavelength and Frequency are Reversed paper is correct. The problem I have with it is that it has an axial Y spin and then an end-over-end Z spin but it is moving in the

I realise that Miles has stated that there are higher axial spins, but he has not proven it. He has not shown

**X**dimension. It is that linear velocity that causes the cycloid motion, not the spin (well, it is both in combination, really). If that was moving in the**Z**dimension, to match the top level spin, then it would be a perfect spiral which, when viewed from the side, would make a perfect sine wave.I realise that Miles has stated that there are higher axial spins, but he has not proven it. He has not shown

**how**it can happen or how it can double the radius. If you read what he says, and ignore the word axial, he is talking about end-over-end spins but calling it axial. That doesn't make any sense. The onus is on Miles to show how any axial spin can double the radius. I am just using the definition of axial spin (to rotate about an axis through the center) and that, in no way at all, doubles the radius. Just look at the wheels on your car. They are rotating about an axis through their center and they do not double the radius of the wheel.**Nevyn**- Admin
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## Re: Photon Sinewave Travel

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The title of this thread is the problem, even if, when viewed orthogonally over a limited interval (after all, you don’t want to spoil the view) you will see what resembles a sine wave. The passing appearance to a sine wave is not a good argument. The cycloid makes more sense than that. Most important, in my view, sine waves are the distinguishing characteristic of E/M field theory. Using the word ‘Sinewave’ because of a passing appearance is confusing and misleading. I started listing the main reasons why a field wave isn't appropriate to compare with photon spin motion in my previous post.

My example, what may be confusing me with this whole business of creating higher axial rotations. I’m riding on my 12 speed bike along an asphalt road as a kid. I can ride without my hands and was casually poking a broomstick on the road alongside the bike. Of course you know what happened, the stick slipped between the spokes, jammed and stopped the wheel’s rotation, causing end-over-end motion allowing me to plant my head on the road while still seated!

We’ve run around this tree before. You honestly believe higher A spins don’t exist. I honestly believe they do. You haven’t convinced me and I gave you mechanisms to support my view. I believe the onus of proof is on you. Please, for all of us, convince Miles that - higher level A spins do not exist! - while he’s still available for discussion.

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The title of this thread is the problem, even if, when viewed orthogonally over a limited interval (after all, you don’t want to spoil the view) you will see what resembles a sine wave. The passing appearance to a sine wave is not a good argument. The cycloid makes more sense than that. Most important, in my view, sine waves are the distinguishing characteristic of E/M field theory. Using the word ‘Sinewave’ because of a passing appearance is confusing and misleading. I started listing the main reasons why a field wave isn't appropriate to compare with photon spin motion in my previous post.

My example, what may be confusing me with this whole business of creating higher axial rotations. I’m riding on my 12 speed bike along an asphalt road as a kid. I can ride without my hands and was casually poking a broomstick on the road alongside the bike. Of course you know what happened, the stick slipped between the spokes, jammed and stopped the wheel’s rotation, causing end-over-end motion allowing me to plant my head on the road while still seated!

We’ve run around this tree before. You honestly believe higher A spins don’t exist. I honestly believe they do. You haven’t convinced me and I gave you mechanisms to support my view. I believe the onus of proof is on you. Please, for all of us, convince Miles that - higher level A spins do not exist! - while he’s still available for discussion.

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**LongtimeAirman**- Admin
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## Re: Photon Sinewave Travel

To me, the problem here is the expectation of a sine wave. As I stated earlier, only the first X spin creates a sine wave and as Jared pointed out, our machines do not measure that low, yet. Also, it is only a sine wave from the side which is not how it would be measured. So I must ask where the expectation comes from. The answer is: The math! Mathematicians love to reduce everything to what they already know. It is much easier to deal with sine waves in the math than it is to deal with real motions when you already have sine wave math. Any wave can be created by a combination of sine waves, but I doubt that extends into 3D. In 2D, yes, you can do that, but we are talking about 3D waves and I don't think the mainstream is aware of that yet. Sine waves are 2D but spin waves are 3D, so we are already dealing with a compression of the true wave.

You can create sine waves in 3D, but if you look at how that is created, you find that there is only 1 dimension that is changing (or waving). We have a 3D wave where all 3 dimensions can be changing but if we limit it to the first X spin, we only have 2 dimensions that are changing, ignoring the linear velocity which is actually a secondary change in one of the previous dimensions. With an X spin, we have motion in the YZ plane, so only the Y and Z values are changing, not the X. When we add a linear velocity to that, and I assume motion in the same direction as the spin axis, then we are changing all 3 dimensions, two with spin (Y and Z) and one with linear velocity (X).

I am using the definition of axial rotation as my proof. Given that definition, there is no way to double the radius. An end-over-end spin is not an axial rotation. It is a rotation about an axis, but that axis is not centered on that which is spinning, so it is not an axial rotation.

Whether a particle can have a linear velocity that is not inline with the top level spin axis is another question. I have asked that question myself on this site. Miles has made statements that seem to limit it to be inline and I am unaware of anything that might change that rule. It seems to me that any collision that could create a linear velocity that is not inline with the top level spin axis would actually cause that top level spin to either decrease in speed (when the collision vectors oppose each other) or it will have no effect (because they are moving in the same direction and therefore there is no difference in velocity to cause a collision).

I have tried to start writing a paper about these higher axial spins but have not gotten very far with it. I want to make sure that I explain it very carefully before I take it to Miles but I haven't put much effort into it.

You can create sine waves in 3D, but if you look at how that is created, you find that there is only 1 dimension that is changing (or waving). We have a 3D wave where all 3 dimensions can be changing but if we limit it to the first X spin, we only have 2 dimensions that are changing, ignoring the linear velocity which is actually a secondary change in one of the previous dimensions. With an X spin, we have motion in the YZ plane, so only the Y and Z values are changing, not the X. When we add a linear velocity to that, and I assume motion in the same direction as the spin axis, then we are changing all 3 dimensions, two with spin (Y and Z) and one with linear velocity (X).

I am using the definition of axial rotation as my proof. Given that definition, there is no way to double the radius. An end-over-end spin is not an axial rotation. It is a rotation about an axis, but that axis is not centered on that which is spinning, so it is not an axial rotation.

Whether a particle can have a linear velocity that is not inline with the top level spin axis is another question. I have asked that question myself on this site. Miles has made statements that seem to limit it to be inline and I am unaware of anything that might change that rule. It seems to me that any collision that could create a linear velocity that is not inline with the top level spin axis would actually cause that top level spin to either decrease in speed (when the collision vectors oppose each other) or it will have no effect (because they are moving in the same direction and therefore there is no difference in velocity to cause a collision).

I have tried to start writing a paper about these higher axial spins but have not gotten very far with it. I want to make sure that I explain it very carefully before I take it to Miles but I haven't put much effort into it.

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## Do Higher A-Spins exist?

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Is that correct? You win by definition? That’s preposterous, one cannot make the observation - A-spins cannot double the radius of the spin state below it - the basis of a proof of the mechanical impossibility of higher level A-spins, especially not while I have a better explanation.

The simplest case. Miles has stated that an axial spin can form on any top level spin - excluding only an existing A-spin. How exactly does an axial spin form on a non-axial spin? I remember you mentioned spinning a coin once, because I was also imagining a toroid (my ring) spinning on a tabletop. The higher level A-spin radius in this case is the same as the radius of the end-over-end, toroidal spin that the A-spin was created from. In other words, you are correct if you believe A-spins cannot be created independently. Instead, an A-spin is transformed from an existing top level non-axial spin.

Imagine Bphoton - A,X,Y,Z,X; Any x direction collision where the Bphoton is at a maximum extent at the time of collision may cause a new top level spin. Collision at the maximum Y or Z (I tried reorienting to top level horizontal – as occurs with a Z-spin, but the description became too confusing) extents would presumably cause a new Y or Z end-over-end spin. I suppose A-spins could result from collisions along the extent of +/- Y / Z diagonals, or in fact, most anywhere in the Bphoton’s YZ plane. This is only a guess of course. I’ve suggested something similar previously, I dimly recall what sounded like howls of despair. I appreciate your patience.

The charged particle result is Bphoton A,X,Y,Z,A. Is each radius double the previous? Yes. How can the new A-spin be considered just twice the energy of Z when we know that that top level’s previous quanta was an X-spin? How exactly is the A-spin axis formed? How are the deeper spins affected? We don’t know, but I’m anxious to find out. Or at least show where I’m wrong.

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**Do Higher A-Spins exist?****Nevyn wrote**. I am using the definition of axial rotation as my proof. Given that definition, there is no way to double the radius. An end-over-end spin is not an axial rotation. It is a rotation about an axis, but that axis is not centered on that which is spinning, so it is not an axial rotation.**Airman**. We know Miles refers to a radius doubling with each new top level spin. How do we go from a Z-spin radius to a new higher axial A-spin at twice the z-spin radius? I agree with you that axial spins cannot increase a radius, only end-over-end spins can increase a radius. From that you seem to infer there can be no higher level A-spins.Is that correct? You win by definition? That’s preposterous, one cannot make the observation - A-spins cannot double the radius of the spin state below it - the basis of a proof of the mechanical impossibility of higher level A-spins, especially not while I have a better explanation.

The simplest case. Miles has stated that an axial spin can form on any top level spin - excluding only an existing A-spin. How exactly does an axial spin form on a non-axial spin? I remember you mentioned spinning a coin once, because I was also imagining a toroid (my ring) spinning on a tabletop. The higher level A-spin radius in this case is the same as the radius of the end-over-end, toroidal spin that the A-spin was created from. In other words, you are correct if you believe A-spins cannot be created independently. Instead, an A-spin is transformed from an existing top level non-axial spin.

Imagine Bphoton - A,X,Y,Z,X; Any x direction collision where the Bphoton is at a maximum extent at the time of collision may cause a new top level spin. Collision at the maximum Y or Z (I tried reorienting to top level horizontal – as occurs with a Z-spin, but the description became too confusing) extents would presumably cause a new Y or Z end-over-end spin. I suppose A-spins could result from collisions along the extent of +/- Y / Z diagonals, or in fact, most anywhere in the Bphoton’s YZ plane. This is only a guess of course. I’ve suggested something similar previously, I dimly recall what sounded like howls of despair. I appreciate your patience.

The charged particle result is Bphoton A,X,Y,Z,A. Is each radius double the previous? Yes. How can the new A-spin be considered just twice the energy of Z when we know that that top level’s previous quanta was an X-spin? How exactly is the A-spin axis formed? How are the deeper spins affected? We don’t know, but I’m anxious to find out. Or at least show where I’m wrong.

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**LongtimeAirman**- Admin
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## Re: Photon Sinewave Travel

LongtimeAirman wrote:Is that correct? You win by definition?

Well, it sounds really bad when you put it like that.

I am saying that there are no higher axial spins because I don't see any collision that could create one and I don't see what use they are even if I could.

**If**anyone thinks that they do double the radius, then the definition of axial rotation puts a stop to that. If you want to theorize about some other spin that does double the radius, then you can't call it an axial spin and you have to explain how it differs from an end-over-end spin.

I think this whole mess of higher level axial spins comes from Miles early papers where he didn't understand that all particles are actually BPhotons with many spin levels. He was still following the mainstream particle list and just used 4 stacked spins on each of them. The electron was a particle that could have 4 spins. The proton was another particle and it could have 4 spins. He later fleshed all of this out into a single particle that just kept stacking spins to give us the particles we have. At that point, I don't think he realised that the axial spins wouldn't work at higher levels.

My main argument against higher axial spins is the very same argument that gives us an end-over-end spin. A single entity can not spin about two axes that go through that entity. The second spin

**must**have an axis that is outside of the gyroscopic influence of the other spin. Any spin that could be called axial must have an axis that goes through what is being spun. I am adhering to the rules of stacked spins and the definition of axial rotation but anyone who uses higher axial spins is not. You have to explain how and why that rule can be broken. If it can be broken, then why is it needed at all?

LongtimeAirman wrote:

Imagine Bphoton - A,X,Y,Z,X; Any x direction collision where the Bphoton is at a maximum extent at the time of collision may cause a new top level spin. Collision at the maximum Y or Z (I tried reorienting to top level horizontal – as occurs with a Z-spin, but the description became too confusing) extents would presumably cause a new Y or Z end-over-end spin. I suppose A-spins could result from collisions along the extent of +/- Y / Z diagonals, or in fact, most anywhere in the Bphoton’s YZ plane. This is only a guess of course. I’ve suggested something similar previously, I dimly recall what sounded like howls of despair. I appreciate your patience.

You have used collisions on the diagonal to build an axial spin, but it does not do so. To a spin, there is no such thing as up, down, left, right, or diagonal. They just don't care about directions. They only care about their own motion. As a collision happens, at the very moment that they touch, you can think of that point as being the top of the spin or the left or bottom or any direction that you want. That collision point is what

**creates the dimension for the next spin level**. We like to talk in solid dimensions like X, Y and Z, but they are actually

**relative**dimensions. They are relative to the previous dimension. Two consecutive spin levels will have dimensions that are orthogonal to each other because the collision must come from a certain direction in order to create the new spin level. Any other direction and it will not create a new spin level but will change the existing spin because it is inline with that spin. So if the particle is spinning about the X dimension, then the incoming particle must be moving in that same dimension in order to create a new spin level. If it is traveling in Y or Z then it is just going to effect the spin itself.

I know that relative dimensions is a hard thing to understand. We tend to create our dimensions and then judge all motion based on that static grid. That is a mathematical abstraction though. Reality does not need to work that way. It only needs to deal with the forces that are created by a collision and the forces that the collision must overcome (pre-existing motion). I'm not sure if I am explaining this very well, it is a difficult topic to describe.

Take a look at SpinSim or Jared's videos. Actually, Jared's latest video is a prime example. It starts with an axial Y spin and it soon adds a Z spin to that. If you think about that Y spin, while it is inside of the Z spin, then its Y dimension

**is traveling**. All of the dimensions inside of the top level spin are moving. They are not static. That is slightly different to relative dimensions that I am trying to explain, but it gives you an idea of how these dimensions can move.

LongtimeAirman wrote:

The charged particle result is Bphoton A,X,Y,Z,A. Is each radius double the previous? Yes. How can the new A-spin be considered just twice the energy of Z when we know that that top level’s previous quanta was an X-spin? How exactly is the A-spin axis formed? How are the deeper spins affected? We don’t know, but I’m anxious to find out. Or at least show where I’m wrong.

.

If it was possible to create an axial spin, then it would have more energy even without doubling the radius. Motion is energy so any motion must increase the energy of the particle. The deeper spins are not affected by higher spins. The motion of the particle is affected, but not the spins themselves. How is the axial spin formed? Well, I leave that for you to describe. I don't see any way for it to be created, but I admit that I could be missing something.

By the way, if you manage to convince me, then it is your patience that is appreciated. If not, then your persistence is still appreciated. I don't want you to just give up. If I am right, then I want you to be able to see why that is, not just accept it. And if I am wrong, and you weren't persistent, then I would be losing valuable information. In my view, we either both come out winners, because we have a more solid understanding, or we just get tired of arguing and agree to disagree.

**Nevyn**- Admin
- Posts : 928

Join date : 2014-09-11

## Re: Photon Sinewave Travel

Perhaps think of it this way: there's no fulcrum in higher-level axial spins to

A screenshot from my last video, for reference:

This is early on, during the Z1 spin stacking. You can see the World axes on the bottom left, X,Y, Z. So we have our initial axial and the firs three stacks.

For the first X1 spin to happen, the A1 photon must be hit by another photon or larger guy. Not just hit, but hit at just the right angle and point, or else it won't spin end-over. It will bounce away, but unless the vector is just right, it can't flip over itself.

This is true of every stacked spin. Not all collisions stack spins; most won't. Almost all will not. It has to be a certain vector and energy to flip the particle over; not any collision at any energy will do. It can't be a head-on collision and it can't be every oblique. It only happens often because there are so very, very many photons creating the charge field.

So back to the reference screenshot: Our B-photon (yellow) is where it is. An incoming B-photon cannot know where it's Volume of Influence (VOI) shell is, since that's just a construct we are using to show it's maximum radius. Remember, the actual radius of our initial B-photon has not changed. Only its

So to create another axial spin, A2, an incoming collider would have to somehow hit our B-photon Z1 and make it rotate not around that B-photon's internal axis at all, but rather the center pivot point of the Z1's VoI sphere - which doesn't exist. The colliding photon nor the original photon don't know anything about that point, at the center of the light blue sphere. Our B-photon may have never even traveled through that point, ever. That point is as much of a construct as the colored VoI spheres I am showing - it does not actually exist.

So the incoming collider would somehow have to force our B-photon to spin about a point that does not exist. It would have to exhibit a torque

However, and end-over spin (X2) is completely possible. If an incoming photon strikes our B-photon example at the right angle from the right direction, bam, it has a fulcrum. It maintains its original spins but feels the torque internally and has no choice but to flip over itself again. That's the most efficient way to transfer the new, incoming motion (energy). It cannot fight the incoming particle head on with tiny swords or lasers, so it must react to its energy in the cleanest way possible. Billiard ball mechanics.

Most colliders would simply redirect our Z1 photon, either because they hit it at an angle not orthogonal to its current Object-Space spin or because they are lower in energy and cannot flip it. In only certain vectors and energies would our baby B-photon get flipped into a higher spin state, or "spun up."

Does that make any sense?

*create*the spin point.A screenshot from my last video, for reference:

This is early on, during the Z1 spin stacking. You can see the World axes on the bottom left, X,Y, Z. So we have our initial axial and the firs three stacks.

For the first X1 spin to happen, the A1 photon must be hit by another photon or larger guy. Not just hit, but hit at just the right angle and point, or else it won't spin end-over. It will bounce away, but unless the vector is just right, it can't flip over itself.

This is true of every stacked spin. Not all collisions stack spins; most won't. Almost all will not. It has to be a certain vector and energy to flip the particle over; not any collision at any energy will do. It can't be a head-on collision and it can't be every oblique. It only happens often because there are so very, very many photons creating the charge field.

So back to the reference screenshot: Our B-photon (yellow) is where it is. An incoming B-photon cannot know where it's Volume of Influence (VOI) shell is, since that's just a construct we are using to show it's maximum radius. Remember, the actual radius of our initial B-photon has not changed. Only its

*maximum*influence size has changed, over a long enough change in time (∆t or dt).So to create another axial spin, A2, an incoming collider would have to somehow hit our B-photon Z1 and make it rotate not around that B-photon's internal axis at all, but rather the center pivot point of the Z1's VoI sphere - which doesn't exist. The colliding photon nor the original photon don't know anything about that point, at the center of the light blue sphere. Our B-photon may have never even traveled through that point, ever. That point is as much of a construct as the colored VoI spheres I am showing - it does not actually exist.

So the incoming collider would somehow have to force our B-photon to spin about a point that does not exist. It would have to exhibit a torque

*outside*of the colliding objects. There's no fulcrum for the particle to spin around. If there was, it would exist*outside*of our particle, which defies all mechanics available. Beyond action at a distance, it would be fulcrum action at a distance.However, and end-over spin (X2) is completely possible. If an incoming photon strikes our B-photon example at the right angle from the right direction, bam, it has a fulcrum. It maintains its original spins but feels the torque internally and has no choice but to flip over itself again. That's the most efficient way to transfer the new, incoming motion (energy). It cannot fight the incoming particle head on with tiny swords or lasers, so it must react to its energy in the cleanest way possible. Billiard ball mechanics.

Most colliders would simply redirect our Z1 photon, either because they hit it at an angle not orthogonal to its current Object-Space spin or because they are lower in energy and cannot flip it. In only certain vectors and energies would our baby B-photon get flipped into a higher spin state, or "spun up."

Does that make any sense?

**Jared Magneson**- Posts : 330

Join date : 2016-10-11

## Re: Photon Sinewave Travel

Here is some evidence that Miles applied axial spin to an electron assuming that it is a solid particle underneath that axial spin.

Emphasis is mine.

That is from the Unifying the Electron and Proton paper.

Miles Mathis wrote:

I will be asked how the electron can show a wave motion with only an axial spin. I have already shown that the wave characteristic of matter and of light is caused by stacked spins. But here we have only the first spin. How is the wave expressed? Well,it isn't expressed by an electron at rest, and we are comparing rest masses here.The electron must be moving to express a wave.If the electron begins moving and expresses a wave,of course it must have a second spin. It must get this spin from collision with photons in the charge field, we assume. And this second spin will add to the energy and therefore the apparent mass of the electron. A moving electron will become a sort of stable meson. As you can see from the math above, we can predict that it will have an energy about 7.2 times (65/9) that of the electron at rest. So in the first instance, the moving electron is not gaining energy only from Relativity. It is primarily gaining energy from x-spin.

Emphasis is mine.

That is from the Unifying the Electron and Proton paper.

**Nevyn**- Admin
- Posts : 928

Join date : 2014-09-11

## Re: Photon Sinewave Travel

Here are some better illustrations which might help. For another axial, we have no connection to the apparent center. For another stack, we have contact connection.

**Jared Magneson**- Posts : 330

Join date : 2016-10-11

## Do Higher A-Spins exist? Continued

.

Thanks for the replies, I hope you all don’t mind being here before. I freely admit to certain mad bull disease-like tendencies, head trauma and all. Keep it simple. Repetition is crucial. Smile a lot.

Please, evaluate only the simplest case. I honestly don’t see how the definitions of axial or end-over-end spins are violated when they are applied as two consecutive actions. Does this simplest case break the rules of stacked-spins or not? I can only guess, the fulcrum is light speed and the lever is the moving Bphoton? The Bphoton always seems to be spinning about points outside itself, that’s a good description of our current spin sims. Does that particular objection apply to the X-spin to A-spin transform alone? Is the simplest case consistent with Miles’ descriptions?

At this point you may slap me and say – How many times do I have to keep telling you - Miles is mistaken – it’s mechanically impossible to include such an additional spin; or, you haven’t figured out a way it could be true, and you’ve pretty much exhausted the possibilities and it's impossible to prove a negative.

How might we give the X-spin Bphoton an additional spin to result in a quanta conversion to an A-spin? The one requirement I can think of is that the motion of the underlying non-axial spins remain intact. There must be a way to give the Bphoton another spin here somehow. I can’t imagine a charge field without A-spins - or it would take some effort.

If we could transform an X-spin into an A-spin, we might assume all non-axial (X, Y or Z) spins must, in turn, develop into full A-spins before new orthogonal non-axial spins can form. Seems quite different from our current model. The Bphoton's top spin surface may then increase incrementally. Any given top-level may be an X, Y, or Z, but the next level down will be an A-spin. Alternating A-spins all the way down.

Thanks for the discussion. You guys make my favorite interest even more rewarding and something to brag about.

.

Thanks for the replies, I hope you all don’t mind being here before. I freely admit to certain mad bull disease-like tendencies, head trauma and all. Keep it simple. Repetition is crucial. Smile a lot.

**Simplest Case**. A-spin creation involves a two-step process. First, a non-axial (X, Y, or Z) spin is created. Second, an axial spin is added to the first non-axial spin.Please, evaluate only the simplest case. I honestly don’t see how the definitions of axial or end-over-end spins are violated when they are applied as two consecutive actions. Does this simplest case break the rules of stacked-spins or not? I can only guess, the fulcrum is light speed and the lever is the moving Bphoton? The Bphoton always seems to be spinning about points outside itself, that’s a good description of our current spin sims. Does that particular objection apply to the X-spin to A-spin transform alone? Is the simplest case consistent with Miles’ descriptions?

At this point you may slap me and say – How many times do I have to keep telling you - Miles is mistaken – it’s mechanically impossible to include such an additional spin; or, you haven’t figured out a way it could be true, and you’ve pretty much exhausted the possibilities and it's impossible to prove a negative.

How might we give the X-spin Bphoton an additional spin to result in a quanta conversion to an A-spin? The one requirement I can think of is that the motion of the underlying non-axial spins remain intact. There must be a way to give the Bphoton another spin here somehow. I can’t imagine a charge field without A-spins - or it would take some effort.

If we could transform an X-spin into an A-spin, we might assume all non-axial (X, Y or Z) spins must, in turn, develop into full A-spins before new orthogonal non-axial spins can form. Seems quite different from our current model. The Bphoton's top spin surface may then increase incrementally. Any given top-level may be an X, Y, or Z, but the next level down will be an A-spin. Alternating A-spins all the way down.

Thanks for the discussion. You guys make my favorite interest even more rewarding and something to brag about.

.

Last edited by LongtimeAirman on Thu Apr 20, 2017 8:44 pm; edited 1 time in total (Reason for editing : Added Title)

**LongtimeAirman**- Admin
- Posts : 749

Join date : 2014-08-10

## Re: Photon Sinewave Travel

No, I don't think you're falling apart or anything here! Your point about pivot centers is a very good one. All our stacked spins except the X1 are pivoting about a point outside the initial B-photon, I'm pretty sure. The X1 pivot point is on the surface of the sphere, so one radius away from the center of that sphere.

I'm not sure how to answer it, so I'll ponder that part.

I also don't know how to address your Simplest Case. Are you thinking that the progression of spins could go in any of several variations, of order? As in, A1, Y1, Z1, A2, X1, or something like that? Or more simply, A1, X1, A2? In that case, how would A2 differ from Y1? Would A2 be spinning about the X1's center point, as opposed to a point on its "surface"? It gets confusing, for sure.

Keep in mind our axes are only relative to the B-photon's initial axial spin. They don't really exist, except relative to A1. In world-space (as opposed to object-space), A1 could be along any of the three, or any angle at all.

I'm not sure how to answer it, so I'll ponder that part.

I also don't know how to address your Simplest Case. Are you thinking that the progression of spins could go in any of several variations, of order? As in, A1, Y1, Z1, A2, X1, or something like that? Or more simply, A1, X1, A2? In that case, how would A2 differ from Y1? Would A2 be spinning about the X1's center point, as opposed to a point on its "surface"? It gets confusing, for sure.

Keep in mind our axes are only relative to the B-photon's initial axial spin. They don't really exist, except relative to A1. In world-space (as opposed to object-space), A1 could be along any of the three, or any angle at all.

**Jared Magneson**- Posts : 330

Join date : 2016-10-11

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