# The Photon - The Mathis' Definition

## The Photon - The Mathis' Definition

This thread is to outline Miles Mathis' definition and description of the "photon".

Cr6

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## Re: The Photon - The Mathis' Definition

3. Unified Fields in Disguise
by Miles Mathis

Now we only need to assign density mechanically. I have given it to E/M, but what part of the E/M field does it apply to? Well, it must apply to the emission. Newton’s equation is not telling us the density of the bodies in the field, it is telling us the density of the emitted field. Of course one is a function of the other. If you have a denser moon, it will emit a denser E/M field. But, as a matter of mechanics, the variable D applies to the density of the emitted field. It is the density of photons emitted by the matter creating the unified field.

Finally, what is G, in this analysis? G is the transform between the two fields. It is a sort of scaling constant. As we have seen, one field--gravity--is determined by the radius of a macro-object, like a moon or planet or a marble. The other field is determined by the density of emitted photons. But these two fields are not operating on the same scale. To put both fields into the same equation, we must scale one field to the other. We are using both fields to find a unified force, so we must discover how force is transmitted in each field. In the E/M field, force is transmitted by the direct contact of the photons. That is, the force is felt at that level. It can be measured from any level of size, but it is being transmitted at the level of the photon. But since gravity is now a function of volume alone, it is not a function of photon size or energy. It is a function of matter itself, that is, of the atoms that make up matter. Therefore, G is a scaling constant between atoms and photons. To say it another way, G is taking the volume down to the level of size of the density, so that they may be multiplied together to find a force. Without that scaling constant, the volume would be way too large to combine directly to the density, and we would get the wrong force. By this analysis, we may assume that the photon involved in E/M transmission is about G times the atom, in size.
Now we continue on to Coulomb’s equation:

F = kq1q2/r2

One hundred years after Newton, we got another unified field equation. Here we have charges instead of masses, and the constant is different, but otherwise the equation looks the same as Newton’s. Physicists have always wondered why the equations are so similar, but until now, no one really knew. No one understood that they are both the same equation, in a different disguise.
...
In scaling up and scaling down, we don’t simply reverse the scale. It is a bit more complex than that, as you have seen. In scaling down, we go from atomic size to photon size. In scaling up, we go from atomic size to our own size.

http://milesmathis.com/uft2.html

Cr6

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## Re: The Photon - The Mathis' Definition

Here’s a question not many ask: why is the velocity squared in the kinetic energy equation, E = ½mv2. Why should the energy depend on the square of the velocity? We have the same question with the equation E = mc2. Einstein was nice enough to provide us with this simple equation, but not nice enough to tell us why the energy depends on the square of the speed of light.

To find out, let us look at how the first equation is derived in textbooks. We start with the constant acceleration equation,

Then substitute a = F/m into that

2Fd/m = vf2 - vi2

If we let the initial velocity equal zero, and define work as force through a distance, we get
W = E = Fd = ½ mvf2

Work is then defined as the change in kinetic energy, in the famous work-energy theorem.

The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will have a force. A car hitting you will apply a force, whether or not it is accelerating.

But can we derive the kinetic energy equation without a force? Can we achieve a square velocity without assuming an acceleration? Not according to classical mechanics. Classical mechanics answers this question by ignoring it completely. QED and string theory do not address it either: they are busy ignoring all sorts of new questions, and would not dream of addressing old ignorance.
...

Even if I didn’t have an answer for this question, I think it would be important to show the hole. These holes should not be covered up, they should be put on the front page: that is the only hope of an answer. But I do have an answer. I can tell you why the kinetic energy is dependent on the square velocity.

In my paper on photon motion, I showed that the measured wavelength and the real wavelength of the photon differ by a factor of c2. This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c2. Energy, like velocity, is a relative measurement. A quantum with a certain energy has that energy only relative to us, since it has its velocity only relative to us. If the wavelength has to be multiplied by c2 in order to match it to our measurements, then the mass or mass equivalence will also. Hence the equation E = mc2. In this way, c2 is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.

The same analysis can be applied to any object. The energy of any object is determined by summing the energies of its constituent atomic and quantum particles, and all these particles also have spins. The quanta will impart this spin energy in collision, so this spin energy must be included in the total kinetic energy.

http://milesmathis.com/kinetic.html

Last edited by Cr6 on Fri Dec 19, 2014 3:35 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

23. HOW NEW TRANSFORMS in SPECIAL RELATIVITY affect MASS, MOMENTUM and ENERGY EQUATIONS

We run the experiment twice. The first time the device in the ball emits a photon toward the mirror at T' = 0s, and then receives the same photon upon its return from the mirror (it does not re-absorb the photon, it simply measures it with an instrument). T' is the time on the ball's clock. At the beginning of the experiment, just before the emission of the photons, T' = T. That is, the clocks of the zero point and the ball are synchronized.

The second time, the device emits a photon at T' = 0s, another at T' = 1s, and then another at T' = 2s. The observer at the zero-point intercepts the second and third photons from the ball in order to calculate where the ball is after T' = T = 0. This observer also intercepts the first photon returning from the mirror.
By the conservation of momentum, the ball must recoil in the opposite direction from the emission of the photon. When the photon returns, the distance the ball has traveled may be measured, and the inertial mass of the ball may be determined.

Question: will the mass of the ball as calculated from the ball be equivalent to the mass of the ball as calculated from the zero point of the experiment? If not, how will they differ?

L = distance from zero-point to mirror
E = energy of photon = 1 x 10-19J (say)
m' = mass of ball, measured by the ball
v' = velocity of ball, measured from the ball

Let us calculate from the ball, first of all. In this case, the ball is the measurer, and the system of the ball is therefore the S' system—the primed system (I make the local system the primed system simply to be consistent with my other paper). What does the ball see?

The simplest thing to do is to let the photon return all the way to the ball. We could let the photon return to the zero-point and then let a signal be triggered, but that seems redundant, since the signal would have to be a light signal.

Let the ball be very tiny, to be sure it travels a nice long distance. But do not assume it reaches velocity instantaneously (this will be important later). When the photon arrives back at the ball, the ball looks at its clock and discovers that 2.5 seconds have elapsed. The ball thinks, "This is very easy. The light took one second to get over to the mirror and one second to get back, and half a second to reach me. If the mirror is 300,000km from the zero-point, then I am 150,000km from the zero-point. I went that far in 2.5s, therefore my average velocity (relative to the system of the zero-point) is:

v'av = 150,000km/2.5s = 60,000km/s

By the conservation of momentum, the momentum of the light must be equal to the momentum of the ball:

E/c = m'v'av
m' = E/cv'av = 1 x 10-19J/(3 x 108m/s)(6 x 107m/s)
= 5.55 x 10-36kg

Now let us calculate from the zero-point. The first photon arrives at the zero-point in 2 seconds, according to the clock at the zero-point. The observer at the zero-point then must measure the distance the ball has appeared to travel. The observer does this by receiving the other photons from the ball. We could use the t-equation from my previous paper, to calculate the difference between the period of the ball and the period of the zero-point. This would be the most direct way to calculate, since the only data the zero-point is receiving from the ball is ticks. [The zero-point is able to calculate velocity simply from receiving ticks, since the zero-point knows the local period of the ball. When the ball was at rest at the zero-point, at the beginning of the experiment, it's period was 1s.] However, since we have already calculated the velocity of the ball according to the ball, I am going to skip this step and use my velocity transformation equation instead. If the ball calculates its own velocity to be 60,000km/s, then the observer at the zero point will calculate (by receiving ticks) the velocity to be:

t = t' + x'/c
t' is a given as 1s. t is incoming data. Therefore x' and v' and v may be calculated.
v =        v'
1 + (v'/c)
vav =        v'av
1 + (2vav/c)
vav =        6 x 104km/s
1 + (12 x 104/3 x 105)
= 42,857km/s

(Again, the zero-point could have arrived at this number without knowing v'av. This is of some importance below.)
m = E/cvav
= 7.77 x 10-36kg

The mass of the ball has appeared to increase, if measured from the zero-point, as compared to measurement from the ball itself. This much is consistent with the findings of Einstein: mass appears to increase as time dilates. But the transformation term is obviously different. I have used a variation of my velocity term alpha rather than gamma.

Now, one may ask, which mass is correct? The mass measured from the zero-point or the mass measured from the ball? Either mass conserves momentum, as long as we keep it in its own equation. But you can see that the mass as calculated by the ball itself must be the correct moving mass, since it is connected to the correct velocity. The zero-point calculates a larger mass only because it has used an incorrect velocity. Its visual data has been skewed by time dilation, making the velocity wrong and then the mass.

Next, one may ask, what was the rest mass in this problem? Well, there must be three calculable rest masses: the rest mass before the emission and two rest masses after (the ball and the zero-point will calculate different rest masses, unfortunately).

mrB = rest mass of the ball, before emission
mrAB = rest mass of the ball calculated by the ball, after emission
mrAZ = rest mass of the ball calculated by the zero-point, after emission
m0B = mass equivalence of the photon, as measured by the ball
m0Z = mass equivalence of the photon, as measured by the zero-point

The photon will have two mass equivalents, since the photon will have a different energy relative to the ball than it will have relative to the zero-point. The ball is moving away from the photon when the photon returns, so that its energy will be redshifted. E'< E.
E = m0Zc2 = the energy of the photon relative to the zero-point
E' = m0Bc2 = the energy of the photon relative to the ball
mrAB= mrB - m0B

We can solve since we also know that mvav = m'vav' = E/c
m = m0Zc/vav
m0Z = m(vav/c)
m0Z = (m - m')/2
= 1.11 x 10-36kg
which agrees with our given value for its energy.
m0Z = αm0B
m0B = 7.9 x 10-37kg

Now all we need is the rest mass. Some will think that is just the mass measured by the ball, since that is the only mass that is truly at rest with regard to its background. But the ball, using the equation above, is calculating with redshifted light. This means that its value for the mass equivalence of the photon is incorrect. In this way my thought problem is not like that of Einstein. In the light planes problem, the body is at rest and the observer is moving away. Therefore the body measures the normal frequency and the observer sees a redshift. But in my thought problem, the observer at the zero-point sees the normal frequency and the ball sees the redshift.

http://milesmathis.com/emc.html

Cr6

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## Re: The Photon - The Mathis' Definition

31. Lorentz Violations

by Miles Mathis

Jeffrey Goldstone is perhaps the ultimate example of this. He invented something called the Goldstone boson. A boson is generally any new particle that fills any old hole, and you get to name the boson for yourself, if you find a proper hole. The Goldstone boson is basically a phonon, and a phonon is a photon that is moving too slow, for reasons unknown in this case. What Goldstone did, among other things, is use the holes in Einstein's equations to build himself a little theory. When an experiment is spitting out less energy than Einstein's equations are predicting, Goldstone tells us this is because phonons are involved instead of photons. So Einstein's equations aren't failing; Einstein just didn't know about the phonon. Of course Goldstone does't say it in this straightforward way. No, we are told that the experimental failure is caused by a "spontaneously broken symmetry". This broken symmetry is used to explain many of the current gaps in physics, and it turns up in gravity models, electroweak models, magnetic models, and so on. Anywhere that the equations aren't working, the events are said to be in a state of broken symmetry. Since the physicist just made up the idea out of nothing as a stop-gap, he has no mechanism for it, hence the added adjective "spontaneous". If something is spontaneous, you don't have to offer a mechanical cause for it, which is convenient. An awful lot of new concepts in physics seem to be spontaneous, for some reason.

I can only touch on the absurdities here, since I cannot waste time addressing all the meaningless new fields, particles, and maths. But I will offer you this as an example of how the theory of violation tends to spin out: Wikipedia tells us that, "The phonons travel at LESS than the speed of light. In general relativistic theories, this leads to a massive graviton (note that this is different from massive gravity, which is Lorentz covariant) which travels at less than the speed of light (because the graviton devours the phonon)".* Lovely, isn't it? The first lie (a slow photon) requires an even bigger lie (a massive graviton), and the second lie is a double lie, since here we have the tweeking of a particle that doesn't exist. The phonon, which doesn't exist here and is only an outcome of a bad equation, requires the recalibration of the graviton—a particle that was manufactured to fill another hole caused by a different bad equation. As if that isn't enough, we have to imagine one fake particle devouring the other; and this meal is spontaneous, we suppose, since there is no possible mechanism for it. The phonon is devoured only because we don't want to have to look at it any more after the experiment.

http://milesmathis.com/phon.html

Last edited by Cr6 on Fri Dec 19, 2014 3:36 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

32. A Critique of General Relativity

Just think of the operator of the merry-go-round. If he were standing in the very center, and if he were not rotating, then he would not see the ball thrown off the ride as curved. He would see like the parents, not like the child.

Even from the point of view of the child on the merry-go-round, I do not see the need for tensors. Einstein says that Euclidean math cannot be used since no Euclidean space can be found in a gravitational field from which to measure the curvature. There is no orthogonal space underlying the curved space. But this is simply false. It is false since curvature necessarily implies a relation to a non-curved background. The meaning of curvature is absolutely dependent upon the existence of a straight line. If the child on the merry-go-round sees a ball thrown off the ride—or calculates a photon emitted—to travel a curved path, he must see it with regard to a Euclidean background. That background is in fact the "space" of the parent off the ride. If the space of the parent were also curved, then the child could not see any curvature. Motion, whether linear or curved, is always defined relative to a background. A curve in curved space is not a curve.

As proof of this, we know that starlight is refracted only by referring it to some Euclidean system. It is curved relative to the background we apply to it in order to see how curved it is. If the ray is refracted some measurable quantity, we know this quantity only by referring it to the path the light would have taken if the field of the sun were not present. This reference field of measurement is precisely the Euclidean field I am talking about. Whether it exists in reality is not pertinent. It can be made to exist as a concept, and indeed it must be in order to measure the degree of aberration of starlight. This field is the field we should choose to do our math in, since it is a Euclidean field and requires no esoteric tensors or slipshod concepts. In fact, the curvature of the light defines the field at that radius. Since a gravitational field is not dependent upon the mass of the test particle, a photon will do as well as any other test mass. But again, the curvature or strength of the field could never be measured without referring it to the path a light ray would have taken without the gravitational tug. That is, a straight light path. This straight light path is precisely the orthogonal field we require: it is the Euclidean space underlying every non-Euclidean field.

Cr6

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## Re: The Photon - The Mathis' Definition

43. A Mathematical Explanation of the Orbital Distance of Mercury

But I have shown in many other papers that the gravitational field is actually a compound field made up of two separate vectors. One of these vectors I continue to call gravity, since it is an apparent attraction. In the Unified Field, this vector points in. I call it solo gravity, since it is the gravitational field without the second field. The second field is the foundational E/M field or charge field. This is the field that underlies both electricity and magnetism, and mechanically it causes the charge between proton and electron. I have shown that although the electrical field may appear to be either negative or positive in interactions, the foundational E/M field is always positive or repulsive. It is caused by simple bombardment, via a particle I have dubbed the B-photon. This B-photon replaces the virtual photon or messenger photon of the standard model.

Since the foundational E/M field is always repulsive, it acts in vector opposition to the solo gravity field. Both fields are active at all levels of size, cosmic and quantum.

I have shown how these fields are both found in Newton’s gravitational equation. I have not added any terms to Newton’s equation: it still stands as it always has. I have simply broken it down into its constituent fields, showing how G works in the equation as a transform between the solo gravity field and the foundational E/M field.

Once the two fields are separated, the solo gravity field turns out to be determined by radius alone. Gravity is an acceleration and nothing more. To be rigorous, it is the acceleration of a length or differential. This means it has nothing to do with density. Density is a part of Newton’s equation, since it is a part of the mass variables, but it turns out that the density buried in the equation is actually a density of the foundational E/M field. It is a density of B-photons. All you have to do is write each mass in Newton’s equation as density times volume, giving the density to the E/M field and the volume to the gravity field. G then acts as the transform between the two fields. I have shown that G is also the relative diameter of the B-photon. In other words, the B-photon is 6.67 x 10-11 times smaller than the hydrogen atom.

Those who have found my theory and math in these other papers to be a bit abstract may be interested to see the field applied to a standing problem of the solar system. I will do that now. I will show that the orbital distance of Mercury is the distance of balance for the three vectors involved.

If, as I have claimed, the solo gravity field is determined by radius alone, we should be able to find an acceleration at the distance of Mercury quite easily, and it should not be the same as the current figure. In another paper I have shown that the solo gravity field of the Earth is not 9.81, but 9.82, and that the foundational E/M field of the Earth is -.009545. If these figures are correct, I should be able to use these figures along with the known parameters of the Sun to find its separate fields. If these two fields do balance at the distance of Mercury, I think I will have gone a long way to convincing skeptics, since only the correct mechanics could hope to solve the problem in this way.

The Sun’s radius is 109 times that of the Earth. If gravity is dependent upon radius alone, then the Sun’s gravitational acceleration at its surface must be 109 times that of the Earth, not 28 times as the standard model tells us. If so, then its surface gravity must be 1070 m/s2, not 274 m/s2. If that is true, then at the distance of Mercury, this acceleration would be:

(1070 m/s2)(696,000/5.8 x 107)2 = .154 m/s2
The standard model currently believes the acceleration at the distance of Mercury is:

Orbital velocity of Mercury = 48,000 m/s
a = v2/r
a = (48,000 m/s)2/(5.8 x 1010m)
a = .04 m/s2

The dirty little secret here is that the standard model believes the surface gravity of the Sun is 274 m/s2 simply because they did this math backwards. NASA has never measured the surface gravity directly, of course. That number 274 is arrived at by working backward from the orbital velocity of the planets. It is not a measured number, it is a calculated number. And it is calculated by assuming that gravity balances the orbital velocity directly. The standard model has never proved that gravity does this, it simply calculates the number required to balance the orbital velocity and calls that number gravity. There is no mechanics involved.

But in my theory, all the numbers are grounded by field mechanics. If gravity is dependent on radius alone, we have a logical field from the beginning. Gravity is an acceleration in the equations, and now it is an acceleration in the field as well. It is an acceleration and nothing else. It does not include density, so it has no mystery to it anymore. Where Newton and the standard model have much to explain, I have nothing to explain. My solo gravity field has nothing to do with mass or density, so most of the historical questions immediately evaporate.

Can my mechanics explain the balance better than the standard model? You be the judge. I have shown that the gravitational strength at the distance of Mercury should be .154 m/s2 and that Mercury’s orbital velocity tends to give it an escaping force of .04 m/s2. Therefore, the E/M field must make up the difference.

That is, I must find that
.154 m/s2 + E = .04 m/s2
E = -.114 m/s2
I must show that, at the distance of Mercury, the Sun is repulsing at that strength. We can do that in two ways. One, use that number to show what the Sun must be repulsing on its surface:

ES (696,000/5.8 x 107)2 = -.114 m/s2
ES = -792 m/s2
Or, go to the surface directly, and figure from the surface numbers:

1070 m/s2 + ES = 274 m/s2
ES = -796 m/s2

To balance the orbit of Mercury we must find that value for ES. Is it possible that this value confirms my value for the Earth of -.009545 m/s2? I found that value for the Earth simply by looking at the parameters of the Earth and Moon, as you can see by returning to that paper. That paper was written several years ago, and when I wrote it I had no idea it would be useful when looking at Mercury or the Sun. Many will have thought I pulled that number out of a hat, by some kind of magic (although the math is quite simple); but if I can use it to balance the orbit of Mercury, many people will be eating their own hats.
Watch this:

I showed that the foundational E/M field of the Earth has a strength at the surface of the Earth of -.009545 m/s2. The Sun has a mass of 333,000 Earths. It has a density ¼ that of the Earth.
(-.009545 m/s2)(333,000)(¼) = -795 m/s2

No magic there, just transparent math and fully defined field mechanics. The Sun is “attracting” Mercury with a straight acceleration of .154 m/s2. It is repulsing Mercury by bombarding it with B-photons at a rate of -.114 m/s2. Mercury is fleeing the Sun at a rate of .04 m/s2. These three vectors balance at the orbital distance of 58 million kilometers. That is the mechanical reason Mercury is at that distance and at no other distance.

http://milesmathis.com/orbit.html

Cr6

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## Re: The Photon - The Mathis' Definition

47. The Ether

by Miles Mathis

Some will say that air facilitates the motion of sound, but this is imprecise as well. The air does not facilitate the motion of sound, the air is in motion itself. Sound is the motion of the air, so nothing is facilitated. There is no medium. There is only the motion of the air. The waveform is only a shape made by the field of air molecules, and sound is this waveform. Talking of a medium implies that the sound is a material thing on its own, without the air, and that the air is a facilitator or medium. But that is not the case. There is not a primary particle with a primary motion and a secondary field of particles with a facilitating motion; there is only one field of particles and one field of motion, so that all talk of a medium is imprecise.

This is why Maxwell's ether has been disproved, and remains disproved. He and the other theorists of the time assumed that light, as an electromagnetic wave, required a medium for transmission. They assumed that light was an analogue to sound. But light is not an analogue to sound. The wave motion physically belongs to the photon. To put it another way, the waveform is created by the motion of a single particle, not by a field of particles. Light is made up of many photons, but each photon moves as a wave. This is not true of air or water waves, where each molecule moves up and down: only the field of molecules creates a waveform.

http://milesmathis.com/ether.html

Cr6

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## Re: The Photon - The Mathis' Definition

[size=32] 47b. Tesla and Einstein Were Both Right[/size]

By Miles Mathis

Tesla was usually not too concerned with theoretical questions like this, but as far as the question interested him, he agreed with Einstein. Tesla was not a supporter of Maxwell’s ether. Tesla found Maxwell to be ham-handed in many ways, and said so. The ether that Tesla believed in was an ether created by the E/M field. In fact, Tesla’s ether has much in common with my foundational E/M field, a real bombarding field emitted by all quanta and all objects. He stated that this field diminished with the square of the distance from Earth (or any spherical object), and my foundational E/M field does this (minus time differentials). He stated that this field combined with the gravitational field, and was often more powerful than it. I have shown this in my Cavendish paper and many other papers.

Einstein was a theorist, not an experimenter like Tesla. He did not know of the foundational E/M field. Almost no one except Tesla has known of it, even among other specialists in electricity. The field I am talking about has concerned only quantum physicists up to now (since it is the field mediated by their ridiculous “messenger photons”). So Einstein could not be expected to have included this field in his theories of the macro-world. But he never denied the existence or importance of the electromagnetic field, and he would never have denied the possibility that other unknown fields existed, even ubiquitous and powerful fields. He would only have denied, based on his theory, that they would be considered the “background of space.” As he showed, space has no background except the motion of light.

He is correct about this, and it is one of two major reasons I refuse to call my foundational E/M field an ether. The other reason is also historical. Tesla called his field an ether, since it was ubiquitous and powerful. It allowed many things to happen, and caused many things to happen. It was fundamental, as fundamental as gravity, or moreso. While admitting all that, I refrain from referring to my foundational E/M field as an ether because it does not fit the even older and more famous definition of ether as the mediator and facilitator of all motion. According to the 19th definitions, the ether was invented to explain the motion of light. It seemed to physicists at that time that light needed a medium through which to propagate, just as sound needed air through which to propagate. Especially as regards the wave motion, it was not understood how light could show this wave without a medium.

Using stacked spins, I have shown how light moves in a wave pattern without the need of any medium. The wave is internal to each photon, and the analogy to sound waves in air completely breaks down. The wave motion of light is not a pattern in a medium, it is real motion of each quantum. You will say, “Motion relative to what?” Motion relative to the previous position, or relative to the void, or relative to a graph you superimpose over the moving quantum. Motion does not require a medium, it only requires a background. That background is automatically created relative to previous positions. You don’t need a medium to describe the motion of quanta. You only need a mathematical or diagrammed background, and previous positions give you that.

In fact, requiring a physical medium for all motion is a reductio ad absurdum. Say that we do define Tesla’s “ether” as the medium. Say that we do define my foundational E/M field as the medium against which the speed of light is calculated. We obviously run into an immediate problem, since my field or Tesla’s is made up of some kind of photon or other emission, fluid or particulate. At that point, you are defining the motion of light against a background of invisible E/M photons. But that brings up many questions: 1) Which photon is more fundamental? The light photon or the photon that transmits the E/M field? 2) How can you measure one against the other? Aren’t they both going c? Or, if they are not going exactly c in all situations, won’t they both vary in the same way for the same reasons? 3) If the light photon is moving relative to the E/M or ether photon, what is the ether photon moving relative to? Don’t we require a sub-ether as a background to the ether photon? 4) It seems we need something that is not moving to be our medium, but Tesla’s ether field, like my foundational E/M field, is made up of moving particles.

http://milesmathis.com/tesla.html

Cr6

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## Re: The Photon - The Mathis' Definition

Cr6, I hope you can chat with us Saturday at 3 pm your time. Let me know, please, if that's a problem. The other 3 guys said it's a good time for them. I plan to write up a list of questions for Michael and invite everyone else to add questions to the list. Maybe Michael will have time to prepare answers before the chat or answer directly during the chat. The chat room link is http://us20.chatzy.com/27985249236108

LloydK

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## Re: The Photon - The Mathis' Definition

LloydK wrote:Cr6, I hope you can chat with us Saturday at 3 pm your time. Let me know, please, if that's a problem. The other 3 guys said it's a good time for them. I plan to write up a list of questions for Michael and invite everyone else to add questions to the list. Maybe Michael will have time to prepare answers before the chat or answer directly during the chat. The chat room link is http://us20.chatzy.com/27985249236108
Hi Lolyd,
That should work for me.

Cr6

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## Re: The Photon - The Mathis' Definition

61. A CORRECTION TO THE EQUATION a = v2/r
......

What this means is that our current equations are just straightforward heuristics. We use them all simply due to the fact that it is easy for us to measure 2πr/t. That is our basic data, data that we like and always have liked, and whether 2πr/t is a velocity, an acceleration, or none of the above, has never really mattered to us. Newton was trying to develop an equation that contained this data, and he did that. He developed a true equation that relates a and 2πr/t. Unfortunately, he labeled 2πr/t as the orbital velocity, and it is not the orbital velocity. Nor is it the orbital acceleration. It is just a relation of two variables and a constant. Because Newton derived a true equation, it has not mattered much (in most situations) that his variable assignments were sloppy. As long as we remember that the v variable in the equation a = v2/r is equal to 2πr/t, we cannot go wrong. But we have not always remembered this, as I will show with Bohr and quantum mechanics.

I have shown that the circle describes not a velocity, but an orbital acceleration. This acceleration is the vector addition of the tangential velocity and the centripetal acceleration.

To find it we use the equation a = aorb2/2r. Using this equation, we find that
(2πr/t)2/r = aorb2/2r
aorb = 2(√2)πr/t
a + r = √(vo2 + r2)
a2 + 2ar = vo2
vo = a√[1 + (2r/a)]

This is another very useful new equation for tangential velocity. It will allow us to calculate velocities and energies that have so far eluded us, such as the energy of a photon emitted by an electron in orbit.

The answer to the question, "which of my equations must replace the current one?" is therefore the first one:

a = √ vo2 + r2 ) -  r.

If we want an equation that relates the velocity of an orbiting object to its centripetal acceleration, we must use this equation, since it is the only equation with a true velocity variable in it.    This works out in other ways, too, since the proportions of the variables in that equation remain the same as the historical equation, while they don't in the equation a = vorb2/2r .

Which brings up another problem. You can see that physics has never had a way to measure tangential velocity.  "Orbital velocity" can easily be measured by circumference per time of one revolution. But tangential velocity must be calculated.  You can calculate using my new equations, but before this paper there was no equation from orbital to tangential velocity.  Neither Feynman nor any of the textbooks were even clear on the difference.  I assume this means that no one was clear on the difference. [Notice that Newton also could not calculate from one to the other, since according to lemma VII they were the same thing.]

http://milesmathis.com/avr.html

Cr6

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Join date : 2014-08-09

## Re: The Photon - The Mathis' Definition

66. Superposition

Update, 2013. I figured it out myself the next time I re-read this paper. To understand why the photon's second spin spins around a point on the original spin surface, we just have to look at the cause of that second spin. I have shown previously it must be caused by collison with another photon. The first photon stacks a second spin on top of the first because it cannot spin any faster on the first axis. It has reached a spin velocity of c, and if it encounters a positive spin collision that would increase its spin energy, it can stack that extra energy on only by creating another spin. Well, since the point of collision is on the outer surface, the photon naturally spins about that point. The second spin must take as its new center that point of collision.

http://milesmathis.com/super.html

67.  SUPERPOSITION AGAIN

I will be told that it is the wave that is polarized, not the particle, but that is just dodging the begged question one more time. Neither the old quantum mechanics nor any of the updates ever bother to tell us how point particles with no radius can create waves, or move in a wave motion. My mechanics explains it, but my mechanics requires a photon with a radius, and with several stacked spins. Without them, mainstream physicists can only rush by this basic question. I have already told you why they do this in about a hundred papers: they are hiding behind the math. If they bring the mechanics back to the front, and let you see all these existing questions in a full light, their famous math begins to melt. Ron Garret calls the squared amplitude in the wave equation a hack, but all the math is hacked from top to bottom, as I have shown.

Again, the thing to take from this addendum is that polarization and superposition are both proof of real photon spin. To create quantum erasers and things like that, each individual photon must have a wavelength. I repeat, not just the wave front, or the wave packet, but each individual photon. This must mean that the polarizers are working upon individual photons, not on wave fronts or fields of photons. And for that to be possible, each photon must have a radius. A photon with no mass and no radius is undifferentiable. In other words, there is no way for a polarizer or other detector or filter to know one photon from another. You cannot tell one point from another. And this means that photons must have mass and radius. And this means that the math of QED, as we know it, comes tumbling down. Ron Garret thinks he deserves a Nobel Prize for noticing that entanglement is a measurement, but he fails to notice that QED needs more than a tweek. It needs a complete overhaul, from the baseboards up. We have to throw out all the math and all the theory and start over from the beginning.

http://milesmathis.com/super2.html

70.  Planck's Constant and Quantization

The standard model states that matter has a dual nature, showing characteristics of a particle in some experiments and characteristics of a wave in others. But it does not tell us how matter expresses these characteristics. It assumes that it is impossible to do so. But it is not impossible. In fact, it is embarrassingly simple. Matter is NOT both particle and wave. Matter is a particle with a wave. That is, the photon, like the electron and the proton and every other incorporation of matter, is a discrete particle, with a radius, a surface area, a mass, and a spin. Fundamentally, it is not a wave. It is a physical sphere, just as much as the Earth or the Moon or a marble or a bowling ball.

So all those smug little PhDs on the forums are wrong. To solve the problem of duality, we must not think of the photon or electron as a probability. We must think of it as a physical sphere. We must do so because doing so is the only way to solve the mysteries of quantum mechanics. I have already proved this in the most direct fashion possible: by solving problems that had stymied them for generations. I have solved problems Bohr could not solve, that Einstein could not solve, that Feynman could not solve, and I have done it by ignoring all the warnings from the physics police—ignoring the warnings of the cult groupies.

They have said there is no way to visualize superposition, but I have done it. They have said there is no way to illustrate wave motions or characteristics, but I have done it. They have said there is no way to explain duality in a physical, mechanical way, but I have done it.

I have done it by showing that a simple sphere, of any size, can have four stacked spins. The first spin is an axial spin, with a size of R (where R is the radius). The second spin cannot be axial, due to gyroscopic rules. It must be outside the influence of the axial spin. Therefore it must be an end-over-end spin, with a size of 2R.

If we give the particle a straight-line motion in any direction, the end-over-end spin will give it a wave. Just monitor the position of the center of the particle, relative to its initial position. On any graph, the motion must create a wave.

I call this first end-over-end spin the x-spin (the axial spin is not an x-spin, since an axial-spinning particle is not moving in the x,y,z field yet). Then we have a y-spin of 4R and a z-spin of 8R. Each spin is orthogonal to inner spins. And each spin creates another wave.

The z-spin creates the electrical field and the y-spin creates the magnetic field. These two outer spins are orthogonal, and the right hand rule applies to them. We achieve all this simply from a mechanical analysis of spin, without any other postulates or theory.

This shows the exact mechanical cause both of the wave and of the duality. Stacked spins cause the wave. The particle is not a wave: the particle moves as a wave against a background, due to stacked spins.

These spins may be caused simply by collisions. We may assume that particles collide all the time. If they survive collision, it is because the collision was indirect. An indirect collision would create a spin on any sphere of any size. Summed over a long period of time, these spins would tend to maximize in one direction or another. That is to say, simply as a statistical matter, we would expect all or most surviving particles to have all possible spins. This applies to photons and electrons as well as to free protons.

http://milesmathis.com/planck.htm

Cr6

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Join date : 2014-08-09

## Re: The Photon - The Mathis' Definition

71. The PLANCK RELATION and the MASS OF THE PHOTON

The Planck relation is just an equation relating the energy of a moving particle to its frequency, via the de Broglie wave. The particle does not have to be a photon; it can be any quantum, like an electron.

E=hf

Where h is Planck's constant. However, since I have shown that Planck's constant is hiding the mass of the photon, we may now unwind this equation, finding much more information buried beneath it. Using simple mechanical postulates, I have calculated that the mass of the infrared photon is h/2,400. To find this, I simply used G to scale down from the proton mass. You can also use the Dalton, 1821, to find this same mass. Simply cube the Dalton and invert it. The Dalton is an outcome of spin mechanics, and the photon is three levels below the proton and two levels below the electron. Using the Dalton, you will get 2,400, but using G will require you multiply by an additional 2.5. This is because G is a scale for size, not mass, and the density of the photon is not equal to the density of the proton.

...

Addendum [February 2010]: we can use the above math to discover what quantum Planck's constant is really hiding. I have said that Planck's constant is hiding the mass of the photon, but since photons have different energies, we may ask which photon it is hiding. Well, we can see from above that h = mc?. So we just make c? = 1, and we will have h = m, you see. The photon that has a wavelength of 1/c is a photon with a wavelength of 3.3 x 10-9m, which is an X-ray. Since current physics is using h as the quantum of action, they must be using the X-ray as a quantum. This is not logical. It is much more logical to use the charge photon as a quantum. We should be using the infrared photon as the field quantum, not the X-ray.

http://milesmathis.com/planck2.html

76. Bohr's First Big Mistake

In my paper on Newton linked above, I predicted a "kinetic energy meltdown" due to the form of F = ma = mv2/r. I said that an improper substitution was being begged by not properly differentiating between the orbital velocity and the tangential velocity. The term mv2/r is a strictly rotational term, but there is nothing but subtle conceptual theory to keep someone from applying it to a translational kinetic energy situation. Bohr has only just avoided this catastrophe, since his electron really is in circular motion. But what if an electron in orbit ejected a photon, and that photon was assumed to be ejected from the tangent? The photon would now be in linear motion, not circular motion. What is its kinetic energy? You can see the problem.

Furthermore, there is actually not even subtle conceptual theory to keep anyone from making this improper substitution, since current theory does not theoretically disallow it. Only my theory disallows it. The subtle conceptual theory is so far mine alone.

Current theory believes that the velocity variable in a = v2/r is an instantaneous tangential velocity. That would make it the same as a translational velocity. That is why current theory uses the same variables for both. And that is why the proofs of Bohr followed the form they did and why they have never been corrected. That is why all modern physics textbooks conflate tangential velocity and orbital velocity.

And that is why I repeat my prediction: this improper substitution will be found to be at the heart of some mathematical impasse in contemporary physics. If I do not root it out of QED very soon, I will be quite surprised.

Update: I have now reworked Bohr's equation completely, finding a new value for the Bohr radius. Coulomb's constant is shown to be an expression of the Bohr radius. See my paper The Bohr Magneton.

77. The Bohr Magneton

I showed that G is a scaling constant that takes us from the size of emitted photons to the size of the atom. Yes, the B-photon is G times smaller than the proton. We find it in Newton’s equation, of all places, because Newton’s equation is actually a unified field equation in disguise. It contains both the gravitational acceleration and the foundational E/M field (or charge field). Well, the same applies to Coulomb’s equation. It looks just like Newton’s equation because it is the same unified field equation in a different disguise. Newton’s equation is hiding the E/M field, and Coulomb’s equation is hiding the gravitational field. Because neither Newton nor Coulomb understood the fields under their equations, they only provided us a math that works. Their equations work because they compress the unified field into one field, and the transform between the two fields is the constant.

http://milesmathis.com/magneton.html

Cr6

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## Re: The Photon - The Mathis' Definition

78. More Problems with Bohr

The change in energy is the energy of the emitted photon, which means that the frequency ν belongs to the photon, and h is a constant that is modifying that frequency, to give us E. That is what the equation means. So why do they then assign mv to the momentum of the electron? Shouldn't they assign it to the momentum of the photon? The author says it “makes sense” to assign the momentum to the electron, but it makes no sense. It is the opposite of sensible. It is an another blatant fudge.

This becomes crystal clear if you read my papers on the photon and Planck's constant, where I show that the constant is hiding the mass of the photon. There is absolutely no reason to apply this fudged momentum to the electron and every reason to apply it to the photon.

I will be told that the energy can apply either to the electron or the photon, since what the photon takes away, the electron loses. But in that case, mv would have to apply to the change in the momentum of the electron during emission, not the momentum of the electron. In the current equations, that is not what the momentum mv stands for. We insert the whole mass of the electron into these equations. We find its entire momentum, not its change in momentum. That one mistake destroys the entire derivation.

To say it another way, since we are finding a change in the energy of the electron ΔE from one level to another, we should also be finding a change in momentum from one energy level to the other. That would be Δp, not p. Notice that Δp ≠ mv.
Since that is true, the other assignment is equally illogical. If the momentum applies to the photon, then it makes no sense to apply the length to the circumference of orbit. The photon is not orbiting, so why would we make that assignment? This derivation is pushed in every line and in every manipulation.
...

Now that I have shown you how to pop a simple unified field equation out of the fudged Bohr equations, we can move on. The first thing to remind you is that I have shown that the electron does not orbit the nucleus. The electron orbits a charge mininum “hole” at the top or bottom of the spin axis of the proton. In alpha particles, the electrons exist inside the alpha, between the stacked protons, orbiting one or other of the axis holes. This explains the first question of this paper, concerning why the electron in any orbit is stable. The current theory can't explain it and must hide it. I can explain it very simply. The charge field existing everywhere—including around the nucleus and in channels through the nucleus—is made of real charge photons. They create a charge pressure that both keeps the orbiting electron from escaping and keeps the orbit energized. The electron is feeding off the charge field all the time. The charge field is causing the orbit in the first place by creating these eddies in the charge field around the nucleus and around the baryons. For this reason the electron can maintain an orbit without any logical problems.

The next thing to answer is how the hydrogen electron can inhabit higher levels, and how these levels create absorption lines. But I will save that for the next paper, in which I recreate the Balmer and Rydberg series with far simpler math and postulates.

http://milesmathis.com/bohr2.pdf

79. Rewriting the Rydberg Formula

Now that we have uncovered the field mechanics of charge, we see that the quantization in the old equations has always applied to the quantization of the photon, not the electron. In other words, you should have noticed by now that we have quantum levels in the photon field, not in the electron orbit. Yes, the electron is “excited” by increased levels in the photon field, and the energy of the electron does change in response to this. And, yes, the energy of the electron does therefore look quantized also, since it hits some levels and not others. But all this is caused by the spin levels of the photon, not the spin levels of the electron. The photon is quantized in this problem, but the electron is not. We aren't studying the electron spins here (although it has them, see my levels above). We have been able to solve by totally ignoring the spin levels of the electron. The electron is not gaining energy by adding spins or changing its wavelength. The photon is. The photon spin is quantized, but only the electron energy is quantized, you see. Not the same thing.

Therefore, the quantum equations have always been tracking levels in the photon. The first quantum number applies primarily to the photon. It is telling us how many stacked spins our charge photon has. The electron then increases its energy in response to bombardment by that field of photons. But (hold on to your hat) the electron does not hop up into another orbit. Since the levels apply to the photon, the electron has no orbital levels. All the levels in QED apply to the photon. QM and QED have been measuring the charge field itself all along, not the electron. This means that the electron is simply increasing its orbital energy by going faster. Since the electron was never going anywhere near c, it is able to take the energy from the photons and put them into velocity. It increases the speed of orbit.

I will be told that we know some electrons have larger orbits than others. Yes, but my nuclear diagrams explain that. Electrons with larger orbits are not quantized larger, they are simply orbiting protons that are in outer shells themselves. The nucleus itself has levels, so of course the electrons will follow those levels like the protons and alphas. This also explains why electrons in those outer levels have more complex spins. They have the spin from circling the hole, and then they have the spin from the proton circling the nuclear center. The nucleus itself spins, so the electron will have that spin, too. The electron will have the angular momentum of the level it inhabits.

http://milesmathis.com/bohr3.pdf

80a. Rewriting the Schrodinger Equation

In the end, the electron may or may not be creating an orbital wave, but it doesn't matter here because that is not where the wave in our data or in the equation is coming from. The waves in quantum wave mechanics are often coming from the photons, not the electrons. If we apply the Schr?dinger equation to solve the emission lines of Hydrogen, the waves belong to the photons. They must be, because it is the photon energy that is quantized in the first instance. This should have been clear from the Bohr formula derivation, where Bohr explicitly assigned the change in momentum of the electron to the photon. The electron velocity and energy are quantized, and the electron does have wave capabilities; but here the waves belong to the photons. The waves in the equations are field changes, and so they must go with the momentum change.

....

This confusion is what has caused the proposed smearing of the electron. The form of Schrodinger's equation makes it look like the electron is indeterminate. But since the wavefunction doesn't even apply to the electron, it isn't the electron that is fuzzy. What is fuzzy is the photon, and that is because none of the quantum equations—including Schodinger's equation—treat the charge field as real or give the charge photon any real variables.

My re-assignment of the wavefunction to the photon rather than the electron solves this problem immediately, because the photon field (charge) IS the system that causes the electron to do everything it does. All motions of the electron are caused by charge photons.

Most readers will not understand what I mean, and I know that. To comprehend my point here, they have to study my rewrite of the Bohr equations. There I show that Bohr conflated the momentum of the electron with the momentum of the photon. He fudged from p to  Δp, p being the momentum of the electron and Δp being the momentum of the photon.

But Schrodinger didn't catch that huge error, and so it infected his equations and analysis as well.

http://milesmathis.com/se.pdf

80b. Maxwell's Equations are also Unified Field Equations

The displacement field equation is also proof of my real charge field, since—as defined by Maxwell— this field is not created by electrons or any other ions. Maxwell's displacement field pre-exists any field created by ions. In fact, if we study Maxwell's use of the displacement field, we find it creates the E/M field. The displacement current has dimensions of density, just as my charge field has, and if we go to Wikipedia, we find this:

The displacement current has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

What does that mean? It means it exists whether or not you have any ions in the field. It is a sub-field to the Electromagnetic Field, and is not equivalent to it. As a matter of straight mechanics, it is the displacement field that creates Electromagnetism, not the reverse. Again, this matches my definition of the charge field. It is the charge field that is primary, and the E/M field that is secondary. Charge is photons, E/M is ions. The photons drive the ions, so they are the fundamental field.

So we already see that Maxwell's displacement field is simply another name for my charge field. And this explains—in yet another way—why the charge field has been undercover for about 150 years. In Maxwell's equations, it has been the electrical field and magnetic field that have gotten all the attention and fame, while the displacement field has been all but hidden. The displacement field has always been seen as little more than a mathematical manipulation, one used to push the equations in line with data. But almost no work has been done in a century and a half to explain the real workings of this displacement field. In fact, the mechanics of the displacement field is purposely hidden to this day, as we can see from this quote at Wikipedia:

There is no free charge in such a material [a bar electret], but the inherent polarization gives rise to an electric field. If the wayward student were to assume the D field were entirely determined by the free charge, he or she would immediately conclude the electric field were zero in such a material, but this is patently not true. The electric field can be properly determined by using the above relation along with other boundary conditions on the polarization density yielding the bound charges, which will, in turn, yield the electric field.

That quote is completely catholic, and no current expert would likely disagree with it. But it is false
nonetheless.

...

You can see the conflation there, and the confusion. Whether the confusion is completely in the minds of those who came after Maxwell, or whether some of the confusion was his own, is not completely clear. But that quoted sentence is true only if you read it correctly. It has not been read correctly. It is true only if by “in the same medium”, Maxwell means in the same general space. But the way it has been read historically is that light is undulations in the E/M field. Light is NOT undulations in the E/ M field. Light is its own field. Furthermore, light is not undulations in anything. The wavelength of light is not a field undulation, it is a spin wave caused by each individual photon. Each photon has a wavelength, as we now know, and the only way it can have that is if the spin of the photon causes the wave. As I have said a hundred times, light is a spin wave, not a field wave.

We have since discovered field waves in the charge field, but these waves are not photons and are not light. They are neutrinos. Neutrinos are “undulations” in the charge field. But they are not thereby undulations in the E/M field. The E/M field and the charge field are completely separate. The charge
field influences the E/M field, but they are not the same.

...

After studying Maxwell's paper closely, I can see that the original fault here is his. He has not been misunderstood or misinterpreted. There has been no misreading, there has simply been a failure to correct him. The central problem here is that he thought and proposed that light was a wave in the E/M field. So he had it upside down from the start. Since electricity and magnetism were discovered before charge and were far easier to study, Maxwell naturally took them as primary. Electromagnetism is the motion of ions, while charge is the motion of photons. Since photons are very much smaller than ions, they hadn't been studied in Maxwell's time. We still know almost nothing about them. For this reason, Maxwell took the E/M field as the foundational field, and tried to fit light into it, explaining light as a field wave in the E/M field. But this is upside down. Light does not move in the E/M field, ions move in the light field. The motions and spin of photons create everything, including ionization, magnetism, current, and so on. The charge field is the fundamental field, and the E/M field is only a creation of it.

This one reversal explains most of the confusion that still exists.

http://milesmathis.com/disp.pdf
http://milesmathis.com/disp2.pdf

Cr6

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## Re: The Photon - The Mathis' Definition

80d. Gauss' Law as a Unified Field Equation

Well, if the mechanics were other than they are, that might be so. If, for instance, the two fields were stacked, with one field existing right on top of the other, then we would expect a product. But that is not the mechanics we have. As it happens—and as we can see by simply studying the real interactions —the fields are not stacked. We do not have charge existing on top of gravity. We do not have a situation that would logically call for integration or multiplication of fields. What we have physically is a charge field that is operating inside of the gravity field. We do “integrate” the charge field into the gravity field, by one meaning of the word “integrate.” We put it in there, which is one meaning of integrate. So some would jump to the conclusion that we should “integrate” with our math as well. To integrate would imply multiplication, not division. The reason we don't is because charge is determined by the real motion of real photons, and like everything else, these photons are moving inside the gravity field.

The cleverest readers will say, “That still isn't enough to decide the question. Inserting one field into another could either give us a product or a ratio. Just because the photons are moving inside the gravity field doesn't automatically give us a ratio.” True. We need to look even closer. We need to look at HOW the photons are moving in the gravity field. In almost all situations, the photons will be moving as a summed vector against the gravity field vector. In both Newton's equation and Coulomb's equation (and in the Lagrangian and Maxwell's equation) charge is moving as a vector against the gravity vector. So, depending on the form of our math, this should either give us a differential (subtraction) or a ratio. If we are calculating single events, or single positions of individual objects, we will often use subtraction, as I have in many of my papers. If we are combining entire fields to get a flux or something like that, we will use a ratio, as here.

So why does the charge vector normally sum opposite to the gravity vector? I have explained this several times before, but I will explain it here again. Photons are actually recycled through matter. They are recycled through the bodies of protons, neutrons, electrons, mesons, and so on. In this recycling, they must go both in and out, so at various times in the cycle they are moving both in and out. This would mean they are moving both with and against gravity, depending on where you measure them. Gravity always points in toward the center of any real object, and if photons are being recycled, they are moving both in and out. But in almost all cases, what we call charge is the photons going out, not the photons going in. Why? Because the greatest results of charge normally happen near the nucleus or proton. Matter tends to take a chaotic and undirectionalized ambient charge field and focus it.

http://milesmathis.com/gauss.pdf

84. Three Problems Solved Mechanically

But very soon these quanta are bumped by other quanta, and they gain both velocity and spin in this way. In most cases, they are bumped by photons, since photons make up the charge field. Photons outnumber everything else by a very large margin, so the odds are very good that this non-spinning electron will gain its new energy from photons. But most of these photons are right-handed photons, so the particle will rebuild as a proton. It gets its spin from the field, and if the field is right-handed, the particle’s new spin will also be right-handed. This means that a normal charge field, as we know it, will turn anti-matter into matter, through this simple process. Over time, all local anti-matter will be turned into matter, and the entire local field and everything in it will become right-handed. Only the fact that the macro-field is not homogeneous, rectilinear, and unidirectional keeps the field from being completely right-handed. Because the field of the Earth is spherical, and receives emission from all directions, it maintains a fraction of anti-matter. It also receives anti-matter via the Solar Wind.
...

As I have shown with my stacked spins, you can build either a proton or an anti-proton from an electron. You do not need to make anti-matter from anti-matter or matter from matter. They are not fundamentally different. Just as you can turn a neutron into a proton, simply by reversing the z-spin, you can turn a proton into an anti-proton simply by reversing the y-spin. You can’t do this just by applying a field to a proton, since you can’t reverse inner spins without affecting outer spins; but you can do it by building the anti-proton from an electron or meson. You simply have to hit it with the right fields in the right order. But my point is, you do not have to build an anti-proton from a positron. So there is no fundamental difference between matter and anti-matter. Anti-matter is not freakish or foreign, it is just emitting photons upside-down compared to “normal” photons, and it is emitting them this way only because of the way its spins stack up.
...

Rather than treat photons as point particles, we will treat them as particles with spin. This spin will immediately provide us with a wavelength, and we will see if that wavelength can explain partial reflection. Feynman tells us that wave theory cannot explain the way our detector works, but that is only in the case that light is a field wave, like sound. If light is a particle with spin, then the wave belongs to the particle itself, and will not diminish with dimming light. In this case, each photon has a wave, and this solves the first big mystery all by itself. The wavelength does not belong to the wavefront or to a large group of photons; it belongs to each photon individually. In this way, the wave can be carried full strength to the detector, even with only one photon.

http://milesmathis.com/feyn3.html

Last edited by Cr6 on Fri Dec 19, 2014 3:31 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

85. The COMPTON EFFECT, DUALITY and the Klein-Nishina Formula

Although current theory on Compton scattering and Thomson scattering is much more filled out than current theory on Rayleigh scattering, it is still very incomplete. It is incomplete because no one has been able to say how the duality expresses itself. It is well understood that Slater's field doesn't work, since the data can't be fit to photons carried that way by field oscillations. Slater was trying to explain interference and polarization and so on with the field, and the Compton effect with the photon, but when it came to explaining how the particle interacted with the field, he was at a loss. Particle physics is still at a loss, which is why they stick with the math and dodge all mechanics.

But I can explain the mechanics. The problem is that everyone from Newton and Huygens to Maxwell and Slater and Einstein has tried to express the wave as a field wave. But the wave of light is not a field wave. The wave belongs to each photon itself, and this is what solves the problem (as I have shown in several papers1). All these theorists could not get out of the rut of thinking of light as an analogy to sound or other field waves. Because all the waves they had been taught in school had been field waves, they naturally thought light must be a field wave, too. So when it was proved by Einstein that light was not traveling via an ether, they were stumped. If there was no field, how could there be a wave? No one has gotten past that apparent dilemma.

The dilemma is a false one, though, since it never required a field to show a wave. Any spinning particle can show a wave, and you do not need a field to create spin. You need collisions, yes, and a field of particles to create these collisions. But once we attach the wave to the spin of the individual particle, we do not need a field to express it. In other words, the field causes the spins on the particles by collision, but the field does not transmit the wave or carry it. The wave is not a shape on a background of particles, as with sound. The wave is the spin of each particle, and is carried by each particle.

To clarify, I will give you an example. In a football stadium, you can find two different types of waves, both caused by people. In the stands, you will see a wave caused by groups of people holding their hands up and then dropping them. That is a field wave, like a sound wave. But in the aisles, you will find individual people creating waves just by walking. Their legs create wavelengths, since each step creates a gap. If you mapped a person walking by following the gap between the legs, you would get a wave. This second example is the analogy to a photon. A photon doesn't have legs, it has spin. If we map the spin over time, this spin will create a wave. The photon has a local wavelength that is determined by the radius of this spin, and that local wavelength is stretched out by its linear motion. That stretched-out wavelength is the one we see and measure.

This immediately explains many things. It explains why individual photons can carry a wavelength, as in the two-slit experiment2. It explains superposition4, since individual photons can stack spins. It explains simultaneous longitudinal and transverse waves, since, again, the photon can stack spins. The electron3 can also stack spins. And it explains the Compton effect because we now have a way to connect the photon to the field. The photon is not carried by the field; the photon is the field. The electromagnetic waves are carried by the photons, not the reverse. The charge field is a field of photons to start with, and the photons tracked by Compton devices and other devices are traveling in a field of other photons. There are photons in the field before the tracked photons are emitted, and that explains all the field mechanics in a direct way.

Particle physicists will say that this doesn't conserve energy, but it does, since quanta larger than photons can recycle photons. We already know that quanta can absorb or emit photons. What we have not understood is that they are absorbing and emitting constantly, and that Compton effects and other effects are just emissions and absorptions above this baseline recycling of the charge field. Even electrons in stable orbits are emitting the charge field. Every existing spinning quantum is recycling the charge field all the time.

This mechanism of spin also explains inverse Compton scattering, since we only have to turn our photons upside down to explain it. Spins are reversed just by a pole reversal. Photons can be spinning CW or CCW relative to electrons, and in one case the angular momenta will add in collision and in the other it will subtract. A subtraction will increase the wavelength of the photon, and an addition will decrease the wavelength.

...

Conclusion: I have shown that my new radius for the electron fits the Klein-Nishina formula like a hand in a glove. If my new number hadn't been very close to correct, we would not have been able to simply move the square constant from the numerator to the denominator.

1 Mathis, Miles. Plank's Constant and Qunatization. 2008.
2 Mathis, Miles. The Double-Slit Experiment. 2008.
3 Mathis, Miles. Unifying the Proton and the Electron. 2008.
4 Mathis, Miles. Superposition. 2005.
5 Mathis, Miles. Hidden Variables. 2008.
6 Mathis, Miles. A Disproof of Asymptotic Freedom. 2008.
7 Mathis, Miles. Unifying the Mesons. 2008.
8 Mathis, Miles. The Bohr Magneton. 2008.
9 Mathis, Miles. The Fine-Structure Constant. 2009.
10 Mathis, Miles. Angular Momentum. 2008.
11 Mathis, Miles. Bye-Bye Pi. 2009.
12 Mathis, Miles. A Refutation of Gamma. 2001.
13 Mathis, Miles. The Mysterious Muon. 2009.
14 Mathis, Miles. The Pound-Rebka Experiment. 2009.
15 Mathis, Miles. Unifying the Photon. 2009.

15 Mathis, Miles. Unifying the Photon. 2009.

http://milesmathis.com/comp.html

Cr6

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## Re: The Photon - The Mathis' Definition

91. BROWNIAN MOTION

But because modern physicists don't understand that, they continue to try to explain all physical phenomena without a charge field. They can't fit the charge field into their equations (a second time), so they are forced to pretend that the charge field is a sort of virtual field. This is precisely why we are taught that charge is mediated by virtual photons. They can't give the field any real presence, because they can't figure out how to fit the field into their equations. But since I have shown that the charge field is already in the old equations of Newton and Einstein, that problem is solved. We can now give the charge field a real presence. We can also give the photon mass and radius and spin without destroying all the old equations. We don't have to unify anything, since it was already unified.

This being so, we can also admit that the photons we know about are capable of forces. We already know that from the photoelectric effect and a lot of other experiments, but modern physicists like to teach their students that photons are real sometimes and not real at other times. So we are taught about the photoelectric effect, and are told that real photons are knocking things about. Then the next week we are taught that all quantum numbers and equations are mathematical only, so that quanta don't have real spin or size or position. And the next week we are taught that charge is mediated by magic, either by little plus signs or by virtual photons that can "tell" electrons to move away or move closer. No one ever seems to see this inconsistency of all this. Modern physicists are right about the photoelectric effect, but they are wrong about the others. The photoelectric effect requires that the photon has position, mass, and size. The charge field requires a mediating particle that also has position, mass, and size. And once we give the charge photon mass and size, it is also capable of creating Brownian motion, from the primary level.

All these tiny particles are knocked about by photons. Everything is awash in a sea of photons, and the photons have real momenta. Because they are coming from all directions, they create random motion. And because they are so tiny, they only create a small displacement with each individual hit.

As with my explanation of heat, you will say that I have only taken the explanation back a step. But once again, that isn't true. It isn't true because photons don't experience Brownian motion themselves. Photons (not in a laser) may have a random trajectory, but they do not wiggle about or zigzag randomly. And light can be lasered, which means it can be made to travel a straight line of choice, without wiggling. So I have not regressed in an infinite line of causes, I have found a first cause. The Brownian motion of dust mites is caused by the Brownian motion of molecules; the Brownian motion of molecules is caused by the Brownian motion of atoms or ions, but the Brownian motion of atoms is NOT caused by the Brownian motion of photons. The Brownian motion of atoms is caused by simply by collision with photons.

We can see that photons are not just another step in an infinite regression by looking at velocity. All these larger particles can be slowed down, which is why they experience Brownian motion. Because they do not have a high linear velocity, and because they have a considerable mass, they can move in a Brownian manner. But light cannot be slowed, it can only be deflected. And because the photon is so small, it can move long distances without deflection, even in air. Light can therefore avoid random motion most of the time. Only in certain (and odd) circumstances would light experience something like Brownian motion, and then only if we could slow down time and shrink space a lot. If we pass light through a dense substance, and slow down time, we might get something like Brownian motion. But even so, it is not this sort of motion of light that causes atoms to vibrate. Atoms move randomly due to a bumping by photons, and it is not necessary that the photons be moving randomly themselves.

http://milesmathis.com/brown.html

92. SUPERCONDUCTIVITY

This solves the superconductivity problem because conductivity is defined as the ability of a substance to let charge pass. Obviously, charge will pass most easily when it is blocked the least, and it is blocked the least when particles aren't getting in the way. In other words, charge photons will pass through still matter more easily than they will pass through vibrating matter. A lack of conductivity is explained by photons colliding with matter, and energetic matter will collide with more photons.

We must also remember that in normal circumstances, the field of charge photons is recycled by all matter. It is recycled via spin. Each particle is spinning, and this spin pulls in photons at the poles and spits them out at the equator. But when heat approaches absolute zero, motions slow down near a stop. When motions slow down, collisions decrease, and when collisions decrease, the spins cannot be maintained. The baryons and electrons slow their spins, and nearly stop recycling the charge field. Since the photons are not being sucked in, they are free to pass. The vortices around all particles are diminished, and the field has less resistance. The substance minimizes its collisions, and the charge field therefore maximizes its efficiency. If the charge field is carrying ions of its own, these ions will pass through the substance with minimal collision.

http://milesmathis.com/conduct.html

93a. WHAT IS HEAT?

What do I mean by that? Let's look at the "neutrality" of molecules and atoms more closely. Are they really neutral or uncharged? No. We know that ions are charged particles, which, according to my theory, means they are recycling the charge field directly: they are taking it in and emitting it, with little or no blockage. So they can transfer their heat or motion to other particles via the charge photons. The charge photons carry energy across space from one particle to the other. But in more complex groups like atoms and molecules, the charge field is not recycled in this way. The charge photons are captured by spins, but then they knock about internally, blocked by electrons or closely neighboring baryons. The charge field causes internal motion or heat or energy, but it is not re-emitted directly. It is either trapped, like with a neutron, going back on itself and creating zero energy pockets, or it is spit out in directionalized streams, between particles.

What this means is that atoms and molecules ARE charged by the field, but they are mostly internally charged. By this I mean they cannot transmit this charge energy by sending out photons, since the photons are blocked. They can transmit this energy only by touch: by collision. You don't have to collide with an ion to feel its energy, since the ion can transmit its energy via the charge photons it is emitting. But you do have to collide with a molecule or an atom to feel its heat or charge.

http://milesmathis.com/heat.html

Last edited by Cr6 on Fri Dec 19, 2014 3:16 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

141. WHY DOES HOT AIR RISE?

According to statistical mechanics, where air has greater kinetic energy it has a greater probability to occupy a higher gravitational potential than less energetic air. That is, the system tends toward maximal entropy.

But that is another report. The statistics are based on observation, and explaining gravity with entropy is a dodge. We don't want an airy statistical mechanics, we want an old-fashioned mechanics that is capable of showing a physical mechanism.

Once again, this question cannot be answered with current theory, which is why the internet and textbooks are full of bad answers. We need the charge field of the Earth to answer it. The charge field is an emission field of real photons, and their (summed) direction is straight up from the surface of the Earth, radially out from the center. This explains rising heat very quickly and simply because we now have a way to bring density into the gravity equations, with simple mechanics. A gas that expands into greater volume will cover more area and encounter a larger section of the charge field. So it will encounter a larger force up, and will rise. It is that simple. An equal mass over a greater volume will rise because it is hit by more photons. This fact underlies all the partial explanations above, on the internet, and in textbooks.
Nor is this an electrical or magnetic answer, since the charge field can exist without any expression of electric or magnetic fields. If ions are present, the charge field will indeed induce electromagnetism, but the charge field is still there when no ions are present. Therefore it is no argument against me that the charge field has not been measured. No one has tried to measure it, independent of the electromagnetic field. Experimenters have so far assumed that measuring the electric or magnetic field is a measurement of the charge field, but it isn't. If the E/M field is present, then the charge field is, too. But lack of an E/M field is no proof of lack of a charge field. It is only proof of a lack of ions.

http://milesmathis.com/hot.html

94. COULOMB'S EQUATION

by Miles Mathis
milesmathis.com
email:mm@milesmathis.com

In this paper, I will continue my analysis of Coulomb’s famous equation. I will show how Coulomb’s constant is connected to the Bohr radius, mathematically and mechanically. I will also show that the current constant, though highly suggestive, is incorrect. But the fact is, the E/M field, like the gravitational field, obeys the inverse square law because the objects in the field are spheres.

Coulomb’s objects were spheres, and electrons and protons and photons are also spheres. The field emitted by the electron is spherical, and the field emitted by Coulomb’s objects is spherical. This determines all the charges and forces. The reason Coulomb’s equation is so simple is because the spherical nature of the field allows us to scale up and down using only a radius or a diameter, as I will show below. Coulomb was measuring a force between two small balls. The force was a force measured at the macro level, between macro (visible) objects.

But this force is actually a summed force caused by a field of forces. It is caused by a bombarding field. Coulomb’s balls are made up of billions of smaller balls (atoms), and the force at the atomic level is a bombardment of these smaller balls by even smaller balls (photons). To find a force on one atom, we need to know the density of the field of photons relative to the size of the atom. How many photons are impacting it each second? To sum up from the quantum field to the macro field requires we sum all these collisions. Since charge in this problem is defined by field density, we sum the field. But there is a problem. Density is a relative term. It is relative to volume, and volume is relative to radius. So a charge measured at the macro level is not the same as a charge measured at the quantum level.

http://milesmathis.com/coul.html

Last edited by Cr6 on Fri Dec 19, 2014 3:29 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

Solid Light?

No, just another bad interpretation of the Charge Field

by Miles Mathis
First published September 19, 2014

Abstract: I will analyze the recent PDF from Princeton and their mainstream report of strongly interacting photons. In doing so, I will show that quantum dimer theory, Cooper pairs, BCS theory, and RVB theory are all non-physical and fudged. I will replace them with a simple mechanical theory based on charge recycling by the nucleus, where all particles, fields, and attributes are real, and where all mathematical variables are assigned to these real particles. This will falsify polaritons and all other quasi or virtual particles, as well as all pseudo-spins and other pseudo-attributes. I will provide diagrams that show directly the cause of the Meissner Effect, and its link to superconduction. Finally, I will explain high-temperature superconduction with the same visuals, showing how it is created mechanically by charge channeling.

As you can probably see already, this explains the Meissner Effect in superconductivity, where interior magnetic lines disappear. We have never been given a simple mechanical explanation for that, but my diagram of Copper supplies it immediately. If this Copper nucleus begins superconducting, that simply means that all photons being recycled are going from pole to pole. None are being recycled out the equatorial or carousel level. As we know, the magnetic field lines are always orthogonal to the electrical field lines. Well, the electrical fields lines go with the conduction. They run south to north here. The magnetic field lines are then orthogonal to that and in a circle, by the old right hand rule. Well, since we have no photons being emitted out the equator in this case, we have no magnetic field being created. Both the electrical field and magnetic field are caused by the charge field, and the charge field is just the recycled photons. Photons that are recycled from south to north in a line create the electrical field, and photons that are recycled through the carousel level create the magnetic field. So if all charge is channeled south to north as through charge, nothing is left to create the magnetic field. It disappears. This disappearance is what we call the Meissner Effect.

...

For me to begin to make sense of any of this, I need an element to start with, so that I can introduce charge in a defined manner. Since these guys are using quantum dimer theory, I will assume they are using high temperature superconductors that use Copper as the main conductor. Dimer theory was introduced through RVB [resonating valence bond] theory, proposed by Anderson and Baskaran back in the late 1980's to fill in big holes in charge theory. What I mean by that is that they had no mechanical way to explain superconductivity, since they had no way to represent charge streams through real material. I have shown that the nucleus is channeling real photons through its interior in defined streams, and that this channeling is what we call charge. But since the mainstream has never realized that, they could only explain superconductivity by pasting complex new maths over the field. That is what this RVB theory and dimer math is. In short, superconductivity is created by valence-bonded electrons forming Cooper pairs, and these Cooper pairs greatly increase the conductivity of the substance.

http://milesmathis.com/solidlight.pdf

Last edited by Cr6 on Fri Dec 19, 2014 3:28 am; edited 1 time in total

Cr6

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## Re: The Photon - The Mathis' Definition

Explaining the Density of Osmium

Those who haven't read my previous nuclear diagramming papers may not understand what I mean by holes. Well, each disk I have diagrammed here is like a compact disk (CD), with a charge hole in the middle. I don't draw all the holes, but they are implied. The hole represents the pole of the proton. Since the proton intakes photons at its poles, the poles act as intake holes. Since each proton is spinning very fast, and since this spin tends to force most of the charge emission to the proton equator, I don't have to draw the proton as a sphere (even those it is one). Since I am diagramming the emitted the charge field of the proton, not the body of the proton itself, I can simplify each sphere into a disk with a hole in the middle. The charge field then goes in the hole and out on the disk's edges. In this way we can follow the charge channeling through the nucleus. And that is why I always have my disks meeting edge to hole.

We can also explain the oxidation states of Osmium with this diagram. Osmium has many oxidation states, but one of them is the rare +8. Osmium has this oxidation state because all eight holes are equally available for bonding. Although all eight positions—the six 4th level positions plus the two inner positions—are filled, none of them are completely filled. Each hole can take six protons, but but none has six protons. Actually, as we see, the eight positions are not quite equal. The two inner positions will not act precisely like the six outer positions. The inner positions are “more open” but harder to get to. They are more open because they are 1/3rdfull instead of ½. But they are harder to get to because they are inner positions. These two factors almost balance out, giving Osmium a potential +8 oxidation state in some instances. We will look more closely at this “balancing out” of factors in subsequent papers.

http://milesmathis.com/osmium.pdf

EVO's and the Charge Field

Since electrons can't do that even with their known charge fields, this looks like a big mystery. This is why Shoulders' research has been used as proof of the zero-point energy field and other exotic fields. But Shoulders never went there himself. He was satisfied with reporting data only.

But with my charge field, you don't need any exotic fields to explain this data. The reason the electron's charge field didn't seem adequate to explain this data was that the charge field was misdefined and misunderstood. The mainstream tells you that electrons and protons are charged, but they aren't. They are in a field that is charged. The particles are not charged, they are simply driven by a charged field. Electrons don't charge the field around them, the field around them charges the electrons. The electron and proton are nothing but buoys in this field, telling us local energy levels. The cause of the energy is the stream of charge photons, which always manages to remain invisible to these researchers. It would be quite simple to switch devices, measuring photons instead of electrons, but somehow that never occurs to anyone in this experiments. It never occurs to them to bring in other machines (which they do have now) to try to measure the photon field fluxes at and around the target. They have been so in love with the electron for so many decades, they cannot dance with another partner.

Actually, the machines they have here are measuring the photon fields quite well, but these researchers and theorists are blind to the data. Even the field numbers are direct confirmation of the charge field, but since they have the wrong equations, they don't see that either. I have tweaked their own equations, giving them back in a form which allows them to solve these things, but no one has accepted my gift.
...

Why would Avogadro's number be showing up here? It is supposed to be the number of molecules per mole of substance. But we aren't looking at molecules or even atoms here. We are supposed to be looking at free electrons. This makes no possible sense until you read my paper re-assigning Boltzmann's constant and Avogadro's constant to the charge field. There I show that both constants are following the charge photons present, not the molecules. Once we understand that, we understand why Avogadro's number is showing up here with electrons. It is because, once again, Shoulders' devices aren't measuring the electrons in the field, they are measuring the photons. As with the molecules, the electrons are just along for the ride. Not only do they not cause any of the major effects, the math— such as it is—doesn't even apply to them. Like all previous researchers, these guys like Shoulders just assume that because they put electrons in (and nothing else), the electrons must be causing the effects. So they apply their field math to the electrons. But since it is the photons that are causing everything, the math should be applied to them.

http://milesmathis.com/evo.pdf

Cr6

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## Re: The Photon - The Mathis' Definition

258. HOW DO PHOTONS TRAVEL?

...

Here we have only the linear velocity and the tangential velocity. I have drawn the linear velocity only a bit larger than the tangential velocity, but let us say we are given a tangential velocity of some magnitude x. We let that magnitude remain constant and we begin increasing the linear velocity. So one line gets very much longer than the other. Obviously, if we make one line very much longer, we no longer need be concerned about the real magnitude of x. We only care about its magnitude relative to the linear velocity. As the linear velocity approaches infinity, the relative tangential velocity would approach zero. This is sometimes called the math of very large numbers, and it allows us to dispense with absolute values. Since velocity is not an absolute value to begin with, this is very easy to do.

You may say that c is not really a “very large number”, but relative to the radius of the photon, it is. We have 31 orders of magnitude between the two, which is a very large number by anyone’s reckoning.

So, as the linear velocity gets larger, the tangential velocity gets smaller relative to it. And the difference between the two is so large that the relationship becomes nearly a perfect inverse relationship. In other words, c2 is so large that the tangential velocity can be estimated very successfully with just the transform 1/c.

This also happens to be the reason we can ditch the ½ in the equation E=mc2. The two equations E=mc2 and E=½mv2 are supposed to be analogous, at least when the first one is applied to a photon and the second one is applied to an electron, say. Nearly all the energy of a photon is kinetic energy, so the equation E=mc2 applied to a photon is an expression of kinetic energy, with m as the mass equivalence of the photon. But the equation has no ½. The reason it doesn’t is that because c2 is so large, ½ becomes negligible. One half of almost nothing is still almost nothing, so we can ignore it as a factor. I explain this in more detail in another paper.

...

To clarify even further, I will show the math for this. All I have to do is import my new equation which takes us from v to ω. I developed this equation by correcting Newton's orbital equation a=v2/r. Current physics believes that the tangential velocity and the orbital velocity are equal, but they are not. Newton assigns them to different vectors or arcs, and then attempts to show they are equal at the limit. I have shown that his math fails.

In my new equations, ω is the angular velocity measured in meters/s, not radians/s, and we can go from v to ω with a simple equation.

I used this equation in my paper on the Bohr magneton to show the real radius of the electron, and to correct the Bohr radius.

ω = √[2r√ v2 + r2) - 2r2]

If we insert our numbers for the photon into this equation, using v=c, we get

ω≈√(2rv) = 4 x 10-8m/s.

http://milesmathis.com/photon2.html

Cr6

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## Re: The Photon - The Mathis' Definition

LloydK

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