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## Antigravity within the photon

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In Modern Physics the photon is considered massless.

But a photon with mass m=0 must have momentum p=m.c=0, and a particle with p=0 cannot transfer any momentum to an electron at rest. Therefore from the concepts of Modern Physics the light is incompatible with the Compton Effect.

Here we will see:

1- how to conciliate the mathematical formalism used by Compton with the concept of a no massless photon, which is proposed in author’s Quantum Ring Theory.

2- The mystery of the photon’s behavior is due to the actuation of repulsive gravity within its structure

#### Introduction

In Compton Effect, a photon with momentum p=h/? collides with an electron at rest. The photon transfers a portion of its momentum to the electron, which gets a momentum pe=me.v, and the momentum of the photon decreases from p to p’, in order that there is a conservation of momentum: p = p’ + pe. The phenomenon is similar to a collision between two billiards balls.

But as in Modern Physics the photon is considered massless, its momentum earlier the collision would have to be p=m.c = 0 , since its mass is m=0. And therefore the equation p=m.c cannot be conciliated with the equation p=h/? from the concepts of Modern Physics.

Besides, from the philosophical viewpoint, it makes no sense to consider that a particle with mass m=0 can be able to supply to an electron at rest a momentum pe=me.v.

One could say that a massless photon has kinetic relativistic energy. But the kinetic energy depends of a particle is a result of its velocity. And we can think about velocity only if we consider a corpuscle constituted by matter, because without matter it makes no sense to speak about velocity, because something that does not exist cannot have velocity. Therefore a particle with no mass cannot have kinetic relativistic energy.

It is known that Compton Effect is incompatible with the wave concept of light. But even if one would wish to say that photon’s relativistic energy is related to its wave nature, nevertheless a wave is a disturbance of a medium, and its propagation would require a massive feature of such medium.

As we realize, there is no way to conciliate the concept of a massless photon with the Compton Effect, from the viewpoint of Modern Physics.

#### Has the light a physical structure according to modern physics?

No, in Modern Physics the theorists consider that light has not a physical structure. Light is described mathematically by Maxwell’s equations, but they have not connection with a physical structure. We can say that light, according to Modern Physics, is like a ghost, an entity of another world different of the physical world where we live.

This is the reason why some philosophers think that there no exist a physical reality, and we live a world of illusion.

#### To suppose that light has not physical structure is obvviously an absurd. How did the theorists arrive to such absurd conclusion?

When Einstein introduced the concept of a photon in the beginning of the 20th Century, he also introduced a new theoretical problem: the concept of the photon (a quantum of light) is not compatible with Maxwell’s Equations. Over several years Einstein tried, in vain, to find a solution to conciliate the concept of a photon with Maxwell’s Equations.

The concept of the corpuscular photon can be conciliated with Maxwell’s Equations through the adoption of the helical trajectory of the elementary particles.

The helical trajectory, known as zitterbewegung, appears in the Dirac equation of the electron, a discovery made by Schrodinger. He did not believe that the successes of the Bohr’s theory could be accidental, and he tried to find the cause of some misteries that appeared in the solution addopted by the theorists in the development of Quantum Mechanics.

#### From the zitterbewegung did the theorists succeed to eliminate the paradoxes of Quantum Mechanics?

The zitterbewegung was interpreted in Quantum Field Theory (the successor of Quantum Mechanics) by a different way of that proposed by Schrodinger, because according to QFT the zitterbewegung is not a helical trajectory.

And the theorisits did not succeed in their attempt to eliminate the inconsistences of Quantum Mechanics from such interpretation of its successor, and several paradoxes persist in the theory.

The zitterbewegung in the sense of Schrodinger is a new attempt that perhaps may be capable to eliminate the paradoxes of Modern Physics.

#### Can the zitterbewegung be applied to the photon?

A new alternative for the explanation of the duality of the wave-particle of particles like the electron and proton has been proposed from a theory developed by considering the zitterbewegung.

Such alternative has been developed by several theorists, like the physicist T. S. Natarajan, the mathematician Waldyr Rodrigues Jr., the physicist David Hestenes, and others.

But can the concept of zitterbewegung be applied to the photon?

The possibility of proposing a physical theoretical model of a photon was rejected by Einstein and Dirac in the beginning of the 20th Century because some experiments have shown that the photon has a statistical behavior in some light polarization phenomena. Their restriction is the most fundamental argument against the viability of a theoretical proposal for a photon model.

Nevertheless, in the paper DEMYSTIFYING THE EPR PARADOX of the book Quantum Ring Theory it is shown that there is a hidden variable in the structure of the photon, responsible for the statistical behavior. So, the fact that in the photon there is a hidden variable implies the following: the statistical behavior can be explained through a physical model of the photon, something considered impossible during the 20th Century.

#### Is the hyphotesis of hidden variable acceptable in physics?

The theorists rejected theories based on the hypothesis of hidden variables because they consider that they violate a theorem known as Bell’s Unequalities.

However, as we will see in the Chapter 2, the Bell’s theorem would be acceptable if all the predictions of Quantum Mechanics were correct. But if the solution based on the zitterbewegung is that one used by Nature for production of the physical phenomena, this imply that Quantum Mechanics is not correct, and so Bell’s theorem cannot be taken in consideration for rejecting a physical model of photon.

A paper on the zitterbewegung of photon was written by D. H. Kobe, published in Physics Letters A in 1999.

#### What premises a model of photon must be able to satisfy?

A model of a photon is unacceptable if it is unable to satisfy the following premises:

1. To be able to yield Maxwell’s Equations.

2. To be compatible with the duality wave-particle of the light.

3. To be able to explain the transverse propagation of light.

4. To be able to explain the light polarization (as we will see in Chapter 2).

5. To be able to conciliate the quantum concept of light with Maxwell’s Equations (they can be reconciled through the present model of a photon, as shown in the Chapter 2).

6. To be able to be fitted to a Lie group. As it’s impossible from a physical model of photon constituted by matter to fit it to any group of symmetry known as Lie Groups, it seems impossible from the mathematical viewpoint the photon to have physical structure.

#### Can light be composed by classical particles in the sense of Newton?

No. When Newton proposed that light is constituted by particles, he was thinking about a classical model of a particle.

On the other hand, Quantum Ring Theory proposes that a non-classical model based on quantum mechanics can explain light: the photon is constituted by two corpuscules, a particle and its antiparticle, in rotation, and they cross the aether.

#### Can the aether be constituted by particles and antiparticles?

Yes. The prediction that the space is fulfilled by particles and antiparticles was made by Dirac. His theory was confirmed by experiments. Thanks to his theory the physicists obtained antiparticles as the positron (antiparticle of the electron).

While a model composed by particle cannot be fitted to a Lie group because it is assymmetrical, unlike a model particle/antiparticle can be fit to a Lie group because it is symmetrical, and therefore it does not brake the symmetry required by the Lie Groups.

#### How to incorporate these ideas to a model of photon?

Let us analyze a physical model of a photon which works through the participation of its interaction with the aether.

Firstly, let us see two macroscopic phenomena which effect we need to consider for the explanation of a photon’s behavior. It is important to note that it is merely a macroscopic example with the objective of remembering the behavior of one particle with electric load crossing a magnetic field.

Figure 2.1 shows that, when a particle with electric positive load +q with rectilinear motion and speed v crosses a magnetic field B situated in an orthogonal plane to the motion, the particle acquires a circular motion with radius r = mv/qB. If a particle with negative load -q has speed v toward the line fluxes of field B, but with a component of v orthogonal to x-axis, the particle acquires a helical trajectory (Figure 2.2). Now let us consider a microscopic particle with load +q emitted by one atom.

Figure 2.3 shows the particle with speed c of light and electric load +q turning by a circular orbit orthogonal to the x-axis, with radius R and angular speed. As the particle has a secondary motion transversal to the x-axis while its main motion peruses the space toward the x-axis, then the particle describes the trajectory shown in Figure 2.4.

Later we will see the reason why the particle turns by an orbit with radius R. Figure 2.4 shows a strong electric field ET toward the z-axis, and a weak magnetic field BT toward the y-axis. The field ET is strong because it is induced by the load +q traveling with the faster speed c, while BT is weak because it is induced by the load +q with lower speed v=?.R (see also Figure 2.5). Figure 2.6 also shows that two other fields are induced: the field EL is toward the x-axis, and it is induced by the tangential speed v= ?.R (and therefore it is weak), and BL also is toward x-axis, but it is induced by the speed c (and therefore it is strong).

Let us verify the relation between the transversal fields ET and BT . Figure 2.7 shows schematically an atom emitting two photons. Let us suppose that the body of the photon “1” is constituted by a quantity “m1” of positive Dirac’s particles e (+) of the quantum vacuum. Thereby m1 is a pure undimensional number. We will call “Phase P” (principal) such a body of the photon with positive electric load. If each particle e (+) of the quantum vacuum has load +q, then m1.q = Q1, which is the load of the photon 1. Using the same method for photon “2”, emitted with greater kinetic energy, m2.q = Q2 is the load, and m1 is the mass. Concerning the relation between the spins of the two photons, it is reasonable to hope that the bigger the mass of a photon, then the angular speed ? of the spin’s rotation is faster. That is, it is reasonable to believe that, if K1 and K2 are the kinetic energies, then

?1 /?2 = K1 / K2 = m1.c2 / m2.c2

Thus,

?1 /?2 = Q1 / Q2 = k

and for any photon:

Q / ? = k

The intensity of the transversal field BT depends upon the values of Q and ?:

BT /(Q.?) = k1

The intensity of the transversal field ET depends upon the values of Q and c:

ET /(Q.c) = k2

Compare the two situations in Figures 2.8 and 2.9: • In Figure 2.8 there is no magnetic field BL crossing the particle’s trajectory. From this condition we would have:

ET /(Q.c) = BT /(Q. ?)

• However, the intensity of the field BL depends upon the speed ?. In addition,, the magnitude of the field BL increases the influence of BT (proportional to ?) in the relation ET versus BT , and therefore in Figure 2.9 we need to consider the influence of the magnetic field BL crossing

the spiral. From this condition we have:

ET / (Q.c) = ?.BT / (Q.?)

and then:

ET / c = BT

The expression above is an old well-known relation between ET and BT in the propagation of electromagnetic waves. Now we can introduce an induced “Phase I” in our model of the photon:

• The atom emits the particle with load +Q (Phase P), as we already have seen.

• However, suppose this particle induces the appearance of an antiparticle with electric load -Q (Phase I).

• Then the antiparticle will have the same fields ET and BT , EL and BL as shown in Figure 2.4 , 2.5, and 2.6 The induced antiparticle with load -Q (with motion toward the x-axis) crosses the field BT induced by the particle with load +Q, and therefore, similar to the macroscopic phenomenon shown in Figure 2.1, the antiparticle acquires rotation around the flux BL. By the same method, the particle with load +Q will cross the field BT (which is induced by the antiparticle), and therefore, similar to the macroscopic phenomenon shown in Figure 2.1, the particle acquires rotation around the flux BL too (but the rotation of the particle has a contrary direction of the antiparticle’s rotation because the particle’s load is positive and the antiparticle’s load is negative).

#### How is the combination of fields?

Let us examine the combination of the fields in Figure 2.10-A. The wavelenght of light is

?=c/?

where c is the light speed and ? its frequency.

The wavelenght ? shown in Figure 2.10-A depends upon the angular speed ? of the spin. Obviously, the lightest photons have a lower ?, and by consequence a longer wavelenght. On the other hand, heavy photons as gamma ray have a faster ?, and their wavelenght are very short.

If T is the time period of one complete revolution of the spin, then when t= 0 we have E=0, B=0 and when t=T/4 we have E=max, B=max and so on..., as shown in Figure 2.10-A.

From these figures we can realize two things:

1- why the field E has a senoidal shape

EZ = EY0 .sen(kx - ?t)

2- and why B with E are in phase:

BZ =(k/?).EY0 .sen(kx- ?t)

where c = ?/k

From the three equations:

ET / c = BT

EZ = EY0.sen(k.x - ?.t)

BZ =(k/ ?).EY0.sen(k.x- ?t)

and starting from the mechanism of the present model of a photon, we can arrive at Maxwell’s Equations.

#### Can such model explain light polirization?

Concerning the polarization of a photon, it depends on the distance between the two particles that constitute the photon, as shown in the paper DEMYSTIFYING THE EPR PARADOX, of the book Quantum Ring Theory.

#### Can such model of photon explain the interference phenomenon??

Look at the Figure 2.11 Concerning the problem of the interference phenomenon of a photon crossing gaps in the wall, when the photon crosses the unique gap 1 of Figure 2.11 we do not have interference between Phase P and Phase I, because the field BL is divided symmetrically by the line center which divides the photon toward the direction of the x-axis (i.e., the direction of the photon’s motion). However, when the photon crosses gap 2 and gap 3, the magnetic field BL is not divided symmetrically, and Phase P interfears with Phase I. This is the reason why a photon can have interference with itself, as experiments have shown, a paradox described by Quantum Physics with the help of complex numbers.

#### What are the conclusions untill now?

CONCLUSIONS

It is possible that the model of a photon proposed herein could be the physical model used by Nature for propagation of light. Actually, it is not only possible, but there is a good chance. This is a very important fact for science, because up to now a “physical” model (able to describe the phenomena of light) was considered impossible by scientists. Indeed, if we consider light as constituted by particles in the sense of Newton (with rectilinear trajectory), such a model cannot survive many theoretical restrictions against its hypothesis. Robert Lindsay and Henry Margenau, authors of the book Foundations of Physics, wrote:

“It is also clear that if the propagation of light is treated as being due to the motion of corpuscles moving with the velocity of light the corpuscles do not move in accordance with the principle of least action. This situation has been discussed by L. de Broglie in his wave mechanics.”

Such an objection is not applicable to two corpuscles, particle and anti-article, moving through the helical trajectory.

#### LIGHT, MATTER, AND ANTIMATTER

When Einstein developed the relativity in the begginning of the 20th Century, nobody cogitated with respect to the existence of antimatter. The positron, that is the antimatter of the electron, was a foresight of the Dirac’s theory, and later it was obtained experimentally.

But what is antimatter ?

Antimatter is not some misterious thing. A particle and its antiparticle have the same mass, the same spin, that is, they are identical particles, but with one difference: they have contrary electric charges. For example, the electron has negative electric charge, and the positron has positive charge.

The basic difference between matter and antimatter is that our Universe is constituted by matter. Out of this, an electron and a positron behave in the same form, except concerning their charges. If we create a positron in the laboratory, it immediatelly goes to interact with an electron, and they will cancel one each other: matter and antimatter, when interact, desintegrate, by producing energy.

When the theorists faced the mistery of the light nature in the begginning of the 20th Century, several theoretical evidences have suggested that the light could not be composed by matter, that is, the light could not be composed by corpuscles. And more than that, Einstein and Diract were even convinced that was impossible to find a physical model for the light, because the statistic behavior of light in the polarization discarded that possibility.

In a certain way Eistein was right in believing that the photon could not be a material particle. That is, the photon could not be matter. Because actually it is not matter, since a corpuscle, constituted by matter and antimatter, is not matter. The combination of particle and antiparticle within the structure of photon confers to it properties that the matter does not have. That is, the photon, because it is not matter, is not submitted to the laws of Classical Mechanics.

#### Does a photon obey to Newton's classical laws? Can a photon be accelerated, as any particle according to classical mechanics?

No, when a force F is applied on a particle with mass m, it accelerates according to Newton’s law expressed by

F=ma

But the photon does not obey this law. The photon cannot be accelerated, because being constituted by matter and antimatter it behaves as if it should not have mass. If we create a magnetic field for accelerating a photon, it appears a force F on the particle, and a force –F on the antiparticle, and the two forces cancell one each other, in order that the total force on the photon is zero. And in the moment of creation of a photon, it is created with the speed of light, it does not accelerate. No particle constituted by ordinary matter can do what the photon does.

For understanding the questions that involve the nature of light, before everything we have to speak about the concept of repose mass. From the most famous Einstein’s equation E=mc2, any energy can be converted to mass, and vice versa. If an electron has a speed V, it has then a total energy that is the sum of two energies:

one is that which is equivalent to the electron’s repose mass

converted to energy through E=mc2,

and the another is the kinetic energy due to its moviment.

In the case of photon, the theorists state that its repose mass is zero, it has its kinetic energy of moviment only, which means that, if a photon collides with a surface and stops, no mass would be deposited on the surface.

Ahead there are the restrictions of the theorists against the hypothesis of being the light of corpuscular nature:

1- Light composed by corpuscles would violate the last action principle.

2- A corpuscular photon would have to have mass. In this case its repose mass could not be zero.

3- According to Relativity Theory the photon is massless.

4- A corpuscular photon would violate a mathematical concept know as gauge invariance.

Let us analyze these restrictions.

1- The least action principle would be violated by a corpusuclar photon if its trajectory should be a classical rectilinear trajectory of a corpuscle. Neverthelesss the photon moves through a helical trajectory, and therefore this restriction is diacarded.

2- A corpuscular photon would have to have mass if it was constituted by ordinary mass. But a photon constituted by matter antimatter has repose mass zero, because when the photon collides with a surface and stops, the particle and the antiparticle interact and desintegrate.

3- A photon constituted by matter antimatter has no mass, and therefore it does not violate the relativity. According to QRT, what the photon has is inertia, due to particle and the antiparticle, but not mass.

4- The gauge invariance is a matematical concept that would be violated if the photon was constituted by ordinary mass (if it was constituted by a particle only, wiithout the antiparticle). This because a photon with this structure of ordinary matter would not have symmetry, and would be impossible to fit it in one of the many groups of symmetry that receive the name Lie Groups. Not having symmetry, it would be impossible to find for this photon a mathematical equation independent on the referential, an indispensable condiction for any model to be accepted. But a photon composed by matter antimatter is symmetrical, and therefore it does not violate the gauge invariance. And the most contusive proof that this model of photon has a equation independent of the referential is the following: in the paper A Model of Photon, of QRT, it is demonstrated that this model of photon generates the Maxwell equations. Thereby the equation of a partnership between a particle and an antiparticle that move through the helical trajectory is the Maxwell equations, and therefore to reject this model of photon would mean to reject the Maxwell equations too.

One could allege: the helical trajectory appears in the Dirac’s equation of the electron, and so we have a good theoretical reason for taking it in consideration.

#### But what about the helical trajectory of a photon? Does it appear in the Maxwell Equations?

Yes, the helical trajectory of photon appears in the Schrödinger-like equation, as shows the paper Zitterbewegung of a photon , by 19- D.H. Kobe , published in Physics Letters A, Vol. 253, No. 1-2, March 1999, where in the absrtract he writes:

“A single photon, satisfying a relativistic Schrödinger-like equation, has velocity operator that undergoes oscillations in a direction orthogonal to its momentum. This Zitterbewegung has a spatial ampliture equal to the classical wavelength. The spin of the photon is the orbital angular mometum due to the Zitterbewegung ”

This model of photon is therefore theoretically viable. And it is not merely viable, because it eliminates paradoxes from the theories of Modern Physics. Indeed, the light beyond the undulatory characteristics, it also has corpuscular characteristic, as dramatically confirmed by Compton’s experiments. In these experiments a photon collides with an electron, and the shock happens exactly as if they were two billiard balls. This does not fit to the undulatory hypothesis. Besides, this photon has momentum. In Physics, the momentum “P” of a body is the multiplication of its mass by its velocity: P=mV. Well, not having mass, how does explain the momentum of a photon ? And how does explain the Compton Effect ?

When an atomic nucleus emits a photon of big energy (gamma ray), the nucleus suffers a with-drawal, as happens with the boat in figure 29, in the instant when the man throws the watermelon to the other man in the other boat. Which indicates that the photon follows the law of actin-reaction, that is applied to the bodies.

#### Is there a controversy on the question the photon is massless, or not?

Yes, there is a controvery on the question of the photon to have, or not, mass.

But mass is a concept created by Newton, who proposed laws for the matter. The photon is not matter, since it is a combination of matter and antimatter. As said, the photon does not follow the law F=ma, because it does not accelerate, just because it is not matter. Then it seems do not be suitable the discussion on whether it has, or not, mass. But there is no doubt that it has something that is equivalent to the mass. Perhaps we have to propose a new concept, of a new mechanics for the alliance matter-antimatter, as Newton proposed the concept of mass for the matter.

Perhaps instead of to call it mass we have simply to call it inertia, since:

a) while the total energy of the conjunct matter-antimatter in repose is zero (because the matter has positive energy, and the antimatter has negative energy),

b) however the inertia of the conjunct matter-antimatter is the sum of the inertias of corpuscle matter and the corpuscle antimatter.

It is not the case of the classical Newtonian inertia, which is applied to matter, because this one refers to the inherent tendency of the bodies of holding their status of motion or repose (Newton’s firs law). It is the case of an inertia tied to the momentum. In the instant of the creation of a photon, it is created with momentum P = i.c, where “i” is the inertia of the photon. Along a collision, the photon transmits this momentum equal to the multiplication of the inertia i by the velocity c of the photon. The momentum of a photon is the multiplication of its frequency by the Planck’s constant h:

P = hf

Therefore the inertia of a photon is

i=h.f/c.

#### Have any experiment detected a mass for the photon?

An experiment led by Jung Luo in 2003, published by the American Institute of Physics. made with a very sensible torsion balance, detected the sediment of mass when the light falls down on a surface. Does this imply that the mass of photon is not zero ?

As already explained, the controversy appears because we want to approach the question of photon through the Classical Mechanics, by the Newtonian concept of mass. Consider the following:

1- A particle of matter has mass “m”, as a particle of antimatter has mass “m” too. So, when they both are within the structure of a photon, each one of them has mass m.

2- But while in the Classical Mechanics the sum of two masses m is 2m, the mass of a corpuscle in which are packed one particle of matter and one particle of antimatter, both with mass m, is not 2m. When they are packed together, the two particles behave as they should have mass zero.

3- When the light falls down on a surface and it is absorbed by it, the structure of the photon is broken, and the two particles are unpacked. Then appears the individual characteristic of each one of them, and as consequence a mass 2m is deposed on the surface.

4- One has to note that, when the photon is absorbed by a surface, with the absorption the particle and the antiparticle leave off of constituting the photon, because the structure of the photon has been broken with the lost of packing. With the unpacking, the two corpuscles lose the electromagnetic properties decurrent from the partnership matter-antimatter inside the structure of photon, and starting from that instant they pass to manifest the individual properties of each corpuscle, which individually has mass.

#### Is the photon accelerated when it passes from the water to the air?

When the light passes from a medium to another one, it changes its velocity, a phenomenon that receives the name refraction. For instance, in the glass the speed of yellow light is 200.000km/s, while in the vacuum the speed is 300.000km/s. In this case is there accleration of light when it passes from a medium to another one?

No, the light does not accelerate. Although the light is faster in the vacuum than in the glass, however it does not accelerate when it passes from the glass to the vacuum. Let’s see why.

First of all we have to understand what is acceleration, which is a Newtonian concept inferred from the relation a=F/m. So, we realize that the acceleration is a property of matter, because a body constituted by matter has mass “m”.

Consider a car moving with constant speed v=50km/h. Suddenly the driver accelerates, in order that in an interval of time “t” the speed increases from 50km/h to 100km/h. Well, along the time interval “t” the car has experienced all the speeds between 50km/h and 100km/h, as for instance 50,1km/h, 50,2km/h...62,8km/h...77,03km/h...100km/h.

This happens because the mass “m” has a Newtonian inertia, and when a force is applied on that body with mass “m” its speed increases continuously. Any body, composed by matter, requires a time “t” to be accelerated. A body composed by matter cannot be accelerated from a speed v to another faster speed V in a time zero.

So, let us suppose that a photon should be a particle with mass “m”. Then it is obvious that (when the speed of such photon passes from 200.000km/s to 300.000km/s) its speed experiences all the intermediary speeds between 200.000km/s and 300.000km/s, because any particle composed by matter suffers acceleration.

But consider a photon composed by particle and antiparticle. Such photon has no inertia in the Newtonian sense. Which means that such photon is not accelerated. When it passes from the glass to the vacuum, its speed changes instantaneously from 200.000km/s to 300.000km/s. In another words, such photon does not experiences the intermediary speeds between 200.000km/s and 300.000km/s, because this change of speed does not waste time. The time of speed changing is zero.

#### Why do theorists state the photon be massless?

In Compton Effect, a photon with momentum

p=h/?

collides with an electron at rest. The photon transfers a portion of its momentum to the electron, which gets a momentum pe=me.v, and the momentum of the photon decreases from p to p’, in order that there is a conservation of momentum: p = p’ + pe. The phenomenon is similar to a collision between two billiards balls.

But as in Modern Physics the photon is considered massless, its momentum earlier the collision would have to be p=m.c = 0 , since its mass is m=0. And therefore the equation p=m.c cannot be conciliated with the equation p=h/? from the concepts of Modern Physics.

Besides, from the philosophical viewpoint, it makes no sense to consider that a particle with mass m=0 can be able to supply to an electron at rest a momentum pe =me.v.

One could say that a massless photon has kinetic relativistic energy. But the kinetic energy depends of a particle is a result of its velocity. And we can think about velocity only if we consider a corpuscle constituted by matter, because without matter it makes no sense to speak about velocity, because something that does not exist cannot have velocity. Therefore a particle with no mass cannot have kinetic relativistic energy.

#### Is the Compton effect compatible with the wave concept of light?

No, it’s known that Compton Effect is incompatible with the wave concept of light, because two waves cannot have collision like two billiard balls.

But even if one would wish to say that photon’s relativistic energy is related to its wave nature, nevertheless a wave is a disturbance of a medium, and its propagation would require a massive feature of such medium.

As we realize, there is no way to conciliate the concept of a massless photon with the Compton Effect, from the viewpoint of Modern Physics.

#### Is it possible to eliminate the controversy on the mass of photon from the model of photon composed by particle & antiparticle?

First of all, we have to understand: what is a massless particle?

A massless particle is that one that does not interact with matter. And if it has no inertia, which means that it is not submitted to Newton’s law F=ma.

For instance, earlier the confirmation of Einstein’s special relativity (where the space is considered empty), the scientists believed that the space would be fulfilled by an aether composed by massless particles. Because if they were not massless, the velocity of the planets about the Sun would decrease slowly, due to the friction of the particles and the matter that constitute the planets.

So now consider the model proposed in Quantum Ring Theory: the photon is constituted by two corpuscles: a particle and its antiparticle. They move with helical trajectory, a motion that confers to light its wave-particle feature. Each corpuscle has mass m/2, in order that photon’s mass is m/2+m/2= m.

The Compton equation

?? = ?C(1- cos?)

has been developed from the energy of the photon given by:

E = h? + mc2 (1)

where he has considered m=0 the repose mass of the photon, in order that eq. (1) becomes:

E = hv (2)

From eq. (2) we get the momentum of the massless photon:

P = E/c = h?/c (3)

As the momentum of a particle is the product of its mass and its velocity, from eq. (3) we have:

p = m.c = h.?/c (4)

and so we get the photon’s mass from eq. (4):

m = h?/c2 (5)

#### How to interpret such mass of the photon?

Let’s see it by considering the model proposed in Quantum Ring Theory.

The photon’s mass is actually constituted by two masses h?/2c2 , one due to the particle (with positive electric charge), and the other due to the anti-particle (with negative electric charge).

Now let’s interpret

E = h? + mc2 (1)

which has been the starting point of Compton’s mathematical development, as follows:

E = h? + [ ( h?/2c2).c2 - ( h?/2c2).c2 ] = h? (6)

and we see that from the model of photon proposed in Quantum Ring Theory we arrive to Compton’s starting point E=h? by considering a photon with mass m, because the positive energy +(h?/2c2).c2 of the particle is cancelled by the negative energy –(h?/2c2).c2 of the antiparticle.

Conclusions:

1) From the viewpoint of the photon’s “energy” (as considered by Compton in his formalism), we see that the energy due to the repose mass of the two particles is cancelled, because they have contrary electric charges.

So we understand that the photon is massless regarding to the “energy aspect”.

2) But from the viewpoint of the photon’s “momentum”, we realize that the photon is not massless, because actually it has a mass m = h?/c2

So, from the aspect of ENERGY, the photon behaves as it should be massless, since it’s null the total energy due to the particle and the antiparticle.

But from the aspect of MOMENTUM the particle with mass

m’ = h?/2c2

and the antiparticle with mass

m’ = h?/2c2

contribute with a total mass

m’ = h?/c2

for the production of the photon’s momentum.

That’s why photon’s momentum is p=mc=h?/c.

#### Is E=mc2 compatible with a massless photon?

The Einstein’s equation

E = mc2

does not depend on the refencial frame. So, we cannot find any referencial frame in which the equation E = mc2 becomes independent of the mass.

Such absolute feature of the equation E = mc2 means that the photon cannot be massless, because the energy of the photon is the equation E = mc2. That is, the photon obeys to Einstein’s equation E = mc2 .

Actually the correct would be to say that the photon is the striking proof of the Einstein’s equation E = mc2 , because:

1) In the instant of light emission, energy (of the atom) is converted to matter-antimatter (photon).

2) In the instant when the light insides on a surface and it is absorbed, matter-antimatter (photon) is converted to energy (heat).

When Newton developed his theory, he could not imagine the existence of antimatter. So, he proposed the fundamental equation F=m.a by considering that all particles in the universe would be constituted by matter, and therefore any particle would have to have mass.

Well, as the photon is not matter, then we cannot apply Newton’s concept of mass to the photon. Actually, when we deal with the photon, we have to consider a new Physics, in which the concept of mass must be changed, by considering that such concept cannot be applied to a particle constituted by matter-antimatter. The concept of mass must be replaced by the concept of inertia. So, the photon is massless when the particle and the antiparticle are packed together in the body of the photon. But when the two particles are apparted (for example when the light falls upon a surface), and the photon is destroyed, the two particles individually behave like no massless. In this sense, the photon is massless while it it is travelling in the space or withing a medium, but the photon becomes no massless when it collides with a surface.

#### Why is not the neutrino massless?

Earlier 2001 the theorists considered the neutrino as massless. So, according to Quantum Mechanics, the neutrino could not have interaction with the matter.

But in 2001 an experiment has shown that the neutrino interacts with the matter. And the theorists had to change their understanding on the neutrino feature.

Look at this incoherence of Modern Physics:

1) The theorists consider that a neutrino cannot be massless, because it interacts with the matter.

2) However the light interacts with the matter too (photo-electric phenomenon, Compton Effect). But in spite of the interaction of the light with the matter, the photon is considered massless, although the neutrino is considered not massless just because it interacts with the matter.

As we see, the criterion applied to the neutrino is different of that applied to the photon.

Why two different criteria?

Incoherences that Modern Physics is unable to eliminate.

#### What would be the new Physics for the behavior of the photon composed by particle & antiparticle?

Let us explain it in another words why Einstein’s relativity requires a massless photon.

Consider an electron with mass “m”, and a positron with the same mass “m”. The positron is the antiparticle of the electron.

Suppose that the electron and the positron perform an atom, with mass 2m. If we apply a velocity “v” to such atom, according to Einstein’s theory its mass increases following the equation:

M = 2m/[ 1- v2 /c2 ] 1/2

When such atom approaches the velocity of light, the mass M tends to infinite.

Now consider a particle and its antiparticle, both with mass “m0” constituting the structure of a photon. So the mass of the photon is 2m0 . As the photon has speed c, the mass of the particle and the antiparticle would have to tend to infinite, because they would have to follow the Einstein’s equation:

M0 = 2m0 /[ 1- v2 /c2 ] 1/2 = ?

And so, according to Einstein’s theory, if the photon would not be massless, it would have to have na infinite mass.

Let’s try to understand why the mass of the photon does not obey to Einstein’s equation, although it is consituted by two corpuscles each one with repose mass m0 .

According to Quantum Ring Theory, all the elementary particles are composed by a body-ring with rotation, crossed by a flux of gravitons, as shown in the figure 1: The aether is constituted by atractive gravitons g(+) & g(-), electric particles e(+) & e(-), magnetic particles m(+) & m(-), particles p(+) & p(-) that promote permeability, and repulsive gravitons G(+) & G(-).

The properties of the partícles G(+) and G(-) are the following:

1. The particles G(+) e G(-) repel each other

2. Two particles G(+) repel each other

3. Two particles G(-) repel each other

4. The particles G(+) attract the particles g(+), and that explains their gravitational nature

5. The particles G(-) attract the particles g(-), and that explains their gravitational nature

6. The particles G(+) do not interact with the particles g(-)

7. The particles G(-) do not interact with the particles g(+)

From the properties of the repulsive gravity, we realize that repulsive gravitons G(+) of the aether are agglutinated about the flux of gravitons g(+) of a particle, as shown in the figure 2, and repulsive gravitons G(-) are agglutinated about the flux of gravitons g(-) of an antiparticle, as shown in the figure 3. Note that there is attraction between two flux of gravitons g(+), and therefore they would have agglutinate one each other. However the repulsive gravitons G(+) avoid such agglutination, since a field of gravitons G(+) is captured by each flux g(+), and as there is repulsion between the gravitons G(+), they avoid the agglutination of fluxes g(+)

As the repulsive gravitons G(+) interact with the gravitons g(+), then the repulsive gravitons G(+) contribute for the inertia of a particle.

And the repulsive gravitons G(-) contribute for the inertia of a antiparticle.

But look at the photon’s motion shown at figure 4. The particle and the antiparticle are packed together, and so in the structure of the photon the repulsive gravitons G(+) and G(-) actuate together.

So, we have to conclude that such agglutination of repulsive gravitons G(+) and G(-) in the structure of the photon inhibits the interaction of the fluxes of gravitons g(+) and g(-) with the free gravitons of the aether. That’s why the photon behaves like a massless particle.

However, we have to note that, individually, the particle and the antiparticle have each one a mass “mo”, but the addition of masses is not applied here, because the actuation of the repulsive gravity inhibits the interaction of the photon with the aether, and so the mass of the photon is not “2m0”, but it actually behaves as a corpuscle of mass zero.

#### New Physics for the photon: m + m = 0

In Physics there is not negative mass. For instance, it makes no sense to write

M + (-M) =0

because a negative mass –M does not exist.

However such arithmetics is not applied to the photon. Although a negative mass does not exist, however into the photon the repulsive gravity cancels the interaction of the mass 2m of a photon with the aether, in order that for the photon the arithmetics is

M + M = 0

#### Conclusions

As we realize, some of the greatest paradoxes in Modern Physics is due to the repulsive gravity. In the new zitterbewegung hydrogen atom we see that the Repulsive gravity within the hydrogen atom also is responsible for many mysteries not solved by the theorists. Here in the present chapter, we have realized that the repulsive gravity is responsible for one of the greatest mysteries of Physcis: the paradoxial behavior of the photon, which, in spite it behaves like a massless particle, it is capable to apply a momentum to an electron in the Compton effect

#### REFERENCES

W. Guglinski, Quantum Ring Theory-Foundations for Cold Fusion, Bauu Institute Press, 2006