| The Rings Model of Elementary Particles by Giulio. C. Cima (Copyright © 2007 Giulio. C. Cima All Rights Reserved) |
Introduction
1.0 The Time-Helix
2.0 Latent and Active Energies
3.0 The Nature of the Rotating Motion
4.0 Curled Space Gradient
5.0 The Minimum Quantum of Energy
6.0 Double Helix Phase Shift Deviations
7.0 The Rings Model
7.1 Classical Electron Radius and Separation between Singlets
7.2 Ring Rotation
7.3 The Effect of Rotational Velocity on Mass of Ring
7.4 Rings Formations
8.0 Particles Decay
8.1 The Mechanics of Particle Decay
9.0 Conclusion
Appendix
1.0 The Time-Helix
2.0 Latent and Active Energies
3.0 The Nature of the Rotating Motion
4.0 Curled Space Gradient
5.0 The Minimum Quantum of Energy
6.0 Double Helix Phase Shift Deviations
7.0 The Rings Model
7.1 Classical Electron Radius and Separation between Singlets
7.2 Ring Rotation
7.3 The Effect of Rotational Velocity on Mass of Ring
7.4 Rings Formations
8.0 Particles Decay
8.1 The Mechanics of Particle Decay
9.0 Conclusion
Appendix
Introduction
The purpose of this paper is to present the concepts of a model for elementary particles that
would resolve flaws in the current Standard Model that are related to the following subjects:
1) Quarks with fractional electrostatic charge. The fractional charge of quarks is due to
the standard model prescription of three quarks for Baryons and two quarks for Mesons
requiring a charge of one third and two thirds (positive or negative) in order to result in a unit
charge (positive or negative) for charged particles and in a zero charge for neutral particles.
This contradicts the fact that the minimum electrostatic charge is that of the electron/positron
which has been named as the unit charge.
2) Non unique particles makeup. There are six quarks and six anti quarks in the standard
model the combinations of which, two or three at a time, define all elementary particles with
the exception of Leptons. This means that the same quark composition could be shared by
different particles with the result that quarks would then acquire the characteristics of the
particles they make up. For instance, in the Baryons family the composition of the proton is
[u u d] which is the same as that of the positive delta particle; in the Masons family the
composition of the positive pion is [u d*] which is the same as that of the positive rho particle.
3) Mass of quarks. In the standard model, because of the non unique particle make up,
the mass of quarks depends on the mass of the particle. In the example above, the three
quarks [u u d] account for a rest mass of 938.3 MeV when they are parts of the proton and a
rest mass of 1232 MeV when they are parts of the positive delta particle: that is, the mass of
quarks depends on the internal dynamics of the particle. That is true; however a 31%
increase in mass, such as the one in the example, requires a considerable increase in the
internal dynamics.
4) Quarks transformation in particles interactions. In the standard model, because of the
six quarks and six anti quarks prescription, quarks transform themselves into other quarks
by the interaction with W vector bosons in order to satisfy particle makeup, as in the beta
decay where a down quark of a neutron transforms itself into an up quark by releasing an
electron and an electron anti neutrino. The interaction is required in order to preserve the law
of charge conservation and to explain the generation of electrons or positrons of unit charge
from quarks of fractional charge.
5) Mass of neutrinos. The assumption in the standard model is that neutrinos are
massless. This has not been shown to be the case although the mass attributed to the
neutrino is very small and is thought to be fluctuating as the neutrino travels through space
at close to the speed of light.
6) Particle/anti particle imbalance. In the standard model, Mesons are made up of quarks
and anti quarks but Baryons are not. This means that the standard model definition of the
proton, which accounts for the bulk of the matter in the universe, disallows the existence of
anti matter. In a universe that is based on probabilities, the probability that most anti matter is
non existent is very small and the same or almost the same quantity of anti matter must exist
together with matter.
The term “model” is defined as a generalized, hypothetical description, often based on
analogy, used in analyzing or explaining something. And so the rings model is not a
mathematical or physical theory about phenomena: it is only an attempt to describe and give
a semblance of reality to events that are outside of it. It is because of this attempt that a
model, any model, has intrinsic flaws. This model, as all models, has its own flaws some of
which are presented in the Conclusion of the paper.
The purpose of this paper is to present the concepts of a model for elementary particles that
would resolve flaws in the current Standard Model that are related to the following subjects:
1) Quarks with fractional electrostatic charge. The fractional charge of quarks is due to
the standard model prescription of three quarks for Baryons and two quarks for Mesons
requiring a charge of one third and two thirds (positive or negative) in order to result in a unit
charge (positive or negative) for charged particles and in a zero charge for neutral particles.
This contradicts the fact that the minimum electrostatic charge is that of the electron/positron
which has been named as the unit charge.
2) Non unique particles makeup. There are six quarks and six anti quarks in the standard
model the combinations of which, two or three at a time, define all elementary particles with
the exception of Leptons. This means that the same quark composition could be shared by
different particles with the result that quarks would then acquire the characteristics of the
particles they make up. For instance, in the Baryons family the composition of the proton is
[u u d] which is the same as that of the positive delta particle; in the Masons family the
composition of the positive pion is [u d*] which is the same as that of the positive rho particle.
3) Mass of quarks. In the standard model, because of the non unique particle make up,
the mass of quarks depends on the mass of the particle. In the example above, the three
quarks [u u d] account for a rest mass of 938.3 MeV when they are parts of the proton and a
rest mass of 1232 MeV when they are parts of the positive delta particle: that is, the mass of
quarks depends on the internal dynamics of the particle. That is true; however a 31%
increase in mass, such as the one in the example, requires a considerable increase in the
internal dynamics.
4) Quarks transformation in particles interactions. In the standard model, because of the
six quarks and six anti quarks prescription, quarks transform themselves into other quarks
by the interaction with W vector bosons in order to satisfy particle makeup, as in the beta
decay where a down quark of a neutron transforms itself into an up quark by releasing an
electron and an electron anti neutrino. The interaction is required in order to preserve the law
of charge conservation and to explain the generation of electrons or positrons of unit charge
from quarks of fractional charge.
5) Mass of neutrinos. The assumption in the standard model is that neutrinos are
massless. This has not been shown to be the case although the mass attributed to the
neutrino is very small and is thought to be fluctuating as the neutrino travels through space
at close to the speed of light.
6) Particle/anti particle imbalance. In the standard model, Mesons are made up of quarks
and anti quarks but Baryons are not. This means that the standard model definition of the
proton, which accounts for the bulk of the matter in the universe, disallows the existence of
anti matter. In a universe that is based on probabilities, the probability that most anti matter is
non existent is very small and the same or almost the same quantity of anti matter must exist
together with matter.
The term “model” is defined as a generalized, hypothetical description, often based on
analogy, used in analyzing or explaining something. And so the rings model is not a
mathematical or physical theory about phenomena: it is only an attempt to describe and give
a semblance of reality to events that are outside of it. It is because of this attempt that a
model, any model, has intrinsic flaws. This model, as all models, has its own flaws some of
which are presented in the Conclusion of the paper.
1.0 The Time-Helix
Let us consider a particle rotating with constant angular velocity W around a center at a
distance R. If its motion is plotted versus time, the resulting form is a three dimensional helix
of two spatial dimensions and one time dimension: a time-helix. This form can be described
by three parametric equations:
X = R cos (W t)
Y = R sin (W t)
Z = (k / 2pi) (W t)
The constant k is the separation between the helix’s loops along the time dimension t. If T is
the rotation period of the particle given by:
T = 2pi / W
Then:
k = T = 2pi / W
And, after substitution:
Z = t
Depending on the direction of the angular velocity of the particle, the helix is either right-
handed if the velocity is in the clockwise direction, or left-handed if the velocity is in the
counterclockwise direction
Let us now consider two cases:
1) Two particles of equal mass rotating with constant and equal (magnitude and
direction) angular velocity W around a center at a distance R: if the two particles enter the
rotation at the same time and at the same point, the X and Y rotational components of one
particle will be in phase with the X and Y rotational components of the other particle
regardless of the direction of the velocity. If both particles enter the rotation at the same time
but at opposite points, or if both particles enter the rotation at the same point but one of them
lags the other by a time delay of half period, then the X and Y rotational components of one
particle will be in opposite phase to the X and Y rotational components of the other particle
regardless of the direction of the velocity.
2) Two particles of equal mass rotating with constant but opposite angular velocities
around a center at a distance R: the X component of particle 1 will also be in phase with the X
component of particle 2 if the two particles start the circular motion at the same point at the
same time; and the X components of the two particles will be in opposite phase if one particle
lags the other by a time delay of half period or enters into circular motion at the same time but
at an opposite point. However, the reverse is true for the Y components. They will be in phase
if one particle lags the other by a time delay of half period or enters into circular motion at the
same time at an opposite point; and out of phase if both particles start the circular motion at
the same point at the same time.
In Case 1, both the X and Y components of one particle are either in phase with or in opposite
phase to the X and Y components of the other particle.
In Case 2, when the X component of one particle is in phase with the X component of the
other particle the Y components are in opposite phase, and when the X component of one
particle is in opposite phase to the X component of the other particle the Y components are in
phase.
In view of the symmetry of Case 1 we will base our discussion on helixes generated by two
particles of equal mass rotating with constant and equal angular velocity around a center at a
distance R. Any deviation from the conditions stated above (same time and point, same time
and opposite points or time delay of half period) will result in helixes with phase shifts of
absolute values greater than zero and less than 180 degrees.
The X and Y components of such helixes for R = 1 are:
X1 = cos (n)
Y1 = sin (n)
X2 = cos (n + S)
Y2 = sin (n + S)
Where: n is the angle of rotation
S is the phase shift
Adding the components:
X3 = X1 + X2 = cos (n) + cos (n + S)
Y3 = Y1 + Y2 = sin (n) + sin (n + S)
At maximum X3, X1 is equal to X2; and at maximum Y3, Y1 is equal to Y2:
X3 = 2 X1 = 2 X2 = 2 cos (nx) = 2 cos (nx + S)
Y3 = 2 Y1 = 2 Y2 = 2 sin (ny) = 2 sin (ny + S)
X3 / 2 = cos (nx) = cos (– nx) = cos (nx + S)
Y3 / 2 = sin (ny) = sin (180 – ny) = sin (ny + S)
S = – 2 nx = – 2 Acos (X3 / 2) = 2 Acos (X3 / 2)
S = 180 – 2 ny = 180 – 2 Asin (Y3 / 2) = 2 Acos (Y3 / 2)
Where: nx is the angle of rotation at maximum X3
ny is the angle of rotation at maximum Y3
Since the phase shift S is the same for the X and Y components, it follows that maximum X3
is equal to maximum Y3, and their values are equal to the ratio of the amplitude of the X or Y
component curve resulting from a phase shift to the amplitude of the X or Y component curve
resulting from zero phase shift:
X3 = Y3 = As / Ao = Ar
The equations above can then be rewritten as:
S = 2 Acos (Ar / 2)
If we now consider the X and Y components of the two rotating particles as simple harmonic
oscillators where the energy is proportional to the square of the amplitudes of the
oscillations:
E = k A^2
we can then say that the energy of the resulting oscillator is null if both the X and Y
components of one of the rotating particles are in opposite phase to the X and Y components
of the other particle; and that the energy of the resulting oscillator is four times the energy of
one of the rotating particles if both the X and Y components of one of the particles are in
phase with the X and Y components of the other particle. The energy Es of the resulting
oscillator is inversely proportional to the phase shift and is given by:
Es = 4 Eo [cos (S / 2)] ^ 2
Where: Eo is the energy of one rotating particle.
From the equations above:
Sqr (Es / Eo) = As / Ao = Ar = 2 cos (S / 2)
For instance, a deviation of 0.057 degrees from a phase shift of 180 degrees will give an Es of
about 1 eV from an Eo of 1 MeV.
Let us consider a particle rotating with constant angular velocity W around a center at a
distance R. If its motion is plotted versus time, the resulting form is a three dimensional helix
of two spatial dimensions and one time dimension: a time-helix. This form can be described
by three parametric equations:
X = R cos (W t)
Y = R sin (W t)
Z = (k / 2pi) (W t)
The constant k is the separation between the helix’s loops along the time dimension t. If T is
the rotation period of the particle given by:
T = 2pi / W
Then:
k = T = 2pi / W
And, after substitution:
Z = t
Depending on the direction of the angular velocity of the particle, the helix is either right-
handed if the velocity is in the clockwise direction, or left-handed if the velocity is in the
counterclockwise direction
Let us now consider two cases:
1) Two particles of equal mass rotating with constant and equal (magnitude and
direction) angular velocity W around a center at a distance R: if the two particles enter the
rotation at the same time and at the same point, the X and Y rotational components of one
particle will be in phase with the X and Y rotational components of the other particle
regardless of the direction of the velocity. If both particles enter the rotation at the same time
but at opposite points, or if both particles enter the rotation at the same point but one of them
lags the other by a time delay of half period, then the X and Y rotational components of one
particle will be in opposite phase to the X and Y rotational components of the other particle
regardless of the direction of the velocity.
2) Two particles of equal mass rotating with constant but opposite angular velocities
around a center at a distance R: the X component of particle 1 will also be in phase with the X
component of particle 2 if the two particles start the circular motion at the same point at the
same time; and the X components of the two particles will be in opposite phase if one particle
lags the other by a time delay of half period or enters into circular motion at the same time but
at an opposite point. However, the reverse is true for the Y components. They will be in phase
if one particle lags the other by a time delay of half period or enters into circular motion at the
same time at an opposite point; and out of phase if both particles start the circular motion at
the same point at the same time.
In Case 1, both the X and Y components of one particle are either in phase with or in opposite
phase to the X and Y components of the other particle.
In Case 2, when the X component of one particle is in phase with the X component of the
other particle the Y components are in opposite phase, and when the X component of one
particle is in opposite phase to the X component of the other particle the Y components are in
phase.
In view of the symmetry of Case 1 we will base our discussion on helixes generated by two
particles of equal mass rotating with constant and equal angular velocity around a center at a
distance R. Any deviation from the conditions stated above (same time and point, same time
and opposite points or time delay of half period) will result in helixes with phase shifts of
absolute values greater than zero and less than 180 degrees.
The X and Y components of such helixes for R = 1 are:
X1 = cos (n)
Y1 = sin (n)
X2 = cos (n + S)
Y2 = sin (n + S)
Where: n is the angle of rotation
S is the phase shift
Adding the components:
X3 = X1 + X2 = cos (n) + cos (n + S)
Y3 = Y1 + Y2 = sin (n) + sin (n + S)
At maximum X3, X1 is equal to X2; and at maximum Y3, Y1 is equal to Y2:
X3 = 2 X1 = 2 X2 = 2 cos (nx) = 2 cos (nx + S)
Y3 = 2 Y1 = 2 Y2 = 2 sin (ny) = 2 sin (ny + S)
X3 / 2 = cos (nx) = cos (– nx) = cos (nx + S)
Y3 / 2 = sin (ny) = sin (180 – ny) = sin (ny + S)
S = – 2 nx = – 2 Acos (X3 / 2) = 2 Acos (X3 / 2)
S = 180 – 2 ny = 180 – 2 Asin (Y3 / 2) = 2 Acos (Y3 / 2)
Where: nx is the angle of rotation at maximum X3
ny is the angle of rotation at maximum Y3
Since the phase shift S is the same for the X and Y components, it follows that maximum X3
is equal to maximum Y3, and their values are equal to the ratio of the amplitude of the X or Y
component curve resulting from a phase shift to the amplitude of the X or Y component curve
resulting from zero phase shift:
X3 = Y3 = As / Ao = Ar
The equations above can then be rewritten as:
S = 2 Acos (Ar / 2)
If we now consider the X and Y components of the two rotating particles as simple harmonic
oscillators where the energy is proportional to the square of the amplitudes of the
oscillations:
E = k A^2
we can then say that the energy of the resulting oscillator is null if both the X and Y
components of one of the rotating particles are in opposite phase to the X and Y components
of the other particle; and that the energy of the resulting oscillator is four times the energy of
one of the rotating particles if both the X and Y components of one of the particles are in
phase with the X and Y components of the other particle. The energy Es of the resulting
oscillator is inversely proportional to the phase shift and is given by:
Es = 4 Eo [cos (S / 2)] ^ 2
Where: Eo is the energy of one rotating particle.
From the equations above:
Sqr (Es / Eo) = As / Ao = Ar = 2 cos (S / 2)
For instance, a deviation of 0.057 degrees from a phase shift of 180 degrees will give an Es of
about 1 eV from an Eo of 1 MeV.
2.0 Latent and Active Energies
We must define the rotating particles and their radius of rotation, but before doing so, we will
review the definition of mass event horizon given in the Author’s “Gravity and Space-Time”:
The velocity V of a body orbiting another body of mass M at a distance d is given by equating
gravitational force to centripetal force, or:
V = sqr (M G / d)
Where: G is the universal gravitational constant
Velocity V is equal to the speed of light C at a distance Rs defined as the critical radius below which
the orbital velocity would be greater than the speed of light. This critical radius defines the event
horizon of mass contracted into a spherical volume of that radius:
Rs = M G / C^2
Orbital velocity equal to the speed of light is a property of the mass event horizon.
If the Planck mass Pm = sqr (h C / G), where h is the rationalized Planck constant, is
contracted to its event horizon then that horizon is equal to the Planck length Pl = sqr (h G /
C^3) = 1.616E-33 [cm]. This is the length below which physical laws may no longer apply, and
is the length that we will use as the radius of rotation. It is important to note,
however, that in our case there is no mass at the center of rotation and, therefore, the Planck
length is simply the radius of rotation and not the event horizon of Planck mass.
Nevertheless, the linear speed of the rotating particle is equal to the speed of light, thus
making the particle a quantum of energy that we will call “ergon”.
For double helixes, we define latent energy as the energy of two ergons rotating
with equal velocities around a center at a distance equal to the Planck length.
We also define active energy as the energy resulting from a deviation from the
180 degrees phase shift of the X and Y components of the motion of the two
ergons rotating with equal velocities around a center at a distance equal to the
Planck length.
For single helixes, we define active energy as the energy of the ergon rotating
around a center at a distance equal to the Planck length.
We must define the rotating particles and their radius of rotation, but before doing so, we will
review the definition of mass event horizon given in the Author’s “Gravity and Space-Time”:
The velocity V of a body orbiting another body of mass M at a distance d is given by equating
gravitational force to centripetal force, or:
V = sqr (M G / d)
Where: G is the universal gravitational constant
Velocity V is equal to the speed of light C at a distance Rs defined as the critical radius below which
the orbital velocity would be greater than the speed of light. This critical radius defines the event
horizon of mass contracted into a spherical volume of that radius:
Rs = M G / C^2
Orbital velocity equal to the speed of light is a property of the mass event horizon.
If the Planck mass Pm = sqr (h C / G), where h is the rationalized Planck constant, is
contracted to its event horizon then that horizon is equal to the Planck length Pl = sqr (h G /
C^3) = 1.616E-33 [cm]. This is the length below which physical laws may no longer apply, and
is the length that we will use as the radius of rotation. It is important to note,
however, that in our case there is no mass at the center of rotation and, therefore, the Planck
length is simply the radius of rotation and not the event horizon of Planck mass.
Nevertheless, the linear speed of the rotating particle is equal to the speed of light, thus
making the particle a quantum of energy that we will call “ergon”.
For double helixes, we define latent energy as the energy of two ergons rotating
with equal velocities around a center at a distance equal to the Planck length.
We also define active energy as the energy resulting from a deviation from the
180 degrees phase shift of the X and Y components of the motion of the two
ergons rotating with equal velocities around a center at a distance equal to the
Planck length.
For single helixes, we define active energy as the energy of the ergon rotating
around a center at a distance equal to the Planck length.
3.0 The Nature of the Rotating Motion
A body can be induced to rotate around a center either if it is physically bound to it or if there
is a strong attractor of such a magnitude that makes the direction of the body’s acceleration
orthogonal to the motion. Excluding the trivial first case, there are only three kinds of strong
attractors that can produce such motion: magnetic fields, electrostatic fields, and gravity
fields, all of which require the presence of mass. In our case, there is no mass in the rotating
ergon nor is there mass at the center of rotation to generate such strong attractors. There
then remains only one other alternative: the high probability that at the Planck level spatial
dimensions are curled.
From the Author’s “Gravity and Space-Time”, it was shown that:
.... a 90 degree deflection of light due to the curvature of space is a property of the mass event
horizon, and the deflection is independent of the angle of incidence.
The above relates to the warping of space by the presence of mass which has nothing to do
with curled dimensions at or below Planck length. It does, however, reinforce the postulation
of the existence of curled dimensions. If mass contracted to its event horizon can warp
space by 90 degrees, so can space warp or curl itself when contracted to some minimum
value. Below this threshold, which should be more properly called the space event horizon,
space folds unto itself to the extent that physical laws, including the general theory of
relativity, no longer apply. In our case, the Planck length is such a threshold.
A body can be induced to rotate around a center either if it is physically bound to it or if there
is a strong attractor of such a magnitude that makes the direction of the body’s acceleration
orthogonal to the motion. Excluding the trivial first case, there are only three kinds of strong
attractors that can produce such motion: magnetic fields, electrostatic fields, and gravity
fields, all of which require the presence of mass. In our case, there is no mass in the rotating
ergon nor is there mass at the center of rotation to generate such strong attractors. There
then remains only one other alternative: the high probability that at the Planck level spatial
dimensions are curled.
From the Author’s “Gravity and Space-Time”, it was shown that:
.... a 90 degree deflection of light due to the curvature of space is a property of the mass event
horizon, and the deflection is independent of the angle of incidence.
The above relates to the warping of space by the presence of mass which has nothing to do
with curled dimensions at or below Planck length. It does, however, reinforce the postulation
of the existence of curled dimensions. If mass contracted to its event horizon can warp
space by 90 degrees, so can space warp or curl itself when contracted to some minimum
value. Below this threshold, which should be more properly called the space event horizon,
space folds unto itself to the extent that physical laws, including the general theory of
relativity, no longer apply. In our case, the Planck length is such a threshold.
4.0 Curled Space Gradient
If at Planck length the properties of space change from what we observe as being linear to
what we describe as being curled, then the question arises whether such change is
continuous or discontinuous. Assuming that the answer is that change is continuous,
space must then have a gradient of “curliness” that varies from a maximum value to a zero
value depending on the distance from the origin.
We must refer again to the Author’s: “Gravity and Space-Time”:
It is intended to show that the deflection of light due to the curvature of space can be obtained … by
using only two premises:
1) light is deflected 90 degrees at the mass event horizon Rs
2) equation: g = M G / d^2
At Rs we have: go = M G / Rs^2
And:
(g / go) = (Rs / d)^2
The ratio (Rs / d) can be considered as the tangent of an angle A equal to:
A = Atan (Rs / d)
And at d = Rs:
Ao = Atan (Rs / Rs) = 45 [deg]
To satisfy the first premise, the angle of light deflection at Rs must be twice the angle Ao, and:
alfa = 2 Atan (Rs / d)
Where: d is greater or equal to Rs.
It should be noted that, when d is less than Rs, the angle of deflection increases and reaches the
value of 180 degrees at d = 0 .... the point must be made that a curvature of space that generates a
light deflection of 180 degrees can only be achieved by a space structure folded unto itself.
As discussed in Paragraph 2.0, the mass event horizon Rs of Planck mass is equal to the
Planck length Pl. The equation above can then be rewritten as:
@ = 2 Atan (Pl / d)
Where: @ is the curled space gradient in degrees
d is the distance from the origin of the curl in centimeters
The Table below shows the values of the curl gradient from Pl = 1.616E-33 [cm] to 100 Pl,
and at proton size of 1E-11 [cm].
[nPl] Curl Gradient
[degrees]
1 90.00
2 53.13
3 36.87
4 28.07
5 22.62
10 11.42
100 1.15
Proton size 2E-20
If at Planck length the properties of space change from what we observe as being linear to
what we describe as being curled, then the question arises whether such change is
continuous or discontinuous. Assuming that the answer is that change is continuous,
space must then have a gradient of “curliness” that varies from a maximum value to a zero
value depending on the distance from the origin.
We must refer again to the Author’s: “Gravity and Space-Time”:
It is intended to show that the deflection of light due to the curvature of space can be obtained … by
using only two premises:
1) light is deflected 90 degrees at the mass event horizon Rs
2) equation: g = M G / d^2
At Rs we have: go = M G / Rs^2
And:
(g / go) = (Rs / d)^2
The ratio (Rs / d) can be considered as the tangent of an angle A equal to:
A = Atan (Rs / d)
And at d = Rs:
Ao = Atan (Rs / Rs) = 45 [deg]
To satisfy the first premise, the angle of light deflection at Rs must be twice the angle Ao, and:
alfa = 2 Atan (Rs / d)
Where: d is greater or equal to Rs.
It should be noted that, when d is less than Rs, the angle of deflection increases and reaches the
value of 180 degrees at d = 0 .... the point must be made that a curvature of space that generates a
light deflection of 180 degrees can only be achieved by a space structure folded unto itself.
As discussed in Paragraph 2.0, the mass event horizon Rs of Planck mass is equal to the
Planck length Pl. The equation above can then be rewritten as:
@ = 2 Atan (Pl / d)
Where: @ is the curled space gradient in degrees
d is the distance from the origin of the curl in centimeters
The Table below shows the values of the curl gradient from Pl = 1.616E-33 [cm] to 100 Pl,
and at proton size of 1E-11 [cm].
[nPl] Curl Gradient
[degrees]
1 90.00
2 53.13
3 36.87
4 28.07
5 22.62
10 11.42
100 1.15
Proton size 2E-20
5.0 The Minimum Quantum of Energy
The rest energy of the electron and of its anti particle, the positron, is the minimum quantum
of energy that we will use for the rotating ergon.
As discussed in Paragraph 1.0, we then have a right-handed helix generated by an ergon of
.511 MeV active energy (the rest energy of the electron or of the positron) rotating around a
center at a distance equal to the Planck length; and a left-handed helix generated by an
ergon of .511 MeV active energy rotating around a center at a distance equal to the Planck
length.
These two opposite single helixes are defined by the rest energy of the
electron (assumed to be the left-handed helix) and by the rest energy of the
positron (assumed to be the right-handed helix), and their electric charge.
Although equivalent in rest energy and charge, single time helixes do not have the quantum
characteristics of free positrons and electrons.
We must also consider a right-handed double helix of null active energy and charge (phase
shift of X and Y components = 180 degrees) generated by two ergons of .511 MeV latent
energy rotating with equal velocities around a center at a distance equal to the Planck
length; and a left-handed double helix of null active energy and charge generated by two
ergons of .511 MeV latent energy rotating with equal velocities around a center at a distance
equal to the Planck length.
Slight deviations from the 180 degrees phase shift of these two opposite double
helixes result in active energies that define neutrinos (assumed to be the
left-handed double helixes) and anti neutrinos (assumed to be the right-handed
double helixes).
The electron and the neutrinos and their counterparts, the positron and the anti neutrinos,
are the minimum energies resulting from the decay of all other particles, excluding the
proton. Together with the proton they are the only universally stable elements. The concept
of form or size for these minimum energies is meaningless: they are the projections on two
spatial coordinates of one quantum or of two quanta of energy rotating in a one-dimensional
curled coordinate.
In a universe that is based on symmetry, matter and anti matter have equal or
almost equal probability for existence. Therefore, equal or almost equal
combinations of single and double right-handed and left-handed helixes are the
fundamental constituents of all particles.
The rest energy of the electron and of its anti particle, the positron, is the minimum quantum
of energy that we will use for the rotating ergon.
As discussed in Paragraph 1.0, we then have a right-handed helix generated by an ergon of
.511 MeV active energy (the rest energy of the electron or of the positron) rotating around a
center at a distance equal to the Planck length; and a left-handed helix generated by an
ergon of .511 MeV active energy rotating around a center at a distance equal to the Planck
length.
These two opposite single helixes are defined by the rest energy of the
electron (assumed to be the left-handed helix) and by the rest energy of the
positron (assumed to be the right-handed helix), and their electric charge.
Although equivalent in rest energy and charge, single time helixes do not have the quantum
characteristics of free positrons and electrons.
We must also consider a right-handed double helix of null active energy and charge (phase
shift of X and Y components = 180 degrees) generated by two ergons of .511 MeV latent
energy rotating with equal velocities around a center at a distance equal to the Planck
length; and a left-handed double helix of null active energy and charge generated by two
ergons of .511 MeV latent energy rotating with equal velocities around a center at a distance
equal to the Planck length.
Slight deviations from the 180 degrees phase shift of these two opposite double
helixes result in active energies that define neutrinos (assumed to be the
left-handed double helixes) and anti neutrinos (assumed to be the right-handed
double helixes).
The electron and the neutrinos and their counterparts, the positron and the anti neutrinos,
are the minimum energies resulting from the decay of all other particles, excluding the
proton. Together with the proton they are the only universally stable elements. The concept
of form or size for these minimum energies is meaningless: they are the projections on two
spatial coordinates of one quantum or of two quanta of energy rotating in a one-dimensional
curled coordinate.
In a universe that is based on symmetry, matter and anti matter have equal or
almost equal probability for existence. Therefore, equal or almost equal
combinations of single and double right-handed and left-handed helixes are the
fundamental constituents of all particles.
6.0 Double Helix Phase Shift Deviations
A perfect double helix would be one where the X and Y components of one of the rotating
ergons is out of phase by exactly 180 degrees with the X and Y components of the other
rotating ergon, resulting in a null active energy and charge. However, slight positive or
negative deviations from the 180 degrees phase shift generate small active energies that are
proportional to the deviations. These deviations may vary to the extent that the flavor of the
neutrinos defined by the resulting active energies may also vary. The probability for the
occurrence of a deviation depends on the amount of the deviation: the higher the deviation,
the lower the probability.
Since the rest mass of the neutrino is very small, the deviation from the 180 degrees phase
shift required to generate it is also very small. Assuming that the rest mass of the heaviest
neutrino (the tau neutrino) is less than .3 eV, the deviation from the 180 degrees phase shift
that is required to generate the tau neutrino energy, according to Paragraph 1.0, is less than
0.044 degrees.
We further define neutrinos and anti neutrinos as the spectrum of active
energies resulting from the variation of slight deviations from the 180 degrees
phase shift of the X and Y components of two rotating ergons of .511 MeV latent
energy.
This definition replaces the standard model concept of neutrino’s flavor with the concept of
active energy spectrum variation.
A perfect double helix would be one where the X and Y components of one of the rotating
ergons is out of phase by exactly 180 degrees with the X and Y components of the other
rotating ergon, resulting in a null active energy and charge. However, slight positive or
negative deviations from the 180 degrees phase shift generate small active energies that are
proportional to the deviations. These deviations may vary to the extent that the flavor of the
neutrinos defined by the resulting active energies may also vary. The probability for the
occurrence of a deviation depends on the amount of the deviation: the higher the deviation,
the lower the probability.
Since the rest mass of the neutrino is very small, the deviation from the 180 degrees phase
shift required to generate it is also very small. Assuming that the rest mass of the heaviest
neutrino (the tau neutrino) is less than .3 eV, the deviation from the 180 degrees phase shift
that is required to generate the tau neutrino energy, according to Paragraph 1.0, is less than
0.044 degrees.
We further define neutrinos and anti neutrinos as the spectrum of active
energies resulting from the variation of slight deviations from the 180 degrees
phase shift of the X and Y components of two rotating ergons of .511 MeV latent
energy.
This definition replaces the standard model concept of neutrino’s flavor with the concept of
active energy spectrum variation.
7.0 The Rings Model
Particles are made up of single right-handed (+) and single left-handed (–) helixes of .511
MeV rest active energy and unit charge; and of double right-handed and double left-handed
helixes of 1.022 MeV latent energy and zero charge named “Z helixes”. Single helixes are
coupled to Z helixes of opposite handedness to form “singlets” of .511 MeV rest active
energy, 1.022 MeV latent energy, unit charge, and null handedness. A pair of opposite
singlets is named a “doublet”. The doublet ensemble has a rest active energy of 1.022 MeV,
a latent energy of 2.044 MeV, and a zero charge, and is represented as:
[+Z –Z*]
Since the handedness of the singlets components of the doublet is null, the overall
handedness of the doublet is also null.
Thus, the structure of a charged particle is composed of a number of doublets and of one
spare singlet of the same charge as the particle (with the exception of the delta ++ particle
which has two spare singlets). The structure of neutral particles is composed of a number
of doublets with no spare singlet. The maximum number of doublets in a particle is
obtained by dividing the rest mass of the particle by the relativistic mass of the doublet
(Paragraph 7.3).
The doublets are strung together, positive end to negative end, to form rings
that are considered “entities” of zero charge with no electrostatic influence
outside their perimeter.
Particles are made up of zero, one, two, or three rings depending on the family of the
particle: Baryons have three rings, Mesons two rings, Leptons (muon and tau) one ring,
and electrons and neutrinos no rings. Baryons and Mesons may have equal number of
doublets in their rings or may have a ring or rings, named “down-rings”, with one less
doublet than the other ring or rings, named “up-rings”. Leptons (muon and tau) have one
ring and are always charged but their spare singlet does not have the Z helix component.
Appendix Tables 1 and 2 show the maximum number of doublets and the type of spare
singlet for some of the particles.
Particles are made up of single right-handed (+) and single left-handed (–) helixes of .511
MeV rest active energy and unit charge; and of double right-handed and double left-handed
helixes of 1.022 MeV latent energy and zero charge named “Z helixes”. Single helixes are
coupled to Z helixes of opposite handedness to form “singlets” of .511 MeV rest active
energy, 1.022 MeV latent energy, unit charge, and null handedness. A pair of opposite
singlets is named a “doublet”. The doublet ensemble has a rest active energy of 1.022 MeV,
a latent energy of 2.044 MeV, and a zero charge, and is represented as:
[+Z –Z*]
Since the handedness of the singlets components of the doublet is null, the overall
handedness of the doublet is also null.
Thus, the structure of a charged particle is composed of a number of doublets and of one
spare singlet of the same charge as the particle (with the exception of the delta ++ particle
which has two spare singlets). The structure of neutral particles is composed of a number
of doublets with no spare singlet. The maximum number of doublets in a particle is
obtained by dividing the rest mass of the particle by the relativistic mass of the doublet
(Paragraph 7.3).
The doublets are strung together, positive end to negative end, to form rings
that are considered “entities” of zero charge with no electrostatic influence
outside their perimeter.
Particles are made up of zero, one, two, or three rings depending on the family of the
particle: Baryons have three rings, Mesons two rings, Leptons (muon and tau) one ring,
and electrons and neutrinos no rings. Baryons and Mesons may have equal number of
doublets in their rings or may have a ring or rings, named “down-rings”, with one less
doublet than the other ring or rings, named “up-rings”. Leptons (muon and tau) have one
ring and are always charged but their spare singlet does not have the Z helix component.
Appendix Tables 1 and 2 show the maximum number of doublets and the type of spare
singlet for some of the particles.
7.1 Classical Electron Radius and Separation between Singlets
The separation between the (+Z) and (–Z) singlets components of a doublet must take into
account the size of the components and, more importantly, the effect of one component on
the other component. Since single and double time helixes do not have size, only the effect
is then considered.
A positron of rest mass m rotating at the speed of light C around an electron at a distance d
generates a centrifugal force equal to:
Fg = (m C^2) / d
Independently of the rotation there is a centripetal electrostatic force equal to:
Fe = (e / d)^2
Where: e is the electron charge of 4.8E-10 [esu].
Fg is equal to Fe at a distance:
d* = e^2 / (m C^2) = 2.82E-13 [cm].
This distance is the so called classical electron radius. In the rings model, the distance
separating adjacent singlets is constant and equal to the classical electron
radius of 2.82E-13 [cm].
The separation between two adjacent singlets in a ring is the chord of the arc subtended by
the two singlets:
sep = 2 r sin (pi / s)
Where: r is the radius of the ring
s is the number of singlets in the ring
From the equation above, the radius of a ring is:
r = sep / {2 sin (pi / s)}
For instance, the radius of a ring of 218 doublets is 1.96E-11 [cm].
The electrostatic force between two adjacent singlets in a ring is given by:
Fe =(e / sep)^2
The force of attraction between singlets separated by a distance equal to the classical
electron radius is 2,905,700 [gr cm sec-2].
The separation between the (+Z) and (–Z) singlets components of a doublet must take into
account the size of the components and, more importantly, the effect of one component on
the other component. Since single and double time helixes do not have size, only the effect
is then considered.
A positron of rest mass m rotating at the speed of light C around an electron at a distance d
generates a centrifugal force equal to:
Fg = (m C^2) / d
Independently of the rotation there is a centripetal electrostatic force equal to:
Fe = (e / d)^2
Where: e is the electron charge of 4.8E-10 [esu].
Fg is equal to Fe at a distance:
d* = e^2 / (m C^2) = 2.82E-13 [cm].
This distance is the so called classical electron radius. In the rings model, the distance
separating adjacent singlets is constant and equal to the classical electron
radius of 2.82E-13 [cm].
The separation between two adjacent singlets in a ring is the chord of the arc subtended by
the two singlets:
sep = 2 r sin (pi / s)
Where: r is the radius of the ring
s is the number of singlets in the ring
From the equation above, the radius of a ring is:
r = sep / {2 sin (pi / s)}
For instance, the radius of a ring of 218 doublets is 1.96E-11 [cm].
The electrostatic force between two adjacent singlets in a ring is given by:
Fe =(e / sep)^2
The force of attraction between singlets separated by a distance equal to the classical
electron radius is 2,905,700 [gr cm sec-2].
7.2 Ring Rotation
Even though rings are entities of zero charge each singlet in a ring is subjected to
electrostatic forces from all other singlets in the ring. The overall force affecting a singlet is
computed by summation of the X and Y components of each force and taking the square
root of the sum of the squares of the components. The resultant Y component of the force
is always zero while the resultant X component is always positive (attractive) and inversely
proportional to the number of singlets in the ring. This attractive force is centripetal and
generates a constrictive stress on the ring unless counteracted by an opposing force
which is provided by a right handed or left handed rotation of the ring:
Fc = (m V^2) / r = m r W^2 = Fxtot
Where: Fc is the rotational centripetal force
m is the relativistic mass of the singlet
r is the radius of the ring
V is the rotation linear velocity
W is the rotation angular velocity equal to V / r
From the equation above we have:
V = sqr (r Fxtot / m)
W = sqr {Fxtot / (m r )}
The rotational velocity V of a ring would make the forces acting between and among
singlets null, resulting in a stress free ring.
These equations, however, can not be solved directly because of the reciprocal
dependency among number of singlets, centripetal force, ring velocity, and singlet
relativistic mass as discussed in Paragraph 7.3.
Since rings are entities that are not affected electrostatically by adjacent rings they can
rotate about any axis and in any directions in addition to their right handed or left handed
spin.
Even though rings are entities of zero charge each singlet in a ring is subjected to
electrostatic forces from all other singlets in the ring. The overall force affecting a singlet is
computed by summation of the X and Y components of each force and taking the square
root of the sum of the squares of the components. The resultant Y component of the force
is always zero while the resultant X component is always positive (attractive) and inversely
proportional to the number of singlets in the ring. This attractive force is centripetal and
generates a constrictive stress on the ring unless counteracted by an opposing force
which is provided by a right handed or left handed rotation of the ring:
Fc = (m V^2) / r = m r W^2 = Fxtot
Where: Fc is the rotational centripetal force
m is the relativistic mass of the singlet
r is the radius of the ring
V is the rotation linear velocity
W is the rotation angular velocity equal to V / r
From the equation above we have:
V = sqr (r Fxtot / m)
W = sqr {Fxtot / (m r )}
The rotational velocity V of a ring would make the forces acting between and among
singlets null, resulting in a stress free ring.
These equations, however, can not be solved directly because of the reciprocal
dependency among number of singlets, centripetal force, ring velocity, and singlet
relativistic mass as discussed in Paragraph 7.3.
Since rings are entities that are not affected electrostatically by adjacent rings they can
rotate about any axis and in any directions in addition to their right handed or left handed
spin.
7.3 The Effect of Rotational Velocity on Mass of Ring
The ring rotational velocity can only be determined indirectly by using an iterative process
that would find the relativistic mass of a rotating ring that is equal to the rest mass of a
target ring. More specifically: given a ring A of x doublets find a ring B of y doublets, where
y < x, so that the relativistic mass of B is equal to the rest mass of A.
The following equations are required:
V^2 = r F / m
m = mo / R
R = sqr (1 – Vc)
Vc = (V / C)^2
Where: r is the radius of the ring
F is the centripetal force on the singlet
m is the relativistic mass of the singlet
mo is the rest mass of the singlet
V is the linear velocity of the singlet
C is the speed of light
From the equations above we get:
(r F)^2 = (Vc mo)^2 C^4 + Vc (r F)^2
Setting: x = Vc
a = mo^2 C^4
b = (r F)^2
We get a quadratic of the form:
a x^2 + b x – b = 0
Solving it for the real root:
x = [ – b + sqr (b^2 + 4 a b)] / 2 a
V = C sqr (x)
m = mo / sqr (1 – x)
Using the neutron as an example:
Rest mass 939.5656 [MeV]
Number of rings 3
Doublets per ring 218, 218, 218
Ring rotational velocity 2.11E10 [cm/sec]
Doublets relativistic mass 1.4361 [MeV]
Ring relativistic mass 313.067 [MeV]
Total relativistic mass 939.201 [MeV]
Mass difference - 0.3646 [MeV]
Rings rotational velocities for rings from 40 to 7000 doublets increase only by a factor of
1.00006 and therefore can be considered independent of the number of doublets and equal
to 2.11E10 [cm/sec] or about 70 % of the speed of light.
We can also calculate a singlet frequency defined as the number of singlets passing
through a point in the ring per unit time:
f = 1 / t
t = 2pi / (W s)
Where: W is the ring angular velocity equal to V / r
s is the number of singlets in the ring
Singlets rotational frequencies for rings from 40 to 7000 doublets increase only by a factor
of 1.0003 and therefore can be considered independent of the number of doublets and
equal to 7.47E22 [singlets/sec].
Using Planck’s equation m = f h / C^2 where m is mass, f is frequency, h is the Planck
constant, and C is the speed of light, that frequency is equivalent to a rest mass of 309.1
[MeV] which is equal to the relativistic mass of a rotating ring of 215 doublets.
This may be an indication of why the proton is the universally stable particle among all
other particles, followed closely by the neutron: their rings compositions are the closest to
a ring of 215 doublets which is the only composition that allows a synchronous frequency
between internal motion and mass. Rings with less than 215 doublets have singlets
rotational frequencies that are greater than their mass frequencies. The reverse is true for
rings with more than 215 doublets. However, individual ring stability does not necessarily
mean stability of the three-ring formation of the proton.
The ring rotational velocity can only be determined indirectly by using an iterative process
that would find the relativistic mass of a rotating ring that is equal to the rest mass of a
target ring. More specifically: given a ring A of x doublets find a ring B of y doublets, where
y < x, so that the relativistic mass of B is equal to the rest mass of A.
The following equations are required:
V^2 = r F / m
m = mo / R
R = sqr (1 – Vc)
Vc = (V / C)^2
Where: r is the radius of the ring
F is the centripetal force on the singlet
m is the relativistic mass of the singlet
mo is the rest mass of the singlet
V is the linear velocity of the singlet
C is the speed of light
From the equations above we get:
(r F)^2 = (Vc mo)^2 C^4 + Vc (r F)^2
Setting: x = Vc
a = mo^2 C^4
b = (r F)^2
We get a quadratic of the form:
a x^2 + b x – b = 0
Solving it for the real root:
x = [ – b + sqr (b^2 + 4 a b)] / 2 a
V = C sqr (x)
m = mo / sqr (1 – x)
Using the neutron as an example:
Rest mass 939.5656 [MeV]
Number of rings 3
Doublets per ring 218, 218, 218
Ring rotational velocity 2.11E10 [cm/sec]
Doublets relativistic mass 1.4361 [MeV]
Ring relativistic mass 313.067 [MeV]
Total relativistic mass 939.201 [MeV]
Mass difference - 0.3646 [MeV]
Rings rotational velocities for rings from 40 to 7000 doublets increase only by a factor of
1.00006 and therefore can be considered independent of the number of doublets and equal
to 2.11E10 [cm/sec] or about 70 % of the speed of light.
We can also calculate a singlet frequency defined as the number of singlets passing
through a point in the ring per unit time:
f = 1 / t
t = 2pi / (W s)
Where: W is the ring angular velocity equal to V / r
s is the number of singlets in the ring
Singlets rotational frequencies for rings from 40 to 7000 doublets increase only by a factor
of 1.0003 and therefore can be considered independent of the number of doublets and
equal to 7.47E22 [singlets/sec].
Using Planck’s equation m = f h / C^2 where m is mass, f is frequency, h is the Planck
constant, and C is the speed of light, that frequency is equivalent to a rest mass of 309.1
[MeV] which is equal to the relativistic mass of a rotating ring of 215 doublets.
This may be an indication of why the proton is the universally stable particle among all
other particles, followed closely by the neutron: their rings compositions are the closest to
a ring of 215 doublets which is the only composition that allows a synchronous frequency
between internal motion and mass. Rings with less than 215 doublets have singlets
rotational frequencies that are greater than their mass frequencies. The reverse is true for
rings with more than 215 doublets. However, individual ring stability does not necessarily
mean stability of the three-ring formation of the proton.
7.4 Rings Formations
1) Baryons
If the separation between singlets in a ring is equal to the electron classical radius, the
separation, or gap, between rings has to be greater than that radius.
Assuming an equal gap between rings of equal number of doublets, the three-ring
formation of Baryons implies either a layered configuration with the centers of the rings
lying on a line of length equal to twice the gap between the rings; a triangular configuration
with the centers of the rings at the vertices of an equilateral triangle with sides equal to
twice the radius of the rings plus the gap between rings; or a linear configuration with the
rings lying on a line of length equal to four times the radius of the rings plus twice the gap
between rings. The rings model as presented in this paper only considers the triangular
configuration.
The circle circumscribing the rings and defining the perimeter of the particle has a diameter
equal to:
diameter = 2 r + 4/3 sin 60 (2 r + gap)
Where: r is the radius of the rings
The triangular configuration is no longer exactly equilateral and the included angles no
longer equal to 60 degrees for formations where rings do not have equal number of
doublets (down-ring and up-ring formations).
For the negative omega particle with three rings of 388 doublets separated by a gap equal
to twice the singlets separation:
rings radius = 3.48E-11 [cm]
included angles = 60, 60, 60 [deg]
particle diameter = 1.51E-10 [cm]
For the proton with two rings of 218 doublets and one ring of 217 doublets separated by a
gap equal to twice the singlets separation:
218 ring radius = 1.96E-11[cm]
217 ring radius = 1.95E-11 [cm]
included angles = 59.95, 60.1, 59.95 [deg]
particle diameter = 8.49E-11 [cm]
The spare singlet of charged Baryons is located at the center of the triangle which is also
the center of the particle (the delta ++ particle has two spare singlets at the center).
The shortest distance between the spare singlet and the rings, assuming an equilateral
triangular configuration, is given by:
d = 2/3 sin 60 (2 r + gap) – r
For three rings of 218 doublets separated by a gap equal to twice the singlets separation
the distance is 3.35E-12 [cm], or 11.9 times the separation between singlets.
2) Mesons
The two-ring formation of Mesons implies either a layered configuration with the centers of
the rings lying on a line of length equal to the gap between the rings; or a linear
configuration with the centers of the rings in a line of length equal to twice the radius of the
rings plus the separation or gap between the rings. The rings model as presented in this
paper only considers the linear configuration. As in the three-ring formation of Baryons,
the gap has to be greater than the separation between singlets. For rings of equal number
of doublets the circle circumscribing the rings and defining the perimeter of the particle
has a diameter equal to:
diameter = 4 r + gap
Where: r is the radius of the rings
For a neutral pion (two rings of 47 doublets separated by a gap equal to twice the singlets
separation): rings radius = 4.22E-12 [cm]; particle diameter = 1.74E-11 [cm]. For the upsilon
(two rings of 3294 doublets separated by a gap equal to twice the singlets separation):
rings radius = 2.95E-10 [cm]; particle diameter = 1.18E-9 [cm].
The location of the spare singlet of charged Mesons is on the center line of the gap
separating the two rings.
3) Leptons
The spare singlet of the one-ring formation of Leptons does not have the Z helix
component and is located at the center of the ring where it is unaffected by the
surrounding singlets but where it has a great influence on the ring.
If a singlet in the ring is of the same charge as the spare singlet then that singlet will be
subjected to a lower centripetal force than the one to which it is subjected without the
spare singlet. If a singlet in the ring is of the opposite charge of the spare singlet then that
singlet will be subjected to a higher centripetal force than the one to which it is subjected
without the spare singlet. This is inducing a torque on the doublets that causes the break
up of the ring.
1) Baryons
If the separation between singlets in a ring is equal to the electron classical radius, the
separation, or gap, between rings has to be greater than that radius.
Assuming an equal gap between rings of equal number of doublets, the three-ring
formation of Baryons implies either a layered configuration with the centers of the rings
lying on a line of length equal to twice the gap between the rings; a triangular configuration
with the centers of the rings at the vertices of an equilateral triangle with sides equal to
twice the radius of the rings plus the gap between rings; or a linear configuration with the
rings lying on a line of length equal to four times the radius of the rings plus twice the gap
between rings. The rings model as presented in this paper only considers the triangular
configuration.
The circle circumscribing the rings and defining the perimeter of the particle has a diameter
equal to:
diameter = 2 r + 4/3 sin 60 (2 r + gap)
Where: r is the radius of the rings
The triangular configuration is no longer exactly equilateral and the included angles no
longer equal to 60 degrees for formations where rings do not have equal number of
doublets (down-ring and up-ring formations).
For the negative omega particle with three rings of 388 doublets separated by a gap equal
to twice the singlets separation:
rings radius = 3.48E-11 [cm]
included angles = 60, 60, 60 [deg]
particle diameter = 1.51E-10 [cm]
For the proton with two rings of 218 doublets and one ring of 217 doublets separated by a
gap equal to twice the singlets separation:
218 ring radius = 1.96E-11[cm]
217 ring radius = 1.95E-11 [cm]
included angles = 59.95, 60.1, 59.95 [deg]
particle diameter = 8.49E-11 [cm]
The spare singlet of charged Baryons is located at the center of the triangle which is also
the center of the particle (the delta ++ particle has two spare singlets at the center).
The shortest distance between the spare singlet and the rings, assuming an equilateral
triangular configuration, is given by:
d = 2/3 sin 60 (2 r + gap) – r
For three rings of 218 doublets separated by a gap equal to twice the singlets separation
the distance is 3.35E-12 [cm], or 11.9 times the separation between singlets.
2) Mesons
The two-ring formation of Mesons implies either a layered configuration with the centers of
the rings lying on a line of length equal to the gap between the rings; or a linear
configuration with the centers of the rings in a line of length equal to twice the radius of the
rings plus the separation or gap between the rings. The rings model as presented in this
paper only considers the linear configuration. As in the three-ring formation of Baryons,
the gap has to be greater than the separation between singlets. For rings of equal number
of doublets the circle circumscribing the rings and defining the perimeter of the particle
has a diameter equal to:
diameter = 4 r + gap
Where: r is the radius of the rings
For a neutral pion (two rings of 47 doublets separated by a gap equal to twice the singlets
separation): rings radius = 4.22E-12 [cm]; particle diameter = 1.74E-11 [cm]. For the upsilon
(two rings of 3294 doublets separated by a gap equal to twice the singlets separation):
rings radius = 2.95E-10 [cm]; particle diameter = 1.18E-9 [cm].
The location of the spare singlet of charged Mesons is on the center line of the gap
separating the two rings.
3) Leptons
The spare singlet of the one-ring formation of Leptons does not have the Z helix
component and is located at the center of the ring where it is unaffected by the
surrounding singlets but where it has a great influence on the ring.
If a singlet in the ring is of the same charge as the spare singlet then that singlet will be
subjected to a lower centripetal force than the one to which it is subjected without the
spare singlet. If a singlet in the ring is of the opposite charge of the spare singlet then that
singlet will be subjected to a higher centripetal force than the one to which it is subjected
without the spare singlet. This is inducing a torque on the doublets that causes the break
up of the ring.
8.0 Particles Decay
The only universally stable particles are the proton, the electron, and the neutrinos, and
the electron and neutrinos anti particles. All other particles decay into those above. In
addition to general conservation laws, of which the law of conservation of charge is of
primary importance, the guidelines for the decay process are:
1) Decays are from high to low rest mass within a family.
2) In the Baryons family, a particle decays either into a proton or a neutron or into a
particle which in turn decays into a proton or a neutron. If enough residual energy is
available then the particle further decays into a pion or a kaon, the particles with the lowest
rest mass of the Mesons family.
3) In the Mesons family, a particle decays into pions or kaons or into a particle which in
turn decays into pions or kaons. Pions decay into electrons and anti neutrinos, or into
positrons and neutrinos; or into particles of the Leptons family which in turn decay into
electrons and anti neutrinos, or into positrons and neutrinos. Kaons decay into pions or
into particles of the Leptons family.
4) In the Leptons family, particles decay into electrons and anti neutrinos or into
positrons and neutrinos.
There are exceptions to the guidelines. In the Mesons family, the decay of the neutral pion
gives an electron and a positron which annihilate into two gamma rays; and the decay of
the J/Psi particle gives an electron and a positron, or a muon and an anti muon.
Appendix Tables 3 and 4 show the modes of decay of some of the particles.
Using the guidelines above and the conservation laws of energy and charge, we can
produce a table of particle decays that predicts modes of decay for any one particle.
However, the probability that a particle decays in a particular mode depends on the rest
mass of the particle and on the amount of energy released in the decay mode.
Using ratios to express probabilities and doublets for mass and energy, we have:
P1 = 1 – Dr / D
P2 = 1 – A / D
Ptotal = P1 P2
Where: Dr is the number of released doublets
D is the number of doublets in the decaying particle
A is equal to 94 for Mesons (the doublets in the neutral pion)
and to 653 for Baryons (the doublets in the proton)
Ptotal is the resulting decay probability
Appendix Tables 5 and 6 show the most probable modes of decay for Baryons and
Mesons.
From Appendix Table 5 we see that the neutron n(0) has the least probability of decay
among all Baryons, followed by the neutral lambda L(0) which decays with almost equal
probabilities into two different modes. At the upper end, we see that the negative omega O
(–) most probably decays into a neutral lambda L(0) and a kaon K(–), and less probably
into either a neutral cascade Xi(0) and a negative pion Pi(–) or a negative cascade Xi(–) and
a neutral pion Pi(0).
The decay process takes place in a stepwise manner, giving preference to the steps that
release the least amount of energy, with the result that the most probable path is the
longest in terms of steps and time.
The only universally stable particles are the proton, the electron, and the neutrinos, and
the electron and neutrinos anti particles. All other particles decay into those above. In
addition to general conservation laws, of which the law of conservation of charge is of
primary importance, the guidelines for the decay process are:
1) Decays are from high to low rest mass within a family.
2) In the Baryons family, a particle decays either into a proton or a neutron or into a
particle which in turn decays into a proton or a neutron. If enough residual energy is
available then the particle further decays into a pion or a kaon, the particles with the lowest
rest mass of the Mesons family.
3) In the Mesons family, a particle decays into pions or kaons or into a particle which in
turn decays into pions or kaons. Pions decay into electrons and anti neutrinos, or into
positrons and neutrinos; or into particles of the Leptons family which in turn decay into
electrons and anti neutrinos, or into positrons and neutrinos. Kaons decay into pions or
into particles of the Leptons family.
4) In the Leptons family, particles decay into electrons and anti neutrinos or into
positrons and neutrinos.
There are exceptions to the guidelines. In the Mesons family, the decay of the neutral pion
gives an electron and a positron which annihilate into two gamma rays; and the decay of
the J/Psi particle gives an electron and a positron, or a muon and an anti muon.
Appendix Tables 3 and 4 show the modes of decay of some of the particles.
Using the guidelines above and the conservation laws of energy and charge, we can
produce a table of particle decays that predicts modes of decay for any one particle.
However, the probability that a particle decays in a particular mode depends on the rest
mass of the particle and on the amount of energy released in the decay mode.
Using ratios to express probabilities and doublets for mass and energy, we have:
P1 = 1 – Dr / D
P2 = 1 – A / D
Ptotal = P1 P2
Where: Dr is the number of released doublets
D is the number of doublets in the decaying particle
A is equal to 94 for Mesons (the doublets in the neutral pion)
and to 653 for Baryons (the doublets in the proton)
Ptotal is the resulting decay probability
Appendix Tables 5 and 6 show the most probable modes of decay for Baryons and
Mesons.
From Appendix Table 5 we see that the neutron n(0) has the least probability of decay
among all Baryons, followed by the neutral lambda L(0) which decays with almost equal
probabilities into two different modes. At the upper end, we see that the negative omega O
(–) most probably decays into a neutral lambda L(0) and a kaon K(–), and less probably
into either a neutral cascade Xi(0) and a negative pion Pi(–) or a negative cascade Xi(–) and
a neutral pion Pi(0).
The decay process takes place in a stepwise manner, giving preference to the steps that
release the least amount of energy, with the result that the most probable path is the
longest in terms of steps and time.
8.1 The Mechanics of Particle Decay
According to the rings model, particle decay is a process of reduction from an unstable
ring structure of high energy to a more stable ring structure of lower energy. The single
attractor in the process is the universally stable three-ring structure of the proton.
Depending on the mode of decay, one or more of the following elementary structures may
also be released: the single left-handed helix of the free electron; the single right-handed
helix of the free positron; the double right-handed Z helix of the free neutrino; and the
double left-handed Z helix of the free anti neutrino.
One of the differences between the standard model and the rings model is the definition of
neutrinos and anti neutrinos. In the standard model, neutrinos and anti neutrinos are
considered particles of the Leptons family even though their masses are assumed to be
zero. In the rings model, neutrinos and anti neutrinos are instead defined by the spectrum
of active energies resulting from variation of the phase shift deviations of Z helixes
(Paragraph 6.0). According to the rings model, a Z helix component of a free doublet (a
doublet that has been released from a ring structure) could be observed as a neutrino or
as an anti neutrino depending on its handedness and on the probability of a phase shift
deviation. Depending on that probability, neutrinos and anti neutrinos can then be
released in any steps of a particle decay process as long as there are free doublets. In
addition, Z helixes with phase shift deviation that are observed as neutrinos or as anti
neutrinos are also released by the decoupling, or splitting, of spare singlets into
components.
When ejected from a ring structure, a doublet looses its identity and separates into its
components: the active energy of 1.022MeV of the two single helixes is released as kinetic
energy; the active energy of a Z helix with phase shift deviation is released as a neutrino or
anti neutrino depending on the helix’s handedness; and a Z helix with no phase shift
deviation is released as free Z helix.
We will use the examples of the decay of neutral lambda and positive sigma particles to
facilitate our discussion.
Neutral Lambda
One of the modes of decay of the neutral lambda particle is:
1) L(0) --> p(+) pi(–) (E)
2) pi(–) --> m(–) v(m)* (E)
3) m(–) --> e(–) v(e)* v(m) (E)
The results are:
L(0) --> p(+) e(–) v(e)* v(m) v(m)* (E)
where E is the amount of energy released in each step and the asterisk denotes anti
particle.
According to the rings model, the neutral lambda decay is written as:
1) [776] --> [653 +Z] [96 –Z*] (26)
2) [96 –Z*] --> [73 –] Z* (23)
3) [73 –] --> (–) (73)
The results are:
[776] --> [653 +Z] (–) Z* (122)
where the number in brackets is the number of doublets in the particle; the number in
parentheses is the number of free doublets; and the asterisk denotes anti neutrino.
There are two stages in the decay shown as Step 1. Firstly, the triangular structure of
neutral lambda restructures itself into that of the proton by releasing 41 doublets from
each of its two up-rings and 40 doublets from its down-ring, and by splitting one doublet
from the down-ring into (+) Z and (–) Z* singlets. The (+) Z singlet stays as the spare singlet
of the proton, while the (–) Z* singlet is released together with the 122 doublets as available
residual energy:
1a) [259, 259, 258] --> [218, 218, 217 +Z] [122 –Z*]
Secondly, since there is enough residual energy available, the released doublets
restructure themselves into a pion of a charge equal to the charge of the available spare
singlet (in this case, a negative pion). The remaining 26 doublets are released as free
doublets:
1b) [122 –Z*] --> [48, 48 –Z*] (26)
In Step 2, the two-ring linear structure of the negative pion restructures itself into the single
ring structure of a negative muon by releasing 23 doublets from its two rings in addition to
releasing the Z* component of its spare singlet as an anti neutrino by decoupling it from
the (–) component. The 23 doublets are released as free doublets:
2) [48, 48 –Z*] --> [73 –] Z* (23)
In Step 3, the one-ring structure of negative muon breaks down completely by releasing
the (–) component as an electron and all of its 73 doublets as free doublets:
3) [73 (–)] --> (–) (73)
Positive Sigma
One of the modes of decay of the positive sigma particle is:
1) S(+) --> n(0) pi(+) (E)
2) n(0) --> p(+) e(–) v(e)* (E)
3) pi(+) --> m(+) v(m) (E)
4) m(+) --> e(+) v(e) v(m)* (E)
The results of the decay are:
S(+) --> p(+) e(–) v(e)* e(+) v(e) v(m) v(m)* (E)
where E is the amount of energy released in each step and the asterisk denotes anti
neutrino.
According to the rings model, the positive sigma decay is written as:
1) [828 +Z] --> [654] [96 +Z] (78)
2) [654] --> [653 +Z] (–) Z* (0)
3) [96 +Z] --> [73 +] Z (23)
4) [73 +] --> (+) (73)
The results of the decay are:
[828 +Z] --> [653 +Z] (–) Z* (+) Z (174)
where the number in brackets is the number of doublets in the particle; the number in
parentheses is the number of free doublets; and the asterisk denotes anti neutrino.
There are two stages in the decay shown as Step 1. Firstly, the triangular structure of
positive sigma restructures itself into that of the neutron by releasing 58 doublets from
each of its rings and the (+) Z spare singlet. The (+) Z singlet and the 174 doublets are
released as available residual energy:
1a) [276, 276, 276 +Z] --> [218, 218, 218] [174 +Z]
Secondly, since there is enough residual energy available, the released doublets
restructure themselves into a pion of a charge equal to the charge of the available spare
singlet (in this case, a positive pion). The remaining 78 doublets are released as free
doublets:
1b) [174 +Z] --> [48, 48 +Z] (78)
In Step 2, the triangular structure of the neutron restructures itself (beta decay) into that of
the proton by splitting a doublet into (+) Z and (–) Z* singlets from one of its rings. The (+) Z
singlet stays with the proton as the spare singlet, while the (–) Z* singlet decouples into (–)
and Z* components which are released as electron and anti neutrino:
2) [218, 218, 218] --> [218, 218, 217 +Z] (–) Z* (0)
In Step 3, the two-ring linear structure of the positive pion restructures itself into the single
ring structure of a positive muon by releasing 23 doublets from its two rings in addition to
releasing the Z component of its spare singlet as a neutrino by decoupling it from the (+)
component. The 23 doublets are released as free doublets:
3) [48, 48 +Z] --> [73 +] Z (23)
In Step 4, the one-ring structure of positive muon breaks down completely by releasing the
(+) component as a positron and all of its 73 doublets as free doublets:
4) [73 +] --> (+) (73)
The Table below shows the end results of the decay process for most Baryons. The first
line identifies the particle and its mode of decay; the second line gives the end results
according to the rings model. The numbers in brackets are doublets; the numbers in
parentheses are free doublets; and the asterisks denote anti neutrinos. The end results
include those of the beta decay of the neutron if a neutron is involved in the process.
n(0) --> p(+) e(–) v(e)*
[654] --> [653 +Z] (–) Z* (0)
L(0) --> p(+) pi(–)
[776] --> [653 +Z] (–) Z* (122)
L(0) --> n(0) pi(0)
[776] --> [653 +Z] 2{(–) Z*} (+) Z (121)
S(+) --> p(+) pi(0)
[828 +Z] --> [653 +Z] (–) Z* (+) Z (174)
S(–) --> n(0) pi(–)
[833 –Z*] --> [653 +Z] 2{(–) Z*} (179)
D(++) --> p(+) pi(+)
[857 +Z+Z] --> [653 +Z] (+) Z (204)
D(+) --> p(+) pi(0)
[857 +Z] --> [653 +Z] (–) Z* (+) Z (203)
D(0) --> n(0) pi(0)
[858] --> [653 +Z] 2{(–) Z*} (+) Z (202)
D(0) --> p(+) pi(–)
[858] --> [653 +Z] (–) Z* (203)
D(–) --> n(0) pi(–)
[857 –Z*] --> [653 +Z] 2{(–) Z*} (203)
X(0) --> L(0) pi(0)
[915] --> [653 +Z] 2(–) Z* (+) Z (260)
[915] --> [653 +Z] 3(–) Z* 2{(+ ) Z} (259)
X(–) --> L(0) pi(–)
[919 –Z*] --> [653 +Z] 2{(–) Z*} (265)
[919 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (264)
O(–) --> L(0) K(–)
[1164 –Z*] --> [653 +Z] 2{(–) Z*} (510)
[1164 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (509)
O(–) --> X(–) pi(0)
[1164 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (509)
[1164 –Z*] --> [653 +Z] 4{(–) Z*} 2{(+) Z} (508)
The following observations can be made:
1) Particle decay is a process of re-formation rather than trans-formation. The three-
ring structure of a Baryon remains intact to the end of the process. The only activity in the
rings is the stepwise ejection of doublets from an unstable configuration to a more stable
one. If the ejection results in the release of enough energy then the ejected doublets
rearrange themselves into the two-ring structure of a Meson. A Meson either rearranges
itself into the one-ring configuration of a Lepton or, if enough energy is available, it splits
itself into two more stable Mesons. The one-ring structure of Leptons breaks down
completely into free doublets releasing the charged component of its spare singlet as free
electron or free positron.
2) The charge of a particle is of primary importance in the process. All particles decay
into a proton regardless of their charge. Therefore, other particles must be generated in the
decay so that the net charge of all resultant particles of the decay is equal to the charge of
the decaying particle (law of charge conservation). The Table below summarizes the
observation using the results above.
[0] --> [+Z] (–) Z*
[+Z] 2(–) 2Z* (+) Z
[+Z] --> [+Z] (–) Z* (+) Z
[–Z*] --> [+Z] 2(–) 2Z*
[+Z] 3(–) 3Z* (+) Z
[+Z] 4(–) 4Z* 2(+) 2 Z
3) The net sum of the handedness of the resultant Z particles (double left-handed
helixes or neutrinos) and Z* particles (double right-handed helixes or anti neutrinos) is
equal to the handedness of the Z component of the spare singlet of the decaying particle.
By net sum of the handedness we mean that right-handedness plus left-handedness equal
zero or, more simply, Z* + Z = 0. This is a handedness conservation law since the
handedness of doublets and free doublets is zero.
4) The number of free doublets released by the process is given by:
y = d – (a + n)
where: d is the number of doublets in the decaying particle
a = 653 (the number of doublets in the proton)
n is the number of doublets that are split in the process (0 to 3)
If d = 1306 and n = 0 then y = 653. This means that there is enough available residual
energy to generate an additional proton. We can define harmonics of Baryon particles by
the number of protons that could be generated in the decay:
1 proton (first harmonic): a = 1 x 653 for d = 653 to 1305 and n = 0
2 protons (second harmonic): a = 2 x 653 for d = 1306 to 1958 and n = 0
3 protons (third harmonic): a = 3 x 653 for d = 1959 to 2611 and n = 0
According to the rings model, particle decay is a process of reduction from an unstable
ring structure of high energy to a more stable ring structure of lower energy. The single
attractor in the process is the universally stable three-ring structure of the proton.
Depending on the mode of decay, one or more of the following elementary structures may
also be released: the single left-handed helix of the free electron; the single right-handed
helix of the free positron; the double right-handed Z helix of the free neutrino; and the
double left-handed Z helix of the free anti neutrino.
One of the differences between the standard model and the rings model is the definition of
neutrinos and anti neutrinos. In the standard model, neutrinos and anti neutrinos are
considered particles of the Leptons family even though their masses are assumed to be
zero. In the rings model, neutrinos and anti neutrinos are instead defined by the spectrum
of active energies resulting from variation of the phase shift deviations of Z helixes
(Paragraph 6.0). According to the rings model, a Z helix component of a free doublet (a
doublet that has been released from a ring structure) could be observed as a neutrino or
as an anti neutrino depending on its handedness and on the probability of a phase shift
deviation. Depending on that probability, neutrinos and anti neutrinos can then be
released in any steps of a particle decay process as long as there are free doublets. In
addition, Z helixes with phase shift deviation that are observed as neutrinos or as anti
neutrinos are also released by the decoupling, or splitting, of spare singlets into
components.
When ejected from a ring structure, a doublet looses its identity and separates into its
components: the active energy of 1.022MeV of the two single helixes is released as kinetic
energy; the active energy of a Z helix with phase shift deviation is released as a neutrino or
anti neutrino depending on the helix’s handedness; and a Z helix with no phase shift
deviation is released as free Z helix.
We will use the examples of the decay of neutral lambda and positive sigma particles to
facilitate our discussion.
Neutral Lambda
One of the modes of decay of the neutral lambda particle is:
1) L(0) --> p(+) pi(–) (E)
2) pi(–) --> m(–) v(m)* (E)
3) m(–) --> e(–) v(e)* v(m) (E)
The results are:
L(0) --> p(+) e(–) v(e)* v(m) v(m)* (E)
where E is the amount of energy released in each step and the asterisk denotes anti
particle.
According to the rings model, the neutral lambda decay is written as:
1) [776] --> [653 +Z] [96 –Z*] (26)
2) [96 –Z*] --> [73 –] Z* (23)
3) [73 –] --> (–) (73)
The results are:
[776] --> [653 +Z] (–) Z* (122)
where the number in brackets is the number of doublets in the particle; the number in
parentheses is the number of free doublets; and the asterisk denotes anti neutrino.
There are two stages in the decay shown as Step 1. Firstly, the triangular structure of
neutral lambda restructures itself into that of the proton by releasing 41 doublets from
each of its two up-rings and 40 doublets from its down-ring, and by splitting one doublet
from the down-ring into (+) Z and (–) Z* singlets. The (+) Z singlet stays as the spare singlet
of the proton, while the (–) Z* singlet is released together with the 122 doublets as available
residual energy:
1a) [259, 259, 258] --> [218, 218, 217 +Z] [122 –Z*]
Secondly, since there is enough residual energy available, the released doublets
restructure themselves into a pion of a charge equal to the charge of the available spare
singlet (in this case, a negative pion). The remaining 26 doublets are released as free
doublets:
1b) [122 –Z*] --> [48, 48 –Z*] (26)
In Step 2, the two-ring linear structure of the negative pion restructures itself into the single
ring structure of a negative muon by releasing 23 doublets from its two rings in addition to
releasing the Z* component of its spare singlet as an anti neutrino by decoupling it from
the (–) component. The 23 doublets are released as free doublets:
2) [48, 48 –Z*] --> [73 –] Z* (23)
In Step 3, the one-ring structure of negative muon breaks down completely by releasing
the (–) component as an electron and all of its 73 doublets as free doublets:
3) [73 (–)] --> (–) (73)
Positive Sigma
One of the modes of decay of the positive sigma particle is:
1) S(+) --> n(0) pi(+) (E)
2) n(0) --> p(+) e(–) v(e)* (E)
3) pi(+) --> m(+) v(m) (E)
4) m(+) --> e(+) v(e) v(m)* (E)
The results of the decay are:
S(+) --> p(+) e(–) v(e)* e(+) v(e) v(m) v(m)* (E)
where E is the amount of energy released in each step and the asterisk denotes anti
neutrino.
According to the rings model, the positive sigma decay is written as:
1) [828 +Z] --> [654] [96 +Z] (78)
2) [654] --> [653 +Z] (–) Z* (0)
3) [96 +Z] --> [73 +] Z (23)
4) [73 +] --> (+) (73)
The results of the decay are:
[828 +Z] --> [653 +Z] (–) Z* (+) Z (174)
where the number in brackets is the number of doublets in the particle; the number in
parentheses is the number of free doublets; and the asterisk denotes anti neutrino.
There are two stages in the decay shown as Step 1. Firstly, the triangular structure of
positive sigma restructures itself into that of the neutron by releasing 58 doublets from
each of its rings and the (+) Z spare singlet. The (+) Z singlet and the 174 doublets are
released as available residual energy:
1a) [276, 276, 276 +Z] --> [218, 218, 218] [174 +Z]
Secondly, since there is enough residual energy available, the released doublets
restructure themselves into a pion of a charge equal to the charge of the available spare
singlet (in this case, a positive pion). The remaining 78 doublets are released as free
doublets:
1b) [174 +Z] --> [48, 48 +Z] (78)
In Step 2, the triangular structure of the neutron restructures itself (beta decay) into that of
the proton by splitting a doublet into (+) Z and (–) Z* singlets from one of its rings. The (+) Z
singlet stays with the proton as the spare singlet, while the (–) Z* singlet decouples into (–)
and Z* components which are released as electron and anti neutrino:
2) [218, 218, 218] --> [218, 218, 217 +Z] (–) Z* (0)
In Step 3, the two-ring linear structure of the positive pion restructures itself into the single
ring structure of a positive muon by releasing 23 doublets from its two rings in addition to
releasing the Z component of its spare singlet as a neutrino by decoupling it from the (+)
component. The 23 doublets are released as free doublets:
3) [48, 48 +Z] --> [73 +] Z (23)
In Step 4, the one-ring structure of positive muon breaks down completely by releasing the
(+) component as a positron and all of its 73 doublets as free doublets:
4) [73 +] --> (+) (73)
The Table below shows the end results of the decay process for most Baryons. The first
line identifies the particle and its mode of decay; the second line gives the end results
according to the rings model. The numbers in brackets are doublets; the numbers in
parentheses are free doublets; and the asterisks denote anti neutrinos. The end results
include those of the beta decay of the neutron if a neutron is involved in the process.
n(0) --> p(+) e(–) v(e)*
[654] --> [653 +Z] (–) Z* (0)
L(0) --> p(+) pi(–)
[776] --> [653 +Z] (–) Z* (122)
L(0) --> n(0) pi(0)
[776] --> [653 +Z] 2{(–) Z*} (+) Z (121)
S(+) --> p(+) pi(0)
[828 +Z] --> [653 +Z] (–) Z* (+) Z (174)
S(–) --> n(0) pi(–)
[833 –Z*] --> [653 +Z] 2{(–) Z*} (179)
D(++) --> p(+) pi(+)
[857 +Z+Z] --> [653 +Z] (+) Z (204)
D(+) --> p(+) pi(0)
[857 +Z] --> [653 +Z] (–) Z* (+) Z (203)
D(0) --> n(0) pi(0)
[858] --> [653 +Z] 2{(–) Z*} (+) Z (202)
D(0) --> p(+) pi(–)
[858] --> [653 +Z] (–) Z* (203)
D(–) --> n(0) pi(–)
[857 –Z*] --> [653 +Z] 2{(–) Z*} (203)
X(0) --> L(0) pi(0)
[915] --> [653 +Z] 2(–) Z* (+) Z (260)
[915] --> [653 +Z] 3(–) Z* 2{(+ ) Z} (259)
X(–) --> L(0) pi(–)
[919 –Z*] --> [653 +Z] 2{(–) Z*} (265)
[919 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (264)
O(–) --> L(0) K(–)
[1164 –Z*] --> [653 +Z] 2{(–) Z*} (510)
[1164 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (509)
O(–) --> X(–) pi(0)
[1164 –Z*] --> [653 +Z] 3{(–) Z*} (+) Z (509)
[1164 –Z*] --> [653 +Z] 4{(–) Z*} 2{(+) Z} (508)
The following observations can be made:
1) Particle decay is a process of re-formation rather than trans-formation. The three-
ring structure of a Baryon remains intact to the end of the process. The only activity in the
rings is the stepwise ejection of doublets from an unstable configuration to a more stable
one. If the ejection results in the release of enough energy then the ejected doublets
rearrange themselves into the two-ring structure of a Meson. A Meson either rearranges
itself into the one-ring configuration of a Lepton or, if enough energy is available, it splits
itself into two more stable Mesons. The one-ring structure of Leptons breaks down
completely into free doublets releasing the charged component of its spare singlet as free
electron or free positron.
2) The charge of a particle is of primary importance in the process. All particles decay
into a proton regardless of their charge. Therefore, other particles must be generated in the
decay so that the net charge of all resultant particles of the decay is equal to the charge of
the decaying particle (law of charge conservation). The Table below summarizes the
observation using the results above.
[0] --> [+Z] (–) Z*
[+Z] 2(–) 2Z* (+) Z
[+Z] --> [+Z] (–) Z* (+) Z
[–Z*] --> [+Z] 2(–) 2Z*
[+Z] 3(–) 3Z* (+) Z
[+Z] 4(–) 4Z* 2(+) 2 Z
3) The net sum of the handedness of the resultant Z particles (double left-handed
helixes or neutrinos) and Z* particles (double right-handed helixes or anti neutrinos) is
equal to the handedness of the Z component of the spare singlet of the decaying particle.
By net sum of the handedness we mean that right-handedness plus left-handedness equal
zero or, more simply, Z* + Z = 0. This is a handedness conservation law since the
handedness of doublets and free doublets is zero.
4) The number of free doublets released by the process is given by:
y = d – (a + n)
where: d is the number of doublets in the decaying particle
a = 653 (the number of doublets in the proton)
n is the number of doublets that are split in the process (0 to 3)
If d = 1306 and n = 0 then y = 653. This means that there is enough available residual
energy to generate an additional proton. We can define harmonics of Baryon particles by
the number of protons that could be generated in the decay:
1 proton (first harmonic): a = 1 x 653 for d = 653 to 1305 and n = 0
2 protons (second harmonic): a = 2 x 653 for d = 1306 to 1958 and n = 0
3 protons (third harmonic): a = 3 x 653 for d = 1959 to 2611 and n = 0
9.0 Conclusion
The litmus test of a particle model is whether or not it can explain the stability of the proton.
Until that explanation is achieved all models, including this one, are surrogates, albeit
necessary surrogates for the understanding of phenomena beyond our physical reality.
The inability of this model to explain the stability of the proton is based on the following:
Z helixes are defined as double right handed or double left handed helixes of null active
energy and zero charge, and singlets are defined as single right handed (positive) or single
left handed (negative) helixes coupled to Z helixes of opposite handedness. There is no
indication of what is keeping the Z helix coupled to the single helix. In addition, rings are
defined as entities of zero charge and considered as having no electrostatic influence
outside their perimeter. The introduction of such an influence would make triangular or
linear rings formations unstable. Formations would then either collapse unto their centers
or break out into separate rings which, in turn, would break down into separate
components. The definition of rings as entities raises the question of what is keeping the
rings in triangular or linear formations. There can not be stability either in rings or in rings
formations without an influence or force that keeps Z helixes coupled to single helixes and
that keeps rings coupled to rings.
There are other flaws in the model related to the rotation of the rings. Even though rings are
defined as entities of zero charge, each singlet in a ring is subjected to electrostatic forces
from all other singlets in the ring. The overall force affecting a singlet is centripetal and
tends to collapse the ring unto its center unless counteracted by an opposing force which
is provided by the rotation of the ring. This poses two questions: 1) what is the influence or
force that initiates the rotation and 2) what is the mechanism that dictates and controls the
velocity of rotation? The model as currently stated leaves those questions unanswered.
The litmus test of a particle model is whether or not it can explain the stability of the proton.
Until that explanation is achieved all models, including this one, are surrogates, albeit
necessary surrogates for the understanding of phenomena beyond our physical reality.
The inability of this model to explain the stability of the proton is based on the following:
Z helixes are defined as double right handed or double left handed helixes of null active
energy and zero charge, and singlets are defined as single right handed (positive) or single
left handed (negative) helixes coupled to Z helixes of opposite handedness. There is no
indication of what is keeping the Z helix coupled to the single helix. In addition, rings are
defined as entities of zero charge and considered as having no electrostatic influence
outside their perimeter. The introduction of such an influence would make triangular or
linear rings formations unstable. Formations would then either collapse unto their centers
or break out into separate rings which, in turn, would break down into separate
components. The definition of rings as entities raises the question of what is keeping the
rings in triangular or linear formations. There can not be stability either in rings or in rings
formations without an influence or force that keeps Z helixes coupled to single helixes and
that keeps rings coupled to rings.
There are other flaws in the model related to the rotation of the rings. Even though rings are
defined as entities of zero charge, each singlet in a ring is subjected to electrostatic forces
from all other singlets in the ring. The overall force affecting a singlet is centripetal and
tends to collapse the ring unto its center unless counteracted by an opposing force which
is provided by the rotation of the ring. This poses two questions: 1) what is the influence or
force that initiates the rotation and 2) what is the mechanism that dictates and controls the
velocity of rotation? The model as currently stated leaves those questions unanswered.
Appendix
Table 1 displays the maximum number of doublets and the spare singlets for several
particles. The number of doublets is obtained by dividing the rest mass of the particle by
the relativistic mass of the doublet (Paragraph 7.3). Depending on the internal dynamics of
the particle, the number of doublets may differ from the one shown; this, however, does
not affect the concepts of the rings model.
Table 2 displays the rings composition for the same particles displayed in Table 1. The
asterisk denotes anti particle.
Tables 3 and 4 have two lines of information: the first line displays the mode of decay of
the particle and the second line displays the final results of the decay using rings model
notation. The rightmost column gives the number of released doublets.
Tables 5 and 6 display the decay probability of Baryons and Mesons. P1, P2, and Ptot
have been defined in Paragraph 8.0 as:
P1 = 1 – Dr / D
P2 = 1 – A / D
Ptotal = P1 P2
Where: Dr is the number of released doublets
D is the number of doublets in the decaying particle
A is equal to 94 for Mesons (the doublets in the neutral pion)
and to 653 for Baryons (the doublets in the proton)
Ptotal is the resulting decay probability
In order to reduce size, Table 5 only displays data for P1 > = 0.865.
Tables 3 through 6 use symbols for particles names, as follows:
Proton p
Neutron n
Lambda L
Sigma S
Delta D
Cascade Xi
Omega O
Lambda c Lc
Pion pi
Kaon K
Rho R
Omega O, W
Phi f
J/Psi J/P
Electron e
Muon m
Tauon t
Table 1 displays the maximum number of doublets and the spare singlets for several
particles. The number of doublets is obtained by dividing the rest mass of the particle by
the relativistic mass of the doublet (Paragraph 7.3). Depending on the internal dynamics of
the particle, the number of doublets may differ from the one shown; this, however, does
not affect the concepts of the rings model.
Table 2 displays the rings composition for the same particles displayed in Table 1. The
asterisk denotes anti particle.
Tables 3 and 4 have two lines of information: the first line displays the mode of decay of
the particle and the second line displays the final results of the decay using rings model
notation. The rightmost column gives the number of released doublets.
Tables 5 and 6 display the decay probability of Baryons and Mesons. P1, P2, and Ptot
have been defined in Paragraph 8.0 as:
P1 = 1 – Dr / D
P2 = 1 – A / D
Ptotal = P1 P2
Where: Dr is the number of released doublets
D is the number of doublets in the decaying particle
A is equal to 94 for Mesons (the doublets in the neutral pion)
and to 653 for Baryons (the doublets in the proton)
Ptotal is the resulting decay probability
In order to reduce size, Table 5 only displays data for P1 > = 0.865.
Tables 3 through 6 use symbols for particles names, as follows:
Proton p
Neutron n
Lambda L
Sigma S
Delta D
Cascade Xi
Omega O
Lambda c Lc
Pion pi
Kaon K
Rho R
Omega O, W
Phi f
J/Psi J/P
Electron e
Muon m
Tauon t
Table 1 Maximum Number of Doublets and the Spare Singlets
Table 2 Rings Composition
Table 3 Modes of Decay: Baryons
Table 4 Modes of Decay: Mesons and Leptons
Table 5 Decay Probability for Baryons (P1 > = 0.865)
Table 6 Decay Probability for Mesons
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Copyright © 2007 Giulio. C. Cima All Rights Reserved
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Copyright © 2007 Giulio. C. Cima All Rights Reserved
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