CHAPTER 5- THE QUANTUM STATES OF THE NEUTRON

SECTION 5-0: INTRODUCTION

In Chapter 3 we used the small differential masses and charges to explain the differences between the angular momentum and the magnetic moment of the neutron. If the charge to mass ratio were a constant, then a balance of the angular momentum would require a balance of the magnetic moment. Therefore for this case the magnetic moment of the neutron would be zero. The differential solution enabled a difference to occur.

Just as the electron has many different states, the same is true of the neutron. In this chapter we will specify the Bohr type equations for the states of the neutron.

In this chapter we will also continue the study of the neutron, from the perspective that the product of charge times mass is a constant. In this case a balanced angular momentum of zero has a net magnetic moment. To justify this solution let us start with the up and down quark.

SECTION 5-1: THE UP-QUARK AND THE DOWN QUARK

According to Wikipedia, the up quark has a charge of 2Q/3 and Energy of between 1.5MEV and 3.3MEV. The down quark has a charge of –Q/3 and energy of between 3.5MEV and 6.0MEV. Therefore for the up-quark, the root mean square value is 2.2248MEV. Using the up quark as the standard and selecting twice the energy for the down-quark, we get:

Charge Up Quark = 2Q/3 (5-1)

Mass Up-Quark = 2.2248MEV (5-2)

Charge Down Quark = -Q/3 (5-3)

Mass Down Quark = 4.4496MEV (5-4)

MaQa = -MbQb (5-5)

In equation 5-5 we find that a lower mass quark has a higher charge and a higher mass quark has a lower charge. For any distribution of quarks we could assume that the charge to mass ratio is a constant. The alternate is to specify that within any pile of masses and charges, the charges and masses will subdivide as per equation 5-5. However the minus is unnecessary if we subdivide the electron within the neutron.

From Chapter 1 we found that the electron only had enough energy to produce one charge Q. The proton could produce up to three charges Q. However if the proton produces all three charges, its mass would be reduced to a very low number.

The proton only has 3.129741 Charge Q’s within it. Repeating Equation 1-29, we get:

# Charge Q’s within proton = 3.129741 (5-6)

The energy available for the production of up-quarks and down-quarks is equal to the difference in charges divided by two, since two charges produces a single bipolar dot. Therefore:

Quark energy = (3.129741 – 3.000)(3/2) 938.272 MEV = 18.26MEV (5-7)

If we broke apart the proton and it only formed up-quarks and down-quarks, we would get three of each.

Charge 3 Up Quarks = 3 (2Q/3) = 2Q (5-8)

Mass 3 Up Quarks = 6.6744MEV (5-9)

Charge 3 Down Quarks = 3 (-Q/3) = -Q (5-10)

Mass 3 Down Quarks = 13.3488MEV (5-11)

Net Charge = 2Q – Q = +Q (5-12)

Net Mass = 20.02MEV (5-13)

Since we got 18.26MEV as the calculated value, the data appears to agree with the proposition that the product of quark mass times quark charge is a constant for a particular construction.

Another way of looking at Equation is to visualize a pile of positive quarks and bipolar quarks. The positive quarks have no mass. If we add negative quarks into the pile, we would have more quark charges. However the addition of more quark charges does not increase the charge of the pile. It causes the charge to go down and the mass to go up. Therefore for any pile of positive charges and bipolar dot-waves, the addition of additional negative dot waves causes the mass to increase and the charge to decrease. Therefore under particular conditions when quarks are being formed:

Charge = B / Mass (5-14)

In Equation 5-14, charge is a constant B over mass. We also know that:

MC = A Q (5-15)

Equation 5-15 tells us that a total package of charge and mass are equal to each other. However Equation 5-14, tells us that a partial breakdown of a structure into quarks does not obey a form of Equation 5-15. Instead it obeys an inverse law.

Equation 5-14 enables us to have a balanced tri-vector for the angular momentum of the neutron and at the same time have an unbalanced magnetic moment. Equation 5-15 enables us to have a proton with a particular angular momentum and a different magnetic moment.

In effect Equation 5-14 enables the universe to exist as it is. Without a difference between the angular momentum vectors and the magnetic moment vectors the universe could not exist.

We can now return to the Neutron to solve for the states of the Neutron’s quarks using equation 5-14.

SECTION 5-2: THE NEUTRON’S ELECTRON EQUATIONS

The electron in the neutron orbit can be defined using quarks, which follow the inverse charge to mass relationship. Thus:

ΔQ ΔM = Constant (5-16)

In equation 5-16 the charge of the quarks times the mass of the quarks is a constant. We know the total charge of the quarks is:

Qa + Qb + Qc = Q = 1.602176E-19 coulombs (5-17)

In addition, we know the total mass of the quarks is:

Ma + Mb + Mc = MNE = 2.305516E-30 Kilograms (5-18)

Also:

QaMa = QbMb = QcMc (5-19)

We also know that the vector sum of the angular momentum of the neutron’s electron quarks is equal to zero. Thus:

Vector sum (MaVaRa + MbVbRb + McVcRc) = 0 (5-20)

We know that the magnetic moment vector has a magnitude and angle with respect to the resultant proton vector of:

Vector Sum (QaVaRa + QbVbRb + QcVcRc) = 4.1168728E-26 (5-21)

Angle = 54.73561 (5-22)

The above equations have many solutions. If the radius of all quarks is identical, then all the quarks see the charge Q of the proton.

All Quarks see: Q

(5-23)

If all the quarks appear at different radii, then the quark nearest the proton sees a charge of +Q. Thus:

Quark a sees: Q (5-24)

Quark b sees: Q- Qa (5-25)

Quark c sees: Q- Qa –Qb = Qc (5-26)

The above equations enable us to look at various solutions. Many solutions are possible. There are a huge amount of stable and unstable solutions with various charge and mass ratios. There are a huge amount of solutions where the magnetic moment of the neutron has different values.

The general solution of the neutron does not require the Bohr/Plank relationship in which the wavelength of the quarks is a multiple of 2πn times the radius. The radius could take any value from a minimum to the maximum of the stable state. The minimum radius of the unstable neutron is found by equating the neutrino energy to the electrical energy gained when the neutron shrinks.

Energy = 939.5653MEV – 938.2720 MEV – 0.510999MEV (5-27)

Energy = KQ/R = 0.78230 MEV (5-28)

R = 1.840678E-15 (5-29)

When the neutron radiates a neutrino, it can return to being a hydrogen atom. It could also become a free proton and electron. However, it can also come closer to the proton.

The equations are:

KQ(Q/3) = (Men/3) V2 R (5-30)

V2 = KQ^2 /Men R (5-31)

Since Men = 2.305516E-30, and R = 1.840678E-15, we get

V = 2.3316210E8 (5-32)

V/C = 0.7777449 (5-33)

From Equation 5-33 we find that the triple quarks can reach a small radius of 1.840678E-15 meters and a velocity of 0.7777459C. For the Q/3 solution, we can have oscillating neutrons, which absorb and radiate any amount of neutrino energy. All these solutions have the same magnetic moment. We can also have many other solutions where the magnetic moment is different. In addition we can have many other solutions where the charges of the three quarks are not basically equal. Finally we can have many other solutions where the radii of the three quarks are different. Therefore each quark can exist at a different energy level.

In effect we can write the stable and unstable solutions based upon the simple Coulomb attraction laws. We now want to know the average state of the neutron on the sun. For the stable and unstable conditions, we can use the RMS value of the neutrino energy as the state of the neutron upon the sun. Thus:

Average Neutron energy level = 0.707106 (0.782301) = 0.55317MEV (5-34)

In general the stable and unstable neutrons will radiate and absorb a difference of energy continuously. Thus:

Differential neutrino energy = 0.782301 – 0.55317 = 0.229131MEV (5-35)

% Differential Neutrino Energy = 29.3 % (5-36)

According to the Internet, the actual measured neutrino energy from the sun is around 33%.

SECTION 5-3 THE MINIMUM RADIUS OF THE NEUTRON’S QUARKS

Let us now understand the limiting factor, which contains the neutron to a simple radius of 1.840678E-15 meters. The radius of the proton is:

Rp = 1.2309626E-15 meters (5-37)

The electric field could bring the neutron quarks down to this radius. Normally the energy that the field would provide would be:

E = KQ/ Rp = 1.169787 MEV (5-38)

The limiting value of the added energy to the electron’s quarks by the electrical field at the proton radius is larger than the 0.78201MEV of the neutron. However it is the proton radius which limits the energy. In order to understand this we must write the Einsteinian equations for the orbital motion.

KQQ/3 = (M/3) V^2 R (5-39)

KQ/R = 0.782301MEV (5-40)

From these equations we obtained:

R = 1.840678E-15 (5-41)

V = 2.331610E8 (5-42)

V = 0.7777449 C (5-43)

The value of V, R, and M are the uncorrected values. We can keep V at 0.7777449C but we must correct both R and M. The Einsteinian mass, M has gone up and the Einsteinian distance, R has gone down. The product MR is a constant.

M = Mo/ (1-(V/C)^2)^0.5 = Mo/Ae (5-44)

R = Ro (1-(V/C)^2)^0.5 = Ro/Ae (5-45)

Ae = (1-(V/C)^2)^0.5 = 1.590888 (5-46)

MR = MoRo (5-47)

Since the product MoRo equals MR equation 5-39 is unchanged. However the limiting value of the neutron quark excursion is the proton radius. Therefore:

R = Ro Ae = Ro/1.590888 = 1.157013E-15 (5-48)

Equation 5-48 tells us that if all the energy came from the electric field, the neutron’s quarks would fall below the proton radius. Therefore a minimum amount of neutrino energy was necessary to change the hydrogen atom into a neutron.

In order to solve this problem we must use successive approximation and select a radius. Then we must calculate the velocity. From that we must calculate the mass increase and the distance decrease. In the end we must end up at the proton radius.

For all solutions:

V^2 R = 100.06671 (5-49)

Equation 5-49 specifies from Equation 5-39, that the square of the velocity times the radius is a constant. We can now select a radius and calculate the solution. As a first try:

R = 2.0 E-15 (5-50)

Therefore:

V^2 = 5.00339E16 (5-51)

V = 2.23682E8 (5-52)

The Einsteinian factor is:

Ae = 1.501939 (5-53)

Therefore we can calculate R:

R = 1.33124E-15 (5-54)

This radius is too large. We are looking for the proton radius. Thus:

R= 1.2309662E-15 (5-55)

The process of successive approximation yields:

R = 1.9077E-15 (5-56)

V^2 R = 100.06671 (5-57)

V = 2.290286E8 (5-58)

V/C = 0.7639572 (5-59)

Ae = 1.549746 (560)

The Einsteinian radius is:

Rein = 1.9077E-15 / 1.59746 = 1.230976E-15 (5-61)

The error is:

Error = 8 PPM (5-62)

This calculation was easy to produce an 8-PPM error rapidly. We see that the neutron quarks reach a velocity of 0.76396C and a radius of 1.9077E-15 meters. This radius looks like an Einsteinian radius at the proton radius. This is the limiting radius that the quarks can achieve.

We can now calculate the energy that the electric field exerts in bringing the quarks to the proton radius from an Einsteinian viewpoint. Thus:

Energy = KQ/1.9077E-15 (5-63)

Energy = 0.754817MEV (5-64)

The neutrino necessary to bring the quarks to this level is:

Neutrino energy = 0.78230 – 0.754817 = 0.027484MEV (5-65)

It would appear that only 0.0274MEV of neutrino energy is required to bring the quarks to the proton radius. However the neutron is quantized. Therefore it will not go to the proton radius but will come as close to the proton radius as possible.

Now that we know the minimum radius, we can solve for the quantum number n. Since n must be a whole number, we must reduce n to the next lowest level. Then we must calculate everything again. This will provide us with the lowest stable neutron state.

SECTION 5-4: THE QUANTUM STATES OF THE NEUTRON

Just as the electron has quantum states using the number n, the neutron also has quantum states using the number 1/n. The Bohr theory specifies that a non-radiating state is one in which the wavelength is:

λ = n(2πR) (5- 66)

In equation 5-65 a stable wavelength is one in which the wavelength is 2 pi times the radius. In addition any whole number n also provides stability.

When we look at the oscillating neutron, we find that it does not have the same stability as the proton or the hydrogen atom. Yet it is partially stable. Therefore it obeys a form of equation 5-64. Thus:

λ = 2πR/ n* (5-67)

n* = 1/n (5-68)

In Equation 5-67 a wave traveling around a radius R has a degree of stability if it can circle R with an exact number of wavelengths. Thus if n*=2, the wave circles twice before meeting itself. If n*=3, the wave circles three times before meeting itself. Therefore any whole number n* will cause a wave to circle a radius a perfect amount of times.

The velocity equation is:

V = 2πKQQ/ n*h (5-69)

We can solve for n when V = 2.290286E8 (5-70)

n* = 1/104.69 (5-71)

Since n must be a whole number, the highest possible number n is 1/104.

n = 1/104 (5-72)

We can then solve for the velocity from Equation 5-68. Thus:

V = 104(2πKQQ)/h = 2.2751975E8 = 0.758924C (5-73)

R = 100.06671/V^2 = 1.933087E-15 (5-74)

KQ/R = 0.744904

Neutrino Energy = 0.78230 – 0.744904 = 0.0374MEV (5-75)

Einsteinian = Ae = 1.535676 (5-76)

Einsteinian radius = 1.933087E-15/1.53676 = 1.2588E-15 (5-77)

We see that the Einsteinian radius is a little greater than the proton radius. This helps prevent the neutron quarks from interfering with the proton quarks.

From this analysis it appears that most of the energy for the production of the neutrino quarks comes from the proton/electron electric field. Only a small amount of neutrino energy (0.0374MEV) is required.