Quark Structure of the Proton

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Offline jerrygg38

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Quark Structure of the Proton
« on: 06/06/2009 13:34:31 »


  In this chapter, we will study the quark structure of the proton. In Chapter 2 we studied the proton from a single quark Bohr/Plank mathematical perspective. We found that the proton had a radius of 1.2309626E-15 meters. At that time we could not readily produce a method of solving both the proton’s magnetic moment and the proton angular momentum.

   A single quark spinning on an axis will tend to have a simple relationship between the magnetic moment and the angular momentum. Over the years no simple relationship has been found. The case is different for the three-quark solution. When the product of charge times mass is a constant for all three quarks, the difference between magnetic moment and angular momentum becomes self-evident.

   We will now study the constant radius solution for the Proton.

  The constant radius solution enables us to use the radius calculated in Chapter 2 and apply this to the proton’s quarks. The radius is:

   R = 1.2309626E-15meters                     (4-1)

  In equation 4-1 we are starting with the radius R derived in Chapter 2. In the proton, the ground state can be considered a stable constant R state. As the proton moves, things will change just like the neutron. However the basic low velocity proton will be unchanged.

   The neutron constantly changes states as will be discussed in Chapter 5. The proton is the most stable structure of the universe. The constant R solution means that the entire proton has a charge Q at the radius R.  The proton could have only positive quarks. It could also have a combination of plus and minus quarks. However as long as the quarks all occupy the same maximum radius, the electrical equations of the proton remain simple. In fact we only need to know the basic Plank equations in chapter 2.

  Therefore we only need to know the quark details to understand the proton. The Proton’s three angular momentum vectors do not add up to zero. Therefore we need to produce a vector diagram for the angular momentum vectors which do not null out.

   It is easy to produce a magnetic moment vector diagram when the angular momentum vectors balance out. It is also easy to produce a magnetic moment vector when the angular momentum does not balance out.

  Let us look at what is known about the proton besides the radius. Thus:

    Angular Momentum = (0.866025) h/2π               (4-2)

    Angle of spin to magnetic vector = 54.7356 degrees            (4-3)

   We can now begin the proton calculations. The latest NIST data is:

   Nuclear magnetron (NM) = 5.0507832E-27               (4-4)

   Proton mag/ Bohr magnetron = 2.792847               (4-5)

   Proton magnetic moment = 1.410606662E-26            (4-6)

      By using above NIST Codata, we have:

   NM = Qh/4πMp = 5.0507832E-26                  (4-7)


    NM = QVpRp = 5.0507832E-26                  (4-8)

  The proton consists of three quarks. These quarks are not necessarily the same charge and mass as the product of the destruction of the proton. When the protons are destroyed there is a lot of photonic energy involved. Therefore the stable proton may only consist of positive charge quarks. Alternatively it could consist of two positives and one negative quark.

  The easiest solution to the proton is the constant radius solution. Thus each quark occupies the same radius. Plus and minus quarks at the same radius produce a singular +Q field external to the proton. The plus/minus solution with different radii is complicated because the negative would be attracted to each positive layer. In addition if the center of the proton looked like a negative charge Q, the negative quark would be repelled from the center and it’s centrifugal force would also repel it. Therefore the only thing holding the negative quark within the proton would be the interaction with the positive quarks. This would produce very complicated equations.

   Since we already know the nuclear magnetron, proton radius, and proton velocity for the total mass of the proton, we do not have to solve force equations for the quarks.

   The proton is very stable whereas the neutron is continually changing states. The neutron will absorb and release energy continuously. Therefore the proton must only obey equation 2-7. Thus:

   n(h/MpVp) = 2πRp                        (4-10)

   Let us assume that all the quarks exist at the same n=1 level. Then for each Mq we have an Mq such that:

    MqVq = MpVp = h/2πRp = 8.7181628E-20               (4-11)
    Vp = 2C/11.706238 = 5.1219267E7                  (4-12)

    Vq = (Vp) Mp / Mq                        (4-13)

    Rp = Rq = 1.2309626E-15                     (4-14)

    Equations 4-11 through 4-14 define the proton quarks for the constant radius solution. In addition all quarks are in the n=1 state. The proton could also move into higher energy states when n is the reciprocal of a whole number.
If we keep adding energy to the proton we could move it to higher and higher energy states. For this chapter we are interested in the ground state of the proton where n=1.

   Let us first start with the angular momentum of each quark.

  MaVaRa = MbVbRb = McVcRc = MpRpVp               (4-15)

  Since Mp = 1.672622E-27, Vp = 2C/11.706238, and Rp = 1.2309626E-15:

   MpVpRp = 1.0545716E-34                     (4-16)

    The angular momentum of the proton is:

  Angular momentum = (0.5 x 1.5)^0.5 h/2π = (0.8660254) 1.0545716E-34                                    (4-17)

   Angular momentum = 0.91328580E-34               (4-18)

  Each quark vector has an angular momentum magnitude as shown by equation 4-16. Equation 4-17 shows that the angular momentum of the proton itself. The proton has the same angular momentum as each quark for the n=1 state multiplied by the square root of three divided by two. 

If we normalize quark value to 1, the vector sum must be:

   Vector sum = 0.8660254                        (4-19)

    We now have the sum of three unit vectors equal to 0.8660254. The first vector is at zero degrees, the top vector is at 93.84097 degrees and the bottom vector is at –93.84097 degrees. The vertical components of the vectors cancel out. The horizontal components are:

   Horizontal vector = 1.0 + 1.0Cos93.84097 + 1.0Cos(-93.8407)      (4-20)

   Horizontal vector = 1.0 - 0.06698737 - 0.06698737 = 0.8660253      (4-21)

  We now have a general vector solution in which the angular momentum has the correct value. The momentum vector is at zero degrees. It is now necessary to select values for the ratio of the proton quark masses, which will produce the correct magnetic moment vector at the correct angle with respect to the angular momentum.
    According to my College physics book, the vectors form a right triangle with the side ratios and angles of:

   Vector Ratios = 1 : 2^0.5 : 3^0.5   = 1 : 1.414214 : 1.732051         (4-22)

  Angles =  54.73561 : 35.264390 : 90.000               (4-23)

  We know that the vectors are defined by Equations 4-22 and 4-33. However we still have to select various mass ratios and see how they fit into the overall solution. Then we may have to correct the ratios to get the desired results.

  As a first try at a solution let us look at the case where the masses are in the same ratio as the quark momentum vectors. We can normalize the mass (a) as 1. The masses will be as follows:

   Ma = 1.00000                           (4-24)

   Mb = 1.414214                           (4-25)

   Mc = 1.732051                           (4-26)

   Sum of mass ratio = 1 + 1.414214 + 1.732051 = 4.146265      (4-27)

 The masses are:

   Ma = Mp(1/4.146265) = 0.2411809Mp = 4.034045E-28         (4-28)

   Mb = Mp (1.414214/4.146265) =0.3410814Mp = 0.57050025E-28   (4-29)

   Mc = Mp(1.732051/4.146265) = 0.4177377Mp = 0.698717267E-28   (4-30)

Check sum = 1.0000000Mp                     (4-31)

  In order to find the corresponding quark charges we must ass the reciprocals of the mass coefficients. Thus:

1/Ma = 1/0.2411809Mp = 4.146265/Mp               (4-32)

1/Mb = 1/0.3410814Mp = 2.931851/Mp               (4-33)

1/Mc= 1/ 0.4177377Mp = 2.393847/Mp               (4-34)

Sum charge ratio = 4.146265 + 2.931851 + 2.393847 = 9.471963      (4-35)

Qa = 4.146265Q/9.471963= 0.4377408Q               (4-36)

Qb = 2.931851Q/9.471963 = 0.3095294Q               (4-37)

Qc = 2.393847Q/9.471963 = 0.2527298Q               (4-38)

   Check sum = 1.0000000Q                     (4-39)

   Also as a further check, the coefficient of:

    MaQa = MbQb = McQc = 0.4377408                  (4-40)

  Therefore the example meets all the initial requirements.

   The corresponding velocities of the quarks are:

  Va = 2(4.146265)C/11.706238 = 0.70838539C            (4-41)

   Vb = 2(2.931851)C/11.706238 = 0.5009040C            (4-42)

    Vc = 2(2.393847)C/11.706238 = 0.4089866C            (4-43)

  From Equations 4-40 through 4-42 we see that the proton quarks all have large amounts of Einsteinian energy within them. Since Mp = 1.672622E-27, Vp = 2C/11.706238, and Rp = 1.2309626E-15, we can now double-check the angular momentum of each quark. Thus:

   MpVpRp = 1.054571E-34                     (4-44)


    MaVaRa = MbVbRb = McVcRc = 1.054571E-34            (4-45)

Angular Momentum = (0.8660254)(1.0545716E-34) = 0.91328579E-34   (4-46)

   We have the correct angular momentum and now we need to calculate the magnetic moment.

  For this 1:2^0.5 :3^0.5 solution we get:

   QaVaRa = (0.43377408Q)(0.70838539C)Rp = 1.805186E-26      (4-47)

  QbVbRb = (0.30952946Q)(0.5009041C)Rp = 0.916710905E-26      (4-48)

  QcVcRc = (0.25272974Q)(0.4089864C)Rp = 0.6111409E-26      (4-49)

    Taking quark a as the plus 93.84097degree vector, quark b as the 0 degree vector and quark C as the –93.84097degree vector, we get:

   Horizontal vector =  0.9167109E-26 + 1.805186E-26 Cos 93.84097
                                   + 0.6111409E-26Cos 93.84097          (4-50)

   Hor. Vector = 0.9167109E-26 – 0.1209247E-26 – 0.0409387E-26   (4-51)

    Hor. Vector = 0.7548475                     (4-52)

Ver. Vector = 1.805186E-26Sin 93.84097 – 0.6111409E-26Sin93.84097   (4-53)

Vertical vector = 1.801131E-26 – 0.6097682E-26 = 1.191393E-26      (4-54)

The resultant vector is:

   Vector sum =  1.410394E-26                     (4-55)

 Vector angle =  57.6423 degrees                     (4-56)

   We are looking for a vector magnitude of 1.4105846E-26 and an angle of 54.7356 degrees.  The magnitude error is:

   Magnitude Error = 190 ppm                     (4-57)

  The angular error is:

    Angular Error = 2.9067 degrees                  (4-58)

       The angular error of three degrees is readily corrected by changing the ratio of masses slightly. Therefore we can get a 3 degree angular correction with minimal change in magnitudes. However there may be other reasons for the angular shift. The velocities are all high in light speed. This might cause an additional phase angle to develop between the magnetic vector and the angular momentum vector.

  In any event the 1: 1.414214: 1.732051 solution produces the correct magnitude and appears to be a small error away from the final answer. The mass vectors by themselves produce the correct angle. Thus:

     Cosine Mass vector angles = Cos 54.735626            (4-59)

    Mass vector angle = 54.73562                     (4-60)

   As we look at the work involved in producing a more accurate answer, every time we increase a mass we will decrease the corresponding velocity and charge. Thus some mass vectors must increase and while others decrease. Without a computer program it is difficult to zero in on a final answer. Therefore for this initial effort, the above answers should be adequate.