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Quark Structure of the Neutron
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Quark Structure of the Neutron
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Quark Structure of the Neutron
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04/06/2009 15:54:07 »
SECTION 3-3: THE QUARK STRUCTURE OF THE NEUTRON
The proton consists of three high-energy quarks. The hydrogen atom contains the same three proton quarks plus and an electron of 0.511MEV. The electron itself is a single quark within the hydrogen atom. Therefore the hydrogen atom is composed of four quarks.
At any split second of time, the proton has three distinct operating planes. The electron has only one plane. Therefore the hydrogen atom has four planes of energy.
The neutron can be considered a high-energy six-plane hydrogen atom. The electron within the neutron split into three planes of energy. Therefore the neutron has six quarks and six planes of energy.
The electron split into three planes and the energy level rose to 1.2933MEV. For the quark mass distribution shown in Table 1, the quark mass/energies became (0.4311+) MEV, (0.4311) MEV, and (0.4311-) MEV. Some of the energy increase came from the proton/electron electric field and some of the energy increase came from the neutrino. Without the neutrino, the electron would not have split into three parts readily. Therefore the electron has a small amount of binding energy which holds the three parts together.
Let us first look at the neutron’s magnetic moment.
The electron has a nuclear magnetron of:
Electron NM = Qh/4πMe = 9.2740065E-24 (3-21)
The proton has a nuclear magnetron of:
Proton NM = Qh/4πMp = 5.0507806E-27 (3-22)
From 2006 NIST Codata, the magnetic moment of the proton is:
Proton MM = 2.792847 NM = 1.410606662E-26 (3-23)
From 2006 NIST Codata, the neutron has a magnetic moment of:
Neutron MM = -1.91304273 NM = -0.96623641E-26 (3-24)
The proton does not change appreciably from the time it was a proton alone to a neutron or to a hydrogen atom. The spin momentum of the electron equals the spin momentum of the proton and this also equals the spin momentum of the neutron. Therefore:
Proton spin = Electron spin = Neutron spin (3-25)
Equation 3-25 tells us that when the electron splits into three parts and comes closer to the proton to form the neutron, the total spin of the split electron equals zero. Thus:
Electron spin within neutron = 0 (3-26)
The vector sum of the three quark electron spins must balance to zero. Therefore we only need to solve the magnetic moment equation to find the ground state of the neutron.
The neutron has many possible states. As it moves it develops even more complex states which include the linear momentum. For the moment we are only concerned with a neutron which is basically stationary in space. The basic equation for the neutron when all three quarks are a common distance from the proton is:
K Qy Qx = Mx V^2 Rx (3-27)
Equation 3-27 is a variation of Bohr force Equation 3-8 with the radius R transposed. We also added the quark charge Qx and the Quark mass Mx to the equation. In addition, the charge Qy is the proton field that the charge Qx sees. When the three quarks are equidistant from the proton, Qy equals Q. Thus:
KQ Qx = Mx Vx^2 R (3-28)
In Equation 3-28 we have the constant proton charge Q and the constant distant R for each quark solution.
For the ground state, all the quark charges and masses are almost equal. Thus the basic solution for the ground state is one in which the radius is:
KQ(Q/3) = (MEN/3)V^2 R (3-29)
Where:
MEN = 2.305516E-30 kg (3-30)
In Equation 3-30 the electron in the neutron orbit has a total mass of 2.305516-30Kg or equivalent energy of 1.2933MEV. When we multiply both sides by 3 we get:
KQ^2 = MEN V^2 R (3-31)
Equation 3-31 is a standard Bohr type equation except that the electron in the neutron orbit has a greater mass.
MEn / Me = 2.530925 (3-32)
For the ground state, the Bohr equations are:
R = nh/2πMV (3-33)
Adding Equation 3-33 to 3-31, we get:
KQ^2 = nhV/2π (3-34)
Solving for the ground state velocity we get:
V = 2πKQ^2/nh (3-35)
Since n=1 for the ground state, the velocity is:
V = 2.18768996E6 = C/137.036 (3-36)
In Equation 3-36, we find that the velocity of all three quarks in the neutron ground state is identical with the Hydrogen atom Bohr orbit velocity ground state. The radius of all three quarks is found from equation 3-31.
R = 2.0908436E-11 (3-37)
R = R(BOHR) / 2.5309268 (3-38)
We see that the electron split into 3 parts while maintaining its velocity. It gained a little electrical energy from the electric field. Thus:
E = ΔKQ/R = 68.87 – 13.58 = 55.29 ev (3-39)
The ground state of the neutron’s electron quarks only involved a differential of 55.29 electron volts of electrical energy. Therefore the basic neutrino energy required to bring the split electron into the ground state is:
Energy neutrino = 1.2933MEV – 0.5109989MEV = 0.7823011MEV (3-40)
When we subtract the electrical energy, we get:
Energy neutrino = 0.7822458MEV (3-41)
Now we know the radius and velocity of the three neutron quarks. We now want to know the exact charge of each quark and the exact mass of each quark. The angular momentum vectors must balance out to zero. Then the magnetic moment must equal the measured value at the measured angle.
The Neutron has a magnetic moment of 2.792847NM while the proton has a magnetic moment of –1.913043NM.
The differential magnetic moment is:
Neutron’s differential MM = -(2.792847 + 1.913043) NM (3-42)
Neutron’s differential MM = -4.70589 NM (3-43)
Neutron’s differential MM = 2.3768430E-26 (3-44)
In Equation 3-44 we see that the neutron’s split electrons must add a negative 4.70589 NM to the proton to produce the neutron’s net magnetic moment.
We now need to calculate the value of the angular momentum vectors. First we can calculate the angular momentum vectors which are balanced to zero. Then we can calculate the magnetic moment. The resultant magnetic moment vector will be a vector of the following characteristics. Thus:
Magnetic moment magnitude = 4.1168128E-26 (3-45)
Vector angle = 54.73561 degrees (3-46)
In Equations 3-45 & 46, we need a magnetic moment vector of 4.1168728E-26 at an angle of 54.73561 degrees with respect to the proton’s magnetic moment vector. This will insure the correct magnitude and angle for the neutron vector.
In order to change a hydrogen atom into a neutron, we must split the electron into three parts. The three quarks will have almost identical masses and almost identical charges. From Equations 3-36 & 37 each quark will have a velocity and radius of:
Velocity = C/137.036 = 2.1876900E6 (3-47)
Radius = 2.0908436E-11 (3-48)
The basic quark masses and charges are:
Qa = Qb =Qc = Q/3 = 5.3405867E-20 (3-49)
Ma = Mb = Mc = MEN /3 = 2.305516E-30/3 = 0.7685053E-30 (3-50)
The angular momentum vectors are all basically equal. Thus:
MxVxRx = 3.515234E-35 (3-51)
The magnetic moment vectors are all basically equal. Thus:
QxVxRx = 2.4428472E-24 (3-52)
In order to solve the exact charges and masses of the three nearly identical quarks, it is necessary to normalize the vectors. We can start with angular vector magnitudes equal to one and spaced 120 degrees apart. The corresponding magnetic vectors will also be equal to one and spaced 120 degrees apart.
We can start with a simple set of vectors such as 0.999 for the plus 120 degree vector, 1.000 for the zero degree vector, and 1.001001001 for the minus 120 degree vector.
● Ma = 0.999000 , Qa =1.0010010
a =
---------------------------●---------------------► Mc = 1.000000 , Qc = 1.0000000
b=
● Mb = 1.0010010, Qb = 0.99900000
Figure 3-1 General Vector diagram
In Figure 3-1 we have an initial vector diagram such that the product of each quark charge times its mass is equal to 1.0000. Thus:
MaQa = MbQb = McQc = 1.0000000 (3-53)
In addition, the ratio of Mc to Ma equals the ratio of Mb to Mc. Thus
Mb/Mc = Mc/Ma = 1.001001001 (3-54)
The ratio of the vectors tends to be very small since the resultant magnetic moment is small compared to each quark magnetic moment. Thus:
Ratio = 4.1168128E-26/2.4428472E-24 = 0.01685252 (3-55)
In Equation 3-54 we find that we need an unbalanced vector sum of only 1.7- percent of the vectors. Therefore we need to write the normalized vector relationships with greater accuracy.
First we write the angular momentum equations. The angular momentum resultant vector is equal to zero. Second we solve for angle b. Third we solve for angle a. Fourth we write the magnetic moment vector equations. Fifth we then find a differential vector, which equals 0.01685252.
The process is successive approximation. Each selected ratio of masses is calculated and compared with Equation 3-55. It takes about 10 tries to get an accuracy of less than 10 PPM using a hand calculator.
The equations for the first test case are as follows.
0.999Sina = 1.001Sinb (3-56)
0.999Cos a + 1.001 Cos b = 1.0 (3-57)
Therefore:
0.999Cos a = 1.0 – 1.001Cos b (3-58)
Squaring Equations 3-56 and 3-58 and adding we get:
Cos b = 1 – (0.999)2 + (1.001)2 / 2 (1.001) (3-59)
Cos b = 50.14985
b =59.9008 degrees (3-60)
Using equation 3-56 we can now calculate angle a. Thus:
a = 60.099 degrees (3-61)
Now that we know the angles, we can calculate the magnetic vectors.
Vertical vector = 1.001 Sin 60.099 – 0.999 Sin 59.901 (3-62)
Vertical vector = 0.867755 – 0.864295 = 0.00346 (3-63)
Horizontal vector = 1.0 – 1.001Cos 60.099 – 0.999 Cos 59.901 (3-64)
Horizontal vector = 1.0 – 0.4990013 – 5.019976 = 0.0009989 (3-65)
In general the horizontal vector is rather small especially when we have to add it at right angles. However when we are looking for about 100 PPM, it is slightly significant. The vector sum of the horizontal and vertical vectors is:
Resultant vector = 0.00035916 (3-66)
The first try produced a normalized resultant vector, which is much too small. The forms of the equations remain the same for all the successive approximation attempts. Therefore the equations can be set up rather fast using the hand calculator. After ten tries, the following was the results.
Ma = 0.995147, Qa = 1.00487666 (3-67)
Mb = 1.00487666, Qb = 0.995147 (3-68)
Mc = 1.0 , Qc = 1.0 (3-69)
Angle a = 60.48436 (3-70)
Angle b = 59.51877 (3-71)
Horizontal vector = 1 – 0.49506366 – 0.5047943 (3-72)
Horizontal vector = 0.000141983 (3-73)
Vertical vector = 0.87446502 – 0.85761309 (3-74)
Vertical vector = 0.016851933 (3-75)
Resultant vector = 0.016852531 (3-76)
Error in resultant vector = 0.65 PPM (3-77)
Magnetic Moment = 4.1168158E-26 (3-78)
The error within 10 tries was 0.65 parts per million. The normalized vectors enable us to calculate the differential to the accuracy of the standard hand calculators. We need to write the numbers down to maximum accuracy. Notice in Equation 3-72, that the difference is very tiny and extreme accuracy is necessary. When we reach the magnetic moment Equation 3-78 the extreme accuracy is no longer necessary.
The above method and calculation enables us to find the differential charge and mass difference between the neutron quarks. The slight angular difference between the angular momentum vectors and the magnetic moment vectors are not important. Only the magnitude of the magnetic moment differential vector is important since this vector must be added to the proton magnetic moment to produce the resultant neutron magnetic moment.
We see that the neutron magnetic moment is the result of a small differential magnetic moment among much stronger vectors. The angular momentum vectors are also stronger but they sum up to zero.
We can now tabulate the mass, charge, radius, and velocity of the three neutron quarks.
Ma = 0.331713 MEN (3-79)
Mb = 0.3349756 MEN (3-80)
Mc = 0.333307 MEN (3-81)
Qa = 0.3349756 Q (3-82)
Qb = 0.331713 Q (3-83)
Qc = 0.333307Q (3-84)
Ra = Rb = Rc = Ro = 2.0908436E-11 (3-85)
Vc = Vo = C/137.036 (3-86)
Va = (Mc/Ma) Vc = 1.00048053Vo (3-87)
Vb =(Mc/Ma) Vc = 0.99501874Vo (3-88)
MEN = 2.305516E-30 kg (3-89)
Q = 1.602176E-19 (3-90)
We see that only small differences occur between the three quarks.
The calculations for Va and Vb would be slightly different if we used Equation 3-31 and solved for the square of V. However a computer program would be necessary to produce a closer fit to all three equations. Equations 3-86 through 3-87 relate the masses and velocities so that a perfect match to the Debroglie wavelength occurs. This insures that the radiuses of all three sub-particles are the same.
Another acceptable solution would have the velocities slightly different and the radii slightly different. Thus there can be a small error band in the solution.
In Chapter 5 we will study the states of the neutron. Right now we will return to the proton and study its quarks.
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