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** CURRENT FILAMENTS:**

Our universe is composed of a large number of charged particles, each containing a closed filament with circulating charge. Each closed charge filament contains one quantum of net electric charge, approximately:

1.60217663 X 10^-19 coulombs

which flows along the filament at the speed of light C, approximately:

2.99792458 X 10^8 m / s.

The net charge is uniformly distributed along the filament. In a stable charged particle at every point along the filament the electric and magnetic forces are in balance. For an isolated charged particle in a vacuum that filament geometry forms the wall of a spheromak. Hence isolated electrons and protons exhibit spheromak geometry.

**SPHEROMAK WINDING:**

The spheromak winding has Np poloidal turns and Nt toroidal turns.
The numbers Np and Nt are positive integers with various special mathematical properties.

The winding length Lh is given by:

Lh^2 = (Np Lp)^2 + (Nt Lt)^2

= [Np 2 Pi Ro]^2 + [Nt Lt]^2

where:

Lp = 2 Pi Ro

is the length of a single poloidal filament winding turn and
Lt is the length of a single quasi-toroidal filament winding turn.

Lh^2 = [(Np 2 Pi Ro)^2 + (K Nt 2 Pi Ro)^2]

or

**[Lh / 2 Pi Ro}^2 = Np^2 + K^2 Nt^2**

where Np and Nt are both integers and:

Nt Lt = K Nt 2 Pi Ro

or

**K = [Lt / (2 Pi Ro)]**

which is a constant common to all normal geometry spheromaks.

**Note that in general the spheromak cross section is not round. Even for a round cross section spheromak, due to solenoid curvature:
K is not exactly equal to one.**

Experimental measurements of the Planck Constant h constrain

(Lh / 2 Pi Ro)

which in turn imposes upper limits on Np and Nt.

The spheromak inner wall and the spheromak outer wall have different radii with respect to the main axis of symmetry. Hence, at the inner wall the current filament to current filament distance is less than at the outer wall. Since the net charge is evenly distributed along the filament this issue causes the average surface charge per unit area at the outer wall to be less than the average surface charge per unit area at the inner wall.

The filament turns are in a single layer that contains no cross overs. The filament path only intersects itself at the filament path closure point. A spheromak will collapse if the number of poloidal turns Np and the number of toroidal turns Nt share an integer factor other than one. The spheromak turns must not overlap or intersect. For the same reason Np can never be an integer multiple of Nt or vice versa unless either:

Np = 1

or

Nt = 1

As a result of prime number constraints a spheromak's Npo value cannot equal the spheromak's Nto value.

Any point on the spheromak winding can be identified by its Theta and Phi values. Theta is the angle about the spheromak minor axis measured with respect to a radial line from the minor axis to the orgin. Phi is the angle about the major axis of symmetry measured with respect to the same radial line. Hence at the current path closure point:

Phi = 0, Theta = 0, or Phi = 2 Pi Np or Theta = 2 Pi Nt.

The range of Phi is:

0 < Phi < 2 Pi Npo radians;

The range of Theta is:

0 < Theta < 2 Pi Nto radians

The average value of dTheta / dPhi is given by:

**dTheta / dPhi** = 2 Pi Nto / 2 Pi Npo

= **Nto / Npo**.

However, in a spheromak dTheta / dPhi varies along the winding and is a function of R.

Symmetry indicates that at the center of the spheromak core the net electric field is zero and the net toroidal magnetic field is zero.

**SUMMARY OF SPHEROMAK WINDING CONSTRAINTS:**

The spheromak winding is governed by multiple spheromak existence constraints including:

1) Np and Nt must both be whole positive integers;

2) Np and Nt cannot be equal;

3) Np and Nt cannot share any integer factors other than unity;

4) If there is a change in spheromak state such that Np and Nt increment-decrement the new values of Np and Nt cannot share any integer factors other than unity;

5) Let dNp be a small change in Np. Let dNt be a small change in Nt. Both dNp and dNt must be whole integers;

6) It can be shown that in order to comply with the aforementioned constraints either:

P = Np + 2 Nt which gives:

dNp = - 2 dNt

or

P = 2 Np + Nt

which gives:

2 dNp = - dNt

7) Each potential governing prime number P has associated with it two families of potential Np and Nt integer values;

The spheromak wall consists of a long filament of charge that forms a complex closed spiral current path with length Lh. This filament contains Nt quasi-toroidal turns and Np poloidal turns. The numbers Np and Nt have no common factors so the filament path never crosses itself. The current flows along this filament path at the speed of light. The length of this filament Lh and the speed of light C together give the spheromak a characteristic natural frequency F where:

Each spheromak has a filament length Lh and a nominal radius Ro where the ratio:

(Lh / 2 Pi Ro)

is a constant greater than one common to all stable normal geometry spheromaks. Thus:

Lh = (Lh / 2 Pi Ro) (2 Pi Ro)

and:

Lh F = C

or

F = C / Lh

= [C /(Lh / 2 Pi Ro)][1 / 2 Pi Ro]

Hence:

**Fh = [C / (Lh / 2 Pi Ro)][1 / 2 Pi] [1 / Ro]**

and

**dFh = [C / (Lh / 2 Pi Ro)][1 / 2 Pi] d[1 / Ro]**

which expression is important in evaluating the Planck Constant and the Fine Structure constant. The parameter:

(Lh / 2 Pi Ro)

is a function of the integers Np, Nt and the parameter (Lt / 2 Pi Ro). The parameter Lt will adopt a value that both meets the boundary condition requirements and that also meets the integer requirements on Npo and Nto.

**PRIME NUMBER THEORY:**

Prime number theory together with the requirement for no common factors in Np and Nt gives two formulae for two families of number pairs Np and Nt that share no common factors other than one. Those formula are:

**Family A:**

**P = Np + 2 Nt**

or

Np = P - 2 Nt;

where:

P = controlling prime number.

Note that Np is always odd and that:

dNp = - 2 dNt.

dNp / dNt = -2

**Family B:**

**P = 2 Np + Nt**

where:

P = prime number.

Note that Nt is always odd and that:

dNt = - 2 dNp

dNp / dNt = - 1 / 2

**STABLE SPHEROMAK:**

When a spheromak has stable relative dimensions its parameter:

Lh / 2 Pi Ro

is stable. Hence a condition of spheromak stability is:

d[(Lh / 2 Pi Ro)] / dNt = 0

Recall that:

Lh^2 = (Np Lp)^2 + (Nt Lt)^2

= (Np 2 Pi Ro)^2 + (Nt 2 (Pie / Pi) Pi K Ro)^2

[Lh / (2 Pi Ro)]^2 = [Np Lp / (2 Pi Ro)]^2 + [Nt Lt / (2 Pi Ro)]^2

where:

Lt = K 2 Pi Ro

where:

K corrects for both the solenoid curvature and for the toroidal winding cross section. Note that if the cross section is circular and the solenoid is straight then:

K = 1,

giving:

[Lh / (2 Pi Ro)]^2= Np^2 + Nt^2

In general:

[Lh / (2 Pi Ro)]^2 = Np^2 + K^2 Nt^2

**FAMILY A:**

**P = Np + 2 Nt**

dNp = - 2 dNt

or

Np^2 = (P - 2 Nt)^2

Thus:

[Lh / (2 Pi Ro)]^2 = Np^2 + K^2 Nt^2

= (P - 2 Nt)^2 + K^2 Nt^2

= P^2 - 4 P Nt + 4 Nt^2 + K^2 Nt^2

= P^2 - 4 P Nt + (4 + K^2 Nt^2

**FIND STABLE OPERATING POINT OF FAMILY "A" SPHEROMAK:**

Find zero slope in [Lh / (2 Pi Ro)]^2 versus Nt curve:

d[Lh / (2 Pi Ro)]^2 / dNt

= d[Np^2 + K^2 Nt^2]

d[(P - 2 Nt)^2 + K^2 Nt^2]

= [2 (P - 2 Nt)(-2 dNt) + 2 K^2 Nt dNt]

= [-4 (P - 2 Nt) + 2 K^2 Nt] dNt = 0

Thus at the stable operating point:

[(P - 2 Nt) - (K^2 / 2) Nt = 0

or

**[Np / Nt] = (K^2 / 2)**

Thus:

Np = Nt + 1

implies that:

(K^2 / 2) = [1 + (1 / Nt)]

BOUNDARY CONDITION:

Bpor = Muo Np I / 2 Ro

In a straight solenoid:

Bto = Muo Nt I / 2 Pi Ro

In a curved solenoid:

Bto = Muo Nt I / 2 Pie Ro

where: Pie is slightly different than Pi.

[Bpor / Bto] = (Np / Nt) Pie

In reality:

[Upor / Uto] = [Bpor / Bto]^2

= (Np / Nt)^2 Pie^2

With a straight solenoid:

[Upor / Uto] = Pi^2 = Pi^2 [Nt / Nt]^2

With a real curved solenoid:

[Upor / Uto] = Pie^2 [(Nt + 1) / Nt]^2

where:

(Pie^2 / Pi^2) = (Nt / Np)^2

= [1 + (1 / Nt)]^-2

or

(Pie / Pi) = [1 + (1 / Nt)]^-1

or

**(Pi / Pie) = [1 + (1 / Nt)]**

Thus to meet the boundary condition Nt must be sufficiently large to compensate for the small difference between Pi and Pie.

Substitution in the boundary condition gives:

Upor / Uto = Pie^2 (Np / Nt)^2

= {Pi / [1 + (1 / Nt)]}^2 {(Nt + 1) / Nt}^2

= {Pi Nt / (Nt + 1)}^2 {(Nt + 1) / Nt}^2

= Pi^2

as expected.

Recall that in a spheromak's stable state:

[Lh / 2 Pi Ro]^2 = Np^2 + K^2 Nt^2

= Np^2 + 2 (Np / Nt) Nt^2

= Np^2 + 2 Np Nt

= Np (Np + 2 Nt)

= **Np P**

A solution to the spheromak appears to be:

Np = Nt + 1

which implies that:

P = 3 Nt + 1

and

[K^2 / 2] = [1 + (1 / Nt)]

which implies that at the stable operating point:

Np = Nt + 1

and

[Np / Nt] = Pi / Pie = K^2 / 2 = [1 + (1 / Nt)]

**INTEGER CONSTRAINTS:**

**[Lh / 2 Pi Ro]^2**

= Np^2 + K^2 Nt^2

= Np^2 + 2 Np Nt

= Np (Np + 2 Nt)

= **[Np P]**

**Note that because Np, Nt and P are integers, [Np P] is an integer.** Hence [Lh / 2 Pi Ro]^2 is an integer. Hence K^2 is a rational number.

F = C / Lh

= {C / [Lh / (2 Pi Ro)}{1 / 2 (Pi Ro)}

= {C / (2 Pi Ro)} {1 / (Np P)^0.5}

=

= (Nt + 1)^2 / [(Nt + 1)(3 Nt + 1)]^0.5

= (Nt + 1)^2 / [3 Nt^2 + 4 Nt + 1]^0.5

Nt^2 / [Np^2 + K^2 Nt^2]^0.5

= Nt^2 / [Np P]^0.5

= Nt^2 / [3 Nt^2 + 4 Nt + 1]^0.5

Np P = 3 Nt^2 + 4 Nt + 1 = M = integer

where both Nt and M are integers. Hence:

3 Nt^2 + 4 Nt + (1 - M) = 0

This is a quadratic equation with solution:

Nt = {-4 +/-[16 - 4 (3)(1 - M)]^0.5} / 6

= {-4 +/-[16 + 4 (3)(M - 1)]^0.5} / 6

= {-4 +/- 2 [4 + (3)(M - 1)]^0.5} / 6

For Nt to be an integer the quantity:

[4 + (3)(M - 1)] = N^2

where N is positive integer,

or

3 M + 1 = N^2

We can step through the possible values of N to find possible values of M, some ow which will correspond to possible values of Nt and some of those will correspond to possible values of:
P = 3 Nt + 1.

One of these combinations should satisfy:

[Pi / Pie] = [1 + (1 / Nt)]

The corresponding Np and Nt values will set the Planck and Fine Structure constants.

N = 1, M = 0, Nt = -1

N = 2, M = 1, Nt = 0

N = 3, M not integer

N = 4, M = 5, Nt not integer

N = 6, M not integer

N = 7, M = 16, Nt not integer

N = 8, M = 21, Nt = 2, P = 7

N = 9, M not integer

N = 10, M = 33, Nt not integer

N = 11, M = 40, Nt = 3, P not prime

N = 12, M not integer

N = 13, M = 56, Nt not integer

N = 14, M = 65, Nt = 4, P = 13

N = 15, M not integer

N = 16, M = 85, Nt not integer

N = 17, M = 96, Nt = 5, P not prime

N = 18, M not integer

N = 19, M = 120, Nt not integer

N = 20, M = 133, Nt = 6, P = 19

We can simplify this process by noting that suitable primes conform to:

P = 3 Nt + 1

or

Nt = (P - 1) / 3

where Nt is an integer. Thus we can list all the relevant primes and delete those that do not conform. The relevant primes are:

**PRIME NUMBERS**

Prime numbers less than 2003 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149,
151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487,
491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593,
599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677,
683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787,
797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883,
887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997,
1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069,
1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163,
1171, 1181, 1187, 1193 , 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249,
1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321,
1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439,
1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511,
1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601,
1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693,
1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787,
1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879,
1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993,
1997, 1999, 2003

**AFTER DELETING NONCONFORMING PRIMES THE REMAINING PRIMES ARE:**

**CONFORMING PRIME NUMBERS**

7, 13, 19, 31, 37, 43, 61, 67,
73, 79, 97, 103, 109, 127, 131, 139,
151, 157, 163, 181, 193, 199, 211, 223, 229,
241, 271, 277, 283, 307,
313, 331, 337, 349, 367, 373, 379, 397,
409, 421, 433, 439, 449, 457, 463, 487,
509, 523, 541, 547,
571, 577, 601, 607, 613, 619, 631, 643, 661, 673, CONTINUE
683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787,
797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883,
887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997,
1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069,
1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163,
1171, 1181, 1187, 1193 , 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249,
1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321,
1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439,
1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511,
1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601,
1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693,
1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787,
1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879,
1889, 1901

**FAMILY B:**

**P = 2 Np + Nt**

or

dNt = - 2 dNp

or

**P = 2 Np + Nt**

or

Np^2 = [(P - Nt) / 2]^2

Thus:

[Lh / (2 Pi Ro)]^2 = Np^2 + K^2 Nt^2

= [(P - Nt) / 2]^2 + K^2 Nt^2

= (P^2 / 4) - (P Nt / 2) + (Nt^2 / 4) + K^2 Nt^2

= (P^2 / 4) - (P Nt / 2) + [(1 / 4) + K^2] Nt^2

Find zero slope in [Lh / (2 Pi Ro)]^2 versus Nt:

d[Lh / (2 Pi Ro)]^2 / dNt = - (P / 2) + [(1 / 2) + 2 K^2)] Nt = 0

Hence slope zero is at:

- P + (1 + 4 K^2) Nt = 0

or

P = (1 + 4 K^2) Nt

= Nt + 2 Np

which indicates that at slope zero:

2 Np = 4 K^2 Nt

or

K^2 = (1 / 2) (Np / Nt)

or

**K = [(1 / 2) (Np / Nt)]^0.5**

Then at the spheromak operating point:

[Lh / 2 Pi Ro]^2 = Np^2 + K^2 Nt^2

= Np^2 + (1 / 2) Np Nt

= Np [Np + (Nt / 2)]

= (Np / 2)[2 Np + Nt]

= (Np / 2)[P]

= {C / [Lh / (2 Pi Ro)]} {1 / (2 Pi Ro)}

= {C /(2 Pi Ro)} {1 / [Np P / 2]^0.5}

= {C /(2 Pi Ro)} {2 / (Np P)]^0.5}

** K^2 Nt^2 = (Np Nt / 2)**

or

K^2 = Np / (2 Nt)

and

**Nt^2 / (Np^2 + K^2 Nt^2)^0.5**

= Nt^2 / (Np [Np + (Nt / 2)])^0.5

= Nt^2 / ((Np / 2) [2 Np + Nt])^0.5

= **Nt^2 / [(Np / 2) P]^0.5**

and

**[Np^2 / (Np^2 + K^2 Nt^2)^0.5]**

= Np^2 / (Np [Np + (Nt / 2)])^0.5

= Np^2 / [(Np / 2)(2 Np + Nt)]^0.5

= 2^0.5 Np^1.5 / [P]^0.5

= Np^1.5 / [P / 2]^0.5

= **Np^2 /[Np P / 2]^0.5**

**We must look at the inside wall boundary condition to find the actual relationahip between Np and Nt.**

**WINDING STABILITY:**

When the spheromak filament length is stable implying that the spheromak contained energy is stable:

d[Lh / 2 Pi Ro]^2 = 0

impling that:

d[Np^2 + Nt^2 K^2] = 0

Hence:

2 Np dNp + K^2 2 Nt dNt = 0

or

Np dNp = - K^2 Nt dNt

or

Np / Nt = - K^2 dNt / dNp

Consider Case A:

If dNt / dNp = -(1 /2) and if K^2 ~ 2 then Np / Nt ~ 1. If K^2 adopts some other value then Npo and Nto will be different. We must determine K^2 from the boundary condition at R = Rc.

Bpor = Muo Np I / 2 Ro

Btc = Mup Nt I / 2 Pi Rc

Hence:
[Bpor / Btc] = (Np / Nt) (Pi Rc / Ro)

or

[Bpor / Btc][Ro / Pi Rc] = (Np / Nt]

However, Np, Nt, dNp and dNt are all integers. For a stable winding prime number theory indicates that:

dNp / dNt = - 2:

P = prime number which is odd

P = (Np + 2 Nt), Np = odd, 2 Nt = even

Recall that for stability:

(dNt / dNp) =-(1/ 2) or - 2

This requirement fixes the ratio of (Np / Nt) in terms of K^2

We need to use the field energy distribution to find K^2. The exact value of K^2 will likely fix P and hence the integers Npo and Nto.

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXAt Np = Nt + 1, Np and Nt have no common factors, which is a condition for spheromak existence. To some extent spheromak sizes are constrained by the availability of suitable prime numbers. Thus a prime number tight group alone is not responsible for the prime number that sets the Fine Structure constant. However, the prime number group selection appears to be primarily the result of electromagnetic theory.

**SPHEROMAK SOLUTION:**

The method of finding the mathematical model for an isolated spheromak is:

1) Find equations for the field energy density both inside and outside the spheromak wall;

2) Solve the field energy density equations simultaniouly to find the equation for the spheromak wall;

3) Find equations for the length Lp of a poloidal turn and for the length Lt of a toroidal turn;

4) Find the equation for the length Lh of the current filament closed loop.

5) Since Lh is inversely proportional to the spheromak energy, at the spheromak stable energy:

dLh = 0

6) At the spheromak stable energy apply the condition that:

dNp = - 2 dNt or 2 dNp = - dNt

7) The resulting equations describe stable spheromak behavior.

8) The solution lies at:

Np = Nt + 1 for CASE A where P = Np + 2 Nt

or at

Nt = Np + 1 for CASE B where P = Nt + 2 Np

9) The current filament length Lh in a spheromak is given by:

[Lh / 2 Pi Ro]^2 = Np^2 [Lp / 2 Pi Ro]^2 + Nt^2 [Lt / 2 Pi Ro]^2

The contained energy of an isolated spheromak is constant. Hence, for an isolated spheromak:

d{[Lh / 2 Pi Ro]^2} = 0.

[Lh / 2 Pi Ro]^2 = Np^2 + (K Nt)^2

Hence:

2 Np dNp + 2 (K Nt) K dNt = 0
or

dNp / dNt = - 2 K^2 Nt / 2 Np

= - K^2 Nt / Np

During these tradeoffs the spheromak must survive, meaning that at all times Np and Nt must have no shared integer factors other than one, which forces either:

dNp / dNt = - 2

in which case:

Np + 2 Nt = P

as in Case A above
or

dNp / dNt = - (1 / 2)

in which case:

2 Np + Nt = P

as in Case B above.

Note that with the contemplated solutions in Case A:

-2 = - K^2 Nt / Np = - K^2 (Nt / Nt +1)

or

K^2 = 2 (Nt + 1) / Nt

= 2 [(P - 1) / 3 + 1] / [(P - 1) / 3]

= 2 (P + 2) / (P - 1)

and in Case B:

- (1 / 2) = - K^2 (Nt / Np) = - K^2 (Np + 1) / Np

or

K^2 = Np / [2 (Np + 1)]

= [(P - 1) / 3] / {[2] [(P - 1) / 3 + 1]}

= (P - 1) / [2 (P + 2)]

which is the inverse of Case A.

Typically:

Lp = 2 Pi Ro

and Lt is calculated from the spheromak wall equation.

Natural spheromaks are very stable, which suggests that an external influence which causes minor incrementation of P should not induce spheromak instability. Hence it is likely that the naturally stable value of P occurs in the middle of a sequence of closely spaced prime numbers. Similarly, the value of Npo or Nto that increments or decrements by 1 likely occurs between two prime numbers and the value of Npo or Nto that increments or decrements by 2 likely occurs within a sequence of closely spaced prime numbers.

It is anticipated that in a stable spheromak Lt may slightly adjust to conform with the constraints on Npo and Nto. Thus an electromagnetic solution of a spheromak only gives an approximate result. The precise result is a function of the actual governing prime number P.

Note that the turn numbers Np and Nt can only take discrete integer values that are further limited by the requirement that Np and Nt share no common integer factors other than unity. This requirement indicates that the Np and Nt values of spheromaks normally are governed by a single prime number P.

**SPHEROMAK FORMATION:**

At the commencement of spheromak formation there is an initial packet of energy. That packet of energy is consistent with a prime number P value. That choice of P value enables a spheromak existence line of constant P of the form:

P = Np + 2 Nt or P = 2 Np + Nt.

Then over time Np and Nt can step along this constant P line in accordance with:

dNp = - 2 dNt

or

dNt = -2 dNp

until the spheromak reaches its stable state. The probability of spheromak deep stability is much higher if the desired value of P is in the middle of a cluster of primes.

**PRIME NUMBER CLUSTERS:**

Each prime number P yields a family of (Np / Nt) integer pairs. The spheromak steps along its existence line:

P = Np + 2 Nt

or

P = 2 Np + Nt

to find its stable state where:

Np = Nt +/-1.

A stable spheromak needs to be tolerant of outside disturbances that cause fluctuations in Np and Nt.

The issue is that the prime number P will not change but an external disturbance can potentially cause Np and Nt to temporarily deviate from their normal values Npo and Nto. A stable spheromak should not collapse under such a condition.

This tolerance to disturbance is improved if P is located within a cluster of prime numbers such as 1087, 1091, 1093 and 1097.

P = 1087 implies:

Nt = 360, Np = 367;

Nt = 361, Np = 365;

Nto = 362, Npo = 363;

Nt = 363, Np = 361;

Nt = 364,Np = 359;

P = 1091 implies:

Nt = 361, Np = 369;

Nt = 362, Np = 367;

Nt = 363, Np = 365;

Nto = 364, Npo = 363;

Nt = 365, Np = 361;

Nt = 366, Np = 359;

P = 1093 implies:

Nt = 362, Np = 369;

Nt = 363, Np = 367;

Nto = 364, Npo = 365;

Nt = 365, Np = 363;

Nt = 366, Np = 361;

P = 1097 implies:

Nt = 363, Np = 371;

Nt = 364, Np = 369;

Nt = 365, Np = 367;

Nto = 366, Npo = 365;

Nt = 367, Np = 363;

Nt = 368, Np = 361;

Note that if Np = 363, Nt can be 362, 364, 365, 367

Note that if Np = 365, Nt can be either 361, 363, 364, 366

Note that if Np = 367, Nt can be 360, 362, 363 or 365

Note that if Nt = 365, Np can be 367, 363, 361

Note that if Nt = 364, Np can be 369, 365, 363, 359

Note that if Nt = 363, Np can be 371, 367, 365, 361

For this group of primes the central point appears to be Npo = 365, Nto = 364, 363

The exact Nto value and hence which P value is dominant is not obvious.
[Lh / 2 Pi Ro] = [Np P]^0.5

P = 1091 or 1093

631.0427878 < [Np P]^0.5 < 631.6209306

**P = Np + 2 Nt**

or

Np = P - 2 Nt

or

Np / Nt = (P / Nt) - 2

Note that as Nt increases (Np / Nt) decreases.

Real spheromaks seem to have a P value of:

P = 1087.

There are prime number clusters at:

223, 227, 229, 233;

and at

821, 823, 827, 829;

and at

877, 881, 883, 887;

and at

1087, 1091, 1093, 1097

and at

1297,1301, 1303, 1307;

and at:

1483, 1487, 1489, 1493;

and at

1867, 1871, 1873, 1877

and at

1993, 1997, 1999, 2003

At Np = Nt + 1, Np and Nt have no common factors, which is a condition for spheromak existence. To some extent spheromak sizes are constrained by the availability of suitable prime numbers. Thus a prime number cluster alone is not responsible for the prime number that sets the Fine Structure constant. The prime number cluster selection appears to be primarily the result of electromagnetic theory.

**ESTIMATED SOLUTIONS:**

P = 1069

P = Np + 2 Nt

Np = Nt + 1

P = 3 Nt + 1

Nt = 356

Np = 357

Lh / 2 Pi Ro = {357 (1069)}^0.5

= 617.76

P = 1087

P = Np + 2 Nt

Np = Nt + 1

P = 3 Nt + 1

Nt = 362

Np = 363

Lh / 2 Pi Ro = [Np P]^0.5

= [363 (1087)]^0.5

= 628.156

P = 1091

P = Np + 2 Nt

Np = Nt + 1

Np = 363

Nt = 364

Lh / 2 Pi Ro = [363 (1091)]^0.5

= 629.311

P = 1093

P = Np + 2 Nt

Np = Nt + 1

Nt = 364
Np = 365
Lh / 2 Pi Ro = [365 (1093)]^0.5

= 631.6209306

Np = Nt +/- 1;

9) The requirement for spheromaks existence by avoiding common factors in Np and Nt demands that:

dNp = -2 dNt,

which in effect causes a spheromak to operate at its relative energy maximum at:

Np = Nt + 1

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

On this web site it is shown that:

[113 / 355] ~ [1 / Pi]

2(113) + 355 = 581

which is not prime

but:

2(355) + 113 = 823

which is prime and is in a dense group of primes 821, 823, 827, 829

**PRIME NUMBERS**

Prime numbers less than 2003 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149,
151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487,
491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593,
599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677,
683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787,
797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883,
887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997,
1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069,
1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163,
1171, 1181, 1187, 1193 , 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249,
1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321,
1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439,
1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511,
1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601,
1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693,
1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787,
1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879,
1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993,
1997, 1999, 2003

**RADIATION AND MATTER:**

A spheromak is a mathematical representation of an isolated charged particle.
Electro-magnetic spheromaks are stable energy states. These stable states are reached by emission or absorption of radiation. During radiant energy emission and absorption total system energy and total system momentum are conserved. Charged particles and radiation, both have characteristic frequencies. During photon emission the emitting spheromak's frequency Fh decreases and the amount of propagating radiant energy increases. During photon absorption the absorbing spheromak's natural frequency Fh increases and the amount of propagating radiant energy decreases.

A spheromak has an electric current which follows a closed spiral path that traces out the shape of the wall of a quasi-toroid. The closed current path has both has both toroidal and poloidal circulation components. The quasi-toroidal surface is referred to as the spheromak wall.

The current circulates at the speed of light. At the spheromak geometry the total field energy density just inside the spheromak wall equals the total field energy density just outside the spheromak wall. Hence the electric and magnetic forces are in balance everywhere on the spheromak wall making the spheromak geometrically stable.

The existence, mass and other properties of each spheromak and hence each real atomic particle is governed in part by the prime number P that simultaneously satisfies all of the spheromak constraint equations. Since P, and the spheromak winding parameters Np and Nt must be integers and Np and Nt cannot share common factors the number such prime numbers and hence the number of real atomic particle possibilities is distinctly limited.

One way of investigating this entire matter is to identify values of parameters Np and Nt at which spheromaks do not collapse and hence particles can exist.If a spheromak's static electromagnetic field energy Ett changes from Ea to Eb and the spheromak frequency Fh changes from Fa to Fb then:

dEtt = (Ea - Eb)

= h (Fa - Fb)

= h dFh

Over time spheromaks in free space will absorb or emit energy until they reach a stable state.

At this stable state the value of (dEtt / dFh) for an electromagnetic spheromak is given by:

(dEtt / dFh) = h,
where:

Fh = the natural frequency of the circulating quantum net charge that forms an electromagnetic spheromak and dFh is the frequency of a radiation emitted or absorbed.

This formula is the basis of quantum mechanics. Spheromaks form the static field structure of charged particles with rest mass. Since spheromaks are the main sources and sinks of radiant energy, spheromak properties in large measure determine the radiant energy absorption and emission properties of matter.

On this web page we are primarily concerned about spheromaks in free space. Spheromaks in an atomic, molecular or crystal environment present additional complications.This web page last updated July 24, 2024.

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