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ELECTROMAGNETIC SPHEROMAK

By Charles Rhodes, P.Eng., Ph.D.

ELECTROMAGNETIC SPHEROMAK:
On the web page titled THEORETICAL SPHEROMAK it is shown that a theoretical spheromak is a stable structure. This web page identifies the circumstances that enable an electromagnetic system to form a stable spheromak.

Spheromak mathematics accounts for the existence of both stable quantum charged particles and semi-stable toroidal plasmas and their properties.

A spheromak is governed by two major equations. There is the equation for spheromak energy content which is derived on the web page titled: SPHEROMAK ENERGY and there is the equation for energy density balance at the spheromak wall, referred to herein as the "boundary condition". These two equations lead to the Planck Constant and the Fine Structure Constant which are the subjects of the web page titled: PLANCK CONSTANT.
 

ROLES OF SPHEROMAK ELECTRIC AND MAGNETIC FIELDS:
The electric and magnetic fields of a spheromak store energy and act in combination to position and stabilize the spheromak wall. It is not sufficient to simply position the spheromak wall. The spheromak wall position must be at a relative energy minimum so that if the spheromak is moderately disturbed it spontaneously returns to its stable equilibrium geometry.
 

DEFINITIONS:
U = with no subscripts indicates total field energy density
Uc = value of U at R = Rc, H = 0;
Us = value of U at R = Rs, H = 0;
Ueo = electric field energy density outside spheromak wall;
Umo = magnetic field energy density outside spheromak wall;
Ueoc = Ueo evaluated at R = Rc, H = 0;

Subscripts for Uxyz are defined as follows:
x = o implies center of spheromak at R = 0, H = 0;
x = e implies that Ue is electric field portion of the field energy density;
x = m implies that Um is the magnetic field portion of the energy density;
y = o implies that the expression for U, Ue or Um is only valid outside the spheromak wall;
y = i implies that the expression for U, Ue or Um is only valid inside the spheromak wall;
z = c implies that the value for U, Ue or Um is only valid at R = Rc, H = 0;
z = s implies that the value for U, Ue or Um is only valid at R = Rs, H = 0; Uo = total field energy density at R = 0, H = 0
Ue = electric field energy density
Um = magnetic field energy density
Ueo = electric field energy density outside the spheromak wall;
Ueoc = electric field energy density outside the spheromak wall at R = Rc, H = 0;
Ueos = electric field enegy density outside the spheromak wall at R = Rs, H = 0;
Uei = electric field energy density inside the spheromak wall;
Ueis = electric field energy density inside the spheromak wall at R = Rs, H = 0;
Umo = magnetic field energy density outside the spheromak wall;
Umoc = magnetic field energy density outside the spheromak wall at R = Rc, H = 0;
Umos = magnetic field energy density outside the spheromak wall at R = Rs, H = 0;
Umi = magnetic field energy density inside the spheromak wall;
Umis = magnetic field energy density inside the spheromak wall at R = Rs, H = 0;

Epsilon = permittivity of free space
Muo = permeability of free space
Bxyz and Exyz by:
B = magnetic field
E = electric field
x = p or t or r subscripts where p indicates a poloidal poloidal magnetic field, t indicates a toroidal magnetic field, r indicates a radial electric field
y = i or o subscripts indicating an inside spheromak wall or outside spheromak wall
z = c or s subscripts indicating core wall or outside wall intersection with the equatorial plane;
 

FIELD DEFINITIONS:
Eroc = electric field outside the wall at R = Rc, H = 0;
Bpoc = poloidal magnetic field outside the wall at R = Rc, H = 0;
Btoc = toroidal magnetic field outside the wall at R = Rc, H = 0;
Bpic = poloidal magnetic field inside the wall at R = Rc, H = 0;
Btic = toroidal magnetic field inside the wall at R = Rc, H = 0;
 

SPHEROMAK CHARGE HOSE PARAMETERS
Define:
Ih = charge hose current;
Lh = axial length of closed spiral of charge motion path;
Nt = Integer number of complete toroidal charge hose turns contained in Lh;
Np = Integer number of complete poloidal charge hose turns contained in Lh;
Lt = length of one purely toroidal charge motion turn
Lp = length of one purely poloidal charge motion turn at R = Rf;
As = outside surface area of spheromak wall
Vi = ion velocity along charge motion path
Ve = electron velocity along charge motion path
Q = proton charge
Qs = net spheromak charge
Rhoh = charge per unit length along the charge motion path
C = speed of light
Theta = angle around the main spheromak axis of symmetry
Phi = angle around the toroidal axis of symmetry measured with respect to the spheromak equatorial plane
 

SPHEROMAK GEOMETRY:
The geometry of a spheromak can be characterized by the following parameters:
R = radial distance from the spheromak's axis of cylindrical symmetry to a general point (R, H);
Rc = spheromak's minimum core radius;
Rs = spheromak's maximum equatorial radius;
Rf = [(Rs + Rc) / 2] = spheromak's top and bottom radius;
Rw = radius of co-axial cylindrical enclosure such as a vacuum chamber;
Z = distance of a general point above the spheromak's equatorial plane;
Zs = height of a point on the spheromak wall above the spheromak's equatorial plane;
(2 |Zf|) = spheromak's overall length measured at R = Rf;
 

EQUATORIAL PLANE:
On the spheromak's equatorial plane:
Z = 0
For points on the spheromak's equatorial plane the following statements can be made:

For R = 0 the radial electric field is zero;
For R < Rc the toroidal magnetic field Btoc = 0
For R < Rc the magnetic field Bp is purely poloidal;
For R = 0 the magnetic field Bpo is along the axis of cylindrical symmetry;

For Rc < R < Rs the electric field Eri is cylindrically radial;
For Rc < R < Rs the electric field Eri is proportional to (1 / R);
For Rc < R < Rs the poloidal magnetic field Bpi = 0;
For Rc < R < Rs the toroidal magnetic field Bti is proportional to (1 / R).

In an experimental apparatus at R = Rs, the field energy density U must meet both the nearly spherically radial requirements of free space and the cylindrically radial requirement imposed by the proximity of a cylindrical metal enclosure wall.

For Rs << R in free space the electric field Ero is spherically radial;
For Rs << R in free space the electric field Ero is proportional to (1 / R^2);
For Rs < R in free space the toroidal magnetic field Bto = 0;
For Rs << R in free space the poloidal magnetic field Bpo is approximately proportional to (1 / R^3);

For Rs < R < Rw at H = 0 in a cylindrical metal enclosure the electric field Ero is cylindrically radial;
For Rs < R < Rw at H = 0 in a cylindrical metal enclosure the electric field Ero is proportional to (1 / R);
For Rs < R < Rw the toroidal magnetic field Bto = 0;
 

SPHEROMAK END CONDITIONS:
The spheromak ends are mirror images of each other. Let Rf be the radius of the spheromak end funnel face at the spheromak's longest point. Then the following boundary conditions apply outside the spheromak end face:
For R < Rc the spheromak has no physical end and the magnetic field is entirely poloidal;

For Rc < R < Rs and |Z| < |Zs| the internal magnetic field Bti is toroidal;
For Rc < R < Rs and |Z| < |Zs| the magnetic field Bti is proportional to (1 / R).

Outside the spheromak wall the magnetic field is purely poloidal;
For Rc < R < Rs and Z^2 < Zs^2 the electic field component parallel to the main axis of symmetry is zero;
For (Z^2 + R^2) >> Hf^2 the electric field is spherically radial;
For (Z^2 + R^2) >> Hf^2 spherical electric field is proportional to (R^2 + Z^2)^-1;
 

When these constraints are properly applied the quantitative agreement between the engineering model and published spheromak photographs is remarkable.
 

CHARGE CIRCULATION:
Assume that the spheromak wall is composed of a charge string of length Lh containing uniformly distributed charge Qs that is circulating around a closed spiral path at the speed of light C. The spheromak inside radius measured from the axis of symmetry is Rc and the spheromak outside radius measured from the axis of symmetry is Rs. From Pythagorus theorm the charge string length Lh and hence the circulating current path length forming the spheromak is given by:
Lh = [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc)Kc / 2)^2]^0.5
= [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc)Kc / 2)^2]^0.5
= {Pi Rc [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2 Kc^2]^0.5}
= {(Pi Ro / So) [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2 Kc^2]^0.5}

Lh / Ro
= {(Pi / So) [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2 Kc^2]^0.5}

Note that Np and Nt are positive integers. The quantity (Lh / Ro) is believed to be a geometric constant for a stable spheromak. The stability of this quantity relies on the stabilities of Np, Nt and So.

Note that as Ro decreases with increasing spheromak energy so does Lh decrease so the ratio (Lh / Ro) remains constant.

The circulating current Ih is given by:
Ih = Qs Fh
= Qs C / Lh = Qs C / {(Pi Ro / So) [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2 Kc^2]^0.5}
= [Qs C So / {Pi Ro Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2 Kc^2]^0.5}]
= [(Qs C) / (Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2 Kc^2]^0.5]

Hence:
(Ih / Rc) = (1 / Rc) [(Qs C) / (Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2 Kc^2]^0.5]
= [(Qs C) / (Pi Ro^2 Nt)] [So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2 Kc^2]^0.5]
 

FIND Ih:
From Pythagorus theorm:
Ih = Qs C / Lh = Qs C / [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc) / 2)^2]^0.5
= Qs C / {Pi [(Np (Rs + Rc))^2 + (Nt (Rs - Rc))^2]^0.5}
= Qs C / {Pi Rc [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]^0.5}
= [Qs C / {Pi Rc Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [Qs C Ro / {Pi Rc Ro Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [Qs C So / {Pi Ro Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [(Qs C) / (Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]

Hence:
(Ih / Rc) = (1 / Rc) [(Qs C) / (Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
= [(Qs C) / (Pi Ro^2 Nt)] [So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
where:
Ih = Q Fh = Q C / Lh
which is referred to herein as the electromagnetic spheromak existence condition. This equation imposes an important relationship between Nr and So in a spheromak.

A critical part of spheromak analysis is finding the functional relationship between Nr and So.
 

NATURAL FREQUENCY:
The natural frequency Fh of a spheromak is:
Fh = C / Lh = C / {(Pi Ro / So) [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2 Kc^2]^0.5}
= So C / {(Pi Ro)[(Np (So^2 + 1))^2 + (Nt (So^2 - 1) Kc)^2]^0.5}

Rearranging this equation gives:
[1 / Ro] = {[(Pi Fh) / (So C)] [(Np (So^2 + 1))^2 + (Nt (So^2 - 1)Kc)^2]^0.5}
= {[(Pi Fh) / (So No C)] [(Np / No)^2 (So^2 + 1)^2 + (Nt /No)^2 (So^2 - 1)^2 Kc^2]^0.5}

Note that this formula applies to all spheromaks
 

SPHEROMAK REVIEW:
Uo can be expressed in terms of known physical parameters by matching the far field functions.

The dimension Ro is the nominal spheromak radius from its axis of symmetry. It is shown on this web site that the near field behavior of Ue leads to determination of the Fine Structure Constant and the Planck Constant.

Inside the spheromak wall the magnetic field energy density is given by:
Umi = Umic [(Rc / R)^2]

At all points on the spheromak wall the total inside energy density equals the total outside energy density. Hence at all points on the spheromak wall:
Umo + Ueo = Umi

At R = Rc and Z = 0:
Umoc + Ueoc = (Umic).

This is a key equation to the boundary condition that accounts for spheromak behavior.

This web page shows that there is a boundary condition which connects spheromak shape factor:
So^2 = (Rs / Rc)
to
Nr = (Np / Nt)

To find the spheromak operating point at particular Np, Nt integer values calculate Nr, then calculate So, and then calculate:
Z^2 = {[Np (So^2 + 1)]^2 + [Np (So^2 -1)Kc]^2}.
and then calculate [1 / Alpha]^2.

SPHEROMAK EVOLUTION:
As shown on the web page titled:THEORETICAL SPHEROMAK, for So < 1.5 the relative depth of a spheromak mutual potential energy well is small which causes a spheromak to be unstable. A spheromak tends to emit static field energy until So^2 increases to a stable value of about 2.0.

As shown on the web page titled:THEORETICAL SPHEROMAK outside a spheromak wall there is an energy density function of the form:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2 and inside the spheromak wall there is an energy density function of the form:
U = Uto (Ro / R)^2

In an electromagnetic spheromak the external energy density function is a result of a combination of radial electric and poloidal magnetic fields. The internal energy density function is a result of a toroidal magnetic field.

There is a complication because A is weakly dependent on:
So^2 = Rs / Rc
When So^2 = 1 then A = 1.0000
However, for plasma spheromaks with So^2 ~ 4.1 the parameter A can be as large as:
A = 1.08
Atomic particle spheromaks seemto exhibit A values of about 1.02. We need to do an energy density analysis along the Z axis to try to determine the exact relationship between A and So.
 

SPHEROMAK WALL:
Under the circumstances of plasma spheromak generation the electrons and ions follow a closed spiral path. This path traces out a three dimensional surface in the shape of a hollow torus, like the icing on a doughnut, known as the spheromak wall. The formation of this wall is discussed on the web page titled CHARGE HOSE SHEET. Inside the spheromak wall the magnetic field is purely toroidal and the net electric field is zero. Outside the spheromak wall the magnetic field is purely poloidal and the net electric field is mainly radial.

The trapped electrons and ions circulate on closed spiral paths within the thin surface known as the spheromak wall. The spheromak wall position corresponds to a system total energy minimum.

In a plasma spheromak the electrons and ions follow similar but opposite spiral paths within the spheromak wall. The positive ions move opposite to the electrons to approximately balance both charge and momentum within the spheromak wall. Within the spheromak wall there is sufficient separation between the opposite flowing electron and ion streams to prevent the energetic electrons being scattered by collisions with the spheromak ions.

A fundamental difference between a plasma spheromak and an atomic particle spheromak is that in a plasma the electrons and ions are of two types and are subject to inertial forces whereas in a quantum charged particle the circulating charge is also coomposed of two types but has no inertial mass. High temperature plasma spheromaks are also subject to relativistic effects.
 

ENERGY DENSITY BALANCE:
For a spheromak wall position to be stable the total field energy density must be the same on both sides of a thin spheromak wall. This requirement leads to boundary condition equation that determines the shape of spheromaks.
 

SPHEROMAK WALL POSITION:
A spheromak is a stable energy state. The spheromak wall positions itself to achieve a total energy relative minimum consistent with the spheromaks natural frequency Fh. At every point on the spheromak wall the sum of the electric and magnetic field energy densities on the outside side of the spheromak wall equals the toroidal magnetic field energy density on the inside of the spheromak wall. This general statement resolves into different detailed boundary conditions at different points on the spheromak wall. This general boundary condition establishes the spheromak core radius Rc on the equatorial plane, the spheromak outside radius Rs on the equatorial plane and the spheromak length 2 Zf.
 

ELECTROMAGNETIC SPHEROMAK STRUCTURE:
1) A spheromak wall is composed of a closed spiral of charge hose or plasma hose of overall length Lh;

2) Spheromak net charge Qs is uniformly distributed over charge hose or plasma hose length Lh resulting in a net charge per unit length:
[Qs / Lh];

3) The spheromak net charge circulates at the apparent speed of light C (constant velocity) along the charge hose path, which gives the spheromak a natural frequency:
Fh = C / Lh

4) The net charge has two orthogonal charge circulation velocity components, a component Vp which contributes to the external poloidal magnetic field and a component Vt which contributes to the internal toroidal magnetic field. Velocities Vt and Vp can each be either positive or negative, so a spheromak has 4 possible quantum states. Hence each spheromak has two orthogonal magnetic vectors (poloidal and toroidal) each of which has two possible orientations. There is poloidal up and poloidal down magnetic vector. For each of the two poloidal magnetic vector directions there is toroidal clockwise (CW) and toroidal counter clockwise (CCW) magnetic vector. If a process generates two spheromaks in which a particular poloidal magnetic state of one spheromak is opposite to the same poloidal magnetic state of the other spheromak the two spheromaks are "weakly quantum entangled". If both the poloidal and the toroidal magnetic states of the two spheromaks are opposite there is "strong quantum entanglement".

Define:
Lp = length of one average charge path poloidal turn
Np = number of charge path poloidal turns
Lt = length of one charge path toroidal turn
Nt = number of charge path toroidal turns.
Rs = toroid outside radius on the equitorial plane
Rc = toroid inside radius on the equitorial plane.

A = elliptical toroidal path [(major radius) / (minor radius)]

5) Lh^2 = (Np Lp)^2 + (Nt Lt)^2 = C^2 / Fh^2

6) Lp = Pi (Rs + Rc)
7) Lt = Pi (Rs - Rc) Kc
8) Ft = Nt Fh
9) Fp = Np Fh
10) Ft / Fp = Nt / Np
11) (Nt Lt / Vt) = (Np Lp / Vp) = (1 / Fh) = (Lh / C)
12) [Nt Pi (Rs - Rc) Kc / Vt] = [Np Pi (Rs + Rc) / Vp] = [Lh / C]
Hence:
13) Vt = C Nt Pi ( Rs - Rc) Kc / Lh
and
14) Vp = C Np Pi (Rs + Rc) / Lh

15) Thus:
Vt^2 + Vp^2
= [C Pi / Lh]^2 [Nt^2 (Rs - Rc)^2 Kc^2 + Np^2 (Rs + Rc)^2] = C^2

16) Hence:
[Pi / Lh]^2 [Nt^2 (Rs - Rc)^2 Kc^2 + Np^2 (Rs + Rc)^2] = 1

17) Thus:
Lh^2 = (Pi)^2 [Np^2 ((Rs + Rc))^2 + Nt^2 (Rs - Rc)^2 Kc^2]
= (Pi)^2 Ro^2 [Np^2 (So + (1 / So))^2 + Nt^2 (So - (1 / So))^2 Kc^2]
= [(Pi Ro) / So]^2 [Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2 Kc^2]

18) Recall that:
C^2 = Vp^2 + Vt^2
= {Fh Lh}^2
= Fh^2 (Pi)^2 [Np^2 ((Rs + Rc))^2 + Nt^2 (Rs - Rc)^2 Kc^2]

19)
20)

21) In a stable spheromak the charge hose current:
Ih = Qs Fh
is constant;

22) The charge hose current Ih causes a purely toroidal magnetic field inside the spheromak wall and a purely poloidal magnetic field outside the spheromak wall;

23) The net charge Qs causes an electric field outside the spheromak wall which is spherically radial;

24) At the center of the spheromak at (R = 0, Z = 0) the net electric field is zero;

25) Define the spheromak nominal radius Ro and shape parameter So by:
(A Rs / Ro) = (Ro / A Rc) = So
or
Ro^2 = A^2 Rs Rc = So^2

26) In the region enclosed by the spheromak wall where:
Rc < R < Rs and |Z| < |Zs|
the total field energy density U takes the form:
Ui = Uio (Ro / R)^2
where:
Uim = toroidal magnetic field energy density inside the spheromak wall;

27) Outside the spheromak wall the total field energy density should take the form:
U = Ue + Um
= Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2

where:
Uo = [(Muo Qs^2 C^2) / (32 Pi^2 Ro^4)];

28) Everywhere on the thin spheromak wall:
U = Ueo + Umo = Uei + Umi
where:

Uei = 0

29) In a electromagnetic spheromak the static electric and magnetic field energy density functions are a result of distributed charge that causes the static electric field and that circulates within the spheromak wall with a characteristic frequency:
Fh = C / Lh
causing the static magnetic fields. The charge circulation pattern is described by five parameters: Np, Nt, So, A and Ro where:
Np = number of poloidal turns per charge circulation cycle;
Nt = number of toroidal turns per charge circulation cycle;
So^2 = (Rs / Rc) = spheromak shape parameter
Rs = maximum radius from spheromak symmetry axis to spheromak wall;
Rc = minimum radius from spheromak symmetry axis to spheromak wall;
Ro^2 = (A^2 Rs Rc) where Ro is the nominal spheromak radius.
A = [2 Zf / (Rs - Rc)]
 

30) In order for a spheromak to exist Np and Nt must not have any common factors. A function which produces integer Np and Nt values without common factors is:
[Np / Nt] = [(N + 1) / (P - 2 (N + 1))]
where P = a prime number where N is an integer in the range:
0 < 2 (N + 1) < P

The total spheromak energy E is of the form:
E = R(Ro) S(So) where:
R(Ro) is a function proportional to (1 / Ro) and hence to natural frequency Fh and S(So) is a function consisting of two orthogonal terms, one proportional to the number of spiral path poloidal turns Np and the other proportional to the number of spiral path toroidal turns Nt.

Existence and boundary conditions impose a mathematical relationship between these two orthogonal energy terms.
 

SPHEROMAK STATIC FIELD ENERGY:
The web page titled SPHEROMAK ENERGY shows that the energy density functions:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
outside a spheromak wall and
U = Uto [(Ro / R)^2]
inside a spheromak wall result in spheromaks with total energy given by:
Ett = [Uo Pi^2 Ro^3 / A^2]
X {1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[(So^2 + 1)^2 - (So - 1)^2 (So^2 + 1) + 2 So (So - 1)^2] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[(So^4 + 2 So^2 + 1) - (So^2 -2 So + 1) (So^2) - (So^2 -2 So + 1) + 2 So (So^2 - 2So + 1)] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[2 So (2 So^2 - 2So + 2)] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2] {4 So (So^2 - So + 1) / (So^2 + 1)^2}
 

CONNECTION WITH ELECTRIC AND MAGNETIC FIELDS:
Outside the spheromak wall:
U = Uo [Ro^2 / (Ro^2 + (A Ro)^2 + Z^2]^2
= Uo [(A^2 Rs Rc)^2 / (A^2 Rs Rc + (A R)^2 + Z^2)]^2

This total energy density function either exactly or very closely approximates physical reality. It represents the total field energy density that arises from the poloidal magnetic field and the radial electric field. At R = 0, Z = 0 due to geometrical symmetry the net electric field is zero and the field energy density:
U = Uo
is entirely due to the poloidal magnetic field.

In the far field where:
[R^2 + (A Z)^2] >> Ro^2
the electric field energy density is given by:
Uo [Ro^2 / (Ro^2 + R^2 + (A Z)^2)]^2
= [Epsilon / 2][Qs / (4 Pi Epsilon (R^2 + (A Z)^2))]^2
 

Let Qs be the net static charge. Hence far field matching gives:
Uo Ro^4 = [Epsilon / 2][Qs / (4 Pi)]^2
or
Uo Ro^4 = Qs^2 / (32 Pi^2 Epsilon)
or
Uo = Qs^2 / (32 Pi^2 Epsilon Ro^4)

Recall that:
C^2 = 1 / (Muo Epsilon)
Hence:
Uo = [Qs^2 Muo C^2 / (32 Pi^2 Ro^4)]

Hence:
Ett = [Uo Pi^2 Ro^3 / A^2] {4 So (So^2 - So + 1) / (So^2 + 1)^2}
 
= [Qs^2 Muo C^2 / (32 Pi^2 Ro^4)] [Pi^2 Ro^3 / A^2]
{4 So (So^2 - So + 1) / (So^2 + 1)^2}
 
= [Qs^2 Muo C^2 / (32 Ro)] [1 / A^2] {4 So (So^2 - So + 1) / (So^2 + 1)^2}

On the web page titled:CHARGE HOSE it is shown that for a wound current ring with poloidal winding current Ip:
Uom = Muo (Np Ip)^2 / 8 Ro^2
where:
Ip = Qs Fh
where Fh = frequency at which Qs cycles around its closed path, giving:
Uom = (Muo / 8 Ro^2) Np^2 [Qs Fh]^2

Bto = Muo Nt It / 2 Pi Ro
= [(Muo Nt) / (2 Pi Ro)] [(Qs Fh)]

Uto = Bto^2 / 2 Muo
= (1 / 2 Muo){[(Muo Nt) / (2 Pi Ro)] [(Qs Fh)]}^2

Thus:
Uto / Uo
= (1 / 2 Muo){[(Muo Nt) / (2 Pi Ro)] [(Qs Fh)]}^2 / (Muo / 8 Ro^2) Np^2 [Qs Fh)]^2
 
= {[( Nt / Pi)]}^2 / Np^2
 
= (Nt / Pi Np)^2

In spheromaks generally Uto << Uo

From the web page titled THEORETICAL SPHEROMAK:

Uto / Uo = (1 / A^2)[So^2 / (So^2 + 1)^2]

Thus:
[Nt / Pi Np]^2 = (1 / A^2)[So^2 / (So^2 + 1)^2]

Uo = (Muo / 8 Ro^2) Np^2 [Qs Fh)]^2 = [Qs^2 Muo C^2 / (32 Pi^2 Ro^4)]
or
( Np^2 Fh^2) = [C^2 / (4 Pi^2 Ro^2)]
or
Fh = C / (2 Pi Ro Np)
Remember that this solution is only valid for a 2 dimensional ring.

The characteristic frequency of a three dimensional particle is:
F = C / Lh
= C / (Pi {[Nt (Rs - Rc) Kc]^2 + [Np (Rs + Rc)]^2}^0.5)
= C / (Pi Ro {[Nt (So - (1 / So)) Kc]^2 + [Np (So + (1 / So))]^2}^0.5)
= C So / (Pi Ro {[Nt (So^2 - 1) Kc]^2 + [Np (So^2 + 1)]^2}^0.5)

For electromagnetic spheromaks at R = 0, Z = 0 the electric field is zero. Hence:
Uo = (Bpo^2 / 2 Muo)
which gives:
Efs = [Uo Ro^3 Pi^2 / A^2]
= [(Bpo^2 / 2 Muo) Ro^3 Pi^2 / A^2]
= [Qs^2 / (32 Pi^2 Epsilon Ro^4)][Ro^3 Pi^2 / A^2]
= [Qs^2 / (32 Epsilon Ro A^2)]
= [Qs^2 Muo C^2 / (32 Ro A^2)]
 

FIELD ENERGY DENSITY INSIDE THE SPHEROMAK WALL:
In the toroidal region inside the spheromak wall where Rc < R < Rs and |Z| < |Zs| the total field energy density is given by:
Ut = Uc (Rc / R)^2

This energy density function arises purely from a toroidal magnetic field.
 

CLASSICAL CHARGED PARTICLE RADIUS Re:(this section is not part of the Planck constant derivation)
The classical expression for particle field only considers the electric field energy and assumes a particle radius of Re. The above expression can be compared to classical electric field energy for an electron given by:
Classical electic field = (1 / 4 Pi Epsilon) Q / R^2
El;ectric field Energy density = (Epsilon / 2)(Electric Field)^2
= (Epsilon / 2) [(1 / 4 Pi Epsilon) Q / R^2]^2
and
Electric Field energy = Integral from R = Re to R = infinity of:
(Epsilon / 2) [(1 / 4 Pi Epsilon) Qs / R^2]^2 4 Pi R^2 dR
= Integral from R = Re to R = infinity of:
[(Qs^2 / 8 Pi Epsilon) / R^2] dR
= [(Qs^2 / 8 Pi Epsilon) / Re]
= [(Qs^2 Mu C^2 / 8 Pi) / Re]
where Re is the classical electron radius.

Equating the two expressions for field energy gives:
[(Qs^2 Muo C^2 / 8 Pi) / Re] = (Muo C^2 / 2)[Qs / 4]^2 [1 / Rs Rc]^0.5
or
[(1 / Pi) / Re] = [1 / 4] [1 / Rs Rc]^0.5
or
Re = (4 / Pi) [Rs Rc]^0.5
= (4 / Pi) Ro

It is shown herein that the ratio of Rs to Rc corresponds to a spheromak energy minimum that sets the Planck constant but the precise reasons why electrons and protons have different mass (rest energy) are uncertain.
 

Inside the spheromak wall the energy density function arises from a toroidal magnetic field and a cylindrically radial electric field.

TOROIDAL MAGNETIC FIELD ENERGY:(this section is not part of the Planck constant derivation)
The toroidal magnetic field is given by:
Bt = [(Muo Ih Nt) / (2 Pi R)]

The toroidal magnetic field energy density is given by:
Bt^2 / 2 Muo = [(Mu Ih Nt) / (2 Pi R)]^2 / (2 Muo)
= (Muo / 8)[(Ih Nt) / (Pi R)]^2

Ih = (Qs C) / Lh
and
Lh^2 = {[Nt 2 Pi (Rs - Rc) Kc / 2)]^2 + [Np 2 Pi (Rs + Rc) / 2]^2}^0.5
= Pi^2 {[Nt (Rs - Rc) Kc)]^2 + [Np (Rs + Rc)]^2}
or
(Ih Nt) = (Qs C Nt) / Lh

Thus:
(Bt^2 / 2 Muo) = (Mu / 8)[(Ih Nt) / (Pi R)]^2
= (Muo / 8)[(Qs C Nt) / (Lh Pi R)]^2

On the spheromak wall:
Zs^2 = A^2 (Rs - R) (R - Rc)

An element of toroidal magnetic volume is:
dV = 2 Zs 2 Pi R dR
= 4 Pi R dR A [(Rs - R) (R - Rc)]^0.5

An element of toroidal magnetic energy dEtm is:
dEtm = (Bt^2 / 2 Muo) dV
= (Muo / 8)[(Qs C Nt) / (Lh Pi R)]^2 4 Pi R dR A [(Rs - R) (R - Rc)]^0.5 = A (Muo / 2)[(Qs C Nt)^2 / (Lh^2 Pi] [(Rs - R) (R - Rc)]^0.5 (dR / R)

Toroidal magnetic energy Etm =
Integral from R = Rc tpo R = Rs of:
A (Muo / 2)[(Qs C Nt)^2 / (Lh^2 Pi] [(Rs - R) (R - Rc)]^0.5 (dR / R)

The characteristic frequency of this particle is:
F = C / Lh
= C / (Pi {[Nt (Rs - Rc) Kc]^2 + [Np (Rs + Rc)]^2}^0.5)

Thus the toroidal magnetic energy Et is given by:
Et = Integral from Xa = (Rc / Rs) to Xb = 1 of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(1 - X)(X - Xa)]^0.5 dX / X)
= Integral from Xa = (Rc / Rs) to Xb = 1 of:
(Muo / 2)[(Qs^2 C Nt^2) Rs / (Lh Pi)] Fh [(1 - X)(X - Xa)]^0.5 (dX / X)
= Integral from Xa = (Rc / Rs) to Xb = 1 of:
[(Muo Qs^2 C) / 4 Pi)] Fh Nt (2 Rs Nt / Lh) [(1 - X)(X - Xa)]^0.5 (dX / X)

This is an accurate expression for the toroidal magnetic energy.


 

CONTINUING WITH THE DEVELOPMENT OF Efs:
Recall that from THEORETICAL SPHEROMAK:
Lp = Pi (Rs + Rc) Lt = Pi (Rs - Rc) Kc Lh = {[Pi (Rs + Rc)]^2 + [Pi (Rs - Rc)]^2 Kc^2}^0.5

Thus:
Fh = C / Lh
= C / {[Pi Np (Rs + Rc)]^2 + [Pi Nt (Rs - Rc)]^2 Kc^2}^0.5
= C / Rc {[Pi Np (So^2 + 1)]^2 + [Pi Nt (So^2 - 1)]^2 Kc^2}^0.5
= (Ro / Rc) [C / Ro Nt](1 / {[Pi Nr (So^2 + 1)]^2 + [Pi (So^2 - 1)]^2 Kc^2}^0.5})
= [So C / Ro Nt Pi] [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}]

Hence:
(1 / Ro) = [Fh Nt Pi {[Nr (So^2 + 1)]^2 + [(So^2 - 1) Kc^2]^2}^0.5}] / So C]

Thus:
Efs = [Muo C^2 Qs^2 / 32 Ro A]
= [ Muo C^2 Qs^2 / 32 A] [Fh Nt Pi {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}] / So C]
= [ Muo C Qs^2 Pi / 32 A] [Fh Nt {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}] / So]
= [Muo C Qs^2 / 4 Pi A] [Pi^2 / 8] Fh Nt ({[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2 Kc^2} / So^2)^0.5

Note that Efs is dependent on Nr^2 and So^2. Since Nr^2 and So^2 are related by the common boundary condition there is a low energy operating value of So^2 and there is a corresponding operating value of Nr^2.

THE SPHEROMAK BOUNDARY CONDITION AT R = Rc. Z = 0:

Outside the spheromak wall the total field energy density is given by:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
At R = Rc, Z = 0 the total field energy density is given by:
Upc = Uo [Ro^2 / (Ro^2 + (A Rc)^2)]^2
= Uo [(A^2 Rs Rc) / ((A^2 Rs Rc) + (A Rc)^2)]^2
= Uo [Rs / (Rs + Rc)]^2

Inside the spheromak walls where the static field energy is entirely toroidal magnetic, at R = Rc, Z = 0:
Btc = (Muo Nt It / 2 Pi Rc)

The corresponding inside field energy density at R = Rc, Z = 0 is:
Utc = Btc^2 / 2 Muo
= (1 / 2 Muo)[(Muo Nt It) / (2 Pi Rc)]^2
= (Muo / 2) [(Nt It) / (2 Pi Rc)]^2

Hence the boundary condition at R = Rc, Z = 0 is:
Upc = Utc
or
Uo [Rs / (Rs + Rc)]^2 = (Muo / 2) [(Nt It) / (2 Pi Rc)]^2

Recall that:
Uo = [Qs^2 Muo C^2 / (32 Pi^2 Ro^4)]

Recall that:
It = (Qs C) / Lh

Substitution for Uo and It in the boundary condition gives:
[Qs^2 Muo C^2 / (32 Pi^2 Ro^4)] [Rs / (Rs + Rc)]^2
= (Muo / 2) [(Nt Qs C) / (2 Pi Rc Lh)]^2
or
[1 / (4 Ro^4)] [Rs / (Rs + Rc)]^2
= [Nt / (Rc Lh)]^2

Recall that:
Ro^2 = A^2 Rs Rc
giving:
[1 / (4 A^4 Rs^2 Rc^2)] [Rs / (Rs + Rc)]^2
= [Nt / (Rc Lh)]^2
or
[1 / (4 A^4)] [1 / (Rs + Rc)]^2
= [Nt / Lh]^2
or rearrange to get:
4 A^4 [Nt / Lh]^2 = [1 / (Rs + Rc)]^2
or
4 A^4 Nt^2 / [(Np Lp)^2 + (Nt Lt)^2] = [1 / (Rs + Rc)]^2
or
4 A^4 Nt^2 / [(Np Pi (Rs + Rc))^2 + (Nt Pi (Rs - Rc) Kc)^2]
= [1 / (Rs + Rc)]^2
or
4 A^4 / {Pi^2 [(Nr (Rs + Rc))^2 + ((Rs - Rc) Kc)^2]}
= [1 / (Rs + Rc)]^2
or
{Pi^2 [(Nr (Rs + Rc))^2 + ((Rs - Rc) Kc)^2]} / 4 A^4 = (Rs + Rc)^2
or
[4 A^4 / Pi^2](Rs + Rc)^2 = [(Nr (Rs + Rc))^2 + ((Rs - Rc) Kc)^2]
or
Nr^2 + [(Rs - Rc)^2 Kc^2 / (Rs + Rc)^2] = [4 A^4 / Pi^2]
or
Nr^2 + Kc^2 [(So^2 - 1) / (So^2 + 1)]^2 = [4 A^4 / Pi^2]

This is the spheromak boundary condition at R = Rc, Z = 0.
 

BOUNDARY CONDITION AT R = Rc, Z = 0 LIMIT ON THE RANGE OF So^2:

Note that:
0 < Nr^2 < (4 A^4 / Pi^2)
which restricts the range of [((Rs - Rc)) / (Rs + Rc)]^2 to:
0 < [((Rs - Rc)) / (Rs + Rc)]^2 < [4 A^4 / Kc^2 Pi^2]
or
0 < [((Rs - Rc)) / (Rs + Rc)] < [2 A^2 / Kc Pi]
or
0 < [(So^2 - 1) / (So^2 + 1)] < [2 A^2 / Kc Pi]

At:
[(So^2 - 1) / (So^2 + 1)] = [2 A^2 / Kc Pi]
or
(So^2 - 1) = [2 A^2 / Kc Pi](So^2 + 1)
or
So^2 {1 - [2 A^2 / Kc Pi]} = {1 + [2 A^2 / Kc Pi]}
or
So^2 = {1 + [2 A^2 / Kc Pi]} / {1 - [2 A^2 / Kc Pi]}

Thus for a spheromak:
1 < So^2 < {1 + [2 A^2 / Kc Pi]} / {1 - [2 A^2 / Kc Pi]}

For A^2 / Kc = 1 This expression simplifies to:
1 < So^2 < {1 + [2 / Pi]} / {1 - [2 / Pi]}
or
1 < So^2 < {Pi + 2} / {Pi - 2}
or
1 < So^2 < {5.14} / {1.14}
or
1 < So^2 < 4.51
 

 
 

SPHEROMAK BOUNDARY CONDITION AT R = Rf, Z = Zf:
Immediately inside the spheromak wall:
Btf = Muo Nt It / 2 Pi Rf
or
Utf = (1 / 2 Muo)[ Muo Nt It / 2 Pi Rf]^2
= (Muo / 2) [Nt It / Pi (Rs + Rc)]^2

Immediately outside the spheromak wall:
Upf = Uo [Ro^2 / (Ro^2 + (A Rf)^2 + Zf^2]^2
= Uo [A^2 Rs Rc / (A^2 Rs Rc + (A (Rs + Rc) / 2)^2 + Zf^2]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + (Zf^2 / A^2)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + (A^2 (Rs - Rf)(Rf - Rc) / A^2)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + ((Rs - Rf)(Rf - ^2 = Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + ((Rs - ((Rs + Rc) / 2))(((Rs + Rc / 2) - Rc))]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2 + 2 Rs Rc) / 4) + ((Rs - Rc) / 2))((Rs - Rc) / 2)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2 + 2 Rs Rc) / 4) + ((Rs^2 - 2 Rs Rc + Rc^2) / 4)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2) / 4) + ((Rs^2 + Rc^2) / 4)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2) / 2)]^2
= Uo [Rs Rc 2 / (Rs + Rc)^2]^2

Utf = Upf
or
(Muo / 2) [Nt It / Pi (Rs + Rc)]^2 = Uo [Rs Rc 2 / (Rs + Rc)^2]^2
or
(Muo / 2) [Nt It / Pi]^2 = Uo [Rs Rc 2 / (Rs + Rc)]^2

Recall that:
It = Qs C / Lh
and
Uo = [Muo C^2 Qs^2 / 32 Pi^2 Ro^4]

Hence:
(Muo / 2) [Nt It / Pi]^2 = Uo [Rs Rc 2 / (Rs + Rc)]^2
or
(Muo / 2) [Nt Qs C / Lh Pi]^2 = [Muo C^2 Qs^2 / 32 Pi^2 Ro^4] [Rs Rc 2 /(Rs + Rc)]^2
or cancelling equal terms:
[Nt / Lh]^2 = [1 / 4 Ro^4] [Rs Rc /(Rs + Rc)]^2
or
[Nt / Lh]^2 = [1 / (4 A^4 Rs^2 Rc^2)] [Rs Rc / (Rs + Rc)]^2
or
[Nt / Lh]^2 = [1 / (4 A^4 (Rs + Rc)^2)]
or
[Nt^2 / (Np^2 Lp^2 + Nt^2 Lt^2)] = [1 / (4 A^4 (Rs + Rc)^2)]
This is the result of the boundary condition at Rf, Zf
 

Rearrange this formula to get:
Nt^2 = [1 / 4 A^4 (Rs + Rc)^2](Np^2 Lp^2 + Nt^2 Lt^2)
or
Nt^2 {1 - [(Lt^2) / (4 A^4 (Rs + Rc)^2]} = [(Np^2 Lp^2) / (4 A^4 (Rs + Rc)^2)]

or
Nr^2 = Np^2 / Nt^2
= {1 - [(Lt^2) / (4 A^4 (Rs + Rc)^2)]} / {Lp^2) / (4 A^4 (Rs + Rc)^2)}
= {4 A^4 (Rs + Rc)^2 - Lt^2} / {Lp^2}
= {4 A^4 (Rs + Rc)^2 - Pi^2 (Rs - Rc)^2 Kc^2} / {Pi^2 (Rs + Rc)^2}
= {(4 A^4 / Pi^2) (Rs + Rc)^2 - (Rs - Rc)^2 Kc^2} / {(Rs + Rc)^2}
 

= {(4 A^4 / Pi^2) - [(Rs - Rc) / (Rs + Rc)]^2 Kc^2}
This is the boundary condition at R = Rf, Z = Zf. Note that it is identical to the boundary condition at R = Rc, Z = 0.

 

Thus: Nr^2 = {(4 A^4 / Pi^2) - [((Rs - Rc) Kc) / (Rs + Rc)]^2}
or
Nr^2 + [((Rs - Rc) Kc) / (Rs + Rc)]^2 = (4 A^4 / Pi^2)
is the boundary condition for an electromagnetic spheromak. Note that Kc is a function of A via ellipse theory.

This equation has the important solution: Nr^2 = Fx(4 A^4 / Pi^2)
and
[((Rs - Rc) Kc) / (Rs + Rc)]^2 = (1 - Fx)(4 A^4 / Pi^2)
where:
0 < Fx < 1
and Fx remains to be determined.

THIS IS THE BOUNDARY CONDITION EQUATION FOR AN ELECTROMAGNETIC SPHEROMAK

Define:
R = [((Rs - Rc) Kc) / (Rs + Rc)]

Then:
Nr^2 + R^2 = (4 A^4 / Pi^2)
 

CHECK THE FOLLOWING SPHEROMAK SHAPE PARAMETER when the magnetic vectors are in balance:
Nr^2 = R^2 = (So^2 - 1)^2 Kc^2 /(So^2 + 1)^2

Hence when the magnetic vectors are in balance:
2 (So^2 - 1)^2 Kc^2 /(So^2 + 1)^2 = [4 A^4 / Pi^2]
or
(So^2 - 1)^2 Kc^2 /(So^2 + 1)^2 = 2 [A^4 / Pi^2]
or
(So^2 - 1) Kc / (So^2 + 1) = 2^0.5 [A^2 / Pi]
or
(So^2 - 1) Kc = 2^0.5 [A^2 / Pi](So^2 + 1)
or
So^2 {Kc - 2^0.5 [A^2 / Pi]} = {Kc + 2^0.5 [A^2 / Pi]}
or
So^2 = {Kc + 2^0.5 [A^2 / Pi]} / {Kc - 2^0.5 [A^2 / Pi]}
or
So^2 = {1 + 2^0.5 [A^2 / Kc Pi]} / {1 - 2^0.5 [A^2 / Kc Pi]}

If (A^2 / Kc) = 1 then:
So^2 = {1 + 0.4501581586} / {1 - 0.4501581586}
= {1.4501581586} / {0.5498418414}
= 2.637409614
and
So = 1.624010349
 

HYPOTHESIS:
The ratio Nr^2 = [(Np / Nt)^2] is locked at an integer ratio which is equal to [2 A^4 / Pi^2]. Hence:
[(So^2 - 1) Kc / (So^2 + 1)]^2 = 2 A^4 / Pi^2
or
[(So^2 - 1) Kc / (So^2 + 1)] = 2^0.5 A^2 / Pi
or
[(So^2 - 1) / (So^2 + 1)] = 2^0.5 A^2 / (Kc Pi)

The relationship between A and Kc is complex and in practise requires a computer program for accurate evaluation.
 

THE NO COMMON FACTOR CONSTRAINT:
An important constraint on the existence of a spheromak is that Np and Nt have no common factors. A mathematical expression of this statement of no common factors is:
[Np / Nt] = [(N + 1) / ((prime) - 2(N + 1))]
 
for
0 <= N <= [(Prime - 1) / 2]

This mathematical function was devised by Heather Rhodes. The quantity (prime) can be any prime number.
 

Prove that the function:
(Np / Nt) = {[(N + 1)] / [prime - 2(N + 1)]}
generates (Np / Nt) values with no common factors.
If (N + 1) = (Fo M)
where Fo is a factor of (N + 1) then the denominator is
(prime - 2 Fo M) which can only be reduced if Fo is a factor of prime, which it is not due to the definition of a prime number.

However, the prime number can in principle adopt a range of values. The value that the prime number adopts will minimize the error in (Np / Nt) where:
Np = N + 1
and
Nt = [prime - 2(N + 1)]

[Np / Nt] = [(N + 1) / [(prime) - 2 (N + 1)]

Thus if A is unique for all spheromaks this formula will pick a particular (prime) that makes (N + 1) an integer. Hence we will have a unique solution.

If we assume that A is unique and is very close to unity (which may or may not be true) then experimental data suggests that:
Np = (N + 1) = 223
and
Nt = 495 and
(prime) = 941

We will test this assumption on the web page titled PLANCK CONSTANT.

For (prime) = 947, N = 222
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 223 / 501 = 0.44510
Error = 50.6 X 10^-4

For (prime) = 947, N = 223
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 224 / 499 = 0.44889
Error = -12.7 X 10^-4

For (prime) = 947, N = 224
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
225 / 497
= 0.4527162
Error = 25.5 X 10^-4

For (prime) = 941, N = 222:
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= (223 / 495
= 0.4505050505
Error = 3.4689 X 10^-4

For (prime) = 937, N = 222
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 223 / 491
= 0.45417
Error = 40.2 X 10^-4

For (prime) = 937, N = 221
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 222 / 493
= 0.45030
Error = 1.42 X 10^-4, seems to be best result

For (prime) = 937, N = 220
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 221 / 495
= 0.44646 Error = - 36.98 X 10^-4

For (prime) = 929, N = 220
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
221 / 487
= 0.45379
Error = 36.4 X 10^-4

For (prime) = 929, N = 219
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
220 / 489 = 0.449897
Error = -2.6 X 10^-4

For (prime) = 919, N = 218
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 219 / 481
= 0.4553
Error = 51.4 X 10^-4

For (prime) = 919, N = 217
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 218 / 483
= 0.45134
Error = 11.8 X 10^-4

For (prime) = 919, N = 216
(Np /Nt) = (N + 1) / [(prime) - 2(N + 1)]
= 217 / 485
= 0.447422
Error = -27.3 X 10^-4

Thus under the assumption that A = 1 the best values for Np and Nt appear to be:
Np = 222 and Nt = 493

Note that experimental measurements of (1 / Alpha) seem to indicate that:
Np = 223 and Nt = 495

Investigate the instability of Np = 221, Nt = 491

Np / Nt = 221 / 491 = 0.450101833 Implies Np = 220, ((Prime) - 442) = 491 or (prime) = 933 which is not prime.
933 = 3 X 311

Hence the values of Np and Nt that best satisfy:
(Np / Nt) = 2^0.5 A^2 / Pi
for A = 1 do not yield a stable solution.

STABLE SOLUTIONS;
In the following work errors are calculated assuming that A = 1.0000000.
Hence:
2^0.5 A^2 / Pi = 0.4501581586

Stable solutions satisfy:
(Np / Nt) = (N + 1) / ((prime) - 2(N +1)).

Consider (223 / 495) = 0.4505050505 ---- (Prime) = (495 + 446) = 941 which is prime and hence (223 / 495) is a stable solution. Error = 4 X 10^-4

Consider (222 / 493) = 0.4503042596 ---- (Prime) = (493 + 444) = 937 which is prime and hence (222 / 493) is a stable solution. Before correcting for A this is the best solution. Error = 2 X 10^-4.

Consider (221 / 491) = 0.450101833 ----- (Prime) = (491 + 444) = 935 which is not prime and hence does not provide a stable solution. It is OK at Nt = 491 but it is unstable under changes in Np / Nt. The ratio (220 / 493) = o.4462 lies on the same unstable line. The ratio (222 / 489) = 0.4540 lies on the same unstable line.

Consider (220 / 489) = 0.4498977505 ----- (Prime) = (489 + 440) = 929 which is prime and hence (220 / 489) provides a stable solution.

Investigate the limited region of stability of Np = 221, Nt = 491

Np / Nt = 221 / 491 = 0.450101833
Implies N = 220, ((Prime) - 442) = 491 or (prime) = 933 which is not a prime number.
933 = 3 X 311

Consider Np = 222, Nt = 489
222 = 2 X 3 X 37
489 = 3 X 163
Note the common factor of 3 causing a spheromak collapse.

Consider Np = 220, Nt = 493: 220 = 2 X 2 x 5 X 11
493 = 17 X 19

Consider Np = 219, Nt = 495
219 = 3 X 73
495 = 5 X 3 X 3 X 11
Note the common factor of 3 causing a spheromak collapse.

Hence the solution Np = 221, Nt = 491 is highly unstable and will not be found in nature.

As (prime) decreases the accuracy with which the ratio (Np / Nt) can represent an arbitrary real number decreases.

Consider (prime) = 919:
Np / Nt = (N + 1) / [919 - 2(N + 1)]
219 / 481 = 0.4553
218 / 483 = 0.4513 Error = 12 X 10^-4
217 / 485 = 0.4474

Consider (prime) = 911:
Np / Nt = (N + 1) / [911 - 2(N + 1)]
217 / 477 = 0.45492
216 / 479 = 0.45093 Error = 8 X 10^-4
215 / 481 = 0.44698

Consider (prime) = 907
Np / Nt = (N + 1) / [907 - 2(N + 1)]
216 / 475 = 0.4547
215 / 477 = 0.4507 Error = 6 X 10^-4
214 / 479 = 0.4467

Consider (prime) = 887
Np / Nt = (N + 1) / [887 - 2(N + 1)]
212 / 463 = 0.4578
211 / 465 = 0.4537
210 / 467 = 0.4496 Error = 5 X 10^-4 209 / 469 = 0.4456

Based on the above list a spheromak will tend to adopt a (prime) value of either 937 or 941. We need to use a computer to explore all the possible prime numbers to ensure that the aforementioned solutions for prime = 937 and prime = 941 are unique. Could there be another constraining factor that has not been taken into account?
 

OK TO HERE*****************************************************
CHECK FROM HERE ONWARDS-- SUSPECT ERRORS

Substitution of Nr^2 into the So^2 dependent term of Efs:
{[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
gives:
{[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
= {[{[8](So^2 + 1)^2 - [Pi^2] [(So^2 - 1)]^2} / {[Pi^2] - [4 / (So^2 - 1)]^2}] + [(So^2 - 1)^2]}^0.5
 
= [({[8](So^2 + 1)^2] - [Pi^2] [(So^2 - 1)]^2 + [(So^2 - 1)^2] [[Pi^2] - [16]}
/ {[Pi^2] - [4 / (So^2 - 1)]^2})^0.5]
 
= [({[8 (So^2 + 1)^2] - [16]} / {[Pi^2] - [16 / (So^2 - 1)^2]})^0.5]
 
= [({[8 (So^2 + 1)^2] - [16]} / {[Pi^2] - [4 / (So^2 - 1)]^2})^0.5] / So
 
= [({[8 (So^2 + 1)^2] - [16]} / {So^2 [Pi^2] - [16 So^2 / (So^2 - 1)^2]})^0.5]
 

Let:
X = So^2

Then the So^2 dependent term of Efs becomes:
[({[8(X + 1)^2] - [16]} / {X [Pi^2] - [16 X / (X - 1)^2]})^0.5]
 
=[({[8(X + 1)^2 (X - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]
 
= [({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]

Hence:
Efs = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]
= [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [(8 {[(So^4 - 1)^2] - [2 (So^2 - 1)^2]} / {So^2 (So^2 - 1)^2 [Pi^2] - [16 So^2]})^0.5]
 
= [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [So^2 - 1] (8 {So^4 + 2 So^2 - 1} / {So^2 (So^2 - 1)^2 [Pi^2] - [16 So^2]})^0.5]
 

DIFFERENTIATION OF Efs:
Differentiate Efs with respect to X = So^2 to get:

d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5
{{X (X - 1)^2 [Pi^2] - [16 X]} {16 (X^2 -1)(2 X) - 32 (X - 1)}
- {[8 (X^2 - 1)^2] - [16 (X - 1)^2]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}

Factor out 8 (X - 1) from both product terms to get:
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{{X (X - 1)^2 [Pi^2] - [16 X]} {2 (X + 1)(2 X) - 4}
- {[(X^2 - 1)(X + 1)] - [2 (X - 1)]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{4 X {(X - 1)^2 [Pi^2] - [16]} {X^2 + X - 1}
- (X - 1) {[(X + 1)(X + 1)] - [2]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{4 X {(X - 1)^2 [Pi^2] - [16]} {X^2 + X - 1}
- (X - 1) {X^2 + 2 X - 1} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qa^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
4 X {(X - 1)^2 [Pi^2]} {X^2 + X - 1} - (X - 1) {X^2 + 2 X - 1} {(X - 1)^2 [Pi^2]} - (X - 1) {X^2 + 2 X - 1} {2 X (X - 1) [Pi^2]}
+ 4 X {- [16]}{X^2 + X - 1} - (X - 1) {X^2 + 2 X - 1}{-16}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X)(X^2 + X - 1) - (X - 1) {X^2 + 2 X - 1} - (X^2 + 2 X - 1) (2 X)}
-64 X^3 - 64 X^2 + 64 X + 16 X^3 + 32 X^2 - 16 X - 16 X^2 - 32 X + 16
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X)(X^2 + X - 1) - (3 X - 1) (X^2 + 2 X - 1)}
- 48 X^3 - 48 X^2 + 16 X + 16
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X^3 + 4 X^2 - 4 X - 3 X^3 - 6 X^2 + 3 X + X^2 + 2 X - 1}
- 16 (+ 3 X^3 + 3 X^2 - X - 1)
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {X^3 - X^2 + X - 1} + 16 {(1 - 3 X^2) (X + 1)}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
{0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8}
{{(X - 1)^2 [Pi^2]} {(X^2 + 1)(X - 1} - 16 {(3 X^2 - 1) (X + 1)}}

CHECK THIS PLOT
A plot of the third line of this expression shows that its zero value is at approximately:
X = 3.765 = So^2

Note that the second line of this expression equals:
4 (X - 1) [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt / Efs]

In a real spheromak, although Efs is the largest single energy term the total energy is significantly affected by Eft and Efst. Hence:
d(Ett) / d(So^2) = 0 at an So^2 value that is somewhat removed from So^2 = 3.765. A photograph of a spheromak plasma suggests that So^2 ~ 4.2. Calculating the So^2 value at which Ett is a relative minimum is a major mathematical exercise. The importance of this exercise is that the steady state value of So^2 determines the steady state value of Nr^2 which in turn determines the Planck constant.
 

ELEMENTAL STRIP ANALYSIS:
Consider an elemental strip of constant radius R. The strip length is:
2 Pi R

The strip width is:
[(Rs - Rc) / 2] dPhi

The strip contains Nt partial windings.

Let L be the length of each partial winding. Then:
L^2 = (R dTheta)^2 + [(Rs - Rc) dPhi / 2]^2

The total charge hose length on the strip is:
Nt L = Nt {(R dTheta)^2 + [(Rs - Rc) dPhi / 2]^2}^0.5
= {(Nt R dTheta)^2 + [(Rs - Rc) Nt dPhi / 2]^2}^0.5
 

SURFACE CHARGE DENSITY:
As shown on the web page titled: THEORETICAL SPHEROMAK the surface charge dQs contained on the elemental strip of constant R is:
dQs = 2 Pi R dLt Sac (Rc / R)
or
Qs = 2 Pi Lt Sac Rc
or
Qs = 2 Pi Lt Sa R
or
Sa = Qs / 2 Pi Lt R

Hence:
dQs = 2 Pi R dLt Sa
= 2 Pi R dLt Qs / 2 Pi Lt R
= dLt Qs / Lt

The corresponding element of area dA is:
dA = 2 Pi R dLt

Hence the charge / unit area Sa on the spheromak wall is:
dQs / dA = (dLt Qs / Lt) / (2 Pi R dLt)
Sa = Qs / {Lt (2 Pi R)]

At R = Rc the charge per unit area Sac is given by:
Sac = Qs / {Lt (2 Pi Rc)]

In general the surface charge density Sa is given by:
Sa = Qs / {Lt (2 Pi R)]

= Sac Ro / R So

This charge distribution equation is required to find the electric field distribution.
 

POLOIDAL TURNS CONTAINED IN AN ELEMENTAL STRIP:
Recall that:
[dTheta / dPhi] = [Np [(Rs + Rc) / 2] / R Nt]

The number of poloidal turns contained in an elemental strip is:
Nt R dTheta / 2 Pi R = [Nt / 2 Pi] [Np (Rs + Rc) / 2 R Nt] dPhi
= [Np (Rs + Rc) / 4 Pi R] dPhi
where:
[Np (Rs + Rc) / 4 Pi R]
is the number of poloidal turns per radian in Phi
 

FIND Bpo:
Now let us attempt a precise calculation of Bpo.

The distance from the spheromak symmetry axis to the toroidal centerline is:
(Rs + Rc) / 2

The distance from the toroidal centerline to the spheromak wall is:
(Rs - Rc) / 2

The radial distance R from the spheromak symmetry axis to a point on the spheromak wall is:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)
where:
Phi = angle between the spheromak equatorial plane and a point on the spheromak wall, as measured a the toroidal center line.

Zs = height of a point on the spheromak wall above the spheromak equatorial plane, given by:
Hs = [(Rs - Rc) / 2] Sin(Phi)

FIX

D = distance from a point on the spheromak wall to the center of the spheromak given by:
D^2 = (R^2 + Hs^2)
= {[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)}^2 + {[(Rs - Rc) / 2] Sin(Phi)}^2
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 [Cos(Phi)]^2
- 2 [(Rs + Rc) / 2][(Rs - Rc) / 2] Cos(Phi) + {[(Rs - Rc) / 2] Sin(Phi)}^2
 
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 - [(Rs^2 - Rc^2) / 2] Cos(Phi)
 
= [(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)

Recall that:
dNp = [Np (Rs + Rc) / 4 Pi R] dPhi

dBpo = (Muo (Qs / Lh) Vp dNp / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5
 
= (Muo (Qs / Lh) Vp [Np (Rs + Rc) / 4 Pi R] dPhi / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5
 
= Muo (Qs / Lh) Vp [Np (Rs + Rc) / 8 Pi] dPhi [R / (R^2 + Hs^2)^1.5

Recall that:
Vp = [C Np Pi (Rs + Rc) / Lh]

Hence:
dBpo = Muo (Qs / Lh) Vp [Np (Rs + Rc) / 8 Pi] dPhi [R / (R^2 + Hs^2)^1.5
 
= Muo (Qs / Lh) [C Np Pi (Rs + Rc) / Lh] [Np (Rs + Rc) / 8 Pi] dPhi [R / (R^2 + Hs^2)^1.5
 
= [Muo Qs C Np^2 (Rs + Rc)^2 / (8 Lh^2)] dPhi [R / (R^2 + Hs^2)^1.5

Recall that:
Lh^2 = (Pi)^2 [Np^2 ((Rs + Rc))^2 + Nt^2 ((Rs - Rc))^2]

Thus:
dBpo = [Muo Qs C Np^2 (Rs + Rc)^2 / (8 Pi^2 [Np^2 ((Rs + Rc))^2 + Nt^2 ((Rs - Rc))^2])] dPhi [R / (R^2 + Hs^2)^1.5
 
= [Muo Qs C / (8 Pi^2)]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
[R / (R^2 + Hs^2)^1.5] dPhi

Recall that:
R = [(Rs + Rc) / 2] - {[(Rs - Rc) / 2] cos(Phi)}

Recall that:
(R^2 + Hs^2)^1.5 = {[(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)}^1.5

Thus:
[R / (R^2 + Hs^2)^1.5] dPhi
 
= {[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)} dPhi
/{[(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)}^1.5
 
= Rc {[(So^2 + 1) / 2] - [(So^2 - 1) / 2] Cos(Phi)} dPhi
/ Rc^3 {[(So^4 + 1) / 2] - [(So^4 - 1) / 2] Cos(Phi)}^1.5
 
= [So^2 / Ro^2] {[(So^2 + 1) / 2] - [(So^2 - 1) / 2] Cos(Phi)} dPhi
/ {[(So^4 + 1) / 2] - [(So^4 - 1) / 2] Cos(Phi)}^1.5
 
= [2^0.5] [So^2 / Ro^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Bpo = 2 X Integral from Phi = 0 to Phi = Pi of:
[Muo Qs C / (8 Pi^2)]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
[2^0.5] [So^2 / Ro^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
 
= [2^0.5 Muo Qs C / (4 Pi^2 Ro^2)]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Compare this integral equal to Bpo as indicated by the electric field distribution.
 

FIND Bpo AS INDICATED BY ELECTRIC FIELD DISTRIBUTION:
Let Qs be the net charge of a spheromak as indicated by an electric field measurement at:
(R^2 + H^2) >> Ro^2.
Note that most field measurements on electrons and protons are far field measurements.

In the far field the energy field density of an electromagnetic system is almost purely electric. The electric field energy density in the far field is:
~ [Epsilono / 2][Qs / (4 Pi Epsilono (R^2 + Z^2)]^2

The theoretical spheromak electric energy density in the far field is:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2

In the far field:
(R^2 + Z^2) >> Ro^2:

Hence for an electromagnetic spheromak in the far field:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
= [Epsilono / 2][Qs / (4 Pi Epsilono (Ro^2 + (A R)^2 + Z^2)]^2
or
Uo Ro^4 = [Epsilono / 2][Qs / (4 Pi Epsilono)]^2
or
Uo Ro^4 = [1 / 2 Epsilono][Qs / (4 Pi)]^2

For an electromagnetic spheromak:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2

At R =0, Z = 0 the electric field energy density Ue = 0 giving:
U = Upo = Bpo^2 / 2 Muo

and hence:
Upo Ro^4 = (Bpo^2 / 2 Muo) Ro^4

The spheromak center energy density projected from the electric field is:
Uo Ro^4 = [1 / 2 Epsilono][Qs / (4 Pi)]^2

Equating these two expressions for Uo Ro^4 gives:
Bpo^2 = = 2 Muo [1 / 2 Epsilono][Qs / (4 Pi)]^2 [1 / Ro^4]
= Muo^2 C^2 Qs^2 / 16 Pi^2 Ro^4
or
Bpo = Muo C Qs / 4 Pi Ro^2
which is the center magnetic field strength of an ideal spheromak.

The corresponding value of Uo is:
Upo = Bpo^2 / 2 Muo
= Muo^2 C^2 Qs^2 / (16 Pi^2 Ro^4 2 Muo)
= Muo C^2 Qs^2 / (32 Pi^2 Ro^4)
 

EQUATE THE TWO EXPRESSIONS FOR Bpo TO SHOW CONSISTENCY:
For an ideal spheromak with an outside energy density defined by:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
the magnetic field at R = 0, Z = 0 is given by:
Bpo = [Muo C Qs / (4 Pi Ro^2)]

Equating this value to our integral gives:
Bpo = [Muo C Qs / (4 Pi Ro^2)]
= [2^0.5 Muo Qs C / (4 Pi^2 Ro^2)]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Cancel out common terms to get:
[1]
= [2^0.5 / Pi]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
or
[Pi / 2^0.5] = 2.22144
= {[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Numerical Evaluation of Integral
So    Integral
1.3xxxx
1.52.1164
1.72.055
1.91.9939
2.01.9638
2.51.8249

Note that the spheromak tends to operate with a central magnetic field strength slightly less than the electric field seems to indicate.

Note that So < 2 and Nt < Np.

SPHEROMAK STATIC FIELD ENERGY:
The web page titled SPHEROMAK ENERGY shows that the energy density functions:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
outside a spheromak wall and
U = Uoc [(Rc / R)^2]
inside a spheromak wall result in spheromaks with total energy given by:
Ett = Uo Pi^2 Ro^3/ A^2
X {1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
 
= Uo Pi^2 Ro^3 {4 So (So^2 - So + 1) / (So^2 + 1)^2}
Note that this formula applies to all spheromaks.
 

The energy density distribution:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
provides an energy density of:
U = Uo [Ro^2 / (Ro^2 + (A Rc)^2)]^2
at R = Rc, Z = 0.

SPHEROMAK ENERGY CONTENT:
Substitution for Uo gives:
Efs = Uo Ro^3 Pi^2 / A^2
FIX A^2 FROM HERE ONWARDS = (Muo / 2) Ro^3 Pi^2 [(Qs C) /(4 Pi Ro^2)]^2
= (Muo / 32) (Qs C)^2 /(Ro)
and
Ett = [Muo (Qs C)^2 / (32 Ro)] {1 -[(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
or
Ett = [(Muo C^2 Qs^2) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}

This equation gives the static electromagnetic field energy content of a spheromak in terms of its nominal radius Ro and its shape factor So where:
So = Rs / Ro = Ro / Rc

FIX A ENTRIES

Thus an electromagnetic spheromak has a net charge Qs, a nominal radius Ro and a theoretical peak central poloidal magnetic field strength given by:
Bpo = [(Muo C Qs) / (4 Pi Ro^2)]

This value of Bpo should equal the value of Bpo obtained by applying the law of Biot and Savart to the circulating charge in the spheromak.
 

MAXIMUM ENERGY VALUE OF So:
An important issue is finding the value of So which maximizes a spheromak's energy content at any particular value of Ro. At that energy maximum:
dEtt / dSo = 0

Recall that:
Ett = [(Muo C^2 Qs^2) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}

The So dependent portion of the function is:
S(So) = {So (So^2 - So + 1) / (So^2 + 1)^2}

dS(So) / dSo = {(So^2 + 1)^2 [(So^2 - So + 1) + So (2 So - 1)]
- [So (So^2 - So + 1)] 2 (So^2 + 1) 2 So}
/ (So^2 + 1)^4
= 0

Hence:
{(So^2 + 1)^2 [(So^2 - So + 1) + So (2 So - 1)]
- [So (So^2 - So + 1)] 2 (So^2 + 1) 2 So}
= 0

Cancelling (So^2 + 1) terms gives:
{(So^2 + 1) [(So^2 - So + 1) + So (2 So - 1)]
- [4 So^2 (So^2 - So + 1)]}

Hence:
(So^2 + 1 - 4 So^2)(So^2 - So + 1) + (So^2 + 1) So (2 So - 1) = 0
or
(- 3 So^2 + 1)(So^2 - So + 1) + (So^2 + 1)(2 So^2 - So) = 0
or
- 3 So^4 + 3 So^3 - 3 So^2 + So^2 - So + 1 + 2 So^4 - So^3 + 2 So^2 - So = 0
or
- So^4 + 2 So^3 - 2 So + 1 = 0
or
So^4 - 2 So^2 + 2 So - 1 = 0
or
So^4 - So^2 = So^2 - 2 So + 1
or
So^2 (So^2 - 1) = (So - 1)^2
or
So^2 = (So - 1)^2 / [(So - 1) (So + 1)]
= (So - 1) / (So + 1)

This equation has a solution of So = 1 which says that at a particular Ro the spheromak's energy is maximum at So = 1, at which point Rc = Ro = Rs corresponding to no volume inside the spheromak wall.

Thus at the maximum energy state: {So (So^2 - So + 1) / (So^2 + 1)^2}
= {1 (1 - 1 + 1) / (1 + 1)^2}
= 1 / 4

At So = 2:
(spheromak energy)
= {So (So^2 - So + 1) / (So^2 + 1)^2}
= {2 (4 - 2 + 1) / (4 + 1)^2}
= {6 / 25}
= which is only slightly less than the spheromak maximum energy.

Thus for modest So values of the order of So < 2 the static field energy content of a spheromak is only weakly dependent on the spheromak's So value. However, at larger So values the spheromak static field energy is proportional to (1 / So).
 

CONFINING THE RANGE OF So:
In the toroidal region: Ut = Utc (Rc / R)^2 at R = Rc the total field is (Eric + Btc).
In general inside the spheromak wall the field is:
(Btc)(Rc / R)
and the field energy density is:
+ (Btc^2 / 2 Muo)(Rc / R)^2
 
= + [(Muo / 2) (Nt Q C / 2 Pi R Lh) ^2)]

Recall that for an ideal spheromak the energy density at R = Ro is (1 / 4) the energy density at R = 0. Hence at R = Ro:
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= [(Epsilono / 2) (Sac / Epsilono)^2 (Rc / Ro)^2]
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh) ^2)]

or
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= [(1 / 2 Epsilono) (Sac / So)^2]
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh) ^2)]

FIX FROM HERE ONWARD
From the spheromak charge distribution:
Sac = Qs So^2 / 2 Ro^2 Pi^2 (So^2 -1)

Hence:
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= (1 / 2 Epsilono) [Qs So / 2 Ro^2 Pi^2 (So^2 -1)]^2
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh)^2)]
or (1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= (Muo C^2 / 2) [Qs So / 2 Ro^2 Pi^2 (So^2 -1)]^2
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh)^2)]
or cancelling terms:
(1 / 4)[1 / (4 Ro)]^2
= [So / 2 Ro Pi (So^2 -1)]^2
+ [(Nt / 2 Lh)^2)]
or
[Lh / Ro]^2 [(1 / 8)^2]
= [Lh / Ro]^2 [So / 2 Pi (So^2 - 1)]^2 + [Nt / 2]^2
or
[Lh / Ro]^2 = [Nt / 2]^2 / {[(1 / 8)^2] - [So / 2 Pi (So^2 - 1)]^2}
 
= Nt^2 / [(1 / 4)^2 - {So / Pi (So^2 - 1)}^2]
 
= Nt^2 Pi^2 (So^2 - 1)^2 / {[Pi^2 (So^2 - 1)^2 / 4^2] - So^2}
 
= Nt^2 Pi^2 (So^2 - 1)^2 / So^2 {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}
 

Comparison with the previous result greatly limits the range of So.

Recall that:
(Lh / Ro)^2 = (Pi^2 / So^2) [Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]
= (Pi^2 / So^2) [Nt^2 (So^2 - 1)^2][{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]
 
= [Nt^2 Pi^2 (So^2 - 1)^2 / So^2][{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]

Comparison of the two expressions for (Lh / Ro)^2 implies that:
[{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]
= 1 / {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}
or
[{Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]
= Nt^2 (So^2 -1)^2 / {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}

Hence:
1 < [Pi^2 (So^2 - 1)^2 / (4 So)^2] < 2
or
1 < [Pi (So^2 - 1) / (4 So)] < 2^0.5

This expression restricts the range of So. We can easily calculate So values at the extreme ends of the possible range of So.

At:
[Pi (So^2 - 1) / (4 So)] = 2^0.5
or
Pi So^2 - 2^0.5(4 So) - Pi = 0
or
So = {2^0.5(4) +/- [32 + 4 Pi^2]^0.5} / 2 Pi
= {5.6568 + [71.4784]^0.5} / 6.2831853
= 2.24588

At:
1 = [Pi (So^2 - 1) / (4 So)]
or
Pi (So^2 - 1) = 4 So
or
Pi So^2 - 4 So - Pi = 0
or
So = {4 +/- [16 + 4 Pi^2]^0.5} / 2 Pi = {4 + [55.47841]^0.5} / 6.2831853
= 1.8220

Thus the possible range of So is restricted to:
1.8220 < So < 2.24588

WRONG

Hence the Fine Structure constant definitely arises from the spheromak relationship and the relative strength of electric and magnetic fields. We need a more exact solution to properly evaluate it.

At R = Rs, Z = 0:
Toroidal magnetic field must approximately match the increase in the electric field. The increase in electric field energy density is given by:
(Epsilono / 2)(Sac Rc / Epsilono Rs)^2

The drop in toroidal magnetic field energy density is:
(1 / 2 Muo)(Muo Nt Q Fh / 2 Pi Rs)^2
= (1 / 2 Muo)(Muo Nt Q C / 2 Pi Rs Lh)^2

Thus, equating these two expressions gives:
(1 / 2 Epsilono)(Sac / So^2)^2 = (Muo / 2)(Nt Q C / 2 Pi Ro So Lh)^2
or
(Sac / So)^2 = (Epsolono Muo) (Nt Q C / 2 Pi Ro Lh)^2
= (Nt Q / 2 Pi Ro Lh)^2
or
(Sac / So) = (Nt Q / 2 Pi Ro Lh)
or
Sac = (Nt Q So / 2 Pi Ro Lh)

Recall that the spheromak surface charge distribution is given by:
Sac = [Qs So^2] / [2 Pi^2 Ro^2 ( So^2 - 1)]WRONG

Equating the two expressions for Sac gives:
(Nt Q So / 2 Pi Ro Lh) = [Qs So^2] / [2 Pi^2 Ro^2 ( So^2 - 1)]
or
(Nt / Lh) = So / [Pi Ro (So^2 - 1)]
or
(Lh / Ro) = Pi (So^2 - 1) Nt / So

This is a tremendously revealing result. It is part of:
(Lh / Ro) = (Pi / So)[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]^0.5

Note that the poloidal and toroidal magnetic fields are orthogonal.

Hence at R = Rs the poloidal magnetic field energy density is given by:
(Epsilono / 2)[Sac Rc / Epsilono Rs]^2 = Bps^2 / 2 Muo
FIX where:
Bps = Muo Np Qs Fh / 2 Pi Rx
= (Bpc Rc / Rs)^2 / 2 Muo
due to the change in perimeter lengths.

The change in radial electric field at R= Rc, H = 0 causes an energy density balance at R = Rc of:
[(Bpc)^2 / 2 Muo] - [(Btc^2 / 2 Muo] = (Epsilono / 2)[Sac / Epsilono]^2
or
[(Bpc)^2 / 2 Muo] = [(Btc^2 / 2 Muo] + (Epsilono / 2)[Sac / Epsilono]^2

If the spheromak was a straight core the poloidal magnetic field at the spheromak wall would be:
Muo Np Q Fh / [2 Pi (Rs - Rc) / 2]

At the outer perimeter this field is reduced by a factor of:
Ro / Rs = 1 / So

At the inner perimeter this field is increased by a factor of:
Ro / Rc = So.

The thesis is that the total energy contained in a spheromak is:
Outside Electric field energy + Outside Poloidal Magnetic field energy + Toroidal magnetic field energy.

From the web page titled: SPHEROMAK ENERGY:
Toroidal magnetic field energy + internal electric field energy
= Poloidal magnetic field energy + external electric field energy = ????????????????

THE INDEX N:
On this web page the total field energy density U is expressed as:
U = Ue + Um
where Ue is the electric field energy density component and Um is the magnetic field energy density component.

At large R and/or large Z, Ue is approximately:
Ue ~ U = Uo [Ro^4 / (Ro^2 + (A R)^2 + Z^2)^2}

In general due to spheromak symmetry at R = 0, Z = 0:
Ueo = 0

At R = Rc, Z = 0:
U = Uo [Ro^2 / (Ro^2 + A^2 Rc^2)]^2

Note that Uoc reflects the reality that R = Rc and Z = 0 there is a radial electric field Eroc pointing toward the spheromak axis of symmetry.

Note that at R = 0, Z = 0:
Umo = Uo

Note that in general outside the spheromak wall:
U = Uo [Ro^2 / (Ro^2 + A^2 R^2 + Z^2)]^2 which is the equation for the total field energy density outside a spheromak wall.

The expressions for Ue and Um in the far field match the well known far field dependences of electric and magnetic field energy densities on distance.
 

STRATEGY:
Find the electric and magnetic field energy densities along the spheromak main axis of symmetry. Compare this function to:
U = Uo [Ro^2 / (Ro^2 + Z^2)]^2 and try to determine A via its effect on the electric and magnetic field energy densities along the Z axis.
 

BOUNDARY CONDITION:
It appears that (1 / Alpha) is primarily set by the spheromak boundary condition which sets:
Nr^2 + R^2 = 1 / [(Pi / 2 A^2)^2]

ELECTRIC FIELD ALONG THE Z AXIS:
Consider an elemental ring of charge on the spheromak wall. The circumference of this ring is:
2 Pi R

If the spheromak cross section is circular the width of this ring is:
[(Rs - Rc) / 2] d(Phi)
where Phi is an angle measured at the toroidal axis from the spheromak equatorial plane to the ring.
At R = Rc, Phi = 0.
At R = Rs, Phi = Pi.

Note that:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] cos (Phi)
and
R^2 = [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 [cos (Phi)]^2
- [(Rs^2 - Rc^2) / 2] cos(Phi)

These expressions for R and R^2 are valid for both the upper and lower halves of the spheromak.

UPPER HALF OF SPHEROMAK:
In the upper half of the spheromak where d(Phi) is positive the area dA of this elemental ring is:
dA = 2 Pi R (Rs - Rc) d(Phi) / 2
= Pi R (Rs - Rc) d(Phi)

At R = Rc the charge per unit area on the elemental ring is Sac. At other radii the charge per unit area on the elemental ring is: Sac (Rc / R)

Thus for the upper half of the spheromak the charge dQs on the elemental ring is:
dQs = Sac (Rc / R) dA
= Sac (Rc / R) 2 Pi R (Rs - Rc) d(Phi) / 2
= Sac Pi Rc (Rs - Rc) d(Phi)

For the upper half of the spheromak the straight line distance from the elemental ring to the point R= 0, H = Rc on the spheromak symmetry axis is given by Pythagoras theorem as:
{{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^0.5

The vertical distance between Z = Rc and the elemental ring is:
{Rc - [(Rs - Rc) / 2] sin(Phi)}
Note that for the upper half of the spheromak this term is positive if the elemental ring is below H = Rc and is negative if the elemental ring is above H = Rc.

For the upper half of the spheromak the element of axial electric field at R = 0, Z = Rc, due to the elemental charge ring is:
dEo = [dQs / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [Sac Pi Rc (Rs - Rc) d(Phi) / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi) / (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 [cos (Phi)]^2 - [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 - [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5
 

SPHEROMAK LOWER HALF:
For the lower half of the spheromak where d(Phi) is negative the area dA of and elemental ring is:
dA = - 2 Pi R (Rs - Rc) d(Phi) / 2
= - Pi R (Rs - Rc) d(Phi)

Thus for the lower half of the spheromak the charge dQs on the elemental ring is:
dQs = Sac (Rc / R) dA
= - Sac (Rc / R) 2 Pi R (Rs - Rc) d(Phi) / 2
= - Sac Pi Rc (Rs - Rc) d(Phi)

For the lower half of the spheromak the straight line distance from the elemental ring to the point R= 0, H = Rc on the spheromak symmetry axis is given by Pythagoras theorem as:
{{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^0.5
Note that for the lower half of the spheromak sin(Phi) is negative.

For the lower half of the spheromak the vertical distance between H = Rc and the elemental ring is:
{Rc - [(Rs - Rc) / 2] sin(Phi)}
Note that for the lower half of the spheromak where sin(Phi) is always negative this term is always positive.

For the lower half of the spheromak where d(Phi) and sin(Phi) are negative the contribution of an elemental ring to the axial electric field at H = Rc is:
dEo = [dQs / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [- Sac Pi Rc (Rs - Rc) d(Phi)/ (4 Pi Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{[(Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 [cos (Phi)]^2
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5

 

TOTAL ELECTRIC FIELD:
Thus the total electric field Eo at R = 0, Z = Rc is:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5

FIX A ENTRIES

Substitute:
Rs = So Ro
and
Rc = Ro / So
to get:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac (Ro / So)^2 (So^2 - 1) d(Phi)/ (4 Epsilono)]
(Ro / So){1 - [(So^2 - 1) / 2] sin(Phi)}
/ (Ro / So)^3 {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (Ro / So)^2 (So^2 - 1) d(Phi)/ (4 Epsilono)]
(Ro / So){1 - [(So^2 - 1) / 2] sin(Phi)}
/(Ro / So)^3 {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
= Integral from Phi = 0 to Phi = Pi of:
[Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 

FIX CHARGE DISTRIBUTION

FIX THE CHARGE DISTRIBUTION EQUATION Substitute the charge distribution expression into the integration for Eo:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
Eo = Integral from Phi = 0 to Phi = Pi of:
[{Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]} (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- {Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]} (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
= Integral from Phi = 0 to Phi = Pi of:
[{Qs So^2 / [2 Ro^2 Pi^2]} d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- {Qs So^2 / [2 Ro^2 Pi^2]} d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5

Factor out:
[Qs / (4 Pi Epsilono Ro^2)] to get:
Eoc = Integral from Phi = 0 to Phi = Pi of:
[Qs / (4 Pi Epsilono Ro^2)][{ So^2 / [2 Pi]} d(Phi)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[Qs / (4 Pi Epsilono Ro^2)][- { So^2 / [2 Pi]} d(Phi)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5

In order to perform these integrations numerically we must first quantify So^2. Thus the order of calculation is to:
a) choose the Np / Nt pair to be investigated;
b) Calculate So;
c) Calculate (1 / Alpha) and the corresponding F value;
d) If (1 / Alpha) is approximately correct perform the this integration;
e) Use the result of this integration to recalculate F;
f) Compare the two calculated F values. If they are almost the same output all the relevant data for this Np/Nt pair;
g) Investigate the stability of adjacent Np/Nt pairs.

Let us assume that the result of this integration is:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] Ic,
where Ic is a unitless integration result.

Recall that:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)

Equate the two expressions for Eoc to get:
[Qs / (4 Pi Epsilono Ro^2)] Ic
= [Qs / (4 Pi Epsilono Ro^2)] [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)
or
Ic = [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)
 

SPHEROMAK ENERGY RATIO:
An issue in electromagnetic spheromaks is the ratio of electric field energy to total energy. On the web page titled SPHEROMAK ENERGY it was shown that the total contained electomagnetic energy of a spheromak before adjustment for the toroidal portion is:
Efs = Uo Ro^3 Pi^2
and the spheromak energy Ett after adjustment for the toroidal portion is:
Ett / Efs
= {1 -[(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}

For R > Rc the electric field energy density of a spheromak outside the spheromak wall is given by:
Ueo = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 [(R^2 + H^2) / (Ro^2 + R^2 + H^2)]^N

Let Z^2 = R^2 + H^2

FIX N

Then:
Ueo = Uo [Ro^2 / (Ro^2 + Z^2)]^2 [(Z^2) / (Ro^2 + Z^2)]^N

The corresponding total electric field energy is given by:
Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ Ro^4 (Z^2)^(2 N) / (Ro^2 + Z^2)^(N + 2)
 

From Dwight XXXX the solution to this integral is:
DO INTEGRATION

The corresponding total magnetic field energy is given by:
Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ Ro^4 [(Ro^2 + Z^2)^N - (Z^2)^N] / (Ro^2 + Z^2)^(N + 2)
 

From Dwight XXXX the solution to this integral is:
DO INTEGRATION

Thus the electric field energy contained in Efs is about ____% of the total field energy of Efs and the magnetic field energy is about ______% of the total field energy of Efs.
 

SPHEROMAK WALL THEORY REVIEW:
On the web page titled CHARGE HOSE PROPERTIES it was shown that for a charge hose (charge motion path) current Ih is given by:
Ih = (1 / Lh) [Qp Nph Vp + Qn Nnh Vn]
= Qs C / Lh
and
Rhoh = (1 / Lh) (Qp Nph + Qn Nnh)
= Qs / Lh
giving:
(Ih / C)^2 = Rhoh^2
= (Qs / Lh)^2
= (1 / Lh)^2 [Qp Nph + Qn Nnh]^2

and that for practical ionized gas plasmas where Ve^2 >> Vi^2 and Ne ~ Ni:
Ih^2 ~ [Q Ne Ve / Lh]^2
and
[(Ni - Ne) / Ne]^2 ~ (Ve / C)^2
and
Qs^2 ~ [Q Ne Ve / C]^2

These equations allow the development of electromagnetic spheromak theory for atomic particles and plasma.
 


BIOT AND SAVART:
The law of Biot and Savart gives an element of axial magnetic field:
d(Bpo) at R = 0, H = 0 due to a current ring with radius R located at height H and having current dI as:
d(Bpo) = [(Muo / 4 Pi) dI 2 Pi R / (R^2 + H^2)] [R / (R^2 + H^2)^0.5]
= (Muo dI R^2 / (2 (R^2 + H^2)^1.5

 

ANGULAR PROGRESSION:
Assume that the cross section of a spheromak in free space is round. This assumption is justified at THEORETICAL SPHEROMAK.
Let Theta = angle about the spheromak major axis
Let Phi = angle about spheromak minor axis.

As the charge moves over length Lh Theta increments by (2 Pi Np)

As the charge moves over length Lh Phi increments by (2 Pi Nt)

On average over the spheromak surface:
d(Theta) /d(Phi)
= (2 Pi Np) / 2 Pi Nt)
= (Np / Nt)
Note that this is an average value. Note that:
[d(Theta) /d(Phi)]|R = Rc < (Np / Nt) < [d(Theta) /d(Phi)]|R = Rs
 

POLOIDAL MAGNETIC FIELD IN THE CORE OF A SPHEROMAK:
Earlier on this web page it is shown that in order to provide the poloidal magnetic field density required for a spheromak to exist the poloidal magnetic field at R = Rc, H = 0 must satisfy:
Bpc = Btc
= Muo Nt Q Fh / 2 Pi Rc
= Muo Nt Q Fh So / 2 Pi Ro
in order to balance the toroidal magnetic field at R = Rc, H = 0.

Recall that:
Fh = C / Lh

Thus:
Bpc = Muo Nt Q Fh So / 2 Pi Ro
= Muo Nt Q C So / 2 Pi Ro Lh

To achieve this magnetic field in the core of the spheromak there must be the required number of poloidal turns in the spheromak. However, in general:
Bpo > Bpc.

Hence in general Bpo must satisfy:
Bpo > Muo Nt Q C So / 2 Pi Ro Lh

Now let us attempt a precise calculation of Bpo.

The distance from the spheromak symmetry axis to the toroidal centerline is:
(Rs + Rc) / 2

The distance from the toroidal centerline to the spheromak wall is:
(Rs - Rc) / 2

The radial distance R from the spheromak symmetry axis to a point on the spheromak wall is:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)
where:
Phi = angle between the spheromak equatorial plane and a point on the spheromak wall, as measured a the toroidal center line.

Hs = height of a point on the spheromak wall above the spheromak equatorial plane, given by:
Hs = [(Rs - Rc) / 2] Sin(Phi)

D = distance from a point on the spheromak wall to the center of the spheromak given by:
D^2 = (R^2 + Hs^2)
= {[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)}^2 + {[(Rs - Rc) / 2] Sin(Phi)}^2
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 [Cos(Phi)]^2
- 2 [(Rs + Rc) / 2][(Rs - Rc) / 2] Cos(Phi) + {[(Rs - Rc) / 2] Sin(Phi)}^2
 
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 - [(Rs^2 - Rc^2) / 2] Cos(Phi)
 
= [(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)

Recall that:
d(Phi) / d(Theta) = (Nt / Np)

Consider two adjacent spheromak windings. Let dTheta and dPhi be the winding spacings in the poloidal and toroidal directions. At any point on a winding the slope of the winding is:
{[(Rs - Rc) / 2] d(Phi)} / {R d(Theta)}

In an elemental slope box with side lengths:
{[(Rs - Rc) / 2] d(Phi)}
and
{R d(Theta)}
contains the fraction [d(Theta) / 2 Pi] of one poloidal winding.

The poloidal windings are equally spaced. Hence the number of poloidal windings per radian in Phi is:
Np / 2 Pi

Hence:
dNp = Np d(Phi) / (2 Pi)

dBpo = (Muo Qs Fh dNp / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5
= (Muo Qs Fh [Np d(Phi) / (2 Pi)] / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5]
 
= [Muo Qs Fh Np d(Phi) / 4 Pi] [R^2 / (R^2 + Hs^2)^1.5]
 
= {(Muo Qs Fh Np d(Phi) / 4 Pi}
{[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)}^2
/ {[(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)}^1.5
 
= {(Muo Qs Fh Np d(Phi) / 4 Pi}
Ro^2 (1 / 2)^2 {[(So + (1 / So))] - [(So - (1 / So))] Cos(Phi)}^2
/ Ro^3 (1 / 2)^1.5 {So^2 + (1 / So)^2 - [(So^2 - (1 / So)^2)] Cos(Phi)}^1.5
 
= (Muo Qs Fh Np d(Phi) / 4 Pi
(1 / 2)^0.5 {[(So + (1 / So))] - [(So - (1 / So))] Cos(Phi)}^2
/ Ro {So^2 + (1 / So)^2 - [(So^2 - (1 / So)^2)] Cos(Phi)}^1.5
 
= (Muo Qs Fh Np So / 4 Pi Ro)(1 / 2)^0.5 d(Phi)
{[(So^2 + 1)] - [(So^2 - 1)] Cos(Phi)}^2
/ {(So^4 + 1) - [(So^4 - 1)] Cos(Phi)}^1.5

Recall that:
Fh = C / Lh

Bpo = 2 X Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 4 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
 
= Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 2 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

 

Set this integral equal to:
Bpc = Muo Nt Qs C So / 2 Pi Ro Lh
to achieve the minimum spheromak core magnetic field strength required to match the toroidal magnetic field strength.

Muo Nt Qs C So / 2 Pi Ro Lh
= Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 2 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
or
Nt / Np = = Integral from Phi = 0 to Phi = Pi of:
(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

This integral can be numerically evaluated at So = 2.0 to get:
(Nt / Np) = 1.12268

This is the maximummum value of (Nt / Np) assuming that the magnetic field in the spheromak core is uniform. If the magnetic field in the spheromak core decays with increasing radius this number will be decreased by the factor:
Ro^2 / (Ro^2 + Rc^2) = So^2 / (So^2 + 1)
which at So = 2 is (4 / 5) giving:
Nt / Np < (4 / 5)(1.12268)
< 0.8981

Thus provided that the electric field is reasonably balanced about the inner spheromak wall and:
(Nt / Np) < 0.8981
the poloidal magnetic field in the spheromak is sufficient to ensure spheromak stability.

However, at R = 0 and H = 0 the peak core magnetic field may not meet the criteria for an ideal spheromak.

For an ideal spheromak with an outside energy density defined by:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2
the magnetic field at R = 0, H = 0 is given by:
Bpo = [Muo C Qs / (4 Pi Ro^2)]

However, at So = 2 it appears that the maximum central magnetic field that we can achieve is:
Bpo attempt = (1.12268) Muo Qs C Np So / 2 Pi Lh Ro)
= (Muo C Qs / 4 Pi Ro^2) (1.12268)(2 Np So Ro / Lh)

Recall that:
(Ro / Lh) = So /{Pi[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]^0.5}
~ So /{Pi[2 Np^2 (So^2 + 1)^2]^0.5}
= ~ So /{Pi (2^0.5) Np (So^2 + 1)}

Hence:
Bpo attempt
= (Muo C Qs / 4 Pi Ro^2) (1.12268)(2 Np So Ro / Lh)
= (Muo C Qs / 4 Pi Ro^2) (1.12268)(2 Np So So /{Pi (2^0.5) Np (So^2 + 1)})
= (Muo C Qs / 4 Pi Ro^2) (1.12268)(2^0.5 So^2 /{Pi (So^2 + 1)})
= (Muo C Qs / 4 Pi Ro^2) (0.4031)

Hence at R = 0, H = 0 the magnetic field for an electromagnetic spheromak is not as strong as in an ideal spheromak.

In terms of spheromak stability this is not a big issue as long as the electric field is balanced at R = Rc, H = 0. However, this non-ideality will introduce a small error into the total spheromak energy calculation.

At So = 1.0:
Bpo = Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 2 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
 
= Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np / 2 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[1 + 1]}^2
/ {[1 + 1]}^1.5
 
= (Muo Qs C Np / 2 Lh Ro)

Set this integral equal to:
Bpo = Muo C Qs / (4 Pi Ro^2)
to achieve the spheromak central magnetic field energy density required to match the electric field energy density.

(Muo Qs C Np / 2 Lh Ro) = Muo C Qs / (4 Pi Ro^2)
or
( Np / 2 Lh) = 1 / (4 Pi Ro)
or
(Lh / Ro) = 2 Pi Np

This is the same result as realized with a simple current ring model.
 

FIELD ENERGY DENSITY BALANCE:
In general, neglecting gravitation and kinetic energy the total field energy density U at any point in a spheromak is given by:
U = [Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]
where:
Bp = poloidal magnetic field strength;
Bt = toroidal magnetic field strength;
Er = radial electric field strength, where Er is

Thus at a spheromak wall:
{[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

 

However, inside the spheromak wall:
Bp = 0
and outside the spheromak wall:
Bt = 0
Hence at a spheromak wall:
{[[Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

This energy density balance (force balance) condition is valid at every point on the spheromak wall.
 

 
 

Define:
R = [((Rs - Rc) Kc) / (Rs + Rc)]

Then:
Nr^2 + R^2 = (4 A^4 / Pi^2)
or
Nr^2 + R^2 = (2 A^2 / Pi)^2

Note that R^2 determines So^2.

Note that since Nr is a ratio of two integers and since Pi is a real number the parameter R is quantized. Hence for spheromak stability, at and near the spheromak operating point there must be a continuous series of stable quantum states.

In general Nr^2 > 0, which causes the upper limit on R to be:
R < [2 / Pi]

To meet spheromak stability criteria Nr < 1.
 

Recall that:
R = (So^2 - 1) / (So^2 + 1)

At So^2 = 4.1:
R = (3.1 / 5.1) = 0.607843
and
R^2 = 0.369473

The form of this function keeps R in the range:
0 < R < 1

Rearranging this equation gives:
R (So^2 + 1) = So^2 - 1
or
So^2 (1 - R) = (1 + R)
or
So^2 = (1 + R) / (1 - R)
and
So = [(1 + R) / (1 - R)]^0.5

Note that since R^2 can only take discrete values So can only take discrete values. The discrete So value actually adopted will be one that is close to the analytic energy minimum So value.

Quantify (Nr^2 + R^2) Use the formula: Nr^2 + R^2 = 1 / {[Pi^2 / 4] }
 

Thus we can readily compute the corresponding [Z^2 / Nt^2] value. We can use this methodology to find [Z / Nt] as a function of So and hence (1 / (Alpha Nt) as a function of So.

We can find the So value corresponding to the stable So value in a plot of (1 / Alpha Nt)^2 versus So. The Alpha value at this point is the Fine Structure constant. However, to determine Alpha we must first determine (1 / Alpha Nt) and then determine Nt.

(1 / Alpha)^2
= (Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2]
or
(1 / Alpha Nt)^2
= (Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2 / Nt^2]

We know (Z / Nt) from the aforementioned computation process. Hence using a computer we can readily determine the So value of the relative minimum in a plot of:
(1 / Alpha Nt)^2 versus So

Plot:
(Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2 / Nt^2] versus So
and find the So value of the stable operating point. At that So value we will know the corresponding value of:
(1 / Alpha Nt)
and we can calculate Nr^2 using the formula:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}

Hence the mathematical procedure is to slowly vary So to find the spheromak lock point in the function:
(1 / Alpha Nt) vs So
where the term:
[Z / Nt}
in the expression for (1 / Alpha Nt) comes from the aforementioned calculation.

Then we recognize that in a spheromak:
Nr = Np / Nt
where Np and Nt are integers with no common factors.
 

Recall that:
Nr^2 = Np^2 / Nt^2
where Np and Nt are integers with no common factors. Thus to find Np and Nt it is necessary to test both the Np and Nt values for integer and factor compliance. This is not a huge task because we know that:
(1 / Alpha) ~ 137
constrains the maximum size of the Np and Nt integer values to less than about 500.

We know that with the simple boundary condition:
(1 / Alpha Nt)^2 = (Pi / 4)^2 [(So^2 - So + 1)^2 / (So^2 + 1)^8] {2 (So^2 + 1)^2 - 4}
/ {- [4 / [(So^2 - 1)^2]] + [Pi^2 / 4]}

We know that:
(1 / Alpha) ~ 137

Hence we can estimate Nt using the equation:
Nt|estimate = (1 / Alpha) / (1 / (Alpha Nt))
= 137 /(1 / (Alpha Nt))

Then we can estimate Np using the equation:
Np|estimate = Nr Nt|estimate
where to calculate Nr we first calculate:
[Z^2 / Nt^2] = {2 - [4 / (So^2 + 1)^2]}
/ {+ [Pi^2 / 4][1 / (So^2 + 1)^2] - [4 / [(So^2 + 1)^2 (So^2 - 1)^2]]} and then calculate Nr^2 using the equation:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}
and the calculate Nr using the equation: Nr = [Nr^2]^0.5

These estimates in combination with a list of prime numbers lead to only a few Np, Nt combinations that need to be fully tested.

Nr = Np / Nt
where Np and Nt are both integers with no common factors and usually either Np or Nt is prime.
 

value of (1 / Alpha) is given by:
(1 / Alpha) = 137.035999

Based on experience with plasmas I expected that the operating So^2 value would be:
So^2 ~ 4.1.

Moving the operating So value requires a change in the electric field energy distribution function. This issue needs more study.

 
 
 

Calculate the corresponding [Z^2 / Nt^2] value.

Calculate the corresponding value of Nr^2 using the formula:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}

Calculate the exact value of Nr using the formula:
Nr = [Nr^2]^0.5

Find Np and Nt which are the smallest integers with no common factors that precisely satisfy the equation:
Nr = (Np / Nt)

Usually Np and Nt have no common factors because one of them is prime.

Calculate the error in Nr base on Np = 223 and Nt = 303.

It is the precision of this ratio of integers coincident with real numbers which are a function of Pi that causes spheromak stability and hence quantization of energy.

Once Nt is precisely determined use the formula:
[1 / Alpha] = [Nt] [1 / (Alpha Nt)]
to determine the calculated value of:
[1 / Alpha].

Check if the calculated value of [1 / Alpha] is close to the value:
(1 / Alpha) = 137.035999
which is published at:
Fine Structure Constant


 
 
 
 

***********************************************************

SUMMARY:
An electromagnetic spheromak is governed by an existence condition and a common boundary condition. After an electromagnetic spheromak forms it spontaneously emits photons until it reaches an energy minimum also known as a ground state .

SPHEROMAK SHAPE PARAMETER:
See the new web page titled: SPHEROMAK SHAPE PARAMETER.
 

CHECK TABLE
So^2Nr^2NrFI(F I)
2.3 16.08516 4.0106 0.69362651.81611.2597
2.5 2.243 1.4976 0.68672 1.6087 1.10476
3.0
3.5
4.0 .549560.7413 0.62185 0.7374 0.45856
4.5
5.0
5.5
6.0

These equations seem to indicate that when a spheromak is initially formed the spheromak will try to operate at So^2 = 2.27, Nr large rather than at its low energy point. This issue needs further investigation. The toroidal region will reduce the spheromak energy. We must figure out how energy minimization affects the existence and/or boundary condition.

NUMERICAL CALCULATIONS:
TRY: So^2 = 2.3
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.155188}] / {1 - 0.959255}
= 0.655382 / 0.0407445
= 16.08516

Nr = 4.0106

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(4.0106)(2.3) / [(16.08516 )(10.89)) + (1.69)]^0.5
= 9.22438 / 13.29877
= 0.6936265

SPHEROMAK EVOLUTION:
The above equations indicate that when a spheromak is initially formed the spheromak will operate at large Nr corresponding to So^2 = 2.27. However, over time the spheromak will lose energy by radiation until its energy falls to a stable minimum energy point at So^2 = 4.0________ where the spheromak can no longer spontaneously radiate. As the spheromak energy decreases, Nr^2 decreases, So^2 increases and the volume of the toroidal region increases.
 

The value of:
So^2 = 4.1
is comparable to the ratio:
So^2 = (Rs / Rc) = 4.2
obtained from a General Fusion plasma spheromak photograph. However, a plasma spheromak may be affected by inertial forces and other issues that do not affect a charged particle spheromak.
 

PLASMA SPHEROMAKS:
Define:
Ih = plasma hose current
C = speed of light
Rhoh = (Qs / Lh)

Recall from PLASMA HOSE THEORY that:
(Ih / C) = Rhoh
= (Qs / Lh)
= Qs / [(Np Lp)^2 + (Nt Lt)^2]^0.5
or
Ih = C Qs / [(Np Lpf)^2 +(Nt Lt)^2]^0.5
= Qs C / {Pi Ro [[(Np (Rs + Rc))^2 / (Ro)^2] + [(Nt (Rs - Rc))^2 / (Ro)^2]]^0.5}
Thus if the charge Q on an atomic particle spheromak is replaced by the net charge Qs on a plasma spheromak the form of the spheromak equations is identical.

However, in a plasma spheromak:
Qs = Q (Ni - Ne)
where (Ni - Ne) is positive.

The web page PLASMA HOSE THEORY shows that for a plasma spheromak:
(Ni - Ne)^2 C^2 = (Ne Ve)^2
where Ve = electron velocity.

The kinetic energy Eke of a free electron with mass Me is given by:
Eke = (Me / 2) Ve^2

Hence:
(Ni - Ne)^2 C^2 = Ne^2 (2 Eke / Me)
or
Qs = Q (Ni - Ne) = Q (Ne / C)[2 Eke / Me]^0.5

Thus in a plasma spheromak Qs can potentially be obtained via measurements of Ne and Eke. However, due to the free electrons being confined to the spheromak wall Ne is not easy to accurately directly measure.
 

CALCULATION OF PLASMA SPHEROMAK Ne FROM THE FAR FIELD:
An approximate expression for the distant radial electric field is:
[Qs / 4 Pi Epsilon] [1 / (R^2 + H^2)]

The corresponding far field energy density is:
Ue = (Epsilon / 2)[Qs / (4 Pi Epsilon)]^2 [1 / (R^2 + H^2)]^2
= [Qs^2 / (32 Pi^2 Epsilon)][1 / (R^2 + H^2)]^2

and that this energy density in the far field must equal
Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2
where:
Uo = (Bpo^2 / 2 Mu)
Equating the two energy density expressions in the far field gives:
[Qs^2 / (32 Pi^2 Epsilon)] [1 / (R^2 + H^2)]^2
= (Bpo^2 / 2 Mu) [Ro^2 / (Ro^2 + R^2 + H^2)]^2
or
[Qs^2 / (32 Pi^2 Epsilon)] = (Bpo^2 / 2 Mu) [Ro^2]^2
or
Qs^2 = (Bpo^2 / 2 Mu) [Ro^2]^2 (32 Pi^2 Epsilon)
= (Bpo^2 / Mu) [Ro^2]^2 (16 Pi^2 / C^2 Mu)
= (Bpo^2 / Mu^2) [Ro^2]^2 (16 Pi^2 / C^2)

Thus:
Qs = (Bpo / Mu) [Ro^2] (4 Pi / C)

Recall that the formula for a plasma spheromak gave:
Qs = Q (Ne / C)[2 Eke / Me]^0.5
or
Ne = Qs C /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Ro^2] (4 Pi / C) C /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Ro^2] (4 Pi /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Rs Rc] (4 Pi /(Q [2 Eke / Me]^0.5)

This equation can be used to estimate Ne in experimental plasma spheromaks.
 

For a spheromak compressed from state a to state b this equation can be written in ratio form as:
(Neb / Nea) = (Bpob / Bpoa)(Rsb Rcb / Rsa Rca) (Ekea / Ekeb)^0.5
or
(Neb / Nea)^2 = (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 (Ekea / Ekeb)
or
(Neb / Nea)^2(Roa^6 / Rob^6)
= (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 (Roa^6 / Rob^6)(Ekea / Ekeb)
= (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 [(Rsa Rca)^3 / (Rsb Rcb)^3] (Ekea / Ekeb)
= (Bpob / Bpoa)^2 [(Rsa Rca) / (Rsb Rcb)] (Ekea / Ekeb)
 

EXPERIMENTAL PLASMA SPHEROMAK DATA:
General Fusion has reported spheromak free electron kinetic energies ranging from 20 eV - 25 eV for low energy density spheromaks at the spheromak generator to 400 ev - 500 eV for higher energy density spheromaks at the downstream end of the conical plasma injector. General Fusion reports a spheromak linear size reduction between these two positions of between 4X and 5X. The corresponding observed apparent electron densities rise from 2 X 10^14 cm^-3 to 2 X 10^16 cm^-3. The corresponding observed magnetic field increases from .12 T to 2.4 T to 3 T. At this time this author does not know for certain: where on the spheromak the electron kinetic energy was measured, where on the spheromak the apparent electron density was measured, where on the spheromak the magnetic field was measured or the absolute dimensions of the measured spheromaks and their enclosure.

Hence:
16 < [Ekeb / Ekea] < 25
20 < (Bpob / Bpoa) < 25
400 < (Bpob / Bpoa)^2 < 625<
4 < (Rca / Rcb) < 5
16 < (Rca / Rcb)^2 < 25
64 < (Rca / Rcb)^3 < 125
[(Nea / Rca^3) / (Neb / Rcb^3)]^2 = 10^-2

It appears that during the plasma spheromak compression Nea decreases to Neb while Qs remains constant. This effect might be due to electron-ion recombination during spheromak compression.
 

This web page last updated April 23, 2019.

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