Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links


XYLENE POWER LTD.

ELECTROMAGNETIC SPHEROMAK

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

ELECTROMAGNETIC SPHEROMAK:
Spheromak mathematics explains the existence of both stable charged particles and semi-stable toroidal plasmas and their properties.

The web page titled SPHEROMAK ENERGY shows that the energy density functions:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^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
X {1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}

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. Outside the spheromak wall these energy density components are believed to be:
Ue = Uo [(Ro^4 (R^2 + H^2)^4) / [(Ro^2 + R^2 + H^2)^6]
and
Um = Uo [Ro^4 [(Ro^2 + R^2 + H^2)^4 - (R^2 + H^2)^4]] / [(Ro^2 + R^2 + H^2)^6]

Note that:
Ue + Um = Uo [(Ro^4 (R^2 + H^2)^4) / [(Ro^2 + R^2 + H^2)^6]
+ Uo [Ro^4 [(Ro^2 + R^2 + H^2)^4 - (R^2 + H^2)^4]] / [(Ro^2 + R^2 + H^2)^6]
 
= Uo [Ro^4 [(Ro^2 + R^2 + H^2)^4]] / [(Ro^2 + R^2 + H^2)^6]
 
= Uo [Ro^4] / [(Ro^2 + R^2 + H^2)^2]
 
= Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2
which is the spheromak energy density function used on the web page titled:
SPHEROMAK ENERGY

In the far field where (R^2 + H^2) >> Ro^2:
Ue = Uo [(Ro^4 (R^2 + H^2)^4) / [(Ro^2 + R^2 + H^2)^6]
 
~ Uo [(Ro^4 / [(R^2 + H^2)^2]

The magnetic field energy density function Um is given by:
Um = Uo [Ro^4 [(Ro^2 + (R^2 + H^2))^4 - (R^2 + H^2)^4]] / [(Ro^2 + R^2 + H^2)^6]
 
= Uo [Ro^4 {[Ro^4 + 2 Ro^2 (R^2 + H^2) + (R^2 + H^2)^2]^2 - [R^2 + H^2]^4}] / [(Ro^2 + R^2 + H^2)^6]
 
= Uo [Ro^4 {[Ro^4 + 2 Ro^2 (R^2 + H^2)]^2 + 2 [(Ro^4 + 2 Ro^2 (R^2 + H^2)] [Ro^2 + H^2]^2}]
/ [(Ro^2 + R^2 + H^2)^6]
 

In the far field where (R^2 + H^2) >> Ro^2 the expression for Um simplifies to:
Um ~ Uo [Ro^4 {2 [2 Ro^2 (R^2 + H^2)] [Ro^2 + H^2]^2}]
/ [(R^2 + H^2)^6]
= Uo [4 Ro^6 (Ro^2 + H^2)^3}]
/ [(R^2 + H^2)^6]
= Uo [4 Ro^6]
/ [(R^2 + H^2)^3]
 

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

At R = 0, H = 0 the electric field energy density Ue is zero as expected and the magnetic field energy density Um is given by:
Um = Uo

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, the Planck Constant.

Inside the spheromak wall the electric field energy density is given by:
Uei = Ueic [(Rc / R)^2]
and 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 + Uei

At R = Rc and H = 0:
Umoc + Ueoc = Umic + Ueic
This is a key equation to the boundary condition that accounts for spheromak behavior.

This web page shows that in order to exist an electromagnetic spheromak should approximately satisfy the existence condition:
or
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

This equation establishes a round about functional relationship between So and the ratio Nr of the number Np of spheromak current flow path poloidal turns to the number Nt of current flow path toroidal turns which is given by:
Nr = (Np / Nt).
However, due to the practical evaluation difficulty involved we develop another method of finding Nr^2 as a function of So.

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

To find the spheromak operating point choose So^2, calculate Nr^2, and then calculate RHS of existence condition. Plot the RHS value of the existence condition versus So^2. Repeat this process to find the So^2 value that gives RHS the closest approach to LHS = (Pi / 2). Use the corresponding value of Nr to find integers Np and Nt.

Typical experimental plasma spheromaks that are sufficiently small compared to their enclosure size so as not to be distorted by the proximity of the enclosure wall have shape factors of about:
So^2 = (Rs / Rc) ~ 4.2
 

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;
H = distance of a general point above the spheromak's equatorial plane;
Hs = height of a point on the spheromak wall above the spheromak's equatorial plane;
(2 |Hf|) = spheromak's overall length measured at R = Rf;
 

SPHEROMAK STRUCTURAL ASSUMPTIONS:
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.
3) At a particular spheromak energy the charge hose current Ih is constant;
4) The charge hose current causes a purely toroidal magnetic field inside the spheromak wall and a purely poloidal magnetic field outside the spheromak wall;
5) The net charge causes a cylindrically radial electric field inside the spheromak wall and an electric field which is spherically radial in the far field outside the spheromak wall;
6) At the center of the spheromak at (R = 0, H = 0) the electric field is zero;
7)Define radius Ro by:
Rs / Ro = Ro / Rc = So
or
Ro^2 = Rs Rc = So^2
8) Inside the spheromak wall where:
Rc < R < Rs and |H| < |Hs|
the total field energy density U takes the form:
Ui = Uei + Umi
= Uio (Ro / R)^2

where:
Uie = radial electric field energy density inside the spheromak wall;
Uim = toroidal magnetic field energy density inside the spheromak wall;
9) Outside the spheromak wall the total field energy density takes the form:
U = Ue + Um
= Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2

where:
Uo = [(Mu Qs^2 C^2) / (32 Pi^2 Ro^4)];
10) Everywhere on the spheromak wall:
U = Ueo + Umo = Uei + Umi;
11) The electric field energy density outside the spheromak wall is:
Ue = Uo [Ro^4 (R^2 + H^2)^4)] / [(Ro^2 + R^2 + H^2)^6]
12)The magnetic field energy density outside the spheromak wall is:
Um = Uo [Ro^4 [(Ro^2 + R^2 + H^2)^4 - (R^2 + H^2)^4]] / [(Ro^2 + R^2 + H^2)^4]
13)Both outside and inside the spheromak wall:
U = Ue + Um
Note that the electric field energy density at R = Rc, H = 0 outside the spheromak wall is:
Uec = Uo [((R^2 + H^2)^4 Ro^4)] / [(Ro^2 + R^2 + H^2)^6]
= Uo [Rc^8 Ro^4] / [(Ro^2 + Rc^2)^6]
= Uo [Rc^8 / Ro^8] [Ro^12 / (Ro^2 + Rc^2)^6]
= Uo [(1 / So^8)] [Ro^12] / [(Ro^2 + Rc^2)^6]
= Uo [(1 / So^8)] [Ro^12] / [(Ro^2)^6 (1 + (Rc / Ro)^2)^6]
= Uo [(1 / So^8)] / [ (1 + (Rc / Ro)^2)^6]
= Uo [(1 / So^8)] / [ (1 + (1 / So)^2)^6]
Uo [(1 / So^8)][So^12] / [ (So^2 + 1)^6]
= Uo [So^4 / (So^2 + 1)^6]
 

EQUATORIAL PLANE:
On the spheromak's equatorial plane:
H = 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 parallel to 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 |H| < |Hs| the internal magnetic field Bti is toroidal;
For Rc < R < Rs and |H| < |Hs| the magnetic field Bti is proportional to (1 / R).

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

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

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;
 

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 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]}

The electric field energy density of a spheromak outside the spheromak wall is given by:
Ueo = Uo [((R^2 + H^2)^4 Ro^4 ) / (Ro^2 + R^2 + H^2)^6]

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

The corresponding total electric energy is given by:
Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ Ro^4 (Z^2)^4 dZ / (Ro^2 + Z^2)^6
 
= Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Ro^4 Z^12 dZ / (Ro^2 + Z^2)^6
 

From Dwight XXXX the solution to this integral is:
FIX

FIXThus the eledtric 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.
 

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 naturally returns to its stable geometry.
 

SPHEROMAK WALL:
Under the circumstances of plasma spheromak generation the electrons and ions form a CHARGE HOSE SHEET. The charge hose sheet is the wall of a hollow torus. Inside the spheromak wall the magnetic field is purely toroidal. Outside the spheromak wall the magnetic field is purely poloidal.

The trapped electrons and ions circulate in the plane of the spheromak wall. The spheromak wall is a constant total energy closed path for the trapped charge(s).

In a plasma the positive ions move opposite to the electrons to approximately balance charge and momentum within the spheromak wall. The electric field component normal to the spheromak wall slightly separates the electron and ion streams, which prevents the energetic electrons being scattered by collisions with the spheromak ions. Hence the electrons and ions follow similar but opposite spiral paths within the spheromak wall.

A fundamental difference between a plasma and an atomic particle is that the electrons and ions in a plasma are subject to inertial forces whereas in a quantum charged particle the circulating charge has no mass. Hence high temperature plasma spheromaks are subject to relativistic effects.
 

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

PLASMA SHEET POSITION:
A plasma spheromak is a semi-stable energy state. The spheromak wall positions itself to achieve a total energy relative minimum. At every point on the spheromak wall the sum of the electric and magnetic field energy densities on one side of the spheromak wall equals the sum of the electric and magnetic field energy densities on the other side of the spheromak wall. This general statement resolves into different boundary conditions for different points on the spheromak wall. These boundary conditions establish the spheromak core radius Rc on the equatorial plane, the spheromak outside radius Rs on the equatorial plane and the spheromak length 2 Hf.
 

On the web page titled THEORETICAL SPHEROMAK it is shown that a theoretical spheromak is a stable structure. The question addressed on this web page is under what circumstances does an electromagnetic system form a stable spheromaks?
 

ENERGY DENSITY MATCHING:
Let Qs be the apparent net charge on 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 + H^2)]^2

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

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

Hence for an electromagnetic spheromak in the far field:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 ~ [Epsilono / 2][Qs / (4 Pi Epsilono (R^2 + H^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:
At R = 0, H = 0 the electric field energy density Ue = 0 giving:
Uo = (Bo^2 / 2 Muo)
and hence:
(Bo^2 / 2 Muo) Ro^4 = [1 / 2 Epsilono][Qs / (4 Pi)]^2
or
(Bpo^2 / 2 Muo) = [1 / 2 Epsilono][Qs / (4 Pi Ro^2)]^2
or
Bpo^2 = [Muo / Epsilono][Qs / (4 Pi Ro^2)]^2

= Muo^2 C^2 Qs^2 / (4 Pi Ro^2)^2

Hence:
Bpo = Muo C Qs / (4 Pi Ro^2)

This equation allows comparison of the poloidal magnetic field of an atomic particle spheromak to externally applied magnetic fields.

The web page titled SPHEROMAK ENERGY gives the maximum possible static field energy Efs of a spheromak as:
B>Efs = Uo Ro^3 Pi^2
and the actual spheromak energy Ett after adjustment for the toroidal portion as:
Ett / Efs = {1 -[(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}

Hence substitution for Uo gives:
Efs = Uo Ro^3 Pi^2
= (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]}

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

Thus an electromagnetic spheromak has a net charge Qs, a nominal radius Ro and a 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.
 

BIOT AND SAVART:
The law of Biot and Savart gives an element of axial magnetic field d(Bpo) at (0, 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

 

POLOIDAL MAGNETIC FIELD AT THE CENTER OF A SPHEROMAK:
At every point on the charge hose:
Ih^2 = Ip^2 + It^2

dL^2 = [R d(Theta)]^2 + [(Rs - Rc) d(Phi)/ 2]^2

Ip / Ih = R d(Theta) / dL
= R d(Theta) / {[R d(Theta)]^2 + [(Rs - Rc) d(Phi)/ 2]^2}^0.5
= R d(Theta) / {[R d(Theta)]^2 + [d(Theta)(Rs - Rc) d(Phi)/ 2 d(Theta)]^2}^0.5
= R / {[R]^2 + [(Rs - Rc) Nt / 2 Np]^2}^0.5
= Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5

Thus:
Ip = Ih Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5

The length of a poloidal current hose element is:
R d(Theta)

In a 360 degree current ring there are Nt such poloidal current elements.

Allowing for poloidal magnetic field contributions from both the upper and lower halves of the spheromak in effect there are (2 Nt) such current hose elements.

The distance from the poloidal current element to the center of the spheromak is:
[Hs^2 + R^2]^0.5

cos(A) = [R / (Hs^2 + R^2)^0.5]

The contribution to the poloidal magnetic field at the center of the spheromak due to the (2 Nt) such elements in the matching rings in the upper and lower halves of the spheromak is:
dBpo = (2 Nt)(Muo / 4 Pi) Ih [Ip / Ih][R d(Theta)] [1 / (Hs^2 + R^2)] cos(A)
= (2 Nt)(Muo / 4 Pi) Ih [Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5][R d(Theta)][1 / (Hs^2 + R^2)][R / (Hs^2 + R^2)^0.5]
= (2 Nt)(Muo / 4 Pi) (Ih Np R^3 d(Theta)) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Hence:
dBpo / d(Phi) = [dBpo / d(Theta)] [d(Theta) / d(Phi)]
= [dBpo / d(Theta)] [Np / Nt]
= (2 Nt)(Muo / 4 Pi) (Ih Np R^3)(Np / Nt) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)
= (2 Np)(Muo / 4 Pi) (Ih Np R^3) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Thus:
Bpo = integral from Phi = 0 to Phi = Pi of:
(2 Np)(Muo / 4 Pi) (Ih Np R^3) d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Recall that:
Hs^2 = (Rs - R)(R - Rc)
= (Rs R - Rs Rc - R^2 + R Rc)

Hence:
Bpo = Integral from Phi = 0 to Phi = Pi of:
(Np Muo/ 2 Pi) Ih R^3 Np^2 d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)

Now find d(Phi) in terms of R:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] cos (Phi)
giving:
dR = [(Rs - Rc) / 2] sin (Phi) d(Phi)

Thus:
dR = [(Rs - Rc) / 2] sin (Phi) d(Phi)

= [(Rs - Rc) / 2] {[Hs / [(Rs - Rc) / 2]} d(Phi)
= Hs d(Phi)

Hence:
d(Phi) = dR / Hs
= {dR / [(Rs - R)(R - Rc)]^0.5}

Thus:
Bpo = Integral from Rc to Rs of:
(Muo / 2 Pi) Ih R^3 Np^2 d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)
 
= Integral from Rc to Rs of:
(Muo / 2 Pi) Ih R^3 Np^2 {dR / [(Rs - R)(R - Rc)]^0.5} / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)
 
= Integral from Rc to Rs of:
(Muo / 2 Pi) Ih R^3 Np^2 dR / ([(Rs - R)(R - Rc)]^0.5 {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)

 
= Integral from Rc to Rs of:
(Muo / 2 Pi) (1 / Rc) Ih (R / Rc)^3 d(R / Rc) Np (Np / Nt)
/ ([((Rs / Rc) - (R / Rc))((R / Rc) - 1)]^0.5 {[(Np / Nt) (R / Rc)]^2 + [((Rs / Rc) - 1) / 2]^2}^0.5
/ [(Rs R / Rc^2) - (Rs / Rc) + (R / Rc)]^1.5)

We can make this integral appear simpler using the substituions:
Z = (R / Rc)
Zo = So^2 = (Rs / Rc)
Nr = (Np / Nt)

Then:
Bpo = Integral from Z = 1 to Z = Zo = Rs / Rc = So^2 of:
(Muo / 2 Pi Rc) Ih Z^3 dZ Np Nr
/ ([(Zo - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(Zo - 1) / 2]^2}^0.5 [Zo Z - Zo + Z]^1.5)

or
Bpo = (Muo Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

which gives the central magnetic field strength for any spheromak.
 

SPECIAL CASE OF So^2 ~ 1:
The special case of So^2 ~ 1 should simplify to the case of a simple current ring. For this special case:
Bpo = (Muo Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Muo Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 [Nr] )
 
(Muo Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
dZ / ([-Z^2 + (So^2 + 1) Z - So^2]^0.5)

From Dwight 380.001 the integral becomes:
= - 1 arc sin[(-2 Z + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5|Z = So^2
+ 1 arc sin[(-2 Z + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5|Z = 1
 
= - arc sin[(-2 So^2 + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5
+ arc sin[(-2 + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5
 
= - arc sin[(1 - So^2) / ((So^2 - 1)^2)^0.5
+ arc sin[((So^2 - 1)) / ((So^2 - 1)^2)^0.5
 
= Pi

Hence for this special case:
Bpo = (Muo Np / 2 Pi) (Ih / Rc) Pi
(Mu Np / 2) (Ih / Rc)

From Biot and Savart the magnetic field at the center of a current ring with Np turns is:
(Muo / 4 Pi) Ih Np 2 Pi Rc / Rc^2 = (Muo / 2) Ih Np / Rc

Hence the integral simplifies as expected for this special case.
 

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]
 

Hence:
Bpo = (Muo Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Muo Np / 2 Pi) [(Qs C) / (Pi Ro^2 Nt)] [So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Muo / 2 Pi) [(Qs C) / (Pi Ro^2)] [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

This value of Bpo should equal the value of Bpo calculated above:
Bpo = [(Muo C Qs) / (4 Pi Ro^2)]
 

ELECTROMAGNETIC SPHEROMAK EXISTENCE CONDITION:
Hence:
[(Muo C Qs) / (4 Pi Ro^2)]
= (Muo / 2 Pi) [(Qs C) / (Pi Ro^2)] [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
and cancelling terms an LHS and RHS gives:
[1 / (4)]
= ((1 / 2 Pi) [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
or
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

which is referred to herein as the electromagnetic spheromak existence condition. This equation imposes an important spheromak relationship between Nr and So.

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

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)

At every point on the spheromak surface:
d(Theta) /d(Phi)
= (2 Pi Np) / 2 Pi Nt)
= (Np / Nt)
 

CHARGE PER UNIT AREA AT R= Rc, H = 0:
On the equatorial plane at the core at R = Rc, H = 0:
Hose rise = (Rs - Rc) d(Phi) / 2
Hose run = Rc d(Theta)

Hose slope = (Rise) / (Run)
= [(Rs - Rc) d(Phi)] / [2 Rc d(Theta)]
= [(Rs - Rc) Nt] / [2 Rc Np]

In the core the distance between adjacent hoses measured along the inner equator is (2 Pi Rc / Nt).

Geometry shows that in the inner core the distance Dhc between hoses measured perpendicular to the hose axis is:
Dhc = (2 Pi Rc / Nt) [(Rs - Rc) d(Phi) / 2] / {[(Rs - Rc) d(Phi) / 2]^2 + [Rc d(Theta)]^2}^0.5
= (2 Pi Rc / Nt) [(Rs - Rc) d(Phi) / 2] / [d(Theta){[(Rs - Rc) d(Phi) / 2d(Theta)]^2 + [Rc]^2}^0.5]
= (2 Pi Rc / Nt) [(Rs - Rc) Nt / 2] / [Np {[(Rs - Rc) Nt / 2 Np]^2 + [Rc]^2}^0.5]
= (Pi Rc) [(Rs - Rc)] / {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5

Hence the net charge per unit area Sac at R = Rc, H = 0 is:
Sac = Qs / (Lh Dhc)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}

Recall that:
Lh = {[2 Pi Np (Rs + Rc) / 2]^2 + [2 Pi Nt (Rs - Rc) / 2]^2}^0.5
= Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5

Hence the charge per unit area at R = Rc, H = 0 is:
Sac = Qs / (Lh Dhc)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}
= (Qs / Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5) {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}
= {Qs / [(Pi^2 Rc) (Rs - Rc)]} {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5
= {Qs / [(Pi^2 Rc^2) (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5

Hence:
Sac^2 = {Qs So^2 / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 

CHARGE PER UNIT AREA AT R = Rs, H = 0:
On the equatorial plane at the outer perimeter where R = Rs, H = 0:
Hose rise = (Rs - Rc) d(Phi) / 2
Hose run = Rs d(Theta)

Hose slope = (Rise) / (Run)
= [(Rs - Rc) d(Phi)] / [2 Rs d(Theta)]
= [(Rs - Rc) Nt] / [2 Rs Np]

On the outer rim the distance between adjacent hoses measured along the outer equator is (2 Pi Rs / Nt).

Geometry shows that at the outer rim the distance Dhs between hoses perpendicular to the hose axis is:
Dhs = (2 Pi Rs / Nt) [(Rs - Rc) d(Phi) / 2] / {[(Rs - Rc) d(Phi) / 2]^2 + [Rs d(Theta)]^2}^0.5
= (2 Pi Rs / Nt) [(Rs - Rc) d(Phi) / 2] / {[d(Theta)]{[(Rs - Rc) d(Phi) / 2 d(Theta)]^2 + [Rs]^2}^0.5}
= (2 Pi Rs / Nt) [(Rs - Rc) Nt / 2] / {[Np]{[(Rs - Rc) Nt / 2 Np]^2 + [Rs]^2}^0.5}
= (Pi Rs) [(Rs - Rc)] / {Np {[(Rs - Rc) Nt / 2 Np]^2 + [Rs]^2}^0.5}
= (Pi Rs) [(Rs - Rc)] / {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5

Hence the charge per unit area Sas at R = Rs, H = 0 is:
Sas = Qs / (Lh Dhs)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / [(Pi Rs) (Rs - Rc)]

Recall that:
Lh = {[2 Pi Np (Rs + Rc) / 2]^2 + [2 Pi Nt (Rs - Rc) / 2]^2}^0.5
= [Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5]

Hence:
Sas = Qs / (Lh Dhs)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / [(Pi Rs) (Rs - Rc)]
= (Qs {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / {[Pi Rs (Rs - Rc)] Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5}
= {Qs / [Pi^2 Rs (Rs - Rc)]} {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5
= {Qs / [Pi^2 Rs Rc (So^2 - 1)]} {[(So^2 - 1) Nt / 2]^2 + [Np So^2]^2}^0.5 / {[Np (So^2 + 1)]^2 + [Nt (So^2 - 1)]^2}^0.5
= {Qs / [Pi^2 Ro^2 (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5

Hence:
Sas^2 = {Qs / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr So^2]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 

SURFACE CHARGE RATIO:
A quantity of interest is the surface charge ratio (Sas / Sac). From above this ratio is given by:
(Sas / Sac) = {Qs / [Pi^2 Ro^2 (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5 / {Qs So^2 / [(Pi^2 Ro^2) (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
 
= {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {So^2} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5
 
= {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5

Note that for large Nr this ratio simplifies to:
(Sas / Sac) = {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5
= {[Nr So^2]^2}^0.5 / {[Nr So^2]^2}^0.5
= 1

Note that for small Nr this ratio simplifies to:
(Sas / Sac) = {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5
= {[(So^2 - 1) / 2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2}^0.5
= (1 / So)
 

Recall that:
Sac^2 = {Qs / [Pi^2 Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 

FIELD DEFINITIONS:
Eroc = electric field outside the wall at R = Rc, H = 0;
Eric = cylindrically radial electric field inside 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;
 

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 for every element of area dA on the spheromak wall.
 

THEORETICAL DETERMINATION OF THE FINE STRUCTURE CONSTANT

BOUNDARY CONDITION AT R= Rc, H = 0:
At Rc = 0, H = 0 field energy density balance gives:
[(Bpoc^2 + Btoc^2) / Mu] + [Epsilon Eroc^2] = [Bpic^2 + Btic^2] / Mu + [Epsilon Eric^2]

Outside the spheromak wall at R = Rc, H = 0 the magnetic field is purely poloidal and the electric field is radial pointing toward the main axis of symmetry of the spheromak. Inside the spheromak wall the magnetic field is purely toroidal and the electric field is cylindrically radial pointing away from the main axis of symmetry of the spheromak. The total field energy densities on both sides of this wall are equal.

For a spheromak at R = Rc, H = 0 the surface charge density Sac causes a step change in the radial electric field which causes the radial electric field to change direction. Hence:
|Eroc| + |Eric| = Sac / Epsilono;
and
Btoc = 0 or no toroidal magnetic field outside the spheromak wall
and
Bpic = 0 or no poloidal magnetic field inside the spheromak wall
giving the simplified boundary condition at R = Rc, H = 0 as:
[Bpoc^2 / Muo] + [Epsilono Eroc^2] = [Btic^2 / Muo] + [Epsilono Eric^2]

Use the spheromak electric field energy density function:
Ueo = Uo [(R^2 + H^2)^4 Ro^4] / [(Ro^2 + R^2 + H^2)^6
which at R = Rc, H = 0 gives:
Ueoc = Uo [Rc^8 Ro^4] / [(Ro^2 + Rc^2)^6]
= Uo [1 / So^8)] [Ro^12 / (Ro^2 + Rc^2)^6]
= Uo [1 / So^8)] [1 / (1 + (1 / So^2))^6]
= Uo [1 / So^8)] [So^12 / (So^2 + 1)^6]
= Uo [So^4 / (So^2 + 1)^6]
to calculate |Eroc| from
|Eroc| = (2 Ueoc / Epsilono)^0.5
and then to calculate Eric from
|Eric| = (Sac / Epsilono) - |Eroc|
= (Sac / Epsilono) - (2 Ueoc / Epsilono)^0.5
= (Sac / Epsilono) - {2 Uo [So^4 / (So^2 + 1)^6][1 / Epsilono]}^0.5
 
which gives:
Eric^2 = (Sac / Epsilono)^2
+ {2 Uo [So^4] / [(So^2 + 1)^6 Epsilono]}
- 2 (Sac / Epsilono){2 Uo [So^4 / (So^2 + 1)^6 Epsilono]}^0.5

Then:
Eric^2 = (Sac / Epsilono)^2
+ {2 Uo [So^4] / [(So^2 + 1)^6 Epsilono]}
- 2 (Sac / Epsilono){2 Uo [So^4 / (So^2 + 1)^6 Epsilono]}^0.5

Ueic = (Epsilono / 2) Eric^2
= (Sac^2 / 2 Epsilono)
+ {Uo [So^4 / (So^2 + 1)^6]}
- (Sac){2 Uo [So^4 / (So^2 + 1)^6] / Epsilono}^0.5
 
and then calculate Umic from:
Umic = Uc - Ueic
where:
Uc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2
= Uo [So^2 / (So^2 + 1)]^2

Hence:
Umic = (Uc - Ueic)
= Uo [So^2 / (So^2 + 1)]^2
- (Sac^2 / 2 Epsilono)
- {Uo [So^4 / (So^2 + 1)^6]}
+ (Sac){2 Uo [So^4 / (So^2 + 1)^6] / Epsilono}^0.5

Recall that:
Uo = [Q^2 / (32 Pi^2 Epsilono Ro^4)]

Hence:
Umic = Uo [So^2 / (So^2 + 1)]^2
- [(Sac^2 / 2 Epsilono)
- {Uo [So^4 / (So^2 + 1)^6]}
+ (Sac){2 Uo [So^4 / (So^2 + 1)^6] / Epsilono}^0.5
 
= [Q^2 / (32 Pi^2 Epsilono Ro^4)] [So^2 / (So^2 + 1)]^2
- (Sac^2 / 2 Epsilono)
- [Q^2 / (32 Pi^2 Epsilono Ro^4)] [So^4 / (So^2 + 1)^6]
+ (Sac){[2 Q^2 / (32 Pi^2 Epsilono Ro^4)] [So^4 / (So^2 + 1)^6] / Epsilono}^0.5
 
= [Q^2 / (32 Pi^2 Epsilono Ro^4)] [So^2 / (So^2 + 1)]^2
- (Sac^2 / 2 Epsilono)
- [Q^2 / (32 Pi^2 Epsilono Ro^4)] [So^4 / (So^2 + 1)^6]
+ (Sac)(Q / 4 Pi Ro^2 Epsilono) {[So^2 / (So^2 + 1)^3]}
 
and factoring out common terms gives:
Umic = {Q^2 / Pi^2 Ro^4 Epsilono}{[1 / 32] [So^2 / (So^2 + 1)]^2
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
- ([1 / 32] [So^4 / (So^2 + 1)^6])
+ (Sac) {Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]]}
 

From the spheromak winding parameters.
Bc = [(Muo Nt Qs Fh) / (2 Pi Rc)]
or
Umic = Bc^2 / 2 Muo
= [(Muo Nt Qs Fh) / (2 Pi Rc)]^2 / (2 Muo)
= [Muo / 8] [(Nt Qs Fh So) / (Pi Ro)]^2

Equating the two expressions for Uimc gives the boundary condition:
[Muo / 8] [(Nt Qs Fh So) / (Pi Ro)]^2
= {Q^2 / Pi^2 Ro^4 Epsilono}{[1 / 32] [So^2 / (So^2 + 1)]^2
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
- ([1 / 32] [So^4 / (So^2 + 1)^6])
+ (Sac) {Ro^2 Pi / 4 Q) {([So^2 / (So^2 + 1)^3])]}
 

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

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

Substitute the expression for (Nt Fh)^2 into the boundary condition to get:
[Muo / 8] [Nt^2 Fh^2][(Qs So) / (Pi Ro)]^2
 
= {Q^2 / Pi^2 Ro^4 Epsilono}{[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}
 
or
[Muo / 8] [(Qs So) / (Pi Ro)]^2 [C^2 So^2] / [Ro^2 Pi^2 {[Nr (So^2 + 1]^2 + [So^2 - 1]^2}]
 
= {Q^2 / Pi^2 Ro^4 Epsilono}{[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}
 
or factoring out Qs ^2 / Pi^2 Ro^4:
[Muo / 8] [So]^2 [C^2 So^2] / [Pi^2 {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}]
 
= {1 / Epsilono}{[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}
 
or since (1 / Epsilono) = Muo C^2:
[Muo / 8] [So]^2 [C^2 So^2] / [Pi^2 {[Nr (So^2 + 1]^2 + [So^2 - 1]^2}]
 
= {Muo C^2}{[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}
 
and cancelling Muo C^2 terms gives:
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}
 
= {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}

 

Recall that:
Sac^2 = {(Qs So^2) / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
or
[Sac Ro^2 Pi / Qs ]^2
 
= {So^2 / [Pi (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
and
[Sac Ro^2 Pi / Qs ]
= {So^2 / [Pi (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
 

Substituting these two expressions into the boundary condition gives:
[1 / 8 Pi^2] [So^4] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
= {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- [(Sac Ro^2 Pi / Q)^2][1 / 2]
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ (Sac Ro^2 Pi / 4 Q) [So^2 / (So^2 + 1)^3]}
 
or
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
= {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- {So^2 / [Pi (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}[1 / 2]
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
- ([1 / 32] [So^4 / (So^2 + 1)^6])
 
+ [1 /4]{{So^2 / [Pi (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5} [So^2 / (So^2 + 1)^3]}

 

Rearrange this equation to get:
[1 / 8 Pi^2] [So^4] / {[Nr (So^2 + 1]^2 + [(So^2 - 1)]^2}
 
+ {So^2 / [Pi (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}[1 / 2]
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
= + {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- ([1 / 32] [So^4 / (So^2 + 1)^2][1 / (So^2 + 1)^4])
 
+ [1 /4][So^2 / (So^2 + 1)^3]{So^2 / [Pi (So^2 - 1)]}
{[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5

 
 
or
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1]^2 + [(So^2 - 1)]^2}
 
+ {So^2 / [2 Pi]}^2 {1 + [2 Nr / (So^2 - 1)]^2}[1 / 2]
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
= + {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- ([1 / 32] [So^4 / (So^2 + 1)^2][1 / (So^2 + 1)]^4)
 
+ [1 / 4][So^2 / (So^2 + 1)^3] [So^2 / 2 Pi]
{[1 + [2 Nr /(So^2 - 1)]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5}

 

Rearrange the boundary condition equation to:
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
+ [So^2 / 2 Pi]^2 {1 + [2 Nr / (So^2 - 1)]^2}[1 / 2]
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
- {[1 / 32] [So^2 / (So^2 + 1)]^2
 
+ ([1 / 32] [So^4 / (So^2 + 1)^2][1 / (So^2 + 1)]^4)
 
= + [1 /4][So^2 / (So^2 + 1)^3]
[So^2 / 2 Pi] {[1 + [2 Nr /(So^2 - 1)]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
 

The terms:
- [1 / 32] [So^2 / (So^2 + 1)]^2
 
+ [1 / 32] [So^4 / (So^2 + 1)^2][1 / (So^2 + 1)]^4
 

Hence the boundary condition equation becomes:
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1]^2 + [(So^2 - 1)]^2}
 
+ {So^2 / [2 Pi]}^2 {1 + [2 Nr / (So^2 - 1)]^2}[1 / 2]
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
- [1 / 32] [So^2 / (So^2 + 1)]^2 [1 - (1 / (So^2 + 1)^4)]
 
- [1 / 4][So^2 / (So^2 + 1)^3][So^2 / 2 Pi]
{[1 + [2 Nr /(So^2 - 1)]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5}
 
= 0
 

From the perspective of the fine structure constant Alpha the parameter that we really need to find is:
Nr^2 + [(So^2 - 1) / (So^2 + 1)]^2

Define:
Z^2 = Nr^2 + [(So^2 - 1) / (So^2 + 1)]^2

Hence:
Nr^2 = Z^2 - [(So^2 - 1) / (So^2 + 1)]^2

Then the frequently reoccurring parameter cluster:
[Nr^2 (So^2 + 1)^2 + (So^2 - 1)^2]
= [{Z^2 - [(So^2 - 1)^2 / (So^2 + 1)^2]}(So^2 + 1)^2] + (So^2 - 1)^2]
= Z^2 (So^2 + 1)^2
 

The frequently reoccurring parameter cluster:
Nr^2 / (So^2 - 1)^2 = [Z^2 / (So^2 - 1)^2] - [1 / (So^2 + 1)^2]
 

The boundary condition equation gives:
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
+ [So^2 / 2 Pi]^2 [1 / 2] {1 + [2 Nr / (So^2 - 1)]^2}
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 
- [1 / 32] [So^2 / (So^2 + 1)]^2 [1 - (1 / (So^2 + 1)^4)]
 
= + [1 / 4][So^2 / (So^2 + 1)^3] [So^2 / 2 Pi]
{[1 + [2 Nr /(So^2 - 1)]^2}^0.5
/ {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
 

Substituting for the frequently reoccurring cluster terms gives:
[1 / 8 Pi^2][So^4] / [Z^2 (So^2 + 1)^2]
 
+ [So^2 / 2 Pi]^2 [1 / 2]{1 + 4 [Z^2 / (So^2 - 1)^2] - 4 [1 / (So^2 + 1)^2]} / [Z^2 (So^2 + 1)^2]
 
- [1 / 32] [So^2 / (So^2 + 1)]^2 [1 - (1 / (So^2 + 1)^4)]
 
= + [1 /4][So^2 / (So^2 + 1)^3] [So^2 / 2 Pi] {[1 + 4 [Z^2 / (So^2 - 1)^2] - [4 / (So^2 + 1)^2]}^0.5
/ [Z (So^2 + 1)]}
 

Now multiply through by Z^2 (So^2 + 1)^2 to get:
[1 / 8 Pi^2][So^4]
 
+ {So^2 / 2 Pi}^2 [1 / 2]{1 + 4 [Z^2 / (So^2 - 1)^2] - 4 [1 / (So^2 + 1)^2]}
 
- [1 / 32] [So^2]^2 [1 - (1 / (So^2 + 1)^4)] Z^2
 
= + (1 / 4)[So^2 / (So^2 + 1)^2] [So^2 / 2 Pi]
{[Z^2 + 4 [Z^4 / (So^2 - 1)^2] - 4 [Z^2 / (So^2 + 1)^2]}^0.5

 

ELEGANT SOLUTION:
The boundary condition can be written in the form:
[1 / 8 Pi^2][So^4]
 
+ [So^2 / 2 Pi]^2 [1 / 2]{1 + 4 [Z^2 / (So^2 - 1)^2] - [4 / (So^2 + 1)^2]}
 
- [1 / 32] [So^2]^2 [1 - (1 / (So^2 + 1)^4)] Z^2
 
= Y
where:
Y = + (1 / 4)[So^2 / (So^2 + 1)^2] {So^2 / 2 Pi}
{[Z^2] + 4 [Z^4 / (So^2 - 1)^2] - 4 [Z^2 / (So^2 + 1)^2]}^0.5}
 

Reordering the terms by Z^2 content gives:
[So^2 / 2 Pi]^2 [1 / 2][4 / (So^2 - 1)^2] Z^2
 
- [1 / 32] [So^2]^2 [1 - (1 / (So^2 + 1)^4)] Z^2
 
- Y
 
= - [1 / 8 Pi^2][So^4]
- [So^2 / 2 Pi]^2 [1 / 2]
+ [So^2 / 2 Pi]^2 [1 / 2] [4 / (So^2 + 1)^2]}
 

Multiply through by 8 Pi^2 / So^4 To get:
{4 [Z^2 / (So^2 - 1)^2]}
 
- [Pi^2 / 4] [1 - (1 / (So^2 + 1)^2)] Z^2
 
- 8 Pi^2 Y / So^4
 
= - [1]
- [1]
+ [4 / (So^2 + 1)^2]}

Solving for Z^2 gives:
Z^2 = {- 2 + [4 / (So^2 + 1)^2]}
/ {4 [1 / (So^2 - 1)^2]}
 
- [Pi^2 / 4] [1 - (1 / (So^2 + 1)^4)]
 
- Y 8 Pi^2 / So^4 Z^2
 
or
Z^2 = {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / (So^2 - 1)^2]}
 
+ [Pi^2 / 4][1 - (1 / (So^2 + 1)^4)]
 
+ Y 8 Pi^2 / So^4 Z^2}

Recall that by definition:
Y = + (1 /4)[So^2 / (So^2 + 1)^2] [So^2 / 2 Pi] {[Z^2 + 4 [Z^4 / (So^2 - 1)^2] - 4 [Z^2 / (So^2 + 1)^2]}^0.5}

Hence:
(Y 8 Pi^2 / (So^4 Z^2))
 
= (8 Pi^2 / (So^4 Z^2)) (1 /4){[So^2 / (So^2 + 1)^2]}[So^2 / 2 Pi]
{[Z^2] + 4 [Z^4 / (So^2 - 1)^2] - 4 [Z^2 / (So^2 + 1)^2]}^0.5
 
= (Pi / (Z^2)) [1 / (So^2 + 1)^2]
{[Z^2] + 4 [Z^4 / (So^2 - 1)^2] - 4 [Z^2 / (So^2 + 1)^2]}^0.5
 
= [Pi / (So^2 + 1)^2]
{[1 / Z^2] + [4 / (So^2 - 1)^2] - 4 [(1 / Z^2) / (So^2 + 1)^2]}^0.5
 
= [Pi / (So^2 + 1)^2]
{[1 / Z^2][1 - (4 / (So^2 + 1)^2)] + [4 / (So^2 - 1)^2]}^0.5
 

Thus we have the itterative relationship:
Z^2 = {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / (So^2 - 1)^2] + [Pi^2 / 4][1 - (1 / (So^2 + 1)^4)] + (Y 8 Pi^2 / (So^4 Z^2))}

in which the term:
(Y 8 Pi^2 / (So^4 Z^2)
= [Pi / (So^2 + 1)^2]
{[1 / Z^2][1 - (4 / (So^2 + 1)^2)] + [4 / (So^2 - 1)^2]}^0.5

The expression for Z^2 is a rapidly convergent itterative relationship. Start at:
Z^2 = 1.
Then for any particular value of So^2 in the range 3 to 6 it should be easy to determine Z^2.

To avoid poles in this function for Z^2 the denominator terms independent of Y and Z should satisfy:
- [4 / (So^2 - 1)^2] + [Pi^2 / 4][1 - (1 / (So^2 + 1)^4)] > 0

Hence:
[Pi^2 / 4][1 - (1 / (So^2 + 1)^4)] > [4 / (So^2 - 1)^2]
or
[Pi^2 / 4] > [4 / (So^2 - 1)^2] / [1 - (1 / (So^2 + 1)^4)]

Hence the condition simplifies to:
[Pi^2 / 4] > [4 / (So^2 - 1)^2]
or
(So^2 - 1)^2 > (16 / Pi^2) / [1 - (1 / (So^2 + 1)^4)]
or
So^2 - 1 > (4 / Pi) / [1 - (1 / (So^2 + 1)^4)]
or
So^2 > 1 + (4 / Pi) / [1 - (1 / (So^2 + 1)^4)] = 1 + 1.273 / [1 - (1 / (2.273)^4)]
= 1 + (1.273 / 0.9625)
= 2.3226

Thus for any particular So value in the range of interest we can readily itteratively compute the corresponding Z^2 value. We can use this methodology to find Z as a function of So and hence (1 / (Alpha Nt) as a function of So.

We can find the So value corresponding to the relative minimum in the plot of (1 / Alpha Nt) versus So. The Alpha value at this relative minimum is the Fine Structure constant.

The formula for Z^2 is used to calculate:
Nr^2 = Z^2 - [(So^2 - 1) / (So^2 + 1)]^2
and
[1 / (Alpha Nt)]
= (Pi / 4)[(So^2 - So + 1) / (So^2 + 1)][Z^2]^0.5

Nr = Np / Nt
where Np and Nt are both integers with no common factors

Subject to the new calculation of Nr^2 my present best estimate of these integers is:
Nr = Np / Nt = (222 / 305)
 

A preliminary BASIC program solution indicates that there is a broad relative minimum in [1 / (Alpha Nt)] located at So = 1.9334. There are two mathematically valid solutions close to that minimum.

Solution #1 located at So = ______ is Nt = 316, Np = 231 which gives:
[1 / Alpha] = 136.8048

Solution #2 located at _________ is Nt = 317, Np = 232 which gives:
[1 / Alpha] = 137.29359

Note that in both solutions either Np or Nt is a prime number, which meets one of the important spheromak existence conditions.

During an experimental measurement of the Fine Structure Constant Alpha both solutions will contribute to the observed result. The exact fraction of contribution will be determined by the experimental environment. A key issue is the ratio of magnetic field energy density to electric field energy density at the spheromak wall.

Based on experience with plasmas I expected that the energy minimum would be at:
So^2 ~ 4.1.
Instead the present theory points to The energy minimum being at:
So^2 = (1.9334)^2 = 3.738
Moving the energy minimum requires a change in the electric field energy distribution function. This issue needs more study.

 
 
 

We do not know the exact So value of the relative minimum in the plot of:
[1 / (Alpha Nt)] versus So.
That value of So must be found numerically. We could attempt to find it by successive manual itterations but that work would be very tedious.

Nr^2 = Z^2 - [(So^2 - 1)^2 / (So^2 + 1)^2]
 

EVALUATION SEQUENCE:

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

Check that with Nt = 305 the corresponding Alpha value satisfies:
[1 / Alpha] ~ 137.03

Do not proceed beyond this point until the computer program correctly calculalates Nr^2, Z^2 and [1 / Alpha].

Increment through So values in 0.1 increments from So = 1.5 to So = 3. Note that some of the function terms will head to infinity at smaller So values. The relative minimum in (Alpha Nt) should occur near
So = 2.026.

At each So value calculate and print:
So, Z^2, Z, Nr, Alpha.

A graph of [1 / (Alpha Nt)] versus So has a relative minimum, near So = 2.0

[ 1 / (Alpha Nt)] is given by the formula:
[1 / (Alpha Nt)] = (Pi / 4) [(So^2 - So + 1) / (So^2 + 1)] Z

I recommend checking the functionality of the computer program at So = 2.0000 at which point the various function terms can easily be verified with a desk calculator.

Find the value of So at the relative minimum in the plot of:
[1 / (Alpha Nt)] versus So for So in the range 1.5 to 3.0

Find the exact value of [1 / (Alpha Nt)] at that value of So.

Calculate the corresponding Z^2 value.

Calculate the corresponding value of Nr^2 using the formula:
Nr^2 = Z^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 integers with no common factors that precisely satisfy the equation:
Nr = (Np / Nt)
Note that the present data indicates that Nt is 316 or 317 and Np is 232 or 233.

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

It is this ratio of integers coincident with real numbers which are functions of Pi that causes quantization of energy.

Use the formula:
[1 / Alpha] = [Nt] [1 / (Alpha Nt)]
to determine the calculated value of:
[1 / Alpha].

An issue that needs more study is does the process of performing the multiplication:
Nt X [1 /(Alpha Nt)]
have the effect of moving the energy minimum up to So^2 ~ 4.1?

What is the dependence of Nt on So? A dependence of the form:
Nt = Nto /(So^2 + 1)
would likely match spheromak theory to experimently observed plasma spheromaks. Possibly the near field approximation of the electric field energy distribution needs to be fixed.

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

 
 
 
 

OLD WORK
The following work is old and is based on a slightly incorrect boundary condition that is much easier to solve with a desk calculator.
 

BOUNDARY CONDITION AT R = Rs, H = 0:
(Epsilon / 2) (Fec Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Epsilon / 2) Eros^2 + (Bpos^2 / 2 Mu)
where:
Eric = Sac / Epsilon
and
Eros = (Sas / Epsilon) + (Fec Rc / Rs)

where from above Sas is given by:
Sas^2 = {Qs / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr So^2]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

This boundary condition can be simplified by noting that:
[(Epsilon / 2) Eros^2 + (Bpos^2 / 2 Mu)]
is the total field energy density at R = Rs, H = 0 outside the spheromak wall. Outside the spheromak wall the total field energy density is given by:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2.

At R = Rs, H = 0 the field energy density outside the spheromak wall is:
U = Uo [Ro^2 / (Ro^2 + Rs^2)]^2
= (Bpo^2 / 2 Mu) [Ro^2 / (Ro^2 + Rs^2)]^2
= (Bpo^2 / 2 Mu)[Ro^2 / (Ro^2 + Rs^2)]^2
= (Bpo^2 / 2 Mu)[Rs Rc / (Rs Rc + Rs^2)]^2
= (Bpo^2 / 2 Mu)[Rc / (Rc + Rs)]^2

Thus the boundary condition at R = Rs, H = 0 becomes:
(Epsilon / 2) (Fec Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(Epsilon / 2) (Sac Rc / Epsilon Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(1 / 2 Epsilon) (Sac Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(Sac Rc / Rs)^2 + (Epsilon / Mu) (Btic Rc / Rs)^2 = (Epsilon) (Bpo^2 / Mu) [Rc / (Rc + Rs)]^2
or
(Sac Rc / Rs)^2 + (Btic^2 / C^2 Mu^2) (Rc / Rs)^2 = (Bpo^2 / C^2 Mu^2) [Rc / (Rc + Rs)]^2
or
(Btic^2 / C^2 Mu^2) (Rc / Rs)^2 = (Bpo^2 / C^2 Mu^2) [Rc / (Rc + Rs)]^2 - (Sac Rc / Rs)^2
or
(Btic^2 / C^2 Mu^2) = (Bpo^2 / C^2 Mu^2) [Rs / (Rc + Rs)]^2 - (Sac Rs / Rs)^2

= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 - (Sac)^2

Recall that:
Btic^2 = (Mu Nt Ih / 2 Pi Rc)^2
= (Mu Nt / 2 Pi Rc)^2 (Ih)^2
= (Mu Nt Qs C / 2 Pi Rc)^2 (1 / Pi^2)[1 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}]
= (Mu Nt Qs C / 2 Pi Rc)^2 [1 / (Pi^2 Nt^2 Rc^2)] [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]
= (Mu Qs C / 2 Pi^2 Rc^2)^2 [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]

Thus the boundary condition at R = Rs, H = 0 becomes:
(Mu Qs C / 2 Pi^2 Rc^2 C Mu)^2 [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]
= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 - (Sac)^2
= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2
- {Qs / [Pi^2 Rc^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

 

Multiply through by (Pi^2 / Qs)^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}.
Then this outer rim boundary condition becomes:
((1) / 2 Rc^2)^2 = (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 (Pi^2 / Qs)^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [Rc^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
((1) / 2)^2 = [(Bpo^2 Rc^4 Pi^4 / Qs^2 C^2 Mu^2)] [So^2 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}

Recall that from energy density function matching:
Bpo^2 = [Mu^2 C^2][Qs / (4 Pi Ro^2)]^2

Hence:
(Bpo^2 Rc^4 Pi^4 / Qs^2 C^2 Mu^2) = [Mu^2 C^2][Qs / (4 Pi Ro^2)]^2 (Rc^4 Pi^4 / Qs^2 C^2 Mu^2)
= Pi^2 [Rc^2 / (4 Ro^2)]^2
= Pi^2 [1 / (4 So^2)]^2

Hence the boundary condition at R = Rs, H = 0 becomes:
((1) / 2)^2 = [Pi / (4 So^2)]^2 [So^2 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
((1)^2 / 4) = [Pi^2 / 16] [1 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
((1) / 4) = [Nr^2 Pi^2 / 16] + [Pi^2 / 16] {[(So^2 - 1) / (So^2 + 1)]^2} - [1 / 4] - {Nr / [(So^2 - 1)]}^2

Collecting Nr^2 terms gives:
[Nr^2 Pi^2 / 16] - {Nr / [(So^2 - 1)]}^2 = (1 / 4) - [Pi^2 / 16] {[(So^2 - 1) /(So^2 + 1)]^2} + [1 / 4]
or
Nr^2 [Pi^2 - Nr^2 (16 / (So^2 - 1)^2)] = 4 (2) - {Pi^2 [(So^2 - 1) /(So^2 + 1)]^2}
or
Nr^2 = (4(2) - {Pi^2 [(So^2 - 1) / (So^2 + 1)]^2})
/ [Pi^2 - (16 / (So^2 - 1)^2)]

This is the boundary condition at R = Rs, H = 0.

Nr^2 = [+ 8 - {Pi^2 [(So^2 - 1) / (So^2 + 1)]^2}]
/ [Pi^2 - (16 / (So^2 - 1)^2)]
or
Nr^2 = (((+ 8) / Pi^2) - {[(So^2 - 1) / (So^2 + 1)]^2})
/ [1 - (4 / Pi (So^2 - 1))^2]

By comparison the boundary condition at R= Rc, H = 0 gave:
Nr^2 = [(+ 8) / Pi^2 - [(So^2 - 1) / (So^2 + 1)]^2} / {1 - (16 / [Pi (So^2 - 1)]^2)}

These two boundary condition formulae are identical and are referred to herein as the common boundary condition.
 

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 (at So ~ 2.026).

The existence condition is:
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

= F I
where:
F = [ Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
and
I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

 

The common boundary condition is:
Nr^2 = (((16 Ni^2 - 16 Ni + 8) / Pi^2) - {[(So^2 - 1) / (So^2 + 1)]^2})
/ [1 - (4 / Pi (So^2 - 1))^2]

 

RANGE RESTRICTION ON So:
The common boundary condition denominator requires that to keep Nr^2 finite:
1 - [4 / (Pi (So^2 - 1))]^2 > 0
or
1 > [4 / (Pi (So^2 - 1))]
or
(So^2 - 1) > (4 / Pi)
or
So^2 > (4 / Pi) + 1

The common boundary condition numerator requires that in order to keep Nr^2 > 0:
[+ {(+ 8) / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] > 0
Compliance with this condition requires that:
So^2 < 19
 

SOLUTION FOR SPHEROMAK OPERATING POINT:
To find the spheromak operating point choose So^2, calculate Nr^2, calculate Nr and then calculate RHS of existence condition. Plot the existence condition RHS value versus So^2. Repeat this process numerous times to accurately determine the So^2 value that gives RHS the closest approach to LHS = (Pi / 2). Use the corresponding value of Nr to find integers Np and Nt that satisfy:
.

Nr = Np / Nt.
 

In the case of an electron it may help to initially choose Ni so that Ett ~ particle mass. That should minimize the number of calculations.
 

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

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

I = Integral from Z = 1 to Z = 2.3 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 4.0106 / ([(1.3)(Z - 1)]^0.5 {[16.08516] + 0.4225}^0.5)
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
(2.3)^3 dZ 4.0106 / ([(2.3 - Z)(1.3)]^0.5 {[16.08516] [2.3]^2 + 0.4225}^0.5 [2.3 (1.5) + 2.3]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 4.0106 / 4.6324((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
dZ 4.0106 / ([(2.3 - Z)]^0.5 {10.51749})
 
= 0.17315  
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ 0.07626
= 0.24941 + 1.5667
= 1.8161

(F I) = 0.6936265 (1.8161)
= 1.2597
 
 

TRY So^2 = 2.5
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.18367}] / {1 - 0.720507}
= 0.62690 / 0.27949
= 2.243

Nr = 1.49766

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(1.49766)(2.5) / [(2.243 )(12.25)) + (2.25)]^0.5
= 3.74415 / 5.45222
= 0.68672

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

I = Integral from Z = 1 to Z = 2.5 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
Integral from Z = 2.49 to Z = 2.5 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 1.49766 / ([(1.5)(Z - 1)]^0.5 {[2.243] + 0.5625}^0.5)
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.49 to Z = 2.5 of:
(2.5)^3 dZ 1.49766 / ([(2.5 - Z)(1.5)]^0.5 {[2.243] [2.5]^2 + 0.5625}^0.5 [2.5 (1.5) + 2.5]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 1.49766 / 2.0514((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.49 to Z = 2.5 of:
dZ 1.49766 / ([(2.5 - Z)]^0.5 {4.6767})
 
= 0.14601  
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ 0.064047
= 0.21005 + 1.3987
= 1.6087

(F I) = 0.6867 (1.6087)
= 1.1047
 
 

TRY So^2 = 4.0
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.3600}] / {1 - 0.180126}
= 0.45057 / 0.819874
= .54956

Nr = 0.7413

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(.7413)(4.0) / [(.54956)(25)) + (9)]^0.5
= 2.96529 / 4.7685
= 0.62185

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

= Integral from Z = 1 to Z = 4.0 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 0.7413 / ([(3.0)(Z - 1)]^0.5 {[0.54956] + 2.25}^0.5)
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
(4.0)^3 dZ 0.7413 / ([(4.0 - Z)(3.0)]^0.5 {[0.54956] [4.0]^2 + 2.25}^0.5 [4.0 (3.0) + 4.0]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 0.7413 / 2.898047 ((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
dZ 0.7413 / ([(4.0 - Z)]^0.5 {5.7557})
 
= 0.05115  
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ 0.025758
= 0.07690 + 0.6605
= 0.7374

(F I) = 0.62185 (0.7374)
= 0.45856
 
 

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 = 4.1 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 August 14, 2018.

Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links