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ELECTROMAGNETIC SPHEROMAK

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

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

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.

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" which is the main subject of this web page. These two equations lead to the Planck Constant which is the subject 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.
 

There is an energy density function outside the spheromak wall of the form:
Up = Upo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 and there is an energy density function inside the spheromak wall of the form:
Ut = Uto (Ro / R)^2

Each of these functions is a combination of electric and magnetic fields.
 

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, like the icing on a doughnut. 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 spirally circulate on a closed spiral path within the spheromak wall. The spheromak wall position causes a system total energy minimum.

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 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 usually only of one type and 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 equations that determine the shape of spheromaks.
 

CHARGE SHEET 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 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 detailed boundary conditions at different points on the spheromak wall. This 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 Hf.
 

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 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 velocity components Vt and Vp.

5) Vt^2 + Vp^2 = C^2

6) At steady state conditions Vt and Vp are each constant along the charge hose path;

7) Vt = [(Rs - Rc) / 2] [dPhi / dT]
= velocity component that forms a toroidal magnetic field inside the spheromak wall;

8) Vp = R (dTheta / dT)
= velocity component that forms a poloidal magnetic field outside the spheromak wall. Since Vp = constant over the path length at small R the quantity:
[dTheta / dT]
is larger than at large R.

;

9) Thus:
[(Rs - Rc) / 2]^2 [dPhi / dT]^2 + R^2 [dTheta / dT]^2 = C^2

10)
C / Lh = Fh = 1 / Th
Vp / Lp = Fp = 1 / Tp
Lp = Pi (Rs + Rc)
Vt / Lt = Ft = 1 / Tp
Lt = Pi (Rs - Rc)
Nt Tt = Np Tp = Th
Nt / Ft = Np / Fp = 1 / Fh = Th
Ft = Nt Fh
Fp = Np Fh
Ft / Fp = Nt / Np
Nt Lt / Vt = Np Lp / Vp = 1 / Fh = Lh / C

11) Nt Lt / Vt = Np Lp / Vp = 1 / Fh = Lh / C
[Nt Pi (Rs - Rc) / Vt] = [Np Pi (Rs + Rc) / Vp] = [Lh / C]
Hence:
Vt = C Nt Pi ( Rs - Rc) / Lh
and
Vp = C Np Pi (Rs + Rc) / Lh

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

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

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

15) Recall that:
C^2 = Vp^2 + Vt^2
= {R [dTheta / dT]}^2 + {[(Rs - Rc) / 2] [dPhi / dT]}^2
= {Fh Lh}^2
= Fh^2 (Pi)^2 [Np^2 ((Rs + Rc))^2 + Nt^2 ((Rs - Rc))^2]

16) Comparing terms gives:
{R [dTheta / dT]}^2 = Fh^2 (Pi)^2 [Np^2 ((Rs + Rc))^2]
and
{[(Rs - Rc) / 2] [dPhi / dT]}^2 = Fh^2 (Pi)^2 Nt^2 ((Rs - Rc))^2

17) Simplifying these two equations gives:
R [dTheta / dT] = Fh Pi Np (Rs + Rc)
and
[1 / 2] [dPhi / dT] = Fh Pi Nt

18) Since:
Fh = C / Lh
R [dTheta / dT] = C Pi Np (Rs + Rc) / Lh
and
[1 / 2] [dPhi / dT] = C Pi Nt / Lh

19) Hence:
[dTheta / dT] / [dPhi / dT] = [C Pi Np (Rs + Rc) / Lh R] / [2 C Pi Nt / Lh]
= [dTheta / dPhi] = [Np (Rs + Rc) / 2 R Nt]

20) This equation is required for determining the spheromak surface charge concentration and the symmetry axis electric and magnetic fields.

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 a cylindrically radial electric field inside the spheromak wall and an electric field outside the spheromak wall which is spherically radial in the far field;

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

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

26) 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 = cylindrically radial electric field energy density inside the spheromak wall;
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 + R^2 + H^2)]^2

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

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

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 fields and that circulates within the spheromak wall with a charactreristic frequency:
Fh = C / Lh
causing the static magnetic fields. The charge circulation pattern is described by four parameters: Np, Nt, So 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 where:
Rs = maximum radius from spheromak symmetry axis to spheromak wall;
Rc = minimum radius from spheromak symmetry axis to spheromak wall;
Ro^2 = Rs Rc where Ro is the nominal spheromak radius.
 

SPHEROMAK STATIC FIELD ENERGY:
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]}
 
= Uo Pi^2 Ro^3 {4 So (So^2 - So + 1) / (So^2 + 1)^2}
Note that this formula applies to all spheromaks, not just minimum energy spheromaks.
 

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:
The surface charge dQs contained on the elemental strip of constant R is:
dQs = [Qs / Lh] Nt L
= [Qs / Lh]{(Nt R dTheta)^2 + [(Rs - Rc) Nt dPhi/ 2]^2}^0.5
 
= [Qs / Lh]{(Nt R[Np (Rs + Rc) / 2 R Nt]dPhi)^2
+ [(Rs - Rc) Nt dPhi/ 2]^2}^0.5
 
= [Qs dPhi / 2 Lh]{([Np (Rs + Rc)])^2 + [(Rs - Rc) Nt]^2}^0.5
 
= [Qs dPhi / 2 Lh][Lh / Pi]
 
= [Qs dPhi / 2 Pi]

The corresponding element of area dA is:
dA = 2 Pi R (Rs - Rc) dPhi / 2

Hence the charge / unit area on the spheromak wall is:
dQs / dA = [Qs dPhi / 2 Pi] / Pi R (Rs - Rc) dPhi
[Qs / 2 Pi^2 R (Rs - Rc)]

At R = Rc the charge per unit area Sac is given by:
Sac = [Qs / 2 Pi^2 Rc (Rs - Rc)]
 
= [Qs So / 2 Pi^2 Ro (Rs - Rc)]
 
= [Qs So^2 / 2 Pi^2 Ro^2 (So^2 - 1)]

In general the surface charge density Sa is given by:
Sa = Sac Rc / R
= Sac Ro / R So
= [Qs So^2 / 2 Pi^2 Ro^2 (So^2 - 1)][Ro / R So]
= [Qs So / 2 Pi^2 Ro R (So^2 - 1)]

Rearranging this equation gives:
Sac = Qs / {2 Rc Pi^2 [Rs - Rc]}
= Qs / {2 (Ro / So) Pi^2 [Ro So - (Ro / So)]}
= {Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]}
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.

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:
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 + H^2)]^2

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

At R =0, H = 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 + R^2 + H^2)]^2
the magnetic field at R = 0, H = 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 + 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]}
 
= Uo Pi^2 Ro^3 {4 So (So^2 - So + 1) / (So^2 + 1)^2}
Note that this formula applies to all spheromaks, not just minimum energy spheromaks.
 

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

SPHEROMAK ENERGY CONTENT:
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]}
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

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:
(Eric + Btc)(Rc / R)
and the field energy density is:
[(Epsilono / 2) Eric^2 (Rc / R)^2 + Btc^2 / 2 Muo)(Rc / R)^2
 
= [(Epsilono / 2) Eric^2 (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)]

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

This is a very important result which indicates that in reality for practical spheromaks So ~ 2.

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, H = 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)]

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.

Note this result suggests that at R = Rs the internal electric field energy density is matched by the external poloidal magnetic field energy density.

At R = Rs the internal electric field is: (Sac Rc / Epsilono Rs)

Hence at R = Rs the poloidal magnetic field energy density is given by:
(Epsilono / 2)[Sac Rc / Epsilono Rs]^2 = Bps^2 / 2 Muo
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:
Electric field energy + 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 H, Ue is approximately:
Ue ~ U = Uo Ro^4 / (Ro^2 + R^2 + H^2)^2

In general outside the spheromak wall the electric field energy density Ueo 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

At R = 0, H = 0:
Ueo = 0
as expected from spheromak symmetry.

At R = Rc, H = 0:
Ueo = Ueoc
= Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2) / (Ro^2 + Rc^2)]^N

Similarly at R = 0, H = Rc:
Ueo =Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2) / (Ro^2 + Rc^2)]^N
 
where N is chosen to match spheromak electric field geometry.

Note that Ueoc reflects the reality that R = Rc and H = 0 there is a radial electric field Eroc pointing toward the spheromak axis of symmetry as well as a radial electric field pointing away from the spheromak axis of symmetry.

The ratio of these two radial electric field vectors expressed via the parameter F determines the spheromak shape factor:
So^2 = (Rs / Rc)

The corresponding magnetic field energy density outside the spheromak wall is given by:
Umo = U - Ueo
= Uo [Ro^2 / (Ro^2 + R^2 + H^2]^2
{[(Ro^2 + R^2 + H^2)^N - (R^2 + H^2)^N] / (Ro^2 + R^2 + H^2)]^N}

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

and at large R and/or large H:
Umo = Uo N Ro^6 / (Ro^2 + R^2 + H^2)^3

Note that in general outside the spheromak wall:
Ueo + Umo = Uo [Ro^2 / (Ro^2 + R^2 + H^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.
 

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

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

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

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

Note that this formula applies to all spheromaks, not just minimum energy spheromaks.
 

PLANCK CONSTANT:
Recall that for a general spheromak:
[1 / Ro] = [(Pi Fh) / (So C)] [(Np)^2 (So^2 + 1)^2 + (Nt)^2 (So^2 - 1)^2]^0.5

Recall that the total static field energy contained in a spheromak is given by:
Ett = [(Muo C^2 Qs^2) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}

Combining these two expressions gives the total spheromak static field energy as:
Ett = [(Muo C^2 Qs^2) / (32)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}
{[(Pi Fh) / (So C)] [(Np)^2 (So^2 + 1)^2 + (Nt)^2 (So^2 - 1)^2]^0.5}
 
which is in the form:
Ett = h Fh where:
h = [(Muo C Qs^2 Pi) / 8]{(So^2 - So + 1) / (So^2 + 1)^2}
{[(Np)^2 (So^2 + 1)^2 + (Nt)^2 (So^2 - 1))^2]^0.5}

and h is termed the Planck constant.
 

FINE STRUCTURE CONSTANT:
In spheromak analysis it is convenient to make the substitution:
Muo C Qs^2 = 2 Alpha h
where Alpha is known as the fine structure constant.

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

Making that substitution gives:
Ett = [(C 2 Alpha h) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}
or
Ett = Alpha h (C / 4 Ro){So (So^2 - So + 1) / (So^2 + 1)^2}

Note that in the region of interest (So < 2) the expression in So decreases slowly with increasing So.

Recall that:
[C / 4 Ro]
= [(Pi Fh) / (4 So)] [(Np)^2 (So^2 + 1)^2 + (Nt)^2 (So^2 - 1))^2]^0.5
= [(Pi Fh) / (4 So)] [Z]

which increases with increasing So. Hence for stability of (1 / Alpha) this function must be stable at the spheromak operating point. This stability arises out of the relationship between Uto (magnetic field energy density at R = Ro) and Upo (the theoretical maximum magnetic field energy density required for the magnetic field to match the electric field.)

This field relationship has the effect of making Ett precisely proportional to frequency Fh and making Alpha constant.

In essence Alpha determines Uto

Combining these two functions gives:
Ett = Alpha h (C / 4 Ro){So (So^2 - So + 1) / (So^2 + 1)^2}
 
= Alpha h [Pi Fh No / 4 So][(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5
{So (So^2 - So + 1) / (So^2 + 1)^2}
 
= Alpha h [Pi Fh No / 4][(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5
{(So^2 - So + 1) / (So^2 + 1)^2}

However:
Ett = h Fh
which gives:
(1 / Alpha) = [Pi / 4][No][(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5
{(So^2 - So + 1) / (So^2 + 1)^2}
 
The magnitude of (1 / Alpha) determines the Planck constant. The ratio:
Nr = ( Np / Nt)
causes quantized spheromak energy changes that stabilize (1 / Alpha).

Thus:
(1 / Alpha) = (Pi / 4)[No][(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5
{(So^2 - So + 1) / (So^2 + 1)^2}
 
= (Pi / 4) Z {(So^2 - So + 1) / (So^2 + 1)^2}
where:
Z = No [(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5
is almost constant.

The expression for (1 / Alpha) is of fundamental importance in quantum physics. Note that Nt, Np and No are positive integers and
Nr = (Np / Nt)
is a ratio of positive integers, which together cause quantization of energy in spheromaks.
 

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

To achieve:
Nr^2 = R^2
the boundary condition will take the form:
2 R^2 = 1 / [(Pi / 2)^2 - (F / R)^2]
or
2 (Pi/2)^2 R^2) - 2 F^2 = 1
or
F^2 = [2 (Pi R)^2 - 1] / 2
or
F = [(Pi R)^2 - (1 / 2)]^0.5

We need an explicit function for F to further understand this situation.
 

ALPHA STABILITY:
Over time a spheromak operating at a constant frequency will reconfigure itself until it reaches a stable energy minimum.

Recall that a general spheromak is described by:
(1 / Alpha) = [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)^2}
{No [(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5}

Hence:
d[(1 / Alpha)] = [Pi / 4] d{(So^2 - So + 1) / (So^2 + 1)}{No}{[(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5}
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)}
d{No [(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5}

 

d{No [(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5}
= (1 / 2 Z) dZ
= (1 / 2 Z) No {(Np / No)^2 2 (So^2 + 1) 2 So dSo + (Nt / No)^2 2 (So^2 - 1) 2 So dSo
+ dNo [(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]^0.5

d{(So^2 - So + 1) / (So^2 + 1)}
 
= [(So^2 + 1)(2 So - 1) - (So^2 - So + 1) 2 So] dSo / (So^2 + 1)^2
 
= [2 So^3 - So^2 + 2 So - 1 - 2 So^3 + 2 So^2 - 2 So] dSo / (So^2 + 1)^2
 
= [So^2 - 1] dSo /(So^2 + 1)^2

Combining these equations gives:
d(1 / Alpha)
= [Pi / 4] d{(So^2 - So + 1) / (So^2 + 1)} {Z}
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)} dZ
CHECK FROM HERE ONWARDS  
= [Pi / 4] [So^2 - 1] {dSo /(So^2 + 1)^2} [Z]
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)}(1 / 2 Z)
{Np^2 2 (So^2 + 1) 2 So dSo + Nt^2 2 (So^2 - 1) 2 So dSo
+ dNo {(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2}
 
= [Pi / 4] [So^2 - 1] {dSo /(So^2 + 1)^2} [Z]
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)}(1 / 2 Z)
{[4 So dSo][Np^2 (So^2 + 1) + Nt^2 (So^2 - 1)]
+ dNo {[(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]}
 
= 0

Multiply through by Z to get:
[Pi / 4] [So^2 - 1] {dSo /(So^2 + 1)^2} [Z]^2
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)}(1 / 2)
{[4 So dSo][Np^2 (So^2 + 1) + Nt^2 (So^2 - 1)]
+ dNo Z {[(Np / No)^2 (So^2 + 1)^2 + (Nt / No)^2 (So^2 - 1)^2]}
 
= 0

Note that:
At a stable spheromak minimum energy operating point:
M dNp = - dNt
where Np and Nt increment or decrement by integers, and M is a small integer, likely 2 or 3.

Thus:
[Pi / 4] [So^2 - 1] {dSo /(So^2 + 1)^2} [Z]^2
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)}(1 / 2)
{[4 So dSo][Np^2 (So^2 + 1) + Nt^2 (So^2 - 1)]
+ [2 Np (- dNt / M) (So^2 + 1)^2 + 2 Nt dNt (So^2 - 1)^2]}
 
= 0

Note that if So is wrong by dSo where dSo is positive then Nt decrements sufficiently to slightly over compensate for the zero error and then Np increments to further minimize the zero error. To achieve this relationship the optimum value of M is 3.
 

ALPHA MAGNITUDE:
Consider the term:
Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2

Introduce factor M = 3 to improve spheromak stability at its operating point.

Choose nominal turn count No such that:
[Np^2 (So^2 + 1)^2 / No^2 So^2] = M [No^2 So^2 / Nt^2 (So^2 - 1)^2]

Then:
Np^2 Nt^2 (So^2 + 1)^2 (So^2 - 1)^2 = M [No^2 So^2]^2
or
Np Nt = M^0.5 [No^2 So^2] / [(So^2 + 1) (So^2 - 1)]

Np = M^0.5 No So / (So^2 + 1)
and
Nt = No So / (So^2 - 1)

Then: Np / Nt = M^0.5 (So^2 - 1) / (So^2 + 1)
and
Nr^2 = M [Np / Nt]^2
= M [(So^2 - 1) / (So^2 + 1)]^2
= M R^2

which is the aforementioned companion equation.
 

d(1 / Alpha) / dSo is always greater than zero so to compensate at least one of d(1 / Alpha) / dNp and d(1 / Alpha) /dNt must be significantly negative.

Thus when Alpha is constant:
d(1 / Alpha)
= [Pi / 4] d{(So^2 - So + 1) / (So^2 + 1)} {[Nr^2 + R^2]^0.5}
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)^3} {{R / [Nr^2 + R^2]^0.5} [4 So] dSo + [dNt][Nr^2 + R^2]^0.5}  
= [Pi / 4] {[So^2 - 1] dSo /(So^2 + 1)^2} {[Nr^2 + R^2]^0.5}
+ [Pi / 4]{(So^2 - So + 1) / (So^2 + 1)^3} {{{R / [Nr^2 + R^2]^0.5} [4 So] dSo + + [dNt][Nr^2 + R^2]^0.5}
= 0

 

Cancel out [Pi / 4] dSo / (So^2 + 1)^2 to get:
{[So^2 - 1]} {[Nr^2 + R^2]^0.5}
+ {(So^2 - So + 1) / (So^2 + 1)} {{R / [Nr^2 + R^2]^0.5} [4 So]
+ [dNt / dSo] [Nr^2 + R^2]^0.5} = 0
or
{[So^2 - 1]} {[Nr^2 + R^2]}
+ {(So^2 - So + 1) / (So^2 + 1)} {{R} [4 So] + [dNt / dSo] [Nr^2 + R^2]}
= 0
or
{[So^2 - 1]}
+ {(So^2 - So + 1) / (So^2 + 1)} {{R /[Nr^2 + R^2]} [4 So] + [dNt / dSo]}
= 0
or
[(So^2 - 1)(So^2 + 1) / (So)(So^2 - So + 1)]
+ {{R /[Nr^2 + R^2]} [4] + [(1 / So)(dNt / dSo)]}
= 0

By inspection we can see that the stability of (1 / Alpha) relies on Nt decreasing as So increases.

Recall that:
[1 / Ro]^2 = [(Pi Fh) / (So C)]^2 [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]
where (1 / Ro) is proportional to the spheromak energy.

Differentiate this expression with respect to So and set the derivative to zero to find the So value at the resulting energy minimum:
d{[1 / Ro]^2} / dSo
= 2 [(Pi Fh) / (So C)][- (Pi Fh) / (So^2 C)][(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]
+ [(Pi Fh) / (So C)]^2 [2 (Np (So^2 + 1)) 2 So + 2 (Nt (So^2 - 1)) 2 So] = 0

Cancelling common terms gives:
[- 1 / (So)][(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]
+ [(Np (So^2 + 1)) 2 So + (Nt (So^2 - 1)) 2 So] = 0

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

Substitute this minimum energy relationship back into the formula:
[1 / Ro]^2 = [(Pi Fh) / (So C)]^2 [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]
to get the minimum energy condition with constant Np and Nt:
[1 / Ro]^2 = [(Pi Fh) / (So C)]^2 2 So^2 [(Np (So^2 + 1)) + (Nt (So^2 - 1))]
 
= [(Pi Fh) / (C)]^2 2 [(Np (So^2 + 1)) + (Nt (So^2 - 1))]

Recall that at the stable spheromak energy minimum:
[Np / Nt]^2 = 3 (So^2- 1)^2 / (So^2 + 1)^2

or
Np / Nt = 3^0.5 (So^2- 1) / (So^2 + 1)

Hence:
[1 / Ro}^2 = [(Pi Fh) / (C)]^2 2 [(Np (So^2 + 1)) + (Nt (So^2 - 1))]
 
= [(Pi Fh) / C]^2 2 {[3^0.5 Nt (So^2- 1)] +[(Nt (So^2 - 1))]}
 
= [(Pi Fh) / C]^2 2 {Nt (So^2 - 1)}{[3^0.5] + 1}

Thus:
[1 / Ro] = [(Pi Fh) / C] {2 Nt (So^2 - 1)[(3^0.5) + 1]}}^0.5

This is an important relationship for enabling numerical analysis of a spheromak at its energy minimum.

Recall that:
[1 / Ro]^2 = [(Pi Fh) / (So C)]^2 [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]
Differentiate with respect to Np and Nt to get:
d{[1 / Ro]^2}
= [(Pi Fh) / (So C)]^2 [(2 Np dNp) (So^2 + 1)^2 + (2 Nt dNt) (So^2 - 1))^2] = 0
or
[(Np dNp) (So^2 + 1)^2 + (Nt dNt) (So^2 - 1))^2] = 0

Recall that at maximum spheromak stability:
2 dNp = - dNt

FIX

Combine these two equations to get:
[Np {(- dNt / 2) (So^2 + 1)^2 + (Nt dNt) (So^2 - 1)^2] = 0
or
[Np {- 1 /2}(So^2 + 1)^2 + (Nt) (So^2 - 1)^2] = 0
or
[Np / Nt] = 2 (So^2 - 1)^2 / (So^2 + 1)^2
This relationship gives the relationship between:
Nr = [Np / Nt] and So.

For numerical analysis purposes this relationship can be rearranged to give So as a function of Nr.

Nr = [Np / Nt]
= 3^0.5 (So^2 - 1) / (So^2 + 1)
or
Nr (So^2 + 1) = 3^0.5 (So^2 - 1)
or
So^2 (3^0.5 - Nr) = (3^0.5 + Nr)
or
So^2 = (3^0.5 + Nr) / (3^0.5 - Nr)

Hence in numerical analysis of spheromaks a computer can choose a Np/ Nt pair, calculate Nr, then calculate So, then calculate (1 / Alpha) and then calculate F. Only a small fraction of the possible Np / Nt pairs will have the approximately correct (1 / Alpha) value and only a small fraction of the remaining Np / Nt pairs remaining will have the approximately correct F value. A desk top computer can do this se?quence of operations in a few seconds with 360,000 possible Np / Nt number combinations.
 

DETERMINATION OF N:
The index N and hence the electric field outside the spheromak wall at R = Rc, H = 0 and hence the spheromak shape factor So^2 can be established by examination of the electric field along the spheromak axis of symmetry.

The functional form of the electric field along the spheromak axis of symmetry is:
Eo = [2 Uo / Epsilono]^0.5 [Ro^2 / (Ro^2 + H^2)] [H^2 / (Ro^2 + H^2)]^(N / 2)

In the far field where H >> Ro this expression simplifies to:
Eo = [2 Uo / Epsilono]^0.5 [Ro^2 / (Ro^2 + H^2)]
~ [2 Uo Ro^4 / Epsilono]^0.5 / H^2
= Qs / (4 Pi Epsilono H^2

Comparing terms gives:
[2 Uo Ro^4 / Epsilono]^0.5 = [Qs / (4 Pi Epsilono)]
or
Uo Ro^4 = Qs^2 / (32 Pi^2 Epsilono)

Hence along the spheromak axis of symmetry the electric field Eo is given by:
Eo = [2 Uo / Epsilono]^0.5 [Ro^2 / (Ro^2 + H^2)] [H^2 / (Ro^2 + H^2)]^(N / 2)
 
= [Qs / (4 Pi Epsilono)] [1 / (Ro^2 + H^2)] [H^2 / (Ro^2 + H^2)]^(N / 2)

At either R = Rc, H = 0 or at R= 0, H = Rc the electric field magnitude is given by:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] [Ro^2 / (Ro^2 + Rc^2)] [Rc^2 / (Ro^2 + Rc^2)]^(N / 2)

Make the substitution:
So = Ro / Rc to get:
 
Eoc
=[Qs / (4 Pi Epsilono Ro^2)] [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)

This expression must be matched to determine N. Note that N must satisfy:
N > 0
to provide the appropriate function response in the far field.
 

The strategy to find N is to evaluate Eo at R = 0, H = Rc by another means.

We can determine Eo at R = 0, H = Rc and hence determine N by integrating all the symmetry axis electric field contributions of the distributed charge on the spheromak wall.

This integration to find Eo and hence N needs to be done from Phi = 0 to Phi = Pi for the top half of the spheromak and from Phi = 0 to Phi = - Pi for the bottom half of the spheromak.
 

Consider an elemental ring of charge on the spheromak wall. The circumference of this ring is:
2 Pi R

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 H = 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, H = 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, H = 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

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
 

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)
 

The index N can be evaluated theoretically by numerically doing the afore described integration or can be evaluated from experimental measurements, or both.

The electric field at R = Rc, H = 0 determines the electric field ratio parameter F which in turn determines the spheromak shape parameter So.
 

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 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 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)]^2}.
and then calculate [1 / Alpha]^2.

The numbers Np and Nt must be integers with no common factors. The correct Np and Nt values are established by using a computer to increment through all the integer value possibilities. For each reasonably possible integer Nt value increment through possible Np values.
Is there an So value that leads to (1 / Alpha) = 137.035999? Do Np and Nt have any common factors?

Are there adjacent solutions allowing incrementation or decrementation of Np and Nt for stability maintenance?

Once Np, Nt and So are determined calculate and print (1 / Alpha)^2.

The spheromak stability condition arises from a rare coincidence between the rational number Np / Nt and a function of the irrational number Pi. Remember that this condition is only strictly valid for an isolated spheromak in a vacuum far from any enclosure walls.

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;
 

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 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 |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]}

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

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.
 

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.
 

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 at every point on the spheromak wall.
 

THE SPHEROMAK BOUNDARY CONDITION:

We have the following constraints:

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= Rc, H = 0 the total field energy density is given by:
Uoc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2
= Uo [So^2 / (So^2 + 1)]^2

At the center of the spheromak where the field energy is entirely magnetic:
Uo = Muo C^2 Qs^2 / 32 Pi^2 Ro^4

Outside the spheromak wall the electric field energy density is given by:
Ueo = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 [(R^2 + H^2)^N] / [(Ro^2 + R^2 + H^2)^N]

At R = Rc, H = 0 the electric field energy density is given by:
Ueoc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2) / (Ro^2 + Rc^2)]^N

Consider the special case of N = 1:
Ueoc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2) / (Ro^2 + Rc^2)]

= Uo [(Ro^4 Rc^2) / (Ro^2 + Rc^2)^3]

For the special case of N = 1 the radial electric field Eroc outside the spheromak wall at R = Rc, H = 0 is given by:
Ueoc = (Epsilono / 2) Eroc^2
or
Eroc = [2 Ueoc / Epsilono]^0.5
= [2 Uo / Epsilono]^0.5 {[(Ro^4 Rc^2) / (Ro^2 + Rc^2)^3]}^0.5
 
= [2 Uo / Epsilono]^0.5 {[(Ro^4 Rc^2 / Rc^6) / (So^2 + 1)^3]}^0.5
 
= [2 Uo / Epsilono]^0.5 [So^4 / (So^2 + 1)^3]^0.5
 
= [2 Muo C^2 Qs^2 / 32 Pi^2 Ro^4 Epsilono)]^0.5 [(So^4) / (So^2 + 1)^3]^0.5
 
= [2 Muo^2 C^4 Qs^2 / 32 Pi^2 Ro^4)]^0.5 [(So^4) / (So^2 + 1)^3]^0.5
 
[Muo C^2 Qs / 4 Pi Ro^2)] [(So^4) / (So^2 + 1)^3]^0.5

At R = Rc, H = 0 the field energy density outside the spheromak wall is balanced by the toroidal magnetic field energy density Umic and the radial electric field energy density Ueic.

From the spheromak winding parameters the toroidal magnetic field Bc at R = Rc, H = 0 is given by:
Bc = [(Muo Nt Qs Fh) / (2 Pi Rc)]
or
Umic = Bc^2 / 2 Muo
= ( 1 / 2 Muo)[(Muo Nt Qs Fh) / (2 Pi Rc)]^2 = (Muo / 8) [Nt Qs Fh) / (Pi Rc)]^2
= (Muo / 8) [Nt Qs Fh) / (Pi (Ro / So))]^2
= (Muo / 8) [Nt^2 Qs^2 Fh^2 So^2) / (Pi^2 Ro^2)]

The radial electric field inside the spheromak wall at R = Rc, H = 0 is:
Eric = F Sac / Epsilono.
where:
0 < F < 1
due to a non-zero Ueoc value.

However:
Ueic = (Epsilono / 2) (Eric)^2
where:
Eric = (Sac / Epsilono) - Eroc
= F (Sac / Epsilono)
 

DEFINITION OF F:

|Eric| + |Eroc| = Sac / Epsilono

|Eric| = F Sac / Epsilono

|Eroc| = (1 - F) Sac / Epsilono

Thus in general:
F = [(Sac / Epsilono) - Eroc] / (Sac / Epsilono)
= 1 - (Eroc Epsilono / Sac)

For the special case of N = 1:
F = 1 - (Eroc Epsilono / Sac)
 
= 1 - [Epsilono Muo C^2 Qs / Sac 4 Pi Ro^2)] [(So^4) / (So^2 + 1)^3]^0.5
 
= 1 - [Qs / Sac 4 Pi Ro^2)] [(So^4) / (So^2 + 1)^3]^0.5

However:
Sac = {Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]}
or
Qs / Sac Pi Ro^2 = 2 Pi (So^2 - 1) / So^2

Thus for the special case of N = 1:
F = 1 - [Qs / Sac 4 Pi Ro^2)] [(So^4) / (So^2 + 1)^3]^0.5
 
= 1 - [1 / 4] [2 Pi (So^2 - 1) / So^2] [(So^4) / (So^2 + 1)^3]^0.5
 
= 1 - [Pi (So^2 - 1) / 2] [1 / (So^2 + 1)^3]^0.5
= 1 - [Pi R / 2] [1 / (So^2 + 1)^0.5]

For the special case of N = 1:
[F / R] = [(So^2 + 1) / (So^2 - 1)] - [Pi (So^2 + 1) / 2] [1 / (So^2 + 1)^3]^0.5
 
= [(So^2 + 1) / (So^2 - 1)] - [Pi / 2] [1 / (So^2 + 1)^0.5]
= [1 / R] - [Pi / 2] [1 / (So^2 + 1)^0.5]

WE MAY BE ABLE TO TEST THESE FUNCTIONS FOR CONSISTENCY WITH GENERAL FORMULAE AND WITH EXPERIMENTALLY MEASURED ALPHA.

For the special case of N = 1 and So = 2:
F = 1 - [Pi (So^2 - 1) / 2] [1 / (So^2 + 1)^3]^0.5
 
= 1 - [Pi (3 / 2)] [ 1 / 125]^0.5
 
= 1 - [Pi (3 / 2)] [0.0894427191]
= 1 - 0.4214888834
= 0.57851

On the web page titled: SPHEROMAK SHAPE FACTOR the general expression for F is found to be:
F = (Pi / 2) [(So^2 - 1) / (So^2 + 1)]
{[(So^2 - 1)^2 / [(So^2 - 1)^2] + [4 So (So^2 - So + 1)}^0.5

Substitution of So = 2.0 into this equation gives:
F = (Pi / 2) [(3) / (5)] {[9 / [9] + [8 (3)}^0.5
= (Pi / 2) [3 / 5] {0.5222329679}
= 0.492192976

Thus the assumption that N = 1 yields a result for F in the right ball park but that result is not sufficiently accurate for our purposes. The F value computed in this manner has about a 15% error. That error is substantial in computations of So and (1 / Alpha.)

The spheromak analysis must be valid for a general F and a general N.

In general:
Ueic = (Epsilono / 2) (Eric)^2
where:
Eric = (Sac / Epsilono) - Eroc
= F (Sac / Epsilono)

Thus in general:
F = [(Sac / Epsilono) - Eroc] / (Sac / Epsilono)
= 1 - (Eroc Epsilono / Sac)

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

Thus:
F = 1 - {[Eroc Epsilono] [2 Ro^2 Pi^2 (So^2 - 1)] /(Qs So^2)}

Assume that Eroc = Eoc

Use the integration result:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] Ic

Hence:
F = 1 - {[Eroc Epsilono] [2 Ro^2 Pi^2 (So^2 - 1)] /(Qs So^2)}
 
= 1 - {[[Qs / (4 Pi Epsilono Ro^2)] Ic Epsilono] [2 Ro^2 Pi^2 (So^2 - 1)] /(Qs So^2)}
 
F = 1 - {[Ic / 2] [Pi (So^2 - 1)] /(So^2)}

Note that Ic comes from the sophisticated electric field integration.

CORRECTED VERSION STARTS HERE:
Ueic = (Epsilono / 2)(F Sac / Epsilono)^2
= (Epsilono / 2) {F Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)] Epsilono}^2
= (1 / 2 Epsilono){(F Qs So^2) / [2 Ro^2 Pi^2 (So^2 - 1)]}^2
= (Muo C^2 / 2) {(F^2 Qs^2 So^4) / [4 Ro^4 Pi^4 (So^2 - 1)^2

The energy balance equation becomes:
Uoc = Umic + Ueic
or
[(Muo C^2 Qs^2) / (32 Pi^2 Ro^4)] [So^4 /(So^2 + 1)^2]
= (Muo / 8) [Nt^2 Qs^2 Fh^2 So^2) / (Pi^2 Ro^2)]
+ (Muo C^2 / 2) {(F^2 Qs^2 So^4) / [4 Ro^4 Pi^4 (So^2 - 1)^2

This equation implicitly assumes that F is sufficiently small that Eroc makes up for the deficiency in central core magnetic field strength.

Cancelling (Muo Qs^2 So^2) / (Pi^2 Ro^2) from all terms gives:
[( C^2) / (32 Ro^2)] [So^2 /(So^2 + 1)^2]
= (1 / 8) [Nt^2 Fh^2)]
+ ( C^2 / 2) {(F^2 So^2) / [4 Ro^2 Pi^2 (So^2 - 1)^2

Multiply through by 32 Ro^2 / C^2 to get:
[So^2 /(So^2 + 1)^2]
= (4 Ro^2 / C^2) [Nt^2 Fh^2)]
+ ( 4) {(F^2 So^2) / [Pi^2 (So^2 - 1)^2
or
[(So^2) /(So^2 + 1)^2]
= [(4 Ro^2 Nt^2 Fh^2) / (C^2)]
+ [4 / Pi^2] [(F^2 So^2) / (So^2 - 1)^2]

which is referred to as the Spheromak Boundary Condition.

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

= Pi^2 (Ro / So)^2 Nt^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

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

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

Thus the spheromak boundary condition becomes:
[(So^2) /(So^2 + 1)^2]
= 4 [So^2 / Pi^2] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
+ [4 / Pi^2] [(F^2 So^2) / (So^2 - 1)^2]
or
[(1 /(So^2 + 1)^2]
= [4 / Pi^2] / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
+ [4 / Pi^2] [(F^2) / (So^2 - 1)^2]

Multiply through by {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2} to get:
Nr^2 + R^2 = [4 / Pi^2] + [4 / Pi^2] (F^2)[(Nr / R)^2 + 1]
= [4 / Pi^2]{1 + (F^2)[(Nr / R)^2 + 1]}

= [4 / Pi^2]{1 + (F^2 / R^2)[Nr^2 + R^2]}
or
(Nr^2 + R^2) [1 - (4 / Pi^2)(F^2 / R^2)] = (4 / Pi^2)
or
Nr^2 + R^2 = (4 / Pi^2) / [[1 - (4 / Pi^2)(F^2 / R^2)]
= 1 / [(Pi^2 / 4) - (F^2 / R^2)]
= 1 / [(Pi / 2)^2 - (F / R)^2]

This is the spheromak boundary condition.

Note that for real spheromaks to exist:
(F / R) < (Pi / 2)
or
(So^2 + 1) / (So^2 - 1) < (Pi / 2 F) or
[(So^2 - 1) / (So^2 + 1)] > [2 F / Pi]
or
(So^2 - 1) > [2 F / Pi](So^2 + 1)
or
So^2 [1 - (2 F / Pi)] > (2 F / Pi) + 1
or
So^2 > [(2 F / Pi) + 1] / [ 1 - (2 F / Pi)]

For F = 1:
So^2 > [(2 / Pi) + 1] / [ 1 - (2 / Pi)]
or
So^2 > [1.6366] / [0.36337]
= 4.5039

Hence for a spheromak to operate at So = 2.0, F must be significantly less than unity.

Hence Eroc > 0

Hence the radial electric field outside the spheromak wall is non-zero at R = Rc , H = 0. This issue was confirmed because the magnetic field in the spheromak core by itself is too weak to support a spheromak.
 

BOUNDARY CONDITION AT R= Rc, H = 0:
At R = Rc, 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 spheromak symmetry axis. Inside the spheromak wall the magnetic field is purely toroidal and the electric field is cylindrically radial pointing away from the spheromak symmetry axis. 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 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]

The spheromak electric field energy density outside the spheromak wall evaluated at R = Rc is:
Ueoc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [Rc^2 / (Ro^2 + Rc^2)]^N

Use this formula to calculate |Eroc| from
|Eroc| = {2 Ueoc / Epsilono}^0.5
 
= {2 Uo [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2)^N] / [(Ro^2 + Rc^2)^N Epsilono]}^0.5
= {[2 Uo / Epsilono]^0.5 [Ro^2 / (Ro^2 + Rc^2)] [Rc^2 / (Ro^2 + Rc^2)]^(N / 2)
 
and then to calculate Eric from
|Eric| = [(Sac / Epsilono) - |Eroc|]
 
= [(Sac / Epsilono)
- [2 Uo / Epsilono]^0.5 [Ro^2 / (Ro^2 + Rc^2)] [Rc^2 / (Ro^2 + Rc^2)]^(N / 2)
 
= [Sac / Epsilono] {1 - (Epsilono / Sac) (2 Uo / Epsilono)^0.5 [Ro^2 / (Ro^2 + Rc^2)] [(Rc^2 / (Ro^2 + Rc^2)]^(N / 2)}
 
= F [Sac / Epsilono]
where:
0 < F < 1
and
F = 1 - (Epsilono / Sac) {(2 Uo / Epsilono)^0.5 [Ro^2 / (Ro^2 + Rc^2)]^2 [(Rc^2 / (Ro^2 + Rc^2)]^(N / 2)
 
= 1 - {(2 Uo Epsilono / Sac^2)^0.5
[Ro^2 / (Ro^2 + Rc^2)] [(Rc^2 / (Ro^2 + Rc^2)]^(N / 2)
 
= 1 - {(2 Uo Epsilono / Sac^2)^0.5
[So^2 / (So^2 + 1)] [1 /(So^2 + 1)]^(N / 2)}

 

Recall that:
Uo = (Muo Qs^2 C^2) / (32 Pi^2 Ro^4)
and
Epsilono = 1 / (Muo C^2)
and
Sac = Qs So^2 / (2 Ro^2 Pi^2 (So^2 - 1))

Hence:
(2 Uo Epsilono / Sac^2)^0.5
= {2 (Muo Qs^2 C^2) / (32 Pi^2 Ro^4)[1 / (Muo C^2](2 Ro^2 Pi^2 (So^2 - 1))^2
/ (Qs So^2)^2
 
= {[(2 Qs^2) / (32 Pi^2 Ro^4)] [4 Ro^4 Pi^4 (So^2 - 1)^2 / (Qs^2 So^4)]}^0.5
 
= {[1 / 4] [Pi^2 (So^2 - 1)^2 / (So^4)]}^0.5
 
= {[1 / 2] [Pi (So^2 - 1) / (So^2)]}

Hence:
F = 1 - {(2 Uo Epsilono / Sac^2)^0.5
[So^2 / (So^2 + 1)] [1 /(So^2 + 1)]^(N / 2)}
 
= 1 - {{[1 / 2] [Pi (So^2 - 1) / (So^2)]}
[So^2 / (So^2 + 1)] [1 /(So^2 + 1)]^(N / 2)}
 
= 1 - {[Pi / 2] [(So^2 - 1)] [1 / (So^2 + 1)]
[1 /(So^2 + 1)]^(N / 2)}
 
= 1 - {[Pi R / 2] [1 / (So^2 + 1)]^(N / 2)}
 
= 1 - {[.9547] [1 / 5.1]^(N / 2)
= F

The present data indicates that:
1.0 < N < 2.0

F = 1 - [Pi R / 2][1 / (So^2 + 1)]^(N / 2)

Then:
Eric^2 = (F Sac / Epsilono)^2

Ueic = (Epsilono / 2) Eric^2
= (F^2 Sac^2 / 2 Epsilono)
 
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
- (F^2 Sac^2 / 2 Epsilono)

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

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

From the spheromak winding parameters the toroidal magnetic field Bc at R = Rc, H = 0 is given by:
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 Nt Qs Fh) / (2 Pi Rc)]^2 / (2 Muo)
= [(Muo Nt Qs Fh Ro) / (2 Pi Rc Ro)]^2 / (2 Muo)
= [Muo / 8] [(Nt Qs Fh So) / (Pi Ro)]^2

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

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
 
- [(F Sac Ro^2 Pi / Q)^2][1 / 2]
 
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
 
- [(F Sac Ro^2 Pi / Q)^2][1 / 2]
 
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
 
- [(F Sac Ro^2 Pi / Q)^2][1 / 2]
 
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
 
- [(F Sac Ro^2 Pi / Q)^2][1 / 2]
 
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
 
- [(F Sac Ro^2 Pi / Q)^2][1 / 2]
 

Recall that:
Sac^2 = {(Qs So^2) / [2 Pi^2 Ro^2 (So^2 - 1)]}^2
 
which gives:
[F Sac Ro^2 Pi / Qs ]^2
= {F So^2 / [2 Pi (So^2 - 1)]}^2

Substitute this expression into the boundary condition to get:
[1 / 8 Pi^2][So^4] / {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}
 
= {[1 / 32] [So^2 / (So^2 + 1)]^2
 
- {F So^2 / [2 Pi (So^2 - 1)]}^2 [1 / 2]
 

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

Cancel out So^4 to get:
[1 / 8 Pi^2] [Nt^2 / Z^2]
 
= {[1 / 32] [1 / (So^2 + 1)]^2
 
- {F / [2 Pi (So^2 - 1)]}^2 [1 / 2]
 

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

 

Rearrange this equation to get:
Z^2 = Nt^2 / {[Pi^2 / 4] [1 / (So^2 + 1)^2] - [F^2 /(So^2 - 1)^2]}

Recall that:
Z^2 = {[Np (So^2 + 1)]^2 + [Nt (So^2 - 1)]^2}

Equate the two expressions for Z^2 to get:
{[Np (So^2 + 1)]^2 + [Nt (So^2 - 1)]^2
= Nt^2 / {[Pi^2 / 4] [1 / (So^2 + 1)^2] - [F^2 /(So^2 - 1)^2]}

Divide through by Nt^2 (So^2 + 1)^2 to get:
{[Np / Nt]^2 + [(So^2 - 1)/ (So^2 + 1)]^2
= 1 / {[Pi^2 / 4] - [F^2 (So^2 + 1)^2 /(So^2 - 1)^2]}

Define R by:
R = (So^2 -1) / (So^2 + 1)

Then:
{Nr^2 + R^2}
= 1 / {[(Pi / 2)^2] - [(F / R)^2]}

THIS IS THE BOUNDARY CONDITION EQUATION FOR AN ELECTROMAGNETIC SPHEROMAK

For particular So, Np and Nt values this equation is readily numerically evaluated, which allows quantification of F and hence N.
 

To find R rearrange this equation as:
[Pi^2 / 4] Nr^2 + [Pi^2 / 4] R^2 - F^2 - F^2 Nr^2/ R^2 = 1
and multiply through by R^2 to get:
[Pi^2 / 4] Nr^2 R^2 + [Pi^2 / 4] R^4 - F^2 R^2 - F^2 Nr^2 - R^2 = 0

or
[Pi^2 / 4]R^4 + {[Pi^2 / 4] Nr^2 - F^2 - 1} R^2 - F^2 Nr^2 = 0
 

This quadratic equation has real solutions for R^2 at:
R^2 = {- {[Pi^2 / 4] Nr^2 - F^2 - 1}
+ {{[Pi^2 / 4] Nr^2 - F^2 - 1}^2
+ Pi^2 F^2 Nr^2}^0.5} / [Pi^2 / 2]

Once F^2 is determined this equation can be used to determine the R^2 value corresponding to any particular Nr^2 value. Note that R^2 determines So^2.

Rearrange the equation:
[Pi^2 / 4] Nr^2 + [Pi^2 / 4] R^2 - F^2 - F^2 Nr^2 / R^2 = 1
to get:
F^2 {1 + Nr^2/R^2}
= {[Pi^2 / 4] Nr^2 + [Pi^2 / 4][R^2] - 1}
or
F^2 = {[Pi^2 / 4] Nr^2 + [Pi^2 / 4][R^2] - 1}
/ {1 + [Nr^2 / R^2]}
 
= {[Pi^2 / 4][ Nr^2 + R^2] - 1}
/ {1 + [Nr^2 / R^2]}

This value of F^2 must equal F^2 derived from the electric field energy distribution where:
F = 1 - [Pi R / 2][1 / (So^2 + 1)^(N / 2)]
or
F^2 = 1 - [Pi R][1 / (So^2 + 1)^(N / 2)]
+ [Pi^2 R^2 / 4][1 / (So^2 + 1)^(N)]

Rearrange the formula:
F = 1 - [Pi R / 2][1 / (So^2 + 1)^(N / 2)]
to find N.

Rearrange terms to get:
(1 - F) = [Pi R / 2] [1 / (So^2 + 1)]^(N / 2)
or
[2 (1 - F) / Pi R] = [1 / (So^2 + 1)]^(N / 2)
or
[(So^2 + 1)]^(N / 2) = {Pi R / [2 (1 - F)]}
or
(N / 2) Ln[(So^2 + 1)] = Ln{Pi R / [2 (1 - F)]}
or
N = Ln{[Pi R / [2 (1 - F)]]^2} / Ln[(So^2 + 1)]

This N value can then be used in the electric field energy density expression. The F value can be calculated from the experimentally measured Fine Structure constant data or can be obtained by analysis of the spheromak core electric field.

DISCRETE VALUES AND RANGE LIMITS:
Recall that:
Nr^2 + R^2 = 1 / [(Pi^2 / 4) - (F / R)^2]

The left hand side of this equation is always positive. Hence:
(F / R) < [Pi / 2]

At So^2 = 4.1, R = (3.1 / 5.1) which gives:
F < 0.9548

Recall that:
Nr^2 + R^2 = 1 / [(Pi / 2)^2 - (F / R)^2]
or
Nr^2 = {1 / [(Pi / 2)^2 - (F / R)^2]} - R^2
= [1 - (Pi R / 2)^2 + F^2] / [(Pi / 2)^2 - (F / R)^2]

Note that since Nr is a ratio of two integers and since F and Pi are real numbers 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]

The lower limit on R is set by:
[Pi / 2] - [F / R] > 0
or
R > (2 F / 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] - [F^2 / R^2]}
to quantify [F^2 / R^2]  

Once F is known this equation allows solution for the R^2 amd hence So^2 values that correspond to particular Nr^2 values and hence integer Np and Nt values.  

Thus for So^2 > 2.4 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.
 

CALCULATION:
We can obtain an approximate value for (1 / Alpha Nt) from the boundary condition equation:
[Z^2 / Nt^2] = {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / [(So^2 + 1)^2 (So^2 - 1)^2]]
 
+ [Pi^2 / 4][1 / (So^2 + 1)^2]}

Plug this expression into the equation for (1 / Alpha Nt)^2:
(1 / Alpha Nt)^2 = (Pi / 4)^2 [(So^2 - So + 1)^2 / (So^2 + 1)^4] [Z^2 / Nt^2]
 
= (Pi / 4)^2 [(So^2 - So + 1)^2 / (So^2 + 1)^4] {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / [(So^2 + 1)^2 (So^2 - 1)^2]] + [Pi^2 / 4][1 / (So^2 + 1)^2]}
 

Thus:
[1 / (Alpha Nt)]^2 = (Pi / 4)^2 [(So^2 - So + 1)^2 / (So^2 + 1)^2] {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / [(So^2 - 1)^2]] + [Pi^2 / 4]}
 

Find the operating point this function numerically. Use the corresponding So value to calculate the corresponding values of:
[Z^2 / Nt^2] = {2 - [4 / (So^2 + 1)^2]}
/ {- [4 / [(So^2 + 1)^2 (So^2 - 1)^2]] + [Pi^2 / 4][1 / (So^2 + 1)^2]}

 
and
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}

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.

Prime numbers less than 774 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773,

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

A preliminary BASIC program solution indicates that there is a broad relative minimum in [1 / (Alpha Nt)] located at So ~ 2.0. When So is precisely:
So = 2.02606822 the corresponding value of (1 / Alpha) is given by:
(1 / Alpha) = 137.035999

The present outstanding issue is to show that the boundary condition causes an energy minimum at:
So = 2.02606822

A key issue is that at R = Rc and H = 0 the electric field Eroc outside the spheromak wall is zero. However, it appears that this is not the case.

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

The theoretical value of So^2 for operation with Np 223 and Nt = 303 is:
So^2 = 4.104952432

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

 
 
 

EVALUATION SEQUENCE:
The definition of Alpha gives:
[1 / (Alpha Nt)] = (Pi / 4) [(So^2 - So + 1) / ((So^2 + 1)^2)] [Z / Nt]
where:
Z = {[Np(So^2 + 1)]^2 + [Nt (So^2 - 1)]^2}^0.5

Check that with Np = 223, Nt = 303 the corresponding calculated Alpha value satisfies:
[1 / Alpha] ~ 137.03

Check 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.

Do not proceed beyond this point until the computer program correctly calculates [1 / Alpha] as a function of So.

Increment through So values in small increments from So = 2.0 to 2.05. The operating point should be near
So = 2.026.

Plot (1 / Alpha) versus So in the region of the operating point.
 

FINDING THE OPERATING So VALUE AT THE OPERATING POINT:
The 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]} gives us a function of So. Plot this function versus So. Focus on the region of So near So = 2.02606822 where it is claimed that:
experimentally:
(1 / Alpha) = 137.035999

Plot this function of So on the same graph sheet as the previous plot. The intersection of the two graph lines will occur at the So operating point. The value of Alpha at this operating point is the true Fine Structure Constant.

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

It is possible that the published value is incorrect or is only applicable to an experimental condition other than the specified 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 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
 

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

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 November 19, 2018.

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