XYLENE POWER LTD.

THEORETICAL SPHEROMAK

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

SPHEROMAK DESCRIPTION:
A spheromak is a stationary mathematical solution to Maxwells electromagnetic equations.

A spheromak has a thin closed wall shaped like the glaze on the surface of a doughnut that divides a three dimensional external region that extends to infinity from a three dimensional quasi-toroidal internal region. This closed wall is known as the spheromak wall. Each region has its own energy density distribution. However, at the spheromak wall the field energy densities in the two regions are identical. Otherwise the position of the spheromak wall would be unstable.

Nature uses spheromaks to store energy in isolated rest mass. The energy fields of a spheromak act together to maintain the spheromak wall geometry.

This web page makes a few simple assumptions about energy density distributions and using these simple assumptions shows the existence of spheromaks and quantitatively derives various spheromak properties.
 

ENERGY DENSITY FUNCTIONS:
Within the internal region the field energy density Ut as a function of position is:
Ut = Uto [(K Ro) / R]^2
where Uto is a constant and Ro is characteristic of the spheromak size. Note that this energy density function is cylindrically radial and that:
Uto = Ut|(R = K Ro)

As shown on the web page titled: Electromagnetic Spheromak in the external region outside a spheromak wall the energy density Up as a function of position is given by:
Up = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
where:
Uo = constant
Z = distance from the center of the spheromak measured parallel to the main axis of symmetry.

Note that at R = 0, Z = 0:
Up = Upo = Uo / K^4

The assumed energy density functions allow perfect matching with known energy densities at both the center of the spheromak and at large distances from the spheromak center.

However, finding the particular energy density functions and boundary condition that match experimental data, although apparently simple, required a lot of work.

This field energy density versus distance assumption is only precisely true for an isolated spheromak in a vacuum. If a spheromak exists in regular atomic array, such as in single crystal silicon, the energy density U does not go to zero at large distances from each silicon atom. Instead at large distances the energy density rises due to the presence of other atoms in the regular crystal. This issue leads to the formation of electron energy band gaps in crystals. However, for now we will focus on isolated atomic particle spheromaks. Atomic nuclei, atoms, molecules and crystals are all more mathematically complicated due to the increased complexity of the energy density versus radial distance function.
 

SPHEROMAK ENERGY CONTENT:
The exact field energy content of a spheromak is derived on the web page titled: SPHEROMAK ENERGY.
 

ELECTROMAGNETIC SPHEROMAK:
Another web page titled: ELECTROMAGNETIC SPHEROMAK shows that the assumed energy density distributions correspond to the combined electric and magnetic field energy distributions in the proximity of a thin filament with Np poloidal turns and Nt toroidal turns and uniformly distributed net charge Q and carrying current I. Within the quasi-toroidal surface formed by this filament the magnetic field is purely toroidal and outside the quasi-toroidal surface the electric and magnetic field energy density distributions are respectively radial and poloidal.

This structure is consistent with the existence of stable charged particles such as electrons and protons, the Planck constant, and particle magnetic resonance. Hence the spheromak properties derived from the assumed energy distributions are the properties of real particles.

This concept can be extended to explain the behavior of electrons around an atomic nucleus. The nucleus provides the central electric field necessary to stabilize the walls of multi-electron spheromaks. As the positive nuclear charge increases the geometry of the electron spheromaks must change to meet the spheromak wall position and stability conditions. It is believed that the electron spheromaks of inert gases are exceptionally stable.

As the atomic number of an atom increases and electrons are added to the spheromaks the poloidal magnetic fields tend to cancel. Note that the conditions for existence of a multi-electron spheromak around an atomic nucleus are similar to the conditions for existence of a discrete electron or proton, so it is not surprising the both systems lead to the Planck constant.

Spheromaks can also form in plasmas, but large plasma spheromaks usually only exist for short times (< 1 ms) due to complicating factors such as neutral particle interactions with spheromak plasma particles.
 

SPHEROMAK WALL CONCEPT:
Assume that three dimensional space is divided into two regions by a thin flexible and moveable closed sheet known as the "Spheromak Wall". Thus a quasi-toroidal shaped internal region "t" exists within a larger external region "p". Assume that region "p" extends to infinity in all directions. The spacial energy density functions in the "p" and "t" regions are different but are dependent on the position of the spheromak wall.

The spheromak wall is free to move. It spontaneously seeks a stable position that results in a total field energy minimum. This requirement leads to boundary condition equations that determine the shape of the spheromak wall.

A stable spheromak wall will be located at a position of force balance (energy density balance). If the spheromak wall deviates outward from its nominal position the local energy density just outside the wall becomes higher that the local energy density just inside of the wall, which results in a net force that pushes the wall inward. Similarly, if the energy density just inside the wall is larger than the energy density just outside the wall there is a net force that pushes the wall outward until the two local energy densities are in balance.

At this stable position the field energy densities on both sides of the spheromak wall are equal so that there is no net force (change in total energy with respect to wall position) tending to move the spheromak wall. Since the spheromak wall is at a position corresponding to a stable total field energy Ett, Ett is minimum:
dEtt / dX = 0,
where X is the wall position in space.

Spheromaks tend to form because there is a certain degree of random energy motion. If random energy motion results in a spheromak configuration with a lower total potential energy and a higher total kinetic energy sometimes part of that kinetic energy becomes a photon that is radiated away. The remaining system is left trapped in a low potential energy (spheromak) state until it absorbs energy (a photon) from an external source. As long as the density of random sufficiently energetic photons in the system environment is small, spheromaks act as particles with stable local accumulations of field energy (rest mass).
 

SPHEROMAK WALL POSITION:
A spheromak is a semi-stable energy state. The spheromak wall spontaneously positions itself to achieve a total energy relative minimum. At every point on the spheromak wall the sum of the electric and poloidal magnetic field energy densities on the outer side of the spheromak wall equals the toroidal magnetic field energy density on the inner side of the spheromak wall. This general statement resolves into different boundary conditions for different points on the wall. These boundary conditions establish the spheromak core radius Rc on the equatorial plane, the spheromak outside radius Rs on the equatorial plane and the spheromak overall length 2 Hm.

In electromagnetic spheromaks the spheromak wall is formed from a long closed filament of circulating current. The form of the equations which describe the spheromak wall sets the ratio of (Np / Nt) where:
Np = number of spheromak filament poloidal turns
and
Nt = number of spheromak filament toroidal turns.
Note that Np and Nt are both large integers. Note also that if the spheromak is disturbed dNp and dNt are both small integers.  

SPHEROMAK INTERNAL ENERGY DENSITY:
A spheromak arises from an electric current circulating around a closed spiral filament path that causes an internal toroidal magnetic field with an energy density Ut of the form:
Ut = Uto (K Ro / R)^2
for:
Rc < K Ro < Rs
where:
Ut = Uto
at
R = K Ro
 

EXTERNAL ENERGY DENSITY:
The sum of the external electric field energy density Ue and the external magnetic field energy density Um outside the spheromak wall is the external energy density function:
Up = Um + Ue

This energy density function is similar in form to the energy density function arising from a thin ring of radius Ro containing circulating current (Np I) and net charge Q.

In the far field:
Ue >> Um
and Ue decreases in proportion to [1 / (R^2 + Z^2)]^2. The external energy density function must equal the known field energy density at R >> Ro and/or Z >> Ro. The external energy density must equal the known field energy density along the spheromak axis of symmetry where R = 0. The external energy density function must equal the internal field energy density at the spheromak walls.

The external field energy density function is assumed to be of the form:
Up = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2,
where:
0 < K < 1.

The internal field energy density function is assumed to be of the form:
Ut = Uto [K Ro / R]^2

The web page titled Electromagnetic spheromak shows that:
Uo = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)]
where:
Muo = permiability of free space;
C = speed of light;
Q = quantum charge;
Ro = characteristic spheromak radius;

These energy density functions yield a practical closed form mathematical model for a spheromak.
 

SPHEROMAK WALL:
The spheromak wall is the locus of points where the energy density is the same for both the internal and external energy density functions. Hence, the spheromak walls occur at the locus of points where:
Up = Ut

Hence on the spheromak wall:
Uo {Ro^2 / [(K Ro - Rw)^2 + Zw^2]]^2 = Uto [(K Ro) / Rw]^2
or
[Uto / Uo] = {Ro^2 / [(K Ro - Rw)^2 + Zw^2]}^2 [Rw / K Ro]^2
or
[Uto / Uo]^0.5 = {[Ro Rw / K] / [(K Ro - Rw)^2 + Zw^2]}

At:
Rw = K Ro:
{[Uto / Uo]^0.5} = [Ro^2 / Zw^2]|(Zw = Ho) = [Ro^2 / Ho^2]

Hence the spheromak wall equation becomes:
[Ro^2 / Ho^2] = {[Ro Rw / K] / [(K Ro - Rw)^2 + Zw^2]}
or
[(K Ro - Rw)^2 + Zw^2] = [Ro Rw / K] / [Ro^2 / Ho^2]
= [Ho^2 / Ro^2] [Ro Rw / K]
or
Zw^2 = [Ho^2 / Ro^2] [Ro Rw / K] - [(K Ro - Rw)^2]

Hence:
Zw(R) = +/- {[Ho^2 / Ro^2] [Ro Rw / K] - [(K Ro - Rw)^2]}^0.5
 

FIND Rm, Hm IN TERMS OF (K Ro), Ho:
A spheromak wall position is defined by:
Zw(R)^2 = Ho^2 (Rw / K Ro) - (K Ro - Rw)^2
or
Zw(R) = [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]^0.5
dZw / dRw = 0.5 [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]^-0.5 [Ho^2 / K Ro + 2 (K Ro - Rw)]

The parameter Zw reaches its peak value Zw = Hm at:
dZw / dRw = 0 at Rw = Rm, where:
[Ho^2 / (K Ro)] + 2 (K Ro - Rm)] = 0
or
Ho^2 + 2 K Ro (K Ro - Rm) = 0
or
Rm = (Ho^2 + 2 K^2 Ro^2) / 2 K Ro
or
[Rm / Ro] = [Ho^2 / 2 K Ro^2] + K
 

SOLUTION CASE:
For the solution case of:
[Ho^2 / Ro^2] = 4 K^2:
[Rm / Ro] = 3 K

Recall that:
Zw(R) = {Ho^2 (Rw / K Ro) - (K Ro - Rw)^2}^0.5
or
Zw(R) / Ro = {[Ho^2 / Ro^2](Rw / Ro K) - [(K Ro - Rw) / Ro]^2}^0.5

The value of (Hm / Ro) corresponding to:
Rm / Ro = 3 K
is given by:
Hm / Ro = {[Ho^2 / Ro^2](Rm / Ro K) - [(K Ro - Rm) / Ro]^2}^0.5
= {[4 K^2] (Rm / Ro K) - [(K - (Rm / Ro)]^2}^0.5
= {[4 K^2] (3) - [(K - (3 K)]^2}^0.5
= [12 K^2 - 4 K^2]^0.5
= 2 2^0.5 K

the spheromak overall length is:
2 Hm = 2 [2 2^0.5 K] Ro
= 4 2^0.5 (K Ro)

Note that Hm is at R = Rm, not at R = Ro and that Hm > Ho and Rm > Ro

For the solution case of:
Ho^2 / Ro^2 = 4 K^2
The spheromak wall equation becomes:
[(K Ro - Rw)^2 + Zw^2] = [Ro Rw / K] / [Ro^2 / Ho^2]
= [Ho^2 / Ro^2] [Ro Rw / K]
or
[(K Ro - Rw)^2 + Zw^2] = 4 K Ro Rw

Expanding this equation gives:
(K Ro)^2 - 2 K Ro Rw + Rw^2 + Zw^2 - 4 K Ro Rw = 0
or
Rw^2 + [- 6 K Ro] Rw + [(K Ro)^2 + Zw^2] = 0

This equation is quadratic with solutions:
Rw = {6 K Ro +/- [(6 K Ro)^2 - 4 (1) ((K Ro)^2 + Zw^2)]^0.5} / 2
= 3 K Ro + / - (1 / 2) [32 (K Ro)^2 - 4 Zw^2]
= 3 (K Ro) +/- [8 (K Ro)^2 - Zw^2]^0.5

At Zw = 0 this equation gives:
Rc = (3 - 2 2^0.5) K Ro
and
Rs = (3 + 2 2^0.5) K Ro

Note that Rc < Ro < Rs, where:
Rs = spheromak wall outside radius
and
Rc = spheromak wall inside radius.

In general:
Rc(Zw) = 3 (K Ro) - [8 (K Ro)^2 - Zw^2]^0.5
and
Rs(Zw) = 3 (K Ro) + [8 (K Ro)^2 - Zw^2]^0.5
 

The spheromak field structure gives an important boundary condition at the inside wall on the equatorial plane which is:
Upc = Utc

SPHEROMAK ENERGY DENSITY GRAPHS:
Examination of the functions Up and Ut shows that at small R values Up < Ut, at medium R values Ut < Up and at large R values Up < Ut.

Shown below is a graph of Up and Ut versus R. On this graph Uto = Uo, K = 0.5 and Z = 0.5. The graph is drawn with two different ordinate scales in order to make the cross over points obvious.

Note that in the region inside the spheromak wall the energy density is lower than it would be if the external energy density function:
Up(R, Z) = Upo {Ro^2 / [(Ro - R)^2 + Z^2}^2
took effect everywhere. Hence the spheromak forms a potential energy well. This potential energy well gives a spheromak its inherent stability.
 

SPHEROMAK GEOMETRY:
The geometry of a spheromak in free space can be characterized by the following parameters:
R = radius of a general point from the spheromak's main axis of symmetry;
Z = distance of a general point from the spheromak's equatorial plane;
Zm = maximum distance of a point on the spheromak wall from the spheromak's equatorial plane;
Zw = a Z value on the spheromak wall; Rm = the value of R corresponding to Z = Zm on the spheromak wall;
Rc = spheromak's minimum equatorial radius;
Rs = spheromak's maximum equatorial radius;
Ro = spheromak's characteristic radius;
Rw = a R value on the spheromak wall;
Ho = positive Zw value on the spheromak wall at Rw = K Ro;
 

SPHEROMAK CROSS SECTIONAL DIAGRAM:
The following diagram shows the cross sectional shape of a spheromak. On this diagram:
K = 0.5
Ho / Ro = 1.0

Note that an ideal spheromak in free space is quasi-toroidal. A real plasma spheromak may be shape distorted by the different properties of a plasma and by proximity of the vacuum chamber walls.
 

POSITION IN A SPHEROMAK:
A spheromak wall has cylindrical symmetry about its main axis of symmetry and has mirror symmetry about its equatorial plane. A position in a spheromak can be referenced by:
(R, Z, Phi)
where:
R = radius from the main axis of spheromak cylindrical symmetry;
and
Z = height above (or below) the spheromak equatorial plane;
and
Phi = angle around the main axis of symetry.
and
Theta = angle around minor axis of symmetry
 

GEOMETRICAL FEATURES OF A SPHEROMAK WALL:
Important geometrical features of a spheromak include:
Rc = the spheromak wall inside radius on the equatorial plane;
Rs = the spheromak wall outside radius on the equatorial plane;
Ro = the characteristic value of R;
Ho = value of Zw at R = K Ro;
Hm = masximum value of Zw;
Rm = value of R at Zw = Hm;
The subscript c refers to spheromak wall "core" surface at the equatorial plane;
The subscript s refers to the spheromak outer "wall" at the equatorial plane;

In order to understand the material on this web page it is essential for the reader to study the spheromak cross sectional diagram and to identify the above mentioned parameters.
 

SPHEROMAK STRUCTURAL ASSUMPTIONS:
1) A spheromak wall is composed of a closed spiral current filament of overall length Lh which contains both poloidal and toroidal turns;
2) Spheromak's net charge Q is uniformly distributed along filament length Lh.
3) The filament current I is:
I = Q C / Lh
4) The filament current causes a toroidal magnetic field inside the spheromak wall and a poloidal magnetic field outside the spheromak wall;
5) There is a net zero electric field inside the spheromask wall and there is a spherically radial electric field for R >> (K Ro) outside the spheromak wall;
6) At the center of the spheromak at R = 0, Z = 0 the net electric field is zero;
7) In the internal region:
the total field energy density U takes the form:
Ut = Uto [(K Ro) / R]^2
8) In the external region the total field energy density takes the form:
Up = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2;
9) Everywhere at the spheromak wall:
Up = Ut
10) An important special case is at R = Rc, Z = 0 where:
Upc = Utc;
11) Np and Nt are positive integers with no common integer factor other than one.
 

SPHEROMAK WALL POSITION:
The spheromak wall is located on the locus of points where the energy density is the same for both the internal and external energy density functions. The spheromak wall exists on the locus of points where:
Up = Ut

If the spheromak wall deviates either inward or outward from its nominal position the total system energy rises causing a net force that tends to restore the spheromak wall to it nominal position where the internal and external energy densities are equal.

Hence a spheromak naturally seeks a minimum energy state.

Differential forces on the flexible spheromak wall due to changes in spheromak energy with respect to wall position will cause the spheromak wall to spontaneously position itself so that the energy densities on both sides of the wall are equal with the wall position at R = Rw, Z = Zw corresponding to a total energy minimum. At this wall position at every point (Rw, Zw) on the spheromak wall:
Up = Ut
or
Uo {Ro^2 / [(K Ro - Rw)^2 + Zw^2]}^2 = Uto [K Ro / Rw]^2
or
[Uo / Uto] = [K Ro / Rw]^2 [(K Ro - Rw)^2 + Zw^2]^2 / Ro^4
or
[Uo / Uto]^0.5 = [K Ro / Rw][(K Ro - Rw)^2 + Zw^2] / Ro^2
= [K / (Rw Ro)][(K Ro - Rw)^2 + Zw^2]
which is the equation for the locus of points that define the spheromak wall.

At Rw = K Ro this equation becomes:
[Uo / Uto]^0.5
= [K / (Rw Ro)][+ Zw|(Rw = K Ro)]^2
= [1 / Ro^2][Ho]^2
= [Ho / Ro]^2
where:
Zw|(Rw = K Ro) = +/- Ho
which is a very important equation in spheromak analysis.

Thus the equation for the spheromak wall given by:
[Uo / Uto]^0.5 = [K / (Rw Ro)][(K Ro - Rw)^2 + Zw^2]
becomes:
[Ho^2 / Ro^2] = [K / (Rw Ro)] [(K Ro - Rw)^2 + Zw^2]

At each intersection of the spheromak wall with the equatorial plane:
Zw = 0
giving:
[1 / K][Ho^2 / Ro^2] = [1 /(Ro Rw)][(K Ro - Rw)^2]
or
[Ho^2 / K] = [Ro / Rw][(K Ro - Rw)^2]

Expanding this equation gives:
Rw Ho^2 = K Ro [K^2 Ro^2 - 2 K Ro Rw + Rw^2]
or
K Ro Rw^2 + (- 2 K^2 Ro^2 - Ho^2) Rw + K^3 Ro^3 = 0

This equation is quadratic in Rw with solutions:
Rw = {(2 K^2 Ro^2 + Ho^2) +/- [(2 K^2 Ro^2 + Ho^2)^2 - 4 (K Ro) K^3 Ro^3]^0.5} / 2 Ro
= (K^2 Ro + (Ho^2 / 2 Ro) +/- (1 / 2 Ro)[4 K^2 Ro^2 Ho^2 + Ho^4]^0.5
= K^2 Ro + (Ho^2 / 2 Ro) +/- (Ho^2 / 2 Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5

Thus:
Rc = K^2 Ro + (Ho^2 / 2 Ro) - (Ho^2 / 2 Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5
and
Rs = K^2 Ro + (Ho^2 / 2 Ro) + (Ho^2 / 2 Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5

[(Rs + Rc) / Ro] = 2 K^2 + (Ho^2 / Ro^2) ~ 1.5

[(Rs - Rc) / Ro] = (Ho^2 / Ro^2)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5 ~ 2^0.5
 

SPHEROMAK SHAPE:
Recall that the equation for the spheromak wall is:
[Ho^2 / Ro^2] = [(K Ro - Rw)^2 + Zw^2] [K / (Rw Ro)]

At Rw = K Ro:
[Ho^2 / Ro^2] = Zw^2 [K / (K Ro Ro)] = [Zw^2 / Ro^2]
as expected.

The spheromak relative shape is set by the value of the ratio (Ho / Ro). As shown on the web page ELECTROMAGNETIC SPHEROMAK there is an important inside wall boundary condition that sets:
Upc = Utc.

The spheromak wall shape defines the surface of a toroid with a distorted cross section having a minimum inside radius Rc (core radius) and a maximum outside radius Rs. The wall position corresponds to a stable total energy minimum.
 

PERIMETER LENGTHS:
A spheromak has four perimeters of interest. Three of them are in the equatorial plane.

The inside perimeter is given by:
2 Pi Rc

The outside perimenter is given by:
2 Pi Rs

The characteristic perimeter is given by:
2 Pi Ro

The fourth perimeter is the length of a purely toroidal filament winding turn along a path tangent to the spheromak wall but which is perpendicular to the minor axis of the spheromak. This perimeter length Lt is found via the line integral around the equation which defines the spheromak cross section.
 

FIND DISTANCE Lt AROUND THE SPHEROMAK WALL:
Recall that on the spheromak wall:
[Ho^2 / Ro^2] = [(K Ro - Rw)^2 + Zw^2][K / (Rw Ro)]
or
[(K Ro - Rw)^2 + Zw^2] = [Ho^2 / Ro^2] (Rw Ro / K)
= [Ho^2 (Rw / K Ro)]
or
Zw^2 = Ho^2 (Rw / K Ro) - (K Ro - Rw)^2
or
Zw = [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]^0.5

Differentiation gives:
dZw / dRw = 0.5 [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]^-0.5 [(Ho^2 / K Ro} + (2 (K Ro - Rw))]
or
(dZw/ dR)^2 = 0.25 [(Ho^2 / K Ro) + 2 (K Ro - Rw)]^2
/ [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]

Hence:
[1 + (dZw / dRw)^2] = {Ho^2 (Rw / K Ro) - (K Ro - Rw)^2 + (1 / 4)[(Ho^2 / K Ro) + 2 (K Ro - Rw)]^2} / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]

= Ho^2 (Rw / K Ro) - (K Ro - Rw)^2 + [(Ho^2 / K Ro)^2(1 / 4) + (Ho^2 / K Ro)(K Ro - Rw) + (K Ro - Rw)^2] / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]

= [+ (Ho^2 / K Ro)^2 (1 / 4) + (Ho^2)] / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]

An element of perimeter distance Lt is given by:
d(Lt / Ro) = [dR^2 + dZ^2]^0.5 / Ro

Thus:
[Lt / 2 Ro] = Integral from Rw = Rc to Rw = Rs of d(Lt / Ro)
= Integral from Rw = Rc to Rw = Rs of:
[1 + (dZw / dRw)^2]^0.5 dRw / Ro
= Integral from Rw = Rc to Rw = Rs of:
(1 / Ro){[(Ho^2 / 2 K Ro)^2 + (Ho^2)] / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]}^0.5 dR

The term:
Integral from Rw = Rc to Rw = Rs of:
{1 / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]}^0.5 dR

= Integral from Rw = Rc to Rw = Rs of:
{1 / [Ho^2 (Rw / K Ro) - (K^2 Ro^2 - 2 K Ro Rw + Rw^2)]}^0.5 dR

= Integral from Rw = Rc to Rw = Rs of:
{1 / [Ho^2 (Rw / K Ro) - K^2 Ro^2 + 2 K Ro Rw - Rw^2]}^0.5 dR

= Integral from Rw = Rc to Rw = Rs of:
{1 / [- Rw^2 + ((Ho^2 / K Ro) + 2 K Ro) Rw - Ro^2]}^0.5 dR

which from Dwight 380.001 (Table of Integrals) equals:
= (-1) arc sin[(2 (-1) Rw + 2 K Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro) + 2 K Ro]^2 - 4 (- 1)(- K^2 Ro^2))^0.5]|Rw = Rs
- (-1) arc sin[(2 (-1) Rw + 2 Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro) + 2 K Ro]^2 - 4 (- 1)(- K^2 Ro^2)]^0.5]|Rw = Rc

= (-1) arc sin[(- 2 Rs + 2 K Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro) + 2 K Ro]^2 - 4 (K^2 Ro^2))^0.5]
- (-1) arc sin[(- 2 Rc + 2 K Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro) + 2 K Ro]^2 - 4 (K^2 Ro^2))^0.5]

= [(-1) arc sin[(- 2 Rs + 2 K Ro + (Ho^2 / K Ro)) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[(- 2 Rc + 2 K Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]

Recall that:

2 (Rs - K Ro) = {(Ho^2 / K Ro) + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5}

2 (K Ro - Rc) = - (Ho^2 / K Ro)) + Ho [(4) + (Ho^2 / K Ro^2)]^0.5}

Hence:
Integral from Rw = Rc to Rw = Rs of:
{1 / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]}^0.5 dR
= [(-1) arc sin[(- 2 Rs + 2 K Ro + (Ho^2 / K Ro)) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[(- 2 Rc + 2 K Ro + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
= [(-1) arc sin[(- 2 (Rs - K Ro) + (Ho^2 / K Ro)) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[(+ 2 (K Ro - Rc) + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
= [(-1) arc sin[(- {(Ho^2 / K Ro) + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5} + (Ho^2 / K Ro)) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[(- (Ho^2 / K Ro)) + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5} + (Ho^2 /K Ro)) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]

= [(-1) arc sin[(- (Ho^2 / K Ro) - Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5 + (Ho^2 / K Ro)) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[(- (Ho^2 / K Ro)) + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5} + (Ho^2 / K Ro)) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]

= [(-1) arc sin[( - Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5) / [(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]
- [(-1) arc sin[( + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5) / ([(Ho^2 / K Ro)^2 + 4 K Ro (Ho^2 / K Ro)]^0.5]

= [(-1) arc sin[( - Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5) / Ho [(Ho / K Ro)^2 + 4)]^0.5]
- [(-1) arc sin[( + Ho [(4) + (Ho^2 / K^2 Ro^2)]^0.5) / Ho ([(Ho / K Ro)^2 + 4)]^0.5]

= [(-1) arc sin[- 1]
- [(-1) arc sin[ 1]

= (Pi / 2) + (Pi / 2) = Pi

Thus:
Lt / 2 Ro = Integral from Rw = Rc to Rw = Rs of:
(1 / Ro){[(Ho^2 / 2 K Ro)^2 + (Ho^2)]^0.5}
{1 / [Ho^2 (Rw / K Ro) - (K Ro - Rw)^2]}^0.5 dR

= (1 / Ro){[(Ho^2 / 2 K Ro)^2 + (Ho^2)]}^0.5 Pi

= (Ho / Ro){(Ho^2 / 4 K^2 Ro^2) + 1}^0.5 Pi
or
[Lt / 2 Pi Ro] = (Ho / Ro) {[(Ho^2 / 4 K^2 Ro^2) + 1]}^0.5

This is a very important equation in spheromak characterization.
 

FILAMENT WINDING:
Now assume that the spheromak wall contains a closed winding with Np poloidal turns and Nt toroidal turns. The average purely poloidal turn length is Lp and the purely toroidal turn length is Lt.

Then to a good approximation the overall winding length Lh is given by:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2
or
[Lh / 2 Pi Ro]^2 = (Np Lp / 2 Pi Ro)^2 + (Nt Lt / 2 Pi Ro)^2

Now assume that:
Lp = 2 Pi Ro

Hence:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Ho^2 / Ro^2] {[(Ho^2 / 4 K^2 Ro^2) + 1]}

Recall that:
[Uo / Uto]^0.5 = [Ho^2 / Ro^2]

Hence:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Uo / Uto]^0.5 {[(1 / 4 K^2)(Uo / Uto)^0.5 + 1]}
= Np^2 + Nt^2 [Uo / Uto][1 / (4 K^2)] + Nt^2 [Uo / Uto]^0.5
= Np^2 + Nt^2 [Uo / Uto]^0.5 + Nt^2 [Uo / Uto]^0.5
= Np^2 + 2 Nt^2 [Uo / Uto]^0.5
= Np^2 + 2 Nt^2 [4 K^2]
 

WINDING STABILITY:
When the spheromak filament length is stable implying that the spheromak contained energy is stable:
d[Lh / 2 Pi Ro]^2 = 0
impling that:
2 Np dNp + 2 Nt dNt [Uo / Uto][1 / 4 K^2] + 2 Nt dNt [Uo / Uto]^0.5 = 0
or
dNp / dNt = - [Nt / Np]{[Uo / Uto][1 / 4 K^2] + [Uo / Uto]^0.5}
= - [Nt / Np]{[Ho^4 / Ro^4][1 / 4 K^2] + [Ho^2 / Ro^2]}

However, Np, Nt, dNp and dNt are all integers. For a stable winding prime number theory indicates that:
dNp / dNt = - 2:
P = prime number which is odd
P = (Np + 2 Nt), Np = odd, 2 Nt = even

With this winding at a nearly constant energy the spheromak can trade off Np and Nt values in whole number increments with certainty of no collapse due to a common factor in Np and Nt.

Hence:
2 = [Nt / Np]{[Ho^4 / Ro^4][1 / 4 K^2] + [Ho^2 / Ro^2]}
or
2 Np / Nt = [Ho^2 / Ro^2] {[Ho^2 / Ro^2][1 / 4 K^2] + 1}

Note that Np and Nt are positive integers where Np is not equal to Nt.

[Ho^2 / Ro^2] {[Ho^2 / Ro^2][1 / 4 K^2] + [Ho^2 / Ro^2] - 2 Np / Nt = 0
or
[Ho^2 / Ro^2] {[Ho^2 / Ro^2] + [4 K^2 Ho^2 / Ro^2] - 8 K^2 Np / Nt = 0

The solution to this quadratic equation is:
[Ho^2 / Ro^2] = {- 4 K^2 +/- [16 K^4 + 4(1) 8 K^2 Np / Nt)]^0.5} / 2
= {- 4 K^2 + 4 K^2 [1 + (2 Np / K^2 Nt)]^0.5 } / 2
= 2 K^2 {- 1 + [1 + (2 Np / K^2 Nt)]^0.5 }

The term:
[1 + (2 Np / K^2 Nt)] must be a perfect square.

FIRST NON-VIABLE SOLUTION ATTEMPT:
Try (2 Np / K^2 Nt) = 3:
Then:
Ho^2 / Ro^2 = 2 K^2
Then:
2 Np / Nt = 3 K^2 = (3 / 2) [Ho^2 / Ro^2]
or
Np / Nt = (3 / 4) [Ho^2 / Ro^2]
Problem because Np and Nt cannot have common factors.
 

VIABLE SOLUTION:
[Ho^2 / Ro^2] = 2 K^2 {- 1 + [1 + (2 Np / K^2 Nt)]^0.5 }

Try:
(2 Np / K^2 Nt) = 8

Then:
Ho^2 / Ro^2 = 4 K^2
Then:
2 Np / Nt = 8 K^2 = 2 Ho^2 / Ro^2
or
Np / Nt = [Ho^2 / Ro^2]
is a viable working solution.
 

SECOND NON_VIABLE SOLUTION ATTEMPT:
Try (2 Np / K^2 Nt) = 15

Then:
[Ho^2 / Ro^2] = 6 K^2
Then:
2 Np / Nt = 15 K^2 = [15 / 6][Ho^2 / Ro^2]
or
Np / Nt = [15 / 12][Ho^2 / Ro^2]
= [5 / 4][Ho^2 / Ro^2]
Problem because Np and Nt cannot have common factors.
 

FIND EXPRESSION FOR Nt:
From above viable solution:
K^2 = [1 / 4][Ho^2 / Ro^2]
= [1 / 4][Np / Nt]
or
Np / Nt = 4 K^2

If Np ~ Nt then K ~ (1 / 2)

Recall that:
Ho^2 / Ro^2 = 4 K^2
Hence:
Ho^2 / Ro^2 ~ 1

Recall that:
[Uo / Uto]^0.5 = [Ho^2 / Ro^2]
Hence:
[Uo / Uto]^0.5 ~ 1
 

If K = (1 / 2) + D
then
K^2 = (1/ 2)^2 + D + D^2
then
4 K^2 = 1 + 4 D + 4 D^2

Thus:
Np / Nt = (1 + 4 D + 4 D^2)

Now assume that:
Np = Nt + 1,
then:
4 D^2 + 4 D = (1 / Nt)
or
D^2 + D - (1 / 4 Nt) = 0
or
D = {- 1 +/- [1 + 4(1)(1 / 4 Nt)]^0.5} / 2
= {-1 +/- [1 + (1 / Nt)]^0.5 } / 2
= {-1 + [Np / Nt]^0.5 } / 2

Thus:
2 D + 1 = [Np / Nt]^0.5
or
Np / Nt = (2 D + 1)^2
The fact that the experimentally measured value of h is constant from system to system indicates that D is constant from system to system.
 

FIND EXPRESSION FOR [Lh / 2 Pi Ro]:
Recall that:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Uo / Uto][1 / (4 K^2)] + Nt^2 [Uo / Uto]^0.5
 

Recall that:
[Uo / Uto]^0.5 = [Ho^2 / Ro^2]

Hence:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Ho / Ro]^4 [1 / (4 K^2)] + Nt^2 [Ho / Ro]^2
 

Recall that:
Ho^2 / Ro^2 = 4 K^2
giving:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Ho / Ro]^4 [1 / (4 K^2)] + Nt^2 [Ho / Ro]^2
= Np^2 + Nt^2 [Ho / Ro]^2 + Nt^2 [Ho / Ro]^4
= Np^2 + 2 Nt^2 [Ho / Ro]^2

Recall that:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Lt / (2 Pi Ro)]^2
which gives:
[Lt / (2 Pi Ro)]^2 = 2 [Ho / Ro]^2

Thus:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Lt / (2 Pi Ro)]^2
= Np^2 + Nt^2 2 [Ho / Ro]^2

Recall that:
[Ho / Ro]^2 = Np / Nt

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

Note that for this stable solution:
Np + 2 Nt = P = prime number
 

FIND ZERO SLOPE IN [Lh / (2 Pi Ro)]^2 VS Nt:
Recall that:
[Lh / (2 Pi Ro)]^2 = [Np Lp / (2 Pi Ro)]^2 + [Nt Lt / (2 Pi Ro)]^2
= Np^2 + 2 Nt^2

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

Thus:
[Lh / (2 Pi Ro)]^2 = Np^2 + 2 Nt^2
= (P - 2 Nt)^2 + 2 Nt^2
= P^2 - 4 P Nt + 4 Nt^2 + 2 Nt^2
= P^2 - 4 P Nt + 6 Nt^2

Find zero slope in [Lh / (2 Pi Ro)]^2:
d[Lh / (2 Pi Ro)]^2 / dNt = - 4 P + 12 Nt = 0
Hence slope minimum is at P ~ 3 Nt

Recall that:
P = Np + 2 Nt
Hence Np ~ Nt.
Thus slope minimum is at Np = Nt + 1
 

CHANGE IN [Lh / 2 Pi Ro] REQUIRED TO ACCESS NEARBY ENERGY STATES:
As shown above, at the slope minimum:
[Lh /(2 Pi Ro)]^2 = P^2 - 4 Nt + 6 Nt^2
or
[Lh /(2 Pi Ro)] = [P^2 - 4 Nt + 6 Nt^2]^0.5

At P = 1087, Nt = 362:
d[Lh /(2 Pi Ro)] = [-2 + 6(362)] dNt / ([1087^2 - 4(362) + 6(362)^2]^0.5)
= 2170 dNt/ [1,181,569 - 1448 + 786,264]^0.5
= 2170 dNt/ 1402.278503
= 1.547481471 dNt
 

STABLE SPHEROMAK:
The difference between excited and ground state spheromaks lies in their contained energies. When a spheromak initially forms its energy is not at steady state. The spheromak must spontaniously absorb or emit photons until it reaches its stable state where:
Np = Nt + 1.
Thereafter the spheromak can absorb or emit photons by changing its value of Ro while keeping (Ho / Ro) constant.
 

QUANTUM NUMBERS:
This relatively simple physical system (one quantum charge) is specified by two quantum numbers. There is the prime number P which is pointed to by D, which is the small deviation of K from (1 / 2), and there are the integers Np, Nt which indicate the energy level within P. The lowest energy level within P is Npo ~ (P / 3) and each increment in Np is 2. Note that Np is constrained by the requirement that Np and Nt have no common integer factors. Note that:
dNp = - 2 dNt

It is reasonable to project that more complex physical systems will have similar mathematical structures. Chemists refer to atomic shells. Each shell will have a P value. Within each shell electron energy states will be subject to comparable Np, Nt common factor constraints. When the energy state (|Np - Npo|) in a shell becomes so large that another P value is more appropriate for that energy the system will adopt the new P value. Note that P values are constrained by available prime numbers.

In essence energy quantization is a consequence of the combination of D > 0 and Np, Nt being integers. Quantization rules are primarily constraints on Np, Nt set by the existence condition that Np and Nt have no common factors. Shells are a consequence of the set of prime numbers.

Thus it appears that the electrons of an atom form multi-electron spheromak(s) around the atomic nucleus. As the atomic number of the atom is raised past an inert gas the operating spheromak adopts a new P value. ie At the transition from He to Li, at the transition from Ne to Na, etc. The multi-electron spheromaks have electrons that move around the atomic nucleus in opposite directions to minimize the net poloidal atomic magnetic dipole moment while maintaining the toroidal magnetic flux. The exact geometry of this multi-electron spheromak structure has yet to be resolved.
 

ESTIMATED SOLUTIONS:
P = 1069
P = Np + 2 Nt
Np = Nt + 1
P = 3 Nt + 1
Nt = 356
Np = 357
Lh / 2 Pi Ro = {357 (1069)}^0.5
= 617.76

P = 1087
P = Np + 2 Nt
Np = Nt + 1
P = 3 Nt + 1
Nt = 362
Np = 363
Lh / 2 Pi Ro = [Np P]^0.5
= [363 (1087)]^0.5
= 628.156

P = 1091
P = Np + 2 Nt
Np = Nt + 1
Np = 363
Nt = 364
Lh / 2 Pi Ro = [363 (1091)]^0.5
= 629.311

P = 1093
P = Np + 2 Nt
Np = Nt + 1
Nt = 364 Np = 365 Lh / 2 Pi Ro = [365 (1093)]^0.5
= 631.6209306

SOLUTION SUMMARY:
[Lh / 2 Pi Ro]^2 = Np P = Np (Np + 2 Nt)

Np / Nt = 4 K^2 = [Ho^2 / Ro^2]

Np = Nt + 1

K = (1 / 2) + D:
where:
D = {-1 + [Np / Nt]^0.5 } / 2

For P = 1087, Np = 363, Nt = 362:
D = 6.901314686 X 10^-4
 

We need to further investigate how D is set by electromagnetic phenomena.

SPHEROMAK POTENTIAL ENERGY WELL:
Outside the spheromak wall the field energy density is:
Up < Ut.

Inside the spheromak wall the field energy density is:
Up > Ut

Hence in an environment where Up extends to infinity, the spheromak causes a local potential energy well inside the spheromak wall.
 

SPHEROMAK CURRENT PATH FILAMENT LENGTH Lh:
Electromagnetic spheromaks arise from the electric current formed by distributed net charge Qs circulating at the speed of light C around the closed spiral path of length Lh which defines the spheromak wall. On the equatorial plane measured from the main axis of symmetry the spheromak inside radius is Rc and the spheromak outside radius is Rs.

Let Np be the integer number of poloidal currrent path turns in Lh and let Nt be the integer number of toroidal current path turns in Lh.

The spheromak wall contains Nt quasi-toroidal turns equally spaced around 2 Pi radians in angle Theta about the main spheromak axis of symmetry.
Each purely toroidal winding turn has length:
Lt
so the purely toroidal spheromak winding length is:
Nt Lt

In one spheromak cycle period the poloidal angle advances Np (2 Pi) radians. In the same spheromak cycle period the toroidal angle advances Nt (2 Pi) radians.
Hence:
(poloidal angle advance) / (toroidal angle advance) = Np / Nt

While a current point moves radially outward from Rc to Rs the toroidal angle advance is Pi radians and the toroidal travel is Lt / 2. The corresponding distance along the equatorial outer circumference is:
Pi (Np / Nt) Rs.
Thus Pythagoras theorm gives the current point travel distance along the winding for the toroidal half turn as:
[(Lt / 2)^2 + ( Pi Np Rs / Nt)^2]^0.5

While the current point moves radially inward from Rs to Rc the toroidal angle advance is Pi radians and the toroidal point travel is Lt / 2. The corresponding distance along the equatorial inner circumference is:
Pi (Np / Nt) Rc.
Thus Pythagoras theorm gives the current point travel distance along this winding toroidal half turn as:
[(Lt / 2)^2 + ( Pi Np Rc / Nt)^2]^0.5

This equation is the result of spheromak geometric analysis.
 

This web page last updated October 19, 2022.