XYLENE POWER LTD.

SPHEROMAK APPROXIMATION:

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

SPHEROMAK APPROXIMATION:
This web page uses relatively simple approximations to show that net charge circulating around a closed path forms the field energy density distribution required for the existence of a spheromak. A more precise mathematical proof of spheromak wall position stability is developed on the web page titled:THEORETICAL SPHEROMAK. The electric and magnetic fields of a spheromak maintain its quasi-toroidal geometry and to hold a stable amount of energy.

A spheromak quasi-toroid has a major axis and a minor axis. The spheromak's circulting current follows a closed spiral path which is characterized by Np turns around the quasi-toroid's major axis and by Nt turns around the quasi-toroid's minor axis. This current path has the quasi-toroidal shape of the glaze on a doughnut and is referred to as the spheromak wall. Inside the spheromak wall the magnetic field is toroidal. Outside the spheromak wall the magnetic field is poloidal. Inside the spheromak wall the electric field for an isolated spheromak is zero. Outside the spheromak wall the electric field for an isolated spheromak is normal to the spheromak surface. Within the core of the spheromak the electric field components parallel to the equitorial plane cancel. In the far field the electric field is nearly spherically radial. The spacial distribution of the circulating charge is constant over time. Hence the electric and magnetic field geometry is constant over time.

The spheromak field structure allows semi-stable plasma spheromaks and discrete stable atomic charged particles to exist and act as stores of energy.
 

ATOMIC PARTICLE SPHEROMAKS:
Atomic particle spheromaks have a quantized charge that superficially appears to be at rest with respect to an inertial observer. Isolated stable atomic particles such as electrons and protons hold specific amounts of energy (rest mass). When these particles aggregate with opposite charged particles the assembly emits photons. This photon emission decreases the total amount of energy in the assembly creating a mutual potential energy well.

In an atomic particle spheromak current moves uniformly around a closed spiral path at the speed of light. The spheromak net charge is uniformly distributed along the current path. The uniform charge distribution along the current path and uniform current cause constant electric and magnetic fields. The time until the current retraces its previous path is (1 / Fh) where Fh is the characteristic frequency of the spheromak.

A real spheromak in free space has a slightly distorted cross section. However, the fields of an atomic particle spheromak may be further distorted by external electric and magnetic fields.
 

SPHEROMAK CONCEPT:
Conceptually a spheromak wall is a quasi-toroidal surface formed from the closed current path of a spheromak. The current path within the spheromak wall conforms to the quasi-toroid surface curvature.

The magnetic field of a spheromak has both toroidal and poloidal components. The current path gradually changes direction over the surface of the quasi-toroid while following a closed spiral.

In a spheromak quantized positive and/or negative charge moves along a closed spiral path embedded in the spheromak wall. A spheromak is cylindrically symmetric about the spheromak major axis and is mirror symmetric about the spheromak's equatorial plane. The net charge Qs is uniformly distributed over the current path length Lh.

For an isolated spheromak in a vacuum, at the center of the spheromak the net electric field is zero. Inside the spheromak wall the magnetic field is purely toroidal and the electric field is zero. In the region outside the spheromak wall the magnetic field is purely poloidal. Outside the spheromak wall the electric field is normal to the spheromak wall and is radial in the far field. The current circulates within the spheromak wall which forms the interface between the toroidal and poloidal magnetic fields.
 

SOLUTION APPROXIMATION:
Note that a neutral spheromak cannot exist because a spheromak relies on the distributed charge on the current filament to balance the attractive magnetic foces between adjacent current filaments. However, to obtain an approximate spheromak solution it may be practical to assume that the field energy contained in the spheromak magnetic fields is much greater than the energy contained in the electric field. Then for spheromak wall geometric stability the poloidal magnetic energy density outside the spheromak wall equals the toroidal magnetic energy density inside the spheromak wall at every point on the spheromak wall. This requirement leads to the approximate spheromak geometry and the ratio of number of poloidal turns Np to the number of toroidal turns Nt. These values in turn lead to the spheromak frequency and the static magnetic energy of the spheromak. The ratio:
(static magnetic energy) / (frequency) = Planck constant.
Note that Np and Nt are both integers which leads to a unique solution. Solve the problem at the inside wall on the equatorial plane. Find that the spheromak cross section is quasi-ellipsoidal.

A real charged spheromak in a vacuum has an external radial electric field which may change the field energy distribution. However, it is instructive to examine the case where the magnetic field energy density is dominant.
A real spheromak case may be an electron spheromak around a positive nucleus. At large distances the electric fields cancel. Inside the spheromak the electric field is zero. At the spheromak inside wall on the equatorial plane the poloidal magnetic energy density outside the wall plus the external electric field energy density equals the toroidal magnetic field energy density inside the wall.
In a plasma or a crystal the electric fields almost cancel so spheromak energy is largely magnetic. Hence it is informative to calculate the theoretical Planck constant for the purely magnetic case.
 

SPHEROMAK GEOMETRY APPROXIMATION:
Assume that th spheromak shape is a round toroid.
Rs = maximum spheromak wall radius in the equatorial plane;
Rc = minimum spheromak wall radius in the equatorial plane:
Ro = Spheromak characteristic radius;
H = Z value of spheromak wall;
Outside zone electric field set by an imaginary charged ring of radius Ro with charge Qs located at (R = Ro, Z = 0);
Outside zone magnetic field set by an imaginary ring of radius Ro carring current Np I located at (R = Ro, Z = 0);
Inside zone magnetic field Bt = Muo Nt I / 2 Pi R

Note that the value of K is a complicated function of Rc, Rs and Ho. Typically:
K ~ 2 Pi.
The exact value of K will in part determine (Nt / Np).

Spheromak existence requirement:
I = Qs C / Lh

To a first approximation spheromak core magnetic field Bpo is given by: Bpo = Muo Np I / 2 Ro

Spheromak core energy density is:
Upo = Bpo^2 / 2 Muo
= [Muo Np I / 2 Ro]^2 [ 1 / 2 Muo]
= Muo Np^2 I^2 / 8 Ro^2

Inside a central sphere of radius Rs the field energy of a spheromak is approximatly:
[Muo Np^2 I^2 / 8 Ro^2][4 Rs^3 / 3] = Muo Np^2 I^2 Rs^3 / 6 Ro^2

Assume that based of plasma spheromak photographs:
Rs ~ 2 Ro

Hence, the field energy inside the central sphere is approximately:
[Muo Np^2 I^2 / 8 Ro^2][4 Rs^3 / 3]
= [Muo Np^2 I^2 / 8 Ro^2] [4 / 3] [2 Ro]^3
= [Muo Np^2 I^2 Ro (4 / 3)]

Outside this central sphere the field energy density is approximately:
[Muo Np^2 I^2 / 8 Ro^2][Rs / R]^4

The total field energy outside the central sphere is:
Integral from R = 2 Ro to R = infinity of:
[[Muo Np^2 I^2 / 8 Ro^2][2 Ro / R]^4] [4 Pi R^2] dR

= Integral from R = 2 Ro to R = infinity of:
[Muo Np^2 I^2 Pi 8][Ro^2 / R^2]] dR

= [Muo Np^2 I^2 Pi 8][Ro^2 / 2 Ro]
= [Muo Np^2 I^2 Ro [Pi / 4]

Hence the total energy content Ett of a spheromak is about:
Ett = [Muo Np^2 I^2 Ro 4 / 3] + [Muo Np^2 I^2 Ro [Pi / 4]
= [Muo Np^2 I^2 Ro] [(4 / 3) + (Pi / 4)]

This estimate of total spheromak field energy is extremely crude and should be improved by spacial energy integration over each zone.

As shown in Filament Charge and Current
I = Qs C / Lh
or
I^2 = Qs^2 C^2 / Lh^2
giving:
Ett = [(Muo Np^2 Qs^2 C^2 Ro) / (Lh^2)] [(4 / 3) + (Pi / 4)]

Recall that:
Lh^2 = Np^2 [2 Pi Ro]^2 + Nt^2 K^2 Ro^2
= Ro^2 {Np^2 [2 Pi]^2 + Nt^2 K^2}

Hence:
Ett = [(Muo Np^2 Qs^2 C^2 Ro) / (Lh^2)] [(4 / 3) + (Pi / 4)]

[(Muo Np^2 Qs^2 C^2 Ro) / (Ro^2 {Np^2 [2 Pi]^2 + Nt^2 K^2})] [(4 / 3) + (Pi / 4)]

= [(Muo Np^2 Qs^2 C^2) /{Np^2 4 Pi^2 + Nt^2 K^2}] [(4 / 3) + (Pi / 4)] [1 / Ro]

= [(Muo Qs^2 C^2) /{4 Pi^2 + (Nt^2 / Np^2) K^2}] [(4 / 3) + (Pi / 4)] [1 / Ro]

Hence:
dEtt / d(1 / Ro) = [(Muo Qs^2 C^2) /{4 Pi^2 + (Nt^2 / Np^2) K^2}] [(4 / 3) + (Pi / 4)]

Recall that:
F = C / Lh
= C / [Ro^2 {Np^2 [2 Pi]^2 + Nt^2 K^2}]^0.5
= C / [{Ro^2 Np^2 [2 Pi]^2 + Nt^2 K^2}]^0.5
= [C / Np] [1 / {4 Pi^2 + (Nt / Np)^2 K^2}]^0.5 [1 / Ro]

Hence:
dF = [C / Np] [1 / {4 Pi^2 + (Nt / Np)^2 K^2}^0.5] d[1 / Ro]

Hence:
dEtt / dF = [dEtt / d(1 / Ro)] [d(1 / Ro) / dF]

= [(Muo Qs^2 C^2) /{4 Pi^2 + (Nt^2 / Np^2) K^2}] [(4 / 3) + (Pi / 4)] [Np / C]{4 Pi^2 + (Nt / Np)^2 K^2}^0.5

= [(Muo Qs^2 C Np) /{4 Pi^2 + (Nt^2 / Np^2) K^2}^0.5] [(4 / 3) + (Pi / 4)]
where the quantity:
[(Muo Qs^2 C Np) /{4 Pi^2 + (Nt^2 / Np^2) K^2}^0.5] [(4 / 3) + (Pi / 4)]
is known as the Planck constant h.

The corresponding value of Ett is:
Ett = [(Muo Qs^2 C^2) /{4 Pi^2 + (Nt^2 / Np^2) K^2}] [(4 / 3) + (Pi / 4)] [1 / Ro]
= [(Muo Qs^2 C^2) /{4 Pi^2 + (Nt^2 / Np^2) K^2}] [(4 / 3) + (Pi / 4)] [Np F / C]{4 Pi^2 + (Nt / Np)^2 K^2}^0.5
= [(Muo Qs^2 C Np) /{4 Pi^2 + (Nt^2 / Np^2) K^2}^0.5] [(4 / 3) + (Pi / 4)] [F]
which gives the same result:
Ett = h F
 

Note that Np, Nt and K must be determined from the boundary and prime number integer constraint conditions.

In order to quantify these conditions an accurate formuation of U(R, Z) outside the spheromak wall is required.

For:
(Nt^2 K^2 / Np^2) ~ 4 Pi^2
Then:
h = [(Muo Qs^2 C Np) /{4 Pi^2 + (Nt^2 / Np^2) K^2}^0.5] [(4 / 3) + (Pi / 4)]

~ [(Muo Qs^2 C Np) /{8 Pi^2}^0.5] [(4 / 3) + (Pi / 4)]
= [(Muo Qs^2 C Np) /{2^1.5 Pi}] [(4 / 3) + (Pi / 4)]

Muo = 4 Pi X 10^-7 ______
Qs = 1.602 X 10^-19 coul
C = 2.997 X 10^8 m / s
giving:
h = Np [(Muo Qs^2 C) / 2^1.5 Pi]
= Np X 4 Pi X 10^-7____ X (1.602)^2 X 10^-38 coul X 2.997 X 10^8 m / s X (1 / 2^1.5 Pi) x [(4 / 3) + (Pi / 4)]
= Np X 2 X 5.429 X 10^-37 joule second X [(4 / 3) + (Pi / 4)]
~ Np X 23 X 10^-37 joule second

The experimental value of h is:
h = 6.62607015 × 10−34 joule second
suggesting that:
Np ~ Nt
~ [6.62607015 × 10−34 joule second] / [23 X 10^-37 joule second]
~ 288

Thus we have developed a crude expression for the Planck constant:
h = (E / F)

However, to get the lead coeeficient right we cannot use the approximations used on this web page. Instead exact expressions must be used. The exact expressions are mathematically much more complex.

We need a more accurate calculation of the field energy distribution outside the spheromak wall to better quantify Np, Nt and K.

The spheromak wall will tend to adopt the fixed geometric parameter values (Rc / Ro), (Rs / Ro) and (Ho / Ro) that result in a stable spheromak wall position. The spheromak wall geometry will determine K. The ratios (Rc / Ro) and Rs / Ro) are set by boundary conditions at (Rc, 0) and (Rs, 0). The integers Np and Nt arise from application of prime number theory to the aforementioned parameters.

The boundary condition at (Rc, 0) generates a factor of about Pi in the ratio of (Np / Nt) which in large measure determines the integers Np and Nt. At any particular controlling prime number P the Np and Nt values increment and decrement together to find the best Np, Nt number pair for meeting this boundary condition. The mechanism by which a spheromak finds its controlling prime number P is not yet fully understood.
 

This web page last updated July 9, 2024.