XYLENE POWER LTD.

SPHEROMAK ENERGY:

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

This web page relies on multiple results derived on the web pages titled CHARGE HOSE PROPERTIES and THEORETICAL SPHEROMAK and ELECTROMAGNETIC SPHEROMAK. The various components of the total spheromak field energy are found in terms of physical constants, the spheromak charge and the peak magnetic field strength Bpo at the center of the spheromak.
 

SPHERICAL FIELD ENERGY DENSITY OUTSIDE THE SPHEROMAK WALL:
Assume that in the region outside the spheromak wall the total field energy density is given by:
U = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2.

Recall that:
Ro^2 = Rs Rc
giving:
U = Uo [(Rs Rc)]^2 / [(Rs Rc + R^2 + H^2)]^2
= Upo [Rs Rc]^2 / [(Rs Rc + R^2 + H^2)]^2
This expression applies everywhere outside the spheromak wall.

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

In the far field where:
(R^2 + H^2) >> Ro^2
the total field energy density is given by:
Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 ~ [Epsilon / 2][Qa / (4 Pi Epsilon(R^2 + H^2))]^2

The ratio of the total energy density in the far field to energy density at the center of the spheromak is:
Ratio = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 / (Uo)
= [Ro^2 / (Ro^2 + R^2 + H^2)]^2
= [(Rs Rc) / (Rs Rc + R^2 + H^2)]^2
 

SPHEROMAK SHAPE FACTOR:
Define the spheromak shape factor So by:
So^2 = (Rs / Rc)
 

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

This energy density function arises from a toroidal magnetic field and a cylindrically radial electric field.

Recall that:
Uc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2
= Uo {[Rc Rs] / {[(Rc Rs) + (Rc^2)]}^2
= Uo {Rs / [(Rs + Rc)]}^2
= Upo {Rs / (Rs + Rc)}^2

Hence in the toroidal region:
Ut = Uo {Rs / [(Rs + Rc)]}^2 (Rc / R)^2
= Uo {So^2 /[(So^2 + 1)]}^2 (Rc / R)^2
= [Upo] {So^2 / (So^2 + 1)}^2 (Rc / R)^2
This expression is valid everywhere inside the spheromak wall.
 

SPHEROMAK WALL POSITION:
As shown on the web page titled THEORETICAL SPHEROMAK the locus of points defining the spheromak wall is given by:
Hs^2 = (Rs - R)(R - Rc)
 

FIELD ENERGY DENSITY SUMMARY:
Thus we have expressions for the field energy density everywhere around and within a spheromak and we have a closed form expression for the spheromak wall position. In principle we can use these expressions to find the total field energy of a spheromak in free space. We will show that this total field energy has a relative minimum at So^2 = Rs / Rc ~ 4.1 and that this relative minimum determines the Planck constant.
 

COMPONENTS OF FIELD ENERGY OF A SPHEROMAK:
The total field energy Ett associated with a spheromak can be expressed in terms of the spheromak peak central energy density (Uo), its nominal radius Ro and its shape parameter So where:
So^2 = (Rs / Rc)
and
So = (Rs / Ro) = (Ro / Rc).

The total spheromak electromagnetic field energy Ett is:
Ett = Efs - Efst + Eft
where:
 
Efs = total spherical field energy if no spheromak wall was present
 
Efst = spherical field energy that would lie inside the spheromak wall if there was no change in the energy density function at the spheromak wall
 
Eft = cylindrical field energy that actually pertains inside the spheromak wall

Each of these spheromak field energy components is expressed below.

The spheromak energy stability is indicated by the ratio:
(Efst - Eft) / Efs
where the larger this ratio is the more stable is the spheromak.
 

Ett FOR STABLE PLASMAS AND ATOMIC PARTICLES:
Note that Ett is the energy corresponding to the available energy in a plasma spheromak or in the rest mass of an atomic particle. An issue of interest is the value of So at a stable relative minimums in:
Ett,
which corresponds to semi-stable spheromak plasmas and stable atomic particle masses. ie What is the value of So where:
d(Ett) / dSo = 0
and
d^2(Ett) / (dSo)^2 > 0
 

Recall that:
Ett = Efs - Efst + Eft
giving:
d(Ett) / dSo = [d(Efs) / dSo] + [d(Eft - Efst) / dSo]

Since:
Efs >> (Eft - Efst)
[d(Efs) / dSo] >> [d(Eft - Efst) / dSo]
giving:
d(Ett) / dSo ~ [d(Efs) / dSo]

Thus at the value of So that makes:
[d(Efs) / dSo] ~ 0
also makes:
d(Ett) / dSo ~ 0

This value of So corresponds to the stable low energy operating point of a spheromak.
 

SPHERICAL FIELD ENERGY Efs:
The upper limit of the total field energy Ett is the spherical field energy Efs which can be obtained simply by integrating the spherical field energy density over all space.
Let:
Z^2 = H^2 + R^2
Then:
Efs = Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ [Ro^2 / (Ro^2 + Z^2)]^2
= Integral from Z = 0 to Z = infinity of:
[Uo Ro^4 4 Pi] {Z^2 dZ / (Ro^2 + Z^2)^2}

Integral is of the form:
Z^2 dZ / (aZ^2 + bZ + c)^2
where:
a = 1;
b = 0;
c = Ro^2

From Dwight 160.22 Integral
= {[(b^2 - 2 a c) Z + (b c)] / [a (4 a c - b^2)(a Z^2 + b Z + c)]} + {[2 c / (4 a c - b^2)]Int [dZ / (a Z^2 + bZ + c)]}
 
= {[( - 2 a c) Z ] / [a (4 a c)(a Z^2 + c)]} + {[2 c / (4 a c)]Int [dZ / (a Z^2 + bZ + c)]}
which from 160.01 becomes:
= {[( - 2) Z ] / [(4 a)(a Z^2 + c)]} + {[2 / (4 a)][2 / (4 a c- b^2)^0.5] [arc tan[(2 a Z + b) / (4 a c - b^2)^0.5]}
 
= {[( - 2) Z ] / [(4 a)(a Z^2 + c)]} + {[1 / (2 a)][1 / (a c)^0.5] [arc tan[(2 Z) / (4 c)^0.5}

At Z = 0 and at Z = infinity the first term disappears. At Z = 0 the second term disappears. Hence the integral becomes:
{[1 / (2 a)][1 / (a c)^0.5] (Pi / 2)}

Hence:
Efs = [Uo Ro^4 4 Pi]{[1 / (2 a)][1 / (a c)^0.5] (Pi / 2)}
= [Uo Ro^4 4 Pi]{[1 / 2][1 / Ro] (Pi / 2)}
= Uo Ro^3 Pi^2

Recall that:
Upo = Uo
or
Uo = Upo
which when substituted into the expression for Efs gives:
Efs = Upo Ro^3 Pi^2

For electromagnetic spheromaks:
Upo = (Bpo^2 / 2 Muo)
which gives:
Efs = (Bpo^2 / 2 Muo) Ro^3 Pi^2

From the web page titled: ELECTROMAGNETIC SPHEROMAK far field matching gives:
Uo [Ro^2]^2 = (1 / (2 Epsilon))[Qs / 4 Pi]^2
or
Upo Ro^4 = (Muo C^2 / 2)[Qs / 4 Pi]^2 or
Upo = (Muo C^2 / 2)[Qs / 4 Pi Ro^2]^2
or
(Bpo^2 / 2 Muo) = (Muo C^2 / 2)[Qs / 4 Pi Ro^2]^2

Hence:
Efs = (Bpo^2 / 2 Muo) Ro^3 Pi^2
= (Mu C^2 / 2)[Qs / 4 Pi Ro^2]^2 Ro^3 Pi^2
= (Muo C^2 / 2)[Qs / 4]^2 [1 / Ro]
= (Muo C^2 / 2)[Qs / 4]^2 [1 / Rs Rc]^0.5
 

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

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

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

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

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

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

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

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

Recall from the web page titled ELECTROMAGNETIC SPHEROMAK that the common boundary condition gives:
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
or
Nr^2 = [+ {8} - {Pi^2 [(So^2 - 1) / (So^2 + 1)]^2}] / {Pi^2 - [4 / (So^2 - 1)]^2}

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

Let:
X = So^2

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

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

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

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

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

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

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

FIELD ENERGY Eft CONTENT OF SPHEROMAK TOROIDAL REGION:
The energy density in the toroidal region is:
Ut = Uc [Rc / R]^2

The field energy content of the spheromak toroidal region Eft is found by integrating over elemental cylinders of constant energy density. Thus Eft is given by:
Eft = Integral from R = Rc to R = Rs of:
Ut (2 Pi R) (2 Hs) dR
= Integral from R = Rc to R = Rs of:
Uc [Rc / R]^2 (2 Pi R) 2 [(Rs - R) (R - Rc)]^0.5 dR
 
= Integral from R = Rc to R = Rs of:
Uc [Rc^2 4 Pi] {(1 / R)[(Rs - R) (R - Rc)]^0.5} dR

Recall that:
Rs Rc = Ro^2
and
Uc = Uo [Ro^2 / (Ro^2 + Rc^2)]^2
= Upo [Ro^2 / (Ro^2 + Rc^2)]^2
= Upo [(Rs Rc) / ((Rs Rc) + Rc^2)]^2
= Upo [Rs / (Rs + Rc)]^2

Hence:
Eft = Integral from R = Rc to R = Rs of:
Uc [Rc^2 4 Pi] (1 / R) [(Rs - R) (R - Rc)]^0.5 dR
 
= Integral from R = Rc to R = Rs of:
Upo [Rs / (Rs + Rc)]^2 [Rc^2 4 Pi] (1 / R) [(Rs - R) (R - Rc)]^0.5 dR
 
= Integral from R = Rc to R = Rs of:
(Upo) [Rs / (Rs + Rc)]^2 [Rc^2 4 Pi] [Ro] (Ro / R) [((So) - (R / Ro)) ((R / Ro) - (1 / So))]^0.5 (dR / Ro)
 
= Integral from R = Rc to R = Rs of:
(Upo) [So^2 / (So^2 + 1)]^2 [(Ro / So)^2 4 Pi] [Ro] (Ro / R) [((So) - (R / Ro)) ((R / Ro) - (1/ So))]^0.5 (dR / Ro)
 
= Integral from R = Rc to R = Rs of:
(Upo) [So / (So^2 + 1)]^2 [4 Pi] [Ro^3] {(Ro / R)[((So) - (R / Ro)) ((R / Ro) - (1 / So))]^0.5} (dR / Ro)

Let Z = (R / Ro)
and
dZ = dR / Ro

Then:
Eft = Integral from Z = (1 / So) to Z = So of:
(Upo) [So / (So^2 + 1)]^2 [4 Pi][Ro^3][1 / Z] [((So) - Z) (Z - (1 / So))]^0.5} [dZ]
 
= (Upo) [So / (So^2 + 1)]^2 [4 Pi][Ro^3] Integral from Z = (1 / So) to Z = So of:
[dZ / Z] [((So) - Z) (Z - (1 / So))]^0.5}

Without the lead constants this integrand is of the form:
[dZ / Z] [(A - Z)(Z - B)]^0.5
= [dZ / Z] (-Z^2 + Z(A + B) - AB)^0.5
= [dZ / Z] (aZ^2 + bZ + c)^0.5
where:
A = (So)
B = (1 / So)
a = -1
b = (A + B) = [(So) + (1 / So)]
c = (- AB) = - (So)(1 / So) = - 1

From Dwight 380.319:
Integral of:
[dZ / Z] [(aZ^2 + bZ + c)^0.5]
= (aZ^2 + bZ + c)^0.5
+ (b / 2) Integral [dZ / (aZ^2 + bZ + c)^0.5]
+ c Integral [dZ / (Z (aZ^2 + bZ + c)^0.5)]
 
= (aZ^2 + bZ + c)^0.5
+ (b / 2) [-1 / (-a)^0.5] arc sin{[(2 a z) + b] / [b^2 - (4 a c)]^0.5}
+ (c) [(1 / (-c)^0.5] arc sin{[(b Z) + (2 c)] / [|Z| (b^2 - (4 a c))^0.5]}

Hence:
Integral from Z = (1 / So) to Z = (So) of:
[dZ / Z] [(aZ^2 + bZ + c)^0.5]
= [a (K So)^2 + b (K So) + c]^0.5
+ (b / 2) [-1 / (-a)^0.5] arc sin{[(2 a K So) + b] / [b^2 - (4 a c)]^0.5}
+ (c) [(1 / (-c)^0.5] arc sin{[(b K So) + (2 c)] / [|K So| (b^2 - (4 a c))^0.5]}
- [a (1 / So)^2 + b (1 / So) + c]^0.5
- (b / 2) [-1 / (-a)^0.5] arc sin{[(2 a (1 / So)) + b] / [b^2 - (4 a c)]^0.5}
- (c) [(1 / (-c)^0.5] arc sin{[(b (1 / So)) + (2 c)] / [|(1 / So)| (b^2 - (4 a c))^0.5]}
 

TERM EVALUATION:
[a (So)^2 + b (So) + c]^0.5 = [-1 (So)^2 + [(So) + (1 / So)](So) - 1]^0.5
= 0

+ [(So + (1 / So))/ 2] [-1 / (-a)^0.5 arc sin{[(2 a So) + b] / [b^2 - (4 a c)]^0.5}]
= + [(So + (1 / So)) / 2] [- arc sin{[(2 (-1) So) + [(So) + (1 / So)]] / [[(So) + (1 / So)]^2 - (4 (-1) (- 1))]^0.5}]
= + [(So + (1 / So)) / 2] [- arc sin{[(- So) + (1 / So)] / [[(So) - (1 / So)]^2]^0.5}]
= + [(So + (1 / So)) / 2] [- arc sin{[(- So) + (1 / So)] / [(So) - (1 / So)]}]
= + [(So + (1 / So))/ 2] [ Pi / 2]
= + (So + (1 / So)) Pi / 4)

+ (c) [(1 / (-c)^0.5] arc sin{[(b So) + (2 c)] / [|So| (b^2 - (4 a c))^0.5]}
= - arc sin{[([(So) + (1 / So)] So) + (-2)] / [|So| ([(So) - (1 / So)]^2)^0.5]}
= - arc sin{[(So)^2 - 1)] / [|So| [(So) - (1 / So)]}
= - arc sin{[(So)^2 - 1)] / [So^2 - 1]}
= - Pi / 2

- [a(1 / So)^2 + b(1 / So) + c]^0.5
= - [(-1) (1 / So)^2 + [( So) + (1 / So)] (1 / So) - 1]^0.5
= 0

- (b / 2) [-1 / (-a)^0.5 arc sin{[(2 a (1 / So)) + b] / [b^2 - (4 a c)]^0.5}
= - ([(So) + (1 / So)] / 2) [- arc sin{[(2 (-1) (1 / So)) + [(So) + (1 / So)]] / ([(So) - (1 / So)]^2)^0.5}
= - ([(So) + (1 / So)] / 2) [- arc sin{[(So) - (1 / So)] / ([(So) - (1 / So)]^2)^0.5}
= - ([(So) + (1 / So)] / 2)[- Pi / 2]
= [(So) + (1 / So)] [Pi / 4]

- (c) [(1 / (-c)^0.5] arc sin{[(b (1 / So)) + (2 c)] / [|(1 / So)| (b^2 - (4 a c))^0.5]}
= arc sin{[([(So) + (1 / So)] (1 / So)) + (- 2)] / [|(1 / So)| ([(So) - (1 / So)]^2)^0.5]}
= arc sin{[((1) + (1 / So)^2) + (- 2)] / [|(1 / So)| [(So) - (1 / So)]]}
= arc sin{[- (1) + (1 / So)^2] / [(1) - (1 / So)^2]}
= (- Pi / 2)
= - Pi / 2

Hence:
Eft = (Upo) [So / (So^2 + 1)]^2 [4 Pi] [Ro^3] Integral from Z = (1 / So) to Z = So of:
[1 / Z] [((So) - Z) (Z - (1 / So))]^0.5} [dZ]
 
= (Upo) [So / (So^2 + 1)]^2 [4 Pi] [Ro^3]
{0 + (So + 1 / So) (Pi / 4) - (Pi / 2) + 0 + [(So) + (1 / So)] [Pi / 4] - Pi / 2}
 
= (Upo) [So / (So^2 + 1)]^2 [4 Pi] [Ro^3] {(So + 1 / So) (Pi / 2) - (Pi)}
 
= (Upo) [So / (So^2 + 1)^2] [4 Pi^2] [Ro^3](1 / 2) {(So^2 + 1) - (2 So)}
 
= (Upo) [So (So - 1)^2 / (So^2 + 1)^2] [2 Pi^2] [Ro]^3

At So = 1:
Eft = 0
as expected.
 

Recall that:
Efs = Upo (Ro)^3 Pi^2

Hence:
(Eft / Efs) = {(Upo) [So (So - 1)^2 / (So^2 + 1)^2] [2 Pi^2] [Ro]^3} / {Upo (Ro)^3 Pi^2}
= [2 So (So - 1)^2 / (So^2 + 1)^2]
 

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

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

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

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

On the spheromak wall:
H^2 + [(Rs + Rc) / 2) - R]^2 = ((Rs - Rc) / 2)^2

Thus:
H^2 = ((Rs - Rc) / 2)^2 - [((Rs + Rc) / 2) - R]^2
= (- Rs Rc / 2) - (Rs Rc / 2) + 2 R ((Rs + Rc)/ 2) - R^2]
= - Rs Rc + R (Rs + Rc) - R^2
= R (Rs + Rc - R) - Rs Rc

Hence:
2 H = 2 [R (Rs + Rc - R) - Rs Rc]^0.5

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

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

Hence:
Et = Integral from R = Rc to R = Rs of:
(Muo / 2)[(Qs C Nt)^2 / (Lh^2 Pi] [R (Rs + Rc - R) - Rs Rc)]^0.5 dR / R
= Integral from R = Rc to R = Rs of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(R / Rs) (1 + (Rc / Rs) - (R / Rs)) - (Rc / Rs))]^0.5 (dR / R)
= Integral from R = Rc to R = Rs of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(Rc / Rs) (- 1 + (R / Rs)) + (1 - (R / Rs))(R / Rs)]^0.5 (dR / R)
= Integral from R = Rc to R = Rs of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(- Rc / Rs) ( 1 - (R / Rs)) + (1 - (R / Rs))(R / Rs)]^0.5 (dR / R)
= Integral from R = Rc to R = Rs of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(1 - (R / Rs))[(R / Rs) - (Rc / Rs)]]^0.5 (d(R / Rs)/ (R / Rs))

Let:
Xo = Rc / Rs
X = R / Rs

Then:
Et = Integral from Xa = (Rc / Rs) to Xb = 1 of:
(Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(1 - X)(X - Xa)]^0.5 dX / X)

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

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

This is an accurate expression for the toroidal magnetic energy. The toroidal electric field energy is additional.

SPHERICAL FIELD ENERGY Efst DISPLACED BY THE TOROIDAL REGION INSIDE THE SPHEROMAK WALL:
The energy related to the spherical field that could potentially be inside the spheromak wall is found by dividing the toroidal region into partial spherical shells of uniform field energy density. The area of these shells is limited by the positions of the spheromak walls.

Consider a shell located at radius Rx on the equatorial plane. An arbitrary point on that shell is:
(R, H)
where:
Ro^2 + R^2 + H^2 = Ro^2 + Rx^2 or
R^2 + H^2 = Rx^2

Let Hs be the height at which the shell intersects the spheromak wall. Let Ri be the corresponding value of R at the intersecton line. The spheromak wall equation gives the radius Ri at H = Hs from:
Hs^2 = (Rs - Ri)(Ri - Rc)

However, the shell equation gives:
Hs^2 = Rx^2 - Ri^2

Equating the two expressions for Hs^2 gives:
(Rs - Ri)(Ri - Rc) = Rx^2 - Ri^2
or
(Rs + Rc) Ri - Ri^2 - Rc Rs = Rx^2 - Ri^2
or
(Rs + Rc) Ri - Rc Rs = Rx^2
or
Ri = (Rx^2 + Rs Rc) / (Rs + Rc)
or
Ri^2 = [(Rx^2 + Rs Rc) / (Rs + Rc)]^2

The value of Hs^2 as a function of Rx is given by:
Hs^2 = Rx^2 - Ri^2
= Rx^2 - [(Rx^2 + Rs Rc) / (Rs + Rc)]^2
= [(Rs + Rc)^2 Rx^2 - Rx^4 - Rs^2 Rc^2 - 2 Rs Rc Rx^2] / (Rs + Rc)^2

Hence:
Hs = [(Rs + Rc)^2 Rx^2 - Rx^4 - Rs^2 Rc^2 - 2 Rs Rc Rx^2]^0.5 / (Rs + Rc)
= [(Rs^2 + Rc^2) Rx^2 - Rx^4 - Rs^2 Rc^2]^0.5 / (Rs + Rc)
= [(Rs^2 - Rx^2) (Rx^2 - Rc^2)]^0.5 / (Rs + Rc)

The area of the elemental shell is given by:
Ashell = Integral from H = 0 to H = Hs of:
2 (2 Pi R) dS
where Rs and dS are functions of H.

Recall that on the shell:
R^2 + H^2 = Rx^2
or
R = [Rx^2 - H^2]^0.5

An element of distance dS along the shell surface line of longitude is given by:
(dS)^2 = (dR)^2 + (dH)^2
where:
(dR) = 0.5 [Rx^2 - H^2]^-0.5 [-2 H] dH
or
(dR)^2 = H^2 dH^2 / [Rx^2 - H^2]
giving:
(dS) = (dR^2 + dH^2)^0.5
= ((H^2 dH^2 + [Rx^2 - H^2] dH^2) / [Rx^2 - H^2])^0.5
= {Rx^2 (dH)^2 / [Rx^2 - H^2]}^0.5
= Rx dH / [Rx^2 - H^2]^0.5

Thus:
Ashell = Integral from H = 0 to H = Hs of:
4 Pi R dS
= Integral from H = 0 to H = Hs of:
4 Pi [Rx^2 - H^2]^0.5 Rx dH / [Rx^2 - H^2]^0.5
= Integral from H = 0 to H = Hs of:
4 Pi Rx dH
= 4 Pi Rx Hs
= 4 Pi Rx [(Rs^2 - Rx^2) (Rx^2 - Rc^2)]^0.5 / (Rs + Rc)
= 4 Pi (Rx / Rc) (Rc^2)[(So^4 - (Rx / Rc)^2) ((Rx / Rc)^2 - 1)]^0.5 / (So^2 + 1)

Let:
Z = (Rx / Rc)
giving:
Ashell = 4 Pi Z (Rc^2) [(So^4 - Z^2) (Z^2 - 1)]^0.5 / (So^2 + 1)

Hence:
Efst = Integral from Rx = Rc to Rx = Rs of:
Uo [Ro^2 / (Ro^2 + Rx^2)]^2 Ashell dRx
 
= Integral from Rx = Rc to Rx = Rs of:
Uo (Rc) [(Ro)^2 / ((Ro)^2 + Rx^2)]^2 [4 Pi Z (Rc^2)] [(So^4 - Z^2) (Z^2 - 1)]^0.5 d(Rx / Rc) / (So^2 + 1)
 
= Integral from Rx = Rc to Rx = Rs of:
Uo (Rc) [(Ro)^2 / (Rc^2 (Ro / Rc)^2 + Rc^2 (Rx / Rc)^2)]^2 [4 Pi Z (Rc^2)] [(So^4 - Z^2) (Z^2 - 1)]^0.5 d(Rx / Rc) / (So^2 + 1)
 
= Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
Uo (Rc / Rc^4) [(Ro)^2 / (So^2 + Z^2)]^2 [4 Pi Z (Rc^2)] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
= Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
Uo (1 / Rc3) [(Ro)^2 / (So^2 + Z^2)]^2 [4 Pi Z (Rc^2)] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
= Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
Uo (1 / Rc) [(Ro)^2 / (So^2 + Z^2)]^2 [4 Pi Z] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
= Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
Upo (Ro^4 / Rc) [1 / (So^2 + Z^2)]^2 [4 Pi Z] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
= Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
Upo (Ro^3) [So / (So^2 + Z^2)^2] [4 Pi Z] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)

Recall that:
Efs = (Bpo^2 / 2 Muo) (Ro)^3 Pi^2
giving:
Efst = Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
[Efs / Pi^2] [So / (So^2 + Z^2)^2] [4 Pi Z] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
= [Efs] Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
[So / (So^2 + Z^2)^2] [4 Z / Pi] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 

CHANGE OF INTEGRATION VARIABLE:
Let:
Y = Z^2
dY = 2 Z dZ
Then:
Efst = [Efs] Integral from Z = 1 to Z = (Rs / Rc) = So^2 of:
[So / (So^2 + Z^2)^2] [4 Z / Pi] [(So^4 - Z^2) (Z^2 - 1)]^0.5 dZ / (So^2 + 1)
 
=[Efs] Integral from Y = 1 to Y = (Rs / Rc)^2 = So^4 of:
[So / (So^2 + Y)^2] [2 / Pi] [(So^4 - Y) (Y - 1)]^0.5 dY / (So^2 + 1)
 
= [Efs][2 So / Pi (So^2 + 1)] Integral from Y = 1 to Y = (Rs / Rc)^2 = So^4 of:
[1 / (So^2 + Y)^2] [(So^4 - Y) (Y - 1)]^0.5 dY BR>  
= [Efs][2 So / Pi (So^2 + 1)] Integral from Y = 1 to Y = (Rs / Rc)^2 = So^4 of:
[1 / (So^2 + Y)^2] [-Y^2 + (So^4 + 1) Y - So^4]^0.5 dY
 

FURTHER CHANGE OF VARIABLE:
Let:
X = So^2 + Y
dX = dY
Then:
Integral from Y = 1 to Y = (Rs / Rc)^2 = So^4 of:
[1 / (So^2 + Y)^2] [-Y^2 + (So^4 + 1) Y - So^4]^0.5 dY
 
= Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [- (X - So^2)^2 + (So^4 + 1) (X - So^2) - So^4]^0.5
 
= Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [- (X^2 - 2 X So^2 + So^4) + (So^4 X - So^6 + X - So^2) - So^4]^0.5
 
= Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [- X^2 + 2 X So^2 - So^4 + So^4 X - So^6 + X - So^2 - So^4]^0.5
 
= Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [- X^2 + X (So^4 + 2 So^2 + 1) - So^2 (So^4 + 2 So^2 + 1)]^0.5
 
= Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [- X^2 + X (So^2 + 1)^2 - So^2 (So^2 + 1)^2]^0.5
 
which is of the form:
Integral from X = (So^2 + 1) to X = So^2 (So^2 + 1) of:
[dX / X^2] [A X^2 + B X + C]^0.5
 
where: A = - 1;
B = (So^2 + 1)^2;
C = [- So^2 (So^2 + 1)^2]
[B^2 + 4 C]^0.5 = {(So^2 + 1)^4 + 4 [- So^2 (So^2 + 1)^2]}^0.5
= (So^2 + 1) {(So^2 + 1)^2 - 4 So^2}^0.5
= (So^2 + 1) (So^2 - 1)

Dwight 380.321 gives the solution of this integral as:
Integral [dX / X^2] [A X^2 + B X + C]^0.5 dX
= - [1 / X] [A X^2 + B X + C]^0.5
+ A Integral{dX / [A X^2 + B X + C]^0.5}
+ [B / 2] Integral{dX / (X [A X^2 + B X + C]^0.5)}
 
= - [1 / X] [A X^2 + B X + C]^0.5
+ A [-1 / (- A)^0.5] arc sin{[2 A X + B] / [(B^2 - 4 A C)^0.5]}
+ [B / 2] [1 / (- C)^0.5] arc sin{[(B X) + (2 C)] / [|X|(B^2 - 4 A C)^0.5]}

Thus:
Integral from X = (So^2 + 1) to X = [So^2 (So^2 + 1)] of:
[dX / X^2] [A X^2 + B X + C]^0.5
 
= - [1 / [So^2 (So^2 + 1)]] [A [So^2 (So^2 + 1)]^2 + B [So^2 (So^2 + 1)] + C]^0.5
+ A [-1 / (- A)^0.5] arc sin{[2 A [So^2 (So^2 + 1)] + B] / [(B^2 - 4 A C)^0.5]}
+ [B / 2] [1 / (- C)^0.5] arc sin{[(B [So^2 (So^2 + 1)]) + (2 C)] / [|[So^2 (So^2 + 1)]|(B^2 - 4 A C)^0.5]}
+ [1 / (So^2 + 1)] [A (So^2 + 1)^2 + B (So^2 + 1) + C]^0.5
- A [-1 / (- A)^0.5] arc sin{[2 A (So^2 + 1) + B] / [(B^2 - 4 A C)^0.5]}
- [B / 2] [1 / (- C)^0.5] arc sin{[(B (So^2 + 1)) + (2 C)] / [|(So^2 + 1)|(B^2 - 4 A C)^0.5]}
 
= - [1 / [So^2 (So^2 + 1)]] [- [So^2 (So^2 + 1)]^2 + B [So^2 (So^2 + 1)] + C]^0.5
+ arc sin{[ - 2 [So^2 (So^2 + 1)] + B] / [(B^2 + 4 C)^0.5]}
+ [B / 2] [1 / (- C)^0.5] arc sin{[(B [So^2 (So^2 + 1)]) + (2 C)] / [|[So^2 (So^2 + 1)]|(B^2 + 4 C)^0.5]}
+ [1 / (So^2 + 1)] [- (So^2 + 1)^2 + B (So^2 + 1) + C]^0.5
- arc sin{[ - 2 (So^2 + 1) + B] / [(B^2 + 4 C)^0.5]}
- [B / 2] [1 / (- C)^0.5] arc sin{[(B (So^2 + 1)) + (2 C)] / [|(So^2 + 1)|(B^2 + 4 C)^0.5]}
 
= - [1 / [So^2 (So^2 + 1)]] [- [So^2 (So^2 + 1)]^2 + (So^2 + 1)^2 [So^2 (So^2 + 1)] + [- So^2 (So^2 + 1)^2]]^0.5
+ arc sin{[- 2 [So^2 (So^2 + 1)] + (So^2 + 1)^2] / [(B^2 + 4 C)^0.5]}
+ [(So^2 + 1)^2 / 2] [1 / So (So^2 + 1)] arc sin{[((So^2 + 1)^2 [So^2 (So^2 + 1)]) + (2 [- So^2 (So^2 + 1)^2])] / [|[So^2 (So^2 + 1)]|(B^2 + 4 C)^0.5]}
+ [1 / (So^2 + 1)] [- (So^2 + 1)^2 + (So^2 + 1)^2 (So^2 + 1) + [- So^2 (So^2 + 1)^2]]^0.5
- arc sin{[- 2 (So^2 + 1) + (So^2 + 1)^2] / [(B^2 + 4 C)^0.5]}
- [(So^2 + 1)^2 / 2] [1 / So (So^2 + 1)] arc sin{[((So^2 + 1)^2 (So^2 + 1)) + (2 [- So^2 (So^2 + 1)^2])] / [|(So^2 + 1)|(B^2 + 4 C)^0.5]}
 
= - [1 / [So^2]] [- [So^2]^2 + [So^2 (So^2 + 1)] + [- So^2]]^0.5
+ arc sin{[(1 - So^2)(So^2 + 1)] / [(B^2 + 4 C)^0.5]}
+ [(So^2 + 1) / 2] [1 / So] arc sin{[((So^2 + 1)^2 [So^2 (So^2 + 1)]) + (2 [- So^2 (So^2 + 1)^2])] / [|[So^2 (So^2 + 1)]|(B^2 + 4 C)^0.5]}
+ [- 1 + (So^2 + 1) + [- So^2]]^0.5
- arc sin{[(So^2 - 1)(So^2 + 1)] / [(B^2 + 4 C)^0.5]}
- [(So^2 + 1) / 2] [1 / So] arc sin{[((So^2 + 1)^2 (So^2 + 1)) + (2 [- So^2 (So^2 + 1)^2])] / [|(So^2 + 1)|(B^2 + 4 C)^0.5]}
 
= 0
+ arc sin{[(1 - So^2)(So^2 + 1)] / [(So^2 + 1) (So^2 - 1)]}
+ [(So^2 + 1) / 2 So] arc sin{[(So^2 + 1)^2 So^2 (So^2 - 1)] / [|[So^2 (So^2 + 1)]|(So^2 + 1) (So^2 - 1)]}
+ 0
- arc sin{[(So^2 - 1)(So^2 + 1)] / [(So^2 + 1) (So^2 - 1)]}
- [(So^2 + 1) / 2 So] arc sin{[((So^2 + 1)^2 (- So^2 + 1))] / [|(So^2 + 1)|(So^2 + 1) (So^2 - 1)]}
 
= 0
+ arc sin{-1}
+ [(So^2 + 1) / 2 So] arc sin{1}
+ 0
- arc sin{1}
- [(So^2 + 1) / 2 So] arc sin{- 1}
 
= - (Pi / 2) + [(So^2 + 1) / 2 So] (Pi / 2) - (Pi / 2) - [(So^2 + 1) / 2 So] (- Pi / 2)
 
= (Pi / 2){-1 + [(So^2 + 1) / 2 So] - 1 + [(So^2 + 1) / 2 So]}
 
= (Pi / 2){-2 + [(So^2 + 1) / So]}
= (Pi / 2){[(So^2 - 2 So + 1) / So]}
= (Pi / 2){[(So - 1)^2 / So]}

Hence:
Efst = [Efs][2 So / Pi (So^2 + 1)] (Pi / 2){[(So - 1)^2 / So]}
= [Efs][(So - 1)^2 / (So^2 + 1)]
 

NUMERICAL EXAMPLE OF RELATIVE SIZES OF Efs, Eft AND Efst:
For:
So = 2
Efs = [Uo Ro^3 Pi^2]
Efst = [Uo Ro^3 Pi^2] (0.20)
Eft = [Uo Ro^3 Pi^2] (0.16)
Ett = [Uo Ro^3 Pi^2] (1 - 0.20 + 0.16)
= [Uo Ro^3 Pi^2 ] (1 - .04)
= 0.96 Efs

Thus at So = 2 the replacement of the spherical energy distribution by the cylindrical energy distribution causes a drop in total spheromak energy Ett of about 4%.

Note that everywhere inside the spheromak wall:
Efst > Eft
which implies that spheromaks can exist.
 

SPHEROMAK EVOLUTION:
For So < 2 the relative depth of the mutual potential energy well is small which causes a spheromak to be unstable. The spheromak energy decays by photon emission until So^2 increases to its stable value of about 4.1.

The initial higher Nr^2 value means that the initial value of So^2 is close to unity and the initial volume enclosed by the spheromak wall is relatively small. The process of replacing poloidal magnetic field energy with less energy dense toroidal magnetic field energy results in photon emission.

The total spheromak energy decreases while the spheromak net charge remains constant. During photon emission:
Nr = (Np / Nt)
decreases and hence So^2 increases until the spheromak stable minimum energy state is reached.
 

CALCULATING TOTAL SPHEROMAK STEADY STATE ENERGY Ett:
Recall that:
Ett = Efs - Efst + Eft
= Efs - [Efs][(So - 1)^2 / (So^2 + 1)] + [Efs] [2 So (So - 1)^2 / (So^2 + 1)^2]
= Efs {[(So^2 + 1)^2 - (So - 1)^2 (So^2 + 1) + 2 So (So - 1)^2] / [(So^2 + 1)^2]}
= Efs {[(So^2 + 1)^2 - (So - 1)^2 [(So^2 + 1) - 2 So]] / [(So^2 + 1)^2]}
= Efs {[(So^2 + 1)^2 - (So - 1)^2 (So - 1)^2] / [(So^2 + 1)^2]}
= Efs {[(So^2 + 1)^2 - (So - 1)^4] / [(So^2 + 1)^2]}
= Efs {1 - [(So - 1)^2 / (So^2 + 1)]^2}
 

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

Ett is a function of Fh and So. Hence:
dEtt = (dEtt / dFh) dFh + (dEtt / dSo) dSo

At steady state Nr = (Np / Nt) adjusts so that:
(dEtt / dSo) = 0
corresponding to a total energy minimum

Plot Ett / {[Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt} versus So to find the value of So that minimizes Ett at constant Fh. At that relative minimum:
(dEtt / dSo) = 0.

This plot shows that the minimum in total spheromak energy Ett occurs at:
So ~ 2.026
at which point:
Ett / {[Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt} ~ 2.2882

Hence at So = 2.026:
Ett = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [2.2882]
which is the approximate operating point of a charged particle spheromak.
 

Hence the change in field energy Ett of a spheromak at steady state is primarily a function of the change in spheromak frequency Fh. However, Fh is a function of the geometrical parameters Ro, Np, Nt and the spheromak's central magnetic field strength Bpo.

Recall that:
So^2 = (Rs / Rc)
where:
(Rs Rc) = Ro^2.
Measurements of Rc, Rs and Bpo of a plasma spheromak allow calculation of Ro and hence So and hence the total spheromak field energy.
 

FIND THE EXACT SPHEROMAK LOW ENERGY OPERATING POINT:
To find the low energy point we need to find the Nr^2, So^2 combination that gives the spheromak its lowest total energy Ett while maintaining Np and Nt as integers.

At this low energy operating point:
d(Ett) / dSo = 0

The plot of Ett versus So shows an energy minimum at approximately:
So = 2.026
or
So^2 = 4.104676

To find the exact low energy point we need to find the Nr^2, So^2 combination that gives the spheromak its lowest total energy Ett while maintaining Np and Nt as integers.

The exact value of Nr^2 corresponding to a particular value of So is given by:
Nr^2 = [+ {8} - {[Pi (So^2 - 1) / (So^2 + 1)]^2}] / {[Pi^2] - [4 / (So^2 - 1)]^2}
 
For the approximate value of So of:
So = 2.026
Nr^2 = [+ {8} - {[Pi (3.104676) / (5.104676)]^2}] / {[Pi^2] - [4 / (3.104676)]^2}
 
= [+ {8} - {3.650860312}] / {[9.869587728] - [1.659920979]}
 
= [4.349139688] / {8.209666749}
 
=0.5297583716

Hence the approximate values are:
So = 2.026
So^2 = 4.104676
Nr^2 = 0.5297583716
Nr = 0.7278450189
Ett = [Mu C Qa^2 / 4 Pi] [Pi^2 / 8] Fh Nt [2.2882]
 

The approximate value of Nr points to the exact integer values:
Np = 222
and
Nt = 305
which give the exact values:
Nr = (Np / Nt)
= 0.7278688525
and
Nr^2 = (Np / Nt)^2
= 0.5297930664

As shown on the web page titled: PLANCK CONSTANT the common boundary condition can then be used to find the precise value of So when the spheromak is at its lowest energy state. This value of So can be used to determine the Planck Constant h which is:
h = dEtt / dFh

Note that historically h was defined as:
h = Ep / Fp where:
Ep = photon energy
and
Fp = photon frequency

In circumstances where the charged particle recoil kinetic energy is negligibly small:
Ep ~ dEtt
and
Fp ~ dFh

Note that in reality there is a small difference between Ep and dEtt due to the charged particle recoil kinetic energy caused by the momentum of the photon. This issue of recoil momentum becomes important in high accuracy measurements of the Planck constant h.
 

This web page last updated February 4, 2018.