XYLENE POWER LTD.

SPHEROMAK ENERGY:

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

This web page relies on results derived on the web page titled THEORETICAL SPHEROMAK. The various components of the total spheromak field energy are found in terms of the spheromak geometry and the energy density at the center of the spheromak.
 

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

This field energy density versus distance assumption is only 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 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 complicated due to the increased complexity of the energy density versus radial distance function.

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

Outside the spheromak wall the ratio of the local energy density to the energy density at the center of the spheromak is:
Ratio = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2 / (Uo)
= [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
= [(A^2 Rs Rc) / (A^2 Rs Rc + (A R)^2 + (B Z)^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
|Z| < |Zs|
the total field energy density is given by:
Ut = Uto (Ro / R)^2

At the spheromak walls:
Ut = Up

Thus at R = Rc, Z = 0:
Uo [Ro^2 / (Ro^2 + (A Rc)^2)]^2 = Uto [Ro / Rc]^2
or
Uto = Uo [Rc / Ro]^2 [Ro^2 / (Ro^2 + (A Rc)^2)]^2
= Uo [Rc Ro / (Ro^2 + (A Rc)^2)]^2
= Uo [(Ro / Rc) / ((Ro / Rc)^2 + A^2]^2
= Uo [A So / ((A^2 So^2 + A^2]^2
= Uo [1 / A^2] [So / (So^2 + 1)]^2

Hence inside the spheromak wall:
Ut = Uto (Ro / R)^2
= Uo [1 / A^2] [So / (So^2 + 1)]^2 (Ro / R)^2
 

SPHEROMAK WALL POSITION:
As shown on the web page titled THEORETICAL SPHEROMAK the locus of points defining the spheromak wall is given by:
Z^2 = (A^2 / B^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 theoretical spheromak in free space.
 

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, the spheromak nominal radius Ro and its shape parameters So and A where:
So^2 = (Rs / Rc)
and
So = (A Rs / Ro) = (Ro / A Rc)
or
Ro^2 = A^2 Rs Rc.

Typical plasma spheromaks produced in a laboratory have shape parameters of about:
So^2 = (Rs / Rc) = 4.2
 

The total spheromak field energy Ett is:
Ett = Efs - Efst + Eft
where:
 
Efs = total outside field energy if no spheromak wall was present so the energy density in all of space was given by:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
 
Efst = field energy that would lie inside the spheromak wall if the energy density function inside the spheromak wall was the same as the energy density function outside the spheromak wall.
 
Eft = cylindrical field energy that actually pertains inside the spheromak wall

Each of these spheromak field energy components is expressed below.
 

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FIELD ENERGY Efs:
The upper limit of the total field energy Ett is the field energy Efs which can be obtained by integrating the outside field energy density over all space.

Use cylindrical elements of volume with energy density that changes along their length. The energy contained in one element of volume is given by:
Integral from Z = - infinity to Z = infinity of:
Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2 2 Pi R dR dZ
 
= [Uo 2 Pi R dR] [(Ro^2)]^2 / B^4
Integral from Z = - infinity to Z = infinity of:
dZ / [(Ro^2 / B^2) + (A^2 R^2 / B^2) + (Z)^2]^2
 

Dwight #160.02 and #160.01 give the solution to this integral as:
{2 Z / [(4 c)(Z^2 + c)]
+ [2 / 4 c][2 / (4 c)^0.5] arc tan[2 Z / (4 c)^0.5]}|Z = infinity
- {2 Z / [(4 c)(Z^2 + c)]
+ [2 / 4 c][2 / (4 c)^0.5] arc tan[2 Z / (4 c)^0.5]}|Z = - infinity
 
= {+ [2 / 4 c][2 / (4 c)^0.5] (Pi / 2)}
- {[2 / 4 c][2 / (4 c)^0.5][- Pi / 2]}
 
= {+ [2 / 4 c][2 / (4 c)^0.5] (Pi)}
 
= {+ [4 / (4 c)^1.5] (Pi)}
 
= [Pi / 2 (c)^1.5]
 
= [Pi / 2 ((Ro^2 / B^2) + (A^2 / B^2) R^2)^1.5]
 

Thus the energy contained in a cylindrical element of volume is:
[Uo 2 Pi R dR] [Ro^4 / B^4][Pi / 2 ((Ro^2 / B^2) + (A^2 / B^2) R^2)^1.5]
 
= [Uo Pi^2] [Ro^4 / B] [R dR / (Ro^2 + A^2 R^2)^1.5]

Thus Efs = Integral from R = 0 to R = infinity of:
[Uo Pi^2] [Ro^4 / B] [R dR / (Ro^2 + A^2 R^2)^1.5]

Y = Ro^2 + A^2 R^2
dY = 2 A^2 R dR

Thus Efs = Integral from Y = Ro^2 to Y = infinity of:
[Uo Pi^2] [Ro^4 / B] [dY / 2 A^2 (Y)^1.5]
 
= [Uo Pi^2] [Ro^4 / (B 2 A^2)]{([- 2 / (Y)^0.5]|Y= infinity) - ([- 2 / (Y)^0.5]|Y= Ro^2)}
= [Uo Pi^2][Ro^4 / (B 2 A^2] [2 / Ro]
or
Efs = Uo Pi^2 Ro^3 / (B A^2)

For the special case of:
A = B = 1
Efs = Uo Pi^2 Ro^3

The value of Ro is dependent on the total amount of energy present. As shown herein integration over all space shows that the total energy present Ett is typically a few percent less than:
Efs = [Uo Ro^3 Pi^2 / (B A^2)]

Thus a spheromak can potentially model a real particle or a real plasma with a total energy Ett and a nominal radius Ro.
 

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

the web page titled: THEORETICAL SPHEROMAK gives:
Zs = (A / B) [(Rs - R) (R - Rc)]^0.5

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 Zs) dR
= Integral from R = Rc to R = Rs of:
Uto [Ro / R]^2 (2 Pi R) 2 (A / B) [(Rs - R) (R - Rc)]^0.5 dR
 
= Integral from R = Rc to R = Rs of:
Uto [(Ro^2 4 Pi A) / B] {(1 / R)[(Rs - R) (R - Rc)]^0.5} dR

Recall that:
A^2 Rs Rc = Ro^2
and
Uto = Uo [1 / A^2] [So / (So^2 + 1)]^2

Hence:
Eft = Integral from R = Rc to R = Rs of:
Uto [(Ro^2 4 Pi A) / B] (1 / R) [(Rs - R) (R - Rc)]^0.5 dR
 
= Integral from R = Rc to R = Rs of:
Uo [1 / (A^2 B)] [So / (So^2 + 1)]^2 [Ro^2 4 Pi A] (dR / R) [(Rs - R) (R - Rc)]^0.5
 
= Uo [1 / (A B)] [So / (So^2 + 1)]^2 [Ro^2 4 Pi]
Integral from R = Rc to R = Rs of:
(dR / R) [(Rs - R) (R - Rc)]^0.5  

Without the lead constants the integral is of the form:
[dR / R] [(Rs - R)(R- Rc)]^0.5
= [dR / R] (- R^2 + R (Rs + Rc) - Rs Rc)^0.5
= [dR / R] (aR^2 + bR + c)^0.5
where:
a = -1
b = (Rs + Rc)
c = - (Rs Rc)

From Dwight 380.311:
I = Integral of:
[dZ / R] [(a R^2 + b R + c)^0.5]
= (a R^2 + b R + c)^0.5
+ (b / 2) Integral [dR / (a R^2 + b R + c)^0.5]
+ c Integral [dR / (R (a R^2 + b R + c)^0.5)]
 
Using Dwight 380.001 and Dwight 380.111 I = (a R^2 + b R + c)^0.5
+ (b / 2) [-1 / (-a)^0.5] arc sin{[(2 a R + b] / [b^2 - (4 a c)]^0.5}
+ (c) [(1 / (-c)^0.5] arc sin{[(b R) + (2 c)] / [|R| (b^2 - (4 a c))^0.5]}

Hence:
Integral from R = Rc to R = Rs of:
[dR / R] [(a R^2 + b R + c)^0.5]
= [a Rs^2 + b Rs + c]^0.5
+ (b / 2) [-1 / (-a)^0.5] arc sin{[(2 a Rs + b] / [b^2 - (4 a c)]^0.5}
+ (c) [(1 / (-c)^0.5] arc sin{[b Rs + 2 c] / [|Rs| (b^2 - (4 a c))^0.5]}
- [a Rc^2 + b Rc + c]^0.5
- (b / 2) [-1 / (-a)^0.5] arc sin{[2 a Rc + b] / [b^2 - (4 a c)]^0.5}
- (c) [(1 / (-c)^0.5] arc sin{[(b Rc + 2 c] / [|Rc| (b^2 - (4 a c))^0.5]}
 

TERM EVALUATION:
Recall that:
a = -1
b = (Rs + Rc)
c = - (Rs Rc)

[a Rs^2 + b Rs + c]^0.5
= [-1 Rs^2 + (Rs + Rc)(Rs) - Rs Rc]^0.5
= 0

[a Rc^2 + b Rc +c]^0.5
= [ - 1 Rc^2 + (Rs + Rc) Rc - Rs Rc]^0.5
= 0
 

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

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

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

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

The sum of the terms is:
0 + 0 + [(Rs + Rc) Pi / 4] + [(Rs + Rc) Pi / 4] + [- (Rs Rc)^0.5 Pi / 2] + [(Rs Rc)^0.5 [ - Pi / 2]
&nbap;
= [(Rs + Rc) Pi / 2] - [(Rs Rc)^0.5 Pi]
= Pi [((Rs + Rc) / 2) - (Rs Rc)^0.5]

Hence:
Eft
= Uo [1 / (A B)] [So / (So^2 + 1)]^2 [Ro^2 4 Pi] Pi [((Rs + Rc) / 2) - (Rs Rc)^0.5]
 
= Uo [1 / (A B)] [So / (So^2 + 1)]^2 [Ro^2 4 Pi^2] [((Rs + Rc) / 2) - (Rs Rc)^0.5]
 
= [Uo Pi^2 Ro^2 / (A B)] [So / (So^2 + 1)]^2 [2] [((Rs + Rc)) - 2 (Rs Rc)^0.5]
 
= [Uo Pi^2 Ro^2 / (A B)] [So / (So^2 + 1)]^2 [2] [Rs^0.5 - Rc^0.5]^2

Recall that:
So = A Rs / Ro
or
Rs = Ro So / A

Then:
[Rs^0.5 - Rc^0.5]^2 = [(Ro So / A)^0.5 - (Ro / A So)^0.5]^2
= (Ro So / A) + (Ro / A So) - 2 (Ro So / A)^0.5 (Ro / A So)^0.5
= (Ro So / A) + (Ro / A So) - 2 [(Ro^2 / A^2)]^0.5
= (Ro So / A) + (Ro / A So) - 2 [Ro / A)]
= (Ro / A)[So + (1 / So) - 2]

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OUTSIDE FIELD ENERGY Efst LOST FROM REGION INSIDE THE SPHEROMAK WALL:
The energy related to the outside field that could potentially be inside the spheromak wall is found by dividing the inside region into concentric vertical cylinders. The length of these cylinders is limited by the Z positions of the spheromak wall. The smallest cylinder radius is Rc. The largest cylinder radius is Rs. This calculation is analogous to the calculation of Efs except there are finite limits on - Zs and + Zs.

Efst = Integral from R = Rc to R = Rs
Integral from Z = - Zs to Z = + Zs
Up (2 Pi R) dZ dR
 
= Integral from R = Rc to R = Rs
Integral from Z = 0 to Z = + Zs
Up (2 Pi R) 2 dZ dR
 
= Integral from R = Rc to R = Rs
Integral from Z = 0 to Z = + Zs
Uo {Ro^2 / [Ro^2 + (A R)^2 + (B Z)^2]}^2 (2 Pi R) 2 dZ dR
 
= Integral from R = Rc to R = Rs
Integral from Z = 0 to Z = + (A / B)[(Rs - R)(R - Rc)]^0.5
4 Pi Uo {Ro^2 / [Ro^2 + (A R)^2 + (B Z)^2]}^2 dZ R dR
 
= Integral from R = Rc to R = Rs
Integral from Z = 0 to Z = + (A / B)[(Rs - R)(R - Rc)]^0.5
[4 Pi Uo Ro^4 / B^4] {dZ / [(Ro^2 / B^2) + (A R / B)^2 + (Z)^2]^2} R dR
 

Let c = (Ro^2 / B^2) + (A R / B)^2

Then:
Efst = Integral from R = Rc to R = Rs
[4 Pi Uo Ro^4 / B^4] Integral from Z = 0 to Z = Zs = + (A / B)[(Rs - R)(R - Rc)]^0.5
dZ / [c + Z^2]^2 R dR
 

Integral from Z = 0 to Z = Zs = + (A / B)[(Rs - R)(R - Rc)]^0.5
dZ / [c + Z^2]^2
 
has a solution given by Dwight #160.01 and 160.02 of:
{[2 Z / (4 c)(Z^2 + c)] + [2 / 4 c][2 / (4 c)^0.5]arc tan[2 Z / (4 c)^0.5]}|Z = Zs
- {[2 Z / (4 c)(Z^2 + c)] + [2 / 4 c][2 / (4 c)^0.5]arc tan[2 Z / (4 c)^0.5]}|Z = 0
 
= {[2 Zs / (4 c)(Zs^2 + c)] + [2 / 4 c][2 / (4 c)^0.5]arc tan[2 Zs / (4 c)^0.5]}
 
= {[Zs / (2 c)(Zs^2 + c)] + [1 / 2 c^1.5] arc tan[Zs / c^0.5]}
 
= [(A / B)[(Rs - R)(R - Rc)]^0.5] / [(2 c)((A / B)^2 (Rs - R)(R - Rc) + c)]
+ [1 / 2 c^1.5] arc tan[(A / B)[(Rs - R)(R - Rc) / c]^0.5]
 
= [(A / B)[(Rs - R)(R - Rc)]^0.5] / [(2 [(Ro^2 / B^2) + (A R / B)^2)]((A / B)^2 (Rs - R)(R - Rc) + [(Ro^2 / B^2) + (A R / B)^2])]
+ [1 / 2 [(Ro^2 / B^2) + (A R / B)^2]^1.5] arc tan[(A / B)[(Rs - R)(R - Rc) / [(Ro^2 / B^2) + (A R / B)^2]]^0.5]

Dealing with this function is a real challenge.

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OUTSIDE FIELD ENERGY Efst LOST FROM REGION INSIDE THE SPHEROMAK WALL:
The energy related to the outside field that could potentially be inside the spheromak wall is found by dividing the inside region into discs. The inner and outer radii of these discs is limited by the R(Z) positions of the spheromak wall. The smallest disc inner radius is Rc. The largest disc outer radius is Rs.

Efst = Integral from Zs = - Zf to Zs = + Zf
Integral from R = Ra to R = Rb
Up (2 Pi R) dZ dR
 
= Integral from Zs = 0 to Zs = + Zf
Integral from R = Ra to R = Rb
 
= Integral from Zs = 0 to Zs = + Zf
Integral from R = Ra to R = Rb
Uo {Ro^2 / [Ro^2 + (A R)^2 + (B Zs)^2]}^2 (2 Pi R) 2 dR dZs
 
= Integral from Zs = 0 to Zs = Zf
Integral from R = Ra to R = Rb
4 Pi Uo dZs Ro^4 {R dR / [Ro^2 + (A R)^2 + (B Zs)^2]^2}
 

Let:
X = [Ro^2 + (A R)^2 + (B Zs)^2]
dX = 2 (A R) A dR
= 2 A^2 R dR

The web page titled: THEORETICAL SPHEROMAK gives:
(B Zs)^2 = [A^2 (Rs - R)(R - Rc)]
or
(B Zs / A)^2 = - R^2 + R (Rs + Rc) - Rc Rs
or
R^2 - R (Rs + Rc) + Rc Rs + (B Zs / A)^2 = 0
which gives:
R = {(Rs + Rc) +/- [(Rs + Rc)^2 - 4 ((Rc Rs) + (B Zs / A)^2)]^0.5} / 2
 
{(Rs + Rc) +/- [(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 2
 

Hence:
Ra = {(Rs + Rc) - [(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 2
and
Rb = {(Rs + Rc) + [(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 2

Thus:
(A Ra)^2 = A^2{(Rs + Rc)^2 + [(Rs - Rc)^2 - 4 (B Zs / A)^2] - 2 (Rs + Rc)[(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 4
 
= A^2 {2 Rs^2 + 2 Rc^2 - 4 (B Zs / A)^2 - 2 (Rs + Rc)[(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 4
 
= {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 - A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2

Then:
[Ro^2 + (A Ra)^2 + (B Zs)^2]
= A^2 Rs Rc + (B Zs)^2 + {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 - A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
 
= (A^2 /2)(Rs + Rc)^2 - (A / 2)(Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5}
 
= (A / 2)(Rs + Rc){A (Rs + Rc)- [A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5}

Recall that:
Rb = {(Rs + Rc) + [(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 2

Then:
(A Rb)^2 = A^2 {(Rs + Rc)^2 + [(Rs - Rc)^2 - 4 (B Zs / A)^2] + 2 (Rs + Rc)[(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 4
 
= A^2 {2 Rs^2 + 2 Rc^2 - 4 (B Zs / A)^2 + 2 (Rs + Rc)[(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 4
 
= A^2 {Rs^2 + Rc^2 - 2 (B Zs / A)^2 + (Rs + Rc)[(Rs - Rc)^2 - 4 (B Zs / A)^2]^0.5} / 2
 
= {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 + A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
 

Then:
[Ro^2 + (A Rb)^2 + (B Zs)^2]
= A^2 Rs Rc + (B Zs)^2 + {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 + A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
 
(A^2 / 2)(Rs + Rc)^2 + {A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
 
(A / 2)(Rs + Rc){A (Rs + Rc) + [A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5}

The common denominator is:
(A / 2)^2 (Rs + Rc)^2 {A^2 (Rs + Rc)^2 - [A^2 (Rs - Rc)^2 - 4 (B Zs)^2]}
 
(A / 2)^2 (Rs + Rc)^2 {A^2 4 Rs Rc + 4 (B Zs)^2}
 
= A^2 (Rs + Rc)^2 {A^2 Rs Rc + (B Zs)^2}

The corresponding numerator is:
[Ro^2 + (A Rb)^2 + (B Zs)^2] - [Ro^2 + (A Ra)^2 + (B Zs)^2]]
 
= (A Rb)^2 - (A Ra)^2
 
= {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 + A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
- {A^2 Rs^2 + A^2 Rc^2 - 2 (B Zs)^2 - A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5} / 2
 
= A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5

Thus the function to be integrated from Zs = 0 to Zs = Zf is:
[A (Rs + Rc)[A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5] / [A^2 (Rs + Rc)^2 {A^2 Rs Rc + (B Zs)^2}]
 
= [A^2 (Rs - Rc)^2 - 4 (B Zs)^2]^0.5 / [A (Rs + Rc) {A^2 Rs Rc + (B Zs)^2}]

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INSERT INTEGRATION FIX UP HERE

The energy related to the outside 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 partial spherical shell or radius Rx which intersects the equatorial plane at R = Rx. An arbitrary point on that shell is:
(R, Z)
where:
R^2 + Z^2 = Rx^2

Let Zs be the height at which the shell intersects the spheromak wall. Let Ri be the corresponding value of R at the locus of intersection points. The spheromak wall equation gives the radius Ri at Z = Zs from:
Zs^2 = (A / B)^2 (Rs - Ri)(Ri - Rc)

However, the partial spherical shell equation gives:
Ri^2 + Zs^2 = Rx^2
or
Zs^2 = Rx^2 - Ri^2

Equating the two expressions for Zs^2 gives:
(A / B)^2 (Rs - Ri)(Ri - Rc) = Rx^2 - Ri^2
OK TO HERE
or
(A / B)^2 [(Rs + Rc) Ri - Ri^2 - Rc Rs] = Rx^2 - Ri^2
or
If A = B
then:
(A / B)^2 [(Rs + Rc) Ri - Rc Rs] = Rx^2
or
Ri = (Rx^2 + (A / B)^2 Rs Rc) / [(A / B)^2 (Rs + Rc)]
or
Ri^2 = {[(Rx^2 + (A / B)^2 Rs Rc)] / [(A / B)^2 (Rs + Rc)]}^2

This expression is only valid if A = B

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

Hence:
Zs = [(Rs + Rc)^2 Rx^2 - (B / A)^2 Rx^4 - (A / B)^2 (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 partial spherical shell is given by:
Ashell = Integral from Z = 0 to Z = Zs of:
2 (2 Pi R) dS
where Rs and dS are functions of Z.

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

Hence:
(dR) = 0.5 [Rx^2 - Z^2]^-0.5 [-2 Z] dZ
or
(dR)^2 = Z^2 dZ^2 / [Rx^2 - Z^2]

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

Thus:
Ashell = Integral from Z = 0 to Z = Zs of:
4 Pi R dS
= Integral from Z = 0 to Z = Zs of:
4 Pi [Rx^2 - Z^2]^0.5 Rx dZ / [Rx^2 - Z^2]^0.5
= Integral from Z = 0 to Z = Zs of:
4 Pi Rx dZ
= 4 Pi Rx Zs
where Zs is the Z value at which the partial spherical shell of radius Rx intercepts the spheromak wall.

Ashell = 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 / (A^2 B)
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)]
 

SPHEROMAK TOTAL STATIC FIELD ENERGY: The total spheromak static field energy is given by:
Ett = Efs - Efst + Eft
= Efs [1 - (Efst / Efs) + (Eft / Efs)]
 
= Efs [1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]
 
= Efs {1 - [(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 - 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}
 

Recall that:
Efs = [Uo Ro^3 Pi^2 / (A^2 B)]

Note that for a theoretical spheromak total energy is proportional to Ro^3.

(Efst - Eft) / Efs
= [(So - 1)^2 / (So^2 + 1)] - [2 So (So - 1)^2 / (So^2 + 1)^2]
= [(So - 1)^2] [(So^2 + 1) - 2 So ] / (So^2 + 1)^2
= (So - 1)^4 / (So^2 + 1)^2

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

NUMERICAL EXAMPLE OF RELATIVE SIZES OF Efs, Eft AND Efst:
For:
So = 2

(Efst / Efs) = [(So - 1)^2 / (So^2 + 1)]
= (1 / 5)
= 0.20

(Eft / Efs) = [(So - 1) / (So^2 + 1)]^2 [2 So]
= [1 / 5]^2 [4]
= 0.16

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

Note that Ett is the total field energy in a spheromak. What is the value of So when:
d(Ett) / dSo = 0?
 

Recall that:
Ett = [Uo Ro^3 Pi^2 / (A^2 B)] {4 So [ So^2 - So + 1] / [(So^2 + 1)^2]}

d{4 So [ So^2 - So + 1] / [(So^2 + 1)^2]} /dSo
= [(So^2 + 1)^2]{4 [ So^2 - So + 1] + 4 So [2 So - 1]}
- 4 So [ So^2 - So + 1] 2 (So^2 + 1) 2 So = 0
or
[(So^2 + 1)]{[ So^2 - So + 1] + So [2 So - 1]}
- So [ So^2 - So + 1] 2 2 So = 0
or
[(So^2 + 1 - 4 So^2)][ So^2 - So + 1] + [So^2 + 1] So [2 So - 1] = 0
or
[1 - 3 So^2][So^2 + 1] + [1 - 3 So^2][- So] + [So^2 + 1][2 So^2 - So] = 0
or
[1 - 3 So^2 + 2 So^2 - So][So^2 + 1] + [1 - 3 So^2][- So] = 0
or
[1 - So - So^2][So^2 + 1] + So [3 So^2 - 1] = 0
or
[1 - So - So^2 + So^2 - So^3 - So^4 + 3 So^3 - So = 0
or
- So^4 + 2 So^3 - 2 So + 1 = 0
or
So^4 - 2 So^3 + 2 So - 1 = 0
which indicates that maximum total spheromak field energy occurs at So = 1 which makes intuitive sense since at So = 1 there is no reduction in field energy inside the spheromak wall. A graph plot of this function shows monotonic increase for So > 1 with no zeros for So > 1.
 

For So > 1 the spheromak total energy decreases but the spheromak becomes more energy stable. The spheromak energy stability is indicated by the ratio:
(Efst - Eft) / Efs
where the larger this ratio is the more energy stable is the spheromak.
 

(Efst - Eft) / Efs = [(So - 1)^2 / (So^2 + 1)]^2
or
(So - 1)^2 / (So^2 + 1)

At So = 1 the value of this function is 0. At So = 2 the value of this function is (1 / 5). At So = 3 the value of this function is 4 / 10. Thus a theoretical spheromak loses energy with increasing So.
 

FINDING THE PLANCK CONSTANT:
Recall that in order to find the Planck constant we need to find:
(dEtt / dRo).

From above:
Ett = [(Uo Ro^3) Pi^2 / (A^2 B)] {4 So [ So^2 - So + 1] / [(So^2 + 1)^2]}

Thus for a constant relative geometry spheromak:
dEtt
= d(Uo Ro^3) [Pi^2 / (A^2 B)] {4 So [ So^2 - So + 1] / [(So^2 + 1)^2]}
 

This web page last updated January 5, 2020.