eb page SPHEROMAK ENERGY

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XYLENE POWER LTD.

SPHEROMAK ENERGY:

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

This web page derives expressions for the total field energy stored inside and outside the spheromak wall for an isolated spheromak in otherwise field free space in terms of the nominal spheromak radius Ro.

This web page relies on results derived on the web page titled Theoretical Spheromak and the web page titled: Spheromak Winding Constraints. The various components of the total spheromak field energy are found in terms of the spheromak geometry and the field energy densities. On other web pages these expressions are used to develop the Planck Constant, the Fine Structure Constant and the magnetic resonance properties of spheromaks.

SPHEROMAK GEOMETRY:
Assume that in the region:
- infinity < Z < + infinity,
0 < R < infinity
the spheromak wall has cross section defined by the function H(R) where:
Rs = maximum spheromak wall radius in the equatorial plane;
Rc = minimum spheromak wall radius in the equatorial plane:
Ro = Spheromak characteristic radius in the equtorial plane;
H(R) = positive Z value of spheromak wall;
- H(R) = negative Z value of spheromak wall;
H(R) = 0 at R = Rc;
H(R) = 0 at R = Rs;
H(R) = Interal from R = Rc to R = R of [dH / dR] as defined on the web page titled: (A HREF="GF Spheromak Winding Constraints.htm"> Spheromak Winding Constraints

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 is set by an imaginary ring of radius Ro carring current (Np I) located at (R = Ro, Z = 0);
Inside zone toroidal magnetic field Bt = Muo Nt I / 2 Pi R

Filament path length Lh given by:
[Lh / 2 Pi]^2 = {[(Rs - Rc) / 2]^2 [Nt^2] +[(Rs^2 + Rc^2) / 2]^2 } {Np^2}
where:
2 Ro^2 = Rs^2 + Rc^2

Winding turns ratio [Np / Nt] is given by:
Np / Nt = - (dNt / dNp)[(Rs - Rc)^2 / 2] / [Rs^2 + Rc^2]
where:
- (dNt / dNp) = (1 / 2), (2).

Experimental plasma spheromak photographs indicate that:
Rs ~ 4 Rc

The outside region has three zones. they are:
R > Rs
R < Rc
and
Rc < R < Rs, |Z| > H(R)

The inside region has only one zone which applies for Rc < R < Rs, |Z| < H(R).

In each zone the total energy is found by integrating over cylindrical elements. Each result is doubled so that it is only necessary to work with Z > 0.

RELEVANT RESULTS FROM THE WEB PAGE TITLED: Theoretical Spheromak

SPHEROMAK WALL SHAPE:
H(R) = H(Rc) + Integral from R= Rc to R = R of [dH / dR].
H(Rc) = 0

From the web page titled: Spheromak Winding Constraints
For Family A
[dH / dR] = +/- { [4 [R - Rx]^2]] + {[(Rs - R)(R - Rc)]}[(Rs - Rc) / Ro]^4
- [2 Rx] [(R - Rx)][(Rs - Rc) / Ro]^4}^0.5
/ 2 [(Rs - R)(R - Rc)]^0.5
and for Family B
[dH / dR] = +/- { [4 [R - Rx]^2] / [(Rs - R)(R - Rc)]] + [(Rs - Rc)/ Ro]^4 [1 / 16]
- [2 Rx] [(R - Rx)][(Rs - Rc)/ Ro]^4 /[16 (Rs - R)(R - Rc)]}^0.5
/ 2

The spheromak wall height is H = 0 at R = Rc and at R = Rs.

The spheromak wall height H = (Rs + Rc) / 2^0.5 at R = Rx = (Rs + Rc) / 2

??????????????? The resulting equation for H = Z(R) is:
H^2 = Ho^2 {1 - [4 (R - Ro)^2 / (Rs - Rc)^2]}
or
Z = Ho {1 - [4 (R - Ro)^2 / (Rs - Rc)^2]}^0.5
or
[Z / Ro] = [Ho / Ro] {1 - [4 (R - Ro)^2 / (Rs - Rc)^2]}^0.5
or
where the parameters Rs, and Rc are as determined on the web page titled:
Theoretical Spheromak

We should also check the value of Ho against the field equations.

?????????????????

Inside the spheromak wall the field energy density is given by:
Ut = Uto (Ro / R)^2

The field energy contained within the spheromak wall is given by: Integral from R = Rc to R = Rs of:
2 H(R) 2 Pi R dR Uto (Ro^2 / R^2)

Recall that: Uto = (1 / 2 Muo)[Muo Nt I / 2 Pi Ro]^2
= (Muo / 8)[Nt I / Pi Ro]^2
= (Muo / 8)[Nt Q C / Pi Lh Ro]^2

Clearly Nt must be large to account for h.

FIELD ENERGY OUTSIDE THE SPHEROMAK WALL:
Outside the spheromak wall the field energy density is given by U(R, Z)

For 0 < R < Rc the field energy is given by:
Integral from Z = 0 to Z = infinity
Integral from R = 0 to R = Rc
2 U(R, Z)2 Pi R dR dz

For Rs < R < infinity the field energy is given by:
Integral from Z = 0 to Z = infinity
Integral from R = Rs to R = infinity
2 U(R, Z)2 Pi R dR dz

For Rc < R < Rs the external field energy is given by:
Integral from Z = H(R) to Z - infinity
Integral from R = Rc to R = Rs
2 U(R, Z)2 Pi R dR dZ

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The maximum overall length of the spheromak occurs at dH / dR = 0 where:

EXPRESSIONS FOR Rc and Rs:
The spheromak wall crosses the equatorial plane at Z = 0 or at:

SPHEROMAK SHAPE FACTOR:
Define the spheromak shape factor [Ho / Ro]^2 by:
[Ho / Ro]^2 = [Uo / Uto]^0.5 = 4 K^2 = Np / Nt

We anticipate finding that a quantum spheromak adopts a shape factor defined by:

SUMMARY OF RESULTS FROM THE WEB PAGE TITLED: Theoretical Spheromak

FIELD ENERGY DENSITY INSIDE THE SPHEROMAK WALL:
In the toroidal region inside the spheromak wall where:
the field energy density is given by:
Ut = Uto (Ro / R)^2

At the spheromak walls:
Ut = Up

Thus at R = Rc, Z = 0:

SPHEROMAK WALL POSITION:
Recall that the locus of points defining the spheromak wall is given by:

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. Via suitable integration we can use these expressions to find the total field energy of a theoretical spheromak in free space.

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COMPONENTS OF FIELD ENERGY OF A SPHEROMAK:
The total field energy Ett associated with a spheromak can be expressed in terms of the spheromak nominal energy density Uo, the spheromak nominal radius Ro and its shape parameter [Ho / Ro] where:
[Ho / Ro]^2 = K [Uo / Uto]^0.5

The total spheromak field energy Ett is:
Ett = Ef - Efp + Eft
where:

Ef = maximum field energy if no spheromak wall was present so the energy density in all of space was given by:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2

Efp = field energy that would be 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 = toroidal field energy that actually exists inside the spheromak wall

Each of these spheromak field energy components is evaluated below.

An important fraction is:
[Ett / Ef] = [Ef - Efp + Eft] / Ef
= [1 - (Efp / Ef) + (Eft / Ef)]
Each of these fractions is calculated herein.
Then:
Ett = [Ett / Ef] Ef

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:
Ef = [Uo Ro^3 Pi^2 /(2 K)]

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

***************************************************************************

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

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

= Uo Ro^4 2 Pi R dR Integral from Z = - infinity to Z = infinity of:
dZ / [(K Ro - R)^2 + Z^2)]}^2

which Dwight 120.2 gives as:
= Uo Ro^4 2 Pi R dR from Z = - infinity to Z = infinity of:
{Z / {(2 (K Ro - R)^2 [(K Ro - R)^2 + Z^2]}}| + {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|

= Uo Ro^4 2 Pi R dR {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|(Z = infinity)
- Uo Ro^4 2 Pi R dR {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|(Z = - infinity)

= {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[Pi / 2]
- {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[- Pi / 2]

= {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[Pi]

= {Uo Ro^4 Pi^2 R dR / [(K Ro - R)^3]}

The energy potentially contained in the entire space is:
Ef = Uo Ro^4 Pi^2 Integral from R = 0 to R = infinity of:
R dR / [(K Ro - R)^3]
which from Dwight 91.3 is:
Ef = Uo Ro^4 Pi^2 {[- 1 / (K Ro - R)] + [K Ro / 2 (K Ro - R)^2]}|(R = infinity)
- Uo Ro^4 Pi^2 {[- 1 / (K Ro - R)] + [K Ro / 2 (K Ro - R)^2]}|(R = 0)

= Uo Ro^4 Pi^2 {[ 1 / (K Ro)] - [K Ro / 2 (K Ro)^2]}

= Uo Ro^4 Pi^2 {[1 / (K Ro)] - [1 / 2 (K Ro)]}

= Uo Ro^4 Pi^2 (+ 1 / 2 K Ro)
= Ef = (Uo / 2 K) Ro^3 Pi^2
This result has been double checked.

**********************************************************

TOROIDAL FIELD ENERGY Eft CONTENT OF SPHEROMAK:
The toroidal field energy content of the spheromak is found by integrating over elemental disks. Each disk has inside radius Rc(Zw), outside radius Rs(Zw) and thickness dZ.

The toroidal energy density function is:
Ut = Uto [K Ro / R]^2

The toroidal energy Eft is:
Eft = Integral from Zw = - Hm to Zw = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Ut 2 Pi R dR dZ

= Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Ut 4 Pi R dR dZ

The web page titled: THEORETICAL SPHEROMAK gives:
Ut = Uto (K Ro / R)^2

Hence:
Eft = Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Uto (K Ro / R)^2 4 Pi R dR dZ

= Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Uto (K Ro)^2 4 Pi dZ (dR / R)
= Integral from Z = 0 to Z = + Hm of:
Uto (K Ro)^2 4 Pi Ln[Rs(Zw) / Rc(Zw)] dZ
= Integral from Z = 0 to Z = + Hm of:
Uto K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from Z = 0 to Z = + Hm of:
[Uto / Uo] Uo K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from Z = 0 to Z = + Hm of:
[1 / 4 K^2] Uo K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro) of:
Uo Ro^3 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)

For the solution:
Ho^2 / Ro^2 = 4 K^2
the web page titled: Theoretical Spheromak gives:
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

Rs(Zw) / Rc(Zw) = {3 + [8 - (Zw / K Ro)^2]^0.5} / {3 - [8 - (Zw / K Ro)^2]^0.5}

This integration needs numerical evaluation. Increment through:
0 < (Zw / Ro) < (Hm / Ro)
where:
(Hm / Ro) = 2 2^0.5 K

At each increment calculate the corresponding value of [Rs(Zw) / Rc(Zw)].
Then use that value to calculate dEft. Sum the dEft values.

***********************************************************************

POLOIDAL FIELD ENERGY Efp 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 elemental disks. Within each disk the Z value is constant. The inside radius of each elemental disk is Rc(Z). The outside radius of each elemental dsik is Rs(Z). The thickness of each dislk is dZ. This calculation is analogous to the calculation of Eft except that the energy density functionis different.

Efp = Integral from Z = - Hm to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Up (2 Pi R) dZ dR

= Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
2 Up (2 Pi R) dZ dR

OK
= Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2 (2 Pi R) 2 dZ dR
OK

Let:
X = (K Ro - R)
dX = - dR
R = K Ro - X

Then:
Efp = Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2 (2 Pi R) 2 dZ dR

OK
= Integral from Z = 0 to Z = + Hm
Integral from X(Z) = K Ro - Rc(z) to X(Z) = K Ro - Rs(Z)
Uo (4 Pi Ro^4) {1 / [X(Z)^2 + Z^2]}^2 ((+ K Ro - X(Z))) (-dX) dZ

OK
= Ro^-1 Uo 4 Pi Ro^4 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
(X(Z) / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

- Ro^-2 Uo 4 Pi K Ro Ro^4 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X(Z) / Ro) = (K Ro - Rc(Z)) / Ro to (X(Z) / Ro) = (K Ro - Rs(Z)) / Ro
d(Z / Ro) d(X / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2
OK

Define:
INT1 = Uo 4 Pi Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
(X / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

and
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

where:
Efp = INT1 - INT2

The above two integrations need to be focused on, one at a time.

INT1 = Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
(X / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

OK

Dwight 121.2 gives:
INT1 = Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X(Z) / Ro) = ((K Ro - Rc(Z)) / Ro) to (X(Z) / Ro) = ((K Ro - Rs(Z)) / Ro)
(X / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{(- 1 / 2[(X(Z) / Ro)^2 + (Z / Ro)^2]|((X(Z) / Ro) = (K Ro - Rs(Z)) / Ro)
- (- 1 / 2[(X(Z) / Ro)^2 + (Z / Ro)^2]|((X(Z) / Ro) = (K Ro - Rc(Z)) / Ro)}
d(Z / Ro)

OK
= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(- 1 / 2 [(K Ro - Rs(Z)) / Ro)^2 + (Z / Ro)^2]
- (- 1 / 2(((K Ro - Rc(Z)) / Ro)^2 + (Z /Ro)^2)}
d(Z /Ro)

OK
= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(- 1 / 2 [(K Ro - Rs(Z)) / Ro]^2)
- (- 1 / 2[(K Ro - Rc(Z)) / Ro]^2)}
d(Z /Ro)

= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(+ 1 / 2 {[Rs(Z)/ Ro]^2 - [Rc(Z)/ Ro]^2)}
d(Z /Ro)
OK

The web page titled: Theoretical Spheromak gives:
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

Hence:
[Rc(Z) / Ro] = 3 K - [8 K^2 - (Zw / Ro)^2]^0.5
and
[Rs(Z) / Ro] = 3 K + [8 K^2 - (Zw / Ro)^2]^0.5

[Rc(Z) / Ro]^2 = 9 K^2 - 6 K [8 K^2 - (Zw / Ro)^2]^0.5 + [8 K^2 - (Zw / Ro)^2]

and
[Rs(Z) / Ro]^2 = 9 K^2 + 6 K [8 K^2 - (Zw / Ro)^2]^0.5 + [8 K^2 - (Zw / Ro)^2]

[Rs(Z) / Ro]^2 - [Rc(Z) / Ro]^2 = 12 K [8 K^2 - (Zw / Ro)^2]^0.5

Thus:
INT1 = Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(+ 1 / 2 {[Rs(Z)/ Ro]^2 - [Rc(Z)/ Ro]^2)}

= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
6 K [8 K^2 - (Z / Ro)^2]^0.5 d(Z /Ro)

Thus INT1 can be numerically evaluated.

Recall that:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = ((K Ro - Rc) / Ro) to (X / Ro) = ((K Ro - Rs) / Ro)
d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

OK

Dwight 120.2 gives:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[(X / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + (X / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan(X / Z)]
|((X / Ro) = (K Ro - Rs) / Ro)
- [(X / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + (X / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan(X / Z)]
|((X / Ro) = (K Ro - Rc)/ Ro)}
d(Z / Ro)

OK
= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rs(Z)) / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/ Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rc(Z)) / Ro)^2] + [1 / 2 (Z / Ro^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

OK

Recall that:
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

or
[8 (K Ro)^2 - Zw^2]^0.5 = 3(K Ro) - Rc
and [8 (K Ro)^2 - Zw^2]^0.5 = Rs - 3 (K Ro)
or
[8 (K Ro)^2 - Zw^2] = 9 (K Ro)^2 - 6 (K Ro) Rc + Rc^2
and [8 (K Ro)^2 - Zw^2] = Rs^2 - 6 (K Ro) Rs + 9 (K Ro)^2
or
Zw^2 + (K Ro)^2 - 2 K Ro Rc + Rc^2 = 4 K Ro Rc
and Zw^2 + (K Ro)^2 - 2 K Ro Rs + Rs^2 = 4 K Ro Rs
or (K Ro - Rc(Z))^2 + Z^2 = 4 (K Ro) Rc(Z)
and
(K Ro - Rs(Z))^2 + Z^2 = 4 (K Ro) Rs(Z)
or
[(K Ro - Rc(Z))^2 + Z^2] / Ro^2 = 4 K Rc(Z) / Ro
and
[(K Ro - Rs(Z))^2 + Z^2] / Ro^2 = 4 K Rs(Z) / Ro

Hence:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rs(Z)) / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rc(Z)/ Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z/ Ro)

= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z))/ Ro) / 2 (Z / Ro)^2 (4 K Rs(Z) / Ro)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/Ro) / 2 (Z /Ro)^2 (4 K Rc(Z) / Ro)] + [1 / 2 (Z /Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z))/ Ro) / (Z / Ro)^2 (4 K Rs(Z))/Ro)] + [1 / (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/ Ro) / (Z / Ro)^2 (4 K (Rc(Z)/ Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

= Uo 4 Pi K Ro^3 Integral from (Z /Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[(K) / (Z / Ro)^2 (4 K (Rs(Z)/ Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [(K) / (Z / Ro)^2 (4 K (Rc(Z) / Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

= [Uo 2 Pi K Ro^3] Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[1 / [(Z / Ro)^2 (4 Rs(Z) / Ro)]] + [(Ro / Z)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [1 / [(Z / Ro)^2 (4 Rc(Z) / Ro)]] + [(Ro / Z)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

INT2 needs numerical evaluation.

STUDY THESE TERMS AND ENSURE THAT THEY CONVERGE. THERE MIGHT BE A PROBLEM NEAR:
(Z / Ro) = 0

Recall that Efp = INT1 - INT2

***********************************************************************

SPHEROMAK TOTAL STATIC FIELD ENERGY: The total spheromak static field energy is given by:
Ett = Ef - Efp + Eft
= Ef [1 - (Efp / Ef) + (Eft / Ef)]

Recall that:
Ef = [Uo Ro^3 Pi^2 / 2 K]

Hence:
Ett = [Uo Ro^3 Pi^2 / (2 K)] { }

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

(Efp - Eft) / Efs
= _____________________________________

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

NUMERICAL EXAMPLE OF RELATIVE SIZES OF Ef, Eft AND Efp:
For:
Ef = [Uo Ro^3 Pi^2 / (2 K)]
Efp = [Uo Ro^3 Pi^2 / (2 k)] (0.20)
Eft = [Uo Ro^3 Pi^2 / (2 K)] (0.16)
Ett = [Uo Ro^3 Pi^2 / (2 K) (1 - 0.20 + 0.16)
= [Uo Ro^3 Pi^2 / 2 K] (1 - .04)
= 0.96 Ef

Thus 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.

This web page last updated October 24, 2022.

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