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By Charles Rhodes, Xylene Power Ltd.


A neutron is believed to be a neutral spheromak. An anti-neutron is identical to a neutron except that the roles of the different charge signs are interchanged.In order to understand nuclear particle bonding we must understand the structure of a neutron, which means we must understand the characteristics of a neutral spheromak.

Consider two spheromak walls, one inside the other, sharing a common toroidal axes, but having opposite net static charges. If these opposite charges are identical the result is a particle with no net charge but with a net external magnetic moment. This is the assumed structure of neutrons and anti-neutrons.

The electric field between the inner spheromak wall and the outer spheromak wall contains part of the neutral spheromak static field energy. Much of the neutron rest mass energy is in a confined photon that circulates within the inner spheromak wall.


A neutral spheromak has an outer spheromak wall with charge -Q and and an inner spheromak wall with charge Q. The following four dimensions are measured on the equatorial plane from the spheromak main axis of symmetry:
Rco = radius to outer wall inside surface
Rso = radius to outer wall outside surface
Rci = radius to inner wall inside surface
Rsi = radius to inner wall outside surface<

The radius of the inner spheromak wall measured from the toroidal axis is:
(Rsi - Rci) / 2
The radius of the outer spheromak wall with respect to the toroidal axis is:
(Rso - Rco) / 2
The radius of the toroidal axis measured from the main axis of symmetry is:
(Rso + Rco) / 2 = (Rsi + Rci) / 2

On the equatorial plane in the region Rsi > R > Rci the field is purely toroidal magnetic

On the equatorial plane in the regions R > Rso and R < Rsi the field is purely poloidal mangnetic.

On the equatoriql plane in the regions:
Rso > R > Rsi
Rci > R > Rco
the field is cylindrically radial electric + toroidal magnetic from the outer spheromak wall + poloidal magnetic from the inner spheromak wall.

The outer spheromak wall is characterized by number of poloidal turns Npo and number of toroidal turns Nto. The inner spheromak wall is characterized by number of poloidal turns Npi and number of toroidal turns Nti.

These turns will theoretically result in two frequencies. At this time we are uncertain as to the relationship of these frequencies.

To move this problem forward we must write the boundary conditions at:
R = Rci
R = Rco
and possibly at:
R = Rsi
R= Rso

This problem is described by three radii (Rco, Rci and [Rso + Rco) / 2] that must spontaneously find a stable energy minimum. Experimental evidence suggests that the energy minimum ismore stabe when there is an external stabilizing field. It appears that the total energy at this minimum may be greater than the total energy contained in separate electron and proton spheromaks.

In general, neglecting gravitation and kinetic energy:
U = [Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]
Bp = poloidal magnetic field strength;
Bt = toroidal magnetic field strength;
Er = radial electric field strength

Thus at a spheromak wall:
{[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside


However, inside the spheromak wall:
Bp = 0
and outside the spheromak wall:
Bt = 0
Hence at a spheromak wall:
{[[Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

This energy density balance (force balance) condition is valid for every element of area dA on the spheromak wall.

On the equatorial plane at the junction between the core region and the toroidal region field energy density balance gives:
[(Bpoc^2 + Btoc^2) / Mu] + [Epsilon Eroc^2] = [Bpic^2 + Btic^2] / Mu + [Epsilon Eric^2]

In the central core at the spheromak equator the magnetic field is poloidal and the electric field is zero. On the other side of the core wall the magnetic field is purely toroidal and the electric field is cylindrically radial. The field energy density on both sides of this wall must be equal.

However, for a spheromak at R = Rc, H = 0 there are the following important conditions:
Eroc = 0 or no electric field in core
Btoc = 0 or no toroidal magnetic field in core
Bpic = 0 or no poloidal magnetic field in toroidal region
giving the simplified boundary condition at R = Rc, H = 0 as:
[Bpoc^2 / Mu] = [Btic^2 / Mu] + [Epsilon Eric^2]
(Bpoc^2 / 2 Mu) = (Btic^2 / 2 Mu) + (Sac^2 / 2 Epsilon)
(Epsilon Bpoc^2 / Mu) = (Epsilon Btic^2 / Mu) + (Sac^2)
(Bpoc^2 / Mu^2 C^2) = (Btic^2 / Mu^2 C^2) + (Sac^2)

FIND Bpoc:
In the core of the spheromak the electric field is zero. Hence Uo is entirely due to the poloidal magnetic field.
Recall that:
Rs Rc = Ro^2

At R = 0, H = 0:
Bpo^2 / 2 Mu = Uo

(Bpoc^2 / 2 Mu) = Uo [Ro^2 / (Ro^2 + Rc^2)]^2
= [Bpo^2 / 2 Mu][Ro^2 / (Ro^2 + Rc^2)]^2
= [Bpo^2 / 2 Mu][Ro^2 / (Ro^2 + Rc^2)]^2
= [Bpo^2 / 2 Mu][Rs Rc / (Rs Rc + Rc^2)]^2
= [Bpo^2 / 2 Mu][Rs / (Rs + Rc)]^2

Bpoc^2 = Bpo^2 [Rs / (Rs + Rc)]^2

This web page last updated July 24, 2018.

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