**Spin Orbit Relationships**

To create a unification between the motion effected by the spin of a particle in the magnetic field (spin orbit coupling), to those associated to the coupling of the gravi(electric-magnetic) fields, we must derive the formulae required to describing the coupling. To do this, we take a new approach to deriving a Greene-like theorem which can be applied in three dimensions in a general sense. While there is a general Greene's theorem called Stokes theorem, this one is preferred because it is richer in information and allows us to understand the fields via their respective components.

We begin with the derivation using a cross product ''×'' and a unit vector ''n'':

∇×F∙n = ∂U/∂r

To get Greenes theorem from this equation, we can distribute the surface

∬ [∇×F∙n] dS = ∮ [∂U/∂r] dS = ∮ F∙dr

The magnetic field is given as

B = 1/emc²∙1/r[∂U/∂r]L

Where L is the angular momentum component, now placing in terms

∬ B∙dS = (1/emc²∙1/r)L ∬ [∇×F∙n] dS = (1/emc²∙1/r)L ∮ [∂U/∂r] dS = (1/emc²∙1/r)L ∮ F∙dr

The magnetic flux is given as

Φ = (1/emc²∙1/r)L∮ F∙dr

These equations will produce the framework of the investigations.

**A Hamiltonian with a Compton Wavelength Term**

We begin with a magnetic field

[1] B = [1/emc²] 1/r [∂U[r]/∂r L]

Take the dot product of the spin on both sides

[2] = [1/emc²] 1/r [∂U[r]/∂r L∙S]

This is the spin orbit equation of quantum mechanics. Take note that (∂U[r]/∂r) is the central potential. In previous work, I proved

[3] erB∙S = [L∙S]/r

Due to Gauge invariance we can state that

[4] A = B × r

Substitution gives

[5] eA∙S = [L∙S]/r

Notice this term ([L∙S] 1/r) exists in the Hamiltonian describing the spin orbit coupling

[6] ΔH = -μ/eħmc² 1/r ∂U[r]/∂r L∙S

So we may simplify terms by substitution giving a new Hamiltonian

[7] ΔH = -[μ/ħmc²]∙[∂U[r]/∂r] A∙S

Using further simplification, where S = ħ

[8] ΔH = -(μA/mc²) ∂U[r]/∂r

A magnetic Hamiltonian can be given by the term

[9] H = eħB/2Mc

or

[10] H = μB

where μ is the Bohr Magneton and using our circular gauge again (A = B × r) and noting that (Er = Gm² = ħc) ''known as the gravitational charge''

we can thus make

[11] ħc = H × r = eħ/2Mc ∙ B × r = μA

Where we have obtained the Bohr Magneton and electromagnetic potential (A) which actually featured in equation [8]

[12] ΔH = -μA/mc² ∂U[r]/∂r

Using some more simplification by replacing μA = ħc , we can obtain

[13] ΔH = -[ħ/mc] ∂U[r]/∂r

A singularity exists when v = c = 0 which makes λ blow to infinity and in which energy inversely blows up. Notice,

λ = [ħ/mc]

Is the Compton wavelength. Essentially, equation 13 is some kind of wave coupling to the potential round some radius and was obtained, simply by obtaining relationships and simplifying the original spin orbit coupling equation.

**In style of the Bohr Theory**

The Newtonian motion (F = ma) for circular orbits is

k Z[e²/r²] = mv²/r

Which turns out to be a centrifugal term. Take note of the interesting dynamics, which is fully equivalent to the term (∂U[r]/∂r) in our derived equation

ΔH = -[ħ/mc] ∂U[r]/∂r

The term (∂U[r]/∂r) is identical to a simple (mv²/r) thus we can take the approximation as a good estimate of the energy required in that equation and it gives simply:

ΔH = -[ħ/mc] k Z[e²/r²]

For hydrogen atoms, this results even simpler, since Z=1

ΔH = -[ħ/mc] k[e²/r²]

**more to be posted**