Preliminary Investigations into Spin Orbit Coupling Fields

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Joined: Sat Nov 07, 2015 7:26 pm

Preliminary Investigations into Spin Orbit Coupling Fields

Postby Bohr'sGhost » Sat Nov 07, 2015 7:54 pm

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


[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

Posts: 2
Joined: Sat Nov 07, 2015 7:26 pm

Re: Preliminary Investigations into Spin Orbit Coupling Fields

Postby Bohr'sGhost » Sat Nov 07, 2015 7:56 pm

Quantization of Magnetic Flux

A spin zitter motion can be given as

1/R = (2/ħ)²p∙S

1/R² = (2/ħ)²p∙S/R

and from previous equations

eA∙S = [L∙S]/r

eA∙p = [L∙S]/r²

we may unite the two definitions by noticing they all reduce to p² or the dot product (p∙p)

eA∙p = p∙S/R = [L∙S]/r²

we can notice just for mentioning it, a kinetic energy term

KE = eA∙p/2m

by uniting some of the equations, one can get the following equations

eA∙p = p∙S/R

eA∙ħ = p∙S

The left hand side has retrieved the electrokinetic momentum

eA = p∙S/ħ

It is integral to rotational equations. Note that a total momentum includes an extra term

mv + eA

and clearly

R × eA = ħ

as a quantization condition. In a sphere of radius R, Stokes theorem allows a relationship between the magnetic flux and the vector potential

Φ = 2πRA

Perhaps interesting to the reader, I did find the appropriate ''couplings'' for the terms required in a spin orbit interaction

Φ = (1/emc²∙1/r)L∮ F∙dr

If we distribute a charge we have

eΦ∙1/2π = R × eA = ħ

From the equations we should note that

eΦ = e ∬ B∙dS = 1/c∙ ∮ [∂U/∂r] dS = ħ

eB = 1/c ∂U[r]/∂r

A Lorenzian force can be obtained

e(c × B ) = ∂U[r]/∂r

ec ∬ B∙dS = ∮ [∂U/∂r] dS = ħc

You can work out the dimensions say of this term ∮ [∂U/∂r] dS by noticing that surface has dimensions of length squared so we eradicate the radius then it it just dimensions of energy times a length or quantum action times a velocity term, which is of course the gravitational charge.

Zeeman Interaction

qr∙E -[eB∙S]/2m

Is the Zeeman interaction term (see references 1.)

Ω =eB/(2m)

is Zeeman frequency

qr∙E - Ω∙S

γ = -e/2m

is the gyromagnetic ratio, a negative charge for an electron, it describes a special case of an electron rotating round a nucleus.

Notice it is a charge to mass ratio, with a coefficient of 2 attached to the mass in the denominator.

The Bohr magneton is therefore

μ = -(e/2m) L

Spin magnetic moment is (differs only slightly)

μ(s) = -(eg/2m) S

A total magnetic moment is the sum of the orbital and spin moments

μ = μ + μ(s) = -(e/2m)[L + gS]

In previous derivations, I discovered that

eB∙S = [L∙S]/r²

(see derivation (2)) Divide the factor of 2m from both sides and we obtain one of the interaction terms

eB∙S/2m = [L∙S]/2mr² = Ω∙S

∵ Ω =eB/(2m)

Interestingly (but perhaps not to unexpected), we have obtained the rotational inertia mr² weighed by the angular momentum dot product with spin

From the ''master equation'' (1) we take the equation

Φ = [1/emc²] 1/r [∂U[r]/∂r] L∙dS

Allow a new notation so not to get confused with the dS term, let the spin vector be S = ћ

= [1/emc²] [∂U[r]/∂r]dS L∙ћ/r

Divide by mr

= (1/emc² [∂U[r]/∂r]dS) L∙ћ/mr²

We have retrieved the inertial rotator mr²

= (1/emc² [∂U[r]/∂r]dS) L∙ћ/mr²

I proved in other work that

eΦ = e ∬ B∙dS = 1/c ∮[∂U[r]/∂r]dS = ћ

thus we can substitute and simplify AND regain the flux (with the dS notation gone, we go back to using the standard sign for the spin vector S)

Φ = (ћ/emc²) L∙S/mr²

The last term is the Zeeman interaction

Φ = (ћ/emc²) eB∙S/2m = (ћ/emc²) Ω∙S

which reduces to a compact equation

Φ = ћ²Ω/emc²

(1) The master equation takes the perspicuous form of

∬ 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

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