Infinitesimal Circular Motion generating Zeeman Effects and Spin-Orbit Coupling

[Shadax]

[math]\int \int \int_V \nabla \times F \cdot dV = \int \int_S F \cdot dS[/math]

Stokes theorem states

[math]\int \int_{S} \nabla \times F \cdot dS = \oint_{\partial S} F \cdot dr[/math]

Greens theorem is a special case of Stokes theorem.

force is

[math]F = \frac{\partial U(r)}{\partial r}[/math]

substitution gives

[math]\int \int_{S} \nabla \times F \cdot dS = \oint_{\partial S} \frac{\partial U(r)}{\partial r} \cdot dr[/math]

As you can see it has dimensions of energy.

Consider the magnetic field now

[math]B = \frac{\hbar}{emc^2} \frac{1}{r}\frac{\partial U(r)}{\partial r}[/math]

Plugging in only part of the magnetic field expression [math]\frac{\hbar}{emc^2} \frac{1}{r}[/math]

[math]\int \int_{S} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dS] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dr[/math]

working from the last equation, performing the integral over the volume this time will yield

[math]\int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]

Since this is just essentially a magnetic field times a surface, we have a magnetic flux

[math]\Phi = \int \int_S B \cdot dS = \int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]

we should obtain a quantization

[math]\hbar = e\Phi = \frac{e}{2\pi} \int \int_S B \cdot dS = \frac{1}{2\pi} \int \int \int_{V} \frac{\hbar}{mc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \frac{1}{2\pi}\oint_{\partial S} \frac{\hbar}{mc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]

It should be noted that the Gauss law for magnetism implies that our flux' date=' if over a closed surface [math']S[/math] is zero:

[math]\Phi = \int \int_S B \cdot dS = \int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS = 0[/math]

From this equation, we can rearrange to find [math]\frac{dS}{dV}[/math] which is a length squared term over a length cubed term, which yields an inverse length we will call the radius, which will imply an equation for magnetism of the form:

[math]\mathbf{B} = \int \int \int_{V} \frac{\hbar}{emc^2} \cdot [\nabla \times F][/math]

(we will show why soon). We can compare the physics of this equation with another magnetic term

[math]\mathbf{B} = \frac{\hbar}{emc^2} \frac{1}{r}\frac{\partial U(r)}{\partial r}[/math]

[math]\int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]

divide volume

[math]\int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot \frac{dS}{dV}[/math]

Surface over volume is equivalent to [math]\frac{dS}{dV} \equiv \frac{r^2}{r^3} = \frac{1}{r}[/math] so we have

[math]\int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r^2} \frac{\partial U(r)}{\partial r}[/math]

We have two terms of the inverse on both sides we can cancel as well to give us a new statement and proves the assertion above of a magnetic field equation:

[math]\mathbf{B} = \frac{\hbar}{emc^2} \cdot [\nabla \times F] = \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

And in fact, by knowing the relationship of the potential [math]\mathbf{A}[/math] with the magnetic field as

[math]\mathbf{B} = \nabla \times \mathbf{A}[/math]

then the inverse operator is acting on the magnetic field in an analagous way to the circular gauge [math]\mathbf{A} = \mathbf{B} \times r[/math], as

[math]\mathbf{A} = (\nabla \times)^{-1}\mathbf{B} = \frac{\hbar F}{emc^2} = (\nabla \times)^{-1}\frac{\hbar}{emc^2} \frac{1}{r}\frac{\partial U(r)}{\partial r}[/math]

The term [math]\frac{\hbar F}{emc^2}[/math] is a bit obscure - what is it? The force over energy can be though of as the inverse of the energy divided by a force which is

[math]\frac{E}{F} = \frac{1}{2} \frac{v^2}{a}[/math]

So

[math]F = \frac{E}{\frac{1}{2} v t} = \frac{2E}{v t}[/math]

So Force is the twice rate of change of Energy with respect to the velocity multiplied by time.

What's the inverse of

[math]\frac{E}{F} = \frac{1}{2} \frac{v^2}{a}[/math]

simplify, rearrange. Simple, gives:

[math]\frac{F}{E} = \frac{2a}{v^2}[/math]

This gives us the relationship

[math]\frac{2a \hbar }{e c^2}[/math]

and you can simplify this all the way now, by noticing the action is an angular momentum times a length term, we finally have

[math]\frac{2Ma c R}{e c^2} = \frac{2F R}{e c} = \frac{2mc^2}{e c} = \frac{2mc}{e} = \frac{2p}{e}[/math]

It may be of interest to note for any further investigations, that this has a dimension of rigidity [math]\mathbf{R}[/math] we can be shown via the equation:

[math]\mathbf{R} = \mathbf{B} r_g = \frac{p}{e}[/math]

where [math]r_g[/math] is the gyroradius (the radius associated to circular motion). It becomes more obvious that

[math]\mathbf{A} = (\nabla \times)^{-1}\mathbf{B} = \frac{\hbar F}{emc^2} = (\nabla \times)^{-1}\frac{\hbar}{emc^2} \frac{1}{r}\frac{\partial U(r)}{\partial r}[/math]

... Is right, when you consider the momentum is in fact the electrokinetic momentum given by [math]e\mathbf{A} = p[/math]. Now it seems a bit strange to me I have one source claiming momentum over charge is a rigidity - but those dimensions work out when you notice [math]\mathbf{A}[/math] has the same dimensions of [math]\frac{p}{e}[/math] in face of the electrokinetic momentum which is known as [math]e\mathbf{A} = p[/math]. Anyway moving on.

[math]]\mathbf{B} = \frac{\hbar}{emc^2} \cdot [\nabla \times F] = \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

rearranging gives

[math]e\mathbf{B} = \frac{\hbar}{mc^2} \cdot [\nabla \times F] = \frac{\hbar}{mc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

relaxing all but primary action terms to the other notations, [math]L[/math] or [math]S[/math], we have distributing the action and then dividing a mass we have now a Zeeman energy interaction term, which will then justify the appearance of a new term in the equation

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

Two equations were of particular interest as:

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

We didn't have to do much to get from

[math]\mathbf{B} = \frac{\hbar}{emc^2} \cdot [\nabla \times F] = \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

We can change the central potential energy [math]U(r)[/math] for a screened Coulomb potential (aka. the Yukawa Potential) and this would give

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \frac{1}{r} \frac{Gm}{r^2} e^{-kr}[/math]

This gives us something alternative to work with. It's not unusual to find a Yukawa term, in a central potential.

A centripetal force is

[math]F = m \omega^2 R[/math]

In which [math]\omega[/math] is the angular velocity.

Equating Newtons classical gravitational force equation with the clasical centripetal force we have

[math]\frac{Gm^2}{R^2} = m \omega^2 R[/math]

simplifying and rearranging we have the term

[math]\omega^2 = \frac{Gm}{R^3}[/math]

which simplifies the term [math]\frac{1}{r} \frac{Gm}{r^2}[/math], so we replace and we have

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \omega^2 e^{-kr}[/math]

we found:

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r}[/math]

And I showed the correct derivation for the second part as

[math]\omega^{2} = \frac{L^2}{m^2R^4} = \frac{1}{mR} \frac{\partial U}{\partial r}[/math]

multiply through by mass we have

[math]m \omega^{2} = \frac{L^2}{mR^4} = \frac{1}{R} \frac{\partial U}{\partial r}[/math]

Now, in other previous equations, I was able to derive the following:

[math]e\mathbf{B}r^2 = mvR = L[/math]

and

[math]e\mathbf{B} = \frac{L}{R^2}[/math]

and to match this with

[math]m \omega^{2} = \frac{L^2}{mR^4} = \frac{\partial U}{r \partial r}[/math]

distribute an action term

[math]e\mathbf{B} \hbar = \frac{L^2}{R^2}[/math]

dividing off [math]mR^2[/math] gives

[math]\frac{e\mathbf{B}\hbar}{mR^2} = \frac{L^2}{mR^4}[/math]

which now fits this term:

[math]\frac{L^2}{mR^4} = \frac{\partial U}{r \partial r}[/math]

Using the correct substitution for [math](1/r \cdot U/r)[/math] now, which is [math]\frac{e\mathbf{B}\hbar}{mR^2}[/math], substitution is now

[math]eR \mathbf{E} - \frac{e\mathbf{B}\hbar}{m} = \frac{L \cdot S}{m^2c^2} \cdot [\nabla \times F] = \frac{L \cdot S}{m^2c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} = \frac{L \cdot S}{m^2c^2}\frac{e\hbar}{mR^2}[ \nabla \times \mathbf{A}][/math]

where [math]\nabla \times \mathbf{A}[/math] is the circular gauge, equivalent of magnetic field.

[imatfaal]

!

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