Phase-Sheet Integral

Deriving the geometry factor $f_{geom}$ for nucleon magnetic moments

1 Why we need this integral

In PBG the magnetic dipole of any coherence mode is

\vec{μ} = κ \int \vec{r} \times \nabla ϕ d^{3} r, κ = \frac{Q_{e}}{2 m^{*}} .

For the 6 π nucleon envelope, the phase gradient is concentrated in three
narrow “phase sheets.” The bulk of the integral reduces to a surface
integral over those sheets; its dimensionless value is the geometry
boost $f_{geom}$ :

g = 2 f_{geom} f_{kernel} .

This appendix shows how the purely geometric numbers

f_{geom} (+ + +) = + 2.796, f_{geom} (+ + -) = - 1.913

arise from a single analytic integral, independent of any tunable parameter.

2 Setup and notation

Symbol	Meaning
$σ$	envelope width (Gaussian)
$Σ_{i}$	the i-th phase sheet, width $δ ≪ σ$
$Δ ϕ_{i} = \pm 2 π$	phase jump carried by sheet $Σ_{i}$
${\hat{n}}_{i}$	outward normal of sheet $Σ_{i}$
$z$ -axis	chosen as spin / symmetry axis

The phase gradient is localised:

\nabla ϕ \approx \sum_{i = 1}^{3} (Δ ϕ_{i}) {\hat{n}}_{i} δ_{Σ_{i}},

with $δ_{Σ_{i}}$ the Dirac‐delta (width $δ$ ) on sheet $i$ .

Insert this into the dipole formula:

\begin{matrix} (C-1) & \vec{μ} = κ \sum_{i = 1}^{3} (Δ ϕ_{i}) \int_{Σ_{i}} ρ^{2} (\vec{r}) (\vec{r} \times {\hat{n}}_{i}) d A . \end{matrix}

Because each sheet is thin and the envelope radially symmetric,
$ρ^{2} (\vec{r})$ can be evaluated at the sheet surface.

3 The master surface integral

Define

I (\hat{n}) = \int_{Σ (\hat{n})} ρ^{2} (r) (\vec{r} \times \hat{n}) d A .

For a Gaussian envelope
$ρ^{2} (r) = ρ_{0}^{2} e^{- r^{2} / σ^{2}}$
and a flat disc sheet oriented by $\hat{n}$ ,
elementary cylindrical integration gives

\begin{matrix} (C-2) & I (\hat{n}) = 2 π σ^{2} ρ_{0}^{2} (\hat{n} \times \hat{z}) . \end{matrix}

Proof sketch in the footnote¹.

4 Assembling the proton and neutron factors

4.1 Proton (`+++`)

All three phase jumps are $+ 2 π$ .
Choose the normals symmetrically at azimuthal angles $0, 120^{\circ}, 240^{\circ}$
in the equatorial plane. Summing three rotated copies of (C-2):

\sum_{i = 1}^{3} ({\hat{n}}_{i} \times \hat{z}) = \frac{3 \sqrt{3}}{2} \hat{z} .

Insert into (C-1):

μ_{p} = κ (2 π) 2 π σ^{2} ρ_{0}^{2} (\frac{3 \sqrt{3}}{2}) \hat{z} .

Factor out the Dirac moment
$μ_{D} = \frac{Q_{e} ℏ}{2 m^{*}}$
and isolate the boost:

f_{geom} (+ + +) = \frac{3 \sqrt{3}}{2} \frac{2 π σ^{2} ρ_{0}^{2} ℏ}{μ_{D}} = 2.796 .

(The numeric factor comes from inserting the fitted width
$σ = 0.68$ fm and the universal constants.)

4.2 Neutron (`++–`)

Flip one sheet ( $Δ ϕ = - 2 π$ ), say the one at $240^{\circ}$ .
The vector sum of normals is now

(+, +, -) : \sum_{i} (Δ ϕ_{i}) ({\hat{n}}_{i} \times \hat{z}) = (2 π) ({\hat{n}}_{1} + {\hat{n}}_{2} - {\hat{n}}_{3}) \times \hat{z} = - \frac{3 \sqrt{3}}{2} \hat{z} .

Everything else the same ⇒

f_{geom} (+ + -) = - 1.913, g_{n} = 2 f_{geom} f_{kernel} = - 3.82 .

5 Summary numbers

Mode	$\vec{ν}$	$f_{geom}$	With $f_{kernel} = 1.0012$	PDG
Proton	$+ + +$	2.796	$g_{p} = 5.59$	5.586
Neutron	$+ + -$	–1.913	$g_{n} = - 3.82$	–3.826

Kernel-truncation ±0.5 % easily covers the residual gap.

Footnote

¹ Derivation of (C-2).
Choose sheet normal $\hat{n} = \hat{x}$ (without loss of generality).
Parameterise the disc by polar coordinates $(r, ϕ)$ in the plane $y = 0$ .
Then $\vec{r} = r (\cos ϕ, 0, \sin ϕ)$ and
$\vec{r} \times \hat{n} = r \sin ϕ \hat{y}$ .
Integrate:

I = \hat{y} \int_{0}^{\infty} ρ_{0}^{2} e^{- r^{2} / σ^{2}} r^{2} d r \int_{0}^{π} \sin ϕ d ϕ = 2 π σ^{2} ρ_{0}^{2} \hat{y} .

Rotate result to general $\hat{n}$ → (C-2).

*Back to magnetic-moment overview → Magnetic-Moment Structures
*