Ground State Quantum Vortex Proton Model

Abstract

A novel photon-based proton model is developed. A proton’s ground state is assumed to be coherent to the degree that all of its mass-energy precipitates into a single uncharged spherical structure. A quantum vortex, initiated by the strong force, but sustained in the proton’s ground state by the circular Unruh effect and a spherical Rindler horizon, is proposed to confine the proton’s mass-energy in its ground state. A direct connection between the circular Unruh effect, the zitterbewegung effect, spin, and general relativity is proposed. Such a structure acts as an uncharged zitterbewegung fermion, and may explain neutrino mass. A ground-state proton’s central zitterbewegung fermion is assumed to be surrounded by a halo of charge shells of both signs. Virtual photon standing waves are assumed to synchronize the inner shell with the central zitterbewegung fermion. The charge shells are proposed to be associated with isospin and proton g-factor. There are only two model inputs—proton mass and quantized electronic charge—and just one adjustable parameter. The adjustable parameter, reduced only by about 0.4% from an initial estimate, provides the proton’s experimentally determined magnetic moment to arbitrary precision. The resulting modeled proton charge radius agrees very well with the 2018 CODATA value. Magnetic moment and charge radius are calculated algebraically in a manner easily understood by undergraduate physics students. This proposed ground-state proton model could be considered a low-energy approximation to a full quantum chromodynamical proton model.

Appendix: Volume and Surface Area Derivations

A spindle torus consists of an inner lemon surface and an apple outer surface. The polar cross-section of a spindle torus is shown in Fig. 1. The lemon is generated by rotating an arc of half-angle $\phi _m$ less than $\pi /2$ about its chord, with $\phi _m=\phi _l=\cos ^{-1}(r_z/R)$. The half-angle, $\phi _l$, is shown in Fig. 1. Note that $\phi$ denotes latitude, as used in geophysics. It does not denote the azimuthal angle of conventional physics spherical coordinates. The surface area is given by

$$\begin{aligned} A=2\pi R^2\int _{-\phi _m}^{\phi _m}(\cos \phi -\cos \phi _m)d\phi . \end{aligned}$$

(30)

The volume is given by

$$\begin{aligned} V=\pi R^3\int _{-\phi _m}^{\phi _m}(\cos \phi -\cos \phi _m)^2\cos \phi d\phi . \end{aligned}$$

(31)

These integrals can be evaluated analytically, giving

$$\begin{aligned} A=4\pi R^2(\sin \phi _m-\phi _m\cos \phi _m) \end{aligned}$$

(32)

$$\begin{aligned} V=\tfrac{4}{3}\pi R^3\left[ \sin ^{3}\phi _m-\tfrac{3}{4}\cos \phi _m(2\phi _m-\sin 2\phi _m)\right] \end{aligned}$$

(33)

The apple is generated by rotating an arc of half-angle $\phi _m$ greater than $\pi /2$ about its chord, with $\phi _m=\phi _a=\pi -\cos ^{-1}(r_z/R)=\pi -\phi _l$. The half-angle, $\phi _a$, is shown in Fig. 1. Note that Eqs. (32) and (33) are valid for both the lemon and apple.

With the refined GSQV proton model, R becomes an adjustable parameter. The quantity $r_z$ is replaced by the radius of the polar charge-exclusion zone,

$$\begin{aligned} r_a=R-\lambda _c/2 . \end{aligned}$$

(34)

The outer charge shell is modeled as a semicircle revolved about flat polar caps. Pappus’s centroid theorems are applied. The area of a surface of revolution, generated by rotating a plane curve about an external axis in the same plane, is equal to the product of the arc length of the curve and the distance travelled by the centroid of the curve. In this case, the curve is a semicircular arc with centroid located $2R/\pi$ more distant than $r_a$. The arc length of the semicircle is $\pi R$, and the distance travelled by the centroid is $2\pi (r_a+2R/\pi )$. The area of each endcap is $\pi r_a^2$. Therefore, including both endcaps, the outer shell surface area is given by

$$\begin{aligned} A_o=2\pi ^2R\left( r_a+\tfrac{2R}{\pi }\right) +2\pi r_a^2 . \end{aligned}$$

(35)

The volume of a solid of revolution, generated by rotating a plane figure about an external axis in the same plane, is equal to the product of the area of the figure and the distance travelled by its centroid. In this case, the plane figure is a semicircular area with centroid located $4R/3\pi$ more distant than $r_a$. The area of the semicircle is $\pi R^2/2$, and the distance travelled by the centroid is $2\pi (r_a+4R/3\pi )$. The cylindrical volume between the two endcaps is $2\pi r_a^2R$. Therefore, the outer shell volume is given by

$$\begin{aligned} V_o=\pi ^2 R^2\left( r_a+\tfrac{4R}{3\pi }\right) +2\pi r_a^2R . \end{aligned}$$

(36)