Building Quark Models
//Ajat basket//, Penan people. Borneo 20th century, 20 (cm) diameter by 35 (cm) height. Hornbill motif. From the Teo Family collection, Kuching. Photograph by D Dunlop.
Ajat basket, Penan people. Borneo 20th century, 20 (cm) diameter by 35 (cm) height. Hornbill motif. From the Teo Family collection, Kuching. Photograph by D Dunlop.

Consider a particle P characterized by some repetitive chain of events

$\Psi ^{\sf{P}} = \left( \sf{\Omega}_{1} , \sf{\Omega}_{2} , \sf{\Omega}_{3} \ \ldots \ \right)$

where each orbital cycle is a bundle of quarks

$\sf{\Omega} = \left\{ \sf{q}_{1}, \sf{q}_{2}, \sf{q}_{3 } \ \ldots \ \right\}$

described by some quark coefficients $n$ and their internal energies $U$. Recall that $H_{chem}$ notes the enthalpy of the chemical quarks in P. The constant $k_{\sf{F}}$ has a positive value, with the units of a force. These quantities are used to define the following radii that describe the form and location of P. The chemical radius is

$\begin{align} \rho_{chem} \equiv \frac{H_{chem}}{k_{\sf{F}}} \end{align}$

The magnetic radius of P is

$\begin{align} \rho_{m} \equiv \frac{ {\Delta}n^{ \mathsf{A}} U^{ \mathsf{A}} - {\Delta}n^{ \mathsf{M}} U^{ \mathsf{M}} }{ k_{\sf{F}} } \end{align}$

the electric radius is given by

$\begin{align} \rho_{e} \equiv \frac{ {\Delta}n^{ \mathsf{G}} U^{ \mathsf{G}} - {\Delta}n^{ \mathsf{E}} U^{ \mathsf{E}} }{ k_{\sf{F}} } \end{align}$

the polar radius of P is defined as

$\begin{align} \rho_{z} \equiv \, \rho_{chem} + \frac{{\Delta}n^{\mathsf{U}} U^{\mathsf{U}} - {\Delta}n^{\mathsf{D}} U^{\mathsf{D}}}{k_{\sf{F}}} \end{align}$

An ordered set defines the radius vector of particle P as

$\overline{\rho} \equiv \left( \rho_{ m}, \rho_{e}, \rho_{ z} \right)$

Components of $\overline{\rho}$ are conserved because they are defined from sums and differences of quark coefficients, and quarks are indestructible. So if some generic particles $\mathbb{X}$, $\mathbb{Y}$ and $\mathbb{Z}$ interact like $\mathbb{ X } + \mathbb{ Y } \leftrightarrow \mathbb{ Z }$ then by the associative propertiesXlink.png of addition and subtraction

$\overline{\rho}^{ \mathbb{X}} + \overline{\rho}^{\mathbb{Y}} = \overline{\rho}^{\mathbb{Z}}$

Also $\rm{\Delta} \it{n} ^{\sf{Z}} \left( \sf{P} \right) = - \rm{\Delta} \it{n}^{\sf{Z}} \left( \sf{\bar{P}} \right)$ so particles and anti-particles have symmetrically opposed radius vectors

$\overline{\rho} \left( \sf{P} \right) = - \overline{\rho} \left( \sf{\overline{P}} \right)$

If P has the same radius vector for every cycle then we call it a rigid particle. We use the foregoing radii to describe particle shape. But some particles are not very solid or exactly located. Then we may use the following radii to describe their range. The inner radius is

$\begin{align} \rho_{\LARGE{\bullet}} \equiv \frac{ \left| \Delta n^{\sf{D}} \right| }{8} \sqrt{ \frac{hc}{ k_{\sf{F}}} \vphantom{\frac{hc}{2 \pi k_{\sf{F}}}^2} } \end{align}$

And the outer radius of P is

$\begin{align} \rho_{\LARGE{\circ}} \equiv \frac{N^{\sf{D}}}{8} \sqrt{ \frac{hc}{ k_{\sf{F}}} \vphantom{\frac{hc}{2 \pi k_{\sf{F}}}^2} } \end{align}$

We say that a particle is perfectly free when $\left| \Delta n^{\sf{D}} \right| = 8$ and $N^{\sf{D}} \! \to ∞$. Note that almost any stable material particle that has absorbed even one photon will be net at least eight down-quarks because quarks are conserved, all photons have $\left| \Delta n^{\sf{D}} \right| \ge 8$ by definition, and our quark models of protons, electrons, neutrons and hydrogen atoms have zero net down-quarks. So, in terms of the radii we can say that a particle is free when its outer radius approaches a large value $\rho_{\LARGE{\circ}} \to {\rho_{\LARGE{\circ}} }_{max}$ while its inner radius is as small as possible $\rho_{\LARGE{\bullet}} \to {\rho_{\LARGE{\bullet}} } _{min}$.

Sensory Interpretation: In these formulae $\Delta n$ means that contributions from sensations on the right side are cancelled by sensations felt on the left. The radius vector depends on their net magnitude. Coefficients of baryonic quarks do not appear in these definitions, only dynamic quarks. So the radius does not depend on thermal sensation, only somatic and visual sensations. Also the null value for energy is referred to down-quarks which are objectified from black sensations. So overall the radius vector is interpreted as a description of the magnitude of somatic and visual sensations, relative to black sensations, net right from left.
Right.png Next step: making room for quark models.
Noun Definition
Chemical Radius $\begin{align} \rho_{chem} \equiv \frac{H_{chem}}{k_{\sf{F}}} \end{align}$ 5-2
Noun Definition
Magnetic Radius $\begin{align} \rho_{ m} \equiv \frac{ {\Delta}n^{ \mathsf{A}} U^{ \mathsf{A}} - {\Delta}n^{ \mathsf{M}} U^{ \mathsf{M}} }{ k_{\sf{F}} } \end{align}$ 5-3
Noun Definition
Electric Radius $\begin{align} \rho_{e} \equiv \frac{ {\Delta}n^{ \mathsf{G}} U^{ \mathsf{G}} - {\Delta}n^{ \mathsf{E}} U^{ \mathsf{E}} }{ k_{\sf{F}} } \end{align}$ 5-4
Noun Definition
Polar Radius $\begin{align} \rho_{z} \equiv \, \rho_{chem} + \frac{{\Delta}n^{\mathsf{U}} U^{\mathsf{U}} - {\Delta}n^{\mathsf{D}} U^{\mathsf{D}}}{k_{\sf{F}}} \end{align}$ 5-5
Noun Definition
Inner Radius $\begin{align} \rho_{\LARGE{\bullet}} \equiv \frac{ \lvert \Delta n^{\sf{D}} \rvert }{8} \sqrt{ \frac{hc}{ k_{\sf{F}}} \vphantom{\frac{hc}{ k_{\sf{F}}}^2} } \end{align}$ 5-6
Noun Definition
Outer Radius $\begin{align} \rho_{\LARGE{\circ}} \equiv \frac{N^{\sf{D}}}{8} \sqrt{ \frac{hc}{ k_{\sf{F}}} \vphantom{\frac{hc}{2 \pi k_{\sf{F}}}^2} } \end{align}$ 5-7
Noun Definition
Radius Vector $\overline{\rho} \equiv \left( \rho_{ m}, \rho_{e}, \rho_{ z} \right)$ 5-8
Noun Definition
Rigid Particle $\overline{\rho} \ \ {\sf\text{has the same value for all events} }$ 5-9
Noun Definition
Free Particle $\lvert \Delta n^{\sf{D}} \rvert = 8 \ \ \ \ \ {\sf{\text{and}}} \ \ \ \ \ N^{\sf{D}} \! \to ∞$ 5-10
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