Excited Particles
 Bead panel from a baby carrier, Kayan or Kenyah people. Borneo 20th century, 27 x 27 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop.

Consider a free particle P described by an ordered chain of events where each orbital cycle $\sf{\Omega}$ can be expressed as a pair of events that are out of phase with each other so that

$\sf{\Omega} ^{\sf{P}} = \left\{ \sf{P}_{\LARGE{\circ}} , \sf{P}_{\LARGE{\bullet}} \vphantom{Q^{2}} \right\}$

and

$\delta _{\theta} \left( \sf{P}_{\LARGE{\circ}} \right) =- \, \delta _{\theta} \left( \sf{P}_{\LARGE{\bullet}} \right) = \pm \rm{1}$

Then $\sf{P}_{\LARGE{\circ}}$ and $\sf{P}_{\LARGE{\bullet}}$ are called phase components of P. Some sub-set of the quarks in $\sf{P}_{\LARGE{\circ}}$ may be matched with quarks of the same type in $\sf{P}_{\LARGE{\bullet}}$. These quarks have phase symmetry with each other, so we use $\mathcal{S}_{\LARGE{\circ}}$ and $\mathcal{S}_{\LARGE{\bullet}}$ to symbolize these sets. A different sub-set of quarks might be matched with anti-quarks of the same type. These quarks have phase anti-symmetry with each other, so we note them as $\mathcal{A}_{\LARGE{\circ}}$ and $\mathcal{A}_{\LARGE{\bullet}}$. Some quarks in $\sf{P}_{\LARGE{\circ}}$ might not correspond with any quarks in $\sf{P}_{\LARGE{\bullet}}$, and vice versa. But such lopsided possibilities seem to be so unstable or superfluous that we do not consider them further. And so P is represented by the union of two entirely symmetric, and two purely anti-symmetric components. Quarks in the symmetric sets may vary independently of the quarks in the anti-symmetric sets, and vice versa. We write

$\sf{\Omega}^{\sf{P}} = \left\{ \left\{ \mathcal{S}_{\LARGE{\circ}}, \mathcal{A}_{\LARGE{\circ}} \right\}, \ \left\{ \mathcal{S}_{\LARGE{\bullet}}, \mathcal{A}_{\LARGE{\bullet}} \right\} \vphantom{{\Sigma^{2}}^{2}} \right\}$

where

 $\mathcal{S}_{\LARGE{\circ}} = \mathcal{S}_{\LARGE{\bullet}}$ $\mathcal{A}_{\LARGE{\circ}} = \overline{\mathcal{A} _{\LARGE{\bullet}} }$ $\delta _{\theta} \left( \mathcal{S}_{\LARGE{\circ}} \right) =- \, \delta _{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right)$ $\delta _{\theta} \left( \mathcal{A}_{\LARGE{\circ}} \right) =- \, \delta _{\theta} \left( \mathcal{A}_{\LARGE{\bullet}} \right)$

This arrangement provides a general way of describing particles that are moved or excited by the absorption of additional quarks. Definition: P is called an excited particle, or said to be in an excited-state, if it contains at least one anti-symmetric pair of quarks

$\mathcal{A}_{\LARGE{\circ}} = \overline{ \mathcal{A}_{\LARGE{\bullet}} } \ne \left\{ {\sf{∅}} \right\}$

These anti-symmetric quark-pairs may be due to the absorption of a photon. Or more generally, to interactions with any field quanta.

## Ground State Particles

Definition: We say that P is in its ground state if it has perfect phase symmetry. Then P has no anti-symmetric quark-pairs

$\mathcal{A}_{\LARGE{\circ}} = \overline{ \mathcal{A}_{\LARGE{\bullet}} } = \left\{ {\sf{∅}} \right\}$

This definition constrains particle-models so that quark-coefficients must all be integer multiples of two when in the ground-state. It is why many nuclear particle models show patterns like 2-4-6 instead of 1-2-3. We can also evaluate the wavevector of P which is given by the sum

\begin{align} \overline{ \kappa }^{ \sf{P}} &= \frac{ k_{\sf{F}}}{hc} \sum_{i=1}^{N} \delta _{\theta}^{\, i} \, \bar{\rho}^{i} \\ &= \frac{ k_{\sf{F}}}{hc} \left[ \delta_{\theta} \left(\mathcal{S}_{\LARGE{\circ}} \right) \overline{\rho} \left( \mathcal{S}_{\LARGE{\circ}} \right) + \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \overline{\rho} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \vphantom{\Sigma^{2}} \right] \\ \\ &= \frac{ k_{\sf{F}}}{hc} \left[ \delta_{\theta} \left(\mathcal{S}_{\LARGE{\circ}} \right) + \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \vphantom{\Sigma^{2}} \right] \, \overline{\rho}^{\mathcal{S}_{\LARGE{\circ}} } \ \ \ \ \sf{\text{because}} \ \ \ \ \overline{\rho} \left( \mathcal{S}_{\LARGE{\circ}} \right) = \overline{\rho} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \\ \\ &= (0, 0, 0) \ \ \ \ \sf{\text{because}} \ \ \ \ \delta_{\theta} \left( \mathcal{S}_{\LARGE{\circ}} \right) = - \, \delta_{\theta} \left( \mathcal{S}_{\LARGE{\bullet}} \right) \end{align}

And so

$\kappa ^{ \sf{P}} \equiv \left\| \, \overline{\kappa}^{ \sf{P}} \right\| = 0$

The wavenumber of a particle in its ground-state is always zero.

 Next step: nuclear events.
page revision: 189, last edited: 21 Aug 2018 21:27