An Electronic Photon |

$\sf{P}_{\! \it{k}}$ | $\delta _{\hat{m}}$ | $\delta _{\hat{e}}$ | $\delta _{\theta}$ | quarks |

1 | +1 | 0 | +1 | $\sf{2u}$ |

2 | 0 | +1 | +1 | $\sf{4\bar{e}}$ |

3 | -1 | 0 | +1 | $\sf{2u}$ |

4 | 0 | -1 | +1 | $\sf{4g}$ |

5 | +1 | 0 | -1 | $\sf{2\bar{u}}$ |

6 | 0 | +1 | -1 | $\sf{ 4e }$ |

7 | -1 | 0 | -1 | $\sf{2\bar{u}}$ |

8 | 0 | -1 | -1 | $\sf{4\bar{g}}$ |

Consider an electronic photon $\sf{\gamma}_ {\sf{e}}$ described by the repetitive chain of events

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

where each orbital cycle is given by

$\sf{\Omega} ( \sf{\gamma}_ {\sf{e}} ) = 4\sf{e} + 4\bar{\sf{e}} + 4\sf{g} + 4\bar{\sf{g}} + 4\sf{u} + 4\bar{\sf{u}}$

Let these quarks be parsed into eight components

$\sf{\Omega} ( \sf{\gamma}_ {\sf{e}} ) = \left\{ \sf{P}_{\! 1}, \sf{P}_{\! 2}, \sf{P}_{\! 3}, \sf{P}_{\! 4}, \sf{P}_{\! 5}, \sf{P}_{\! 6}, \sf{P}_{\! 7}, \sf{P}_{\! 8} \right\}$

as shown in the accompanying table. The compound event $\sf{\Omega}$ cannot fully satisfy the conditions for being a space-time event because there are not enough quarks to properly distinguish components by their magnetic polarity $\delta _{\hat{m}}$. So instead we just assert that the photon's quarks are distributed as shown, as if the photon was part of a well-defined atom. This convention is called a three dimensional model of an electronic photon.

A three-dimensional quark model of an electronic photon. |