A Three Dimensional Model of a Photon
 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.