A Three Dimensional Model of a Graviton
 The Electronic Graviton
 $\sf{P}_{\! \it{k}}$ $\delta _{\hat{m}}$ $\delta _{\hat{e}}$ $\delta _{\theta}$ quarks 1 +1 0 +1 $\sf{ 2u } \ \ \sf{2\bar{u}}$ 2 0 +1 +1 $\sf{4e} \ \ \sf{4\bar{e}}$ 3 -1 0 +1 $\sf{ 2u } \ \ \sf{2\bar{u}}$ 4 0 -1 +1 $\sf{ 4g} \ \ \sf{4\bar{g}}$ 5 +1 0 -1 $\sf{ 2u } \ \ \sf{2\bar{u}}$ 6 0 +1 -1 $\sf{4e} \ \ \sf{4\bar{e}}$ 7 -1 0 -1 $\sf{ 2u } \ \ \sf{2\bar{u}}$ 8 0 -1 -1 $\sf{ 4g} \ \ \sf{4\bar{g}}$

Consider an electronic graviton composed from an electronic photon and an electronic anti-photon

$\sf{\Gamma} _{\! \sf{e}} \equiv \left\{ \gamma_{\sf{e}} , \overline{\gamma}_{\sf{e}} \right\}$

This graviton is described by a 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

$\sf{\Omega} ( \sf{\Gamma} _{\! \sf{e}} ) = 8\sf{e} + 8\bar{\sf{e}} + 8\sf{g} + 8\bar{\sf{g}} + 8\sf{u} + 8\bar{\sf{u}}$

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 satisfy the conditions for being a complete space-time event because components are not distinguished by their magnetic polarity $\delta _{\hat{m}}$. So instead we just assert that the graviton's quarks are distributed as shown, as if the graviton has been emitted from a well-defined atom. This convention is called a three dimensional model of an electronic graviton.

 A three-dimensional quark model of an electronic graviton.