Bead Panel, Kayan people. Borneo 20th century, 31 X 27 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop. |

Consider a photon $\gamma$ defined by an out of phase pair of components

$\sf{\Omega}^{ \gamma} = \left\{ \mathcal{A}_{\LARGE{\circ}} , \mathcal{A}_{\LARGE{\bullet}} \right\}$

that are anti-symmetric so that

$\mathcal{A}_{\LARGE{\circ}} = \overline{\mathcal{A}_{\LARGE{\bullet}} }$

Many photon attributes are nil because phase anti-symmetry guarantees that every quark is matched with a corresponding anti-quark somewhere in the photon. So for any type of quark Z, the net number of quarks is zero

${\Delta}n^{\sf{Z}} \equiv n^{\sf{\bar{z}}} - n^{\sf{z}} = 0$

Substituting this condition into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

$q^{\, \gamma } =0$ $S^{\, \gamma } =0$ $L^{\gamma } =0$ $B^{\, \gamma } =0$ $H^{ \gamma } =0$

Recall that the lepton-number, baryon-number and charge are conserved, so a particle may freely absorb or emit countless photons without altering its own values for these quantum numbers. Moreover no work is required to assemble a photon because the two phase components $\mathcal{A}_{\LARGE{\circ}}$ and $\mathcal{A}_{\LARGE{\bullet}}$ have radius vectors with the same norm $\rho^{\mathcal{A}}$ but in opposing directions

$\overline{\rho} \left( \mathcal{A}_{\LARGE{\circ}} \right) =- \, \overline{\rho} \left( \mathcal{A}_{\LARGE{\bullet}} \right)$so that$\overline{\rho}^{\, \gamma} = \overline{\rho} \left( \mathcal{A}_{\LARGE{\circ}} \right) + \overline{\rho} \left( \mathcal{A}_{\LARGE{\bullet}} \right) = (0, 0, 0)$

However photons have one important characteristic that is not zero; their wavevector which is found from the sum

$\begin{align} \overline{ \kappa }^{\, \gamma} \equiv \frac{1}{k_{\sf{A}}} \sum_{i=1}^{N} \delta _{\theta}^{\, i} \, \bar{\rho}^{i} = \frac {\delta_{\theta} \left(\mathcal{A}_{\LARGE{\circ}} \right) \, \overline{\rho} \left( \mathcal{A}_{\LARGE{\circ}} \right) }{k_{\sf{A}}} + \frac {\delta_{\theta} \left( \mathcal{A}_{\LARGE{\bullet}} \right) \, \overline{\rho} \left( \mathcal{A}_{\LARGE{\bullet}} \right) }{k_{\sf{A}}} \end{align}$

Photon Type | $\lambda$ (m) |

a gamma-ray |
$\lesssim 10^{-12}$ |

an X-ray |
$10^{-11} \sim 10^{-8}$ |

an ultraviolet photon |
$\sim 10^{-8}$ |

a visible photon |
$\sim 10^{-7}$ |

an infrared photon |
$10^{-6} \sim 10^{-3}$ |

a microwave |
$10^{-3} \sim 1$ |

a radio-wave |
$1 \sim 10^{8}$ |

But $\mathcal{A}_{\LARGE{\circ}}$ and $\mathcal{A}_{\LARGE{\bullet}}$ are out of phase so $\delta_{\theta} \left( \mathcal{A}_{\LARGE{\circ}} \right) =- \, \delta_{\theta} \left( \mathcal{A}_{\LARGE{\bullet}} \right) = \pm 1$ and

$\begin{align} \overline{ \kappa }^{\, \gamma} = \frac { 2 }{k_{\sf{A}}} \overline{\rho} \left( \mathcal{A}_{\LARGE{\circ}} \right) = - \frac { 2 }{k_{\sf{A}}} \overline{\rho} \left( \mathcal{A}_{\LARGE{\bullet}} \right) \end{align}$

The wavenumber is given by the norm of the wavevector and recall that $k_{\sf{A}} \equiv hc / 2 \pi k_{\sf{F}}$ so that

$\begin{align} \kappa^{\gamma} = \frac{ 2 \rho ^{\mathcal{A}} }{ k_{\sf{A}} } = \frac{4 \pi}{hc} k_{\sf{F}} \rho ^{\mathcal{A}} = \frac{4 \pi}{hc} W^{\mathcal{A}} \end{align}$

where $W^{\mathcal{A}} = k_{\sf{F}} \rho ^{\mathcal{A}}$ is the work required to build one phase-component. Then finally, the wavelength of a photon can be expressed as

$\begin{align} \lambda = \frac{2 \pi}{\kappa} = \frac { hc }{ 2 W^{\mathcal{A}} } \end{align}$

Photons may be classified by their wavelength as noted in the accompanying table.