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

Consider a photon $\large{\gamma} \,$ defined by an out of phase pair of components

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

that are almost perfectly anti-symmetric so that

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

for all quarks except down-quarks. Many photon attributes are nil because phase anti-symmetry implies that almost every quark is matched with a corresponding anti-quark somewhere in the photon. So for most types of quark Z, the net number of quarks is zero

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

As for the down-quarks, ${\Delta}n^{\sf{D}}$ is not zero. But recall that by convention, the internal energy of the down-quarks is zero. $U^{\sf{D}} \! =0$, so any imbalance between ordinary quarks and anti-quarks can often be ignored. Substituting these conditions into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

$q ( {\large{\gamma}} ) =0$ $S ( {\large{\gamma}} ) =0$ $L ( {\large{\gamma}} ) =0$ $B ( {\large{\gamma}} ) =0$ $H ( {\large{\gamma}} ) =0$

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. No work $W$ 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, but in opposing directions

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

Then $W(\gamma) \equiv k_{\sf{F}} \Vert \, \overline{\rho}^{\, \gamma} \Vert =0$ too. But not all photon characteristics are null; the outer radius $\ \rho_{\LARGE{\circ}}$ and the inner radius $\, \rho_{\LARGE{\bullet}} \,$ may be greater than zero. Also consider the wavevector $\, \overline{ \kappa }$ which is found from the sum

$\begin{align} \overline{ \kappa } &\equiv \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \sum_{\sf{q} \, \in \, {\large{\gamma}} } \delta _{\theta}^{\sf{q}} \, \overline{\rho}^{\sf{q}} \\ \ \\ \ \\ &= \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \left[ \, \delta_{\theta}^{\mathcal{A}_{\LARGE{\circ}}} \! \! \! \sum_{\sf{q} \, \in \, \mathcal{A}_{\LARGE{\circ}}} \! \overline{\rho}^{\sf{q}} \ \ + \ \ \delta_{\theta}^{\mathcal{A}_{\LARGE{\bullet}}} \! \! \! \sum_{\sf{q} \, \in \, \mathcal{A}_{\LARGE{\bullet}}} \! \overline{\rho}^{\sf{q}} \, \right] \\ \ \\ \ \\ &= \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \left[ \, \delta_{\theta}^{\mathcal{A}_{\LARGE{\circ}}} \, \overline{\rho}^{\mathcal{A}_{\LARGE{\circ}}} \ \ + \ \ \delta_{\theta}^{\mathcal{A}_{\LARGE{\bullet}}} \, \overline{\rho}^{\mathcal{A}_{\LARGE{\bullet}}} \, \right] \\ \ \\ \ \end{align}$

But $\mathcal{A}_{\LARGE{\circ}}$ and $\mathcal{A}_{\LARGE{\bullet}}$ are out of phase, so $\delta_{\theta}^{\mathcal{A}_{\LARGE{\circ}}} =- \, \delta_{\theta}^{\mathcal{A}_{\LARGE{\bullet}}} = \pm 1$. And radius vectors are symmetrically opposed, so $\ \overline{\rho}^{\mathcal{A}_{\LARGE{\circ}}} \! =- \, \overline{\rho}^{\mathcal{A}_{\LARGE{\bullet}}}$. These two negative factors cancel each other so that

$\begin{align} \overline{ \kappa } = 2 \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \, \delta_{\theta}^{ \mathcal{A}_{\LARGE{\circ}} } \overline{\rho}^{\mathcal{A}_{\LARGE{\circ}}} = 2 \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \, \delta_{\theta} ^{\mathcal{A}_{\LARGE{\bullet}}} \overline{\rho} ^{\mathcal{A}_{\LARGE{\bullet}}} \end{align}$

The wavenumber is given by the norm of the wavevector, so

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 $1 \sim 10^{8}$

$\begin{align} \kappa = 2 \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \left\| \, \overline{\rho}^{\mathcal{A}} \right\| \end{align}$

We can drop the subscript on $\mathcal{A}$ because both phase-components have the same norm. And recall that $W^{\mathcal{A}} = k_{\sf{F}} \Vert \, \overline{\rho}^{\mathcal{A}} \Vert$ is the work required to build one of these phase-components. So in energetic terms, the photon's wavenumber can be written as

$\begin{align} \kappa = \frac{2W^{\mathcal{A}} }{ k_{\sf{F}}} \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$

Then the wavelength of a photon is

$\begin{align} \lambda = \frac{1}{\kappa} = \frac{ k_{\sf{F}} }{ \, 2W^{\mathcal{A}} } \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right)^{\! -1} \end{align}$

Finally recall that by definition the inner radius $\, \rho_{\LARGE{\bullet}} \,$ is constrained such that for all photons

$\begin{align} \rho_{\LARGE{\bullet}} \ge \sqrt{ \frac{hc}{ k_{\sf{F}}} \vphantom{\frac{hc}{2 \pi k_{\sf{F}}}^2} } \end{align}$

Then if $\gamma$ is a free particle where $\, \rho_{\LARGE{\circ}} \to ∞ \,$ and $\rho_{\LARGE{\bullet}}$ is as small as possible, the wavelength will be

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

Photons are classified by wavelength as noted in the accompanying table.

Right.png Next step: fields of photons.

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