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 $\gamma$ defined by an out of phase pair of components

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

that are 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, it may be that ${\Delta}n^{\sf{D}} \ne 0$. But recall that by convention the internal energy $U$ of the down quarks is zero. So any imbalance between down quarks and anti-quarks can often be ignored. Substituting these conditions into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

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

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 $W$ 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

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

Then $W^{\gamma} \equiv k_{\sf{F}} \left\| \, \overline{\rho}^{\gamma} \right\| =0$. However photons have some important characteristics that are not zero, starting with their wavevector which is found from the sum

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

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 radius vectors are symmetrically opposed; $\ \overline{\rho} \left( \mathcal{A}_{\LARGE{\circ}} \right) =- \, \overline{\rho} \left( \mathcal{A}_{\LARGE{\bullet}} \right)$. So these two negative factors cancel each other and

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

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

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 $1 \sim 10^{8}$

$\begin{align} \kappa^{\gamma} = \frac{ 2 }{ k_{\sf{A}} } \left\| \, \overline{\rho}^{\mathcal{A}} \right\| = \frac{4 \pi k_{\sf{F}} }{hc} \left\| \, \overline{\rho}^{\mathcal{A}} \right\| \end{align}$

Recall that $W^{\mathcal{A}} = k_{\sf{F}} \left\| \, \overline{\rho}^{\mathcal{A}} \right\|$ is the work required to build one phase-component. So in energetic terms, the photon's wavenumber can be written as

$\begin{align} \kappa^{\gamma} = \frac{4 \pi}{hc} W^{\mathcal{A}} \end{align}$

We can use this to express the momentum of a photon in a perfectly inertial reference frame as

$\begin{align} p ^{\gamma} = \frac{ h \kappa }{ 2 \pi } = \frac{2 k_{\sf{F}} }{c} \left\| \, \overline{\rho}^{\mathcal{A}} \right\| = \frac{2 W^{\mathcal{A}} }{c} \end{align}$

Then, 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.

Right.png Next step: fields of photons.
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