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 down-quarks is so small that it is usually taken to be zero. Then any imbalance between ordinary quarks and anti-quarks can 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
$\begin{align} \lambda = \frac{2 \pi }{\kappa} = \frac{2 \pi k_{\sf{F}} }{ \,2 W^{\mathcal{A}} } \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right)^{\! -1} \end{align}$
Finally recall that by definition the inner radius $\, \rho_{\LARGE{\bullet}} \,$ is constrained such that for all photons
$\begin{align} \rho_{\LARGE{\bullet}} \ge \sqrt{ \frac{hc}{ 2 \pi k_{\sf{F}}} \vphantom{\frac{hc}{2 \pi k_{\sf{F}}}^2} } \end{align}$
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
$\begin{align} \lambda = \frac{ hc}{\,2 W^{\mathcal{A}} } \end{align}$
Photons are classified by wavelength as noted in the table above. A more general treatment considers that a photon's wavelength may also depend on its surroundings. Then the symbol $\lambda _{\sf{o}}$ is used to indicate a wavelength where any such environmental effects are negligible.
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Types of Photons |
Summary |
Noun | Definition | |
Gamma-ray | $\lambda \; \lesssim \; 10^{-12} \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
X-ray | $\lambda \; = \; 10^{-11} \sim 10^{-8} \; \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
Ultraviolet Photon | $\lambda \; = \; \sim 10^{-8} \; \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
Visible Photon | $\lambda \; = \; \sim 10^{-7} \; \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
Infrared Photon | $\lambda \; = \; 10^{-6} \sim 10^{-3} \; \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
Microwave | $\lambda \; = \; 10^{-3} \sim 1 \; \; \sf{\text{(m)}}$ | 7-XX |
Noun | Definition | |
Radio-wave | $\lambda \; = \; 1 \sim 10^{8} \; \; \sf{\text{(m)}}$ | 7-XX |