Consider finding the mechanical energy $\, E$, of a photon $\, \large{\gamma}$, that is specified by a pair of phase components $\mathcal{A}_{\LARGE{\circ}}$ and $\mathcal{A}_{\LARGE{\bullet}}$
$\sf{\Omega} \left( {\large{\gamma}} \right) = \left\{ \mathcal{A}_{\LARGE{\circ}} , \mathcal{A}_{\LARGE{\bullet}} \vphantom{\left( {\large{\gamma}} \right)} \right\}$
As discussed earlier, the wavenumber of this photon can be written as
$\begin{align} \kappa \left( {\large{\gamma}} \right) = 2 \left\| \, \overline{\rho}^{\mathcal{A}} \right\| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$
where $\rho_{\LARGE{\bullet}}$ is the photon's inner radius, $\rho_{\LARGE{\circ}}$ is its outer radius and $\overline{\rho}$ is a radius vector. The subscript on $\mathcal{A}$ is dropped because both phase-components have the same norm. We can use this wavenumber to express the momentum of the photon, in a perfectly inertial reference frame, as
$\begin{align} p \left( {\large{\gamma}} \right) = \frac{h}{2\pi} \kappa = \frac{h}{\pi} \left\| \, \overline{\rho}^{\mathcal{A}} \right\| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$
Writing-out the norm in terms of the radial components of $\mathcal{A}$ gives
$\left\| \, \overline{\rho}^{\mathcal{A}} \right\| = \sqrt{ \vphantom{ {{\rho_{m}^{\mathcal{A}}}^{2}}^{2} } \; \left| \; k_{mm} {\rho_{m}^{\mathcal{A}}}^{2} + k_{ee} {\rho_{e}^{\mathcal{A}}}^{2} + k_{zz} {\rho_{z}^{\mathcal{A}}}^{2} + 2 k_{em} {\rho_{m}^{\mathcal{A}}} {\rho_{e}^{\mathcal{A}}} + 2k_{mz} {\rho_{m}^{\mathcal{A}}} {\rho_{z}^{\mathcal{A}}} + 2 k_{ez} {\rho_{e}^{\mathcal{A}}} {\rho_{z}^{\mathcal{A}}} \; \right| \; }$
This expression can be simplified if the photon is not a gamma-ray. For long-wavelength photons the coefficients of leptonic quarks must all be zero. Then the electric and magnetic radii are also null; $\rho_{m}^{\mathcal{A}}= \rho_{e}^{\mathcal{A}} = 0$. And recall that $k_{zz} = 1$. Then
$\left\| \, \overline{\rho}^{\mathcal{A}} \right\| = \left| \, \rho_{z}^{\mathcal{A}} \vphantom{\left( {\large{\gamma}} \right)^{9}} \right|$
and so
$\begin{align} p \left( {\large{\gamma}} \right) = \frac{h}{\pi} \left| \, \rho_{z}^{\mathcal{A}} \vphantom{\left( {\large{\gamma}} \right)^{9}} \right| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$
The photon's momentum is proportional to the absolute-value of a polar radius as defined by
$\begin{align} \rho_{z} \equiv \, \frac{ H_{\sf{chem}} + {\Delta}n^{\mathsf{U}} U^{\mathsf{U}} - {\Delta}n^{\mathsf{D}} U^{\mathsf{D}}}{k_{\sf{F}}} \end{align}$
where $H_{\sf{chem}}$ is the enthalpy due to any chemical quarks, and $n$ is a quark coefficient. This expression can be simplified too because by convention $U^{\mathsf{D}} \! =0$. Also, $\Delta n^{\sf{U}} \left(\mathcal{A}\right) = 0$ or else the photon would be a gamma-ray. Thus
$\begin{align} \rho_{z}^{\mathcal{A}} = \, \frac{ H_{\sf{chem}}^{\mathcal{A}} }{k_{\sf{F}}} \end{align}$
and so
$\begin{align} p \left( {\large{\gamma}} \right) = \frac{ h }{ \pi k_{\sf{F}} } \left| \, H_{\sf{chem}}^{\mathcal{A}} \vphantom{ {H_{\sf{chem}}^{\large{\gamma}}}^{9}} \right| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$
The photon is ethereal so its mechanical energy is
$\begin{align} E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \; } = cp = \frac{ hc }{ \pi k_{\sf{F}} } \left| \, H_{\sf{chem}}^{\mathcal{A}} \vphantom{ {H_{\sf{chem}}^{\large{\gamma}}}^{9}} \right| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$
Substituting-in definitions for the radii gives the energy in terms of quark coefficients as
$\begin{align} E \left( {\large{\gamma}} \right) = 2 \left| \, H_{chem}^{\mathcal{A}} \vphantom{{H_{chem}^{\large{\gamma}}}^{9}} \right| \cdot \left[ \frac{64}{\left( \Delta n^{\sf{D}} \right)^{2} } - \frac{64}{ \left( N^{\sf{D}} \right)^{2} } \right] \end{align}$
For a perfectly free particle, $\left| \Delta n^{\sf{D}} \right| = 8$ and $N^{\sf{D}} \! \to ∞$. So a free photon's energy approaches