Template: Photon Theorem

Consider finding the mechanical energy 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 for photons that are not gamma-rays. For these longer wavelength photons $n^{\sf{q}} = 0$ for all leptonic quarks. This condition nullifies the electric and magnetic radii such that $\rho_{m}^{\mathcal{A}}= \rho_{e}^{\mathcal{A}} = 0$. Also, 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

$\begin{align} \rho_{z} \equiv \, \frac{ H_{chem} + {\Delta}n^{\mathsf{U}} U^{\mathsf{U}} - {\Delta}n^{\mathsf{D}} U^{\mathsf{D}}}{k_{\sf{F}}} \end{align}$

where $H_{chem}$ is the enthalpy due to any chemical quarks. This expression can be simplified too because by convention $U^{\mathsf{D}} \! =0$. Also, we know that $\Delta n^{\sf{U}} \left(\mathcal{A}\right) = 0$ or else the photon would be a gamma-ray. Thus

$\begin{align} \rho_{z}^{\mathcal{A}} = \, \frac{ H_{chem}^{\mathcal{A}} }{k_{\sf{F}}} \end{align}$

and so

$\begin{align} p \left( {\large{\gamma}} \right) = \frac{ h }{ \pi k_{\sf{F}} } \left| \, H_{chem}^{\mathcal{A}} \vphantom{ {H_{chem}^{\large{\gamma}}}^{9}} \right| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$

The photon is ethereal so its mechanical energy is

$\begin{align} E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \ } = cp = \frac{ hc }{ \pi k_{\sf{F}} } \left| \, H_{chem}^{\mathcal{A}} \vphantom{ {H_{chem}^{\large{\gamma}}}^{9}} \right| \left( \frac{1}{\rho_{\LARGE{\bullet}}^{2}} - \frac{1}{\rho_{\LARGE{\circ}}^{2}} \right) \end{align}$

Substituting-in definitions for the radii gives the energy in terms of quark coefficients as

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