Why? Why? Tell Me Why?

From the post "Less Mass But No Theoretical Mass" and "Two Types Of Photons" we have,

$m_{po}=m$ or equivalently,

$m_{\rho\,po}=m_{\rho}$

$m_{\rho\,e} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dx=m_{\rho\,po} c^2-\int^{x_a}_{0}{\psi}dx=m_{\rho\,po} c^2-\int^{2x_z}_{0}{\psi}dx$

and

$m_{\rho\,ph} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dx=0$

$m_\rho c^2=\int^{x_a}_{0}{\psi}dx$

In general from the post "We Still Have A Probelm",

$m_{\rho\,particle}c^2+\int^{x_a}_{0}{\psi}dx=m_\rho c^2=m_{\rho\,po} c^2$

which is of course an assumption that all particles is due to a common manifestation.  In which case, there is one value of $m_\rho$ and the value of $x_a=2x_z$, determines what type of particle we have.

$m_{\rho\,particle}c^2+\int^{2x_z}_{0}{\psi}dx=m_{\rho\,po} c^2$

Why $m_{\rho\,po}$ and why $x_z$?  Why do these parameters (mass densities, $m_{\rho\,x}$), take on the values that they do?

$m_{\rho\,particle}+\cfrac{1}{c^2}\int^{2x_z}_{0}{\psi}dx=m_{\rho\,po}$

From symmetry about $x=x_z$

$m_{\rho\,particle}+\cfrac{2}{c^2}\int^{x_z}_{0}{\psi}dx=m_{\rho\,po}$

When this mass is fully manifested as $\psi$, we have a photon,

$E_{\rho\,ph}=\int^{2x_z}_{0}{\psi}dx=m_{\rho\,po} c^2=constant!$

where $m_{\rho\,po}$ is the mass density of a proton and $E_{\rho\,ph}$, the energy density of the photon.

Which, strangely, is consistent with the photons being particles in a helical path, where their kinetic energy is,

$KE=\cfrac{1}{2}m_{ph}c^2+\cfrac{1}{2}m_{ph}c^2=m_{ph}c^2$

What happens to photoelectric effects?  $r$ the radius of the helical path changes inversely with frequency $f$.

$2\pi rf=c$,    $2\pi r=\lambda=\cfrac{c}{f}$

A smaller $r$ pushes the electron further towards the nucleus and is ejected with greater velocity after the photon passes.  This means $r$ is inversely proportional to $E$, the energy of the ejected electron (from the post "Miss e- Miss e- Not").  And so,

$E\propto f$

Still, why $m_{\rho\,po}$ and why $x_z$?  Why do these parameters (mass densities, $m_{\rho\,x}$), take on the values that they do?