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?
