and from the post "Not Exponential, But Hyperbolic And Positive Gravity!",
\(\int { 1 } d\, x=-i2{ mc^{ 2 } }\int { \cfrac { 1 }{ F^{ 2 }-A } } dF\)
\(x=i2{ mc^{ 2 } }\cfrac { 1 }{ \sqrt { A } } archtanh(\cfrac { F }{ \sqrt { A } } )+x_z\)
why would \(x_z\) take on different values of opposite sign?
A plot of log(cosh(x)) with \(x_z=0\) is shown below,
For the case of \(x_z\le0\), \(\psi\lt 0\) for \(x\gt 0\), and \(\int^x_{0}{\psi}dx\rightarrow-\infty\) as \(x\rightarrow \infty\), this would require infinite negative energy to realize. So \(x_z \gt 0\), and \(0\lt x\le x_a\), where at \(x=x_a\), \(\psi=0\) such that \(\int^{x_a}_{0}{\psi}dx=+finite\) and so does not require infinite amount of energy.
If this is the case than,
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A more negative value in \(F\) appears as \(x_z\) increases and
\(\int^{x_a}_{0}{\psi}dx\), the total \(\psi\) over the extend of \(x_a\) increases. Could it be that,
\(m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dx\)
that the energy, \(m_{\rho\,particle}c^2\) associated with the mass density of the particle, \(m_{\rho\,particle}\) in real time is the result of the total \(\psi\) being subtracted from the total kinetic energy, \(m_\rho c^2\) along the \(t_c\) (or \(t_g\)) dimension. This reduction in mass density is to provide for the total \(\psi\). Such that when the particle is annihilated the total amount of energy released is,
\(m_{\rho\,particle}c^2+\int^{x_a}_{0}{\psi}dx=m_\rho c^2\) (per unit volume)
where \(m_\rho\) is the mass density of the particle along time \(t_c\) (or \(t_g\)), and \(m_{\rho\,particle}\) is the mass density of the particle in \(t\) time dimension.
This means, when a particle is a wave in two space dimensions and one time dimension, together with a fourth time dimension along which it exists, part of its mass \(m\), is fuzzed into an energy density field in 3D space and real time \(t\) around the particle. The remaining energy manifest its self as \(m_{p}\), the mass of the particle in 3D space and real time \(t\).
This is the reason why an electron has lower mass, part of its mass is manifested as \(\psi\) around it.
What then is this value of \(m\) to start with? If \(m_{\rho}\) is a positive value, that means the constrain on \(x_a\) is,
\(m_{\rho}c^2-\int^{x_a}_{0}{\psi}dx\ge0\)
ie. as long as \(m_{\rho\,particle}\ge0\). And given that \(F\) has rotational symmetry about the origin,
\(F(x)=-F(-x)\),
\(x_a=2x_z\)
We have one constrain on the value of \(x_z\).
This however does not restrict the values of \(x_z\) and the corresponding \(x_a=2x_z\) for specifically an electron or the analogous gravity particle.
