\(m_{\rho\,p} c^2=m_\rho c^2-\int^{2t_{g\,z}}_{0}{\psi}\,d\,t_{g}\)
Let light speed, \(c=v\), a variable velocity \(v\) along \(t_g\),
\(m_{\rho\,p} v^2=m_\rho v^2-\int^{2t_{g\,z}}_{0}{\psi}\,d\,t_{g}\)
Differentiating with respect to \(t_g\),
\(m_{\rho\,p} 2v\cfrac{\partial\,v}{\partial\,t_g}=m_\rho 2v\cfrac{\partial\,v}{\partial\,t_g}-\psi\)
where it is understood that \(\psi\) is valid for \(0\le t_g\le 2t_{g\,z}\). Differentiating again,
\(m_{\rho\,p} 2\left\{(\cfrac{\partial\,v}{\partial\,t_g})^2+v\cfrac{\partial^2v}{\partial\,t^2_g}\right\}=m_\rho 2\left\{(\cfrac{\partial\,v}{\partial\,t_g})^2+v\cfrac{\partial^2v}{\partial\,t^2_g}\right\}-\cfrac{\partial\,\psi}{\partial\,t_g}\)
We know that,
\(-\cfrac{\partial\,\psi}{\partial\,t_g}=F_\rho\)
is the force on the system to affect a change in \(\psi\). Let
\(\cfrac{\partial\,\psi}{\partial\,t_g}=F_{co}\)
be the reaction force from the system as \(\psi\) collapses or increases.
\(m_{\rho\,p} 2\left\{(\cfrac{\partial\,v}{\partial\,t_g})^2+v\cfrac{\partial^2v}{\partial\,t^2_g}\right\}=m_\rho 2\left\{(\cfrac{\partial\,v}{\partial\,t_g})^2+v\cfrac{\partial^2v}{\partial\,t^2_g}\right\}-F_{co}\)
\(F_{ co }=2\left( m_{ \rho }-m_{ \rho \, p } \right) \left\{ (\cfrac { \partial \, v }{ \partial \, t_{ g } } )^{ 2 }+v\cfrac { \partial ^{ 2 }v }{ \partial \, t^{ 2 }_{ g } } \right\} \)
Let \(\cfrac { \partial \, v }{ \partial \, t_{ g } } =a_g\)
\(\cfrac{\partial\,\psi}{\partial\,t_g}=F_{ co }=2\left( m_{ \rho }-m_{ \rho \, p } \right) \left\{ a^{ 2 }_g+v\cfrac { \partial \,a_g }{ \partial \, t_{ g } } \right\} \)
This is the expression for the force that would drive us to and fro \(t_g\) from a collapsing or increasing \(\psi\).
The acceleration along \(t_g\), \(a_g\) is given by,
\( a^{ 2 }_{ g }=\cfrac { F_{ co } }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } -v\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } \)
If we achieve a steady force, \(F_{co}\), then
\(\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } =0\)
\(a_{ g }=\pm \sqrt { \cfrac { F_{ co } }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } } \)
The ambiguity from the square root is most troubling. Furthermore, for a negative \(F_{co}\),
\(a_{ g }=\pm i \sqrt { \cfrac { |F_{ co }| }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } } \)
In this case \(t_g\) has been rotated to one of the other time axes, \(t_c\) or \(t_T\), although,
\(|t_{now}|=\cfrac{1}{\sqrt{3}}|t_g|=\cfrac{1}{\sqrt{3}}|t_c|=\cfrac{1}{\sqrt{3}}|t_T|\)
is still true for all time axes.
To deal with the plus and minus sign ambiguity after taking the square root, we consider,
\(a^{ 2 }_{ g }=\cfrac { F_{ co } }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } -v\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } \)
Differentiating wrt \(t_g\),
\( 2a_{ g }\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } =\cfrac { 1 }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } \cfrac { \partial \, F_{ co } }{ \partial \, t_{ g } } -v\cfrac { \partial ^{ 2 }a_{ g } }{ \partial \, t^{ 2 }_{ g } } -a_{ g }\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } \)
\( 3a_{ g }\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } =\cfrac { 1 }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } \cfrac { \partial \, F_{ co } }{ \partial \, t_{ g } } -v\cfrac { \partial ^{ 2 }a }{ \partial \, t^{ 2 }_{ g } } \)
\( 3a_{ g }\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } } =\cfrac { 1 }{ 2\left( m_{ \rho }-m_{ \rho \, p } \right) } \cfrac { \partial \, ^{ 2 }\psi }{ \partial \, t^{ 2 }_{ g } } -v\cfrac { \partial ^{ 2 }a }{ \partial \, t^{ 2 }_{ g } } \)
If we ensure that the acceleration profile at the on start is such that,
\(\cfrac { \partial \, a_{ g } }{ \partial \, t_{ g } }\gt 0\) and
\(\cfrac { \partial ^{ 2 }a }{ \partial \, t^{ 2 }_{ g } }\lt 0\)
an example of such a profile f(x)=1-e^(-x) is given below,
In which case the sign of \(a_g\) depends on the sign of \(\cfrac { \partial \, ^{ 2 }\psi }{ \partial \, t^{ 2 }_{ g } }\), the second rate of change of \(\psi\), near the steady state of \(a_g\) as
\(\cfrac { \partial \, ^{ 2 }a_g }{ \partial \, t^{ 2 }_{ g } }\rightarrow0\).
As long as there is some value of \(\psi\),
\(m_{ \rho }-m_{ \rho \, p }\gt0\).
It is unlikely that we leave all our protons behind when we time travel. Similarly on deceleration, if we have the deceleration profile as show below, f(x)=e^(-x)-1,
The sign of \(a_g\) depends on the negative of \(\cfrac { \partial \, ^{ 2 }\psi }{ \partial \, t^{ 2 }_{ g } }\), the second rate of change of \(\psi\), near the steady state of \(a_g\) as
\(\cfrac { \partial \, ^{ 2 }a_g }{ \partial \, t^{ 2 }_{ g } }\rightarrow0\).
It is possible that because of the negative sign before \(m_{ \rho \, p }\) that, a collapsing \(\psi\) creates a positive time force and an increasing \(\psi\) generates a negative time force. In both cases, the magnitude of the force is proportional to \(\cfrac{\partial\,\psi}{\partial\,t_g}\).
So, by manipulating \(\psi\) we can generate a time force that will propel us through time. Hello Nobel Prize.
