Time Travel By Manipulating $\psi$

Consider $\psi$ along $t_g$,

$m_{\rho\,p} c^2=m_\rho c^2-\int^{2t_{g\,z}}_{0}{\psi}\,d\,t_{g}$

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.