where \(r\) is very small. \(t_g\) is at light speed, \(v=c\). If it is possible to slow down time,
\(\cfrac{\partial\,t_{gf}}{\partial\,t}\lt c\) (in ticks per second)
such that,
\(\cfrac{\partial\,t_{gf}}{\partial\,t}=r\omega\)
\(r=\cfrac{1}{\omega}\cfrac{\partial\,t_{gf}}{\partial\,t}\lt\cfrac{c}{\omega}\)
where both \(\omega\) and \(r\) are physically manageable number, consider the equation from the post "Time Travel Made Easy" again,
\(\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}\) ---(*)
\(\cfrac{\partial\,t_{gi}}{\partial\,t}=c\)
\(\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\cfrac { 1 }{ mc^{ 2 } } (cE_{ \Delta h }+t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } )+c\)
\( t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =mc^{ 2 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } -cE_{ \Delta h }-mc^{ 3 }\)
\( t_{ gi }=\cfrac { 2\pi a_{ \psi \, n } }{ c } \)
\( a_{ \psi \, n }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =\cfrac { 1 }{ 2\pi } \left\{ mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } -c^{ 2 }E_{ \Delta h }-mc^{ 4 } \right\} \)
\( \begin{equation*}a_{ \psi \, n }=\cfrac { 1 }{ 2\pi } \left\{ -mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right) ^{ -1 }+mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right) ^{ -1 }-c^{ 2 }E_{ \Delta h }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right) ^{ -1 } \right\} \end{equation*}\)
Since,
\( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0\)
\( \begin{equation*}a_{ \psi \, n }=\cfrac { 1 }{ 2\pi } \left\{ mc^{ 4 }\left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }-mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } \left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }+c^{ 2 }E_{ \Delta h }\left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 } \right\} \end{equation*}\)
replacing \({a_{ \psi \, n }}_o=\cfrac { 1 }{ 2\pi } mc^{ 4 }\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }\)
\(a_{ \psi \, n }={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t } +\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t \right\} \)
From (*)
\(\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\cfrac { 1 }{ mc^{ 2 } } (\cfrac { \partial \, t_{ gi } }{ \partial \, t } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t+\cfrac { \partial \, t_{ gi } }{ \partial \, t } t\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } )+\cfrac { \partial \, t_{ gi } }{ \partial \, t } \)
\( \cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t } \) --- (**)
So,
\(a_{ \psi \, n}={ a_{ \psi \, no } } \left\{ 1-\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\}+\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t \right\}\)
\(a_{ \psi \, n }={ a_{ \psi \, no } } \left\{-\cfrac { t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right\}\)
\(a_{ \psi \, n }=\cfrac{1}{2\pi}c^2.t\)
The particle travels at a forward velocity of,
\(v_{tele}=\cfrac{1}{2\pi}c^2\)
Is this speed possible? Yes, time speed has been slowed down, that implies greater than light speed space travel. However we are NOT traveling through space this way.
At time \(t=0^{-}\), just before \(t=0\),
\(\require{cancel}\)
\(a_{ \psi \, n, \,t=0}={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t } +\cancelto{0}{\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t } \right\} \)
\(a_{ \psi \, n,\,t=0 }={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t } \right\}\)
in this case, \(\cfrac { \partial \, t_{ gf } }{ \partial \, t }\ne0\), and the teleported distance is \(a_{ \psi \, n\,t=0 }\). As with the case considered in the post "Teleportation", with
\(\cfrac { \partial \, t_{ gf } }{ \partial \, t }\)=constant
the particle return velocity is,
\(a_{ \psi \, no } \cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t=-\cfrac{1}{2\pi}c^2.t=v_{tele}.t\)
\(v_{tele}=-\cfrac{1}{2\pi}c^2\)
At \(t\gt0\) however,
\(v_{tele}=\cfrac{1}{2\pi}c^2\)
It seems paradoxical that the particle can be at different locations and have different velocities just before and after \(t=0\). The trick here is,
\( \cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t } \) = constant
The expression \(\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }.t \) is held constant such that
\( \cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t } \)
from (**) is a constant. This means
\(\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t}\)
where \(A\) is a constant.
This way, only the situation for \(t=0^{-}\) applies and as will be discussed below \(t\ngtr0\).
Simple, we beam the particle through another portal that induces \(-\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }\). After passing through the second portal, \(\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=0\). Equation (*) collapses to,
\(\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{\partial\,t_{gi}}{\partial\,t}\)
as at \(t=0\), \(E_{\Delta h}=0\).
And the particle does not return from \(x=a_{ \psi \, n,\,t=0 }\), \(t\ngtr0\).
Since time, \(t_g\) is curled around \(x\) where \(it_g=x\) as required in the post "Time Travel Made Easy". \(x\) is perpendicular to the circular plane containing \(t_g\). The portal is directional. This sort of teleportation unfortunately is restricted to line of sight. Unless we can draw a straight line between the two points, teleportation is not possible.
And the billion dollar question is, how to induce \(\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t}\) in a particle?
