Teleportation

Why did the Philadelphia Experiment traveled into the future time?  From the post "Rotating Al Again", if they have just rotated the ($t_c$-$t_T$) plane to the ($t_g$-$t_T$) plane by rotating the projection of $t_{now}$ on the ($t_c$-$t_T$) plane by 60^o onto the ($t_g$-$t_T$) plane, and the space dimensions are coiled up along the time axes such that any space dimension is perpendicular the time dimension and at the same time orthogonal to all other dimensions, space and time cannot cross nor superimpose no matter what rotation or transformations.

In our 3 space dimension world time flows freely, continuously.  If we are able to rotate/transform a space dimension on to a time dimension then space along that dimension will flow freely.  The time dimension that has been swapped with a space dimension will stand still unless kinetic energy is applied.

From the post "Eternal Embrace",  time and space loop around each other.  As we shorten time, space elongates.  It might be possible that using the equation from the post "Time Travel Made Easy",

$\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}$

and setting

$\cfrac{\partial\,t_{gf}}{\partial\,t}=0$

$\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}=0$

$t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =-\left( E_{ \Delta h }+mc^{ 2 } \right) \cfrac { \partial \, t_{ gi } }{ \partial \, t }$

but,

$\cfrac { \partial \, t_{ gi } }{ \partial \, t } =c$

So,

$t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =-c\left( E_{ \Delta h }+mc^{ 2 } \right)$

since,

$t_{ gi }\gt0$,   $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0$

Furthermore,

$t_{ gi }=\cfrac { 2\pi a_{ \psi \, n } }{ c } =-c\left( E_{ \Delta h }+mc^{ 2 } \right) \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } c^{ 2 }\left( E_{ \Delta h }+mc^{ 2 } \right) \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

Initially when, $E_{ \Delta h }=0$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

As $\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right)$ is negative,

${a_{ \psi \, n }}_o=\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$ --- (*)

So, depending on the value of $\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$,  the particle is transported to a distance defined by equation(*).  Afterwards,

$E_{ \Delta h }=\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } c^{ 2 }\left\{ t+mc^{ 2 } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 } \right\}$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  }mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }-\cfrac { 1 }{ 2\pi  } c^{ 2 }t$

$a_{ \psi \, n }={a_{ \psi \, n }}_o-\cfrac { 1 }{ 2\pi  } c^{ 2 }t$

After the initial jump, if the particle is still subjected to $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0$, as time passes, the particle travels back to its origin with a velocity $v_{tele}$,

 $v_{tele}=-\cfrac { 1 }{ 2\pi  } c^{ 2 }$

If $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }$ is set to zero immediately at ${a_{ \psi \, n }}_o$ then the particle remains at ${a_{ \psi \, n }}_o$.

Teleportation!  Warp Speed!  When the particle is allowed to return, on a smaller scale, Brownian motion.