Strained And Witchcraft

Consider the relation between time and space,

$\cfrac{2\pi.t_c}{t}=c$

$2\pi.t_c=ct$

$2\pi\cfrac{\partial\,t_c}{\partial\,t}=c$

$\cfrac{\partial\,t_c}{\partial\,t}=\cfrac{c}{2\pi}$

In a similar way and considering the direction of these dimensions,

$\cfrac{\partial\,t_T}{\partial\,t}=-i\cfrac{c}{2\pi}$

$\cfrac{\partial\,t_g}{\partial\,t}=i\cfrac{c}{2\pi}$

And

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, t } +\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t } +\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$ --- (*)

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}\left\{\cfrac { \partial \, x }{ \partial \, t_{ c } }+i\cfrac { \partial \, x }{ \partial \, t_{ g } }-i\cfrac { \partial \, x }{ \partial \, t_{ T } }\right\}$

If all time dimensions are equivalent,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }=\cfrac { \partial \, x }{ \partial \, t_{ g } } =\cfrac { \partial \, x }{ \partial \, t_{ T } }$

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}\cfrac { \partial \, x }{ \partial \, t_{ c } }$

We also have

$\left(\cfrac { \partial \, x }{ \partial \, t } \right)^2=\left(\cfrac { \partial \, x }{ \partial \, t_c } \right)^2+\left(\cfrac { \partial \, x }{ \partial \, t_g } \right)^2+\left(\cfrac { \partial \, x }{ \partial \, t_T } \right)^2$

 $\cfrac { \partial \, x }{ \partial \, t } =\sqrt{3}\cfrac { \partial \, x }{ \partial \, t_{ c } }$

This is possible only if,

 $\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c }= 0$

At this point we can swap $x$ for $t_c$, since they are both dummy variables and conclude by symmetry that when $\cfrac{\partial\,x}{\partial\,t}=\cfrac{c}{2\pi}$,  $\cfrac { \partial \,t_c}{ \partial \, t } =0$

However, for the fun of it,

When $x$ wraps around $t_c$,

$\cfrac{2\pi.x}{t}=c$

$2\pi.x=ct$

$2\pi\cfrac { \partial \, x }{ \partial \, t } =c$

and,

 $2\pi\cfrac { \partial \, x }{ \partial \, t_c } =c\cfrac{\partial\,t}{\partial\,t_c}$

So,

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c } \cfrac{\partial\,t_c}{\partial\,t}$ --- (***)

Since,

$\cfrac { \partial \, x }{ \partial \, t_{ c } } =\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t_{ c } } +\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t_{ c } }$

As changes in $x$ and $t_c$ does not effect $t_g$ and $t_T$, and as far as $x$ is concern $t_g$ and $t_T$ are equivalent, !?!

$\cfrac { \partial \, x }{ \partial \, t_{ c } } =\cfrac{c}{2\pi}\cfrac{\partial\,t}{\partial\,t_c}=2\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t_{ c } } =2\cfrac { \partial \, x }{ \partial \, t_{T } } \cfrac { \partial \, t_{ T } }{ \partial \, t_{ c } }$

ie.

$\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac{\partial\,t_c}{\partial\,t} =\cfrac{c}{2\pi}$ --- (**)

and

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }=2\cfrac { \partial \, x }{ \partial \, t_{ g } }$

From which we formulate,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t } =2\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }$ --- (1)

Similarly,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }=2\cfrac { \partial \, x }{ \partial \, t_{ T } }$

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t } =2\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$ --- (2)

Therefore (1)+(2),

$\cfrac { \partial \, x }{ \partial \, t_{ c } } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\left\{\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}$

Substitute (**) into the above,

 $\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\cfrac{\partial\,t_c}{\partial\,t}\left\{\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}$

But,

 $\cfrac{\partial\,x}{\partial\,t}=\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac { \partial \, t_{ c } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$

So,

  $\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\cfrac{\partial\,t_c}{\partial\,t}\left\{\cfrac{\partial\,x}{\partial\,t}-\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac { \partial \, t_{ c } }{ \partial \, t }\right\}$

But from (***),

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c } \cfrac{\partial\,t_c}{\partial\,t}$

So,

$\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=0$

But,

$\cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t } +\cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$

Thus,

$\cfrac {c }{ 2\pi } \cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

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

Blasphemy, I know.  The point is,

when $\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}$, $\cfrac { \partial \, t_{ c } }{ \partial \, t } =0$ and when $\cfrac { \partial \, t_c }{ \partial \, t } =\cfrac{c}{2\pi}$, $\cfrac { \partial \, x }{ \partial \, t } =0$ are consistent.

The factor of $\cfrac{1}{2\pi}$ appearing before $c$ is the result of circular motion around the orthogonal axis.  A particle traveling in circular motion in space about the time axis is not traveling in time, so time speed equals zero.  Similarly, the same particle traveling in circular motion in time about the space axis has zero space speed.

This relationship can be rewritten as,

$v^2+v^2_t=\left(\cfrac{c}{2\pi}\right)^2$ --- (+)

where $v$ is the particle velocity in space, $v_t$ is the particle velocity in time and $c$ a constant.  Pythagoras' Theorem applies because $v$ is perpendicular to $v_t$.  Obviously expression (+) satisfies the boundary value conditions.

More importantly, the time speed limit occurs at $\cfrac{c}{2\pi}$ not at $c$, because the particle goes into circular motion.