From previously, looking at the work done equation along the time dimension,
\(E=\int { { F }_{ v }dx } \)
\({ F }_{ v }=\cfrac { d{ P }_{ v } }{ d{ t }_{ v } } =\cfrac { mdv }{ d{ t }_{ v } }\)
Therefore,
\(E=\int { \cfrac { mdv }{ d{ t }_{ v } } } dx\) and since \(\cfrac { c }{ v } .\cfrac { 1 }{ d{ t }_{ c } } =\cfrac { 1 }{ d{ t }_{ v } }\)
We have
\(E=\int { m } \cfrac { c }{ v } \cfrac { dx }{ { dt }_{ c } } dv\quad ,\quad but\quad \cfrac { dx }{ d{ t }_{ c } } =v\)
\( E=mc\int { dv }\)
And so for time speed from 0 to c,
\(E=mc\int _{ 0 }^{ c }{ dv }\)
\(E=m{ c }^{ 2 }\)
This suggests then, mass m travelling along the time dimension at speed c has kinetic energy \(E=m{ c }^{ 2 }\) , and when m stops, we will recover this amount of energy. What does it mean to stop in the time dimension? One moment it is here another moment it disappears completely. Total annihilation! As if its mass, m is converted completely to energy.
And so, Einstein's equation \(E=m{ c }^{ 2 }\) can be interpreted as kinetic energy along the time dimension. With one assumption, that c is light speed. Are we travelling along the time dimension at light speed? At this point there is something fishy about how Einstein got his expression...
