A Mass In Time And In Mind

From the post "Into A Pile Of Deep Shit" dated 09 Jun 2016,

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } =2\cfrac { \partial \, T }{ \partial \, t_{ c } }$

For $\cfrac { \partial \, x }{ \partial \, t_{ c } } \ne 0$, multiply both sides by $\cfrac{ \partial \, t_{ c } }  { \partial \, x }$

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, x } =2\cfrac { \partial \, T }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, x }$ -- (*)

The reason (*) is still possible is because, the wave $\psi$ is stationary in space, that $\cfrac{\partial\,\psi}{\partial\,t}|_{x=x_o}=0$.

A wave $\psi$ in circular motion, radiating energy is analogous to circular motion in mechanics.  In circular motion,

in turning $v$ into a circular path a centripetal force directed to the center the circle acts on the body.  As $\Delta \theta\to0$ in the diagram, $F_{m}$ is directed at the center of the circle.  If the body encounters resistance along its circular path but remains in circular motion, a resistance force pointing away from the center of the circle acts on the body.  Work done against this resistance force is the energy needed to overcome the resistance along the circular path.  This force acts on the mass of the body, $m$ and is a Newtonian force.

In the case of $\psi$, in circular motion around the center of a particle, a similar "force" develops as $\psi$ encounters resistance along its path.  "Work done" against this resistance is radiated outward away from the center of the particle along a radial line.  This force acts on $\psi$ (not on mass $m$ but on energy density, $\psi$) and results in a rate of change of kinetic energy,

$F.c=\cfrac{dKE}{dt}$

where $\dot{x}=c$.

$F=\cfrac{1}{c}\cfrac{dKE}{dt}$

Image $\psi$ along an energy number line mark consecutively with increasing energy magnitude.  $\cfrac{dKE}{dt}$ is the equivalent of velocity along the energy number line. $\cfrac{1}{c}$ is the equivalent of mass.  $F$ is then the momentum of energy density, $\psi$.  $\psi$ is a point of energy denoted as $\cfrac{1}{c}$ just as a point mass is notated as mass density, $m$.

We can have,

$F.c=ma.c=mc.a=\cfrac{dKE}{dt}$

but $F$ does not act on $m$ but on $\psi$.  If the equivalent mass is $mc$ then $mc.a=\cfrac{dKE}{dt}$ would make $a$, a velocity.  However, $m$ is a space concept of inertia and does not exist in time.  However we can view the time dimension from the perspective of space and obtain,

$E=mc^2$

where $c$ is the speed limit in the time dimension or

$E=32\pi^4c^2$

where $c$ is the speed limit defined in the space dimension.

$32\pi^4v_x^2+v_t^2=32\pi^2c^2$

both of which deals with the time dimension at the terminal speed $c$ in the time dimension from the perspective of the space dimension.

$m$ marks a point in space, $x$ along a space number line.  $\psi$ marks a point of energy in time along a time number line.  In the time dimension, the first rate of change is the second rate of change in space; Acceleration (in space) is the analogue of velocity (in time) and the change of energy, not over time and stationary in space is the analogue of the change of time (position) along the time number line.

Energy $\cfrac{1}{c}$ in the time dimension is the equivalent of mass $m$ in the space dimension.  In the space dimension,

$F=ma$

$m$ is the inertia, resistance along a space number line.

In the time dimension,

$F.c=\cfrac{dKE}{dt}$

$\cfrac{dF}{dt}.c=\cfrac{d^2KE}{dt^2}$

assuming $c$ is a constant.

$\cfrac{dF}{dt}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}$

A force in the time dimension is the rate of change of $F$ acting on $\psi$.

The first rate of change in time is the second rate of change in space; the change of position in time is the first rate of change in space (change in velocity; change in $KE$).  Inertia in the time dimension is $\cfrac{1}{c}$.

This is just consistent coincident.