Like Doppler

Another view of the torus photon's time trip backwards is,

baring in mind that the kinetic energy along the time dimension is $\color{red}{mc^2}$.  As we account for total energy across space and time dimensions,

$\Delta_{\small{forward}}= \cfrac{1}{2}m(c+v_{boom})^2+\color{red}{mc^2}-\left(\cfrac { 1 }{ 2 } mv_{ boom }^{ 2 }+\cfrac { 1 }{ 2 } mc^{ 2 }+\color{red}{mc^2}+\color{red}{mc^2}\right)$

$\Delta_{\small{forward}}=mcv-mc^2=mc(v-c)$

which is just the required energy to to accelerate a particle from ($v_i=v$) to ($v_f=c$) in the expression,

$E=mc\int ^{v_f}_{v_i}{ dv }$

from the post "No Poetry for Einstein" dated 06 Apr 2014.

If this discrepancy is solely compensated for by one particle,  $mc^2$ along one timeline of existence,

$\Delta t=\cfrac{\Delta_{\small{forward}}}{mc^2}=\cfrac{v}{c}-1$

As $v\lt c$, $\Delta t\lt 0$,  a negative value indicates backward in time at the instance $T$ when the torus collapses.  In term of $\tau$,

$-\Delta \tau=\cfrac{1}{\Delta t}=\cfrac{1}{\cfrac{v}{c}-1}$

$\Delta \tau=\cfrac{1}{1-\cfrac{v}{c}}$

Both expressions are different from the one in the previous post "Light Speed Back In Time" dated 15 Jan 2018,

$\Delta t=\cfrac{v}{c}$

$\Delta \tau=\cfrac{c}{v}$

because previously the focus is on one particle along its timeline.  In this post, the total energy across both timelines are considered with the discrepancy compensated solely by one particle.  Both timelines are kept in view.

$v_{boom}$ is not the speed of the torus photon source.  It is assumed that $v_{boom}$ is in the same direction as $c$.  If $v_{boom}$ acquires a negative value.

$\Delta \tau=\cfrac{1}{1-\cfrac{-v}{c}}=\cfrac{1}{1+\cfrac{v}{c}}$

$v_{boom}$ can acquire a negative value when the photon ray is observed from the side moving at right angle to the line of sight.  To establish a common reference given two $\Delta \tau$ values, we use the geometric mean,

$\Delta \tau_{+}=\cfrac{1}{1-\cfrac{v}{c}}$

$\Delta \tau_{-}=\cfrac{1}{1+\cfrac{v}{c}}$

$\Delta \tau=\sqrt{\Delta \tau_{+}\Delta \tau_{-}}=\sqrt{\cfrac{1}{1+\cfrac{v}{c}}*\cfrac{1}{1-\cfrac{v}{c}}}$

This geometric mean is the ray $mc^2$, that appear previously in this derivation,

$\Delta t=\cfrac{\Delta_{\small{forward}}}{mc^2}$   or

$\Delta \tau=\cfrac{mc^2}{\Delta_{\small{forward}}}$

So,

$\Delta \tau_1=\cfrac{\Delta \tau_{+}}{\Delta \tau}=\cfrac{\sqrt{1-\cfrac{v^2}{c^2}}}{1-\cfrac{v}{c}}=\cfrac{mc^2}{\Delta_{\small{forward}}}=\cfrac{f_o}{f_1}$

$f_1=\cfrac{1-\cfrac{v}{c}}{\sqrt{1-\cfrac{v^2}{c^2}}}*f_o$

and

$f_2=\cfrac{1+\cfrac{v}{c}}{\sqrt{1-\cfrac{v^2}{c^2}}}*f_o$

which look like the transverse Doppler effect but has nothing to do with a moving source.  It is a ray moving in a medium that collapses the torus photons to $v_{boom}$ that in turn eject torus photons upon further collisions at $v_{boom}$.  A ray through a region of gas that emitted the torus photons.

OK??