If \(\psi\) is conservative, that it does not lose energy impart onto it; that the deformation from a spherical particle to a torus as the result of \(v_{max}=v_{boom}\), stores the kinetic energy and that this energy is completely recovered when the particle collapses back into a sphere.
\(E_{torus}-E_{absorbed}\rightarrow E_{sphere}+\cfrac{1}{2}m_av^2_{boom}\)
where \(-E_{absorbed}\) is the energy absorbed by the photon when it collapses.
If the torus is at light speed, \(v_{boom}\) will kick the particle beyond light speed. It travels back in time. As far as our perception is concerned, the particle appears first at a distance, with \(v_{boom}\) speed (as energy is conserved between space and the time dimension) causing other photon torus to be ejected. From there, the light emitted traces backwards to its point of origin.
If \(v_{boom}\) is slow, the light path can be seen growing at \(v_{boom}\) speed back to its origin. We are not observing the speed of light, but the emitted light (photon torus) from colliding particles at \(v_{boom}\) that returned from the time dimension, at a time before its emergence from the source. The path of light traces the movement of the source as Earth moved.
Given,
\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
When
\( \psi _{ n }-\psi _{ max }\lt0\)
\( \psi _{ n }\lt\psi _{ max }\)
\(v^{ 2 }_{ { max } }\lt 0\)
Since we need the remaining particle to repel the ejected torus, \(a_{\psi\,n}=a_{\psi\,ne}\) is the best candidate with \(\psi_{max}=a_{\psi\,\pi}\). If not \(a_{\psi\,ne}\lt a_{\psi\,n}\lt a_{\psi\,\pi}\), will also eject a photon torus.
The trick is to have \(a_{\psi\,ne}\) at \(v_{boom}\) speed when they are formed. In a discharge tube, such particles in the Faraday Dark Space is above \(v_{boom}\) speed. Furthermore, to observe the arc outside of the discharge tube, \(v_{boom}\) is of the air at atmospheric pressure and not of the low pressure gas.
Why an arc outside of the discharge tube? The particles time traveled there. The apparent displacement in space is the result of Earth's movement. The path of light observed traces the discharge tube movement as Earth moved.
Time travel at \(c+v_{boom}\) speed returns to the space dimension with \(v_{boom}\) speed with work done in the time dimension. A particle exist because of its speed in the time dimension at \(c\). The total energy in the time dimension due to this speed is \(E=mc^2\) (post "No Poetry for Einstein" dated 6 Apr 2016). When the particle goes back in time, it starts off a second line of existence in space that requires this energy of \(E=mc^2\) in the time dimension. So, the work required in the time dimension is \(E=mc^2\). Leaving this quantum of energy behind, at particle at speed \(c+v_{boom}\) exit the time dimension with speed \(v_{boom}\).
Where \(v_{boom}\) is rate of delivery of kinetic energy of particles at light speed that would trigger a boom. Boom generates more photons that are visible as the particles return to the space dimension. And we see this boom light from afar tracing backwards to the source of the particle. It is not the speed of light in reverse.
The path of the boom light traces the relative movement of the source through time as Earth moved. The particle enters into the time dimension just as it is ejected at the source. It appears earlier in time, before the source moved into place due to Earth's motion. On switching off the discharge, the boom light appears to return to the source as it traces moment less and less earlier to the present. So, a pulse train that switches on and off periodically will provide a persistent beam of light from a point (in space) forward in time, to the source at present.
Like a light saber but you cannot fight with it.
