Reinterpreting Spectral Lines

From the post "Peanuts And Peanut Butter" dated 09 Dec 2017, where $\psi$ can disperse in a circular fashion around the point of impact.

$v^{ 2 }_{ { max } }\lt0$

and so,

$v_{ { max } }=i\sqrt{\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\}  e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }}$

What happens when $a_{\psi\,c}$ collides with one of these rings?

What happens to fragments of $\psi$ less than $a_{\psi\,c}$?

If ${a_{\psi\,ne}}=182.29\,nm$ exist and that $a_{\psi}$ changes polarity beyond ${a_{\psi\,ne}}$ then all emission spectral series ends at $n=n_e$  ie, no energy transition between energy states $n_1\gt n_e$ and $n_2\gt n_e$.  Energy transitions between particles of $n\ge n_e$ do not occur as these particles are repulsive.  A particle of $n\ge n_e$ can still receive particle of size $n\lt n_e$, because these particles are still attractive.

An absorption line occurs when a big particle splits into smaller particles of higher energy.  But if the split involves negative particles $n\lt n_e$, these particles will collide again and coalesce with an equal emission of energy.  Both absorption and emission will occur at the same time within same confinement of particles.  The effects will cancel and neither an absorption nor an emission line will be seen.  Absorption spectral lines can be seen only if subsequent collisions and coalescence do not occur.  This is when a big particle splits into particles of $n\ge n_e$.  The first absorption line is when $n=2n_e$ splits into two particles of $n=ne$.  Two particles of $n=n_e$ will not coalesce and emit a spectral line as they are repulsive (at the limiting border line between positive and negative).  This is the lowest absorption spectral line visible.

We have to reinterpret the data presented in the post  "Sizing Them Up" dated 3 Dec 2014.  The data are energy transitions from $n_2=2n_e$ to $n_1=n_e$ and not of the particle $(n+1=2)\rightarrow (n=1)$ where the photon emitted is the particle itself.

$a_{\psi}$ (nm)	$f$ (GHz)	$\lambda$(nm)
19.34	2466067.5	121.57
16.32	2922728.6	102.57
15.48	3082568.8	97.25
14.77	3230699.3	92.79

These are data for the photon absorbed when particles of each type of size $n=2n_e$ split into two particles $n=n_e$.  No corresponding emission line occurs because the resultant particles does not coalesce; there is no energy transition back from $n_1=n_e$ to $n_2=2n_e$.

What is the value of $n_e$?  Is $2*n_e$ possible given that $a_{\psi\,\pi}$, $n=77$.

Why would radiating at $2.4660675\,Hz$ that encourages the breakup of particles, heat up the matter?  If $a_{\psi}=19.34\,nm$ is not the temperature particle, then it should not resonate to $2.4660675\,Hz$.  It is still possible that the resultant smaller particles at $n=n_e$ absorb particles of $n\lt n_e$ and become positive.  These particles will not split again until $n=2n_e$.  The result is an increase in the number of positive particles.  It is still possible to conclude that $a_{\psi}=19.34\,nm$ is due temperature particles.

All similar radiations, $\cfrac{f}{10^{6}}$ will increase the respective number of positive particles.

Negative particles are generated when big particle split into particles of $n\lt n_e$, the smallest of which is $a_{\psi\,c}$.

So, a stream of negative particles will first converge and coalesce, when they are of size $n=n_e$ they are neutral.  When these coalesce further with particles of $n\lt n_e$ they become positive, and the stream of particles begins to diverge.

So, a stream of negative electric charges turns neutral then positive.  When the stream is neutral, all deflection mechanisms using electric fields and magnetic coils will fail.

And phosphorous luminescence due to positive charges on a CRT screen comes to the rescue...There is no need for electrons knocking out electrons to create positive ions for phosphorous luminescence; the electrons turned positive.

This blog is definitely not for the faint heart-ed!

Note:  Negative time is associated with negative temperature particle; cold.  We look for negative time particles in $v_{boom}$, especially, $2n_e\rightarrow n_e$ and then $n_e\rightarrow (n_c=1)+(n_e-1)$ that reduce positive particles and generates negative particles.

So, to generate $T^{-}$, first, radiation at $2.4660675\,Hz$ breaks the $n=2n_e$ positive particles up, then a stream of particles at $v_{boom}$ pops $a_{\psi\,c}$ from the neutral $a_{\psi\,ne}$.  Both $a_{\psi\,c}$ and the resultant particle are negative.

If time particles exist, neutral time particle stream stops time, negative time particle stream reverses time and positive time particle stream carries us into the future.

$TRT$: Time Ray Tunnel