Spread Beyond Light Speed

From the graphs of emission and absorption of a fluorescence material, the right tail of the absorption graph is similar to the left arm of the emission spectrum.

The graphs were taken from the site: http://www.iss.com/resources/research/technical_notes/PC1_LWPolarizationStandards.html

This give rise to the relation at the cross point,

$f_{cross}=f_{max}+\cfrac{1}{2}\Delta f$

where $f_{max}$ is the preceding maximum point, $\Delta f$ is the difference between the two peaks and $f_{cross}$ at the cross point.

On the absorption graph the right arm plots the the energy of the absorbed photons that bring the energy of the particle beyond $\small{\cfrac{1}{2}mc^2}$.  On the emission graph both arms graph the spread of energy around $\small{\cfrac{1}{2}mc^2}$.  The theoretical emission graph is given in the post '"Not Exactly A Fluorescence Polarizer" dated 28 Jul 2015.  Both the right arm of the absorption spectrum and the left arm of the emission spectrum graph the spread of energy where the particle is at/around light speed $c$.

The similarity between the two graphs suggest that the underlying mechanism that store energy when the particle is at light speed are of similar nature.  This energy beyond the particle's $\small{KE=\cfrac{1}{2}mc^2}$, is wholly emitted.  We see this from the emission graphs that begin immediately after the local maxima on the absorption graph.  At the absorption maxima, the particle is at light speed.

This suggests that the particle is first driven to light speed through photons absorption and then it oscillates about the $\psi$ cloud center.  Photons emission occurs at the center.

What happens when the particle swaps velocities $\small{v_{cir}\to v_{shm}}$, $\small{v_{shm}\to v_{cir}}$ as the post "Color!" dated 25 Jul 2015 suggests?

Mathematically this swap occurs with certainty.  At the center both $\small{v^2_{cir}}$ and $\small{v^2_{shm}}$ attains a negative sign, and so both $\small{v_{cir}}$ and $\small{v_{shm}}$ are multiplied by a factor $i$.  This changes $\small{v_{cir}}$ to be along $x$, the radial line and $\small{v_{shm}}$ to be perpendicular to $x$.  The particle now oscillates with velocity $\small{v_{cir}}$ and performs circular motion with velocity $\small{v_{shm}}$.  The change in circular motion frequency results in the emission or absorption of a photon.  (Paradoxically, both $\small{v^2_{cir}}$ and $\small{v^2_{shm}}$ are zero, at the center, after that they are both negative.)

Does the change in oscillation frequency also result in an emission/absorption of a quantum of energy? This question arises because the projection of circular motion onto the diameter of its circular path is SHM oscillation.  Oscillations can be viewed as projected circular motion along the diameter.

The particle's motion is the superposition of circular motion and oscillation.  One energy packet is enough to account for energy gain or loss from its resultant motion.  But where did this packet of energy originate from?  Circular motion or oscillation?  The initial delay when the particle first absorbs a photon then emits one for values of $\small{\theta\gt45^o}$ suggests that the packet of energy is from circular motion.