It's Fluorescence

The emitted photons frequencies are less than the absorbed frequencies, emitted after a time delay; this is fluorescence.

Stokes shift is the direct result of,

$E_p=h.\Delta f_{cir}$

as

$\Delta f\lt f_{cir}$

And the spectra overlap occurs exactly at the cross point,

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

The second lower hump on the left corresponds to secondary photons absorbed that drive the particle beyond $v^2_{min}$ up the two arms of the $v^2$ vs $x$ curve.  That is the Delta Spectrum.

The graph above allow us to obtain $\theta$,

$\Delta f=f_{blue\,max}-f_{red\,max}=\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}=\left\{1-\sqrt{{tan(\theta)}}\right\}.f_{cir}$

Maximum absorption corresponds to an energy level of $\cfrac{1}{2}mc^2$, where the particle is orbiting at speed $c$ afterward (from the post "Initial High Energy Photon" dated 26 Jul 2015).  The majority of the particle is at a nominal energy state, $v^2_m\lt c^2$ given temperature, after absorption of photons (at  $\cfrac{1}{2}m(c^2-v^2_m)$) most particle will be in circular motion at speed $c$.

So,

$f_{cir}=\cfrac{c}{2\pi x_r}$

$\Delta f=f_{blue\,max}-f_{red\,max}=\left\{1-\sqrt{{tan(\theta)}}\right\}.\cfrac{c}{2\pi x_r}$

where,

$x_r=x_v$  or  $x_c$

Since other particles along the two arms of the $v^2$ vs $x$ curve around $v^2_{min}$ can absorb smaller packets of energy (which results in the Delta Spectrum) and reach $c^2$, the emitted peak is higher than the absorption peak.  There is more particle with circular energy $\cfrac{1}{2}mc^2$ than those that absorbed an energy packet $\cfrac{1}{2}m(c^2-v^2_m)$ directly.

When particles can travel along both the left and right arms of the $v^2$ vs $x$ curve, the emission spectrum will have twin peaks as shown,

When the particles received higher energy photons and is moved to orbit of light speed, $c$, from the post "Initial High Energy Photon",

$cos(\theta_v)=\cfrac{x_p}{x_r}$

where $x_p$ is the initial orbital radius and $x_r$ is the orbital radius when the particle has speed $c$ according to the $v^2$ vs $x$ curve,

with

$x_v\lt x_c$

$cos(\theta_v)=\cfrac{x_p}{x_v}\gt cos(\theta_c)=\cfrac{x_p}{x_c}$

The change in frequency that results in photon emission is,

 $\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}=\left\{ 1-\sqrt { \frac { \sqrt { \left( \cfrac { x_{ p } }{ x_{ v } }  \right) ^{ 2 }-1 }  }{ \cfrac { x_{ p } }{ x_{ v } }  }  }  \right\} \cfrac { c}{ x_{ v } }$

$= \left\{ 1-\sqrt { \cfrac { x_{ v } }{ x_{ p } } \sqrt { \left( \cfrac { x_{ p } }{ x_{ v } }  \right) ^{ 2 }-1 }  }  \right\} \cfrac { c }{ x_{ v } }$

$=\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ v } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c}{ x_{ v } }$

as $x_p\gt x_c\gt x_v$

$\cfrac{x_v}{x_p}\lt \cfrac{x_c}{x_p}\lt 1$

From the following plot,

when $x_v\lt x_c$

$\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ v } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c }{ x_{ v } } \lt\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ c } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c }{ x_{ c } }$

$E_{p\,v}\lt E_{p\,c}$

Photons emission from the left arm has less energy than photon from the right arm.  Particles along the right arm result in energy emission closer to the absorption spectrum, at energy higher than the emission due to particles moving along the left arm.  In the case when the left arm is truncated,

less particles on the left arm absorb photons to the elevated state.  At the same time, since

$v^2_l\lt c^2$

$v_l\lt c$

$E_{p\,v}\lt\lt E_{p\,c}$

The second emission peak due to the left arm is lower in fluorescence than the right arm, and it is also of greater wavelength.