Light Dispersion?

From the post "Like Wave, Like Particle, Not Attracted to Electrons", the energy required to move from  $r_{eo}$  to its final position  $r_{ef}$,

${E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\ge PE_e$    ----(*)

The energy required to move from  $r\rightarrow\infty$ to  $r_{eo}$,

$PE_e(r_{eo})={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ eo}-{ r })\}}$

The energy required to move from  $r\rightarrow\infty$ to  $r_{ef}$,

$PE_e(r_{ef})={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r }}{{r}_{ef}})}+C({ r }_{ ef}-{ r })\}}$

And so,

${E}_{s}=\lim _{ r\rightarrow \infty  }\left\{PE_e(r_{ef})-PE_e(r_{eo})\right\}$

${E}_{s}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +C(r_{ef}-r_{eo})}\right\}$

The electron is in circular motion about the nucleus, at  $r_{eo}$ it experiences the centripetal force,

$F_c=\cfrac{m_ec^2}{r_{eo}}=\cfrac{\partial\,{E}_{s}}{\partial\,r_e}={m_ec^2}\left\{ \cfrac{1} {{r}_{ef}{r_{eo} }}-C\right\}$

$C=\cfrac{1} {{r}_{ef}{r_{eo} }}-\cfrac{1}{r_{eo}}=\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }$

Therefore,

${E}_{s}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }(r_{ef}-r_{eo})}\right\}$

For the passing photon, if we consider only the direction along a radial line from  $r_{eo}$  to  $r_{ef}$,  since its longitudinal velocity does not effect net changes in energy in this direction.  (Actually if the interaction between the photon and electron is electrostatic than, the photon slows as it approaches the orbiting electron but speeds up again as it leave the vicinity of the electron; energy is conserved along its direction of travel).

The repulsion between the electron and the photon reduces the centripetal force on the photon.  It is expected that the photon path radius increases, .

$r_{po}\rightarrow r_{pf}$

together with a decrease in circular velocity and so, a drop in $KE_p$.  It is this drop in  $KE_p$  that accounts for the increase in $PE_e$  of the electron.  (The electron orbiting at light speed around the nucleus.)  The centripetal force, $F$  in account for photon/electron repulsion is,

$F=F_c-F_r=\cfrac{m_pc^2_{pf}}{r_{po}}-F_r=\cfrac{m_pv^2_{pf}}{r_{pf}}$

where $F_c$  is due to drag at light speed,  $F_r$  is the repulsion force,  $v_{pf}$  is the velocity of the photon at the furthest point in its orbit right over the atom.  The relative position of the photon and electron is unknown at this point, but  $F_{r\, max}$  is,

$F_{r\,max}=\cfrac{q_pe}{4\pi\varepsilon_o (r_{po}-r_{eo})^2}$

$\Delta E_p=\Delta KE_{p}=\cfrac{1}{2}m_p\left\{v^2_{pf}-c^2 \right\}=-{E}_{s}(r_{eo},\,r_{ef})$ --- (*)

From this we have,

$\cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } } =\cfrac { m_{ p }v^{ 2 }_{ pf } }{ r_{ pf } }$

$m_{ p }v^{ 2 }_{ pf }={ r_{ pf } }\left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\}$

substitue into  (*)

$\cfrac { 1 }{ 2 }{ r_{ pf } }\left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\} -\cfrac { 1 }{ 2 } m_{ p }c^{ 2 }=-{E}_{s}(r_{eo},\,r_{ef})$

${ r_{ pf } }=A \left\{ \cfrac { 1 }{ 2 } m_{ p }c^{ 2 }-{ E }_{ s }(r_{ eo },\, r_{ ef }) \right\}$

where  $A$  is,

$A=\cfrac { 2 }{ \left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\}  }$

This is the maximum value of  $r_{pf}$, simply because we set  $F_r$  to be maximum.  But since,

$c=2\pi .r .f$

there is a corresponding maximum decrease in frequency.

$f_{pf}=\cfrac{c}{2\pi.r_{pf}}$

In fact, since interaction between the photon and electron at different distances are possible,  there is a spread of frequency from  $f_{pf}$  to  $f_{po}$.  That photons emerging from the material has now a spread of energy, and the frequency range is,

$f_{po}≥f≥f_{pf}$

where  $f_{op}$  is the frequency of the incident photon passing through without interacting with any electrons.  If subsequently, these photons with a spread of frequencies are to eject more packets of energy,  the resulting quanta will show a spread of colors as each photon interacts with different electron at different  $r_e$,  $r_e$  vs  $T$  profile and kink point.

This might explain the spread of colors we see when white light enters a prism.  The resulting spectrum is continuous.  However, this model indicates that monochromatic light will also spread continuously down the frequency spectrum.  More importantly,  in this model, photons are self propelling dipoles that will regain the lost  $KE_p$  and returns to a circular velocity of light speed by itself.  So if we recombine the spread of light,  we will see white light further down its path with the same intensity!

Where do photons go to die?  Where's the photon graveyard?