a particle in orbit about the center of a \(\psi\) cloud at light speed \(c\). On impact of a photon, this particle is displaced from its initial position. Since it is at the speed limit, \(c\), its speed remained unchanged at the new position and so neither does its radius of circular motion change.
From the post "Nature Has Orbits" dated 19 Jul 2015,
\(v^2=-\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
and the energy stored is,
\(E_{xd}=\cfrac{1}{2}mv^2=\cfrac{1}{2}m\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
\(E_i=E_{xd}=\cfrac { 1 }{ 2 } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_d}\)
and the particle is orbiting at speed \(c\),
\(c^2=\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_v}\)
\(cos(\theta)=\cfrac{x_v}{x_d}\)
\(x_v\) is fixed given light speed, but \(x_d\) changes with photon energy. Because \(v^2\) has a minimum value, there is also a minimum photon energy below which the particle does not move on impact.
\(E_{i\,min}=\cfrac{1}{2}mv^2_{min}\)
Above this threshold all energy from the photon is emitted EXCEPT when,
\(c^2=\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_c}\)
at the other solution of \(v^2=c^2\). The particle at point this will continue to orbit around the center with this higher value radius and does not oscillate It is not possible to excite the particle beyond this radius because of the light speed limit. At values less than this upper limit, the lower solution of \(x\) to a given \(v^2\) is occupied first than upon further absorption of photons the next higher solution above the minimum point is occupied.
It is counter intuitive that a smaller packet of energy at \(v^2_{min}\) can cause the particle to go further out along the radial line. In this situation, it is not the greater energy going the furthest distance, but the appropriate energy state at the appropriate location in a \(\psi\) cloud.
The range of energy over which the photon is absorbed and emitted is delimited between \([x_v,\,x_c)\), not including \(x_c\).
Since,
\(x_c\gt x_d\gt x_v\)
\(cos(\theta)=\cfrac{x_v}{x_d}\lt 1\)
It is then possible that when,
\(\sqrt{2}.x_v=x_d\)
\(cos(\theta)=\cfrac{1}{\sqrt{2}}\)
\(\theta=45^o\) at this point no photons are emitted (from the post "Polarization And Invisibility" dated 25 Jul 2015), the particle continues to oscillate, otherwise the oscillation is damped with emission of photons. At \(\theta=45^o\), we do not have resonance, as the amplitude of oscillation cannot increase further, but we have no loss from the oscillation. It could be that the particle begins to perform circular motion around a second axis.
The range of energy that the particle can absorb and emit is,
\(\cfrac{1}{2}mv^2_{min}\lt E_p\lt\cfrac{1}{2}mc^2\)
except a sharp bandwidth, when \(v^2(x=x_v\sqrt{2})\) at which \(v^2=v^2_{v}\), absorption of this photon is not emitted.
The diagram above shows three issues. A particle at \(c^2\) receive a photon \(E_p\) and transit to higher radius. It is still in circular motion at \(c\). This transition is a gain of energy by \(E_p\) over and above \(\cfrac{1}{2}mc^2\). The other solution of \(x\) also at an energy level of \(E_p\) is accessible only through \(v^2_{min}\) upon a further absorption of energy at \(v^2_{min}\). When the photon is at \(\cfrac{1}{2}mv^2_v\), which drive the particle to the point \(x=x_v\sqrt{2}\), this photon is not emitted.
The particle absorb a photon and finds its appropriate location in the \(\psi\) cloud. Absorbing a photon with less energy can propel the particle to a further location. The minimum energy absorbed is \(\cfrac{1}{2}mv^2_{min}\). Further absorption at any position on the curve below \(c^2\), pushes the particle further along the curve. This results in a Delta Spectrum of lower frequencies, that is the difference of the energy between the final and initial positions on the curve. (Remember the little hump on the left).
The initial \(E_{p}\) finds the position \(x\) on the \(v^2\) vs \(x\) curve, subsequent absorption of lower energy photons move the particle along the curve.
While on the curve, the particle is in circular motion at light speed \(c\). Once displaced the particle is no longer on the curve above, it is in oscillation about the center of the \(\psi\) cloud (from the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015.
The particle is oscillating between point \(a\) and \(b\). This graph is plotted with the requirement that \(x=0\), \(v^2=0\). As the particle move along the radial line, it is at zero radius with respect to its circular motion, when \(x=0\), this requires that \(v=0\), thus \(v^2=0\). The graph above is both \(v^2_{cir}\) and \(v^2_{shm}\); these two differ by a constant factor of \(\cos(\theta)\) and \(sin(\theta)\), respectively.
As the particle passes through \(x=0\), it emits a photon. This lost of energy, damps the oscillation.
The photon emitted is given by,
\(E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}\)
from the post "Color!" dated 25 Jul 2015. The next emission as the particle passes through \(x=0\) again is,
\(E_{p}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.f_{shm}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}\)
and the next emission,
\(E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}\)
and
\(E_{p}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}\)
...
Here we have a problem that was set aside when we have the expression,
\(\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\)
It is unbounded, \(cos(\theta)=0\), and is not guaranteed to be smaller than one. But from the four expressions above for \(E_P\), two will result in a positive emission (ie. lost in energy).
What is then the physical interpretation of the the remaining pair? In this case, photon absorption is not a passive process but an active one, where energy from the surrounding \(\psi\) is absorbed at a determined place \(x=0\), in an determined amount, \(E_p\).
One set of expressions for \(E_p\) emits a photon. The reciprocal set of expressions for \(E_p\) absorb packets of energy from the \(\psi\) cloud. In this case, \(E_p\) is best express as,
\(E_{p}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}\)
and
\(E_{p}=h.\left\{1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right\}.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}\)
where a negative \(E_p\) is an absorption of a photon. And we get rid of the absolute value bar by allowing for active spontaneous photon absorption from the surrounding \(\psi\).
A pump!
Note: We still need a constraint on \(\theta\).
