From the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015,
\(v^{ 2 }_{ cir }=cos(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x))+\cfrac { \psi _{ d } }{ m } \right\}\)
and
\(v^{ 2 }_{ shm }=sin(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x))+\cfrac { \psi _{ d } }{ m } \right\}\)
where a particle is both in circular motion and oscillating along a radial line. A change in sign of \(v^2_{shm}\) occurs at \(v_{shm}=0\).
When both \(v^2_{cir}\) and \(v^2_{shm}\) are negative (ie. \(v_{cir}\) and \(v_{shm}\) are complex), their role switches.
\(v_{cir}\to v_{shm}\) and
\(v_{shm}\to v_{cir}\)
We have a change in the radius of the circular motion and a change in the velocity along the radial line at the same time. This is so because both expressions for \(v_{cir}\) and \(v_{shm}\) instantaneously attains zero at the same time. \(KE\) is still conserved, but the frequency of oscillation changes abruptly. The latter result in a abrupt change in \(PE\), from \(E=h.f\).
This could be the mechanism by which a particle releases or gains packets of energy.
When the switch occurs, \(v^2_{cir}\) changes by a factor,
\(\cfrac{sin(\theta)}{cos(\theta)}\)
and the corresponding change in frequency is by a factor,
\(\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\)
If \(\Delta E= h. \Delta f\), then, the photon released is,
\(E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}\)
since \(v^2_{cir}\) is changing, we have only a specific \(v^2_{min}\) and expect the frequency, \(f_{cir}\) to spread to higher values. So,
\(E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{v_{min}}{2\pi x_{min}}=\hbar.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{v_{min}}{x_{min}}\)
where \(x_{min}\) is the value of \(x\) at which \(v^2\) is minimum and is also the radius of the circular motion around the center of the \(\psi\) cloud. From the post "Two Quantum Wells, Quantum Tunneling, \(v_{min}\)" dated 19 Jul 2015,
\(v^2_{min}=-\cfrac{\psi^2_c}{m}.tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }\)
Considering only magnitude,
\(E_{p}=\hbar\cfrac{\psi_c}{\sqrt{m}}\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.\sqrt{tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }}.\cfrac{1}{x}\)
We have an expression for the range of energy for the photon emitted/absorbed by a particle orbiting at a radius \(x\).
The particle is in motion by an initial excitation, afterwards the emission and absorption of such packets of energy occurs naturally when both \(v_{cir}\) and \(v_{shm}\) attains zero instantaneously.
In the derivation for \(v^2_{cir}\), \(\theta\) is the angle between the plane of the circular motion and the radial line joining the particle with the center of the \(\psi\) cloud. The particle in effect move on the surface of two cones pointed tip to tip at \(x=x_z\) , with \(\theta\) held constant.
Now that \(\theta\) has gained greater significance, what is \(\theta\)?
When the particle is at light speed,
\(E_{p}=\hbar.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{c}{x_{orbit}}\)
If \(\theta\) is a constant, then this is a sharp photon emission/absorption from the particle in orbit.
What is \(\theta\)?
