Two For TwoTwo

Continued from the previous post "Spread Beyond Light Speed" dated 07 Aug 2015,

For a oscillating system the maximum speed at the center is,

$v_{max}=\omega.A$

where $A$ is the amplitude of oscillation, and $\omega$ the angular frequency.

Either $\omega$ or $A$ can vary to accommodate the change in $v_{max}$.

But neither happens util the end of the oscillation at $x=A$, where the particle returns towards the center of the $\psi$ cloud.

The change in circular motion frequency occurs instantaneously at the center, when the circular velocity is swapped.  The frequency change, $\small{\Delta f}$,

$\Delta f=\cfrac{v_{cir}}{2\pi x_c}-\cfrac{v_{shm}}{2\pi x_c}=\cfrac{1}{2\pi x_c}(v_{cir}-v_{shm})$

and the energy change $\small{\Delta E}$,

$\Delta E=\cfrac{1}{2}m({v^2_{cir}}-{v^2_{shm}})=\cfrac{1}{2}m(v_{cir}+v_{shm})(v_{cir}-v_{shm})$

$\Delta E=m{\pi x_c}(v_{cir}+v_{shm})\Delta f$

Since $v_{cir}$ and $v_{shm}$ are both driven to light speed at $x=A$ (at the beginning),

$v_{shm}=c$  and  $v_{cir}=c$

(these were the values we started with when the particle in circular motion at light speed, absorbs a photon and is driven to $x_d=A$ from the center at light speed.  The $v^2$ vs $x$ graph was used twice, once to find the radius of circular motion and then again to find the position of the particle (from the center of $\psi$), when it has light speed in a direction always perpendicular to the circular velocity.)

$\Delta E={2\pi x_c}mc.\Delta f$

From the posts "de Broglie Per Unit Volume" and "de Broglie Per Person" both dated 20 Dec 2014, where Planck's constant is,

$h=2\pi a_{ \psi  }mc$

We obtain,

$\Delta E=h\Delta f$

which is Planck's relation if $x_c=a_\psi$.

This means, the particle's motion extends to the edge of the $\psi$ cloud at the radius $a_\psi$.  Light speed is the first constraint on the particle's motion, the extend of $\psi$, $a_\psi$ serves as the second constraint on the particle's motion when it is bombarded with photons.

Two constraints for two superimposed "linear" motions.

Note:  If $v_{shm}=c$  and  $v_{cir}=c$ then why is $(v_{cir}-v_{shm})$ not immediately zero?  The answer is in the difference between instantaneous values and time averaged values.  In an oscillatory system you would pick the point at which energy is maximum to be the total energy of the system.  In this case, $\small{\Delta f}$ is calculated from $\small{(v_{cir}-v_{shm})}$ and the latter is substituted away.   And energy change is calculated with $x_c=A$, at the beginning of the particle's superimposed motion.  In other words, energy change with reference to the energy of the system at $x_c=A$, in the beginning.