Cracking Balls

If this is the torus photon,

we can approximate,

$E=hf=h*\cfrac{2\pi a_{\psi\,l}}{v}=\cfrac{1}{2}mv^2_{boom}$

that all energy due to $v_{boom}$ is stored in $a_{\psi\,l}$ and the initial energy before impact and also the energy of the photon upon collapse remains unchanged in $a_{\psi\,s}$.

We are not going to get $m$, because,

$h=2\pi a_{\psi\,l}mc$

So,

$2\pi a_{\psi\,l}mc*\cfrac{2\pi a_{\psi\,l}}{v}=\cfrac{1}{2}mv^2_{boom}$

$v^2_{boom}=2(2\pi a_{\psi\,l})^2\cfrac{c}{v}$

$v_{boom}=\sqrt{\cfrac{2c}{v}}*2\pi a_{\psi\,l}$

or

$a_{\psi,l}=\cfrac{v_{boom}}{2\pi\sqrt{2}}*\sqrt{\cfrac{v}{c}}$

We may also define a $k_{\psi}$ such that,

$\cfrac{1}{2}ka^2_{\psi\,l}=\cfrac{1}{2}mv^2_{boom}=h*\cfrac{2\pi a_{\psi\,l}}{v}=(2\pi a_{\psi\,l})^2m*\cfrac{c}{v}$

$k=8\pi^2\cfrac{c}{v}m$

as if the photon is a spring extended to accommodate and store the energy in $v_{boom}$.  Naturally,

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{8\pi^2\cfrac{c}{v}}=\sqrt{\cfrac{2c}{v}}$

This means a collision frequency at $f_{res}$ will stretch a torus photon along its larger radius.  Given $E_p$ as the energy per photon, a power setting of,

$P_{res}=f_{res}E_{p}\, Js^{-1}$

will create a halo around the ray impacting a blocking medium.  Since $E_p$ is very small, $P_{res}$ is low, the effect of which is small.

$f_{res}$ is the frequency to resonate a torus photon and may also be the frequency to grow a basic particle, because if the torus collapses,

$a_{\psi\,l}\rightarrow a_{\psi\,s}=a_{\psi\,c}$

$v\rightarrow c$

$f_{res}\rightarrow \sqrt{2}$

Given a simple string-ball pendulum,

$f=\cfrac{1}{2\pi}\sqrt{\cfrac{g}{l}}=\sqrt{2}$

$l=\cfrac{g}{8\pi^{2}}$

$g=9.806\,ms^{-1}$

$l=0.1242\,m$

Do the balls heat up, start to glow, become positively charged and explode?