Amnesia's Dream: Cracking Balls

Tuesday, January 16, 2018

If this is the torus photon,

we can approximate,

E = h f = h * \frac{2 π a_{ψ l}}{v} = \frac{1}{2} m v_{b o o m}^{2}

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

that all energy due to

v_{b o o m}

$v_{boom}$ is stored in

a_{ψ l}

$a_{\psi\,l}$ and the initial energy before impact and also the energy of the photon upon collapse remains unchanged in

a_{ψ s}

$a_{\psi\,s}$ .

We are not going to get

m

$m$ , because,

h = 2 π a_{ψ l} m c

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

So,

2 π a_{ψ l} m c * \frac{2 π a_{ψ l}}{v} = \frac{1}{2} m v_{b o o m}^{2}

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

v_{b o o m}^{2} = 2 (2 π a_{ψ l})^{2} \frac{c}{v}

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

v_{b o o m} = \sqrt{\frac{2 c}{v}} * 2 π a_{ψ l}

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

or

a_{ψ, l} = \frac{v_{b o o m}}{2 π \sqrt{2}} * \sqrt{\frac{v}{c}}

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

We may also define a

k_{ψ}

$k_{\psi}$ such that,

\frac{1}{2} k a_{ψ l}^{2} = \frac{1}{2} m v_{b o o m}^{2} = h * \frac{2 π a_{ψ l}}{v} = (2 π a_{ψ l})^{2} m * \frac{c}{v}

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

k = 8 π^{2} \frac{c}{v} m

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

as if the photon is a spring extended to accommodate and store the energy in

v_{b o o m}

$v_{boom}$ . Naturally,

f_{r e s} = \frac{1}{2 π} \sqrt{\frac{k}{m}} = \frac{1}{2 π} \sqrt{8 π^{2} \frac{c}{v}} = \sqrt{\frac{2 c}{v}}

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

This means a collision frequency at

f_{r e s}

$f_{res}$ will stretch a torus photon along its larger radius. Given

E_{p}

$E_p$ as the energy per photon, a power setting of,

P_{r e s} = f_{r e s} E_{p} J s^{- 1}

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

will create a halo around the ray impacting a blocking medium. Since

E_{p}

$E_p$ is very small,

P_{r e s}

$P_{res}$ is low, the effect of which is small.

f_{r e s}

$f_{res}$ 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_{ψ l} \to a_{ψ s} = a_{ψ c}

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

v \to c

$v\rightarrow c$

f_{r e s} \to \sqrt{2}

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

Given a simple string-ball pendulum,

f = \frac{1}{2 π} \sqrt{\frac{g}{l}} = \sqrt{2}

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

l = \frac{g}{8 π^{2}}

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

g = 9.806 m s^{- 1}

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

l = 0.1242 m

$l=0.1242\,m$

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

Tuesday, January 16, 2018