Emmy Noether II

From the post "Emmy Noether" dated 6 Jul 2015,

Consider again energy $mv^2$ imparted onto a particle, a portion, $p$ of which is loss due to entanglement,

$E_{loss}=mv^2p$

Since this particles exist in the time dimension $t$, and are travelling along time $t$, the hypothetical force resulting in $E_{loss}$ is given by,

$F_{drag}=-\cfrac{d\,E_{loss}}{d\,x}=-\cfrac{d\,E_{loss}}{d\,t}=-2vmp\cfrac{d\,v}{d\,t}$

If,

$F=m\cfrac{d\,v}{d\,t}$

then

$F_{drag}=-2vpF$

At terminal velocity $v=c$, the total force on the particle is zero,

$\sum{F}=F-F_{drag}=F-2cpF=0$

this implies,

$2cp=1$

The portion $p$ of energy loss is,

$p=\cfrac{1}{2c}$

which is,

$p=0.0000000016678205$

This is essentially the results are before.

If the particles are uniform, all have a fractional entanglement of $p$, then the total number of particle entangled is, $N$

$N=\cfrac{1}{p}=2c$

But if the particles are not uniformly distributed, but Gaussian

but we adjust the total population of the random variable $p_x$ as,

$N_{p_{x}}=N.\sigma\sqrt{2\pi}$

Each instance of $p_x$, in the Gaussian distribution represents entanglement between two particles, so

$N_{\small{G}}=N_{p_{x}}*2=2N.\sigma\sqrt{2\pi}$

This is the total population of $p_x$ along one direction, the total population of $p_x$ in a sphere, in 3D space is,

$A_v=\cfrac{4}{3}\pi (N_{\small{G}})^3=\cfrac{4}{3}\pi(2N.\sigma\sqrt{2\pi})^3$

$A_v=\cfrac{32}{3}(2)^{3/2}(\pi)^{5/2}(N.\sigma)^3$

But what is $\sigma$??

If  $A_v$ is the Avogadro's constant in $kg$ then,

$A_v=6.022140e26$

in $1000g$.  So,

$6.022140e26=\cfrac{32}{3}(2)^{3/2}(\pi)^{5/2}(2*299792458.\sigma)^3$

where $N=2c$

$\sigma=0.1743$

What is the significant of $\sigma$?  Firstly, $0\le p_x\le1$ but $p\lt \sigma$, the spread of $p_x$ is within
one standard deviation, $1\sigma$.  More correctly, Gaussian is a bad fit for the distribution of $p_x$, Maxwell-Boltzmann distribution or Poisson distribution would be considered next.

Win some, Lose some.