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.
