The Fuzz With $n_{\lambda\,e}$

A Poisson Distribution Plot with $\lambda=1.6678205$ is time-scaled by ($10^9$),

The good thing about Poisson is its sum property which states that,

"Suppose that $N$ and $M$ are independent random variables, and that $N$ has the Poisson distribution with parameter $c\gt0$ and $M$ has the Poisson distribution with parameter $d\gt0$. Then $N+M$ has the Poisson distribution with parameter c+d."

and

"Suppose that N has the Poisson distribution with parameter $c\gt0$. Thenfor $n\in N^+$, $N$ has the same distribution as $\sum_{i=1}^n N_i$ where ($N_1,N_2,…,N_n$) are independent, and each has the Poisson distribution with parameter $\cfrac{c}{n}$."

which lead us to conclude that, as $(c+d)\rightarrow 1$, $(N+M)\rightarrow A$, where $A$ is the total number of particles in the population presenting the Poisson statistics. ie,

$\sum_i^{A}n_{\lambda\,e\,i}=1$

That simply,

$A.n_{\lambda\,e}=A.p\cfrac{mc^2}{\Delta Q}=A.\cfrac{1}{2c}\cfrac{mc^2}{\Delta Q}=A.\cfrac{mc}{2\Delta Q}=1$

$A=2\cfrac{\Delta Q}{mc}$

As $p$ is derived from $c$ along one direction, for all direction in 3D space inside a sphere,

$A_v=\cfrac{4}{3}\pi A^3=\cfrac{32}{3}\pi \left(\cfrac{\Delta Q}{mc}\right)^3$

$A_v$ is the Avogadro Constant (an assumption), referring to mass measured in $kg$.

If we replace,

$\Delta Q=m_{\small{Q}}c^2$

that the entanglement quantum packet has an equivalent mass (inertia), $\small{m_{\small{Q}}}$ in the time dimension.

$A_v=\cfrac{32}{3}\pi c^3\left(\cfrac{m_{\small{Q}}}{m}\right)^3$

If $A_v$ is a constant then, the ratio $\cfrac{m_{\small{Q}}}{m}$ is also a constant.

$6.022140e26=\cfrac{32}{3}*\pi*(299792458)^3\left(\cfrac{m_{\small{Q}}}{m}\right)^3$

$\left(\cfrac{m_{\small{Q}}}{m}\right)^3=\cfrac{6.022140}{9.029022}$

$\left(\cfrac{m_{\small{Q}}}{m}\right)^3=0.66698$

$\cfrac{m_{\small{Q}}}{m}=0.8737$

which is large.  This lead to the postulation that $A_v$ is wrong, that in fact,

$\cfrac{m_{\small{Q}}}{m}=1$

that entanglement is collision in the time dimension of similar particles of equal mass.  The time dimension became accessible at light speed.  In which case, the new constant,

$A_{\small{D}}=\cfrac{32}{3}*\pi*(299792458)^3=9.029022e26$

shall be called the Durian constant.

Have a nice day.