How to make a hollow particle? When \(a_{\psi}\lt a_{\psi\,\pi}\) set off a perturbation in \(\psi\) towards the center of the particle. Pinch, pull and let go.
When is \(a_{\psi}\lt a_{\psi\,\pi}\)? When \(n\lt 77\), when the particle is smaller.
We can direct \(\psi\) only if \(\dot{x}\) retains its direction when the particle returns from the time dimension. We point ourselves in space at Jupiter, travel to light speed in space and exit into the time dimension, travel in time by the comparative age of Earth and Jupiter (Jupiter is older by 50000 years; a guess); travel back in time by 50000 years and pops back into the space dimension, in the same direction. The gain is in the higher speed limit in the time dimension of \(\small{4\sqrt{}2.\pi^2c}\) and possibly a short travel distance of 50000 years in time.
If these details are true, only places of about the same age as the starting point is worth travelling to this way.
But this enunciation does bring out a point; the need to travel through the distance in time between the starting and destination points.
Which lead us to "time particles".
Not only do the entangled particles in space be hollow, they have to exist in the time dimension as one big particle. One big time particle. Only then, there is entanglement.
Which bring us back to the Durian constant...
If applying the Durian constant is correct, then this big time particle contains \(\cfrac{4}{3}\pi(2c)^3\) number of basic particles. One big lump of entanglement.
Why? My guess is it takes an equivalent distance of \(2c\) to reach light speed under the action of the field in the time dimension. At \(2c\) from the center of a particle, a test particle \(\small{\cfrac{1}{c}}\) returns to the space dimension.
Unfortunately,
Mass \(m\) is the proportional constant in the space dimension \(x\). If a point of unit mass move through \(x\). A point mass of \(m=2\) moves through by \(\cfrac{1}{2}\). In a very strained but analogous way, in the time dimension at light speed \(c\) (\(F.c\),energy is flowing out of the particle at light speed), a time particle moves through time by,
\(n_c=\cfrac{c}{\left(\cfrac{1}{c}\right)}\)
number of time particle of inertia \(\cfrac{1}{c}\) per second. But since we are in the time dimension,
\(n_{c\,t}=\cfrac{d}{dt}\left\{n_c\right\}\)
we differentiate \(n_c\) in time again to obtain \(n_{c,t}\) the number of particles per unit time in the time dimension.
\(n_{c\,t}=2c\)
And so in the time dimension a sphere of entangled particle is,
\(A_D=\cfrac{4}{3}\pi (2c)^3\)
Why do we consider a windows of "per second" over a distance of \(c\) in time? It is for the same reason we differentiate \(n_c\) in time again to obtain \(n_{c,t}\). In the post "A Mass In Time And In Mind" dated 04 Jul 2016, the expression
\(F.c=\cfrac{dKE}{dt}\)
was differentiated with respective to time to arrive at,
\(\cfrac{dF}{dt}=\cfrac{1}{c}\cfrac{d^2KE}{dt^2}\)
the force in time.
In the time dimension displacement is measured in time, in order to obtain a dimensionless count in number, we differentiate once with time. In order to obtain a per unit time count, (count per second), we differentiate again with respective to time. So a dimensionless count is consider over a per second window. The time dimension is the equivalent of the frequency domain where relevant parameters are in per unit time.
Intuitively speaking...
