The Universe Particle

From the post "Pound To Rescue Permittivity" dated 30 May 2016,

$\cfrac { q }{ \varepsilon _{ o } } =4\pi a^{ 2 }_{ \psi  }m.2c^{ 2 }ln(cosh(\pi ))$

but,

$\Delta q=m.4\pi r^{ 2 }.\Delta r$

$\cfrac { dq }{ dr } =m.4\pi r^{ 2 }$

so,

$\cfrac { q }{ \varepsilon _{ o } } =\cfrac { dq }{ dr } |_{ a_{ \psi  } }.2c^{ 2 }ln(cosh(\pi ))$

If we let,

$\varepsilon _{ o }=\cfrac { 1 }{ 2c^{ 2 }ln(cosh(\pi )) }$

$q=\cfrac { dq }{ dr } |_{ a_{ \psi  } }$

Since,

$\cfrac { q }{ 4\pi r^{ 2 }\varepsilon _{ o } } \equiv \cfrac { G^{ ' }M }{ r^{ 2 } }$

as both types of field, electrostatic and gravitational are equivalent force fields due to particles with $psi$ wrap around a center.

$q=4\pi \varepsilon _{ o }G^{ ' }M=G_oM=\cfrac { dq }{ dr } |_{ a_{ \psi  } }$

where,

$G_o=4\pi \varepsilon _{ o }G^{ ' }$

With this out of the way, from the same post "Pound To Rescue Permittivity" dated 30 May 2016,

$F=\int{F_{\rho}}dx$

where,

$F_{ \rho  }=i\sqrt { 2{ mc^{ 2 } } } \, G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right)$

So,

$F=i2{ mc^{ 2 } }.ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }))$

for $0\le x \le a_{\psi}$, within the boundary of $\psi$.  There is a problem with $i$, but we know that $F$ is towards the center of the particle, an attractive force between like particles interacting as wave (the opposite of repulsive force between like particles when they are interacting as particles).

If the universe is one big particle,

within the universe, there is a force directed towards the center of the universe that increases with distance from the center.

Objects further from the center is accelerated greater towards the center.

This is not an expanding universe but a collapsing one.  $a_{\psi}$ however is a constant; the universe itself is not collapsing.

From a perspective other than the center, objects on the same side as the center are accelerating away from the observer and objects on the opposite side as the center are accelerating towards the observer.  Since, the force is greater at greater distance from the center, objects approaching the observer will always have greater acceleration.

Where do we stand?