Correcting The Correction

Maybe,

$\cfrac{q^2}{4\pi \varepsilon_o r^2}=\cfrac{q}{4\pi \sqrt{\varepsilon_o} r^2}.\cfrac{q}{\sqrt{\varepsilon_o}}$

that, the change in $KE$ per unit test charge is,

$\cfrac{q}{4\pi \sqrt{\varepsilon_o} r^2}$

This would change the definition of $\varepsilon_o$ in this model, and the symbol $\sqrt{\varepsilon}$ replaces $\varepsilon_o$. The correction factor to $c$ considering $\sqrt{\varepsilon}$ remains,

$\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }$

the same.

What Change $KE$?

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016,

$q|_{a_\psi}=2\dot{x}F_{\rho}|_{a_\psi}$

and at a distance $r$, $r\gt a_{\psi}$ from the center of the field,

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=\dot{x}F_{\rho}|_{r}$    defined this way

where $\dot{x}=v$.

This was Coulomb's inverse square law, as a constant power is divided over the surface area of a sphere at radial distance, $r$ from the center of the field.

So, with

$\cfrac{1}{\varepsilon_o}=2c^2ln(cosh(\theta_{\psi}))$ and,

$F_{\rho}|_{r}=\cfrac{F_{\rho}|_{a_\psi}}{4\pi r^2}.\cfrac{2}{\varepsilon_o}$

here the factor $\cfrac{2}{\varepsilon_o}$ behaves like surface area,

${F_{\rho}|_{a_\psi}}.\cfrac{2}{\varepsilon_o}=\Psi_c$

is the total flux emanating from a sphere of surface area of $4\pi a_{\psi}$, at a distance $a_{\psi}$ from the center of the field.  So that, $F_{\rho}|_{r}$ is simply,

$F_{\rho}|_{r}=\cfrac{\Psi_c}{4\pi r^2}$

And when $\dot{x}=v=c$, we have,

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{\rho}|_{r}.c$

This is different from Coulomb's

$\cfrac{1}{4\pi r^2}\cfrac{q}{\varepsilon_o}=F$

the force in a field per unit charge.

From Newton force,

$F=ma=m\cfrac{dv}{dt}$

An incremental work done by this force,

$F.\Delta x=m\cfrac{dv}{dt}\Delta x$

If we allow $m$ to change with velocity, ie to associate the factor $32\pi^4$ in the expression,

$32\pi^4v^2_x+v^2_t=32\pi^4c^2$

that originated from considering the different in speed limits in the time dimension and the space dimension and that energy is conserved across the two dimensions.  At the speed limit, a particle passes over to the orthogonal dimension; in space, light speed brings a particle to the time dimension at $v_t=0$.  So, we have,

$F.{ \Delta x }=\left\{ m(v_{ p })+\Delta m(v_{ p }) \right\} \cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

where $m=m(v_p)$, the particle mass is a function of its velocity, and the particle is displaced by $\Delta x_p$ instead.  $\Delta m(v_p)$ is due to a change $v_p$, $\Delta v_p$ with $\Delta x_p$,

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\Delta m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\cfrac { dm(v_{ p }) }{ dv_{ p } } \Delta v_{ p }.\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\cfrac { dm(v_{ p }) }{ dt } \Delta v_{ p }.{ \Delta x_{ p } }$

Over time $\Delta t$,

$F.{ \cfrac { \Delta x }{ \Delta t }  }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \cfrac { \Delta x_{ p } }{ \Delta t }  }+\cfrac { dm(v_{ p }) }{ dt } \cfrac { \Delta v_{ p } }{ \Delta t } .{ \Delta x_{ p } }$

as $\Delta t\rightarrow 0$

$\cfrac{\Delta x_p}{\Delta t}=v_p$,  $\cfrac{\Delta v_p}{\Delta t}=a$

$F.{ \cfrac { \Delta x }{ \Delta t }  }=m(v_{ p })v_p\cfrac { dv_{ p } }{ dt } +\cfrac { dm(v_{ p }) }{ dt } a.{ \Delta x_{ p } }$

since $\Delta t$ is small and $\Delta x$ is small, $a$ is a constant, basically,

$v=u+at$

$\Delta x=ut+\cfrac { 1 }{ 2 } at^{ 2 }$

$\Delta x=u\cfrac { (v-u) }{ a } +\cfrac { 1 }{ 2 } \cfrac { (v-u)^{ 2 } }{ a }$

$2a\Delta x=2(uv-u^{ 2 })+(v-u)^{ 2 }=v^{ 2 }-u^{ 2 }$

when $u=0$, the particle is subjected to the field from rest,

$2a\Delta x=v^{ 2 }$

$a\Delta x=\cfrac { 1 }{ 2 } v^{ 2 }$

So,

$F.{ v }=m(v_{ p }).v_{ p }\cfrac { dv_{ p } }{ dt } +\cfrac { 1 }{ 2 } v^{ 2 }_{ p }\cfrac { dm(v_{ p }) }{ dt }$

$F.{ v }=m(v_{ p })\cfrac{1}{2}\cfrac { dv_{ p }^2 }{ dt } +\cfrac { 1 }{ 2 } v^{ 2 }_{ p }\cfrac { dm(v_{ p }) }{ dt }=\cfrac { d }{ dt }\left\{\cfrac{1}{2}m(v_p).v_P^2\right\}=\cfrac { dKE }{ dt }$

$v$ is independent of $v_p$, it is a parameter associated with the field only.

From previously,

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

$n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

$c$ is the flow of energy from  the center of the field, irrespective of the velocity of a test charge in the field.  $c$ is a constant between all interacting particles with the same $n$ number of constituent basic particle.

When $v=c$, the idea of $F$ changing $KE$ directly and $F.c$ changing $KE$ directly differ by a constant.

$F.c=\cfrac { dKE }{ dt }$

$F.c$ is decreasing with distance $r$ from the center of the field because the fixed total amount of energy from the field passes through a bigger surface area $4\pi r^2$ at $r$.
...

This is still prelude to an expression for mass $m$, but correct for $F.c$ that change $KE$ directly.

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{\rho}|_{r}.c\ne F_{newton}$

Two things have changed, a field is quantifies by its power output $F.c$ and the force in a field is different from the Newtonian force.

Prelude To "What Is Inerita?"

From the previous post "Entanglement Made It So" dated 29 Jun 2016,

$v^2_x+v^2_t=c^2$

$v_x$ is in the space dimension, $v_t$ is in the time dimension, both are of different reference frame.  In the time dimension the unit of measure for velocity, displacement per unit time, changes with the unit of time that lengthens with increasing velocity in time.  We can, however, compare the maximum points in each of the dimension and expresses both in the space dimension.  From the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016, for example, the maximum kinetic energy of a particle in the time dimension, with reference to the space dimension was,

$E_t=32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$ --- (**)

compared to

$E_{x}=\cfrac { 1 }{ 2 } mc^{ 2 }$

in the space dimension.

So, (*) above with reference to the space dimension only, could be,

$32\pi^4v^2_x+v^2_t=32\pi^4c^2$ --- (*)

that would satisfy the two extremema, $v_x=0\,\,\,v^2_t=32\pi^4c^2$ and $v_t=0\,\,\,v^2_x=c^2$

But in between these extremema, the transition of $v_x$ to $v_t$ is not in linear proportion.

If we associate the factor $32\pi^4$ with the mass of the particle, then as the velocity $v_x$ of a particle of mass $m$, increases towards light speed $c$, its mass increases to $32\pi^4m$ and its $KE$ is given by (**) with the understanding that the particle is now in the time dimension.

The point of this is prelude to find the particle's mass, $m$ from $\psi$ and light speed $c$.

For the sake of...

$\cfrac{v^2_t}{c^2}=\gamma^2=32\pi^4\left(1-\cfrac{v_x^2}{c^2}\right)$

$\gamma=4\sqrt{2}.\pi^2\sqrt{\left(1-\cfrac{v_x^2}{c^2}\right)}$

Einstein, or if you travel through time.

$F=m_ia=m_i\cfrac{dv}{dt}$

An incremental work done by this force,

$F.\Delta x=m_i\cfrac{dv}{dt}\Delta x$

over time $\Delta t$,

$F.\cfrac{\Delta x}{\Delta t}=m_i\cfrac{dv}{dt}\cfrac{\Delta x}{\Delta t}$

as $\Delta t\rightarrow 0$

$F.v=m_iv\cfrac{dv}{dt}=\cfrac { 1 }{ 2 } m_{ i }\cfrac { dv^{ 2 } }{ dt } =\cfrac { dKE }{ dt }$

where $F$ changes along the distance between the particle and the center of the field.

$F.v =\cfrac { dKE }{ dt }$

is consistent, where the input power (rate of change in energy) results in a change in kinetic energy.  From,

$F.v=\cfrac { 1 }{ 2 } m_{ i }\cfrac { dv^{ 2 } }{ dt }$

$m_{ i }=\cfrac { 2F.c }{ \cfrac { dv^{ 2 } }{ dt }  }$

From (*),

$32\pi ^{ 4 }\cfrac { dv^{ 2 }_{ x } }{ dt } +\cfrac { dv^{ 2 }_{ t } }{ dt } =0$

since $c^2$ is a constant.

$\cfrac { dv^{ 2 }_{ x } }{ dt }=-\cfrac{c^2}{32\pi^4}\cfrac { d\gamma^{ 2 } }{ dt }= \cfrac { dv^{ 2 } }{ dt }$

and we are not getting anywhere.

$m_i=-\cfrac{64\pi^4}{c}F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}$

This particle is not at rest in the field so, $\cfrac { dv^{ 2 } }{ dt }\ne0$ and $\cfrac { d\gamma^{ 2 } }{ dt }\ne0$.  As the particle speed increases in the field $F$, time speed decreases and so $\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)\lt 0$.

At light speed,

$\lim\limits_{v\to c}{F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}}\rightarrow-1$

as both $F$ and $\cfrac{d\gamma^2}{dt}$ is $\cfrac{dv^2}{dt}$ dependent.

$\lim\limits_{v\to c}m_i=\cfrac{64\pi^4}{c}$

This is the mass of the particle at light speed, $c$.

And boldly we are at where no rational man has gone before, because

$\lim\limits_{v\to c}{F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}}\rightarrow-\cfrac{c}{64\pi^2}m_i$

and

$\lim\limits_{v\to c}m_i=m_i$

It seems that rest mass is the mass at light speed.  The speed limits in the space dimension and the time dimension are different; $c$ and $4\pi^2\sqrt{2}$ respectively.

Still, how to obtain $m_i$ from

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

$n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

???

Once $c$ is defined and given a unit, $ms^{-1}$, $kg$ is also defined.

Entanglement Made It So

We know that because of entanglement,

$max(v_x)=c$

Entanglement shares energy and acts like a drag force proportional to $v_x^2$, that give rise to a terminal velocity, $max(v_x)$.  And

$max(v_t)=c$

since all dimensions are equivalent, swapping $x$ for $t$ is arbitrary.

Since, $v_t$ and $v_x$ are oscillating and orthogonal,

$v_x=max(v_x)sin(\omega t)=c.sin(\omega t)$

$v_t=max(v_t)cos(\omega t)=c.cos(\omega t)$

With the trigonometrical identity,

$sin^2(\omega t)+cos^2(\omega t)=1$

$c^2sin^2(\omega t)+c^2cos^2(\omega t)=c^2$

So,

$v^2_x+v^2_t=c^2$ --- (*)

where $c$ is the speed limit in both dimensions.

We have conservation of energy across two dimensions for a single particle.  As $v_x$ increases in a field, $v_t$ decreases and we have time dilation.

Good night.

$c$ Is For Circle

Is light speed still a constant for all particles without the adjustments?

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

$n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

so, given $n$, $\theta_{\psi}$ is determined and $c$ is fixed.  $c$ is the same for all particles with $n$ number of basic particles.

$c_{n=i}\ne c_{n=j}$

for $i\ne j$

But this does not proof that $c$ is a constant even for particles with the same $n$.  $c$ was assumed to be a speed limit when $\theta_{\psi\,c}=0.7369$ was obtained from

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}$

in the post "New Discrepancies And Hollow Earth" dated 23 Jun 2016.

We are going in circles otherwise...

Adjustments And Light Speed...

But does it make sense to apply the correction,

$c_{ adj }=c.\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }$??

If we generalize,

$16*3.135009*{ \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(3.135009 )) }tanh(3.135009)\cfrac { 1 }{ 4\pi a_{ \psi }^{ 2 } }=\\\cfrac{1}{2}.(32\pi ^{ 4 }-1).\cfrac { c }{ 2\pi a_{ \psi \, c } } \cfrac { 1 }{ 4\pi a_{ \psi \, c }^{ 2 } } *\cfrac{77}{2}$

to

$16.\theta _{ \psi  }{ \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(\theta _{ \psi  })) }tanh(\theta _{ \psi  })\cfrac { 1 }{ 4\pi a_{ \psi  }^{ 2 } } =\cfrac { 1 }{ 2 } .(32\pi ^{ 4 }-1)\cfrac { c }{ 2\pi a_{ \psi \, c } } \cfrac { 1 }{ 4\pi a_{ \psi \, c }^{ 2 } } \cfrac { n }{ 2 }$

${ \dot { x } c }.\theta _{ \psi  }{ ln(cosh(\theta _{ \psi  })) }tanh(\theta _{ \psi  })=n\cfrac { (32\pi ^{ 4 }-1) }{ 128\pi  } \left( \cfrac { a_{ \psi  } }{ a_{ \psi \, c } }  \right) ^{ 3 }$

since,

$\left( \cfrac { a_{ \psi  } }{ a_{ \psi \, c } }  \right) ^{ 3 }=n$

and $\dot{x}=c$

$c^2 .\theta _{ \psi  }{ ln(cosh(\theta _{ \psi  })) }tanh(\theta _{ \psi  })=n^2\cfrac { (32\pi ^{ 4 }-1) }{ 128\pi  }$

does it make sense to multiply $\left(\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }\right)^2$ to the right hand side?

Nooooooooooooooo!

Or, why should $\left(\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }\right)^2$ be divided from the left hand side?

Simple, there is a mistake, the actual force due to Coulomb and Newton was,

$F_e=\cfrac{q_1q_2}{4\pi\varepsilon_o r^2}$  and

$F_g=G\cfrac{m_1m_2}{ r^2}$

but neither $\varepsilon_o$ nor $G$ were necessary in this formulation for $c$, that consider the flow of energy from the basic particles through the surface of the manifested particle.  Both constants should not be involved in the final expression for $c$ explicitly.

However, if we define,

$\varepsilon_{d}=\cfrac{1}{2c^2ln(cosh(\theta_{\psi}))}$

$c^2=\cfrac{1}{\varepsilon_{d}.2ln(cosh(\theta_{\psi}))}$

then

$c^2.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}=\cfrac{1}{\mu_o\varepsilon_{d}}=c^2_1$ --- (*)

But we defined,

$c^2=\cfrac{1}{\mu_o\varepsilon_o}$

which leads to,

$\varepsilon_o=\cfrac{1}{\mu_oc^2}$

$\varepsilon_o=\varepsilon_d.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

the model $\varepsilon_d$ defer from the definition $\varepsilon_o$, by a factor of

$\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

So,

$\cfrac{1}{\mu_o\varepsilon_d}=\cfrac{1}{\mu_o\varepsilon_d\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}}\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}=\cfrac{1}{\mu_o\varepsilon_o}\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

and from (*),

$c^2_1=\cfrac{1}{\mu_o\varepsilon_o}\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

However, the definition,

$c^2=\cfrac{1}{\mu_o\varepsilon_o}$

is a mistake, but if this model is correct then when we measure $\varepsilon_o$, denoted by $\varepsilon_m$, we are actually measuring $\varepsilon_d$ ie,

measuring $\varepsilon_o$ obtain $\varepsilon_m$ but $\varepsilon_m=\varepsilon_d$

so,

$\varepsilon_o\rightarrow \varepsilon_d$

and the expression for $c^2_1$ becomes,

$c^2_{adj}=\cfrac{1}{\mu_o\varepsilon_d}\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

but from (*),

$c^2.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}=\cfrac{1}{\mu_o\varepsilon_{d}}$

$c^2_{adj}=c^2.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

$c_{adj}=c.\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}$

The adjustment made to the calculated $c$ in the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016 was for the fact that the measured $\varepsilon_o$ is $\varepsilon_d$.

Note:  This post is for those who keep substituting

$c^2=\cfrac{1}{\mu_o\varepsilon_o}$

back into (*) and obtain,

$c^2=c^2$

Enjoyed yourselves?

All For One...Flow Rate

Two manifested particles of $n$ constituent basic particles does not interact at the total $c_{adj,\,n}$ energy flow rate level.  This is because the total energy flow is divided among $n$ basic particles.  Thus two manifested particles of $n$ constituent basic particles interact at,

$\cfrac{c_{adj,\,n}}{n}$

which is a constant.  For this reason, the maximum attained velocity is also a constant, in a field of constant flow rate.  Given sufficiently large $n$, this limit is the same constant for all $n$.

One For All.

Big Particle Exists

The following plot shows the change in $\psi$ along a radial line from the center of a particle,

A force $F_r$ on $\psi$ along a radial line, derived from the force density, $F_{\rho}$

$F_{\rho}=-\cfrac{\partial\,\psi}{\partial x}$

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

act opposite to the direction of increasing $\psi$.  If $\psi$ moves to higher value than a positive force is required in the positive radial direction to perform work on $\psi$, for $\psi$ to gain energy.  This force $F_{pinch}$ in the positive direction is countered by $F_r$ above in the negative direction, as $\cfrac{\partial\,\psi}{\partial x}$ is positive.

Work is done against $F_r$.  This work done increases the potential of $\psi$.

$F_r$ is subjected to the inverse square law.  When \cfrac{\partial\,\psi}{\partial x}\) at $a_{\psi}$ and beyond, is a constant or does not increase enough,

$\cfrac{\partial\,\psi}{\partial x}|_{x\gt a_{\psi}}\lt\cfrac{2}{r^3}$

where

$-\left(\cfrac{1}{r^2}\right)^{'}_r=\cfrac{2}{r^3}$

then $F_r$ decreases with distance from the center of the particle.  Any non zero force $F_{pinch}$ that displaces $\Delta \psi$ away from the center of the particle, pulls $\Delta\psi$ away to infinity with greater and greater acceleration.

$\Delta\psi$ is removed from the particle.

In this way, $a_{\psi}$ the probable size of a particle was arbitrarily set at,

$\theta_{\psi}=\pi=\cfrac{G}{\sqrt{2mc^2}}a_{\psi}$

where $tanh(\pi)\approx1$, for $F_r|_{a_{\psi}}$ to increase at greater distance from the particle center, that the particle remain intact with application of small $F_{pinch}$.  $\psi$ has minimum resistance from being pinched apart.

This does not mean that big particles of higher values of $a_{\psi}$ beyond the $\theta_{\psi}=\pi$ limit do not exist.

Such big particles will still interact at a constant (slightly higher) light speed limit.

Speed Limited For A Different Reason

Obviously,

$\left( \cfrac { c_{ adj\,n } }{ n=77 }  \right) ^{ 2 }\ne\left( \cfrac { c_{ adj\,n } }{ n=78 }  \right) ^{ 2 }$

but all particles with $77$ basic particles interact at the same light speed limit, and all particles with 78 basic particles interact at a different but constant light speed limit.  Only for,

$\theta_{\psi}\gt40$

$n\gt159939$

using $n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

is the light speed limit about the same for values of $n$ greater than $159939$.

That's a big particle!

Further more, with the Durian constant in mind, is $n=A_D$?  (The particle size limit, $a_{\psi\,\pi}$ is still an assumption only based on the need for an inverse square relation of the force in the field over distant, and so a decreasing or constant $F_{\rho}$.)  That entangled particles sharing energy in the time dimension is also in close proximity in space, so much so that a manifested particle is made up of $n=A_D$ number of basic particles as far as elementary charges are concerned.

An electron is a basic particle, but free charges are much bigger; maybe a coalesce of $n=A_D$ number of basic particles.

In this way, entanglement is not observed at the macro level of manifested particles, but only at the basic particle level.  Manifested particles are entangled within themselves.  Basic particles entangled as part of a manifested particle, pried from the manifested particle will display entanglement, and are entangled to each other.

Gas molecules cannot be entangled to each other, but as big manifested particles, they are subjected to a common light speed limit irrespective of their size, $n\gt\gt159939$.

Gas molecules and the like are subjected to the light speed limit not because of entanglement but for $\cfrac{c_{adj,\,n}}{n}=constant$ in a field.

May light speed be with you.

Light Speed As Flow Of Energy Per Basic Particle

We have attempted to generalize the expression for $c$,

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

$\theta_{\psi}=\cfrac{a_{\psi\,c} G}{\sqrt{2mc^2}}\sqrt [  3]{n  }$

A more useful expression for $n$ and $\theta_{\psi}$ is,

$n.a^3_{\psi\,c}=a^3_{\psi}$

by consider $n$ particles of radius $a_{\psi\,c}$ forming into one particle of radius $a_{\psi}$.

$\theta_{\psi\,c}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi\,c}=0.7369$

$\theta_{\psi}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi}$

$\cfrac{\theta_{\psi}}{0.7369}=\cfrac{a_{\psi}}{a_{\psi\,c}}$

$n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

we will do it again with,

$c_{ adj }=c.\cfrac {2ln(cosh(3.135009)) }{ \mu _{ o }  }$

that in general,

$c_{ adj }=c.\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o }  }$

$c_{ adj\,n }=n\sqrt { \cfrac { (32\pi ^{ 4 }-1).ln(cosh(\theta _{ \psi  })) }{ 32\pi .\theta _{ \psi  }tanh(\theta _{ \psi  }) }  } .\cfrac { 1 }{ \mu _{ o } }$ --- (*)

where $\mu _{ o }=4\pi \times 10^{ -7 }$.

So,

$\left( \cfrac { c_{ adj\,n } }{ n }  \right) ^{ 2 }=\cfrac { (32\pi ^{ 4 }-1).ln(cosh(\theta _{ \psi  })) }{ 32\pi .\theta _{ \psi  }tanh(\theta _{ \psi  }) } .\cfrac { 1 }{ \mu _{ o }^{ 2 } }$

A plot of ln(cosh(x))/(xtanh(x)),

and $c_{adj\,n}$ can be a constant with changing $n$, provided that,

$\theta_{\psi}$ is large and so $n$ is large.

$c$ was the rate at which energy leaves the boundary of the particle at $a_{\psi}$.  This boundary was made up of $n$ number of basic particles (each of radius $a_{\psi\,c}$), reformed into a sphere of radius $a{\psi}$

After $c$ is adjusted by expression (*), $c\rightarrow c_{adj\,n}$ we find the rate of flow of energy $E_{excess}$ out of its boundary $a_{\psi}$ increase proportionally with $n^2$, irrespective of the final boundary $a_{\psi}$ as denoted by a changing $\theta_{\psi}$.

$c_{adj\,n}=n.D$

$c_{adj\,n}$ is the flow of energy from the particle made up of $n$ basic particles and,

$D=\sqrt { \cfrac { (32\pi ^{ 4 }-1).ln(cosh(\theta _{ \psi  })) }{ 32\pi .\theta _{ \psi  }tanh(\theta _{ \psi  }) }  } .\cfrac { 1 }{ \mu _{ o } }$

is a constant.

$c_{adj\,n}$ per basic particle,

$\cfrac{c_{adj\,n}}{n}=constant$

 Irrespective of increasing $a_{\psi}$ and increasing total excess energy $E_{excess}$.  So, the total excess energy of $n$ basic particles is,

$E_{excess,\,n}=n.E_{excess,\,1}$

and the total energy flow (times per sec) out of the manifested particle boundary at $a_{\psi}$,

$c_{adj\,n}.E_{excess,\,n}=c_{adj\,n}.nE_{excess,\,1}=n^2\cfrac{c_{adj\,n}}{n}.E_{excess,\,1}$

$\because$ $E_{excess\,1}$ is a constant, $\cfrac{c_{adj\,n}}{n}$ is a constant.

$c_{adj\,n}E_{excess\,n}\propto n^2$

the total energy flow out of the manifested particle boundary at $a_{\psi}$ is proportional to its number of basic particles squared, $n^2$.

In this case, the light speed limit is interpreted as the flow of energy in a field.  Energy is imparted upon a particle until it achieve the same speed as this flow of energy.  When the particle is at less than light speed, energy flows into the particle increasing its kinetic energy.  The particle speed increases.  When its speed is at the energy flow rate, energy does not enter into the particle and its speed is constant at light speed.

No matter what the manifested particle size is (in terms of $n$, $a_{\psi}$ or $E_{excess}$), this flow rate (flow per $n$) is the same.  In other words, interactions between pairs of manifested particles of equal $n$ will produce the same light speed limit, irrespective of $n$.

But interactions between manifested particles of unequal $n$ results in different rate of transfer of energy, corresponding to particles of different charge experiencing different force in a field.  These interacting particles will still be at same constant flow rate per basic particle relative to each other when the transfer of energy between them stops.

So, an alternate view of the light speed limit is the constant flow of energy per basic particle provided the number of basic particles in the manifested particle is large.

Eventually this flow rate traces back to the flow of $\psi$ at light speed back to the time dimension at the center of each basic particle.  What is surprising is that after $n$ number of basic particles reform into a sphere of radius $a_{\psi}$ as they coalesce, the resulting energy flow rate out of the new boundary at $a_{\psi}$, is a constant per basic particle.

Light speed is mine!

Note:  Since energy flow per basic particle traces back to the flow of $\psi$ at the center of each basic particle back to the time dimension, the presentation here does not answer the question why is there a speed limit.  That was answered by entanglement as energy sharing in the time dimension, which manifests as a drag force proportional to speed squared in the space dimension.  Here, we see how manifested particles might share the same speed limit as energy flows through them at a constant flow rate per basic particle.

A manifested particle is made up of $n$ basic particles

In the starting equation equating energy emanated and excess kinetic energy, $E_{excess}$ from the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016, in expression (*) repeated,

$16\pi { \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(\pi ))tanh(\pi) }=\cfrac{1}{2}(32\pi ^{ 4 }-1).\cfrac { 1 }{ T }$

flow rate was given by the term $\cfrac{\dot{x}}{a_{\psi}}=\cfrac{c}{a_{\psi}}$ in per sec.  In the discussion above flow rate is $c$ and $c_{adj\,n}$ in times per sec.  To fully define light speed would require the definition of a meter.  That happens when adjustments was made to $c$,

$c\rightarrow c_{adj\,n}$

by the factor

$c_{ adj,\,n=77 }=c.\cfrac {2ln(cosh(3.135009)) }{ \mu _{ o }  }$

specifically when $n=77$.

Have a nice day.

And Light Speed Has A Pulse...

From the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016,

$c=\sqrt { \cfrac { 38.5*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  *\left( \cfrac {3.135009}{ 0.7369 }  \right) ^{ 3 }}$

Since,

$\left( \cfrac { 3.135009 }{ 0.7369 }  \right) ^{ 3 }=77$

$c=\sqrt { \cfrac { \frac { 77^{ 2 } }{ 2 } *(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  }$

We generalize the above expression,

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

where given $n$, $\theta_{\psi}$ is fixed.

$n.a^3_{\psi\,c}=a^3_{\psi}$

as $n$ basic particles reform into one particle, a sphere of radius $a_{\psi}$.

$a_{\psi}=\sqrt [  3]{n  }.a_{\psi\,c}$

and

$\theta_{\psi}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi}=\cfrac{a_{\psi\,c} G}{\sqrt{2mc^2}}\sqrt [  3]{n  }$

$G$ is not the gravitational constant, it is still unknown.

We have,

$\left(\cfrac{c}{n}\right)^2=\cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }$

where $\theta_{\psi}$ is proportional to $\sqrt [  3]{n  }$

A plot of xln(cosh(x))tanh(x) gives,

and a plot of 1/(xln(cosh(x))tanh(x)) and 3/x^2 gives,

Both curves coincide only at $x=1$.  In the expression for $c$, $c$ is not a constant but changes with $n$.  It is an expression for $c$ that is valid only at the point $n=77$ with the assumption that,  

$\theta_{\psi}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi}\approx\pi$

which delimits the particle size to be less than or equal to,

$\theta_{\psi}=\cfrac{G}{\sqrt{2mc^2}}a_{\psi}=\pi$

$n=77$ is not a more fundamental constant, changing $n$ does not change $c$ in reality, but changes the magnitude of the charge.

And light speed has a pulse...

Still No Free Lunch

The truth of the matter is,

$n=77$

and

$c_o=77*\cfrac { 2ln(cosh(3.135009)) }{ 4\pi\times10^{-7}  }$

$c_o=300283934$

where $n$ is the number of fundamental/basic particles that coalesce into one manifested particle.

However, when $n\gt77$, $c \ngtr c_o$ but the force field around the manifested particle is greater, the flux from the charge is greater and the magnitude of the charge associated with the field is greater; all of which are equivalent.  $c_o$ remains the speed limit of the wave.

If,

$KE_{t}=32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$

where $c$ is time speed,

is the peak value of kinetic energy in the time dimension, where did the energy come from?

Such energy must eventually return to the time dimension by attaining light speed in space.

From the perspective of the time dimension,

$KE_{x}=32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$

where $c$ is "space" speed.

In the time dimension, a similar force field surrounds a particle suspended in time.  The energy that flows in the field will also eventually return to the space dimension when it attains light speed in the time dimension.

$KE_{t}$ and $KE_{x}$ are like two equal scoops of energy being passed in the opposite direction between time and space.  A doorway opens to the orthogonal dimension at light speed.  The passage through this doorway is a wave at the speed limit.

Between time and space, energy is conserved.

Just When You Think $c$ Is The Last Constant

The kinetic energy of a particle at speed $c$ in time is given by,

$E=mc^2$

it is doubled the KE in space because as speed in time increases, the time unit increases and speed tends to decrease.  The post "No Poetry for Einstein" dated 6 Apr 2016, shows that energy required to accelerate to light speed in time is twice the normal expression for KE in space.

Since, the time dimension wrap around the space dimension, the kinetic energy of the particle in time is adjusted for by $4\pi^2$ as shown by the previous post "May The Issue Rest" dated 25 Jun 2016,

$E=4\pi^2.mc_{t}^2=8\pi ^{ 2 }.\cfrac { 1 }{ 2 } mc_{t}^{ 2 }$

In the dimension space,

$KE=\cfrac { 1 }{ 2 } mc_{ x }^{ 2 }$

Since, in circular motion,

$v\rightarrow 2\pi v$

time and space do not share the same speed limit,

$c_{t}=2\pi c_{x}=2\pi c$

The kinetic energy in the time dimension is,

$E=8\pi ^{ 2 }.\cfrac { 1 }{ 2 } mc_{t}^{ 2 }=8\pi ^{ 2 }.\cfrac { 1 }{ 2 } m(2\pi c)^{ 2 }$

$E=32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$

Are we double counting here?  No, in the first instance at time speed $v_t$ in time because of circular motion we have an factor of $4\pi^2$. In the second instance, compared to speed in space $v_x$, $v_t$ in time has a higher speed limit also because of circular motion in time, that introduced another factor $4\pi^2$.  Together with the fact that acceleration to $v_t$ in the time dimension requires twice ($\times2$) the amount of energy compared to acceleration to the same speed in the space dimension, there is a combined factor of $32\pi^4$ greater than the energy required in the space dimension.

$\psi$ is oscillating, $E$ is sinusoidal squared.  When $E$ is averaged over one period of oscillation,

$E_{ave\,t}=\cfrac{1}{2}E_t=\cfrac{1}{2}.32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$

in the time dimension and

$E_{ave\,x}=\cfrac{1}{2}E_x=\cfrac{1}{2}.\cfrac { 1 }{ 2 } mc^{ 2 }$

in the space dimension.

Where did the excess energy go as $\psi$ oscillates between time and space?

$E_{excess}=E_{ave\,t}-E_{ave\,x}=\cfrac{1}{2}.(32\pi ^{ 4 }-1).\cfrac { 1 }{ 2 } mc^{ 2 }$

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016,

${ q_{a_{\psi}} } =2\dot { x }F_{\rho}|_{a_{\psi}}$

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

when, $x_z=0$ and

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{\psi})=\pi$

$q_{ a_{ \psi  } }=4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }tanh(\pi)$

In this model,

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

So,

$\cfrac { q_{ a_{ \psi  } } }{ \varepsilon _{ o } } =4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }.{ 2c^{ 2 }ln(cosh(\pi )) }tanh(\pi)=16\pi { \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(\pi )) }tanh(\pi).\cfrac { 1 }{ 2 } mc^{ 2 }$

$\cfrac { q_{ a_{ \psi  } } }{ \varepsilon _{ o } }$ is the power emanating from the particle at the boundary $x=a_{\psi}$

If the energy discrepancy between the time and space dimension is emanated, over one period as $\psi$ oscillates between time and space,

$16\pi { \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(\pi ))tanh(\pi) }=\cfrac{1}{2}(32\pi ^{ 4 }-1).\cfrac { 1 }{ T }$

$T=\cfrac{2\pi a_{ \psi  }}{c}$

as the factor $2\pi$ to $c$ is applied per revolution.

$16\pi { \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(\pi ))tanh(\pi) }=\cfrac{1}{2}(32\pi ^{ 4 }-1).\cfrac { c }{ 2\pi a_{ \psi  } }$ ---(*)

$\dot{x}c=\cfrac { 32\pi ^{ 4 }-1 }{ 64\pi ^{ 2 }ln(cosh(\pi ))tanh(\pi) }$

as $\dot{x}=c$

$c=\sqrt{\cfrac { 32\pi ^{ 4 }-1 }{ 64\pi ^{ 2 }ln(cosh(\pi )) tanh(\pi)}}$

$c=1.42156133$

This value for $c$ is too far off to be considered further.

If instead $77$ particles made up the big particle, from the post "New Discrepancies And Hollow Earth" dated 23 Jun 2016,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}))=\cfrac{1}{4}$

and

$\varepsilon _{ o }=\cfrac { 2 }{ c^{ 2 }}$

for one particle.  After the $77$ particles coalesce into one big particle however,

$\varepsilon _{ o }=\cfrac { 1 }{ 2c^{ 2 }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}))}$

where,

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}=3.135009$

So,

$\pi\rightarrow 3.135009$

$tanh(\pi)\rightarrow tanh(3.135009)$

$\cfrac { q_{ a_{ \psi  } } }{ \varepsilon _{ o } } =16*3.135009*{ \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(3.135009 )) }tanh(3.135009).\cfrac { 1 }{ 2 } mc^{ 2 }$

on the particle side of the expression (*) not the $KE$ discrepancy and so from (*),

$16*3.135009*{ \cfrac { \dot { x }  }{ a_{ \psi  } }  }{ c^{ 2 }ln(cosh(3.135009 )) }tanh(3.135009)\cfrac { 1 }{ 4\pi a_{ \psi }^{ 2 } }=\\\cfrac{1}{2}.(32\pi ^{ 4 }-1).\cfrac { c }{ 2\pi a_{ \psi \, c } } \cfrac { 1 }{ 4\pi a_{ \psi \, c }^{ 2 } } *\cfrac{77}{2}$
--- (**)

Only half the power is radiated outwards.

At the center, $\psi$ is at light speed and is returned to the time dimension.

Since, the surface area of the big particle and its constituent $77$ particles are different, we equate power per unit area, ie intensity instead.  This is done by dividing both sides of the expression (**) by the spherical areas of radii $4\pi a_{ \psi }^2$ and $4\pi a_{ \psi \,c }^2$, respectively.

As $\dot{x}=c$,

$c=\sqrt { \cfrac { 38.5*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) } *\left( \cfrac { a_{ \psi } }{ a_{ \psi \, c } }  \right) ^{ 3 } }$

but,

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}=3.135009$ and $\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}=0.7369$

$\cfrac{a_{\psi}}{a_{\psi\,c}}=\cfrac{3.135009}{0.7369}$

$c=\sqrt { \cfrac { 38.5*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  *\left( \cfrac {3.135009}{ 0.7369 }  \right) ^{ 3 }}$

$c=77.5871223$

In this model, however,

$\varepsilon_o=\cfrac{1}{2c^2ln(cosh(3.135009))}$

$c^2=\cfrac{1}{\varepsilon_o.2ln(cosh(3.135009))}$

By definition,

$c^2_{defined}=\cfrac{1}{\mu_o\varepsilon_{old}}$

So,

$c^2*\cfrac { 2ln(cosh(3.135009)) }{ \mu _{ o } }  =\cfrac{1}{\mu_o\varepsilon_{o}}$

This is with $\varepsilon_o$ derived in this model.  In the conventional definition,

$\varepsilon_{old}=\cfrac{1}{\mu_o c^2}$

And we adjust for this definition of $\varepsilon_{old}$,

$\cfrac{1}{c^2.2ln(cosh(3.135009))}\rightarrow\cfrac{1}{\mu_oc^2}=\cfrac{1}{4\pi\times10^{-7}c^2}$

applying the factor,

$\cfrac{2ln(cosh(3.135009))}{\mu_o}$

to $\varepsilon_o$.  So,

$c_{ adj }^{ 2 }=\cfrac { 1 }{ \mu_o\varepsilon _{ o }.\cfrac { 2ln(cosh(3.135009))}{ \mu _{ o } }  } .\cfrac { 2ln(cosh(3.135009)) }{ \mu _{ o } } .\cfrac { 2ln(cosh(3.135009)) }{ \mu _{ o } }$

the first $\cfrac { 2ln(cosh(3.135009)) }{ \mu _{ o }  }$ factor adjust for the absence of $\mu_o$.  The second $\cfrac {2ln(cosh(3.135009))}{ \mu _{ o }  }$ factor adjust for $\varepsilon_o\rightarrow \varepsilon_{old}$

Therefore,

$c_{ adj }^{ 2 }=c^2.\left(\cfrac { 2ln(cosh(3.135009))}{ \mu _{ o }  }  \right)^2$

$c_{ adj }=c.\cfrac {2ln(cosh(3.135009)) }{ \mu _{ o }  }$

$c_{ adj }=c.\cfrac { 2ln(cosh(3.135009)) }{ 4\pi\times10^{-7}  }$

$c_{ adj }=301763665$

Compare this with the definition for $c=2.99792458e8$, we are off by a factor of $1.0066$.

And I killed light speed.

Note:  If we use $\left\lceil\cfrac{77}{2}\right\rceil=38$

$c=\sqrt { \cfrac { 38*(32\pi ^{ 4 }-1) }{ 64*3.135009*\pi .ln(cosh(3.135009))tanh(3.135009) }  *\left( \cfrac {3.135009}{ 0.7369 }  \right) ^{ 3 }} .\cfrac { 2ln(cosh(3.135009)) }{ 4\pi\times10^{-7}  }$

$c=299797757$

Jackpot!  More than half the emanated energy is returned to the center.

$G$ here is not the gravitational constant.

May The Issue Rest

If in circular motion the momentum of a particle is adjusted by $2\pi$,

$mv\rightarrow 2\pi mv$

then the kinetic energy of the particle,

$\cfrac{1}{2}mv^2\rightarrow \cfrac{1}{2}m(2\pi v)^2$

$KE\rightarrow 4\pi^2KE$

is adjusted by a factor $4\pi^2\approx 39.478$, and

$\cfrac{1}{2}mv^2\rightarrow \cfrac{1}{2}(4\pi^2m)v^2$

$m\rightarrow4\pi^2m$

$m_{or}=4\pi^2m_r$

This could account for the difference between "rest" mass, $m_r$ and the mass of a particle in orbit around another, $m_{or}$.

A Small Point Missed...

When mass measurement are apart from the gravitational constant $G_o$,

$q_e=\cfrac{1}{9.029022e26}=1.107540e-27=m_e$

is also the elementary gravitational mass, $m_e$.

$m_e*77=\cfrac{1}{9.029022e26}*77=8.5281e-26$

$m_e*38=\cfrac{1}{9.029022e26}*38=4.2087e-26$

and with the maximum attractive force, considering shielding,

$\left\lfloor\cfrac{77}{2}\right\rfloor=\left\lfloor38.5\right\rfloor=39$

$m_e*39=\cfrac{1}{9.029022e26}*39=4.3194e-26$

We could be talking about the atomic mass constant or the neutron rest mass.  A proton as an elementary charge particle does not experience a force in a gravitational field and have no gravitational mass.  But it has inertia under an electric field.

Genius And Creative Bullshit

Cont'd from the previous post "The Elementary Electron Charge" dated 25 Jun 2016.

It is possible that we do not have to account for 3D space using the factor $\sqrt{3}$, but must account for the fact that the single charge is a lump of $77$ particles.  In any one direction half of these particles on one side shield the other half of the particles on the other side.  In effect, a test charge only faces half of the total number of particles in a single charge.  The effective number of particles, $n_e$ is,

$n_e=\cfrac{n}{2}=\cfrac{77}{2}$

half the total number of particles in a charge, $n$.

So,

$q_{adj}=\cfrac{1}{9.029022e26}*\cfrac{2ln(cosh(3.135009))}{4\pi\times10^{-7}}*\cfrac{77}{2}=1.658431e−19$

in any one direction.

Good night...

If we really fuss about it,

$\left\lceil\cfrac{77}{2}\right\rceil=\left\lceil38.5\right\rceil=38$

when the experiment to find the elementary charge $e$ is using repulsive force and the situation is presented with the least repulsive force.

$q_{adj}=\cfrac{1}{9.029022e26}*\cfrac{2ln(cosh(3.135009))}{4\pi\times10^{-7}}*\left\lceil\cfrac{77}{2}\right\rceil=1.636893e-19$

and we edge closer to the quoted value of $e=1.602 176 565e-19$

Good morning!

The Elementary Electron Charge

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

$\varepsilon_o=\cfrac{1}{2c^2ln(cosh(\pi))}$

after applying the $\sqrt{3}$ factor to account for 3D space,

$\varepsilon_o=\cfrac{1}{2c^2ln(cosh(\pi))}*\sqrt{3}$

$\varepsilon_o=\cfrac{1}{2*299792458^2*ln(cosh(\pi))}*\sqrt{3}=3.9325e(-18)$

Why was it not necessary to divide by the Durian constant $A_D=\cfrac{4}{3}\pi(2c)^3=9.029022e26$?

Because the elementary charge $q_e$ has absorbed the constant,

$q_e\rightarrow \cfrac{q}{A_D}$

Given one particle, $q=1$

$q_e=\cfrac{1}{9.029022e26}=1.107540e-27$

since by definition,

$\varepsilon_o=\cfrac{1}{\mu_oc^2}=\cfrac{1}{4\pi\times10^{-7}c^2}$

If we adjust $q_e$ by the factor,

$\cfrac{2ln(cosh(\pi))}{4\pi\times10^{-7}*\sqrt{3}}$

to return to the definition of $\varepsilon_o$. ie,

$\cfrac{1}{2c^2ln(cosh(\pi))}*\sqrt{3}\rightarrow\cfrac{1}{\mu_oc^2}=\cfrac{1}{4\pi\times10^{-7}c^2}$

As both denominator and numerator of $\cfrac{q}{\varepsilon_o}$are multiplied by the same factor, we have,

$q_{adj}=q_e*\cfrac{2ln(cosh(\pi))}{4\pi\times10^{-7}*\sqrt{3}}$

$q_{adj}=\cfrac{1}{9.029022e26}*\cfrac{2ln(cosh(\pi))}{4\pi\times10^{-7}*\sqrt{3}}=2.493676e-21$

and we further account for the fact that $77$ particles coalesce up to the limit $tanh(\cfrac{G}{\sqrt{2mc^2}}a_{\psi\,\pi})=1$

$\pi\rightarrow 3.135009$

$ln(cosh(\pi))\rightarrow ln(cosh(3.135009))$

that is,

$\varepsilon_o=\cfrac{1}{2c^2ln(cosh(3.135009))}$

and so,

$q_{adj}=\cfrac{1}{9.029022e26}*\cfrac{2ln(cosh(3.135009))}{4\pi\times10^{-7}*\sqrt{3}}*77=1.914991e-19$

we compare this with the quoted value of $e=1.602 176 565e-19$, we are about $1.195$ times off.

If we do not adjust for the factor $\sqrt{3}$,

$q_{adj}=\cfrac{1}{9.029022e26}*\cfrac{2ln(cosh(3.135009))}{4\pi\times10^{-7}}*77=3.316861e-19$

we are about twice off the quoted value,

$\cfrac{q_{adj}}{2}=\cfrac{3.316861e-19}{2}=1.658431e-19$

What happened?  If the experiment to obtain $e$ is with point charges then we may not have to factor in $\sqrt{3}$ for charged bodies in 3D space, but this results in a calculated elementary charge of twice the quoted value.

When we do factor in $\sqrt{3}$, the discrepancy seems to reflect the ratio between the Durian constant and Avogadro constant,

$\cfrac{A_D}{A_v}=\cfrac{9.029022e26}{6.02214e26}=1.49930$

there is however no reason to adjust the Durian constant here.   Furthermore, swapping Durian for Avogadro increases the calculated charge, $q_e$ that is already too high.

Otherwise, another constant hangs on my wall!

Note:  Dividing by the Durian constant was necessary to account for entanglement.

One Too Many...

Actually,

$tanh(\pi)\ne1$

but,

$tanh(\pi)=0.996272$

and

$\cfrac{1}{tanh(\pi)}=1.033741$

If we apply this correction to $G_o$, the gravitational constant calculated before,

$G_o=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi*9.029022e26 }*\sqrt{3}*{tanh(\pi)}$

$G_o=6.69855e-11$

and the quoted value of $G_o$ is $6.67259e(-11)$.

Without the correction,

$G_o=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi*9.029022e26 }*\sqrt{3}$

$G_o=6.72361e-11$

Which suggests that, the particle is below the limit set by $tanh(\cfrac{G}{\sqrt{2mc^2}}a_{\psi\,\pi})=1$ and it is likely that when many particles coalesce into one,

$\left\lceil n \right\rceil =\left\lceil 77.486 \right\rceil =77$

instead of $\left\lfloor n \right\rfloor =78$.

From the post "Sticky Particles Too...Many" dated 24 Jun 2016,

$n\left(\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}\right)^3=n\left(\cfrac{0.7369}{x}\right)^3=1$

where $\pi$ has been replaced by an unknown $x$ and $n=77$.

$77\left(\cfrac{0.7369}{x}\right)^3=1$

$x=3.135009$

So,

$G_o=\cfrac{299792458^2*ln(cosh(3.135009))}{2\pi*9.029022e26 }*\sqrt{3}$

$G_o=6.70562e-11$

with the correction factor, $tanh(3.135009)$,

$G_o=\cfrac{299792458^2*ln(cosh(3.135009))}{2\pi*9.029022e26 }*\sqrt{3}*tanh(3.135009)$

$G_o=6.68029e-11$

And before we get obsessed with decimal number, lunch time!

Sticky Particles Too...Many

Unfortunately for this model of waves and particles, when

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}$

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}=0.7369$

the particles stick into lumps without end.

The particles in black and red surround a central piece.  The green dotted particles is a second layer on top of the black and red.

The problem is, the particles continue to lump without end.  Only if $\psi$ merge into one and the particles "grows" to

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,\pi}))=ln(cosh(\pi))=2.450311$

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,\pi}=\pi$

the size at which the force around the particle decreases with distance from the center of the particle as Coulomb's inverse square law applies, does this sticky issue resolve itself.

How many particles must merge?

$\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}=\cfrac{0.7369}{\pi}$

$n\left(\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}\right)^3=n\left(\cfrac{0.7369}{\pi}\right)^3=1$

$n=77.486$

$\left\lfloor n \right\rfloor =78$

which would make the resulting big particle slightly over the limit,

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,\pi}=\pi$

This implies, it is possible to split a big particle into a smaller fractions.  The smallest of which is $\cfrac{1}{78}$ of the original big particle.

The magic number to look for is $78$.

Sticky Particles...

From the previous post "New Discrepancies And Hollow Earth" dated 23 Jun 2016, where we now denote,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}$

$ln(cosh(0.7369))=\cfrac{1}{4}$

and

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,\pi}))=ln(cosh(\pi))=2.450311$

what is between $a_{\psi\,c}$ and $a_{\psi\,\pi}$, where the force increases with distance away from the center?

An accumulation of other like particles.  At distances greater than $a_{\psi\,c}$, up to $a_{\psi\,\pi}$, the attached particle experiences a greater attractive force due to the increasing value of $F_{\rho}$.

$a_{\psi\,c}$ is then interpreted as the minimum value for $a_{\psi}$ that holds up a particle.  Values of $a_{\psi}$ less than $a_{\psi\,c}$ collapses the particle.

And we have a lumpy issue...Particles of various sizes that stick together.  Interestingly,

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

at $x=a_{\psi\,c}$,

$F=m\cfrac{c^2}{2}$

This is the attractive force that holds the particles in the lump together.

New Discrepancies And Hollow Earth

From the previous post "$\psi$ Gets Inverted And Fourier Xformed" dated 22 Jun 2016,  it was suggested that the extend of $\psi$ is such that from $x=a_{\psi}$ to the center of the particle, the acceleration due to $F_{\rho}$ achieve light speed at the center.  That is to say,

$\cfrac{1}{2}mc^2=\int_{0}^{a_{\psi}}{F_{\rho}}dx$

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

with $x_z=0$,

$\cfrac{1}{2}mc^2=i2{ mc^{ 2 } }.ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi }))$

this implies,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}))=\cfrac{1}{4}$ --- (*)

If this is so, there is a new discrepancy in the calculations for $\varepsilon_o$ and $G$ that had previously assumed,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}))=ln(cosh(\pi))=2.450311$ --- (**)

Since, the function $ln(cosh(x))$ is monotonously increasing, the assumed value of $a_{\psi}$ given by (**) results in light speed before reaching the center under the action of $F_{\rho}$.  The particle is then hollow at the center.

If (*) is correct, then immediately after $a_{\psi}$, the force in the field around the particle does not obey coulomb's inverse square law but increases until $ln(cosh(\pi))$ or $\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}=\pi$, where (**) holds true.  Furthermore, if (*) is true,

$\varepsilon _{ o }=\cfrac { 2 }{ c^{ 2 }}$

and

$G_o=\cfrac { c^{ 2 } }{ 8\pi  }.\cfrac{3}{4\pi(2c)^3}.\sqrt{3}=6.86e-12$

A lower value for $a_{\psi}$ as given by (*) allows for two particles to interact as waves without merging below the particle limit,

$a_{\psi}=\pi\cfrac { \sqrt { 2{ mc^{ 2 } } }  }{ G }$

had (**) been assumed.  But for the case of $G_o$, the gravitational constant, (**) seems more appropriate.

Have a nice day.

Note: $\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi} =acosh(e^{\cfrac{1}{4}})=0.736904590621$

Eternity At A Black Hole

If this particle is a black hole, the event horizon where an object at light speed orbits the black hole is also where the object makes a splat AFTER being sucked into the center by the belly.  This splat is all over the spherical event horizon surface; the object is being Fourier transformed at the point in the center and is evenly distributed over the entire surface area of the event horizon that defines the black hole.

Perpetually...

Empty On The Inside?

The center of the particle is not a black hole because $\psi=0$ at $x=0$.  Or is it?

How big SHOULD a particle be?  The bound of the particle is where, from the surface of the particle, the total acceleration that a test particle experiences on moving inwards due to $F_{\rho}$, brings it to light speed $v_x=c$ along a radial direction.  In this way, the particle is able to recycles itself.

$\psi$ is on a Möbius strip, drawn at its belly towards the center, accelerating to light speed only to return to the same point with light speed, once again drawn at its belly further along the strip towards the center.

For energy density, $\psi=0$ at $x=0$ but kinetic energy of the test point, $KE_{max}=\cfrac{1}{2}mc^2$ at $x=0$.

Have a nice day.

$\psi$ Gets Inverted And Fourier Xformed

If $\psi$ flows inwards everything collapses, but,

$\psi$ is energy oscillating between two dimensions.  At the center of the particle, $\psi$ departs from space returns to the time dimension ($v_x=c$ and $v_t=0$).  On the surface of the particle where $iv_x=v_{x\bot}=c$ and $iv_t=v_{t\bot}=0$, $\psi$ returns to the space dimension.  This space dimension is orthogonal to that at the center of the particle.  Space at the center and the surface of the particle are connected through $v_{t}=0$, but twisted.  $\psi$ emerge through an orthogonal dimension in space at the surface, as if it has been Fourier transformed at the center of the particle.

$\psi$ recycles itself.  $\psi$ need not return to the same particle.

And so, $\psi$ is not delimited to $a_{\psi}$ but extends to infinity.

The correct graph of $\psi$ should be,

This does not mean $\psi$, energy density is unbounded,  $\psi$ is instead bounded by $v_{x\bot}=c$ where it "drops" into an orthogonal time dimension.

Whether $\psi$ "drops" at the edge from the space dimension and appear at the center in the time dimension or $\psi$ "disappear" at the center from the space dimension and "emerges" at the edge in the time dimension, are equivalent orthogonal view of the same situation.

$\psi$ is oscillating between two dimensions, when it is maximum in one dimension it is minimum in the other dimension.

If $\psi$ is subjected to $F_{\rho}$, $\psi$ is also subjected to the $\psi$ vs $x$ graph above.  As $\psi$ approaches the universe center, $\psi$ reduces to zero.  $\psi\rightarrow 0$ as $x\rightarrow 0$.

Is the size of an electron the same anywhere in the universe?

The Red And Blue Divide

If the universe has a center,

there is a plane, perpendicular to the line joining such a center and Earth;  on the same side as the center there is red shift and one the opposite side of the center, there is blue shift, given local conditions.

Good night.

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?

$c$ And $\pi$

$\psi$ is circular  It is wrapped around a circle of perimeter $2\pi r$, where $r$ is the radius from the particle center, travelling at light speed $c$.  On every cycle of period $T=\frac{1}{f}$, it traveled once along the perimeter and has rotated by $2\pi$ rad.  It is then better to redefine light speed with consideration for the value of $\pi$,

$c=\pi*10^8$

instead of

$c=2.99792458*10^8$

such that $r$ is defined/measured without the uncertainty associated with the value of $\pi$.

$\pi$ at light speed will be quite a splat.

Pressure And Anger Management

As only positive temperature particles $T^{+}$ are captured in the electron orbit, it is the negative temperature charge particle that exert a pressure on the containment wall initially.  When temperature increases by driving away $T^{-}$, pressure drops.  When all $T^{-}$ particles are driven away but one (per constant area) pressure is at the minimum.  Pressure due to one $T^{+}$ and one $T^{-}$ per constant area is the same.  Pressure increases as temperature increases with more positive temperature particles added afterwards.  At this point and higher temperature beyond, pressure on the containment wall is due to $T^{+}$ particles distributed on the surface of the containment wall.

There is no temperature point at which there is zero temperature particle (per constant area) distributed on the containment.  The transition from one negative temperature particle to one positive temperature particle, per constant area, on the containment share the same pressure.  This is assuming that positive particles repel each other equally as negative particles repel each other.

If this scenario is not a dream, pressure has more to do with the containment wall's ability to hold and distribute temperature particles than the gas being contained.  A containment of insulating material that does not accumulate temperature particles irrespective of the temperature of the gas inside and redistribute temperature particles within it uniformly, will not experience increase in pressure on its inside wall (both locally and the general interior) as temperature of the gas increases.

Un-fuzzed when hot.

But, No Gas, No Pressure

If pressure is solely due to temperature particles, why would a vacuum not exert a pressure on the walls of a containment maintained at a constant temperature?

Temperature particles are charged particles.  Temperature particles of the same charge repel each other.  Pressure on the containment wall is due to $T$ particles, attracted by the $T$ particles (of the opposite charge) on the gas, distributed on the inside wall of the containment.

Without a gas inside the containment, the charged particles move to the outside surface of the containment.

There is no pressure inside the containment, but the $T$ particles on the outside, still causes an outward stain on the containment.  The associated stress acts against the pressure crushing the containment inwards.

Reducing temperature on the containment reduces this outward force and allows outside pressure to collapse the containment.

How educational can "Myths Busters" get?

Death Experiment With Moving Blades

With this setup we increase pressure without increasing volume nor temperature,

as the gas molecules gain velocity from the rotating fan.  Measurements of pressure is taken after the fan has stopped completely.

What will happen?  Does the contained gas retains its pressure given constant temperature and volume?

Guess law?

It is likely that all kinetic energy imparted by the fan is lost through collisions with the containment walls in time with temperature held constant.

In that case, we postulate that the total kinetic energy of the gas is made up of,

$KE_{total}=KE+KE_{T}$

where $KE_{T}$ is Thermal Kinetic Energy.

Thermal kinetic energy is the kinetic energy associated with temperature particles $T^{+}$ caught in the orbit of electrons around the nucleus.

$KE$ is the difference of $KE_{T}$ from $KE_{total}$.  The origin of this part of kinetic energy is left opened for the time being.

A plot of $P$ over time after the fan is switched off would indicate the effect of collisions on kinetic energy.
Why would the gas retain any kinetic energy?  Is has always been assumed that a gas left to stand will have only kinetic energy associated with its temperature.  Is this true?

A plot of $P$ vs $v^2$ with the fan at different speed settings might allow the pressure due solely to $KE_{T}$ to be found when the graph y-intercept is extrapolated.

If a solid can exert pressure on the containment then pressure is not due to $KE$ and collisions.

How much of the pressure is due to collisions and how much of the pressure is due to the distribution of temperature particles on the inner surface of the containment?  If temperature particles exist.

My guess is, collisions are not involved in pressure without agitations.  Pressure is due solely to the distribution of temperature particles in the rested state.

Good night...zzz..

Contained System And Entanglement Loss

The signifigance of $A_D$ the Durian constant,

$A_D=\cfrac{4}{3}\pi (2c)^3=9.029022e26$

is, a contained system, where the system has integer multiples of $A_D$ number of particles in it,

$N=nA_D$

$n=1,2,3...$

where $N$ is the total number of particles in the system.

Entanglement results in energy loss from the system.  In a contained system, it can be assumed that entanglement is within the system that no energy is loss to outside the system.

A non-contained system is when the number of particles in the system, $N$ is not an integer multiples of the Durian constant, where part of the entanglement is outside of the system.

There is no certainty that all entanglement is within a contained system, only greater assurance that the particles inside, subjected to the same physical conditions, are entangled within the system.

Since, entanglement loss is not a defined problem, this is the undefined solution to a unappreciated problem.

When Ego Is Permitted To Go...

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

$\varepsilon_o=\cfrac{1}{2c^2ln(cosh(\pi))}=\cfrac{1}{2*299792458^2*ln(cosh(\pi))}=2.2704e(-18)$

Should the $\sqrt{3}$ factor also be applied?

$\varepsilon_o=2.2704e(-18)*\sqrt{3}$

$\varepsilon_o=3.9325e(-18)$

In case which, compared to the quoted value of $\varepsilon_o=8.8542e(-12)$,  we have

$\cfrac{8.8542e(-12)}{3.9325e(-18)}=2.251969e6$

Interestingly it is 2.20462 pound to a kg.

Maybe the pound is natural units and imperial; that $\varepsilon_o=3.9325e(-18)$ is per pound instead of per kg.

$\varepsilon_o$ can be redefined!

Gravitational Constant At Last!

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

$q=m.4\pi a^2_{\psi}= \cfrac { dq }{dr}|_{ r=a_{\psi} }$

This points to surface property, that the behavior of $q$, $\frac { dq }{ dr }$ at the surface boundary $r=a_{\psi}$ determines the field extended by the particle.  But it leads to high values for gravity and the gravitational constant, when the particles are large gravity particles (because there was a mistake).

If we reformulate more clearly,

$M=m.4\pi a^2_{\psi}= \cfrac { dM }{dr}|_{ r=a_{\psi} }$

Because,

$F_{ { G } }\cfrac { 1 }{ 4\pi a^{ 2 }_{ \psi  } } =G\cfrac { M }{ a^{ 2 }_{ \psi  } } =2{ mc^{ 2 } }.ln(cosh(\pi ))$

We note that $GM$ takes the place of $\cfrac{q}{4\pi\varepsilon_o}$.  We have

$G\cfrac { M }{ a^{ 2 }_{ \psi  } } =2{ mc^{ 2 } }.ln(cosh(\pi ))$

$G\cfrac { m.4\pi a^2_{\psi}}{ a^{ 2 }_{ \psi  } } =2{ mc^{ 2 } }.ln(cosh(\pi ))$

$G=\cfrac { c^{ 2 }ln(cosh(\pi )) }{ 2\pi  }$

$G=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi}$

$G=3.504958e16$

To account for entanglement, we divide by the Durian constant.

$G=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi*9.029022e26 }$

$G=3.88188e-11$

which is just a coincident.

Compare this with the quoted value of $G=6.67259e(-11)$, we missed by a factor of $\sqrt{3}$.

$G*\sqrt{3}=6.7236e-11$

What happened?  It could be,

where one dimensional space along $r$ is elevated to three dimensional space.  A particle in one dimension is now a body with extends in two other dimensions.  The body "feels" the force in these additional two dimensions and the result is a factor of $\sqrt{3}$ ( used to be called Boltzmann).

Good night.

And Gravity Is Five Hundred!

Combining both results from the post "Where Did The Pound Come From?" dated 15 Jun 2016,

from

${ q_{ a_{ \psi  } } } =i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }$

and

$\cfrac{1}{4\pi a^2_{\psi}}.i4\pi{ \cfrac { \dot { x }  }{ a_{ \psi  } }Mc^{ 2 } }=G\cfrac{M}{a^2_{\psi}}$

$G={ \cfrac { c^{ 3 }  }{ a_{ \psi  } }  }$

$F= \cfrac { c^{ 3 }  }{ a_{ \psi  } }.\cfrac{M}{x^2}$ --- (1)

From,

${ q_{a_{\psi}} }=2\dot { x }F_{\rho}|_{x}=4\pi GM$

and

$F=\cfrac{c^3}{a^2_{\psi}}.\cfrac{M}{x}$ --- (2)

$G_{l}=\cfrac{c^3}{a^2_{\psi}}$

At $a_{\psi}=6371000$ and $M=5.972e24$, both expression (1) and (2) is the same,

$g=\cfrac{299792458^3}{6371000^3}*5.972e24$

$g=6.222417e29$

If we divide by the Durian constant considering entanglement,

$g=\cfrac{6.222417e29}{9.029022e26}=6.891573e2$

Even if we consider $g$ as a sinusoidal at frequency of $7.489 Hz$ and we take the rms value,

$g_{rms}=\cfrac{6.891573e2}{\sqrt{2}}$

$g_{rms}=487.30$

This is still a large number, given the error in Earth's mass, $M$.

Note:  The unit dimension of the expression for $F$ is a mess because after the integration the $tanh(x)$ function carries a unit and is not dimensionless.

Where Did The Pound Come From?

This post is defunct the post "Into A Pile Of Deep Shit" dated 09 Jun 2016, from which this discussion originates has been corrected.

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016,

$q_{ a_{ \psi  } } =i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }$

If we compare the force due to $\psi$ and Newton's Gravitational force,

$m=M_e$

$\cfrac{1}{4\pi a^2_{\psi}}.i4\pi{ \cfrac { \dot { x }  }{ a_{ \psi  } }M_ec^{ 2 } }=G\cfrac{M_e}{a^2_{\psi}}$ --- (*)

we have,

$G={ \cfrac { i\dot { x }  }{ a_{ \psi  } } c^{ 2 } }$

The Gravitational constant, $G$ depended on the size of the planet.

if $i\dot{x}=c$,

$G={ \cfrac { c^{ 3 }  }{ a_{ \psi  } }  }$

For earth, $a_{\psi}=6371000$,

$G=\cfrac{299792458^3}{6371000}$

$G=4.229164e18$

If we consider entanglement by dividing by the Durian Constant,

$\cfrac{G}{A_D}=\cfrac{4.229164e18}{9.029022e26 }=4.683967e(-9)$

In the calculation for $\varepsilon_o$ we did not divide the result by the Durian Constant because the particle involved is not interacting as wave and is entangled to other particles.  In the case of gravity above where like particles attract, they are interacting as waves and are entangled to each other.

When we correct for the fact that the quoted $G$ is actually $\pi G$ because of the spin of the planets,

$\pi G=\pi*4.683967e(-9)=1.4715116317e-8$

What happened?  Maybe it should have been,

$m.2c^2ln(cosh(\pi))=G\cfrac{m}{a^2_{\psi}}$ --- (*)

at $x=a_{\psi}$.

$G=2a^2_{\psi}c^2ln(cosh(\pi))$

$G$ is dependent on $a^2_{\psi}$ and I am in deeper....

$G=2*6371000^2*299792458^2*ln(cosh(\pi))$

$G=1.787754e31$

If we replace $c^2$ with $\dot{x}^2$ and interpret it to be the spin of the planet,

$\dot{x}=\cfrac{2\pi a_{\psi}}{24*60*60}$ 

$G=2a^4_{\psi}\left(\cfrac{2\pi }{24*60*60}\right)^2ln(cosh(\pi))$

$G=a^4_{\psi}*2.5916926e(-8)$

$G$ is dependent on $a^4_{\psi}$, for earth $a_{\psi}=6371000$,

$G=4.269863e19$

We are wrong; on both counts, expression (*) and $i\dot{x}=v_s$.

And lastly, if we have,

${ q_{a_{\psi}} }=2\dot { x }F_{\rho}|_{x}=4\pi GM$

as

$4\pi x^2.G\cfrac{M}{x^2}=4\pi GM$

where the attractive force on the gravity particle interacting as wave is due only to the flux emanating below it.

$G=\cfrac{\dot { x }F_{\rho}|_{x}}{2\pi M}$

where,

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

and so,

$G=\cfrac{\dot { x }}{2\pi M}i\sqrt { 2{Mc^{ 2 } } } \, G_o.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

with $x_z=0$

$G=\cfrac{i\dot { x }c}{\pi \sqrt{2M}}\, G_o.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

$\cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } a_{\psi}=\pi$

$G=\cfrac{i\dot { x }}{a_{\psi}}c^2.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

and $G$ becomes a running target...

Because at $x=a_{\psi}$,  

$tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (a_{\psi}) \right)=1$

we can approximate,

$tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)\approx\cfrac{x}{a_{\psi}}$

$G\approx\cfrac{i\dot { x }}{a_{\psi}}c^2.\cfrac{x}{a_{\psi}}$

$F=G\cfrac{M}{x^2}=\cfrac{i\dot { x }}{a^2_{\psi}}c^2.\cfrac{M}{x}$

when $i\dot{x}=c$,

$F=\cfrac{c^3}{a^2_{\psi}}.\cfrac{M}{x}$

$G_{l}=\cfrac{c^3}{a^2_{\psi}}$

$G_{l}=\cfrac{299792458^3}{6371000^2}=6.638148e11$

$\cfrac{G_{l}}{A_D}=\cfrac{6.638148e11}{9.029022e26 }=7.352e(-16)$

And deep blue is brown...

Note:  What if $i\dot{x}$ is interpreted as the spin of the planet?  Instead of  $i\dot{x}=c$, we have $i\dot{x}=v_s$ and we replace $c$ with $t\dot{x}$. So,

$F=\cfrac{c^3}{a^2_{\psi}}.\cfrac{M}{x}$

$F=\cfrac{v_s^3}{a^2_{\psi}}.\cfrac{M}{x}$

$G_{l}=\cfrac{v_s^3}{a^2_{\psi}}$

$G_{l}=\cfrac{465^3}{6371000^2}$

$G_{l}=2.4771006228e(-6)$

$\psi$ at light speed wrap around a sphere manifest as gravitational mass that spins independently of $\psi$.  $i{x}\ne v_s$ the speed of the wave is not the spin of the mass.

  

Guessing My Way Through Physics

From the previous post "Equating To A Indefinite Integral" dated 10 Jun 2016,

$\dot { x } ^{ 3 }=-\cfrac { 1 }{ 4\pi  } \cfrac { q }{ m } a_{ \psi  }$

where the negative sign signifies an attractive force.

$\dot { x } =-\sqrt [ 3 ]{ \cfrac { 1 }{ 4\pi  } \cfrac { q }{ m } a_{ \psi  } }$

$\cfrac { \partial \, T }{ \partial \, x } =m\dot { x } \cfrac { \partial \, \dot { x }  }{ \partial \, x }$

$\cfrac { \partial \, T }{ \partial \, x } =-\sqrt [ 3 ]{ \cfrac { q }{ 4\pi  } a_{ \psi  } } .\cfrac { \partial \, \dot { x }  }{ \partial \, x }.m^{ 2/3 }$

Only a decreasing $\dot{x}$ along $x$ is,

$\cfrac { \partial \, T }{ \partial \, x }\gt 0$

that would make,

$q_{ a_{ \psi  } } =3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\lt0$

where

$\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k$

If $\dot{x}$ increases along $x$ or decreases less along $x$,

 $\cfrac { \partial \, T }{ \partial \, x }\lt 0$

or

 $\cfrac { \partial \, T }{ \partial \, x }_1\lt \cfrac { \partial \, T }{ \partial \, x }_2$

such that,

$q_{ a_{ \psi  } } =3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\gt0$

the particle switches sign.

This seems to mark the boundary when two particles with interacting $\psi$ turn from attractive (interacting as waves) to repulsive (interacting) as particles.  It is not.  The particles' $\psi$s are already interacting and they are interacting as waves.  After the sign reversal, the particles remains together.

$3k=2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}$

and is neutral (neither attractive nor repulsive) to each other.  We can simplify to obtain,

$3 \cfrac { \partial V\,  }{ \partial \, x }|_{a_{\psi}} =-\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}$

this is valid only at the neutral point as $\cfrac { \partial V\,  }{ \partial \, x }$ changes with $\cfrac { \partial \, T }{ \partial \, x }$ along $x$.  $k$ given $\psi$ is a constant.

It is not likely that $\dot{x}$ increases.  In the case when $\dot{x}$ decreases less along $x$, $\psi$ spreads itself away from the center of the particle as it reduces speed along $x$.  In free space, a particle is negative, in a medium where $\psi$ is encouraged to spread, the particle is positive.  eg. an electron in free space and an electron on glass ($Si$), respectively.

Axiomatic but a good guess.

If electrons can turn positive, do we still need protons?

Here's A Small Pile, $P$...

This post is wrong, the math in the post "Into A Pile Of Deep Shit" dated 09 Jun 2016 is wrong.

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016

$\cfrac{\Delta E}{\Delta t}=P=\dot{x}F_{\rho}$

as $\Delta t\rightarrow 0$

but from the post "Into A Pile Of Deep Shit" dated 09 Jun 2016,

$\cfrac { dq }{ dx } =2\dot { x }\cfrac { \partial \, \psi  }{ \partial \, x }$

and

${ q_{a_{\psi}} }=2\dot { x }F_{\rho}|_{a_{\psi}}$

Why is,

${ q_{a_{\psi}} } =2P|_{a_{\psi}}$

??

Why is the total flux through the spherical surface at $x=a_{\psi}$ twice the power $P$ when we cross $x=a_{\psi}$ with velocity $\dot{x}$?

$P$ is in the space dimension only, ${ q_{a_{\psi}} }$ of the particle oscillates in one space dimension and one time dimension.  When time and space dimensions are equivalent, then to account for the time dimension, twice the power is needed.  Another way to see this is that, energy input to the system in one dimension is shared equally with the other dimension, eventually only half the energy remains in the dimension receiving energy input.

Otherwise the partial differential maths in the post "Into A Pile Of Deep Shit" dated 09 Jun 2016, is wrong.

It is likely that the maths is wrong.

$\psi$ Is Attractive

From the post "Not Exponential, But Hyperbolic And Positive Gravity!" dated 21 Nov 2014,

$F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

If $G=D.e^{ i\pi /4 }$ is real then

$F=e^{i\pi/2}\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

$F=i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

If however we allow the augment to the function $tanh(x)$ to be complex, and interpret $e^{ i\pi /4 }$ as rotation about the origin anti-clockwise by $\pi/4$ from $e^{ i\pi /4 }$ , and then a further rotation by $3\pi/4$ from $e^{ i3\pi /4 }$, $F$ is rotated by $\pi$ about the origin in total.

$F=-\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

$F$ is then along the negative $x$ direction, towards the center of the particle which is consistent with the expression

$4\pi { \cfrac { \dot { x } ^3 }{ a_{ \psi  } } m }=-q$

that a negative charge is normally associated with particle.

But, this interpretation of $e^{ i\pi /4 }$ in the augment of $tanh(x)$ is odd because $tanh(x)$ does not accept complex augments.  In this way, the modulus of a complex augment serves as the real input to the function and the argument of the complex augment rotates the x axis of the graph of the function about the origin anti-clockwise.

If this is true,

$F=-\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

without $i$.  Force density $F$, points towards the center of the particle, along the negative $x$ direction.

And we have a new class of real functions with complex augments.

Good night.

Note: This suggests $e^{ia}$ has immunity, that

$F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

$F=e^{ i3\pi /4 }.e^{ i\pi /4 } D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

$F=e^{ i\pi } D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

$F=- D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

 $e^{ia}$ passes through a function as a constant coefficient passes through an integral.

Warning, over generalization!!