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?