\(c_{ adj }=c.\cfrac {2ln(cosh(\theta_{\psi})) }{ \mu _{ o } }\)??
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 (*),
\(c^2_1=\cfrac{1}{\mu_o\varepsilon_o}\cfrac{2ln(cosh(\theta_{\psi}))}{\mu_o}\)
\(c^2=\cfrac{1}{\mu_o\varepsilon_o}\)
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?
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?
