\(E=mc^2\)
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,
\(\pi\rightarrow 3.135009\)
\(tanh(\pi)\rightarrow tanh(3.135009)\)
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.
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.
