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