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