\(F_{\rho}=i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (x-x_z) \right)\)
and so the energy density, is given by,
\(-\cfrac{\partial\,\psi}{\partial\,x}=F_{\rho}\)
\(\psi=-\int{i\sqrt { 2{ mc^{ 2 } } }\,Gtanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (x-x_z) \right)}dx\)
\(\psi=-i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }(x-x_z)))+c\)
when \(x=0\), \(\psi=0\)
\(c=i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }x_z))\)
A plot of tanh(x-1) and -log(cos(x-1)) is given below,
\(x=x_a\) where \(\psi=0\) delimits the physical extend of \(\psi\) as we consider only positive energy, \(\psi\) cannot be negative. The plot is illustrative only, the value of \(x_a\) is scale by the factor,
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }\)
\(F_{\rho}\) is valid up to when \(\psi=0\) at \(x=x_a\). Since,
\(\psi(x)=\psi(-x)\) and \(\psi\) is symmetrical about the \(y=0\) axis.
The shape of \(\psi\) depends on \(x_z\). In this case, it is a double bell centered at \(x\) with \(x_z=1\)
The 3D plot is generated using scilab
o=0.0;
for n=1:(1/0.02)
t = 0:0.01:(2.0*%pi);
x=o*ones(1,size(t,'c'));
r=-(log(cosh(o-0.5))-(log(cosh(-0.5))));
y = r*sin(t);
z = r*cos(t);
param3d(x,y,z,45,60,"X@Y@Z",[2,3]);
o=o+0.02;
end
o=0.0;
for n=1:(1/0.02)
t = 0:0.01:(2.0*%pi);
x=-1*o*ones(1,size(t,'c'));
r=-(log(cosh(o-0.5))-(log(cosh(-0.5))));
y = r*sin(t);
z = r*cos(t);
param3d(x,y,z,45,60,"X@Y@Z",[2,3]);
o=o+0.02;
end
Two collapsed views are also plotted,
collapsed along the y-axis.
collapsed along the x-axis, the plot is a perfect circle. A plot corresponding to a smaller value of \(x_z=0.5\) is also given,
Except for smaller average radius, it is essentially the same shape as the previous plots. These are plots of energy density along the \(\pm\) x-axis of a single particle. These are not electron clouds in 3D. If we plot the contour of maximum \(\psi\) around the particle, it is just a sphere.
There is a problem, \(F_{\rho}\) beyond \(x_a\) is zero. In the old formulation of gravity or electrostatic force, \(F\rightarrow0\) as \(x\rightarrow\infty\), the associated field stretches to infinity, and does not abruptly decrease to zero. In the case of gravity, not only does the field density drop to zero abruptly, it is positive, pushing mass outwards just before disappearing.
The fact is \(F_{\rho}\) is not \(F\). At \(x=x_a\),
\(F_{\rho}(x_a)=F_{xa}\)
For \(x\gt x_a\),
\(F=F_{xa}\cfrac{4}{3}\pi x^3_a\cfrac{1}{4\pi x^2_a}\cfrac{1}{4\pi x^2}=\cfrac{1}{12\pi x^2}F_{xa}x_a\)
where \(F_{xa}\) is a constant. This formula is valid for \(x\ge x_a\) and since both \(F_{xa}\) and \(x_a\) are constants, it obeys Coulomb's Law.
So, the force field around a particle can be divided in two, a near field \(x\lt x_z\) and a far field \(x\ge x_a\).
The fact is \(F_{\rho}\) is not \(F\). At \(x=x_a\),
\(F_{\rho}(x_a)=F_{xa}\)
For \(x\gt x_a\),
\(F=F_{xa}\cfrac{4}{3}\pi x^3_a\cfrac{1}{4\pi x^2_a}\cfrac{1}{4\pi x^2}=\cfrac{1}{12\pi x^2}F_{xa}x_a\)
where \(F_{xa}\) is a constant. This formula is valid for \(x\ge x_a\) and since both \(F_{xa}\) and \(x_a\) are constants, it obeys Coulomb's Law.
So, the force field around a particle can be divided in two, a near field \(x\lt x_z\) and a far field \(x\ge x_a\).
