Double-Bell, Balls

Does this orbits give raise to the double bells electronic cloud energy density?

$\psi_T=\psi_1+\psi_2$

$\psi_T=\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x-a_e)^{ 3 } }$

where $2a_e$ is the orbital separation.  $\psi_1$ and $\psi_2$ are due to the two orbiting electrons individually.

We formulate

$\psi_T \ge \psi_{min}$

$\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x-a_e)^{ 3 } }\ge\psi_{min}$

$\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x-a_e)^{ 3 } }\ge \cfrac{mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x_{min})^{ 3 } }$

and so,

$\cfrac { 1 }{ (x+a_{ e })^{ 3 } } +\cfrac { 1 }{ (x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } }$

Consider  $x \gt a_e$,

$\cfrac { 1 }{ (x+a_{ e })^{ 3 } } +\cfrac { 1 }{ (x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } }$

$\cfrac { (x-a_{ e })^{ 3 }+(x+a_{ e })^{ 3 } }{ (x+a_{ e })^{ 3 }(x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } }$

Since $(x-a_{ e })^{ 3 }\gt 0$,

$\cfrac { (x+a_{ e })^{ 3 }(x-a_{ e })^{ 3 } }{ (x-a_{ e })^{ 3 }+(x+a_{ e })^{ 3 } }\le x^3_{ min }$

This implies that given a $\psi_{min}$ and a corresponding $x_{min}$, the expression above for $x$ must be smaller or equal to $x_{min}$.

The constrain on $x$ due to $x_{min}$ is plotted in red.  Given this solution for $x$, with reference to the diagram below, we will find $r$ such that $|\psi|\gt|\psi_{min}|$.

$\psi=\cfrac{1}{2\mu_o}B^2=-\cfrac{1}{2}\varepsilon_oE^2=\cfrac { mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ x^{ 3 } }$

We shall consider the B field,

$B=\sqrt{2\mu_o\psi}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi  }  \right) ^{ 1/2 }\cfrac { 1 }{ x^{ 3/2 } }$

In this diagram, the electrons in parallel orbits are travelling out of the paper at $\psi_{-}$ and $\psi_{+}$.  Their respective $B$ fields are circular with centers at $\psi_{-}$ and $\psi_{+}$.  Without lost of generality we will set the angle on the side of $B_{+}$ at 90^o and plot for values of $r\ge a_e$, given $x$,

$B_{//}=B_{+\,//}+B_{-\,//}$

$B_{//}=B_{+}sin(\theta)+B_{-}sin(\phi)$

$B_{//}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi  }  \right) ^{ 1/2 }\left( \cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{r^{5/2}}+\cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{[(2a_e+x)^2+(r^2-x^2)]^{5/4}}\right)$

and

$B_{\bot}=B_{+\,\bot}+B_{-\,\bot}$

$B_{\bot}=B_{+}cos(\theta)+B_{-}cos(\phi)$

$B_{\bot}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi  }  \right) ^{ 1/2 }\left( \cfrac { x }{ r^{ 5/2 } } +\cfrac {2a_e+x  }{ [(2a_{ e }+x)^{ 2 }+(r^{ 2 }-x^{ 2 })]^{ 5/4 } }  \right)$

Therefore,

$B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi  } \left[\left( \cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{r^{5/2}}+\cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{[(2a_e+x)^2+(r^2-x^2)]^{5/4}}\right)^2+\\\left(  \cfrac { x }{ r^{ 5/2 } } +\cfrac {2a_e+x  }{ [(2a_{ e }+x)^{ 2 }+(r^{ 2 }-x^{ 2 })]^{ 5/4 } }\right)^2 \right]$

If  $2a_e\lt\lt x$,

$B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi  }  \left[ \left( \cfrac { \sqrt { r^{ 2 }-x^{ 2 } }  }{ r^{ 5/2 } } +\cfrac { \sqrt { r^{ 2 }-x^{ 2 } }  }{ r^{ 5/2} }  \right) ^{ 2 }+\left( \cfrac { x }{ r^{ 5/2} } +\cfrac { x }{ r^{ 5/2 } }  \right) ^{ 2 } \right]$

$B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi  } \cfrac { 4 }{ r^{3 } }$

So, we have

$\cfrac{1}{2\mu_o}B^2= \cfrac{1}{2\mu_o}\cfrac { \mu_omc^{ 2 } }{ 6\pi  } \cfrac { 4 }{ r^{ 3 } }\ge|\psi_{min}|$

The LHS is just the expression for $\psi$ of two electron charges at a distance $r$.

$\cfrac { 4 }{ r^{ 3 } }\ge\cfrac { 1 }{ x_{ min }^{ 3 } }$

$r\le\sqrt[3]{4}\,x_{ min }$

Since, $x_{min}$ is from the expression,

$\psi_{min}=\cfrac{mc^{ 2 } }{ 12\pi  } \cfrac { 1 }{ (x_{min})^{ 3 } }$

around a single particle, so provided that $x_{min}\gt\gt a_e$, we have approximately,

The shape of the graph is not actually what we expected.  The approximation to $\psi$ at $r$ may need improvement.