\(B=\left\{ \cfrac { \mu _{ o }qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 2 }_{ e } }{ a^{ 2 }_{ e } } \right\} \\ \left\{ { \left( \cfrac { 3\sqrt { 2 } }{ 4 } +\cfrac { \sqrt { 3 } a_{ e } }{ r_{ e } } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 3 } r_{ e } }{ a_{ e } } }+\left( -\cfrac { \sqrt { 2 } }{ 2 } -\sqrt { 2 } \cfrac { a_{ e } }{ r_{ e } } +\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 2 } r_{ e } }{ a_{ e } } } } \right\} \)
of post "Too Hot! Keep Your Distance." date 16 May 2016 which graph the \(B\) field, is shown below,
\(B\) does not decreases monotonously. For increasing values of \(r_e\), \(B\) dip below zero (the direction of spin is reversed of that from the right hand screw rule) and is negative.
When the other \(B\) field due to the other electron is drawn in and we zoom in to the resultant sum from the plot,
((3*(2)^(1/2)/4+(3)^(1/2)*a/x-a^2/x^2)*e^(-(3^(1/2)*x/a))+(-(2^(1/2)/2)-(2^(1/2))*a/x+a^2/x^2)*e^(-(2^(1/2)*x/a)))+((3*(2)^(1/2)/4+(3)^(1/2)*a/(pi-x)-a^2/(pi-x)^2)*e^(-(3^(1/2)*(pi-x)/a))+(-(2^(1/2)/2)-(2^(1/2))*a/(pi-x)+a^2/(pi-x)^2)*e^(-(2^(1/2)*(pi-x)/a)))
When the other \(B\) field due to the other electron is drawn in and we zoom in to the resultant sum from the plot,
((3*(2)^(1/2)/4+(3)^(1/2)*a/x-a^2/x^2)*e^(-(3^(1/2)*x/a))+(-(2^(1/2)/2)-(2^(1/2))*a/x+a^2/x^2)*e^(-(2^(1/2)*x/a)))+((3*(2)^(1/2)/4+(3)^(1/2)*a/(pi-x)-a^2/(pi-x)^2)*e^(-(3^(1/2)*(pi-x)/a))+(-(2^(1/2)/2)-(2^(1/2))*a/(pi-x)+a^2/(pi-x)^2)*e^(-(2^(1/2)*(pi-x)/a)))
We see that the \(T^{+}\) particle rest in a non-zero minimum resultant \(B\) field, at \(\phi=pi/2\) between the two particles for high values of \(a=1\) down to \(a=0.7\). Below \(a=0.7\) the minimum lifted and is now a maximum.
Below is a plot of (((3*(2)^(1/2)/4+(3)^(1/2)*a/x-a^2/x^2)*e^(-(3^(1/2)*x/a))+(-(2^(1/2)/2)-(2^(1/2))*a/x+a^2/x^2)*e^(-(2^(1/2)*x/a))))^2 which graph the energy curve \(U_B\) with varying \(r_e\).
\(\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2 } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } . \right\} ^{ 2 }\cfrac { r^{ 4 }_{ e } }{ a^{ 4 }_{ e } } \\ \left\{ { \left( \cfrac { 3\sqrt { 2 } }{ 4 } +\cfrac { \sqrt { 3 } a_{ e } }{ r_{ e } } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 3 } r_{ e } }{ a_{ e } } }+\left( -\cfrac { \sqrt { 2 } }{ 2 } -\sqrt { 2 } \cfrac { a_{ e } }{ r_{ e } } +\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 2 } r_{ e } }{ a_{ e } } } } \right\} ^{ 2 }\)
\(U_B\) near zero, due to one electron, first increases then decreases monotonously. A peak values occurs before the descends to zero again.
The resultant \(U_B\) between two orbiting electron is (illustratively), from the plot
(((3*(2)^(1/2)/4+(3)^(1/2)*a/x-a^2/x^2)*e^(-(3^(1/2)*x/a))+(-(2^(1/2)/2)-(2^(1/2))*a/x+a^2/x^2)*e^(-(2^(1/2)*x/a)))+((3*(2)^(1/2)/4+(3)^(1/2)*a/(pi-x)-a^2/(pi-x)^2)*e^(-(3^(1/2)*(pi-x)/a))+(-(2^(1/2)/2)-(2^(1/2))*a/(pi-x)+a^2/(pi-x)^2)*e^(-(2^(1/2)*(pi-x)/a))))^2
where \(0.1\le a\le1\) in steps of \(0.1\).
We have a minimum at the middle of the two electrons when \(a\) is small, \(a\le\approx 0.6\). We see that given a displace from this minimum, a force,
\(F=-\cfrac{\partial\,U_B}{\partial\,r_e}\)
\(F=-\cfrac{\partial\,U_B}{\partial\,r_e}\)
always pushes the \(T^{+}\) particle back to the minimum.
\(a\le\approx 0.6\) set up an oscillatory system, where the \(T^{+}\) particle can swing closer to one electron then return to approach the other electron, in Simple Harmonic Motion.
For large values of \(a\), \(a\ge\approx 0.6\), \(U_B\) due to two electrons presents a maximum, the system is not oscillatory, because a displacement from the maximum creates a force that pushes \(T^{+}\) further away from the maximum.
\(a\approx0.6\) marks the transition of elements able to absorb and retain heat (\(T^{+}\) particles) and elements that do not absorb heat. Elements that do not have \(T^{+}\) particles oscillating along the electron orbits do not have an emission spectrum. Emission spectrum will be discussed next...
This is provided that the expression for \(B\) and \(U_B\) is correct.