\(r_e=\cfrac { m_{ e }\pm \sqrt { m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) } }{ 2A } \)
two positive roots implies,
\(m_{ e }\gt\sqrt { m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) } \)
\(m^2_{ e }\gt m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) \)
But for non complex solutions,
\(m^2_e\gt\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } )\gt 0\)
\(\cfrac { q^{ 2 } }{ \varepsilon _{ o } } \gt\cfrac { T_{ e }T_{ n } }{ \tau _{ o } }\)
This condition was also seen when \(r_e\) is a double root. As the y-intercept increases from zero, \(r_e\) has two positive values, then a double root, followed by two increasing positive roots. As \(r_e\) is symmetrical about the minimum point, both roots will have the same absolute gradient.
\(\cfrac{d\,r_e}{d\,x}|_{r1}=-\cfrac{d\,r_e}{d\,x}|_{r2}\)
Also, at these roots all forces sum to zero,
\(\sum F=0\), \(\cfrac{d\,PE_{re}}{d\,r_e}=0\)
So, \(\cfrac{d\,PE_{re}}{d\,x}=\cfrac{d\,PE_{re}}{d\,r_e}\cfrac{d\,r_e}{d\,x}=0\) for both values of \(r_e\).
Notice that the kink point expounded in previous posts still exist due to the square root in the expression for \(r_e\). The kink point is given by,
\(r_e=\cfrac{m_e}{2A}\) at the double root.
The following plot shows the situation as the y-intercept increases with decreasing \(T\),
\(y_{intercept}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } }\)
This is the reason why electrons can jump from the valence band to the conduction band across the kink point. The two valid values for \(r_e\) straddle the kink point, one in the valance band and one conduction band.
The plot suggests that there is a limit to which decreasing temperature can increase conductivity; right at the double root where \(r_e\) has only one solution. Beyond this point, \(r_e\) has no valid solution. Does this mean that the electron start falling off? Yes, a case of Bose-Einstein condensate. Luckily, \(T\) is restrained by \(T\gt0\). The y-intercept cannot increase beyond,
\(y_{intercept}\le \cfrac{q^2}{4\pi\varepsilon_o}\)
At the double root, the y=intercept is,
\(v^{ 2 }A=\cfrac { 1 }{ 4\pi r_{ e }^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } )\quad \)
\(v^{ 2 }Ar^{ 2 }_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } \) from the post "Electrons In Orbit Again"
Furthermore, increasing \(T\) shifts the y-intercept down and leaves a single positive solution above the kink point. Does this mean that the material is more conductive? If decreasing \(r_e\) widens a conduit for conduction electrons that results in a decrease in transverse velocity, then conductivity of the material decreases and vice versa.
Noteworthy is that there is a temperature point, \(T_c\) where one of the solutions to \(r_e\) is zero. This occurs when the y-intercept is zero,
\(\cfrac { q^{ 2 } }{ \varepsilon _{ o } } =\cfrac { T_{ e }T_{ n } }{ \tau _{ o } }\) and \(r_e=\cfrac{m_e}{A}\)
Where \(r_e\) is exactly twice the distance to the kink point. At this temperature point the material may also display high conductivity. It is expected that conductivity increases suddenly against a decreasing trend as temperature rises.
This value of temperature also suggests that the electrons can be easily pushed into the nucleus (at \(r_e=0\)) and create matter/anti-matter interactions. A critical matter/antimatter fusion temperature, \(T_c\). Image that at this temperature, electrons are pushed into the nucleus with photons or magnetic pulses to trigger plasma formation.
