Critical Matter/Anti-Matter Fusion Temperature, $T_c$

What if  $r_e$ has two possible positive values?  From the post "Electrons In Orbit Again",

$r_e=\cfrac { m_{ e }\pm \sqrt { m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) }  }{ 2A }$

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"

This is a small value,  to achieve a double root,  $T$  has also to be small (a few Kelvins).  At which point, since the electrons are at the conduction band the material will exhibit high conductivity.

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.