Wanted: Attractive Force. Electrostatic Need Not Apply

There is no other way.  The electron is attracted to the nucleus, it is accelerated to light speed and begins curl in the direction perpendicular to its forward motion.  It transcribe a helical path.  It is the drag force that provides the centripetal force that rotates the electron in the direction perpendicular to its direction of travel.  The magnetic repulsive force in the vicinity of the nucleus and thermal gravity pushes the electron way, and the new centripetal force is given by,

$\cfrac { m_{ e }c^{ 2 } }{ r_{ e } } =Ac^{ 2 }+G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } } -\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o }(r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \tau_{ o }r^{ 2 }_{ e } }$

where $Z$ is the atomic number, $m_a$ is the atomic mass, and $r_p$ is the proton spin radius.

The equation does not have the electrostatic term,

$+\cfrac { Zq^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } }$

as the charges are in constant motion.

The gravitational term,

$G_Z=G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } }$

plays a much more important role.

$Ac^{ 2 }r^{ 2 }_{ e }-m_{ e }c^{ 2 }r_{ e }+\left(Gm_{ a }m_{ e }-\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r^{ 2 }_e }{ (r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \varepsilon _{ o } } \right)=0$ --- (*)

we see that the equation for $r_e$ is still essentially a quadratic equation.  For heavy nucleus elements, it is expected that the gravitation term is high enough to give a positive y-intercept to equation (*).  A positive y-intercept will allow the parabola to intercept the x-axis twice and give two positive solutions to $r_e$.  The lower solution corresponds to the an orbit in the valence band and the higher solution is in the conduction band.  An electron can make the transition from the valence band to the conduction band, both valid $r_e$ solutions to the force equation (*).  This behavior is expected of heavier elements but not all elements.

Increasing temperature can bring the y-intercept down, and eventually a possible zero valued solution to $r_e$.  This temperature would correspond to $T_c$, the temperature for the formation of plasma.

Decreasing temperature or increasing the nucleus mass, ie the $G_Z$ term, moves the curve upwards and moves the two roots closer until a double root is created.  At the double root, the kink point, the root and the minima all coincide.  The element is very conductive, very shinny as the electron readily crosses kink point and emit a packet of energy.

But it is commonly believed that $G_Z$ is small and does not play a part in this situation.  Unless there is another attractive force in the equation, I believe common opinion is wrong, gravity plays a much more important role and cannot be ignored.  Electrostatic attraction between the charges cannot account for a positive y-intercept even if it is admitted because it is already smaller than the magnetic repulsive force by a factor of $\pi$.

Let's look for another attractive force that hold the electron to its orbit of $r_e$.