\(r_e=\cfrac{m_e}{2A_of(lnT)}\left\{{1\pm \sqrt{1-4\cfrac{A_of(lnT)}{m_e}{r_{ec}}}}\right\}\)
and
This is a graphical explanation for bandgap, as a result of drag of free space near terminal velocity, \(2c^2\).
Band gap energy is given by,
\(\Delta v^2_{BG}={\cfrac{d(r_e)}{d\,t}|_{r_{e1}}}^2-{\cfrac{d(r_e)}{d\,t}|_{r_{e2}}}^2\)
\(E_{BG}=\cfrac{1}{2}m_e\Delta v^2_{BG}\) this energy is released, moving from \(r_{e1}\) to \(r_{e2}\).
Band gap can be calculated theoretically from the difference of the square of the gradient,
\(\cfrac{d(r_e)}{dT}=\cfrac{d(r_e)}{d\,t}\cfrac{d\,t}{dx}\cfrac{dx}{dT}\)
If we assume a conservative field where the energy changes are independent of the path taken, the time taken, and where only the starting and ending points mattered, we let,
\(\cfrac{d x}{d\,t}=1\) --- (*)
when we move from \(r_{e1}\) to \(r_{e2}\), this path is also constrained by Hamilton’s principle of stationary action. (This principle and the differential equation (*) above fully describe the path from \(r_{e1}\) to \(r_{e2}\).) And so we have,
\(\cfrac{d(r_e)}{dT}=\cfrac{d(r_e)}{dt}\cfrac{dx}{dT}\)
\(\cfrac{d(r_e)}{dt}={\cfrac{d(r_e)}{dT}\over\cfrac{dx}{dT}}=\cfrac{d(r_e)}{dT}\cfrac{dT}{dx}\)
where \(\cfrac{dT}{dx}\) is the postulated temperature gradient around the nucleus. Obviously \(x\) is in the direction of increasing \(r_e\) and the unit dimension of the last expression is consistent given that it is multiplied by 1 ms-1 from \(\cfrac{d x}{d\,t}=1\).
Both \(\cfrac{d(r_e)}{dT}\) and \(\cfrac{dT}{dx}\) are negative which makes \(\cfrac{d(r_e)}{dt}\) positve. If \(T\) is not defined as energy (Joules) then a multiplicative constant is needed for a consistent unit dimension on both sides of the equation.
A derivation using the Lagrangian is presented in the posts "Energy Band Gap....Gap...Gap.Gap" and "Pag.Pag...Pag....Pag Dnab Ygrene".
