What? Both,
\(\cfrac{d(r_e)}{d\,t}|_{r_{e1}}=0\)
and
\(\cfrac{d(r_e)}{d\,t}|_{r_{e2}}=0\)
at \(r_{e1}\) and \(r_{e2}\).
The truth of the matter is, \(r_e\) is not fixed. The electron is performing SHM along the radial line joining the nucleus and the electron.
Not only is \(\cfrac{d(r_e)}{d\,t}|_{r_{e}}\ne0\), it is varying within a small band of values, performing SHM.
This is totally analogous to an orbiting moon. A force similar to that derived in the post "The Force is Strong with You... Too Strong", acts against a change in \(r_e\) and a SHM situation occurs. This force is,
\(F_{shm}=-\cfrac{d\,PE_{re}}{d\,r_{e}}\)
where \(PE_{re}\) is the potential energy of the electron. The negative sign was not included in the post previously because the direction of the force was discussed separately. In general, given a local potential minima,
a displacement \(\Delta x\) from the minimum requires a gain in potential energy. This implies there is work done against a force in the opposite direction of \(\Delta x\). This force per unit inertia (in this case, the electron charge. \(q\)) is given by the negative of the potential gradient.
This is equivalent to \(E=-\cfrac{d\,V}{dx}\), where \(E\) is the electric field and \(V\) the electric potential.
Moreover, an initial velocity \(\cfrac{d\,r_e}{d\,t}\) sets the system into SHM about a mean orbital radius \(r_e\). This force, \(F_{shm}\), must oppose the change in \(r_{e}\) taking place, otherwise the electron will not be in stable orbit in the first place.
Compare this force with that of the simplest oscillator,
\(-\cfrac{d\,PE_{re}}{d\,r_{e}}=-\cfrac{k}{m}r_e\)
\(\cfrac{k}{m}=\cfrac{d}{d\,r_e}(\cfrac{k}{m}r_e)=\cfrac{d}{d\,r_e}(\cfrac{d\,PE_{re}}{d\,r_{e}})=\cfrac{d^2 PE_{re}}{d\,r^2_{e}}\)
where the system resonance frequency is given by,
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{d^2 PE_{re}}{d\,r^2_{e}}}\)
This approximation for resonance is the same as taking the Taylor expansion of \(F_{shm}\) and ignoring all terms higher than \(r_e\), which is valid when the displacement from the minimum potential point is small.
\(f_{kink}=\cfrac{1}{2\pi}\sqrt{\cfrac{d^2 PE_{kink}}{d\,r^2_{e}}}\)
The question is what is the nature of the dissipated energy when the electron jumps over the kink? When this SHM is centered about the kink, huge number of such energy packets will be emitted. And given wide swings in the temperature profile around the nucleus, the resonating material will also cool rapidly. Please note this resonance is specific to a local, \(r_e\).
Have a nice day.