Earnshaw's theorem says "no stable equilibrium for any $\frac{1}{r}$ potential field in charge-free space". Now I am confused in some aspects, and I would like some helping hands.
General physics Textbooks draw a picture of potential field versus distance about electrons around nuclei, there is a place for local minimum (and therefore, it is nice to use a "spring approximation" around that point). That sounds kind of contradiction to me, would anyone be kind enough explaining that to me?
- It seems exactly same reason below listed - where nuclei much massive than electrons, so to a good approximation - a stable circular motion.
Does this theorem work on varying-potential-field as well?
- It seems not. considering planetary motion's reduced mass effective potential energy $V(r)_{eff}=U(r)+\frac{L^2}{2\mu r^2}$, which reduce a two-bodies motion to one-body-motion, then there is a $r$ such that the one-body-motion along r direction at a stable equilibrium. But if back to an inertial frame, then it also seem we do NO have such stable equilibrium (if $m_1$~$m_2$ )?
Do we need quantum theory to explain why we have a solid object (in a reasonable time scale ~10^3 years) due to no stable equilibrium for static inverse square force?
- Even if above effective potential energy works, it doesn't save that, since to my knowledge, there is no way to reduce N-bodies-motion to one-body-motion, as well as atoms in a solid object just don't do "planetary motion" to each other!
- Not sure if it is a physics question: Does "stable circular motion" count as "stable equilibrium" in r direction?
