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How do we know there is only 1 stable position?

(A perfect circle would have many, but you could consider that a degenerate configuration.)



There isn't necessarily just one. A rectangle-ish form could have 4.


Yes, rectangle would also be degenerate. Basically any configuration where adding an epsilon weight somewhere generates a non-epsilon shift so to speak.


To me, "degenerate" means that it has a symmetry so that the different states are somehow identical (e.g. equal energy in a quantum system). If it's rectangle-ish (say, an imperfect hand-drawn rectangle) then it has 4 distinct equilibrium positions. Not sure if we're thinking of the same with "degenerate".


A plus shape "+" has four stable equilibriums which all stable optimums, resistant to weight and self correcting.


You mean unstable?




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