Equilibrium — Question 9

Question 9

A local minimum need not have nonzero curvature

A particle moves along $x$ with potential $U(x)=cx^4$, where $c>0$. At $x=0$ both $U^{\prime}$ and $U^{\prime\prime}$ vanish.

1. Classify the equilibrium without relying only on the second derivative test.

2. Compare the restoring force at displacements $d$ and $2d$, with $d>0$.

Question 9 - Solution

A local minimum need not have nonzero curvature

(a) Derivation and calculation

The force is $$F(x)=-U'(x)=-4cx^3.$$ For every $x\ne0$, $$U(x)-U(0)=cx^4>0,\qquad xF(x)=-4cx^4<0.$$ Thus the origin is a strict energy minimum with a restoring force: $$\boxed{x=0\text{ is stable}}.$$

(b) Result and interpretation

At the two positive displacements, $$F(d)=-4cd^3,\qquad F(2d)=-4c(2d)^3=-32cd^3.$$ $$\boxed{|F(2d)|/|F(d)|=8}.$$ The vanishing second derivative makes the quadratic test inconclusive; the leading potential and force terms are quartic and cubic, respectively.

Physical check

The restoring force is nonzero for every nonzero displacement, unlike a flat neutral potential.