Nature of stability in a conservative system

Question

In CM, a conservative system can be described by a potential energy function, $V(x)$

The states of the system which will be in equilibrium will be found at the extrema of this potential where:

• $\frac{dV(x)}{dx}|_{(x=x_0 )}=-F(x_0)=0$ or

• Where the classical force vanishes as suggested by Newton’s Law

The stability of the equilibrium point is governed by sign of the second derivative:

• $\frac{d^2V(x)}{dx^2}|_{(x=x_0 )}>0$ Implying stable, bounded, and periodic motion
• $\frac{d^2V(x)}{dx^2}|_{(x=x_0 )}<0$ unstable (in fact exponentially increasing) and bounded motion

I was wondering where they derived the bottom two conditions form? Is it just a convention or is there a place they derived it form

Answer 1

It is a rather fundamental mathematic concept. The roots of first derivative are points of local extrema, where forces are vanishes. But these extrema could be local minimum (a stable equilibriium point), or local maximum (a unstable equlibrium point). The criterion of checking local maximum or minimum or saddle point (in higher dimension space) is to check the second derivatives.

Let use Taylor expansion to illustrate the potential near an equilibrium point $x_0$:

$$ V(x) = V(x_0) + 0 (x - x_0) + a (x - x_0)^2 + O(\Delta x)^3; \tag{1}\\ \text{ where the coefficient } a = \frac{d^2V}{dx^2}|_{x=x_0}. $$

The force near $x_0$ is $$ F(x) = -\frac{dV}{dx} = - 2 a (x - x_0). \tag{2} $$

From Eq.(2), the force resembles a restoring force of simple harmonic oscillation if $a > 0$, and $x_0$ is the equilibrium point.

When $a < 0$, the force is positive pushing away from the equilibrium points.

From mathematic point of view,, the $\frac{d^2V}{dx^2}|_{x=x_0} > 0$, the curve has a positive curvature. The potetial curve has a local minimum at $x_0$, and potential bends upwards in both side. On the other hand, $\frac{d^2V}{dx^2}|_{x=x_0} < 0$, the potential has a local maximum at $x_0$. When $\frac{d^2V}{dx^2}|_{x=x_0} = 0$, it is a turning point.