I Kleppner Classical mechanics: Question about stability (p.217)

[ThreeCharacteristics]

I have a question understanding the reasoning in the book.

The book says in one dimension F=-dU/dr(p.185). From this, the system is stable at distance a when U'(a)=0 and U''(a)>0 where U is differentiated with respect to r.(p.217)

My question arises from the instance of a pendulum where a massless rigid rod of length l and a mass m are considered(p.217).

The book says the system is stable if the pendulum hangs downward and unstable if it hangs upward, both vertically.

I also agree the result, but I can't understand the reasoning in the book since the derivatives of potential energy in the reasoning are with respect to the angle theta, not r.

I have tried to solve it with the chain rule, indeed, we have dU/dr=dU/d(theta) × d(theta)/dr and d²U/dr²=dU²/d(theta)² × (d(theta)/dr)² + dU/d(theta)×d²(theta)/dr². But stuck due to the trouble with identifying d(theta)/dr.

To try again, at this time, I made a guess that generalises the stability of the system to higher dimension. The system is also stable at the point of minimum potential energy, by investigating small displacement along every axis. Is this OK?

 

[vanhees71]

The math in both cases is the same. It doesn't matter whether your dynamical configuration variable is a Cartesian vector component or an angle.

The the example of the mathematical pendulum. The Lagrangian is
$$L=T-V=\frac{m}{2} R^2 \dot{\theta}^2 + m g R \cos \theta,$$
i.e. the potential is
$$V=-m g R \cos \theta.$$
The equation of motion is easily derived from the Euler-Lagrange equations,
$$p_{\theta}=\frac{\partial L}{\partial \dot{\theta}} = m R^2 \dot{\theta} \; \Rightarrow \; \dot{p}_{\theta}=mR^2 \ddot{\theta}=\frac{\partial L}{\partial \theta}=-V'(\theta)=-mgR \sin \theta.$$
Now we can look for stationary solutions, $\theta=\text{const}$. Obviously for such solutions you must have $V'(\theta)=m g R \sin \theta =0$. That's fulfilled for all $\theta=n \pi$ with $n \in \mathbb{Z}$. Of course only the two cases $\theta=0$ and $\theta=\pi$ are really different.

Now you may ask which of this stationary solutions are stable under small deviations. To that end one linearizees the equation of motion. So let's write $\theta=\theta_0 + \epsilon$, where $\theta_0$ is a stationary solution and $\epsilon$ a small deviation from it. If $\epsilon$ stays small, it's allowed to linearize the equation of motion, i.e., writing
$$m R^2 \ddot{\theta}=mR^2 \ddot{\epsilon} = -V'(\theta_0+\epsilon)=-V'(\theta_0) - \epsilon V''(\theta_0) + \mathcal{O}(\epsilon^2) =- \epsilon V''(\theta_0) + \mathcal{O}(\epsilon^2).$$
Neglecting the higher-order terms you get
$$\ddot{\epsilon}=-\frac{V''(\theta_0)}{m R^2} \epsilon.$$
If now $V''(\theta_0)>0$, i.e., if the potential has a minimum at $\theta=\theta_0$, you get an equation of motion of a harmonic oscillator,
$$\ddot{\epsilon}=-\omega^2 \epsilon \; \Rightarrow \; \epsilon(t)=\hat{\epsilon} \cos(\omega t -\varphi),$$
which means that the deviation from the equilibrium limit stays small for all times, provided the amplitude $\hat{\epsilon}$ is small. Here $\omega^2=V''(\theta_0)/(m R^2)>0$ and thus $\omega \in \mathbb{R}$.

If on the other hand $V''(\theta_0)<0$, i.e., if the potential has a maximum at $\theta=\theta_0$, the equation of motion reads
$$\ddot{\epsilon}=+\Omega^2 \epsilon,$$
where $\Omega^2=-V''(\theta_0)/(m R^2)>0$. Now the general solution of our linearized equation reads
$$\epsilon=A \exp(\Omega t)+B \exp(-\Omega t).$$
This means that the stationary solution is not stable of all small perturbations since if $A \neq 0$ the perturbation doesn't stay small but grows exponentially, and the linear equation of motion is not valid at later times.

All this generalizes to the multidimensional case.

 

[wrobel]

Theorem(little bit informally)

If in a Hamiltonian system with a smooth Hamiltonian $H(p,x)=T+V$ the function $V$ has a local isolated minimum at a point $x_0$ then $x_0$ is a stable equilibrium.

If $H$ is an analytic function then the inverse is also true.