To think about potential wells, we often imagine little beads rolling on the plot of the potential well (as a wire) under gravity without friction. How precise is this analogy?

Let’s consider a system with one degree of freedom being described by $$\ddot x = f(x), \; x \in \mathbb{R}$$ with kinetic energy $$T = \frac{1}{2} \dot x^2$$ and potential energy $$U(x) = - \int_{x_0}^x f(\xi) d \xi$$. Alternatively we could write

$\ddot x = -\frac{dU}{dx}$

## Potential wells in reality

Suppose we have some function $$U(x)$$. Imagine a bead of mass 1 moving along a wire $$\{(x, U(x)) \,\vert\, x \in \mathbb{R}\}$$ under gravity with zero fiction. The lagrangian of the system is hence $$\frac{1}{2} (\dot x^2 + \dot y^2) - gy$$. Imposing the constraint $$y = U(x)$$, the lagrangian is now

\begin{align*} L&=\frac{1}{2} (\dot x^2 + \dot y^2) - gy\\ &= \frac{1}{2} \bigg(\dot x^2 + \dot x^2 \bigg(\frac{dU}{dx}\bigg)^2\bigg) - g U(x) \\ &= \frac{1}{2} \dot x^2 \bigg(1 + \bigg(\frac{dU}{dx}\bigg)^2\bigg)- gU(x) \end{align*}

The Euler-Lagrange equation would now be

\begin{align*} 0&= \frac{d}{dt}\bigg(\frac{\partial L}{\partial {\dot x}} \bigg) - \frac{\partial L}{\partial x} \\ &= \frac{d}{dt}\bigg[\dot x \bigg(1 + \bigg(\frac{dU}{dx}\bigg)^2\bigg)\bigg] \\ &\phantom{= } -\bigg[ \frac{1}{2}\dot x^2\frac{d}{dx}\bigg(\bigg(\frac{dU}{dx}\bigg)^2\bigg) - g \frac{dU}{dx}\bigg] \\ &= \ddot x \bigg(1 + \bigg(\frac{dU}{dx}\bigg)^2\bigg) + \dot x \bigg(\dot x \frac{d}{dx}\bigg(\bigg(\frac{dU}{dx}\bigg)^2\bigg)\bigg) \\ &\phantom{= }-\bigg[ \frac{1}{2}\dot x^2\frac{d}{dx}\bigg(\bigg(\frac{dU}{dx}\bigg)^2\bigg)- g \frac{dU}{dx}\bigg] \\ &= \ddot x \bigg(1 + \bigg(\frac{dU}{dx}\bigg)^2\bigg) + \dot x \bigg(\dot x \times 2 \frac{dU}{dx} \frac{d^2U}{dx^2}\bigg) \\ &\phantom{= }-\bigg[ \frac{1}{2}\dot x^2\times 2 \frac{dU}{dx} \frac{d^2U}{dx^2}- g \frac{dU}{dx}\bigg] \\ &= \ddot x \bigg(1 + \bigg(\frac{dU}{dx}\bigg)^2\bigg) + \frac{dU}{dx} \bigg(\dot x^2\frac{d^2U}{dx^2} + g\bigg) \end{align*}

Hence

$\ddot x = \frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2} \bigg(g + \dot x^2 \frac{d^2 U}{ dx^2} \bigg)$

This isn’t particularly illuminating. Let’s understand the above term by term.

First Term: $$\frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2} g$$

Let $$\theta$$ be the normal angle of $$U(x)$$ to the $$x$$-axis, then we could write $$-\frac{dU}{dx}$$ as $$\tan \theta$$. Now we would have $$\frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2} = \frac{- \tan \theta}{1 + \tan^2 \theta} = -\sin \theta \cos \theta$$. As such we have

$\frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2} g = \cos \theta \times (-g \sin \theta)$

This could be thought of as the horizontal component of the normal component of gravitational acceleration.

Second Term: $$\frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2}\dot x^2 \frac{d^2 U}{ dx^2}$$

We would like to connect the second derivative of $$U(x)$$ with its curvature as follows.

\begin{align*} &\dot x^2 \frac{-\frac{dU}{dx}}{1+(\frac{dU}{dx})^2} \frac{d^2 U}{ dx^2} \\ &= \frac{\dot x^2 + \dot y^2}{1 + (\frac{dU}{dx})^2} \times \bigg(-\frac{dU}{dx} \sqrt{1+\bigg(\frac{dU}{dx}\bigg)^2}\bigg) \times \frac{\frac{d^2 U}{dx^2}}{(1+(\frac{dU}{dx})^2)^{3/2}} \\ &= v^2 \times - \cos \theta \times \kappa \\ &= \cos \theta \times (- v^2/r) \end{align*}

where $$v$$ is the magnitude of velocity, $$\kappa$$ is the signed curvature and $$r$$ is the signed radius of curvature. $$\frac{v^2}{r}$$ could be understood as the centripetal acceleration, $$a_c$$, necessary to keep the bead on the wire. $$a_c$$ is positive if the curve opens upwards and negative if the curve opens downwards.

Combining the two terms, we have

$\ddot x = \cos \theta ( - g \sin \theta - a_c)$

$$-g \sin \theta - a_c$$ could be thought of as the magnitude of the net acceleration on the bead. The multiplier $$\cos \theta$$ nets us the horizontal component.

Comparing the two potentials

Setting $$g = 1$$, if $$a_c \ll 1$$ (e.g. initial velocity is $$0$$ or the curvature of $$U(x)$$ is close to $$0$$) then $$\ddot x \approx - \cos \theta \sin \theta$$. This is strikingly similar to the case of the abstract potential well for low $$\theta$$ where $$\ddot x = - \frac{dU}{dx} = \tan \theta$$. We could see this by plotting $$-\tan \theta$$ in red and $$-\sin \theta \cos \theta$$ in blue.

For the abstract potential wall, the acceleration it could capture is unbounded. For the potential well “in reality”, the acceleration is bounded by one half of gravitational acceleration.

## Pendulum potential

If $$U(x) = -\cos x$$, for an abstract potential well we could plot the phase diagram as follows: (x-axis is $$x$$ while y-axis is $$\dot x$$)

\begin{align*} \dot x &= y \\ \dot y &= -\sin x \end{align*}

For a potential well “in reality” we have

\begin{align*} \dot x &= y \\ \dot y &=\frac{-\sin x}{1 + \sin^2 x} \big( 1 + y^2 \cos x \big) \end{align*}

There are now local minimums of $$\dot x$$ around $$x=\pi/2$$, that’s because the curvature of $$-\cos x$$ changes sign and the vertical velocity starts converting back into horizontal velocity due to the constraint force. The peaks of $$\dot x$$ at $$x=\pi$$ are also lowered as there’s less gravitational potential energy. Below is a plot of $$U(x) = - \cos x$$ for reference.

If $$U(x) = x^2$$, for an abstract potential well we could plot the phase diagram as follows: (x-axis is $$x$$ while y-axis is $$\dot x$$)
\begin{align*} \dot x &= y \\ \dot y &= -2x \end{align*}
\begin{align*} \dot x &= y \\ \dot y &= \frac{-2x}{1+4x^2}( 1 + 2 y^2 ) \end{align*}