61.1: Examples

Example $\PageIndex{1}$ Simple Harmonic Oscillator

As an example, we may again solve the simple harmonic oscillator problem, this time using Hamiltonian mechanics.

Solution

We first write down the kinetic energy $K$, expressed in terms of momentum $p$ :

$$K=\frac{p^{2}}{2 m}$$

As before, the potential energy of a simple harmonic oscillator is

$$U=\frac{1}{2} k x^{2}$$

The Hamiltonian in this case is then

$$\begin{align} H(x, p) & =K+U \\ & =\frac{p^{2}}{2 m}+\frac{1}{2} k x^{2} \end{align}$$

Substituting this expression for $H$ into the first of Hamilton's equations, we find

$$\begin{align} \frac{d x}{d t} & =\frac{\partial H}{\partial p} \\ & =\frac{\partial}{\partial p}\left(\frac{p^{2}}{2 m}+\frac{1}{2} k x^{2}\right) \\ & =\frac{p}{m} \end{align}$$

Substituting for $H$ into the second of Hamilton's equations, we get

$$\begin{align} \frac{d p}{d t} & =-\frac{\partial H}{\partial x} \\ & =-\frac{\partial}{\partial x}\left(\frac{p^{2}}{2 m}+\frac{1}{2} k x^{2}\right) \\ & =-k x \end{align}$$

Equations $\PageIndex{5}$ and $\PageIndex{8}$ are two coupled first-order ordinary differential equations, which may be solved simultaneously to find $x(t)$ and $p(t)$. Note that for this example, Eq. $\PageIndex{5}$ is equivalent to $p=m v$, and Eq. $\PageIndex{8}$ is just Hooke's Law, $F=-k x$.

Example $\PageIndex{1}$ Plane Pendulum

As with Lagrangian mechanics, more general coordinates (and their corresponding momenta) may be used in place of $x$ and $p$. For example, in finding the motion of the simple plane pendulum, we may replace the position $x$ with angle $\theta$ from the vertical, and the linear momentum $p$ with the angular momentum $\mathscr{L}$.

Solution

To solve the plane pendulum problem using Hamiltonian mechanics, we first write down the kinetic energy $K$, expressed in terms of angular momentum $\mathscr{L}$ :

$$K=\frac{\mathscr{L}^{2}}{2 I}=\frac{\mathscr{L}^{2}}{2 m \ell^{2}}$$

where $I=m \ell^{2}$ is the moment of inertia of the pendulum. As before, the gravitational potential energy of a plane pendulum is

$$U=m g \ell(1-\cos \theta)$$

The Hamiltonian in this case is then

$$\begin{align} H(\theta, \mathscr{L}) & =K+U \\ & =\frac{\mathscr{L}^{2}}{2 m \ell^{2}}+m g \ell(1-\cos \theta) \end{align}$$

Substituting this expression for $H$ into the first of Hamilton's equations, we find

$$\begin{align} \frac{d \theta}{d t} & =\frac{\partial H}{\partial \mathscr{L}} \\ & =\frac{\partial}{\partial \mathscr{L}}\left[\frac{\mathscr{L}^{2}}{2 m \ell^{2}}+m g \ell(1-\cos \theta)\right] \\ & =\frac{\mathscr{L}}{m \ell^{2}} \end{align}$$

Substituting for $H$ into the second of Hamilton's equations, we get

$$\begin{align} \frac{d \mathscr{L}}{d t} & =-\frac{\partial H}{\partial \theta} \\ & =-\frac{\partial}{\partial \theta}\left[\frac{\mathscr{L}^{2}}{2 m \ell^{2}}+m g \ell(1-\cos \theta)\right] \\ & =-m g \ell \sin \theta \end{align}$$

Equations $\PageIndex{15}$ and $\PageIndex{18}$ are two coupled first-order ordinary differential equations, which may be solved simultaneously to find $\theta(t)$ and $\mathscr{L}(t)$. Note that for this example, Eq. $\PageIndex{15}$ is equivalent to $\mathscr{L}=I \omega$, and Eq. $\PageIndex{18}$ is the torque $\tau=-m g \ell \sin \theta$.