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Example: simple pendulum

The Hamiltonian for the simple pendulum is:

\begin{align} H &= T + V \tag{1} \\ &= \dfrac{ml^2 \dot{\theta}^2}{2} + mgl\sin{\theta}. \tag{2} \end{align}

Problem is, however, that this is written in terms of $q_i$ and $\dot{q}_i$ not $q_i$ and $p_i$ . First we need to determine what our canonical momentum is. Equations (??) yields:

\begin{align} p_{\theta} &= \dfrac{\partial L}{\partial \dot{\theta}} \notag \\ p_{\theta} &= ml^2\dot{\theta}. \tag{3} \end{align}

Rewriting Equation (3) with $\dot{\theta}$ as its subject:

\begin{align} \dot{\theta} &= \dfrac{p_{\theta}}{ml^2}. \tag{4} \end{align}

Substituting Equation (7) into Equation (2) yields:

\begin{aligned} H &= \dfrac{ml^2}{2} \left( \dfrac{p_{\theta}}{ml^2} \right)^2 + mgl\sin{\theta} \\ &= \dfrac{ml^2}{2} \dfrac{p_{\theta}^2}{m^2l^4} + mgl\sin{\theta}. \end{aligned}

Cancelling out common factors in the numerator and denominator of the kinetic energy term yields:

\begin{align} H &= \dfrac{p_{\theta}^2}{2ml^2} + mgl\sin{\theta}. \tag{5} \end{align}

Hence the equations of motion are:

\begin{align} \dot{p}_{\theta} &= -\dfrac{\partial H}{\partial \theta} \notag \\ &= - mgl\cos{\theta} \tag{6} \\ \dot{\theta} &= \dfrac{\partial H}{\partial p_{\theta}} \notag \\ &= \dfrac{p_{\theta}}{ml^2}. \tag{7} \end{align}

As we can see, Hamiltonian mechanics yields a series of first-order ordinary differential equations (ODEs) instead of second-order ODEs. Which is important because Lagrangian mechanics and Newtonian mechanics both yield second-order ODE systems. That being said, if you rewrite Equation (7) to make $p_{\theta}$ its subject and substitute it into Equation (6) and rearrange it to make $\ddot{\theta}$ the subject, you will obtain Equation (??).

© Brenton Horne. I do not claim to be infallible, so feel free to open issues or pull requests on GitHub if you notice flaws or things that can be improved.
I favour Oxford spelling, so if you notice I generally use different spelling to Australian spelling, that is probably why.
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