6.9: A Simple Example

For a particle moving in a potential in one dimension, $\begin{equation} L(q, \dot{q})=\dfrac{1}{2} m \dot{q}^{2}-V(q) \end{equation}$.

Hence

$$\begin{equation} p=\dfrac{\partial L}{\partial \dot{q}}=m \dot{q}, \quad \dot{q}=\dfrac{p}{m} \end{equation}$$

Therefore

$$\begin{equation} \begin{array}{c} H=p \dot{q}-L=p \dot{q}-\dfrac{1}{2} m \dot{q}^{2}+V(q) \\ =\dfrac{p^{2}}{2 m}+V(q) \end{array} \end{equation}$$

(Of course, this is just the total energy, as we expect.)

The Hamiltonian equations of motion are

$$\begin{equation} \begin{array}{l} \dot{q}=\dfrac{\partial H}{\partial p}=\dfrac{p}{m} \\ \dot{p}=-\dfrac{\partial H}{\partial q}=-V^{\prime}(q) \end{array} \end{equation}$$

So, as we’ve said, the second order Lagrangian equation of motion is replaced by two first order Hamiltonian equations. Of course, they amount to the same thing (as they must!): differentiating the first equation and substituting in the second gives immediately $\begin{equation} -V^{\prime}(q)=m \ddot{q}, \text { that is, } F=m a \end{equation}$, the original Newtonian equation (which we derived earlier from the Lagrange equations).