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).