Find the Lagrangian, the generalized momentum, and the Hamiltonian for a free particle (no forces at all) confined to move along the x axis. (Use x as your generalized coordinate.) Find and solve Hamilton's equations.

Solution

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Step 1

1 of 2

Since there is no potential Lagrangian is equal to kinetic energy.

\begin{align} \mathcal{L} &= T = \frac{1}{2}m\dot{x}^2 \end{align}

Conjugate momentum is given by:

\begin{align} p &= \frac{\partial \mathcal{L}}{\partial \dot{x}} = m \dot{x} \Rightarrow \dot{x} = \frac{p}{m} \end{align}

Using (2) we can find the Hamiltonian of the system:

\begin{align} \mathcal{H} &= p\dot{x} - \mathcal{L} = p \dot{x} - \frac{1}{2}m\dot{x}^2 = p\frac{p}{m} - \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{m} - \frac{1}{2}\frac{p^2}{m} \\ \mathcal{H} &= \frac{p^2}{2m} \end{align}

Using Hamilton's equations on (3) we find:

\begin{align} \dot{x} &= \frac{\partial \mathcal{H}}{\partial p} = \frac{p}{m} \\ \dot{p} &= -\frac{\partial \mathcal{H}}{\partial x} = 0 \end{align}

From (5) we see that $p = p_0 = const.$ . Inserting this into (4) we have:

\begin{align} x(t) &= \frac{p_0}{m}t +x_0 \\ p(t) &= p_0 \end{align}

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