Chapter 13, Problem 13.1P

To determine

The Hamilton’s equation for a free particle.

Answer to Problem 13.1P

The Hamilton’s equation for a free particle is $x=v_{0}t+x_0$ which represents a straight line.

Explanation of Solution

Write the expression for kinetic energy for a free particle moving along $x$ direction.

  $T=\frac{1}{2}m{\dot{x}}^2$        (I)

Here, $m$ is the mass..

Write the expression for potential energy for a free particle moving along $x$ direction.

  $U=0$        (II)

Write the lagrangean by substituting expression (I) and (II),

  $\begin{array}{l}L=T-U\\ L=\frac{1}{2}m{\dot{x}}^2\end{array}$        (III)

Write the generalized momentum by substituting expression (III),

  $\begin{array}{c}p=\frac{\partial L}{\partial \dot{x}}\\ p=\frac{1}{2}m\left(2\dot{x}\right)\\ =m\dot{x}\\ \dot{x}=\frac{p}{m}\end{array}$        (IV)

Substitute expression (IV) in expression (I), re-write the expression for kinetic energy

  $T=\frac{1}{2}m{\left(\frac{p}{m}\right)}^2$        (V)

Write the Hamiltonian, by substituting expression (V) and (II),

  $H=\frac{p^2}{2m}+0$        (VI)

Generalize the momentum

  $\begin{array}{c}\dot{p}=-\frac{\partial H}{\partial x}\\ \dot{x}=\frac{2p}{2m}\\ \dot{x}=\frac{p}{m}\\ \frac{dp}{dt}=0\\ p=p_0\end{array}$        (VII)

Conclusion:

Substitute expression (VII) in $\dot{x}=\frac{p}{m}$

  $\begin{array}{l}\frac{dx}{dt}=\frac{p_0}{m}\\ dx=\frac{p_0}{m}dt\end{array}$        (IX)

Integrating the expression (IX) on both sides

  $\begin{array}{l}{\displaystyle \int dx}=\frac{p_0}{m}{\displaystyle \int dt}\\ x=\frac{p_0}{m}t+C\end{array}$        (X)

.

Consider the condition in with ar $t=0,x=x_0$ in initial displacement.

  $\begin{array}{l}x=\frac{p_0}{m}t+x_0\\ x=v_{0}t+x_0\end{array}$        (XI)

This equation (XI) represents a straight line.