SICM Exercise 1.04
Solution to exercise 1.4 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Lagrangian actions
🛈 Note
For a free particle an appropriate Lagrangian is
\[ L(t, x, v) = \frac{1}{2} m v^2 \]Suppose that \(x\) is the constant-velocity straight-line path of a free particle, such that \(x_a = x(t_a)\) and \(x_b = x(t_b)\). Show that the action on the solution path is
\[ \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a} \]
The position of the particle at time \(t\) can be written as
\[x(t) = x_a + (x_b - x_a) \frac{t - t_a}{t_b - t_a}\]The velocity is the time derivative of the position
\[v(t) = Dx(t) = \frac{x_b - x_a}{t_b - t_a}\]The equation of the Lagrangian action is
\[S[q] (t_1, t_2) = \int_{t_1}^{t_2}L(t, q(t), Dq(t))\]Which in this case is
\[S[q] (t_a, t_b) = \int_{t_a}^{t_b}\frac{1}{2}mv^2dt\]Substituting for \(v\) we get
\begin{equation*} S[q] (t_a, t_b) = \int_{t_a}^{t_b}\frac{1}{2}m{\left(\frac{x_b - x_a}{t_b - t_a}\right)}^2dt \ = \frac{1}{2}m{\left(\frac{x_b - x_a}{t_b - t_a}\right)}^2\int_{t_a}^{t_b}dt \ = \frac{1}{2}m{\left(\frac{x_b - x_a}{t_b - t_a}\right)}^2(t_b - t_a) \ = \frac{1}{2}m\frac{{(x_b - x_a)}^2}{t_b - t_a} \end{equation*}