Computing Actions

Lagrangians in generalized coordinates

The function \displaystyle{S_\chi} takes a coordinate path; the function \displaystyle{\mathcal{S}} takes a configuration path.

\displaystyle{\begin{aligned}  \mathcal{S} [\gamma] (t_1, t_2) &= \int_{t_1}^{t_2} \mathcal{L} \circ \mathcal{T} [\gamma]  \\   S_\chi [q] (t_1, t_2) &= \int_{t_1}^{t_2} L_\chi \circ \Gamma [q]  \\   \end{aligned}}

\displaystyle{\begin{aligned}  \mathcal{S} [\gamma] (t_1, t_2)  &= S_\chi [\chi \circ \gamma] (t_1, t_2) \\  \end{aligned}}

Computing Actions

\displaystyle{\texttt{literal-function}} makes a procedure that represents a function of one argument that has no known properties other than the given symbolic name.

The method of computing the action from the coordinate representation of a Lagrangian and a coordinate path does not depend on the coordinate system.

Exercise 1.4. Lagrangian actions

For a free particle an appropriate Lagrangian is

\displaystyle{\begin{aligned}  L(t,x,v) &= \frac{1}{2} m v^2  \\   \end{aligned}}

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

\displaystyle{\begin{aligned}  \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a} \\   \end{aligned}}

— Structure and Interpretation of Classical Mechanics

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\displaystyle{\begin{aligned}  L(t,x,v) &= \frac{1}{2} m v^2  \\   \end{aligned}}

\displaystyle{\begin{aligned}  S_\chi [\gamma] (t_1, t_2) &= \int_{t_1}^{t_2} L_\chi (t, q(t), Dq(t)) dt \\  &= \int_{t_2}^{t_1} \frac{1}{2} m v^2 dt \\  &= \frac{1}{2} m v^2 \int_{t_2}^{t_1} dt \\  &= \frac{1}{2} m v^2 (t_2 - t_1)  \\  &= \frac{1}{2} m (\frac{x_2 - x_1}{t_2 - t_1})^2 (t_2 - t_1)  \\  &= \frac{1}{2} m \frac{(x_2 - x_1)^2}{t_2 - t_1}   \\   \end{aligned}}

— Me@2006-2008

— Me@2019-03-10 11:08:29 PM

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2019.03.10 Sunday (c) All rights reserved by ACHK