# 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

.

\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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# Distinguishing hardware and software

If

X is a pattern of Y,

then

Y is hardware

and

X is software.

— Me@2019-03-10 12:10:20 AM

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