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202112021037 Exercise 1.4 Lagrangian actions

For a free particle an appropriate Lagrangian is

Eq. (1.8):

\mathcal{L}(t,x,v)=\displaystyle{\frac{1}{2}mv^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

Eq. (1.9):

\displaystyle{\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}}.

Extracted from Structure and Interpretation of Classical Mechanics, SICP, 2e

Background. (Lagrangians and Lagrangian actions)

The function \mathcal{L} is called a Lagrangian for the system, and the resulting action,

Eq. (1.4):

S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}\circ\Gamma [q]},

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

Eq. (1.5):

S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}(t,q(t),\mathrm{D}q(t))\,\mathrm{d}t}.

(Section 1.3 The Principle of Stationary Action)

Working. (roughly)

\begin{aligned} S & = \int_{t_a}^{t_b}\frac{1}{2}mv^2\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}(\mathbf{v}\cdot\mathbf{v})\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}\bigg( \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\cdot\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\bigg) \,\mathrm{d}t \\ & = \frac{m}{2}\int_{x_a,t_a}^{x_b,t_b}\frac{(\mathrm{d}x)^2}{(\mathrm{d}t)} \\ & = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} \\ \end{aligned}