Optimizing Another Path

Classical Mechanics Level 2

A particle of mass $m$ travels along a one-dimensional path from $x = 0$ to $x = D$ . The potential energy of the particle $V(x) = -F x$ . Suppose the position of the particle varies over time as follows (where $a$ is a number to be determined):

$x = \frac{1}{2} a t^2$

The action $S$ for the path is:

$S = \int_0^{\sqrt{2 D /a}} \Big(\frac{1}{2} m \dot{x}^2 - V(t) \Big) \, dt$

The quantity $S$ is a function of the variable $a$ . Determine the value of $a$ such that the following is true:

$\frac{\partial{S}}{\partial{a}} = 0$

\frac{F}{m}

F D

F m^2

F

1 solution

Steven Chase
Feb 14, 2021

The premise of the question is that we have guessed the form of the solution, which includes an unknown parameter $a$ that must be solved for using the stationary action principle.

$x(t) = \frac{1}{2} a t^2 \\ \dot{x} (t) = a t$

Kinetic energy, potential energy, and action:

$T = \frac{1}{2} m \dot{x}^2 = \frac{1}{2} m a^2 t^2 \\ V(t) = - F x(t) = - \frac{F a}{2} t^2 \\ S = \int_0^{\sqrt{2 D / a}} \Big( \frac{1}{2} m a^2 t^2 + \frac{F a}{2} t^2 \Big) dt = \frac{1}{2} (m a^2 + F a) \int_0^{\sqrt{2 D / a}} t^2 dt \\ = \frac{1}{6} (m a^2 + F a) \Big( \frac{2 D}{a} \Big)^{3/2} \\ = \frac{(2D)^{3/2}}{6} (m a^2 + F a) a^{-3/2} \\ = \frac{(2D)^{3/2}}{6} (m a^{1/2} + F a^{-1/2})$

Applying the stationary action principle:

$\frac{\partial{S}}{\partial{a}} = 0 \implies \frac{1}{2} m a^{-1/2} - \frac{1}{2} F a^{-3/2} = 0 \\ m a^{-1/2} = F a^{-3/2} \\ F = m a \\ a = \frac{F}{m}$

The value of $a$ that makes the action stationary corresponds to the acceleration value from Newton's Second Law.

Optimizing Another Path

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