Revisiting "Least Action"

Classical Mechanics Level 4

I've had some misconceptions about the concept of "least action" in classical mechanics, largely due to taking that phrase too literally. As it turns out, a physical path is one for which small changes to the path have essentially zero impact on the action. It is therefore better to call it the "principle of stationary action".

$\frac{\partial{S}}{\partial{\text{(path)}}} = 0 \,\,\,\, (\text{physical path})$

Let us investigate this empirically. Consider a projectile of mass $m$ being launched from ground level at speed $v_0$ at an angle $\theta$ with respect to the horizontal, under the influence of ambient gravitational acceleration $g$ . This results in the well-known parabolic trajectory. Let the horizontal and vertical coordinates for this path be $(x(t), y(t))$ . Let $t_f$ be the time at which the particle lands again on the ground. The kinetic energy, potential energy, and action for the path are:

$E(t) = \frac{1}{2} m \Big(\dot{x}^2 (t) + \dot{y}^2 (t) \Big) \\ U(t) = m g \, y(t) \\ S = \int_0^{t_f} \Big(E(t) - U(t) \Big) \, dt$

Consider five other trajectories. Trajectory $1$ is nearly the same as the physical trajectory. Trajectories $2$ and $3$ are nearly identical, with the $y$ component scaled down relative to the physical trajectory. Trajectories $4$ and $5$ are nearly identical, with the $y$ component scaled up relative to the physical trajectory.

$\Big(x_1 (t), y_1 (t) \Big) = \Big(x(t), 0.999 \, y(t) \Big) \\ \Big(x_2 (t), y_2 (t) \Big) = \Big(x(t), 0.500 \, y(t) \Big) \\ \Big(x_3 (t), y_3 (t) \Big) = \Big(x(t), 0.499 \, y(t) \Big) \\ \Big(x_4 (t), y_4 (t) \Big) = \Big(x(t), 2.000 \, y(t) \Big) \\ \Big(x_5 (t), y_5 (t) \Big) = \Big(x(t), 1.999 \, y(t) \Big)$

Let $(S, S_1, S_2, S_3, S_4, S_5)$ be the actions for the physical path, and for the five non-physical paths, respectively. Per the "stationary action principle" we expect small changes in the path to have the least effect on the action when the path is close to the physical one.

In light of that expectation, determine the following sum of ratios:

$\frac{|S_2 - S_3|}{|S - S_1|} + \frac{|S_4 - S_5 |}{|S - S_1 |}$

Details and Assumptions:
1) $m = 1$
2) $v_0 = 10$
3) $\theta = \frac{\pi}{4}$
4) $g = 10$
5) The answer I provided is an approximation

The answer is 3000.0.

1 solution

A Former Brilliant Member
Feb 18, 2020

Let $x_i(t)=x_0(t)=v_0t\cos \theta=10\times t\times \cos {\dfrac{π}{4}}=5√2t$ , $y_i(t)=α_iy_0(t)=α_i(v_0t\sin \theta-\dfrac{1}{2}gt^2)=α_i(5√2t-5t^2)$ . Then $E_i(t)=25(1+α_i^2)-50√2α_i^2t+50α_i^2t^2$ and $U_i(t)=50√2α_it-50α_it^2$ . (Here $i=1,2,3,4,5, α_1=0.999, α_2=0.500, α_3=0.499, α_4=2.000, α_5=1.999$ ). We have $t_f=\dfrac{2v_0\sin \theta}{g}=√2$ . Hence, $S=\dfrac{25√2}{3}\times 2$ ,

$S_i=\dfrac{25√2}{3}[(1-α_i) ^2+2]$ , which yields

$S_1=\dfrac{25√2}{3}[2+(0.001)^2]$ ,

$S_2=\dfrac{25√2}{3}[2+(0.5)^2]$ ,

$S_3=\dfrac{25√2}{3}[2+(0.501)^2]$ ,

$S_4=\dfrac{25√2}{3}[2+(1)^2]$ , and

$S_5=\dfrac{25√2}{3}[2+(0.999)^2]$ .

So $|S-S_1|=\dfrac{25√2}{3}\times (0.001)^2, |S_2-S_3|=\dfrac{25√2}{3}\times 0.001001, |S_4-S_5|=\dfrac{25√2}{3}\times 0.001999$ , and the required sum is $\dfrac{0.001001+0.001999}{0.000001}=\boxed {3000}$

Revisiting "Least Action"

The answer is 3000.0.

1 solution

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