Action 9-8-2020

Classical Mechanics Level 3

A particle of mass $m = 1$ undergoes motion according to the following Lagrangian:

$\mathcal{L} = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - x - y$

At time $t = 0$ , the particle's position and velocity are $(x,y) = (0,0)$ and $(\dot{x}, \dot{y} ) = \Big( \cos (\pi/3), \sin (\pi/3) \Big)$ . The particle moves until it once again reaches $y = 0$ at time $t_f > 0$ .

The action for this path is (the parentheses indicate that the Lagrangian varies over time):

$S = \int_0^{t_f} \mathcal{L} (t) \, dt$

If the coordinates for this path are $\Big(x(t), y(t) \Big)$ , define two altered paths:

$\Big(x_1 (t), y_1(t) \Big) = \Big(x(t), 0.9 \, y(t) \Big) \\ \Big(x_2 (t), y_2(t) \Big) = \Big(x(t), 1.1 \, y(t) \Big)$

Note that these altered paths preserve the starting and ending coordinates in space and time. Let $S_1$ and $S_2$ be the actions for these paths (calculated over the same time interval as before). Determine the quantity $Q$ :

$Q = \frac{S_1}{S} + \frac{S_2}{S} - 2$

Bonus: What is the physical significance of the action ratios?

The answer is 0.01866.

2 solutions

Mark Hennings
Sep 8, 2020

Solving Lagrange's equations yields $x \; =\; \tfrac12t - \tfrac{1}{2m}t^2 \hspace{1cm}y \; = \; \tfrac{\sqrt{3}}{2}t - \tfrac{1}{2m}t^2 \hspace{2cm} 0< t < t_f = m\sqrt{3}$ and hence we calculate $\begin{aligned} S & = \; \big(\sqrt{3} - \tfrac{3}{2}\big)m^2 \\ S_1 = S_2 & = \; \big(\tfrac{801}{800}\sqrt{3}-\tfrac32\big)m^2 \end{aligned}$ so that $\frac{S_1}{S} + \frac{S_2}{S} - 2 \; = \; \tfrac{1}{200}(2 + \sqrt{3}) \; = \; \boxed{0.0186603}$ The fact that $S_1,S_2 \ge S$ is an example of the Principle of Least Action.

Lol I used the same method... weird. I got an error someway... well done! I upvoted :)

Krishna Karthik - 9 months ago

Karan Chatrath
Sep 8, 2020

Code-based solution provided.

I remember seeing your previous problems where you speak of the principle of 'stationary' action and not that of 'least' action. In this case, the solution of stationary action corresponds to that of least action, so each of the action ratios are greater than unity.

clear all
clc

% Numerical resoloution:
dt    = 1e-5;

% Initial conditions:
dy(1) = sin(pi/3);
dx(1) = cos(pi/3);
x(1)  = 0;
y(1)  = 0;
t(1)  = 0;

% Time index:
k     = 1;

% Action initialisation:
S     = 0;
S1    = 0;
S2    = 0;


% Loop runs till the paticle stays at or above ground level (y=0)
while y >= 0

    % Equations of motion after plugging into the Euler-Lagrange equation:
    ddx     = -1;
    ddy     = -1;

    % Lagrangian for each case:
    L       = 0.5*(dx(k)^2 + dy(k)^2) - x(k) - y(k);
    L1      = 0.5*(dx(k)^2 + 0.9^2*dy(k)^2) - x(k) - 0.9*y(k);
    L2      = 0.5*(dx(k)^2 + 1.1^2*dy(k)^2) - x(k) - 1.1*y(k);

    % Numerical integration for stationary action trajectory:
    dy(k+1) = dy(k) + ddy*dt;
    dx(k+1) = dx(k) + ddx*dt;

    y(k+1)  = y(k) + dt*dy(k+1);
    x(k+1)  = x(k) + dt*dx(k+1);

    % Numerical integration for action:
    S       = S + L*dt;
    S1      = S1 + L1*dt;
    S2      = S2 + L2*dt;

    % Time and index increment:
    t(k+1)  = t(k) + dt;    
    k       = k + 1;

end

ANSWER = (S1/S) + (S2/S) - 2
% ANSWER = 0.0187

First to solve. I did a code based solution as well; however at first I tried integrating action analytically ; that took me about 20 minutes, then, I did a code based solution, which took me about 20 minutes. So, after 40 minutes of going round a merry-go-round, I finally got the answer. Lmao.

Nice one btw. Thanks for posting the solution :)

Krishna Karthik - 9 months ago

Nice job. Did your analytical approach yield the right answer?

Steven Chase - 9 months ago

Nope. I decided to go with a numerical approach instead. My theory was right, with the analytical computations of the integral and the ratios, but yeah. I may have computed something wrong. That's the tricky part with an analytical method.

Krishna Karthik - 9 months ago

Action 9-8-2020

The answer is 0.01866.

2 solutions

0 pending reports