Force Field and Action

@Karan Chatrath has shown the details of how to solve this problem. I will comment on the physical meaning of the results. "Least action" is a misleading name for the principle being considered here. A much better name is "the principle of stationary action". The physical path is one for which a small deformation in the path results in essentially zero change to the action. In this problem, we have introduced a "deformation parameter", which is the multiple out in front of $y(t)$ .

$y' (t) = \alpha y(t)$

Suppose we make a plot of the action $S'$ vs $\alpha$ . The physical path corresponds to an $\alpha$ value at which the following is true:

$\frac{d S'}{d \alpha} = 0$

So on the graph, there should be a local extreme for $\alpha = 1$ . This means that a small deviation from the physical path (and correspondingly, a small deviation in $\alpha$ ) has no effect on the action (because the slope is locally zero). And since the two other paths have $\alpha$ values just below and just above unity, we expect the actions for those paths to either both be less than (for a local max) or both be greater than (for a local min) the action for the physical path. As it turns out, both action values for the deformed paths are smaller than the action for the physical path, and it is a local maximum.

So the results of this exercise agree with the "stationary action" principle. Looking for "least action" would have led one astray.

The external force acting on the particle is:

$F_y = -y^2$ The potential energy associated with this force is governed by:

$\frac{dV}{dy} = -F_y$ $\frac{dV}{dy} = y^2$

Given that the reference is $y=0$ , the PE of the system is finally:

$V = \frac{y^3}{3}$

Extra details:

Applying conservation of energy:

$T +V = T_{init} + V_{init}$ $\frac{\dot{y}^2}{2} + \frac{y^3}{3}=\frac{25}{2}$

As time progresses $\dot{y}$ decreases: $\dot{y} = -\sqrt{25 - \frac{2y^3}{3}}$

The particle comes to rest when kinetic energy is zero. This happens when $y$ is governed by: $0 + \frac{y_f^3}{3}=\frac{25}{2}$ $y_f = \left(\frac{75}{2}\right)^{1/3}$

Separating variables and integrating:

$2\int_{0}^{y_f}\frac{dy}{\sqrt{25 - \frac{2y^3}{3}}}=t_f$

The time taken is multiplied by two to account for the return journey of the particle to $y=0$ . Having found the time of motion, the next step is to compute the action which is done as such:

$S = \int_{0}^{t_f} \left(T-V\right) \ dt$

This integral is solved numerically. In fact, all the calculations are performed numerically. The above time duration computation is some extra detail for the problem.

Simulation code attached bolow:

$\boxed{Q = 0.06}$

It is strange that the actions $S_1$ and $S_2$ are lower than $S$ . I would have expected $S$ to be the least. This result is counter-intuitive.

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clear all
clc

% Initial conditions:
y(1)  = 0;
dy(1) = 5;

% Time step:
dt    = 1e-4;

% Time initialisation:
t(1)  = 0;

% Time index initialisation:
k     = 1;

% Actions initialisations:
S     = 0;
S1    = 0;
S2    = 0;

% Simulation till the particle returns to y = 0:
while y(k)>=0  

    % Acceleration:
    ddy      = -y(k)^2;

    % Kinetic and potential energies:
    T(k)     = 0.5*dy(k)^2;
    V(k)     = (1/3)*y(k)^3;

    % Numerical integration:
    dy(k+1)  = dy(k) + ddy*dt;
    y(k+1)   = y(k)  + dy(k+1)*dt;

    % Numerical action integral:
    S        = S + (T(k) - V(k))*dt;

    % Kinetic and potential energies when y1 = 0.9*y:
    T1(k)     = 0.5*(0.9*dy(k))^2;
    V1(k)     = (1/3)*(0.9*y(k))^3; 

    % Numerical action integral when y1 = 0.9*y:
    S1        = S1 + (T1(k) - V1(k))*dt;

    % Kinetic and potential energies when y1 = 1.1*y:
    T2(k)     = 0.5*(1.1*dy(k))^2;
    V2(k)     = (1/3)*(1.1*y(k))^3;  

    % Numerical action integral when y1 = 1.1*y:
    S2        = S2 + (T2(k) - V2(k))*dt;

    % Time and time index increments:
    t(k+1)   = t(k)  + dt;
    k        = k + 1;
end


Q = (1 - S1/S) + (1 - S2/S)