Ch 2. Hamilton Jacobi EQ Examples


Example 1: Free Particle

For a Free Particle in 3 dimensions, V = 0 and the HJE reduces to

(1)   \begin{equation*} \frac{\partial S}{\partial t}+\frac{1}{2m} ((\frac{\partial S}{\partial x})^{2}+(\frac{\partial S}{\partial y})^{2}+(\frac{\partial S}{\partial z})^2)=0} \end{equation*}

or, by assuming uniform energy,

(2)   \begin{equation*} -E+\frac{1}{2m}(\nabla S)^{2}=0} \end{equation*}

where we’ve replaced \frac{\partial S}{\partial t} with -E as discussed at the end of chapter 1. I’m not going to go through the motions of solving this thing, as it’s been done elsewhere (usually using an additive Separation of Variables technique)(Details can be found here and elsewhere). We’ll just skip right to the solution:

(3)   \begin{equation*} S(x,y,z)= p_xx+p_yy+p_zz\ \ (which\ implies\ E=\frac{p_x^2+p_y^2+p_z^2}{2m}) \end{equation*}

You can verify the solution by substituting 3 into 1 and churning it out, if you are so inclined. Note that the Gradient of S is obviously the momentum {p_x, p_y, p_z}. We are going to plot the vector field that we’ve obtained though, just to be sure everything is clear.

Figure 2.1.1: Plot of vector field for a free particle with mass=1 and momentum ={3,1,2}

Figure 2.1.1: Plot of vector field for a free particle with mass=1 and momentum ={3,1,2}

As expected, the vector field points in one direction with a constant magnitude, representing a particle moving with constant velocity, with arbitrary initial position.


Example 2: Central Potential

For the next example, we’ll be working in Spherical Coordinates, with a single particle under the influence of a central potential, V(r) = -1/r. The Hamilton Jacobi Equation is

(4)   \begin{equation*} -E+\frac{1}{2m} ((\frac{\partial S}{\partial r})^{2}+\frac{1}{r}(\frac{\partial S}{\partial \theta})^{2}+\frac{1}{r Sin(\theta)}(\frac{\partial S}{\partial \phi})^{2})-\frac{1}{r}=0 \end{equation*}

where we’ve assumed uniform energy again. And again, we’ll just jump to the solution:

    \[ S(r,\theta ,\phi )= \pm \int \sqrt{2m(E+\frac{1}{r})-\frac{C_\theta}{r^{2}}}dr \]

(5)   \begin{equation*} \pm \int \sqrt{C_\theta-\frac{C_\phi}{(Sin(\theta))^{2}}}d\theta +C_{\phi}\phi \end{equation*}

However, we are mostly interested in the partial derivatives, since these are what we use to generate our velocity field (and hence the particle’s motion).

(6)   \begin{equation*} \frac{\partial S}{\partial \phi }=p_{\phi }=C_{\phi } \end{equation*}

(7)   \begin{equation*} \frac{\partial S}{\partial \theta }=p_{\theta }=\pm \sqrt{C_{\theta }{}^2-\frac{C_{\phi }{}^2}{(Sin(\theta ))^2}} \end{equation*}

(8)   \begin{equation*} \frac{\partial S}{\partial r}=p_r=\pm \sqrt{2m(E+\frac{1}{r})-\frac{C_{\theta }}{r^2}} \end{equation*}

The 2 constants of integration, C_\phi, and C_\theta, can be interpreted as the angular momentum about the polar axis and the total angular momentum, respectively.

Let’s look at a particular example and see how everything behaves. We’ll start by setting our constants to:

(9)   \begin{equation*} m=1 \end{equation*}

(10)   \begin{equation*} C_{\phi }=C_{\theta }=1 \end{equation*}

(11)   \begin{equation*} E=-1/8 \end{equation*}

Equation (10) means that the angular momentum about the polar axis equals the total angular momentum (i.e. all of the angular momentum is about the “z” axis). Setting the energy to a negative value (equation (11)) means that the particle is bound to the central potential.

Now, check out equation (7), the component of the momentum in the “\theta” direction. Since this expression generates the a component of the momentum of our particle, we are going to demand that it resolve to a real number. This implies that

(12)   \begin{equation*} C_{\theta }{}^2-\frac{C_{\phi }{}^2}{(Sin(\theta ))^2}=1-\frac{1}{(Sin(\theta ))^2}\geq 0 \Longrightarrow Sin(\theta )=1 \end{equation*}

In other words \theta is \pi/2. Which means that our solution is only valid in the x-y plane! There is no velocity field and hence no particle flow outside of this plane, which agrees with the fact that central force motion is always confined to a plane.

But there is another restriction. Check out equation (8). The same idea applies there:

(13)   \begin{equation*} 2m(E+\frac{1}{r})-\frac{C_{\theta }}{r^2}=2((\frac{-1}{8})+\frac{1}{r})-\frac{1}{r^2}\geq 0 \end{equation*}

Let’s see a plot of this function:

Figure 2.2.1: The discriminant of Equation (8)

Figure 2.2.1: The discriminant of Equation (8)

Numerically, we find that the function is only positive (and hence we only have valid solutions) when ‘r’ is between .536 and 7.464. Let’s go ahead and plot our velocity field and a sampling of possible trajectories:

Figure 2.2.2: Velocity Field and Trajectories of the Central Potential System

Figure 2.2.2: Velocity Field and Trajectories of the Central Potential System

Here we can see the power of the HJE and one of it’s practical shortcomings. On one hand, solving the HJE automatically provides the constants of the motion and tells us all possible trajectories consistent with a given set of those constants. On the other hand, because S is single valued (and hence so is the velocity vector at any location), different possible paths cannot cross each other. This explains why we only see a half elliptical path, instead of a full ellipse as expected. Completing the loop would cross over other possible trajectories, causing the vector field to be multiple valued. It turns out that the other halves of the elliptical paths are hiding–check out the +/- sign in equation (8). Let’s re-plot using the minus sign instead of the plus sign:

Figure 2.2.3: The "missing" part of the Trajectories

Figure 2.2.3: The “missing” part of the Trajectories

While we’re at it, let’s check out ‘S’ itself to get a view of the generating functions. Again, the Gradient (the vector that points in the direction of steepest ascent and has magnitude equal to the slope) of these functions produces the 2 velocity fields shown above.

Figure 2.2.4 Plots of the generating functions 'S';

Figure 2.2.4 Plots of the generating functions ‘S’

As an exercise, see if you can work out the example of m = 1, C_\theta=2, C_\phi=1, E = -1/8.

Chapter 3

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