Ch 1. Hamilton Jacobi EQ

Suppose there are ‘n’ particles inhabiting 3 dimensional space. Let’s describe the positions of the particles as a single point in 3n dimensional position coordinate space


(Note: We’re working in Cartesian coordinates unless otherwise noted.)
Now imagine an arbitrary vector field in this coordinate space (\vec{p} is just the coordinate space position variable)

(1)   \begin{equation*} \vec{u}(\vec{p}\text{,t)=}\{u_1(\vec{p},t),u_2(\vec{p},t),\text{...},u_{3n}(\vec{p},t)\} \end{equation*}

If our particles are in some arbitrary motion, \vec{Q} will be a function of time, and the coordinate space point will trace out a path through the vector field \vec{u}. The particular vector at point Q(t) is

(2)   \begin{equation*} \vec{u}(\vec{Q}\text{(t),t)=}\{u_1(\vec{Q}(t),t),u_2(\vec{Q}(t),t),\text{...},u_{3n}(\vec{Q}(t),t)\} \end{equation*}

Figure 1.1: Schematic representation of an arbitrary vector field and a path through it.

Figure 1.1: Schematic representation of an arbitrary vector field and a path through it.

We can take this a step further by assuming that \vec{u}, is a velocity field and that \vec{Q} is constrained to move along the streamlines of the field. We can then relate the velocity field with the particle trajectory like this:

(3)   \begin{equation*} \frac{d\vec{Q}(t)}{dt}=\vec{u}(\vec{Q}(t),t) \end{equation*}

This is a form of Burgers’ Equation. Now, an interesting question that we might ask: “Is it possible to find a \vec{u} such that \vec{Q} will trace out paths that match those traced out by physical systems? In particular, can we find a velocity field whose streamlines match Newtonian Mechanics?”

Let’s start by letting the masses of the particles equal m_1, m_2, ..., m_n and define a new set of variables for convenience (let’s call them “extended mass variables”)

    \[ M_1=M_2=M_3=m_1, \]

(4)   \begin{equation*} M_4=M_5=M_6=m_2,..., \end{equation*}

    \[ M_{3n-2}=M_{3n-1}=M_{3n}=m_n \]

The total acceleration vector of our phase space point is going to be:

(5)   \begin{equation*}  \vec{a}=\frac{d^2\vec{Q}(t)}{dt^2}=\frac{d}{dt}(\vec{u}(\vec{Q}\text{(t),t))} \end{equation*}

We are going to concentrate on the i^{th} component of this vector for simplicity. Newton’s 2nd law, equation (5), and letting V be a Potential Energy function will lead to

(6)   \begin{equation*}  F_i=M_ia_i=M_i(\frac{d}{dt}(\vec{u}(\vec{Q}\text{(t),t))})_i=M_i(\frac{\partial u_i}{\partial t}+\sum _{j=1}^{3n} u_j\frac{\partial u_i}{\partial q_j}\text{)=-}\frac{\partial V}{\partial q_i} \end{equation*}

Switching over to momentum coordinates by letting p_i = M_i u_i, we get

(7)   \begin{equation*} \frac{\partial p_i}{\partial t}+\sum _{j=1}^{3n} \frac{p_j}{M_j}\frac{\partial p_i}{\partial q_j}\text{=-}\frac{\partial V}{\partial q_i} \end{equation*}

One problem we might encounter with this equation is that streamlines may run into each other and cause “shock” (the vector field will become multi-valued at the point of intersection). To avoid this, we are going to restrict our velocity field to being Conservative, meaning that it is the Gradient of a Scalar Potential (remember, the Gradient of a Scalar function is a vector that points in the direction of steepest ascent). This is going to wind up ensuring that our streamlines never run into each other. We’ll call our Scalar function ‘S’ and let

(8)   \begin{equation*} \vec{p}=\nabla S \end{equation*}

In other words,

(9)   \begin{equation*} p_i= \frac{\partial S}{\partial q_i} \end{equation*}

A neat trick this allows us to do is swap indices on the partial derivatives of \vec{p}.

(10)   \begin{equation*} \frac{\partial p_i}{\partial q_j}=\frac{\partial p_j}{\partial q_i} \end{equation*}

This is a standard theorem from Calculus III; in 3-dimensions it means that the Curl of \vec{p} is zero if and only if \vec{p} is Conservative (See here for more info). This will allow us to transform equation (6) to

(11)   \begin{equation*} \frac{\partial p_i}{\partial t}+\sum _{j=1}^{3n} \frac{p_j}{M_j}\frac{\partial p_j}{\partial q_i}=-\frac{\partial V}{\partial q_i} \end{equation*}

A bit more manipulation is going get us to an interesting place:

(12)   \begin{equation*} \frac{\partial p_i}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{\partial }{\partial q_i}(p_j{}^2)=-\frac{\partial V}{\partial q_i} \end{equation*}

(13)   \begin{equation*} \frac{\partial }{\partial t}(\frac{\partial S}{\partial q_i})+\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{\partial }{\partial q_i}((\frac{\partial S}{\partial q_i})^2)=-\frac{\partial V}{\partial q_i} \end{equation*}

(14)   \begin{equation*} \frac{\partial }{\partial q_i}(\frac{\partial S}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(\frac{\partial S}{\partial q_i})^2=-V) \end{equation*}

(15)   \begin{equation*} \frac{\partial S}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(\frac{\partial S}{\partial q_i})^2+V=0 \end{equation*}

This is the Hamilton Jacobi Equation (HJE), in some ways the most powerful equation in Classical Mechanics. We will sometimes use an alternate notation for the HJE. Let’s define

    \[ \{x_j,y_j,z_j\} \]

    \[ (or \ \{r_j,\theta _j,\phi _j \} \ in \ Spherical \ Coordinates) \]

as the spatial coordinates for the j^{th} particle and let

(16)   \begin{equation*} \nabla _j=\{\frac{\partial}{\partial x_j},\frac{\partial}{\partial y_j},\frac{\partial}{\partial z_j}\} \end{equation*}

(17)   \begin{equation*} (or \ \nabla _j=\{\frac{\partial}{\partial r_j},\frac{1}{r_j}\frac{\partial}{\partial \theta _j},\frac{1}{r_j(Sin(\theta _j))}\frac{\partial }{\partial \phi _j}\}\ in\ Spherical\ Coordinates) \end{equation*}

be the Gradient with respect to the j^{th} particle. Then the HJE looks like this:

(18)   \begin{equation*}  \frac{\partial S}{\partial t}+\sum _{j=1}^n \frac{1}{2m_j}(\nabla _jS)^2+V=0 \end{equation*}

To recap, if you solve this for S, you will have a function whose Gradient gives streamlines in phase space that agree with Newton’s Laws. This can be quite difficult to visualize, especially for multi-particle systems. One way to think about it is that we have a very large number (so large that it appears as a continuous fluid in phase space) of sets of ‘n’ particles. Each particle interacts with the other particles in it’s own set, but doesn’t know about the existence of the other sets of particles. The examples in the next section should help with visualizing solutions of the HJE.

By the way, the HJE is usually the last stop (or close to it) in most Classical Mechanics Texts. As such, the typical path to it goes through: Calculus of Variations, Lagrangian Mechanics, Hamiltonian Mechanics, Canonical Transformations, (and possibly other stuff). That is a long way to go and makes it seem more obscure than it actually is. The derivation shown above is the simplest I know of.

Note that since \nabla _j S is the momentum of the j^{th} particle, equation (18) can be rewritten as

(19)   \begin{equation*} \frac{\partial S}{\partial t}+\sum _{j=1}^n \frac{(p_j){}^2}{2m_j}+V=\frac{\partial S}{\partial t}+T+V=0 \end{equation*}

where ‘T’ is the Kinetic Energy at a given point in phase space. Evidently, \frac{\partial S}{\partial t} equals -1 times the total Energy at each point. This turns out to always be true if the Potential is Conservative. In that case \frac{\partial S}{\partial t} with -E and worry only about the spatial part of S.

Chapter 2


Baby needs diapers!
If you find anything on this site interesting or valuable, please consider making a donation!
Your charity will keep this project alive and growing.

An essay on free content by Sam Harris


  1. How did you get from Equation 5 to Equation 6? I think I follow everything up until then but the stuff after the du_i/dt term confuses me. Where did the sum come from? I understand where -dV/dq_i came from, because that’s from familiar, basic intro mechanics physics. It’s a definition that relates a force to a potential. I’ve honestly never liked V for potential, because it’s the same symbol for voltage, but I know why you used it.


    1. I’m just taking the total derivative of u, which is a function of Q(t) and t. But since Q(t) is a vector, you could write this as u(q1(t),q2(t),q3(t),t). Now take the total derivative. You should get what I got. Peace.

      Edit: Also note that we’re letting derivative of qj(t) equal uj, since u is assumed to be Q’s velocity field. If you expand out the partial derivatives, then make this substitution, it should make sense.

Leave a Reply

Your email address will not be published. Required fields are marked *