15.4: Hamilton-Jacobi Theory

Hamilton used the Principle of Least Action to derive the Hamilton-Jacobi relation (chapter $15.3$)

$$H(\mathbf{q},\mathbf{p}, t) + \frac{\partial S}{\partial t} = 0 \label{15.11}$$

where $\mathbf{q}, \mathbf{p}$ refer to the $1 \leq i \leq n$ variables $q_i, p_i$ and $S(q_j (t_1), t_1, q_j (t_2), t_2)$ is the action functional. Integration of this first-order partial differential equation is non trivial which is a major handicap for practical exploitation of the Hamilton-Jacobi equation. This stimulated Jacobi to develop the mathematical framework for canonical transformation that are required to solve the Hamilton-Jacobi equation. Jacobi’s approach is to exploit generating functions for making a canonical transformation to a new Hamiltonian $\mathcal{H}(\mathbf{Q}, \mathbf{P}, t)$ that equals zero.

$$\mathcal{H}(\mathbf{Q},\mathbf{P}, t) = H(\mathbf{q},\mathbf{p}, t) + \frac{\partial S}{\partial t} = 0 \label{15.90}$$

The generating function for solving the Hamilton-Jacobi equation then equals the action functional $S$.

The Hamilton-Jacobi theory is based on selecting a canonical transformation to new coordinates $(Q, P, t)$ all of which are either constant, or the $Q_i$ are cyclic, which implies that the corresponding momenta $P_i$ are constants. In either case, a solution to the equations of motion is obtained. A remarkable feature of Hamilton-Jacobi theory is that the canonical transformation is completely characterized by a single generating function, $S$. The canonical equations likewise are characterized by a single Hamiltonian function, $H$. Moreover, the generating function $S$, and Hamiltonian function $H$, are linked together by Equation $\ref{15.11}$. The underlying goal of Hamilton-Jacobi theory is to transform the Hamiltonian to a known form such that the canonical equations become directly integrable. Since this transformation depends on a single scalar function, the problem is reduced to solving a single partial differential equation.

Time-dependent Hamiltonian

Jacobi’s complete integral $S(q_i, P_i, t)$

The principle underlying Jacobi’s approach to Hamilton-Jacobi theory is to provide a recipe for finding the generating function $F = S$ needed to transform the Hamiltonian $H(\mathbf{q}, \mathbf{p}, t)$ to the new Hamiltonian $\mathcal{H}(\mathbf{Q}, \mathbf{P}, t)$ using Equation $\ref{15.90}$. When the derivatives of the transformed Hamiltonian $\mathcal{H}(\mathbf{Q}, \mathbf{P}, t)$ are zero, then the equations of motion become

$$\dot{Q}_i = \frac{\partial \mathcal{H}}{ \partial P_i} = 0 \label{15.91}$$

$$\dot{P}_i = − \frac{\partial \mathcal{H}}{ \partial Q_i } = 0 \label{15.92}$$

and thus $Q_i$ and $P_i$ are constants of motion. The new Hamiltonian $\mathcal{H}$ must be related to the original Hamiltonian $H$ by a canonical transformation for which

$$\mathcal{H}(\mathbf{Q}, \mathbf{P}, t) = H(\mathbf{q}, \mathbf{p}, t) + \frac{\partial S}{ \partial t} \label{15.93}$$

Equations $\ref{15.91}$ and $\ref{15.92}$ are automatically satisfied if the new Hamiltonian $\mathcal{H} = 0$ since then Equation $\ref{15.93}$ gives that the generating function $S$ satisfies Equation $\ref{15.90}$.

Any of the four types of generating function can be used. Jacobi chose the type 2 generating function as being the most useful for many practical cases, that is, $S(q_i, P_i, t)$ which is called Jacobi’s complete integral.

For generating functions $F_1$ and $F_2$ the generalized momenta are derived from the action by the derivative

$$p_i = \frac{\partial S}{ \partial q_i} \label{15.4}$$

Use this generalized momentum to replace $p_i$ in the Hamiltonian $H$, given in Equation $\ref{15.93}$, leads to the Hamilton-Jacobi equation expressed in terms of the action $S$.

$$H(q_1, ...q_n; \frac{\partial S}{ \partial q_1 }, ..., \frac{\partial S}{ \partial q_n} ;t) + \frac{\partial S}{ \partial t} = 0 \label{15.94}$$

The Hamilton-Jacobi equation, $\ref{15.94}$, can be written more compactly using tensors $\mathbf{q}$ and $\boldsymbol{\nabla}S$ to designate $(q_1, ..q_n)$ and $\frac{\partial S}{ \partial q_1 }, ..., \frac{\partial S}{ \partial q_n}$ respectively. That is

$$H(\mathbf{q}, \boldsymbol{\nabla}S, t) + \frac{\partial S}{\partial t} = 0 \label{15.95}$$

Equation $\ref{15.95}$ is a first-order partial differential equation in $n + 1$ variables which are the old spatial coordinates $q_i$ plus time $t$. The new momenta $P_i$ have not been specified except that they are constants since $\mathcal{H} = 0$.

Assume the existence of a solution of $\ref{15.95}$ of the form $S(q_i, P_i, t) = S(q_1, ..q_n; \alpha_1, ..\alpha_{n+1};t)$ where the generalized momenta $P_i = \alpha_1, \alpha_2, ....\alpha$ plus $t$ are the $n + 1$ independent constants of integration in the transformed frame. One constant of integration is irrelevant to the solution since only partial derivatives of $S(q_i, P_i, t)$ with respect to $q_i$ and $t$ are involved. Thus, if $S$ is a solution of the first-order partial differential equation, then so is $S + \alpha$ where $\alpha$ is a constant. Thus it can be assumed that one of the $n + 1$ constants of integration is just an additive constant which can be ignored leading effectively to a solution

$$S(q_i, P_i, t) = S(q_1, .....q_n;\alpha_1, .....\alpha_n;t) \label{15.96}$$

where none of the $n$ independent constants are solely additive. Such generating function solutions are called complete solutions of the first-order partial differential equations since all constants of integration are known.

It is possible to assume that the $n$ generalized momenta, $P_i$ are constants $\alpha_i$, where the $\alpha_i$ are the constants. This allows the generalized momentum to be written as

$$p_i = \frac{\partial S(\mathbf{q}, \boldsymbol{\alpha}, t)}{ \partial q_i } \label{15.97}$$

Similarly, Hamilton’s equations of motion give the conjugate coordinate $\mathbf{Q} = \boldsymbol{\beta}$, where $\beta_i$ are constants. That is

$$Q_i = \beta_i = \frac{\partial S(\mathbf{q}, \boldsymbol{\alpha}, t)}{ \partial \alpha_i} \label{15.98}$$

The above procedure has determined the complete set of $2n$ constants $(\mathbf{Q} = \boldsymbol{\beta}, \mathbf{P} = \boldsymbol{\alpha})$. It is possible to invert the canonical transformation to express the above solution, which is expressed in terms of $Q_i = \beta_i$ and $P_i = \alpha_i$, back to the original coordinates, that is, $q_j = q_j (\alpha , \beta , t)$ and momenta $p_j = p_j (\alpha , \beta , t)$ which is the required solution.

Hamilton’s principle function $S_H(\mathbf{q}_i, t; \mathbf{q}_o t_o)$

Hamilton’s approach to solving the Hamilton-Jacobi Equation $\ref{15.95}$ is to seek a canonical transformation from variables $(\mathbf{p}, \mathbf{q})$ at time $t$, to a new set of constant quantities, which may be the initial values $(\mathbf{q}_0, \mathbf{p}_0)$ at time $t = 0$. Hamilton’s principle function $S_H(q_i, t; q_ot_o)$ is the generating function for this canonical transformation from the variables $(\mathbf{q}, \mathbf{p})$ at time t to the initial variables $(\mathbf{q}_0, \mathbf{p}_0)$ at time $t_0$. Hamilton’s principle function $S_H(q_i, t; q_ot_o)$ is directly related to Jacobi’s complete integral $S(q_i, P_i, t)$.

Note that $S_H$ is the generating function of a canonical transformation from the present time $(\mathbf{q}, \mathbf{p}, t)$ variables to the initial $(\mathbf{q}_0, \mathbf{p}_0, t_0)$, whereas Jacobi’s $S$ is the generating function of a canonical transformation from the present $(\mathbf{q},\mathbf{p}, t)$ variables to the constant variables $(\mathbf{Q} = \boldsymbol{\beta}, \mathbf{P} = \boldsymbol{\alpha})$. For the Hamilton approach, the canonical transformation can be accomplished in two steps using $S$ by first transforming from $(\mathbf{q}, \mathbf{p}, t)$ at time $t$, to $(\boldsymbol{\beta}, \boldsymbol{\alpha})$, then transforming from $(\boldsymbol{\beta}, \boldsymbol{\alpha})$ to $(\mathbf{q}_0,\mathbf{p}_0, t_0)$. That is, this two-step process corresponds to

$$S_H(\mathbf{q}, t; \mathbf{q}_ot_o) = S(\mathbf{q}, \boldsymbol{\alpha}, t) − S(\mathbf{q}_0, \boldsymbol{\alpha}, t_0) \label{15.99}$$

Hamilton’s principle function $S_H(\mathbf{q}, t; \mathbf{q}_ot_o)$ is related to Jacobi’s complete integral $S(\mathbf{q}, \boldsymbol{\alpha}, t)$, and it will not be discussed further in this book.

Time-independent Hamiltonian

Frequently the Hamiltonian does not explicitly depend on time. For the standard Lagrangian with time-independent constraints and transformation, then $H (\mathbf{q}, \mathbf{p},t) = E$ which is the total energy. For this case, the Hamilton-Jacobi equation simplifies to give

$$\frac{\partial S}{ \partial t} = −H( \mathbf{ q}, \mathbf{ p}, t) = −E (\boldsymbol{\alpha}) \label{15.100}$$

The integration of the time dependence is trivial, and thus the action integral for a time-independent Hamiltonian equals

$$S(\mathbf{q}, \boldsymbol{\alpha},t) = W (\mathbf{q}, \boldsymbol{\alpha}) − E (\boldsymbol{\alpha})t \label{15.101}$$

That is, the action integral has separated into a time independent term $W (\mathbf{q}, \boldsymbol{\alpha})$ which is called Hamilton’s characteristic function plus a time-dependent term $−E (\boldsymbol{\alpha})t$. Thus using equations $\ref{15.97}$, $\ref{15.101}$ gives that the generalized momentum is

$$p_i = \frac{\partial W(\mathbf{q}, \boldsymbol{\alpha})}{ \partial q_i} \label{15.102}$$

The physical significance of Hamilton’s characteristic function $W (\mathbf{q}, \boldsymbol{\alpha})$ can be understood by taking the total time derivative

$$\frac{dW}{ dt} = \sum_i \frac{\partial W(\mathbf{q}, \boldsymbol{\alpha})}{ \partial q_i} \dot{q}_i = \sum_i p_i\dot{q}_i \nonumber$$

Taking the time integral then gives

$$W (\mathbf{q}, \boldsymbol{\alpha}) = \int \sum p_i\dot{q}_i dt =\int \sum p_idq_i \label{15.103}$$

Note that this equals the abbreviated action described in chapter $9.2.3$, that is $W(\mathbf{q}, \boldsymbol{\alpha}) = S_0(\mathbf{q}, \boldsymbol{\alpha})$.

Inserting the action $S (\mathbf{q}, \boldsymbol{\alpha})$ into the Hamilton-Jacobi equation $(15.2.1)$ gives

$$H(\mathbf{q}; \frac{\partial W(\mathbf{q}, \boldsymbol{\alpha})}{ \partial \mathbf{q}} ) = E (\boldsymbol{\alpha}) \label{15.104}$$

This is called the time-independent Hamilton-Jacobi equation. Usually it is convenient to have $E$ equal the total energy. However, sometimes it is more convenient to exclude the $k^{th}$ energy $E(\alpha_k)$ in the set, in which case $E = E(\alpha_1, \alpha_2, ...\alpha_k−1)$; the Routhian exploits this feature.

The equations of the canonical transformation expressed in terms of $W (\mathbf{q}, \boldsymbol{\alpha})$ are

$$p_i = \frac{\partial W(\mathbf{q}, \boldsymbol{\alpha}) }{\partial q_i } \quad \beta_i + \frac{\partial E(\boldsymbol{\alpha}) }{\partial \alpha_i} t = \frac{\partial W(\mathbf{q}, \boldsymbol{\alpha})}{ \partial \alpha_i} \label{15.105}$$

These equations show that Hamilton’s characteristic function $W (\mathbf{q}, \boldsymbol{\alpha})$ is itself the generating function of a time-independent canonical transformation from the old variables $(q, p)$ to a set of new variables

$$Q_i = \beta_i + \frac{\partial E(\boldsymbol{\alpha})}{ \partial \alpha_i } t \quad P_i = \alpha_i \label{15.106}$$

Table $\PageIndex{1}$ summarizes the time-dependent and time-independent forms of the Hamilton-Jacobi equation.

Table $\PageIndex{1}$: Hamilton-Jacobi formulations

Hamiltonian	Time dependent $H(q, p, t)$	Time independent $H(q, p)$
Transformed Hamiltonian	$\mathcal{H}= 0$	$\mathcal{H}$ is cyclic
Canonical transformed variables	All $Q_iP_i$ are constants of motion	All $P_i$ are constants of motion
Transformed equations of motion	$\dot{Q}_i = \frac{\partial \mathcal{H}}{ \partial P_i} = 0$, therefore $Q_i = \beta_i$ $\dot{P}_i = − \frac{\partial \mathcal{H}}{ \partial Q_i} = 0$, therefore $P_i = \alpha_i$	$\dot{Q}_i = \frac{\partial \mathcal{H}}{ \partial P_i} = v_i$, therefore $Q_i = v_i t + \beta_i$ $\dot{P}_i = − \frac{\partial \mathcal{H}} {\partial Q_i} = 0$, therefore $P_i = \alpha_i$
Generating function	Jacobi’s complete integral $S(\mathbf{q}, \mathbf{P}, t)$	Characteristic Function $W(\mathbf{q}, \mathbf{P})$
Hamilton-Jacobi equation	$H(q_1, ...q_n; \frac{\partial S}{ \partial q_1 }, ..., \frac{\partial S} {\partial q_n} ;t)+\frac{\partial S}{ \partial t} = 0$	$H(q_1, ...q_n; \frac{\partial W }{\partial q_1} , ..., \frac{\partial W}{ \partial q_n} ) = E$
Transformation equations	$p_i= \frac{\partial S}{ \partial q_i}$ $Q_i= \frac{\partial S}{ \partial \alpha_i} = \beta_i$	$p_i=\frac{\partial W}{ \partial q_i}$ $Q_i=\frac{\partial W}{ \partial \alpha_i} = v_i t + \beta_i$

Separation of variables

Exploitation of the Hamilton-Jacobi theory requires finding a suitable action function $S$. When the Hamiltonian is time independent, then Equation $\ref{15.101}$ shows that the time dependence of the action integral separates out from the dependence on the spatial variables. For many systems, the Hamilton’s characteristic function $W(\mathbf{q}, \mathbf{P})$ separates into a simple sum of terms each of which is a function of a single variable. That is,

$$W(\mathbf{q}, \boldsymbol{\alpha}) = W_1(q_1) + W_2(q_2) + \cdots \cdot \cdot W_n(q_n) \label{15.107}$$

where each function in the summation on the right depends only on a single variable. Then Equation $\ref{15.100}$ reduces to

$$H(q_1, ...q_n; \frac{\partial W }{\partial q_1} , ...,\frac{ \partial W}{ \partial q_n} ) = E \label{15.108}$$

where $E$ is the constant denoting the total energy.

Hamilton’s characteristic function $W( \mathbf{ q}, \mathbf{ P})$ can be used with equations $\ref{15.101}$, $\ref{15.102}$, $\ref{15.91}$, $\ref{15.92}$, and $\ref{15.93}$ to derive

$$p_i = \frac{\partial W( \mathbf{ q}, \boldsymbol{\alpha}) }{\partial q_i} \quad Q_i = \frac{\partial W( \mathbf{ q}, \boldsymbol{\alpha}) }{\partial P_i} \label{15.109}$$

$$\dot{Q}_i = \frac{\partial \mathcal{H}}{ \partial P_i} = 0 \quad \dot{P}_i = \frac{\partial \mathcal{H}}{ \partial Q_i} = 0 \label{15.110}$$

$$\mathcal{H} = H + \frac{\partial S}{\partial t} = H − E = 0 \label{15.111}$$

which has reduced the problem to a simple sum of one-dimensional first-order differential equations.

If the $i^{th}$ variable is cyclic, then the Hamiltonian is not a function of $q_i$ and the $i^{th}$ term in Hamilton’s characteristic function equals $W_i = \alpha_iq_i$ which separates out from the summation in Equation $\ref{15.107}$. That is, all cyclic variables can be factored out of $W( \mathbf{ q}, \boldsymbol{\alpha})$ which greatly simplifies solution of the Hamilton-Jacobi equation. As a consequence, the ability of the Hamilton-Jacobi method to make a canonical transformation to separate the system into many cyclic or independent variables, which can be solved trivially, is a remarkably powerful way for solving the equations of motion in Hamiltonian mechanics.

Example $\PageIndex{1}$: Free particle

Consider the motion of a free particle of mass $m$ in a force-free region. Then Equation $\ref{15.93}$ reduces to

$$H(q_1, ...q_n; \frac{\partial S}{ \partial q_1} , ..., \frac{\partial S}{ \partial q_n} ;t) + \frac{\partial S}{\partial t} = 0 \nonumber$$

Since no forces act, and the momentum $\mathbf{p} = \boldsymbol{\nabla}S$, thus the Hamilton-Jacobi equation reduces to

$$\frac{1}{ 2m } \nabla^2S + \frac{\partial S}{\partial t} = 0 \tag{A}\label{A}$$

The Hamiltonian is time independent, thus Equation $\ref{15.101}$ applies

$$S(\mathbf{q}, t) = W(\mathbf{q}, \boldsymbol{\alpha}) − E(\boldsymbol{\alpha})t \nonumber$$

Since the Hamiltonian does not explicitly depend on the coordinates $(x, y, z)$, then the coordinates are cyclic and separation of the variables, $\ref{15.107}$, gives that the action

$$S = \boldsymbol{\alpha} \cdot \mathbf{ r} − Et \tag{B}\label{B}$$

For Equation $\ref{B}$ to be a solution of Equation $\ref{A}$ requires that

$$E = \frac{1}{ 2m} \boldsymbol{\alpha}^2 \tag{C}\label{C}$$

Therefore

$$S = \boldsymbol{\alpha} \cdot \mathbf{r} − \frac{1}{ 2m} \boldsymbol{\alpha}^2t \tag{D}\label{D}$$

Since

$$\mathbf{\dot{Q}} = \frac{\partial S }{\partial \boldsymbol{\alpha} } = \mathbf{r}− \frac{\boldsymbol{\alpha}}{ m }t \nonumber$$

the equation of motion and the conjugate momentum are given by

$$\mathbf{r} = \mathbf{\dot{Q}} + \frac{\boldsymbol{\alpha}}{ m} t \quad \mathbf{p} = \boldsymbol{\nabla}S = \boldsymbol{\alpha} \nonumber$$

Thus the Hamilton-Jacobi relation has given both the equation of motion and the linear momentum $\mathbf{p}$.

Example $\PageIndex{2}$: Point particle in a uniform gravitational field

The Hamiltonian is

$$H = \frac{1}{ 2m} (p^2_x + p^2_y + p^2_z) + mgz \nonumber$$

Since the system is conservative, then the Hamilton-Jacobi equation can be written in terms of Hamilton’s characteristic function $W$

$$E = \frac{1}{ 2m} \left[\left(\frac{\partial W}{ \partial x} \right)^2 + \left(\frac{\partial W}{ \partial y} \right)^2 + \left(\frac{\partial W}{ \partial z} \right)^2 \right] + mgz \nonumber$$

Assuming that the variables can be separated $W = X(x) + Y (y) + Z(z)$ leads to

$$p_x = \frac{\partial X(x)}{ \partial x} = \alpha_x \nonumber$$

$$p_y = \frac{\partial Y (y)}{ \partial y} = \alpha_y \nonumber$$

$$p_z = \frac{\partial Z(z) }{\partial z} = \sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y} \nonumber$$

Thus by integration the total $W$ equals

$$W = \int^x_{x_0} \alpha_x dx + \int^y_{y_0} \alpha_ydy + \int^z_{z_0} \left(\sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y }\right) dz \nonumber$$

Therefore using $\ref{15.106}$ gives

$$\beta_z = t − t_0 = \int^z_{z_0} \frac{mdz}{ \sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y } } \nonumber$$

$$\beta_x = \text{ constant }= (x − x_0) − \int^z_{z_0} \frac{\alpha_xdz}{ \sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y } } \nonumber$$

$$\beta_y = \text{ constant } = (y − y_0) − \int^z_{z_0} \frac{\alpha_ydz }{\sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y }} \nonumber$$

If $x_0, y_0, z_0$ is the position of the particle at time $t = t_0$ then $\beta_x = \beta_y = 0$, and from $\ref{15.106}$

$$x − x_0 = \left(\frac{\alpha_x}{ m }\right) (t − t_0) \nonumber$$

$$y − y_0 = \left(\frac{\alpha_y}{ m} \right) (t − t_0) \nonumber$$

$$z − z_0 = \left( \frac{\sqrt{ 2m(E − mgz) − \alpha^2_x − \alpha^2_y }}{ m} \right) (t − t_0) − \frac{1}{ 2} g(t − t_0)^2 \nonumber$$

This corresponds to a parabola as should be expected for this trivial example.

Example $\PageIndex{3}$: One-dimensional harmonic oscillator

As discussed in example $15.3.5$ the Hamiltonian for the one-dimensional harmonic oscillator can be written as

$$H = \frac{1}{ 2m} ( p^2 + m^2\omega^2q^2) = E \nonumber$$

assuming it is conservative and where $\omega = \sqrt{\frac{k}{m}}$.

Hamilton’s characteristic function $W$ can be used where

$$S (q, E, t) = W (q, E) − Et \nonumber$$

$$p_i = \frac{\partial W}{ \partial q_i} \nonumber$$

Inserting the generalized momentum $p_i$ into the Hamiltonian gives

$$\frac{1}{ 2m} \left(\left[ \frac{\partial W }{\partial q} \right]^2 + m^2\omega^2q^2 \right) = E \nonumber$$

Integration of this equation gives

$$W = \sqrt{ 2mE} \int dq \sqrt{1 − \frac{m\omega^2q^2}{ 2E}} \nonumber$$

That is

$$S = \sqrt{ 2mE } \int dq \sqrt{ 1 − \frac{m\omega^2q^2}{ 2E}} − Et \nonumber$$

Note that

$$\frac{\partial S(q, E, t)}{ \partial E} = \sqrt{\frac{2m }{E}} \int \frac{dq}{\sqrt{1 - \frac{ m\omega^2q^2}{ 2E}}} − t \nonumber$$

This can be integrated to give

$$t = \frac{1}{ \omega }\arcsin \left( q \sqrt{\frac{m\omega^2}{ 2E}}\right) + t_0 \nonumber$$

That is

$$q = \sqrt{\frac{2E}{m\omega^2}} \sin \omega (t − t_0) \nonumber$$

This is the familiar solution of the undamped harmonic oscillator.

Example $\PageIndex{4}$: The central force problem

The problem of a particle acted upon by a central force occurs frequently in physics. Consider the mass $m$ acted upon by a time-independent central potential energy $U(r)$. The Hamiltonian is time independent and can be written in spherical coordinates as

$$H = \frac{1}{ 2m} \left( p^2_r + \frac{1}{r^2} p^2_{\theta} + \frac{1}{r^2 \sin^2 \theta} p^2_{\psi} \right) + U(r) = E \nonumber$$

The time-independent Hamilton-Jacobi equation is conservative, thus

$$\frac{1}{ 2m} \left[\left(\frac{\partial W}{ \partial r }\right)^2 + \frac{1}{ r^2} \left(\frac{\partial W }{\partial \theta} \right)^2 + \frac{1}{ r^2 \sin^2 \theta} \left(\frac{\partial W}{ \partial \phi} \right)^2 \right] + U(r) = E \nonumber$$

Try a separable solution for Hamilton’s characteristic function $W$ of the form

$$W = R(r) + \Theta (\theta ) + \Phi (\phi ) \nonumber$$

The Hamilton-Jacobi equation then becomes

$$\frac{1}{ 2m} \left[\left(\frac{\partial R} {\partial r} \right)^2 + \frac{1}{ r^2} \left(\frac{\partial \Theta}{ \partial \theta} \right)^2 + \frac{1}{ r^2 \sin^2 \theta} \left(\frac{\partial \Phi}{ \partial \phi } \right)^2 \right] + U(r) = E \nonumber$$

This can be rearranged into the form

$$2mr^2 \sin^2 \theta \left\{ \frac{1}{ 2m} \left[\left(\frac{\partial R}{ \partial r} \right)^2 + \frac{1} {r^2} \left(\frac{\partial \Theta}{ \partial \theta} \right)^2 \right] + U(r) + E \right\} = − \left(\frac{\partial \Phi}{ \partial \phi} \right)^2 \nonumber$$

The left-hand side is independent of $\phi$ whereas the right-hand side is independent of $r$ and $\theta$. Both sides must equal a constant which is set to equal $−L^2_z$, that is

$$\frac{1}{2m} \left[\left(\frac{\partial R}{ \partial r} \right)^2 + \frac{1}{ r^2} \left(\frac{\partial \Theta}{ \partial \theta} \right)^2 \right] + U(r) + \frac{L^2_z }{2mr^2 \sin^2 \theta} = E \nonumber$$

$$\left(\frac{\partial \Phi}{ \partial \phi} \right)^2 = L^2_z \nonumber$$

The equation in $r$ and $\theta$ can be rearranged in the form

$$2mr^2 \left[ \frac{1}{2m} \left(\frac{\partial R}{ \partial r} \right)^2 + U(r) − E \right] = − \left[\left(\frac{\partial \Theta}{ \partial \theta} \right)^2 + \frac{L^2_z}{ \sin^2 \theta} \right] \nonumber$$

The left-hand side is independent of $\theta$ and the right-hand side is independent of $r$ so both must equal a constant which is set to be $−L^2$

$$\frac{1}{2m} \left(\frac{\partial R}{ \partial r} \right)^2 + U(r) + \frac{L^2}{ 2mr^2} = E \nonumber$$

$$\left(\frac{\partial \Theta}{ \partial \theta} \right)^2 + \frac{L^2_z}{ \sin^2 \theta} = L^2 \nonumber$$

The variables now are completely separated and, by rearrangement plus integration, one obtains

$$R(r) = \sqrt{2m} \int \sqrt{ E − U(r) − \frac{L^2 }{2mr^2}} dr \nonumber$$

$$\Theta (\theta ) = \int \sqrt{ L^2 − \frac{L^2_z}{ \sin^2 \theta}} d\theta \nonumber$$

$$\Phi (\phi ) = L_z \phi \nonumber$$

Substituting these into $W = R(r) + \Theta (\theta ) + \Phi (\phi )$ gives

$$W = \sqrt{2m} \int \sqrt{ E − U(r) − \frac{L^2 }{2mr^2}} dr + \int \sqrt{ L^2 − \frac{L^2_z}{ \sin^2 \theta}} d\theta + L_z \phi \nonumber$$

Hamilton’s characteristic function $W$ is the generating function from coordinates $(r, \theta , \phi , p_r, p_{\theta} , p_{\phi} )$ to new coordinates, which are cyclic, and new momenta that are constant and taken to be the separation constants $E, L, L_z$.

$$p_r = \frac{\partial W}{ \partial r} = \sqrt{2m} \sqrt{ E − U(r) − \frac{L^2 }{2mr^2}} \nonumber$$

$$p_{\theta} = \frac{\partial W}{ \partial \theta} = \sqrt{ L^2 − \frac{L^2_z}{ \sin^2 \theta}} \nonumber$$

$$p_{\phi} = \frac{\partial W}{ \partial \phi} = L_z \nonumber$$

Similarly, using $\ref{15.109}$ gives the new coordinates $E, L, L_z$

$$\beta_E + t = \frac{\partial W}{ \partial E} = \sqrt{\frac{m}{ 2}} \int \frac{dr}{\sqrt{ E − U(r) − \frac{L^2}{ 2mr^2}}} \nonumber$$

$$\beta_L = \frac{\partial W}{ \partial L} = \sqrt{2m} \int \frac{dr}{\sqrt{E − U(r) − \frac{L^2}{ 2mr^2}}} \left( \frac{−L }{2mr^2} \right) + \int \frac{Ld\theta}{\sqrt{ L^2 − \frac{L^2_z }{\sin^2 \theta}}} \nonumber$$

$$\beta_{L_z} = \frac{\partial W}{ \partial L_z} = \int \frac{d\theta}{\sqrt{ L^2 − \frac{L^2_z}{ \sin^2 \theta}}} \left( \frac{−L}{ 2mr^2} \right) + \phi \nonumber$$

These equations lead to the elliptical, parabolic, or hyperbolic orbits discussed in chapter $11$.

Example $\PageIndex{5}$: Linearly-damped, one-dimensional, harmonic oscillator

A canonical treatment of the linearly-damped harmonic oscillator provides an example that combines use of non-standard Lagrangian and Hamiltonians, a canonical transformation to an autonomous system, and use of Hamilton-Jacobi theory to solve this transformed system. It shows that Hamilton-Jacobi theory can be used to determine directly the solutions for the linearly-damped harmonic oscillator.

Non-standard Hamiltonian:

In chapter $3.5$, the equation of motion for the linearly-damped, one-dimensional, harmonic oscillator was given to be

$$\frac{m}{ 2} [ \ddot{q}+ \Gamma \dot{q} + \omega^2_0 q ] = 0 \tag{a}\label{a}$$

Example $10.5.1$ showed that three non-standard Lagrangians give equation of motion $\alpha$ when used with the standard Euler-Lagrange variational equations. One of these was the Bateman[Bat31] time-dependent Lagrangian

$$L_2 (q, \dot{q}, t ) = \frac{m}{ 2} e^{\Gamma t} [ \dot{q}^2 − \omega^2_0 q^2] \tag{b}\label{b}$$

This Lagrangian gave the generalized momentum to be

$$p = \frac{\partial L^2}{ \partial \dot{q}} = m\dot{q} e^{\Gamma t} \tag{c}\label{c}$$

which was used with equation $(15.1.3)$ to derive the Hamiltonian

$$H_2(q, p, t) = p\dot{q} − L_2(q, \dot{q}, t ) = e^{−\Gamma t} \frac{p^2}{ 2m} + \frac{1}{ 2} m\omega^2_0 q^2e^{\Gamma t} \tag{d}\label{d1}$$

Note that both the Lagrangian and Hamiltonian are explicitly time dependent and thus they are not conserved quantities. This is as expected for this dissipative system.

Hamilton-Jacobi theory:

The form of the non-autonomous Hamiltonian $\ref{d1}$ suggests use of the generating function for a canonical transformation to an autonomous Hamiltonian, for which $H$ is a constant of motion.

$$S(q, P, t) = F_2(q, P, t) = qPe^{ \frac{\Gamma t}{ 2}} = QP \tag{d}\label{d2}$$

Then the canonical transformation gives

$$p = \frac{\partial S}{ \partial q} = P e^{\frac{ \Gamma t}{ 2}} \label{e}\tag{e}$$

$$Q = \frac{\partial S}{ \partial P} = qe^{\frac{ \Gamma t}{ 2}} \nonumber$$

Insert this canonical transformation into the above Hamiltonian leads to the transformed Hamiltonian that is autonomous.

$$\mathcal{H}(Q, P, t) = H_2(q, p, t) + \frac{\partial F_2}{ \partial t} = \frac{P^2}{ 2m} + \frac{\Gamma}{ 2} QP + \frac{m\omega^2_0}{ 2} Q^2 \tag{f}\label{f}$$

That is, the transformed Hamiltonian $\mathcal{H}(Q, P, t)$ is not explicitly time dependent, and thus is conserved. Expressed in the original canonical variables $(q, p)$, the transformed Hamiltonian $\mathcal{H}(Q, P, t)$

$$\mathcal{H}(Q, P, t)= \frac{p^2}{ 2m } e^{−\Gamma t }+ \frac{\Gamma }{2} qp + \frac{m\omega^2_0 }{2} q2e^{\Gamma t }\nonumber$$

is a constant of motion which was not readily apparent when using the original Hamiltonian. This unexpected result illustrates the usefulness of canonical transformations for solving dissipative systems. The Hamilton-Jacobi theory now can be used to solve the equations of motion for the transformed variables $(Q, P)$ plus the transformed Hamiltonian $\mathcal{H}(Q, P, t)$. The derivative of the generating function

$$\frac{\partial S}{ \partial Q} = P \label{g}\tag{g}$$

Use Equation $\ref{g}$ to substitute for $P$ in the Hamiltonian $\mathcal{H}(Q, P, t)$ (Equation $\ref{f}$), then the Hamilton-Jacobi method gives

$$\frac{1}{2m} \left( \frac{\partial S }{\partial Q} \right)^2 + \frac{\Gamma}{ 2} Q \frac{\partial S}{ \partial Q} + \frac{m\omega^2_0}{ 2} Q^2 + \frac{\partial S}{\partial t} = 0 \nonumber$$

This equation is separable as described in $\ref{15.107}$ and thus let

$$S(Q, \alpha , t) = W(Q, \alpha ) − \alpha t \nonumber$$

where $\alpha$ is a separation constant. Then

$$\left[ \frac{1}{2m} \left(\frac{\partial W }{\partial Q} \right)^2 + \Gamma Q\frac{\partial W}{ \partial Q} + \frac{m\omega^2_0}{ 2} Q^2 \right] = \alpha \tag{h}\label{h}$$

To simplify the equations define the variable x as

$$x \equiv \sqrt{m\omega_0} Q \label{i}\tag{i}$$

then Equation $\ref{h}$ can be written as

$$\left(\frac{\partial W}{ \partial x} \right)^2 + Ax\frac{\partial W}{ \partial x} + (x^2 − B ) = 0 \tag{j}\label{j}$$

where $A = \frac{\Gamma}{ \omega_0}$ and $B = \frac{2\alpha}{\omega_0}$. Assume initial conditions $q(0) = q_0$ and $\dot{q}(0) = 0$

For this case the separation constant $\alpha > 0$, therefore $B > 0$. Note that Equation $\ref{j}$ is a simple second-order algebraic relation, the solution of which is

$$\frac{\partial W}{ \partial x} = −\frac{\alpha x}{ 2} \pm \sqrt{B − \left[ 1 − \left(\frac{A}{ 2} \right)^2 \right] x^2} \label{k}\tag{k}$$

The choice of the sign is irrelevant for this case and thus the positive sign is chosen. There are three possible cases for the solution depending on whether the square-root term is real, zero, or imaginary.

Case 1: $\frac{A}{ 2} < 1$, that is, $\frac{\lambda }{2m\omega_0 }< 1$

Define $C = \sqrt{\left[ 1 − ( \frac{A}{ 2} )^2 \right]}$ Then Equation $\ref{k}$ can be integrated to give

$$S = −\alpha t − \frac{Ax^2}{ 4} + \int \sqrt{(B − C^2x^2)}dx \tag{l}\label{l}$$

and

$$\beta = \frac{\partial S}{ \partial \alpha} = −t + \frac{1}{ \omega_0 } \int \frac{ dx }{\sqrt{(B − C^2x^2)}} \nonumber$$

This integral gives

$$sin^{−1} \left( \frac{Cx}{ \sqrt{B}} \right) = C\omega_0 (t + \beta ) \equiv \omega t + \delta \nonumber$$

where

$$\omega = \omega_0 C = \omega_0 \sqrt{1 − \left( \frac{\Gamma }{2\omega_0} \right)^2} = \sqrt{ \omega^2_0 − \left(\frac{\Gamma }{2} \right)^2} \label{m}\tag{m}$$

Transforming back to the original variable $q$ gives

$$q(t) = Ge^{−\frac{ \Gamma t }{2}} \sin (\omega t + \delta ) \tag{n}\label{n}$$

where $G$ and $\delta$ are given by the initial conditions. Equation $\ref{m}$ is identical to the solution for the underdamped linearly-damped linear oscillator given previously in equation $(3.5.12)$.

Case 2: $\frac{A}{ 2} = 1$, that is, $\frac{\Gamma }{2\omega_0} = 1$

In this case $C = \sqrt{\left[ 1 − ( \frac{A}{ 2} )^2 \right]} = 0$ and thus Equation $\ref{k}$ simplifies to

$$S = −\alpha t − \frac{Ax^2}{ 4} + x \sqrt{B} \nonumber$$

and

$$\beta = \frac{\partial S}{ \partial \alpha} = −t + \frac{x}{ \omega_0 \sqrt{B}} \nonumber$$

Therefore the solution is

$$q(t) = e^{− \frac{\Gamma t}{ 2}} (F + Gt) \label{o}\tag{o}$$

where $F$ and $G$ are constants given by the initial conditions. This is the solution for the critically-damped linearly-damped, linear oscillator given previously in equation $(3.5.15)$.

Case 3: $\frac{A}{ 2} > 1$, that is, $\frac{\Gamma }{2\omega_0} > 1$

Define a real constant $D$ where $D = \sqrt{\left[ ( \frac{A}{ 2} )^2 - 1\right]} = iC$, then

$$S = −\alpha t − \frac{Ax^2}{ 4} + \int \sqrt{(B + D^2x^2)}dx \nonumber$$

Then

$$\beta = \frac{\partial S}{ \partial \alpha} = −t + \frac{1}{ \omega_0 } \int \frac{dx}{\sqrt{(B + D^2x^2)}} \nonumber$$

This last integral gives

$$\sinh^{−1} \left( \frac{Dx}{ \sqrt{B}} \right) = D\omega_0 (t + \beta ) \equiv \omega t + \delta \nonumber$$

where

$$\omega = \omega_0C = \omega_0 \sqrt{\left( \frac{\lambda }{2m\omega_0} \right)^2 − 1} \nonumber$$

Then the original variable gives

$$q(t) = Ge^{− \frac{\Gamma t}{ 2 }} \sinh (\omega t + \delta ) \tag{l}\label{l2} \nonumber$$

This is the classic solution of the overdamped linearly-damped, linear harmonic oscillator given previously in equation $(3.5.14)$​​​​​​. The canonical transformation from a non-autonomous to an autonomous system allowed use of Hamiltonian mechanics to solve the damped oscillator problem.

Note that this example used Bateman’s non-standard Lagrangian, and corresponding Hamiltonian, for handling a dissipative linear oscillator system where the dissipation depends linearly on velocity. This nonstandard Lagrangian led to the correct equations of motion and solutions when applied using either the time-dependent Lagrangian, or time-dependent Hamiltonian, and these solutions agree with those given in chapter $3.5$ which were derived using Newtonian mechanics.

Visual representation of the action function $S$.

Figure $\PageIndex{1}$: Surfaces of constant action integral S (dashed lines) and the corresponding particle momenta (solid lines) with arrows showing the direction.

The important role of the action integral $S$ can be illuminated by considering the case of a single point mass $m$ moving in a time independent potential $U(r)$. Then the action reduces to

$$S(q, \alpha , t) = W(q, \alpha ) − Et \label{15.112}$$

Let $q_1 = x, q_2 = y, q_3 = z, p_1 = p_x, p_2 = p_y, p_3 = p_z$. The momentum components are given by

$$p_i = \frac{\partial W(q, \alpha ) }{\partial q_i} \label{15.113}$$

which corresponds to

$$\mathbf{p} = \boldsymbol{\nabla}W = \boldsymbol{\nabla}S \label{15.114}$$

That is, the time-independent Hamilton-Jacobi equation is

$$\frac{1}{2m} |\boldsymbol{\nabla}W|^2 + U(r) = E \label{15.115}$$

This implies that the particle momentum is given by the gradient of Hamilton’s characteristic function and is perpendicular to surfaces of constant $W$ as illustrated in Figure $\PageIndex{1}$. The constant $W$ surfaces are time dependent as given by Equation $\ref{15.101}$. Thus, if at time $t = 0$ the equi-action surface $S_0(q, t) = W_0(q, P_i)=0$, then at $t = 1$ the same surface $S_0(q, t)=0$ now coincides with the $S_0(q, t) = E$ surface etc. That is, the equi-action surfaces move through space separately from the motion of the single point mass.

The above pictorial representation is analogous to the situation for motion of a wavefront for electromagnetic waves in optics, or matter waves in quantum physics where the wave equation separates into the form $\phi = \phi_0 e^{\frac{ iS}{ \hbar }} = \phi_0 e^{i(\mathbf{k} \cdot \mathbf{r}−\omega t)}$. Hamilton’s goal was to create a unified theory for optics that was equally applicable to particle motion in classical mechanics. Thus the optical-mechanical analogy of the Hamilton-Jacobi theory has culminated in a universal theory that describes wave-particle duality; this was a Holy Grail of classical mechanics since Newton’s time. It played an important role in development of the Schrödinger representation of quantum mechanics.

Advantages of Hamilton-Jacobi theory

Initially, only a few scientists, like Jacobi, recognized the advantages of Hamiltonian mechanics. In 1843 Jacobi made some brilliant mathematical developments in Hamilton-Jacobi theory that greatly enhanced exploitation of Hamiltonian mechanics. Hamilton-Jacobi theory now serves as a foundation for contemporary physics, such as quantum and statistical mechanics. A major advantage of Hamilton-Jacobi theory, compared to other formulations of analytic mechanics, is that it provides a single, first-order partial differential equation for the action $S$, which is a function of the $n$ generalized coordinates $\mathbf{q}$ and time $t$. The generalized momenta no longer appear explicitly in the Hamiltonian in equations $\ref{15.94}$, $\ref{15.95}$. Note that the generalized momentum do not explicitly appear in the equivalent Euler-Lagrange equations of Lagrangian mechanics, but these comprise a system of $n$ second-order, partial differential equations for the time evolution of the generalized coordinate $\mathbf{q}$. Hamilton’s equations of motion are a system of $2n$ first-order equations for the time evolution of the generalized coordinates and their conjugate momenta.

An important advantage of the Hamilton-Jacobi theory is that it provides a formulation of classical mechanics in which motion of a particle can be represented by a wave. In this sense, the Hamilton-Jacobi equation fulfilled a long-held goal of theoretical physics, that dates back to Johann Bernoulli, of finding an analogy between the propagation of light and the motion of a particle. This goal motivated Hamilton to develop Hamiltonian mechanics. A consequence of this wave-particle analogy is that the Hamilton-Jacobi formalism featured prominently in the derivation of the Schrödinger equation during the development of quantum-wave mechanics.