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# 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.

 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 $$S$$.

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.

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.