Hamilton–Jacobi Equation
The Hamilton–Jacobi formulation rewrites dynamics in terms of a single function $S$, called Hamilton’s principal function. Instead of solving the equations of motion directly, one solves a first-order partial differential equation for $S$ and then extracts the trajectory from it. This method is powerful because it connects variational mechanics, Hamiltonian mechanics, and the integration of motion in one framework.
Hamilton’s principal function and the Hamilton–Jacobi equation
For a system with generalized coordinates $q_i$, Hamilton’s principal function is defined as the action evaluated along the true path from a fixed initial point $(q_{i0}, t_0)$ to the final point $(q_i, t)$:
\[S(q_i,t)=\int_{t_0}^{t} L(q_i,\dot q_i,\tau)\,d\tau\]Since the upper endpoint is $(q_i,t)$, $S$ is a function of the endpoint variables. Its differential along the true path is
\[dS=\sum_i p_i\,dq_i-H\,dt\]where
\[p_i=\frac{\partial L}{\partial \dot q_i}, \qquad H=\sum_i p_i\dot q_i-L\]Comparing this with the ordinary differential of $S(q_i,t)$,
\[dS=\sum_i \frac{\partial S}{\partial q_i}dq_i+\frac{\partial S}{\partial t}dt\]we obtain the fundamental relations
\[\frac{\partial S}{\partial q_i}=p_i, \qquad \frac{\partial S}{\partial t}=-H\]Since the Hamiltonian is originally $H(q_i,p_i,t)$, substituting $p_i=\partial S/\partial q_i$ gives
\[H=H\left(q_i,\frac{\partial S}{\partial q_i},t\right)\]Hence the Hamilton–Jacobi equation becomes
\[\frac{\partial S}{\partial t}+H\left(q_i,\frac{\partial S}{\partial q_i},t\right)=0\]This is the Hamilton–Jacobi equation. It replaces Hamilton’s $2n$ first-order equations by one first-order partial differential equation for $S$.
For time-independent Hamiltonians, one usually writes
\[S(q_i,\alpha_i,t)=W(q_i,\alpha_i)-Et\]where $W$ is Hamilton’s characteristic function. Then the Hamilton–Jacobi equation reduces to
\[H\left(q_i,\frac{\partial W}{\partial q_i}\right)=E\]Origin of the constants $\alpha$ and $\beta$
A complete solution of the Hamilton–Jacobi equation for one degree of freedom contains one arbitrary parameter. Write it as
\[S(q,\alpha,t)\]The constant $\alpha$ enters because the Hamilton–Jacobi equation is a first-order partial differential equation, and its complete integral must contain an arbitrary constant parameter. For time-independent systems, this parameter is usually chosen to be the energy, so $\alpha=E$.
The motion of one degree of freedom is governed by a second-order differential equation, so the final trajectory must contain two independent constants. One of them is already $\alpha$. The second arises from the quantity $\partial S/\partial \alpha$, whose constancy can be proved directly.
Start from the Hamilton–Jacobi equation
\[\frac{\partial S}{\partial t}+H\left(q,\frac{\partial S}{\partial q},t\right)=0\]Differentiate it with respect to $\alpha$:
\[\frac{\partial^2 S}{\partial t\,\partial \alpha}+\frac{\partial H}{\partial p}\frac{\partial^2 S}{\partial q\,\partial \alpha}=0\]because $H$ depends on $\alpha$ only through
\[p=\frac{\partial S}{\partial q}\]Now along the actual motion, Hamilton’s equation gives
\[\dot q=\frac{\partial H}{\partial p}\]Therefore
\[\frac{\partial^2 S}{\partial t\,\partial \alpha}+\dot q\,\frac{\partial^2 S}{\partial q\,\partial \alpha}=0\]But the total derivative of $\partial S/\partial \alpha$ along the motion is
\[\frac{d}{dt}\left(\frac{\partial S}{\partial \alpha}\right)=\frac{\partial^2 S}{\partial q\,\partial \alpha}\dot q+\frac{\partial^2 S}{\partial t\,\partial \alpha}\]Hence
\[\frac{d}{dt}\left(\frac{\partial S}{\partial \alpha}\right)=0\]So $\partial S/\partial \alpha$ is conserved along the orbit. Therefore we set
\[\frac{\partial S}{\partial \alpha}=\beta\]where $\beta$ is the second constant of motion. Thus for one degree of freedom, the two constants are $\alpha$ and $\beta$. In the time-independent case with $\alpha=E$,
\[\beta=\frac{\partial S}{\partial E}=\frac{\partial W}{\partial E}-t\]This relation yields the trajectory.
Simple harmonic oscillator
Consider the one-dimensional harmonic oscillator with Hamiltonian
\[H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 q^2\]The Hamilton–Jacobi equation is
\[\frac{\partial S}{\partial t}+\frac{1}{2m}\left(\frac{\partial S}{\partial q}\right)^2+\frac{1}{2}m\omega^2 q^2=0\]Since the Hamiltonian is time-independent, take
\[S(q,t)=W(q,E)-Et\]Then
\[\frac{\partial S}{\partial t}=-E, \qquad \frac{\partial S}{\partial q}=\frac{dW}{dq}\]Substitution gives
\[\frac{1}{2m}\left(\frac{dW}{dq}\right)^2+\frac{1}{2}m\omega^2 q^2=E\]Thus
\[\left(\frac{dW}{dq}\right)^2=2mE-m^2\omega^2 q^2\]and hence
\[\frac{dW}{dq}=\sqrt{2mE-m^2\omega^2 q^2}\]Therefore
\[W(q,E)=\int \sqrt{2mE-m^2\omega^2 q^2}\,dq\]Let
\[A^2=\frac{2E}{m\omega^2}\]Then
\[\sqrt{2mE-m^2\omega^2 q^2}=m\omega\sqrt{A^2-q^2}\]so
\[W(q,E)=m\omega\int \sqrt{A^2-q^2}\,dq\]Using the standard integral,
\[\int \sqrt{A^2-q^2}\,dq=\frac{1}{2}\left[q\sqrt{A^2-q^2}+A^2\sin^{-1}\left(\frac{q}{A}\right)\right]\]we obtain
\[W(q,E)=\frac{m\omega}{2}\left[q\sqrt{A^2-q^2}+A^2\sin^{-1}\left(\frac{q}{A}\right)\right]\]Hence the principal function is
\[S(q,t)=\frac{m\omega}{2}\left[q\sqrt{A^2-q^2}+A^2\sin^{-1}\left(\frac{q}{A}\right)\right]-Et\]Now the second constant is obtained from
\[\beta=\frac{\partial S}{\partial E}\]Since $A^2=2E/(m\omega^2)$, differentiation gives
\[\beta=\frac{1}{\omega}\sin^{-1}\left(\frac{q}{A}\right)-t\]Rearranging,
\[\sin^{-1}\left(\frac{q}{A}\right)=\omega(t+\beta)\]so
\[\frac{q}{A}=\sin(\omega t+\phi)\]where $\phi=\omega\beta$ is a constant phase. Therefore the trajectory is
\[q(t)=A\sin(\omega t+\phi)\]with
\[A=\sqrt{\frac{2E}{m\omega^2}}\]Thus the Hamilton–Jacobi method reproduces the familiar harmonic oscillator motion. The constant $\alpha$ is the energy $E$, and the second constant $\beta$ appears from the conserved quantity $\partial S/\partial \alpha$.
Main points
- Hamilton’s principal function is the action evaluated on the true path.
- From $dS=\sum_i p_i\,dq_i-H\,dt$, one obtains $\partial S/\partial q_i=p_i$ and $\partial S/\partial t=-H$.
- Substituting $p_i=\partial S/\partial q_i$ into the Hamiltonian gives the Hamilton–Jacobi equation.
- A complete solution $S(q,\alpha,t)$ contains a parameter $\alpha$.
- The quantity $\partial S/\partial \alpha$ is constant along the motion because its total time derivative vanishes.
- Hence the two constants of motion for one degree of freedom are $\alpha$ and $\beta$.
- For the harmonic oscillator, this procedure leads directly to $q(t)=A\sin(\omega t+\phi)$.
Practice questions
- Starting from $dS=p\,dq-H\,dt$, derive the Hamilton–Jacobi equation for one degree of freedom.
- Prove directly that $d(\partial S/\partial \alpha)/dt=0$ along the actual motion.
- Solve the Hamilton–Jacobi equation for a free particle in one dimension.
- For the harmonic oscillator, show that $\beta=\partial S/\partial E$ leads to the sinusoidal trajectory.
- Explain clearly the difference between Hamilton’s principal function $S$ and Hamilton’s characteristic function $W$.