Hamilton–Jacobi Equation with Example of Harmonic Oscillator
Learning Objectives:
- Understand the formulation and significance of the Hamilton–Jacobi equation in classical mechanics.
- Learn how to reduce the problem of solving equations of motion to solving a partial differential equation.
- Apply the Hamilton–Jacobi method to solve the harmonic oscillator problem.
Key Concepts / Definitions:
- Hamilton–Jacobi Equation: A first-order partial differential equation for the action function $S(q, t)$ derived from Hamilton’s equations.
- Action Function ($S$): A function whose complete solution generates the equations of motion via canonical transformation.
- Harmonic Oscillator: A system in which a particle experiences a restoring force proportional to its displacement.
Theory and Explanation:
The Hamilton–Jacobi equation (HJE) is an alternative formulation of classical mechanics that can be used to solve the equations of motion by solving a partial differential equation. It is especially powerful because it reduces the problem of dynamics to integration.
The central idea is to find a generating function $S(q, t)$ such that the transformation to new coordinates results in constant generalized momenta. This function $S$ satisfies the Hamilton–Jacobi equation:
\[\frac{\partial S}{\partial t} + H\left(q, \frac{\partial S}{\partial q}, t\right) = 0\]If $S(q, t)$ is known, then the equations of motion can be obtained by simple differentiation, and the trajectories can be determined directly.
Mathematical Formulation:
Let the Hamiltonian of a system be:
\[H(q, p, t)\]We define the action function $S(q, \alpha, t)$, where $\alpha$ are constants of integration. The canonical momentum is:
\[p_i = \frac{\partial S}{\partial q_i}\]Then the Hamilton–Jacobi equation is:
\[\frac{\partial S}{\partial t} + H\left(q, \frac{\partial S}{\partial q}, t\right) = 0\]For time-independent systems, we use separation of variables:
\[S(q, t) = W(q) - Et\]Substituting into HJE gives:
\[H\left(q, \frac{\partial W}{\partial q}\right) = E\]Solved Examples:
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Example 1:
Problem: Derive the Hamilton–Jacobi equation for a 1D harmonic oscillator and find the action function.
Solution:The Hamiltonian of a 1D harmonic oscillator is:
\[H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 q^2\]Assume $S(q, t) = W(q) - Et$, then the time-independent HJE becomes:
\[\frac{1}{2m} \left( \frac{dW}{dq} \right)^2 + \frac{1}{2} m \omega^2 q^2 = E\]Solving:
\[\left( \frac{dW}{dq} \right)^2 = 2m \left( E - \frac{1}{2} m \omega^2 q^2 \right)\] \[\frac{dW}{dq} = \sqrt{2mE - m^2 \omega^2 q^2}\]Integrating:
\[W(q) = \int \sqrt{2mE - m^2 \omega^2 q^2} \, dq\]Let $A^2 = \frac{2E}{m \omega^2}$, then:
\[W(q) = \frac{E}{\omega} \arcsin \left( \frac{q}{A} \right) + \frac{m \omega}{2} q \sqrt{A^2 - q^2}\]Hence, the full action is:
\[S(q, t) = W(q) - Et\]
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Example 2:
Problem: Using the Hamilton–Jacobi method, find the trajectory of a particle in a harmonic oscillator potential.
Solution:From the previous example, we know $S(q, t) = W(q) - Et$.
From the Hamilton–Jacobi method:
\[\frac{\partial S}{\partial E} = \text{constant} = \beta\]The statement “$\partial S / \partial E$ is constant” means:
Along a trajectory governed by the Hamilton-Jacobi equation, if you consider S = W - E t, then the change in S with respect to E is linear in time and independent of q, so its partial derivative with respect to E is constant.
So,
\[\frac{\partial W}{\partial E} - t = \beta\]Using $W(q)$ from before:
\[\frac{\partial W}{\partial E} = \frac{1}{\omega} \arcsin \left( \frac{q}{A} \right)\]Therefore:
\[\frac{1}{\omega} \arcsin \left( \frac{q}{A} \right) - t = \beta \Rightarrow \frac{q}{A} = \sin(\omega t + \phi)\]Thus, the trajectory is:
\[q(t) = A \sin(\omega t + \phi)\]which is the expected solution for a harmonic oscillator.
Important Points / Summary:
- The Hamilton–Jacobi equation provides a powerful method to solve mechanical problems using partial differential equations.
- It can simplify finding trajectories, especially for integrable systems.
- For time-independent systems, separation of variables is often applicable.
- In the harmonic oscillator, the HJE approach reproduces the sinusoidal motion.
Practice Questions:
- Short Answer:
- What is the physical interpretation of the action function $S(q, t)$ in the Hamilton–Jacobi theory?
- How does the Hamilton–Jacobi equation relate to canonical transformations?
- Numerical:
- Derive the Hamilton–Jacobi equation for a free particle in 1D.
- Find the generating function $S(q, t)$ for a particle in a uniform gravitational field.
- MCQs:
- The Hamilton–Jacobi equation transforms the problem of motion into solving:
- a) A linear equation
- b) A second-order ODE
- c) A first-order PDE
- d) A matrix equation
Answer: c)
- In the Hamilton–Jacobi method, if $S = W(q) - Et$, the function $W(q)$ is known as:
- a) Hamiltonian
- b) Characteristic function
- c) Action-angle function
- d) Phase function
Answer: b)
- The Hamilton–Jacobi equation transforms the problem of motion into solving:
Few more examples
🧭 Example 1: One-Dimensional Free Particle
Hamiltonian
\[H = \frac{p^2}{2m}\]Hamilton–Jacobi Equation
\[\frac{1}{2m} \left( \frac{\partial S}{\partial q} \right)^2 + \frac{\partial S}{\partial t} = 0\]Solution
Assume a separable solution: $S(q, \alpha, t) = W(q, \alpha) - \alpha t$ where $\alpha$ is the separation constant (i.e., energy).
Then:
\[\frac{1}{2m} \left( \frac{dW}{dq} \right)^2 = \alpha \Rightarrow \frac{dW}{dq} = \sqrt{2m\alpha}\]Integrating:
\[W(q, \alpha) = \sqrt{2m\alpha} \cdot q\]Hence,
\[S(q, \alpha, t) = \sqrt{2m\alpha} \cdot q - \alpha t\]Equation of Motion
To obtain the trajectory:
\[\beta = \frac{\partial S}{\partial \alpha} = \frac{q}{\sqrt{2m\alpha}} - t\]Solving for $q(t)$:
\[q(t) = \sqrt{2m\alpha}(t + \beta)\]This represents uniform motion: $q(t) = v t + q_0$ where $v = \sqrt{2\alpha/m}$ and $q_0 = \sqrt{2m\alpha} \cdot \beta$.
🧲 Example 2: Particle in a Central Potential
(Coulomb potential: $V(r) = -\dfrac{k}{r}$)
Hamiltonian (in spherical coordinates)
\[H = \frac{1}{2m} \left( p_r^2 + \frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2 \sin^2\theta} \right) - \frac{k}{r}\]Hamilton–Jacobi Equation
Assume:
\[S(t, r, \theta, \phi) = -Et + S_r(r) + S_\theta(\theta) + S_\phi(\phi)\]Then:
\[\frac{1}{2m} \left[ \left( \frac{dS_r}{dr} \right)^2 + \frac{1}{r^2} \left( \left( \frac{dS_\theta}{d\theta} \right)^2 + \frac{1}{\sin^2\theta} \left( \frac{dS_\phi}{d\phi} \right)^2 \right) \right] - \frac{k}{r} = E\]Let: • $\dfrac{dS_\phi}{d\phi} = p_\phi = l_z$ • Introduce a constant $l$ such that the total angular part becomes $l^2$
Then:
\[S_\phi = l_z \phi, \quad S_\theta = \int \sqrt{l^2 - \frac{l_z^2}{\sin^2\theta}} , d\theta\]For radial part:
\[\left( \frac{dS_r}{dr} \right)^2 = 2mE + \frac{2mk}{r} - \frac{l^2}{r^2}\]So:
\[S_r = \int \sqrt{2mE + \frac{2mk}{r} - \frac{l^2}{r^2}} , dr\]Final Form of Action
\[S(t, r, \theta, \phi) = -Et + \int \sqrt{2mE + \frac{2mk}{r} - \frac{l^2}{r^2}} , dr + \int \sqrt{l^2 - \frac{l_z^2}{\sin^2\theta}} , d\theta + l_z \phi\]Equation of Orbit
Solving the above gives elliptical orbits:
\[r(\phi) = \frac{a(1 - e^2)}{1 + e \cos \phi}\]with • $a$ = semi-major axis • $e$ = eccentricity
This recovers Kepler’s laws.