The Hamilton-Jacobi Equation
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You should be comfortable with the Hamiltonian formulation of mechanics, the definition of canonical transformations, and the basic properties of Poisson brackets. Familiarity with solving first-order partial differential equations is helpful but not strictly required, as we will focus on the physical interpretation and structure.
Why this matters
The Hamilton-Jacobi equation transforms the problem of solving equations of motion into the problem of finding a single generating function.
While Newton's laws and Lagrange's equations require solving coupled second-order differential equations for each coordinate, the Hamilton-Jacobi approach recasts the entire dynamics of a system into a single first-order partial differential equation. This shift is profound because it allows us to leverage the powerful methods of PDE theory and, more importantly, it reveals the deep connection between symmetries and conservation laws in a way that is often obscured in the standard Hamiltonian formalism. By finding a complete solution to this equation, we effectively solve the system's motion entirely, reducing the dynamical problem to a set of algebraic relations.
The Core Idea: From Dynamics to Geometry
Imagine you are trying to find the path of a particle moving through a landscape. In standard mechanics, you track the particle's position and momentum at every instant, updating them continuously based on forces. This is like navigating by constantly checking your speedometer and steering wheel. The Hamilton-Jacobi approach is different. Instead of tracking the instantaneous state, we seek a "map" of the entire landscape that tells us the optimal cost or action to reach any point from a starting point. This map is described by a single function, Hamilton's principal function, \( S \). Once you have this map, the actual path is just the gradient of this function. The key insight is that if we choose our coordinates wisely—specifically, if we transform to a new set of coordinates that are constant in time—the "energy" or Hamiltonian in this new frame vanishes. This condition leads directly to the Hamilton-Jacobi equation. It is a geometric statement: the dynamics are encoded in the shape of the action surface in phase space.
The Hamilton-Jacobi Equation
Recall that a canonical transformation is defined by a generating function. If we choose a generating function \( S(q_1, \dots, q_n, \alpha_1, \dots, \alpha_n, t) \) that depends on the old coordinates \( q_i \) and a set of new constants \( \alpha_i \), the transformation is given by:
Here the constants \( \alpha_i \) play the role of the new momenta and the \( \beta_i \) are their conjugate new coordinates. We demand that both are constant in time: \( \dot\alpha_i = -\partial K/\partial \beta_i = 0 \) and \( \dot\beta_i = \partial K/\partial \alpha_i = 0 \). Both hold automatically if the new Hamiltonian \( K \) is identically zero. The relationship between the old Hamiltonian \( H \) and the new one \( K \) is given by \( K = H + \frac{\partial S}{\partial t} \). Setting \( K = 0 \) yields the fundamental equation of this lecture:
This is a first-order partial differential equation for \( S \). A "complete integral" is a solution that contains \( n \) independent arbitrary constants \( \alpha_i \). The existence of such a solution is guaranteed under mild conditions, and it provides the complete solution to the mechanical problem.
Solving the Dynamics
Once we have found a complete integral \( S(q, \alpha, t) \), the equations of motion are obtained by simple differentiation. The new momenta \( \beta_i \) are given by \( \beta_i = \frac{\partial S}{\partial \alpha_i} \). Since \( \alpha_i \) are constants, \( \beta_i \) are also constants of motion. The original coordinates \( q_i \) and momenta \( p_i \) as functions of time are found by inverting the relations:
This inversion is algebraic, not differential. We do not need to integrate any further differential equations. The constants \( \alpha_i \) and \( \beta_i \) are determined by the initial conditions. For example, if we know the initial position \( q_i(0) \) and momentum \( p_i(0) \), we can solve for the \( 2n \) constants. This reduction from \( 2n \) second-order ODEs to the solution of a single PDE and subsequent algebraic inversion is the primary computational advantage of the method.
Time-Independent Systems
Many physical systems have a Hamiltonian that does not depend explicitly on time. In this case, \( \frac{\partial H}{\partial t} = 0 \). We can look for a solution \( S \) that separates into a spatial part and a temporal part: \( S(q, \alpha, t) = W(q, \alpha) - Et \), where \( E \) is a constant. Substituting this into the Hamilton-Jacobi equation gives:
This is the time-independent Hamilton-Jacobi equation. The constant \( E \) is identified as the total energy of the system. The function \( W \) is known as Hamilton's characteristic function. This separation is particularly powerful because it reduces the PDE to a single equation involving only the spatial coordinates. The constant \( E \) becomes one of the \( n \) constants of integration. This form is closely related to the WKB approximation in quantum mechanics, where \( W \) plays the role of the phase of the wavefunction.
The derivation
The Hamilton-Jacobi Equation
For a system with Hamiltonian H(q, p, t), there exists a canonical transformation to new coordinates that are all constants of motion, generated by Hamilton's principal function S(q1,…,qn, α1,…,αn, t). Requiring the transformed Hamiltonian to vanish yields the first-order partial differential equation H(q, ∂S/∂q, t) + ∂S/∂t = 0, whose complete integral reduces the dynamics to the algebraic inversion of 2n constants.
A common pitfall is assuming that any solution to the Hamilton-Jacobi equation is a complete integral. A complete integral must contain \( n \) independent constants. If the solution depends on fewer constants, it is an incomplete integral and does not provide the full solution to the equations of motion. Furthermore, the constants \( \alpha_i \) must be independent; if they are related, the transformation is not canonical in the required sense. Always verify the independence of the constants by checking that the Hessian matrix \( \frac{\partial^2 S}{\partial q_i \partial \alpha_j} \) is non-singular. C
Worked example
Problem. A particle of mass \( m \) moves in one dimension under a constant force \( F \), so \( V(x) = -Fx \). Solve the Hamilton-Jacobi equation and find \( x(t) \) for \( x(0) = x_0 \), \( p(0) = p_0 \).
- \( H = \frac{p^2}{2m} - Fx \) has no explicit time dependence, so try \( S = W(x) - \alpha t \) with \( \alpha = E \). The HJ equation becomes \( \frac{1}{2m}\left(\frac{\mathrm{d}W}{\mathrm{d}x}\right)^2 - Fx = \alpha \).
- So \( \frac{\mathrm{d}W}{\mathrm{d}x} = \sqrt{2m(\alpha + Fx)} \) and \( W = \frac{2\sqrt{2m}}{3F}(\alpha + Fx)^{3/2} \).
- The new coordinate is constant: \( \beta = \frac{\partial S}{\partial \alpha} = \frac{\sqrt{2m}}{F}(\alpha + Fx)^{1/2} - t \). (Forgetting the \( -t \) from \( -\alpha t \) gives a static, wrong answer.)
- Invert: \( \alpha + Fx = \frac{F^2 (t + \beta)^2}{2m} \), so \( x(t) = \frac{F(t + \beta)^2}{2m} - \frac{\alpha}{F} \), and \( p = \frac{\mathrm{d}W}{\mathrm{d}x} = F(t + \beta) \).
- Initial conditions: \( p_0 = F\beta \) gives \( \beta = p_0/F \), and \( \alpha = \frac{p_0^2}{2m} - Fx_0 = E \). Substituting and expanding the square, the constant terms cancel to leave the familiar result.
Units check. \( p_0 t/m \) is \( (\mathrm{kg\,m/s})(\mathrm{s})/\mathrm{kg} = \mathrm{m} \) and \( F t^2/m \) is \( \mathrm{N\,s^2/kg} = \mathrm{m} \); \( \beta = p_0/F \) is a time, as it must be to be added to \( t \).
Three quick tests
- What is the physical meaning of the constant \( E \) in the time-independent Hamilton-Jacobi equation?
Answer
It is the total energy of the system, which is conserved because the Hamiltonian does not depend explicitly on time. - If the Hamiltonian depends explicitly on time, can we still use the separation \( S = W - Et \)?
Answer
No, because \( E \) would not be a constant of motion. We must solve the full time-dependent PDE. - How many constants of integration does a complete integral of the Hamilton-Jacobi equation for an \( n \)-degree-of-freedom system contain?
Answer
It contains \( n \) independent arbitrary constants.
Problem set for PU-201 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.