Variational principles in mechanics

Hamilton's principle, symmetries and conservation laws, and the canonical formalism.

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Hamilton's principle and its consequences

Hamilton's principle and its consequences

Application

Hamilton's principle

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Newton's equations for particles moving in a potential are exactly the Euler–Lagrange equations of the action, the time integral of kinetic minus potential energy, and they keep this form in any generalized coordinates, including moving ones.

Prerequisites: Several functions and higher derivatives

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §21, pp. 83–84; Arfken, Weber & Harris (2013), §22.2, pp. 1097–1098

Consider $N$ point particles with masses $m_1, \dots, m_N > 0$ and positions $r_1, \dots, r_N \in \mathbb{R}^3$ . Collect the positions into one configuration vector $x = (r_1, \dots, r_N) \in \mathbb{R}^{3N}$ .

The forces come from a potential if there is a $C^2$ function $V(t, x)$ on $\mathbb{R} \times G$ , with $G \subset \mathbb{R}^{3N}$ open, such that the force on particle $k$ is $F_k = -\nabla_{r_k} V(t, x)$ . Time dependence of $V$ is allowed.
The kinetic energy is $T(\dot x) = \sum_{k} \tfrac12 m_k |\dot r_k|^2$ and the Lagrangian is $L(t, x, \dot x) = T(\dot x) - V(t, x)$ .
The action of a $C^2$ path $x : [t_0, t_1] \to G$ is the number

\mathcal{S}[x] = \int_{t_0}^{t_1} L\bigl(t, x(t), \dot x(t)\bigr)\, dt .

A path $x$ is a stationary path of the action if $\frac{d}{d\varepsilon}\mathcal{S}[x + \varepsilon h]\big|_{\varepsilon = 0} = 0$ for every $C^2$ function $h : [t_0, t_1] \to \mathbb{R}^{3N}$ with $h(t_0) = h(t_1) = 0$ .
Generalized coordinates are given by a $C^2$ map $X : \mathbb{R} \times U \to G$ , $U \subset \mathbb{R}^{3N}$ open, such that for each fixed $t$ the map $q \mapsto X(t, q)$ is injective and its Jacobian matrix $X_q = (\partial X_i / \partial q_j)$ is invertible. The coordinates may move with time (a rotating frame, for instance). A path $x(t) = X(t, q(t))$ has velocity $\dot x = X_t + X_q \dot q$ , and its Lagrangian in the new coordinates is

\tilde L(t, q, \dot q) = L\bigl(t,\ X(t, q),\ X_t(t, q) + X_q(t, q)\,\dot q\bigr).

Application(Hamilton's principle)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §21, Theorem, pp. 84–85; Arfken, Weber & Harris (2013), §22.2, eqs. (22.37)–(22.38), p. 1098

Let $V$ be $C^2$ and let $x : [t_0, t_1] \to G$ be a $C^2$ path. The following are equivalent.

$x$ obeys Newton's equations $m_k \ddot r_k = -\nabla_{r_k} V(t, x)$ for $k = 1, \dots, N$ and all $t \in [t_0, t_1]$ .
$x$ is a stationary path of the action $\mathcal{S}$ .
For one, and then for every, system of generalized coordinates with $x(t) = X(t, q(t))$ and $q$ of class $C^2$ , Lagrange's equations hold:

\frac{d}{dt}\frac{\partial \tilde L}{\partial \dot q_j} - \frac{\partial \tilde L}{\partial q_j} = 0 \qquad (j = 1, \dots, 3N).

Idea of the proofOriginal

In Cartesian coordinates the Euler–Lagrange system of $L = T - V$ is Newton's law term by term, so 1 and 2 are the same statement. For 3, a direct computation shows that the Euler–Lagrange expressions in the new coordinates are the old ones multiplied by the transposed Jacobian $X_q^{\mathsf T}$ . An invertible matrix sends a vector to zero only if the vector is zero, so the equations hold in one system exactly when they hold in the other.

ProofAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §21, pp. 84–85; Arfken, Weber & Harris (2013), §22.2, p. 1098

Write $p_k = \partial L / \partial \dot r_k$ and define the Euler–Lagrange expression of $L$ along the path,

E_x(t) = \frac{d}{dt} L_{\dot x}\bigl(t, x, \dot x\bigr) - L_x\bigl(t, x, \dot x\bigr) \in \mathbb{R}^{3N},

and $E_q(t)$ in the same way for $\tilde L$ along $q$ .

Stationary paths are the solutions of $E_x = 0$ . Because $L$ is $C^2$ , differentiating under the integral and integrating the $\dot h$ term by parts gives
$\frac{d}{d\varepsilon}\mathcal{S}[x + \varepsilon h]\Big|_{\varepsilon=0} = -\int_{t_0}^{t_1} E_x(t)\cdot h(t)\, dt ,$
with no boundary term because $h(t_0) = h(t_1) = 0$ . If $E_x \equiv 0$ the right side vanishes for every $h$ . Conversely, if it vanishes for every admissible $h$ , then $E_x \equiv 0$ by the Euler–Lagrange system for several functions (the fundamental lemma applied to one component at a time).
$E_x = 0$ is Newton's law. For $L = \sum_k \tfrac12 m_k|\dot r_k|^2 - V$ we have $\partial L/\partial \dot r_k = m_k \dot r_k$ and $\partial L / \partial r_k = -\nabla_{r_k} V$ . The $r_k$ -block of $E_x$ is therefore $m_k \ddot r_k + \nabla_{r_k} V$ . This proves $1 \Leftrightarrow 2$ .
How the Euler–Lagrange expression transforms. Let $x(t) = X(t, q(t))$ . We claim
$E_q(t) = X_q\bigl(t, q(t)\bigr)^{\mathsf T}\, E_x(t), \qquad\text{i.e.}\qquad (E_q)_j = \sum_i \frac{\partial X_i}{\partial q_j}\,(E_x)_i .$
- Since $\dot x_i = \partial_t X_i + \sum_j (\partial X_i/\partial q_j)\, \dot q_j$ , the velocity $\dot x_i$ depends on $\dot q_j$ with coefficient $\partial X_i / \partial q_j$ . By the chain rule, $\partial \tilde L / \partial \dot q_j = \sum_i L_{\dot x_i}\, \partial X_i/\partial q_j$ .
- The velocity also depends on $q_j$ : $\partial \dot x_i / \partial q_j = \partial_t \partial_{q_j} X_i + \sum_k \partial_{q_k}\partial_{q_j} X_i\, \dot q_k$ . Because $X$ is $C^2$ its mixed partial derivatives commute, so this is the time derivative of $\partial X_i / \partial q_j$ along the path. Hence
  $\frac{\partial \tilde L}{\partial q_j} = \sum_i \Bigl( L_{x_i}\, \frac{\partial X_i}{\partial q_j} + L_{\dot x_i}\, \frac{d}{dt}\frac{\partial X_i}{\partial q_j} \Bigr).$
- By the product rule, $\frac{d}{dt}\, \partial \tilde L/\partial \dot q_j = \sum_i \bigl( \frac{d}{dt} L_{\dot x_i}\bigr)\, \partial X_i/\partial q_j + \sum_i L_{\dot x_i}\, \frac{d}{dt}\, \partial X_i/\partial q_j$ .
- Subtracting the last two lines, the terms containing $\frac{d}{dt}\,\partial X_i/\partial q_j$ cancel and the claim remains.
Conclusion. The matrix $X_q$ is invertible by the definition of generalized coordinates, so $E_q(t) = 0$ if and only if $E_x(t) = 0$ . Thus 3 holds in a given coordinate system exactly when 2 holds, and then it holds in every system. $\square$

Remark(Where the hypotheses are used)Original

Forces from a potential are used in step 2, where $\partial L/\partial r_k$ must equal the force. Friction is not of this form, and $L = T - V$ cannot describe it.
$C^2$ regularity of $V$ and of the path makes $E_x$ continuous, which the fundamental lemma needs in step 1. $C^2$ regularity of $X$ is what lets mixed partial derivatives be exchanged in step 3.
Invertibility of $X_q$ is used only in step 4. If there are fewer coordinates than $3N$ , for example a pendulum described by one angle, then $X_q$ is a tall matrix and $E_q = X_q^{\mathsf T} E_x = 0$ says only that the vector $E_x$ , the mass-weighted acceleration minus the applied force, is orthogonal to every direction the constraint allows. That is precisely the statement that the constraint force does no work in any allowed displacement (an ideal constraint). This is why Hamilton's principle, written in coordinates on the constraint surface, gives the correct motion of pendulums and rigid bodies; constraints of this kind are treated in Pointwise constraints.

Remark(Stationary, not least)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §21, Remark 2, p. 85; Gelfand & Fomin (1963), Ch. 7, §36.2, pp. 159–160

The older name "principle of least action" claims more than is true. Take a unit-mass oscillator, $L = \tfrac12(\dot q^2 - \omega^2 q^2)$ , on $[0, T]$ with $q(0) = q(T) = 0$ . The rest state $q \equiv 0$ solves Lagrange's equation and has action $0$ . For the competitor $\varepsilon h$ with $h(t) = \sin(\pi t/T)$ ,

\mathcal{S}[\varepsilon h] = \frac{\varepsilon^2}{2}\int_0^T \bigl(\dot h^2 - \omega^2 h^2\bigr)\, dt = \frac{\varepsilon^2 T}{4}\Bigl(\frac{\pi^2}{T^2} - \omega^2\Bigr),

which is negative as soon as $T > \pi/\omega$ . Over such long intervals the true motion is a stationary point of the action, not a minimum. The threshold $\pi/\omega$ is the first conjugate time, studied in Conjugate points and Jacobi's condition.

Example(A free particle seen from a rotating frame)Original

A particle of mass $m$ moves in a plane with no forces ( $V = 0$ ). Describe it in coordinates $q = (\xi, \eta)$ that rotate with constant angular velocity $\Omega$ :

x = R(\Omega t)\, q, \qquad R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.

Lagrangian. Since $\frac{d}{dt}R(\Omega t) = \Omega R(\Omega t) J$ with $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ , we get $\dot x = R(\Omega t)(\dot q + \Omega J q)$ . Rotations preserve length, so

\tilde L = \frac m2\, |\dot q + \Omega J q|^2 = \frac m2\Bigl[(\dot \xi - \Omega \eta)^2 + (\dot\eta + \Omega \xi)^2\Bigr].

Lagrange's equations. For $\xi$ : $\partial\tilde L/\partial\dot\xi = m(\dot\xi - \Omega\eta)$ and $\partial \tilde L/\partial \xi = m\Omega(\dot\eta + \Omega\xi)$ , so $m(\ddot \xi - \Omega\dot\eta) - m\Omega(\dot \eta + \Omega \xi) = 0$ . Doing the same for $\eta$ ,

\ddot \xi = 2\Omega\dot\eta + \Omega^2 \xi, \qquad \ddot \eta = -2\Omega\dot\xi + \Omega^2\eta .

In vector form this reads $\ddot q = -2\Omega J\dot q + \Omega^2 q$ : the term $\Omega^2 q$ is the centrifugal acceleration and $-2\Omega J \dot q = (2\Omega\dot\eta,\ -2\Omega\dot\xi)$ is the Coriolis acceleration. Both came out of the theorem, with no separate argument about non-inertial frames.

A check with numbers. Release the particle at rest in the rotating frame at $q(0) = (a, 0)$ . In the inertial frame it starts at $(a, 0)$ with velocity $R(0)\,\Omega J q(0) = (0, a\Omega)$ , so it moves on the straight line $x(t) = (a,\ a\Omega t)$ . Hence

q(t) = R(-\Omega t)\, x(t) = \bigl(a\cos\Omega t + a\Omega t \sin\Omega t,\ \ -a\sin\Omega t + a\Omega t\cos\Omega t\bigr).

Substituting into the two equations above makes both residuals vanish identically. At $\Omega t = \pi/2$ the particle is at $q = (\pi a/2,\ -a)$ , which for $a = 1$ is $(1.5708, -1)$ : seen from the rotating frame, the particle has drifted outward and backward against the rotation.

Exercise(The elastic pendulum)Original

What it teaches: choosing generalized coordinates adapted to the forces. A bob of mass $m$ hangs from a fixed point on a spring of stiffness $k$ and natural length $\ell_0$ and moves in a vertical plane under gravity $g$ . Using polar coordinates $(r, \theta)$ , with $\theta$ measured from the downward vertical, show that

\tilde L = \frac m2\bigl(\dot r^2 + r^2\dot\theta^2\bigr) + mgr\cos\theta - \frac k2 (r - \ell_0)^2,

derive $m\ddot r = mr\dot\theta^2 + mg\cos\theta - k(r - \ell_0)$ and $\frac{d}{dt}(mr^2\dot\theta) = -mgr\sin\theta$ , and linearize about the hanging equilibrium. For $m = 0.5$ , $k = 50$ , $\ell_0 = 0.4$ and $g = 9.81$ (SI units) you should find $r_{\mathrm{eq}} = \ell_0 + mg/k = 0.4981$ , a radial frequency $\sqrt{k/m} = 10$ and a swinging frequency $\sqrt{g/r_{\mathrm{eq}}} \approx 4.438$ , in radians per second.

ApplicationOriginal

Hamilton's principle is the standard way to set up equations of motion in mechanics and engineering: one writes down $T$ and $V$ in whatever coordinates suit the problem (joint angles of a robot arm, Euler angles of a spacecraft, normal modes of a molecule) and reads off the equations, with constraint forces never appearing. In celestial mechanics and geophysical fluid dynamics, the rotating-frame Lagrangian of the example above is the starting point for the restricted three-body problem and for the Coriolis terms in atmosphere and ocean models. The same principle with a Lagrangian density gives field equations such as the wave equation (see Several independent variables), and the symmetries of $L$ give conservation laws through Noether's theorem.

Theorem

Noether's theorem

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If the action is unchanged by a one-parameter family of transformations of time and coordinates, an explicit combination of momenta and energy is conserved along every motion; time shifts, translations and rotations give conservation of energy, momentum and angular momentum.

Prerequisites: Hamilton's principle

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §20, Definition and eqs. (45)–(48), pp. 80–81

In this unit $L(t, q, v)$ is a $C^2$ Lagrangian on $\mathbb{R}\times G\times\mathbb{R}^n$ , $G\subset\mathbb{R}^n$ open, and along a curve $q(t)$ we write

p_i = \frac{\partial L}{\partial v_i}\bigl(t, q, \dot q\bigr), \qquad E = \sum_{i=1}^n \dot q_i\, p_i - L\bigl(t, q, \dot q\bigr)

for the momenta and the energy function. For a mechanical Lagrangian $L = T - V$ with $T$ quadratic in the velocities, $E = T + V$ .

A one-parameter family of point transformations is a pair of functions $\theta(\varepsilon, t, q) \in \mathbb{R}$ and $Q(\varepsilon, t, q) \in G$ , of class $C^2$ in $(\varepsilon, t, q)$ for $|\varepsilon|$ small, that reduce to the identity at $\varepsilon = 0$ : $\theta(0, t, q) = t$ and $Q(0, t, q) = q$ . Its generators are

\tau(t, q) = \frac{\partial \theta}{\partial \varepsilon}(0, t, q), \qquad \xi(t, q) = \frac{\partial Q}{\partial \varepsilon}(0, t, q) \in \mathbb{R}^n .

Given a $C^2$ curve $q : [a, b] \to G$ , put $s_\varepsilon(t) = \theta(\varepsilon, t, q(t))$ . For small $|\varepsilon|$ its derivative is close to $1$ on $[a, b]$ , so $s_\varepsilon$ is increasing and we may define the transformed curve $q^\ast_\varepsilon$ on $[s_\varepsilon(a), s_\varepsilon(b)]$ by $q^\ast_\varepsilon\bigl(s_\varepsilon(t)\bigr) = Q\bigl(\varepsilon, t, q(t)\bigr)$ .

The action is invariant under the family if, for every $C^2$ curve $q$ in $G$ , every interval $[a, b]$ in its domain and every sufficiently small $|\varepsilon|$ ,

\int_{s_\varepsilon(a)}^{s_\varepsilon(b)} L\Bigl(s,\ q^\ast_\varepsilon(s),\ \frac{dq^\ast_\varepsilon}{ds}(s)\Bigr)\, ds = \int_a^b L\bigl(t, q(t), \dot q(t)\bigr)\, dt .

Theorem(Noether's theorem)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §20, Theorem (Noether) and Remark, pp. 81–83

If the action of $L$ is invariant under a one-parameter family of point transformations with generators $\tau$ and $\xi$ , then along every $C^2$ solution $q(t)$ of the Euler–Lagrange equations of $L$ the quantity

I = \sum_{i=1}^n p_i\, \xi_i\bigl(t, q\bigr) - E\, \tau\bigl(t, q\bigr)

is constant.

Idea of the proofOriginal

Write the transformed action as an integral over the original time $t$ and differentiate it with respect to $\varepsilon$ at $\varepsilon = 0$ . Invariance on every subinterval forces the integrand of that derivative to vanish at each instant. This gives a pointwise identity, valid on any curve, involving $L$ , its partial derivatives and the generators. On a solution of the Euler–Lagrange equations, $\partial L/\partial q$ is $\dot p$ and $\partial L/\partial t$ is $-\dot E$ , and the identity collapses to $\frac{d}{dt} I = 0$ .

ProofAdaptedAdapted from Evans (2010), §8.6.2, proof of Theorem 1, pp. 482–483; Gelfand & Fomin (1963), Ch. 4, §20, pp. 81–83

Fix a $C^2$ curve $q$ and an interval $[a, b]$ . Dots denote derivatives along the curve. Write $\theta_\varepsilon(t) = \theta(\varepsilon, t, q(t))$ and $Q_\varepsilon(t) = Q(\varepsilon, t, q(t))$ .

Pull the transformed action back to $[a, b]$ . Substitute $s = \theta_\varepsilon(t)$ , so $ds = \dot\theta_\varepsilon\, dt$ . By the chain rule $\frac{dq^\ast_\varepsilon}{ds} = \dot Q_\varepsilon / \dot\theta_\varepsilon$ . The invariance identity becomes
$\int_a^b \Lambda(\varepsilon, t)\, dt = \int_a^b L\bigl(t, q, \dot q\bigr)\, dt, \qquad \Lambda(\varepsilon, t) = L\Bigl(\theta_\varepsilon,\ Q_\varepsilon,\ \frac{\dot Q_\varepsilon}{\dot\theta_\varepsilon}\Bigr)\, \dot\theta_\varepsilon .$
Invariance on every subinterval gives a pointwise identity. The left side does not depend on $\varepsilon$ . Because $\theta$ and $Q$ are $C^2$ and $q$ is $C^2$ , the functions $\dot\theta_\varepsilon = \theta_t + \theta_q\cdot\dot q$ and $\dot Q_\varepsilon$ are $C^1$ in $(\varepsilon, t)$ , and so is $\Lambda$ , since $L$ is $C^2$ . Differentiating under the integral sign therefore gives $\int_a^b \partial_\varepsilon\Lambda(0, t)\, dt = 0$ . This holds for every subinterval of the curve's domain, and $\partial_\varepsilon\Lambda(0, \cdot)$ is continuous, so $\partial_\varepsilon \Lambda(0, t) = 0$ for every $t$ .
Compute the derivative. At $\varepsilon = 0$ we have $\theta_0 = t$ , $Q_0 = q$ , $\dot\theta_0 = 1$ and $\dot Q_0 = \dot q$ . Their $\varepsilon$ -derivatives are $\tau$ , $\xi$ , $\dot \tau$ and $\dot \xi$ , where $\dot\tau = \frac{d}{dt}\tau(t, q(t))$ ; for the last two we exchanged $\partial_\varepsilon$ with $\frac{d}{dt}$ , which is allowed because the mixed second derivatives of $\theta$ and $Q$ are continuous. The $\varepsilon$ -derivative of the velocity slot $\dot Q_\varepsilon/\dot\theta_\varepsilon$ is $\dot\xi - \dot q\,\dot\tau$ . Hence
$0 = \partial_\varepsilon\Lambda(0, t) = L_t\,\tau + L_q\cdot\xi + p\cdot\bigl(\dot\xi - \dot q\,\dot\tau\bigr) + L\,\dot\tau = L_t\,\tau + L_q\cdot\xi + p\cdot\dot\xi - E\,\dot\tau . \tag{$\ast$}$
So far $q$ was any $C^2$ curve.
Use the Euler–Lagrange equations. Now let $q$ solve them, so $L_q = \dot p$ . Then the energy function satisfies $\dot E = -L_t$ : differentiating $E = \dot q\cdot p - L$ gives
$\dot E = \ddot q\cdot p + \dot q\cdot\dot p - L_t - L_q\cdot\dot q - p\cdot\ddot q = \dot q\cdot\bigl(\dot p - L_q\bigr) - L_t = -L_t .$
Assemble. Substituting $L_q = \dot p$ and $L_t = -\dot E$ in $(\ast)$ ,
$0 = -\dot E\,\tau + \dot p\cdot\xi + p\cdot\dot\xi - E\,\dot\tau = \frac{d}{dt}\bigl(p\cdot\xi - E\,\tau\bigr) = \frac{dI}{dt} . \qquad\square$

Remark(Where the hypotheses are used)Original

Invariance on every subinterval is what turns an integral identity into the pointwise identity $(\ast)$ in step 2. Invariance on a single interval would give only $\int_a^b \partial_\varepsilon\Lambda(0,t)\,dt = 0$ .
The Euler–Lagrange equations enter only in step 4. The identity $(\ast)$ holds on every curve; conservation of $I$ holds only on motions.
A failure of invariance. The Lagrangian $L = \tfrac12 e^{\gamma t}\bigl(\dot q^2 - \omega^2 q^2\bigr)$ , $\gamma > 0$ , has Euler–Lagrange equation $\ddot q + \gamma\dot q + \omega^2 q = 0$ , a damped oscillator. It depends on $t$ explicitly, so the action is not invariant under time shifts. Its energy function $E = \tfrac12 e^{\gamma t}(\dot q^2 + \omega^2 q^2)$ obeys $\dot E = -L_t = -\tfrac{\gamma}{2} e^{\gamma t}(\dot q^2 - \omega^2 q^2)$ , which is not zero in general.

Remark(Symmetries up to a total derivative)Original

The proof gives slightly more than the statement. Suppose that, instead of invariance, $\partial_\varepsilon \Lambda(0, t) = \frac{d}{dt}\Phi(t, q(t))$ for some $C^1$ function $\Phi$ and every $C^2$ curve. Then step 5 becomes $\frac{d}{dt}(I - \Phi) = 0$ , so $I - \Phi$ is conserved. Galilean boosts are the standard case. For $N$ particles with masses $m_k$ , total mass $M$ and a potential that depends only on the differences $r_k - r_l$ , take $\theta = t$ and $Q_k = r_k + \varepsilon u t$ for a fixed vector $u$ . Then $\xi_k = u t$ , $\tau = 0$ , and $\partial_\varepsilon\Lambda(0,t) = \sum_k m_k\dot r_k\cdot u = \frac{d}{dt}\bigl(M R\cdot u\bigr)$ , where $R$ is the centre of mass. The conserved quantity is $u\cdot(P t - M R)$ , where $P$ is the total momentum. Since $u$ is arbitrary, $R(t) = R(0) + P t/M$ : the centre of mass moves uniformly.

Example(Energy, momentum and angular momentum)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §20, Example 3, p. 83; Gelfand & Fomin (1963), Ch. 4, §22, items 1–3, pp. 86–88

Take $N$ particles with $q = (r_1, \dots, r_N)$ and $L = \sum_k \tfrac12 m_k|\dot r_k|^2 - V(t, r_1, \dots, r_N)$ , so that $p = (m_1\dot r_1, \dots, m_N\dot r_N)$ and $E = T + V$ .

Time shifts: $\theta = t + \varepsilon$ , $Q = q$ , so $\tau = 1$ and $\xi = 0$ . Shifting a curve in time leaves its action unchanged for every curve exactly when $L$ does not depend on $t$ , that is when $\partial V/\partial t = 0$ . Then $I = -E$ : energy is conserved.
Translations: $\theta = t$ , $Q_k = r_k + \varepsilon e$ for all $k$ , with $e$ a fixed vector. Velocities are unchanged, so the action is invariant if $V(t, r_1 + \varepsilon e, \dots, r_N + \varepsilon e) = V(t, r_1, \dots, r_N)$ . Here $\xi_k = e$ , $\tau = 0$ and $I = \sum_k p_k\cdot e$ : the component of total momentum along $e$ is conserved.
Rotations: $\theta = t$ , $Q_k = R_e(\varepsilon) r_k$ , where $R_e(\varepsilon)$ is the rotation by angle $\varepsilon$ about the unit axis $e$ . Rotations preserve lengths, so $T$ is unchanged, and the action is invariant if $V$ is unchanged by rotating all particles together. The generator is $\xi_k = e\times r_k$ , and $I = \sum_k p_k\cdot(e\times r_k) = e\cdot\sum_k r_k\times p_k$ : the component of total angular momentum along $e$ is conserved.

Example(A screw symmetry)Original

Here the conserved quantity mixes angular and linear momentum, so it is not one of the familiar three. A particle of unit mass moves in $\mathbb{R}^3$ in the potential

V(x, y, z) = \tfrac12 w^2, \qquad w = -x\sin(z/\kappa) + y\cos(z/\kappa), \qquad \kappa = 0.5 .

The potential vanishes on the helicoid $w = 0$ and grows away from it, so the particle is confined near a twisted ribbon, like a bead in the groove of a screw.

The symmetry. Let $\theta = t$ and let $Q(\varepsilon, \cdot)$ rotate $(x, y)$ by angle $\varepsilon$ about the $z$ -axis while shifting $z$ by $\kappa\varepsilon$ . The direction $(\cos(z/\kappa), \sin(z/\kappa))$ turns by the same angle $\varepsilon$ as $(x, y)$ , and $w$ is the component of $(x, y)$ perpendicular to it, so $w$ is unchanged. The kinetic energy is unchanged because the map is a rigid motion. So the action is invariant, with $\tau = 0$ and $\xi = (-y,\ x,\ \kappa)$ .

The conserved quantity. Noether's theorem gives

I = -y\,p_x + x\,p_y + \kappa\, p_z = \ell_z + \kappa\, p_z ,

where $\ell_z = x p_y - y p_x$ is the $z$ -component of angular momentum. As a direct check, Newton's law gives $\dot\ell_z = -(x V_y - y V_x)$ and $\dot p_z = -V_z$ , so $\dot I = -\xi\cdot\nabla V$ . This is minus the derivative of $V$ along the screw motion, which is $0$ .

Numbers. Start at $(x, y, z) = (1, 0, 0)$ , where $w = 0$ , with velocity $(0,\ 0.5,\ 0.3)$ . Then $\ell_z = 0.5$ , $p_z = 0.3$ and

I = 0.5 + 0.5\times 0.3 = 0.65 .

Integrating the equations of motion numerically (relative tolerance $10^{-12}$ ) gives, at $t = 10$ , $\ell_z \approx 0.661788$ and $p_z \approx -0.023576$ . Neither is conserved separately ( $\ell_z$ ranges over $[0.5, 0.677]$ on $0 \le t \le 10$ ), but $\ell_z + 0.5\,p_z = 0.650000$ to the accuracy of the integration. The energy $E = \tfrac12(0.5^2 + 0.3^2) = 0.17$ is also conserved, by item 1 of the previous example.

ApplicationOriginal

Noether's theorem is how conservation laws are found in practice: one lists the symmetries of the Lagrangian instead of hunting for first integrals. In the Kepler and other central-force problems, rotational symmetry gives angular momentum and hence planar orbits and Kepler's second law. In an axisymmetric magnetic-confinement device (an idealized tokamak), the invariance of a charged particle's Lagrangian under rotations about the symmetry axis conserves its canonical angular momentum about that axis, a mixture of mechanical momentum and the vector potential much as $I$ mixes $\ell_z$ and $p_z$ in the screw example, and this is what keeps the particle's orbit close to a magnetic surface. In field theory the same argument, applied to a Lagrangian density, turns space-time translations into conservation of energy and momentum densities (see Several independent variables). On the Hamiltonian side, a quantity is conserved exactly when its Poisson bracket with the Hamiltonian vanishes (see The canonical equations).

Theorem

The canonical equations

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When velocities can be solved for in terms of momenta, the Legendre transformation turns the n second-order Euler–Lagrange equations into Hamilton's 2n first-order canonical equations, and first integrals are then recognized by a vanishing Poisson bracket.

Prerequisites: Hamilton's principle, Existence and uniqueness of solutions

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §16, pp. 67–69; Evans (2010), §3.3.1(b), eq. (9) and Definition, p. 114

Let $L(t, q, v)$ be a $C^2$ function on an open set $\Omega \subset \mathbb{R}\times\mathbb{R}^n\times\mathbb{R}^n$ ; $v$ is the slot into which the velocity $\dot q$ is substituted.

The canonical momentum belonging to $(t, q, v)$ is $p = L_v(t, q, v) = (\partial L/\partial v_1, \dots, \partial L/\partial v_n)$ .
Regularity hypothesis (R). There is an open set $W \subset \mathbb{R}\times\mathbb{R}^n\times\mathbb{R}^n$ (the phase domain) and a $C^1$ map $\hat v : W \to \mathbb{R}^n$ such that $(t, q, v) \mapsto (t, q, L_v(t, q, v))$ is a bijection from $\Omega$ onto $W$ with inverse $(t, q, p) \mapsto (t, q, \hat v(t, q, p))$ . In words: the velocity can be recovered from the momentum, smoothly.
The Hamiltonian is the function on $W$
$H(t, q, p) = p\cdot \hat v(t, q, p) - L\bigl(t, q, \hat v(t, q, p)\bigr).$
Passing from $(t, q, v, L)$ to $(t, q, p, H)$ is the Legendre transformation in the velocity variables.
The canonical equations (Hamilton's equations) for a curve $(q(t), p(t))$ in the phase domain are
$\dot q_i = \frac{\partial H}{\partial p_i}(t, q, p), \qquad \dot p_i = -\frac{\partial H}{\partial q_i}(t, q, p) \qquad (i = 1, \dots, n).$
The Poisson bracket of $C^1$ functions $f, g$ on $W$ is $\{f, g\} = \sum_i \bigl(\partial_{q_i} f\, \partial_{p_i} g - \partial_{p_i} f\, \partial_{q_i} g\bigr)$ . Gelfand–Fomin and Lee use this sign; some physics texts use the opposite one.
A first integral of the canonical equations is a $C^1$ function $f(t, q, p)$ on $W$ that is constant along every solution.

TheoremAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §16, eq. (11), p. 70; Gelfand & Fomin (1963), Ch. 4, §17, eq. (13), p. 71; Evans (2010), §3.3.1(b), Theorem 2, p. 114

Assume $L$ is $C^2$ on $\Omega$ and satisfies (R). Then:

$H$ is $C^2$ on $W$ , and for $(t, q, p) \in W$ , with $\hat v = \hat v(t, q, p)$ ,
$H_p = \hat v, \qquad H_q = -L_q(t, q, \hat v), \qquad H_t = -L_t(t, q, \hat v).$
A $C^2$ curve $q(t)$ with $(t, q, \dot q) \in \Omega$ solves the Euler–Lagrange equations $\frac{d}{dt}L_v - L_q = 0$ if and only if $(q(t), p(t))$ , with $p(t) = L_v(t, q, \dot q)$ , solves the canonical equations. Conversely, if a $C^1$ curve $(q, p)$ in $W$ solves the canonical equations, then $q$ is $C^2$ , $p = L_v(t, q, \dot q)$ , and $q$ solves the Euler–Lagrange equations.
Along every solution of the canonical equations, $\frac{d}{dt}H(t, q, p) = H_t(t, q, p)$ . In particular, if $L$ does not depend on $t$ , then $H$ is a first integral.
A $C^1$ function $f(t, q, p)$ satisfies $\frac{d}{dt}f = f_t + \{f, H\}$ along every solution. Hence $f$ is a first integral if and only if $f_t + \{f, H\} = 0$ at every point of $W$ . If $H$ does not depend on $q_k$ , then $p_k$ is a first integral.

Idea of the proofOriginal

The defining relation $p = L_v$ makes the Hamiltonian insensitive, to first order, to how the velocity $\hat v$ moves when $(t, q, p)$ changes. Every term in which $\hat v$ is differentiated has the factor $p - L_v$ , which is zero. So the partial derivatives of $H$ are read off as if $\hat v$ were constant. With these formulas, the definition of $p$ becomes the first canonical equation and the Euler–Lagrange equation becomes the second. The time derivative of any function along a solution is then a chain-rule computation, which is where the Poisson bracket comes from.

ProofAdaptedAdapted from Evans (2010), §3.3.1(b), proof of Theorem 2, pp. 114–115; Gelfand & Fomin (1963), Ch. 4, §16–17, pp. 69–71

Derivatives of $H$ . By (R), $L_v(t, q, \hat v(t, q, p)) = p$ on $W$ . Differentiate $H = p\cdot\hat v - L(t, q, \hat v)$ by the chain rule, with $\hat v$ evaluated at $(t, q, p)$ :
$H_{p_i} = \hat v_i + \sum_k \bigl(p_k - L_{v_k}\bigr)\,\partial_{p_i}\hat v_k = \hat v_i ,$ $H_{q_i} = -L_{q_i} + \sum_k \bigl(p_k - L_{v_k}\bigr)\,\partial_{q_i}\hat v_k = -L_{q_i}, \qquad H_t = -L_t + \sum_k \bigl(p_k - L_{v_k}\bigr)\,\partial_t\hat v_k = -L_t .$
Each bracket vanishes by (R). The right-hand sides $\hat v$ , $-L_q(t, q, \hat v)$ and $-L_t(t, q, \hat v)$ are $C^1$ , because $L$ is $C^2$ and $\hat v$ is $C^1$ , so $H$ is $C^2$ .
Euler–Lagrange implies canonical. Let $q$ be a $C^2$ solution and $p = L_v(t, q, \dot q)$ . By (R), $\dot q = \hat v(t, q, p)$ , so part 1 gives $\dot q = H_p(t, q, p)$ . The Euler–Lagrange equation says $\dot p = L_q(t, q, \dot q)$ , and by part 1 again $L_q(t, q, \hat v) = -H_q(t, q, p)$ . So $\dot p = -H_q$ .
Canonical implies Euler–Lagrange. Let $(q, p)$ be a $C^1$ solution in $W$ . Then $\dot q = H_p = \hat v(t, q, p)$ , so by (R) $p = L_v(t, q, \dot q)$ . Since $\hat v$ is $C^1$ and $q$ , $p$ are $C^1$ , $\dot q$ is $C^1$ , so $q$ is $C^2$ . Finally $\frac{d}{dt}L_v(t, q, \dot q) = \dot p = -H_q = L_q(t, q, \dot q)$ , which is the Euler–Lagrange equation.
Time derivative of a function along a solution. For $f \in C^1(W)$ and a solution $(q, p)$ , the chain rule and the canonical equations give
$\frac{d}{dt}f = f_t + \sum_i\bigl(f_{q_i}\dot q_i + f_{p_i}\dot p_i\bigr) = f_t + \sum_i\bigl(f_{q_i}H_{p_i} - f_{p_i}H_{q_i}\bigr) = f_t + \{f, H\}.$
- With $f = H$ the bracket $\{H, H\}$ is zero by antisymmetry, so $\frac{d}{dt}H = H_t$ . If $L$ does not depend on $t$ , then $H_t = -L_t = 0$ by part 1. This proves statement 3.
- If $f_t + \{f, H\} = 0$ on $W$ , then $f$ is constant along every solution. Conversely, suppose $f$ is a first integral and let $(t_0, q_0, p_0) \in W$ . Since $H$ is $C^2$ , the right-hand side of the canonical equations is $C^1$ , hence locally Lipschitz, and by the existence and uniqueness theorem there is a solution through $(q_0, p_0)$ at time $t_0$ . Along it $f$ is constant, so $f_t + \{f, H\} = 0$ at $(t_0, q_0, p_0)$ .
- If $H$ does not depend on $q_k$ , the second canonical equation reads $\dot p_k = 0$ . $\square$

Remark(Where the hypotheses are used)Original

Hypothesis (R) is used in every step: it makes $H$ well defined and it is the cancellation $p - L_v = 0$ in part 1. By the inverse function theorem, (R) holds near any point where $\det\bigl(\partial^2 L/\partial v_i\partial v_j\bigr) \ne 0$ , so the canonical equations always exist locally there. It fails for Lagrangians that are linear in the velocities, where $p = L_v$ does not involve $v$ at all, and for the length Lagrangian $L = |v|$ of a parametrized curve, where every $v$ in the same direction gives the same $p = v/|v|$ and $p\cdot v - L$ is identically $0$ .
$C^2$ regularity of $L$ makes $H$ of class $C^2$ , which is what part 4 needs to apply the existence theorem. Without existence of a solution through each point, a first integral need not satisfy $f_t + \{f, H\} = 0$ at points no solution visits.

Remark(The Legendre transformation as convex duality)AdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §18.1, eqs. (14)–(18), (23), pp. 72–75; Evans (2010), §3.3.2(a), Theorem 3, pp. 116–117

Part 1 shows that the construction can be run backwards. The "velocity" belonging to $H$ is $H_p = \hat v$ , and $\hat v\cdot p - H = L(t, q, \hat v)$ . So applying the same recipe to $H$ returns $L$ : the transformation is its own inverse. Write $\Omega_{t,q} = \{v : (t, q, v) \in \Omega\}$ . If moreover each $\Omega_{t,q}$ is convex and $\partial^2L/\partial v^2$ is positive definite there, then for fixed $(t, q, p)$ the function $v \mapsto p\cdot v - L(t, q, v)$ is concave on $\Omega_{t,q}$ with a critical point at $\hat v$ , so

H(t, q, p) = \max_{v \in \Omega_{t,q}}\bigl(p\cdot v - L(t, q, v)\bigr), \qquad\text{hence}\qquad p\cdot v \le L(t, q, v) + H(t, q, p)

for all $v \in \Omega_{t,q}$ (Young's inequality). Evans takes the supremum of $p\cdot v - L$ as the definition of $H$ , which also makes sense when $L$ is convex but not differentiable.

Example(The quadratic Lagrangian)Original

Most Lagrangians of mechanics have the form

L = \tfrac12\, v\cdot M(t, q)\, v + A(t, q)\cdot v - V(t, q),

with $M$ a symmetric positive definite matrix. Then $p = Mv + A$ , so (R) holds on all of phase space with $\hat v = M^{-1}(p - A)$ . Substituting,

H = (p - A)\cdot M^{-1}(p - A) - \tfrac12 (p - A)\cdot M^{-1}(p - A) + V = \tfrac12 (p - A)\cdot M^{-1}(p - A) + V .

For a charged particle, $A$ is the charge times the magnetic vector potential. In a rotating frame, $A$ comes from the frame's rotation, as below.

Example(A free particle in a rotating frame)Original

In Hamilton's principle a free particle of mass $m$ seen from axes rotating at angular velocity $\Omega$ has

\tilde L = \frac m2\,|\dot q + \Omega J q|^2, \qquad J = \begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}, \qquad q = (\xi, \eta).

Legendre transformation. This is the quadratic case with $M = mI$ , $A = m\Omega Jq$ and $V = -\tfrac{m}{2}\Omega^2|q|^2$ , because $|\dot q + \Omega Jq|^2 = |\dot q|^2 + 2\Omega\,\dot q\cdot Jq + \Omega^2|q|^2$ . So $p = m(\dot q + \Omega J q)$ and

H = \frac{|p - m\Omega Jq|^2}{2m} - \frac m2\,\Omega^2|q|^2 = \frac{|p|^2}{2m} - \Omega\,\ell, \qquad \ell = \xi p_\eta - \eta p_\xi ,

using $p\cdot Jq = \ell$ and $|Jq| = |q|$ .

Canonical equations. $\dot\xi = p_\xi/m + \Omega\eta$ , $\dot\eta = p_\eta/m - \Omega\xi$ , $\dot p_\xi = \Omega p_\eta$ , $\dot p_\eta = -\Omega p_\xi$ .

First integrals. $H$ does not depend on $t$ , so it is conserved, although the coordinate change to the rotating frame depends on time. Moreover $\{\ell, |p|^2\} = p_\eta\cdot 2p_\xi - p_\xi\cdot 2p_\eta = 0$ and $\{\ell, \ell\} = 0$ , so $\{\ell, H\} = 0$ : $\ell$ is conserved, and therefore so is $|p|^2 = 2m(H + \Omega\ell)$ .

Check with numbers. Take $m = \Omega = 1$ and the motion $q(t) = (\cos t + t\sin t,\ -\sin t + t\cos t)$ found in the Hamilton's-principle example. At $t = 0$ : $q = (1, 0)$ , $\dot q = (0, 0)$ , $p = (0, 1)$ , so $\ell = 1$ and $H = \tfrac12 - 1 = -\tfrac12$ . At $t = \pi/2$ : $q = (\pi/2, -1)$ , $\dot q = (t\cos t,\ -t\sin t) = (0, -\pi/2)$ , so $p = \dot q + Jq = (0, -\pi/2) + (1, \pi/2) = (1, 0)$ . Then $\ell = \tfrac{\pi}{2}\cdot 0 - (-1)\cdot 1 = 1$ and $H = \tfrac12 - 1 = -\tfrac12$ , as predicted. The value $-\tfrac12$ is the inertial energy $\tfrac12$ minus $\Omega$ times the inertial angular momentum $1$ .

ApplicationOriginal

The canonical equations are the form in which mechanics is studied as a dynamical system: phase space $(q, p)$ carries a volume preserved by the flow (Liouville's theorem), which underlies equilibrium statistical mechanics, and the Poisson bracket $\{f, H\}$ becomes the commutator in the passage to quantum mechanics. Numerically, long-time simulations in celestial mechanics and molecular dynamics use integrators that respect the canonical structure (see Symplectic integrators). The rotating-frame Hamiltonian $H = |p|^2/2m - \Omega\ell + V$ is the setting of the restricted three-body problem, where its conservation is known as the Jacobi integral.

Theorem · advanced

The Hamilton-Jacobi equation

Adapted

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The action, viewed as a function of the endpoint, solves the first-order Hamilton–Jacobi equation; conversely, a complete integral of that equation produces all solutions of the canonical equations by differentiation and elimination alone (Jacobi's theorem).

Prerequisites: The canonical equations, Implicit and inverse function theorems, Differentiation under the integral sign, First-order equations and characteristics, Existence and uniqueness of solutions

DefinitionAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §23, eqs. (65)–(72) and Theorem 2, pp. 88–92

Let $H(t, q, p)$ be a $C^2$ function on an open phase domain $W \subset \mathbb{R}\times\mathbb{R}^n\times\mathbb{R}^n$ , for example the Hamiltonian of a Lagrangian $L$ satisfying the regularity hypothesis (R) of The canonical equations. Write $\nabla S = (\partial S/\partial q_1, \dots, \partial S/\partial q_n)$ .

The Hamilton–Jacobi equation is the first-order partial differential equation
$\frac{\partial S}{\partial t} + H\bigl(t, q, \nabla S\bigr) = 0$
for a $C^1$ function $S(t, q)$ on an open set $D \subset \mathbb{R}\times\mathbb{R}^n$ with $(t, q, \nabla S(t, q)) \in W$ for all $(t, q) \in D$ .
A complete integral is a family of solutions $S(t, q, \alpha)$ , with parameters $\alpha$ in an open set $A \subset \mathbb{R}^n$ , such that $S$ is $C^2$ in all $2n + 1$ variables on an open set $\mathcal{D} \subset \mathbb{R}\times\mathbb{R}^n\times A$ and the mixed Hessian is nonsingular:
$\det\Bigl(\frac{\partial^2 S}{\partial q_i\,\partial\alpha_j}\Bigr) \ne 0 \quad\text{on } \mathcal{D}.$
Action as a function of the endpoint. Fix a starting event $(t_0, q_0)$ . A field of extremals from $(t_0, q_0)$ is a $C^2$ map $(t, a) \mapsto \mathsf{Q}(t, a)$ on $[t_0, t_1)\times A$ , $A \subset \mathbb{R}^n$ open, such that each $\mathsf{Q}(\cdot, a)$ solves the Euler–Lagrange equations of $L$ with $\mathsf{Q}(t_0, a) = q_0$ , and $\Psi(t, a) = (t, \mathsf{Q}(t, a))$ is injective on $(t_0, t_1)\times A$ with $\det \mathsf{Q}_a \ne 0$ there. By the inverse function theorem $D = \Psi\bigl((t_0, t_1)\times A\bigr)$ is open and $\Psi^{-1}$ is $C^1$ , so every $(t, q) \in D$ is reached by exactly one extremal of the field, with label $a(t, q)$ . The action function is
$S(t, q) = \int_{t_0}^{t} L\bigl(s,\ \mathsf{Q}(s, a),\ \partial_s\mathsf{Q}(s, a)\bigr)\, ds, \qquad a = a(t, q).$

TheoremAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §23, eqs. (69)–(72), p. 90, and Theorem 2 (Jacobi), pp. 91–93

(a) The action solves the Hamilton–Jacobi equation. Let $L$ be $C^2$ and satisfy (R), with Hamiltonian $H$ , and let $S$ be the action function of a field of extremals from $(t_0, q_0)$ . Then $S$ is $C^1$ on $D$ ,

\nabla S(t, q) = L_v\bigl(t, q, \partial_t\mathsf{Q}(t, a)\bigr) \quad (\text{the momentum of the extremal arriving at } (t, q)), \qquad \frac{\partial S}{\partial t} = -H\bigl(t, q, \nabla S\bigr).

(b) Characteristics. Let $S \in C^2(D)$ solve the Hamilton–Jacobi equation. If a $C^1$ curve $q(t)$ in $D$ satisfies $\dot q = H_p\bigl(t, q, \nabla S(t, q)\bigr)$ , then $q(t)$ together with $p(t) = \nabla S(t, q(t))$ solves the canonical equations, and $\frac{d}{dt}S(t, q(t)) = p\cdot\dot q - H$ .

(c) Jacobi's theorem. Let $S(t, q, \alpha)$ be a complete integral, and fix $\alpha \in A$ and $\beta \in \mathbb{R}^n$ .

If a $C^1$ curve $q(t)$ , with $(t, q(t), \alpha) \in \mathcal{D}$ , satisfies the $n$ equations $\frac{\partial S}{\partial \alpha_j}\bigl(t, q(t), \alpha\bigr) = \beta_j \qquad (j = 1, \dots, n),$ then $q(t)$ and $p(t) = \nabla S\bigl(t, q(t), \alpha\bigr)$ solve the canonical equations.
Every solution of the canonical equations whose value at some time $t_\ast$ satisfies $p(t_\ast) = \nabla S(t_\ast, q(t_\ast), \alpha_\ast)$ for some $\alpha_\ast$ with $(t_\ast, q(t_\ast), \alpha_\ast) \in \mathcal{D}$ is, near $t_\ast$ , of the form in 1, with $\alpha = \alpha_\ast$ and $\beta = \partial_\alpha S(t_\ast, q(t_\ast), \alpha_\ast)$ .

So the $2n$ constants $(\alpha, \beta)$ give a $2n$ -parameter family of motions, obtained without integrating the canonical equations; by 2 this family contains every motion whose momentum at some time $t_\ast$ equals $\nabla S(t_\ast, q(t_\ast), \alpha)$ for some admissible $\alpha$ .

Idea of the proofOriginal

For (a), vary the endpoint of the extremal. Because the path is an extremal, the change in action comes only from the endpoint, through the momentum: moving the endpoint by $\delta q$ at fixed time changes $S$ by $p\cdot\delta q$ , and waiting a time $\delta t$ changes it by $(L - p\cdot\dot q)\,\delta t = -H\,\delta t$ . For (b), a solution $S$ assigns a momentum $\nabla S$ to every point; moving along the velocity field $H_p$ and differentiating the equation in $q$ shows that this momentum evolves by $\dot p = -H_q$ . So the canonical equations are the characteristic equations of the Hamilton–Jacobi equation. For (c), the quantities $\partial S/\partial\alpha_j$ do not change along such characteristics, because the equation holds for every $\alpha$ . Fixing their values pins down the curve.

ProofAdaptedAdapted from Gelfand & Fomin (1963), Ch. 4, §23, pp. 89–93

Part (a). Put $\sigma(t, a) = \int_{t_0}^t L\bigl(s, \mathsf{Q}, \partial_s\mathsf{Q}\bigr)\, ds$ on $[t_0, t_1)\times A$ and $P(t, a) = L_v\bigl(t, \mathsf{Q}, \partial_t\mathsf{Q}\bigr)$ , the momentum along the extremal with label $a$ .

Derivatives of $\sigma$ . By the fundamental theorem of calculus, $\partial_t\sigma = L\bigl(t, \mathsf{Q}, \partial_t\mathsf{Q}\bigr)$ . The integrand is $C^1$ in $a$ because $L$ is $C^2$ and $\mathsf{Q}$ is $C^2$ , so we may differentiate under the integral sign:
$\frac{\partial\sigma}{\partial a_j} = \int_{t_0}^t \Bigl( L_q\cdot \partial_{a_j}\mathsf{Q} + L_v\cdot \partial_s\partial_{a_j}\mathsf{Q} \Bigr)\, ds ,$
where we used $\partial_{a_j}\partial_s\mathsf{Q} = \partial_s\partial_{a_j}\mathsf{Q}$ ( $\mathsf{Q}$ is $C^2$ ).
Use that each curve is an extremal. Along $\mathsf{Q}(\cdot, a)$ the Euler–Lagrange equation gives $L_q = \partial_s P$ . The integrand is therefore $\partial_s\bigl(P\cdot\partial_{a_j}\mathsf{Q}\bigr)$ , and
$\frac{\partial\sigma}{\partial a_j}(t, a) = P(t, a)\cdot\partial_{a_j}\mathsf{Q}(t, a) - P(t_0, a)\cdot\partial_{a_j}\mathsf{Q}(t_0, a) = P(t, a)\cdot\partial_{a_j}\mathsf{Q}(t, a),$
because every extremal of the field starts at $q_0$ , so $\mathsf{Q}(t_0, a)$ does not depend on $a$ and $\partial_{a_j}\mathsf{Q}(t_0, a) = 0$ .
Transfer to $S$ . On $(t_0, t_1)\times A$ we have $S\bigl(t, \mathsf{Q}(t, a)\bigr) = \sigma(t, a)$ , and $S = \sigma\circ\Psi^{-1}$ is $C^1$ . The chain rule gives
$\nabla S\cdot\partial_{a_j}\mathsf{Q} = \partial_{a_j}\sigma = P\cdot\partial_{a_j}\mathsf{Q} \quad (j = 1, \dots, n), \qquad \partial_t S + \nabla S\cdot\partial_t\mathsf{Q} = \partial_t\sigma = L .$
The vectors $\partial_{a_j}\mathsf{Q}$ form the invertible matrix $\mathsf{Q}_a$ , so the first set of equations forces $\nabla S = P$ . The second then gives $\partial_t S = L - P\cdot\partial_t\mathsf{Q}$ . By (R), $\partial_t\mathsf{Q} = \hat v(t, q, P)$ , so the right side is $-H(t, q, P) = -H(t, q, \nabla S)$ .

Part (b). Let $p(t) = \nabla S(t, q(t))$ ; it is $C^1$ because $S$ is $C^2$ .

Differentiating along the curve, $\dot p_i = \partial_t\partial_{q_i}S + \sum_k \partial_{q_k}\partial_{q_i}S\;\dot q_k$ .
Differentiating the Hamilton–Jacobi equation with respect to $q_i$ (allowed since $S$ is $C^2$ and $H$ is $C^1$ ), with $H$ and its derivatives evaluated at $(t, q, \nabla S)$ :
$\partial_{q_i}\partial_t S + H_{q_i} + \sum_k H_{p_k}\,\partial_{q_i}\partial_{q_k}S = 0 .$
Mixed partials of $S$ commute, and $\dot q_k = H_{p_k}$ by hypothesis. Subtracting the two lines gives $\dot p_i = -H_{q_i}$ . Together with $\dot q = H_p$ these are the canonical equations. Finally $\frac{d}{dt}S(t, q(t)) = \partial_t S + \nabla S\cdot\dot q = -H + p\cdot\dot q$ .

Part (c). All derivatives of $H$ are evaluated at $(t, q, \nabla S(t, q, \alpha))$ .

Differentiate the equation in the parameters. For each fixed $\alpha$ the function $S(\cdot, \cdot, \alpha)$ solves the Hamilton–Jacobi equation, so the identity holds on $\mathcal{D}$ and may be differentiated in $\alpha_j$ :
$\partial_{\alpha_j}\partial_t S + \sum_k H_{p_k}\,\partial_{\alpha_j}\partial_{q_k} S = 0 .$
Differentiate the defining relations along the curve. Since $\partial_{\alpha_j}S(t, q(t), \alpha) = \beta_j$ is constant in $t$ ,
$\partial_t\partial_{\alpha_j}S + \sum_k \partial_{q_k}\partial_{\alpha_j}S\;\dot q_k = 0 .$
Use the nondegeneracy. Subtracting step 7 from step 8, and exchanging mixed partials of the $C^2$ function $S$ ,
$\sum_k \frac{\partial^2 S}{\partial q_k\,\partial\alpha_j}\,\bigl(\dot q_k - H_{p_k}\bigr) = 0 \qquad (j = 1, \dots, n).$
The matrix is nonsingular, so $\dot q = H_p\bigl(t, q, \nabla S(t, q, \alpha)\bigr)$ . Part (b), applied to $S(\cdot, \cdot, \alpha)$ , now shows that $(q, p)$ solves the canonical equations. This proves 1.
Every solution arises this way. Let $(\bar q, \bar p)$ be a solution with $\bar p(t_\ast) = \nabla S(t_\ast, q_\ast, \alpha_\ast)$ , where $q_\ast = \bar q(t_\ast)$ , and set $\beta = \partial_\alpha S(t_\ast, q_\ast, \alpha_\ast)$ . The map $q \mapsto \partial_\alpha S(t, q, \alpha_\ast)$ has invertible Jacobian (the transpose of the mixed Hessian), so by the implicit function theorem the equations $\partial_\alpha S(t, q, \alpha_\ast) = \beta$ define a $C^1$ curve $q(t)$ near $t_\ast$ with $q(t_\ast) = q_\ast$ . By 1, it gives a solution $(q, p)$ of the canonical equations with $p(t_\ast) = \nabla S(t_\ast, q_\ast, \alpha_\ast) = \bar p(t_\ast)$ . The right-hand side of the canonical equations is $C^1$ because $H$ is $C^2$ , so by uniqueness of solutions the two solutions coincide near $t_\ast$ . $\square$

Remark(Where the hypotheses are used)Original

The field condition in (a), that each $(t, q)$ is reached by exactly one extremal of the family with $\det\mathsf{Q}_a \ne 0$ , is used in step 3 to solve for $\nabla S$ . It fails where neighbouring extremals cross. For the oscillator $H = p^2/2m + \tfrac12 m\omega^2 q^2$ with $q_0 = 0$ , $t_0 = 0$ , the extremals $\mathsf{Q}(t, a) = (a/\omega)\sin\omega t$ form a field for $0 < t < \pi/\omega$ , and the action function is $S = \tfrac12 m\omega\, q^2\cot\omega t$ , which solves the Hamilton–Jacobi equation there. At $t = \pi/\omega$ every extremal returns to $q = 0$ and $S$ blows up. These crossing points are the conjugate points of Conjugate points and Jacobi's condition.
The fixed starting point is used in step 2. If the extremals started from different points, a boundary term at $t_0$ would remain.
The nondegeneracy $\det(\partial^2 S/\partial q\,\partial\alpha) \ne 0$ is used in step 9 to cancel the matrix, and in step 10 for the implicit function theorem. A family of solutions with fewer than $n$ effective parameters, such as $S$ not depending on some $\alpha_j$ , gives too few equations to determine $q(t)$ .
$C^2$ regularity of $S$ is used in step 6 and step 9, where mixed partial derivatives of $S$ are exchanged, and in step 7, where the Hamilton–Jacobi equation is differentiated in $\alpha$ . (Step 1 exchanges mixed partials of the field map $\mathsf{Q}$ , which is $C^2$ by hypothesis; $S$ plays no part there.) Under the hypotheses of (a) the action function is in fact $C^2$ , so (b) applies to it directly: $\nabla S = L_v\bigl(t, q, \partial_t\mathsf{Q}\circ\Psi^{-1}\bigr)$ is $C^1$ , because $L$ is $C^2$ , $\partial_t\mathsf{Q}$ is $C^1$ and $\Psi^{-1}$ is $C^1$ ; and then $\partial_t S = -H(t, q, \nabla S)$ is $C^1$ because $H$ is $C^2$ .

Remark(Relation to the method of characteristics)AdaptedAdapted from Evans (2010), §3.2.5, Example 6, pp. 109–110

When $H$ is linear in $p$ , say $H = b(t, q)\cdot p + c(t, q)$ , the Hamilton–Jacobi equation is the linear equation $S_t + b\cdot\nabla S = -c$ treated in First-order equations and characteristics. Its characteristic curves solve $\dot q = b = H_p$ , and along them $S$ changes at the rate $-c = p\cdot H_p - H$ . Part (b) is exactly this, extended to a general nonlinear $H$ . The new feature is that the characteristic speed $H_p$ depends on the unknown slope $\nabla S$ , so the slope must be carried along the characteristic too, and it evolves by $\dot p = -H_q$ . In this sense the canonical equations are the characteristic system of the Hamilton–Jacobi equation, and conversely a solution of the PDE can be built by solving the canonical equations.

Example(Vertical throw)Original

A ball of mass $m$ moves vertically in uniform gravity $\gamma$ , so $H(q, p) = p^2/2m + m\gamma q$ , with $q$ the height.

Action function (part (a)). The extremals from $(t_0, q_0) = (0, 0)$ are $\mathsf{Q}(t, a) = a t - \tfrac12\gamma t^2$ , and for $t > 0$ the label of the extremal through $(t, q)$ is $a = (q + \tfrac12\gamma t^2)/t$ . Integrating $L = \tfrac12 m\dot{\mathsf{Q}}^2 - m\gamma\mathsf{Q}$ along it gives

S(t, q) = \frac{m q^2}{2t} - \frac{m\gamma q t}{2} - \frac{m\gamma^2 t^3}{24}.

Then $\nabla S = m q/t - m\gamma t/2 = m(a - \gamma t)$ , which is the momentum $m\dot{\mathsf{Q}}$ at time $t$ . A direct computation gives $S_t + (S_q)^2/2m + m\gamma q = 0$ , as part (a) asserts.

A complete integral (part (c)). Try a solution affine in $q$ , $S = (\alpha - m\gamma t)\,q + f(t)$ . Then $S_q = \alpha - m\gamma t$ , and the equation becomes $f'(t) + (\alpha - m\gamma t)^2/2m = 0$ , so

S(t, q, \alpha) = (\alpha - m\gamma t)\, q + \frac{(\alpha - m\gamma t)^3}{6 m^2\gamma}, \qquad \frac{\partial^2 S}{\partial q\,\partial\alpha} = 1 \ne 0 .

Jacobi's theorem gives the motion by differentiation alone:

\beta = \frac{\partial S}{\partial\alpha} = q + \frac{(\alpha - m\gamma t)^2}{2m^2\gamma} \quad\Longrightarrow\quad q(t) = \beta - \frac{(\alpha - m\gamma t)^2}{2m^2\gamma}, \qquad p(t) = \frac{\partial S}{\partial q} = \alpha - m\gamma t .

So $\alpha$ is the momentum at $t = 0$ and $\beta$ is the height at which $p = 0$ , the top of the flight.

Numbers. Take $m = 2\ \mathrm{kg}$ and $\gamma = 9.8\ \mathrm{m/s^2}$ , and throw the ball upward from $q(0) = 1\ \mathrm{m}$ at $4.9\ \mathrm{m/s}$ . Then $\alpha = 9.8\ \mathrm{kg\,m/s}$ and

\beta = 1 + \frac{9.8^2}{2\cdot 2^2\cdot 9.8} = 1 + 1.225 = 2.225\ \mathrm{m}.

The top is reached when $p = 0$ , at $t = \alpha/(m\gamma) = 0.5\ \mathrm{s}$ . At $t = 1\ \mathrm{s}$ the formulas give $q = 2.225 - 1.225 = 1\ \mathrm{m}$ and $p = -9.8\ \mathrm{kg\,m/s}$ : the ball is back at its starting height, moving down at its launch speed, as energy conservation requires.

ApplicationOriginal

In geometrical optics the Hamilton–Jacobi equation becomes the eikonal equation $|\nabla S| = n(x)$ for the optical path length $S$ ; its level sets are wavefronts and its characteristics are light rays, which is how ray tracing and wavefront propagation are related. In quantum mechanics, inserting $\psi = e^{iS/\hbar}$ into the Schrödinger equation and keeping the leading order in $\hbar$ gives the Hamilton–Jacobi equation, the starting point of the WKB approximation. In celestial mechanics, finding a complete integral by separating variables (for example $S = -Et + W(q)$ when $H$ does not depend on $t$ ) is the classical route to the Kepler orbits and to action–angle variables. In optimal control, the value function of a control problem satisfies a Hamilton–Jacobi–Bellman equation of the same form.