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Topic 8 of 11

Canonical Transformations

Canonical transformations change coordinates in phase space while preserving Hamilton's equations — they are the symplectomorphisms of classical mechanics. The right transformation can trivialize a problem: action-angle variables reduce any integrable system to uniform motion.

Key Concepts

Canonical Transformation

A coordinate change

(q,p)\to(Q,P)

that preserves the form of Hamilton's equations. Equivalently, it preserves the Poisson brackets:

\{Q_i,P_j\}=\delta_{ij}

\{Q_i,Q_j\}=0

\{P_i,P_j\}=0

Generating Functions

A function

F(q,Q,t)

(or one of three other mixed representations) defines a canonical transformation via

p=\partial F/\partial q

P=-\partial F/\partial Q

K=H+\partial F/\partial t

Action Variable

J_i=\oint p_i\,dq_i

(integral over one period of the

i

th degree of freedom). Action variables are adiabatic invariants: they are conserved when the system changes slowly.

Action-Angle Variables

The canonical transformation to action-angle variables

(J,\theta)

makes

H

depend only on

J

H=H(J)

. Then

\dot J=0

(J is constant) and

\dot\theta=\partial H/\partial J=\omega(J)

is constant — uniform motion in

\theta

Key Equations

Symplectic condition

M^T J M = J, \quad J = \begin{pmatrix}0&I\\-I&0\end{pmatrix}

The Jacobian M of a canonical transformation satisfies this matrix equation.

Action variable

J = \oint p\,dq

Integral over one complete cycle; equals the enclosed area in phase space.

Angle variable EOM

\dot{\theta} = \frac{\partial H(J)}{\partial J} = \omega(J)

In action-angle variables, the angle increases uniformly at rate ω(J).

Worked Example

Action Variable for the Harmonic Oscillator

Problem

Compute the action variable $J=\oint p\,dq$ for a harmonic oscillator ( $H=p^2/(2m)+\frac{1}{2}m\omega_0^2 q^2=E$ ) and express $H$ in terms of $J$ .

Solution

The phase space orbit is an ellipse: $q=A\cos\phi$ , $p=-m\omega_0 A\sin\phi$ , with $A=\sqrt{2E/(m\omega_0^2)}$ .

J = \oint p\,dq = \oint (-m\omega_0 A\sin\phi)(-A\sin\phi)\,d\phi = m\omega_0 A^2\int_0^{2\pi}\sin^2\phi\,d\phi = \pi m\omega_0 A^2

Since $E=\frac{1}{2}m\omega_0^2 A^2$ , we have $A^2=2E/(m\omega_0^2)$ :

J = \pi m\omega_0\cdot\frac{2E}{m\omega_0^2} = \frac{2\pi E}{\omega_0} \implies H = E = \omega_0 J/(2\pi) \equiv \omega_0 J

(Using $J$ defined without the $2\pi$ .) Then $\dot\theta=\partial H/\partial J=\omega_0$ , confirming uniform angular motion.

Answer

J=2\pi E/\omega_0

; the Hamiltonian becomes

H=\omega_0 J/(2\pi)

Practice

Exercises

7 problems

1 of 7

For a harmonic oscillator with $\omega_0=4.0$ rad/s and energy $E=20$ J, find the action variable $J=2\pi E/\omega_0$ (in J·s).

Your answer

J·s

2 of 7

A free particle (mass $m=2.0$ kg) bounces between walls at $x=0$ and $x=L=1.0$ m with speed $v=5.0$ m/s. Find the action $J=2mvL$ (in kg·m²/s).

Your answer

kg·m²/s

3 of 7

What is the Poisson bracket $\{J,\theta\}$ for action-angle variables? (It equals 1 by the canonical condition.)

Unlock Exercise 3