advanced30 min studyLesson 132 in Physics

Hamilton's equations and phase flow

Read a Hamiltonian as a vector field on phase space and distinguish a conserved contour from motion along it.

Before you begin

On this page

The question for today

What you’ll understand

Derive the signs in Hamilton's equations.
Predict oscillator motion from phase-space coordinates.
Separate conservation of H from the speed and direction of phase flow.

Position alone is not enough to predict a particle's future: a cart at the same location may be moving left, moving right, or momentarily stopped. Phase space records position and momentum together, so a state is one point and evolution is a curve through those points. Hamilton's equations attach a velocity vector to each state. The Hamiltonian therefore supplies both a family of contours and a precise rule for traveling through them.

Two first-order equations replace one second-order equation

Differentiate $H=p_i\dot q_i-L$ before eliminating the intermediate velocity differentials. The terms $p_i\,d\dot q_i$ cancel because momentum is the velocity derivative of L. What remains identifies the momentum derivative of H with velocity and the position derivative with minus the force-like momentum rate. The minus sign is not a convention that can be removed independently; it is tied to the canonical pairing and the standard bracket convention used throughout this module.

A regular system with n configuration coordinates has 2n phase-space coordinates and 2n first-order evolution equations. It still needs the same physical amount of initial information as n second-order equations. A fixed point requires both kinds of derivatives to vanish. A position turning point only requires the velocity to vanish, so a particle can pass through a turning point in phase space without stopping its phase-space motion. Its momentum is changing there.

\dot q_i=\frac{\partial H}{\partial p_i},\qquad \dot p_i=-\frac{\partial H}{\partial q_i}

Derivatives hold the other canonical variables and time fixed. These equations assume a regular canonical description.↔ Scroll wide equations horizontally.

An oscillator moves around an energy ellipse

For a harmonic oscillator, the potential is quadratic in position and the kinetic term is quadratic in momentum. Positive energy contours are ellipses. At positive position and zero momentum, the state moves toward negative momentum first: the spring starts pulling toward the origin. At zero position and positive momentum, it moves toward positive position. These two local arrows determine the orientation of the entire orbit without solving a trigonometric equation.

Rescaling the axes can make the ellipse look circular, but position and momentum ordinarily have different units. The visual shape depends on the plotting convention; the physical period does not. In normalized coordinates, equal increments of time need not correspond to equal ordinary polar-angle increments on an unscaled ellipse. Action-angle coordinates later provide a phase variable with uniform angular motion. That variable should not be confused with the angle read from an arbitrary screen plot.

H=\frac{p^2}{2m}+\frac{m\omega^2q^2}{2},\qquad q=A\cos(\omega t),\quad p=-m\omega A\sin(\omega t)

This solution starts at q = A and p = 0 at t = 0. It assumes m > 0 and a constant positive angular frequency ω.↔ Scroll wide equations horizontally.

A conserved value does not mean a frozen state

Along a Hamiltonian trajectory, the contributions from position change and momentum change cancel in the total derivative of H. Only explicit time dependence remains. Consequently a time-independent Hamiltonian is conserved, even while both position and momentum change rapidly. Energy specifies which contour contains the trajectory, but the full derivative structure of H specifies the rate and direction along that contour. Multiplying H by a positive constant preserves its contours while rescaling the speed of the flow.

Damping provides an instructive boundary. The familiar friction force proportional to velocity makes oscillator energy decrease and contracts areas in its ordinary position-momentum plane. That two-dimensional damped system is not generated by the same autonomous canonical Hamiltonian. One can introduce a larger environment or alternative formulations, but those are different descriptions. When inspecting an animation, first determine whether it displays conservative Hamiltonian motion or a dissipative comparison rather than assuming every smooth trajectory is canonical.

\frac{dH}{dt}=\sum_i\left(H_{q_i}\dot q_i+H_{p_i}\dot p_i\right)+H_t=H_t

The cancellation follows from Hamilton's equations. Conservation requires no explicit time dependence in the chosen canonical Hamiltonian.↔ Scroll wide equations horizontally.

Work it through

A quarter-cycle in phase space

Let $H=p^2/2+2q^2$ in normalized units, with $q(0)=1$ and $p(0)=0$ . Find the state at $t=\pi/4$ and check its energy.

Identify unit mass and $\omega=2$ , since $\omega^2q^2/2=2q^2$ . Hamilton's equations are $\dot q=p$ and $\dot p=-4q$ . Initially the flow points downward in the q-p plot.
The solution is $q=\cos(2t)$ and $p=-2\sin(2t)$ . At $t=\pi/4$ , the phase is $\pi/2$ , so position is zero and momentum is −2. The state has not stopped.
Initially $H=2$ . At the requested time $H=(-2)^2/2=2$ . Potential energy has become kinetic energy while the conserved total remains unchanged; the full period is $2\pi/2=\pi$ .

The result

The state is q = 0, p = −2 after one quarter of a period, with H = 2 at both times.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Follow a conservative phase-space state

Set damping ratio to zero and natural frequency to 2. Predict the first zero crossing near π/4, then use playback and inspect displacement and velocity around that time. Unit mass makes velocity equal to momentum in this lab. Increasing damping produces a dissipative comparison, not the same canonical Hamiltonian system. The plot is displacement versus time; use its readouts to reconstruct the phase-space point rather than mistaking it for a phase portrait.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA conserved oscillator changes position and momentum continuously while its Hamiltonian stays fixed.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Hamilton's equations give q̇ = p/m = 0 and ṗ = −mω²A < 0. A turning point in position is not a fixed point in phase space.

Your turn

Try it yourself

Compare H and $K=3H$ for the same autonomous oscillator and identical initial canonical coordinates. Do their trajectories have different shapes or different periods?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Take derivatives of K and compare the complete vector fields, not only the energy contours.

Compare with the solution

Every phase-space derivative generated by K is three times the corresponding derivative generated by H. The unparameterized orbits are the same ellipses, with the same orientation, but K traverses each orbit three times faster. Its period is one third as large. This comparison treats the canonical coordinates and time unit as fixed; it is a mathematical rescaling of the dynamical generator.

Keep the idea

Take it with you

Hamilton's equations define a vector field on the full phase space.
A position turning point is usually not a phase-space fixed point.
Conserved contours and the speed along them are distinct information.

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