Hamilton's equations of motion: Question bank — Physics deep dive

Before we start, three words must be earned so no line uses an undefined symbol:

Spot the error

H = \frac{1}{2} m \overset{x}{˙}^{2} + V

, so I just plug in and I'm done."

The error is leaving

\overset{x}{˙}

H

. You must substitute

\overset{x}{˙} = p / m

to get

H = p^{2} /2 m + V

; writing

H

with

\overset{x}{˙}

mixes the Lagrangian and Hamilton pictures.

\overset{p}{˙} = + \partial H / \partial q

by symmetry with

\overset{q}{˙} = + \partial H / \partial p

The plus is wrong. Matching

- \overset{p}{˙} d q

against

+ (\partial H / \partial q) d q

d H

forces

\overset{p}{˙} = - \partial H / \partial q

; the asymmetry is essential, not a symmetry to be restored.

"For the magnetic Lagrangian I set

p = m \dot{r}

and continue."

Wrong momentum. Here

p = \partial L / \partial \dot{r} = m \dot{r} + q A

; using the kinetic momentum instead of the canonical one gives incorrect equations.

\frac{d H}{d t} = \frac{\partial H}{\partial q} \overset{q}{˙} + \frac{\partial H}{\partial p} \overset{p}{˙}

, and that's the full time derivative."

It omits the explicit

\partial H / \partial t

term. The chain rule for

H (q, p, t)

has three pieces; the first two cancel via the canonical equations, leaving exactly

\partial H / \partial t

"Since

\partial H / \partial t = - \partial L / \partial t

, and energy is conserved,

L

must be time-independent too."

Conflates two things.

H

can be conserved (no explicit

t

H

) while

L

has explicit

t

only if the transform is arranged so, but generically

\partial H / \partial t = - \partial L / \partial t

— the reasoning "energy conserved therefore

L

time-independent" skips that

L

time-dependence directly produces

H

time-dependence.

"In phase space the trajectory can cross itself since it's just a curve."

For a time-independent

H

the flow is deterministic: each point has one unique tangent, so trajectories cannot cross — a crossing would mean two futures from one state.

"The SHO orbit is a circle in

(x, p)

It is an ellipse in general.

H = p^{2} /2 m + \frac{1}{2} m ω^{2} x^{2} = E

has different scales on the

x

and

p

axes; only after rescaling variables does it become a circle.

Why questions

Why trade

\overset{q}{˙}

for

p

at all, rather than keep the Lagrangian's second-order equations?

First-order equations make the state a single point in phase space with one unique flow line, exposing the symmetric

(q, p)

structure behind Poisson brackets and Liouville's theorem.

Why does the

d \overset{q}{˙}_{i}

term vanish when we take

d H

Because

\partial L / \partial \overset{q}{˙}_{i} = p_{i}

by definition, so the term

p_{i} d \overset{q}{˙}_{i}

exactly cancels

\frac{\partial L}{\partial q ˙ _{i}} d \overset{q}{˙}_{i}

— that cancellation is the entire purpose of the Legendre transform.

Why do the two chain-rule terms cancel when computing

d H / d t

Substituting

\overset{q}{˙} = \partial H / \partial p

and

\overset{p}{˙} = - \partial H / \partial q

gives

\frac{\partial H}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial H}{\partial p} \frac{\partial H}{\partial q} = 0

; the minus sign is precisely what makes them annihilate.

Why is the minus sign called "area-preserving"?

The antisymmetric pairing

(\overset{q}{˙}, \overset{p}{˙}) = (+ \partial_{p} H, - \partial_{q} H)

makes phase-space flow divergence-free, so volumes (areas in 2D) are conserved as the system evolves (Liouville).

Why must

\overset{q}{˙}

be expressed through

(q, p)

before writing the canonical equations?

Otherwise

\partial H / \partial p

is ambiguous — you cannot differentiate

H

with respect to

p

H

still secretly depends on

\overset{q}{˙}

, since

\overset{q}{˙}

and

p

are not independent.

Why does Newton's second law reappear in Example A?

Combining

\overset{x}{˙} = p / m

and

\overset{p}{˙} = - V^{'} (x)

gives

m \overset{x}{¨} = - V^{'} (x)

; Hamilton's structure is equivalent to Newton, just written in doubled first-order form.

Edge cases

What happens to the formalism if

L

is independent of a velocity

\overset{q}{˙}_{j}

(no

\overset{q}{˙}_{j}

term)?

Then

p_{j} = \partial L / \partial \overset{q}{˙}_{j} = 0

is a constraint, not an invertible relation — the Legendre transform is singular and requires constrained (Dirac) treatment.

For a free particle (

V = 0

), what does the phase-space trajectory look like?

\overset{p}{˙} = 0

p

is constant, and

\overset{x}{˙} = p / m

is constant; the trajectory is a horizontal line in

(x, p)

traversed at uniform speed.

What is the limiting phase-space orbit of the SHO as energy

E \to 0

The ellipse shrinks to the single fixed point

(x, p) = (0, 0)

— the particle at rest at the potential minimum, a degenerate orbit of zero enclosed area.

\partial H / \partial t \neq = 0

but the trajectory happens to sit at a point where

\partial H / \partial t = 0

momentarily, is

H

conserved?

No. Conservation requires

\partial H / \partial t = 0

along the whole trajectory; an instantaneous zero does not stop

H

from changing before or after.

What does

H

equal in a uniformly rotating frame where a bead slides on a spinning wire?

H

is conserved (if the setup has no explicit

t

) but

H \neq = E

; it is energy minus a term

\propto ω^{2}

, the classic case where the "energy =

H

" intuition fails.

For a coordinate that is cyclic, what does the reduced system look like?

p_{j}

becomes a fixed parameter (constant of motion), so the effective dynamics live in fewer dimensions — the seed of the reduction used in Hamilton-Jacobi Theory and Noether's Theorem.

Recall Self-check before moving on

Can you, without looking, state the one reason $H \neq = E$ in general? ::: The kinetic energy is not quadratic in velocities and/or the constraints depend explicitly on time (rheonomic). Can you state why the minus sign cannot be flipped? ::: It is forced by matching $- \overset{p}{˙} d q$ with $+ \partial_{q} H d q$ in $d H$ ; flipping it breaks $d H / d t$ cancellation and thus energy conservation.

See also: Legendre Transform, Phase Space and Liouville's Theorem, Poisson Brackets, Canonical Transformations.

Question bank — Hamilton's equations of motion

True or false — justify

Spot the error

Why questions

Edge cases