Cyclic coordinates — corresponding conservation law: Question bank

Before you attack the traps, we set up the two pictures every question below leans on, so no symbol is a stranger.

Definition The Lagrangian and the coordinates we will keep reusing

The Lagrangian $L$ is a single number built from a system's positions and velocities; think of it as the "score sheet" $L = (kinetic energy) - (potential energy)$ that Nature reads to decide the motion.

$m$ is the particle's mass.
In the plane, $(x, y)$ are the usual left–right / up–down coordinates. A dot means "rate of change", so $\overset{x}{˙}$ is the sideways speed.
Polar coordinates $(r, ϕ)$ describe the same plane differently: $r$ is the distance from the origin and $ϕ$ is the angle measured round from a reference direction. See the picture below.

Two Lagrangians recur throughout this page:

Free particle in the plane: $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2})$ .
Central force in polar form: $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)$ , where $V (r)$ is a potential depending only on the distance $r$ .

Figure — Cyclic coordinates — corresponding conservation law

Look at the figure: the same point $P$ is pinned by $(x, y)$ or by $(r, ϕ)$ . The angle $ϕ$ is the thing that "cycles" — walk it from $0$ round to $2 π$ and you land back where you started. That periodicity is what makes $ϕ$ a natural cyclic coordinate, and it also warns us that $ϕ$ and $ϕ + 2 π$ are the same physical place (a branch subtlety we return to in Edge cases).

Now hold these four anchors in your head — every trap below pokes at one of them:

Recall The four anchors (open if you forgot the topic)

Cyclic means the coordinate $q_{k}$ is absent from $L$ : $\partial L / \partial q_{k} = 0$ . Its velocity $\overset{q}{˙}_{k}$ may still appear.
What is conserved is the conjugate momentum $p_{k} = \partial L / \partial \overset{q}{˙}_{k}$ — not the coordinate itself.
The engine is the Euler–Lagrange equation: $\overset{p}{˙}_{k} = \partial L / \partial q_{k}$ . Zero on the right ⇒ zero rate of change on the left.
Absence of time in $L$ is a different theorem — it conserves energy (the Hamiltonian), see Conservation of Energy and Hamiltonian Mechanics.

True or false — justify

L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)

: the coordinate

ϕ

is cyclic even though

\dot{ϕ}

appears in

L

True. "Cyclic" is about the coordinate

ϕ

itself being absent; here

ϕ

never appears, only

\dot{ϕ}

does, so

\partial L / \partial ϕ = 0

and

p_{ϕ}

is conserved.

q_{k}

is cyclic, then

q_{k}

itself stays constant in time.

False. It is the conjugate momentum

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

that stays constant. In central-force motion

ϕ

keeps increasing forever, yet

m r^{2} \dot{ϕ}

is fixed.

If the velocity

\overset{q}{˙}_{k}

is missing from

L

, then

q_{k}

is cyclic.

False. Cyclic requires the coordinate

q_{k}

to be absent, not its velocity. A missing

\overset{q}{˙}_{k}

is a different (degenerate) situation entirely.

A cyclic coordinate always has a constant conjugate momentum, regardless of what the other coordinates are doing.

True.

\overset{p}{˙}_{k} = \partial L / \partial q_{k} = 0

holds identically once

q_{k}

is absent, no matter how wildly the other coordinates evolve — that constancy is exactly the content of the theorem.

L

does not depend explicitly on time, then some linear momentum is conserved.

False. Time-independence conserves the energy/Hamiltonian, a separate theorem. It says nothing directly about any conjugate momentum

p_{k}

For a free particle

L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2})

, both

p_{x}

and

p_{y}

are conserved.

True. Neither

x

nor

y

appears in

L

, so both are cyclic and both

m \overset{x}{˙}

and

m \overset{y}{˙}

are constant — full translational symmetry in the plane.

Adding a potential

V (x)

that depends on

x

can never destroy the conservation of

p_{x}

False. Once

V

depends on

x

\partial L / \partial x = - V^{'} (x) \neq = 0

in general, so

x

is no longer cyclic and

p_{x}

changes. The symmetry (and its law) is broken by the potential.

The conjugate momentum of a cyclic angular coordinate is always the physical angular momentum

L_{z}

Mostly, but read carefully. It equals angular momentum for a genuine rotation in a plane (

p_{ϕ} = m r^{2} \dot{ϕ}

); in general

p_{ϕ}

is whatever

\partial L / \partial \dot{ϕ}

evaluates to, which coincides with the physical angular momentum only when

ϕ

really parametrizes a rotation.

For a charged particle in a magnetic field,

\partial L / \partial q_{k} = 0

guarantees the mechanical momentum

m \overset{q}{˙}_{k}

is conserved.

False. With a velocity-dependent potential the conserved conjugate momentum is

p_{k} = m \overset{q}{˙}_{k} + q A_{k}

(mechanical plus field part), not

m \overset{q}{˙}_{k}

alone. See Edge cases for the full story.

Spot the error

"Since

ϕ

is cyclic in

L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)

, the angular velocity

\dot{ϕ}

is constant."

Error: the conserved quantity is

p_{ϕ} = m r^{2} \dot{ϕ}

, not

\dot{ϕ}

. As

r

shrinks,

\dot{ϕ}

must grow to keep

m r^{2} \dot{ϕ}

fixed (this is precisely Kepler's second law).

"Gravity acts on the projectile, so no momentum can be conserved."

Error: gravity acts only vertically, so

x

stays cyclic and

p_{x} = m \overset{x}{˙}

is conserved. Only the vertical momentum changes; horizontal momentum is untouched.

q_{k}

is cyclic means the generalized force

\partial L / \partial q_{k}

is nonzero along

q_{k}

Error: it is exactly the opposite. Cyclic means

\partial L / \partial q_{k} = 0

, i.e. no generalized force along

q_{k}

— that vanishing force is why the momentum has nothing to change it.

"Because

x

is cyclic, we can drop the Euler–Lagrange equation for

x

; it gives no information."

Error: it gives the most useful information of all — a first integral

p_{x} = const

. We don't drop it; we use it to reduce the problem (this is the basis of Routhian reduction in Hamiltonian Mechanics).

"Time

t

is absent from

L

, so

t

is a cyclic coordinate and its conjugate momentum

p_{t}

is conserved."

Error:

t

is the evolution parameter, not a generalized coordinate — there is no

\dot{t}

to differentiate by. Time-independence conserves the Hamiltonian instead, via the energy-function argument, not the cyclic-coordinate one.

"The bead-on-sphere

L = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2})

has

θ

cyclic because it's an angle."

Error:

θ

appears explicitly through

sin^{2} θ

, so

\partial L / \partial θ \neq = 0

and

θ

is not cyclic. Only

ϕ

(which never appears) is cyclic. Being an angle is neither necessary nor sufficient. (See the bead-on-sphere figure below in Edge cases.)

Why questions

Why does "the Lagrangian not caring about

q_{k}

" translate into "something is conserved"?

L

is unchanged when you shift

q_{k}

, there is no generalized force

\partial L / \partial q_{k}

to push along

q_{k}

; with nothing to change it, the momentum

p_{k}

along that direction is frozen. Symmetry ⇒ conservation (see Noether's Theorem).

Why is the conserved quantity the momentum conjugate to

q_{k}

rather than some other momentum?

Because Euler–Lagrange for coordinate

q_{k}

reads

\overset{p}{˙}_{k} = \partial L / \partial q_{k}

with

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

. Setting the right side to zero freezes exactly that

p_{k}

and no other; the pairing is built into the equation.

Why is

p_{ϕ} = m r^{2} \dot{ϕ}

(with an

r^{2}

) instead of just

m \dot{ϕ}

Because

p_{ϕ} = \partial L / \partial \dot{ϕ}

, and the only term with

\dot{ϕ}

\frac{1}{2} m r^{2} \dot{ϕ}^{2}

; differentiating gives

m r^{2} \dot{ϕ}

. The

r^{2}

carries the units and geometry of a genuine angular momentum.

Why is time-independence a different theorem from a cyclic spatial coordinate?

A cyclic coordinate is a symmetry under shifting a coordinate and conserves its conjugate momentum. Time-independence is a symmetry under shifting the clock and conserves the energy function

h

— different symmetry, different conserved object, proved by a different manipulation.

Why does a cyclic coordinate still leave a fully solvable equation of motion, even though

q_{k}

"vanished"?

The conserved

p_{k} = const

lets you solve algebraically for

\overset{q}{˙}_{k}

in terms of the constant and the other coordinates, then feed it back into the remaining equations — the ignored coordinate reappears as a source of a first integral, reducing the problem's degrees of freedom.

Why can two coordinates be cyclic at once, and what does that buy you?

Nothing forbids

L

from being independent of several coordinates; each independent absence gives its own conserved conjugate momentum. Each conserved momentum is a first integral that lowers the effective dimension of the problem by one.

Why might a coordinate look non-cyclic yet still carry a conserved momentum (the gauge trap)?

With velocity-dependent potentials (magnetic fields), a change of gauge

A \to A + \nabla χ

adds a total time derivative to

L

and can make

q_{k}

appear explicitly, hiding the symmetry. The conserved

p_{k}

is unchanged; only its split into mechanical and field parts, and its visibility, shifts with the gauge.

Edge cases

What happens if

\partial L / \partial q_{k} = 0

holds only at one instant, not for all time?

Then

q_{k}

is not cyclic and

p_{k}

is not conserved. "Cyclic" is an identity in

q_{k}

—

L

must be free of

q_{k}

everywhere, not just where

\overset{p}{˙}_{k}

happens to vanish momentarily.

Free particle in the plane: is angular momentum conserved even though its trajectory is a straight line, and about which point?

Yes, about the chosen origin. Angular momentum is origin-dependent; in polar coordinates centred at that origin the free-particle

L

has

ϕ

cyclic, so

p_{ϕ} = m r^{2} \dot{ϕ}

is conserved. A straight line has constant angular momentum about any fixed point, but the value depends on which point you pick.

q_{k}

is cyclic but the corresponding

\overset{q}{˙}_{k} = 0

t = 0

(particle momentarily still along that direction), does the coordinate stay put forever?

Yes for that special case:

p_{k} = 0

is conserved, so

\overset{q}{˙}_{k}

stays zero (assuming

p_{k}

is a simple multiple of

\overset{q}{˙}_{k}

). But this is a boundary value, not a general feature — nonzero initial

p_{k}

keeps

q_{k}

moving forever.

Suppose

L

has no coordinates at all appearing, only velocities (e.g. free particle). What does the theorem say?

Every coordinate is cyclic, so every conjugate momentum is conserved. This is the maximally symmetric case: all momenta are constants of motion and the trajectory is straight-line, uniform.

In central motion, at the turning points of

r

(where

\overset{r}{˙} = 0

), is

p_{ϕ}

still conserved?

Yes, absolutely.

p_{ϕ} = m r^{2} \dot{ϕ}

is conserved for all times regardless of what

\overset{r}{˙}

does; the turning points of

r

are irrelevant to the

ϕ

-symmetry. This is why the particle sweeps equal areas even as it speeds up and slows down radially (see Central Force Motion).

Since

ϕ

is periodic (

ϕ

and

ϕ + 2 π

are the same physical point), does its cyclicity or the conservation of

p_{ϕ}

break at the "branch cut"

ϕ = 2 π

No. Cyclicity is a statement about

L

not containing

ϕ

, which is unaffected by how we label the branch;

p_{ϕ} = m r^{2} \dot{ϕ}

depends on the velocity

\dot{ϕ}

, which stays smooth as

ϕ

winds past

2 π

. The branch cut is a bookkeeping artifact, not a physical discontinuity — indeed the periodicity is precisely why

ϕ

is cyclic.

Velocity-dependent potential (charge

q

in magnetic vector potential

A

L = \frac{1}{2} m \dot{r}^{2} + q \dot{r} \cdot A (r)

. If

x

is absent from

L

, what is conserved?

The canonical momentum

p_{x} = \partial L / \partial \overset{x}{˙} = m \overset{x}{˙} + q A_{x}

, not the mechanical

m \overset{x}{˙}

. The field part

q A_{x}

carries the difference. Choosing a gauge where

A

depends on

x

can make

x

appear explicitly and hide the symmetry, yet the physics (the conserved

p_{x}

) is gauge-invariant up to the added total derivative.

What if

L

depends on

q_{k}

only through a total time derivative, e.g.

L \to L^{'} = L + \frac{d}{d t} f (q_{k}) = L + \frac{\partial f}{\partial q _{k}} \overset{q}{˙}_{k}

The equations of motion are unchanged by a total time derivative, so a coordinate appearing only this way is still effectively cyclic. Concretely the new momentum is

p_{k}^{'} = \partial L^{'} / \partial \overset{q}{˙}_{k} = p_{k} + \partial f / \partial q_{k}

; taking

d / d t

and using

\partial L^{'} / \partial q_{k} = \partial L / \partial q_{k} + \frac{d}{d t} \frac{\partial f}{\partial q _{k}}

, the extra pieces cancel in Euler–Lagrange, so

\overset{p}{˙}_{k}^{'} = \overset{p}{˙}_{k}

. The conserved quantity is just shifted by the constant-along-motion term

\partial f / \partial q_{k}

; the conservation law survives.

A coordinate is cyclic in one set of generalized coordinates but appears explicitly after a change of variables. Which conclusion is correct?

Both are consistent: cyclicity is a property of the coordinate choice, not an absolute. A clever choice that makes a coordinate cyclic reveals a conserved momentum; a poor choice hides it. The underlying symmetry (via Noether's Theorem) is coordinate-independent even when the "cyclic" label is not.

The bead-on-sphere

L = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2})

: which angle is cyclic, and what does the geometry show?

The azimuthal

ϕ

(spin round the polar axis) is cyclic — it never appears — so

p_{ϕ} = m R^{2} sin^{2} θ \dot{ϕ}

is conserved. The polar

θ

is not cyclic because

sin^{2} θ

multiplies the

\dot{ϕ}^{2}

term. See the sphere figure: rotating the whole picture about the vertical axis changes nothing, which is exactly the

ϕ

-symmetry.

Connections

Euler–Lagrange Equations — every "why" here traces back to $\overset{p}{˙}_{k} = \partial L / \partial q_{k}$ .
Generalized Momentum — the object that gets sealed.
Noether's Theorem — the coordinate-free reason the symmetry survives changes of variable.
Central Force Motion — home of the $ϕ$ -cyclic, $p_{ϕ}$ -conserved traps.
Conservation of Energy / Hamiltonian Mechanics — the time-translation analogue and reduction machinery.